~CD-ROM
~INClUDED
Brian Vicente / Max Nori / H. Scott Fogler
Soludons Manual for
Elements
of Chemical Reaction Engineering Fourth Edition ~
Prentice Hallitaternational Series in the Phys!ptl and Chemical E.ngineering Sciences
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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<2 Not done 2 < D:5 4 Dependences 4 < D:5 6 Parameter values 6
A even: score> 85 % z = random numbers
Perf. No. = 53~07 Score> 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 <1 10
x
P2-9 (c) Volume of second CSTR added in series to achieve 80% conversion
__l!.AO(X2- Xl). V2(-rA )
V2
= 500000 X (0.8"- 0.5)
V2 = 150000m3 x
2-17
P2-9 (d) Volume of PFR added in series to first CSTR to achieve 80% conversion
1 VPFR = (-x 400000 x 0.3) + (100000 X 0.3) 2 VPFR = 90000m3
P2-9 (e) ForCSTR, VI = 60000 m3 (CSTR) Mole Balance
V=
x
FAOX
(-rA )
x == 0.515 ForPFR, V2 = 60000 m3 Mole balance
V=FAO
X2
dX
Xl
(-rA )
f---
nd
Eqn . Of 2 line (sloping upwards)
y -100000 =1.3 x10 6 (x - 0.5) X2
60000 = f (l.3x10 6 (x -- 0.51) + 100000)dX
x
Xl
=> X2 = 0 ..746
P2-9 (0 Real rates would not give that shape. The reactor volumes are absurdly large .
P2-10 Problem 2-10 involves estimating the volume of three reactors from a picture., The door on the side of the building was used as a reference., It was assumed to be 8 ft high. The following estimates were made:
h = 56ft
d = 9 ft
V = nr 2h = 11:(45 fti(56 ft) = 3562 ft' = 100,865 L
2·18
Length of one segment = 23 ft Length of entire reactor = (23 ft)(12)(II) = 3036 ft D=lft V = m 2h = n(O.5 ft)2(3036 ft) = 2384 ft3 = 67,507 L Answers will vary slightly for each individuaL
P2-11 No solution necessary. P2-12 (a) The smallest amount of catalyst necessary to achieve 80 % conversion in a CSTR and PBR connected in series and containing equal amounts of catalyst can be calculated from the figure below.
CSTR PBR
Conversion~
X
The lightly shaded area on the left denotes the CSTR while the darker shaded area denotes the PBR This figure shows that the smallest amount of catalyst is used when the CSTR is upstream of the PBR
See Polymath program P2-12.pol.
P2-12 (b) Calculate the necessary amount of catalyst to reach 80 % conversion using a single CSTR by determining the area of the shaded region in the figure below .
2-19
60
2
,4
6 Conversion, X
8
10
The area of the rectangle is approximately 232 kg of catalyst.
P2-12 (C) The CSTR catalyst weight necessary to achieve 40 % conversion can be obtained by calculating the area of the shaded rectangle shown in the figure below.
60
10 _____ ~ __-LI____~_____ 4 10 6 8
~
2
Conversion, X
The area of the rectangle is approximately 7,6 kg of catalyst
P2-12 (d) The catalyst weight necessary to achieve 80 % conversion in a PBR is found by calculating the area of the shaded region in the figure below,
2-20
6
Conversion, X
The necessary catalyst weight is approximately 22 kg .
P2-12 (e) The amount of catalyst necessary to achieve 40 % conversion in a single PBR can be found from calculating the area of the shaded region in the graph below.
60
f
1;;
40
If
<;J,-1...~30
u.
I
20
10
2
.4
.8 6 Conversion. X
1.0
The necessary catalyst weight is approximately 13 kg.
P2-12 (1) 2-21
1.0
~ ~
0. 8 0.6
04 0.2 0.0
0. 0
6.-1
12..8 \V 19. 2
25 . 6
32.0
P2-12 (g) For different (-rA) vs . (X) curves, reactOIs should be arranged so that the smallest amount of catalyst is needed to give the maximum conversion.. One useful heuristic is that for curves with a negative slope, it is generally better to use a CSTR Similarly, when a curve has a positive slope, it is generally better to use a PBR
CDP2-A(a) Over what range of conversions are the plug-flow reactor and CSTR volumes identical?
We first plot the inverse of the reaction rate versus conversion. 10" r
8
A /
/' /
/
/
o ,. -.........-..-.---.-,.-.. . -.. --.-------,--.. .--.--.-----,-.. .-.-.. ----..-.. . .,.----.---,-.---.. o
0.2
OA
,)_6
0 .. 8
1.0
Conversion X
Mole balance equations for a CSTR and a PFR: CSTR: V = FAOX -- r A
PFR: V=
x dX
S-
0-- r A
Until the conversion eX) reaches 0.5, the reaction rate is independent of conversion and the reactor volumes will be identical. 2-22
. vPFR = 05f -dX- = F-AO- 05fdX --
1.e.
o - rA
-
rA
0
F X AO
-
rA
- VCSTR '-
CDP2-A(b) What conversion will be achieved in a CSTR that has a volume of 90 L?
For now, we will assume that conversion (X) will be less that 0.5. CSTR mole balance:
V = !AO X = voC AO X - rA - rA
V X =---=. V C 3 o AO 5 m - rA S
0.09 m 3 =3xlO-13 l 3 X 200 mo X 3 X 108 !'1 .~ m3 mol
CDP2-A (c) This problem will be divided into two parts, as seen below: 8
l()~
-'A (m3 •• '\ ,-
, ---··--1 ! mot 1 )
0.2
{I.4
0,6
1.0
Conven,ion X
•
The PFR volume required in reaching X=O.5 (reaction rate is independent of conversion).
•
The PFR volume required to go from X=O..5 to X=o.,7 (reaction rate depends on convelsion).,
2·23
lO
= 8*10 m
3
Finally, we add V 2to VI and get: Vtot =VI + V2
=2.3 xlO 11 m 3
CDP2-A (d) What CSTR reactor volume is required if effluent from the plug-flow reactor in part (c) is fed to a CSTR to
10' r
l'
1113
1/
/\
~,~
ruol
f
2
I)
C;onversion X
raise the conversion to 90 %
We notice that the new inverse of the reaction rate (I/-rA) is 7*108. We insert this new value into our CSTR mole balance equation:
CDP2-A (e) If the reaction is canied out in a constant-pressure batch reactor in which pure A is fed to the reactor, what length of time is necessary to achieve 40% conversion? Since there is no flow into or out of the system, mole balance can be written as:
_dNA
Mole Balance: rA V
- --
Stoichiometry: N A
= N AD (1 -- X)
dt
2-24
dX
Combine: r AV = N AO - -
dt
=
From the stoichiometry of the reaction we know that V Vo( 1+eX) and e is 1. We insert this into our mole balance equation and solve for time (t):
Vo
dX
N AO
dt
-rA -(1+X)=-tSdt - C XS __d_X__ o
-
AO 0 -
rA (1 + X)
After integration, we have:
1 t =--CAoln(l + X) - rA
Inserting the values for our variables: t = 2.02 X 1010 s That is 640 year'S.
CDP2·A (f) Plot the rate of reaction and conversion as a function of PFR volume.
The following graph plots the reaction rate (-rA) versus the PFR volume: Ii' ~..
~"~"".'"
"''''''''''' " '" ""''''''''''''''' '"'''''' '
il
Reaction Rate (--rA) ver'sus Reactor' Volume (V)
4 . 0*10-
'I
II-'A 3O'10_9"--"--------~ 11[~o:s) 2 .. 0*10-9_
------
-
.'-..._.•. ......_,"'_.","",.
1 . 0*10-9
o o
- - - - t - - -___----,>----_,--_, 1 . 0*10"
2 . 0*10"
3 . 0*10"
4 .. 0*10"
5 . 0*1011
V (m 3 )
Below is a plot of conversion versus the PFR volume. Notice how the relation is linear until the conversion exceeds 50%.
2·25
Conversion (X) versus Reactor Volume (V) 1 .. 0
0 .. 8
x
0 .. 6
0 .. 4
0. 2
0--1<----_--;,----..-,-----t,----o, o
1 . 0*10"
2.0*10"
3.0*10"
4 .. 0*10"
5 .. 0*1011
V (m 3 )
The volume required for 99% conversion exceeds 4*1011 m 3 .
CDP2-A (g) Critique the answers to this problem. The rate of Ieaction for this problem is extremely small, and the flow rate is quite large. To obtain the desired conversion, it would require a reactm of geological proportions (a CSTR or PFR approximately the size of the Los Angeles Basin), or as we saw in the case of the batch reactor, a very long time .
CDP2-B Individualized solution CDP2-C (a)
o
0.1 O~ a.3 0.4 ~ e.G 0.1 0.1 o.~ COfivtrll01l. X
!a.}
For an intermediate conversion of 03, Figure below shows that a PFR yields the smallest volume, since fOI the PFR we use the area under the curve . A minimum volume is also achieved by following the PFR with a CSTR In this case the area consideIed would be the rectangle bounded by X =0.3 and X = 0 . 7 with a height equal to the CAr/-rA value at X = 0.7, which is less than the area under the curve .
CDP2-C (b) 2-26
so t/min v ... Vo I. 'Where
v o ....
,0 .3
I
,.
-C
J_
+ <0.1-0.3) -
o
AO
-%A
*"
-+
1
2'
(0
X-O.1
.3-
(15)
"" IS :::c.in So V .. v .. 1:
III
(SO l.J:ad.n) (15 lJdn) .. 75,0 1
III
750
=.t.o
CDP2-C (C) The $~allo$t ~:O& caa be achieved by asi~, ouly Qua CSTR with this systom. C"I'I I ... (O.1--{).Ol....u::.;.
* (0.7-0)(15) - 10.5
-r).
mi~
'1-0.7 So
V - v
o
I .. (SO) Ihniu (10.5) lAiu -
525 1
We Yo~ld further roduco the tot~l vQl~e by usin& a PFR at first up to the
CDP2-C (d)
2-27
to t11(1
of tlu rcctan.;lc up to tho specified c;ouvcr:Jion.
&::C&
I '" (O.45-0)
V ~ v
So
o
eM -r
(0.45-0)(37) - 16.55
~in
A :X ..O.1
I . (50 I/min}{16.65 min.)
832.5 1
u
For thePFR,
o. 4S
I ..
fo
4(48) ... 2(43) + 31)
.. So V
..9.....Sli 3
(10 ... 4(1.5 ... 3S ... 43 ... 48) ... 2(20 + 43 ... 50 + 43) ... ]1) ... :1.5. n
v I -
p
o
There is also
(50 I/min} {15.12 miu) - 186 1 ~
solution at au X>O.1
Tr:r I. .. O.S
For the PFR
+ 4( 15) .... 33)
""
For
.!hl. 3
tl:a
UO ... 4'\ 20 +- SO +- 3Z .... 1.3) ....
~H43 .... 43 .... 1.1} ... 33) ...
CSU
I • (O.19)(30) - 13.7 min So V ...
'V
c
t ... (SO
1/;:dn)(23.7 min) ... 11S! 1 Cdm~)
2·28
Z3.9
For t:.h.a PFR.
I - 23.9 -
~ ... (O.S-Q.79)(30+33) - 23.9 - 0.315 - 23.58 mi:
So V .. v I '" (50 I/mill)(23.SS mal -1179 1 o o.~
diffc:rcu:u:o
CDP2-C (e)
x CA0/-:r A{J:).inJ -;(minJ
0.0
0.1
0,,2
0.3
0.4
0.5
0 .. 6
0.1
0.8
10
20
43
SO
43
32
11
15
33
0.0
l.O
S .. t)
15.0
17.2-
16.0
10,.2
10 .. 5
2,6.4
3CI
1$ 20
o
0.1
02
0.4
X
v
"= . - , .
7'00 1.
50 If::r.i.a. "" 14 m.in.
2-29
0.7
CDP2-D Data taken at WI.;; kPa (10 atrn) and 227"C (500.2 K)
y/\O
(0.333XlO) ==. 00811"-)'d-' c~ ; ; ; .y,,(}p __ ...-;;:;; ---------,J gInO I m
=0333
Ao
RT
.. -...
100000
~
(0,082X500.2)
~"'--"""----'-
.
-l
~,-- -------.~-~-.
! 60000
I
40000
... -···········-·······----1
I I
·-----·--···------·-·-··-~i
zoooC)
t
() -~---.- .... -.- ......- ... --... ---.~..-----..-.-.- ....-•... -.---.-...
o
0.2
x
l
--.-.
0,6
0.4
CDP2-D (a) 30% conversion in PFR: , -C pfR ;:;:
C AO
0.3
cLX
5--....... :;: ;: 4,664 . 84
s
V --
=?
f
~64 .84 s,
V o "£ -- (4 \' 0
0'-' fA
1
min }2 m '/mm .) ····6······.:·-.·--
, Os
- - m3 = L5 ).)
CDP2-D (b) 30 to 50% conversion in CSTR:
c
(X,·-·X)
1:CS'IR ;;;:; .... _Au, .. : ............... !...: .... IAl
:;::;: 12,169.2 s
=?
"1 Y
CSIR ::::::
CDP2-D (C)
2-30
1 (12 ,169 min ]2 In 3/..) - 64 m 3 . _')S{' -....... IIlln =40 3. 60s f
Total Volume: VTouu == 155.5+405.6 == 561.1 m
J
CDP2-D (d) 60% conversion in PFR: 1:PFR ::-:.
C At>
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
<.,.------,--."'~""."""
Q,.8
1.0
0.0 0.2 0.4
0.6 0.8
x:
. "". ."""" -"""'' ' -' ' ' ' ' ' -' ' ' ' -' 2-33
1.0
Case 2: PFR --+ CSTR PFR:
Area under curve:: 1667 From the graph - Xl ::;; 0.259
CSTR. :
/\rea::: 66.67 From the grapb . X 2
::;;
0.515
CDP2-F (b) Two 400 L CSTR's in series.
CSTRl:
V;;:;: FAQ(A.rea) mea;;:;; 66.67 From the graph - XI ::: 0.36
CSTR2 :
Area = 66.67 From the graph . X z = 0.595
(b) Two CSTR's in Series
(c) Two CSTR's in Parallel 500 1"'----------:Jr-. 400
500·~·--------------r~
400
. 30G
. 300
~
~
::: 200
::.; 200
100
100
o
o 0,0
0.2
0.4
0.6
0.8
LO
0.0
X
Two 400 L CSTR's in parallel. To each CSTR goes half of the feed :::;:;
6/2 ;:::: 3 moUmin
V::: FAo(Area) V
400
Area ::: .._,-- :: ------ = 13 3...3 FA" 3 From the graph:
0.6
X
CDP2-F (C)
FAo
0 . 2 0.4
X = 0.52
2-34
0 .. 8
1.0
CDP2-F (d) PFR : V ::: FAo (Area under curve) From the graph we can find the area under the curve for a conversion of 0.60: Area == (0 . 60)(300) :.::: 90 2 V = (2 mol1min)(90) = 180 L ~--~.-
-------
..
...................
-.-,-----..
".,.--~
- - -.......-.-.. .. .. ...........
............
~
,
"
(d) Single .PFR 500 T-----------.---.---.--~._, 400
.
.::: 300 -
200
1010
o 0.0
0.2
0.6
OA
0.8
1..0
X
CDP2-F (e) Pressure reduced by a factor of 10. A decrease in pressure would cause a decrease in the overall concentration which would in turn cause a decrease in CAo and F.-\,o" By looking at the design equation:
v::: !.",,,X "'[A
it is apparent that to compensate for the decrease in F.Aa there would be and increase in X.
CDP2-F (1) Use the graph of 1/.,(... VS. X to find values for all volumes. (Assume a flow rate of 1 mol/min.) Generate the following table and graphs:
...... "
X
-
0 0.1 ... ,,0.4
- - ; : ,........ "
i----
0. 7 0. 9
-r"
0.2
"·"'0.'01'67
V 0 -;-3 . 494
··0.0048·S-- -"4i984
=+. .
0.00286 0.00204
125.878 225. 088
2-35
1"""'"""" .. """'-'----,.,-'"'-'-'-"'t
! ~
Conversion
(f) i
""'---~-~~---'--"
-
"'S.
-------,
!
''flolume
(0-1.""
!
_",~ _ _ _ _ _ •
'IS.
Volume
DO! -,--.---'-------~
0&
0.015
0.,6
.; 0,01
0. 4
o
.,.,,_,_.,.,_...,__,-..-J
IL-_, _____0 ,__ , ,_,~ 100 V 200 __ "' CDP2-F (g)
0005
I
0,2.
300 ____
o ,..___ ,_..,._.. Q
,J
Individualized solution
2-36
~
100
__,___..J V
ZOO
300
Solutions for Chapter 3 - Rate Law and Stoichiometry
P3-1 Individualized solution. P3-2 (a) Example 3-1 0 . 008 0 . 007 - t - - - - - - - - - - - - 0 . 006 - t - - - - - - - - - - - - - - - - - - - - - - - # - - - - j 0 . 005
~ ~ 0 . 004 + - - - - - - - - - - - - - - - - - - F .iOi
0 . 003 0 . 002
+-----------~~-------
0 . 001 - t - - - - - - - - - =.......~---
o
.•....
. T
310
315
.....,.-++.... _................,
"
320
325
330
335
T(K)
For E = 60kJ/moi
k
For E1 = 240kJ/moi
= 1.32 X 1016 expC-- 6~~OJ )
kl = 1.32 X 1016 exp( -
2~~OOJ)
E=60 kj/mol E = 240 kj/mol 6000000 35E-22 3E-22
5000000
t---+------------i 4000000
25E-22 t---~._---------___j
g
2E-22
:;; 1 5E'22 1E-22
. . ...............................-_ •.•
"'2.
t----\--------------I t------\-----
t----~-=--
3000000 t - - - - - - - - " . . : : : - : - - - - - - - - j
-'"
2000000 t - - - - - -
t-------'b------1000000
5E-23 t--------"'c--~
t---.----------~-__l
o o 00295
0003
0 00305 0 0031
000315 00032 000325
000295 0003
0.00305 00031 000315 0.0032 000325
1fT (11K)
1fT (11K)
P3-2 (b) Example 3-2 Yes, water is alJeady considered inert.
3-1
P3-2 (c) Example 3-3 The solution to the example at a conversion of 20% would remain unchanged. For 90 % conversions of the caustic soda, the final concentration of glyceryl sterate is 0 instead of a negative concentration. Therefore 90 % of caustic soda is possible. bed P3-2 (d) Example 3-4 A + - B ~ -c + - D a
So, the minimum value of @B
a
a
113
= b/a = -- = 0.33 1
P3- 2 (e) Example 3-5 For the concentration of N2 to be constant, the volume of reactor must be constant. V = Vo. Plot:
1
0.5(1-0.14x)2
-rA
(I-X)(0.54--0.5X) 1/(-ra) vs X
180
160
140
120
f"100
::!:...
80
60
40
20
02
04
08
06
12
X
The rate of reaction decreases drastically with increase in conversion at higher conversions.
P3-2 (0 Example 3-6 For a given conversion, concentration of B is lower in flow reactor than a constant volume batch reactor. Therefore the reverse reaction decreases . C ro = constant and inerts are varied. N 2 0 4 *"72N0 2 A
~
2B
C
2
Equilibrium rate constant is given by: K c =.~ CA,e
Stoichiometry: £ =
YA0 6 = YAO (2 -1) = YAO
Constant volume Batch:
3-2
CB -- 2N AO X --2CAO X
Vo
Plug flow reactor:
= _~AO(l- X) = C Ao(l- X)
C
v
A
o
(1 + eX)
and
B
(l + EX)
Y;o;0o = YAo(O.07176)mol! dm
CAO =
C
=
2FAO X v o(l + eX)
= 2CAO X (1 + eX)
3
Combining: For constant volume batch:
C 2 4C 2 X2 -~Ao . K cCAe C AO (1 - X)
4CAO
For flow reactor:
C
4C 2
2
=_._~=
K C
K C (1 - X e)(1 + eX e)
X2
Ao
CAO (1 - X)(1 + eX)
CAe
4C AO
See Polymath program P3··2-LpoL POLYMA TH Results
NLES Report (safenewt) Nonlinear equations [1] f(Xeb) = Xeb - (kc*(1-Xeb)/(4*Cao»I\() . 5 = 0 [2] f(Xef) = Xef - (kc*(1-Xef)*(1 +eps*Xef)/(4*Cao»A0 . 5 = 0 Explicit equations
yao = 1 kc = 0.1 [3] Cao 0 . 07174*yao [4] eps yao
[l) [2]
= =
1 .....•.••..................
<:
0.9
+-------------------------:.7-~---i
0. 8
+------------------~--------......,-.mIIIJIIl''----------i
.§~ ~;~~-~:i -~~ -;-;:;~~~~~~_=--
-Batchl
:::J
-FIOW.J
Qi 0 . 6
oE
____
0.7
~
0. 5
~ 04
'3
Jr
0. 3 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0. 2 + - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1
0. 1 - 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1
o
...
o
·······1· .. ·........ ".. ," ... 1 .... · ........ , ... ·.. T.... · ................... ·.. T"·
0. 1
0. 2
0.3
04
.. ... " .. ' ..... "1' ........... " ....... ,•• 'r~ ........ ".' ..' ".. ·.. ·T·~ .. ' ..·.. · " .... n~ ..·T
........ '1.'''...
0. 6
0. 5
y inerts
3-3
0. 7
0. 8
0.,9
Yinert
Xef
Xeb
Yao
0.44 0.458 0.4777 0.5 0.525 0.556 0.5944 0.6435 0.71 0.8112 0.887 0.893
1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.05 0.044
0 0.1 0.2 0.3 0.4 - 0.5 0.6 0.7 0.8 0.9 0.95 0.956
0.508 0.5217 0.537 0.5547 0.576 0.601 0.633 0.6743 0.732 0.8212 0.89 0.896
P3-2 (g) No solution will be given P3-2 (h) 1
1
2
2
A+-B-7-C
Rate law: - r A =
kAC A2C S and
dm
2
3 1 kA = 25- - s ( mol J
~=.~.= rc
-1
-112
112
(dm 3
J2 kc =kB =12.5-1 s mol P3-2 (i) A+3B -7 2C Rate law: -
rA
= k A C A CB
at low temperatures.
At equilibrium,
C
K = ..---- Ce' C C I12C A,e
3/2
B,e
At "Iuil;b,;um, -, A = 0 , so we can sugge" th,t -
rA
= k A ( CA 1/2 CB 112 -
But at t = 0, Cc = 0 So the rate law is not valid at t = 0 .
3-4
~~ J
Next guess:
P3-3 Solution is in the decoding algorithm available separately from the author. P3-4 (a) Note: This problem can have many solutions as data fitting can be done in many ways. Using Arrhenius Equation For Fire flies: --- r;:::-_. T(in K) 294
298 303
-
11T
0 . 00340 1 0.00335 6 0.00330 0
Flashes/ min 9
In(flash es/min) 2.197
12.16
2.498
16.2
2 . 785
2.66
2.52
2.38
2.24
Plotting In(flashes/min) vs lIT, we get a straight line .
See Polymath program P3-4-firetlics.pol.
210L---~·----~----~--~
3.30E-:3 .332E-3 3.34E
h 3.36E-3
1
3.38E-3 3.40E-3
For Crickets:
11T
~2
x103 3.482
chrips/ min 80
In(chirPsi min) 4 . 382
293 . 3
3.409
126
4.836
300
3.333
200
5.298
T(in K)
5.3
5.1
-49
Plotting In(chups/min) Vs lIT, we get a straight line . -+ Both, Fireflies and Crickets data follow the Arrhenius Model. In y = A + BIT , and have the same activation energy.
·~.5
See Polymath program P3-4-crickcts.pol. 4..3
3-5
'---~--~--~--
3.HE-3 1.36E-3 3.39Eih 1.42E-3 3.45E-3 3.48E-.3
P3-4 (b) F or H oney!bee: T(in K) 1rr x10 3 3.356 298
3.,300 3,,247
303
rsoa-
20 , - - - - - - - - - - - - - - - - ,
V(cm/s)
In(V)
0.7
-0.357
1.4
1.8 3
0 . 588 1.098
o. s
Plotting In(V) Vs liT, almost straight lme. In(V) = 44.,6 - 1.33E4/T At T = 40°C(313K) V = 6Acmls At T = -5°C(268K) V = 0,005cmls(But bee would not be alive at this temperatme)
See Polymath program P3-4-bces.pol.
0.2
-0.4
-1.0 L-_ _ _ _ _ _ __ _ ___' 3.25E-3 ,'.2'7E-3 3.29E HIE-, 3.3.3E-3 3.36E-3 ~
~
1lr
~
~
P3-4 (c) For ants' 1rr x1Q"
V(cm/s)
In(V)
3,53
0,,5
-0,,69
293
3,41
303-
3,,30
2 3,4
0 . 69 1,22
6,,5
1.,87
T(in K)
-,---
283
---,,-,------'-
3,,21
311
~--
'------
20
r---------------,
1.4
--
o.S
'------"
0.2
Plotting In(V) Vs liT, almost straight line,
See Polymath program P3-4-ants.pol. -10L--~-----So activity of bees, ants, crickets and fireflies follow 3.22E-J .uSE-J J,J,tE1tr 3.41E-J Arrhenius modeL So activity increases with an increase in temperatme, Activation energies fOl fireflies and crickets are almost the same,
---:-c-
Insect Cricket Firefly Ant Honeybee
3.47E-33.53E-3
--
Activation Energy 52150 54800 --95570 141800
P3-4 (d) There is a limit to temperature for which data for anyone of he insect can be extrapolate, Data which would be helpful is the maximum and the minimum temperatme that these insects can endme before death., Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperatme it could be useless.,
P3-5 There are two competing effects that bring about the maximum in the corrosion rate: Temperatme and HCN-H2S04 concentration. The corrosion rate increases with increasing temperatme and increasing concentration of HCN-H2S04 complex, The temperatme increases as we go from top to bottom of the column and consequently the rate of conosion should increase" However, the HCN concentr'ations (and the
3-6
HCN-H2S04 complex) decrease as we go from top to bottom of the column. There is virtually no HCN in the bottom of the column, These two opposing factors results in the maximum of the cOIrosion rate somewhere around the middle of the column.,
P3-6 Antidote did not dissolve from glass at low temperatures. P3-7 (a) If a reaction rate doubles for an increase in lOoC, at T = T 1 let k = kl and at T = T 2 = T 1+ 10, let k = k2 = 2k • Then with k = Ae-EIRT in general, kl = Ae-E1R1i and k2 = Ae-E'RT2 , or l
k2-_e -~(*-*) kl
E
or
R
Therefore:
In (k2 J(T. (T. +10)) E = R--
(In 2)(7; (7; +10)) " 10
kIll
= R·---,-·
(7;-7;)
T. (T. + 10) = lOE 1
1
RIn2
which can be approximated by T
!OED 5
= ---RIn2
P3-7 (b) E
Equation 3-18 is
Dividing gives
k = Ae
RT
E( 1 I)
~2, = e -Ii Tz""-"1;"
, or
1
3-7
1.99 cal ] [ 273 K][ 373 K] E= mol K In(·050) = 7960 cal lOOK .001 mol [
E
A = k1e RT, = 10-3 min -I exp .
P3-7 (c)
= 2lO0min- 1 Cal 1.99_ ](273K) ( molK
Individualized solution
P3-8 When the components inside air bag are ignited, following reactions take place, 2NaN3 --7 2Na + 3N2 ... ·.. . , ' , ............. ' .. ' ,,(1) , ,(2) lONa + 2KN03 --> K20 + 5Na20 + N2 , K20 + Na20 + Si02 --7 alkaline silicate glass" ",.,.,(3)
5x rxn(l) + rxn(2) + rxn(3) = rxn(4) NaN3 + 0.2KN03 + 01Si02 -7 04Na20 + 1.6N2 + complex/l0., ... , ,(4) Stoichiometric table: -
r - - ' - - - ,--_.
Species Symbol NaN3 _ A KN0 3 B Si02
C
Na20 N2
D E
Initial NA
NABe
Change Final ---NAX NA(1-X) -O.2XNA N A( BB - 02X) -··OIXNA N A ( Be - 01X)
0 0
O.4XNA 1.6XNA
NA~B
Given weight of NaN 3 = 150g Therefore, no., of moles of NaN3 = 2 . 3
O.4XNA 1.6XNA
M wt of NaN3 = 65
1 moles of NaN3 requires 02 mole of KN0 3
=> Moles ofB, KN0 3 = 02(2..3) = 0.46 moles
Mwt ofKN03 = lOLl Therefore, grams ofKN03 required = 046 x lOLl = 46..5 g 1 moles ofNaN3 requires 01 mole of Si02, Moles of C, Si02 = 0.1(2.3) = 023 moles M wt of Si02 = 60,08 Therefore, grams of Si02 required = 0,23 x 60,,08 = 13,8 g Following proposals are given to handle all the un-detonated air bags in cars piling up in the junkyards: • Store cars in cool, dry, ventilated areas" • Avoid Physical damage of the bag in cal. • It is stable under ordinary conditions of storage" Decomposes explosively upon heating (over 221 0 For 105 0 C), shock, concussion, or friction, 3-8
•
Conditions to avoid: heat, flames, ignition sources and incompatibles.
P3-9 (a) dX
From the web module we know that - - = k (1 - .x) and that k is a function of temperature, but not a
dt
linear function. Therefore doubling the temperature will not necessarily double the reaction rate, and therefore halve the cooking time .
P3-9 (b) When you boil the potato in water, the heat transfer coefficient is much larger, but the temperature can only be 100°C. When you bake the potato, the heat transfer coefficient is smaller, but the temperature can be more than double that of boiling water.
P3-9 (C) No solution will be given
P3-10 (a) 1) C2H6
~
C2H 4 + H2
Rate law: -fA = kCC2H6 Rate law: -fA = kC C2H4 C~:2
2) C2H4 + 11202 -+ C2H40
3) (CH3)3COOCCCH3)3 ~ C2H 6 + 2CH3COCH3 ~
A
4) n-C4HIO
+
B
Rate law: -fA = k[C A- CBCc2IKcJ Rate law: orA = k[ CnC4 H 10 --CiC4 H IKe]
1- C4H lO
+-+
2C
lO
5) CH3COOC 2H s + C4H 90H +-+ CH3COOC4H9 + C2H sOH
+
A
B
+-+
C
+ D Rate law: -fA = k[CACB - CcCnlKcJ
P3-10 (b) 2A (1) (2)
+ B-- C orA orA orA orA
(3) (4)
= kC ACB 2 = kC B
= k = kCACB - 1
P3-10 (C) (1) C 2H 6
~
C2H 4 + H2
3-9
(2) H2 + Br2
(3) H2 + 12
---->
---->
2HBr
2HI
P3-11 (a) Liquid phase reaction,
o /"-
CH2-·-OH
I
CH2 - CH2 + H2 0·~ CHT-OH A +B--~ C 3 C AO = 1 Ibmol/ft C BO = 3.47 lbmol/fe
Stoichiometric Table· Species Symbol Ethylene A oxide -B Water
Initial C Ao=1Ibmol/ft3 C BO = 347lbmol/ft1,
----
Change -CAOX
Remaining C A= CAO(l-X) = (l·X) Ibmol/ft3
-CAOX
CB= C AO ( BB -X)
CAOX
=(3.47-X) lbmol/fe Cc = CAOX = X Ibmol/ft3 ....-
BB=347 f--
Glycol
'----
C
0
----
- - - - _.. _-----
Rate law:
-rA = kCAC B
Therefore,
-rA = k C~o (1-X)( BB -X) = k(l-X)(347-X)
At 300K
E = 12500 cal/mol,
X = 0 ..9,
0.1 3 3 Ilbmol.s = 0.0035ft Ilbmol.s 1000 CAOX (1)(0.9) -1 =--- = - - 2 - - - - - - - - - = 1000..56 s -rA (0.0035)(1) (1- 0.9)(3.47 - 0.9) 3
k = O..ldm ImoLs =--x35.315ft
'rCSTR
At 350K, k2 = k exp«E/R)(lIT-lIT 2»= O.0035exp«125001l.987)(11T-IIT2»= 0.071 dm3/moLs Therefore, 'rCSTR
CAOX -rA
= -- =
(1)(0.9) (0.071)(1) (1- 0.9)(3.47 - 0.9) 2
P3-11 (b) Isothermal, isobaric gas-phase pyrolysis, CZH 6 C2H 4 + H2 A ~ B + C Stoichiometric table:
3-10
_-1
-
49.3s
0 0 FTO=FAO
B C
C 2H 4 H2
+FAOX +FAoX
FB=FAoX Fc=FAOX F~FAO(1+X)
e =Yaob= 1(1+1-1)= 1 v = vo(1+ eX) => v = vo(l+X) CAO = YAO CTQ = YAO -
P
RT
(I)(6atm)
=
( C = A
0.082 m3atm
FA v
K.kmol
3
= 0..067 kmol/m3 = 0..067 mol/dm
J(1100K)
= FAO(l- X) = C
(1- X) AO (1 + X)
v0 (l + X)
mol/dm3
FAO(X) X 3 =CAO - - - mol/dm vo(l+X) (I+X)
FB v
CB = - = -
Fe v
Cc = -
=
FAO(X) vo(l+X)
=CAO . -X- - mol/dm3 (I+X)
Rate law: -fA
(I-X)
(I-X)
(I+X)
(I+X)
=kC AO - - - =o..o.67k---
= kCA
If the reaction is carried out in a constant volume batch reactor, =>( e = 0.) 3 C A= C AO (1-X) mol/dm3 C B = CAO X mol/dm C c = CAOX mol/dm3
P3-11 (C) Isothermal, isobaric, catalytic gas phase oxidation,
1
C2H4 + - O2 ~ C2~0
2
A
1 2
+ -B
~
C
Stoichiometric table: Species
--- .-::::------,--"------
Entering F AO F Bo
Change -FAOX
Leaving FA=FAO(l-X)
O2
Symbol A B
-BBFAOX
FB=FAO ( BB -X)
C 2H 4O
C
0.
+FAOX
Fc=FAoX
C2~
3-11
_ _ P _2 ( 6atm ) _ mol CAO - YAOCro - YAO-- ( 3) -0.092--3 RT 3 0.082 atm.dm (533K) dm mol.K FA
C - -
A-
V -
FB
C - -
B-
FAo(1-X) CAO (1-X) 0.092(1-X) - --~---.:.. Vo (1 + eX) - (1-0.33X) - (1-0.33X)
-
V
FAo(eB-~) - --'-----'0.046(1-X)
-
vo(1+eX) - (1-0.33X)
-
Fe C = -- = e
FAoX 0.092(X) = ---'---"vo(1+eX) (1-0.33X)
V
If the reaction follow elementary rate law
_ {0.092(1- X)}{0.046(1- X)}O.5 => -r -k .
A
(1-0.33X)
(1-0.33X)
P3-11 (d) Isothermal, isobaric, catalytic gas phase reaction in a PBR C6H6 + 2H2 ~ C6HlO A + 2B ~ C Stoichiometric table: Change Species Symbol Entering FAO Benzene A -FAOX B FBO=2FAO -2FAOX H2 C
C6H lO
1
-- --
0
FAOX
(1)
P RT -
3
CA
=(
Fc=FAOX
-,-
3)
vo(1+eX)
-1)
6atm ( 3 = 0.055mol / dm 0.082 atm.dm (443.2K) 3 mol.K
_ FA _ FAo(1-X) _ CAo (1-X) _ 0.055(1-X) ------
---
V
FB=FAO( BB -2X)
2
e=y, AO J=-(1-2-1)=-,3 3 CAO = CroYAO =
--------
Leaving FA=FAO(1-X)
(1-~X)
(1-~X)
3-12
= FB = FAO (BB -2X) = O.l1(I-X)
C
v
B
C
V
= Fe =
c
V
(1-~X)
o (1+£X)
FAoX __ = CAOX
=
0.055X _
(1-~X) (1-~X)
vo (1+£X)
If the reaction follow elementary rate law. Rate law:
·-rA'= kCAC;
(1
X)3
-rA'= 0.0007k----3
(1- ~ X)
For a fluidized CSTR:
W
= FAoX. -rA '
W
=
FAoX
_
(1- X)3
0.0007k--"-3
(1-~X )
mol " - - at300K kgcat min atm 3
k=53 -
!.J)
k = kj exp(E(l.. __ R ~ T
=
53exp(80~9.Q_(_I_""_-I-_)J= 1663000 mol 8.314 300 443 kgcatminatm
FAO = CAO * Vo 3 Vo = 5 dm /min
atX= 0.8 W = 0.0024 kg of catalyst
3-13
3
P3-12 1
+ -B
A
2
-7
C
Stoichiometric table for the given problem will be as follows A ssummg gas plhase
Species
Symbol
C2H 4 O2 N2 C2H 4O
A B I C
8B
-
1 2 FAO
1
e
=---=-
FAO
Entering FAO FBo = eBFAO FI = elFAO 0
2
I
Leavin~
Change -FAOX -112 FAOX
FAoCI-X) F AOce B- x/2) FAOel FAOX
-----_.
FAOX
=_FIO F
= 0.79 F
FAD'
0.21
10
BD
=?
e =e I
F
YAO = -.:iQ.. = 0.30, FTO
C AO
= YAO P = 0.041!!1013 RT
F
C
=_AO A
dm
=C
V
'AO
(1-X) (1 + EX)
0.041(1- X) 1-- 0.15X
(!--~-X)
_ FB _ 2 2 0.020(1---- X) CB ---CAO ---------V 1-----0.15X 1-0.15X C =J:c = CAO~ 0.041X c V 1- 0.15X 1- 0.15X
P3-13 (a) Let
C = Nibroanaline D = Ammonium Chloride
A=ONCB B=NH3
A+2B
~
C+D
3-14
B
0.79 0.21
= 1.88
P3-13 (b) Species A
Entering FAO F BO = 0 BFAO =6.6/1.8 F AO 0 0
--
B
c--:::------C D
---
Change -FAOX -2 FAOX
FAOX FAOX
---
Leaving FAO(1-X) FB= FAO C0B-2X) Fc=FAOX Fo=FAOX
P3-13 (C) For batch system, CA=NAN
P3-13 (d) -rA = kCACB FA = _~A = _~A = ~:O (1- X) = CAO (1- X) , CA = !A = FA = CAO (1- X) Vo Vo V Vo FB
=-; =-v,B =- T:O(BB-2X)=CAO(BB-2X), Vo N
N
N
o
F
CB =-1L=CAo (BB- 2X ) Vo
-rA = kC~o (1- X)( BB -2X) BB = _~BO_ = 6.6 = 3.67 CAO 1.8
CAO = 1.8i'!!lol 3
m
FrA =k(1.8r(1-X)(3.67--~~)] P3-13 (e) 1) At X = 0 and T = 188°C = 461 K
kmOl)( 6.6kmOl)
3
2 -rAO = kCAoB B = kACAOCBO = 0.0017
E -rAO -
= 0.0202
m . ( 1.8--kmolnun m3
krnol 3
.
In nun -
2) At X = 0 and T = 25C = 298K
3-15
-
m3
k~ ko exp( !(:o + nJ m3
k =0.0017
[ 1.987
kmol.min
.~ (461
112735!il 1 1 mo - + -
exp
1
298)
mol.k
= 2.12xlO-6 -lAO
m3 kmol.min
= kCAOC BO = 2.52 X W
3)
k
5
3
kmoJ/m min
t- nJ
~koexp[ !(
11273~ ( 1
m3
k =O.0017---exp kmolmin 1.987 cal
461K
1 561K
J
molK m
3
k=0.0152--kmolmin -rAO = kCAOCBO -rA
=0.0152-~~-. (1.8 kmol)(6.6 kmol) 3 3 kmolrmn
m
m
kmol -rA = 0.1806-3- . m rmn
I' - - - - - - - - P3-13 (f) lA
= kC AO\1-X)(e B-2X)
At X = 0.90 and T = 188C = 461K l)att=188C
- _. r
A
= (0.0017
--.!!!~-J(1.8 km~!_)2 (1- 0.9 )(3 . 67 - 2(0.9)) kmol.rmn m
= 0.00103 kmol 3 m min 2)
At X = 0.90 and T = 25C = 298K
3-16
- rA = (2.12XlO-6
-m~J(1.8 km~1)2 (1-0.9)(3.67 - 2(0.9)) kmol.rmn m
= 1.28 X10-6 kmol m 3 min
3)
At X =0..90 and T
- rA
=288C = 561K
= (0.0152
km~1)2 (1-- 0.9 )(3.67 - 2(0.9))
m3. J(1.8 kmol.rmn m
=0.00333 __ kmol 3
m min
P3-13 (g) 3 3 Vo = 2m /min = 0..002m /min 1)Fof CSTR at 25C -fA =
= 1.28 X10-6
kmol 3 . mmm
- r A : X =09
.9.0.02m3 /m~nX1.8km~!/m3 xO.l
= 281.25m 3
1.28xlO- 6 _ kmol_
m3 min kmol 2)At 288C, -rA = 0.00333- 3 . m rmn
v = v o CAo (I-· ~) - rA :X =O.9
V = 0.002!11 /m~?<1.8kmollm3_~JJ...:..! = 0.108m 3 3
0.00333 kmol m 3 min
P3-14 C 6H 120 6 + a02 + bNH3
---+
C(C44H7 3N0860L2) + dH20 + eC0 2
To calculate the yields of biomass, you must first balance the reaction equation by finding the coefficients a, b, c, d, and e.. This can be done with mass balances on each element involved in the reaction. Once all the coefficients are found, you can then calculate the yield coefficients by simply assuming the reaction proceeds to completion and calculating the ending mass of the cells.
P3-14 (a) 3-17
Apply mass balance For C 6 = 44c + e For N b = o..86c Also for C, 6(2/3)
6 + 2a = L2c + d + 2e 12 + 3b = 7.,3c + 2d
ForO ForH
= 4,4c which gives c = 0.909
Next we solve for e using the other carbon balance 6 = 44 (0.,,90.9) + e
e=2 We can solve for b using the nitrogen balance b = o..86c = 0..86* (0.,90.9) b = 0.78 Next we use the hydrogen balance to solve for d 12 + 3b = 7..3c + 2d 12 + 3(0..78) = 7.3(0.,,90.9) + 2d d =3.85 Finally we solve for a using the oxygen balance 6 + 2a = 1.2c + d + 2e 6 + 2a = L2(o..9o.9) + 3.85 + 2(2) = 1.47
a
P3-14 (b) Assume 1 mole of glucose (180. g) reacts:
Y e/s= mass of cells / mass of glucose = mass of cells /180. g mass of cells = c*(molecular weight) == 0.,90.9 mol* (9L34g1mol) mass of cells = 83,.12 g
Y e1s Y els
= 81.12 g /180. g
= 0.46
Y e/02 = mass of cells / mass of O2 If we assume 1 mole of glucose reacted, then 1.47 moles of O2 are needed and 83 . 12 g of cells are produced"
mass of O2 = L47 mol * (32 glmol) mass of O2 = 47,0.4 g
Y e/02 = 83.12 g /47,0.4 g Y e/o2 =1.77
3-18
P3-15 (a) Isothermal gas phase reaction.
1 3 --N +-H2 ---7NH 3
2
2
2
Making Hz as the basis of calculation:
1 3
2 3
H2 +-Nz -'?- NH 3
1 3
2 3
A+-B-'?-C Stoichiometric table' Species
H2 N2
Symbol A
Initial F AO
B
' - - - - - 1----
NH3
C
Leaving FA=FAO(1-X)
FBO=BBFAO
change -FAoX -F AoXl3
0
+2FAoXl3
Fc=(2/3)FAOX
FB=FAO( 0B -Xl3)
--
P3-15 (b)
6=(~ -i-1)=--~ e= YAo 6 =0.5X( _
CAO
-
~)=-i
(16.4atm)
0.5 (
J 0.082 atm.dm~ (500K)
3
= 0..2 mol/dm
mol.K C H2
=C = A
CAO
(~_~~_Xl = 0.2(1- X) = O.lmol/ dm 3
(l+eX)
(1-})
2 CAo (l-X)
C
3
= C =--x NH3
C
(l+eX)
2
O.2(X)
3 (1-{)
=-x
=O.lmol/ dm
P3-15 (C) kNZ = 40 dm3/mol.s (1) For Flow system:
3-19
3
(1- X)
(I-X)
(1--})
P3-16 (a) Liquid phase reaction -? assume constant volume Rate Law (reversible reaction):
-rA =k[CA CB
_ Stoichiometry:
-~] K c
CA =CAO (1-X), CB =CAO (1-X), Cc =CAOX To find the equilibrium conversion, set -rA = 0, combine stoichiometry and the rate law, and solve for Xe
CACBKc =Cc
C~o (1- Xe
r
Kc
X: -(2+ CAoKC 1
= CAOX e JX e +1=0
Xe =0.80 To find the equiliblium concentrations, substitute the equilibrium conversion into the stiochiometric relations.,
3-20
mol mol CA = CAO (I-X) = 2 -3 (1-0.80) =0.4-3 dm dm mol ) mol CB =CAo (I-X)=2-3 (1-0.80 =0.4-3 dm dm CA = CAOX = 2 mo! *0.80 = 1.6 mo!. dm dm
P3-16 (b) Stoichiometry:
e= YAo t5=(1)(3-1)=2 and Be =0 NAN AO (I- X) (1- X) C ---C A - V - Vo(1+eX) - AO (1-t-2X) C = Ne = __.3NAO X = C 3X e V Vo (1 +eX) 'AO (1+2X) Combine and solve for x.,.
K C C
.Q..-- Xe) =[c 3Xe ]3 AO (1+ 2X e) AO (1+ 2X e)
Kc (1- Xe)(1 + 2Xe)2 = 27C~oX:
J
27C~o X 3 +3X +1=0 - 4+--K e e ( e
Xe = 0 ..58 Equilibrium concentrations:
Po lOatm mol C =--=--------------=0.305--AO RTo (400K) 0.082 m 3 atm m3 molK
(d
CA
_ -
C = e
J
d
(1-0 ..58) _ mol 0.305 ( ( )) - 0.0.59--3 1+2 0.58 dm 3(0 ..58)(0.30.5) mol =0246-·3 (1+2{0.58)) . dm
P3-16 (C) Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction.,
3-21
Stoichiometry:
C =NA =NAO (1-X)=C (I-X) A V V AO o
C =Nc =3NAO X =3C X c
V
AO
V
o
Combine and solve for Xe
KCCAO (1- Xe) = (3C Ao X e )3 Xe
=0.39
Equilibrium concentrations
CA
mol =(0.30.5)(1-0.39) =0.193
Ce
=(0.30.5)(0.39) =0.36--3
dm mol
dm
P3-16 (d) Gas phase reaction in a constant pressure, batch reactor Rate law (reversible reaction):
-rA =k[CA ._
C~]
K
e
Stoichiometry: £=
YAoO=(I)(3-1)=2 and Be =0 NAN AO (1- X)
(1 - X)
C ---C A - V - Vo(1+£X) - AO (1+2X) C c
= Nc =_l!'AO X --=3C V
Vo (1+£X)
X
AO (1+2X)
Combine and solve for Xe:
KCCAO (1- Xe) 1+ 2X e Xe
= [3CAO X e ]3 1+ 2X e
= 0.58
Equilibrium concentrations: 3-22
C =
0.305(1·-0.58) 1+2(0.58)
A
C =
3(0.305)(0.58)
c
1+2(0.58)
mol =0.059--3 dm mol =0.246-3 dm
P3-17 Given: Gas phase reaction A + B -7 8C in a batch reactor fitted with a piston such that V = O.IPo
(ft3f
k=l.O---· lbmol 2 sec -rA =kC~CB N AO = N Bo at t = 0 Vo = 0.15 ft 3 T = 140°C = 600 0 R = Constant
P3-17 (a)
=
N AO =0.5 N AO +NBO 8=8-1-1=6 e = YAo8=3 YAO
T
and .-.- = 1,
Now
1'0
Therefore
v = lOVo2 (1 + eX) lOV
Po = lOVo ' and P = lOV
or
N
Therefore
3-23
AO
= ( YRT AOPo ) v:0
= 5.03*10-9 A
[1
-r
-
X]3
3
Ib mol 3
(1+3X)2 ft sec
P3-17 (b) V2 =Vo2(1+eX) 0.22 = 0.15 2 (1 + eX) X
= 0.259
-rA = 8.63*10- 10 }b mol ft 3 sec
P3-18 No solution will be given. P3-19 No solution will be given. P3-20 No solution will be given . --_..._ - - - - - - - - -
CDP3-A
3-24
w 6.5
InW 1871802
13 18
2.564949 2.890372
r
1fT'
300 310
0.003333
313
0. 003195
..........-.--..
---.--.--.-..i~w;;1rr
3.5 , . . - - -..- .._ ...-._.
0.003226
. ··-·---················__·· . ···1 •......_ .... _ I
3 2 ..5 ~
e
-
2 1.S
I
y '" -7120.4x + 2:5.593 Fr.O.S8$3
0.5
0+-,----.-o..(XX:I1S
' - - . - -.."'.. _ _.. , - . , - - - - - 1
0.0002
0..0032S 1trK
~." ....... ~,- .. " •••• , .. "" ..... ,
"".,.
~ ...... -~~ ................... - , - - " -
"......... ".".~ .... -"--"-,,~ .......... ,.... _ ............... ,....7"" .............. -. __ ........ .
From the graph: E= 7120 In W = --7120.4 W(4L5<>C)
*(_._ . -!. ,,,.--) 1- 25.593 == 2.95 41.5+273
=19.2cm I s
__._--------
---------------------------_.-_. CDP3·B Polanyi equation: E = C - a(-.1HR) We have to calculate E for the reaction CH3 0 + RBI' -7 CH3Br + Ro Given: dHR = - 6 kcaUmol From the given data table, we get 6.8 = C - a(17.5) and 6,.0 = C -- a(20) => 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<0.151:)
0 .. :18- f<0.15X)
!!O
c
0
+(l.15X
0.151:
H2 O
D
3 ... t' 0 .. lSI:)
N 1
I
°0.67
3 + t'0.151:)
0
0 .. 67
Totill
l'
1.00
+ 1/4(0 .. 15,I,)
1
In.itiol N2 ... 0.19(1-0.15) "'" 0.1.9(0.8S). 0.67
Initial 02 = O.a5-0.61~O_lg
lJ
~i
Pi ... YiP ...
ttr
~i P -
8.2
;J.t::l
(1+(0.151411; )
3-27
+ I/4( O .. lSl'.)
i.
P.
Catm)
l.
A
0.15(1-1.) 5 4'
0.18 -
'-------,
C.
l.
'''''-'''''''--
I-I 0.03 1+(0.15/411
1.23 1+{O.15/4]X
(0 036 - .Q~ X) /(1 + .Q...ll I)
(O.lSI.)
4
c
4
OJ1311+ O~5 X
O.I5X
1. ::t CO.HI) 1
a.lll!(l~ 1}
0.67
v ...
Ci
d
Vo (l..,O+B I) , ....
0 .. 1 1\1
Pi .... CiI::I ...
1
'Ill 1
S .,
0.1.1 - -:; (O .. Ul:)
0.15 X
~ I
8.2 ~:L
Ci
o .:l.,${ 1;-1:) II
Ill!
i
Pi
1 .. 23 (I-X)
0 .. 03 Ct.-X) Q
....... 6 n
"".I
_
0.03
o. '15, :x
"
1 .. 1(0 .. 11 ....
:x
(0.151:)
0.67
1.%3 X 1"","
0 .. 1.33 '0 .. 1.99 + 0.0011 ... 0 .. 2 (l ... 0 ~1.5 .x)
3-28
:r5 {Q .. UX) X
P'tO't'" 1.2 - 0 ... 313 I .. taI. 3)
Suu; u
d)
(l)
4A+5B-1'4C+6D
Mole balance for a PFR
elF • A dV
dF "... b . . = dV dF,c ...... dV
-r,
= -rc
dF v dV
Combine with equation 3,,40 and 3-48
3-29
CDP3-E A -+ 2B --)
C
-+-
D
a.)
Species
Symbol Initi..a1 moles
Ben:royl chloride
A
A.:rom.ooia
B
Be!".zy~e
C
Ammonium C1l1oride
D
b)
CAO
4!'!-
2.
D
2 gma-i/l AO IV "'"
N
A ... n _)
-
NAG NAoSs N Ao6 c NA o6 D
D
+""'" 2 :2
3-30
(llange
Final moles
Conceruration
--·NAoX
NAo(l-X)
N Ai)( 1-X).fV
-2.'""lAQX
NAO{~·2X)
NAoX NAoX
NAo(8 c +X) NAQ(8 D -+-X)
NAoC&g-1XW N Ab(6c.--+X)!V N A o(6D+XW
Species
Ammooia Benzo chloride
Symbol B
Entering
Olange
Exit
Fao
-FsoX
A
SAFBO
~FooXI2
Fao{1-X) F}3O(S KXI2)
... C
Conceruration F:ao(l-X)/vo F:ao(9 a .JlY'z,}/v0
9cFBO
FBO(ec+XI2)
FBo(ac+~/vo
0
eoFBO
FMXI2 Fsr:XJ2
fBO<9 n+X/2)
~B
Benzylamide Ammorrlum Chloride •
1)
:Moh.r floy :l:1.tcs considered rll.ther tha.n numb..,r of moles.
2)
e.1 ""
3)
Concentration found by dividing the e~i1:ing molar floy rate by the
F. IF.", , as opposed to N. INA 10.;:>0
~olnmetric
10
flo. rate
'IT
.0
•
"
CDP3-F Given. : A + B -) C taking placo. in .. sqU&xo: duct ..
T
F AO .. loS Ib molel sec. (~)
If B is at equilibrium in the gAS phase throughout the reActor Since 13 m&int&ins its oquilibriUll'l v&por prouure
throughout the re&ctor. it
is replaced by
&
&$
soon as 1 moloculo of B is consumed by the re&ctioll.J
lIIalccuh of B in tha liquid.
3-31
Ronco, 0 ... 1 - 1 • 0, ~ '" TAOQ
. .. .. (1:1)
.At 1: ,..
0.5
[- ( .25 a'tm) ( .75
.em f~3 ( .730 . .~ 11:1 lnole OR
3.'t1ll) ..........
-1
2 % lOOOOR)
1 'b.mo 1 e {O.S]" .114'; 3 •~t: ~An~ .W~_
CDP3-G a)
3 5i HC13 .+ 3Hz ~ 15i (S) + 7HCl + 5i H2 Ch Take Si HC13 as basis
Spede~
Symtx:!l
Emering
ClJani.'C
Si HO] (g) Hz Cg) H2HCl (g)
A
-FACiX
B
FAC f=Bo =&s FAo
C
0
+2FAoX
Si lhC1 z (g)
D
0
Si (s)
S
0
+}FAoX 2.FA oX·
~F}.D
3
3-32
L&mdDr; FA
n
= FAO (I-X)
=FAo(ea .. X}
Fe :::2FAOX
3 Po =lFAoX 3
Assume isothermal and constant pressure. Neglect the vapor pressure of Si(S}
as = 1 Stoichiometric feed o only involves the changes in gas phase E
= YAO 0 =:; .1 (1. -+- L - 1 - 1) 233
e=}tt)=t
CAO - Y 0 Po . - 1 ( 2,Q ) . A R Tn - 2 (O_OR2HI127\ ""
c" = £~,
:= CAO
\)
:=
Cc
= Fc.
1,)
tl
{l- ifL
(1 + EX)
g. ::::
Ca
"" _.
IJ
= 0.0088 (1- X)
C {Ss-=-X} :::: A (1 + EX)
== 7 .,.' F AoX _. 3 \)0 (1 +. EX)
I \ .
1'1- X/3
O.~88 (1·X) 1 + X/3
= 2 CAO __ ,_,X____ 3
(1
CD == F D .::::} l!~9.~ -=0.OO31 ....- ~-.-. v 3vo(1+c:X) 1+~
...
1he SOlUllon to parts (b) and (c)
and e)
a...""e similar to Part (a).
CDP3-H
3-33
'1-
£X)
::: 0 021 •
X (1 + XJ3)
$0 vc can use the stoichiome~ric tAble. i
FAO(l-X)
CAO{l-X)
B
4F AO(l--X)
4C
4F AO .l.
D
_ F 00 5
titer
~--
A
C
whor~
C.,.
F,
... ...
CAO
=
:x
FAC
3ubstit~ting C's for F's
AO 4C
{l"-X) AO
eAQ
:! X
5 CAO
0.0658 g~ol/l aDd FAD - 0.02631 Bruol/s
cQ~gell~a~ion
F t V"'v o -""'V F Q t.o
Not~:
at X - 0.5 (beginning of condensation) C.; c· i
1.
A B C
5FAirF A (}X FAO(5-X) 4FAOX ..... __"_--1?. ___. . . . _........._. 9:.1FAO(5·~?f)!Q. 9. Total FAO(5-X)+FAO(5-X)/9
3-34
CAo(l-X)(4.5)/(5-X) 4C A0(1-X)(4.5)/(5-X) 4C,,,oX,(4.5)J(S. -X) C Ao (5-X)(4_5)/(5-X)/9=O.5C ._-...,._..- .... .. A{) . C A o(5Xt/40.5
__ _-_
"
....
Pv 95 .--::w-P 9S0
u
0.10
Scoich Tl.ble
Spocies
DoCore Co ndel1s;1I.l:.i on Pp < P y r4111&ining
After COl1dens&tiol1
C1:l&ngo
FAOU-IJ
___ -._,----_ -.
In
Sy:ubol
........ ,.. ....
........
....
CB:4
A
FAe
,·,F X ,Ac
FA - FActI-X)
C1
2
13.
4F AO
-4FAcf.
Fa
liC1 4
C
+4FAOl:
Fe
eel.",
.0
0
D(g)
DO)
+FAOI
...
-
Fn
AO (1-X)
4F
4FAOA
Pp - Py rem&il1il11
AO (1-lJ
4F
4FAQ1O.10FT
.. FAOX
0
---" . ....
"'.","',,"-_
Fr ...
SF AO
Fro
When conde •• atioD first begins
Fr
=
5 FAa
.
'r
0.50
1)
AI' "'" AP "" .0
""..,.)
p
-
.0
--} e ,.. 0
Total concentration is consta,llt
o
" ,RL' -"" o
C.0820S atm lit X 348 K
gmol K
3-35
,----.
..
FT-SFAO-FAOX+O.10FT , F (S'-X) AO FT ... .0.90
CDP3-I
T+
473 K
200~C ~ tl
10E
:;:: "v::
P
506,S;:; 0203 2500 •
)toich t~bl/!l
In
Species
'" .....,..... __ ~ •... "~ '''_~""'''''''''
C2H6 BI2,
HBr C2F14B1 l
A B C D(g)
FAll 2FAO
o o o
Fi before F'i After condensation PD ::::: condensation PD Pv remaining < P remaining v .. _.... -.-' ...... _ ....... "....... " -...... •. _.... w'"....... __,....... ...-, .. _.,.". ........,_....... ,,,FA(;X FA : .:.: FAO( 1"X) FAO( IX) 2FAO( IX -2FAoX Fu ;:::: 2FAO( I-X)
Change
_~~
2FAoX FAOX
Fc == 2FA()..'X FD
Bafar. coadeas&tioa ---) e '" 0
3-36
=FAaX
,4. _ _ _ . _..... _ .. _ . o K "
2FAoX O.203F'r
P 'Clt
_._0
a-I
CAO •
o .. 0820S
40
1
3' <=-rIO
0.673 :d
••
'A
t
prAll.
.
t
24. • 7 . . '/... ttti -
= -
'JOn~u
3
&t=
.
lit
~
l:
1=01 I:.
0.673 gcol/l
413 I
,mol/l,. 0,.212 ,moll1
'.
pAll. l '~ll.-X)
'0 ...11.
IM1I.
a-lJ 1
11 J.&ll
';41
lPA.O
<::.40 u -%)
Fe
&IN".
C 4
'r
0.H16
.0.:1%
0
0
0.11.1
0.111
G.1
a,au
Of .Ur.)
0.041
Q.OU
Il.ua
0,4
13.1>44
0.111
0.o"
0.041
0.311
0.U1
o.'¢~
0.041
O.Ul
(I.U'
Q.Q'"
o.JU
0.01l
t
, .AQ(1.-11
l).fO!!)
a.IML
t.' toO
IF>\I.'l'thIl
o-n
.u:n
IFClIl;l'l
l'>\I.'l'l
F.w-r:Ti
'J.J::IJ.1;1
a.ou
O.t.1f
0 • .,(.5
O.ltl
O.OU
0.011
0.641
0.110
a.aH
0,037
1)
0
Q.lll
O.OH
Q.l"
\)
3-37
CAl)
O-t) \I
lCjD(l-Xl ~."1'.4
Q.llt
"'PC .10
A,Q
Q.134
0.:110
0.1"
O.lll
1"';-1211: \l' J ll ... X)
1e
"
Ml
d.l"
ttl~
i).1J'
,t"
e
U-l)
(Q ..
iq71
J..Q
0.1,:$ ,
O.Hl
F AO
=
v
CAO
=<
Q
F.
C
:0=
l.
-
(0.212) gmal/1 (0.5) lIs
F.
1-
V
'If
0
0.106 gmal/s
F.:l
1.
:::t
n
(l+f:l.)
... v
0
So we ca~ first use t~e stoichia~etric t~ble. SUbstituting C's for F's i
....
F.
C __ .~ _._:..L
(l.-:n
C el-I.) AO
._.__...•,.•..l...._ _
A
F
B
2F AO (1-·,X)
c
2F AO
-
D
Total
AO
.!
___:.10 1_ 3 F
2C
AO :lC
(1-I.} AO
-- ehQ
X X
.--
.3 CAO
AO
3-38
v .. v
F 'C '
o F
to
F.
C
1
i
""
". V
-::Ill
--1.
F.1. (2.392... v C3-IJ
2.
"Y
o
,.,. C(A) .... e(8)
C(9710111)
.. C(Cl C(O)
0.
OQ
0.1
Q.l
as
1),4
os
04 01 04 09
"L j --.--. OJ
I.Q
____~. .
-~
+
FCA}
<>- F(6)
F ((1no1lS) OJ.
,.. F(C)
""'Fm
0.1
O.O~~-::--
____-=::::~~ Qj
Q.4
V
04 U
1.0
X
CDP3-J
3-39
,13-X) o 2.39
Fso as =,,--=1
F AO
YAo :::;; 0 . 5 E
== Y ADO =0.5(1+ 12,.-3--4)= 3
_ C
C.~o(I
A --
- X) _ 6.18*IO·-&(1-X)
l+&Y- -~'-'i+3X
= 1.68 *~-a(~_ ~~
C
)
1+3X
8
C = 2.0~:'10-a X c • 1 +- 3X . 2.47 * 10-7 X C = -.D 1+3X CAO
= Y .... O(::1'O
=
o_i'It)= o.'){.'.···.---.·P;;-~~;3.'.... ".J= --l * 8:314··_··········
mol*k
_. _ C.-IO (1-, x) _ 6.18" 1{)8 (1 ' x) C. A. - -"--"'-"'" ................ -. - . _....................... _ ...... _ ......"........ -'-' l+EiX
• ('1J
1+3X
J
1.68 * 10-$(1 . . . ~~:. = --_.. _.-........................ . . . .:"..........................1+3X
2.06'" lO·-a__ ....x-C c = ............... - ..................... • 1·+ 3X c... = .2.47*107X. . . . . . . . . . . . . . . . . . _.-.. . . . . . . .. '0 1+3X 3·40
97.3K
6.18"" lO-lt
CDP3-K CA)
C QUllZ Q :t\l;;:a. t
~D:tIQl
el2
A
bilid m!.lh~ 1.0
CH4
B
0 • .5
c
0
Rel
D
0
l'ot31
T
1.5
CIUel2
(b)
0 "" 1 C
'"
-I-
YAO
2 - 1 I) '"
Q:r:~'.
~ 7(0)
A
~A
:m
:lI!
1-1
O.S(l-X)
-0.5:1
.5.1
0.51.
+.1
X
1.S
0
c
-400 m::dl,
.. <400
k C 1
A
CAO (l-l.AJ
~;:I. .. l.. :2
-:1
+{)
..... X" =- 1.5189 - } C1I el docs 1 z
-r
~
2 '" 0 • YAO ... 1.0
s,.S1:O': is lao,s ph"'$c Ulltil F
(c)
Chu,U
C (l-ll) AO
3-41
U,Q
t
cgn.d~n.:u
!. t
1 a 'bI ..
(d)
c.AO 3E....!...."" 1.5
.1.
O~og206
= 0.02724 (1-O.6) ~ 0.01090
,mol
3
W:zt -r
A
Klmpl ~3
(f)
A e::t:.p
k
-
A
"" 2
.:t
.10
(-
.$
Lz Rr"
~2.
d
s
gm.ol
2.
at:
gJ:llol
(--
-E E.
(
373.2
3-42
2S"'C
Solutions for Chapter 4 - Isothermal Reactor Design P4-1 Individualized solution, P4-2 (a) Cooking food (effect of temperature), removing of stains with bleach (effect of bleach cone,,), dissolution of sugar in coffee or tea.
P4-2 (b) Example 4-1 There would be no enur! The initial liquid phase concentration remains the same"
P4-2 (C)
Example 4-2
For 50% conversion, X = 0.5 and k = 0.311min- 1
. F AO =Fe - =6.137 - - = 12 .27 lb rna If mm
X
Also,
0.5
FAO
VAO
=C=
12.27 ft3 min
AO
Vo = VAo + VBo = 2VAo and also, VAo = VBo Vo= 24.52 fe/min Using Mole Balance,
~oX _ = 24.52xO,5 k(l- X) 0.311(0 ..5) = 592gal
V
=
=
78.93
This is less volume than Example 4-2 because the rate is higher..
P4-2 (d) Example 4-3 For P = 60atm, C AO = 0.0415 lbmol/fe
(C
= YAO Po = AO
RTo
60 ) 0.73x1980
Using equation E-4-3,,6, for X = 0.8 We see that the only thing that changes is CAO and it increases by a factor of 10, therby decreasing the volume by a factor of 10"
VOO
1
P
P4-2 (e) Example 4-4 New Dp = 3DoI4 Because the flow is turbulent
4-1
1)p2
fJ2
=A-
1
=0.0775- =0.lO33 0.75
1)PI
I
2.0.lO3 atm . 60ft 1ft 10atm
2fJo L Po
Now 1 - - - ~
0
,so too much pressure drop P =
2 I
= (-0.24)2
°
and the flow stops,
(0
P4-2 Example 4-5 For without pressure drop, conversion will remain same as example X = 0,,82. With Pressure drop, anew
= a o = 0.0037 kg- I lO
For decrease in diameter, Po = 2xO.01244[266.,9x2+ 12920.,8] = 33475lbtlft3 = 334.75(lbr/ft')xlI144(ft2/in2)xll147(atml( Ibtlin2» = 52.71 kPalm For turbulent flow: fJ ,_.'
_1_ and
a-.L => a 2 = aI to!
Poz
Po
1)p
X -
kCAOW va
(1 aw) 2
1)pI Dp2
_
4,6(0.,949)
-~~C:W(l_.9'W)- - + 4~6(0.,949) 1
Vo
2
X = 0,,8136·- virtually the same (2) Optimum diameter would be larger
a = 0.037x 52.71 = 0.0756kg- 1 25,,8
4-2
= 0. 15814atmlft
Dl a2 =a -P-=2a1 1D p2
aW = 1-a W = 1-(0.037)(27.5) = -0.0175 1
1 __ 2 2
Now l-uW < 0, too much pressure drop due to higher superficial velocity.
P4-2 (g) Example 4-6 For turbulent flow
1
a-Dp
a,
and
1
a·---
Po
~~( ~:-)( ;':)~~[ilG)~~
Therefore there is no change.
P4-2 (h) Exmple 4-7 For pressure doubled and temperature decrease ero = 2*PJRT and T == 688K See Polymath program P4-2--h.poJ. POLYMA T!l Results Calculated values of the DEQ variables Variable V Fa Fb Fe E T Cto Ft Ca k
ra Fao rb vo re X
Tau rateA
initial value ---0 2.26E-·04 0 0 2 . 4E+04 688 0.573773 2.26E-04 0.573773 213.40078 -70 . 254837 2 . 26E-04 70 . 254837 3 . 939E-04 35 . 127419 0 0 70.254837
minimal value 0 1.363E-·05 0 0 2.4E+04 688 0.573773 2.26E·-04 0.0236075 213 . 40078 --70 . 254837 2.26E-04 0 . 1189309 3 . 939E-04 0 . 0594654 0 0 0 . 1189309
maximal value -1-. 0E--~2.26E-04 2.124E-04 1.062E-04 2.4E+04 688 0 . 573773 3.322E-04 0 . 573773 213.40078 -·0 . 1189309 2.26E-·04 70 . 254837 3.939E-·04 35.127419 0.9395277 0 . 253044 70.254837
ODE Report (RKF45) 4-3
final value 1. OE-04 1. 363E-05 2 . 124E-04 1.062E-04 2 . 4E+04 688 0 . 573773 3.322E-04 0.0236075 213.40078 -0.1189309 2.26E-04 0.1189309 3.939E-·04 0.0594654 0.9395277 0 . 253044 0.1189309
Differential equations as entered by the user [1 J d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb [3 J d(Fc)/d(V) = rc Explicit equations as entered by the user [1] E = 24000 [2] T=688 [3J Cto = 2*1641/8.314rr [4] Ft = Fa+Fb+Fc [ 5] Ca = Cto*FalFt [6 J k = 0.29*exp(EI1.,987*(1/500-1rr)) [7] ra = -k*Ca"2 [ 8] Fao = 0,,000226 [9 J rb -ra [10] vo = Fao/Cto [11] rc -ral2 [12] X= 1-FalFao [13] Tau Vivo [14] rateA=-ra
= = =
3,Oe-4..-------
.2 .4e-4
O.OHO OOHO
2.0<,-5l.Oe-5
v
............. .., .........
_--".......... .
60e-5
S,Oe-5
""
1.0e-4
P4-2 (i) Example 4-8 Individualized solution, P4-2 (j) Example 4-9 Using trial and error, we get maximum feed rate ofB Q.,Olmol/dm3 ,
= O,,0251dm3/s to keep concentration ofB
See Polymath program P42-j.,pol. POLYMA TH Results Calculated values of the DEQ variables Variable -t ca
cb
initial value
o
minimal value
o
0,,05
0,0063485
o
o
maximal value 500 0.05 0.009981
4-4
final value 500 0.0063485 0,,009981
0 0 0 0 0 . 22 0.22 k 0.0251 vOO 0.0251 0.025 0.025 cbO 5 5 vO 0.05 0.05 caO 0 rate 0 5 5 v 0 0 x equations as entered by the user [1] d(ca)/d(t) :::: -k*ca*cb-vOO*ca/v [2] d(cb)/d(t) :::: -k*ca*cb+vOO*(cbO-cb)/v [ 3] d(cc)/d(t) :::: k*ca*cb-vOO*cc/v [4] d(cd)/d(t) :::: k*ca*cb-vOO*cd/v
0 . 0078965 0 . 0078965 0.22 0.0251 0 . 025 5 0.05 3.91E-05 17.55 0 . 5543321
cc cd
Explicit equations as entered by the user [1] k:::: . 22 [2J vOO:::: 0.0251 [3J cbO:::: 0 . 025 [4J vO:::: 5 [5J caO:::: 0.05 [6 J rate:::: k*ca*cb [7J v:::: vO+vOO*t [8] x:::: (caO*vO-ca*v)/(caO*vO)
If the concentration of A is tripled the maximum feed rate becomes 0..064 dm3/s
0.0078965 0.0078965 0.22 0 . 0251 0.025 5 0.05 1. 394E-05 17.55 o . 5543321 Differential
0.010 0.008
GJ
0.006 0.004 0.002 0.000
0
100
200 t
300
400
500
P4-2 (k through r) Individualized solution. P4-3 Solution is in the decoding algorithm given with the modules. P4-4 We have to find the time required to cook spaghetti in Cuzco, Peru.
Location Ann Arbor Boulder Cuzco
Elevation (km) 0.21 -- 1.63 3.416
--
Pressure (mm Hg) 739 625 493
~-
Assume reaction is zero order with respect to spaghetti conversion: -E
de
-r =k=Ae RT = ___ A A
dt
so that 4-5
Boiling Point (0C) .. 99.2 94.6 88.3
Time (min) 15 17 ..?
For complete conversion (i.e . : well cooked) CA = 0 at time 1. Therefore -E
CAO =tAe RT C -E ·~=teRT
A
In ( CAO ) A
= In k = In t _ E ~
R~ Now, plot the natural log of the cooking time versus liTb and get a linear relationship . Extrapolation to T b = 88 ..3°C = 36145 K yields t = 21 minutes.
I 2l-L I I
I
Ull t}
11~1
I lS-t
I I 2.69
::'.72
2.767
P4-5 (a)
4-6
e Aa :;:;:; coo;;;; 2: !il;mtld:ft~1>
kl!l!i,
VA;;;; VB =;S drt1:'ltttif~
E. . . ;;;; ZO..COO cillmol
0;01
TI=:lOOK
'LTsi'll.g the
t\lirheniu:s
eqUi.lillc·l} .!i. tll¢
CSTR rernpeli'il'i.WRof 3,SOK y~~!ds, tbe: new sp!l!tifi(:
reactio'll rate.
CSTR tleslf:,";o Btruatilll1::
.".
" \' y\ - r,<,,}('r:: ;"',1'
rt r'lI
_~=I,
X = V ( kC~o (1_. X)2) 10
From the :quadratic ~qu~tion:
PBR
V=800 dm3
T=300K
Design Equa.tion:'
dX
F"o'-=-rA dV
~___ kC~oV 1-- X
FAD
X=O,,85 So, considering the above results, we will choose a CSTR.
4-7
P4-5 (b) Batch Reactor V=200dm' NAO=NBO=200 moles X=0.90 Assume Isothermal Design Equation:
t=NA O
rX dX
-lJ -rAV
r
t
=
t
dm'. J(lmal)2(200dm3) mal * nun dm 3 = 1.06 min
(200males)
(
_dX
8.4.5
(l-XY
P4-5 (c) T=273K Find the specific Ieaction rate at the new temperature of 273 K using the Anhenius Equation.
k
= 2..54 X 10-3
(200)(9) . k = -r-(- - ' - - 3'~) = 3.543 nun 2.,.54 x 10- A200)
= 2..5days
P4-5 (d) 1) CSTR and PFR rue connected in seIies:
X
. = (200dm 3 )(007dm' / mal.min)(1mall dm 3 )2(1-- X)2 CSJR lOmal / min
Solving the quadratic equation, X CSIR = 0 . 44 ForPFR,
dX = (0.07dm
1.
dX 044 (1- X)3
3 /
mall min)C,40(1- X)(l- X)2 dV lOmale/min
= (0.07dm 3/mallmin)(lmall dm 3)2(800dm 3 ) lOmale/ min
X =0.736 2) when CSTR and PFR rue connected in parallel,
= (200dm 3)(0.07dm 3/ mal,min)(lmal / dm3)\1- X)2
X CSTR
•
.5mal/nun.
X CSIR = 0.56
4-8
ForPFR,
1
dX = (0.07dm 3 / mol.min)(1moll dm 3 )2(800dm 3 ) o (1- X)2 5moll min
XPFR = 0.92 Hence, final conversion X =
0.56+0.92 =0.74 2
P4-5 (e) To process the same amount of species A, the batch reactor must handle
2M
(5d~~J(60min)(24hJ = 14400~lOl hr
mm
day
day
If the reactants are in the same concentrations as in the flow reactors, then
v = (14400-mo.lJ(~dm3J = 14400 dm day
mol
3
day
So the batch reactor must be able to process 14400 dm3 every 24 hours" Now we find the time required to reach 90% conversion" Assume the reaction temperature is 300K.
tR
=
N AO X . N Ao 2 - - , and smce - - = CAO VkCAO I-X V
1 X 1 0.9 tR =-_._--= 3 --=2.14hr kCAO I-X 4.2 dm_*I(mol) 0.1 mol·hr dm 3 Assume that it takes three hours to fill, empty, and heat to the reaction temperature. tf= 3 hours
ttotal = 2. 14hours + 3 hours = 5.14 hours., Therefore, we can run 4 batches in a day and the necessary reactor volume is
14400dri = 3600dm3 4 4-9
Refening to Table 1-1 and noting that 3600 dm3 is about 1000 gallons, we see that the price would be approximately $85,000 for the reactor.
P4-S (f) The points of the problem are: 1) To note the significant differences in processing times at different temperatures (i . e. compare part (b) and (c)). 2) That the reaction is so fast at 77°C that a batch reactor is not appropriate. One minute to react and 180 to fill and empty. 3) Not to be confused by irrelevant information. It does not matter if the reactor is red or black.
+
'rA=kC",CB k (F Ai,,) (F NU)
":A ""
butanoi
+-
..."
eiernentary reaction for liquid systems volumetric flow u =uo
4-10
water
V
F~oX Fe - --.------ = -.--.-----.--2
CSTR -
- ..
kC
rA
£7\: =4
A
_~o
(l---XXS--X)
106~.
6 , Fe = 4. x 10 Ib / yr , 30d / yr operation
F := 4. x 106 It!.. x ..!:x! _1.£.... x 1 Ib mol .C yr 30d 24 hr 278 Ib V
•.
CSIR
-lOC'fl ~1 I'I.J ga.
=20.0 !t'~1
1ft" 1 ..j.) ~_",-3 x. ------:::: J u. 7.48 gal "
133.7 h 3 = . ~ ___ .._._ ..... 20.1bmole/hr_ . ._._ ._._ J
12-;--..fE.--_.-;-··... lomol<:::·lll
Xi. • 6X,· 1.89
=:
2
(.Q:~lbmOle) (1 ·X) (5-X)
f?
0
X"" 0.33
P4-6 (b) To increase conversion, use PFR, higher temperature, or use better catalyst.
P4-6 (C) ME
Rl.
s c: E
P4-6 (d) PFR Design Equation d)(
dV
dX
F Ac =:
IcC!o (1- X XS"'_· X )
W-= ___ ~_ , _-
.. ""." •• " .• "...... ~" .." . - , _ _ _.~
dV
V = 535dm 3 4-11
P4-6 (e)
MOle baiance.;
The above equation relates the reaction time for a batch and the conversion achieved during that hatch.. There is a trade···off bct\vcen high conversion and few batches and low conversion but many batches per day_ What conversion will result in the smallest nurllber of reactors?
N AO
:::;;;
133.7 ft;'
* C AO
C AO
:=:
__ .!2.(?:!!!:.c..~.__ . . _'!:lQ ;y*-_~~~C = .!!~AOX ~~. = . . . . . . reactor * day
t
.
+3
t bau:h
r=
O.2lbmoll ft>
'Y~QJ!:!:_~_. ____ = f{X ) 1
(
<; ...
X \
'A:i":~~; In ,.~.... '5X
J1- 3
J~I'?l:pr.5:c!~:!. =2~}~:!?~19J,!q! = 480.~q!. day
30days 278lb
day
480moll f(X)moll dayJ reactor
n :::;:.-._--...
o
0.4
0..2
U.O
0.8
x---->
The minimum occurs at X
= 0.82 and corresponds to 4.192 or 5 reactors
P4-6 (f) Individualized solution P4-6 (g) Individualized solution P4-6 (h) Individualized solution P4-7 (a) 4-12
Elementary gas phase reaction, A-4B+2C
c
",c (1-X} ...0 (1+
r\
£X)
P,IQ 10 mol CAO '" RT ::: (0.0$2){400) '" O.3'dml
V;;:967
P4-7 (b) v", FAc X(l+!!:l CAJ(l-X)
P4-7 (c) For a = O. OOldm-3
See Polymath program P4,. 7 . c.pol. POLYMA TH Results Calculated values of the DEQ variables Variable v x
initial value 0 0
Y
1.
Co
0.3 2 0_001
esp alfa
minimal value 0 0 0_1721111 0_3 2 0.001
maximal value 500 0 . 656431 1 0.3 2 0_001
4-13
final value 500 0_656431 0 . 1721111 0_3 2 0_001
C k
0.3 0.044 -0" 0132 2.5
r
fo
0. 3 0 . 044 -3.422E-04
0.0077768 0.044 -0 . 0132 2.5
0 . 0077768 0 . 044 -3 . 422E-04
2. 5
2.5
Differential equations as entered by the user [1 J d(x)/d(v) = -r/fo [2 J d(y)/d(v) = -alfa*(1 +esp*x)/(2*y) Explicit equations as entered by the user [1] Co 0 . 3 [2] esp = 2 [3] alfa = 0 . 001 [4] C Co*(1-x)*y/(1 +esp*x) [5] k = 0 . 044 [6] r = -k*C [7] fo=2 . 5
1.0 . . . . . . : : : - - - - - - - - - - - - - - - - - - ,
=
0.8
=
0.6
~. LrJ
0.-:1
AtV= 500, x = 0..66, Y = 0..17
0.2
P4-7 (d)
Individualized solution
0.0
o
P4-7 (e) A
If--_ _-'--_ _--'-_ _- '_ _ _-'--_ _--I
~
B + 2C
Law:
100
200
v
300
400
)::: 0
X
Stoichiometry:
1.+ eX
esc:!
Kc =---~.=
C"
s : : ; 2 and C Ao
;;;;;
0 ..3
X "1:1 --O-? .31_
X:::; (O,,90)X~ :::: 0,.47
4-14
500
3
Using these equations in Polymath we get the volume to be 290 dm
,
P4-7 (0 ~
A
B+2C
Rate Law: Kc == 0,,025
C ::::C"'Q(L:.~l .. \ l+eX
Stoichiometry:
CBC~ _ (_C,-oX 1(2~~oX)!(
K -
C ...
C -
-
heX) l+eX
C
,-:a
=C",,,2S. 1+- ex'
Cc=2C...o X l+eX
__C".,(l··X) . !,7 ex );; ,(1+eX}2(1-X) __ ,,_4C~oX3..__
e::; 2 and CA,o::: 0.3
Xeq::: 0.52 X::: (O.90)Xeq::: 0.47
,E.EB. ,... t"
M
orA
dX r dV'"
- - ::: -
0r
dV F -,,,..,,.' ;;;; """.ia. dX -r. .
=(i~~~)((1- Xl - (1:~fd
Using Polymath to solve the differential equation gives a volume of 290 dm3
See Polymath program P4-7-f.pol. POLYMATH Results Calculated values of the DEQ variables Variable -X V Kc Faa Caa k
e ra
initial value 0 0 0.025 2,,5 0.3 0.044 2 -0,0132
minimal value 0 0 0,,025 2.5 0.,3 0.044 2 -0,,0132
maximal value 0.47 290,,23883 0.025 2 "5 0,,3 0.044 2 ··9" 391E--04
ODE Report (RKF45)
4·15
final value 0,,47 290.23883 0,,025 2,,5 0.3 0.044 2
-9.391E-04
Differential equations as entered by the user [1 J d{V)/d{X) = Fao/{-ra) Explicit equations as entered by the user [1] KC= . 025 [2] Fao = 2 . 5 [3] Gao= . 3 [4] k = . 044 [5] e=2 [6] ra = -(k*Gao/{1 +e*X))*{{1-X)-{4*Gao/\2*X/\3)/{{1 +e*X)/\2*Kc))
V:::: 1300 dm) PFR with pressure drop: Alter the Polymath equations flom part (c)..
See Polymath program P4 . T. fpressure.pol. POLYMA TH Results Calculated values of the DEQ variables Variable v x Y
Kc alfa Cao k
esp fo r
initial value 0 0
----"---
1
0 . 025 0 . 001 0.3 0 . 044 2 2.5 -0.0132
minimal value 0 0 0 . 3181585 0.025 0 . 001 0.3 0 . 044 2 2.5 -0.0132
maximal value 500 0.5077714 1 0.025 0 . 001 0. 3 0 . 044 2 2. 5 -1.846E-04
ODE Report (STIFF) Differential equations as entered by the user [1] d{x)/d{v) = -r/fo [2] d{y)/d{v) = -alfa*{1 +esp*x)/{2*y) Explicit equations as entered by the user [lJ Kc =.025 [2 J alfa = 0 . 001 [3] Gao = 0 . 3 [4J k=0.044 [5J esp = 2 [6J fo = 2 . 5
4-16
final value ---500 0 . 5077714 0.3181585 0.025 0.001 0. 3 0.044 2 2. 5 -1.846E-04
[7]
r = -(k*Cao/(1 +esp*x»*(4*Cao"2*xI\3/«1+esp*x)1\2*Kc»
At V = 500 dm3 X = 0.507 and y = 0.381
P4-7 (g) Membrane reactor: A -+ B + 2C Cc = CoFCIFI CA= CoFAlFr CB = CoFBIF1 FT = FA + FB + Fe and orA = rB = rel2 Using polymath, ForPFR,
See Polymath program P4·7···g.pol. POLYMA TH Results Calculated values of the DEQ variables Variable v Fa Fb Fc Kc Ft Co K kc ra X
initial value 0 2.5 0 0 0.025 2.5 0.3 0.044 0.08 -0.0132 0
minimal value 0 1. 3231889 0 0 0.025 2. 5 0.3 0 . 044 0.08 -0.0132 0
maximal value 1040 2.5 0.3635477 2.3536223 0.025 3 . 7452437 0.3 0.044 0 . 08 -3.827E·-04 0.4707245
Differential equations as entered by the user [ 1 J d(Fa)/d(v) = ra [2] d(Fb)/d(v) =ora - kc*Co*Fb/Ft [3 J d(Fc)/d(v) -2*ra
=
Explicit equations as entered by the user [1] Kc =0 . 025 [2] Ft = Fa+ Fb+ Fc [3J Co =0.3 [4J K = 0.044 [5J kc = . 08 [6 J ra = - (K*Co)*(Fa/Ft· CoI\2*Fb*FcI\2/(Kc*FtI\2» ['7 J X = 1 - Fa/2 . 5
Solving for when X = 0 . 47, we get V = 1040 dm3
4-17
final value 1040 1.3231889 0.0684325 2.3536223 0.025 3.7452437 0.3 0.044 0.08 -3.827E-04 0 . 4707245
3.0 2.4
-
Fa Fh
1.8 1.2
0.6 0.0 0
208
416 v
832
624
10-10
P4-8 (a) The blades makes two equal volumes zones of 500gal each rather than one 'big' mixing zone of IOOOgal.
Si71.g1e B12c:'e
Predicted 'V =
F
.~.
AO kll \., _.
X)-
X
;:;; a ~-'-"' __ 7
(1·· X).
rnal
[aj
=:;SL. :.-.----..-:" CAok
-2:
~~~;;,;l ~~l)
1···· 3X1 ·+
Xl
xf =0
=_ .... -::;:::. .38 2
Reac:or 2
a = [gal] X ··?. = a . -....... .5 :::::. 2a 1000 gal = a---..
(l··-Xt
.25
1-· 2X-:> + X~:;:;:X, ··38 "" :..,
a=500 gal
.,
./;.,.
.
X1--3X + 1.38=0 4-18
2
Measured
500 gal == 5U{10 ~~!;~~L .!.-. X.·I·~ ... '"
. -~ £.."
P4-8 (b) A CSTR is been created at the bend due to backmixing, so the effective arrangement is a PFR is in series with aCSTR.
-c PF~ 8-]-~__
----------~-~.--
CSTRzone created due to backmixing
A--+B 3 k = 5 min- 1 Vo = 5 dm /min . x.,xpected = 0.632 Xactual = 0.586
V = xfFAOdX = va In [ _ 1 ] o -rA k 1- X Expected
=1 In ( 5
1 ) =1.0 1- .632
Now,
4-19
=
Vp
ForPFR,
Vc
For CSTR,
Also,
In (_1 J. I-Xl
. ..1
= FAO(Xactual -
Xl)
-rA
Vp
Vc_
V
V
= (Xactual - Xl) (1-· Xactual)
............. .2
........................... ..3
-+--}".
Solving I, 2 and 3 by using polymath,
See Polymath program P4-8-b.pol. POLYMA TH Results NLES Solution Variable Xl Vc
Value 0 . 350949 0.,567756
V
1
X2 Vp
0 . 586 0 . 432244
f(x)
lni Guess
--~3~,~1~4~8E~--cl~0~--0~·
-3 . 297E-14
1
NLES Report (safenewt) Nonlinear equations [ll f(X1) = In(1/(1-X1))-Vp = 0 [21 f(Vc)
=(X2-X1 )/(1-X2) -, Vc =0
Explicit equations l11 V = 1 [21 X2 = .586 [3] Vp = V - Vc Vc = 0..57 dm3 ; Vp = 0.43 dm3
;
Xl = 0.,35
P4-8 (c) CAO = 2 mol/dm3 A~B
Assuming 1st order reaction, For CSTR,
'r"= CAOX ~ -rA -rA = kC A = kCAO(l-X)
=> rk
ForPFR,
V
= -~ == ~.4 = 0.,67 I-X
=F AO
=> X PFR
0.6
X f o kC
dX
_ rk
(I-X)' AO
= Xf~_ 0
I-X
=1- exp(-rk) =1-exp( -0 . 67) = 0.486
Now assuming 2nd order reaction, For CSTR, Now, assuming 2nd Older reaction,
4-20
ForCSTR,
CAOX
1:=--
-rA -rA =
kC~=kC~o (1- X)2
=> 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 A<=>B 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 <\..1
:VIol.! Babnce :
-
F ...-r ' "v :~
,\
V~~~. = C vdt'"
C v·· kC., V
.
~
';0...
0
.~s:.~ :::5: A. V.e,. _ S::{~.2.. :~~1 dt
V
V
Use POLYMATH: to plot C" vs. t .... sc:;:
(b) C,a lIS t
1., Eqgal::ions '. d(ca) Jd (t.) '" (cao·vo/v)·· (ca"" (k"v+vo) Iv)
, ,
cao=4 .. S vo=70000 v",15000000 k"O.0025 t.O '" O.
~
I:: f
1000 ':.&.aoC
I
.
~""~"'''"''''''''- ..f-.-.~",.""~- .. ~""",,,,,,,,,,,,,,,._,, .• ~ . ........... .,._. :3..'%00 c,~ f:I.':X
o"coo
t
For steady-state: t ::.:: 4.6 -...:.......
1 + 1:'k:
::::: 641 !:us
CA ;:;; .CADV. ::::2.930mgldm3 kV+v~
99% of this is 1.905 mg/dm \. which l5 below rhe standard of 3.0 mg/dm) . Polymath solution(Ans CDP4-B-a) b)
dN dt
Mole Bal::Ulce :
Use POLYMATH
...: ....i..:=
[0
F - F -+ r V .-I"
."
gener.ate plot of C ... and N"
Em;"- £.~2.!1:!:.
.~Ii¢l
d(nal/d( cl .. cao"vo·~ca"VOt;::;;-k~cil cao",4.5
Ke.!
yo,,,,70000
...
'IS. L
T
- I ,.=
. ~
.._~ Na
vouc=50000 11:=0.0025 "=20000"1:: ca=na/v t
Na vs t
f '" '150'
t
4-69
Ca vs t
.Ku
Ca ),000
•• /XX)
I,:)',)(l
'T I
1, i .1 i
Oc,_
L---'
D..COO
----·---+---··---"-,-';O,,-,,,··,-"-+----+-I lfPl,Q(:)Q :n.a...:;oo "iC...'C:OO 'tu.coo
Polymath solutionCAns CDP4·B·b) c)
60 61
tjsc same equaIio[;s as in pan: (a),
07
but
0,3 4,9
40
d (ca 1 fd
(t:)
= (C;;to~".ro/v) -
lea· (}C"_'.-vo) Iv)
41
37
vo •• 7 OfJ-() 0
3,3
v'" ,:,5 ,,0 noO()
29 20
}<=0_0025 1:
f
.EQlymath solutionCAns CDP4·B·c) d) This part is almost same as part(b) with minor changes: V = 15000000 - 10000t Vo= 80000 and Vout = 70000 The reason the graph looks so different from(a) is that pure water is evaporated, but water with atrazine is coming in, 45
Polymath code: d(ca)/dt=cao*vo/v)(ca (kv+vout) Iv) ca(O)=4.5 # vo = 80000 # V = 15000000 - 10000*t 0.0025 k = # 4.5 cao # vout = 70000 # t(O)=O # t(f)=1000
44 43 42 41 40 39 3,8
3.7 36 3,5 100
200
300
400
500
600
700
800
900
101
Polymath solutionCAns CDP4-B-d)
CDP4-C a)Find the number of moles of receptors: 10 cells ;5 receptors 1 mole receptors l000mL -10 I x 10 "'----" * 1 x 1O-m,--,,_- *,'''---------------''""- '*' '--'-'--'-"-.- == 1.66 x 10 M mL cell 6.022 x 1023 receptors L.
1_66.',. XW'lO .~£!. + .OlL:: 1.66 x 1O- 12 moies L' " . Design equation:
dX
NAO ---::-r dt A
Rate Law: Stoichiometry:
CA = CAo(l- X) Cs =CAO(aB-X)
Where:
Total number of moles; 1.56 x 10-12 + 1 X 10-9 == 1.002 X 10-9 hence;
Combining and solv41g:
dX k<:;'odt ·(i.-:x)(e~:~:X)' == o '
ii:
-_.}:"--, In . _.~!!::L._ = ~5;;o~ 8'/1 ··,·1 aB{l" X) N AO
t=O.071mm b) Assume en == egO Design Equation:
ax
N,1.0 dt-- == "rA Rate Law: Stoichiometry: CA
=C;\O(1····· X)
CB = eRG
4-71
Combine and solve: 1-· X
N Ao
1. --. :::: . kCAOCBOt In .---. .-.. .----.. -.1 -- X N A.O
t==O.069 min A very good approximation .
Design Equation:
c)
dX
N ·---=:.-r AO dt A RateL1.w:
Stoichiometry:
CA == C"o(l- X) Co == CSO Cc =:' C"oX Combine and solve:
N AO~;:::: kjCAo(l··· X)C so
·····kJ~·AOX
dX kjCAOCSO
dt
kfCAOCBO -krCIlOX
x
dX
5
D
k/'BO--X(kjCslJ"f k,)
0
N AO
CAodt
f-:---·-·---·~"·:--·--·-·-:;;: S. .
X'-
kfC
1.
_ .sC
no -- -..................-:-. - .---AO -.. . . . . ---.. -.-- -........--- In ...-----...... ........... _. .-...--.... kfC SO +k, kjClJ()--X(kjC so +k,) N Ao
...L. --.ln -.--. -. 1+ ,1
::~
,1···· XC.l
-.
+ . 1)
= 500
X =0.5
CDP4-D Batch reaction: 2A + B -) 2C k j = L98 feJIbmoLmin, k2 = 9..2 X 103 (feJIbmolilmin V = 5 gal =0. . 67 fe, x = 0..65 . CAO = 0. . 0.0.2 IbmollO.67ft3 = 2 . 98 x 10-3 Ibmol/ft3 CBO = 0..0.0.18 IbmollO.67fe =2.69 x 1O-3 lbmollft3 Mole Balance: Rate law: Stoichiometry:
dNA dt
---
= -rAV 2
-rA = kjCAC B+ k2C ACB CA = CAO (1-X) CB = CAo (8 B-(b/a)X) = C Ao(D . 9 - D5X) 4-72
Combine:
dX
CAD -
dt
=CAD 2(1- X)lkr l (0.9 -
0.5X) + k 2 C AD (1- X)(0.9 - 0.5X)
Integrating between X=O to X=O.65 for t = 0 to t = t gives t =24.1 min
CDP4-E Ha..--_ _-..
e.
i o - - - -.........;> A,.
Finally F
AF
, FH2f
FBF
CBF (ete) YBF "" 10- 5
"'r
a
=
k C!:!A Cn
"'
Liquid feed is a mixture of A and B YAo = 1 - YSo= 0.999 Because of the low concentration of B in the feed, such properties as SG, the specific gravity, Mw, the molecular weight, and the solubility of H2 are essentially that of component A. assume that any H2 depleted via reaction is instantaneously replaced via absorption of H2 gas over the liquid reacting mixture. Then at 500psig,
4-73
2]
·~> l
1
;:(.
[~2b':'1 :r [ __ ..Ll~.le._.) .x (.L...1l!.ll..J 2&
l
454 g:tloles
g!'.l
.UfL.:U!L.aI. ll.L..1.~
.020 81 b::.t91~
zal
3 .72 Ibmo1eJ~.::
1 b::::.o~le
v
o
constant ..
P Jl
~
2.
or
0
r.'
,··5 10 Ibmolel;;a.l
• !:to' 1.'0 ----V .;l
U:.ing ec;,nll.tio:;. for
3.
CSTR
(89_3 zallhrHO.:.21L.......__ _ 50 .. d (Q .QZQ8 11,r'9U (0.01) ,11.1
k
= 0.49
x 10 3 ;;alf(l~ole - hr) is:
:::
C
a2
(
0.G127 IbwoleJgal
Let
4-74
300+14.7) 500+14 * 1
314.7 1""'01., {.020l0 l'Si'4:71 Ed
~ 1-1
= 2
F
(SO gal)(S.49x10 3 gal/lbmol-hrHO.0127 Ibmol/g-all 89.3 gal/h.r
= en
B2 ,F
<';2.F
= (.0127 Iba o1e )(S9.4 g al) = 1.134 Ibmole/hr
v
gal
0
hr
3 (3.72x10- ) (1-0.984) = 1.23xlO-S 1-'1.34+3.72
CDP4-F NAo
~ = VpC (.rJ "" VPC kC A
Uqtlid Phase C A
"'" CAO
OX) "'" ~~. (IX)
QX= Pc k {I Xl de • ,
t
X In--L -. (I-X)
o
o
o
5
.3 .36
10
20 .73 1.31
.5
.69
4-75
30 .85 1.89
40 0.93 2.66
50 .97
3.:5
A. plot In [[i:~} i:; lir:ear v.n rime so we concllmc me reaction is filsi order in 5. 6 benzz:roqumoline with k'=O.o,,¥ th'''':'
f '" .___. ,. ".ru ......_ '" 6248 oK .30.__ (353{383 ))
E ;;; 12,414 .;.:;;i mol
k (363)
k (36.3)
=:
0,63 k (373)
~ =~. £'.:2.,5.. "" (.63) (2) k: -
kl
Pel
:: 126 k' 1
1
In --L..."" k't i-X
o o
't
X
30 0.42
20 0.82
10
057
40
50
0.97
0,,99
CDP4-G Develop a design equation Mole balance: FACr) - FACr + L1r) + rAC2nrMh) = 0
_ FA (r) FA (r + ~r) _
~
-~
~r
Rate law:- r A
=
(2
m h)
_ dFA
~
-~
dr
(2
m h)
k1C ACB
Assuming 8B=1 and E = 0 Stoichiometry: Also FA
CA
= CAO (1- X) and CB = CAO (eB -
= F AO (1- X)
dFA
=> - -
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--
wherek',pc' and
3
Stoichiometry:
4-85
=0.78
Pressure Drop: when neglecting the turbulent contribution to pressure drop, ~ 0 is given by
= G[- l~S)~~_, ,p}~~-l
r< Po
'
(T
'[)"
~3
p
' (J
=
ill
Ac The ma.ss flow rate is also and unkrlOwn constant Thus Po can be written as a con~t,~t~ E, over cross·· sectional area.
where
Where E is a fUllction of the mass flow rate, feed properties, and catalyst PIOpclties-'--
(70'om )2 :::: -'14d J mZ A (,,;;;;; It'_
-
4
B and leo must be arbitrarily chosen al first Then, depending on wheLher POLTh1ATH gives high or Im·v values of '/ and X, one Cffi1 converge on the true values of}~ and B. In order to do so efficiently. hm.vevcr, \Jue must rnakc usc of the foHm'iing trends: (Note that tile tollO\vmg taNe IS completely tnj(~ on.1 YwIlen c > (J.)
a) from POL YMATIi: B==S43 k==O.21 x=.67 P=yPo=·973* 1500=1459..5
4-86
£qua&.ions .:. dey) /d(z) ",···beta! (po"y)' (~+eps*x)
d( x1 1.4fzl=ra,·Ac/Fao po,"2500
x and y vs z
'eps;O ,.5 !.QCQ
l:'ao"'950 Re10
Cao"'O.4 k=O.:21 B",8~3 ~~
AC;=3 .1416" (R H 2)
Ca;Cao"Y*
(l
....--....._..
"
.......... ~~~~.....-" .... ". "',
·x) I (l.+eps·x)
beta;B/Ac ra=··k~Ca·"2
220
~!:12E..L
x and y vs z
d (y) Id(z) ""'oeta! (po"y) .. (l+eps·x) d(x} Id(z) =·'·ra-}l.c!Fao po=1500
;
fu:.l -x
eps=0.5 l:ao==950
~iIOQ
i
....
... y
R==25
Cao"'O.4. k=0.21
ll=84.3 L=24
Ac",3 .. 14lli*{R""2··(z"'L) ''''2)
Ca"'Cao"y· (l"'xJ 1 (1 + eps"x) :::'a=-k"Ca .. " 2 beta.="B/AC
Zo :
0,
zf
= 48
b) From polymath: Maximum flow rate = 1750 molls aqua cio!!!!:, dey} Id(zl= .. oota/ (po·y) .. n,·eps·x)
(b)
d(xl/d(z) ",··ra"Ac/Fao
Fao ... 1750
",0=,,1.500 T;
ep8=0.5 a..DC::'
'''ao=>:1750
"",25 cao==(j ..
4
1<;=0.21
3=843
L"24 Ac==3.1.41.6 w (a""2 .. · (z .. L) .";;l)
Ca""Cao"y· (l'-x)! (I+eps"x) ~:;.:eo
ra='"k-Ca*-,2 beta"B/Ac
Zo '" o.
43
4-87
.......-1"...-.-.•• --.•. "C.;3CO
." c) From POLYMATH
Nllnimum pressurc""gO kPa Eg:,.:aci~!:!~:.
d
(y)
Id (zj ",···beta! (po ~y)
• (l~·eps·x)
(c)
d (Xl Idl z) ::~ra -Ac/l"'ao
Po;; 80
po"'80 e?so:O 5
x:
Fao",SSO
y
....."""
R:=26
',.
Cao",O . 4.. k'=O .. 21.
13"'843 L::24
he""3 .1.41.6* (R"'·2··· (z;· ..·Ll·".2) Ca::Cao"Y" U ..~x) I fl+eps*xl
:::30= . ·'k"'Ca '" '"2
1:
bee ta=:S , I AC
d) Flom POLYMATI1: d(y)
Id{:d " ..bet.a/ (po~y)· (l+eps~x)
Cd)
d(x) Id (z)" .... );.a*Ac/23.0
Kex
pO=I.5CO
eps"O S
-x
:;;'ao=950
y
...
1<.=25 Cao .. O.4
k=0 . 21 13",843
he=3 141&*CR··2 .. (z-L).'Z) Ca"'Cao*y· (1. ·xl f (l~eps ~x) ;::'a~""k""Ca1/("*2
De';:a=B/Ac 43
4-88
~".,..~
..•".
CDP4-L (a) C Kc::: ._l! C'"
C. . "" C"'o(l- X) , c.s == C",oX Kc =_.C . . oX.._.::;: .~ C",.(l .. X) 1·.. · X X"l "" _~s..". '" _..2:~._. 1-+ Kc 1 -+ 0.5
"" 0.33
For a conventional PFR:
(b)
dX FAo--;::;:-rA dV
"'r" =k(C .... -Ca/K,}=kCAQ{(l-.X) -XlKc ) For the IMRCF: dF -"'::::::rv dt
"'.,
~.!!. == "J:'"v. '"
v"kcCs
··rA := k{C" ... CafKc)
CA
:::
en =F~N (1'. . + FrJ ._._-....•. __ .
FAN ,
V, =:: v
•
FA"
Use these equations in POLYMATH to generate plots of the conversion profile. ::1 .. 3::0
8. lac
0. :20
8.:.JOC
::leo
4-89
",
l::::.CC
!,
-===
............
o .. ooc 0.000
(c)
-~.-+
...
18.,::)00
'''-'--1'-'''-'-'-'~'''-"-'1
2"1.000
3CLOGC
Use the same equations for the IMRCF in POLY¥ATIi to generate the desired plot.
2.
t::c
... fb
tL9GG
..;..
_.....
1 s~ ceo
4-90
:3~J, 8(J::~
(d)
By varying one parameter at a time we can see the effect of each:
Increasing the specific reaction rate causes changes in conversion. concentration. and molar flow rate [0 occur more quickly . Lowering the transpon coefficient (k,) causes an increase in both C a and Fa. which causes a decrease in conversion . By raising the equilibrium constant (K.,). we cause a decrease in the molar flow rate of A and an increase in conversion.,
(e)
A signitlcant increase in temperature for an exothermic reac1.ion would drive the reaction in the reverse This would cause a decrease in X and C s and an increase in the CA' A significant dect-ease in temperarun.': for an exothermic reaction would cause increase in the rate of the forward reaction. This would drive up X and Cll• while lowering C.... An increase in temperature would drive an endothermic reaction forward, raising X and Ca. While lowering C" A decrease in temperature would cause an increase in the reverse reaction for an endothermic reaction . This would raise CA' and lower both X and C II .
CDP4-M No solution
CDP4-N
4-91
~A4
Y F A3
F F A2
F
,.1,.0
.- y 1',.1,.3
FAl (1·-X) "nere X is con.version defined with respect to • .'1.1.
A1
Also Stoichiometric :able
v
2
t"Al
CAl
v.
1
v
lDO 1
_~~l_
F!~1
t.A1
100
CDP4-0 No solution .-----------------------------------_.----------------
4·92
CDP4-P
A
(n)
At equilibtimll
Initial Condition
ki~lle = k"CCe
e no "" CUe
k 2C nO
C ;\(Ii
.'"
k 3C AO
(1)
(b)
= --k[C"
elt
=> C A == C AO e
klt
dec
,cit = +k;2C' llO ... k3 C(, is it constant
U::::. 1"1 C-AO
I:~ ··k1t
,.
Problems fwn, I'robjot
From (I)
de
ell
>0
C:::;
for a minima in ell
4-93
vtoto " L~ocle 0,' t
'''0
CCe ""Cc\O
+ CCe
CDP4-Q CDP4-R CDP4-S CDP4-T
CDP4-U CDP4-V
4-94
Solutions for Chapter 5 - Collection and Analysis of Rate Data P5-1 P5-1 P5-1 P5-1 P5-1 P5-1 P5-1
(a) Individualized solution (b) Individualized solution
(C) Individualized solution (d) Individualized solution (e) Individualized solution (0 Individualized solution (g) Individualized solution
P5-1 (h)
Example 5-1 The graphical method requires estimations of the area under and above curves on a plot as well as in reading the intersection of lines on the plot. This can lead to small inaccuracies in each data point The Finite differences method uses mathematical estimates to calculate the rate. It is only possible to use this when the time interval of each data point is uniform, The graphical method uses polynomial regression to approximate CA as a function of time. The derivative of the polynomial is then used to calculate the rate,
P5-1 (i) Example 5-2 Assuming zero order reaction: Rate law:
_.4C
A
dt
= k'
~~~__~+-____~__~_______~~__~~__-+~~___+~3~OO~~-J
.-L..:='::'~--L.::0:.:.:::'O 17r=J
zero order reaction
1st order reaction 4.5
35
~ ~
25
2 15
05
50
dC
Rate law: ___ A =k'C
dt
100
150
t(min,,)
t(mln . )
A
5-1
200
250
300
TI27 1 3.49
1 3.66
13.94
1 3.81
14.05
Since none of these plots are straight lines, its is not 1st order reaction or second order reaction.,
P5-1 (j) Example 5-3 Because when a is set equal to 2, the best value of k must be found. P5-1 (k) Example 5-4 -C (moVdm'i) HC10
:
2
-rHel,O(mol / em .s)xlO
7
4.0 2.,0
1.0 L2
2.0 1.36
0.5 0.,74
0.1 0..36
<--.
See Polymath program P5···1··k.poL POLYMA TH Results Nonlinear regression (L-M) Model: r = k*(Callalfa)
Variable k alfa Precision R"2 R"2adj
lni guess 0.1 0.5
Value 1.0672503 0.4461986
95% confidence 0.0898063 0 . 076408
= 0,,9812838 = 0.9750451
= 0.0341709 Rmsd Variance = 0 . 0097304
Rate law: _r
P5-1 (I)
n
= 1.1C~:;:mol/ em 2 .s
Example 5-5
rate law: rCH4
= kP:!oP~
Regressing the data
~oiCHJgcat.min) 5.2e-3 13.2e-3 30e-3 4.95e-3 7.42e-3 5.25e.. 3
Pco(atm)
P H2 (atm)
-
1
1.8 4.08 1 1
--
--
See Polymath program PS .. I-Lpo1. POLYMA Tn..&§.!!lts Nonlinear regression (L-M) Model: r = k*(PCO"alfa)*(PH2 I1beta)
Variable k alfa beta Precision R"2
lni guess 0. 1 1 1
Value 0 . 0060979 1 . 1381455 0.0103839
95% confidence 6.,449E-04 0.0850634 0,1951217
= 0.,9869709 .5-2
I 1 I
0.1 0.5 4
---
-
-
R"2adj = 0.9782849 Rrnsd = 4.17BE-04 Variance = 2 . 093E-OB
Therefore order of reaction = 1.14 Again regressing the above data putting
fJ =1
POLYMATH Resu1ts Nonlinear regression (L-M)
Model: r = k*(PCOAO.14)*(PH2) Variable Ini guess
Value
o.:r-'---
k Precision
95% confidence
0-:.0040792
0.0076284
R"2 = -0.8194508 R"2adj =-0.8194508 Rrnsd = 0.0049354 Variance = 1.754E-04
Therefore, k = 0..0.0.4 (gmoICHJ(gcat.min . atm I14»
P5-2 Solution is in the decoding algorithm given with the modules. P5-3 Individualized solution P5-4 (a) The kinetics of this deoxygenation of hemoglobin in blood was studied with the aid of a tubular reactor. Hb02 -7 Hb + O2 -IA=ke~
Rate law:
FAO dX =ke~o(1-Xr
Mole balance:
dV dV = Acdz , where Ac is the tube cross sectional area
= ke~O~.L (1 - X
dX
dz
FAo
r
In( :~)=lna+nln(l--X) therefore,
ken A where, a = __ ....&Q._L FAo ••
~
1
0.
Position (cm)
rA~'(clIl)
!
..........
'-'
,.
!.5
,......... ..
IConversion ofHb02 (XA)
I 0..0.0.0.0.
(l-XA)
lioOOD
.ax "Ii xl Ii z-(cm: i )
r
<
Electrode Position
I I I
2
3
I
.5
10
,1
.5
.5 IT I 0..0.382 I
r f
0..0.193
0..0.568
r 0..9618', 0..9432 'f 0. . 0.193" r 0..0.189 I 10.980.7
0. . 0.0.386
I
I 0..0.0.378
5-3
0..0.0372
,
'r
.5 20.
5 f···· I 0..0.748
I
I,
'r
6 ....
!"
I
7 3D
5
Electrode Position dXldz'(cm- l ) A hi~~t.ogr~~ plot ~TLi XIii ~ vs . z is then p~oduced Th~values of dXldz are ~~~iuated using~qual-area graphical differentiation: 0 . 00395 0.0039 0 ..00385 0 . 0038 00375 g 0 . 0037
dX/dz vs. distance
EO
~0.00365
~ 0 . 0036
0.00355 0.,0035 0.,00345 0.,0034
+--+--~+--~+--~t--~f--~r--l~--+~~f.----l
o
25
5
75
10
125 15
175 20
225
25 275
30 325
35 375 40
z(cm)
Using the values obtained above, a plot ofln(dXA/dz) vs. In(l-XA) is produced and a line is fit to the data
.
In(dXldz) V$, In(1.Xj
-
~~~~~~~~~~~~~~~
In ( y = 1 0577x- 5 5477 •
~~ ) = In a + n In (1 - X )
where a = ,
kCn A Ao c F Ao
In(1-X)
In(
~~) = -S.S477 + l.OS77In (1- X)
n=l
In(a) = -5.S a = exp(-5.S)
= 4 . 1 x 10- 3 em.
Concentration of blood is 150g hemoglobin per liter of blood Molecular weight of hemoglobin = 64500 Ac=D., 0 196cm
= 2.3x1W6 mol I em 3 FAo = 45.7 X 10-6 moles I s CAO
k
=
FAoa C1of\:
=
6
4S.7x1O- moles Is . ____ ( 4.1x1O-3 em) 2.3 X 10-6 moles I em 3 xO.0196em 2
Hence rate law is,
-rA =4.1CA
mol 3
dm ·s
5-4
= 4.1s-1
P5-4 (b) First we fit a polynomial to the data. Using Polymath we use regression to find an expression for X(z)
See Polymath program P5--4-b.pol. POLYMATH Results Polynomial Regression Report Model: X = aO + a1*z + a2*z"2 + a3*z"3 + a4*z"4 + a5*z"5 + a6*z"6 Variable aO a1 a2 a3 a4 a5 a6
Value 2.918E-14 0.0040267 -6.14E-05 7.767E-06 -5.0E-07 1. 467E-08 -1.6E-10
95% confidence 0 0 0 0 0 0 0
General Order of polynomial = 6 Regression including free parameter Number of observations = 7 Statistics R"2 = Rl\2adj= Rmsd= Variance =
1
o 1.669E-10 1. 0E+99
Next we differentiate our expression of X(z) to find dXldz and knowing that
In ( :~) = In a + n In (1·- X) where a = ,
ken A Ao
r
F
Ao
Linear regression of
In (.:;-)
as a function of
In (1- X)
as in the finite differences . POLYMATH Results Linear Regression Report Model: In(dxdz) = aO + a1*ln(1·X) Variable aO a1
Value
-~. 531947
1..2824279
95% confidence 0 . 0241574 0 . 3446187
General Regression including free parameter Number of observations = 7 Statistics R"2 = R"2adj= Rmsd= Variance = n= 128
0.9482059 0.9378471 0.0044015 1.899E-04
5-5
gives us similar vaules of slope and intercept
In a = -5.53, a = 0.00396
4;.7XlO-6m~les/s
k= FAOa = C~oAc
2
(3.96xlO-3 em)= 4.0s- 1
2.3xlO- moles / em xO.0196em
Hence rate law is,
_rA = 4 •0 CAl 28
mol 3 dm ·s
P5-5 (a) Liquid phase irreversible reaction: A -7 B -+- C; c AO = 2 mole/dm3
CAO-CA = kc a A
T
In ( CAO ; CA) _. Space
=In k + a In CA j
time ( T )min.
~. 38 100 300 1200
-
CA(molldm 1.5 1.25 1.0 0.75 0.5
)
In(CA) 0.40546511 0.22314355 0 -0.28768207 -0.69314718
In«CAO-CA)/T) -3.4011974 -3.9252682 --4.6051702 -5.4806389 -6.6846117
By using linear regression in polymath:
See Polymath program P5-S·a.pol.
In((Cao-CaO/tau) vs In{Ca)
POL YMA TH, Results Linear Regression Report Model: y = aO + a1*lnCa
-q.s
-0.6
-0.4 -1 -2
In ( CAO ;- CA) Variable
= In k + a In CA
y = 2 . 9998x - 4 . 6081
-4
Value
95% confidence 0 . 0162119 2.9998151 0.0411145
":"a=O=-=-=--'--'~--47.-:6:-::0-=8-=-0-=5:79
a1 Statistics RI\2 = RI\2adj = Rmsd= Variance =
. . . . .,. ~. ,_._"'.-. . . '. .-.'.'. ::s. ...............
0 . 9999443 0.9999258 0 . 003883 1.256E··04
··8
In(Ca)
a= slope::3 In(k) = intercept = -4.6 therefore, k = 0.01 mole- 2min- 1 .
dC dt
-6 -7
Hence,
Rate law:
-3
3
A --= O.OlCAmol / dm
3·
mm
P5-5 (b) Individualized solution P5-5 (C) Individualized solution
5-6
....................... ; ....
PS-6 (a) Constant voume batch reactor: Mole balance:
A-7B+C
_ dCA =kC a dt A Integrating with initial condition when t = 0 and CA = C AO for a =t 1.0 1 C(l-a) - C(1-a) 1 (2)(1-a) _ C(l-a) t = AO A =_ A ............. ,' ... ,' "substituting for initial concentration C AO = 2
k
k
(1- a)
(1- a)
t (min.)
0 5 r---" 9 IS r---. 22 30
--
~-
60
CA(m~Vdm3) 2 1.6 1.35 1.1 0.87 0.70 0.53 0.35
See Polymath program P56··a.pol. POLYMATH Results Nonlinear regression (L-M) Model: t = (1/k)*((2A(1-alfa».. (CaA(1"alfa)))/(1·alfa) Variable Ini guess Value 95% confidence k 0.1 0.0329798 3.628E-04 alfa 2 1.5151242 0.0433727 Precision RA2 = 0.999'7773 RA2adj = 0.9997327 Rmsd 0" 1007934 Variance = 0,,0995612
=
K= 0..03 (moVdm3yo5"s·1 and Hence, rate law is
dC A dt
a = 1.5
-._-
= O.03CA15 mol / dm 3 .s
PS-6 (b) Individualized solution PS-6 (C) Individualized solution PS-6 (d) Individualized solution PS-7 (a) Liquid phase reaction of methanol and triphenyl in a batch rector . CH30H + (C6Hs)3CCI -7 (C6Hs)3COCH3 + HCI A + B -7 C + D Using second set of data, when CAO = 0 . 01 mol/dm3 and CBO = 0.1 mol/dm3
5-7
------
CA (molJdm3) 0.1 0.0847 0.0735 0.0526 0.0357
t (h) 0 1 2 5 10 -' Rate law:
-rA =kC:C~ «
For table 2 data:
C AO
Using eqn 5-21, t
= _.
1
k
CBO
=>
C~l;m)
-
-Y'A
= kI
= kC;o
-
(1- m)
I
I
1 (O.01)(l-m) _ C(l-m)
cCl-m) A
C; where k
A
k
(1 - m)
I
See Polymath program P5··7-a-l.poL POL XMA l]!.!l.esnlts Nonlinear regression (L-M) Model: t = (1/k)*((O.1 !\(1-m))··(Ca!\(l·m)))/(1·m) Ini guess
Variable k
1
m
2
- -Value ----
95% confidence 0.0109025 0 . 0021115
L 815656 2.0027694
Nonlinear regression settings Max # iterations = 64 Precision R"2 R"2adj Rmsd Var iance
= = = =
1 0 . 9999999 3268E-04 8 . 902E-07
Therefore, m = 2 For first set of data, equal molar feed => 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<:::;:;
0.129 ~ 0.01 ::::: .119
3.0 •. t 19 = 2.SS1 K == .! i2 : : : J...I] torr.. oL "? (.01)(2.381) P:ac
-
~...
<=
:: 90 :< 10-)
tg :;.: x;: (.119 torr.l (3.0 ~orr)
k: :::; 0.23 ({or. c~tt
5-30
CDP5-G From given data, find the rate law" Given: Oxidation of propene to acrolein Rate law:
rA = kPpapo
b
2
~W=O.5g Using Polymath non linear regression, the following results were obtained: Nonlinear regression (L-M) Model: ra = k*PpA a*P02A b Variable k a b
lni guess 5
95% confidence 2.685E-05 0.0046367 0.001358
Value 0.006609 0.9948724 0,,2034299
1
1
Precision RA2 = 0.9999969 RA2adj = 0 . 9999953 Rmsd 5.722E-07 Variance 4.011E-12
= =
CDP5-H 1)
A ..- ..!~-7 prod
B
.,,-.!~..~
- f1
= kAC ... +kaCa
-fA
prod
=kAC A
- fa = kaCll
_ d~·~I =kAC",,(l-X ... )+kBC:ao(l-X a )
dt
C1(t) = C A (t)t-Ca(t:):= C ...o(l-X ... ) t CaQ(l-X a) C1(t) = CAo(l- X ... + 9
g ......
XB)
=C
Ao
(1.75 - X ... - Xs)
.~~L=CAO(kA -kAX", +k a9 a -k aXB) dt
_.~~:l = C.Ao (k A dt II)
tn
n -) prod Crr(t)
crt}
kal/.3
Xt t. • • 14
de II
-
... Cn(t)
So, ten.)
~ ~
kAX", + O.75kaea •. kaXa)
2
2:
d;- '" -r II '" knC n '" kn eno (l-ln)
"" eno
7;
(I-In) 0
0
10
lO
30
40
.50
'0
0.01,.4
o.ou~
0.0011
0.00'"
0.0074
11 ,00i40
o.ooso
O.31J
I).SlI
0.700
0 ••1$
1.00
1.lU
0
5-31
If
n. 10
0
~ Alt}
-~)
cio (1-10,2
If II.
-:LhlO
~l.hJ.O-4
.. l.hlO- 4 • Z.h10'" 1.9hlO LJ7S
Ln
~.
-4
0.400
0.301
0.179
-1.ldO
+1.h10-4
-4
0.571
0.411 -ldO
"1.1:tlO-4
~.
.
-1.4110
-4
-
·l.ltlO
Q.6.43
-ldO
-4
+l.ltlO·'
..0,1110 ....
1.3UdO
9.Uhl.O- 3 7.06dO-'
5.41b10-'
l.'UtlO-5
1.'Odo~$
1.60.1
1.700
l.lU
3.311
l.lO]
'~4
1.701
If I! were true .. all kO's should 0(:: the same .._-) II is not true:
rime (s) .------,---------------~
CDP5-I (a)
Experimental Plan to tind the rate law for the hydrogenation of cyclopentane on a PtIAll)l catalyst:
1. Since this is a stable catalyst we don't have to worry about catalyst decay and an Integral React(l" will be used. 2, Perform several different nms, holding CAO and W constant while from run to run.
3., Plot X A•OUI
V$,
FAG
is valied
WIF;>.o for all IUllS,
4. Fit a curve through all points which passes through the origin,. The slope at any point is the reaction late Record the slope and COII'cspouding CAl} for many different X A values. 111cse data can be used to determine the rate law.,
(b)
Experimental Plan to find the rate law for the liquid-phase production of methyl bromide from an aqueous solution ofmerhyl amine and bromine cyanide: L Fot a liquid-phase reaction without a catalyst. use a batch reactor.
2. Vlhile running the reaction record both CA and en at equal time intervals 3.. Repeat to ensure a.ccurate data.
CDP5-I (c) No solution CDP5-J No solution will be given.
5-32
CDP5-KNo solution will be given.
5-33
Solutions for Chapter 6 - Multiple Reactions
P6-1 Individualized solution P6-2 (a) Example 6-2 ForPFR,
dC A - dr
= -kJ -
dCB dr
=k
kzC A
Z
-
dClL - k J dr
k3 C A
dCy dr
C z
A
=k
CZ 3
A
3
In PFR with V = 1566 dm we get X = LO and SB/XY = 0.394 also at V = 533 dm 3 SBIXY is at its maximum value of 0 . 625 0.04 . - - - - - - - - - - - - . - - - - - - ,
0'.7r-----
o.J
0. . 6
0..2
0..4
0.1
0.",3
0.,,0.
0.,1
-0..10.
313
626 tau 940.
1253
1566
0.",,0. . - - - " - - - "-----~----' 1253 1566 0. 313 626 tau 940.
See Polymath program P6--2·a.poJ. POL YMATH Result" Calculated values of the DEQ variables Variable tau Ca Cx Cb Cy Cao X
k1 k2 k3 Sbxy
initial value 0 0.4 1.. OE-07 0 1.0E-06 0,,4 0 1.0E-04 0 . 0015 0.,008 0
minimal value 0 -0 . 0439962 1 . OE·-07 0 1. OE-06 0.4 0 1,OE-·04 0,0015 0,,008 0
maximal value 1566 0,,4 0,1566001 0.1490534 0 . 1617665 0..4 1.1099906 1 . OE-04 0.0015 0,,008 0,,6436717
ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(tau) = -k1-k2*Ca-k3*CaI\2 [2] d(Cx)/d(tau) = k1 [3] d(Cb )/d(tau) = k2*Ca [4] d(Cy)/d(tau) k3*Ca1\2
=
6-1
final value 1566 -0 . 0439962 0 . 1566001 0 . 1256308 0,,1617665 0.4 1 . 1099906 1,OE-04 0 . 0015 0,008 0.3946104
Explicit equations as entered by the user [1] Cao = 0 . 4 [2] X = 1-Ca/Cao [3] k1 = 0.0001 [4] k2 = 0 . 0015 [5] k3=0.008 [ 6] Sbxy = Cb/(Cx+Cy)
(2) Pressure increased by a factor of 100. Now CAO = PIRT = 0.4 x 100 = 40 mol/dm3 For single CSTR, C A ' does not change but
r - -v - CAO -
Vo
CAO - CA - ----'=----=..:...- rA - klA + k 2A C A + k3AC~ T=21217sec -
CA
40-0.112 ---------sec 0.0001 + 0.00168 + 0.0001
P6-2 (b) Example 6-3 (a) CSTR: intense agitation is needed, good temperature control. (b) PFR: High conversion attainable, temperature control is hard - non-exothermic reactions, selectivity not an issue (c) Batch: High conversion required, expensive products (d) and (e) Semibatch: Highly exothermic reactions, selectivity i.e. to keep a reactant concentration low, to control the conversion of a reactant. (t) and (g) Tubular with side streams: selectivity Le . to keep a reactant concentration high, to achieve higher conversion of a reactant
(h) Series of CSTR's: To keep a reactant concentration high, easier temperature control than single CSTR. (i) PFR with recycle: Low conversion to reuse reactants, gas reactants
(j) CSTR with recycle: Low conversions are achieved to reuse reactants, temperature control, liquid reactants (k) Membrane Reactor: yield i.e. series reactions that eliminate a desired product (1) Reactive Distillation: when one product is volatile and the other is not
P6-2 ( c) Example 6-4 For kl = k2' we get C A =CAoexp(-klT')
6-2
and
X opt
= 1- exp·( k(l' ') = 1- e -1 =0.632
ForaCSTR:
W= FAo-FA = vO(CAO-CA) -r' A
r
-r' A
, CAO -CA CAO -CA = = ---""'---'-"-r'A kjCA
CA = CAO (r'kj +1) C CB , C -CB r = BO = - - -B- - = . -r'B -kjCA+k2 CB -kjCAO(r'kj +1)+k2 CB 1. 8 0 . . - - - - - - - - - - - - - - - . .
1.64
to find the maximum concentration of C, differentiate CB with respect to .' and set it equal to 0 .
148
See Polymath program P6·2··c.pol.
simplifying we get
320
Then use the quadratic formula to solve for .'. (2) Operating temperature = 325 K
6-3
340
r
360
380
400
Equation (E6-6.8):
P6-2 (0 Example 6·7 For equal molar feed in hydrogen and mesitylene. 3 CHO = YHOC ro = (0.5)(0.032)lbmollft =0.016Ibmollfe 3 CMO = 0.016lbmollft Using equations from example, solving in Polymath, we get T opt = 0.38 hr., At1' = 0.5 hr all of the H2 is reacted and only the decomposition oiX takes place . , XH CH .. CM C !-, - - ! .x l'
----_.
Ex 6-7 0.50 0.0105 0.0027 0.00507 0.2hr 0 . 596
--
This question 0.99 -0.00016 -0.0042 0.0077 --_._---0..38hr ._-1.865
SXIT
See Polymath program P6·-2-f.pol. POL YMATH Results Calculated values of the DEQ variables Variable tau
CH CM CX k1 k2
rIM r2T
rlH
initial value -----
minimal value
o
o
0 . 016 0.016
1.64E-06 0.0041405
o
55.2 30.2 -0.1117169
o -0.1117169
r2H
o
rlX r2X
0_1117169
o
o
55.2 30 . 2 -0.1117169
o -·0.1117169 -0 . 0159818 2.927E-04 -0.0159818
maximal value 0.43 0.016 0 . 016 0.0077216 55.2 30 . 2 -2.927E-04 0.0159818 -2.927E-04
o
0.1117169
o
6-4
final value
0.43--1. 64E-06 0.0041405 0.0077207 55.2 30.2 -2.927E-04 2.986E-04 -2.927E-04 -2.986E-04 2.927E-04 -2.986E-04
ODE Report (RKF45) 0.020,.-----------------,
Differential equations as entered by the user [1] d(CH)/d(tau) = r1 H+r2H [2] d(CM)/d(tau) = r1 M [3] d(CX)/d(tau) = r1 X+r2X
0.016 0.012
Explicit equations as entered by the user [1] k1 = 55,.2 [2] k2 = 30.2 [3] r1 M = -k1*CM*(CHI\.5) [4] r2T = k2*CX*(CHI\. 5) [5] r1 H = r1M [ 6] r2H = -r2T [7] r1 X = -r1 M [8] r2X = -r2T
Increasing 9H decreases 'opt and
- CH .. Ci\I
ex
0.008 0. 004
00000.000
0.086
O.I72 0.258 tau
0.344
0.430
S xrr··
P6-2 (g) Example 6-8 Using equation from example 6-8: Polymath code:
See Polymath program P6·2··g.poL rOLYMA TUResults NLES Solution Variable CH CM CX tau K1 K2 CHo CMo
Value 4.783E-05 0.0134353 0.0023222 0.5 55.,2 30.2 0,,016 0.016
f (x) -4,,889E-ll -1. 047E--ll -9.771E-12
Ini Guess 1 . 0E-04 0 . 013 0.OQ2
NLES Report (safenewt) Nonlinear equations [1] [.'2] [3]
f(CH) = CH-CHo+K1*(CM*CHA.5+K2*CX*CHA.5)*tau = 0 f(CM) = CM-CMo-tK1*CM*CHI\,,5*tau = 0 f(CX) = (K1*CM*CHI\.5-K2*CX*CHI\0,,5)*tau-CX = 0 0.020,-·-----
Explicit equations tau = 0,,5 K1 = 55 . 2 K2 = 30 . 2 [4] CHo=0016 [5] CMo=0 . 016 [1] [2] [3]
0.016 '
'- -- -.,-,
........ ..........'".........
......
.. -'""...~.--........... " ....
--.,...'"'"--........ ,~...
0.012
Df] CH
0.008
A plot using different values of T is given . For T =0.5, the exit concentration are CH = 4 . 8 xW5 lbmoVfe CM =00134IbmoVfe Cx =0.00232 Ibmollft3
" Cl\,,1
- ex
0.004 0.000 lL:"'-_ _~-=:::==:::====:d 0.00 0. 04 0.08 tau 012 0.16 0.20
6-5
The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for T =0..5:
YMX
Fx = --.:..:...-F MO -FM
Cx CMO
= - -0.00232 ---0.016-0.0134
-eM
0.89mole· xylene· produced mole· mesitylene . reacted
The overall selectivity of xylene relative to toluene is:
=.Fx = 8.3mole· xylene· produced
S
FI
x IT
mole· toluene· produced This Question 4.8 x 10-5
Ex 6-8
0..0.0.89 0..0.0.29 0..0.0.33 0..5 0..41 0..7
CH CM Cx T
Y MX SXff
0..0.134 0..0.0.232 0..5 0..89 8.3
---
-_c--.
P6-2 (h) Example 6-9 (1) SDIU
Original Problem-
P6-2 h 1.0.1 0..20.8 SDIU
Membrane Reactor 2.58 PFR 0..666 .. Doubling the mcommg flow rate of specIes B lowers the selectIvIty.
(2) The selectivity becomes 6.52 when the first reaction is changed to A+2B
-~
D
P6-2 (i) Example 6-10 Original Case -. Example 6-10 15
P6-2 i
---_..
---
15
- FA 12
.. FA
9
.. .. ..
6
-
FC FD FE
..
IfIf
- Fe - FD
FB
9
6
3
o0
FB
12
3
2
4
V
6
8
10
0
0
1
2
V
3
5
The reaction does not go as far to completion when the changes are made. The exiting concentration of D, E, and F are lower, and A, B, and C are higher.
See Polymath program P62i.pol.
P6-2 (j)
6-6
O.Oe+O
140
O
280 t
420
560
700
At the beginning, the reactants that are used to create TF-VIla and TF-VIIaX are in high concentration. As the two components are created, the reactant concentration drops and equilibrium forces the production to slow. At the same time the reactions that consume the two components begin to accelerate and the concentration of TF-VIla and TF-VIlaX decrease . As those reactions reach equilibrium, the reactions that are still producing the two components are still going and the concentration rises again . Finally the reactions that consume the two components lower the concentration as the products of those reactions are used up in other reactions.
P6-2 (k) Equal-molar feed
Base case 10
-
8
10
..
Fx
..
Ii'!
-
8
6
4
4
2
2
40
80
FlU Fh Fx
FlUe
6
o0
-
FlU Fh
V
120
160
200
o
o
40
80
V
120
160
200
Increasing YMO will increase the production of m-xylene and methane, but will result in a large amount of un-reacted mesitylene .
P6-2 (I) Individualized solution P6-3 Solution is in the decoding algorithm given with the modules ( ICM problem) P6-4 (a) Assume that all the bites will deliver the standard volume of venom. This means that the initial concentration increases by 5e-9 M for every bite .
6-7
After 11 bites. no amount of anttvenom can keep the number of free SIteS above 66 7% of total Sites, ThIs means that the imttal concentration of venom would be 5 5e-8 M The best result ocelli'S when a dose of anttvenom sucb that the Iruttal concentration of antivenom In the body IS 5.7e-8 M, wIll result in a mmimum of 66 48% free sites, wincb IS below the allowable nurumurn.
See Polymath program P6-4-a.pol.
P6·4 (b) The VICtim was bitten by a hatmless snake and anttvenom was Injected. ThiS means that the tmttal concentration of venom 18 O. From the program below. we see that if an amount of antivenom such that the tmttal concentration in the blood is 7e-9 M, the pattent Will dIe.
See Polymath program P6-4-h.pol. POLYMATn Results Calculated valges of the DEQ variables Variable t
initial value
o o
.
fsv fs Cv
o
Ca
7 OE-09
1
fsa Cp
kv ksv ka kia eso ksa
kop 9 h m
o o
5
final value 0.5
o
0.6655661
0.6655661
1
o
o
o
4.503E-·09 0.3344339
4 503E-09
7.0E-09 0.3344339
o
o
2 OE+OS 6.0E+OS 2.0E+08 5.0E·-09 6.0E+08 1.2E+09
2.0E+08 6.0E+08 2 OE+OS 1 S.OE-09 6.0E+08 1.2E+09
2.0E+08 6.0E+OS 2.0E+08 1 5 OE-09 6 OE+OS 1 2E+09
2.0E+08 6.0E+08 2.0E+08 1 5.0E-·09 6.0E+08 1. 2E+09
0.3 o 3
0.3 0.3
3 0,,3
0.3 o 3
o
o o
o o o
j
maximal value
o o
o
kov koa
o o
o o
1
kp
minimal value
-2.1E-09
o o
o
o
o
o o
o
-2.1E··09
-1 351E-- 09
o o
-1. 3S1E-09
ODE Report (STIFf) Differential equations as entered by the user [ 11 d(fsv)/d(t) kv * fs * Cv • ksv * fsv * Ca [2 J d(fs)/d(t) -kv*fs*Cv - ka * fs • Ca + kia * fsa + 9 [3] d(Cv)/d(t) Cso • (-kv * fs • Cv - ksa * fsa * Cv) + h [4 J d(Ca)/d(t) Cso*(-ka * fs • Ca + kia * fsa) + j [ 5 ] d(fsa)/d(t) ka * fs * Ca • kia • fsa - ksa * fsa • Cv [6] d(Cp)/d(t) Cso • (ksv· fsv· Ca ,+ ksa * fsa * Cv) + m
= = = =
=
=
100..-------ExpliCit equations as entered by the user [1] kv=2e8 [2J ksv=6e8 [3J ka=2e8 [4J kia=1 rSJ Cso 5e-9 [6] ksa=6e8 [7J kp=12e9
0.92 ---"-- -- - .. -.------.-.-----.-._----- .----- ... --.-.-0.84 '-'--'--'--
=
Q
--.----.-------..-----.--.. --
0.76 ------ .--------.068 ---------------------.-.-..---""'-....::-...... 0.60L----------0.0 0.1 02 t 03
6··8
0.4
0.5
[8] kov= 0 [9] koa=O . 3 [10] kop=O.3 [11] 9 = ksa * fsa * Cv + ksv * fsv * Ca [12] h = -kp * Cv * Ca - kov * Cv [ 1 3 J m = kp * Cv * Ca - kop * Cp [14 J j = -Cso * ksv • fsv * Ca - kp * Cv * Ca - koa * Ca
P6-4 (C) The latest time after being bitten that anti venom can successfully be administerd is 27.49 minutes. See the cobra web module on the CDROMlwebsite for a more detailed solution to this problem I
09 'rI :2.-:: I:: <::>
'J:
"'" 0.8 .;::
.:::... '"g::
.... l-<
Amivcnom Injected at t
2749 min
OJ
o ()
tO
02
04
()6
t· ·4'!
OJl
1.4
16
Time (houri
P6-4 (d)
Individualized Solution
P6-5 (a) Plot of C A , CD and C u as a function oftime (t):
See Polymath program P6·Sa.poJ. POL YMA Tn Results Calculated values of the DEQ variables Variable t Ca Cd Cu k1
initial value
o
1
minimal value
maximal value
final value
15 1 0.7995475 0 . 5302179 1
15 0.0801802 0 .. 7995475 0.1202723 1
o 0 . 0801802
o o
o o
1
1
6-9
.1 8
2
k2 K1a K2a Cao
100 10 1.5
100 10 1.5
100 10 1.5
100 10 1.5
1
1
1
1
0.9198198
0.9198198
o
o
X
ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -(k1 *(Ca-Cd/K1 a)+k2*(Ca-Cu/K2a)) [2) d(Cd)/d(t) = k1 *(Ca-Cd/K1 a) [3] d(Cu)/d(t) k2*(Ca-Cu/K2a)
=
Explicit equations as entered by the user [1] k1 1.0 [2] k2 = 100 [3] K1a = 10 [4] K2a = 1.5 [5] Cao = 1 [6) X = 1-CaiCao
=
1.0
1.0
0.8
C~l
- Cd
0.8
0. 6
0. 6
0.4
0.4
0.2
0.2
0.0
J--_~
o
__
3
~_.
____
6
9
0.0
~.
12
15
l~
o
----'
---~-~-,-.--'-.
3
6
9
12
15
To maximize CD stop the reaction after a long time . The concentration of D only increases with time
P6-5 (b) Conc. Of U is maximum at t = 0.31 min.(C A= 0.53)
P6-5 (C) Equilibrium concentrations: 3 CAe = 0.08 mol/dm 3 CDe = 0 . 8 mol/dm CUe = 0.12 mol/dm3
P6-5 (d) See Polymath program P6-S-d.poL POLYMATH Results NLES Solution Variable Ca
Value 0.0862762
f(x) -3.844E-14
lni Guess 1
6-10
0 . 7843289 0 . 1293949 1 1 100 10 1.5 100
Cd Cu CaO k1 k2 K1a K2a t
-2 . 631E-14 6 . 478E-14
0 0
NLES Report (safenewt) Nonlinear equations [1] f(Ca) = CaO-t*(k1 *(Ca-Cd/K1 a)+k2*(Ca-Cu/K2a))-Ca = 0 [2] f(Cd) = t*k1 *(Ca-Cd/K1 a)-Cd = 0 (3] f(Cu) = t*(k2*(Ca-Cu/K2a))-Cu = 0
T
Explicit equations [1] [2]
[3] [4] [5J [6] [7]
10 min
100min
0.295
0.133
0..0862
CDexit
0.2684
0.666
0.784
CUexit
0.436
0.199
0.129
X
0 ..705
0.867
0.914
C Aexit
=
1----
CaO 1 k1 = 1 k2 = 100 K1a=10 K2a = 1 . 5 t=100 X = 1-Ca/CaO
1 min
P6-6 (a) kl
= 0.004(
;;3 J1I2 I
k2 =0.3min-1
A·~B
dm 3 k3 =0.25----I . mO.mm
Sketch SBX, SBY and
SBIXY
as a function of CA
See Polymath program P6. ·6-a.poL
1)
6-11
min
40~----------------------~
32
24 16 8
o
0.0001 0.0326
0.065\.~.p.0976
0.1301 0.1626
2)
120o.1-----------------------, 960 720
480 240
o
0.0001 0.0326
0.065\~.P0976
0..1301 lU626
0.065\~J).0976
0.1301 01626
3)
.-1
2
o
0.0001 0.0326
P6-6 (b) Volume of first reactor can be found as follows
6-12
We have to maximize SBIXY 3 From the graph above, maximum value of SBXY = 10 occurs at CA' = 0 . 040 mol/dm So, a CSTR should be used with exit concentration CA* 3 Also, C AO = P AIRT = 0.162 mol/dm 1/2
+rB+ry = (k 1C A
And-rA=YX
=
=
>V
2
+k2 C A +k 3C A )
*
*
vO(CAO -CA ) = vO(CAO -CA ) = 92.4dm 3 * 112 * * 2 -rA (k 1(C A ) +k 2C A +k3 (C A ) )
P6-6 (C) Effluent concentrations:
.
CB
We know, 't = 9..24 mm => 1: = -
k2 CA
rB Similarly: C X
*
mol = 0.007 -"3 dm
mol = -C B- ~ CB * = 0.11-3 dm
*
mol
Cy = 0.0037 --3
and
dm
P6-6 (d) Conversion of A in the first reactor:
CAO -C A = CAOX
~
X =0.74
P6-6 (e) A CSTR followed by a PFR should be used. Required conversion = 0.99
dV
FAO
=> 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<1 )then(-k1 *Ca+2*t)else( -k1 *Ca) [2] d(Cb )/d(t) = -k2+k1 *Ca
=
Explicit equations as entered by the user [1) k1 = 10 [2) k2=O.192
US: C B never rises above 0 . 8 gIL so the is no time that it would be illegaL Sweden: t = 26 hours Russia: t = 5.2 hours
P6-7 (0 60 g of ethanol immediately -7 C A = 1.5 gIL C B = 08 gIL at 0 . 0785 hours or 4 . 71 minutes. So the person has about 4 minutes and 40 seconds to get to their destination.
P6-7 (g) A heavy person will have more body fluid and so the initial concentration of C A would be lower. This means a heavier person will reach the legal limit quicker. The opposite is true for a slimmer person. They will take longer to reach the legal limit, as their initial concentration will be higher.
P6-8 (a)
Let A be the tarzlon in the stolllach and B be the tanion in the blood.
iVlole Balances: dC~
_ .. _ _t_ ..
dt
=r A
deB
-·-=r dt B
6-17
Rate Laws: ·.. rA ra
;;;;;;
k/-~A
-+- kZC A
= kiC A -
k, _., k~Cs
All k values .are given in the proble m statement. It must be noted, however, that
for CB < 0, k, must be equal to O. These equations when entered in POLTh!ATH generate the following results:
See Polymath program P6-8·,·a.po1. POLYMATH Results Calculated values of the DEQ variables Variable t Ca
Cb k1 k2 k4 k3
initial value
o
6,,25
o 0 . 15 0.6 0.2 0.1
minimal value
o
0.3111692
o 0.15 0.6 0.2 0.1
maximal value 4 6.25 0.5977495 0.15 0.6 0.2 0.1
ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = -k1 *Ca-k2*Ca [2] d(Cb)/d(t) = k1 *Ca-k3-k4*Cb Explicit equations as entered by the user [1] k1 =0,,15 [2] k2 = 0.6 [3] k4 = 0,,2 [4] k3 = if(Cb
P6-8 (b)
6-18
final value 4 0.3111692 0.4057018 0.15 0.6 0.2 0.1
From the following graph generated using the above program in POLY1v1ATH, We can see the proper doses of the drug: . 1. First take two doses of the crug. 2, Six hours later take one dose. ), Take one dose every four hours from then on.
Tr\
6·8b
ii \
~:1 \(~YVf~V~~~V\ ! I
. . . . . ;- __ .. "!. ~~
£~.Gcri1
~'""_'o. _ _ _ _ .~_....
2D.CCG
.
+.
'3lJ.,CGG
P6-8 (c) If one takes initially two doses of Tarzlon, it is not recommended to take another dose within the first six hours Doing so will result in build up of the drug in the bloodsueam that can cause harmful effects .
P6-8 (d) If the drug is taken on a full stomach most of it will not reach the wall at all. The processed food can also drag the drug to the intestines and may limit its effectiveness . This effect can be seen in the adsorption constant kl and elimination constant k2 values If kl decreases this means that the adsorption process is slow
6-19
and if k2 increases means that the rate of elimination of Tarzlon increases. The next graph shows the concentration profiles for kl = 0.10 h- I and k2 = 0.8 h- I " Note that the maximum amount of the drug in the bloodstream is reduced by two.
O:lncentration Profiles -+- Ca (1l§'dm3) 8~--------------------------~Cb(Il§'dm3)
0.6
'IiIm(h) Concentration profile for Tarzlon in the stomach (A) and bloodstream (B).. The maximum amount of Tarzlon in the bloodstream is 03 mg/dm 3•
P6-9 (a) Reactor selection A+B~D
=1Oexp(-8000K / T)C A CB 112 31Z r ZA = lOOexp(--lOOOK IT)C A C B
'iA
A+B-~U
rD SDU= - =
ru
lIZ
exp(-8000K IT)C A
lOexp(--8000K IT)CAC B .
lOOexp(--lOOOK IT)CAI/2C/IZ
lOexp(-lOOOK IT)C B 112
AtT =300K kl = 2.62
X
10- 11 &
k2 = 3.57
At T = 1000K kl = 335
X
10-3 &
k2 =36.78
9.2xlO-S SD/U =
CB
C/
IZ
IIZ-
Hence In order to maximize SDU, use higher concentrations of A and lower concentrations of B . This can be achieved using: 1) A semibatch reactor in which B is fed slowly into a large amount of A 2) A tubular reactor with side streams of B continually fed into the reactor 3) A series of small CSTR's with A fed only to the first reactor and small amounts of B fed to each reactor.. 8 '--_._.... _..._.._. ···-T........... __.....A·······_·__···_
B~b/'J -j
·"l·-·
cLJLll
'-
~~;' ___ Pure A intitially Semi batch
f
T
Tubular reactor with side streams
-
l
T
l:"'-
·_··--·-l
i
"d--'::..::0 lll I
Series 01 small: CSfRs
6-20
T
Also, since ED> Eu, so the specific reaction rate for D increases much more rapidly with temperature. Consequently, the reaction system should be operated at highest possible temperature to maximize SDU Note that the selectivity is extremely low, and the only way to increase it is to keep
6 c )Yz ( C: < 10- and
add B drop by drop.
P6-9 (b) A+B~D
and
A +B ~U
and
r
D SDU= --=
ru
rIA
=100exp(-1000KIT)CA C B
r2A = 10 6 exp( -8000K IT)C A CB
100exp(-1000K IT)CAC B 6
10 exp(-8000KIT)C A CB
=
exp(-1000K IT)
10 4 exp(-8000KIT)
AtT= 300K 6 kl = 3.57 & k2 = 2.623 SDU = 1.14 x10 AtT = 1000K kl = 36.78 & k2 = 3354,.6 SDU = 0.103 Hence we should keep the temperature low to maximize SDU but not so low that the desired reaction doesn't proceed to a significant extent
P6-9 (C) A+B-·-~D and
B+ D
S
-7 U
and
r2A = 109 exp(-lO,OOOK IT)CBCD
= 'iA = lOexp(-8000K I1)CA\~ DU r2A 109 expC-10000K IT)CBCD
=
S DU
exp(-8000IT)CA 10 exp(-10000IT) CD 8
Therefore the reaction should be run at a low temperature to maximize SDU, but not too low to limit the production of desired product. The reaction should also take place in high concentration of A and the concentration of D should be limited by removing through a membrane or reactive distillation
P6-9 (d) A~D
and
'iA = 4280exp(-12000K IT)C A
D~Ul
and
r2D = 1O,100exp(--15000K IT)C D
A~U2 and
s
DIUIU2
r 3A
= 26 exp( --·10800
KIT)C A
= _~_ = 4280exp(--:12000K IT)C A -1O,100exp(-150~OK IT)C D rVI + rV2 1O,100exp(-1.5000K IT)CD + 26exp(-10800K IT)C A
AtT= 300K 14 kl = 1.18 X 10- &
k2 = 194 X 10- 18 & k3 = 6.03
X
10- 15
If we keep C A > 1000CD
S
=~~XlO-I4CA -L94xlO- CD ""~~~=1.96 D/UIU2 1.94 X 10-18 CD + 6.03x 10-15 CA .603 18
6-21
At T= 1000K k j = 0.026 &
k2 = 3.1 X 10-3 & k3 = 5.3
X
10-4
If we keep C A > 1000CD
S
= DIUlUZ
3
0.026C A -3.1xlO- C D 3.1xlO- 3 CD +5.3xlO-4 C A
.026 =49
""
.00053
Here, in order to lower U j use low temperature and high concentration of A But low temperature and high concentration of A favours U z So, well have to optimize the temperature and concentration of A. Membrane reactor in which D is diffusing out can be used.
P6-9 (e) A+B ~ D and D ~ A + Band A +B ~U rD
=ru
SDIU
and
'iA
= 10 9 exp(-10000K IT)CAC B
rzv = 20 exp( -2000 K IT)C D r3 A
= 10 3 exp( -- 3000 KIT) cAe B
9
10 exp(-10000K IT)CACB -20exp(-2000K IT)C D 10 3 exp(-3000K IT)CACB
AtT =300K k j = 3.34 X 10-6 &
k2 = 0.025
& k3 = 0.045
The desired reaction lies very far to the left and CD is probably present at very low concentrations so that:
SDIU
",,0
AtT= 1000K k j = 4.5399 ..9 & k2 = 2.7 & k3 = 49.7 If we assume that CACB > 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) Spect<::5 A:
de
--"-:r dt
-\
-ro. = ktC,-\ Species B:
Species C:
,
de c _ .. --·c dt
,
. . fC::::::
klC B
Pluggmg mto POLYMATH: gets the following,
dice) Id{:::) =rc
cb
k,1-=.4
cc
!E!b,8:~1:.Y~!~ MaximUlll ..~ Minimum.'y~~ !'i~!. . ~ a 100 1.00 0 6' .. 789Be .. 18 5.7898e-18 1.6 1.6 () n.6037 1.4556 0 () .9963 0 .. 9963 0 0
k2"'.{)1
kl
0.4
0.4
04
0.4
ra", . kl"ca rc"k2"cb
k2
0.01
ra
·0 64 0 0.54
0.01 ·2.'71592et8 0.014556
0.01 -0.54
0 .. 01 -2.71.S92e-18
0
0.00603'1
O.6~
-O.OU2417
ESI::!=a.ti~:§!:
var~!!:
d(calldP::) "'rat ci{co) /0.( t) =z·o
t
. '"
ca
rb"'k:;:'·ca· .. k2 "cb ·0
0,
t"...
'"
I'C
1.00
rb
key ~Ca
. Cb •.. ·Cc
P6-10 (b) For CSTR, T = O.Sh First calculate kJ and k2 :
6-26
0.006037
k =O.Olex (20'OOO(_1__ ~)J 2 P R 373 T
See Polymath program P6-1O-b.pol. POLYMA)'H Results
NLES Solution Variable Ca Cb Cc tau T
k1 ra k2 rb rc
Value 1 . 5268515 0.0319385 0 . 04121 0.5 760 0 . 0958161 -0.146297 2 . 580579 0 . 0638771 0.0824199
lni Guess 1.5
f (x)
-6.446E-13 7.28E-14 0
o o
NLES Report (safenewt) Nonlinear equations [1] f(Ca) = tau*(··ra)-(1 . 6-Ca) = 0 [2] f(Cb) [ 3] f(Cc)
=tau*(rb)·Cb =0 =tau*(rc)-Cc =0
Explicit equations [1] tau = . 5 [2] T = 760
[3] k1 = 72*exp(-10000/(1 . 987*T)) ra = -k1 *Ca k2 = 1457152*exp(-20000/(t.987*T)) [6] rb k1 *Ca-k2*Cb [7] rc = k2*Cb [4] [5]
=
Cb vs" temperature
350E-02 3 . 00E-02 + - - - - . - - - - - -
I
2.50E-02 ~ 2.00E02 II
~
tl 1.50E·02
---f------\--~ I
100E-02
.0...._1
5.00E-03 +-_________-+--_________
o.OOE+OO--·-·-----.-·----·· o 200
600
400
------.-----------,-----J 800
temperature (K)
Therefore, CB is maximum at T=760K.
6-27
1000
1200
P6-10 (c) (e) ?art is similar to part b except for two rate laws: fA -k _II *('·s
*(' - kI 'A
rB =k 1 *C-k -,-\ -·1 *C B -k,;. *C a Using those rate laws inPOLYThtlATH produce the following:
___
... .......-..,., Eauations:
l!ti t ~a 1 ..~1::~
d{ca) /d(t) ;::::r:-a
1.6
d(cb)Jd(tl=rb d(cc) Id(tl ::::e'C
a
o
klr·=8.33e·S
2.JC::
-_.
k1£"'.OOOl k2=2.78e-5
\ ..ti::::
-.
!.,2IJC
1\
rc:::k2"cb ra=klr""cb··klf"ca
!\
o.oeo c"cco
var:iable
0.808
J.
-_-.-. _
_
----
t
value In,itial value Maximu.m value Minim1.4'11 ....... ...........value _........ ............... -:inal .,. .......... _-, ...... ....... -------.-'"'--350000 0 350000 0
ca
1.6
1.6
0.436316
0.4.16316
cb
0
0.833237
0
0.5159
cc
(J
0.647'784
0
0.64'1784
k.lr
8 3.3e·"05 0.0001 2.78e06
S.33e··05
13. Be-OS
8 . 33e-OS
0.0001
0.0001
0.0001
2.78e·06
2.78e-06
2.78e-·06
rc ra
0
2 _3164e-- 06
a
1.4342e·-06
0.00016
··0.00016 --1.20632e··06
-6. 57168e-- 07
:r:b
klf
k2
~--
-0.00016
6 _S7168e--07 0.00016
P6-10 (d) Th isis similar to
part d except for one rate law: rc -k' --"1 *C.... 8: -_ok . -2 *C"'c
Using that in POLTh1ATH produces the following:
6-28
. 0' .7703 4e-... 07
EqlJ.§!ci~~.: 2
d{ca)/d(t)=ra d(cb)/d(t)=:::b
ce:
~ 5.:lC
T i
.!..
d(cc)/d(t);r;c k1:::=8 _33E!-'S k1.f= _, 0001 k2:=2,78e--6 k2r=1.. 3 %-6
ra"k1:r:-·cb-· kIf'cd
rc'"'k2f*cb-·k2:::*cc r'b=kl f"ca '" klr 'cb-- k2 f "cb+k2.::'cc
to '"
(),
t:l! .L
Variable --t
'"
350000 ~~..s~,~! __ ::-al':l:~ 1:!~im1.l!:LvallJ.e Minim~ya1ue Final._y~!ue 350000 0 0 150000
Cd
1. 6
1 5
a,490306
0 490306
cb
0
0 833769
0
0.583662
cc
0
0.525032
0
0,526032
8,33e'OS
8.33e05
k1r
8,,33e,05
8 33e-05
kif
0.0001
()
kl:
2,788 06
k2:
1. 3%06
oootS
00001
0.0001
2 730 06
2. ii3e' 06
2.,78e-·06
1. 3ge06
1 39806
L '3%,,1)6
0001
··4 115738·-0'1
'·0 ,00016
--,4 .11573e .. 0 7
l:C
0
2 25569806
0
8.91396,0·-07
r;b
0, 00Gl6
0 GOO16
0
::::d
-1..124.93806
"4,, 79824eO)
P6-10 (e) (e) Vlhen kl>l00 and k2
When k2 ;:;;; 1 then the concentratlon of B spikes again and remains high, while very little of C is formed.. This is because after R is fonned it will not got to C because the reverse reaction is faster. Vlhell k-2 ::::: 0.25, B shoots up, but does not stay as high because the second reverse reaction is a slightly slower than seen before, but still faster than the forward reaction..
P6-11 (a) InteImediates (primary K-phthalates) are fOlmed flOm the dissociation of K-benzoate with a CdCh catalyst reacted with K-terephthalate in an autocatalytic reaction step:
C AD
:::::~ = RT
A-~R~S
Series
R + S -~ 2S
Autocatalytic
1l0kPa =0 02moll dm' (8.314 kPa.dm')( 683K) . mol.K
6-29
Maximum in R occurs at t = 880 sec. See Polymath program P6-11-a. pol, POLYMATH Results Variable t A
initial value 0 0,02
minimal value
o 0.003958
S
0
o o
k1 k2 k3
0.00108 0.00119 0.00159
0,,00108 0.00119 0.00159
R
0
maximal value 1500 0.02 0,0069892 0,0100382 0.00108 0.00119 0.00159
final value 1500 0.003958 0.005868 0.0100382 0.00108 0.00119 0.00159
ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1 *A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [3) d(S)/d(t) (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 108e-3 [2] k2 = 1,1ge-3 (3) k3 = 1 ,,5ge-3
=
P6-11 (b) 1) T= 703 K 3 CAO = 0,,019 mol/dm
Similarly,
k~ =3.3xlO- s- and k~ =3.1xlO-3 dm 3 Imol.s 3
1
Maxima in R occurs at around t =320 sec" See Polymath program P6·11·b i pol POLYMATH Results Calculated values of the DEQ variables Variable
initial value
minimal value
t O O
A
0,019
3,,622E-04
R S
0 0
o
kl k2 k3
0 00264 0.0033 0.0031
0,,00264 0.0033 0.0031
o
maximal value 1500 0,,019 0,,0062169 0,0174625 0.00264 0,,0033 0,0031
ODE Report (RKF45) Differential equations as entered by the user [1] d(A)/d(t) = -k1 *A [2] d(R)/d(t) = (k1 *A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 2,,64e-3 [2) k2 = 3,,3e-3
6-30
final value 1500 3,622E-04 8,856E-04 0,0174625 0.00264 0.0033 0.0031
[ 3 J k3 = 3,1 e-3
2) T = 663 K 3 CAO =0,19 moIJdm
(_I_-_I_)J
k; = (1.08xlO-3 s-l)ex p ( (42600cal / mol) (1.987callmol.K) 683K k~
3
=0.4xlO- s-
663K
1
k~ = 0.78xlO- dm 3
= 0.42 X10-3 S-1
3
/
mol.s
See Polymath program P6-11-b2.pol. POLYMA Tn Results Calculated values of the DEQ variables Variable t A
initial value 0 0.019
R
0
S k1 k2 k3
0 4.2E-04 4.0E-04 7.8E-04
minimal value
o
2.849E-04
o o
4.2E-04 4.0E-04 7.8E-04
maximal value 10000 0.019 0.0071414 0,016889 4.2E-04 4.0E-04 7.8E-04
final value 10000 2.849E-04 0.0012573 0.016889 4.2E-04 4,OE-04 7.8E-04
ODE Report (RKF45) Differential equations as entered by the user l l ] d(A)/d(t) = -k1 *A l2] d(R)/d(t) = (k1 *A)-(k2*R)-(k3*R*S) [3] d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = OA2e-3 [2] k2 =OAe-3 [3] k3 =0,,78e-3 Independent variable variable name: t initial value: 0 final value: 10000 0.020.------,-·------·-·-------,
0,004
2000
4000 t
6000
8000
10000
Maxima in R occurs around t = 2500 sec,
P6-11 (C) Use the Polymath program from part Ca) and change the limits of integration to 0 to 1200, We get: 3 CAexit = 0.0055 moIJdm 3 CRexit = 0.0066 mol/dm
6-31
CSexit = 0,0078 mol/dm'
P6-12 (a)
P6-12 (b) . mol == -_(.7)(.1) .....__ ... _:;;:: 0.023-""--,,, 3.3 dm 3 _$
-riA fIB :;;:: ....-
r"ll = 0 * ro_D =: 0-dm' m<:t._ ... _s 1:,B ::: 0 * fj€
P6-12 (c) _ -riA _
""
0
m~l_..
dm' .s
P6-12 (d) (,7)(.1)
f lC - "-""'- - -,,--- :::
3
j
_-2*r
,
mol
0,023 "·-1"
_,2(3)(50 1 (.I)
2D f:c - - ...... -' -_..._--,. __.,_.-
3
_
mol
== -"Q,O(h2·"t--
3
dm' *5
lic =rJ £ ;:·-(.2)("049)( .51) = ··O.005-!~I_., dm'
*5
P6-12 (e) Ii" ::::: 0 '" liA .
mol =O..dm ·~-3......*5 mol
'iE ::: O~· liD::: 0,:: -". dm'·s r,E ::: (.2)(.049)(.51):::;: O.OOS ....!!!.ol 3
dm .s
P6-12 (I) fA :::
P6-12 (g) F,F V=, . M-, . ~ .
--0.07 - 0.0026 : : : -0.0726 mol
.-~"
::::::~!{~·~~_: . ~:.~2 Ie:::
dm-
fD
·"·fA
0.023 . . ,0.0052· 0.005 = O.0128 .... ~~~.
_ 100(3 ,0.1) _ 3 - ..,_. . _""' ...."........ ::::: 4000dm 0,0726
.5
= 0.0078 ······0.033 == 0 OOll"""il:..?L .. 1
dm _s
P6-12 (h) Mole balance:
Rate law,
CAD
·-
= (_. r A)r Cc = (rc}r CD = (rD)r
CA
rA = -[k"CA
+~kWCAC~ ]
6-32
Solving in polymath: SBID
=I'B/rD = 247
See Polymath program P6-12··h. pol. POL YMATH Results NLES Solution var'iable Ca Cb Cc Cd Ce kd ka rb ra ke rc rd re tau Cao
Value 0.0068715 0.9620058 0.5097027 0.0038925 0,2380808 3 7 0.0160334 -0.0498855 2 0,008495 6,,488E-05 0,003968 60 3
fIx)
-2.904E-10 -1 . 332E-15 --1,67E-08 -2.391E-08 1.728E-08
lni Guess 3
o o o o
NLES Report (safenewt) Nonlinear equations [1] f(Ca) = Cao-Ca+ra*tau = 0 [2] f(Cb) = Cb - rb*tau = 0 [3] f(Cc) = Cc-rc*tau =0 [4] [5]
=
=
f(Cd) Cd-rd*tau 0 f(Ce) = Ce - re*tau = 0
Explicit equations [1] kd = 3 [~2]
ka=7 rb = ka*Ca/3 ra = -(ka*Ca+kd/3*Ca*CcJ\2) ke =2 rc = ka*Ca/3 - 2/3*kd*Ca*CcJ\2 ,. ke*Cd*Cc rd = kd*Ca*CcJ\2 - 4/3*ke*Cd*Cc re = ke*Cd*Cc 19] tau = 60 [10] Cao = 3 [3] l4] [5] [6] [7] [8]
P6-12 (i) For PFR and gas phase: Mole balance:
dF dV
- - A-
= rA
dF dV
- -B= r B
dF dV
-.i:..·=r
6-33
C
dF dV
_D_=r D
dFE =r __ dV
E
Rate law:
Stoichiometry:
FT =FA +FB +Fc +FD +FE
dy dV
= -a
FT
2y FTO
Plot of CB and Cc are overlapping.
See Polymath program P6··12·i pol. POLYMATH Results Calculated values of the DEQ variables Variable
initial value
minimal value
V
o
Fa Fb Fe Fd Fe
20
9 . 147E-04
o o
o o o o
o
o
o
y
1
Ft Cta Ce
20 0.2
0.9964621 13.330407 0.2
o
o
ka kd ke
7 3
7 3 2
Ca rb ra Cd Fta re rd re alfa X
2 0.2 04666667 -1.4
o 20 0.4666667
o
o
1.0E-04
o
maximal value 100 20 6 . 6638171 6 . 6442656 0 . 02012S8 0.0043322 1
20 0.2 0 . 0993605 7 3 2
1.367E-05 3.191E-05
0.2 o 4666667 -9.S86E-05 3 . 0E-04 20 0.4666667 8 . 653E-04 S . 908E-OS 1.0E-04 0 . 9999543
-1..4
o 20 -1.923E··05 -7 . 012E- 05
o 1.0E-04
o
ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb [3] d(Fc)/d(V) = rc f 4] d(Fd)/d(V) = cd [5] d(Fe)/d(V) = re
6-34
final value 100 9.147E-04 6.6638171 6 . 6442167 0 . 0171261 0 . 0043322 0 . 9964621 13 .330407 0.2 0.0993325 7 3 2
1 . 367E-OS 3 . 191E-OS -9.S86E-05 2.S6E-04 20 -1 . 923E-05 -6.742E-OS 5 . 087E-·05 1.OE-04 0.9999S43
d(y)/d(V):::: -alfa*Ft/(2*y*Fto)
[6]
Explicit equations as entered by the user [1] Ft:::: Fa+Fb+Fc+Fd+Fe [2] Cto:::: 0.2 [3] Cc:::: Cto*Fc/Ft*y [4] ka:::: 7 [5] kd:::: 3 [6] ke:::: 2 [7] Ca:::: Cto*Fa/Ft*y [8] rb:::: ka*Ca/3 [9] ra:::: -(ka*Ca+kd/3*Ca*CcA 2) [1. 0 1 Cd:::: Cto*Fd/Ft*y [Ill Fto::::0 . 2*100 [12J rc:::: ka*Ca/3 - 2/3*kd*Ca*CcA 2·' ke*Cd*Cc [13] rd:::: kd*Ca*CcA 2 - 4/3*ke*Cd*Cc [lA] re:::: ke*Cd*Cc [15J alfa:::: 0.0001 [16] X::::1-Fa/20
·40. . V 60 ..
12
!l
>4
\I
0
(I.. !l$(f
#.,1124
(I,out iJAltZ 1).• IlM
n.IiOO
Il
41l
V
H)O
6(1
P6-12
G) Changes in equation from PaIt (i):
dFc dV
rc -Rc
=
Rc
=
kdiffilSeCC
k diffilSe
= 2 nun . -1
See Polymath program P6···12-j.po1.
6-35
20r----------------------------,
0,020.--------------------------,
16
0.016
F~
lBJ
12
8
4
40
V
60
80
100
40
V
60
P6-13 (a) m-xylene --> benzene + methane A -->B +M m-xylene --> p-xylene A --> P
See Polymath program P613 . a.pol. POLYMA TH Results Calculated values of the DEQ variables Variable
'-V Fa Fb Fp
Fm Fi Ft k1 k2 Cta Ca rl Cb r2
Cp Spb tau y
X
initial value
o
75
o o o
minimal value
o
6.1072984
o o o
25 100 0.22 0.71 0.05 0.0375 -·0.00825
25 100 0.22 0 . 71 0.05 0.0026257 -·0.00825
-0.026625
-0 . 026625
o
o o o o o
o
o o o o o
maximal value 6000 75 16.297198 52.595503 16.297198 25 116.2972 0.22 0.71 0.05 0.0375 -5 . 777E-04 0.0070067 -0.0018643 0.0226125 3.2272267 3 0,7634409 0.9185694
ODE Report (RKF45) Differential equations as entered by the user [ 1) d(Fa)/d(V) = r1 +r2 [2) d(Fb)/d(V)=-r1 [3) d(Fp)/d(V) = -r2 Explicit equations as entered by the user [1) Fm = Fb [2) Fi =25 [3) Ft = Fa+Fb+Fp+Fm+Fi [4J k1 = ,,22
6-36
final value 6000 6.1072984 16 . 297198 52.595503 16 . 297198 25 116.2972 0.22 0.71 0.05 0.0026257 -5.777E-04 0.0070067 -0.0018643 0.0226125 3.2272267 3 0 . 7634409 0.9185694
80
100
[5] k2 = . 71 [6] Cto = . 05 [7] Ca = Cto*Fa/Ft [8] r1 = -k1*Ca [9] Cb = Cto*Fb/Ft [ 10 ] r2 =-k2*Ca [11] Cp = Cto*Fp/Ft [12] Spb = Cp/(Cb+ . 0000001) [13] tau = V/2000 [14] Y = Fp/(75 . 00000001-Fa) [15] X = (75-Fa)/75
a l' =2.8 is necessary to achieve 90% conversion 10r--------------------------,
4.0 .-----------------------------,
0.8
3.2
0. 6
2t
. - - - . - - - -.....- - - . - - - -•.•.---.•- - - - -
I~ ~Pj
L6
r........ .
0. 8
2tOO V
4800
3600
6000
0. 0
0
1200
2400 V
3600
4800
P6-13 (b)
CSIB \
FM =1"MV
Mole Balances:
:c ~.? == rPV
Rate La\vs.· .,," "E .r -."1 - -.I ". . . ') k :;:; k. exp.· ,,_.d l ,0 R IT 946· \. ./
Using these equations and Polymath we find the optimal temperature is 1194 K. The maximum concentration of p-xylene is 0.013 mol/dm3
See Polymath program P6·13b.pol. pOLYMA THResul!§
NLES Solution Variable Ca
Cb Cm Cp Cao tau k10 E1 R
k20 E2
Value 0 . 012197 0 . 0122301 0 . 0122301 0 . 0130729 0.0375 0.5 0 . 22 2 . 0E+04 1.. 987 0.71 10000
f(x) 7 5 5 -3
. 15E-ll . 284E-12 . 284E-12 . 069E-ll
Ini Guess 0.0375
o
o o
6-37
6000
Yp
0.5166557 1194 2.0054175 0.0244601 0.0244601 2.143628 0.0261459 -0.050606
T
k1 rb rm k2 rp ra
NLES Report (safenewt)
Nonlinear equations [1] [2] [3] [4]
f(Ca) = Ca-Cao-ra*tau = 0 f(Cb) = Cb-rb*tau = 0 f(Cm) = Cm-rm*tau = 0 f(Cp) = Cp-rp*tau = 0
Explicit equations [1] Cao = .0375 [2] tau = ,,5 [3] k1o=,,22 [4] E1 = 20000 [5] R = 1.987 [6] k20 = .71 [ 7] E2 = 10000 [8] Yp = Cp/(.03750000001-Ca)
[9] T= 1194 [10] k1 = k1o*exp((E1/R)*(1/946-1/T»
rb = k1*Ca rm = k1 *Ca k2 = k2o*exp((E2/R)*(1/946-1/T» rp = k2*Ca [ 15] ra = -k1 *Ca·k2*Ca [11] [12] [13] [14]
-_._-----------
---------------------
P6-14(a) 50dm 3 PFn Mole balance:
dCA
f';\
Vo
dCc rc dll Vo 11(:£ ./ B = dV Vo [
deB dl T dev dV
.:::::::
f'13 ,,~
Uo
tlOr
Fp
dV'
=~~;
6-38
Rate laws: r . ' 1 ' 1 'U 1-- ~li
Fa
·2rVI
£:;2
l'F:"
l'c
rlJ 1
I' [::2 .
"21'1' 3
tn
TO]
2TB2
I TF:)
'IE
TE2
Til(} equation for I.he conversion
ur A is :
C-"I )( = .G. AU ----.
C AO
See Polymath program P6-14··a. pol. POLYMATH Results No Title 08-01-2005, RevS 1.233 Calculated values of the DEQ variables Variable t Fa Fb Fe Fd Fe Ff va Caa Cba Ft Cta kd1 ke2 kf3 Ce Cd Cb Ca rd1 re2 rf3 re rf rd ra rb re
initial value 0 15 20 0 0 0 0 10 1..5 2 35 3.5 0 . 25 0.1 5 0 0 2 1.5 1.5 0 0 0 0 1..5 -.1. 5 -3 1..5
minimal value ----0 0 . 2090606 1. 3440833 0
0 0 0 10 1.5 2 15 . 582463 3.5 0 . 25 0.1 5 0 0 0 . 3018965 0 . 0469574 0 . 0010699 0 0 0 0 -0.042376 -1.5 -3 -0 . 0962952
maximal value 50 15 20 1 . 9655663 7 . 2554436 2.5920934 4.6265981 10 1.5 2 35 3.5 0 . 25 0. 1 5 0 . 228291 1.. 4503322 2 1..5 1.5 0.1004639 0.3632767 0.1004639 0.3632767 1..5 -0 . 0215011 -0 . 0116593 1..5
ODE Report (RKF45) Differential equations as entered by the user
6-39
final value 50 0.2090606 1.. 3440833 0.3535564 6 . 4570707 2 . 5920934 4.6265981 10 1.5 2 15 . 582463 3.5 0.25 0. 1 5 0.0794128 1 . 4503322 0 . 3018965 0.0469574 0.0010699 0.0068104 0 . 0095194 0 . 0068104 0 . 0095194 -0 . 0030314 -0 . 0215011 -0 . 0116593 -0.0111585
[1] [2]
[3] [4] [5] [6]
d(Fa)/d(t) = ra d(Fb)/d(t) = rb d(Fc)/d(t) = rc d(Fd)/d(t) = rd d(Fe)/d(t) = re d(Ff)/d(t) = rf
Explicit equations as entered by the user [1] vo= 10 [2] Cao = 1.5 [3] Cbo = 2 [4] Ft = Fa+Fb+Fc+Fd+Fe+Ff [ 5] Cto = Cao+Cbo [6] kd1 = 0.25 [7] ke2 =.1 [8] kf3=5 [ 9] Cc = Cto*Fc/Ft [ 10] Cd = Cto*Fd/Ft [ 11] Cb = Cto*Fb/Ft [12] Ca = Cto*Fa/Ft [13] rd1 = kd1 *Ca*CbA 2 [14] re2 = ke2*Ca*Cd [15] rf3 = kf3*Cb*CcA2 [16] re = re2 [17] rf = rf3 [18] rd = rd1-2*re2+rf3 [19] ra = -rd1-3*re2 [20] rb=-2*rd1-rf3 [21] rc = rd1 +re2-2*rf3 20 -
16
12
Fa Fh
Fe Fd
Fe
8 -4 ()
o
10
20
V 30
40
50
P6-14 (b) (b) Det.ermine the effiuellt concentration and eOllversion from a. 50dm:3 CSfI'll. IVlo]e Balanee: }~iO·-- f:4 -TA V F HO ·PH
Fc' Fl.)
~~
THV TC V
TDV
6-40
F1:;
J'E V
FI"
1'FT-l
== £. combining'J rate law and 110'
I
-
fleA)
CA
.t(C H )
CB
f(Cd f"'\ C... /),) f(C FJ ) f(CF)
--,Cc:
C.fto
--
lllUlc
halancc,.
rAi
C'BO-- rBl
+ reT
--Co I rv T
,
+ t ET --(/p + 'fFT
--C'B
Polyumth code,
(ca)=ca-caO""xa*tau f(cb)-cb-cbO-rb*tau f (cc) =r c*tau-cc f(cd)=rd*tau-cd
f
f(ce)=tau*re-ce f(cf)=tau*rfcf tau""V/vo V=50 \'0=10 caO=1,5 cbO=2.0 rd1=kdl*ca*cb"2 re2"'ke2*ca*cd rf3=kf3*cb*cc'2 kdl"'0,25 ke2=O.1 kf3=5 ra=rdl·3*re2 rb=2*rdl·rf3 rC"'l'd1.+re2·2*rf3
rd=rdl 2*re2+rf3 re=re2
rf=rf3 ca(O).,;;l cb(O)=! cc(O);l cd(O)=l ce(O)=l cf(O)=1 (Ans) C/I = 0,6 t,
eu =
0 79, C c
() 11, CD
= 045, CEO:; 0 14, C p
= 025
P6-14 (C) (c) V(J = ilOdlll:; S(,wj·,Batch tendor, (l) A is fed to B, (2) B h" fed to A (Catle 1) A itl fod to 13,
6-41
dCA elt
CAO
-
rA
+ tiO
···a£"· -
t'jj
--voV' Ce, .... vo·_V
deu
dCC) ·~lt
=
1'C
deD dt
.
l'J)
dCs dt
=
rp;
--
l"f'
-~
deF elt .t'{
CA
en
.. ."yo.
CD
vo··v· CE
tJo··V
OF .... vo--·
V (C80 (:8) _._...._----CBO
alld mudifying eorrespondiug ]>OlYlllHt.h code.
d(ca)!d(t)=ra·vo!V*ca d(cb)!d(t)"'.rb+vo!V*(cbO·cb) ca(O)=1.5 cb(O)"O x"'(caO·ca)!caO Diffol cntes (1) Becallse Cs o is highel than CAO (i.e ;10% higher molar flow rate), case (2) fenches (2) \Vitll the snme reason, C(ISC (2) produces D and F mote (3) With the same reason,
eB
(in
tilSC
X~:C
2) increases more dwstkally(exeessivcly) than GA
1 ill e!ulier time. (ill
ease 1)
(<.:) case I concentration vs., lime G1dl)h litle
"'"
'''' 1.<10
\ ,\ \
110
.
.,
. ",
""
.
••!••~.~~~~~~~.s~.~!m~~!;~~~~!flI'(~Q '''' .~..~~;::"oo~'~"~~"'~.c~.~~t~;¢~. oo~~~~""~"'~~~".~o",~~~t~t~ ,
6-42
conversion
Case
VS,
time
HIO
1
041}
case
ttl
016 ,__.",.,_"-____..._ _ _-..<......_ _ _....._ _ _-'-_ _ _ _' -_ _ _..1
woo
0.00
2000
3j) flO
41} 00
5000 t
Case 2 concentration vs. time lW
111
!"S2'
lH
IH
\ \ \I \
:P
~-j
\
\
;I
d
"
6-43
&0 DO
70.00
eliDa
90
on
100 llil
Case 2 conversion vs. time
--_.._._--
1.00
090
OSO Q 70
01>0
050
case (2) 040 0.30
02f1
D.li! 0.00
' - - -.........- - - ' - - - . - - ' - - - - - ' ! ' - -..-.-... --......---~.-........- - - ' - - - -.......- - - - '.....- -.... 0.00 1000 2DOO 3000 40.00 50.00 60.00 701)0 80.00 gaoa 10000
P6-14 (d) As aB increases the outlet concentration of species D and F increase, while the outlet concentrations of species A, C, and E decrease . When aB is large, reactions 1 and 3 are favored and when it is small the rate of reaction 2 will increase .
P6-14 (e) When the appropriate changes to the Polymath code from part (a) are made we get the following.
See Polymath program P6-14·e.poL POLYMATH Results Calculated values of the DEQ variables Variable V Fa Fb Fe Fd Fe Ff vo Ft Cto kd1 ke2 kf3 Ce Cd Cb Ca rd1 re2 rf3 re rf
initial value 0 20 20 0 0 0 0 100 40 0.4 0.25 0.1 5 0 0 0. 2 0. 2 0 . 002 0 0 0 0
minimal value 0 18.946536 18 . 145647 0 0 0 0 100 38.931546 0.4 0.25 0.1 5 0 0 0.1864364 0.1946651 0.0016916 0 0 0 0
maximal value 500 20 20 0.9342961 0.8454829 0.0445942 0.0149897 100 40 0. 4 0.25 0. 1 5 0.0095994 0.0086869 0.2 0.2 0.002 1. 691E-04 8.59E-05 1 . 691E-04 8 . 59E-05
6-44
final value 500 18.946536 18.145647 0.9342961 0.8454829 0.0445942 0.0149897 100 38.931546 0.4 0 . 25 0.1 5 0.0095994 0 . 0086869 0.1864364 0 . 1946651 0.0016916 1.691E-04 8.59E--05 1.691E·-04 8.59E-05
rd ra rb re Sed Sef
0.002 -0 . 002 -0.003469 0.002 1.1734311 83 . 266916
0 . 0014393 -0.0021989 -0 . 004 0 . 0016889 1 0
0 . 002 -0.002 -0 . 004 0 . 002 1 0
0 . 0014393 -0.0021989 -0.003469 0 . 0016889 1 . 1734311 1.9686327
ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(V) = ra [2] d(Fb )/d(V) = rb [3] d(Fc)/d(V) = rc [ 4] d(Fd)/d(V) = rd [5] d(Fe)/d(V) = re [6] d(Ff)/d(V) = rf Explicit equations as entered by the user [1] vo = 100 [2] Ft = Fa+Fb+Fc+Fd+Fe+Ff [3] Cto = A [4] kd1 = 0..25 [5] ke2 = . 1 [6J kf3=5 [ 7 J Cc = Cto*Fc/Ft [8] Cd = Cto*Fd/Ft [ 9 J Cb = Cto*Fb/Ft [10] Ca = Cto*Fa/Ft [11] rd1 = kd1 *Ca*CbJ\2 [12] re2 = ke2*Ca*Cd [ 13] rf3 = kf3*Cb*CcJ\2 [14] re = re2 [15] rf=rf3 [16] rd=rd1-2*re2+rf3 [ 17] ra = ··rd1-3*re2 [18] rb = -2*rd1-rf3 [19] rc = rd1+re2-2*rf3 [20] Scd = rc/(rd+..0000000001) [21] Sef = re/(rf+ . 00000000001) 1.20 ~~~ .... ~...-".~...."...~
.---------
90 , - - - - - - - - -
_.__
0.96
72
0.72
54
048
36
C-SI" ~
0.2-1
18
(100
o
o
100
400
200 V 300
500
P6-14 (1) dF
--.---~
The only change from part (e) is: d~-
= rD -
kcDCD
6-45
.............................~".- ...... 0
100
21)0 V 300
400
500
See Polymath program P6-14·-f.pol. 1.15
-----------" -.•
..•.-~--•..
-
125,.--------------,
100
~~
1-
0.69
Scid
J
75
046
50
0,23
25
0.OOO'-----1~0-0--20-0-V~3-00--4~0-0---'500
o0
100
200 V 300
400
500
P6-14 (g) dF dV
The only change from part (e) is: _---1L
=r
F VT
--1rQ.. B
3
where V r = 500 dm and F BO = 20 moVmin
See Polymath program P6--14-g,pol. 1.2 , . - - - - - - - - - - - - - .
700.------
1.0
560
07
420
0. 5
280
0.2
140
flO ' - - - - - - .
o
100
200 V
300
400
500
o
P6-1S (a)
6-46
I...---=~--.---
I)
100
......-..•.-.-..".-""......"--..-., . . -.. . .
200 V
300
400
500
(a) Enter the given program into POLY?vlATH Equations for the concentrations must be added. The following maximums can be seen in the graph given (More exact values can be found in (he corresponding table in POLY1VIATH ) CCm.u.
== 0 . 0434 and CDtrcu ;:;: 0.0033
~gu<,t:: i o~s .:
.!~~. S.~ al_.,!~.J:!:::~
d{fc)!d(v)=kl.{fa/ft}*(fb/ft).·ll!2)~k3.(fc/ft}+k4·{fw1ft
0
,*(fd/ft) d(fal/d(v)=-kl O (fa/fcl*lfb/ft)"(1!21-k2*(fa/ftl*W2
9.8.3
d{fbl/dlv'=-kl/2*'fa/ft)-,fb/fcj··11/2)
4.91
o o o o
d(fw}!d(v)=k3*lfc/ftl-k4*(fw/fc)*(fd/it) d(fd)!dlv)=k2IZ*lfa/fc'··Z-k4· (fd/ft)* (fwiftl d(fe) /d{v} ::ok)'" (re/ft) d(fg) Id{v} ",k4* (fw/ft.)
* (Ed/ft)
kl=O . 04 f t=fa ·,·fb+ fc+fd+ fetfw+fg
~ ..
k2=O.007 k3=O 01.4 k4=O 45
vo=lOO cto",O . 147 ca"'cto" (fal it)
C.2ca
cb"'cto*(fb/it) cc,,"cto*(fc/ftl
c.aca
cci=cto*(fd/ft)
ce=cto*(fe/ft) cw=cto' (fw/ft) cg",ct:o~(fg/ft)
VI)
""
0,
1000
."'., ...•......•.... ,..... ;
a.oco o
CDO
P6-15 (b)
6-47
C:.2:::'C
:::. :soc
(b) Overall yield of HCOOH:
Selectivity of HCHO to CO:
- - F S .. ~~.. ,\E ~ F, E
Selectivity of HCOOCH 3 to CH.;OH: Selectivity of BCaOH to HCOOCH J
8 02
:
ell =C-· A
Add these equations to the previous program and use it to generate the desired plots .
o. acc
"S.tioo
1 OSCJ
o. "CO
O'10C
f. t
i
r I
l
?.·ne
T
-:--¥ C ... S~
1.
1 LL __ . ___ . /".~.;
~,sa;
)tp2
P6-15 (c)
6-48
Modify the original POLY1-IATH program by adding y to each of the concentration terms Also add (he followil1g equation:
(c)
1~~~~(~~) a
::= 0002 Fro"" 15 The graphs of concentration down the reactor are very similar to those generated in Palt (a),. The only major difIerence is that with the change in pressure, the maximum reactor volume is significantly smaller.
E~~£ion~-.: d{ fe)/d (vl =kl* (fa/fe) .. (fb/fe.) "* (1/2 )*y k3" (fe/ft) "'y+·k4" (f
:!.9::\:~i~~, ,. Y~lue 0
w/ft)"(fd/ft)"y
d(fa) Id(v) '=-kl* (fa/ft) * (fb/ft) ** O/2)·y·k2* (fa/ft) 'H2*y
10
d(fb} Id(v) ;;:·kL/2* (fa/ft) * (fb/n.) ** (1/2) *y
5
d (fw) Id (v) =k3'" (fel ft) *y .. k4" Cfw/ ft) .. (fd! ft) *y d (fd) Id (vl ;k2/2" (fa/ft.) '"*2 *y··k4" (f'.... /ft)
* (fd/ft)
0 *y
a
d(fe)/d(v)=k3*Cfc/ft)*y
0
d(fgl/d(vl=k4*(fw/ft)"(fd/ft)*y
0 1
d(y)/d(v)=-O.002/2/y*(ft/fto) k2=O,,007
kl",Q,04
COlICentralion profile with pre •• Ul'C .ha:nge
ft=fa+fb+tctfd+fe+fw+fg
k3=O.014 k4;;;O.45 vo=1011
ca",fa/vQ cb=fblvo cc=fc/vo
cd::;ofd/vo -J.fJCO
ce=fe./vo
c.coo
cg=fg/vo
Vo '"
COlICentIalion profile with pre"Ul'C .hange
0,
P6-15 (d) 6-49
·
r fJ ---"-""' 1 1 \'\ . l.)
k"., "'" k,,n eXt11. ~I \ R\ T
1',.)
Subsdtute this equation in for all of the k values. Vary 'r and find Dut wbat temperature maximizes the yield of C, The 'best temperature at \vhich to ron the reactor is 523 K or 250"(;.
P6-16 (a) (a)
Mole Balances:
dFc _·,"-'-=r dV
C
dFp dV
dFA ·-···-·=f
··,-·=f p
dV
A
dF
o --,-=t
dV
0
Ie ::;:: --kl C c -- k 2 C e
Rate Laws:
fA
;;;;:;k1C C +kJC p ,--k,;C" "kJC A
c o Cro (.FF{.>, )
., (FI') C" =L!'o F~'
Stoichiometry:
=:
'I'
l'
FI =F:+F C P+F A +F0 Use these equations in POLYMATH
to
generate aplol of the flow rates vs.
't ..
;:r:;.,-:::;t; .. \
15.:::<: . \ '.
f,e
\
ip fa fo
..
.. -" '
P6-16 (b) (b)
For a CSTR
Mole Balances:
Fc
=Fco + Ie V
Fo =ro V !'p = ktC c -- k3Cp+ k 4 C A
Rate Laws:
fo
6-50
== kSC A
Swichiometry : f(C c) =: 0 = CcC e" -;-
Combine:
fCC?) = 0 = C p
f(C A ) =O=C A
(k 1 + kzTc'"
(kiC c- K)C p + k ,eA)r:
-
.....
(ktC c ... kJC p --k4CA-ksC~:,Jr
f(C o )= 0= Co ...... kSCAr Use these equations in POLYMATH to generate values for the flow lates at different
values of 1: Use these values to generate the desired etHve. E;~~.Sior~~.:. f (eel "'CC "000+ (kltk2)
I;niS1:a1. vah:e 2
~cc*t:au
o
£ (cp) "'cp" (kl *co-kJ"cp+k4"'cal ""tau £ (ca} =ca·· (kl""cc-·k3 ·cp·k4··ca·-·kS "'cal' tau
o
f (co) =co··-kS"ca"tau
o
cco=2
kl",O.12 Sol wt 1 0<1
k2=O .046 k3",O. {)2
ll. 039:370 1 0 .. 211711 O. Df.J633865 O. Cl?S0638
cc
k~",O.O:J4
c:p
vocolO
Cil
co cco
kS=() . 04
V",3DOO fc;:,:vo·cc fp=vo*cp fa-:vo*ca £o;vo"co
2 D.t2 (] 046 C. D2 CU134 10
k 1 k .? k ., .J .~
k4 \/0
k5
tat~=V/vo
U
3COU
fc ip fa
C. '393701 2 .. ,171 t 0.(;63:3865
fa
D.76()638
tau
300
(b) Flow rates vs. 1; 200 !80 Wl 14()
. Fe
120
. . . FP,
100 80 60
-FA' Fe) .
:j()
20 ()
0
50
leo
150 1;
200
350
250
(nJill)
P6-16 (C) Individualized solution
6-51
··8.54':5e"·'7 .- 1. ()78e--16
·.. 2 .. :30ge··15 '8,518e
18
P6-17 Individualized solution P6-18 (a) Blood coagulation living example See Polymath program P6-18.po[ OLYMA TH Results Example CD Solved Problems - Blood Coagulation 08-25-2005, Rev51233 Calculated values of the DEQ variables Variable initial value t 0 TF 2. SE-ll VII 1. OE-08 TFVII 0 VIla 1. OE--10 TFVIIa 0 Xa 0 IIa 0 X 1.6E--07 TFVIIaX 0 TFVllaXa 0 IX 9.0E-08 TFVllaIX 0 IXa 0 II 1.. 4E-06 VIII 7 . 0E-I0 VIlla 0 IXaVllla 0 IXaVlllaX 0 VIIlalL 0 VIIla2 0 V 2 . 0E-08 Va 0 XaVa 0 XaVaII 0 mIla 0 TFPI 2.SE-09 XaTFPI 0 TFVIIaXaT 0 ATIII 3 . 4E-06 XaATIII 0 mllaATIII 0 IXaATIII 0 TFVlIIaAT 0 IIaATIII 0 kl 3.2E+06 k2 0.0031 k3 2 . 3E+07 k4 0.0031 kS 4 . 4E+OS k6 1. 3E+07 k7 2 . 3E+04 k8 2.SE+07 k9 1. OS k10 6 kll 2 . 2E+07 k12 19 k13 1.0E+07 k14 2.4
minimal value 0 8.24E-14 3.S13E-10 0 1. OE-10 0 0 0 1. 426E-07 0 0 8.994E-08 0 0 -3.41E-24 -2.024E-28 0 0 0 0 0 -1.SSE-S2 0 0 -3.938E-26 -8 . 77E-2S 2.094E-09 0 0 2 . 001E-06 0 0 0 0 0 3.2E+06 0.0031 2.,3E+07 0.0031 4 . 4E+OS 1. 3E+07 2.3E+04 2.SE+07 1. OS 6 2.2E+07 19 1.OE+07 2.4
maximal value 700 2 . SE-ll 1. OE--08 2.027E-ll 9.724E--09 3.361E-13 1.481E-09 2.487E-07 1. 6E-07 1.869E-13 S.673E-14 9.0E-08 7.2E-14 3 . S79E-ll 1. 4E-06 7.0E-IO S.3S2E-I0 2.988E-12 S.372E-12 6.S8SE-10 6.S8SE-10 2.0E-08 1.943E-08 1 . 492E-08 2.281E--10 3.788E-07 2.SE-09 3.867E-I0 1 . 881E-ll 3.4E--06 6.073E-IO 8 . 247E-07 1.301E-ll 8.,3S4E-14 S.,734E-07 .3 . 2E+06 0.0031 2.3E+07 0.0031 4.4E+OS 1 . 3E+07 2 . 3E+04 2.SE+07 LOS 6 2.2E+07 19 1.0E+07 2.4
6-52
final value 700 8.24E-14 3.S13E-10 S.71E-12 9.724E-09 1.. 66SE-13 1.481E-09 1. 846E-09 1.426E-07 8.423E-14 2.608E-14 8.994E-08 3.S68E-14 3. S79E-ll -1. OSE-2S -1.026E-38 3.366E-ll 2.873E-12 4.99SE-12 6.S8SE-10 6.S8SE-10 2.793E-90 S.077E--09 1.492E-08 -6.977E-27 1. 663E-2S 2.094E-09 3.867E-IO 1. 881E-ll 2.001E-06 6.073E-10 8.247E-07 1.301E-ll 8.,3S4E-14 S.,734E-07 3 . 2E+06 0.0031 2.3E+07 0 . 0031 4.4E+OS 1. 3E+07 2 . 3E+04 2.SE+07 1. OS 6 2.2E+07 19 1.0E+07 2.4
k15 k16 k17 k18 k19 k20 k21 k22 k23 k24 k25 k26 k27 k28 k29 k30 k31 k32 k33 k34 k35 k36 k37 k38 k39 k40 k41 k42 r1 r2 r3 r4 r5 r6 r7 r8 r9 r10 r11 r12 r13 r14 r15 r16 r17 r'18 r19 r20 r21 r'22 r23 r24 r25 r26 r27 r28 r29 r30 r31 r32 r33 r34 r35 r36 r37 r38 r39
1.8 7500 2.0E+07 1.0E+07 0.,005 1 . 0E+08 0.001 8.2 0.006 2.2E+04 0 . 001 2.0E+07 4.0E+08 0,,2 1.0E+08 103 63 . 5 1.5E+07 9.0E+05 3.6E-04 3.2E+08 1 . lE-04 5 . 0E+07 1500 7100 490 7100 230 8,OE--13 0 5 . 75E-14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1..8 7500 2 "OE+07 1 . 0E+07 0.,005 1. OE+08 0.001 8.2 0.006 2.2E+04 0,,001 2 . 0E+07 4.0E+08 0.2 1.0E+08 103 63.5 1 . 5E+07 9 . 0E+05 3.6E-04 3 . 2E+08 1.1E-04 5 . 0E+07 1500 7100 490 7100 230 8 . 0E-13 6.283E-14 5.75E-14 1 . 042E-15 9 . 923E-16 1_444E-ll 4 _073E-ll 1_318E-12 1.962E-13 1 . 121E-12 5 . 381E-15 1.078E-12 3.024E-13 1,728E-13 1 . 296E-13 2.883E-13 1.159E-ll 4 . 674E-14 1 . 494E-14 4.406E-ll 5,372E-15 4.405E-ll 3 . 211E-12 9 . 527E-15 5 . 372E-15 3 . 312E-10 2,999E-09 2,,974E-09 3 . 764E-08 2,35E-08 1,449E-08 7.132E-09 2.762E-12 1 . 372E-13 4 . 492E-14 2.065E-15 3 . 19E-15 4,387E-12 7 . 708E-09
1..8 7500 2"OE+07 1 . 0E+07 0 . 005 1.0E+08 0.001 8.,2 0 . 006 2 . 2E+04 0 . 001 2 . 0E+07 4.0E+08 0.,2 1. OE+08 103 63.5 1 . 5E+07 9 . 0E+05 3.6E--04 3.2E+08 1.1E-04 5 . 0E+07 1500 7100 490 7100 230 9 . 728E-17 0 L 735E-14 0 0 0 0 0 0 0 0 0 0 0 0 -1. 6E-29 -9 . 831E-28 0 0 0 0 0 0 0 0 -2 . 108E-52 0 0 -3.763E-24 -4 . 056E-24 -2.5E-24 -1 . 733E--25 0 0 0 0 0 0 -1,248E-26
6-53
1..8 7500 2 . 0E+07 1 . 0E+07 0 . 005 L OE+08 0.001 8.2 0.006 2 . 2E+04 0 . 001 2.0E+07 4.0E+08 0.2 1 . 0E+08 103 63.5 1.5E+07 9.0E+05 3.6E-04 3.2E+08 1.lE-04 5.0E+07 1500 7100 490 7100 230 9.728E-17 1. 782E-14 1. 855E-14 5.189E-16 2.696E-17 6 . 953E-12 1_602E-14 5 _971E-13 8.896E-14 5.083E-13 5 . 381E-15 4 . 982E-13 1_ 506E-13 8.609E-14 6.456E-14 4 . 511E-30 -4 . 231E-40 1.217E-14 1. 439E-14 4.106E-ll 5.006E-15 4 _105E-ll 2 . 044E-13 9 . 527E-15 5.006E-15 1 . 152E-91 2 . 999E-09 2.974E-09 6 . 12E-25 1,,105E-24 6,81E-25 7.062E-26 2,762E-12 1.372E-13 1.762E--14 2.065E-15 3.19E-15 4.387E-12 4 . 499E-27
r40 r4I r42
Total
o
o o
o
o o o o
3 . S02E-14 4.224E-09 1. 78E-16 S.749E-07
3.S02E-14 2.704E-ll 7 . 70SE-17 1.903E-09
ODE Report (STIFF) Differential equations as entered by the user [1] d(TF)/d(t) = r2-r1-r3+r4 [2] d(VII)/d(t) = r2-r1-r6-r7-r5 [ 3] d(TFVII)/d(t) = r1-r2 [4] d(Vlla)/d(t) = -r3+r4+r5+r6+r7 [5] d(TFVlla)/d(t) = r3-r4+r9-r8-r11 +r12-r13+r14-r42-r37 +r15 [ 6] d(Xa)/d(t) = r11 +r12+r22-r27+r28-r33+r34-r38 [7] d(lIa)/d(t) = r16+r32-r41 [8] d(X)/d(t) = -r8+r9-r20+r21 +r25 3.40E-13 [ 9] d(TFVllaX)/d(t) = r8-r9-r10 [ 10] d(TFVllaXa)/d(t) = r10+r11-r12-r35+r36 [11] d(IX)/d(t) = r14-r13 2.72E-13 [12] d(TFVllaIX)/d(t) = r13-r14-r15 [ 13] d(lXa)/d(t) = r15-r18+r19+r25-r40 2 . 04E-13 [14] d(II)/d(t) = r30-r29-r16 [15] d(VIII)/d(t) = -r17 [16] d(Vllla)/d(t) = r17 -r18+r19-r23+r24 136K·13 [17] d(lXaVllla)/d(t) = -r20+r21 +r22+r18-r19 [18] d(IXaVlllaX)/d(t) = r20-r21-r22-r25 [19] d(Vllla 1L)/d(t) = r23-r24+r25 6.80E-14 [20] d(Vllla2)/d(t) = r23+r25-r24 [21] d(V)/d(t) = -r26 O.OOEO 0. 00 [22] d(Va)/d(t) = r26-r2TH28 [23] d(XaVa)/d(t) = r27-r28-r29+r30+r31 [24] d(XaVall)/d(t) = r29-r30-r31 [25] d(mlla)/d(t) = r31-r32-r39 [26] d(TFPI)/d(t) = r34-r33-r35+r36 [27] d(XaTFPI)/d(t) = r33··r34-r37 [28] d(TFVllaXaTFPI)/d(t) = r35·r36+r37 [ 29] d(ATIII)/d(t) = -r38-r39-r40-r41-r42 [30] d(XaATIII)/d(t) = r38 r31] d(mllaATIII)/d(t) = r39 [32] d(IXaATIII)/d(t) = r40 [33] d(TFVlllaATIII)/d(t) = r42 [34] d(lIaATIII)/d(t) = r41
r;:::====:;--r,----,
Explicit equations as entered by the user [1] k1 = 3 . 2e6 [2] k2 = 3.1 e-3 [3] k3=2.3e7 [4] k4 = 3 . 1e-3 [ 5] k5 = 4..4e5 [6] k6 = 1 . 3e7 [7] k7 = 2.3e4 [8] k8=2.5e7 [9] k9 = 1.05 [10] k10=6 [11] k11 = 2 . 2e7 [12] k12 = 19 [13] k13=1.0e7 [14] k14 = 2 . 4 [15] k15 = 1.8 [16] k16=7 . 5e3
6-54
138.93
277.8~
416.79
555.72
694.64
[17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [ 30] [31] [32J [33] [34J [35] [36] [37] [38] [39J [40] [ 41J [42J [43J [44] [45J [46] [47] [ 48J [49] [50J [51J [5'2J r 5 3] [54] [55] [56] [57] [58] [59] [60] [61] [62] [63 J [64] [65J [ 66] [67] [68] [69] [70J [71J [72] [73 J [74] [75] [76]
k17=2e7 k18 = 1..0e7 k19 = 5e-3 k20 = 1e8 k21 = 1e-3 k22 = 8,.2 k23 = 6e-3 k24 = 2 . 2e4 k25 = 1e-3 k26 = 2e7 k27 = 4e8 k28 = 0,.2 k29 = 1e8 k30 = 103 k31 = 63 . 5 k32 = 1 . 5e7 k33 = ge5 k34 = 3,.6e-4 k35 = 3.2e8 k36 = 11e-4 k37 = 5e7 k38 = 1 . 5e3 k39 = 7..1e3 k40 = 4 . ge2 k41 = 7.1e3 k42 = 2 . 3e2 r1 = k1 *TF*VII r2 = k2*TFVII r3 = k3*TF*Vlla r4 = k4*TFVlla r5 = k5*TFVlla*VII r6 == k6*Xa*VII r7 = k7*lIa*VII r8 = k8*TFVlla*X r9 = k9*TFVllaX r10 = k10*TFVllaX r11 = k11 *TFVlla*Xa r12 = k12*TFVllaXa r13 = k13*TFVlla*IX r14 = k14*TFVllaIX r15 = k15*TFVllaIX r16 = k16*Xa*1I r17 = k17*lIa*VIIi r18 = k18*IXa*Vllla r19 = k19*IXaVllla r20 = k20*IXaVllla*X r21 = k21*IXaVlliaX r22 = k22*IXaVlliaX r23 = k23*Vllla r24 = k24*Vllla1L*Vllla2 r25 = k25*IXaVlllaX r26 = k26*lIa*V r27 == k27*Xa*Va r28 k28*XaVa r29 = k29*XaVa*1I r30 = k30*XaVall r31 = k31*XaVall r32 = k32*mlla*XaVa r33 k33*Xa*TFPI r34 = k34*XaTFPI
=
=
6-55
[77] [78]
[79] [80] [81] [82] [83] [84] [85]
=
r35 k35*TFVllaXa*TFPI r36 =k36*TFVllaXaTFPI r37 =k37*TFVlla*XaTFPI r38 =k38*Xa*ATIII r39 = k39*mlla*ATIII r40 = k40*IXa*ATIII r41=k41*lIa*ATIII r42 = k42*TFVlla*ATIII Total lIa+1.2*mlla
=
P6-18 (b) No solution will be given P6-19 C2H4 + 11202 E + 1120
(1)
FlO
~ ~
C2H40 D
C2H4 + 302 E + 30
(2)
~ ~
2C02 + 2H20 2U I + 2U2
= 0.82Fm = 0.007626
P6-19 (a) Selectivity of D over CO 2
s= FD FUl
See Polymath program P6·19···a.po1. POLYMATH Results Variable W
Fe Fo Fd Ful Fu2 Finert Ft Kl K2 Pto Pe Po kl k2 X S
rle r2e
initial value -0----5 . 58E-04 0.001116 0 1.. OE-07 0 0.007626 0.0093001 6.5 4.33 2 0_1.199987 0.2399974 0.15 0.088 0 0 -0.0024829 -0.0029803
minimal value 0 1. 752E--I0 4.066E--·05 0 1 . OE--07 0 0 . 007626 0.0091804 6.5 4.33 2 3.817E-08 0.008858 0.15 0.088 0 0 -0.0024829 -0.0029803
maximal value 2 5.58E-04 0.001116 2.395E-04 6.372E-04 6.371E-04 0.007626 0.0093001 6.5 4.33 2 0.1199987 0 . 2399974 0.15 0.088 0_9999997 0.,4101512 -3.692E-I0 -8.136E-IO
Differential equations as entered by the user [1] d(Fe)/d(W) = r1 e+r2e [ 2] d(Fo)/d(W) = 1/2*r1 e + 3*r2e [3] d(Fd)/d(W) = -r1 e [4] d(Fu1 )/d(W) = -2*r2e [5] d(Fu2)/d(W) = -2*r2e Explicit equations as entered by the user [1 J Finert = 0.007626 [2] Ft = Fe+Fo+Fd+Fu1 +Fu2+Finert [3J K1 = 6 . 5
6-56
final value 2 1. 752E-10 4,,066E-05 2 . 395E-04 6.372E-04 6.371E-04 0 . 007626 0.0091804 6.5 4.33 2 3.817E-08 0.008858 0.15 0.088 0.9999997 0_3758225 -·3.692E-I0 ·-8 _136E-·I0
[4] K2 == 4 . 33 [5] Pto == 2 [ 6] Pe == Pto*Fe/Ft [7] Po == (Pto*Fo/Ft) [8] k1 == 0.15 [9] k2 == 0.088 [10] X == 1 - Fe/0 . 000558 [11] S == Fd/Fu1 [12] r1e == -k1*Pe*Po"O . 58/{1+K1*Pe)"2 [13] r2e == -k2*Pe*Po"O.3/{1 +K2*Pe)/\2
x == 0..999 and S == 0.376(mol of ethylene oxide)/(mole of carbon dioxide)
P6-19 (b) Changes in equation fr'Om part (a): a dF1 +3r +R =-r and Fa () a =0 a IE 2E dW 2 R = 0.12xO.0093 = 0.001116 mol a W 2 kg.s
From Polymath program: X == 0..71 S = 0..0.4 (mol of ethylene oxide)/(mole of carbon dioxide) See Polymath program P6- t9-b . pol
P6-19 (C) Changes in equation from pmt (a):
d~ - = rlE dW
+r 2E +R E
and
FE () a =0
R = .0.06 x 0.0~9~ = 0.000558 mol E W 2 kg.s X = 0. . 96 S == o.Al(mol of ethylene oxide)/(mole of carbon dioxide) See Polymath program P6-19·c .pol
From Polymath program:
P6-19 (d) No solution will be given . P6-20 Solved on web Go to http://www.wits.ac.zalfac/engineering/procmatiARHomepage/frame.htm
P6-21 (a) Isothermal gas phase reaction in a membrane reactor packed with catalyst.
A
~-7
B +C
lic,
1
c = lsc [ CA - -CBC _.
KIC
A-7D 2C+D -7 2E
6-57
CAD
=~ = RT
24.6atm (O.082dm atmlmol.K) (500K) 3
= O.6mol I dm 3
See Polymath program P6··21-a.po] POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value WOO
Fa Fb Fe Fd Fe
10 0
0.349438 0
0 0 0
0 0 0
y
1 0.4 0.2 10 0.6
0" 3404952 0.4 0.2 9.7581913 0.6
k2d K1e Ft Cta Cb Ca Cd Ce kb k1e r2d k3e rIc ra r3e rd rb re re Ce alfa Fto
0
0
0.6 0 0
0.0073158 0 0
1
2 0.24 5 1.2 -1.44 0 0.24 1..2 1.2
maximal value 100 10 3,,2375418 4.9873025 2.7304877 1.3722476 1
0.4 0,,2 13.220737 0.6 0.1403618 0.6 0.1019635 0.2117037 1
1
2 0,,0029263 5 0,,0051635 -1,,44 0 0.0014457 0,,0051635 -0.0042625
0 0
0 0
0.008 10
0.008 10
2
final value 100 0.349438 0,,4443151 4.8617029 2,,7304877 1.3722476 0.3404952 0.4 0.2 9.7581913 0.6 0.0093022 0.0073158 0.057l654 0.1017844 1 2
0,,24
0.0029263
5
5
1.2 -0.0080898 0,,0216828 0.24 1.2 1.2 0,,0216828 0.0398819 0.008 10
0.0051635 ··0.0080898 0.0029612 0.0014457 0.0051635 0.0022023 0,,0029612 0.0287293 0.008 10
ODE Report (RKF45) Differential equations as entered by the user [1] d(Fa)/d(W) = ra [2] d(Fb)/d(W) = rb·(kb*Cb) (3) d(Fc)/d(W) = rc [ 4] d(Fd)/d(W) = rd (5) d(Fe)/d(W) = re [6) d(y)/d(W) = -alfa*Ft/(2*Fto*y) 10
Explicit equations as entered by the user [1J k2d = 0..4 [2J K1c=0 . 2 [3] Ft = Fa+Fb+Fc+Fd+Fe [4] Cto = 0 . 6 [5 J Cb = Cto*(Fb/Ft)*y [6] Ca = Cto*(Fa/Ft)*y [7] Cd = Cto*(Fd/Ft)*y [8] Cc = Cto*(Fc/Ft)*y [9] kb=1 [10] k1c=2 [11] r2d = k2d*Ca [12) k3e 5 [13) r1 c = k1 c*(Ca-(Cb*Ce/K1 e» [14] ra = -r1 e-r2d
- Fa 8
•. li'b
.. Fe - Fd 6
4
2 -"', .....".........
=
_
...........
o "."..•... -====--~-.----~----I 80 100 o 20 40 \V 60
6-58
[15] [16] [17]
[18] [19] [20] [21] [22]
r3e = k3e*(CcA 2)*Cd rd r2d-(r3e/2) rb = r1c rc=r1c-r3e re = r3e Ce = Cto*(Fe/Ft)*y alfa = 0 . 008 Fto = 10
=
P6-21 (b) The interesting concentrations here are species C and D, both of which go through a maximum Species C goes through a maximum for two reasons: (1) it is an intermediate product which is formed and then consumed, and (2) there is pressure drop along the length of the reactor and as pressure drops, so does concentration. The concentration of species D goes through a maximum because of reason (2) above . Species D is formed but then the pressure drops, which causes the concentration to falL
P6-21 (e) Individualized Solution P6-22 (a) What factors influence the amplitude and frequency of the oscillation reaction? Ans: k and the initial conditions
P6-22 (b) Oscillations eventually cease because the CA is decreasing and becomes the limiting factor. P6-22 (ae) Observation 1: 'tl and 't2 decreased Observation 2: 'tpI increased Observation 3: 'tp2 decreased Now, Tl =
~£ In( JLlJL~ J
and T 2 =
~£ In(PO* J JL2
2n Tpl
=(JLl*2- + KU )112
_ ko
£--,
2n and
Tpz
= (JL2*2-; K
-)l/Z-
u
* _[(1-2Ku )±f!=8Ku ]112 JL12 -
k ' 2
2
From observation 2 and 3, we get Decreasing III * and increasing 112* => Ku = (ku/k2) will increase Also, from observation 1, E increased => koIk2 should be increased Now, P -7 A k = ko ....... (1) A-7B k=ku. (2) A+2B -7 3B k = kl ....... (3) B-7C k=k2 ............ (4) Hence the reaction (1) and (2) are more temperature sensitive than reaction (4)
P6-22 (d) Individualized solution
6-59
P6-23 Individualized solution CDP6-24 t).. ::
-u() Ca + ra W
•
·kl CA - k3 CA
ra = k1 CA
=0
..
k2 CB
rc:: k2 CB
-Uo Cc + rc W ;:: 0
"Uo CD +rD W =0 113' • .9'.9'
KEY: 1 -
ca
2 - cb
:] .. cc ..... cd
CDP6-25 (a) PFR:
6-60
r-,.'lo1e balance:
dC},t = r --_._.
dr
M
Rate laws
e KC ::::: 0 . 021
C MO ;;;;; 00105 T ==.5 Plugging those into POLYMATH gels the following:
6-61
CSTR:
Mole balance:
FHO '" Ffj == (.- rtB -. f;H···· FMO-FM
r,H) V
=-"I'I:-I V
Fx =(rtx+f1X)V FT =(r21+ f;r) V F;-'1< == (£'1 M~ + fz ),10 t·
r.l.\k )
V
Fa =: f'll Rate Law:
Stoichiometry: FH ::::: 'lOCH
~"1::::
VOCM,
Fx == voCx.
FT == voC r F:\1e :::: VOC Mc == vo(C l10
...•
ell)
ell
FEI :: \'oC 3= vo[(C)"m '" eM)" ex·' Combining all of these: ·' CH::;;;;.·l (k COH~(''M + k-:: (~l)5C ) C . -HO ·,tt 'K t'. k ; c,Q,$c" HI r
CM
¢
::::
eHG -C H
ell = (C;"iO - eM ) . . C x .- CT The following i:s the POLTh1.-'\TH program and tIle concemrar,ions. E:'i'!a;.i 0E!::!!:'
me summary table showing all of
f (en] :eh-eno+ (k1""ch ~ . 5 ·c:n-·k2 ~cn" . 5 'cx+kl "ct: 'ch~ . 5) "cau £ (c:n) =cm· 'crne i kt ~c~ - ~ 5 .... c:tr'" 'C.a.!l Uartaolt? ~(cCI;lk2·ch-.5·cxk3·c:·chA.5)·~aa-ct
c:'0=.021
ch em ex
k1.=5S.2 k2",3a.2
Solutl.O["'f
--
Value ........... . "
.......,.,~
..... .......
0.00776519 0.00301658 0.00:317467 0.00286611
0.021
ld=l1.2
1;8.u"".51. erne"',
.......
kl
55.2
k2
"30.2 11.2
1<:3
alaS
0.5l erne cmt?
6..62
0 .. 0105 0.01323-;0 0.00114254
~"".--
..
~"
-
.. fO .. ..... --~
"
2.993.-15 I .463",
14-
2 .. 79t? -·14 1.495e- i"
The con\.crston of Hydrogen and i'v[esitylene are (hen: X. = S'.:HQ .. 0 . 11 -~g.9gLS. == 0.63 H C O O '. )- 1 HO
_Sll : ;: .2:
X "'1
_ .....e --......-C- - . =--.0010)··,0.0030 . . . . . _._ . . . . . . . . -.. . ----- -_ .~.f 1
-
MO "
SI
O.OW5
'C!'-IO
C H = 0,0078 Ib mol/frl C r = 0.0029 ib mollft'
0.0030 lb molJfr\ Crr.::;::;;; 0.Q13 Ib mollft' eM:;:;
C x = 0.00.32 Ib mol/fr' C~ ::::: 0,,0014 Ib mollfr'
CDP6-25 (b) When eM is reduced to 1.5, it now takes a L of 0.24 h to achieve a maximum of xylene, Increasing eM to 10 now requires a L of only 0 . 08 h
CDP6-25 (C) To find out the reactor schemes needed, use the attainable region to get these graphs
Using fi PFR wouldmaxirnize C,. If ---'Ie used at rl1ri{) of mesil:ylt!ne to hydrogen of 10, tben we l.vQuld only have to have a t of .08 hours. So our voulme of the reactor would bt; only 38,08 ft\ So our entering cOl1centraiol'ls would be .0105 Ib mollir' of Hydrogen and, 105 111 moUff of Mt;s[~yler;e.
CDP6-25 (d) Ftfst fmd the protJortionality ';;OEst;;;mts of the rate consr.ants usin2 (he Anheniu$ equation.' • 's
A, :::: 12674
Once that is done, by t.rial fu'1d erw[ corne up w;th the temperature where SXT ""STB and th~~ r.msw~r is then 862K or 155 L6°R.
6-63
~qu~9.!.::!.: dIem) Id (t);:Dn
0.0105 0.021
d(chl/ci(cl =!h
o o o
d(cx)/d (t) =rx d(eme) Id(t) ;;rme
d(ccl/d(c)=rt: dlcb}/d(t)=r'b
Scale: 10
rm=-55 . 2 "cm"ch". 5 rb::::l1 . 2 "ct"chA. S
o
2 2.:::
K&'£.;. ,,""
em.
!x:;:~rm-30.2*cx*ch~.S
ch
rt;.=30 ,,2"'c:X:"'ch". 5-rb rh:::rm-30.2"c:x:"ch".5-rb
e'X
..... -"
cme 1.:lCC
.
rme;::·-.t1l\+30 . 2"cx"'ch" .S+rb
,,
'
..
~.
ct
cb
O.JCO
t
!E:bE!~l . Y~l~ Max!~ Ye.!:!~ ~:h..t!l:!!!:!!!!.Y!:1~s: () 51 (} 0
em
{}.O!.OS
0.01.05
(}
en
o. 021.
0.021
0 . 00381.84"1
YeE~~e.!.~
. 0001599836
~i::~! .. ~!:l~!: 0 .51
a
000699836
0 00381847
ex
0
()
0
0.00372332
erne
0
0.0111.815
0
0,,017181.5
c::.
Q
0.OO4772.U
0
0.00477233
cb
a
Q . 00130452
0.00130452
00506614.
::m
--0 .. 083992
.. Q.OO:n871S
0 -Q .083992
::b
0
000.3521.27
0
0,00330298
::x
0.083992
(}.08.1992
.• 1.1 _. 00523484
-0.0045612
-~
0
0 01.46868
°
0 00364547
.. 0 . 0126384 0.083992
0.0126384
0.0126384
!-h
0.083992
l::me
0.08,3992
-0.083992
The maximum concentration of xylene occurs at I: ;:;;: 0 . 19 h .
6-64
···0.00238715
-0 .0126384
CDP6-26 (a) Starr wicb ihe mok balances:
dF,
_dEl .....,,"- ;;;;;: C
dV
,"',","'-::: r
It
dF,
'
If.!::::
9
dF dV
H
H ---,-,:;;;: r
-''''''''.- ::: r,
dV
dV
(" «""'} +- k" ('"}j'("~3 + k 5 C'"ti " C-, k'~ C-ii«".~H + k ~ (~'c' "f~i() -+ k-1'fl ?
Finally ,the stoichiometry:
Putting all of (hose together and put it into POLYMAnr and get the following program and answers d(cll)/d(V) = rl/vo # cll(O)=0.l37
d(c9)/d(V) = (-r2+r3)/vo
# c9(O)=0
d(clO)/d(V) = (-rl+r2)/vo
# c10(O)=0
d(c8)/d(V)
(--r3+r4) /vo
# c8(O)=O
d(c7)/d(V)
(-r4+rS) /vo
# c7(O)=O
d(c6)/d(V)
-rS/vo
# c6(O)=0
d(ch) Id(V)
(rl+r2+r3+r4+rS)/vo
# ch(O)=0.389
890 c9/(cl0+c8+c7+c6+0.0000001) 887 c8/(c7+0.0000001) # 889 c8/(c9+0.0000001) # 8910= c9/(cl0+0.000000l)# kS*10 kl kS*17.6 # k2 kS*2.7 kS*4.4 # k4 k3 2.1 vo 1 # kS rl -kl*ch A O.S*cl1 # r2 -k2*ch A O.S*cl0 # r3 -k3*ch A O.S*c9 # r4 -k4*ch A O.S*c8 # rS -kS*ch A O.S*c7 # X = l-cll/cllo # cll0 = 0.1 V(O)=O # V(f)=0.8
#
6-65
006 0.05 0.05
0.04 0.03 0.03 0.02 0.02 0 . 01 0.01
0.00 0 . 00
0 . 03
0 . 16
0 . 24
0 . 32
0.40
0.43
0.56
0.64
072
0
V
• The ratio of hydrogen to pentamethylbenzene is 2.83 and the volume is 0.8 m polymath solution
3
•
CDP6-26 (b) The polymath program is the same as the first, we see that the value of c11(o)=0.092 and ch(o)=0.434 and the ratio now becomes 4 . 8 to 1 and the volume increase to 6.8 m3 to maximize SS9' To maximize SS7 it follows that the volume would be smaller because earlier the reaction ends the less C 7 is formed. 67.0 60.3 53 . 6
46 . 9 40 . 2 33 . 5 26.8 201 13.4 6 ..7 0.0 0.00
0.67
1.35
2 . 02
2.70
3 . 37 V
CDP6-26 (C) Individualized solution CDP6-27
6-66
4.05
4.72
5 . 40
6.07
674
I;:;.~'::' tftJ 1<=-09
o. S
~(fca)==ca·V-fca
V·"" 10
no
fho=lOOOO
fb=4320 fbo:..;:7200
k::",,2 . 7
Kl=.0264 k2=:.07 T=40J
Poo""1400 R;:::8.]09
PO;::: I?bo" , 6.
vo=2 .. 41e7/Poo
ph=fh-S.309*Q03,vo
Y"" fc / ( fbc.)~ fb} pc_fc*8.309-T/vQ
~h=(:b)-k2·K2*Pc·Ph/(1.K2-Pc)
rc::-irb) ~k2"'K2"'Pc*Ph! (1+K2".l;>C)
Val UE?
f C:,8
5:3.5618 82:32.08 €l52.1S:
U
umu
: ~1 0
1 Ul)ClO
ib
4:320
,ba
72DO 2,,7
k 1 Kl
k2 K2 Pbo Pb va
Pt,
Pho
5~826@
1] L8",7\'? 12 28-12"" 14
rc
O. ':1264 U .. 07
0.04
4U3 :400 8 .. 105 84(] '.7214.3
2.3541 2.85836 16Ul.31
The highest yield occurs at pressures: P flO:::: 1400 kPa
PHa
::::
19452 kPa
6-67
19";5,2 ,9. OS-j 26 0 .. 852181 -9.93644 8.232U8
CDP6-28 (a)
Mole Balances: dF + r2A dV D ..dF _=r. dV ZA --~ ::: riA
Rate Laws:
kl(c",Ci -- CcIK t ) -I'y. := k1 (c A CD - C1i;C SIK2 ) -riA::
-f3C ;;;
k)Cc:
Stoichiometry : C,
=(:,.(~J;)
Fy == FAt- Fat- Fe + Fo + FE + Fo
p
e'G =R'i'a Use these equations in POLThIATII- Vary Po and To to fmd the optimal conditions. We determine these to be:
T" =315.8 K
Po ::;: 160 atm <:;:<1;(:to" (fa/:;;;) • (';,'o/'rl
cb$Cto . . {fblf':)·(Tol:') cc=Cto"IEc/!c) "tTo/T)
al!.]
Id{v).~r2&
6
d (:0) lcl(v) "'4*rla· :::-2e· ::-3c
14
d( fa) Id(v) "r1a .. ;:2a
:5
dl!clJd(v).~:la·:Jc
Q
c.(:gJid(v):.;;-rJc
0
co.::;C-;,;o'(fd/f-;,;) "(To/T) ce~C~o~(fejf~)V(To!T)
:::3c"" .. kJ"cc rla=-k:*(ca~cb~2~CC/~1)
r'2a=-k2'" (ca."cd-ce/:~2)
4
1'0:=31.5
a
T"''I:o
k1.:=. 9::n"e;<;;i (2.5' (3140011. 5a7~ C./J30 "liT)))
K:,.13J.567" (0. DOl 98 "1:)"" 2 "'e;-cp LHl520.i L 987'" (1/1>1.,1 298) i
k2=. 6J6"t?:xpi 1300011.98'7" (U :'SOD-liT) l K2 =103943 *el':,;;J 198J4/!. . 987'" n. iTo, 1/298) ) ~:J::;
.. 244 *@.xp (1.5 w 28956 /! . 98 '{,. (1./ 32S-1./T) }
6-68
v
G
200
o
6
3.99S't2
5
J.A
1:.3038
1.4.
o
"!. .. 13754
£d it 1;'Q
k1
o o
O.68201H 1.13754
[} . 00:'0827
G.00441.766
30
27 A837
27,';:933
315.8
].15.8
'H58
:t60
l50
1.00
315.3
")
0.00428571
0.00,128571
O .. 0042857'\.
0.0042857:.
279::'12
279:112
2791.,12
2.SH122
2 881.22
2,831.22
4076: . S
40761.5
D.034 ..l899
0 .. 0.14]399
(} .0.HJ899
6.17366
6 ]:,865
o
O.I:l4:S214
15 g
') L5 .. S
40761..5
kJ
ca
l) ..
}lS 3
2.831.22
Ct.o
lB4792
0.:84792
3D
K1. X2
9 . 99558
6 178,,5 1. . B'573
1 23$73
2,88337
041.5214
2.88337
O.20249.B
L 48::'36
n.1S32S6
cd
O~82Ja2l
0.323821
0.000236652
o
cO:
1.23573
2.24735
1.23S73
2.24593
·c
.. 0.00596389
·"·,0.0:)22,5647
-{j _ :~440299
ria
-1.9JJ0'5
08099:251.5
.(1. [HJ527:J54
-0 . 1)02:564'] 4 GOG·He·OS MeChanQl Syncbesig
v ................. _._._ Ij
()
20 40
O.6357();an
60
0.Si:334782
0.4<1820445
ao
O . 8'l()427)
100
O.90B2S74j O.sn76642
o .8554.1262 o . 30470055 (} . 74650071 20a
(b)
o . 5a:S".l95
Use the same POLYMATH program as aoove and vary the ratio of entering reactants. The optimal ratio would be :
fs
hydrogen gas,
TI
carbon monoxide, and ~ carbon dioxide
These results are similar to those in paxt (a) in tllat the optimal volume is still 100 dm\ and the cQucentrarion profile is very similar in shape. 'nle primary difference is that the Fe values are more thar) doubled.
6-69
Mechanol synthesis
equations:
:tnitl.al value
d{Fa)/d(V)=~la+r2a
8
d{Fb)/d(V)=2·rla-~2a-r3c
16
~(Fc)/d(V)=-rla+r3c
o
d(Fe) IdCV) =-::2a
6
d(Fg}/d(V)=-r3c
o o
;1(Fd) Id (V)
~::2a
1:0=315.8
Po=160 TeTo J.CCC
r , I
2.~
t
t
t1 o.~o
1_.__._. .___
+__.___ _
. - , - - - + _ _ , -........;._.__
1().:C~
ea a:c
12~.a:{;
I ~.oc;:
2:)0::
u
kl=.933-exp«(2.S-(31400/l.987-(1/330-1/T})) Kl=131667-(O.00198·T)~2-exp(30620/1.9a7·(1/T-1/298})
k2=O.636-~(18000/1.9a7-(1/300-1/T})
3:2::.10] 943"ex;,') (9834/1. 987· (1/T-1129B) I ~J=O.244·exp(1.S·289S6/1.9a7~(1/325-1/T»
Cto=?0/(.082-To) Ca=C~o·(Fa/F~)·(TO/T)
C~=Cto·(!,b/Ftl·(To/T)
.Kee.h4l>ol. ~i.:I
.,
~val ...
!!:i.u-
0
200
0
lOa
.f-
a
a
1 94105
r.>
U
li
7 IIS7.a
1 U~Q5 7_aanS
Ce=Cto*(Fe/ft}·(To/TI
Fe
0
2 flU
Q
Y.
5
41
5 .. 9''''1
::3c=-kJ ·Cc
f'1i
I)
4 00756
\)
N
Q
n.
J,
o OGOStoln lO
0 11
To
llS I
Po
15.:1
l'
lLS i CI 0041ISU 2191 12
CC=Cto*C?c/ft}·(To/T) cd=Cto·(Fd/:~I·(To/T)
rla=-kl·(Ca·Cb~2-Cc/~)
r2a=-k2-(Ca*Cd-Ce*Cb/Kl)
~;',:;i&l
vsi.able
kl ~~
value
1.5 l60
a
5.""1
4 D07S4I
II 0005'JOala
lasS
US.
:1:...as
3l.S •• 0.004215'11
US.S
115 .. 8
O.. OO41IS71
2791.12 2 !la~22 40161 5 Q OJ4J."
nn
o 00421511. 2791..1l 2 IBU2
Ct.Q
, 178i&
, nan
C'""
<:t>
!. o4'15~ 1.29528
Cc:
0
C
0 ~.Zl'7)
•. U7i4 J lf528 o 7243" o Oool.Snn ~ Sil5l
.:lc: .:l..t.'
-0
-0
"0.Q7S07a
-".Ol.lUQ8
6-70
lOU"
ll.Sa 160
IeJ
El
F~l_l...
laO
l '41.l2 4~'Gl S o O14lU'
lcl
v.he
12 2 .. 111.22 401&1..5 o 0143899 , %.7866 0 . 5(7917
2 0
22U~
,0751.S
o OlUU9
, ::raSi Q
547'H7
:z 11,;a
0
o 57nJ3 o OOOUi7a.
I. :U571
1 S93'S!.
-0 Oa.uU -0 .. 076578
-a.Ol".,.
.. 0 . Qll.l .. oe
Met:hanol Synthesis
v -.-.--... ....... ..... .. .. ~
~
~"
"
Fc
o 20
1.2H3568
40
:: .. 0129151
60
2.4406221
30
2 6365715
100
2.680932
120
2.6292203
1'1.0
2 .. 5193659
160
2 376518
180
2.2171.669
2DO
2.0519865
CDP6-29 No solution will be given
CDP6-A a) A + B-1> C + D C-+B~E+D
First, find 'r. To do this use the original desigl1 equation for a CSTR:
V == .~AO~ -··rA Then since F . . a ==
CAOVO and 1;;:::
Vivo then the design equation becomes: . _ CAQX 1--' ·-rA Using the rate law and stoichiometry we find:
m·-rA = k i CAO .(1 .... X) Combining aU these and solving for't when X = .3, CAQ::::: .1 and kl ::::: .412 1: :;:::
~M!'F:.. . . . :;: : -.. --.~-~. ~)~-. : : : 1.04b
k\CAQ(l-X) .412*.1*.7 Once that has been calculated, redo all the mole balances: CAl) -., C A
::;;; --rA 1:
C BO · · · · C B ::;;; -rB1:'
Cc
==rc1:'
CD == Io1:'
6-71
Then do the rate laws: -fA
= klC A
,,·rB :::: klC A + k 2CC fC ;:;;;;;
fD
ktC A
-
k 2C C
== klC A -1- kzC c
fg =: k 2 C C Combining and rearranging into a function: fCC A );:;:: 0 == C A - CAO + k,cA-r
f( Ca):::: 0 == Cn ..... Coo + (k\C ... + k 2Cc)-r f(cd==O=(k,C ... -"kzCc)-r-Cc
feeD) = 0:;;: (k1C ... + klCC)-r~· CD f(C g )::::. a
=k C 2
C "f .....
CE
Plug those into POLyp.,IATII: 6-F ESFlations; f(ca)=ca-cao+kl·ca"cau f (cb) =cb·'-cbo";. (kl"'ca..-k2 "cc) .... tau f(ccJ;(kl*ca-k2·cc)*tau-cc
Solution Uanab!e
ca cb
cc
f(cd}·(kl·ca~k2·cc)·tau-cd
cd
f{c.J=k2~cc·tau-ce
Cf;'
cao=.l kl=.412 tau=1.04
C.ao
kl
__
U
0.0700015 0.10B382 0.0283726 Il03i6ia" O.OO16229i III 0.112 l.tM
cbo==.14
tau r.be
O,ii
k.2=.055
t2
o.ffi5
Cc ;::: .028 need Fco == 10 v
o
= Fe
V=
C·c
'tVa
;;::;: -~~0"8 .....
=357 gall h
= 1.04 *' 357 = 371.3gal
b)No solution will be given.
CDP6-B kt ~/I D A---76/'
~C
BatCh r~:.or: YAO == 1
6-72
3.J6le-15 -t.02Je-! I 9. 112e--13 "' 9. iOSe-· i 3 LSi'S!?! !
kl
(a)
;;
s.:C- 1
O.Ot
T
t
== 1..5 ruin::: 90 s
A-7B
NAO!2X = -fA V dt
CAO~= kl
CA
::::
ki CAo(i.X}
in Loo.1__);;:;; ktt
\l-X t
.--1- =: ct,t I-Xl
CA = C.-...n{l-Xd = CAO e-k\t
.fA. = cook,t ::;; c-{O.Ol}(90) ::: 0.41 CM
B--7 D
rSl=-r",=kIC",
XI
fC
=:
k:z CB
X2
fC>
: :; ; 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 <.~+ k2.,'C+ k3 t) kl t
6-74
~; __1_ z:< CB (1
+ \::2 1: + \::3 1:) CM \::1 1:
CAo
1 + \::1
~=
\::11: (l + kl "C)(1 + kl1: + kJ 1:)
CAQ
1:
t(s)
10
Ca/C,Ml
0.08
50 0.26
100 0.33
150 0.34
200 0.33
CDP6-C First~ find the values for k. kl :::: 298ge("·IO,~7/(OO19S7"100»
k1, ::;;: 9466e(-l5ll1(OOI9&"7-100}} In this problem
= L2 ::::
k J = 11127eC'1500(OOI9117°100l) Isoburylene :::: I Merhacrolein :::: ,M
.18
:::;; .22
CO,::::D 4
C0 ::::C
Oxygen:::: 0 The mole balances for these species are as follows: F~o ... FI :::: ('I'll ,.- fu - IJr)V
F.'>i = riM V Fo =r:m V
Fe = flO V Oxygen is in excess so we will assume that F" ::: F 00' The rate laws for these reactions are as follows: . "IiI :::::--110 = riM = kLC1C OO -lil::;;:
25
(" ('""'00 ::;;:41 rm:;;;: k 2"1
6-75
300 0.3
-··r = 4rJO = . ~.r. == k C C
-17 4 3C 3 1. 00 Combine all of these and come up with the following: Cxo -C 1 = (kIC[C oo +. kZC1C oo + k 3C rC oo )'r J!
Coo - Co = (k1C1C OO +~t-'k.tC!COO
+~}k3CIc()(,)'r
c,1.f == kIC/Cort' CD =4 * kzC\Coo't C c == 4'" k3CrCoo't Before plugging into POL YMATH, evaluate the parameters and derive equations for conversion. C . == ...P. . : ;: : __........3__.__ : ;: : 0.034 1 RT .082 * 700 C/O == YAoCr .1 *' .034::;::: 0.0034 Ccq ::::; 0.031
=
X [ ::::: S;lQ:;:5':1 C[O
X
u
= .c . . :.\':! ('
'[0
X ;;; _~::!L 21
4('
'\.o
X){::;::: XI ······X 11 ······X2! Then plug into POL Y~'1ATH to get the following: !n!ti~!-....y!!!.~~
Equa.t!~.:..
f (ci.) ""c:i·"ci.o+ (kl. -c:i. "c:oo+k2 ~ci. ·.:;:oo+1<:.J "ci "coo) ·cau
0 . OU1
f (cmJ =kl *ci. ~coo"c:au· . cm
'0,.001
f (cd) ='1 "k2 "'ci "coo'"::;au.··cd f(cc)=4,"ci"coo"tau-cc f {col =co···coo+- (lel. ·ci."c:oo+25/ 4 "k2*c:i"coo+17/2 "'k:3 "ci "'cool "cau
O.Q04. 0.004 0.01
c:i.o=.0034 kl.=l. . :2
coo=.03l. }<2;; .. 1.8
k3== .22
t;.au=lO xl.=cm/cio
x2"".2S"cd/ci.o x=={cio'ci)/ci x.3=x ··:x1 .... x2
6-76
valuE' Cl
0.00227273
-1_775e-12
ern
0.Q0084-5"155
Cd
0.000507273
1" 0 11 e--12 -! 38:&-12 J.B3::'€' i:3
C.Cl028:818
D.CC3"'
1.2 O~C31
Coo k2
0, L8
C.22 10 D.2"8663 Cl.03?29S5 C . 495 0.21 DC]?
x2 x
xJ
Xl;;; 0.496
X2! ::;::
0.037
X~l
::;:: 0.21
CDP6-D A-··tB C--3<[)+E Fr :::: 10 Ibmollsec dF
.,••• ~, :::::: fA
dV
;;;·k j C A
F.~::= CAV o
"" -.. -g:::?---.--.
C 10
:. Vo
0.73*900 = 13140
::: 7,.61 * l{r{i
V == l005ft'
See POLYMATH solution below d{ fal Jd (VI z;ra
5
d (fe) /d(V) =rc
5
d (fb) /d {V) "'ra
d{fd) /d(V)",z:-c
o o
d(fe)/d(V)::="rc
Q
kl"'lQ k2"" .. OJ
6·77
ft=fa~·fc t·fb+fd+ fe
fao=5 cto=S/(900*.73) fto==1.0
rc::;-k2 X'" (fao·~ f.a)
J fao
avo=fto/cto ca=cto'" (fal ft)
ra=-kl*ca y~~
:nJ:Sial"
V
0
fa.
fc
MinimU\~,,_~~b.~
F i~!Y£!h\±~
lOS
0
5
5
:2.49621
lOS 2.49621
5
5
1. 65
l.8S
fb
0
2.50379
0
2.50379
Ed
0
3.15
0
fa
0
3.15
0
3.15 3.15
kl
10 0.03 10
10
10
10
0 . 03
0.03
0.03
13 15
10
13.15
1<2
....... ""~ :ao
y~~ ~~i~~,,,.~~l lie
5 0 . 00761(}35
5
::;
5
cto ft.o
0 . 00761035
O. 00761035
10
10
0.00761035 10
rc
··0 03
·0.03
-0.03
-0.03
x avo ca
0
0.500758
0
0.500758
1314 {) .. 00.3 8051.8
1314
131.4
1314
0.00330518
0.00144464
0OOl444€4
CDP6-E a) Using the equation for the equilibdum constants:
We can come up with the equations for C A , CD, and ex.
CA =. ~~·f.q D . KelCH CD :::: .!!!~ACB.
Cc
ex = !;!.1.Cc~fL Cy
The test can be found with stoichiometry. Cy =Cx CD:;:: RO _·CD,·Cy
e
Cc =cAO·······cA -ex
6-78
10
POLYlv1ATH 6--5
6--5
.-:1.
~ffildt.~.
Sotl".)':lon
nc 'cb)
f (ca) "'ca-··cc 'cdl
Var!abl~ _ _ ·_···N.' ____
E (cdl :cd-'K: ·ca.·cb/cc t ~cx) -=cx"'K2""cc "cb/cy
.~_._
.•
_~"_.
ca
K2=1 c.y=cx
__
Ualue.•" ••.
.,,_,~_._
iO
_~
cx
Cl.CJ306182 0.109652 0.714386
Kl
4
cd
Kl=4
3.~J4ge-19
!.08 4 e-,t8 L 1 05e-' 16
K2
cbo=LS c'iO~l
d
.. :;
cc=cao"'ca--c:x
Cl.J
0.714'386
cbo
1.5 1.5
};;"(ca.o-"c3.1 Jcao
CoO
cb"'coo-cd-,cy
cc
0,754995 0.979588 0.675961
x cb
C,4, =0_0306
ex
=0,71
C e =0,68 C y =0.'11
We also fmd that X :::: 0.98 b) With the new equatjon we must fmd the new equilibrium equations.
ex =Cy _···Cz C,A
= CAli ······CD·CZ
E:g;uat:i£s~ ;, flc4)=c~-k3·ca·cx
10
f (c:yl :=cy--k2 ·Cc: .. c:b/c.x
cz
flcd)=cd-kL·ca-cbicc
eLJ
k;3",5
cd cx
cx:;.r;:;:y~-cz
CdC
1<2:..1,,51.
ct:>Q
T
sdz
cao=l
S
cbo=,L 5
sdz;-;tcdicz
0 . 990725 t .07:384 CL 2681 0 1 0.0831 135 ! .5 1.5 :300 U.27C6 i 1
1.08389 k3
c"
k2
49.4255 0241174 D . 480271
cb
0. ; 5806
ca=cao'-cd--cz.
k 1
cb:;:cbo-c:d-cy
x cc
8.26888 0.839217 : . 17571 1
1.
<~
-r '.
6-79
......
~
1"" ......
'''':;J:;:'
! 09212--13 Lt"''1e-12 2 .s07c'-15
Wefmd that Cz =0.65
C y =0.91 C a =0.22
C/o. = 0.48
ex =0.27
CD =0.37
C c =0.76
X =0.68
Sex =2.8
SDZ
Syz =1.41
=057
c) When the temperature is raised from 300 K to 500 K, Sex goes down,
SDZ
goes
up and SyZ goes up. d) First find the proportionality constants fwm the A. rrhenius equation. 0.002 ::::: A~
A1e-n()Q(X)/1 911YHX})
== 38603
O.. 06 -
A
'''"\l
e
-(1(,(,'00/1.987' ;(0)
A2 == 2..24 x IOu 0.3 ::::: AJe,Pt)()OO/1 987".1(0) AJ
::::
2.16 X 10 21
Once those are known, come up with equations for the equations in terms of space-time 'to .w . -. . . . .C A . .,..._ 'r: :::: ..............,C . . ._ .. _ .. kj + k 2C,;, +- k3C~ rkl + rkJCA t· ti:~C~ - C1{)
+ C,4 :::: 0
r'·· . . ' '· . ''''''', . . ;...
.
.... "." ............,.....................
··. ··(;:/{,+1}+ I(rk, +1)-·4(rkJ )(rk·C ) . ,·. ··. . ·\J '. 1 AO . __···. . . _-..·. (.'~,\ = _ . . . ··_.. _
;;·;..i··. ·. . ·. . . _. ·. . . ·. _. -.. . . . . . . . . . . . . . . --.-. - \.. 3
··c
I"= .,..... ZL
·. --k 1
-.. en
r == ...k1CA
'i
=--. --c . <~;
··k>C;
ex ;:; kl r
CiJ :::: k:.1.CA Cy = k,'t't:1 Use EXCEL solver in order to find the temperature that maximizes Ca T
Ca
312.5336841
0.016513937
k1 0,Q03919233
Gx 0.039192325
k2
Cb
0.230891512
k3 2.260443602
0.038129279 Cy
0 . . 006164458
CDP6-F (a) Mole balances:
6-80
dFA/dV = -rD-IU dFn/dV =rD dFu/dV =IU Rate laws:
= klC~ ru = k 2 CA
kl = 15 fellbmol.s
rD
k2 = 0.015
S·I
Stoichiometry: C A = CTo(FA/FT), F1 = FA + FD + Fu Cost = 60FB - 15Fc -lOFAo Using these equations in polymath to find the necessary volume to maximize Cost:
v = 1425 fe POLYMATH Results Calculated values of. the DEQ variables Variable initial value minimal value V
0
0
fu fa
0 0.06705
0 0.0018652
fd
0
kl k2 faa ft cao ca Cost rd ru
0
15 0.015 0.06705 0.06705 0.00447 0.00447 -0.6705 2.997E-04 6.705E-05
15 0.015 0.06705 0.06705 0.00447 1.243E-04 -0.6705 2.319E-07 1.865E-06
maximal value 2000 0.0237312 0.06705 0.0414537 15 0.,015 0.06705 0.06705 0.00447 0.00447 1. 46684 2.997E-04 6.705E--05
final value 2000 0.0237312 0.0018652 0.0414537 15 0.01.5 0.06705 0.06705 0.00447 1..243E-04 1. 4607519 i.319E-07 1.865E-06
ODE Report (RKF45) Differential equations as entered by the user [1] d(fu)/d(V) = ru [ 2] d(fa)/d(V) = -ru-rd [ 3] d(fd)/d(V) = rd Explicit equations as entered by the user [1] k1 = 15 [2] k2=0 . 015 [ 3] fao = 0 . 06705 [4] ft = fu+fa+fd [5] cao = 0 . 00447 [6] ca = cao*(fa/ft) [7] Cost = (60*fd-15*fu-1 O*fao) [8] rd = k1 *(caI\2) [ 9] ru = k2*ca Independent variable variable name: V initial value: 0 final value: 2000
CDP6-F (b) Mole balances: FA =FAo + IAV
FD =rDV
Rate laws: fA = -ID - IU
ki
Fu = IUV
rD = klC~
= kio exp ( ~ )(~-
ru = k 2 CA
91;.67)
F; = VOCi
Cost = v o(60FB -- 15Fc ·,lOFAo ) Use these equations in Polymath and vary T from 860 0 R to 1160 0 R and find maximum value of cost.
6--81
Cost steadily rises with temperature and reaches a maximum at 970 0 R(51O.33 0 F).
pOLYMATH Results NLES Solution Variable ca cd cu cao E1 R T k20 E2 k10 vo V tau cost k1 k2 rd ru ra
Value 0.0023213 0.0016225 5.262E-04 0.00447 10000 1.987 970 0.015 2.OE+04 15 15 400 26.666667 0.6713603 11.292187 0.0085009 6.085E-05 1.973E-05 -8.058E-05
f(x)
5.354E-08 -5.354E-08 -2.76E-15
Ini Guess 0,,00447
o o
NLES Report (safenewt) Nonlinear equations [1] f(ca) = ca-cao-ra*tau = 0 [2] f(cd) = cd-rd*tau = 0
[3] f(cu) = cu··ru*tau = 0
Explicit equations [1] cao = 0.00447 [2] E1 = 10000 [3] R = 1..987 [4] T= 970 [5] k20 = 0,,015
[6] E2 = 20000 [7 J k10 = 15 [8J vo=15
[9J V = 400
[ 10 J [11] [12] [13]
tau = Vivo cost = vo*(60*cd-15*cu-1 O*cao) k1 = k1o*exp«E1/R)*(11T-1/919.67)) k2 = k2o*exp«E2/R)*(11T-1/919,,67)) [14] rd = k1*(caA 2) [15] ru = k2*ca [16] ra = -k1 *(caA 2)-k2*ca
-----_._---------------------CDP6-G3
6-82
Mole Balances:
dC,
at
:= f\
de ....: ..,.::;: I'f dt
dt Rate Laws:
C
=.wk,C. .". ,. . C~
..:.,.-~.:.."
It·KAC~
Stoichieme try :
-4 k,meJ)'rlurn'' C~'Eo == (';'Y
41
:=)...
0 mo1Id . m3
from Henry's Law: C ~,' ';;:;; 5.9 MPa(O.058 kmol/m J . MPa)= 0,3422 mol/dm! Use these equations in the following POLYMATH program to generate graphs of CA' C,,_ and C F as a fUIlction of time
eL'
d(ea) /d(e) "":co.
o
d(cb) .ld(e) "'1:0
540
o o o
di,ec) Id(:.:.) "':::c d(ce)/d(t)=.:;e d(c:::) /d(e)=;::
kl",O . 000468
Ka=:22 . 76 k2",O . 000227
k4=O.00147 k3=;O .00282
rl:w*kl*ca-cb/(l.Xa-ca) r2=w*k2*ca*cc/ll+Ka*ca) :4=w~k4·ca·ce!(1+Ka·cal
ro"'· r1
20
6-83
3422
20 ll.3422
o
ca
o o 3422
9 l.B51'7e-20
20 9 1851-7",--20
c~
540
540
539 658
SJ9
cc
o
0,311641
o
0.205615
ce
o
II 136398
o
cf
o
6 .. 37928e-05
o
0116398 6 ,.3 7926e-·05
10
10 0 .. 000468
10
10
kl
0,000468
o
Ita
22,76
k2
22.76 0.000227
0.000468 22.76
0.000227
0 . 00022'1
0.000227
k4
0 .. 001.4.7
O.00H'7
0.00147
0.00147
k3
0.00282
0.00282
0.00282
0.00282
o
o
0.00580002
2.319S1e,,19
2,31981",-l.9
o o
1 .. 84167e--22
o
r3 r1
o
000468
00873821 0.0984025
0984025
22,,76
1::'2
o
1. 621e-·05
::'4
o
4.96363e .. 06
rb
··0 .. 0984025
"2 .. 31981e-1.9
.. 0 09811025
-231981.", .. 19
re
o a
000878695
0.00560002
.. 2.32208e .. 19
n o ··n .091H025
0 .. 09811025
·'0.,00834598
-0.00580002
rf
2 .. 07778e··05 0,0984025
:::a
0.0984025 ~.
-;:
~.~.
:.n:
ca ec eEl
Ca, Ce. Ce vs
t
;:, ;,,::
:l.,lea
cf vs
! /"" t / 'rj
t
j
/
/
1.iL/..
• •• ~ .... _.~._ ••• ~ ••• B ...... _ _ _ _ ............ .
~.
ago::
2,27051e··22 ·2,32208e·l'}
Cb vs
t
::Lt-:
cf
4.2S838e·.;n
B ::lC::
CDP6-H
6-84
t
The equilibrium constant for the reaction
+::.
trOL
cs--OL
can be csrimaccd as a function of tcmpcratuIe from (he mole fraccion data below t(X)0c. _ Yes K --_. P Ya:
T
323
375
Y:t
.8
. 75
Yc'!;
.2
. 25
Kp
.,-)
,
33
In Kp
1
lCooff
3.09
-, -
39 ··1 11
2.63
}
__ ~39J::J:.!.n.x 1000 -:= -683 "" ~~it 3_09 - 268
R
6.H '" . . I 3 :51 _.S£l rool
CDP6-I 2,9-17 CDP6-J CDP6-K 2, 9-13 CDP6-L Given: liquid feed
to CSTR
L1tttid Peed. CAll ... 0.4 tm.clJI Cl'O ...
0 ..4
[ttlOlil
V"", 1201
6-85
With the following re:lCcon s.equence )::1
A --~ C
.. rl
k2
:;;:
k! C A
)::1 :::
0 . 0 t min- 1
=- 0.02 min- t
A·--,1- B k3 B -..,. C
-£'z ;: )::2 C A
)::2
·r3 ::: );:3 CB
k3 = 0.07 min-I
F -..,. B + D i
-r4 ::: k... C~
k.. ::: 0 ..50 Vgmol - min
k."
1<2 k3 (a) Since C is an ~ product. formed through intermediate B by either A -...,r B -7 Cor k.". 1::] ~l F -...,r B + 0 i -.;. c. or din:ctJ.y from A ~ C. the m.aximum concenmuion of C occurs when all A and F have b:.:en convened to C: (b)
Cm.u;: CAO + CFOr (with E == 0).
Both A and F an: only decomposed by the above scheme:
Bahnee on A: ~ CAO· 'U() CA",:(.rA) V;: (-rl • fZ) V ={kl + k~d c" V
CAO:.C~ CA
;: (kl + kl)·Y.. "" {kt + kt} 1: Uo
C~ __ ==
or C" ;:
1 ... (k! :
)::2) 't
0.4 gmol!! = __0.4 gmol/l 1 + (0.01 + 0.02) min~l (21J~') t -+- (0.03) (60)
C,,;: 0.143 gmat /1 Balance on F: \10 Cr.J· Uo C;; ;: (-fF) Y ;: ~ C~ V . V.:.r. . wnen:: 1: :::--::; \N" rrun Uo
Solving we get: CF = O.lgmoVI Balance forB:
O-voCB
= (-rB)V = (k3CB -k 2 C A +k4 C;)V
(k 2 C A + k 4 C;)T (0.01min"lxO.143gmo1/l + 0.51.mor l min-'l 0.lgmo1 /l2)60min CB = 1 + k3T -= 1 + 0.07min-l 60 min
C B = 0.0907 gmo1 / 1
= (--r~.)V = (-k3CB -klCA)V = (k3C B + klCA)T = (O.07min- l xO.0907gmoll1 +O.01min-1 xO.143gmo11l)60min
Balance for C:
Cc
O-voCe
C c = 0.467 gmol/ I
=
Mole fraction of C =
CA+CB+CC+C F
0.467 0.143 + 0.09 + 0.467 + 0.1
6-86
= O.~~z. = 0.583 0.8
CDP6-M 4A + 5B -:7 4C + 6D
-ljA
= k1ACAC;
2A +1.5B -:7 E + 3D
-rZA
= kZACACB
2C + B -:7 2F
-r3B
=k3BC~CB
4A + 6C -:7 5E + 6D
-Y4C
= k 4C CCC1 13
Rate laws:
=fIA + f2A + (2/3)f4C -fC = -fIA + 2f3B + f4C
-fB
-fA
-fE
= -O.Sf2A -
=
1.2SfIA + O.7Sf2A +f3B
= -l.SfIA -fp = -2f3B
-fD
(S/6)f4C
Using these equations inpolymath to find the exiting molar flow rates. POLYMATH Results NLES &.lutioll NLES Re,Qort {safenewtl Value Variable Nonlinear equations 0.998927 ca [ 1 ) f(ca) ::::; vo*ca-fao-ra*W ::::; 0 [ 2)
[3) [4) [5) [6)
f(cb) ::::; vo*cb-fbo-rb*W ::::; 0 f(cc) ::::; vo*cc-rc*W ::::; 0 f(cd) ::::; vo*cd-rd*W ::::; 0 f(ce) ::::; vo*ce-re*W ::::; 0 f(cf) ::::; vo*cf-rf*W ::::; 0
Explicit equations [1) vo::::; 10 fao::::; 10 [ 3) W::::;3 [4] fbo::::; 10 [5) rho::::; 00012 [6] k1 ::::; 5 [7] k2 ::::;2 [8) k3::::; 10 [9] k4::::;5 [10J fa::::; vo*ca [ 11] fb::::; vo*cb [12] fc::::; vo*cc [1.3 J fd::::; vo*cd [ 14] fe::::; vo*ce [ 15] ff::::; vo*cf [16] 1'1 ::::; rho*k1 *ca*(cb"2) [17] r2 ::::; rho*k2*ca*cb [18J r3::::; rho*k3*cb*(cc"2) [19J r4::::; rho*k4*cc*(ca"(2/3» [20] rf::::; 2*r3 [21] re ::::; O. 5*r2+(5/6)*r4 [22J ra ::::; -r1 +r2-(2/3)*r4 [23J rb ::::; -1 .25*r1-0. 75*r2-r3 [24] rc ::::; r1-2*r3-r4 [25] rd ::::; 1 . 5*r1 + 1 . 5*r2+r4 [2)
cb cc cd ce cf vo fao
0.997227 0.0017849 0.0037612 3.613E-04 1. 285E-12 10 10
W
3
fbo rho k1 k2 k3 k4 fa fb fc fd fe ff rl r2 r3 r4 rf re ra rb rc rd
10 0.0012 5 2 10 5 9.98927 9.9722697 0.0178487 0.0376119 0.0036129 1. 285E-ll 0.0059604 0.0023908 3.812E·-08 1.07E-05 7.625E-08 0.0012043 -0.0035767 -0.0092436 0.0059496 0.0125374
6-87
l.Sf2A - f4C
f(x) -1. 05 6E-15 1. 08 2E-15 -1.422E-16 -1.041E-16 -1. 73 5E-18 1. 254E-13
Ini Guess 1 1 0 0 0 0
3
(b)
Y AE
FE 3.6x100336 = = = . FAa -- FA 10-9.989
Y BF
=
7
FF F Ba - FB
YAC
=
_
2.29xl010-9.973
= 8.25xl0-6
Fc _ 0.0178 = 1.663 FAa - FA 10-9.989
CDP6-N 3, 6-21 CDP6-0 3, 6-25
6-88
Solutions for Chapter 7 - Reaction Mechanisms, Pathways, Bioreactions and Bioreactors P7-1 (a) Example 7-1 The graph of loll will remain same if CS 2 concentration changes. If concentration of M increases the slope of line will decrease .
P7 -1 (b) Example 7-2 For t = 0 to t = 035 sec, PSSH is not valid as steady state not reached . And at low temperature PSSH results show greatest disparity. See Polymath program P7-1-h.pol. POL YMA TH Results Calculated values of the DEQ variables Variable t C1 C2 C6 C4 C7 C3 C5 C8 CP5 CP1 k5 T
k1 k2 k4 k3
initial value
minimal value
0 0.1 0 0 0 0 0 0 0 0 o. 1 3.98E+09 1000 0 . 0014964 2.283E+06 9 . 53E+08 5.71E+04
0 2 . 109E-04 0 0 0 0 0 0 0 0 2.166E-04 3 . 98E+09 1000 0 . 0014964 2 . 283E+06 9.53E+08 5 . 71E+04
maximal value
final value
12
12 2 . 109E-04 L311E-09 3.602E-09 1.276E·-08 0 . 0979179 0.0012475 0 . 0979179 6.237E-04 0 . 0979123 2 . 166E-04 3 . 98E+09 1000 0 . 0014964 2 . 283E+06 9.53E+08 5 . 71E+04
o. 1
L311E-09 3.602K-09 2 . 665E-07 0.0979179 0 . 0012475 0.0979179 6 . 237E-04 0 . 0979123 0.1 3 . 98E+09 1000 0.0014964 2 . 283E+06 9.53E+08 5.71E+04
ODE Report (STIFF) .t.Oe-9 , . . - - - - - - - - - - . - - . - - - ,
Differential equations as entered by the user [1 J d(C1 )/d(t) = -k1 *C1-k2*C1 *C2-k4*C1 *C6 [2] d(C2)/d(t) = 2*k1 *C1-k2*C1 *C2 [3] d(C6)/d(t) = k3*C4-k4*C6*C1 [4] d(C4)/d(t) k2*C1 *C2-k3*C4+k4*C6*C1·· k5*C41\2 [5] d(C7)/d(t) k4*C1 *C6 [6] d(C3)/d(t) = k2*C1 *C2 [7] d(C5)/d(t) = k3*C4 [8] d(C8)/d(t) Q..5*k5*C41\2 [9] d(CP5)/d(t) = k3*(2*k1/k5)A05*CP1I\O.5 [ 101 d(CP1 )/d(t) = -k1 *CP1-2*k1 *CP1(k3*(2*k1 /k5)1\Q . 5)*(CP 11\Q . 5)
32e-9
=
~
2Ae-9
=
16e-9
•
(''l
..
('6
=
S.lle-HI
OOe+O
P7-1 (C) Example 7-3 The inhibitor sho'NS competitive inhibition .
See Polymath program PT··l·e.poL
7-1
"'---~------
0.0
24
·tS t
72
9.6
120
0.40..--------------.. 0.32
- I'_inhibitor
0. 08
r
-
_.-
--'
O.OOL------~--~--------l
S.Oe-3
7.0e-~~llr~ge-3
6"Oe-3
9.0e-3
LOe-2
P7-1 (d) Example 7-4 l)Now Cure. = 0.001mol/dm3 and t = 10 min = 600 sec. t
= '~ln(_I_)+.CureaX VMAX
I-X
_ 600 sec _
VMAX
3
3
0.0266mol / dm 1n ( -1-) + --L (O.OOlmol / dm )X _ ____ _..L-_ 3 3 0.000266mol / s.dm 1- X 0.000266mol / s.dm
Solving, we get X =0,,9974. 2) For CSTR, r = t = 461.7 sec
CureaX
T=----_... r
.- rurea
-
urea = CureaX
r~rea
VMAXCurea
=K
C
M + urea
T
0.000266mol / s.dm 3 xO.lmol / dm 3 (1- X) -_ -(O.lmol / dm )X . -,---._-3
~
O.0266mol/ dm 3 + O.lmol/ dm 3 (1- X)
Solving, we get X = 0.,675
See Polymath program PT, I·e.pol. POLYMA TH Results
NLE Solution Variable X
Value 0.6751896
fIx)
-8.062E-10
Ini Guess 0.5
NLE Report (safenewt) Nonlinear equations [1]
f(X)
=O,,0000266*(1-X)/(O.0266+0.1 *(1-X))-O" 1*Xl461. 7 = 0
3) ForPFR,
T=
CO
dC
Cum
-
f
urea r
and
Curea
7·2
= CureaO (1- X)
461.7 sec
=>
Curea X = -k M- In ( -1-) + --"--""--
r
VMAX
VMAX
1- X
x=o ..8
Same as batch reactor, but t replaced by r
P7-1 (e) Example 7-5 Y
= - t,.cs = - (238.7 - 245) = 2 18 /
SIP
5.03-2.14
J'j.C p
. g g
1 1 YplS =--=--=0.46g/g YSIP 2.18
1 YSIC +P = y - C+PIS
1
0.075+0.46
= 1.87 g / g
Yes there is disparity as substrate is also used in maintenance .
P7-1 (0 Example 7-6 1) if we go for 24 hIS, fermentation will stop at 13..2 hIS as Cp
= Cp '
.
See Polymath program PT-l-f-1.pol. POLY'!yIATH Results Calculated values of the DEQ variables Variable . --t Cc Cs Cp r'd Ysc Ypc Ks
m
urnax rBm kobs rg
initial value
minimal value
maximal value
final value
o
o
13 . 2
13 . 2
1
1
250
39 . 292786
0 . 01 12.5 5.6 1.7 0.03 0.33 0.03 0.33 0.3277712
0.01 12.5 5.6 1.7 0 . 03 0.33 0.03 0 . 0039386 0 . 0624825
16 . 558613 250 92 . 981376 0.16559 12.5 5.6 1.7 0 . 03 0.33 0 . 4967701 0 . 33 2.1455962
16.550651 39 . 292786 92.981376 0 . 1655065 12.5 5.6 1.7 0.03 0 . 33 0.4965195 0.0039386 0.0624825
o
o
ODE Report (RKF45) Differential equations as entered by the user [1] d(Cc)/d(t) = rg-rd [2] d(Cs)/d(t) = Ysc*(-rg)-rsm [3] d(Cp)/d(t) = rg*Ypc Explicit equations as entered by the user [1] rd =Cc*0 . 01 [2] Ysc = 1/0.08 [3] Ypc = 5 . 6 [4] Ks = 1 . 7 [5] m =0 .. 03 [6] umax = 0 . 33 [7] rsm = m*Cc [8] kobs = (umax*(1-Cp/93)"O.52) [9] rg =kobs*Cc*Cs/(Ks+Cs)
7-3
2)Semi-Batch reactor:
See Polymath program P7-1··f-2.pol. POLYMATH Results Calculated values of the DEQ variables Variable t Cc Cs Cp
rd Ysc Ypc Ks m umax rsm kobs rg Cso vo Vo V
initial value 0 1.0E-04 1.. OE-04 0 1. OE-06 12.5 5.6 1.7 0.03 0.33 3.0E-06 0 . 33 1.941E-09 5 0.5 1 1
minimal value 0 1.0E-04 1. OE-04 0 1. OE-06 12 . 5 5. 6 1.7 0.03 0.33 3.0E-06 0.3294925 1.941E-09 5 0.5 1 1
maximal value 24 0.0474697 12.206266 0.2748298 4.747E-04 12.5 5.6 1.7 0.03 0.33 0.0014241 0.33 0 . 0137289 5 0.5 1 13
ODE Report (RKF45) Differential equations as entered by the user [1] d(Cc)/d(t) = rg-rd [2] d(Cs)/d(t) = vo*CsoN + Ysc*(-rg)-rsm [3] d(Cp)/d(t) = rg*Ypc Explicit equations as entered by the user [1] rd = Cc*0 . 01 [2] Ysc = 1/0 . 08 [3] Ypc = 5 . 6 [4] Ks = 1.,7 [5] m 0 . 03 [6] umax 0.33 [7] rsm =m*Cc [8] kobs = (umax*(1-Cp/93)1\().52) (9) rg =kobs*Cc*Cs/(Ks+Cs) r10] Gso = 5 [11] vo = 0 . 5 [12] Vo = 1 [13] V = Vo+vo*t
=
=
7-4
final value 24 0.0474697 12.206266 0.2748298 4.747E-04 12.5 5.6 1.7 0.03 0.33 0.0014241 0.3294925 0.0137289 5 0.5 1
13
0,30
15
0,24
12
018
9
~
0.12
6
- C)
0.06 (100
0.0
3
4.8
96
t
14.4
19,2
0
140
0,,0
4.8
9.6 t
144
19.2
24.0
3)Changes from part(2)
See Polymath program PT-l·f·3.po1. POL YMA TH Results Calculated values of the DEQ variables Variable -t Cc Cs Cp rd Ysc Ypc Ks m umax rsm kobs Ki Cso vo Vo V rg
initial value
minimal value
maximal value
final value
0 1 . 0E-04 1.0E-04 0 1.0E-06 12 . 5 5. 6 1.7 0.03 0.33 3 . 0E-06 0.33 0. 7 5 0.5 1 1 1,941E-09
0 1. OE-04 1.. OE-04 0 1.0E-06 12,5 5.6 1.7 0.03 0.33 3,OE-06 0,,3299991 0. 7 5 0.5 1 1 1,941E-09
24 1.. 514E-04 12 . 823709 4.669E-04 1.. 514E-06 12.5 5 "6 1..7 0.03 0,,33 4.541E-06 0 . 33 0.7 5 0.5 1 13 8 . 215E-06
24 1.514E-04 12.823709 4.669E-04 1 .. 514E-06 12.5 5.6 1..7 0.03 0.33 4.541E-·06 0.3299991 0.7 5 0. 5 1 13 2 . 568E-06
ODE Report (RKF45) Differential equations as entered by the user [1] d(Cc)/d(t) = rg-rd [2] d(Cs)/d(t) = vo*CsoN + Ysc*(-rg)-rsm [3] d(Cp )/d(t) = rg*Ypc
15
Explicit equations as entered by the user [1] rd = Cc*Q.,o.1 (2] Ysc=1/o.,o.8 [ 3] Ypc = 5,,6 [4] Ks = 1 . 7 [5] m=o."o.3 [6] umax = 0.,,33 [ '7 1 rsm = m*Cc
9
12
6 3 0
7-5
~---~----~----~----~----~
0,0
48
9,,6 t
14 4
19,2
240
[8] kobs:;:;: (umax*(1-Cp/93)J\().52) [9] Ki 0.7 [10) CSO 5 [11) vo = 0.5 [12] Vo=1
=
=
[13] V = VO+VO*t [14] rg kobs*Cc*Cs/(Ks+Cs+CSA2/Ki)
=
1.00e-4
O.Oe+O
9.6 t
0.0
14.4
19.2
IL-_~_~
0.0
24.0
4. 8
4)After 9.67 rus, C s = O. 300,--------· 240 180
c~c
-
c.s
120
C)
60
oL-=:::::;;;;;;;~====::::::::::=U
o
P7 -1 P7-1 P7-1 P7 -1 P7-1 P7-1 P7-1 P7 -1 P7-2
2
4
t
6
8
10
(g) Individualized solution (h) Individualized solution (i) Individualized solution
(j) Individualized solution (k) Individualized solution (I) Individualized solution (m) Individualized solution (n) Individualized solution
Solution is in the decoding algorithm given with the modules.
7-6
_ _ _ _~
9.6 t
14.4
19.2
24 . 0
P7-3 Burning:
GO' '~OH' y~
-+i,)H·
then
-reI.
= 0 =ks(HCl)(H) -- k6(CI)(H)
7-7
Cl _ k} (Hell • k6
.+kj (o!};. 3k4 (H·) (O:d' kz (B- }(02l' 2ks (HCl) (H-)
fH =:
.:_ .
r'-------.. . . . . .,,----·,---··,'''--.. ,,·,,·----lr,I_!,:_:, 4k ~~=~ ),.~._~,~,~ ~.~.) _~.~_{~.1~: l.~ .( f:.:.U ConstamVoiurne
e'.btdCH·· -' (""H *b*"hle -ae··hl dt Using the integraring f:lctor d(e-!:l!C ) ---"........-..H,-,-:::: a e-,be dt
C£+ t ::
0
CH
=: ..
:::;
'~' +. KI eot
therefore K 1 == ~
0
,--'_.,,-:' ..
....
' ' -]
tc~~ :. t ~~~.~~l~ '
'''
> ks {HCI)
) then "b" is positive
> kd02}
, then"b" is negative
If
k4 (C'}z)
If
kj (HCt)
,
..".
,,,e"
1/ t .........". "" . ,."."_..".....,,'-.. _ .. , ,............, '"''''''
7-8
=·a··p(C\-{.)·
p(CH )= 3k,! (02 '[:f{
= CCJo
3k4 (OJ a::::: 4ls (0 2 ) p:::::
t "'"
····..
0
Ceo
·····~··-·-··························-··········~··························-·-······
r,.. ~~ i~·..CO :::; Cc.O-o -
.1
at·· p _aLt· ..; c" .... pi'1
._.~_~.__ . . . . . ._._J
L. . __.. .
Reaction Pathway with HGL Reaction Pathway 'lifo HCL
[• . J. . . . . . . . . . . . . . . . . • ~)~
[~;~~
1-_~_ . 1 J:1 [1_ . ]
I2:,
-lH-
-
'
..
[ OR
--T
.I
[c~J [J~J
····I ] l CO
P7 -3 (d) Individualized solution P7-3 (e) Individualized solution P7-4 The re::!crion sequence is
I"l
= k!
CH3CBO-.~···>CHJ --teBO .
!2
= kz CAt: COh-
CHI·+CH,CHO-~CH3· +-CO+CH4
I)
= kl
\4
= k4 ('.J;., · ..... 1J"
C'HO·· +(JIi~'HO---·~·-~ CIl3 -+2CO + H2
7CH ..···_ .. k ,·.·--4C H ~'3 ' "2 '6
7-9
CAe
eCHO- CAe
,2: =CAC (k !... k'2 (~.. G~;· + ,KJ( ,-CHO.i
Active imc:n:ned.iates:
(1)
Cfh·, CHO·
·-rc:h;:;:; -r! .... r2 .. r2" r3 +l..r ::: 0 2 4 1
"rCHo- :::: -rl
:=·k1C,4C + k2CACC(11)
-
== ..,kle"AC +- k2 CAC('"CH,
"•. "
+ 1'3
:::: .. k!
k3C CHO C,\C .,' k;.CACCCHl
k ('"C'flO ('~'-'{C+ 1 k4 (,,2 'eH,'
CAe + k3 eCHO. CAe;::: 0
2
+2' k C '4
2
CH}
(2)
(3) Substituting (3) in (2) gives:
C2Hs
P7-4 (d) Individualized solution P7-4 (e) Individualized solution P7-5 (a) 7-10
Gas phase reaction
with third order kinetics and an apparent negative activation energy . Consider the following mechanism, in which NO} is an active intermediate.
fj "'" kll NOJ[02J -k 1[N03 ]
==kzINOlfNOJ]
T2
"" k![NO][ O2 ] k 1!N03 ]· ·ls[N0J[NO,] [N03 ] =5I~<.?le~1 kl ·+~[NO] "T NO
== . . k iNO][O,] + (k 1
k
~l..2
.
[NOJ)
~~P,'V.~(?l(?:J l k [NO]
k
1
1
== ... ~!5I~~)tL?2 ]::~5~j\.foy 19.21
"_I
+.s[NO]
_ 2k1k2 [NOY[Oz] -"--k~;·-t-iSfi~iorkl »k2 [NO)
. ····r
No
2k1k, = _.- . . ·k· ··-"-[NO]41] .()z ·1
For the overall activation energy to be negative,
···(E1 + EJt =::!;>
E~l == EDt, < 0
E -
A.s long as all energies are positive values.
P7-5 (b)
k3
Cl+CO~COCl ks
COCl + Cl z -7 COCl z + Cl
7-11
= 0 = k3 (Cl)( CO)-k4 (COCI)-ks (COCI)( CIJ
r COCt
(COC!) = k3 (Cl)( CO) k4 + k5 (Cl2 ) rcoCt2
r ct
= ks ( COCI ) ( Cl2) =
klk3 ( CO) ( Cl) ( Cl 2)
()
k4 + ks Cl2
=0 = kl (C1 2)-k2 (CI)2 -k3 (Cl)( CO)+ k4 (COCI) + ks (COCI)( C12)
add I'COCI to rCI r Ct
+ r eoct = 0 + 0 = kl ( C12) - k2 ( Cl)2
( Cl) = ~ ( Cl2 ) k2 2
~(CIJ
(Cl)=
k2
:1 (C12t (CI2) 5
klk3 (CO) rCOCt2
=
k4 + ~s ( Cl ) 2
kl k3
:12 (CO)( CI2)%
.- --k-=4'--+-k-(·-C-1-)s 2
k4 » k5 (Cl 2 )
2k _ k1 3 ( CO )( Cl )~2 2 k2k4 P7-5 (C) Individualized solution rCOCt2 - - -
P7 -5 (d) Individualized solution P7-6 (a) K[
N02 + hv
-~
NO + 0 K2
O2 +O+M -7 0 3 +M K3
0 3 +NO
-~
N02 + O2
UsingPSSH,
P7-6 (b)
7--12
0"..... M and O} aonear in [he denoI11inacot 2c,;ve ~r~·e"";~s .. ... .... ~
,,~
~Jt"
~.....
Sl,,,,,,,,.r(M1 (') ' ".;:,.;:::,'-';' ..... .....;" ...:> ~
fl..pplying rule '3 of t2.b!e 71 to 0::. and 0,:
0]
-i-
(h';-
0··.,. 202
0;.0 3
0 3 and M appear in the numerator. Applying rule 1 of Table 7·1 to 0;: 0,.-7 0;: + 0 If the second and third equations are combined, and M is added to each side of the equation:
A mechanism is proposed which satisfies all the rules of thumb: ro,::::' r l t r2 frJ
[01
'i '''2' :;:o_ ..
kz [()JO 1(M] k; [0 ][03 ]::::' 0 0 ·][M j'-k;,lo '10, ]= 0 rJ ::::. k, [lrf][Ol]···· k2 [0 2 1 rJ
::::. .•
kl [lvll[C?11
j.
k,[MJ[~~L._
kz[OJM]+'k;,[OJ
P7 -6 (d) Individualized solution P7 -6 (e) Individualized solution P7-7(a)
7-13
LOW TEMPERATURES • NO ANTIOXIDAm'
~] = ri c(~
==
c(R~-]
= 2ko(I2] - ki[I-J[RH1
(A)
-~(I-][RH] - kpl[ROz-J[RH]
(B)
= lep, [R-] [Oil - IcI'1[ROze][RH] - k, (ROz-P
(C)
c:{R-]
Cit = ki(RHJ(I-] - kp, [R-] [021 + kpl[ROz-][RH] dRADlCALS]
PSSA. ...
(A)
=0
(D)
= 0
dt
2ko[h]
rj
= 0 [I-] = ki [RH]
c(!.J = 0 = ki [RH][I-] - kpdR-J[021 + kpl(R
c(~~-] = 0 = kp: [R-][Ch] - kp%[R02-][RH] - kt [R02·j~ Substitute for [R·] :
2ko [Iti - k t [R~·F =
0
[ROrJ = _!lko [Ill
'V
kt
Now substitute the expressions for [he radicals into d[:~1 the expression for the degradation of the oil.
c{RHJ = _k"[2ko[hJl(RH] _k [2ko[hn llZ[RHl cIt I ki [RH]J P:z kl J .
J. [1-]
= 2ko (I:zJ- [2ki~koJr[I:zJ1n [RH] 7-14
ODE to be solved- low temperarures~ no antioxidant:
_c{RH] = cit
.
.,1,._ "I?1 + [2~1 [koJ e.AQ
L -..
kt
llnf'lI ]112 [RHl J
2
..
P7-7(b) Low temperatures with anti-oxidant
<{~e]
= ki[RH](J.]-kp1 lR.][Oll+kp:(R02.][RH] (same)
dfAe] dt =
kAl [AH][ R02-] - kA 2[A·][ ROr]
c(RHJ = _ k.l [1-1.[RH1kp.. [Ro,·] rRH] dt .... _~
ApplyPSSH:
d[I-]
-=0 cit
fIe] = 2ko[h] .. kj [RH]
d(R·l = 0
----0:.
dt
d(
2
R
2 .]
== 0
2ko[h] - kt[ROr]2 .. 2(l
:. - kt [R0:2·p- 2kAl [AH] (RO:z·] + 2ko [12] = {} k,[Reh·P+ (2kAl [AH]) [ROr] - 2ko[h]
=0
[ROr] = • 2kAl [AH] ± ../(2kAl [AH])2 + 8kt ko [h] 2k t
Now (FINALLY!)~ let'S substitute into d(:~] :
7-15
Quadradtic in [Rehe ]
[R~.] MUST be positive
t{RH] = _ 2ko [12] _kp [RH] - 2kAl [AM] + ../{2k A1 [AHJ)2 + 8kt
dt
?k t
1
leo (Ill
P7-7(c) If the radicals are formed at a constant rate, then the differential equation for the concentration of the radicals becomes:
d ~ e] and
=ko _ ki [Ie] [RH] = 0
[Ie]=-~-o-
ki [RH]
The substitution in the differential equation for R" also changes. Now the equation is:
d
~~e] = ki [Ie][ RH] -kpi [Re][ 02]+ kP2 [R02e][ RH] = 0 .
..,
and solvmg and substltutmg gIves:
[Re] =
ko+kP2[R02e][RH] [ ] kpi 02
Now we have to look at the balance for R02
d[R0 e ] .
2
_. _ -2- = kpi [Re][02]--k p [R02e][RH]-kt [R02e] = 0
dt
and if we substitute in om expression for [R] we get
0= ko - kt [R02 e
[R02 e] =
t
which we can solve for [R02·]·
I~!L
Vkt
Now we are ready to look at the equation for the motor oiL
d [~~e] = -ki [Ie][RH]-k P2 [R02e][ RH] and making the necessary substitutions, the rate law for the degradation of the motor oil is:
d[RHe] ~~ di= r = --ko - k [RH] p2VT RH
P7-7(d)
With antioxidants
Without antioxidants
7-16
-----------------------------------------------
-- -- --------------------------
----- --------------
- ----- ---- - - -
-'I } \1(".
it
-+R H )I-
I
Ii·
H1.
P7 -7 (e) Individualized solution P7 -7 (f) Individualized solution P7 -7 (g) Individualized solution
P7-8 (a) Given: Illness mechanism
H
-7
r
k:2 r + E-, 21
{Healthy person g<:rs (healthy per;;on cormac:s from ill oerson i (Hl person getS v. cH)
I ···t D
(Ill person dies)
7-17
rD:::
k.t [lJ
Applying pseudo steady state hypothesis
l:O 1:
Ii = fl + f: ~
I:"} .•
ft=;:C 0
P7-8 (b) Jt
b+k.& (H] = - - the demh i.H.e becomes infinite . k2
P7-8 (c) It is enlightening fH "'" -rl - f2
to
calculate rtf:
+ f3 =:·k 1 [H) ~ k2 [H] [IJ + kJ [fj ::: ·k! [H] . {kz [H] .. kJHI]
(!!4--:-- ;;: _
rH;;: --- -k t k.t (1<3 + k.t ~ k2 tH)}
fl)
<0
This expression states rhat for every person who is ill ilnd dies. a healthy person becomes ill For the population as a whole, "illness is but a 'way station' on [he road to 'death'." Note fur..her that IR < 0, and therefore the population will die off evenmally_ Initially, the death rate will be slow. until [H].,,. {k3 + ~} I k2. I11e model neglects the possibility of birth. In practice, it would appear to be useful in describing epidemic-like diseases, which occur over a short time so that the birth :rate can be neglected.
P7-8 (d) See Polymath program P7-8-d.pol. 1.0\"+9
1. 00E9
fi"======:: .... ,
8. 0e+8
8.00E8
6,OH8
6.00E8
_.---...._--_.
CJ - I
". n
4.0e+8
400E8
2,Oe+8 .
2.00E8
O"Oe+O
O.OOEO l_"""",===O. OOEO 2.00E9 4.00Et9 6.00E9 800E9 1.00EIO
--- -
-~~-.~.-~~.....
0.01'+0
2.0H9
4.01'+f
6.01'+9
8.0e+9
LOe+10
.....
Everyone becomes ill rather quickly, and the rate at which an ill person recovers to a healthy person is much slower than the rate at which a healthy person becomes ilL Eventually everyone is ill and people stru:t dying.
7-18
P7 -8 (e) Individualized solution P7 -8 (0 Individualized solution P7-9 (a) Starting with the design equation for a batch reactor
de
--.. -.£.
dt
== r P
Tp=kC
c= ((~o -(~{l"""~i~J de . r C 1 ·····d. . ·r·!:· =kfC so . . C ] 1--·£ • 1.: r __ .~ 1 4.
I,'
integrating
Finding k, l:~40
y '" 0 C02x + 1. ],697
1
.......
o
~
.. -
20
--~.-:-
"
60
40 tlme ------r'- ...... , .. .
Using the data at 40"F and 45"F the following graphs ll!~l!1:ade:
Finding k
o o
'5 y :O.CG2x t· l.l697
o
20
40
60
80
time L:::~.:.:.:::::::::::: :::::::::::::::T:.:::.:.::::: .......... _." . . '".........!.."... ~~::~~.~:::.-~.~.:.~:_'"_ .. ~.. ,~""~ ... ".. ,..
., ... _.. ,_~....... "" ........ ,.. _:...: .. ...::::=~~.:"_"
7-19
. . . . ..
k
from these:
1n
(
C -1.5) ::;::7.14kt + C
.933..;..a.·~. ···--
C p ··-1.4
E = _Rln(k.2lkll :::: 108120 Bt~ lfT1, --'l/1~
mol
P7-9 (b) k3&
:=
( -E
k4s1 0'" ~ 2.3R
(1"4-9;7 -'- 5041)) =.18'
104;
'I'
Using interpolation, C:::: 1.18086 Using the same equation used on the graphs we can solve for t when C p :;':: 1.1 t = 49 days.
P7-9 (c) kn = k4s lO'"
(2'~fR (5·16' ,_. ':5'64)) = .0023
C::::: 1.211 t::::: 2 days
P7-9 (d) The data appears that it may fit the Monod equation for substrate consumption at the stationary phase.
P7-9 (e) Individualized solution P7-9 (I) Individualized solution P7-10 (a) E+S
kJ
)E-S
E-S
E·S-E..~P+S rE • S =
P+S'-~E-S
kJ(E)( s) - k2 (E-S) -k3 (E-S) + k4 (p)(s)
since S is not consumed:
rE • S
k2 )E+S
Sy
= S + E-S
or
S = Sy - E-S
= kJ(E)[( ST) -( E-S)]-k2 (E-S) -k3 (E-S) + k4 (P)[( Sy ) -( E-S)]
(E-S) = kJ(E) (SrJ+ k4 (P)(ST) kJ( E) + k2 + k3 + k4 ( p) -rE =kJ(E)(S)-k 2(E-S)
_
( )()[ S 1-
-r -k E E
J
7
kJ(E)+k4(P)] - kJ k2(E)(Sy)+k2k4(P)(ST) kJ(E)+k2+k3+ k4(P) kJ(E)+k2+k3+ k4(P)
7-20
P7-10 (b) E+S~E-S
E-S
E-S E-P
k3 )E-P
k2 )E+S k4 )E-S
E-P~P+E
rEop = 0 = k3 (E-S)- k4 (E-P) - k5 (E-P) (E-P)
= k3 (E-S) k4 -+-ks
rEoS = ~ (E)(S)-k2 (E-S)-k3(E-S)+k4 (E-P) since E is not consumed:
ET
= E + E-S + E-P or E = ET -
k3) E = ET - E-S - - - E-S ( ~+~
E-S - E-P
. (1+-""----k3) E = E[ - E-S ~+~
Insert this into the equation for rE s and solve for the concentration of the intermediate:
(E-S)
r
=
=k
kl (S)(Ey) _ __ k (S)+k2 +k3 - k4 3 k4 + ks k4 + ks
[ l+~-]kl
(E-P) = k3 k5(E-S) k 4 +ks
s
P
klk3k5 (S)( E[ )
=
r P
--.--=-.::......:::..~~::.-!.---
[k3 +k4 +k5]kl (S)+k2k4 +k3k5
P7-10 (C) No solution will be given P7-10 (d)
k3
E-S1 -+- S2
p
k4
E-S1S 2
k5
E-S1S2 --7 P+ E (1) (2)
= 0 = kl (E)( SI) - k2 (E-S) - k3 (E-S)( S2) + k4 (E-S 1S 2) rEoS,sz = 0 = k3 (E-S) (S2 )_. k4 (E-S 1S2 ) - k5 (E-S 1 S2) rEoS,
If we add these two rates we get: (3)
rEoS, + 'Eos,s2
= 0 = kl (E)( S) -
k2 (E-S) - k5 (E-S 1S2)
Plug this into equation 3 and we get:
7-21
P7-10 (e) kl
E+Sp(E.S) ~
j
k3
(E.S)j p( E.S)2 + ~ k4 ks
(E.S)2 -7Pz +E (1) ~E.S)1 =O=k j(E)(S)-k 2(E.S)j-k 3(E.S)j +k4(E.S)2(1n (2) ~E.S)2 = 0 = k3 (E.S)j - k4 (E.S)2 (p)j - k5 (E.S)2 k3 (E.S)j (EoS)2 4 (R)+k5
="
j
Add (1) and (2) to get ~E.S)1
+ ~E.S)2
= 0 = kj
(E)(S)-k2 (E.S)j -k5 (E.S)2
kj(E)(S)
7-22
k 2
kSk3
+ kS +k4(In
=
(EoS)
k/S(E)(S) 2 k2k4 (Pr) + k2 kS + kSk3 r =k (EoS) = kr k3kS(E)(S) 2 P S 2 k2k4 (Pr) + k2kS + kSk3
(Ey) =(E)+(EoS)r +(EoS)2
P7-10 (0 ko
k,
E+S~EoS-~P k2 k4
E+P~EoP k5 k6
EoS
+P~EoSoP k,
ks
EoP+S~EoSoP
"" r=k(EoS) P 3 (1)
~EoS)
(2)
~EoP) =O=k4(E)(P)--k s (EoP)-k s (Eop)(S)+k9(EoSoP)
(3)
~EoSoP) = 0 = k6 (EoS)( p) - k7 (EoSoP) + kg (Eop)( S) - k9 (EoSoP)
=0 =kr (E)(S) -
(2) + (3): '(EoP)
+ ~EoSoP)
k2 (EoS) - k3 (EoS) + k6 (EoS)( p) + k7 (EoSoP)
= 0 = k4 (E)( p) - ks (EoP) + k6 (EoS)( p) - k7 (EoSoP)
now add (1) to this:
7-23
~E.S) + ~E'P) + ~E.S.P) = 0 = kl (E)( S) - k2 (Ee S )- k3 (Ee S )+ k4 (E)( p) - k5 (Ee p )
(Ee ) = kl (E)(S)+k4 (E)(P)-k5(Ee p ) S k2 +k3 (Ee P) = k4 (E)(P)+k9 (EaSa P) k5 + kg (S) (EeSe p ) = k6 (EeS)(p)+kg (Eep)(S) k7 +k9 k4 (E)(P)+k9 (k6 (EeS)(p)+k g(Eep)(S)] (Ee P) =
k7 + k9 ._--"-k5 + kg (S) (Ee ) = k7k4 (E)(P)+k9k4 (E)(P)+k9k6 (EeS)(P)+k g(EeP)(S) P k5 + kg (S) (Ee P ) = k7 k4(E1( p) +k9 k4(E)( p) +k9k6 (Ee S )( p). k5 kl (E)( S) + k4 (E)( p) - k5 (k7 k4(E)( p) + k9 k4(E)(P) + k9 k6(EeS)( P)] (EeS)=
_____
k5 k2 +k3
(EeS) = kl (E)(S)+k4 (E)(P)·-k 7k4(E)(~)+k9k4 (E)(P) k2 +k3 -k6 k9(p) rp = k3 (Ee S ) = k3 (E) kJ (S) + k4 (p) -. k7 k4(p) + k9~4 (p) k2 + k3 - k6 k9(p) (Er )=(E)+(EeS)+(EeP)+(EeSeP) (E] ) = (E)+ kJ (E)(S)+k4 (E)( p)- k7 k4(E)( P)+k9 k4(E)(P) k2 +k3 ·-k6k9(p) k7 k4(E)( p) + k9 k4(E)( p) + k9 k6(kJ (E)(S) + k4 (E)( p) -_. k7 k4(E)( p) + k9 k4(E)( P)J( p) +__ . _ k2 +k3 --k6~9 (p) .-'----~
+
k (kJ (E)( S) + k4 (E)(P) - k7 k4(E)( p) + k9k4 (E)( P)] 6 k2 +k3 -k6 k9(P) (p) -
7-24
-
kJ +k9 all of the terms in the numerator have (E) in it and so the (E) can be factored out and an expression for (E) in terms of (El)' (P), and (S) can be made and plugged back into the equation for rp.
P7-10 (g) kl
k3
E+SpEeS~p k2
k4
E+ppEep k5
~E.S)
=O=kl(E)(S)-k2(Ee S )-k3(Ee S )
(Ee S) =5lS)(E) k2 +k3 ~E.P) =O=k4(E)(P)-k5(Ee P) (Ee p )= k4(E)(P) k5 rp = k3 (EeS) - k4 (E)( p) + k5 (Eep) rp
= klk~ (~~( ~1 __ k4 (E)(P)+k4 (E)(P) 2
3
Isk3 (S)(E) k2 +k3 (ET ) = (E) + (EeS) + (EeP) rp
=--'--:"""":
(ET )=(E)+ kl(S)(E)+ k4(E)(P) k2 + k3 k5 (ET )=(E)[1+_ k1 (S) + k4(P)] k2 + k3 k5 Er ) r - - - - klk3 (S)( -'--'--c-"-'--_ _ P - (k2 +k )[1+ kl. (S) + k4(P)] 3 k2 + k3 k5 _ k3 (S)( ET ) r p (k2 +k3 +(S)+ k4 (k z +k3 )(P)) kl klk5
--------'-'--'--'--:"-'---~
7-25
P7 -10 (h) No solution will be given P7-10 (i) No solution will be given P7-10 (j) No solution will be given P7-10 (k) No solution will be given P7-11 (a) The enzyme catalyzed reaction of the decomposition of hydrogen peroxide . For a batch reactor:
~ ~~s
;;
~.~~ = fS
""
·:Ky-m~x~~ ·r LS
at t
til
=:
0, Cs "" CS o
Rearranging and integrating:
Cs .,.. •Km In -....
eS o
or J In t
S"t) '" \-5
Cs· CSo
"" • V . t max
.~Sil:~l+- ':' m1!~.. Km t
Km
A plot of L I CSt) . Cso - Co;; .' t n C vs _. -i........c- should be lwear with s:ope ..•J ... s
Km CS!L:CS t
10
.01775
1.1268
.01193
20
.0158
.01179
50
.0106 .0050
12654 1.8867 4.0000
.01386
100
OI27()
.00225 .0042 .0094
.000225 .00021 . 000188
.0150
.00015
00140
0.0135
IIn~Jl t Cs
0.0130
0.0125
00120 0.0115 15
16
1 7 18
19
20
21
22 23
(C{so) - C(S»)lt X 1 E5
7·26
24
25
· 013· 012 ( min .j "~ F rom t h e gnarl, slope "" ...-~--. -'--.--.--, -,,.-------"1 -
(17.5 ,. 206) x 10 5 ,g mol I min; J
,
:::: ,.0310 g mot!l
Km
C Cs
C
C
1 75 x 10-4 I·min
mol
Art In ,~. '" ..Q~l,. ~.Q, . ,,,.2 ::;: """'''''' t
rom
mu = (01" V 5 64. 'K; , .> +- ,,'
Vm.z:s.
_·
x
10"»
'I mlD
=, O'8 l ' 0) mm r
-,.\
- ,,1){031·)gmOl\ -. ( --'l""'-j
:::;:; (..018 6J X Hun
=:;
-78 X '0 ), I
4
gmol ---':-"'I,nnn
P7-11 (b) V m= a. (EJ. If the enzyme concentration is incrc:;lscd by a factor of th:ee. then t/
-'3(-"78*10'4 gmol \-1'" gmol 0-4 'L-;-;~ri;~' .) "l*~n' ) - 7,.>4 * 1
v rna=< - .
r
I _ , _ * 20 min L' mm
km In·~·L +- Cs '" Cso :::;:; '·-1734 * 10 4,,~!pq.
C so
=:
,.-34.7 * 10
-ntis equation shouid be sOlved by ni;),l and error, Rearr;:mging, in .Cs., :::: CSO,:::, Cs_.::...:.Q?:~" Cso Km
:. Cs
=:
C so exorCSQ.~:~,,~,,~.J)3~JJ '" ,,02 cxo f~:Q!47,~~d .L Km 1 "l, .0310 1
Assume a Cs • caicut:lte a new (lme from the RHS of the :l.t~ove el1!luation. Assume Cs New Cs % difference O.0l5·-"1r.r;OO't\'7"'7r---;-;~:4"""l'"8;----
O.®077 0.0097 0.0091 0.®®93 Q,.Cl®92
O.®®97 O.®®91 O.®®93
20.~ ~.,
2.2
0.0092
L1
0.0092
0
P7 -11 (c) Individualized solution P7 -11 (d) Individualized solution P7-12 (a)
7-27
4F.".,!::)1 L
Given me reacrion sequence:
E+S =£·5
The plat of ·r$ vs
~
is s.'lown below.
the rnec:banism for the aI:xwe :reaction is: -r,. ::::; k Cs lit 1 + Kl Cs + K,
low values ofCs.
r.;
; For ('-S « I: -rs'" k C Et in qualitative agreement with the graph at
fur Cs »
1: -rs ""
;2~ .
This is also in agreement with the gr:aph.
=0
I 2I'l'fs ) -= _..._. -_...... ·.·.·~_)k._El~S !<-.1..... _. ___ or C s2 == -K-·widi: ~- - - ld~(~~ ( C' -, 1 +- Kl s~' Kl ('\2 ~:~I
2kE t
r- Kl - 3{K~' +-:-i K11 < 0,,', ~.
.~~_?_dL:_~1.~ni~L.:.~K2~~1. (1 • K j Cs + Kz c~f
(-r,) goes through:l maximum.
2•
This observation also agrees with (ile above graph
P7-12 (b)
7-28
For a CSTR operatbg with V.::; 1000 I; Uo ;:; 3 1. IJmin
v Do f""
with CAD"" 50 m mo\es/l .
= 50~~i~':::: this (linear) equacion is plotted on the accompanying ,)
.
.)
graph, the equari?n imersects the (.!,) v s C s curve from .he rare of renccion at
1~2 mID.Qks. ...• ('-~ "" --. ? 1 ll.rr.ill.e.s. -(5 ··r,; -- 0 .;) . I .. Il1ID
·1
Stability of points: J.Ssume u'1ar a pe:,urbllrion
as
==
c so . Cs occurs where the overb:;r designarcs
the steady-state condition. ~1ated:ll balance for any time:
V ~~2. == rsV + Cso \)0' Cs dt
or
1:
·c
~-~. cit
:: IS:
+
eso . . Cs
at Steady stare
r5 ,.
rs
{ d rs \
\:-l
\d Cs IC-·.;:,
(Cs"
Uo
.'\
}
Combining
-
csl "" "1
11 6s + 6s[)
:1:. t ""
0.
i ,1
For this solution of the 3.0ove eqtw.rion
!.~,r~~. I , I.
\
~ C' ~\.d -s Ie:;
10
< I
as
-= 0
be stable:
or
t"L ~-,~:~ ('" ~
d,S
r > ·1
les
At C. = 2. 1 m moies/!"fs "" O. tS~ 111 moles! I·min. rhe ~doce of rhe re:lction rare curve is posithe 'Therefore, rhis oper-:J.ring poim is stable. For the other (WO poi:1ts,. the s(;),bi!icy may be exarnined by estimating the derivative gr;J.phically:
L\CS
7-29
1 '" • ~L5..Q 2. min· '
:. t.'1e point Cs :::; 9.4.
''is::=
(~.L.::~!.\
At Cs"'" 2.1 m molesll.
( ~i::~lL aCs les At
,
f(
Cs
1: =: ( ..
0132 is unsrable ..
aCs
fE.s
=2.1
.=
~L::.~1. ;:;: D"Cs
Q,-Qfi.12.. :.D_J218
= - Q~~.
24 - 44
min-I
20
min'1) (3 n min) ::: - 0.775> .,! ; s.abk 2"
u,.D!!25.
=5.1 m moleSJ1.
S el(or o -_ .....- )).IS necessary
(1 C s '
-rs :::; 0.154 rrLrTI.Q~~~. appears to be sc:!ole. but more accurate calculation
~o
I-mm
bl" .. l' d ,... , esra Isn tillS conC,USIon ennmvelV. .
•
X = Cso -Cs = 50-2.1 =.958 Cso 50
P7-12 (C) -rs =--z 1+ K1C, + KzCs !fEr is reduced by 33%, -rs will also decrease by 33%. From the original plot, we see that if the curve -rs is decreased by 33%, the straight line from the CSTR calculation will cross the curve only once at approximately Cs = 40 mmol/L
X=O..2
P7 -12 (d) Individualized solution P7 -12 (e) Individualized solution P7-13 (a) Data on Ba.ker's Yeast at 13.4 °c no
su Ha:1i b.midc.
23.5 33.,0 "'!,-7 .5 42..0 43.0 ·:+3.0
~\~ 2C mg
._L
$u[[';mllarnidc!rrH adccc to medium
17.::t
Q~ no $ulfan!l.;mlidc
2..0
30.8 36439.6
.0425 .0:303 .02.666 02.38 .02.33
. ""~)
;lOJ)
(n,3)
.200
25.6
7-30
1.0 .66
AO '1"'~
.0575 .0391 .03246 .02747 .0'2.53 .0250
...
.1.
::;
fp
_K!!!_ + ___L __ Vma.x S YmJ.x
;::;;
K IIL \(1' ___ ____ "_)
+ ______.1! - ._____ Po, V ma;x
V m.;u
::
_1_ QC'z
I versus -p1---- wt'11 h Km .., anu-an !' lave a sI ope ' or ---;.;'... Imercept ot--Un P l~ot ot~ n:,-~
V ron
O;l
'-<\.h
0.06
.... WItt!
ODS
O/tlln (C!5ClI C{:s»
0.04
OM
om 0.01 0.00 0.0
1..0
t5
1/1){O) (rnm HQ)
From the graph, slope'" JllQJ_:_J1J1. =0165 IlHcrCcpt;;;:: 0.019 1 ... V m.:u = 5263, ml
nr rng cells
Km "" 0.0165 Vm:1X ::; (0_,0165)(52.63)
~
0
mrnHg
P7-13 (b) Now, with competitive lnhibidon:
E +S
C;::l
E•S
Km( 1 + Rare law becomes:
f
K It In this C:lse, (he slope is :
(1) + "S \mu
111
~
+_.L} K"
----"---""--. '.vhile (he imer::epr is the S;;lme as in c;).se fa) V ma ;>;
7-31
For the case of uncomperitive inhibition:
E+ S
= E· S
E·$+1=1-E-5 E'SC;)P+E
Rate law becomes:
1
In this case, the slope is the same. bur the inrercept is
+-'
·v.JSL maA
And for the case of non-competitive inhibition:
E+I=E-1 E"I+S~I·E·S E+S~E·S
I·E·S¢;:)E·S+I E"S¢;:)P+E
In this case both the slope and intercept change. Plotting the data of
t- in mmlig versus r-!:~
'
with sul1anilamide on the same plot as was ploned the data for the case with no sulfanilamide. it is seen that the slopes are different. but the intercept is the same. 11lerefore the inhibition is
cornparari ve.
P7 -13 (c) Individualized solution P7 -13 (d) Individualized solution P7-14 For No Inhibition, using regression, Equation model:
---.!= aO + a1(.l..) -r S s
aO = 0 . 008
al = 0.0266
For Maltose, Equation model: _1_
-rs
= aO + a1(.l..)
aO= 0.0098
S
al = 0.33
For a-dextran,
7-32
Equation model: _1-
-rs
=
aO + al(!) S
aO = 0008
al = 0.0377
=> Maltose show non-competitive inhibition as slope and intercept, both changing compared to no inhibition case.
=> a-dextran show competitive inhibition as intercept same but slope increases compared to no inhibition case.
P7-15
=k(EHS) (EHS) = KM (EH)(S) rp = kKM (EH) (S) (EH;) = K2 (H+ )(EH) (EH) = Kl ( H+ ) ( E- ) rp
(E-)=~EH)
Kl (H+)
(E] )=(E-)+(EH)+(EH;) (EH) + (Er)=-K1(H+) +(EH)+K2(H )(EH) (EH) = _ 1+ K,
(E] )
_
(~') + K2 ( H' )
Now plug the value of (EH) into rp
kKM (E] ) Kl ( H+ ) ( S ) rp = kKM (EH) (S) =· · - - - - - - - - - - 2 1+ Kl (H+) + KIK2 (H+) At very low concentrations of H+ (high pH) rp approaches 0 and at very high concentrations of H+ (low pH) rp also approaches 0 . Only at moderate concentrations ofH+ (and therefore pH) is the rate much greater than zero.. This explains the shape of the figure .
P7 -15 (a) Individualized solution P7 -15 (b) Individualized solution
P7-16 (a) For batch reaction,
7-33
dCs dt
rs --
- - = rs
&
See Polymath program P7-16-a.poI. POLYMA TH Results Calculated values of the DEQ variables initial value 0 20 20 0.25 0.5 0.1 0.1 1 -0.0987654 0.0493827
Variable t Cs eso KIn
Yes Ceo Ce umax rs re
minimal value 0 6.301E-ll 20 0.25 0.5 0.1 0.1 1 -8,,0781496 1.273E-09
maximal value 10 20 20 0.25 0.5 0.1 10.1 1 -2.546E-09 4.0390748
final value 10 6.301E-ll 20 0.25 0.5 0.1 10.1 1
-2.546E-09 1.273E-09
ODE Report (RKF4S) Differential equations as entered by the user [1) d(Cs)/d(t) = rs Explicit equations as entered by the user [1) Cso = 20 [2] Km:::: 0 . 25 [3] Ycs = 0 . 5 [4] Cco=O.1 [5] Cc = Cco+Ycs*(Cso-Cs) [6] umax:::: 1 [7] rs = -umax*Cs*Cc/(Km+Cs) [8] rc -Ycs*rs
=
0.0
20
··1.8
16
c:J
-3.6 -5.4
12 8
-7.2 -9.0
.:I
0
2
.:I
6
8
0
10
P7-16 (b) For logistic growth law:
dCc dt
---=r g
7-34
0
2
.:I
6
8
See Polymath program P7-16--b.pol. POLYMATH Results Calculated values of the DEQ variables initial value 0 0. 1 1
Variable t Cc umax Coo rg
1
0 . 09
minimal value 0 0.1
maximal value 7 0 . 9918599
1
1
1 0 . 0080739
1 0_2499857
final value 7 0.9918599 1 1 0.0080739
ODE Report (RKF45) Differential equations as entered by the user [1] d(Cc)/d(t) == rg Explicit equations as entered by the user [1] umax == 1 [2] Coo == 1 [3] rg == umax*(1-Cc/Coo)*Cc
030.---------------
1.00
0.82
0.64
0.46
0.28
0.10
0. 0
1.4
28
4_2
t
5.6
O. OOL---~-------0.0 1.1 2_8 t 4.2 5.6
7. 0
7.0
P7-16 (C) ForCSTR, rg
=_Y
r
CIS s
= YClSf.imaxCsCC
- rate at Wh-IC h was hout occurs = D I-I utlOn
K
M
+C
s
YClSf.imaxCSO KM +Cso
1
3
O.5xlhr- x20g / dm O.25g / dm 3 + 20g / dm 3
= 0.494hr-1
P7-16 (d)
~ -J = O.5xlhr- X[l- ~O.25g/dm r O.25~ +20g/dm / dm ~~o 3
Dmaxprod = YClsf.imax[lDmaxprod
1
= 0.44hr- 1
7-35
]
3
P7-16 (e) Ifrd= k.tC c
rh Divide by CcV,
= Cc v0 = (rg
- rd )v
(YC/s,umaxCs~_kd
D= (rg -rd ) = Cc
KM +Cs
C -_.JD+kd)~_~ Yc/sf.Jmax -- D ~s
~
- rs = Ys/crg
Now
Cc = YC,sD(Cso - Cs ) D+kd For dilution rate at which wash out occur, ~ Cso=Cs ~
DMAX.
Cc = 0
- (D+kd )K M_. C so-
YC/sf.Jmax - D
1 3 O.02hr- xO.25g / dm 20g/dm 3 +O.25g/dm 3
= YC/S,umaxCSO -- kdKM = ~:5xlhr-l X 20g / dm 3 C so +KM
= 0.493hr-1
There is not much change in Dilution rate while consideling cell death to one where cell death is neglected., 2
DC =YC/SD (c:so- Cs) C D + kd d(DC c ) For D max .prod' . =0 dD D max prod = 0,446 hI-I
Now
C =_(D+kd)K M s
P7-16 (0 Now -rm=mCc
DCc=rg rh
&
D(Cso-Cs)=-rs-rm
= Ccvo = (rg)v = f.JCcV 7-36
YC/S,umax - D
D
Divide by CcV,
Solving
= Ji = (YC/sJimaxCs) KM +C s
DKM YC/sJi max .- D - r"s = Ys1Crg C -
~
S -
Now
~
= [D(C so -
Cc
Cs )] YclsD+m
For dilution rate at which wash out occur,
Cc = 0
Cs
~
C
~
so
=
Now
For
YCISJimax C so C so +KM
D max
D max
DK--=M__ YC/sJimax - D
DKM YC/sJimax- D
DMAX
~
=
prod'
= 0.494hr-1
d(DC c ) =0 dD
---.-
= 0.4763 In·
prod
3
1
O.5xlhr- X 20g / dm 20g/dm 3 +O.25g/dm 3
l
P7 -16 (g) Individualized solution P7 -16 (h) Individualized solution P7-17 Tessier Equation,
. - Jimax (1 - e-Cs I k )c
rg
-
C
(a) For batch leaction,
dC_ _ s =r dt
cc
. - Jimax (1 - e -Cslk)c
rg
s' = Ceo
C
-
+ Yc I 5 ( C so - C 5 )
See Polymath program P7-17·a.poL POLYMATH Results Calculated values ofthe DEQ variables Variable t Cs Ceo Yes Cso Ce k
umax Yse rg rs RateS
initial value
o 20 0. 1 0. 5 20 0.1 8 1 2
0 . 0917915 -0.183583 0 . 183583
minimal value -0-----
maximal value
final value
7
7
0 . 0852675
20
0. 5
0.1 o. 5
20 10.057366
0 . 0852675 0.1 0.5 20 10.057366
8 1
8 1
o. 1
20 0.1 8 1
2 3.8563479 -0.183583 7.7126957
2
0.0917915 -7 . 7126957 0 . 183583
7-37
2
0.1066265 -0 . 213253 0.213253
ODE Report (RKF45) Differential equations as entered by the user [1] d(Cs)/d(t) = rs Explicit equations as entered by the user [1] Cco =0.1 [2] Ycs = 0.5 [3] Cso = 20 [4] Cc = Cco+Ycs*(Cso-Cs) [5] k= 8 [6] umax= 1 [7] Ysc=2 [8] rg =umax*(1-exp(-Cs/k))*Cc [9] rs = -Ysc*rg [ 10 1 RateS = -rs 8.0
16 12
6.4
F-~
W
4.8
8
3.2
4
1.6
0.0
o 0. 0 (c)
2.8
1.4
DCc
4.2
5.6
b::::2::::::'--'_~
0. 0
7.0
14
_ _~ __~.
2.8
t
5.6
4.2
D(Cso--C s )= rs
= rg
m= Ccvo = {rg)v = ,uCcV Divide by C cV,
D = ,u = ,urnax (1-. e -C s I k )
Cs = -kln(l-
,u:-)
Now For dilution rate at which wash out occur,
(d)
=>
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<--_ _ _ _ 0.0 02 ~
1.0 00
0. 2
04 t
0.6
P7 -20 (b) When we increase D, C s
0.8
1.0
~
__
~
0.4 t
__
0. 6
~
_ ___'
0.8
increases, C X ! decreases, and C X2 has very little decrease.
P7-20 (C)
When C so decreased, Cs and C X ! both decreases, C X2 has no noticeable change . When C Si increased, C X ! increases, C X2 has no noticeable change for large t
P7 -20 (d) Individualized solution P7 -20 (e) Individualized solution P7-21 (a and b) Run
#1 #2
No Yeast Extract Yeast Extract
TIle percem volume of the gro'W1h product H 2S collected above [he broth was reponed as a function of time:
cell t· nutrient." more cells + plOducr 1 ......... _......... ,........ ,; "j~.;..'"" ....._
...
.-
o
... . • ..
(a and b)
ex ;::: Cxo&(t .1<....)
or
ex ""
CXQ ¢ill
e
j.l. 1;..
curve fit exponential CHIve;
7-46
o~
...... -._.,.
1.0
where A ::::: CXQ e iJ. ~
Run 1 (between 15, 20, 30 hrs)
Run 2 (poims 10, 15,20 ill'S)
= 0.2125 hrl
!lmu
!lOla;; ::
A ::: 7492.6 tlog
== In
tlag
::=
(~{4§2~r.)
A
0.3124 hr· l
= 557L7
: : 0.21252
6.0 hr
:;: 5.1 hr
il ag
P7-21 (C) Stationary tstationary between
TIme
45 to 55
length of time
25 to 15 20hr
10 ill
P7-21 (d) Production struts at the end of the exponential (for both runs) P7-21 (e) dCc :::: D {.(.~-co'· C' -_. -{;J" dt
jJ.Cc•
= 0
D{C'..co· Cd + !lee;:;:: 0 -D Cc + jJ.f=<: "" 0
Cc :::
0 or D :::::
jJ.
wash om occurs when D :>
gm;u
P7 -21 (0 Individualized solution P7-21 (g) Individualized solution P7-22 (a) dC..- ::: 0 (Cr . . . C..r) + fg dt ....... C ' . . . . . . _
--~
=
}lma;;,
1.5 hrl
Cs~ == CeQ ::: 0.5 g/dm3 r __ d_'-l. '-cit
"" ()
~o £'.; = )
p'
f' :; g.,em
Ks == 1 gJdm3 D == 0.75
eso
=:
.30
YCiS
== v.08
7-47
o=
!-lrnax Cs Cc \ .. DC'-c + ---.-·....--.----·--l (
\ KS .,.
CS) JI (' (l . 1<:-1 ,,$
..,.
[----·----------------·---1 ' ~l:r.ax Cs 'I' 1,D = / ~~:~~$ (~-.-~-~~-\). I
\
'- .......... w_" ............ _ ••• _
••
D
,Ku,
_.-L._w. _ _ ...... w~ _ _ ._._._•• ,
C:;'Q..
12_5 Cc
pick Cs , caiculate D and
eso
=::
::0
YclS
(Cso' ('51
(~:
30.0 g.!dm->
D= (;:~~5( ;~;;jl
Cc "" (300··, C$)(0.08)
For D = 0 . 876hr-! (C s = 2..5g/dm\ production rate is maximum
P7-22 (C) Cc
"'-'-- Cs
For D = 0.27hr-! , Ce = 0 if Ceo = 05 g1dm3 . For D = 0.J14hr-! , Ce = 0 if Ceo = 0 g1dm3 , And for maximum production rate, D = 0,,876hr- l
P7-22 (d) For batch reactor,
7-48
dC dt
- -c= r g
See Polymath program P7-22-d.pol. POLYMA TH Results Calculated values of the DEQ variables Variable t Ce Cs Ki umax Ks rg Yes Yse rateS
initial value
o
minimal value
o
final value 2
maximal value 2
2. 9
2. 9
30 50 1.5
2.01E-07 50 1.5
1
1
1
8.744E-07 0.08 12 . 5 1.093E-05
2 . 9019168 0.08 12.5 36.27396
8.744E-07 0.08 12.5 1.093E--05
0.5 30 50 1.5
0.5 2.01E-07 50 1.5
1
0.4591837 0.08 12.5 5.7397959
ODE Report (RKF45) Differential equations as entered by the user [1] d(Cc)/d(t) = rg [2] d(Cs)/d(t) = -Ysc*rg Explicit equations as entered by the user [1] Ki=50 [2] umax= 1.5 (3] Ks= 1 [4] rg = umax*Cs*Cc/(Ks+Cs*(1+Cs/Ki» (5] Ycs=O.08 [6] Ysc=1lYcs [7] rateS = Ysc*rg 30 , . . . ; : : - - - - - - . - - - -
40
c:J CC
24
..
18
. s.
('
24
12
16
6
8
o
~==========:=::!:=l 2. 0 c..0 04 0.8 t 1.2 1.6
0
P7-22 (e) For semi-batch reactor,
dCc dt
=
r g
=r g
~
~.~
32
_ VaCc_ V
JimaxCsCc Ks+Cs(I+C s / K/)
--.---=~~...:=..~-
See Polymath program P7--22-e.pol.
7-49
..
0.0
04
0.8
-, t
.. -.~~
1.2
1.6
2.0
POL YMA TH Results Calculated values of the DEQ variables variable t Ce Cs Ki va Va V Csin umax Ks rg Yes Yse rateS VCe
initial value 0 o. 5 2 50 50 10 10 30 1.5 1 0.487013 0 . 08 12.5 6.0876623 5
maximal value 4. 5 2 . 2593341 24 . 016878 50 50 10 235 30 1.5 1 2 . 2464622 0.08 12.5 28.080778 530 . 94351
minimal value 0 0.2329971 0 . 8327919 50 50 10 10 30 1.5 1 0.2329022 0 . 08 12.5 2.9112771 5
final value 4 .. 5 2 . 2593341 0.8327919 50 50 10 235 30 1.5 1
1.5283423 0 . 08 12.5 19 . 104279 530.94351
ODE Report (RKF45) Differential equations as entered by the user [1) d(Cc)/d(t) = rg-vo*Cc/V [2) d(Cs)/d(t) = -Ysc*rg+vo*(Csin-Cs)/V
30
Explicit equations as entered by the user [ 1 ) Ki =50 [2] vo= 50 [3] Vo = 10 [4] V = Vo+vo*t [5] Csin = 30 [6] umax = 1.5 [ 7] Ks = 1 [8] rg = umax*Cs*Cc/(Ks+Cs*(1 +Cs/Ki» [9] Ycs = 0 . 08 [10 J Ysc = 1IYcs [11 J rateS = Ysc*rg [12J VCc = V*Cc
12 6 0
600
24
480
18
360
12
240
6
120
C'?" _,
0.0
~J
18
30
0
~.c
-
24
119
0.0
18
t
2.7
-------------------~
[vcc]
I)
09
1.8 t
2.7
36
-1.5
0. 0
7-50
0.9
1.8 t
2.7
3.6
4.5
P7-22 (I) Individualized solution P7-22 (g) Individualized solution P7-23 (a) Fit the data to the equation: D ; ___ 5:!it: Ks+C s Using POLYMATH, find the values for fi.,rw: and
Ks.
TIley are L98 and 0.97 respectively.
$·29
.--~,
CS l i .OOC
......."., ._.-._;---~'~"'''''':'''~'----'''-''1
7.::lC:;J
9 G::JC
P7-23 (b) Using this equation, solve for Y C,'5:
With the given inf<.umation Yc;s"'" 0.099, therefore, YS1C is equal to the inverse of that, lOJ)'lS.
P7-24 No solution wiIl be given .
.--~---------------
P7-25 No solution will be given
P7-26 See Professional Reference Shelf 7..5 on the website for a sample solution. CDP7-A
No solution wiIl be given.
----------------------------------------------------------CDP7-B
7-51
Given the following re:rc:lon scr:e::le: i + 00 . _} <
or··
+- C! .
with the following f:!te: law: Active intermediates assumed to be HOCl and HOI From table I, the: first rule of thumb suggests:
OH'
'"<
kl HOCI .,.., oCt
< ;.
H20. -r\:.::: kl COH CWXI- k.\ Coo' CH:O
kl
kz OH" + HOI <:..., 01· +- H20 ,
I +- HOC! ., HOI + Cl<
,r3 ""
k1 C!· CHOC1
This step makes the overa.ll reaction sequence 1-.;- OCl·· ....~ OI·· + C1 ' possible: k
... TO! ;;;;:: r2 :.::::
[HOI ::=
--r2
k)C{ C HOCI
2COI
C[f~O
..
k;;,C[fO CHor
+ ') "" k:,.(-'OfJ C iJOI
...
k. 2C or ('H
k.2 Cor CfJp
:.:::: k2CO[fCHOJ
1
0 .~
k 3C
1
Cd
,
CHOCl :;;;; k.! C{Xl CHiJ ;
+ k t C{ . CHCX:J == 0
. k -1 Ccx::l CH.,Q i.e.,, (HOC1 ::: ;-------<..- ... Kl CoB· ·t k3 CI
.1 X3 ('1 CO::X Cn,o . 1'ben:
r Of
;;;;;; KC. ......,.-_COC' . ......!-. C. Oll
0
;:;;;;"()f
·lllClC..1 ::::: -rl - 1.3 :;;: kl COH· CH(X.l< k ·1 C(Xl CrhO
or (kl ('..OW .;- k3
CHOC! =:
k3
k_rkl K :::;----. k1
WheI~
An alternative approach assumes that reaction 1 quickly attains equilibrium, then:
• fhen
<.
rO! ::;;;::
k
'3'-- r ~
(-t
'HOCI
','k 3 k .. C., . o
"' ;;;;'l -k ...r., 1
"j Ct· C,C!
_(~_ _ . 1../ '''',
O-..
..
,
.
OR·
TIlcse two approaches are basically equivalent
CDP7·C
7-52
(a) Assumptions:
• Transfer rate from bubble buLle to f1 uid interface is not rate limiting, i.e., Ci is tneequilibrium oxygen concemration. • System is:ar pseudo steady-state with regard to the particle size, i.e., panicle growth is slow compar'ed to oxygen transfer. • Rate of oxygen consumption is direcl1y proponional to the cell growth rote:
ro,•
-=
_yl rc where •vo,• 0,
=•vield of cells on oxv£en (.-.mmoe g ce1lls ) o1 .~
This implies t.~ac any oxygen utilization required to maintain the cells is negligible, and there is no significant metabolic product being synthesized.
Oxygen balance:
where:
kl ilb :: ove.all mass transfer lesiSlance from the bubble to the bulk.
ac
= surface area per gra."11 of cells
k
:: mass tronsier of cells on 02
yo,
::: yield of cells on 02
Rearrange {'}z balances: Ro, C·'i - (''u --.-"::;
(1)
k;:~c.c ""
(2)
kl ab
,lo,.Ro.. = Tlk C'.c
Cb - C"
c .-$
(3)
Add equations 1, 2 and 3
~~. ::; k/ab + J-c (k(;I;c+ ~~)
(4)
0) When oxygen consumption by the cells is slow, the process is reaction rate limited. C =: - .......... I +. ..:..-. Yo, Thus Tl··~ 1 and: -1. Ro. k! ab <=<: k
{2} When oxygen consumption is much faster than In3.SS tr.lnsier. the mass transfer becomes [he limiting factor. ..,9,,, "" .. J.... + --........ L._.. Ro, kl ab kc ac Cc
7-53
(b) To increase the growth rate, you could: • Ifl(,,-rease kl ab by increasing fem1cntor agitarion, • Increase the concemration of cells (since this is an autoca::alyuc reaction) • Increase clump surface area and the effectiveness factor by decreasing the particle size (also by increasing fermentor agitation)., (c) Re "" YOl R(h From equation (4), have:
~c:
:=1
Ci
Yo., [ ____L
"'lkl
.;. ..L {:__ L,
<.tb
CC
+
Yfh.'ll r;kt
KC 3.(:
yO;-~C; [k~;~ ,~.;: ~tl dec " Ldl
[
+
{1 _yo,) In Ceo Cc "" Y0 Cit
Cc .,. CeQ ~kl-ab -+ ~~;;; -t-- ~1k
~
2
(d) Assumptions: • There is a constant number of particles • Each pellet is roughly spherical and has const::!.nt density. We do not know which resistances are controlling, so we know theIe are no reaction limitations. but. may be either internal or external diffusion limitations.
~:
:=
Yo.
C[j;}a~ + de' (~~l~; + ~~)j
1
As particle growth increases, kc. ac and II will change as functions of the particle diameter. Thus need to fmd particle diameter as a function of Cc Cc "" npe V c Vc "" TI;, d 3
,
where n "" number concennation of particles (Ill)
6 P
Pc ;;::: the density of the parTicles (gil) Ve :::: particle volume (1)
Intemal diffusional resistance can be modded as: ;,
T)z(
:;::; (Xi
do.
External diffusional resis::aflce with or 'Withom shear is;
.. _L. kc
3{:
where cr:! :;;:: at
iCn Pc
Dropping the primes and simplifying:
7-54
Cc,,~Cco k[
+ 3
,
(r.....c113 . CcoI(3) + 2£1 (C" b!3. Ccobn) "" b'c
Y01
C
i t
(e) From part (c), we have:
For a vigorously stirred fen-nemor, assume [hac tluid shear is sufficiently high, that transpOrt to the edge of the floc is negligible(l):
_. . 1.. . . .
--~
0
kcac The mass transfer resistance from air bubble to bulk liquid depends on me fermentor design, air flow rate, agitation rate and a number of other factors. For a 10 1 laoonuOlY scale fermentor. kl ab was f()und to be~· 150 mMz. during the growth phase of the I hI arm fermentation . (2) Dividing through by Hemy's low constant:
k1 ab ;;
(ISO mMQ!~)(0.88;Ur.n.L . .}(;:;JJl!....) 1 hr atm
mMole ..,600 s
"'" 3.67 x 10.2 S ·1 Effectiveness factor:
Microbial growth on multiple substrates (here oxygen and glucose)
is typically modelled using Moncxi type kinetics:
Rc ::
),l.(':c
By representing the reaction as first-order with respect to oxygen, we are essentially assuming a low oxygen concentration, relative to the intrinsic rate parameter. Ko:
7-55
__,.1_
--t
0
kcac The mass transfer resistance from air bubble to bulk liquid depends on the fermentor design, air flow rate, agitation rate ::md a number of other factors. For a 10 1 laboratOIY scale fennemoI', kl ab was found to be -- 150 mJ,1Qk during the growm phase of the 1hr atm fermentation", (2) Dividing through by Henry's low constant: kl ab
:=
~~;;~HO.88 '~;~~i~']{3A~LJ
{1.50
= 3. 67 x 10.2 S·l
Effectiveness factor.
Microbial growth on multiple substrates (here oxygen and glucose)
is typically moocHed using Monoo type kinetics:
Rc ;;: ).lee
By representing the tt:action as first"order with respect to oxygen, we are essentially assuming a low oxygen concentration, relative to lbe intrinsic rate parameter, Ku:
For a first· order reaction. the effectiveness factor is;
",~. ($ cosh ¢ .. 1) ¢(
n
11 ::: OA5 Reaction rate constant:
k ""
83 x 10
tI
5 S ,1
,.-.max "" ..... ,,,_.,, ................, ..,,,,,,._ .•
-K a
()
3.2 x 10.4 ~,-.~
:: 0.26·,··..1·, . ,. g0 2 S
1
Finally, aSSume an initial cell concentration of 0.25 g cell/l, L-'le cell concem:."";;!.tlon equation now becomes:
r
(Cc
367 x 10.2 s'
I
-
!
1.5
~~!~.)! 7:l x 10 ;) ~9.:d::) ; In 4Cc ;:;: (1. 5 2.g02 \ 1 i.{(45)/O.26 0 "J,I '. \ g. 2
'" I
..
t
$,
2.7,25 (C-=C; ··0.2.5)
f
1281n4Cc;:;: 0.012t
Clearly, mass trilllsfer from the gas to lhe liquid phase ond internal diffusion play UnpOItilllt roles in determining the cell growth rote.
7·56
Cell mas vs time, Stan ::n 0.25 gil t
{hr)
Cc (g/l)
Cc (g./l)
1 thr)
-0--" 0.25 "'---"'8-113 2.37 3.74 S.17 6.64
1 2 3 4 5
'm
1 LIS 12.67 14.21 15.74 17.28 0.62 1..72
9 10 11 12 05
6 8. 13 L5 1 9.63 From the graph it can be seen that growth startS our exponentially and becomes linear as the fennentation becomes limited by gas· liquid mass tmnsfer. Sensitivity analysis:
TIle gas~liquid mass transfer coefficient is related to the agitation rate to the 0.95 powerO):
k!
au u
NO.95
What is the effect of increasing the agitation by 50%?
kl
ti2 ""
kJ
al
(1.5°·95)
::: (3.67 x 10 .2}s -I{ LSO.95}
;; {S.39 x !D,lls·1
(see graph)
Since cell growth has an exponential ponion. another way to increase the growth mte would be to increase the innocuous size. C<:o. What happens when Ceo is quadrupled?
C'-cu
=:
1,0 gil
(see graph)
From the results shown in the graph. a relatively small increase in the agitation nne leads to
a significant increase in the cell growth rate. while an inCIe:lSe in innoculum size means that the fennentation leaches a gas·liquid ttansfer,limited state more quickly, but the growth rate remains the same.
Cell Mass 'IS Time for a STR P("" ........ fermentation
18 '"
I
I
16 .. I 14 .... •
cell mass
12 ..
(gIl)
10·..
I /
I
lq a ~ 3. 67 x 10-1 5 1 Ceo = O. 25 g/i kl a = 5 39 x 10.1 S .;
/
I
Ceo " 0 25 g.'1 kl a '" 3 6.7 x 10 ('':;C '"
1.0 gil
"-j . -T--r--"'r·T'-r·-"'·T' 2 4 fi 8 10 12 14
t (ttr)
7-57
Z $:
References: (I) James E. Bailey and David F. Ollis, 5iQfh~rIlicalElJ..giDeerjng, FllndamemaIs, (NY: McGraw-Hilf. 1977), Chapters "1 and 8
(2) (iF. Payne, PhD Thesis, University of Michigan, (1983)..
(3) D.Le Wang, et at, F~r:rm;.!l!illiQQ_l!nQ."Gnn'lIIt.TechD.Q102"y, (NY: John Wiley & Sons, 1979), Chapter 9
CDP7-D No solution will be given. CDP7-E Since the deninification f6liow$ Michaelis Memon kinetics, first detemrine V mall; and Km from Lineweaver BUlk plot.
Initial
. .:]
r-N-;:'ol' __ "-J' t
.:2
Time for 50.6°
Rate of reaction
,
I .
1f!
_ N 01_fl?E!!~2..._. ___.._._.-:duc~!.~:~.: . (min )_r:...{E£!l1!E.~_~_._. . __.___._ [
25 50 7.5 100
100 (given)
.04 .02
.013
35 38 44
.010 .005
50 66 (given)
lIr
2.0··
{~~¥;;)
1 ..5 ..·..
852
1.173
1000 l.515
1.000 0660
rime for 50% reduction
'.j
2.5 .
2.800 1.520
...._J:::~.~al [N0;;.·l!.~_ ._...
where the n.He of reaction is fbl."lnd by the r:mo: 3.0
.357 .658
f
10
0. 5 0 .. 00
7-58
LineweaveI· BUlk: From the plot, V ma>; :: _ _L_.... "" 2.73 EE~. intercept min
:. Km = 165 ppm Next, need conversion as a function of rime.
NSO~· m
Design equation:
;:;:
··rs V
_ V max S
Rate Law:
·IS -
R:+s s
Stoichiometry:
== Sop··X)
Finally, VInax t =- So X • Km In (1 .. X) Know V max • Km, t =- 68 hrs :; 4080 min So "" 0..2.5 • Iteration
[0
(!DOI.) (46.JL.) (.~Qg~rr:~) (!.PP!.::) i
mol
,.
J
mgil
=:
11,500 ppm
find the conversion obtained after 68 hrs gives:
X:;: 0.930
The [N( 2 ) level is: (1·· 0.930) (11.500 ppm) == 805 ppm Since t, Km. and So are fixed by the system. the change must be made in Vnu:x'
Desired conversion:
... Desired V r;1:U;
Since Vmax Ct [El ] Vmu;
,
x=
(.9565)1651n (l ... 9565) _. _. __ ..._........ _. __ •40"80 . '-'..................... _. __. = = 11,500
2.823 E~~ mm
increasing the conceno·ation of whole cells in the emulsion will increase
(2...8.ll) 50 ... ,r:.:~.~en.~.... 273
1 ... 200 _. :::: . 9565 11.500
ml emulsion
mg c:!~ . . _.
= Sl.7-
ml emulsion
Therefore, increasing the cell loading to 52 •. .::::£.~~n.s-- would results in a level of [NO£J < ml emulslOn 500 ppm after 68 hours .
CDP7-F No solution will be given.
7-.59
CDP7-G >folt bala.nces in a CSTR DC so ..- LJ(:$ -;- fs
··DC p
C
~
?
·' II'
=:::;
0
C
r
,r.
D
the Lite la\-v as given is: r~
;;;;; "Is
r
o=.Y~.~~ I(1· Ee. 1 (,,)
r
11'
• ("
1.\.5 '~"S \
'p )
Plugging tho$e into POLYMATH and using different values of D and C t , come up with the following; Using a volume of 200 dm' and a cell concentration of 50, get the best production
of the Lmalic acid. f~cs]=cs ·cso~rp:D
c50=2
Sol ut:
Q"
f( )
cs
CJ ,35: 657 L 64834
cp
esc
LJ
cc vrndx
76
krn
C. :.><;8
cpst ar
CDP7-H Michaelis·l\·fentoll Kinetics:
E=40rng
7-60
3. 222E''" i 5 <3.2772·': 5
Mole Balance: __ dC.s. == VWil.CS% dt K,.. +Cs~ CS~
== CSo%(1--· X)
=>
1.4 --0.2 14
](==-··. -==0.86
Where C S $ is the percent of fish oil. For most oils: density 0 . 9 3 xl 0.-3 mo1Jm1 :;:: 3 x 103 p.mo11m 1 C., :;::._---;;;:_._:;:: 1 MW 300
Cs : ;: ; CrCs%
f . .K. . .
+C
{)2
=>
m(..
~-.~"(iC s%::::::
14"S%
(1.4)
Jtv(';'" dt == f 5.6xlO3 dt 0''1
0
5.6x 10 3 /3
CDP7-J No solution will be given.
(b)
1 X ;;;;: . _-_. n 1.- p
m
-
t
t:::::: 6.21n(7)+~:~ min == 7106 min == 118.4 hrs
CDP7-1 No solution will be given.
(a)
.
5.6xlO··3
=> ~ln 0.2 +- (1.4- 0.2);;;: -3--· =>
3
t
For Xn :;:; 5:
p :;:: 0 . 80
X,=dO:
p:;::0..90
X" == 20 :
p ::;;: 0 95
7-61
Use these equations to generate the desired graphs: Yl 5
o2
YS.
j for X. = 5, 10, and 20
.I
r-····--·--·'··'-·--'·'··-·~='~~~~~·~.. "~::~-=-- j
020I l '- -Xn;;;!O ;:-XneeZO {I
---.-.-..-----.--... ---.-.--.
o 10 ~ "
;:: 040
!
030
1\\
S, 10, and 20
0,60
I !
: t
=:
070
:'
o 15 i' ;>-~
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
-.
<)$
::'.1,i>4e·-,oS
0 .. .')01-
O.S 5,,·01 S.·06~ne·4~
·2 9g:5Bi:
n
:: :;c;;
1
~EJ~
-
.-,;
: i
J ZOO
0 .. ':00
CSTR:
I 20 -·· I 2:::: ····r1: *"...
M .- M :::: ..... r I *r
o )<.. POI YM'ATH the followino- !rraphs are The rate laws are the same so agam usrng . to to J
generated. Initiator Concentratlon ..-" Space Tlme CSTR
Monomer vs space time CS TR
·········. 1
•
'50000
) OOooc-
1 SO
...... ~---.-""----+
2(101)(10
S:)O;)O
Tau 1J'::t
lQOOCO
l$C·OOO
20000-0
Z5~oao
Tau, H:,.
2.5 a~oc1.
:mo=]
t:.du=5el1 io.:=.Ol
kp=:10
ko"'le.·· 3
1':=.5 k:::·=5e"J
b)
Fm two CSTRs, the design equations change just a bit '1"
M··M = .. ! lvi, *.::-2 o 1
M .... ,. 1\1, :::: ,. L j
•
.,,:
* 1:2
1'1 ."" I,.., :::: -.
*' "T2'. .
The rate laws are the same with the exception that instead of just 12 or M, 12 [> Iw MI' or M;) are used depending on which reactor they came from and the following graphs ale generated
7-66
..
..'.
.l •... ".,.'" ...., .. ~ ... " ...
., ..
I
1:_'1!1
I
I
.;2:
a
•
.. . . a ..
•••
000.
r
.
o ,:)os L 000. [
{) 00" ~• o{)'c;i: ~
r ~~
!loa t
Q,jQl :.
*!: ••
2. [
>.••
Q
Ii ..
0>01
roC
o ._1. ...................,._...•,........... _
........... . 'S-IKIQq
Q
lOOOQO
T--., ,,.
.-
c) Making kg bigger causes [2 to decrease rapidly and M does not get formed as much. Increasing kt> causes M to decrease sl.ightly but not by t.hat much. Increasing ~, causes M nOt to decrease by very much staying very close to 3. 1
CDP7-M No solution will be given CDP7-N I +iH ..5_~ R1 R.J t·i.V!_.!:J:. ····7R.~J --'I ::::; k/vlI
a)
, ..... ..,
!
~ OQ~ ~
Balance on I
L, .. ' I
10 .. I
. (1
k;iYll
'f;;;;;; .-...-----:;;;; -..... ..
10 => 1;;;;;;",-· l11k,M ~
'-I~1
;;;:;:
kiMI + kp!v(2':Rj
ASLR
J=l
j
-==
10 -- I
j~!
'-r~l :::: kMI + k PM(I 0 .- I) 1
7.. 67
........ ..
Balance on M: r
_ l'tlfo-;lV! A-fo - iYf .,,','-----"" == ,----"------,, .-.---.---_.._.
~
·r;,!
kiA-1f + k/d( If)-
=> rk/vll +rkplYf( 10
-
!)
I) "'" Mo - M
rk"ly[Jo -rk/oM( rk,lvt) => ----------'-,-- + .'"'_~,._.,_.,___ c, = 11,11) .- fH I + 'rk,lv/ 1 + rk,114 i . => rkJiJAl +-r2klJoj'yf2 ::;;: (J"10 =>
-
M)(l+ rk,M)
,k, (1 + rk,?~) )M" + (1 + 'lk'!a -, rk,i'vio )1'.1 ,,',. A10 :::: 0
b) --If<: :::::: --'kiJ\U + kl,;'vlRI Balance on R1
Similm.!y, ,<
R,:::;
R21
Rj
:::;
1 -~'rk!
lv/X;'-
1
I
I + k l v1-'(·
If
ri< \/j 1 + rk p lrf)
\
k
,
p \
p
)
k"
Initiation Rate constant ~ < < propagation rate constant Hence, nearly no change in the concentration of Initiator (I).
c)
Mo 10
1
t
xi
0015
kp
1QC()
7-68
......... w._ ••••• ~ ••••••••
0.5
+-. __.......
1.5
1 tau
d)
7-69
w ........ ~~ ........... w . . . . . . . ...
2
'~2
x(1+x)
j
= ·(-l:=~~-)T
As 2., J.X r=l
CDP7-0 a)With the reaction self catalyzed the mole balance and rate law becomes: dlCOOHl , _.. _J.:_:_.:..:.J
at
=k[COOHY
We can then get [COOH 1as a function of time. The following graph shows both the given values of p and the calculated value as 11 function of time where p ::= t CO,Ql!Jn:1CC!..l?!!J
.... __ ...,l~:~~~!J1. ,. ,. P
1
•+
0,8
••
'IS
",.",,'"''
time
..... 'T...... ".""""""' .. "''',...--, ...... ""..""."","" .... - ••.. ,..• " ..•.•.".-._,,,.,,., ... ,"". """ •. " ....,""" ...."" .. """" .... "
41\ 41\
06 0.4 02
o
41\ .."'',.""----•• '
o
t , , , , , , , , " " , · ·.. ··- , ....•.
t"""" 15(:0
500 time
It appears to fonow this above 500 min"
7-70
2000
b)
The new mole balance and late law is:
-,!!l£.~f!!J == k[COOH][OHJ[lI+] [OH]=[COOH] COOH:~CO(r +H+
=Icq,~~~:J
K
[COOH]
eq
[COO"
]= [Bt ]
[Rtf == K~q[COOH] .. .:1~q,g{:!J:= kICOOH]12
dt ' Solving for [COOH] as a function of time gives the following graph:
p vs time ·~--;-'-t"--'-·t·-····"'I'···'-.··-··-'····
0.8 • •
•
0.6 0.
0,4 ' 0,2
o
500
1500
1000 time
2000
It follows the data above 200 min. e) This mechanism can be made to fit either rate law, depending on whether UA dissociates before or after the fust reaction.
CDP7-P CDP7-Q
7-71
I + M--·~--7 RI R.) +M ·--~---'tR-1+1
_. _~i!_ = kOMI =
(k M }~2! P k
dt
"
dI
k
."'-_._- == .._S!. I k/r1.dt
k,v'
d8:::::k p Mdt .. ~
dI
k
dB
kp
--- =.__ . '2. I => I "'" I e x, 0
dR --_..l=-·-k MI +k MR dt 0 p I dR
•. _.J.. ==
de
);,
~.
.-;-'~ 1 e[ ',) ._- R
k
. . . '2.. kp
0
I
_~.{<~!) == .~Q.I e(l ~'J dB
kp
dR de
0
-- J. :;;;;
R.. ..- R ' :!
CDP7-R
7-72
1 -+ ivf'~'--7 R1 R.)
-+ lvf "-'~"-7 R./+1
r:::: .I.11.._-... __[ ko/vfI
8' ::::. kpi'vFr:
8' :::: ~f!!L,::::
. !.l
koI
k/ I == -............ _"..o_.......-. k o8' +kp l"tl ..,·lvl k/,;flo
l' :::: ....JL ...... ,... --
A'fo --A1 ;:;; -----.. kl,lor + 1 9';:;; _..~e!~!?~. " k/o'f+ 1 R,;.. r :::: -----.--...-..... kolvlI --- k,)rfR,
Ie' R ;::: ...k_Q,.... ! k1',(1+ fJ') 1.'
&_ . . ,. ,_.-..
== -.. _.-.---.-.. k/,fR{- k/V{R2
R 8'
R2 :::
k k 1 8'2
k I 8 1z
i:~,{i; == k)i~~;1-~~e';':k~); (j . ~-e;Y~~e;·~-~k:)
7-73
Fogler 7-24 Solution Problem Statement: In biotechnology industry, E. coli is grown aerobically to highest possible concentrations in batch or fed-batch reactors to maximize production of an intracellular protein product. To avoid substrate inhibition, glucose concentration in the initial culture medium is restricted to 100 grams/liter in the initial charge of 80 liter culture medium in a 100 liter capacity bioreactor. After much of this glucose is consumed, a concentrated glucose feed (500 gil) will be fed into the reactor at a constant volumetric feed rate of 1.0 liter/hour. When the dissolved oxygen concentration in the culture medium falls below a critical value of 0.5 mg/liter, acetic acid is produced in a growth-associated mode with an a of 0.1 g acetlg cellmass. The by-product acetic acid inhibits cell growth linearly, with the toxic concentration (no cell growth) at Cp * oflO g/liter. Find the optimum volumetric flow rate that will maximize the overall rate of cell mass production when the bioreactor is filled up and if the feed is turned on after glucose falls below 10 gil. Inoculum concentration is 1 g cellslliter. Additional parameters: f.lmax
= 1.2 hr,-I, KG = 1.0 gil, Ko = 1 mg/l, Y xls = 0.5 gig, Yp/s = 0.3 gig,
qo/x
= 1000 mglg
Oxygen mass transfer rate kLa = 500 hr- I, Saturation oxygen concentration C02*= 7.5 mg/liter
Increase the value of mass transfer rate (up to 1000) or the saturation oxygen concentration (up to 40 mg/liter) to see if higher cell densities can be obtained in the fed·· batch reacator" (a) list ways you can work this problem inconectiy" (b) How could you make this problem more difficult? (Contibuted by Prof. D. S. Kompala, University of Colorado)
Solution: This problem is solved numerically in three parts, using the following equations on Berkeley Madonna package: 7-74
The first time period covers the simple batch culture, when glucose and dissolved oxygen are being consumed for cell growth. METHOD Stiff STARTTIME = 0 STOPTIME = 3.4 DT=0.02 INITG =100 INIT X = 1.0 INITO =7.5 INITP=O mumax = 1.2 KG = 1.0 KO = 1.0 Yxs = 0,,5 Yps = 0.3 q = 1000 kLa = 1000 alpha = 0 SATG = G/(KG+G) SATO = O/(KO+O) SATP = 1.0 - (PIlO) mux = mumax * SATG*SATO*SATP*X d/dt(X) = mux d/dt(G) = - muxlYxs d/dt(O) = kLa*(7.5 - 0) - q*mux d/dt(P) = alpha*mux The numerical simulation results shown below identifies the time at which the dissolved oxygen concentration falls below the critical value of 0.5 mg/liter, triggering the formation of the by-product acetic acid. From the simulation results, we find that the dissolved oxygen concentration falls below the critical values of 0.5 mg/l at the batch culture time of 3.64 hours. At that time, the glucose concentration has fallen to 67.1 gil and cell mass concentration has growth to 17.45 gil. The by-product acetic acid concentration remains zero through the early batch culture, as the dissolved oxygen concentration is above the critical level throughout this time. In the program above, the parameter alpha is set to zero to ensure that no acetic acid is produced. 7-75
Run 1: 43 steps in 0 seconds
18
100
16
80
14
70 12 60 10 0
..
~
.... ...- ...
50
,
1><
~
8
40 6 30 4 20
2
10
o o
0.5
1
1. 5
2
2.5
3
3.5
4
TIME
In the second part of the batch culture, acetic acid is getting produced and glucose is still above its set point of 10 gil, when the concentrated glucose feed is added to the bioreactor.
7-76
The program equations are given above slightly modified to change the alpha value to the given value of 0.1 g acetic acidlg g cell mass and integrated from the end of first part of batch culture. RlJn 1~ 2t; steps
in 0 s~90nds
~~~~~~~~~~~~~~~~-r~~~~,r45
.
c.o
20
3.5
4
4. 5
5
5 ..5
6
6.5
7
7.5
8
TIME From the simulation results, we see that glucose concentration reaches the predetermined value of 10 gil (for turning on the feed) at 7.65 hours of batch culture. At that time, the cell mass concentration has reached 41.02 gil and the by-product acetic acid concentration has reached 2.85 gil. Using these values as the initial conditions for the third part of culture, a glucose feed is added and the balance equations are therefore modified to include the dilution of all bioreactor contents with the fresh nutrient medium. The modified program equations are shown below: 7-77
METHOD Stiff STARTTIME = 7.65 STOPTIME = 209.5 DT=0.02 INIT G =10.0364 INIT X = 41.8957 INIT 0 = 0.1978 INIT P = 2.85834 INIT V = 80 mumax= 1.2 KG = 1.0 KO = 1.0 Yxs = 0 . 5 Yps = 0.3 q = 1000 kLa= 1000 alpha = 0.1 vin=O.1 SATG = G/(KG+G) SATO = O/(KO+O) SATP = 1.0 - (PIlO) mux = mumax
* SATG*SATO*SATP*X
d/dt(X) = mux - X*vin/V d/dt(G) = - muxlYxs +(vinlV)*(500 - G) d/dt(O) = kLa*(7.5 - 0) - q*mux d/dt(P) = alpha*mux - P*viniV d/dt(V) = Yin The constant value for yin, the volumetric feed rate can be systematically varied to find the highest cell mass concentration, when the reactor volume gets filled, i.e. becomes 100 liters. Simulation results for different Yin values ar'e tabulated below:
yin
time, hrs
Volume
X cell con cent
0.,,0.5
40.7.8
10.0.
87,53
0.1
10.0.
8758
10.0.
87.,6
0.,3
20.8 10.7.,7 75,,3
0.4
574
0.,2
10.0.
87.,5
10.0.
87.,3
7-78
0.5
47.,7
0,6
41
0.7
36,2
100 100 100
87,5 87.,6 87,,5
It is clear from these simulation results that volumetric feed rate does not make a strong difference in the final cell mass concentration. The time for filling up the reactor volume to 100 liter is of course strong affected by the volumetric feed rate It is expected if the kLa is smaller, then the acetic acid production will be higher. In that case, the volumetric feed rate will have a significant effect on the maximum cell mass concentration achieved in the fed batch reactor.
These simulations nevertheless provide a useful introduction to the concepts of fed-batch culture.
7-79
Solutions for Chapter 8 - Steady-State Nonisothermal Reactor Design P8-1 Individualized solution
P8-2 (a) Example 8-1 ForCSTR
v = FAOX = __X__ Dok(l-X)
--fA
'tk 'tAe -E/RT X=---=--, l+'tk 1+ Ae- E/ RT One equation, two unknowns Adiabatic energy balance
T=To_MIRXX
CpA In two equations and two unknowns In Polymath form the solution
'tAe- E/ RT f(X)= X --E/RT l+Ae f(T) = To -
~HRxX_ CPA
Enter X, A, E, R, C p A , To and i1HRx to find '( and from that you can find V.
P8-2 (b) Example 8-2 Helium would have no effect on calculation %Error =
-~Cp (T - TR ) -[MI~x +~Cp(T-TR)]
1270 =----x100 = 5.47% 23,210
P8-2 (c) Example 8-3 (a)
V =0.8 m
3
See Polymath program pg-2-·c.pol.
8-1
POLYMA TH Results Calculated values of the DEQ variables Variable
initial value
o
V
o
x Cao Fao T
Kc k Xe ra rate
minimal value
o o
9.3 146 . 7 340 2.4595708 8.5452686 0.7109468 -110.4184 79.470998
9.3 146.7 340 2.8783812 8.5452686 0.7421605 -79.470998 79.470998
maximal value 0.8 0.5403882 9.3 146.7 363.39881 2.8783812 38.191248 0.7421605 -79.470998 110.4184
final value 0. 8 0.5403882 9.3 146.7 363 . 39881 2.4595708 38.191248 0.7109468 -85.208593 85.208593
ODE Report (RKF45) Differential equations as entered by the user [ 1) d(X)/d(V) = -ralFao Explicit equations as entered by the user [1] Cao=9.3 [2] Fao = . 9*163 [3] T = 340+43.3*X [4] Kc = 3 . 03*exp(-830 . 3*«T-333)/(T*333))) [5] k = 31..1*exp(7906*(T-360)/(T*360» [6] Xe = Kc/(1+Kc) [7] ra=-k*Cao*(1-(1+1/Kc)*X) [ 8] rate = ora
PFR T
330
X
0.26
340 054
350 0.68
370 0..66
390 0.65
420
0 . 62
450 0.59
500 0.55
600 0.48
----------_._-
0.7 0.6 0.5
[~J
0.4
0.3 02
330
384
-438 T
492
546
600
CSTR has the same trend.
P8-2 (d) Example 8-4
Counter-Current: Guess Ta at V =0 to be 330 and it will give an entering coolant temperature of 310 K. See Polymath program P8-2-d.pol.
8-2
POL YMATH Results No Title 08-17-2005, Rev51 233 Calculated values of the DEQ variables initial value
Variable V X T Ta Cao Fao Kc
0 0 310 330,,7 9,,3 14.67 3,6518653 0.9004084 0.7850325 -8,,3737978 -6900 5000 159 8.3737978 50 75
k
Xe ra dHrx Ua Cpo rate m Cpc
minimal value 0 0 310 310,16835 9.3 14,,67 2.7812058 0,,9004084 0,7355341 -27,114595 -6900 5000 159 0.0460999 50 75
maximal value 5 0,,7797801 344,71423 335,79958 9,,3 14.67 3,6518653 11.. 763976 0.7850325 -0.0460999 -6900 5000 159 27.114595 50 75
final value 5 0.7797801 310.83085 310.16835 9.3 14.67 3,,6255777 0,,9639302 0,7838108 -0,0460999 -6900 5000 159 0.0460999 50 75
ODE Report (RKF45) Differential equations as entered by the user [11 d(X)/d(V) = -ralFao [2] d(T)/d(V) = «ra*dHrx)-Ua*(T-Ta))/Cpo/Fao [3] d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc Explicit equations as entered by the user [1 J Cao = 9.3 [2J Fao = ,,9*163*,,1 [3 J Kc = 3,,03*exp(-830,,3*«T-333)/(T*330))) [4] k = 31,,1 *exp(7906*(T-360)/(T*360)) [5J Xe = Kc/(1+Kc) [6] ra = -k*Cao*(1-(1+1/Kc)*X) [ 7 J dHrx = -6900 [8] Ua = 5000 [9J Cpo = 159 [10] rate ,·ra [11] m = 50 [12] Cpc = 75
=
P8-2 (e) Example 8-5 At V = 0, Ta = 995.15 and gives a counter current entering temperature of 1250 K. See Polymath program P8-2·e.poL POL YMATI{Results Calculated values of the DEQ variables variable V X T Ta Fao Cpa delCp Cao To
initial value 0 0 1035 995,,15 0,,0376 163 -9 18.8 1035
minimal value 0 0 972.39417 986.00676 0,,0376 163 -9 18.8 1035
maximal value 0,,00;1. 0,,3512403 1035 1249,,999 0.0376 163 -9 18,,8 1035
8-3
final value -----cf:001 0,,3512403 1034,,4748 1249,,999 0.0376 163 -9 18.8 1035
7 . 414E+04 -67 .. 304 1 . 65E+04 0.111 34 . 5
7 . 414E+04 -67.304 1 . 65E+04 0.111 34.5
dHrx ra Ua me Cpe
7 . 47E+04 -6.3363798 1.65E+04 0.111 34.5
ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) =-ra/Fao [2 j d(T)/d(V) = (Ua*(Ta-T)+ra*dHrx)/(Fao*(Cpa+X*deICp» [3 j d(Ta)/d(V) = -Ua*(T-Ta)/mc/Cpc Explicit equations as entered by the user [1] Fao = .0376 [2] Cpa = 163 [ 3] delCp = -9 [4] Cao = 18 . 8 [5] To = 1035 [6] dHrx =80770+deICp*(T-298) [7] ra = -Cao*3.58*exp(34222*(1!T0-1/T»*(1-X)*(TolT)/(1 +X) [ 8] Ua = 16500 [9] mc = . 111 [ 10 1 Cpc = 34.5
(b)
P8-2 (f) Example 8-6
Energy balance will remain the same X EB = 2 xlO-3 (T - 300) for 2A-"72B
.... , \ \ 1st Order
2nd Order \
\ \
(c)
, ....
.... ........
P8-2 (g) Example 8-7 8-4
7.414E+04 -31.792345 1.65E+04 0.111 34.5
Both Xe and XEB will change. The slope of energy balance will decrease by a factor of 3.
x
T
Also Xe will be more temperature sensitive
K =K e
e
exp~HRX (~--~J R T T 1
The dotted line in the plot below shows an increase in -AHRx
\ \ \ \
x
\ \ \
, " '" ....
T
(d) (1) (2)
P8-2 (h) Example 8-8 CAO will decrease but this will have no effect 't will decrease 't'=
401.1 ft 3 466.1 ft3
(3)
Is
In the energy balance the slope of the energy balance of X vs . T will be greater
8-.5
l:8 i C PC = 35 +(18.65X18)+4 x (1.67X19.5)= 35 + 335.7 + 130.2 =501 BTU kmoloR XM
,/
-- _.
XM, T
I I I I
I I I
I /
I I
Basecase
I I I
I I
_
_ _
Change QM
I I I I I l
8-6
Less Conversions
P8-2 (i) Example 8-9 Change C p =29 and -~H = 38700 POLYMA TH Results
NLES Solution va:r::·iable
Value
x
fix) 2 . 444E-11 1. 2E-09
0.7109354 593 . 6885 0 . 1229 1.696E+13 3.24E+04 1. 987 20.01167
T
tau A
E R
k
Ini Guess 0.367 564
NLES Report (safenewt) Nonlinear equations [1 J f(X) = X-(397.3*(T-535)+92 . 9*(T-545))/(38700+7*(T-528)) = 0 [2J f(T) = X-tau*k/(1+tau*k)
=0
Explicit equations [1] tau = 0 . 1229 [2J [3] [4] [5 J
A=16.96*10A12 E = 32400 R = 1.987 k = A*exp(-E/(R*T))
Vary the heat exchanger area to find the effect on conversion.
P8-2 (j)
=
3
a 1.05 dm See Polymath program P8·2··j.pol. POL YMA TH Results
Calculated values of the DEQ variables
8-7
Variable V Fa Fb Fe T Y
kla k2a Cto Ft To Ca Cb Cc rla r2a Fto alpha
initial value 0 100 0 0 423 1 482.8247 553.05566 0.1 100 423 0. 1 0 0 -48.28247 -5.5305566 100 1. 05
minimal value 0 2.738E-06 0 0 423 0.3120454 482.8247 553 . 05566 0. 1 77.521631 423 2.069E-09 0 0 --373.39077 -848 . 11153 100 1. 05
maximal value 1 100 55.04326 22.478369 812.19122 1 4.484E+04 1.48E+07 0.1 100 423 0.1 0.0415941 0 . 016986 -5.019E-05 -1.591E-ll 100 1. 05
final value 1 2.738E-06 55.04326 22.478369 722.08816 0 . 3120454 2.426E+04 3.716E+06 0.1 77.521631 423 2.069E-09 0.0415941 0.016986 -5.019E-05 -1. 591E-1l 100 1. 05
ODE Rel!ort (RKF45} Differential equations as entered by the user [1] d(Fa)/d(V) r1 a+r2a [2] d(Fb)/d(V) = -r1 a [3] d(Fc)/d(V) =-r2a/2 [4] d(T)/d(V) (4000*(373-T)+(-r1 a)*20000+(-r2a)*60000)/(90*Fa+90*Fb+ 180*Fc) [5] d(y)/d(V) =-alpha/21y*(FVFto)
=
=
Explicit equations as entered by the user [1] k1a 10*exp(4000*(1/300-1/T» [2 J k2a = 0 . 09*exp(9000*(1/300-1/T» [3 J Cta =0.1 [41 Ft = Fa+Fb+Fc [5J To 423 [6] Ca Cto*(Fa/Ft)*(To/T) ['7 J Cb =Cto*(Fb/Ft)*(To/T) [8] Cc =Cto*(Fc/Ft)*(TolT) [9] r1 a = -k1 a*Ca [ 10] r2a = -k2a*Ca"2 [11] Fto = 100 [12] alpha = 1.05
=
= =
(e)
P8-2 (k) Example 8-11
VaryUA
J/m
3 VA= 70,000 -s-K only the lower steady state exists at T =318 K SBC = 0.05
J/m 3
VA=60,000 -s-K only three steady states exist T =318, 380 (about) and 408 (about) depending how you read the intersection on the graph. 3 VA= 700 -s-K only three steady states T =300 (about), T =350 (about) and one are a very high temperature off the scale of the R (T) and G(T) plot. In all cases SBC remains low at 0.05, meaning that the reaction has neared completion to form species C therefore reactor is too large.
J/m
8-8
To
= 275, very little effect.
Vary l' 't = 0.001 only the lower steady state at T = 316 about and other off scale SBC = 0.05 z =0.0001 only are steady state at T =316 and others off scale SBC =0.05 't = 0.00001, SBC =5 However, the upper steady state is off the graph and needs to be studied (f)
P8-2 (1) Example PRS P8-4 ..l 1
(J.----
dpPo
U=U 2
j :::::
=[~:2)(~)J=1
No effect for turbulent flow if both dp and P changed at the same time.
P8-2 (m) Example TS-3 fie = 200 g/s
See Polymath program P8·2·rn.pol. POLYMATH Results Calculated values of the DEQ variables variable W
Ta Y
T X
alpha To Uarho Me Cpme Hr Fao theta I CpI CpA the taB CpB Cto Ea Ke ka yao xe Cao sumep Ca Cb Ce ra
initial value 0 320 1 330 0 2 . 0E-04 350 0.5 200 18 -2.0E+04 5 1 40 20 1 20 0.3 2.5E+04 66.01.082 0.046809 0.3333333 0 . 8024634 0. 1 80 0.1060606 0 . 1060606 0 -·5 . 265E--04
minimal value 0 320 0 . 3044056 330 0 2.0E-04 350 0.5 200 18 -2 . 0E+04 5 1 40 20 1 20 0.3 2.5E+04 0 . 8247864 0 . 046809 0.3333333 0 . 3122841 0. 1 80 0 . 0137198 0 . 0137l98 0 -0 . 0143957
maximal value 4500 334.77131 1 385 . 31436 0.5645069 2 . 0E-04 350 0. 5 200 18 -2.0E+04 5 1 40 20 1 20 0.3 2 . 5E+04 66.01082 11 . 205249 0 . 3333333 0 . 8024634 0.1 80 0 . 1060606 0.1060606 0.0724316 -1 . 745E-05
8-9
final value 4500 334 . 77l31 0.3044056 338.18498 0.5645069 2 . 0E-04 350 0.5 200 18 -2 . 0E+04 5 1 40 20 1 20 0.3 2.5E+04 31.. 551036 0.1177827 0 . 3333333 0 . 7374305 0. 1 80 0.0137l98 0 . 0137l98 0.0355685 -1 . 745E-05
ODE Report (RKF45) Differential equations as entered by the user [Ii d(Ta)/d(W) = Uarho*(T-Ta)/(Mc*Cpmc) [2] d(y)/d(W) = -alpha/2*(T/To)/y [3] d(T)/d(W) = (Uarho*(Ta-T)+(-ra)*(-Hr»/(Fao*sumcp) [4 J d(X)/d(W) = -ra/Fao Explicit equations as entered by the user [ 1] alpha =.0002 (2) To=350 [3) Uarho = 0.5 [4] Mc = 200 [5] Cpmc = 18 [ 6 J Hr = -20000 [7) Fao=5 [ 8] thetal = 1 [9] Cpl = 40 [10J CpA =20 [l1J thetaB = 1 [12] CpB=20 [13J Cto = 0.3 [14] Ea = 25000 [15 J Kc = 1000*(exp(Hr/1.987*(1/303··1/T))) [16J ka = .004*exp(Ea/1.987*(1/310-1/T» [17 J yao = 1/(1 +thetaB+thetal) [18] xe = Kc"O.5/(2+Kc"O.5) [19] Cao = yao*Cto [20 J sumcp = (thetal*Cpl+CpA+thetaB*CpB) [21 J Ca = Cao*(1-X)*y*To/T [22] Cb = Cao*(1-X)*y*To/T [2.3] Cc = Cao*2*X*y*Torr [ 24] ra = -ka *(Ca *Cb-CcJ\2/Kc)
P8-2 (n) (1) The concentration of A near the wall is lower than in the center because the velocity profile is parabolic. This means near the walls the velocity is much lower and therefore the time space near the wall is much larger than in the center. This means the reaction has longer to take place and conversion will be higher near the wall. Thus the concentration is lower. Below is the FEMLAB solution. 1. Parameters in simulation on the tubular reactor from Example 8-12 (First Order reaction): Reaction: A + B -7 C
A- propylene oxide; B- water; C- propylene glycol (1) operating parameters Reactants • Feed rate of A FAO = 0.1 molls •
Inlet flow rate of A
VAO
=
FAOM A
PA
0.lx58.1xlO-3 -6 3 = =7x10 mls 830
8-10
• • • • •
• •
Inlet flow rate of B v BO Inlet total flowrate
Vo
= 2.5 x 2 x V AO
6
= 35 X 10- rrhs
= 2v AO +v BO = 14xlO-6 +35xlO-6 = 49xlO-6
Inlet concentration of A C AO
FAO = -Vo
0..1 49xlO-
. F v P Inlet concentIatIOn of B C BO = ---.!!.Q.. = ~- = Vo
Inlet temperature of the reactant To
6
MBVO
= 20.40. .8
3
mol/m
35xlO-6 x1000 3 = 39682.5 mol/rn 6 3 18xlO- x49xlO-
= 312K
Coolant flowrate, m] = 0.,,01 kg/s Inlet temperature of the coolant, Tao = 298 K
•
(2) properties of reactants • Heat of reaction, I1HRx , dHrx = -525676+286098+ 154911.6=-84666.4 J/mol • Activation energy, E = 75362 J/mol • Pre-exponential factor, A = 16.96x1012 13600 lis • Specific reaction rate ko = 1.28/3600 lis @300K • Reaction rate k = ko exp[ E
R
• • • • • •
m 3/s
(-.~ - ~)] or k = A exp[- ~] ~ T RT
Gas constant, R = 8.314 J/mol·K Rate law - r A = kC A Thelmal conductivity of the reaction mixture, ke = 0..559 W/mK Average density of the reaction mixture, p, rho = WOo. kg/m3 Heat capacity of the reaction mixture, Cp = 4180 J/kg"K Diffusivity of all species, Diff = 10-9 m%
(3) properties of coolant 2 • Overall heat transfer coefficient, Uk = 1300 J/m "s"K • Heat capacity of the coolant, Cp] = 4180 J/kg-K 2. Size of the Tubular Reactor • Reactor radius, Ra = 0.1 m • Reactor length, L = 1.0 m
2. Femlab screen shots (1) Domain
(2) Constants and scalar expressions - Constants
8-11
~
tw·· . ··-·-··-·-··-···--····--··· f--··----····-·-----···--
-+----.---{"
.314 12
- - - - - - - - - - -..-------------.-
~---.-+------------~.Q................ -.-- . . . . . - ...... .
fBD __ ._ . __. _. __ . +_____._____._.__....__._..._......__..._._.....__...__.__.. _.._{----_._..._.... _....__...•._-_._...._----_........ -...__.._. ----·F=-=-··---------·------·-----+-=
r=-=---_.-.--.-.-.. -f'.---.--..-----.----------.-------..--.---fc:-::..----.--..-
- Scalar expressions
(3) Subdomain Settings - Physics (mass balance) Equation V.(-DVcA+cAu) =R, cA =concentration
Subdomain selection
Library material: .[ - - ; ..•.-
.~ .C~:.~j , ,='
~"'-,-,-
T
Valueil'xpressionDestriptiOI1 . [ ... ····---~~~:l~ime..~~~fing coefficient
0'5
G) D isotropic
~!!"
.•'..
-. IDiffusion coeffic~nt ~ ~;-i -I'
'
o D anisotropic._ .. --:::l Pilfus!9n (;oefficient R u
10=~===-:JJ'{ilaCiion rat!'
8-12
(Energy balance) V*C-kVT + ~hiND,j) =Q - PCpu.VT, T=temperfrture
i PhYSics i Init ; Element ' Thermal properties and heat sourcesfsinks
~~
Library material Quantity
ok
1\5
Valueibpression
Description
tC~~~=~-==~.~-~~--~]
Time-scaling coefficient
Thermal conductivity
(isotropic)
(':' k (anisotropic)
Thermal conductivity
Heat capacity
C
p
Q
hiND.1 l~p'~~~~~~i!~~~~?~ !~_~~~~ ! [~iCial~fUSio~~:_J
(Cooling Jacket) Equation
v.r = F c Subdomain
selection
i Coefficients
In~
Element Weak
PDE coefficients
COefficient VilluelExpression
o
Qescription
..l§~..~'=] Flux vector
F;~g~:iT~!~j1~e:;~;;;~~l
Source term
da
Mass coefficient
[ ] Select by group
~ Active in this domain
(Source Term) F = Taz-2*pi*Ra*Uk*(T-Ta)/(CpJ*m1) - Initial values (Mass balance) cA(tO) = cAO (Energy Balance) T(tO) = TO (Cooling Jacket) Ta(tO) = TaO - Boundary Conditions @ r = 0, Axial symmetry @ inlet, cA = cAO (for mass balance) T = TO (for energy balance) @ outlet, Convective flux @ wall, Insulation/Symmetry (for mass balance) qO = -Uk*(T-TaO) (for energy balance)
8-13
(4) Results (Concentration, cA)
.." ?,&' 1~3·aaa
1782.os5
660222
O.6~
0.6
0,4
0.2
'-0.2
- -..- - - - . - - . - - - - - - - - - - -.. ..Ql5
..01
-0.05
-------~-=-
0.05
0.6
0.4
""
0.2
---
-0.2
·015
~====--,---.----
..0.1"
-0.05
0.05
8-14
01
0,,15
0.2
..--".--..025"
Second order reaction [1] Domain
o Axis equal
[2] Constants & Scalar expressions (1) Constants
, ~l'l.t'rlE;! .•..
• !;;::;Pf~ssiQn
uz
2·uO·C1-CrlRa).A2) .t-.... _.-._-...." ..--_ ..._ ...
FkO;;""",.._~r°1e~~~~~2)
0--
txA
cB
'cC
1-- ~." V"·,.,_,
rf!:
J
...__ .
(cAO-cA)/cAO
"-"-TBO~~~.Q~L~-·-·······
_".,_"._'"+2'cAO'xA ___ ..
.~._~" ~._o.
v""''',.'''" __ •• _._ .. ~ .. ,,~'"_, .. " .•,, __
·_,,",v~·_,_·
__
..rf!:'~~!1~~'!r0.9.~~·(c~_"c_B-E~~~9J...... . 1303··1
-~
. . . . . -.. . .-.. .-.. . . .-.-.. . . . .
8-15
[3] Subdomain Settings (1) Physics
(Mass balance) Equation-' --,,V'(-D'lcA+cAu) =R, cA =concentration
i,Subdomain selection--
Ifilii:! Iii ffl 1111 imll1lllll ,±; [1
library material:
i 1
Quantity
II
lh
II ~
!
ii I1
o D isotropic Diffusion cQefficienl o D anisotropic ii,ij,Q,:'.".=, _::.c._.J'DiffuSlon coefficient
II !
ValuelExpressiofi Description;
[ ____ ..~~-~=] Time-s"aling coefficieoi
I
IL
R
~=--~-=-=='=:l Reacti':'Q;/'ate
v
~~, __7_-Jz~vel~qty
lo--·-·-------J,r,veloc~y
u
EJ Select by group
~rlifiCial Di!fu,~ion_,
G:'.I Active illIbis domain
]
(Energy balance) I Equalion
,
I v.(:kVT ",:" Ilhi~D,I) = Q., pCp.....VT. T=t~mperatL!re I Subdomain selection ~'. r'Physj~ in~ !Element 1 Tl:lermal properties and heat sourcesfsinkS
~ibr8ry material;
".~'.'..',~~J L.LOad ....
J
Value&press;OO De.cription
Quantity
h' ---- mm:"':::::':':.i.5;t:';-T Tjme-SC~ling coefficient
0kQsolroplC)
Therrr:taLcon~ivily
Ok (.niSoiropic)
Tberma.l.concWclivilY:-
~:c:;::::::c::::::::::::::::;
Dens~¥
~::,::,==or;===::::,~ Heal Cop'CKY
Cp Q
'''''''';;'''''''=c,;''''':::!
Healsoi1rce r..velo~iiy z-vel
o Select by group
~~!!~~.~~
hiND ,I
[tJ Acti\ie in t~is '~6main
[ ~rtJ!ici~i~t~s~;]
(Cooling Jacket) ;Equation
lv·r =
F
L EI~ment I W~~k:j PDE coefficients-
II
Coefficient
I
r
ValueiEHpre~siQfJ
Des"ription
lQ_~_=-fo'cc,; _~-J Flux~ecto;
rl~Uk;(T~T~)~ceE~J~, ~o(.lr~e term ~-- -_,=:'~':~~:~~ :5' ~,ass coef!iCi~nt
F
_
I da E]Selecl by group ~ Aclivein this dd!'lj~i!'l
(Source Term) F = Taz-2*pi*Ra*Uk*(T-Ta)/(CpJ*m1) 8-16
(2) Initial values (Mass balance) cA(tO) = cAO (Energy Balance) T(tO) = TO (Cooling Jacket) Ta(tO) = TaO (3) Boundary Values @ I' = 0, Axial symmetry @ inlet, cA = cAO (for mass balance) T = TO (for energy balance) @ outlet, Convective flux @ wall, Insulation/Symmetry (for mass balance) qO = -Uk*(T-TaO) (for energy balance) [4] Results
:~
_. 2
·.Q15
-01
··0 os
0.05
--0-1- - 0 - 1 5 ---0-'-2-·~-0.25-- 03 Min: 81735
xA is undefined at (--0 .. 1,0,,6)
8·-17
0.6
0:4
02
.0,05
·0.15
,.......
0.05
015
0.1
. 02
0.25 .
,;d''i~ ~d~~i~~d ~t t-9,,~,O.6j
P8-2 (0) Individualized solution
P8-3 Solution is in the decoding algorithm available in the beginning of this manual. P8-4 Find the reactor temperature at steady state (prior to shutdown) Let M = mass of the NH4N0 3 in reactor. F AO =lbs/hr of NH4N0 3 fed to the reactor. Mass balance: FAD - FAD (1- X) = -rAV = kM X
= k(T)M FAD
Energy Balance: -FAD LBJHw - Hi (T)] + FADXMl Rxn(T) + FAD (1- X)MlVA
=0
Where Hj(T) is the enthalpy of i at the temperature of the reaction, Ml VA is the heat of vaporization of A, and Ml Rxn (T) is the heat of reaction at the outlet temperature.
8-18
NH 4NO(l) ~ 2H2 0(g) + N 2 0(g) A~2B+C
The last term FAO (1- X)t:Jl VA accounts for the unreacted N&N03, which exits as vapor rather than the liquid. Now, we can make some substitutions BB
= .17 ,BA =l,Bc =0
.83 Heat capacity of A is given, and the enthalpy change for water 200 0 F (I) ~ 500 0 F (g) is also given. So, after dividing by F AO, we obtain CpA (T -200)+ BB[H B(500° F) + C P1 (T - 500)]+ t:Jl RxnX + (1- X)t:Jl VA = 0
The previous equation assumes that the heat capacities are constant over a reasonable temperature range. the phase change NH4NO(aq) ----7 NH 4NO(1) is isenthalpic.
In addition, we must account for the effect of the temperature dependence of Ml Rxn . 44 Ml Rxn(T) = Ml Rxn(TR) + (T-- TR)[80 C P1 -CpA] = Ml Rxn(Tr) + (T -Tr)~Cp Let Ml Rxn = HB (500) -- HB (200), we have CpA(T -200)+OB(MlW +Cps(T -500)+ X [MlRxIl +i1Cp(T -500)]+(1- X)Ml VA =0 or,T[CpA -t0BCp> ]+[-CpA (200) +OB(Mlw --500Cp,)] + X[Ml Rxll + i1Cp(T -500)]+ (1-- X)MlVA
Numerical Substitution with K(560) =:; 5.03 and k(51O) = .53 In(k2)
kj -::::;.E
R
=-
(i_ -~J
E R T2
~
=-In(k2)(~_~J-j =44499 kj
T2~
E Ao = kj exp(-) = 4.51424xlO19 R~
Btu CpA =.38--;Cp1 =.466 Ih.R (P = 1 atm over 450 - 500Dp Himmelblau) Btu C pc = .2521_ (Himmelblau, App E, over 230-265°C) Ih.R
8-19
=0
8 =.:!2. = .205 .83
B
Ml = 1034 Btu .03 w lb F
= 257.3 lb NH 4 N03 hr
AO
ACp = 44.02 (.2521)+ 36.03 (.466)-.38=-.0316 Btu 80.05 80.05 Ib.R MlRxn = C pA (41O-T)
Substituting all these into the mass and energy balances:
)M
4 ..51x1019 ex ( -44499 P T+459.67 Mass balance: X = ---------''----........:..257.3 Energy balance: 0= .48T +88.21 + X(-320.20-.03T) + (1- X)(1.55.80"-.38T) Assume X = .96 and M=500. Then, from mass balance, T = 5100P
P8-5 A+B~2C
_.-
A
B
C
F,. ( lb -h~Ole )
10
10
0.0
Tio(F)
80
80
C~Cb::e"F )
51
44
MW'(~--) lbmol
128
p;,(; )
63
-,
'
47 ..5
94
222
67.2
65
, - - - + - ------- --
-----------'--
R
-----
-- f - - - - - -
-"""-----,--_t__
Ml = 20 000
--
-
Btu lb mol A'
Energy balance with work term included is:
8-20
F 10 BA =1' B =~=-=1 X B F 10 ' AF =1 AO
Q=UA(Ts-T) Substituting into energy balance, UA(Ts - T) - Ws - FAOM-J RX AF = FAO [ CpA + CpB ] [T - To]
~UA(Ts -T)--Ws -FAOM-JR ={FAO [ CpA +CpB]+UA}[T-Yo] T
= T, + UA(Ts -T)- Ws o
-W
FAOM-J R FAO [ CpA +CpB]+UA
=63525 Btu hr
S
:. T
= 199°F
P8-6 A+B~C
Since the feed is equimolar, C AO = CBO =.1 mols/dm 3 CA = CAOO-X)
C B =CBo (I-X) Adiabatic:
T =To + {[-_M!R(To)l rACp; + X8Cp 8Cp =CpC-CpB-CpA =30-15--15=0 All R(T) = Hc - H B - H A = --41000-(-15000-(-20000) = -6000 cal/mol A "L./i;C-
i
cal = CpA +BBCpB =15+15=30--
mol K
T = 300 + ~~~~ X = 300 + 200 X
30 -rA = k C~o(l- X)2 = .0Ik(1- X)2
P8-6 (a)
8-21
For the PFR, F AO = CAovo = (.1)(2) =.2 mols/dm
3
See Polymath program P8-6-a.pol. Calculated values of DEQ variables.., ..·-·--··t "-""""." "--''''---'1 .... ! IVariable nitial value Minimal value Maximal value i Final value
,---r-'---"
~---~~~----
--.~,-,--."""
I I -·-··--r·-·.·-.------·-·-----..·..
""-J:;'--'-""'--'
I
··T", •• ·-.., ·.. ··_ _·_-·····'·T--- .. -----.-. --,
illX . l
0
i V
0
.,', .•
10
10.85
_"."_~.,,._._'"
...1--".,.. , ,-. .. ,,,.,,.,, '-""-'
0
0.85
'" ". .......___.. _~ __ ...
1308.2917
1
_.JI
08.2917
1
'E-~~~f~=~]6~~===~-JF===:J~t=-=]I
151T 1300 . _.~v~~.~'~",¥m~"'w""'~"___ 16ik 10.01 f-""'T"'"' ,. ,-.~,~ "~.,, ""~·'I~~~'"-'~'~''''' j71ra 1-0.0001 •. .,,, ..•. ,, ..• _ " rw'-""~~
300 .
1
h
""'~.w '~~~N~'~'O 0
___
~~~~~~C
1470.1470.. ~" _,~'w,_=~ ~,_"," ww,~_,, 14.150375 14.150375
""''01" , ,.
10.01 w·,t·" '~'W~ 'W~ ""."~,-" "'~l ~ 1-0.0018941 1-0.0001 .. . ... ___ " ' ¥ ' ¥ ¥ _ N __
"'''"
"'W
',f'
"
,=,
YW
___
,~
.~~y,.,_~w~,~· ~ .•.." -----, ,~"W' ~., '~--"--'-' ••
1-0.0009338 , . . . '" .......
I ;
I 1
Differential equations :g,: d(V)/d(X) = -FaO / ra Explicit equations Cao = .1
FaO
= .2
T = 300 + 200 * X k = .01 *exp((10000 / 2) * (1 / 300 - 1 / T» $ ra = -k * (CaO /\ 2) * ((1 - X) /\ 2)
v = 308.2917dm3 For the CSTR, X = .85, T = 300+(200)(85) = 470 K. k =4.31 (Using T =470K in the formula). -rA = .000971 mol/dm3/s V=_l!AO X = .lx2x.85 =175 dm3
-rA 9.71xlO-4 The reason for this difference is that the temperature and hence the rate of reaction remains constant throughout the entire CSTR (equal to the outlet conditions), while for a PFR, the rate increases gradually with temperature from the inlet to the outlet, so the rate of increases with length.
P8-6 (b) X[-Mi ] T = T + - · -Ro
~)1;Cp;
For boiling temp of 550 k, 5.50 =To + 200 To = 3.50K 8-22
P8-6 (c) 0.90
500
0.72
460
GJ
0.54· 0,.36·
380
0.18
340
O.(lU
62
(I
[!]
:120
123 V 185
246
3U8
300
0
P8-6 (d) FAoX -rA
VCSTR = - -
=> X
= VCS7R (-rA ) FAO
=
3
For V 500 dm , F Ao=.2 -rA = k C~o(1- X)2 = .Olk(l- X)2 T =300+ 200 X Now use Polymath to solve the non-·linear equations. See Polymath program P8·6. d . ·1. pol. Calculated values of NlE variables I' lvariablejvalue if(x) ;Initial Guess'
liT
1484.4136 :0
21X
!0.9220681j-2.041E-09i o.9
1480.
iVariable t!Value >
lr~
2 :ra
n
16.072856
I,(). 0003688
Nonlinear equations 1;ci f(T) = 300 + 200 * X - T
f(X)
= 500 - .2 * X /
=0 ra = 0
Explicit equations
1 k = .01 * exp(10000 /
1.98
* (1/300 - 1/ T)) 8-23
62
123 V 185
246
308
= 0.01 * k * (1 - X)
ra
Hence, X = .922 and T
A
2
=484.41 K
For the conversion in two CSTR's of 250 dm3 each, 3 For the first CSTR, using the earlier program and V =250 dm ,
I rV~"~iable Ivalue 11 ]k""
15.105278
]~F~
I6.0007~59
Nonlinear equations f(T) = 300 + 200 * X - T i~ f(X)
= 250 "" .2 * X /
ra
=0
=0
Explicit equations k = .01 * exp(lOOOO / 1.98 ra
= 0.01 * k * (1 - X)
A
* (1/300 - 1 / T))
2
T = 476.48 ad X = .8824 Hence, in the second reactor, V = FAO(X - Xl). CSTR
X
~
-rA
V
=_CSTR
F AO
(-r )+ X A
I
= Tou/,CSTRl + 200 (X _. Xl)
T
See Polymath program P8-6"d·2 . pol. Calculated values ,.-of"'rNlE variables : IVariable;Value if(x) IInitial Guess ;1 1493.8738 fO" ... 1480. .. ..... f"-~Y'"
~~.-~".-,
"~>-"-'f
~,-.~~-.~~.
~
."",~,~"
~~'t'·~~~-
IT "
121~
JO.9693688 t-1.359E-09 j0.8824
; lVariablelval~e " ~.
t,
'~"N '~ __ ·.'''''N·.·_"" T---"
jllk
r2r~a ~
."
,7.415252 '16.958E-05
.(,
8-24
31X1
;0.8824
Nonlinear equations J~ f(T) = 476.48 + 200 * (X - Xl) - T
i
f(X) = 250 - .2
* (X - Xl) /
Explicit equations ij k = .01 * exp(lOOOO / 1.98
= 0.01 * k * (1 - X) Xl = .8824
::~; ra
A
=0
ra = 0
* (1 / 300 .. 1/ T))
2
Hence, final X = .9694
P8-6 (e) Individualized solution P8-6 (0 Individualized solution
P8-7 (a) For reversible reaction, the rate law becomes
-orA =k(CA
C- KCcJ B
c
x =.~ l+Ke
e
T =300+200X k = k(300)ex p
(ER (_l300___ .!.)J T
(-.l___ .!.)]
K = K (450)exp [Ml Rxn C C R 450
T
Stoichiometry: Cc = CAOX C A = CAo(l- X)
CB
= CAo(l- X)
See Polymath program P87-a . pol. Calculated values of DEQ variables "'lv~;i~bi'~lInitial val~~TMini;;':~1 value Maximal value Final valuel ,
l
i
'"
,.,
",...
".
i
" , .
'l1V
iOO
110.
10.
l?Ix.
\0
10.0050806
0.0050806
10
8-25
-;
!
Differential equations d(X)/ dey) = ora / FaO Explicit equations T = 300 + 200 * X
= .2 i~i Kc = 10 * exp(6000 / 2 * (1 / 450 - 1 / T)) k = .01 * exp((10000 / 2) * (1/ 300 - 1/ T)) FaO
CaO =.1
i
= -k * (cao 2) * ((1 - X) Xe = Kc / (1 + Kc) ra
A
A
2 .. X / Kc)
302.0
0.30 , - - - - - - - - - - - - - - ,
30L6
0.24
--..-,----
...,.--......., . -,."., ...~"-... ,......
301.2
0. 18
300. 8
0..12
300.4
0.06
300.0
0
2
4
V 6
8
10
{tOO
IT] - x .,. ~e
(I
2
...
V
P8-7 (b) When heat exchanger is added, the energy balance can be written as dT _ Ua(Ta ·-T) + (-rA ) [-MIRxn (T)]
dV -
FAO
(I BiCpi +~Cp)
So with ~Cp= 0, I~CPi =30, MIRxn= -6000 cal/mol dT
Ua(Ta -T) + (-rA ) [6000]
=-~---~~-~
dV FAO (30) Where Va = 20 cal/m3/s/K, T a = 450 K 8-26
6
8
10
See Polymath program P8-7-b.pol. Calculated values of DEQ variables ""W'"r IVariable !Initial value l Minimal value: Maximal value I Final value
~"~~""r-~AA~~_vn_'~_''''~~"l
~~C~~l~''''''''''-~-~vw-''~'
,-~.-4-~~"'''''.'~'~~''''''-''''~!''
"~,,
"'f'
1 Iv:O :2
lxJo·
10
... fo
3Jr
i450.
i4 ICaO 's·IFao
!0.1 iO.2
10.1 /0.2
'110.
Tio. i ..
16TKc .
I?~Ik
..1
,,-~
... ·1~50.
12.586706. i~·58~?06 f-0.0258671 ;-0.027946
19 lxe
"]0.909090910.9090909
j10. . 1450. ~
1
'463.99741452.3087
.1~:1
10.1 ····]0.2
10.2 ···-T1.2.22765
.-'
450.
1
flO.
10.9244008
TO.9118643 . lio .
[450. i 10 ... 1
110 .. ,i
Differential equations J, d(X)/d(V) = -ra / FaO
d(T)/d(V)
!10.34614
13.616753i2.737647 i-0.0035089 T-0.0035089
!1 0 . !10. 1
10 .
,. +~"."
".~!" ~-~~,
;10. 110. ···_····:0.5717112··10.5717112
18 ira 10 fUfo.
,"~'~
= (UA * (Ta - T) + (--ra) * 6000) /30/ FaO
Explicit equations it CaO = .1
= .2 Kc = 10 * exp(6000 /2 * (1 / 450 - 1/ T)) k = .01 * exp((10000 / 2) * (1/ 300 - 1/ T)) ~l ra = -k * (CaO /\ 2) * ((1 - X) /\ 2 - X / Kc) Xe = Kc / (1 + Kc) UA = 10 FaO
I~JTa
= 450 ;al A = 10
8-27
470
1.0
466
0.8
462
0. 6
EiJ
458
0.4
454
0.2
450
2
0
4
V
6
2
10
8
4
V
6
8
P8-7 (C) For a co-current heat exchanger,
Q=mcCpc(Tal-T)[l-.exp( . . UA mcCpc CpC = lcal/glK, Ta1 =450 K,
J]
m=50 L
sec
See Polymath program P87·c.pol. Calculated values of DEQ variables r~.~!variab'~li~~itia' v~i~~] Minimal value rMa~imal ~~lu~l~il1~i val~~]
11 12 13 :r"O"
. . 10 ·····10
IV
Ix IT
1450.
1"
T"
14 ICaO 1'15 I FaO
10.1 o
]0
[0.575571
]450.
,,,,
!0.2
11.
·1~·586706
1'.,•: .8•.. ·.•.. k. · ra .9 '1-0.025867~
110jXe 11.1 jUA
112lT~ [131A
10.9090909
1141mc "150.
[lslq
I~
!0.1 '1' 10 .2
"ce,,,
10.1
!
;0.2 .\12.46524
11.
11.
[2.586706. 1-0.0281568 10.9090909
13.734633 1-0.0034309 16.9257347
Tlo. 1450./450. 110. lio. 150.
•... ~O.
.0·····
!0.1 f 10 .2
.<..
hO.37812
150.
. ···..
I I
·1
1
/1.1 I2.751.765 J
]-.0.0034309 J !0.912112 ····[10. 1 1 1450.
I
:10.
.. 150. .
1-139.420110
I,·
J.
I
J-22.833091
Differential equations
= -ra j FaO 12; d(T)jd(V) = (Q + (-ra) * 6000) j
I j
..... 10.575571
/465.3828' 1452.51.92" .]
lio·rio. 1450. 110. ....
110 .
..... ·········r
··.Jl0.JlO.
17 kpc rll.,· . .
110 .
~
l' . ".,. r
1611
jo
d(X)jd(V)
30 j FaO 8-28
Explicit equations
= .1 FaO = .2 3. Kc = 10 * exp(6000 /2 * (1/450 - 1/ T)) ,fl· Cpc = 1 k = .01 * exp((10000 / 2) * (1/ 300 - 1/ T)) 6" ra = -k * (CaO /\ 2) * ((1 - X) /\ 2 - X / Kc) Xe = Kc / (1 + Kc) 8. UA = 10 1, CaO
9. Ta
= 450
= 10 .1l mc = 50 12 Q = mc * Cpc * (Ta - T) * (1 - exp(-UA / 10 A
mc / Cpc))
470
1.0
466
tl8
462
0. 6
458
0.4
454
lL2
450
0
2
"
V
8
6
-------------,
~ l~
2
10
4
V
6
8
Next increase the coolant flow rate and run the same program to compare results.
P8-7 (d) For counter-current flow,
Q=
meC,e (I;, -T)[l +exp ( - m~~)]
See Polymath program P8,,:7 ·d.po!. Calculated values of DEQ variables !Variable'Initial value:Minimal valueiMaximal value ' Final value ,
r~-l"--'"
1 iV ! 2
3
Ix
::r
<
'r~<'
~~.
~."
'~/'i
.'0'"
10
:450.
'
;~
;
, '~l
10 {
,~~"
~
[10.
io
1"450.
1
ilO.
'0.5395082 1451.6471
.. ,
8-29
10.5395082
... ~
.450.2745
10
f4'" r~o'''''-lo~i is !FaO 10.2
" .......·'1-· ..·'.. · .. , ........ " Y l
--'''''''
, .. '
.. ,
'"''''''''''''''''''''~'''
, , - , .. ,
.""TO:1-" . . . """.:_:~j,:~~"(,:.,., . ._,~~. ~~:'" ~.'",I~~~~. ,~.~ . 10.2..........." ......" . "110.2...." ......... .. ,t10.2, . -, ". . ."."""('., ; 10. 110.24606 110.04072 , .............., , ."...," "', 'T'......· . . . , ... , . , ....... ""'''1'''''''' . .,,--,. . ............. "
,. . . . . . .
.
"-, ''''-
I....Kc,. ._. . . . . . . .".-1110. ,....6,. . + .......,'-, . , 17 ICpc 11. 11. 11. :1. 18'Tk-''-'''T2:s86706' "T2:586'706"'''''''T2~6936S7''-'''''' 1"2.604286 19' "F~'" . " ···'T~0.02"58671 r~0':0260956 ,." "T~o~"o04123i""" r~o:o04i231 lioTXe ""'10.9090909 10.9090909"'10:91108" 10.9094262 (' A 'Eo.'" "liD."" ···rio.'Tio. 1"12 !Ta ....t450.145o. 1450. l4S0.
ii3IA"Tio.
,..,',. Fo: .
;10.
flO.
1i:!~~ro-o, --- Ji~.75n-~f~o,.
W~.96003
Differential equations d(X)/d(V) = -ra / FaO d(T)/d(V)
= (Q + (-ra) * 6000) / 30/ FaO
Explicit equations CaO = .1
;2,'
= .2 Kc = 10 * exp( 6000 / 2* (1 / 450 - 1 / T» Cpc = 1 is,~;; k = .01 * exp((lOOOO / 2) * (1 / 300 -1/ T» ra = -k * (CaO /\ 2) * ((1 - X) /\ 2 - X / Kc) Xe = Kc / (1 + Kc) FaO
UA
= 10
Ta
= 450
10 50
l~1I A = j,J, mc = Q
= mc * Cpc * (Ta ,. T) * (1 + exp(-UA /
mc / Cpc»
8..30
452.0,.---------------,
10
0.8
~
4512
0.6
450.8
O·.j
450.4
02
2
4
V 6
00
10
8
U;J
0
2
.:j
V 6
8
P8-7 (e) We see that it is better to use a counter-current coolant flow as in this case we achieve the maximum equilibrium conversion using a lesser volume of the PFR.
(0
P8-7
If the reaction is irreversible but endothermic, we have -rA =k C~o(1- X)2 =.Olk(1- X)2 as obtained in the earlier problem.
MI Rxn = 6000cal / mol For co·-current flow of coolant,
Q~mcC,c (7;, -T)[l-exp ( - m~~J See Polymath program P8-7·f·co.pol. we use 8- 7f cocurrent.pol Calculated values of DEQ variables
i"'fVariablehniti~1 v~i~'~l, Mini~al value I, Maximal value 1I Final value , ,
, r i '., 1 IV 10 ~--;
i2 IX I
13 ,IT !4 CaO 15 FaO 16 Cpc 7 i!k 8 IIra
,
T
:l~lmc
.,
f
,.
:
.
10 ,
1 10.10. ; fO.5227896
;0.5227896
;0
10
"l
1
1450.
1439.9908
10.1
10.1
10.1
;0.1
10.2
10.2
"io. 2
11.
11.
·2.586706
:2.352492
1-0.0053573 !110
1-0.0053573
llO.
1450.
:450.
SO.
50.
i
:0.2
1.
11. .,
12.586706
[2.008972
1-0.0258671
1-0.0258671
:446.1887
1
'
~~
110. ,
9 !UA
10ha
Y
'
1450. , 150.
A50. '50.
I
8-31
c~····
T'·· ·······,····,······'··T·'· ... 10
i 12 iQ 1 ' ....,!., ..
~.,
.........,.. ',
~~, .. ,
~.
.'.......---..' ...
,. ,).".¥~ .." .. ~ ...." " ..
'···~r
.
.. 'c-'-
~
¥ ' •• ' ¥ ••••••
··-···-··-·'····-····~·--'~··'-T····'
...... ........ - .. ' .. --......, ~
J~O'~.?~779J~4.?4357
10
'¥.~.".¥ . .,-. •• ~~,: ..•. ~~.
Differential equations d(X)jd(V) = -ra j FaO d(T)jd(V)
= (Q + (-ra) * (-6000)) j
30 j FaO
Explicit equations Cao = .1 FaO =.2
~a~ Cpc = 1 .41 k = .01 * exp«10000 j 2)
= -k * (CaO UA = 10
ra
~Ta , ,<
A
2)
* (1 j
* «1 - X)
A
300 - 1 j T)) 2)
= 450
¥~: me
= 50 fii Q = me * Cpc * (Ta - T) * (1 .. exp(-UA j 450
me j Cpc))
-.------
0.60
OAS 0.36 ·US
0.2-4
[J]
434
0. 12
430 01...--~2---4--\-!-6~-·-··8
~---'--~--~--~--~
10
2
4
V 6
For counter-current flow,
Q=rhcCpc(Tal--T)[l+exp(-
.UA
mcCpC
J]
See Polymath program P8--7 -·f-counteLp0l. POLYMATH program 8-7f countercurrent.pol Calculated values of-. DEQ variables -, '-.. ., "'1 .... ....... IVariablelInitial value!Minimal value IMaximal valuelFinal value ---r'~~ ~~.~
r'
1 IV
~,~~.
... r
'~"~--"-
10 '·10 }150.
2lx IT' . . .
3
'1V~""
~~.
,-~~ ~.
··r· 10
fa
f448.4634
.~
.
r'
:10. 10.5594826 ..... 1450.
8-32
..
~
~«'~,
......
r~
,
0
~:.
...
110. fo:s594826 l449.669
8
10
lcao
iO.1
iO.1
'5~1~~~
:0.2
0.2
:6 ICpc
1.
1.
'4
'ill<,
iO.1
11.
i
!2.586706
12.586706
2.4901
8T~~
1-0.0258671
-0.0258671
[ -0.0049788
9juA
110. 1450. J
10.
110.
450.
A50.
i50.
:50.
50.
10
10
139.7022
101Ta o
"
lli mc 12jQ ,
"
" [30.09601
Differential equations ~.•
d(X)/d(V)
~,d(T)/d(V)
= -ra /
FaO
= (Q + (-ra) * (-6000)) /30/ FaO
Explicit equations cao = .1
= .2 Cpc = 1 k = .01 * exp((lOOOO / 2) * (1/ 300 _. 1/ T)) :5 ra = -k * (CaO /\ 2) * ((1 - X) /\ 2) FaO
q. UA =
10
= 450
") Ta
8 me = 50 ~. Q
= me * Cpc * (Ta - T) * (1 + exp(-UA /
450.0.--------
0.60 ..-'~~~"
449.6
0,48
0.36
449.2
Q
1),.24 0,12
448,4 448.0
me / Cpc))
-.-~----~ --~
o
2
4
Y
6
8
10
000
()
P8-7 (g) For a runaway reaction, the following must be true: RT2
T, -TC >-'E
8-33
2
4
V
6
8
10
and T =
Yo + KTa
= 300+3*450 = 412.5
I+K' 1+3 So if we plug this value into the original equation we get: c
1.987 T2 -T +450> 0 10000 Tr= 499 K T
1
P8-8 (a) A+B~C
Species Balance: I
dX
rA
dW
FAO
--=--
= 20dm 3 / s Po = lOatm Vo
Stoichiometry: CA =CAO ( I-X )I--,where £=1 1+£X To
~ C, =
CAO ( : :
~-):o
Rate Law is:
(-~- !)] T
--r = kC with k = 0.133 exp [ E A A' R 450 E=31400
!VIRxn = -20,000 J / mol Energy Balance: T
X[--·M!R(To)]_ o + '" ~B;Cp; + XI1Cp
_T -.I
=15 + 24--40 =0 T = 450+ 20,0~0 X =450+ 500X I1Cp
40
See Polymath program P8-8·-a.pol. Calculated values of DEQ variables
r~lv~~i~bi~I~~iti~i;~i~~Tfv,i~i;;i:~~ lue fMaxim~lvalue ['~in~1 V~.llI~J i1 1X
12.j~
10
. . I~.....
10
10
10.8
10.8
143.13711'"j43.1371.1 8--34
I
1
r-~
'3:T
1450.
'450.
,:20.
:20.
~ ..
~~
'~t~~
- '.~
1450 .
,5iTO
'850. ..
20.
,.. 1
450.
1450.
'0.133
'6.904332
A'·l
'0.133
6ik
~
Differential equations ~ d(W)/d(X) = vO * (1 + X)
*T /
k / (1 - X) / TO
Explicit equations
:lin = 450 + 500 * X
= 20 3TO = 450 :4 k = .133 * exp(31400 / 8.314 * (1/ TO - 1/ T)) 2. vO
0.8 0.6 0. 5 0. 3 0. 2 0.0
9
0
18
W 27
36
P8-8 (b) Species Balance for CSTR: FAOX WeHR
=--,'-rA
T
= 450 + 500 X = 450 + 500(.8) = 850 K
k
= .133exp [}1400(_1__._1_)] = 6.9 8.314 450
WCSTR
850
= 39.42 kg
P8-8 (C) Individualized solution P8-8 (d) For pressure drop, an extra equation is added
8-35
dP dW
CA
=_ a(~J-~(I+8X) 2 1'0 (PlPo)
= CAO
(I.-X) + I
T P X To Po
See Polymath program P8-8-··d.pol. Using POLYMATH program CRE_8_8d.pol For a=.019 Calculated values of DEQ variables
i var;bi~tn;itii;l;';I~~ rMi~imal,;;'I~~lMaximal valuejFinal val~~1 ?'W'l'~"
10
J~.8t~·8J
.
,
G*I~.o13E+06 ji.002E+06 ·r850.····· K~~::~~ .. j~::::~~ r41i··1450. .. 1450. 1850. ...... 151vo
"'120:120:"
161To
1450. '1450. ',·"10.133'10.133
l!Jk,
,8IpOI1.013E+06 r
T .
,9lalpha
10.019
'fio:·120.·'1
1
450 . J 1450. 16.904332f6.904332
11.013E+06h.013E+06 r 10.019 10.019
Differential equations l d(X)jd(W) = k j vO * (1- X) j (1 + X)
d(P)jd(W)
= -·alpha j
* TO j T * P j PO 2 * (T j TO) * PO A 2 j P * (1 + X)
Explicit equations T = 450 + 500 * W
vO ~:TO
= 20
= 450
iJ k =
.133 * exp(31400 / 8.314 * (1 / TO - 1/ T)) PO = 1013250 alpha = .019
8-36
I
11.013E+06 . ···T··· i 10.019 .J
0 . 060
900
0.0-18
800
Q
0.036 I}, 02-1
600
0 . 012
500
0.000
000
016
032 W 0.48
G
700
400
0.80
0.6-1
0.00
0.16
0.32 W 0....8
0.64
0.80
P8-9 (a) We use the same equations as problem P8-8, except that the energy balance changes as:
dT
I -Ua ( Ta -T ) +(-rA)(-Mi Rxn ) =p ---------
dW
FAOC pA
Where -Mi Rxn = 20,000 J/mol, T a=323 K, C pA =40 l/mollK See Polymath program P8"·9"-a.pol. Calculated values of DEQ variables
lV~riablellniti~1 ~~I~e~l Minimal~alue Maximal value 1Final value i
Iw ,
;1
3
,
l
'
1010
.. ".
:50.
150."
'2jX
10
10
iO.1376181
;0.1376181
'3 IT
i450.
138 1.1888
1450.
)381.1888
1450.
i45O .
1450.
i450.
14
ITO
15 ivO 16
r
i20.
jk
iO.133
'{
1:
120 . [0.0292331
i7 IUarho
10.08
0.08
is" ITa
1293.
1293.
,
i
19
ipo
1101cAO
1.013E+06 1270.8283 f""
1111CA 1121rA
I
20.
!20. , ,/0.0292331
:0.133
r l
08 1°. i
10.08 " 1293.
j293.
r
11.013E+06
i1.013E+06
!1.013E+06
1270.8283
270.8283
12.42.3648 t
270.8283
!270.8283 i 1242 .3648
1·-36.02017
'-7.085084
..
1270.8283 1 " 1-36.02017
I
-7.085084
Differential equations d(X)/d(W) = k * (1 - X) / (1 + X)
* TO / T / vO d(T)/d(W) = (Uarho * (Ta - T) + rA * 20000) / vO /
CAO / 40
8-37
Explicit equations TO = 450
= 20 ljl k = 0.133 * exp(31400 /8.314 * (1/ TO - 1/ T)) vO
~J Uarho
= 0.08
= 293 PO = 1013250 CAO = PO / 8.314 / TO 1m CA = CAO * (1 - X) / (1 + X) * TO / T rA = -k * CA Ta
450.----- - - - - - - - - - - - ,
0.20
436
0.16
422
0.12
40S
O.OS
394
0.1)4·
10
20
\V 30
40
50
0.00
0
10
20
W 30
40
50
If !!.-!2. was increased by a factor of 3000, we use the same program with the new value. The profiles
Ph are in the graphs below.
----
450
0.20 0.16
434
EiJ
US 402
0. 08
386
0,,04
370
Q
0.12
_.
0
10
20
W 30
40
50
10
P8-9 (b) 8-38
20
W 30
40
50
For non-constant jacket temperature, the equation for incorporating the flow needs to be introduced. co-current: Tao = 50°C
~50r-----------------------~
0.20,----------------------------,
436
016
~22
012
408
0,,08
394
0,U4
380 L.-_ _
~
o
_ _ _ _~_ _ _ _ _ ,_~_,-=:S
10
20
W 30
40
10
50
20
W 30
40
50
countercurrent: dTa dW
= mCCpC
Taf= 50°C guess and check Tao until T a = 323 K at W = 50 Tao = 438.8 K 0.2U 016
422
0,12
.:J08
0,U8
394
n 114
380 L--_ _ _----~-----o 10 20 W 30
40
5U
ouo
10
0
P8-9 (C) For a fluidized CSTR with W = 80 kg, VA = 500 J/s/K,
8-39
20
\V 30
40
50
W -CAO
Species balance: X MB = ~,r = ..:..P-",b_ _ l+rk FAO UA (T-Ta) +C (T-T)
F
Energy balance: X EB
0
pA
=...:.P.-!:b:.----"A~O- - - - - - -MIRxn
X EB =XMB Solving, X = .95, T
=323 K
P8-9 (d) For a reversible reaction, we have all the previous equations, but the rate law is modified as: --rA = kfCA -k,CBCC C =C =C B
C
AO
~1'0 l+X T
Plugging the equation for kl> and solving using POLYMATH program, we get the plots. Only the co··current program and plots are shown. See Polymath program P8-9-d.pol. Calculated values of DEQ variables
!~~]ly~~i~~I~'lll~~i~ial v~'~~l Minimal valuE!1 Ma~,im~1 va'u~IFi~~' "-~IU,!!JI' j1 _.,,,." IW:O f"
f-''''~'
!~-l~
13 IT j..•... \... J4
JTa
10
180.
1 .. ,
J~ . 1°. . \450. 1420.7523 . t'< . " ' " .,
,P?~:
!~2~:
!?~!~" .. J450·l4?0. i6 iVO
120.
l~f~~:hot::'3 191po . li.013E+06 110'fCAo·'127o.82'83
42131
li·013E+06 [270.8283
flila!270.8283" "f 258.1071 fi21k'r 102"10.076962
(like"10" r141'c8 ",'.'][0. j15FA
'{
jO.057593 1450. j
,1 4?6.1627 ]450. ,20.
l20.
. ]it:.
180.
1
'"
.
1~·057~93
'"
1420.7523 ... .
,
'v
,1
J
!
J
,I
j42.0.7565. , 450 . 1 120. I
1
··1~~~31~:'42131i
h:013E+06 '11.613E+06 1 1270.8283'12708283'
'j b70.8283h58~io711
10.2
'10.076962
"'1
10t15.77362
h5.77362'j
[0'115.77362
115.77362"'1
1-36.02017' 1-36.02017
1-0.0062421
Differential equations d(X)jd(W) = -rA j vO j CAO
8-40
. . . . ]-0:0062421]
= (Uarho * (Ta - T) + rA * 20000) / vO / 3 d(Ta)/d(W) = Uarho * (T - Ta) / .2/ 5000 2 d(T)/d(W)
+ sign = cocurrent, -ve sign
CAO / 40
=countercurrent in RHS of eqn
Explicit equations
1
TO
= 450
= 20 k = 0.133 * exp(31400 /8.314 * (1 / TO - 1/ T))
vO
~ Uarho == .08
4-
* 3000
S •PO = 1013250 6
CAO = PO / 8.314 / TO
= CAO * (1 - X) / (1 + X) * TO / T aE'ili kr = 0.2 * exp(51400 /8.314 * (1/ TO - 1/ T)) CC = CAO * X / (1 + X) * TO / T ~m: CB = CAO * X / (1 + X) * TO / T 1;:~i rA = -(k * CA - kr * CB * CC) "l...
CA
450..--------------.,
OU60
.:14.:1
I)
.:138
0.036
432
OJ124
.:126
U.II12
16
32
vV.:I8
64
8U
U.:l8
o.ouo
0
16
32
W 48
6.:1
80
P8-9 (e) Individualized solution
------------------P8-10 (a) A-7B+C
C
A
=c_FFA I
T
e[ = FF[
A
Cr
= CA +C[
Fr =FA +F[
8-41
C
=
CAO +CIO
BI + 1
AOI
P8-10 (b) Mole balance: dX = -rA
dV
FAO -rA = kCA
Rate law:
. h· C - C 1-- X 1'0 StOlC lOmetry: A AOI 1+£ X T £ =
YAO
c5
8=1+1-1=1 Y = FAO .=
Fro
,AO
FAO
__ 1_
FAO + F;o
1+ B;
1
£=--.
1+8.I T=
-XMIRX
+ (CPA +B;CPi )1'o
CpA +B;CPi
Enter these equations into Polymath See Polymath program P8--lO-b.pol. POLYMATH Results Calculated values of the DEQ variables Variable V X
Cao Cio theta Fao Cao1 e To dRrx Cpa Cpi T k
ra
initial value 0 0 0.0221729 0.0221729 100 10 4.391E-04 0.009901 1100 8 . 0E+04 170 200 1100 25.256686 ···0 . Oll0894
minimal value 0 0 0 . 0221729 0.0221729 100 1.0 4.391E-04 0.009901 1100 8.0E+04 170 200 1098.3458 24.100568 -0.0110894
maximal value 500 0.417064 0.0221729 0.0221729 100 10 4.391E-04 0.009901 1100 8.0E+04 170 200 1100 25.256686 -0 . 0061524
ODE Report (RKF45) Differential equations as entered by the user [l] d(X)/d(V) = -ra/Fao
8-42
final value 500 0 . 417064 0.0221729 0.0221729 100 10 4.391E-04 0.009901 1100 8.0E+04 170 200 1098.3458 24.1.00568 -0,0061524
Explicit equations as entered by the user [ 1] Cao =2/.082/1100 [2] Cio = Cao [ 3] theta = 100 [4) Fao = 10 [5 J Cao1 = (Cao+Cio)/(theta+ 1) [6] e 1/(1 +theta) [7] To = 1100 [8] dHrx 80000 [9j Cpa = 170 [10] Cpi = 200 [11] T = (X*(-dHrx)+(Cpa+theta*Cpi)*To)/(Cpa+theta*Cpi) [12] k exp(34 . 34-34222/T) [13] ra =-k*Cao1 *(1-X)*To/(1 +e*X)/T
=
=
=
Conversion
1.0
--
1060
0.8
-
0.6
No inert i\fodelate Larae
980
0. 2
940
0
LOU
100
200 V 300
400
-
1020
004
0.0
Tell1per~lture
1100
500
900
--_.. 100
(}
No inert Moderate Larae
-~---200 V 300
400
500
--_.
E~ 040 0.20 0..0°0
3
6
Ihe~a
12
15
P8-10 (C) There is a maximum at e =8. This is because when e is small, adding inerts keeps the temperature low to favor the endothermic reaction. As e increases beyond 8, there is so much more inert than reactants that the rate law becomes the limiting factor.
P8-10 (d) The only change to the Polymath code from part (b) is that the heat of reaction changes sign. The new code is not shown, but the plots are below. See Polymath program P8··1O-d.pol.
8-43
1.0
Conversion
rr------=.......- - - - - - - ,
Tempel"nture
1600..-------'---------, 1500
0.8
- No inerts
- No inerts l\Iodcrate - Larue
0.6 0.4
1300
0.2
1200
0,0
w:::;..._~
o
__
100
~
_ _ _ _ _. _ _
200 V 300
- Moderate - Laroe
1400
400
500
1100 ...----:::=.==~-.,-.-.,~-=,~=~,.,:~ 0 100 200 V 300 400
.-.---'
500
1.00.-----~--------_,
0.92 0,,84
0.76 0.68 ,~--~-~-,---'-.-.
10
20
TheiN
40
so
The maximum conversion occurs at low values of theta (8 < 8) because the reaction is now exothermic. This means heat is generated during the reaction and there is no advantage to adding inerts as there was in the endothermic case.
P8-10 (e) We need to alter the equations from part (c) such that -rA
= kC~
and CAo =1
A plot of conversion versus theta shows a maximum at about 8 =5. See Polymath program P81O··e.pol.
8-44
0.90.----------------,
-
0500' - - - - 2 - - - 4
rheIN
10
8
P8-10 (0 We need to alter the equations from part (c) such that We already know that CA
-'A
~~c ]
= k [ CA _ C
I-X T
= CAO ----......Q.. Now weed need expressions for CB and Ce. From
I+EX T stoichiometry we can see that CB =Ce. In terms of CAO we find that: C =C =C B e · AO
X To I+EX T
We also need an equation for Ke: Kc = KCl exp
[MiR
RX
!J]
(~T.. T
= 2exp
When we enter these inot Polymath we find that the maximumn conversion is achieved at approximately e = 8. See Polymath program P8-1 o Lpol. 100
0.30 ' - - - - : 3 - - 6 - - - - 9 - - - 1 2 0 Iheta
15
P8-10 (g) See Polymath program P8-1 Og.pol.
8--45
!)]
[8000~ (_1_ 8.314 1100 T
Fb
10
-~
8
- Part(d) .. Part (e)
6
4
2
o0
100
200
V
300
400
500
P8-11 (a) Start with the complete energy balance:
I
I
dE " s -"LEiF; in-"LEiF; out -=Q_·W dt The following simplifications can be made: It is steady state. In part a, there is no heat taken away or added There is no shaft work That leaves us with 0= -"LEiF; lin -"LEiF; lout
Evaluating energy terms: In: H AOFAO + H BOFBO + H eoFeo Out: HA(FA + RAV) + HB(FB+RBV)+He(Fe + ReV) Simplifying, HA(FA+RAV)+HB(FB+RBV)+He(Fe + ReV)-HAoFAO -HBOFBO -HeoFeo = 0 If only C is diffusing out of the reactor we get: HAFA +HBFB +He (FB +RBV)-HAOFAO -HBOFBO -HeoFeo =0 Now we evaluate Fi FA = FAO .- FAOX
= FBO + FAOX Fe = Feo + FAOX -
FB
ReV Inserting these into our equation gives: HAFAO -HAFAOX +HBFBO +HBFAOX + HcFeo + HeFAoX -HAOFAO -HBOFBO -HeoFeo =0 and note that FBo =Feo =0 H AFAO - H AFAOX + H BFAOX + H eFAoX - H AOFAO = 0 and combining and substituting terms gives: FAO ( H A _. H AO ) + FAo X MJ.RX
=0
8-46
Differentiating with respect to V with i1Cp
=0
dT dX FAoC -+FAo-(LViRx(T))=O P dV dV dT
(rA)[ LVi Rx (T) ]
dV
'i.~Cp
Combine that with the mole balance and rate law: dFA =r dFB =-r dFe =-r -k C dV A dV A dv Ace -rA =kCA
C =C A
TO
FA To F T
=C
C B
FB
TO
T
Yo
C
FTC
=C
Fe
TO
T
Yo
F T I
Forkc =10 See Polymath program P8-11a.pol. Calculated of DEQ variables -- .,-.-.---..values ,. - _._ ..,; lVariable :Initial value iMinimal value Maximal value Final value
..
1 IV & "'
i2
Fe
\3
Fb
:0
:0
50.
50.
0 10
;0
0.0012968
0.0005261
:0
0.1978837
10.1978837
5.42
15.222116
548.9418 5.421297
:5.420526
;
I
i5 ;T
,450.
16 1Ft
,5.42
15.222116 , 1450 . 15 .42
i7
dHrx
'-2.0E+04
:-2.0E+04
-2.0E+04
;-2.0E+04
i8
lk
iO.133
iO.133 '10.
0.6036997
10.6036997
10.
i1O.
2.710027
12.710027
2.710027
12.610831
10.0006905
10.0002635
10
;0.0006482
10.000263
i4 lFa {
9 ike
i5.42
~,
,
(
il0.
1
"r
'loiCto
12.710027
2.710027
:l1iCa
12.710027
;2.610831
f121Ke ,
10.0006905 ,
10.0002635
1
;548.9418
+
~l
i13 ICe
,0
114icb
12.710027
12 .61 0831
12.710027
12.610831
l15lra ;
1-0.3604336
1-·0.3604336
1-0.0026249
1-0.0026249
116iCpa :
~40.
140.
140.
140.
i
1
1
,,~
~
Differential equations d(Fe)jd(V) = -ra - ke * Ce d(Fb)jd(V)
= -ra 8-47
!
~J d(Fa)/d(V) = ra ~J d(T)/d(V) = ra * dHrx / Cpa
Explicit equations Ft = Fa + Fb + Fc dHrx = -20000 k = .133 * exp«31400 / 8.314) * (1 / 450 - 1/ T»
kc
= 10
Cto = 100 / .082 / 450 Ca = Cto * Fa / Ft Kc = .01 * exp«dHrx / 8.314) * (1 / 300 - 1/ T» Cc=Cto*Fc/Ft Cb = Cto * Fa / Ft ;~Il ra = -k * (Ca - Cb * Cc / Kc)
Cpa = 40 6.0 4.,8
~ F~ [j
3,,6
Fa
2.-4
1.2
10
20
V 30
40
50
vary kc to see how the concentration profiles change.
P8-11 (b) Now, the hear balance equation needs to be modified. dT
dV
= Ua(Ta-T) + FAO(rA)[ LVIRx (T)] FAO'X-B;Cp
See Polymath program P8·,1l-b.pol.
8-48
6.0.----------------, 4.8
[ill FC
~ Fh
3. 6
Fa
2.4
1.2
10
20
40
V 30
50
P8-12(a) To find the necessary heat removal, we start with the isothermal case of the energy balance
.lL_ Ws F F AO
AO
--X[Ml RX +!J.CP (T-T,R )]="ac .(T-T) L,.IPI 0
Because there is no shaft work W s = O.
!J.Cp =60--25-35=0 And for isothermal operation T =To If we simplify the energy balance using this information we get: Q
--·--XMl RX =0 FAO or Q = FAOXMl RX
= CAOvXMl RX
We now know everything except the heat of reaction to solve for the heat removed term. To find the heat of reaction consider the adiabatic case:
!J.Cp = 60- 25-35 = 0 Q =0 and Ws = 0 --XMlRX = IB;CPi(T-To) Because feed is equal molar in A and B, 88 - X Ml RX = (CpA + CPB )( T- To )
Ml RX
(25+35)(350-300)
kJ
-0.4
mol
=1
= ----.--.------ = -7500-
Now go back to the isothermal case:
.
Q=CAOvXMlRX
(mal) mJ(0.2) (-7500~ol kJ ) 1000 m (.5 min 3
=
3
8-49
· kJ Q = -750000-. mm
P8-12(b) We start with the energy balance for the second CSTR (already simplified):
VA
-(Ta -T)-( X 2 -
XI)MI RX
FAO
= (CPA + CpB )(T --To)
This equation has two unknowns (T and X 2) and so we need another equation. Now wee need the mole balance for the second reactor FAO ( X 2- X I) FAO ( X 2- X I)
v: -
- ---'-'--'---
--rA2
2 -
-
kC~o (1- X 2 )2
This equation also brings in another unknown: k. We know that the specific reaction rate is dependant on temperature and if we have the activation energy, we can make an implicit equation for k as a function of T. To calculate the activation energy we will use the isothermal and adiabatic information for reactor 1 and the mole balance for reactor 1.
=
V. I
FAoX I 2 (
kCAO I-XI )2
k=
FAoX VC~o (1- X)2
vX -------VCAO (1- X)2
Solving for kat 300 and 350 K gives: k(300 K) 0.00015625 k(350 K) = 0.0005555
=
If we plug these numbers into the Arrhenius equation we get
(kJ E(I IJ we get EIR = 2664.
In - 2 = -- -- -- kl R 'Fr T2
Which means k(T) = k, ex p [
iU -~ J]
If we use a nonlinear equation solver to solve the energy balance and mole balance for reactor 2 we find that the exit concentration is 0.423.
See Polymath program P8-12-h.pol. POLYMATH Results
NLES Solution Variable
T X2
UA
Value f (xl Ini Guess --~3~2~7~,,~6~87~1~2~~-2~.~.2~7~4E~--:1~3~~3~4~0~==
0.4214731
-6.666E-12
0.4
4
8-50
350 0. 5 0. 2 -7500 300 1 3 . 309E-04 1000 500 -110.73657
Ta vo Xl dHrx To V k
Cao Fao ra
NLES Report (safenewt) Nonlinear equations f(T) = (UA)*(Ta-T)/Fao-(X2-X1 )*dHrx·60*(T-To) = 0 [2] f(X2) = V-Fao*(X2-X1 )/(-ra) = 0 [1]
Explicit equations [1] UA= 4 [2J Ta == 350 [3] vo=D..5 [4J X1 0.2 [5] dHrx = -7500
=
[6J To = 300 V =1 [8] k = . 00015625*exp(2663 ..8*(1/300-1/T» [ 9] Gao = 1000 [101 Fao = Gao*vo [11 J ra = -k*GaoJ\2*(1-X2)"2 [7J
P8-12(c) Now we need the differential form of the energy balance dT Ua(Ta -T)-rAMlRx Ua(Ta -T)-rAMlRx
dV-
FAOL~CPi
-- FAO(CPA+CPB)
we also need the mole balance for a PFR. For this case it simplifies to: dC _ dCB _ - -A- - - - - r dt dt A with -rA = kCACB and we can use the same equation for k as in part (b). When we put these equations into Polymath we get an outlet conversion of X See Polymath program P8--12-c.pol. POLYMATH Results Calculated values of the DEQ variables Variable --V
T X
Cao Ua Ta dHrx Cb
initial value 0 300 0.2 1000 10 300 -7500 800
minimal value 0 283 . 98681 0.2 1000 10 300 -7500 671.87016
maximal value 1 300 0 . 3281298 1000 10 300 -7500 800
8-51
final value 1 283 . 98681 0 . 3281298 1000 10 300 -7500 671 . 87016
= 0.33
0.5 500 25 35 1.563E-04 800 -100
v Fao Cpa Cpb k
Ca ra
0.5 500 25 35 1 . 563E-04 800 -42 . 749596
0.5 500 25 35 9.47E-05 671. 87016 -100
0.5 500 25 35 9.47E-05 671.87016 -42.749596
ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb)) [2] d(X)/d(V) = -ralFao Explicit equations as entered by the user [1] Cao = 1000 [2] Ua = 10 [3] Ta=300 [4] dHrx = -7500 [5] Cb = Cao*(1-X) [6] v =0 . 5 [7] Fao = Cao*v [8] Cpa = 25 [9] Cpb 35 [10] k 0.00015625*exp(2664*(1/300-1/T)) [11] Ca Cao*(1-X) [12] ra = ··k*Ca*Cb
= = =
P8-12(d) In this case we need to replace the rate law we used in part (c)
-rA
=k
[c
A
CB _ K Cc ] C
We also need an equation to calculate Kc at different temperatures. K
C
= K CI exp[~RX.(.! __~J] R T.T 1
be careful of the units when entering Kcl into Polymath. Also note that the initial temperature is different than in part (c) We get an outlet conversion of X
=0.48
See Polymath program P8-12-d.pol. POLYMA III Results Calculated values of the DEQ variables Variable V T X R
Ua Ta dHrx Cao
initial value 0 350 0.2 0.0083144 10 300 -7500 1000
minimal value 0 314.93211 0.2 0.0083144 10 300 -7500 1000
maximal value 1
350 0.4804694 0.0083144 10 300 -7500 1000
8-52
final value 1 314.93211 0.4804694 0.0083144 10 300 -7500 1000
0. 5 500 25 35 5.556E-04 800 800 200 0.002 -300 . 01803
¥
Faa
Cpa Cpb k
Ca Cb Cc Kc ra
0. 5 500 25 35 5.556E-04 800 800 480.46942 8.63E+121 -64 . 253241
0. 5 500 25 35 2.381E-04 519 . 53058 519.53058 200 0 . 002 -335.38132
0. 5 500 25 35 2 . 381E-04 519.53058 519.53058 480.46942 8.63E+121 -64 . 253241
ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb» [.2] d(X)/d(V) = -ra/Fao Explicit equations as entered by the user [1] R = 8 . 3144/1000 [2] Ua = 10 [3J Ta = 300 [4] dHrx=-7500 [ 5 j Cao = 1000 [6J v 0 . 5 [7] Fao = Cao*v [8J Cpa = 25 [9] Cpb = 35 [10] k = 0.00015625*exp(2664*(1/300-1/T» [11] Ca = Cao*(1-X) [12] Cb=Cao*(1-X) [13] Cc=Cao*X [14] Kc = .002*exp«dHrxlR)*(1/350-1/T» [15] ra = -k*(Ca*Cb-Cc/Kc)
=
P8-12( e) Individualized solution P8-12(f) For the gas phase the only the stoichiometry changes.
C A
and
=C (~_)(To) l+EX T AO
E=
YAoJ=0.5(1-1-1) =·-0.5
From Polymath we see the exiting conversion is X = 0.365 See Polymath program P8-12fpol. POLYMATH Results Calculated values of the DEQ variables Variable V T X
To Ua Ta
initial value 0 300 0. 2 300 10 300
minimal value 0 279 .3717 0.2 300 10 300
maximal value
final value
1
1
300 0.3650575 300 10 300
279 . 3717 0.3650575 300 10 300
8-53
-7500 1000 0.5 500 25 35 1.563E-04 -0.5 888.88889 888.88889 -123.45679
dHrx Cao vo Fao Cpa Cpb k
e Ca Cb ra
-7500 1000 0.5 500 25 35 1.563E-04 -0.5 888.88889 888.88889 -56 . 423752
-7500 1000 0. 5 500 25 35 8.ll1E-05 -0.5 834.06666 834 . 06666 -123.45679
ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb» [2] d(X)/d(V) = -ra/Fao Explicit equations as entered by the user [1] To = 300 [2] Ua = 10 [3] Ta= 300 [4] dHrx = -7500 [5] Cao = 1000 [6] vo 0 . 5 [7] Fao = Cao*vo [8] Cpa = 25 [9] Cpb = 35 [10) k = 0 . 00015625*exp(2664*(1/300-1/T)
=
e =-.5
[11)
[12) Ca = Cao*(1-X)*To/(1+e*X)/T [1.3] Cb = Ca [14] ra = -k*Ca*Cb
P8-13
X;
= CCCD =
K
CACB
C
~X
.jK;
= e
(1-. Xe)2
+.JK
1
T =To -
c
MIRX CPA
+ CP
B
=300- (--30000). X = 300 + 600X ( 25 + 25)
See Polymath program pg··13.pol. Calculated values of NLE variables Tv~;i~b-I~lv~I"~;f(x) . ---- -~-r i~-itial Guess
-I
r~.999764~13.si8E-lilo:s .~ 6 <-x~·<
llxe ,,-
1. )
T-·
i IVariable IValue !ll! ]300. r'~'''"f A"_'''''~~'N~'
,,'
~","
.. " 1""0_",,,,.,,,.
,¥,,·,~.w_,
8-54
-7500 1000 0.5 500 25 35 8.1llE-05 -0.5 834.06666 834.06666 -56.423752
2 Kc
li8E+07'
.J
Nonlinear equations f(Xe) = Xe - (1 - Xe) * Kc
A
0.5
=0
Explicit equations
= 300 Kc = 500000 * exp(-30000 /1.987 * (1/323 - 1/ T))
1. T
2
T
X
300
1
320
0 ..999
340
0 . 995
360
0.984
380
0 ..935
400
0.887
420
0.76
440
0 ..585
460
004
480
0.26
500
0 ..1529
520
0.,091
540
0.035
560
0.035
8-55
Xe 1
,-------~~-------.----------.----------.---------~
0.9
C 0.8 o 'en ... 0.7 ~ 0.6 c 0.5
FxeJ
8
§. 0.4 !:!:!.. 0.3 Q)
><
0.2 0.1
o
·------·--···--···-·--·T--···--····--··-·----·-,--··--·-----·---,··-·-·-------·--,------·--··---····---1---·
300
350
400
450
500
550
Temp (K)
P8-14 For first reactor, X Kc K = e l . or X c I-X el el I+KC For second reactor B X K -B K =2~orX = C BZ C I-X eZ eZ I+KC For 3rd reactor K =.BB3+ X e3 orX _ Kc -BB3 C I-X e3 e3 I+KC 1st reactor: in first reaction Xe =0.3 So, FB = F A01 (.3) 2nd reactor: Moles of A entering the 2nd reactor: F A02 = 2FA01 - F A01 (.3) = 1.7FA01 BBZ
= .2FA01 =.12 1.7 FAO
-FAOZ'LCp;~ (T -To) + FAOZX (-/ili R) = 0 X=
(CPA +BBCpB )(T -T 0) -/ili R
Slope is now negative 3rd reactor:
8-56
X e2
=0.3
(.2FAOl ) + .3FA02 = FAOl (.2 + .3x1.8) FA03 = FAOl + FA02 (1- XeJ = FAOl + 1.8FA01 (1- X e2) = 1 + 1.8(1- .3)FAOl FA03 = 2.26FAOl Feed to the reactor 3: (2FA01 ) + .3FA02 = FAOl (.2 + .3x1.8) = 0.7FAo1 (}B =
Say
(} =
.74 2.26 Feed Temperature to the reactor 2 is (520+450)12 =485 K Feed Temperature to reactor 3 is 480 K Xfinal = .4 Moles ofB = .2FA01 + .3FAo 2 + .4FA03 =F Aol (.2 + .54 + (.4)(2.26)) X = FB/3FAOI = .54 B3
= 1.64 F Aol
Xe.
,
-----~~
............ __ -
1-/13
/"
.'f .~
P8-14 (b) The same setup and equations can be used as in part (a). The entering temperature for reactor 1 is now 450 K and the outlet is 520 K. When the two streams are joined prior to entering reactor 2 the temperature is (520+450)12 =485 K Say that the outlet temperature for reactor 2 is 510 K. Then the entering temperature for reactor 3 would be (510+510+450)/3 = 490 K For any reactorj, -FAOl,CFj~ (T-To)+ FAOjX (--MI R ) X
=0
= (CpA + (}B C pB ) (T - To) -MIR
and 9B for reactor 1 =O. For reactor 2, 8B > O. This means that the slope of the conversion line from the energy balance is larger for reactor 2 than reactor 1. And similarly 8B for reactor 3> 8B for reactor 2. So the line for conversion in reactor 3 will be steeper than that of reactor 2. The mass 8-57
balance equations are the same as in part (b) and so the plot of equilibrium conversion will decrease from reactor 1 to reactor 2, and, likewise, from reactor 2 to reactor 3.
x
P8-1S (a) Substrate -7 More cells + Product S-7C+P G(T) =
X*-~HRX
To solve for G(T) we need X as a function of temperature, which we get by solving the mass balance equation.
_ FAOX _ FsoX V - - - - - - - and since -rs -~
rg
= jiCc
-~
and ji
= -rg
~/!
then,
= ji(T) __C. _ 5 _ Ks +Cs O.0038*Tex p( 21.6- 6~O)
where peT) = J4max
------
1+exP (153- 48~OO)
if we combine these equations we get:
v=
~/sFsoX ji(T)CcCs
Ks +Cs
8-58
V=~/sFsoX(Ks+Cs) p(T)CcCs
V= ~/sFsoX(Ks+Cso(1-X)) p(T)CSO~/l XCso (1- X) Canceling and combining gives:
Now solve this expression for X:
Now that we have X as a function ofT, we can plot G(T). To get R(T) we must calculate the heat removed which is the sum of the heat absorbed by reactants to get to the reaction temperature and the heat removed from any heat exchangers . The heat gained by the reactants =
Cpo (T - To )
The heat removed by the heat exchanger = UA(T-Ta)/Fso
VA
R(T) = Cps (T --Yo)+-F-(T -Ta) so
Now enter the equations into polymath and specify all other constants_ The adiabatic case is shown below_ The nonadiabatic case would be with explicit equation [12] as A = 11.
See Polymath program P8-15--a.pol. Differential equations as entered by the user [ 1] d(T)/d(t) = 1
Adiabatic Case
Explicit equations as entered by the user [lJ mumax=_5 [2J Ycs = . 8 [3] vo=1 [4J Ta = 290 [5] mu = mumax*( . 0038*T*exp(21 . 6-6700/T))/(1 +exp(15348000/T)) [6] Ks=5 [7] V = 6 [8] Cso = 100 [9 J Fso = vo*Cso [10] Cps = 74
20000 . - - - - - - - - - - - - - - - , 10000
o -10000 -20000 -30000
L-_~
294
8-59
_ _~_~_ _~_---I
299
303 T 308
312
317
[ 11] dH = -20000 UA = 0*300 kappa = UA/(Cps*Fso) [14] To = 280 [15] X = 1-«Fso*Ks)/«mu*V*(Cso"2))-Fso*Cso)) [16] Gt = X*(-dH)*Ycs [17] Rt = Cps*(T-To)+UA*(T-Ta)/Fso [12] [13]
Non-Adiabatic Case
20000 . . . . - - - - - - - - - - - - - - - - , 1-4000
Independent variable variable name: t initial value : 0 final value: 30
8000 2000 -4000 -10000
P8-1S (b)
_ _~_--,-_ _~__--l 299 303 r 308 312 317
Il-_~
2 4
To maximize the exiting cell concentration, we want to maximize the conversion of substrate . If we look at G(T) from part A, we see that it is at a maximum at about 310 K. This corresponds to the highest conversion that can be achieved. By changing the values of UA and Il1c we can change the slope of the R(T). What we are looking to do is get R(T) to intersect with G(T) at 310 K. Since we now have a limited coolant flow rate we will use a different value for Q.
and so,
-u~J) + Cps (T -To)
R(T) = mcCpc (T -Ta {1-exp( \ mcCpc
Now we set R(T) equal to the maximum value of G(T) which is 15600 Jlh
f
u~.J) + FsoC
G(T) = 15600Fso = mcC pc (T - Ta 1- ex p(- \ mcCpc
PS
(T ._- To)
And now plug in the known values . Assume the maximum coolant flow rate and that will give the minimum heat exchange area.
15600{(100 L ) g hr
=
8··60
1560000
~. = (88800000 ~) hr hr
-UA
1-- exp [
4440000
_J_)
+ (162800
~) hr
hr K
J VA=70415-hrK
P8-1S (C) There are two steady states for this reaction" There is an unstable steady state at about 294.5 K and a stable steady state at 316 K.
P8-1S (d) Increasing To enough will eliminate the lower temperature steady state point. It will also lower the outlet concentration of cells. Decreasing To will increase the outlet concentration of cells., Increasing me increases the slope of R(T) and will increase the exit concentration of cells, Increasing T a will lower the exit concentration of cells.
P8-16 (a) G(T) = illi R X
X=
Tk l+Tk'
-3
T=6.6xlO exp
[E[ 1- T1)] Ii 350
= cpo (1 + K)(T - Tc) cpo = I. BP p; =BACpA +B1CpI =20+30=50
R(T)
K=
VA =_8000_=2 CpoFAO (50)(80)
= ToFAOCpO +UA~ = KT,,_+To
=350 VA+CpoFAO l+K To find the steady state T, we must set G(T) =R(T). This can be done either graphically or solving the equations. We find that for To =450 K, steady state temperature 358 K. Tc
P8-16 (b) First, we must plot G(T) and R(T) for many different To's on the same plot. From this we must generate data that we use to plot Ts vs To.
8-61
I
G('l) \S. R(1)
I
1;,'5. T.
10000 ,-------;-~,.........,,_r"7"__.
425
;----~t-)'#~(;C.:;;;;;I
400
I; ~+-------~"~~~~~--~ i-
8000
375
CD
E~r-----~~~~~~-----~
~ ~r---~~~~~~~----~ O~"
350
325
__~~U--L~~____~~
Z15
325
375
TOO
300
425
200
250
300
350
-----"
P8-16 (C)
400 T. (X)
450
500
550
For high conversion, the feed stream must be pre-heated to at least 390 K. At this temperature, X = .98 and T =380 K in the CSTR. Any feed temperature above this point will provide for higher converSIOns.
P8-16 (d) For a temperature of 375 K, the conversion is .968
P8-16 (e) The inlet extinction temperature is 375 K.
P8-17 The energy balance for a CSTR: -FAO XllH Rx (T)
UA = FAO ["'£.e.c pi.(T -.r,0 )+--(T -.r,)] F 0 I
AO FAOX =-'rAV
Cps isindependentofT.
Cpo (T -- To ) = C p'
G(T)=(-llHRx)(-~VJ F AO
R(T) =
"f'l'
FAOCpS
+UATr -UAT
".
dR(T) dT dG(T) _ -llHRx d(-rA) dT FAO dT
D 1 lerentlatmg: ---- = UA
where
d( -r ) A
E = - - 2 ( -rA )
dt RT Setting these two equations equal to each other and manipulating, 8-62
In 1.421 1.127
=[E(323-313)] R 323 *313
E=19474 FAOCpS + (T _ T ) > RT,2 ,andit will be runaway reaction UA T a E 2 At A < 2 ..5 m it will become a runaway.
P8-18 (a) Mol Balance:
~ _ FAOX _ -
-rA
voCAO X
-kTC
A
-CB / Ke1
(:'}[(l-X)-X I K,] X [1+rk(l+ 11 Ke)] X
=
= rk
rk l+rk(I+11 KJ
G(T) = -M/RxX = 80000X
k=lmin-- 1 T=lOmin
Ke =100 X =. 10 =.901 1+ 10(1.01) G(400) = 72080 cal/mol
P8-18 (b) UA R(T)
= 3600 =9 10*40
= CpA (1 + K)(T -·Tc ) = 400(T -
Tc)
Tc =ToI+K +KT .=31O a
R(T)
= 400(T -
310)
The following plot gives us the steady state temperatures of 310,377.5 and 418.5 K See Polymath program P8-18--b.pol.
8-63
-r-----
-
1_
OF------~~-_+_
-
_ _,_-__o_--+___...:=._ 390
P8-18 (C) 3 10K and 418.5 K are locally stable steady-state points
P8-18 (d) R{418.5) = 400(418.5-310)=43400 G(418.5} = 43400 = 80000 * x
43400 x = 80000 = .54
P8-18 (e)
-
The plot below shows Ta varied. ..
'
.....
...... ._,.".".~." .."'''''-''''./''.''
t ....
..
Ta~
___._____.c=..__
• 100:--_ _-_~=----_, _~
m
_
_
~
m
m
_
GO
The next plot shows how to find the ignition and extinction temperatures. The ignition temperature is 358 K and the extinction temperature is 208 K.
8-64
...
~5S
...
...
.----'---~---.-.---.---
1501.----150
-
P8-18 (f) This plot shows what happens if the heat exchanger quits. The upper steady-state temperature now becomes 431.5 K.
-, 70000
_50000 15 .l!.
i..
4DOOO
E
" 30000
20000
1DOOO
351 0
,.0
330
3IiO
TIKI
P8-18 (g) At the maximum conversion G(t) will also be at its maximal value. This occurs at approximately T =404 K. 0(404 K) =73520 cal. For there top be a steady state at this temperature, R(T) =OCT). See Polymath program P8-18-g. pol.
8-65
VA
K=--FAOC pA
R(T) =
CpA
where T
(1 + K)(T - TC) = 73520
= T0 +KTa I+K
C
If we plug in the values and solve for VA, we get: VA 7421 cal/miniK
=
P8-18 (h) Individualized solution P8-18 (i) The adiabatic blowout flow rate occurs at T = 0.0041s V T = 0.0041s =-Vo Vo
= TV = 0.0041 *10
Vo
dm 3 =0.041.. mIll
See Polymath program P8-18·-i.pol. 6000
4600 321.11.1 1801.1
-11.10tll-------~-
o
30
60
t
90
120
151.1
P8-18 (j) Lowing T 1.1 or Ta or increasing VA will help keep the reaction running at the lower steady state.
P8-19 Given the first order, irreversible, liquid phase reaction: A·-7B VA = l.Ocal! minl°C
8-66
C 0, = CPo == 2eal/ g - mol/DC
Pure A Feed = o...5g mol / mm I1CP
M1 R
= I1CpA -- ilCpB = 0
= -200
cal gmofA
= canst
_ FA oX _ Vo Designqn· V -----.---E -rA k(l---X) RateLaw..
-rA
StolchlOmetry
= kCA
C A = CAO(l- X)
EnergyBafance -UA(T -T,1)-- FA oX MR
Simplifying, -rk
= FAOL~Cpi(T --1'0)
1
X=--=-l+-rk
1+_1
-rk
o
~g
=F X(--M ) == F40 (--M-!.R)_ AO
R
}
1+rk ThIS IS the curve plotted m the prohlem statement From the equatIOn for heat generatIOn curve, we get X == __ Jl~_-_ =-~ = Qg F 4o (--MI R ) 5*200 100 The equatIOn for heat removal curve IS· QR =FAoCp4.(T-To)+UA(T-T4)==·5*2*(T--I'u)+1 O*(T-T j )
QR = 2T--l;-·} 00 Plot thl:' along with the heat generatlon curve for vanous To This I:' shown m the figure
r=:::~:.---
Qg------1
I 1- -
I .•- T0=15O I I, . . . . . 1·0= 160 Ii
1-' -
·To=170
I
i-- ...- TO=180 \
l-
r 0=1901 . - -T0=200 I.____- ._.-___ To=210 L ._--"I
120
140
160
180
200
220
T fe)
The intersectlOn between Qg and QR can be used to plepme the igmtion - extinctlOn CUlve :.hown m figure P8-22-:2 The value~ fOl T~ a~ a flJnction of mlet temperature, To, are tabulated below.
8-67
To (degree C) 150 157 160 170 190 202 210
Ts (degree C) 132 135, 172 137, 168,176 142.5, 167, 181 154.5, 165, 193.5 162, 199 204.5
P8-19 (a) To obtain high conversion, the reactor must operate at or beyond point 14 in figure. So the minimum inlet temperature for high conversion is To 2 202°C.
P8-19 (b) The temperature of the fluid in the reactor corresponding to temperature in part 1 is Ts > 199°C (point 14 in figure).
P8-19 (c) The inlet temperature of the fluid is To = 202 + 5 = 207°C. This will be somewhere between points 14 and 15 in figure. Once the fluid is cooled from this temperature, it will follow the line formed by points 15, 14, 12,9, etc .. Now To =207 - 10 = 197°C. From figure, Ts = 195.5 °C. From equation (5), Qs = 2*196.5-19-100 =95 call mol
QR
=Q =96 cal/mol
So X ,
g
= 100 Q =0.96 g
P8-19 (d) Extinction temperature is the temperature corresponding to point 3 in the figure = 157 °c
P8-19 (e) Individualized solution P8-20 (a) The following are the explanations for the unexpected conversion and temperature profiles Case 1: Broken preheater or ineffective catalyst Case 2: The eqUilibrium conversion was reached due to a problem with the heat exchanger Case 3: Broken preheater or ineffective catalyst Case 4: The equilibrium conversion was reached due to a problem with the heat exchanger Case 5: Ineffective catalyst Case 6: The equilibrium conversion was reached due to a problem with the heat exchanger
P8-20 (b) 8-68
The following are the explanations for the unexpected conversion and temperature profiles Case 1: Broken preheater or ineffective heat exchanger Case 2: Ineffective catalyst Case 3: Broken pre heater or ineffective heat exchanger Case 4: Ineffective catalyst
P8-21 Below is the FEMLAB solution. 1. Parameters in simulation on the tubular reactor in Problem 8-6: Reaction: A + B ~ C
(1) operating parameters Reactants • Inlet concentration of A C AO
= 100
3 mol/m
•
Inlet concentration of B C BO = 100 moilln'
•
Inlet total flow rate
•
Inlet temperature of the reactant To = 300K
Vo
= 2xlO-3
m3/s
(2) properties of reactants • Heat of reaction, t:.HRx , dHrx = -41+20+15=-6 kcallmol=-25100 J/mol • Activation energy, E = 41840 J/mol • Specific reaction rate ko = 0.0IxlO-3 m 3/mol.s @300K • Reaction rate k
= ko exp[ E (~-!-)] R To
T
• Gas constant, R = 8..314 J/mol·K • Rate law - fA = kC A CB
Assumption: • Thermal conductivity of the reaction mixture, ke
= 0,,68 W/mK
(needed in the mass halance and the energy balance) • Average density of the reaction mixture, p, rho = 1000 kg/m3 (needed in the enelgy balance)
• Heat capacity of the reaction mixture, Cp = 4200 J/kgX (needed in the energy baJance) 9
• Diffusivity of all species, Diff= 10- m% 2. Size of the Tubular Reactor (1) Volume of reactor sized by a PFR = 0.317 m' (2) From FEMLAB • Reactor radius, Ra = 0 . 1 m • Reactor length, L = 10.0 m
8-69
3. Femlab Screen Shots (1) Domain
[ ] Axis equal
:\
[
OK ,,'
!I [
cane;]
,
[A;i¥___J
(2) Constants and Scalar Expressions - Constants
- Scalar Expressions
[~'"~-:J r'~-~ OK Cancel j
'"
(3) Subdomain Settings - Physics
8-70
j
~z~·····,··,
.
-,
['---'''-J Apply _'_'
(Mass balance) rEquation
I Vo(-DVcA+cAu) =R, cA =concentration
I
rSubdomain selection"'~"
'_-13]:1
Species Library material:
ValueJExpression
Quantity
Time-scaling ...,.".ff;,~;",-,t
°ts
Diffusion coefficient
D isotropic
Diffusion coefficient
D anisotropic
R
[J Select by grou!"
__ "'" ___._N.____ ._"_____
~===;;;;;=~
Reaction rate
u
'=::;;;;;:;;:;;;;;;;;;;;;:;;==":: r-velocity
v
z-velocity
------.---J ._--_._-_._--
[ .__Artificial Diffusion ...
~ Active in this domain
(Energy balance)
Description
¥ _ _ _ _ _ · _ · _ _ _ , , _ _ _ _ • _ _ _ ' , · _ _ _ _ _· _ - - - - - . - - - - - - - - - - . - - -
;-Equation
Ii V.(··kVT +
~;h;HDJ) = Q - PCpu.VT, T=temperature
; rsubdomain selection
In~.: Element
Physics
Thermal properiies and heat sources/sinks
I
library material: i.
ValueiEKpression
Quantity
[1
(V k (isotropic)
~k~
ok
Dllscription
1 Time-scaling coefficient
01S
•. , Thermal conductiv!y Thermal conductiv~y
(anisotropic)
Dens!y Heat capac!y
Heat source
u
.........................
o
hj"O,i
Select by group
~ Active in this domain
[
r-veloc~y
z.veloc!y
v
l~Eecie: ~~.!~~i"~~~~~~~ ¥~
:J
Artifici;oiff';-sio~ ..
- Initial Values
(Mass balance) cA(tO) = cAO (Energy Balance) T(tO) = TO - Boundary Conditions
r = 0, Axial symmetry @ inlet, cA = cAO (for mass balance) T = TO (for energy balance) @ outlet, Convective flux @ wall, InsulationiSymmetIy (for mass balance) Thermal Insulation (for energy balance) (4) Results @
(Concentration, cA)
8-71
File Ed~ Options Draw Physics Mesh SOlve Poslp"ocessing Mulliphyslcs Help
_ti~ ~_ .iL~_~~I~I~~~J~_~~~_li[~~~_~L~~~~~~_~~l !. ____ _
____________R_________
[!]
___________________._____._
Max: 97.38
Contour: Concentration, cA
10
&~,
--------------- -------;===:;==:::;::~~;:;=;::;:':;
>--,
~
467
/
_~-1==:;=========~ 01 02 -------0-.3- - - - - ' - 0 - _ 4 - - - - - -
2--~--
.05"
'~
[SNAP i
(0,4.~)
10~'--'"----·----..····-'"--·-----·-··------378
333.361 331 344. ;.' 329 321 •• 327 31
4
03
8-72
P8-22 (a) Liquid Phase: A + B ---7 C
= -rA
dX dV
FAO
-rA =kCACB
k=.01*exp [- RE(~.T __ l )] 300 CA = CAo (l-- X) = CB The energy Balance: dT _ Ua(Ta -T)+(-rA)(-LlliR ) FAo [CPA
dV' -
+ CpB ]
4 D
a=-
AssumeD =4 U
=-~x2 m Ks
2
leal x 1m = .0120 cal 4.1841 lOdm 2 dm 2 Ks
~ee Polymath program pg·22·a.poJ. Calculated values of DEQ variables
'-lv~~i;bleli~itial-~~I~e-)Mini~al valu~!Maximal valuelFinal value ,,1
i
i1 IV f ,2 IT '3
Y"~
"
.fo
IX
;300.
SR
11.988 ,
16'E ; I! r 17 leao
. "~~,,
~
0
1300.
:4 "Ta
;8
,.
10
'0
~
i1000.
11000.
1434.7779
1348.2031
10.9620102
10.9620102
r
-"t'"
300.
~
;300.
i300. i
110000. 10 . 1
110000. ,
1.988 !10000.
10 . 1
iO.1
10.1 :0.1
1'6.003799 ;0.003799
10.1 101 l •
10.003799
'0.01 . '-0.0001
,0.01
11.808628
iO.1018746
-0.0010427
,-1.47E-06
15.
i1S.
,,i-1.47E-06
15.
15.
0.2
10.2
1
:1.988
11.988 , 110000.
1
t .
l'"
10.1 , iO.003799
", r-
'[
lea
19 'f~b .
I
iO. 2
~
r~6000.
.1
,
-6000.
.. ...
i
i
115. , 15 .
1
10.2 r-~
1-6000. ~
-
.1=6000: 8-73
Differential equations d(T)jd(V) = (U * a * (Ta - T) + (-ra)
* (-Dhrl)) j
(faa
* (cpa + cpb))
2; d(X)jd(V) = -ra j faa Explicit equations Ta = 300 R = 1.988 E = 10000
caa
= .1
ca = caa
* (1 - X)
cb = caa
* (1 - X)
* exp(-E j R * (1 j T - 1 j 300)) ra = -k * ca * cb
k = .01
cpb = 15 ~~r~l cpa j~~;M
= 15
faa = .2
'~l' Dhrl = -6000 J3~a=1
1l;~.; U = .0120
500
1.0
460
0.8
EiJ
420
0.6
380
0.4
340
0. 2
300
0
200
400 V 600
800
1000
200
P8-22 (b)
8-74
400 V 600
800
1000
Gas Phase:A f=Z B + C dX
= -rA
dW
FAD
= kjCA -
-fA
krCBCC
X Yo
C =C =C - - B e A D 1+ X T
k = .133*exp [- E(J:___ 1 )] R T
450
k. =.2exp [E, (_I__ !)],Er ,
R
450
T
= 51.4kJ Imol
The energy Balance: dT Ua(Ta -T)+(--rA)(-Mi R ) dW
=
Mi R
= -40 -
50 + 70=-20 kllmol
U=5 See Polymath program P8-22-b.pol. Calculated values of ,-, DEQ variables .. ", .....,. .T ""T' ,-- i ' :Variable! Initial value IMinimal value, Maximal value! Final value ,1 jW . :0 " " ' 1 0 150.' . " "/50. 12 ,X
10
10
10.0560855' \0.0560855
,3 T
[400.
/371.902
1400.
,
1
IJl1.902 1
>
'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
In(kl c(500)
)]
J =7628
kIC (400)
(4~0 5~0)
=15,158
! (~ -5~O
cal mol K
P8-24 (a) See the additional homework problems in chapter 8 at bttp://www.engin.umich.edu/~cre for the full solution. ,-----------------, Iso1hermal: T = 321 K RMAt\. = I, Q = 40 m~lIL 1
1 ~
:
II
~.
S4-
~-~
!:~--~
.---"" ..,---~
.___'..-
i
-
o -t=----,-.--,.- --,----
o -.2MIB
02
04
._-_....-,,2M2B
x
06
-Me0H
I
i
-l
0.8 --"'TAME
P8-24 (b) See the additional homework problems in chapter 8 at http://w\yw.engin.umich.edu/~cre for the equations to enter into Polymath. Then vary to see its effect.
8-77
P8-24 (c) See the additional homework problems in chapter 8 at http://www.engin.umich.edu/-cre for full solution Non-Isothermal: To = 353 K, Tw = 298 K RMIIA = 1.0, Q= 200 Llntin 7
-----------------
6
-
5
$ 4 o .,5,3 U
2
0. -1"'"'----,.-----.- ---.------,-----1 o 0.2 08 0..4 X 0.6
P8-24 (d) No solution will be given
P8-25 Mole balance: dF - -A= r dW A Rate Laws: rA = --rZB + 'iA
dF dW
- -B= r B
dFc --=r. dW
C
+ ljA
rc = -r3A -'iA = klCA -rZB = kBCB
--r3A = k3 CC Stoichiometry: C A
=C 1
_FA'Fo
FT 1
C =C FBTO B T FT T
Energy balance: _dT = Ua(Ta -- T) + (---'iA)( -LV!RIA) + (-rRZB ) + (-ljA)(-LV! R3A) dW
FAC pA + FBC pB + FcCpc 8-78
·dT = 16(500 - T) + (-ljA)1800 + (-rR2B )1800 + (-'3A)1100 100(FA + FB + Fe)
dW
kJ =0.5exp[2(l-320IT)] kJ k =--
K
2
C
k3 = 0.005 exp [4.6(1-460IT)] Kc =lOexp[4.8(430IT-1.5)]
See Polymath program P8"25.pol. ~alculated
"alues of D~Q variables " r IInitial value; Minimal value Maximal value i Final value , Variable , < l
""
1
iW
.~
'2 ,fb
o
11. I
13 Ifa r ' 4 :fc ,5 IT
i1. :0 330.
;100.
0
1
1.368476
10.9261241
jO.9261241
I
iO.6296429 iO
i1.
~
1
j
10.8694625
10.2044134
10.2044134
330.
1416.3069
;416.3069 :16."" :1
,6 jua
16.
,16.
]16.
7 ITa
500.
500.
]500. ,
}
500.
8 /Dhrla
i- 1800.
·-1800.
1-"1800.
)-1800.
9 iDhr3a
i-llOO.
i-llOO.
:-1100.
'100.
100.
1~1l00. ,i100.
,100. 1100 .
<1
I 100
'100.
100.
1100.
. 1100. ,
;100.
1
1
'100.
1
:0.5312401
0.5312401
:0.7941566
:0.7941566
iO.0008165
'0.0008165
;0.0030853
:0.0030853
12.
2.
'2.
i2.
2.
Ir'2 .""""
:2.
'"2. 1 1330 .
;330.
17!To
1330.
181Kc
I1~·885029
r
11.062332
3.885029
19:k2
,0.1367403
10.1367403
0.74756
i21 cb
1.
~0.5682599
1.
i1.
)0.7341242
1.253213
Io.?341242
'22[rla
1-0.5312401
-0.5748799
-0.362406
" ;-0.5473402
-0.002138
]-0.0007594
i" .~,
f··
".~.
!201ca J
f
,
f
!-0.0008165
231r3a T'
241rc
;0.0008165
i25 jr2b
1-0.1367403 I 10.3944998
j
!261rb
~
i
127lra
,
,;0.002138 -0.5770243 -0.0522707
~<-~
i-0.3953164
!-0.1367403 , 10.3944998 ,
-0.3953164
)~~~
--
""p30. :1.062332 10.74756 "" 10.6892094
!-0.0021264
··T'
10.0021264 1-0.5488019 T
" --
1-0.0014617
.~"
10.0510521 8"79
1-0.0006647
Differential equations d(fb)jd(w) = rb
= ra d(fc)jd(w) = rc d(T)jd(w) = (Ua * (Ta - T) + (-r1a) * (-Dhr1a) + (-r2b) * (Dhr1a) * (-r3a) * (-Dhr3a)) j d(fa)jd(w)
cpb
(fa
* cpa + fb *
+ fc * cpc)
Explicit equations Ua = 16 Ta
= 500
Dhr1a
= -1800
Dhr3a
= -1100
= 100 cpb = 100 cpc = 100 k1 = .5 * exp(2 * (1 - 320 j T)) k3 = .005 * exp(4.6 * (1 - (460 j T))) l,~,~ ct = 2 !U:ift = 2 ~Z~To = 330 m~ Kc = 10 * exp(4.8 * (430 j T - 1.5)) r~~; k2 = k1 j Kc !t~~ ca = ct * fa j ft * To j T cb = ct * fb j ft * To j T r1a = -k1 * ca [,1,;11 r3a = -k3 * ca l~~: rc = -r3a cpa
~~ r2b
= -k2 * cb rb = -r1a + r2b a~ ra = -r2b + r1a + r3a
P8-25 (a) As seen in the above table, the lowest concentration of o-xylene (A) = .568 mol/dm3 P8-25 (b) The maximum concentration of m-xylene (B) = 1253 molldm3 P8-25 (c) The maximum concentration of o-xylene = 1 molldm3 P8-25 (d) The same equations are used except that F BO =O.
=
The lowest concentration of o-xylene 0.638 mol/dm3. The highest concentration of m-xylene = 1.09 mol/dm3. The maximum concentration of o-xylene = 2 mol/dm3.
8-80
P8-25 (e) Decreasing the heat of reaction of reaction 1 slightly decreases the amount of E formed. Decreasing the heat of reaction of reaction 3 causes more of C to be formed. Increasing the feed temperature causes less of A to react and increases formation of C. Increasing the ambient temperature causes a lot of C to be formed.
P8-25 (0 Individualized solution
P8-26 (a) AH B+C
A-7D+E A+C-7F+G We want the exiting flow rates B, D and F Start with the mole balance in PFR: dF dFc dFA --=ra - ·B= r --=r. B dV dV c dV dF dFE dFG _F·=r -= r --=r. dV E dV F dV G
dF dV
- -D= r
Rate Laws:
8-81
D
Fe FT
Pe =--Pro ~=~+~+~+~+~+~+~+~
F[ = steamratio x .0034
Energy Balance: dT - ( 'is /lll RIA + r2B /lll R2A + r3T /lll R3A ) dV FA *299+ FB *283+ Fe *30+ FD *201 + FE *90+ FF *249+ Fo *68+ F[ *40
Kpi =exp(bi +
i
+b3 In(T)+[(b4T+b5 )T+b6 ]T)
See Polymath program P8-26.pol. For TO =800K Calculated values of DEQ variables
111~ariabl~~ti~1 val;;elrini;;;;;'~alu~lrjximal valu;l:~a! value! "...
If
]2 a 13 14 15 Ifd
...,
.,
10.00344
Ifb '10 IreTa 1
16 rf~ 17'
Iff
Is'l1=g 1 -['
;9;1
16 !
\0 "'[0
nip " 1,41p,~i
lsll
~~!:
118rr~,
j19 1Pa
.~
jO
10
.,.
h.078E-05 i3.588E-05 h.588E-05 """1800'."
r1.18E~051i.18E+05
J
j
I
j
1
I !1.078E-OSj j3.588E-05j
]3.588E-05 ',1765.237
I
[-s.39E+041-5.39E+041 12137. J
hBl.
a.4
,jc):41 10:0459123 ... 10.0196554 10.0459123
1~:!988· ·1~:!988 ·16:!988
., "'1!~·05332 ,10.05332 10.154838710.1104652
]i:991E-06
"... ....
jl.18E+051 k052E+05T1.052E+C)sji.oS2E+05!
1~.4
ri~I;;b j~.991E-06 Ed
10.00344 iO.0 02,496 10.0008974 1o.Q()08974 ' jO.0008615 10.000861S' i 11.078E-05';l 1.078E-05
f-5.39E+04' '!-5.39E+04 12137. .,". 12137.
i20lpbI~
]23
......
10.002496 /0 10 10
1010 ":,i800. T1765.237
10TH1al1.18E+05 11IH2a . Ti:052E+05
121 H3~
, .
~
[0.0542282 10.1548387
lO.4
J
10.0196554 1
1~:!988· j
fO:c)542282 , j
10.1104652 J
fo
10.039715510.0397155 I
I~.16E-07
li.991E-06
tLE-07
!~:~!~~:1~:~!:~:;71 8-82
]5.16E-07
i24ire f
2.991E-06
!2.991E-06Is.16E-07
2SF3t--Io
10
10-10
i26T~·To
10
271rg
J4.196E-06
,!4.196E-06 14. 196E-06
:?8.r~~-
ig·_0002138
-i2.481E-OS
!0.0002138
)2~! rb
10.0002138
12.481E-OS
:0.0002138
:.!3301._.•l..i rr.ac: _
10.0002138
12.066E-OS
~.
10.0002138
'0;
1-2.948E-OS
1-0.0002167]-0.0002167
... L
i2.481E-OS
L
~. t_.
12.066E-OS '+ ... j-2.948E-OS
Differential equations
= ra 7. d(fb)jd(v) = rb ,J d(fc)jd(v) = rc d(fd)jd(v) = rd ~! d(fe)jd(v) = re ;p; d(ff)jd(v) = rf d(fg)jd(v) = rg J
d(fa)jd(v)
l
d(T)jd(v) = -(rls
•.. . * 68 + fi * 40)
* Hla + r2b * H2a + r3t * H3a) j
(fa
* 299 + fb * 273 + fc * 30 + fd * 201 + fe * 90 + ff
Explicit equations Hla
= 118000
H2a
= 105200
H3a
= -53900
P = 2137
l5
phi
= .4
J§t. KI = exp(-17.34 fi
* In(T) + «-2..314e-1O * T + 1.302e-6) * T + -0.004931) * T)
= 14.5
7~: sr 8i~
1.302e4 j T + 5.051
= sr *
.00344
ft=fu+fu+k+~+fu+ff+~+fi
f~ Pa
= fa j
ft
* 2.4
= fb / ft * 2.4 re ··l pc-l'cjft*24 r2b = p * (1 - phi) * exp(13.2392 ... 25000 j i~~] Pb
'<:.':~
II
•
T)
* Pa
rd = r2b
it$! re = r2b !~qi r3t = p * (1 - phi) * exp(.2961 .- 11000/ T) * Pa * Pc ll' rf = r3t ~~ rg
= r3t rls = p * (1 - phi) * exp( -0.08539 - 10925 j T) * (Pa - Pb * Pc j 8-83
KI)
= rls .~1.j rc = rls - r3t
;~Qi rb
= -rls - r2b - r3t
.~! ra
= 0.0008974 Fbenzene = 1.078B-05 Ftoluene =3.588B-05 Fstyrene
SS/BT
= 19.2
P8-26 (b) To= 930K
=0.0019349 Fbenzene =0.0002164 Fstyrene
Ftoluene
SS/BT
= 0.0002034
=4.6
P8-26 (c) To= 1100K
= 0.0016543 Fbenzene =0.0016067 Ftoluene =0.0001275 Fstyrene
SS/BT
=0.95
P8-26 (d) Plotting the production of styrene as a function of To gives the following graph. The temperature that is ideal is 995K
1. i=b (S~) VI. Tol
0.0023
r------:=--------.
0.0022, C.0021
t G.~02 f
0.001' C.0:U8 i' C.OO17
i
0.0018 T 0.001& ,.
.,/~
f/
.\
0.0014 -'-----...,...--~-__J
lOa
100
tooo
1100
1200
8-84
P8-26 (e) Plotting the production of styrene as a function of the steam gives the following graph and the ratio that is the ideal is 25:1
Q.OO2
--..:::=====::-___
0 . 0019 +0001' •
!
00017-
o OGIa0.0015 ....- - - -_ _ _ _ _ _ _ _,.j
to
20
40
P8-26 (0 When we add a heat exchanger to the reactor, the energy balance becomes: dT
=
Ua(Ta -!)-('isLViRIA +r2BLViR2A +r3TLViR3A)
dV FA *299+ FB *283+ Fe *30+ FD *201 + FE *90+ FF *249+ FG *68+ F, *40 With Ta = 1000 K Va = 100 kJ/min/K = 1.67 kJ/s/K The recommended entering temperature would be To = 440 K. This gives the highest outlet flow rate of styrene. 40e-3
300
3.2e-3
24£1
24e-3 1.6e-·3 80e-4
- fh - fe - fd - fe
180 120
tt
60
2
4
v
6
8
10
4
P8-26 (g) Individualized solution P8-26 (h) Individualized solution P8-27 (a) 8-85
v
6
8
10
Adiabatic exothermic, adiabatic endothermic, exothermic with cooling, endothermic with heating. All the profiles show the rate of reaction dropping toward the end of the reactor.
PS-27 (b) The non-adiabatic profiles show an increase and a decrease in temperature profile, and the adiabatic profiles do go from increasing temperature to decreasing temperature (or decreasing to increasing).
PS-27 (c) Figure E8-5.3 shows a decrease in temperature while the reaction rate is large because the reaction is endothermic. Once the reaction rate drops, the heat exchanger increases the temperature profile because the reaction is no longer removing much heat. Figure E8-3.1 shows a steady increase in temperature until the reaction rate drops to near zero. The reaction rate increases at the beginning of the reactor because of the increase in temperature affecting the specific reaction rate. At too high a temperature the equilibrium constant gets vbery small and the reverse reaction becomes more prominent and thus the rate decreases as the temperature rises above 350 K.
PS-27 (d) In Figure E8-1O.1, the temperature increases quickly until the reactants are used up. Then there is no more heat generated from the reaction, and the heat exchanger lowers the temperature. Figure E810.2 shows that the flow rate of A drops to about zero at about the same time the temperature reaches a maximum. Once there is no reactant, the reaction ceases and the flow rates of the products remains constant.
PS-2S (a) 2A-,'m"-'~ B
a) Design equation:
Rate law: Stoichiometry:
Energy balance:
.~, == ,f!q,[!1.-:: T) ~L~!1X:,t:!!M..2 dV
FAOCpA
,4:£ = 5(7OQ.:: T )~ ~~?,~~' :gJ3{~~:~?~)X::~E1} dV
5*122
Plugging those into POLYMATH gets the following program and the following graphs. The conversion achieved is 0.36.
8-86
See Polymath program P8-28-a.pol. POLYMATH Results Calculated values of the DEQ variables variable V
T X
Faa Ua Ta dHr Cpa k
Caa Ca ra
initial value 0 675 0 5 5 700 -235.524 0.1222 0.0734336 1 1 -0 . 0734336
minimal value 0 675 0 5 5 700 -236.01067 0.1222 0 . 0734336 1 0 . 6419118 -0 . 3335975
maximal value 10 715 . 55597 0.3580882 5 5 700 -235 . 524 0 . 1222 0.3658299
final value 10 704 . 76882 0.3580882 5 5 700 -235 . 88123 0.1222 0 . 243008 1 0.6419118 -0 . 1001316
1
1 -0.0734336
ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(V) = (Ua*(Ta-T)+(·ra*(-dHr)))/(Fao*Cpa) [2] d(X)/d(V) = -ra/Fao Explicit equations as entered by the user [1] Fao=5 [2] Ua = 5 [3] Ta=700 [4] dHr=-231-0 . 012*(T-298) [5] Cpa = . 1222 [6] k = 1A8e11*exp(-19124fT) [7] Cao = 1 [8] Ca=Cao*(1-X) [ 9 ] ra = -k*CaA 2
0.32
700
0.24
690
016
680 _ ._ " _ ._ _ _ _ _ _ _
670 '----~-------------.o 2 4 V 6 8 10
2
P8-28 (b)
8-87
4
V
6
~_--.l
8
10
Using the same POLYMATH program we were able to change the entering temperature and come up with this graph.
r
Conversion vs temperature
I
1.2.---------...-;..-----..., )(
1· 0 .8
___J
5
eO.6
~0.4
8 0 .2
Or---~--~----~----~-~
o
200
400 600 Temperature, K
800
1000
P8-28 (c) Again using the same POLYMATH program, we can vary the ambient temperature until the reaction runs away. As the following sununary table will show the maximum temperature is 708 K. POLYMA TH Results Calculated values of the DEQ variables Variable -V T X
Fao Ua Ta dHr Cpa k
Cao Ca ra
initial value
-0-----675 0 5 5 708.2 -235.524 0.1222 0 . 0734336 1 1 ·-0.0734336
minimal value
maximal value
0 675 0 5 5 708.2 -·236 . 5196 0.1222 0.0734336 1 0.5137156 -1 . 078707
10 758 . 02032 0 . 4862844 5 5 708.2 -235.524 0.1222 1.6320169 1 1 -0 . 0734336
final value 1()---712.2666 0.4862844 5 5 708.2 -235 . 9712 0.1222 0 .. 3233503 1 0.5137156 -0.0853333
P8-28 (d) When the reaction becomes adiabatic the energy balance will then become:
X-All ) T = 1: + _(-r ..._A_,., __ ..,.L, o
CpA
However, the heat of reaction is a function of temperature. This is a circular reference, so we need to find T as a function of just X, 1~CpA -231X -3.576X
T =._-"'''--_.".,,'''''--_.._,.,'-' ' ',. .-
CpA··O.012X Plugging that into POLYMATH gets the following program and graphs.
8-88
See Polymath program P8-28-d.pol. POLYMA TH Results Calculated values of the DEQ variables initial value 0 0 5 5 708.2 675 0.1222 675 1 1 --235 . 524 0 . 0734336 -0.0734336
Variable V X
Fao Ua Ta To Cpa T Cao Ca dHr
,<
ra
minimal value 0 0 5 5 708 . 2 675 0.1222 618 . 33416 1 0.9695163 -235 . 524 0.0054738 -0 . 0734336
final value 10 0 . 0304837 5 5 708.2 675 0.1222 618.33416 1 0.9695163 -234.84401 0.0054738 -0.0051452
maximal value 10 0 . 0304837 5 5 708 . 2 675 0 . 1222 675 1 1 -234 . 84401 0 . 0734336 -0.0051452
ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(V) = -ra/Fao ' Explicit equations as entered by the user [1] Fao = 5 [2] Ua = 5 [31 Ta=708.2 [4] To = 675 [ 5] Cpa = .1222 [6] T = (To*Cpa-231*X·3 . 576*X)/(Cpa-0.012*X) [7] Cao = 1 [8] Ca = Cao*(1'X) [9] dHr = -231-0012*(T-298) [10] k= 1.48e11*exp(-19124/T) [11] ra = -k*CaI\2
----
0.040
680
0.032
666
0.1)24
652
0 . 016
I
638
Q
0.008
[J
624 "-
0.000
0
2
4
V
6
8
10
610
0
2
4
V
6
S
P8-28 (e) When it becomes reversible with inerts the two equations that change are the rate law and the energy balance.
-r.=\c!-; )
8-89
I
10
We can use those equations in POLYMATH and the following graph is made:
1 ~ 0.8 ! 0.6 ~ 0.4 02
8
Conversion vs. temperature, reversible
~------~~~~----------,
_ _J
O+-----------~~----------~
o
500
1000
Temperature
There is no maximum. because the reaction is a runaway at a certain temperature and the conversion goes to close to one at that point.
P8-28 (I) No solution will be given
P8-29 A+B
The elementary, reversible, gas phase teaCIion
~
2C
Feed: FAO =Fso =20 moVs =1200 mol/min. Po
=580.5 kIno! =5.74 atm.
To = noc =3SOOK YAO=YBO=0.5 CAO
=YAO Po = RTo
Rate
(05) (5.74 atm.) = 0.1 gmo1J1 10.082 1attn ) (3SOOK) \ gtmlOK
conswlt: k = kl exp[f(i1 - ill =0.035 exp [7~3~ (2j3 -i)] :: 0.035 exp [8419.5
(2i3 -il]
(1)
8-90
Equilibrium constant: Mia (Ta) =2Hc - HA - Hs =2(45.000) + 40,000 + 30.000 =-20.000 J/mole
= 2C?c - CpA - c;,. =2(20) - 25 -IS =0
ac;,
SO .MiR
en = constant = -20,000 l/mole
(2)
Kc=Kcl exp[~R{*_+)1 ICc =25,000 exp [-i~3~ (2~8 -
Kc = 25,000 k and
i)]
exp [-2405.6 (~- i)]
(3)
Kc can be calc:ulatcd by equation (1) and (3) ifT is given.
Kc=~ CACB CA=CA«)(l-X)f wheref=]:;}. T
~=CAQ(l-X)f
Cc=2CAQXf Substitute CA. Ca. and Cc into Kc
Kc =
4cio X2 f2
cio (1 • X)2 f2
=
4Xl (1 - X)2
(4) (5)
-JKc
Calculate as a function of temperature from equation (3)~ substitute in equation (5) to get Xe.g. as a function ofT_ Energy balance for adiabatic condition(6)
9A
=9B =
1 • 9c = 0 • To = 3SOK
Substitute: + 20,OOOX = (25 + 15) (T - 350)
or X
=
T = 500X + 350
=0.002 (T - 350)
(7)
Equations (3) = (4): 4X2 = 25000 (1 - X)2
exp [-2405•6 (....L _ 1 )~ 298 350 + SOOX ~ 8-91
Xeq = 0.8667 0.85Xeq =0.7366
=0.7
(a) Plug flow reactor design equation: Rate law:
orA =
(8)
Ic[CACa 4~1
-rA = Ie clAO f2 [(1 -X-y. -
(~~
(9)
(10)
To evaluate the integral, we need to evaluate fOO as a function of X. This is done in the table below:
I!K
Kc
k
f.£Xl
(eqn.7)
(eqn. 3)
(eqn. 1)
(cqn.9)
350
7534.8 3191.1
30.96 626.• 1
1635.9 958.5
6491.5 4.216 x 1()4
0.0323 2.58 x 10-3 3.98 x 10-4
618.9 429.9 315.8 192.8
1.949 x lOS
A 0 0.1
400 450 500 550
0.2 0.3 0.4
600
0.5 0.6 0.7
650 700
6.978 x lOS 2.054 x 1()6 1.155 x 107
9.89 x 10-5 3.53 x 1.70 x 1.08 x 4.98 x
10-5 10-5 10-5 1(}-6
The integral can be evaluated using Simpson's rule for the fust six segmentS and trapeziodal rule for the last segment.
1=[ f(X) dX '"
or-
+ 2(3.53 x 10..5) +
[0.0323 + 4(2.58 x
4( 1.70 x
10-') + 2(3.98 x 10-4) + 4(9.89 x 10.5)
10..5) + 1.08 x 1O.S] +
1- L5x 10-3 VPFR
= (1.2
x
HP)(1.5 x 10-3)= 180L 8-92
DzL[ 1.08 x 1O's + 4.98
X
10-6]
,Exothermic. adiabatic
T ~-------X
distance down the reactor
(e) CSTR. design equation:
v=
V _ FAO XA -rA
FAO X
kcio f1 [(l-Xf -ifC] X
_ =
(mol] 1 - 001"5 1200:: - . -
1500L X (0.1)2
kf2 [(l-Xf-~J
{l-xf· ~ = 80...x...
Kc
f2k
1-2X + X2 • 4X2 = .8.QX.
Kc
f 2k
(1. ic)X2-{2+ f~Qk)X+ 1 =0 Let bi
~
= 1 - -.!...and i>2 =2 +...&L 2 Kc
X =
f k
(11)
b2±.J~ - 4b} 2b2
(12)
Nonadiabaric energy balance:
-lO--YL. x 2 m2 K x 1200
mm.
lkW Ihr 3.6 x 1()6 J lOOOW x 60min x lKW.hr (T - 290) + 20000X
=40(T - 350)
min
-T + 290 + 20.000X =40T - 14,000
T = 348.5 + 487.8 X
(13)
8-93
So the procedure to calcularc X is as follows: 1. Choose increments in T and calculate X as a function of T from equation (13). This is the value given by energy balance. 2. Choose increments in T and calculate bl and b2 from equation (11). 1 or X < O. This is the value of X given by material balance.
4. Plot X vs. T given by equation (13) and (11) on the same graph. The intcrSeC1ion'lives the conversion in the reactor. A typical graph looks like the fonowing: "
x
Operating paint
The aaual calculation gives: X - 0 which is the conversion in CSTR. d) The same equations can be used except that ~ = 20000 and To following graph shows the equilibrium. conversion for this case.
=SSOK. The
Conversion vs temperature 1 ·r-------------~----------_=======I go.s·
. :!O.6 !
•
~O.4
gO.2
o ~--------------~~----~------~------~ o
50
100
150
Temperature
200
250 '
The following POLYMATH program gives the PFR. volume necessary to get a
conversion of .65. See Polymath program P8··29.pol. POL YMA TH Results Calculated values of the DEQ variables Variable V X
Fao T
initial value 0 0 1200 550
minimal value 0 0 1200 225.22683
maximal value 1. OE+08 0.6495463 1200 550
8-94
final value 1.0E+08 0.6495463 1200 225.22683
0 . 064 550 1 . 949E+05 1 . 01E+06 1 0 . 064 0.064 0 -798.17344
Cao To k
Kc f
Ca Cb Cc ra
0 . 064 550 5.047E-05 1841 . 4832 1 0 . 0547713 0.0547713 0 -798 . 17344
0 . 064 550 1 . 949E+05 1 . 01E+06 2 . 4419826 0 . 064 0 . 064 0.2030311 -1 . 503E-07
0 . 064 550 5 . 047E-05 1841 . 4832 2.4419826 0.0547713 0.0547713 0 . 2030311 -1 . 503E-07
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(X)/d(V) = -ra/Fao Explicit equations as entered by the user [1] Fao = 1200 [2] T = -500*X+550 [3] Cao= . 064 [4] To = 550 [5] k = . 035*exp(8419 . 5*(1/273-1fT» [6] Kc = 25000*exp(2405.6*(1/298-1fT» [7] f = TofT [ 8 j Ca = Cao*(1-X)*f [9] Cb=Cao*(1-X)*f [ 10 1 Cc = 2*Cao*X*f [11] ra = -k*(Ca*Cb-Ccl\21Kc)
The CSTR conversion can be found similarly to the equilibrium conversion. The following graph was made to find the conversion. Conversion vs temperature
=:j
. 8f ~0 0 . 6 r-------------_ ~
:0.4'
::==::::_
8o.~. 230
280
330
380
Temperature
The graph shows a conversion of .39 at a temperature of 351.8K e) If the reaction is left in a large enough volume it will "tuIlaway·. If any of the quantities get bigger. then it will run away even faster.
f) The ambient temperature around the CSTR has little effect on the conversion in theCSTR.
P8-30 (a) P8-6 is adiabatic so the radial reactor has no effect on it. P8-30 (b)
8-95
dX
-r
dW
~o
_=--:1..
-rA =kCA CA
1-- XL = C".40 ·---·.Q +X T.'V 1
-a
dy dW
---=""--
dT
_
-
2y
u(r)~(TA ........
T)+ (-rA)(-MR ) ._-
,-~--.-
._.,
dW
Since ~ and p are unknown, we will assume that they are both equal to one. We wIll also assume that r varies asW 5 varies. so: dT _ U(r)(~~ - T) + (-,:~)(-L\HR) -------~-.:..-~~-~ dW F"oCiM
- -
w..s )'S( ToT)S
U = U{lQ);' (
See Polymath program pg·JO·b.pol. POLYMA Til Results Calculated values of the DEQ variables Variable W Y
X T Fao alpha Ta dHr1 WO To E R
Cao Xl y1 U
k
initial value 0 1 0 450 5 0,.007 300 -·2.0E+04 0,,01 450 3.14E+04 8,,314 0.2498645 0 1000 5"OE+04 1
minimal value 0 0.8062258 0 450 5 0.007 300 -2.0E+04 0.01 450 3,,14E+04 8.314 0.2498645 0 806,,22577 0.9244844 1
maximal value 50 1 0.7845538 756.08452 5 0.007 300 -2.0E+04 0,,01 450 3.14E+04 8.314 0.2498645 784.55376 1000 5"OE+04 29.893648
8-96
final value 50 0.8062258 0.7845538 756.08452 5 0.007 300 -2.0E+04 0.01 450 3.14E+04 8.314 0.2498645 784.55376 806,,22577 0.9244844 29.893648
Ca Cc Cb Kc ra
0 . 2498645 0 0 . 2498645 1635 . 8981 -0 . 0624323
0.2498645 0.2333455 0 . 2498645 1635 . 8981 -0 . 0220235
0 . 0320395 0 0 . 0320395 187.88715 -0 . 1234947
0.0320395 0.2333455 0 . 0320395 187.88715 -0 . 0220235
ODE Report (RKF45) Differential equations as entered by the user [1] d(y)/d(W) = -alphal(2*y) [2] d(X)/d(W) = -ralFao [ 3] d(T)/d(W) = (U*(Ta-T)+(-ra)*(-dHr1 »/(Fao*40) Explicit equations as entered by the user [1] Fao=5 [2] alpha = .007 [3J Ta= 300 [ 4 ] dHr1 = -20000 [5] Wo = . 01 [6] To=450 [7 J E = 31400 [8] R=8.314 [9] Cao = 9 . 22/(..082*450) [10] X1 = 1000*X [11] y1 = 1000*y [12] U = 5*(Wol\.5/(WI\.5+ . 000000001 »1\.5*(TlTo)I\ . 5*yl\ . 5*(1 +X)I\.5 [13] k = exp((E/R)*(1/450-1ff) [14] Ca = Cao*(1-X)*TolT [15] Cc = 2*Cao*X*TolT [16] Cb = Ca [17] Kc = 25000*exp((dHr1/R)*(1/298-11T» [18] ra = -k*(Ca*Cb-CcI\2/Kc) 0.80
800
0.64
720
0.48
64U
Q
0.32
20
W 30
·to
[2J
560
400
'---~-.--~----------I
o
P8-30 (C)
8-97
10
20
W 30
40
50
= p v = p7tr2h
W
=
dW
= 2n:prhdr
Material balance:
Taking the limit as tJ.r ~ 0
~ FA
=
FAO (l-XA)
=
:: (rA) 21tprh
ar = A
-FAO
~"
. dXA _ -2milp (rA) FAO
(1)
so. dr
or: dXA :: _...!A..
. dW
(2)
FAO
Assuming ptessure drop is negligible. The rate equation is:
-rA = k(~)l/2 [P(h _{ KpPS(h Pso, )2] Pso, The stoichiometry :
A+.LB=C 2
o=
-0.5, YAO
PsQz = PsOu Then:.r
= 0.11 :. £ = YAOO = -0.055 ; ec = 0 ; 1- XA ; Pso, 1 +£XA
~k(t)(l-XA)lJ2[p
A
XA
SOu
= PsOu
XA 1 +£XA
; Paz
Sa
= ¥t = 0.91
A) = Pso,.o (O.91-0.5X ~--..;..;.;..;;..;.:.::. 1 +£XA
(O.91-0.SX A )_ xl. 1 1-O.05SXA (l.XJ K~l
(3)
Equation (3) is aue for X A < 0.05 for X A ~ 0.05 - fA
- fA
= k(t) (4.3S>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-<:oncentration phase plane. These 31'e the four graphs.
9-2
~:,lCOT
19'5_OOOi KEY:
-T
'~-t ;::I5.oo0.-!-
'~-t
~.
~--- - - ..
~-----+I----------'i~·-------+.--~--~t
75.000
c..:~
~•
0.1iI'tlO
0.000
.soo
v.¢Q
:l..zou
'1 . oot).
T\ T ..!..
.:), ~2'C
T
KEY:
-Ca
0.::190
!
Q.co;a
1
;
t
o...~JO
Q.ooa
t
---~'-'---+--'--.--";os-coo l~QOO l::J~.QOQ tiS;!i.COO ..,..~
2:2tl. to"Uu~·
Startup of a CSTR
!
1\
1'--!
'--i
~~.ac:
Tl =40)'
To;120~
.._.- -.. . . .--.----.. -.. . . . . . . . .
~-----.---
-
..
15':>.000
~---"--
....
f
1.0.0CO+ };
KEY:
-T
t H,0.::I0011 60.000
-¥ r
zo..,oco
-1. .--------+,-= . __ . . -.. ..".,.,.,..-+--....-~. ~ . .....,..-------t---.,...~"--Q.ooo O. !lOa :. q:a::I :2. "\00 loa ".000 ::I.
9.. 3
a CSTR To=120 Tl;40 Cal=.2 (L;;60
t· 1
t/~'y. q.):;!O
_"'_1:._::._
-Ca
T
P9-2 (0 Example 9-6 Using the code flom Example 9-5, we could produce the following graphs either by changing To and finding the steady state conversion or changing the coolant flow rate and finding the steady state conversion and temperatme. These are the graphs of those:
Conversion 'IS To 0,,9
------,-".,"',"---"--"'-,-
. ...--........................._ _ .............
0.8 0.'7
c o 0.6
f
(1)
0.5
:> 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<:c:=1
(i.90c
t
P9-2 (i) Example RE9-2 Using the code from Example RE9-2 ,we can change the values of kc and '1:[ and find values that produce the lowest oscillations and the quickest return to steady~ state and getting kc =:; 150 and L J ;;;: • L The following graph shows the result.
.....3.:1:0(1 -+--.. t
P9-2 (j)
No solution will be given
P9-3 9-6
Find time to explosion
For the unsteady~su[e. we assume the following: (i) The inlet is closed. but the ourlet is not closed (ii) Operation is adiabatic, and PV terms are negligible so that H "'" U
Equarion 8-61 reduces to: (-611R) (-fA V}= NAC'-PA
c;J
(-6HR) (kNA)": NACpA ~ ill. -; AHlhl'1 d! Cp".
k(T)
k ::: .53 exp (44499 (9jO -
+)} / 60 min.
See Polymath program P9-3.pol POLYMA TH Results Calculated values of the DEQ variables Variable t
initial value -0----
T
dH Cpa k
970 -336 0 . 38 0.53
minimal value 0 970 -336 0.38 0 . 53
maximal value 2.9 1.119E+23 -336 0.38 4.438E+19
ODE Report (RKF45) Differential equations as entered by the user [lJ d(T)/d(t) = -dH*k/Cpa Explicit equations as entered by the user [1 J dH =-336 [2] Cpa = .38 [3] k = .53*exp(44498*(1/970-1/T»
P9-4 Mole balance:
9-7
final value ---2.9 1.119E+23 -336 0.38 4.438E+19
There is no A leaving the reactor so F" :::; O. 111ere is no Bore entering the reactor so Fao and FQ) ::: 0.. The amount of B and C leaving is equal to the reaction rate so Fa::: VIa and Fe::: VIc' Simplifying:
4:~!l.. == 0 dt
rf!!c
:=
0
dt Rate law:
• Stoichiometry:
Simplifying:
Energy balance:
Evaluate the parameteIS:
InCji )=~(8(~j' .. ~£~) E::: 3467
k = .19 exp
[i1i: (4}m . . -})1
tVi(400) ::8kjlmol = LVf(8(X)) = till (1 200) = MI(T) Assume a heat capacity for A of 20 J/molK and assume there is no A already in the reactor. We can plug these equations into POLYMATH and get this answer See Polymath program P94 .pol
POLYMATH Results No Title 08-11-2005, RevS I 233 Calculated values of the DEQ variables Variable -t Na T
Fao
dHr Cpa
initial value
-0-----
1.0E-09 400 100 -8000 20
minimal value
o
1.. OE-09 400 100 -8000 20
maximal value 100 326,,10947 801.. 376 100 -8000 20
9-8
final value 100 312 . 51175 800 100 -8000 20
To Vra
400 0.3202745 -1.9E-10
400 0.19 -101.76832
400 0.19 -1. 9E-10
k
400 0.319988 -100
ODE Report (STIFF) Differential equations as entered by the user [1] d(Na)/d(t) = Fao+Vra [2] d(T)/d(t) = «Vra*dHr)-Fao*Cpa*(T-To»/(Na*Cpa) Explicit equations as entered by the user [1] Fao = 100 [2] dHr = ··8000 [3] Cpa = 20 [4] To= 400 [5] k = . 19*exp(3467/8 . 314*(1/400-1/T» [ 6] Vra = -k*Na
NA
=326.125
When the flow is turned off, F AO :::: 0 and we get this graph of T and N.... 2000
1.0e··9
1600
Q
1200
6.0e-10
800
·t.Oe-10
400
2.0e.. l0
0
0
2
4
t
6
8
O.Oe+O
10
(I
2
4
t
8
6
10
P9-5 (a) Mole balance : dNa --_. = ra.V dt
dNb - - - := fa.
dt
dNe
. . . . .---- =
Vr Fha
Rate law:
-ra:: k(T).Ca.Cb
Stoichiometry :
Ca::: .t:!a
dt
Cb
V
Nb
:=--....
V
IF (t<50) nffiN (V:;;;; Yo + u.t) ELSE (V=Vo + 50) dm31miu
9-9
-Ta. V
Ne_. Cc:: .-... V
where Yo:;;:; 50 dm3 , '\) = 1
Combining:
Na Nb
-- ra ;:;: k (T) . ------
V V
k(T);:;:
ko.exp[-~(.L -"'~)\l . R To
where E::: 10000 cal/mol
T_.
R == 1.987 callmoLK,
ko ;:;: 0.01 drn3/mol.min , To= 300 K
Energy balance: dT _ Q. WS d~-
'LFio.Lpi.(T-Tio)+(--IlHTX(T)).(--ra.V)
- ----····-·----··--·----··!Ni ..(;pT--.. -..- - - - · -
where Q=Ws=O AHn: (f) == MIrx(fref) + ACp(T-Tref) ACp = !:." Cp -"~'" C'pb·· Cpa a a
= 30 ·······15 . ·····15"'" 0
Mlrx(273);;:;; [(-41)-(-20}{15)]. lO J =6000 cal/mol
ilHrx(f) :::: -6000 cal/mol
2:Fio.Cpi.(T-Tio) =: fl)o.(15).(T-323) LNLCpi:::::: NaCpat· f'.c'b.Cpb -+- Nc.Cpc = 15Na + 15Nh + 30Nc See Polymath program P9-5-a.pol £OLYM~:rI!.Resul!§
08-11-2005, RevS. 1 233
Calculated values of the DEQ variables Va:r:iable -t Na Nb Nc
T X
Fbo Nao Cbo k
vo V ra
initial value 0 500 0 0 298 0 10 500 10 0,,0089352 1 50 0
minimal value 0 0.1396707 0 0 298 0 0 500 10 0 . 0089352 0 50 -0.3365379
maximal value 120 500 42.43357 499.86033 510.44411 0.9997207 10 500 10 10 . 085112 1 100 0
ODE Report (RKF45)
9-·10
final value 120 0.1396707 0.1397745 499,,86033 510.44411 0,,9997207 0 500 10 10" 085112 0 100 -1.969E--05
Differential equations as entered by the user [1] d(Na)/d(t) = ra*V [2] d(Nb)/d(t) = ra*V+Fbo [3] d(Nc)/d(t) = -ra*V [41 d(T)/d(t) = «6000*(-ra*V»-(Fbo*1S*(T-323)))/(1S*Na+ 1S*Nb+30*Nc) [5] d(X)/d(t) = -ra*V/Nao Explicit equations as entered by the user [1 J Fbo = if(kSO)then(10)else(0) [2J Nao = SOO [3] Cbo = 10 [4 J k = .01 *exp«10000/1 . 987)*(1/300-1/T» [ 5 J vo = Fbo/Cbo [6 J V = if(kSO)then(SO+(vo*t»else(100) [7 J ra = -k*Na*Nb/(VI\2) 600.------
1.0 0.8
Q
0.6 04 280 200
0.2
o
24
48
t
96
72
24
120
48
t
72
96
120
P9-5 (b) This is the same as part (a) except the energy balance .
Energy balance : dT _ Q_.. Ws -
L Fio. Cpi.(T ... Tio)--::-...+ (--Mirx(T)), (-ra. V) -.-.-.....-..-""-..--..
-;j;" ---." ..-.--.-... ...............
,,·~~-~-C·
.£..; lVZ.
where Q :::: UA(Ta . T)
VA;;:: 100, 'fa::::: 323
See Polymath program 1'95 b .pol
9-11
pI.
-~-
1.0
r - - - - . -____---~.~.........- .......".............- .....--."-....... ..
500 r - - - - - - - - - - - - - - ,
0. 8
-t-t0'
0,6
380
0.0'1
320
0,2
260· 200
36
72
t
108
I-t4
L...-_~
o
180
_ _~_ _ _ _ _ ' _ _ _ - - l
36
72
t
108
144
180
P9-5 (C) This is the same as part (b) except the reaction is now reversible
Rate law:
-ra:::: kI(1').Ca.Cb·· k2(T).Cc
Combining:
'·'ra::::
kl(T){~'.'~'}' k2 C D.( i-)
k2(T)
= koexp[ ~{};, -
¥)]
where E:::: 16(X)(), R::;:; 1.987 ,
ko :::: 10 , To ::;: 300 See Polymath pIOgram P9··5-cIJol 0.9 ..
500
--,...,.,...---.,,~.---------
-260
72
t
108
H-t
180
200
o
36
72
t
108
P9-6 (a) Mol balance: Rate law: dNa
dNa ···········'·;:;;;;raV dr . . ra:::: k .. Ca
But Ca.V:::: Na
-_._-;:;;;; -kNa dt
9-12
14-t
180
Na In--;;;;-,k.t Nao
Na:::: Nao.e·kt Energy balance :
where Nao:::: 0,5 x 50 == 25
dT, ::;:; Q- Ws- LFio.Cpi(T - Tio) + (---Mlrx)(-ra.v)- :::: 0 "._"""--",-,-.,-,-,,,, ..,,.,,,,-,,,,,-.,-,--.. dt LNi.Cpi constant T
Q=Ws=O
LFi.Cpi(T - Tio):::: (-AHrx)(-ra.V) FCQ,Cpc.(T- Tio)::;:; (-Mfrx)(k.Na)
:::: ("L~Hrx)(k.Nao.ekt )
(-Mlrx).(kNao.e b ) Feo =.-.'-...-'----~-"''''''-Cpc.(T - Tio)
::;:; .~SOO'9,~ (WOO ~ 2 x ~?,,~::!
:::: 3.16 Ibis
P9-6 (b)
Av ;;; l000Btu Ilb Nao::;:; 25 Ib/mol
Vo = 300 ft3 of which 250 if is solvent dT ·-=0
dt
Fs.
Av = ::;:; (-Lllh'x) (-fa.V)
(-··AHrx). (kNao.e- b F s ;;; _.,_.".__.......
)
,_,0""""""" .. ", •• _",-,,,
25000 x 0.00012 x 25 x exp( -.().OOO12 x 2 x 36{)O) ::;:; _
.. ,·..
_--"_··,,--'_·""""'''''·'''''''''1000--'··'--,·_..·. . ._-_·."""""""""'-.-".-.'.'.
/I." Fs ::;:; 0.0316 Ibis
P9-7 9-13
Batch problem dX
Nao~---
Mol balance:
' dX C00--':::: '-Ta dt
= ··ra V
dt
Rate law:' Ca ~ Cao(lX) :::: 0 . 1(l-X)
Stoichiometry :
.t1(D =
Cb
:=
Cao( l25-X) :::: O.1(1.2S-X)
Cc
=:
Cao(O+X) :::: 0 IX
ko.expl'~(:/~' -" '~J]
E =: 100000 llmoi., R:::: 8.314 J/moLK
where
ko ;: ; : 0.002/$ > To;::;;: 373 K E::= 150000 J/mol., R == 8.314 J/mol.K,
ko :::: O.O
where INi.Cpi L":.C"p Y' 7
=:
=:
Nao(I8iCpi + L\CpX)
.,~." Cpc -.,.~" Cpb -, Cpa :;: 40 a
a
25 "". 25 = "'·10
Cbo, b Ceo Cpt ="Cao .. '-, ('_pa +-.---. Cpt--·· __·, Cpc Cao Cao Cao
Q:"""
01..
~
= 25 +- 0 125/01 25 + 0 =56..25 L\Hrx(f) = .6Hrx(Tref) + L\Cp(T-Tlef) ::: '." 40000", lO,.(1'298)
Nao
V=="""""",· Cao
dT
subs
NaO') [ 40000 + 10" (T .,., 298)] ,,' ( ra, ,.-:;--..
""'' ' ' ' ' :::: '''' ,."",,",,. ,,,._._... "., ..._..."" .... __ ..._. __......______"....,..Cao.:.... dr Nao(56.25 """ lOX)
9-14
>
(-ora) r40000 + 10. (T -·298))...·· l
dT
.". . , " . . = . dt
cancelling
._.__."".""".."",,". . . . . ,,_.. 0.1 __..
(56.25··· lOX)
See Polymath program P9-7.pol POLYMA Tn Results Calculated values of the DEQ variables minimal value
initial value
Variable t
o o
°°
x
373 0.002 0.0749517 0 . 0999517 3.0E-05
373 0.002 0. 1 0.125 3.0E-05
T
k1 Ca Cb k2
o
o
Cc ra
-·0.1344598
-2.236E-04
final value
maximal value 10 0.2504829 562.91803 106.13627 0.1 0.125 366.75159 0.0250483 8.644E-06
~--
0 . 2504829 562.91802 106.13612 0.0749517 0.0999517 366.75083 0.0250483 1.3E-07
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(X)/d(t) = -ral . 1 [2 J d(T)/d(t) = «40000+(1 0*(T-298)))*(-ra)*(1/.1 »/(56 . 25-(10*X» Explicit equations as entered by the user [1] k1 = .002*exp«100000/8.314)*(1/373-1/T» [2] Ca = .1*(1-X) [3] Cb = .1 *(1 . 25-X) [4 J k2 = . 00003*exp«150000/8 . 314)*(1/373-1/T» [5] Cc = .1*X [ 6] ra = -«k1 *(Cal\.5)*(Cbl\.5»-(k2*Cc»
----
600
0.15,..-----·---· - - -
[J
480
0.09
300
DD - eb
0.06
420 360
r-----_'. . -. .-
0 . 12 - - . -.. -..- - - - -..- - -
540
-""'- ....-.... 0
-
c, ,
...-"'"
2
Check answer:
0.03
4
8
6
Cc .~ Ca JCi;
-
10
0.00
2
0
kl
-'ki
0.02504 'JrE'0"?07""'4:=::9;;:::;;S';;..JO.09995
_",_',',
,~
tOO
=3.6~j == 0.289
9-15
4
t
~=~
..<¥=-~¥M
6
8
_J
10
P9-8 (a) Use Polymath to solve the differential equations developed from the unsteady state heat and mass balances. See Polymath program P9-8·a.pol
POLYMATH Results No Title 08·11-2005. Rev5 1233 Calculated values of the DEQ variables Variable t Ce Cs T
Iprime KIn
mu1max Yes mu Q Cps Hrxn V
rho rg Cpe rs
initial value 0 0,,1 300 278 0.0866522 5 0.5 0,,8 0.0426159 0 74 -2"OE+04 25 1000 0,,0042616 74 -0.005327
minimal value 0 0,,1 43.080524 278 6.499E-04 5 0.5 0,,8 2" 912E·-04 0 74 -2"OE+04 25 1000 0.0042616 74 -30 . 908876
maximal value 300 205.63558 300 333.55016 0,,3593146 5 0.5 0. 8 0" 1725264 0 74 -2"OE+04 25 1000 24.727101 74 -0.005327
ODE Report (STIFF) Differential equations as entered by the user [ 1 J d(Cc)/d(t) = rg [2 J d(Cs)/d(t) = -rg!Ycs [ 3] d(T)/d(t) = (O+(-Hrxn)*(rg))/(rho*Cps) Explicit equations as entered by the user [1] Iprime = (0,,0038*T*exp(21.6-6700/T))/(1 +exp(153-48000fT)) [2] Km =5.0 [3] mu1max= 0,,5 [4] Ycs=0 . 8 [5 J mu = mu1 max*lprime*(Cs/(Km+Cs)) [6] 0=0 [7J Cps = 74 [8] Hrxn = -20000 [9J V = 25 [10] rho = 1000 [111 rg = mu*Cc [12] Cpc = 74 [13] rs=-rg!Ycs
9-16
final value 300 205.63558 43.080524 333,,55016 6.499E-04 5 0.5 0.8 2,,912E-04 0 74 -2.0E+04 25 1000 0,,0598648 74 -0 . 074831
340.--------------------------,
300 ...
326
240
312
180
298
120
284
60
60
120 t
180
240
300
o
--_=-----------,
~
W
---'-'o
2-40
60
300
0.40..---------------------------,
40 .--------------------------,
0 ..32
32
0.24
~
~l
24
0.16·
16
0.08\,..·----
8
0.000 -·---6~0----1-2(~)::'t-I 80
240
300
LJ
oo
----~~--
60
120 t
..-.--.- .•......
180
240
300
P9-8 (b) When we change the initial temperature we find that the outlet concentration of species C has a maximum at To = 300 80
---------.--
64
48 32 16
o 280
300 To 310
320
330
P9-8 (C) Cc can be maximized with respect to To (inlet temp), Ta (coolantiheating temperature), and heat exchanger area. Therefore, if we are to find the optimal heat exchanger area the inlet temperature and coolantiheating temperature needs to be specified. If we take To = 310 and T a = 290 we find that the optimal heat exchanger area is an infinite amount of area . As A increases Cc increase without a maximum in 24 hows. Cc = 118..
9-17
P9-9 First order liquid phase, CSTR
First solve the steady state problem fin the heat exchange area A for notmal operation T= 358 K. Co
.- ......... J:\
Mol balance:
1:
V :::::-.dCs __ ... =: 0 dt
In(}1·)
== ..-·-·-··l····--..·. ·-'~·-···-r . -·-··--......•- ..........................._..... _---
R.T2
8.314(323)
R.n
8.314(313)
94852 ::.: JI Inol K k
= Llexp[ ?~~??(_l 8314
.......
313
l)t"l07A Imin J
T
Steady state solution:
CAo =2 M::;: 2*90 "" 180 g/dm'
C A :;;;; 180 107.4 (0.4) CA 180 H (0.4* 101.4)
9-18
for T;;;:.; 358 K
CB :: -
Energy balance:
( -
440) 04 = 175 . 9 gldm3
Q.- Ws- LFio.Cpi(T····Tio)+(-AHrx)(-··ra.V) .dT . . .-=----. . . . . . . .- . . . . .-.. .-.-.-.--.. . . . . . . . . . . . . ._. . . . -.. . . . . . . . . . . . __.... == 0 LNi.Cpi
dt
Q == UA(fa - T) =(120*60)A(273-1) FAOCpA.(f - TAO)
=90000*2*(T - 313)
J/min J/min
A :::: .-180000.(358··313) . . . . . _. . . . . _. . . . . . ._.•.--_. . -.. . .-..- + -(250)( - - .44Ox200) - . - - = 22.7m2 7200(273 - 358)
Use the unsteady state equations to detennine what heat exchange area A will give a l1maway reaction. where
I
Ni. Cpi = Cpsol Psol V =2 x 900 x 200 =360000 11K
See Polymath program P9-9 . pol
POLYMATH Results Calculated values of the DEQ variables Variable t Ca Cb T
tau A k
V ra Na Nb
initial value 0 180 0 313 0.4 22.696 1..1 200 -198 0. 9 0
minimal value 0 4.1007565 0 313 0.4 22.696 1..1 200 -439 . 75367 0.0205038 0
maximal value 10 180 175 . 89924 357.97719 0.4 22.696 107.23721 200 -198 0.9 0.8794962
final value 10 4.1007565 175.89924 357.97719 0.4 22.696 107 . 23721 200 -439.75367 0.0205038 0.8794962
ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(t) = «180-Ca)/tau)+ra [2] d(Cb)/d(t) = -ra-(Cb/tau) [ 3] d(T)/d(t) = «7200*A*(273·T»-(90000*2*(T-313) )+( (-250)*ra*V) )/(2*900*200) Explicit equations as entered by the user [1 j tau =.4 [2] A = 22 . 696 [3] k=1.1*exp(11409*(1/313-1/T» [4J V = 200 [ 5 1 ra = -k*Ca [6] Na = Ca/V [7] Nb = CbN
9-19
360,-----=======1
310 OI---2~--4--t-6---·8~--l10
A sensitivity analysis for different values of A show that a new steady state is reached at every new A but at an increasingly higher T. The required A depends on the definition of reactor runaway.
P9-10 (a) X
=L 8;CPi [(T - To )/(-Illi Rx)]
(1)
X=1 7 T=T f gives
P9-10 (b) CA
= CAO (1- X)
(2)
Insert (1) in (2)
(3)
For x= 1, T f =T expression (l) becomes
(4) Insert the inverse of (4) in (3)
9-20
(5) Remember that C BO is not constant here. Similar to C A, use equations (1) and (4)
IeicPi
CB
= CBO -- CAO • (_ Ml Rx) . (T - To) = CBO -
C
[eB ·T! -T + (To -eB 'To)]
(T-To) _ CAO • (Tj _ To) - CBO
•
e
B •
(T-To) (T-To )_ (Tj _ To) - CAO • (Tj - To) -
T -T,
AO
f
0
For developing -rA use
(6)
8
C =C B
AO
[
B
. T f- O T + (T, B - 8 . T, )] 0_ T -T, f
(7)
0
(8)
Insert (7) and (8) in (9)
P9-10 (C) The unsteady energy balance for a batch reactor is showed in equation 9-11.
dT
dt'=
Q+Ws + (-Ml Rx )· (-rA .V) INtCPi
Adiabatic operation (
.
Q=0), neglecting W, and assuming ~Cp=O gives the following using the expression for -rA:
9·-21
P9-10 (d) P= 1, E>B = 3 in the equation from part (c)
Put a = 1,
P9-10 (e) Use the reaction constant expression from the great Swedish chemist Arrhenius and develop expression from 9-lOd.
dT [(--MI ---= - - " -)] Rx
dt
IepPi
.
k·CAO (Tj
-T,)
2 [(
Tj -T ) . ( 3·Tf -T+ ( Yo -3·Yo ) ) ] ~
0
P9-10 (0 First use a plot ofT vs . t to get To (329 K) and Tf (439 K).. Checking the concentration of species B gives E>B=3. Make a table showing t, T, dTldt, left hand side of equation P9-10 ..7 and liT.. Plot left hand side of equation P9-10.7 vs. liT in Polymath and use linear regression to get E from the slope and kJ from the intercept. Regression equation as shown in Polymath:
y = aO+ al· T _inverse =-8944.2· T _inverse + 17.534 Activation energy and Arrhenius constant from slope and intercept:
slope =
-% => E = -8944..2·8.314 = -74. 36
kJ/mol
9-22
~ ·C J+~' = In(kl .6.7) => kl = 12100 ·exp T-T) R·Yo 12100 6.7 AO
aD =17.534 = In(
(
f
2
0
17534
= 1.298.1011
L-.,..--'
neglect
P9-10 (g) Follow the procedme from example 9-3 gives heat of reaction. X=l and T=T f gives
-MIRx
= I8 CPi .(T-To)=(CPA +8 B .CPBXTj-To) j
=(189.7+3.75.4. J )(439-329K)=-45.75kJlmO[ mo[·K
P9-11 (a) Mol balance:
. dNa '..' ...•. ::;: ra.V dt
dX
Nao--·-·· ='raV dt
dNb
.__.-.. :::;: 2. Ta. V + Fha dt
Fba::;;: CbO.1)o=
dNc
. . --:::: -raY dt
Rate law:
Stoichiometry: .,
- ra == k.Ca.Cb
4.1)0
V==Vo +1)o .t 2
CQ="~c:..
Ne C.C = ". . . . _. O
V
V
Energy balance:
\\ hc:re Q:= UA.(Ta - T) 'vis::::: 0
= 250.(390 - T)
t~Hrx(T)
::::: -55000
2:Fio.Cpi ;;;; Fbo.Cpb :::: Cbo. Do.Cpb = 4*1)0*20:::: SODa 9-23
dT 250.(290 ----::: ---_ ..
T) -80uu.{T -- 32S)~· (-55000). (-ra. V)
.. _.....,..",-_ ........"""""'".
- .- .
~,.,..,...,.......-.-----
35. Na + 20. Nb +- 75 ,vc
See Polymath program P9-11-a.pol POLYMA TH Results Calculated values of the DEQ variables Variable t Na Nb Nc X T
vb k
V Fbo Ca Cb Cc ra
initial value 0 50 0 0 0 300 1.5 5.0E-04 10 6 5 0 0 0
minimal value 0 0 . 3324469 0 0 0 283 . 60209 1.5 2.3E-04 10 6 2.202E-04 0 0 -0.0028611
maximal value 1000 50 5900 . 6649 49.667553 0 . 9933511 300 1.5 5.0E-04 1510 6 5 3.9077251 0 . 0718765 0
final value 1000 0 . 3324469 5900.6649 49.667553 0.9933511 298.76383 1.5 4.73E-04 1510 6 2.202E-04 3 . 9077251 0 . 0328924 -1. 59E-06
ODE Report (RKF45) Differential equations as entered by the user [ 1 j d(Na)/d(t):::; ra*V [2 J d(Nb)/d(t):::; 2*ra*V+Fbo [3 J d(Nc)/d(t):::; -ra*V [ 4 J d(X)/d(t):::; -ra*V/50 [5 J d(T)/d(t):::; ((250*(290-T)) .. (SO*vb*(T·325))+(-55000*(-ra*V)))/(35*Na+2O*Nb+75*Nc) Explicit equations as entered by the user [1J vb:::; 1.5 [2 J k:::; . 0005*exp((SOOO/1 . 9S7)*(1/300-1/T)) [3] V:::; 1O+(vb*t) [4] Fbo:::; 4*vb [5] Ca:::; Na/V [6] Cb:::; NbN [7] Cc:::; Nc/V [8 J ra:::; -k*Ca*CbI\2
-l []
5
3
2
-..- - - -
296 292 288 28-t
1
o0
300
200
-ton t
600
800
1000
281.1
'----~---~---~--~---'
o
9-24
200
400 t
600
800
1000
1.0,------------==""'" 0,8
0.6 0.4 0.2
200
400 t
600
800
1000
T"ble ! Iterating \\ It Uo for X:::O.8 and T<403K and daily Nc " 110 mole
--'''·''·''-'-1. ~()
T
dm'/min ;
f"-·"'-·--_····_·,·"·,·",,·,,··,,_·,"-_··_'-""_·,,,,·, .••-"-", .._.",,Nc (X.::::Q 8) Daily Nc
(X=O.8) t (X:O 8) K 'min
mol
.. Nc(24*601t+30)
A flowfate of 1.5 mol B/min would produce 120 mol C/ day, with X ;;;;: 80 % and T< 130°C at all times
P9-11 (b) If the max. coolant rate falls to 200 molimin, then it may not be prudent to assume that the coolant leaves at the entering arnbient temperature, Ta. It should be assumed that the coolant temperature varies spatially along the heat exchanger pipes and the required tenn for the heat exchange would be :
Q =: l'11;:; Cp.:ool (Ta 1 - Tal)
Ta2 = l' -
(1" -
where Tal:= ambient T coolant entering Tal = ambient T coolant leaving
Tal).ex p(-_U:4 ) rTkCpcoJ/
The reduced flowrate and hence heat exchange. may increase the reactor temperature to approaching 130o.C, the upper limit, at conversions approaching 80 %, and so more caution is required. The incorporation of temperature control would be prudent.
P9-12 CSTR startup Need steady-state values at To;::::: 75~
9-25
dNa Mol balance: -;;;-;::: F ao . . . Fa + ra.v dCa
(Cao - Ca).vo
dt
V
--=
-
deb
(Cbo - Cb)v.,
-=:
dt
V
+ra
+ra
dCc (Ceo·- Cc)u, --.dt V
dCM ---
...-.".
ra
__ V
(CMO -. CM)U, ....._. -
dt Fio
C:ao == Faq. = 0.18157
Cia=:'Va
Ib moll ft3
1)"
Cbo = .!bo = 2.2696 Ib moll ff u.. Cco=o CMO
=
.!:.~?. ;: : 0.22696 v.,
Ib moll ftl
1 V ;::: ....--...;. X 500 7.484 Fat) pao
Fba F.tfO 80 1noo 100. 3 .+ ._._....... ;::: ---'-- + - - + ---..- :::: 440 6 ft / hr pbo PMO 0.932 3.45 L54 .
Vo ;::: .-..... _- + .,_....-
Rate law:
- fa;:::
k.ea
k = 16.96eI2.exp
(Co in excess)
(-i9s?1:ir460))
Energy balance:
dT
Q- Ws·- LFio.Cpi(T.- Tio) + (-·-t:.Hrx)(-ra_V)
°d;- = .---................ ··_········· ..--·I)ifi. C;i--·-·----·-·.·. . . --·
9-26
Ws=o Q = IDc.Cpcool.(Tal - Ta2) == 1000 x 18 x (60 - Ta2) l:Fio.CpL(T ~ Tio) == [Fao.Cpa + Fbo.Cpb + FMO.C,PM ].(f - 75)
AHrx ==
~36000
Btullb mol
I,..Ni.Cpi == Na.Cpa + Nb.Cpb + Nc.Cpc + NM'CPM == 53Ca. V + 18Cb.V + 46Cc.V + 19.5CM.V
Ta2::: T- (T-- Tal).exJ _ V A y~ l1kCpcool == T -- (T - 60) exp( . .
J
l~_)
lOOOx18
Ta2 == T - 0.41111(T - 60) tIT -dt
18000(60- (T - 0.41111(T- 60)- 2275O(T -75) + (36000)(-ra.V) .. .--........... ...-.35Ca.V + 18Cb.V + 46Cc.V + 195Cv.V
Initial conditions: To::.:; 75, T == 138.5 o:F. Ca:::: 0.03780 • Cb :;::: 3.3062 • Cc;;: 0.0144,
If
eM::::
O.22691b moll ft
To drops from 75 to 70 OP
P9-12 (a) P-control only:
manipulated variable:;: lIlc controlled variable = T where Illco = lOOOlb mollh Tsp == 138.5 'F
~::: IIlco + kc.(T - Tsp)
kc= 10 See Polymath program 1'9· 12a..pol
POLYMA TH Results Calculated values of the DEQ variables Variable
initial value
minimal value
t O O
maximal value 4
9-27
final value 4
3
Ca Cb Ce Cm T I
FaO TO V
Tsp UA
Tal ke k FbO FmO meO ra NCp ThetaCp vO CaO CbO CmO tau X
me Ta2 Q
0.03789 2 . 12 0 . 143 0.2265 138.53 0 80 70 66.809193 138 . 5 1.6E+04 60 10 24.990212 1000 100 1000 -0.9468791 3372 . 5882 284.375 441 . 46403 0 . 1812152 2 . 2651902 0.226519 0 . 1513355 0.7909116 1000 . 3 106 . 23674 8.325E+05
0.03789 2 . 12 0 . 1227539 0.2265 125 . 70694 -39 . 636625 80 70 66 . 809193 138 . 5 1 . 6E+04 60 10 13.770535 1000 100 1000 -0.9468791 3372.5882 284.375 441.46403 0.1812152 2 . 2651902 0.226519 0 . 1513355 0 . 6775218 872 . 17351 102 . 00022 6.594E+05
0 . 0584442 2 . 1423214 0.143 0.226519 138 . 53 0 80 70 66 . 809193 138.5 1.6E+04 60 10 24 . 990212 1000 100 1000 -0.7913918 3385 . 3399 284.375 441 . 46403 0 . 1812152 2 . 2651902 0.226519 0 . 1513355 0.7909116 1000.3 106.23674 8.325E+05
ODE Re~ort {RKF4S} Differential equations as entered by the user [1] d(Ca)/d(t) = 1/tau*(CaO-Ca)+ra [2J d(Cb)/d(t) = 1/tau*(CbO-Cb)+ra [ 3 J d(Cc)/d(t) = 1/tau*(0-Cc)-ra [4J d(Cm)/d(t) = 1/tau*(CmO-Cm) [5] d(T)/d(t) = (·Q-FaO*ThetaCp*(T-TO)+(·36000)*ra*V)/NCp [6j d(I)/d(t) = T-Tsp Explicit equations as entered by the user [11 FaO = 80 [2] TO = 70 [3] V = (1/7484)*500 [4] Tsp = 138.5 [ 5 ] UA = 16000 [6j Ta1 =60 [7] kc= 10 [8J k = 16. 96e12*exp(-32400/1 . 987/(T+460)) [ 9 1 FbO = 1000 [10] FmO= 100 [11] mcO = 1000 [ 12] ra =·k*Ca [13] NCp = Ca*V*35+Cb*V*18+Cc*V*46+Cm*V*19 . 5 [14] ThetaCp = 35+FbO/FaO*18+FmO/FaO*19 . 5 [15] vO = FaO/0.923+FbO/345+FmO/1 . 54 [16] CaO = FaOlvO [ 17] CbO = FbOlvO [18] CmO = FmO/vO [19] tau = V/vO [20] X = (CaO-Ca)/CaO [21] mc=mcO+kc*(T-Tsp) [22] Ta2 = T-(TTa1)*exp(-UA/(18*mc))
9-28
0 . 053671 2.1376461 0.1275442 0 . 226519 128 . 49299 -39.636625 80 70 66.809193 138.5 1.6E+04 60 10 15.702753 1000 100 1000 -0.8427832 3383.2355 284.375 441.46403 0 . 1812152 2 . 2651902 0.226519 0.1513355 0.703827 899.92991 102.98479 6 . 963E+05
[23] Q
= mc*18*(Ta2-Ta1)
140
1100
136
1040
E!J
132
980
-
128
920 860
124 120
Q
0.0
0.8
1.6
t
2,4
3.2
4.0
800
0.0
0.8
1.6 t
3.2
2.4
4.0
P9-12 (b) I-control only :
manipulated variable ::: .~ controlled variabLe::: T
kc
mc:::mc... + - ]
where IDeo::: 1000 Ib mollh
1.1
kc ::: 10. 'tI::: 1 dI - : (T-Tsp)
Tsp= 138_5 ~
tit
See Polymath program P9··12·b.pol
---
10001,.,...,-----------·----, 940
[mJ
880 820 760 110 '---~---~0.0 0.8 1.6 t 2.4
P9-12 (C)
3. 2
4. 0
700
0. 0
0. 8
L6 t
2.4
3.2
4.0
manipulated variable == me
PI-control only:
controlled variable = T rrk
==
kc
meo
+
kc(T - Tsp) + ..- I
where nlco:;;:;: 1000 Ib molfh
'tl
kc ;;;; 10, 1:1 :;;:;:1
_.dl,'... = (T dt
Tsp ;;;; 138. 5 "F
Tsp)
See Polymath program P9-12·c.pol 1-'10,.------- ' - - - - ,
1100 1020
132
9-'1U
128
860
12-'1
780
120
'------~----------'
0.0
0,,8
1.6 t
24
3. 2
-'10
700
0,0
0,8
1.6 t
2.4
32
---,------,----
P9-13 dCa
-,-
Mol balance:
(Cao . . ,. Ca) =~.
dt
,,_._. - - + ra
Cao ::: 0 ..1 kmollm 3
!
Cao :::::: 0.1 mol/dm3
deb (Cbo-- Cb) --""= '--'_.'."'' ' ' ' .-'' ' ' ' -' -+ ra
dt
Cbo= Cao
1"
dec _ (0- Cc)
"
- ._ .. - -,-----,...... ra
dt Rate law:
1:;;;;
t
50 s
,',' ra ;;;; k.Ca.Cb k(T) ;;:;; 0.01 expr",,~_oooo.(_. 1 __ ..~)]
L1.987
9··30
300
T
-'1.0
Energy balance:
dT
Q-
w.~
. . . . L Fio, epi(T -
'dr" = .------.--......
Tio) + C,-t:JirxX-ra. V)
L~vi:-
Cpj---,··---·-----------
V == VI) • 't :: 2 x 50 == 100 dm3
=0.1 x 2 =.:: 0.2 molls:: Fbo (equimolar feed) LFio.Cpi.(r - Tio) =2 x 0.2 x (15)(r - 300) =6(T- 300) Fao:: Cao . 1)0
tJfrx(f)
=Lllirx(To) + ACp.(T - To)
but ACp;;;: 0
= - 41000 - (- 20000) - (-. 15(00) = .' 6000 cal/mol
f!T =~(T.,::- 300) + (6000).(-ra. V) dt
15.Ca.v+ 15.Cb. V +30.Cc. V
No control: See Polymath program P9·13 .. a.po!
POLYMA Til Results Calculated values of the DEQ variables Variable t Ca Cb Cc T
V k
ra
initial value 0 0,,001 0,,001 0 300 100 0.01 -·1" OE-08
minimal value 0 1.027E-04 1.027E-04 0 300 100 0.01 -0.0088037
maximal value 400 0.0732572 0.0732572 0.099864 2.574E+05 100 1.893E+05 -1.0E-08
final value
400---1.027E-04 1.027E-04 0,,099864 2.574E+05 100 1.893E+05 -0.001998
ODE Report (RKF45) Differential equations as entered by the user [1 J d(Ca)/d(t) = «.1-Ca)/SO)+ra [2] d(Cb)/d(t) = «.1-Cb)/SO)+ra [3 J d(Cc)/d(t) = (-Cc/SO)-ra [4 J d(T)/d(t) = «6*(T'300»+(6000*(-ra*V)))/(1S*Ca*V+ 15*Cb*V+30*Cc*V) Explicit equations as entered by the user [1] V = 100 [2] k = .01*exp«10000/1 . 98'7)*(1/300-1rr» [3] ra = -k*Ca*Cb
9-31
300000
rn
2-40000
.. ell
180000
- Cc
120000 .
60000 0.00
0
80
160
t
240
320
400
0
0
80
160 t
240
320
Tbe results without control, indicate a runaway reaction, as T continues to increase after the concentrations have approached their steady-state values Control aspects:
assume that the operating T should not exceed 550K, the boiling point of the liquid.
Without given data for beat exchange: Try
manipulated variable::::: Tio (iulet feed T) conuolled variable::::; T
P-control
Tio::;: IF (T<550)'I1IEN (300) ELSE (Tioo +- kc.(f - Tsp)) where Tsp:::: 550 K, Tioo:::: 300 K, kc =-10 ~
Tio rnanipuJation not feasible fc)[ T control for any kc (requires Tio .. ···2000 K)
See Polymath program P9··13-b.pol 8001'----·------------, 700 600 500
400 300 L-_""""'==-__
o
Try
80
160 t
240
320
400
manipulated variable::::: Va (inlet volumetric flowrate) controlled variable::::: T
9-32
400
P-control
U o ::::
IF (T<550) THEN (2) ELSE ('0 00 + kc.(f '- Tsp» where Tsp :::: 550 K,
U OO
::::
2 dm 3, kc :::: 10
See Polymath program P9-13-c.pol 600~----------.....,
540 480 420 360 300
L...--~"---
o
_ _ ~_~_ _
80
160 t
240
320
400
P9-14 (a) CSTR startup Initial conditions: To:::: 75, T = 138.5 "f. Ca::: 0.03780 , Cb:::; 2.12 • Cc::: 0.0144, eM:::: O.22691b mol f ft3
If
T drops froIn 138.5 to 133.5 OP
As in P9-12B except:
dCa dt
Fao- Ca. Va V
..---- = - - - - - . + ra
Fao
1000 3.45
100 1.54
Va ::::: ......... ,..................f. --"'. ._- .+ ... _-
0.932 dT
.--- :::;:;::
dE
_.
18000(60·(T ····0.41111(1'- 60) -(35Fao + 19950).(1'-75) + (36000)(·-ra. V) ............................ -...-.."-.--. "'.".~.- ....----.-.......................-~ ..,-_., ._.__ ...,,_._...... --_..._.. __............. ~- .. ,,- -. .-.~ .. - .............. _"._--.
".,.. ",,--~-
'""
35Ca. V+ 18Cb. V
+ 46Cc..1' + 19..5C~. V
manipulated variable = Faa
I-·control only:
controlled variable = T Icc
Fao = Faoo+-I 11
where Faoo::::; 80 Ib roollh.
say kc = -0.2 ) 'tt == D~I
9--33
-----
dI :::: (T - Tsp) dt
-
Tsp = 138.5 OP
See Polymath program P9-14-a pol
POLYMATH Results Calculated values of the DEO variables Variable t Ca
Cb Cc
em T I
FaOo TO V
Tsp UA
Tal ke k
FbO FInO
meO
ra NCp FaD ThetaCp vO CbO CmO
tau CaO me
Ta2 Q X
initial value
o
0.03789 2.12 0.143 0.2265 138.53
o
80 70 66.809193 138.5 1.6E+04 60 0.2 24.990212 1000 100 1000 ·-0.9468791 3372.5882 80 284 375 441 . 46403 2.2651902 0.226519 0.1513355 0.1812152 1000 106.24535 8.324E+'05 0.7909116
minimal value
o
,,·27 . 922973 2.12 '.. 1. 7469378 0.2265 17.999228 -307.34656 80 70 66.809193 1385 1.6E+04 60 0.2 0.0260082 1000 100 1000 ·-0.9468791 3372.5882 ·-534.4229 -3993.8255 ·,,224.21625 -183,_ 76896 -18 376896 -12 277456 --118 36922 938.55771 34 . 305432 --4.341E+05 -65.857622
maximal value 4 0.0898852 95.168593 0.143 9.3421656 138.53
o
80 70 66.809193 138 5 1.6E+04 60 0.2 24.990212 1000 100 1000 0.723285 5.574E+04 80 9.686E+04 441 46403 364.28299 36.428299 24.337453 61.10206 1000 106.24535 8.324E+05 12.667586
final value 4 -27 922973 95.168593 -1.7469378 9.3421656 17.999228 -307.34656 80 70 66.809193 138.5 1 6E+04 60 0.2 0.0260082 1000 100 1000 0 . 723285 5.574E+04 -534.4229 -2.3299873 -224 21625 -4.45998 -0.445998 -0.2979677 2 3835154 938.55771 34.305432 -4.341E+05 12.667586
ODE Report (STIFF) Differential equations as entered by the user (1) d(Ca)/d(t) 1/tau*(Cao-Ca)+ra (2) d(Cb)/d(t) = 1/tau*(Cbo-Cb)+ra [3 J d(Cc)/d(t) = 1/tau*(Q-Cc)-ra [4] d(Cm)/d(t) = 1/tau*(Cmo-Gm) [5] d(T)/d(t) =(-Q-FaO*ThetaCp*(T-TO)+{-36000)*ra*V)/NCp [6 J d(I)/d(t) =T-Tsp 150 -----.----,-----.--
=
144 138
132 126 120
9·14
0.0
08
1..6 t
2.4
3.2
4.0
FmO = 100 [ 11] meO = 1000 [12] ra = -k*Ca [13] NCp = Ca*V*35+Cb*V*18+Cc*V*46+Cm*V*19,5 [14] FaO = FaOo+(kc/,,1)*J [15] ThetaCp = 35+FbO/FaO*18+FmO/FaO*19.5 [16] vO = FaO/0.923+FbO/3.45+FmO/1 ,,54 [17] CbO = FbO/vO [18] CmO = FmO/vO [19] tau = VivO [20] CaO = FaOlvO [21] mc = mcO+kc*J [22] Ta2 = T-(T-Ta1)*exp(-UA/(18*me)) [23] Q=mc*18*(Ta2-Ta1) [24] X = (CaO-Ca)/CaO [10]
90
88 86 ...
82 80
P9-14 (b) Tal
""""---.-- -----
84 .
0.8
0.0
1.6
t
3.2
2.4
4.0
= 55°P
I-control onl y :
nUl11ipulated variable ;;::: To controlled variable::::: T
To;;::: Tooo
+-~~1
where Too
=75 OP ,
say kc,:;:; ·,0.2. 't'I
dl
=0.1
Tsp;:: 138.5 OP
dt ::::: (T - Tsp) See Polymath program P9-14-b.pol 150
85
144
82
138
79
132
76
126
73
120 '---~--~
0.0
0.,8
1.6 t
2.4
3.,2
4.,0
70
01)
P9-14 (C) No solution will be given P9-15 9-35
08
L6 t
2.4
3.2
4.0
ammonium nitr'ate A Feed, rno
nitrous oxide
stearn C
B
= mao +- meo ; ; ;: 0,,83 mo + 0.17 mo dfllfa
Mol balance:
(rnco
=::
0 "I- l'a. V
.._-:::;; mao"
dt
17% liquid water in feed)
Ma == Ib A, mao = lblb A
dMb
. _.._. :::: 0- mh···, rQ" V dt leaving:
m};;;::: mao. Gb
(
"\ t,;;b X )::::: mao; Sb:::: 0, b/a::;;:: 1
(X;;;;; 1, i.e. all of A entering reacts, but there is some initially inside) dlvlc
."'-..-~, :;:;;: mea- me -- 2. ra. V
dt
leaving: me:::: mao. ( 6t + ~-
x) :; ;:
mea + 2mao ;
e.
.;;t;(},
cia;;;;: 2
assuming all water entering with the ammonium nitrate leaves as steam.
Rate law:
-ra. V :;:;:; k.Ca. V ::::: k.Ma
Ln':)) 0 -''') ( 5.03.
R,T2
1 1.987(560+460)
R.n =::
1 1.987(510·+460)
88500 Btu Ilb mol
Energy balance : aT Q- Ws~· ')' mio . Cpi(T·, Tio) ." .., meo. (Hi~ Hio)-+- (--Mlrx)( -"'la. V)
d;' = --,..--"-..~,,,-..where
....... ,-..,.---
I. l~fi:'L:;;i-''''-'''''''''''''''''''''''-'---'--'~ . .". . . . . . .". _.
Ws=o Q =UA.(Ta - T) 9-36
Imio.Cpi.(T ~ Tio) == mao_CpaCT - Tio) ::::: 0.83 x 310 x 0.38 x (T - 200) mco.(Hi - Hio) =:: mco.[Hg(T)* HI (Tio)] =:: 0.17 x 310 x [(1202 + 0.47(f - 500)) - 168J assumes all liquid water in feed leaves as steam. Hg(T);;:; Hg(Tref) + Cpc.(T - Tref):::: 1202 + OA7.(T ~ 500)
HI (200);;;:: 168 Btu fIb 2:MLCpi ;::; 0.38 Ma + Cpb.Mb + 0.47 Me
mphysical properties systems: 1 Cpb (N,O. 516"F) = 1065 J/kg.K. (10 55 Btu ). ( 0.454 (~.) From ChemCAD
:J
::: 044 Btullb N 20, K
P9-15 (a) att=:O Ca=:5001b, T=516OP
__ ___
dT 10000.(515- T) ···,97..77(1' -, 200) ...... 52.7[(1202 + 0.47(T.,- 500»)·· ··1681 +- 336.(-ra V) ... .. :;;:;::: .... .... .................. .......... ......... ..... ............-,-... ......... ...... .............. .,..................... ... ....... .. ......__ ....... _... ................... _..__.._..... ... ....... . ..... .................... OJ8Ma ·-j·OA4.Mb + OA7Mc dt
"'" .. .,
~
~
,.,
,
.,
"
~-
"
-
-~-,-
"'
,
,
,
,
,
See Polymath program P9, IS-a. po]
POLYMATH Results Calculated values of the DEQ variables Variable -t Ma Mb
Mc T
mco mao k
r'aV rob mc
initial value
o
500
o
o 516 52.7 257.3 0 . 7028594 -351..4297 257.3 567.3
minimal value
o
---
399.06704
o o
514.15817 52.7 257.3 0 . 6447579 -366.56963 257.3 567.3
maximal value 10 500 100.93296 201.86591 517.83877 52.7 257.3 0 . 7677849 -257.30162 257.3 567.3
final value 10 399 . 06704 100.93296 201.. 86591 514.15817 52.7 257.3 0.6447579 -257 . 30162 257 . 3 567.3
ODE Report (RKF45) Differential equations as entered by the user [1] d(Ma)/d(t) =31 0*. 83+raV [2] d(Mb)/d(t) = -mb-ra V [3] d(Mc)/d(t) = mco-mc-2*raV [4] d(T)/d(t) = ((1 0000*(515-T))-(99 . 77*(T·200))-(52.,7*(1202+(.47*(T-500)))-168)+(336*(-raV)))/(.,38*Ma+.44·' Mb+.47*Mc) Explicit equations as entered by the user [1] mco = 310*.,1] [2 J mao = 310*.83 [31 k = .,53*exp(44540*(1/970-1/(T+460)))
9-37
[4 [5 [6
raV = -k*Ma mb = mao me = meo+2*mao
soo ,------------------------,
S18~--~-------------------.
::I~
S17 S16
------------------~
S16
380
515
340
514 '------.- 2 4 0
t
6
8
10
300 '---~----------------------' o 2 4 t 6 S 10
P9-15 (b) lJA = 10000 Btu lh.ft2 P-{:ontrol only:
manipulated variable;:: Ta controlled variable = T
Ta:;;;;; Tao + kc.(T - Tsp) where Tsp == 516
of,
kc = -.5, Tao == 97.5 ~ == 515 ~
See Polymath program P9J5-b pol
SOO 460 420
380
--..._- _._ . _., .........
SIS.60 0
2
4
t
6
S
340
.................... .
10
300
0
P9-15 (C)
9-38
2
4
t
6
8
10
UA :::: 10000 Btu lh.ft2 PI-control:
manipulated variable :::: Ta controlled variable:::: T
7'0.::::
Tao+kc.(T,-Tsp)+ kr:...I 1J
dI dt
- ' ::::(T-Tsp)
where Tsp;;;: 516 OPt kc;;;: -5, 'tI
=:-
1. Tao =975 ~
=515 OP
See Polymath program P9·15·c,.po] 516.20.-----------------,
500
516.12
460
516.04
420
515.96
380
51588
340
515.80
2
0
4
6
t
8
10
300
0
---~---'---~
2
4
t
6
8
P9-15 (d) UA;;;: 10000 Btu lh.ft2 PI-control1oop 1 : manipulated variable = Ta controlled variable:::: T To.
= Tao + kcL(T -
kcl Tsp) +--.11 1:11
dll dl
-=(T-Tsp)
where Tsp PI-cOIlu'olloop 2 :
=516 <>p, kcl =-5, crd
;:;;; 1. Tao;:;: 975 oR;:;;; 515 ~
manipulated variable;:;;;; mao eontmlled variable == M (= Ma + Mb + Me) 9-39
10
kc2 I". . . mao ;;;;; maoo + kc2 . (1\.1 - Jyfsp) +~.' t/2
dI2 dt
~-
= (Ivf -
NIsp)
where Msp::::: 500 Ib kc2::;: 25, 't12:::: I, maoa :::: 310 Iblh 1
See Polymath program P9-15-d.pol 516.20...--------------,
500
516.12
460
51604
420 380 340
515.80 L -_ _ _
o
~
2
_ _~_~_----I
4
t
6
8
10
300
0
2
4
t
6
P9-16 (a) Plot R(T) vs G(T):
R(T) = U.4(T ,,-···1:) + fJVoCp(T . . 1~)
G(T) =
-'L'\H~VkCA
Evaluat.e the parameters in those equations: k
= 2 * 7.08;;: lOti exp( -30{}OOI 1.9871 T)
MI RX
;;;;;
-30000 BTU Ilbmol
p ::::: 50lb I .ftJ [IA:;:: 150'* 250;;;;; 37500 See Polymath program P9-16-a pol
POLYMA TH Results Calculated values of the DEQ variables Variable t T
dHr V Caa va VA
initial value 0 520 -3 . 0E+04 48 0. 5 400 3.75E+04
minimal value 0 520 -3 . 0E+04 48 0.5 400 3 . 75E+04
maximal value 140 660 -3.0E+04 48 0. 5 400 3 . 75E+04
9-40
final value 140 660 -3 . 0E+04 48 0.5 400 3.75E+04
8
10
530 50 530 3.0E+04 0.3478362 0.12 0.0241085 0.75 2.404E+05 -5.25E+05
530 50 530 3.0E+04 0.3478362 0.12 0.479966 0.75 2.404E+05 -5.25E+05
Ta rho To E k
tau Ca Cp Gt Rt
530 50 530 3.0E+04 164.49672 0 . 12 0.479966 0.75 5.711E+06 6.825E+06
530 50 530 3.0E+04 164.49672 0.12 0.0241085 0.75 5.711E+06 6.825E+06
ODE Report (RKF45) 7.00E6 , . - - - - - - - - - - - - - - - . . . ,
Differential equations as entered by the user [ 1] d(T)/d(t) = 1
5.40E6
Explicit equations as entered by the user [ 1] dHr = -30000 [2] V=48 [3] Cao = ,,5 [4] vo = 400 [5] UA = 250*150 [6] Ta=530 [7] rho = 50 [8] To=530 [9] E = 30000 [10] k=2*7.08e11*exp(-E/1.987/T) [11] tau = Vivo [12] Ca = Cao/(1 +tau*k) [13] Cp = .75 [ 14] Gt = -dHr*V*k*Ca [15] Rt = UA*(T·Ta)+rho*vo*Cp*(T-To)
1-
G;T)
- Rn
1
3.80E6 2.20E6 6.00E5
-1.00E6
L-_~
520
_ _~_~_ _ _~_....J
548
576 T 604
P9-16 (b) From part (a), we find the concentration and temperatures at the points where G(T) and R(T) intersect.
- .-
=547.t<'R
., - 04·'15 (;/\ C•A ::: 0.319 .
T
CA,= 0.068
T::: 628.6°R
T::: 57L3°R
The extinction temperature is To = 506°R. At this point R(T) is tangent to G(T) at the upper steady state.
P9-16 (C) Using the unsteady-state equations for a CSTR we get the following Mole balance:
d~;'_ = .~o{ C AO ....•. <:A} _ kC dt
V
A
Energy balance:
See Polymath program P9-16-·c ..pol 9-41
632
660
,?Ol:-YI\llp'TH Report Calculated values of DEQ variables """1'''' """'''' ; ' i , ,Variablellnitial valuelMinimal value IMaximal value/Final value "'~-"r"~'"
"" " " ' ' ' ' ' r ' '
'ifi"-"lo ,,","'," [21Ca
'1'6"'16.
,~l
i3~J:('
)~~8.
A ICaO '5 {FaO
'0.5 '200.
6 ITO
i530.
J
;J
J638.1791 0.5 ,200. '530.
11.416E+12
IR
11.987
j
1150 .
;150.
!1.416E+12
1.987
!1.987
1400. 530. j48. 14iE
j3.0E+04
15/k
151.27669
16hau
;0.12
:17;rA
i
1150. ,
I
1400.
1250. 1400 i .
1530 .
'530.
148. i3.0E+04
'48.
TO~4253139 ,j547.0711 10.5 '[200. l530.
1.416E+12
i250.
1250.
"
<
10.4290427
JAr
"lu
"" '16.
1
11.416E+12 '11.987 t 1150 . I " 1250 . :400. :530. 148. ,
13.0E+04
r
!75.24179.
!3.0E+04 11.463353
:0.12
:0.12
10.12
!-3.491943
-3.971954
-0.4433316
[-0.6223844
i18 JI delH
j-3.0E+04
-3.0E+04
'-3.0E+04
i-3.0E+04
i19lNao
124.
24.
24.
24.
175.
'75.
~
1.165967
~
"f' , 1
20 lCpa
;75.
Differential equations d(Ca)jd(t) = rA + CaO j tau - Ca j tau
121 d(T)jd(t)
= (U * A * (Ta - T) - FaO * Cpa * (T -, TO) + (rA * V) * delH) j
Explicit equations CaO = 0.5 2;1 FaO = 200
3/F TO
= 530
= 1.416 * 10 R = 1.987 U = 150 "7 A = 250 vO = 400 Ar
A
12
Ta = 530
'fijj V = 48
9-42
(NaO
* Cpa)
~tJ E = 30000
1'4) k = Ar * exp( -E / R / T) i£~l tau = V / vO
= -k * ca 't~ delH = -30000 ,161NaO = CaO * V Cpa = 37.5 * 2 rA
""""""d
0.5 0.'1 0.3 0.2 0.1
IV 0.0
O.
0
1.2
204 t
3.6
48
When the upper steady state is used as the initial conditions, the unsteady-state mole balance shows that this steady state is actually unstable. The concentration increases and the temperature drops to the lower stable steady state.
P9-16 (d) When To or Ta is increases slightly, the upper steady state becomes stable . At these elevated inlet and coolant temperatures, the lower steady state is no longer stable.
P9-16 (e) Starting at the lower steady state, if To is increased to 550 OR, the lower steady state is no longer stable, but the upper steady state is 0.50..---·-----------------,
800
0040
740
0.30
680
0.20
620
0.10
560
--6.0
0. 00 '---.-.-'----~-----....- .:1.8 0.0 1.2 2.4 t 3.6
500
""""""...........---------_._.-
0.0
1.2
2.4
t
3.6
48
6. 0
If we plot the concentration-temperature phase-plane trajectory, we see that increasing Ta will shift the trajectory to the left. However, the final steady state is shifted to the right. This means that from the initial conditions, at any temperature CA is lower for the larger Ta until the minimum in CA is reached. 9-43
0.5
0.5
0..-1
04
GJ
03
n.3
02
0. 2
0.1
tlJ
0.0
UO
'--------~-=-----'
500
560
P9-16 (f)
620 T 680
740
800
'--_~_~_._...._C:::...._
500
560
620
r
680
Individualized solution
P9-16 (g) The following Polymath program gives the linear analysis of the plOblem A = 1.175, B = 6.03, J = 1600, 1" =12 See Polymath program P9-16g.pol
~,~~¥ 1VIJ\~,f:!R,~portt, Calculated values of lVariable:lnitial value Minimal value Maximal value:Final value ''''
'm~'""l''~'~'''
i
1 't
;
1
0
0
6
12.
0.0176788
A:159719
iO:0176788
10.2
-0.0225558
0.2
!-0.0001258
:4 jea
!~.0681
0.04847
;5 IT
1628.
;542.6046
0.4290427,0.4253139 1 :638.1791 :547.0711
"6 )cpa
175.
175.
175.
~ [y t,
ix
3
17k~0
6
l
)75.
18 ;Fao
fo.5 .1200.
9 lTO
1530.
1530.
/530.
i1.416E+12
h.416E+12
:1.416E+12h.416E+12
i 1.987
11.987 150 .
1.987
, f
1150. :250.
10.50.5 . '.. 1200. ;200.
1
1400.
1250 . 1400.
(530.
1530.
:48.
17lE
i3.0E+04
181k f 19ltau
151:27669 10.12
,
15!Ta
;~61v
.
;
L
'
:0.5 j200. 1530. 11.987
! l 5 0 · l i 5 0.. 1250. : 250. :400. 1400. /530.
t530.
J48.
t
+
J,
j
13.0E+04
i3.0E+043.0E+04
i i
148.
148.
1.165967
175.24179
1.463353
0.12
!0.12
0.12
9-44
740
800
Differential equations d(y)jd(t) = (-J * (1 - A)
* x + (B - C) * y)
= (-A * x - B * Yj J) j tau d(Ca)jd(t) = rA + CaO j tau - Ca j tau :ft d(T)jd(t) = (U * Area * (Ta - T) - FaO * Cpa * (T - TO) + (rA * V) * delH) j d(x)jd(t)
Explicit equations Cpa = 37.5 * 2
cao = 0.5
= 200 TO = 530 Ar = 1.416 * 10 /\ 12 1§~;;~i R = 1.987 FaO
U = 150
Area
= 250
= 400 :,~ITa = 530 V = 48 E = 30000 t~~l k = Ar * exp( -E j vO
~4; tau
=V j
R j T)
vO
rA =-k * Ca
= -30000 NaO = CaO * V !~~iJ J = -delHjCpa j ;~'~! delH
CaO
f~!C=6 ~1:f;l) t~ B = 6.03 A = 1.175
P9-16 (f) Only the lower steady state plot of xl and yl will be shown.
9-45
(NaO
* Cpa)
0.50 , - - - - - - - - - - - - - - - - ,
0.50
038
0.38
r--------------,
~
l&J
0.26
0.26
OJ-I
(1.1-1
0.02 . +--~- .........::;;:=""""'.......--~-_1
0. 02 .
-0.10'---~----~--~-----'
0.0
12
24 t
3. 6
.:I S
60
·0.10
O.Oe+O 1.6e-3
3.2e-~1
.:I.Se-3
6Ae-3
ROe-3
P9-17 Batch reactor· series reaction dea -.._.":::: ral dt
Mol balance on A :
···.ral ::: kl.Ca
Ca ::::: Cao.exp( ·k! .t)
deb .... ;;;:; rbi + rb2 dt
Mol balance on B :
rbl :::;; . . ral
= kLCa
and
l'b2;:::; k2.Cb
deb
""ii·· = kLCao.exp( k1.t)· k2Cb
dCc
Mol balance on C;
dt
k2 ==
Energy balance :
·····,-b2
4.58.exp(-~::iif{·~) . . })) Q+ (AHrxal)(-·raIY)+ (i.VJrxb2)(·rb2.v) 'dT dt'" == '-'".. ..... ......- ..... 2..Ni:cpT· . ··_· . ··-_·_··. · , _· -, · . ·· . ·_·
9-46
Q = UA.(ra ... T) LNi.Cpi = Ca V.Cpa + Cb.V.Cpb +Cc.V.Cpc dT dt
UA.{330- T)+55000(-ral.V)+ 71500(-rb2.v) 200.v,(Ca+ Cb+ Ge)
P9-17 (a) dT
UA·:;:;O
•......•.... :::::
dt
V[55000{-"'raL) _._-_._-_ __.- +__71500{-rb2.)] _-.....
..•...
...........
200.v.(Ca+ Cb + Ge)
See Polymath program P9-17-a . pol
POLYMATH Results Calculated values of the DEQ variables Va:riable t Ca Cc Cb T UA
v k1 k2 ra1
initial value
minimal value
o
o
0. 3
1. 34E-65
283
o o o
10 1.1172964 4.081E-09 -0.3351889
:rb2
o
rb1
0.3351889
maximal value --
final value
-2.02E-44 283
0.2 0.3 0.3 0.2895784 915.5
0.2 . 1. 34E-65 0.3 -1.864E-65 915.5
10 1..1172964 4.081E-09 -35.016552 -27.963241 -6 . 345E-60
10 2.141E+05 1.04lE+06 6 . 345E-60 2.103E-38 35 . 016552
10 2.141E+05 1.041E+06 6.345E-60 -3.974E-59 -6.345E-60
o
o
o
o
ODE Report (STIFF) Differential equations as entered by the user [1] d(Ca)/d(t) = ra1 [2 J d(Cc)/d(t) = -rb2 [3J d(Cb)/d(t) = rb1+rb2 [4] d(T)/d(t) = ((UA*(330-T))+(55000*(-ra1 *V))+(71500*(-rb2*V)))/(200*V*(Ca+Cb+Cc)) Explicit equations as entered by the user [ll UA= 0 [2] V=10 [3] k1 = 3 . 03*exp((9900/1 . 987)*(1/300-1/T)) (4) k2 = 4.58*exp((27000/1.,987)*(1/500-1/T)) [5] ra1 = -k1*Ca [ 6 ] rb2 = -k2*Cb [7] rb1 = -ra1
9··47
1000,-------------, 8-10
0.22
- Ca
0.14
680
- Cc Cb
0.06
-0.02
-0.10
520
1--==---.-"'----------.--
0.00
0. 0-1
0. 08 t
0.12
360
016
020
200
'---~--~----~----'
0. 00
0.04
0.08 t
0.12
0.16
0.20
P9-17 (b) UA: 10000 llmin.K 0.30 '""""'---,--".....-.--.."....'--.'--.." ......".........---~...."' ....
1000,----------·---,
0..22
840
t
0.1-1
ll
680
("c
-.1D:. -_ Cb
0.06
520
··0.02 -O . lH
360
0.00
0.04
0.08 t
fU6
0.12
0.20
200
000
00-1
008 t
0 . 12
0.30
900
0.22
760
Q
ell
-
('c
620
CI>
0.06
·480
--... .....-... -"--. ... ...."-. --". ...... .......
...."."" ... ... ~"
-002 -0.10
0.00
0. 04
"' .. ,,"",'
....
,
0 . 08 t
" '"'
0.12
UA: 100000 l/min.K
020
V::::::. 10 drn 3
UA :=: 40000 JJmiu.K
0.1-1
016
"
0 ..16
-
,
0.20
3·40 200
000
V;;;;;. lOdm3
9-48
0.04
008 t
. 012
016
020
030
I,
500 440
0.24
Em - Cc
0.18
Q
380
- Cb
0.12
0.06 ) 0.00 0.00
320 260
, 0.04
0.12
0. 08 t
0.16
0.20
200 o. . 00
0.04
0.08 t
0.12
0.16
0.,20
At UA = 100000 J/min.K the 2nd reaction for C is totally suppressed
P9-17 (c) UA = 40000 Jhnin.K
To =320 K 900 r--;--.-------------,
0040 0,,30
\
780
.
0.20
Ca
660
Cc
. ell 540
0.10 0.,00
lJ
-0.10 0.00
420
0.04
0.08 t
0.12
0.16
0.20
300 0.00
0.04
0.08 t
0.12
P9-18 Semibatch with parallel reactions Mol balance;
dNa
-'"d;'; Fao+(rl+r2).V dNd dt
~,= Foo+ (rl +r2). V
.
dNu -dt =-r2. V
-·-=-rl.V
Rate laws:
-rl:;:;:kLCa
- r2:;:;: k2.Cb
9-49
0.16
0,,20
Na
Stoichiomeu), :
•
=_Nb . --
Cb
Ca::::: ..- V
V
Nu
Nd Cd:;::;-,V
Cu=V
v:::: Vo + OO.!+ vb.!
assume Vo=O
vcz:::;; . .Faa ....... -
w == .--.Fho . . . . _Cbo
Cao
Ca6 : : ;: 5 molJdm3
Cbo == 4 mol/dm? where
V =uO.t
But Fao == Fba (equimolar feed)
Fao Fba + ....... Cao Cbo
1)0;;;;;:; -.-.--
no
=GA5.Fao
v == 0.45Fao.t dT
Energy balance:
Q" WS "LFio.Cpi.(T - Tio) + (-AHr:x(D).(-ra. V)
'dt' == -.--.-'----.- ..."., ....... ':2: Ni:C;;i---'·_ . . ···,·. ·. -'--_. ··_·,-' where Q==Ws==O
MIrx(T) == .6.Hrx(Tref) + 1.\Cp(T-Tref)
1.\Cp1 == :!.. Cpd'" ,?.,Cpb- Cpa.:::: 50-20--30 = 0 a a 1.\Hrx 1(T)
:=; '.,
3000 cal/mol A
1.\Gp2;:::: -Z:.Cpu ..~.. Cpb·· Cpa == 40- 20 -30:::: --10 a
a
&Irx2(T) ::: . , 5000, IOcr ... 3(0)
2.:Fio"Cpi.(r·Tio):::: 20 FaO.(T-Tao) + 30.Fbo.(T-Tbo) :L.Ni,Cpi == NaCpa + Nb,.Cpb + Nd,Cpd + Nu.Cpu
.
dT. :::= _._20 . Fao(T ., Tbo ),f:,?5~~::t!'L Y.l~J?~.::}.gQ~.:: ~.25J1K~~~:YJ .. ... -,-........ ...... . Tao) -+ 30. Fba. dt 20. Na. + 30Nbt 50.1Vd + 40" Nu
".
".~
~
-~
Selectivity
rl s= _.-.,
72
P9-18 (a) 9-50
The selectivity is proportional to CAIC B and therefore a small concentration of B in the reactor is preferable. To maintain a low concentration of B it would be beneficial to run a semibatch reactor where B is slowly added to A.
P9-18 (b) Let the rate laws be :
- r1 == kl.c'''b
-r2 ==k2.Ca
Now it would be best to slowly add A to B in a semibatch reactor .
Let the rate laws be :
-d == kLCaCb ····rl == kLCa - r1 ==k1.Cb
- 1'2 ;;;;; k2.CaCb - 1'2;;;;; k2.Ca -r2= k2.Cb
For these rate law combinations a semibatch reactor will not improve selectivity.
Let the rate laws be :
. r1 :::: kl.Ca.Cb .. 1'1 =kl.Cb
-1'2 =k2.Ca • r2 :::: k:2.CaCb
Now it would be best to slowly add A to B in a semibatch reactor to maintain high concentrations ofB and low concentrations of A.
Let the rat.e laws be :
- 1'1 == kl.Ca - rl ; : : lel .Ca.Cb
- 1'2 :::: k:2. CaL"b - r2=k2.0>
Now it would be best to slowly add B to A in a semibatch reactor to maintain high concentrations of A and low concentrations of B .
P9-18 (C) AHrxl
callmol A
aHlX2(300K) ca1lmol A
Temp. K
3000
If AIh-xl is constant, then if .1.Hrx 1 increases, T increases and S decreases. If AIm! is constant, then if L1Hrx2(300K) increases, T increases and S decreases. S is dependent on r1 and 1'2. whicb depend on kl and k2, which depend on the temperature. The greater the temperature in the reactor the smaller the ratio between kl and k2, hence reduced S.
P9-19 Plotting the data oft vs. T gives the initial and final temperature of the reaction, To = 52.5 °C and T t =1668
9-51
0c.
200~------------------------~
80 -10
oo
s
16
t
24
32
40
Recalling equation (8-29) with LlCp = 0, X = 1 and T == T f gives:
cal ( cal !ill R = -18166.8 - 52.5 )K = -2057.4x mol.K mol Make a table oft (min), T CC), dT/dt, liT (K\ In(dT/dt) and plot In(dT/dt) vs 11K (K) in Polymath. According to equation (E9 . 3-14) the slope of the curve is E over R Linear regression gives: See Polymath program P9-19. pol
~~.~'f\1!~,..tll~~p'otl·' Model: InTdot
= aO + a1 *T_inverse_kelvin
VariableiValue
;95% confidence
aO
18.88162 . 2.425175
al
-6224.846 880.596
Statistics RA2.
0.9755663;
:RA 2adj
0.9720758
•Rmsd
0.0597988
Variance 0.0413783 Remember that T s is the self· heating 1 ate, which occur s after the onset temperature is reached.
. --6224 InT, =---+18.88 T
E
6224=R
9--52
E=6224K ·1.987
cal cal =-12367mol·K mol
P9-20 No solution will be given CDP9-A
CDP9-B
CDP9-C Adiabatic. batch. reversible. I~sign equation
:
Rate law: Stoichiometry : Combining :
kl Ln" Parameter evaluation: E =-1' b 1 ............ ".
,~.,".
R.T2
_R.TI
0217 I.n--·--·::::: -'--T'-~. . __ 03_2_4··-·T···-· = 8498J I mol
...............,......
,
...., ..... -,."' ..........
"
8314(340)
......., ...............,.... ... ... "
8.314(300)
(I 1)
Ke(T) In' ................. -....... ::::: _AHr:x ................ _. . ................. -....
Ke(300)
R
300
T
LlCp == 2 Cpa" CPA::::: 20 .. 12 ::::: 8 ]lmol.K
AHrx ::::: LlHIX (300) + LlCp(T· 300) == ., 75000 + 8 (1' " 3(0)
9-53
* 900 '" 254 = 272 127 mol A
N AO == 0 . 6667
Nro = 0..3333 * 900 *454 = 136064 moll N AO I V
CAO :::=
:::=
T
Energy balance: ~
(1)(12)
=272127 I (50 *28.12) :::= 192 moll dm3
+ 0 + (136064/272027)*(15):::= 195 J/mol
• [.- Lllirx(3OO).X] (.- 75(00). X = 10+-···-·-·--··-----·== 300+·······································__•·········
CPA + tJ.Cp.X
195+8.X
POLYMATH
?~a t.i ons_,;.,
d {Xl / d ( c)
Ini::ial
""kl~
val>:i~
o
( ( 1-x) .. {4' cao· (x.... 2} I l:(e:} l
TQ",300
cao=192 TzTQ+11500Q·x/{19,5+{8~x}))
kl=. 217 "eX? (11322" Dhrx=-75000 ...
(a
w
(l/HO~
llT) )
(T-3UO) )
Ke:z7QOaO·exp(Dh:::::-x/S.31.4."" /1J300-1,/T))
c:,.,
~
0,
" 9-8 variable
To
;::<:
:ni;:;ia,l._'~~
Final value
()
2
-----~~.-
o
o .O~':;9S73
JOQ
300
192
1.92
JOO 191
300
469."731
Joe
'io69.1S1
0_145345
0.49785
a . 145345
0 . 49735
·-7SQiJO
-7364.1 .8
--'75000
-73641.8
70000
loooa
1. 62'532
1..62532
o
O.04.:j,9S7]
300
.~.:;::::::
;,)I. . : : . : :
9-54
9-·3
"I:~"
,,,,,'r
"".---i'--"""'--' O.~OO
"',.,"',.+".,-"·,,-'.",,..,,-,..,,+"'''··,....'''-'''. ,,-,''·. ''i'' o,.;;)QO
:,,200
1.600
:2 COC
Equilibrium conversion. Xe :: 4.5 % 90 % of Xe ::: 0.9
* 4.5 ;:: 4.05 % cOIlversion
Time for this conversion"" 0.11 min:: 6,,6 s
Check: at equilibrium: (4.CAo2).Xe 2
using the solution to a quadratic:
+ (Ke.CAo). Xe- Ke.(7Ao:: 0
_. Ke,. Cw ± ..JK~2~C~~i':.i6.K~:(:;'~3· Xe = .--_._. """""""'''''''''''?'''-----''''.''''.. '''''''-''-
where at T ;:: Te ::: 470 K,
8.CAO-
Ke:: 1.625 moll drIl3
From POLYM.A.."TII, Xe:: 0.045
CDP9-D
9-55
CSTR startup CSTR startup, gas phase, isobatic - no pressure drop
Part (a)
dFa
Mol balance:
._-.- =:
dt
(Fao-· Fa) + ra ..v
'dF'b --" = (0- Fb) .. 2.m.V dE
Fb Cb= .. ···
Fa
Ca =, ...
VI
iJr
Fro ::::: Cro.vlQ =:
1)]" :;;;;:
P - ' - : " . VIa
R.l0
Fr T Fw To
111'0 .-.-. -...•..
Fr;:::; Fa + Fb
Fro::=: Fao + Tho V=: lrr.'t
1:
=500 s
Energy balatlce :
UA --- == 51 I kgcat.s. K
peat UA
=5 x 50
=:
but kg catalyst =: 50 kg
2.."50 J I s.K
ACp=O
dt
40. Ca. V+ 20 Cb. V
Ta= 300 K, Tio:::: 450 K.; Fao = 5 moUs
No control:
!~itial
E9.::t,at.ions: difa)!d(~l~{:ao-fal·(ra·VI
Ie-OS
d(::b)/d(c)"'(~:a)-(2"ra"V)
0
c',T) Id{t.};( {250" (Ta-T}} -t40"fao'" (T-Tio)" (20000" (-::::a"V})) / ((
450
40wCa·V)+{20~Cb·Vll
':'a=3Qa 7~.o=4.50
1:0.:;450
e,a.u=500 1:au:l;SQQ k.,.;exp( (31400/8.3U}
,,< !1I450) -(lIt1]}
cao=O.2S vo<.;fao/cao Ca""fa/v Cb;:fb/v V=v"t:au ra~--k"Ca
9-57
value
Mininr..:.tn value final value
o
30
a
30
Ie-OS
0_0106221
le-OS
0.00969749
o
299.Q98
a
299.098
450
451.548
445.634
451.548
s
s
5
'1'a
300
300
5 300
'1':.0
450
450
450
450
le-OS
299.107
le·-08
29:9.107
450
450
450
4sa
fa
300
taU
soo
500
sao
50{)
Tau
500
500
sao
500
1
1-0292
0 . .921054-
1.0292
0.25
0.25
0.25
Q.25
5
5
5
5
20
20
20
20
4e-08
1200.55
4.e~08
1200.55
va
Ca
0,25
0.251187
S .07756e··06
f1. Q77S6e-06
Cb
o
0,250078
a
-0.249135
V
la-OS -0.25
600273 -8.31]]ge-06
2e-05
600273
-0.25
-8 ..HJ3ge·06
ra
.,
.. ;: \ . 200.i. I
_....-...-~-
~
-"..,...=-...--~~.. .-"'" -_.,..,.."--- ..-....-----.--
~-""'-------"""'-'" ... --~"" .
.....e.scc.l(-"' .:..;
""e,:;oo·I 4.·:i..,,-t-CO·"·~
j ~
•t
-4H
ao~ ~oa-"-'-"-'-"""+-""'-"---"-"'-+-"~""-""'-'+'---.-.~-- .....t--~-'----'1 ........
6.000
::<.::00
19.000
2 ..... 000
30.00e
Part (b)
Mol balance:
dCa == (eao···· .................. ...................,.Ca) _. . . ._. + ra dt 1:
Cao == 0.1 kmollm3
.~-,.
9·58
:::::
0.1 111olldm
3
P9-10 cont'd
Rate law:
=:;;
Cbo=Cao
dCc (O··Cc) _ . = -_ ..._ ........ Ta dt 1.
t=50s
"ra:: k.Ca.Cb
EneIgy balance:
v
deb :;: ---._ (Cbo'''' -_. .......-Cb) + ra dtz
t)o • 1" =
or
d~
Q_.• Ws·"LFio.Cpi(T··· Tio) + (-Mlrx)(·ra. V)
:;: -_...........__. . . . .-.._-LNi.Cpi··---·········_··········
m •• -
2 X 50 ::: 100 dm3
Fao =:;; Cao " 'Uo = 0.1 x: 2 ::: 0.2 molls::: Fbo (equimolar feed)
IFio.Cpi.(T - Tio) =2 x 0.2 x (15)(1' - 300) = 6(1' - 3(0) AIlrxCD;::: .6.H.Ix(To) + .6.Cp.(T - To)
but ACp ::: 0
::: - 41000 - (- 20000) - (- 15000) ::: - 6000 cal/mol
dT 6(T - 300) (6000) . ( -ra. V) .. - = ------_ . . ._ + .. _ _.. _ -_
dt
15.Ca.v+15.Cb.V·t30.Cc.V
E;g;::ations:
Initial value
d(ca)/d(cl-«O.1-ca)/50)+ra
0.001
d(cb}/dICJ={(O.l-cbI/SD)+~a
>1),001
d(cc}fd(C)={(O-cc)/50}-ra
a
d(T) Id{c) =( (6" (T-JOO)) +{6000" (-ra*V) 1 } It (IS"ea"''\!)'" (lS"'cb*
)00
V) ... DO"cc"V) ) '\!",100
1<:=0.01 *exp { {1000011. 987) .. ( (lIJ 00 I,,· (liT) ) }
raz- k"ca",cb
to '" O.
t
f == 40.0
V
~al v"alue
~im\!ft\
t:-
O 0.001
400 0.0732615
0.00010274
O. aOOlO::n4
O. OQ01021'4
0.00010274
ca cb
value Mini.mum
v~lye
0
!ind 400
v~lu~
0.001
0.0132615
CC
I)
Q.0:99$64
{)
0.099864-
T
300
25.7:380
V
251380 100
300
100
100
100
k
0.01
lenSO
0.01
la9280
-Ie-OS
-ie-OS
--O.Oll(HHl
-0.00199796
:t'a
9·59
$c.lf':
10-- 52 ,;>00
~£!:.:... .~.
T
-rT
2.\00
"i-
TI
t .. 5CO
T
o_.seo
"I'
0.300
I-
T i
...... +--_...... _.............. ". ..;.-"" ._ . . -.------+ . ,,-,. __. _-_ . ---\ 16ti,cca
2~O.:lOO
• .:fL!.:o.
9-60
320.0CC
400 or;
Solutions for Chapter 10 - Catalysis and Catalytic Reactors
PIO-I Individualized solution PIO-2 (a) Example 10-1 (1) Pentane isomerization nP -pt--7 iP
Assume that Pt is the catalyst used. Maximum f = 5 molecules/site/sec Minimum f =3e-3 molecukes/site/sec Maximum rate: --rp
= fD(_l __ ) %Pt 100
MWpt
-rp = 5(0.5)(_1_)_~ = 1.28*10-4 mo~ 195 100 s geat Minimum: --rp = 3*10-3 (0.5)(_1_)_~_ = 7.69*10- 7 mol__ 195 100 s geat 1
(2) SO2 +--0 2 2
--7
SO3
No turnover frequency is given so this rate law cannot be determined by this method (3) H2 + C2 H 4 -~ C2H 6
Assume Cobalt is the catalyst. =
--r H2
fD(_l _) %C~ MWCo
100
Maximum: f =100 molecules/site/sec 1 mol --rH =100 ( 0.5 )( - 1 ) ---=0.00849--2
58.9 100
s geat
Minumum: f= 0.01 1 ) 1 -7 mol ( 58.9 -----=8.49*10 100 s geat
-rH =0.01 ( 0.5 ) 2
PIO-2 (b) Example 10-2 10-1
(1) Cv
=
1
Cr 1+ KrPr + KBPB Ks = 1.39 Kr = 1.038 Pro = YroProtal = 0.3*40 = 12 For 60% conversion
Pr = Pro (1- X) = 12*0.4 = 4.8atm PB =PTOX =12*0.6=7.2atm Cv = 1 =_1_=0.063 Cr 1+(1.038)(4.8)+(1.39)(7.2) 15.99
6.3% of the sites are vacant (2) X
=0.8
Cros = .CvKrPr _ KrPr CT Cr 1+ KrPr + KBPB eros = (1.038)(1)(1-0.8) = 0.2076 =0 09 CT 1+(1.038)(1)(1-0.8)+(1.39)(1)(0.8) 2.3196 .
9% of the sites are covered by toluene (3) Linearize the rate law to:
PH?Pr 1 KBPB KrPr ----=-+--+---r7 k k k
PIO-2 (c) Example 10-3 Increasing the pressure will increase the rate law.
p?
-r---T--R A 7 l+KrPr If the flow rate is decreased the conversion will increase for two reasons: (1) smaller pressure drop and (2) reactants spend more time in the reactor. From figure ElO-3.1 we see that when X =0.6, W
=5800 kg.
PIO-2 (d) Example 10-4 With the new data, model (a) best fits the data (a) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/(1 +Kea*Peat·Ke*Pe)
10-2
Ini guess 3 0.1 2
variable k
Kea Ke
Value 3.5798145 0.1176376 2.3630934
95% confidence 0.0026691 0.0014744 0.0024526
Precision RA2 = 0,,9969101 R A2adj = 0 . 9960273 0 . 0259656 Rmsd variance = 0.0096316
=
(b) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/(1+Ke*Pe) Ini guess 3 2
variable k
Ke
Value 2.9497646 1.9118572
95% confidence 0.0058793 0.0054165
Precision RA2 0 . 9735965 RA2adj = 0.9702961 Rmsd = 0.0759032 Variance = 0.0720163
=
(c) POLYMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/((1 +Ke*Pel"'2) Ini guess 3 2
Variable k
Ke Precision RA2 RA2adj Rmsd vari.ance
Value 1.9496445 0.3508639
95% confidence 0.319098 0 . 0756992
= 0.9620735
= 0 . 9573327
= 0 . 0909706 =
0 . 1034455
(d)
POL YMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pel\a*Phl\b Variable
Ini guess
a
3 1 1
-k---b
Precision RA2 R A 2adj Rmsd Variance
Value 0.7574196 0 . 2874239 1.1747643
95% confidence 0 . 2495415 0.0955031 0.2404971
= 0 . 965477 = 0.9556133 = 0 . 0867928 = 0.107614 10-3
Model (e) at first appears to work well but not as well as model (a). However, the 95% confidence interval is larger than the actual value, which leads to a possible negative value for Ka. This is not possible and the model should be discarded. Model (f) is the worst model of all. In fact it should be thrown out as a possible model due to the negative RII2 values. (e) POLYMA TH Results Nonlinear regression (L·M) Model: ReactionRate = k*Pe*Ph/((1 +Ka*Pea+Ke*Pe)ll2) Variable k
Ini guess -3---
Ka Ke
1 1
Value 2.113121 0,,0245 0 . 3713644
95% confidence 0 . 2375775 0.030918 0 . 0489399
Precision RA2 RA2adj Rmsd Variance
= 0 . 9787138 = 0.9726321 0 . 0681519 = 0 . 0663527
=
(f)
POLYMATH Results Nonlinear regression (L·M) Model: ReactionRate
= k*Pe*Ph/(1+Ka'Pea)
Variable k
Ini guess 3
Ka
1
Precision RA2 RA2adj Rmsd Variance
Value 44.117481 101. 99791
95% confidence 7 . 1763989 16 . 763192
=·0 . 343853 = -0 . 5118346 = 0.5415086 = 3 . 6653942
PIO-2 (e) Example 10-5 (l)X =1-
1 klk
(l+kd t )
d
As t approaches infinity, X approaches 1.
10-4
large klta
x
t (2) Second order reaction with first order decay. dX ,W - = - rA
-
dt
NAD 2
-rA '=ak'CA
a = exp[-kdt] dX = Wk' C2 (1- X)2 exp[-·k t] dt N AD d AD
dX =k(I-X)2 exp[-ki] dt X k 1-- X = k(1-exp [-kd t ]) d
~. t... X =kas t -y m Imty -.-l--X kd
X=J{d
1+ k/
/kd (3) First order reaction with first order decay dX Wk' di- = CAD (1- X)exp[-kd t ]
""N AD
dX
-
dt
= k(l- X)exp[-kdt]
InC~x)= ~ (I-exp[-kdtj) t -7 infinity X = l-exp [ -
~]
PIO-2 (0 Example 10-6
10--5
Increasing the space time makes the minimum disappear. Decreasing the space time moves the minimum to the left and the concentration is higher. Increasing the temperature so that the rate constants are higher will cause the catalyst lifetime to be shorter.
If tau =0.005 the minimum C A=0.607 If tau = O.Olthe minimum CA = 0.5088
PIO-2 (g) Example 10-7 (1) If the solids and reactants are fed from opposite ends,
da
kda
dW
Us
atW=We ,a=l
--=-
C 1
= kd~e U
S
This gives the same expression for conversion as in the example. (2) Second order decay 1 a=----kd W 1+--Us
_~_= kC~oUs In(I+'5dW J I-X
kdFAO
Us
1.24= (0.6)(0.075)2 US_In
(0.72)(30)
(1 + (0.7~)2~~?OJ Us
Solve for Us by trial and error or a non-linear equation solver. Us 0.902
=
(3) If G =2
dX dW
(l-x)2 AO (1+c X 2
r
F --- = kC ------a AO
10-6
2 ( 2£ (1 + £ ) In (1 - X ) + £2 X + 1+£)2 X = kC AOUS ( 1- exp [-kd W]J e
I-X
Us
kdFAO
9X
12In(I-X)+4X +--=1.24 I-X
X =0.372
PIO-2 (h) Example 10-8 Uo = 0.25 1 . 0 -,- - - - - - - - - - - - - - ,
Uo = 0.025
0.8
~.'."" L~J
0.6
0.6
0.4
0.4
0.2
0.2 "-",
4 1.0
z
8
6
10
Uo =2.5
0.0 1.0
0
2
4
z
6
8
10
Uo=25
.~
0.8
0.6 0.4 0.2
~-
4
z
6
8
10
0.0
~--'"".--=~
I)
2
4
z
6
PIO-2 (i) For EA = 10 and Ed = 35, for first order decay we rearrange Eq 10-120 to:
In(l- kdotEd EA
J
= Ed (!-'~J R
T
To
=Ed - ( ._--1 1
R I'o T
J
lO-7
8
10
1
R -In Ed
l-
T=
kdOtEd
1 1 =--1'a T
EA 1'a
400 =--------~--------~
1+0.07948In(
1 ) 1-0.00286t
PIO-2 (j) Individualized solution PIO-3 Solution is in the decoding algorithm given with the modules PIO-4
W+S,·.,.w·s
TBA.S,(.,.TBA + S
PIO-4 (a) Surface Rm Limited
r
".Wl.
=0
kA (~ ~7EA·S
C1,,.,,C,,.
'k, V {" t ==-'j('-" == £\.1BAI..'nl.·1 L y, ¥
.
1
K
l '
SInce
D
K" .
';;;;;
D
10-8
'TBA
PIO-4 (b) Adsorption of isobutene limited rADl ;kAl
[Cr - ::~] Cv •
Bey Ridea1 Kinetics
PIO-4 (d)
10-9
W .. S + 19S1HTBA+S2+S1 2 r
-= k
S
S
C _.~1.BAKSC:~~'l=Y1l
[' C
'W·S2 1.$1
en == C V1 -+- CloSl CT2 =c'V2'W-S2 +C
PIO-4 (e) Individualized solution PIO-4 (0 Individualized solution PIO-5 (a) H..,+ C..,H .. ~ 4 H +E-4A Bey Rideal. E·S~4>A+S
E + S ii"~ E .. S • CE1OS :::: KE. PE. Cv "
TS "" ks[ CE .. S PH]
CT""'CV+CE"S
f~ =C~~{-l:~~;l PIO-5 (b) Individualized solution PIO-5 (c) 02 +2S
p
Az +2S p
20·S
+ O·S ---) C3H 5 0H.S C3 H 60H.S P C3 H 5 0H + S C3H6
B + A·S ---) C·S C·S---)C+S
-riJ = r's = k 3 PB C A • S rAD =kA [ PAzCV2
-
2A·S
C~'S] KA 10-10
rAD =0 kA CAoS = CV~KAPA -rB = rS
= k3PBCV~KAPA
re.s = kD [ Ce•s -
P~~v] = kD [Ce.s - KePeCv 1
rcos =0
kD CCoS
= KcPcCv
Cr = Cv + CAoS + CCoS -rB = rs
k 3 Cr PB
= Cv [1 +~KAPA + KcPc ]
.JK:P:
= l+~KAPA + KcPc
PIO-6 (a) A (butanol) ;: B (butene) + C (warer) Possible meChanism: 7 .. :::: ,~..
K (PC . . . .K~·AS_:) AA
,4
S
AA.
rs "" ks (CA.s Cs • CaoS CcoS Ks) rDB
C·S
t:! c+s
=:
knB
rDC;: klX
Assume stuface reaction controlling:
( ··a-s ' - PB Cs. -- PB K AS ('··S -K DB
' == -KPc Cs '" PC KAC (~-5 ( ,<:'5 rx:
10-11
(esoS . PB Cs/Kos) {Ccs - Pc
Cs/Kr:d
'-
-
(
.,!
..
pgpc C;._.) -_ kS KAi·s -'Zl'pA
·r,J - rs - k s! P,tK,Li·s -- KKK \
S
site bala:lce: :. ···rA
;:: _.......
DB
DC
_.
}!ePc h . K - KKK K K Jl were eq S AA DC DE eq
('T == Cs +- CAS +- CBS + Ccs :;: Cs (l + KAA PA+ KAB PB + KAc Pc)
ks. KAA cf (P.A ~......... F1l_ ......P~) _ .. _ _ ._.
{I + P A K.A.A + PB K.A.B + Pc K A c)2 If PBO::: 0 and Pco == 0 , then .;
:=
AO
~2.!MS1!:M.;:: {I -+ PAn KAA )2
Is1 PM 1 -+ k2
_.
Pio +- k3 PAO
This is consistent with the observation.
PIO-6 (b) From r..iJ.c figure. 2
3
4
:;
6
0
0.275
0.5
0.77
0.77
0.5
0
4.5
27
54
112
229
405
6 . 45
8..14
L2.06
21.4
Point numb::::r
~I:
(_lmr~~)
AO hr'!b cat
PAC (arm)
.J.p-; . . ·"fAO
10-12
At large PAl):
-r~O =~.Ll'.~Q :;; ¥1... k2
At small P AO: .. rAO
. ;AO:= kl
=..... _... 1T
P~o
K2
(_l_) ; using point 6: k, PAO
=(229) (0.5) ;; 114
k2"
P AO ; using point 2: kJ :: 0061
=
k2:::: 5.34xlO-4
0. 061. ~ AO ............... ; so 5.34x 10-4 PAo + k, P AO
Using point 3:
k3 :: 7.05xIO· 2
Using poim 4;
k3
Using rJOim 5;
k3 "" L05x 10.2
:=
3. 19x 10,,2
= }
k3 = 3.19xlO-2 (The reason for the different values of k, is from reading the graph)
25.0·~ - - y "" 4.3103 + O.073937x R= 099583 20.0 -
lntercepb1/ko. 5 k::: 0.054 Slope
o.0
J
'~'j"'.:"""'~
o
KA
:::
:=
KA/kO'!
:::
4.31
= 0.074
0.32
····-~~i· .. ·~~ .. ·"':"·····':"· .. "':"."""';.... "1"..""":"...... ~ .... ~-.;" ..~-:"'--o:"' •• ~~\.' "';'-'''':''$''~''''"'~''''-i
50
100
P
ISO
200
250
a
PIO-6 (C) Find the percent of vacant sites
Pb and Pc "" 0 so that red llces to:
%vacant ::::: ._-_ 1..... - .. _.:;::: .........._....1__ ...__. =: 0.41 l+K AAPAO 1+0.01596*90
Find the percent of sites occupied by A and B. No B will have occupied any sites at X : : : O. So; KAAP ~90 -- 0-9 A i''''1:." -_ .-.......... _ . . _-._. -_ ...... 0.01595 ,. . . . . ,....._, ........,..................... . ,J 1 + K,<\,APA "1 -+- 0.01595" 90
0/
PIO-6 (d) Individualized solution PIO-6 (e) Individualized solution
---------------------------------------------------------------------PIO-7 10·13
ME
--'+
D}(fE + rhO
. -- ... _-----
~
100
300
The r.ue o~ formation ofDME is gream:r inirially. 11ris is a result of IJl(){e vacant siteS being inirially available for reaction because water is not adsorbed on the sites. As time goes on
the equilibrium concenll:u:ion of Water sites is reached. WaI.C:J: is strongly adsorbed on this catalyst.
Probable Mech;mism ME+5
:;= ~fE·S
2ME· S""""1> W· S + DME
t
S
Assume Surface Reaction Controls
CME-S "" K/.u:. PME Cv
Cw.s == Kw Pv". Cv
PIO-8 Given: Kinetic r.:ue expression for the reduction of NO over :l solid catalyst: kPNPC
t" ;: """"""'""""---"--------
{I +- Kl PN+- K z Per
~
== partial pressue NO
Pe =: partial pressure CO
Assume that overall reaction is of the form
NO +CO
"....,.t·N2 +COz
PIO-8 (a) 10-14
It is seen that neither Nz or C~ appear in the denominator. This infeIs that neither is adsortxd on the catalyst, On the other hand. it can be infered that both NO and CO are
adsorbed on the surface. The squared denominaror suggests a dual site surface reaction of the adsorbates of NO and CO. Therefore the following mechanism is proposed.
P,.. ;;;: P,vo
kA."i
NO(g) + S
<=:
NO· S
Pc::::: Pea
keN CO(g) + S
<=:
CO· S
'-1"0/ ;;;:
k:cN [Pc Cs - Cco.s/KcN]
ks NO • S + CO • S -1- ~ N2 + Cth + 25 With the surface reaction controlling
CNQ-S ;;;: KAN' PN Cs
4:o-S :: keN Pc Cs Tuen CT :: Cs ,. CNQ.S + Cco.s :: Cs (1 +- KAN PN + Kc."'N Pc]
and therefore reaction is -rs:: ks CNo-S Cco-s:;; ks c..~ PN Pc KAN Ik"N or.r$ ""!ss..!C~,~,,f~?s:.,gL [1 +- K.~"I fIN + N.:N PcP
with
k\ :: ks
ct Km KeN
Kl :;;KAN K1=KcN
kJ Pc PN'
"'1"s :::: ---,.",,,-,,,--'- ' "--'-.--~,,-..-.
[l + K t PN + K2 PcJ2
PIO-8 (b) Assume chat Pc
»flN. Then Pc chattges very little during the course of the reaction
and remai.'1s consu.m,. A maximum in (-rs) rhen occurs. for a fixed value of PN at:
10-15
The rate of n::.action will. increase ..vim an inC're:l.Se in Pc until me: above value is reached. • after which it '..vil.! decrease. It appears tbat there is an excess pressure which will mini:miz.e reactOr YoiU!l'le .. Operating at excess pressure greater than t..'"tis value will. decrease (-rs), and hence increase V. This analysis is exact if the catalytic reaCtor is a CSTR. If me reactor is
t:re:ued as PFR. the Cthical value of Pc is only approximarc. but the general observation is qualitatively the same. This analysis further assumes that the excess CO can be elimitmcd easily and economically
downstream from the NO converta'.
PIO-8 (C) The conditions for which the rate law and mechanism are consistent are the following.. DIe CO S surface reaction must be the rate limiting. P cdP NO must be small. The mechanism must be a dual site mechanism (which it is).
PIO-9 Methyl ethyl ketone (MEK) is an important industrial solvent that can be plOduced {roll1lhc dehydrogenation of outall·2-ol (Btl) over a zinc oxide catalysl
Bu·+MEK+ Hz /i) 11 t C The following data giving the reaction rate for MEK were obtained in a difterentialreactO! at 490°C, _., - - - - , • - .,.... _,-
l~\Iu:
(atm)
PH, (atm)
_~;[J!:~ tIY1f:l!h x
-¥~~"'"
0.5
PBIt (atm)
cat)
5
o
o
o
2 1
0.044
O(W)
0069
0060
0 0
10
0.043
0.059
PIO-9 (a) Suggest a rate law that is consistent with the experimental data From data sets 2 and 5 PBIi (aim) l~HBK (atm)
()
PII , (atm)
0
(molf}! x g cat)
. __.:c:= ...._. __.__.
0 . 040
0..043
-... _ .._ ....._..._......__ ._._ .........---.. - .... _ ... .
we can say that an increase in BlI partial pressure slightly increases the reaction rate.
10-16
From data sets 1 and 5 1
5
l~"le-K (aIm)
5
Pit, (aim)
0
o o
Data Set Pllu (aIm)
0.043 we can see that the MEK partial pressure has little if no effect 011 the rate law. From data sets 4 and 6 ...
DataSet.. -.-,.--,PSlI (arm)
,,~ ~~
P.WEK
..
.,-.-~,.-~~.~"
.. ..",.. .... ,~.~
"
'"~
..
6 1
4 1
(atm)
~/1 (aim)
r;~~!i .(~rl(){/ft~~JL ca~L
...
o
1 1
10
0.060
0.059
Is seems that the pattial pressure of Hz has no effect on the reaction rate. If MEK and Hz are weakly adsorbed (or not adsorbed at all) we can proposed initially the following
.- r : : : r ::::: . . . . . kiP, . ---.:.......... J
,
A
B
ItkP 2 A
But, from the complete data sel "'
Data Set
2
2 0.1
0.044
0.040
'" ..•..•... __._ _ _ _ •.•
cat)
_~
.•• _•• _...................... _-...-_._.'_.'
_~
.......... __ ............. .
4
3 05
___ ••... __ ........................... _.• .·.M .. .. _____ ·_ ••·..." ~
2
0.069
0..060
We can see that the reaction rate goes through a maximum 008 0. 07 006 «i ~ 0.05
S0
,5, ~ - lii
...
004 003 002 001 0
0
15
0.5
,
~
2
PUu (lltm)
10-17
0.043
.........
-.-~-
....... 6 . -.- .... .
-
1
0.059
PIO-9 (b) Suggest a reaction mechanism and rate limiting step consistent with the late law. One possible mechanism is the following one
<-> A·S (2) A S + S ,) B . S + C . S (1) A r-S
(3) B S' "7 Bt, S
(4)C·S<)C+S If the limiting step is #2 (irreversible surface reaction) and the others me at Pseudo Steady State (# L, 3and 4)
,
kC A.) ("/V
.... fA ;;:: '2
A site's balance wHl yield
TherefOlC,
Solving tm Cv
C"
(1
=
1 + P.K + "
I
K3
+ f~.
K .!
Substituting the expressions for Cv and CAS into the equation for -r' A ". r~
=.:
k 2 C;\ sC"
= k, K IPr\C..~ =,_,._~f..f
1+ p. K + J>!L il
I
K,
+!~.lK) 4
which for the case of weak adsorplion of MEK and Hz reduces to
PIO-9 (c) Individualized solution
10-18
PIO-9 (d) First we need to calculate the rate constants involved in the equation for -r' A in part (a). We can rearrange the equation to give the following
fE;*+ jtp, YJk:' and intercept equal to Y.Jk:·' Shown below is the linear
which is a linear equation with slope equal to regression we did using the problem data
g
"'-1..
7
6 V1
0 <; ~
I
~
4
e::.'"
3
~
I
------i
5
I I
2
•••
(PaJr'~"Y{}5
I
'.H
= 2 7298l>flu +
I 3362,
,,': 0 .9991
1
I
I
I
0 0
05
f 15
2
2..5
Thus from the slope and intercept data mol
kl ::::: 0.56· .--._. h· gcat·cum
and
k2
:::::
I 2.04 ---_. atm
Thus,
The design equation for t.he PBR is
F ..cJ?i_. :::::--,' AO
dW
II
From stoichiometry (gas phase)
C CA({i~-~~-\J¥ ~t1 :::::
From the reaction E:= 1+ L·· 1 _Assuming isobaric and isothermal operation and using ideal gas law
p,\
=PA{Li-)
Using equations 10 ..8-13,10-8-14 and 10-8·-16 together with Polymath we can solve for W at X::::: 90%..
10-19
See Polymath program P 109-d..pol. POL YMA TH Results Calculated values of the DEQ variables initial value
Variable
o o
W X
maximal value
o o
10 10 560 2.04 600 -12.228142 12.228142
final value
23
23
0.9991499
0.9991499
1
1
1
e Paa Pa k1 k2 Faa ra rate
minimal value
1
10 10 560 2.04 600 -2 . 3403948 68 . 5622
10 0.0042521 560 2 . 04 600 -68 . 5622 2.3403948
10 0.0042521 560 2.04 600 --2 . 3403948 2 . 3403948
ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -ralFao Explicit equations as entered by the user [1] e=1 [2] Pao = 10 [3] Pa = Pao*(1-X)/(1 +e*X) [4] k1 = 560 [5] k2 = 2 . 04 [6J Fao = 600 [7] ra = -k1 *PaI((1 +(k2*Pa))A2) [8] rate = -ra ~ . mol mol it should be mentioned that I, . :::: to . . ;;;: 600--tlU min hr
---
1.0
70
56
0.8
Q
06
-12
0.4
28
0.2
1-1 '---_ _ _ _ _ _ _ _ "_ _ _ _ _ _
"~
.0.0
(I
0.0
-1.6
9.2
\"
13.8
184
23.0
0.0
PIO-9 (e) Individualized solution PIO-9 (0
10-20
4.6
9.2
W 13.8
_ _ .J
18.4
23.0
Now consider the change in pressure: Stoichiometry:
P =C RT=C
Pressure:
. "".dy_. . . =.·'n ".". - (1+ X )
AA
dW
A"
(1-X }P
._--, -·..,RT l+iX Po
2y
Use these new equations in the Polymath program from part (d). See Polymath program P 1O-9-f.pol.
POLYMATH Results Calculated values of the DEQ variables Variable W
initial value
o
X
o
Y
1 1
e Pao Pa
kl k2 Fao ra rate alpha
10 10 560 2.04 600 -12.228142 12.228142 0.03
minimal value
o o
0.0746953 1
10 7.771E-05 560 2.04 600 -68.584462 0.0435044 0.03
maximal value 23 0.9997919 1 1
10 10 560 2.04 600 -0.0435044 68.584462 0.03
ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(W) = -ra/Fao [2] d(y)/d(W) = -alpha*(1 +X)/2/y Explicit equations as entered by the user [1 J e = 1 [2J Pao=10 [ 3 J Pa = y*Pao*(1-X)/(1 +e*X) [4] k1 = 560 [5J k2 = 204 [6J Fao = 600 [7 J ra = -k1 *Pa/( (1 +(k2*Pa) )1\2) [ 8 J rate = ora [ 9] alpha = ,,03
10-21
final value 23 0.9997919 0.0746953 1
10 7.771E-05 560 2.04 600 -0.0435044 0.0435044 0.03
1.0
70
08·
56
0. 6
42
04
/,
[;J
./!
r:f.f{-
28
/\
I
/'
I
\
<",,"'
0.2
14
0.0
(I
9.2
0.0
W
13.8
1804
23,0
,,>P'
~ '"
...
'}
/
~#'V,-/t7
\
" ",
\
0.0
4.6
9.2
\V
13.8
18.4
"
"
23.0
PIO·IO (a) Iso,ocrene +. Hydro gcn --? isCH)Ctanc
A
B
-t.
Discrirninarion of mcdds: Assume
'fA ;:
k
C1 q
For runs 2 and J, 0 <
q: a: < 1 ; 2 and 4, 0 < f)
< 1 ; and 2 and 5 , ·1
< r < O. From
Perry's handbook, 5th cd". p. 4-8; the reaction is probably surface reaction rate con'Q'"olling. Mechanism I (H. Alvord):
A+S+!A.S B2+2S ~ 2B·S
A·S,2B·S;:! C·S+2S C·S ;:! C+S Hence.
k fp;.. - PcfKuJ .... _-,._.. -rA, "'" _._., ..•,..._. '--,•..PB ,_.,-""-,
[1 +'KAP;..+KfPft+ KcPcP Mechanism II (S. L Mullick): A+S +! A.S
B+S ~ B.S A·S+B·S f:!: C·S+S C·S ~ C+,S Hence,
, _
k[P Ps - Pc/l<.,qJ ...... _ ••... , . , ....•.
"1;.. - ...." ._..._, ...• "...• _...A ,••.•.- ." •.._
[1 ,. K ... FA t· KB Ps -+- Kc PcP
10·22
From runs 2. 9
11. 12. P A:::: PB
=Pc "" P. a plot of ~rA vs. P shows a parabollic
behavior. therefore we Mil drop the second term in the denominator for easy linearization. Tne readers can calculate ~ value by Gibbs free energy change 4'1 this equation (up to 650
K. the reverse reaction is negligible) Tne linearized regression model is:
Using given 12 data pointS to solve for these four unknowns: y :::: 3.0 -+ 1.42 P A
.;.
0.97 Pa
-t.
1.42 Pc
The tinal results are:
P A Fg .... ". 0..1113 _._-,,--,,-,,_. - _ ............... _ ...._" (1 + 0.475 PAt 0322 PB .;. 0 . 414 Pcr
-fA:: ...-
The comparison of the pen:emage error between the model and the experimental data are: r (exp.)
(~A.!:lit .5
r (calc.)
0
0..0362
1 1 1
0..0.239 0..0390
:'5.26 647
0.0345 0.0227
8.77 9.25
0..0410
3
O.D114 0.0534
Run
PA
Pa
Pc
2 3
1 1 1 3
1 1
4
5 6 7 8 9 10. 11 12
1
:3 1 1 10 1
0 0 10.
2-
2
2
0.2 0.1 5
0.2 0.1 5
0.2 0.1 5
I 10 1
1
o...Q351
0..0310 0.0033 0..0380 0.0032 0..0008 0.0566
expo
9.37 13,,69 17.96
lo.Al 10.26 3.54 3.54 21.02
% elmr
-4.8 -4.7 +5.0
A.a 0.0334 +5..7 0.0120 -5.4 0 . 0.505 -2.6 0.0302 -4.5 0.00315 0.0380 0 0.00288 ·9.8 +10.7 0.00089 0.0599 ....:.1:2 _ Isuml =59.2 avg.
=4.9
PIO-IO (b) Discussion: The readers may check the validity of mechanism L To reduce the accumulation of error in calculations, the readers should have used ail data points and solved all unblowns simultaneously. To get the maximum informacion of complex kinetics of a reaction from the least runs, it is advantageous to do planned experiments such as factOrial design .. (W. G . Humer. and A. C. Atkinson. Olem1cai Engin~~rin!!, p . 159. June
6. 1966). A paper discussing chemical reaction rare equations from experimental data is in: C. n. \Vare 11., Summer Compmer Simulation Conference. Proceedings. 1975, Pa.."t 1. p .
368. 10-23
PIO-IO (c) 0.2223 C~o (RTr {l-Xr / (l-O.SXY ··rA =.-....- - -..- - . - - . - - . - . - -................
--.-~.-.--'J-
r1 + CAOO ~T {(0.475 + 0..322 (I-X)} + OA14x}11-
l
.)x
.
0 ..1 113 (RT CAot -r!\ = ----.
(l·.xF I {l-O.5Xr ..--.--.------..
~------
.
FAO:::
r 1 + c.:..o RT fO 79 7 .
2.5
1-0.SX \.
=:::
,
•
0383 .
x'\j2
'!
i50 I~l
CSTR:
w;;;;;; 21.380 g ::::: 21.4 kg
x
w::; 1501 o
2:1l13J1.5J:i!·X(LL0.5 x)Z ~
1 1 - 0 ..5 X
..,..
11+' _. . . . . .,..2. . . . . . {O.797- 0..383 X)I~ l.
'j
10-24
1 x
w= . .- . ..J~Q
G (X}dX
0.1113 x 2.25
0
. 1..5 (0..797' - 0..383 Xl]2 (1 - 0.5 xjZ [...1-.. + -.-----------. G(X):= L.:Q~.X ___ . __.._._. . . _. . . . . . _. . . where (1-
xf
x
G{X)
0. 0 ..1
4.82
~l
5.38
~t
0.2
6.13
0..3
7.16 8.66
OA 05 0.6 0.7 0. 8
COO
11.00
:1 0
IS.03
0':'
a.l
I)
23.15
Using Simpson s rule. area. under the curve ~ i 0
w:;: ..._____.J2!1_____. __ x.
10:;: 6000 g "" 6 kg
0.1113 x. 2.25
PIO-IO (d) Consider the differential section between L and L
+ dL
In - Out + Generation"" Accumulation 4
,L)Ar Pc-H······ t/J)dL ==
-.dl{"1· Arv{L)CJL)]
But: A1 v(L)C4 (L):= N)L).:::: f~ll(I ... X) :. FAD
f·:: TA pdl ..
$) rA {Pd
Relating the expression fA to X and L :
;; lFo (l·X) 2
1
:;: }Fo (l-X)
R, =-(2······ X) 2 X_•. P __F ...:1.. _I...........
Fc=FAOX
l-Pr -2· . ·X
== tFox
PA == PA ... 1:.:X. Pr 2·X
0.4
"
44 . 62
r)C
c..J
Pc=-X-Pr 2-X
LetP=Pr,
10-25
as cu.
0::
lUI
From th.e Ergun equation. we: have:
All the things on R11S arc: constant. except far p
p=p,((L!x) . :i = ~2X(~:)I~)gc%;;(1;1)l~50 ~-$)l: + [75 GJ df. == 2.:X dL
2P
~o where 6 .
0
=.!?Q.Q.. . . ~~:~?r.~. ~?,..0.,:d» 1: +Poge D? Q> L Dp
Let L" ",., L..... : p. =..P. . then: dE"", l-X. L-des Po dL 2p·
1.75 G
\3:
where
$=04
G :;:: t~ An fiA + Fso fiB) == AT '"'''At: 9 G == J.,~ T
P :; (3 atm) o
T.......
. ,.u.a.ll.•
frZ hr
(14.1 ?E~,·) (1441 frZm:.) =63487 .lb. ft2 attn
p ;; C r m == (5ZJll.\ L 0-.'·
( I'sO gmol.) (2 + 112}~.-,{---.JL"-.) hr gmol 4)".6.g, T K. {fr..25:I\2 f~ 4 12}
lhmal J RT
(57 _Jh.,_.1 (3 ann) := ._ _ _ _ •_ _ _
(0,7301 \
.Jbmcl.!........"." . . . ,....:.....__ .. _ _...
ft~..rurrL\1 [(lOO + 273.15)( 1.8 jJ OR
IhmoloR,
10-26
IiJ ;:: 0.4
Dp ;: (_1 in) U . . fL} :: _1 ft 16
\12in
192
&<: = 32.174 lli.!IL.JL Ibr sect
-JL-)
(150 grnol) (2 + 112) .....1..-.. {.. G;:: (FAO rnA +- Fae mB) =. hr gmol ~5.).6 g AT
T
1L{n..2llj f~ 4 12
G=~ J.bm.. f~hr
T
Po =(3 atm)(l1:r:iHl~f~~-) = 6348.7!i' __
(57
,.J.b-:-) (3 atm)
={l~:n~lJh= (0,,7301 ..fy3 9trU~IL(20~:;3.15H~ .all oR
Po:; Cr m
IbmoloR,
Po:: 0.2750 !wt ft)
Ii (3 atID. 200°C);: 9.4719 x 10,3 cp
(t Cg,
~. He) }
-~.
:: 8.6211
=
X 10.3
PPROP
cp (all Ca)
(94719 X 1O·3j- 8.6211 x 10.3 ) -4 Ibm " !J. ;; ................- -..- -.........- - - - - cp 6.7197 x 10 ----2 ft sec cp x JQQQ..z..;;. hI" Il "" 002188~n.
ft ru"
10·27
where
• rAT p~ (1 . 1j» K P; Let: a. "" - . -.. . ------FAo Then: dX:= ell
(1.- _ ••.....
K~
=(KA + Ks) Po
J!.:::..~_
~K', K' xl1 lP:-+-l'1' 2 J
f2-X
For a given T. we
can
solve the
twO
ODEs to get X(L *)., We need to guess T until X :::; 0,8.
For the rate cOnStaIUS. we use the solution of 6.10 (a) K :::; 0 . 1118 - .... ~:::~! . .-.. .gear'" hr • ~~.
;
KA"" 0.475 amr 1 ; KB
Kc "" 0.414 atm l
- (0 ..~ TAO K ·2" ..... ,4"7.)
a:
:=
() .~~")} - • 149 .... ,.,.- ~ .)-"'1..
85.958 T
10-28
= 0.322 atnl 1
~. =(5...5.ll x 1(J4 f2..hr.) (378.09 + .ll2.Ql) J1'm.. o
T
ll:xn .
T
f~
hr
~: == _Q 2191 _ 8,~;5:i
A FORTRAN program is written to solve the equations. The results show that any numbers 6f tubes !oufiicient to allow the given flow rates with a positive pressure provides more than enough catalyst for the desired conversion The problem as stated, therefore has no solution. However, we can choose a ditIerent L, and it only changes the dimensionless parameters.. With L = 20, the problem is still unsolvable For L ::;: 10 ft
, T:::: 2.16 rubes
=
1-::::
0.46
In
Note: Using the modified program with
Dp ::;i~ in • II "" 9 . 05 x 10"3 cp
L""' 10 ft
A:; 2.15 il'l2
=> X", 0.80 • p. == 0.7531
Using 1 1/2 in schedule 80 pipe (I.D. := 1..5 in) • A = 1.76715 in2 and me length to get X "" 0 . 8 is impossible (p( < 0). With 2 in schedule 80 pipe U.D. "" 1.939 in). A =- 2.952877 in1
This gives L "" 6.6Jft , X == 080 and p:::: 0.9113 1 1/2 in schedule 40 => A "" 2.03580 in2
L '"
1O.9~857
, X = 0 . 8 and P "" 0.6903
PI0-ll (a)
10·29
Assume a rate.,ljmiting step; start with surface reaction A·S+S.....-··BS+H S
H·S ..... ·'H+S Assume a rate·lit,wting step; start with surface reaction ~-~~~
.~4D.
:::
0
.:fl!2.. ::;;
0
kAD
knv
Find the expression for Cy
Combine all of those to get the following rale la\v suggestion.
CheckinO' to see if it fit>;, we see that for high P A' increases in P A cause decreases in the rate. \Ve see that if PB or PH increase the rate will go down, which is consistent with the rate law .
PI0-ll (b) Now nsing POLYMATH's nonlinear equation regression we can find the values for the parameters. We find that k=O ..o0137 K·a = 4'16 • I
0 ")-9 K B --._':>
Kc::= 0.424 In the problem it is given that K.. . is 1 or 2 orders of magnitude greater than KB and ~: which is tme so this is a good auswc{.
10·30
"aXl:~OO 3.200
T
o
~Ii'c;;r "li;;;~
on
:;lAta
J. .
[J Cdl;::~i
vat de
2 .... 00
t.600 0.8CC
0.000
Model: ~
,. azi<.""Pa..... ' 1 ~Kil ... P.a·Kb.Pb··l(h)(Pt')-2
"
0.00137022 Ka .. '+.7608'11 ? pOsatlv'@ r€'su::l,~ .. ls.
ICb .. 0.259382 1<1'\ '" O. "t2359
:3 "eqatlv~ .... 5lC:luals.
Sum o! squa r -E1'S '" 5.27033~-15
PIO-ll (C) The estimates of the rate law parameters were given to simplify the search techniques to make sure that it converged on a false minimum. In real life, one should make a number of guesses of the rate lw parameters and they should include a large range of possibilities
PIO-12 (a) Assume that the second reaction is the rate-limiting step.
Using PSSH, we know that rS"'·H ",= 0 I 4'"
= k1CS·u .
lD4·
sC·'V
10-31
Perform a site balance:
Combining all of these we find:
This rate law is consistent with the data. As the concentration gets larger, the rate change gets srnaller which is consistent with the rate la\v as given.
PIO-12 (b) No answer is right or wrong, but the points will probably be higher than the ones given to see that the change in rate becomes even smaller.
PIO-13 Assume the rate law is of the form rD
ep
= kP::rl~O
1+ KPVI1PO
At high temperatures K J,. as T i and therefore KP::rIPO «1 '[Yep
= kPy~ [PO
'[Yep
--=k
Py~IPO
Run 1
0.028 ----2
=11.2
( 0.0.5) Run 2 0.45 = 11.28
(0.2)2 7.2
Run .5 - - 2 = 11.25 (0.8) At low temperature and low pressure '[Yep
= kP~TlPO
r
--.!l!L = k
Py~IPO
Run 1
.2. 004 = 0.4 (0.1)2
10-32
Run 2 0.015
(0.2)2
= 0.375
These fit the low pressure data At high pressure KPv~1PO »1 rDep
_ kP~1PO _ k KR2 - K
-
VT1PO
This fits the high pressure data At PVTIPO = 1.5, r =0.095 and at P VTIPO =2, r =0.1 Now find the activation energy At low pressure and high temperature k = 11.2 At low pressure and low temperature k =0.4
In(k2) = E(~._._L) = E(!2 -7;) kl R 7; T2 R 7;T2 E(~3-~~)
In(I1.2) _ 0.4 - R (473)(393) E =7738 R
E
cal = 15375----
mol
PIO-14 r1"iD,. :::::
ks.~.s
1;;:; Iv + ft.s;::: iv(l+
~1;
:
Kp
K,fj)
• Ii "" Kp Pfnp/PPI
_ ks Pr Kr 1 + PI Kr
I'iiOt----···
10-33
h-s:::; ~K[j~
f~.~, == 1:1:~' ks
Pr Kl
I'Ti(~ ;:; 'r::;:: P~ Kr
kPirIP I pp! rno, =i~r~~;?-I-'i~)K Low
High
kP;'TP
=1>;1'" ~:P:;;;K
Pnn>:
Rxn is second orrier
Since
1»
K PfnpfPPl
PIT!P: Reac:tion is ze:ro order
High Tempermure Kr very smail such that
PIO-15 (a) Using Polymath non-linear regression few can find the parameters for all models: See Polymath program PI 0 15. po I. (1)
POLYMA TH Results Nonlinear regression (L-M) Model: rT = k*PMl\a*PH2I\b Variable k a
1 0,,1
b
0. 1
PreciSion RA2 RA2adj
Ini guess
Value 1 . 1481487 0 . 1843053 -0,,0308691
95% confidence 0.1078106 0.0873668 0,1311507
= 0,,7852809 = 0,,7375655 = 0,,0372861
Rmsd Variance = 0,,0222441
a = 0.184 ~ = -0.031
k = 1.148
(2)
POLYMATH Results Nonlinear regression (L-M)
10-34
Model: rT
= k*PMI(1+KM*PM)
Variable
Ini guess
k
1 2
KM
Value 12.256274 9,,0251862
95% confidence 2.1574162 1.8060287
Precision R"2 = 0,,9800096 R"2adj = 0.9780106 = 0,,0113769 Rmsd variance = 0,,0018638
=12.26
k
KM = 9.025
(3) OLYMATH Results Nonlinear regression (L-M) Model: rT = k*PM*PH2/((1+KM*PM)A2) Variable
Ini guess
k
1
KM
2
Value 8.4090333 2.8306038
95% confidence 18.516752 4.2577098
Precision R"2 =-4.3638352 R"2adj = -4.9002187 Rmsd = 0,,1863588 Vari.ance = 0.5001061
k
=8.409
KM=2.83
(4) POL YMATH Results Nonlinear regression (L-M) Model: rT = k*PM*PH2/(1+KM*PM+KH2*PH2) Variable
Ini guess
k KM
1
2 2
KH2
Value 101,,99929 83.608282 67.213622
95% confidence 4.614109 7.1561591 5.9343217
Nonlinear regression settings Max # iterations = 300 Precision R"2 = -3,,2021716 R"2adj =-4 . 1359875 Rmsd = 0,,1649487 variance = 0.4353294
k =102
KM =83.6
KHZ =67.21
PIO-15 (b)
10-35
We can see from the precision results from the Polymath regressions that rate law (2) best describes the data.
PIO-16 Using Polymath non-linear regression few can find the parameters for all models: See Polymath program P1 0-·16.pol. (1)
POL YMATH Results Nonlinear regression (L·M) Model: r = k*KNO*PNO*PH2/(1 +KNO*PNO+KH2*PH2)
variable
Ini guess
k--
-1---
KNO KH2
1 1
Precision R"2 = R"2adj = Rmsd = Variance =
k = 0.0031
Value 0 . 0030965 57_237884 10l. 9967
95% confidence 3 . 702E-05 1.0353031 2 . 2870513
0 . 9709596 0 . 9645062 5 . 265E·07 4 . 436E-12
KNo = 57.23
KH2 = 102
(2)
• POL YMATH Results Nonlinear regression (L·M) Model: r = k*KNO*PNO*KH2/(1 +KNO*PNO+KH2*PH2)
Variable k
KNO KH2
Ini guess
0:1-10 1
Value ----4.713E-06 -108.42354 1 . 046E-05
95% confidence l.297E-05 4.9334604 2 . 878E-05
Nonlinear regression settings Max # iterations = 300 Precision R"2 = -9 . 6842898 R"2adj = -12 . 058576 Rmsd = 1. 01E-05 Variance = 1.632E-09
k = -0.00000471
KNO = -108.4
KH2 = 0..00001046
(3)
POLYMATH Results Nonlinear regression (L·M)
10-36
Model: r = k*KNO*PNO*KH2*PH2/((1+KNO*PNO+KH2*PH2)"2)
KNO
Ini guess 0.1 10
KH2
1
Variable k
Value 5.194E-04 13 .187119 18.487727
95% confidence 2 . 242E-04 7.659298 7.7652667
Nonlinear regression settings Max # iterations = 300 Precision
= 0,,9809761
R"2 R"2adj Rmsd Variance
= 0,,9767486
= 4 . 262E-07 = 2 . 906E-12
k = 0.000519
KNO
= 13.19
KH2
=18.49
The third rate law best describes the data.
PIO-17 (a) Mole balance:
-dX. . . . . =,_(-.,~)a . .__ . . . . " ~
dW
FAC
Rate law:
Decay laVor:
.4.':. =._ kdc:.~:A. dW
Us
Stoichiometry:
Evaluate the parameters:
8=.8
POLYMA'I11
m tial ..y!!!!!
~at.4:2E!.l.
o
d(x) la{w} :::a* (··ra) Ifao
d lal/d(w) ",··kd*a "ca!Us
1
kd:::9
fao=4000COO
10·37
Us=250000
(d;
Nov 1:::.g ,. Sed Reac cor
k::::90 cao-= .. 8
Ca
eps'" 8 ca=cao* (1. "xl f (l+eps*x)
-
x
50000
PIO-17 (b) Mole balance:
{!~:,',4 == _(~dL dlZ
Everything else is the same and we !Jt:cd to know"
r
V = ,., = 0.004 Ii;)
5 CS'ffi' s in Series !~~!:~51 d(a)/d(c)~-kd·a·ca
d (ca)
fa (t) ;ca4!t.au ( ( (1*ya4) I
1 (l +'ca/eto)
) t tau'a *k;) 'ca/t.au
d(ca41/d (t) "ca.,3/tau'" ( ( (1+ya3) I (1+ca4/eto) ) • tau· a*k) *ca4/t
O. 8 Q 8
d(call/d(tl"cao/tau'" « (l'-yao) I {l+callcto» ttau*a"k}*caJ./t
0.8
d(ca2) lalt) "'cal.ita\'!- « (l+yal) I (lica2/cto»
0,8 o. S
+tau*a"k) 'ca2/t d (ea}) hilt) "ea2/tau- ( ( (1"'ya2ll (1+ca3/cto) ) +tatl'a"'k) 'ca3/t ko"'9
10-38
value
tau=O 004 cto;:l k"'45
.... q
Ca
ca,0=0.8
--- x .
ya4=ca4/cto yal;cal/cto ya,2;::ca2/cco ya3:;ca3/cto
x=Ccao .. ca)/cao yao=cao/cto to = 0, t
0.5
f
PIO-17 (C) The only change from part (a) is the decay law:
'~='''k aCA dt d r == . W··W _. _.llll!L.•••.•_ .• Us cit =: ;~~~. Us
da dW Integrating:
w· _ _ _ •
=kd(~a. _ _""'--~"_._
Us
In a = .k.CW . .!L_.:i............+k Us W=:WM.AX@a=l
k
=:
_._kdCA!~~AX.
Us
a = exp(~l;~{~~~~~~J) .Ini:::!!::!. y~
Equa!::~£Ef!:l
o
d(x) Id(w) =a* ( . :raj Ifao k=.90
fao=40000DO kd;;;;S Us",250000 ==50000 cao"" .. 8
eps'='.8 ca=cao*(1.-x)/(l+eps"x)
tc) LOOO
~
a Ca
0"""
.. -- x Q.6CO
Q,100
'Moving
TI
"T"",
t
l
.....
:ra",-·k"ca a"'e:)Cp (kd*ca/Us" (w~=) )
PIO-17 (d) 10-39
(Count:er:::Ul:re.nt)
To find the Time-Temperature Trajectory we need to use the following equation for first·,order decay,
r [ E' 1'1 I 'IT] E ;;:k~~C~E:t 1 exp;_~RlT- 300JJ. ·r
t
"'d
Since no initial temperamre \vas given, we assumed one of 300K This is the graph
of thar equation, Temperature- Time trajectory 1000 -800 ::,;: ~ ..... 600 E f: 400 ~ ;:) 200
iii...
o
o
0,1
0.2
Time
0,3
(h)
PIO-I7 (e) The two energy of activations are switched and this is {he new graph made,
Temperalure- Time Trajectory
. " ., ......... "'- J
1000 -..800 ¢i ~oo :::J
iii
E w400 ...
0. ill
I-
-
200"
- ------------
-..
()
The graph looks the same just the time is much smaller.
PIO-I8 (a)
10-40
For all of the parts, the mole balances and rate laws are the same, 'They are: de", ---. =r
v
dvV '" 0 'A::::: --·kaC",
Find the equation needed f()f a.
da
··--::::::kdaCA
dt
a:::::: exp(-.-kdC",t)
a:::::l when t=O
Assuming values for Vo. k, and ~ come up with the following graphs according to the cases described, y:!~ t.ia~.. ,.Y~!~
E:...quat.i.23~!.
o
d(cb)/d(w)~rb*vo d{ca) Id{w) =ra*vo
vo=lO (a)
a=l.
ko=l
..... Ca
t=20
Cb
_. a
k=l
0 a-oC
!Ll!\CO
ra=,,·k*a*ca
50 3 .. 200'
T
Q.~OO t:~0000
".. ., l "*':"
KJ;;,X. .......
C,~
,
Cll
-
Cas. I @ t-1000
1.000
ill
10-41
~ o~c
-
'-"-'., -..... ••..........~ --.--Q."'OO ~cc
-~.'"'~"'t-
Q
...... - ...
-.-.~
o.aoo
..... -.-"'"'!
PIO-IS (b) Find the new equation for a:
,_!la...;;;:; k dt
a:;;;::"···'·······
a2 d
e
i\
1
l+k d C A t Using same values come up with these graphs: g~.t:::i~l,Y~l~~
~~.i,.?E.E_-'
o
d(cbl/d{w)~xb'vo
1
dlcal/dlw)=ra*vo vo",lO
(!:Ii
kd"'.OOl t",lOOO k"'l a=11
e1;:; -.~".
(l+kd~ca*t)
a
1
(h)
Cas,"" II @ t::;:80
Ca Cb a
10-42
Cc.:so I
@ t~
100e
{b)
Case IE
@
t=lG-
lSU Ca Cb - a
PIO-18 (C) Find the new equation for a;
__ .da = k aC
ilt
d
B
a;;;;; expC kd Cst)
The following graphs are made:
-_.__
Eouat.ions:.-
Ini~~~±,. y~:!~~
d(ca) Id(w) =:r:'a*vo
1.
d(cb) Id(w) '"l:'b*Vo
o
vo=lO
kd=.OOl Ie)
t;;;;1000
k:::l a=;exp( ··kd*cb*t)
z:b;;;;:k"a"ca
IOO::
..... Ca
. cb
-a
J:·a",··k*a*ca
10-43
Case I @ :-1000
l.Oc:;
K~
ea en a
~
-----
.. ooc
2.0ae
(e)
1.0':::
.......
-~-.,-'''.-
".000
.......
,.
'.'---.~.~;-,.-~-.---;
6.~CC
Case lXI @
B,.ooe
~-lO
KE'(
Ca cb
:}"iJ!:n'!
Q.
0... 600
c,.lIce -:d.2{;!:!
+
J ;)GG
PIO-18 (d) ~'iaking
a inlO a differential equation we come up with this:
, ,...4!!.. == kdCAG dt W
t :=" Us
. . . .,~a..", == .~:!,~~~.. dW
U,
10-44
JO .. OQC
Equati.~".
lnitia!
d{ca)/d{wj=ra~vo
1
d(cb)/d(w)=rb*vo
Q
d(a)!dtw)=·kd~a·ca/Us
1
va1.u~
vo",10
kd=.OOl Us;1.0
(d)
Case I
~
k=l .th::::::k*a*ca
Ca
Cn
.....- a
ra.t=:'""k*a 1rr ca
ld}
CaSe: II
KEY Ca C.b
a
-+-~.~o~ $P~ 1 .00(/
(d) -~.- .•.. K~X
Ca Cb
T r
t
a
(j.. QOO
1.
0 .. 000
-
Case III ....
--.-.--.- ~-.- -.... ..
..
-
'............
~- ,
~~----
,,,,m~-"'r'''''' .... ::.000
.., ..... ""~
.... ... -,
.....
...
::..:.+=:-...;. ··· ..-..;·~r..-·.-----i-';--'.oeo
..
~
~.,
~ .-~ ~
a.ooo
000
PIO-18 (e)
10-45
,.",.,,~
=i
lo,ooe
Everything from part (d) is the same except for the decay law..
.~~ =-k.aC dt 4'A _ l~ttAX _. HI t--_········,···,·
U, dW dt= - ..... Us da ==: kJC, a ... dW Us .,..".".~
'-'-~'-
Integrating: In a =: ~:L~:'A t~ + k
Us \'11 =- \V iv1A.'.: @ a == 1
~2::~t:iS'!?§~ d{ca)fd(w)"'ra~vo
o
d(cb)!d(w)"'rb~vo
vo"d,O kd"l k=,001 '..mlOlx;;10
a"'exp(kd.*ca/UsX (W'wmax) )
:t:bmlk'-,a,*ca
"'0 '" 0,
10
......... --- ... ".. _- +-- . .. .. .. --1 1> coo G: ace lC,;;;:~<';: ~.
...
10-46
~ac
~.
Case II
(€:)
EEX
_. Ca
", Cb
-
a
~-~~""il
a.ooo
'c_ooC
f;EY Ca
... Cb
_. a
t,'T
O~2cn
I
-'
i
o~oo.c
~~:...:.:.::,;.~.:..:.r:~.:.::.:~::.:::.:
0,000
~ .. boo
2 .. CO(;
., .
,.,+
'¥ ....
~" ........ ' .. ~" .. -.'. ~~ ... ~, .......
... ~ _ . . . ,...". . ~ ~. . ~_ ....-.__ ._......., a. 00;] tc,.o~~
:i" 10-:;
PIO-19(a) da =-k dt
W-Ut - s
D
~__4..f!.. = =kD dW dt
a
dW=Usdt
=1-- kD~ Us
Us
kDW t hen W =---=-=2.5 U s ·5 kg If 0=1--Us kD .2
!X = -TA =~(-rA (0)) = akC:~o (l_X)2 dW dX dW
f
FAO
=
FAO
(1- .~D J (1-
dX (I_X)2
W Us
= kCAO Vo
kCAO
FAO X)2
Vo
fl- kDW dW Us
Activity is zero for W > 2.5 kg, so the catalyst weight only goes to the effective weight.
10-47
2
_kCDWAO [ e ]_1(0.2)[ -x- W -k2.5- 0.2(2.5)2]_ -0.25 1- X Vo e 2U s 1 2 *0.5 X =0.2
PIO-19(b) 1
o ________ ______ ~
~
2.5
o
5
PIO-19(c) For infinite catalyst loading a = 1.
dX
= kCAO (1- X)2
dW
Vo
~=kCAoW=1 I-X
Vo
X=0.5
PIO-19(d)
~= kC
AO
I-X
Vo
[W _ kDW2] _
2U,
-~=0.2[5-- 0 . 2*25] 1-0.4
2U,
_
kg
Us =1.5s
PIO-19(e) kDW Us
a=I----
0=1- kDW_ Us
10-48
Us =kDW=0.2*5=I.
kg s
PIO-19(f) a
=0 means there is no reaction is taking place. Activity can never be less than o.
PIO-19(g) U=-U s . da = kD dW Us kW a=-D-+C Us _kDW; C I ---+ Us
when W =W;, a =1
a = kDWe +1- kDW; Us Us Now find We. 0= _kDWe.+ I·_.kDW; Us Us We
= Us [kD"i. __ I]=.:~[:3*5 kD
.2.5
Us
-IJ
a
We =2.5
\Ve
dX.=(I_ kDW; +wJkCAo(I-xt dW Us Vo
~=~~AO I-X
Vo
\\'
1 - kDW; +W I-lW Jl 1 Us r
~ =~~AO [(W; -W.)(I- kDW; J+.W;2 --We
2
I-X
Vo
_
Us
]
2.
~ = 0.2 [( 5 - 2.5) (1- 0.2 *~.) + 25 - 6.25]
I-X X --=1.875 I-X X =0.65
0.5
2
PIO-19(h) $ = I60FAo X -lOU s
10-49
\Vt
kDW a=l---· Us dX =ka=k(l-!'DW] dW Us 2 X =kW _ kk DW 2U s To maximize profit, a maximum in profit is reached and so we set the differential of profit equal to
O. . d$ =O=160F dX_ lO dU s AO dU s dX kk D W 2 dUs - 2U~ 2 160F kk DW = 10 AO 2U2 S
2
8FAO kk DW
Us
=U~
=~8FAOkkDW2 =~8(2)(.2)(.2)(25)
Us =4
k~
mm
PIO-19(i) No solution will be given
PIO-20 (a) Stan with the mole balance for a balch reactor:
Rare lavv;
Decay law:
da
..-._-:z:
(i t
k J a"'
Stoichiomeny:
Plugging those into POLYrvlATH gets the fc)llov,.ing program and the following graph
10-.50
2~ti2.!~:L' dlx)/dlc)=-rt"w!cto
o
d{a}/d{ti~'kd~a~2
1
w=5 k;:=20 "'~.-~ ....
kd"'.l...6
r
v=l
:).&00
-1- /
T=735
p1:=pto·
(1 ··x)
rt"'·-kc·pt"a
II
Tl
lltO=pt.o"v/ (R"T)
=
//
T/ tI Tf
2.;.082
::0
-
/,/
-!---..... -- -".--.-----,.~-~ ....." ...............- ...................... f - - - - - - . - - j
n.ctio
0,
a. ::u;;;::
.,...~
..
PIO-20 (b) For the moving··bed reactor the mole balance now becomes:
dX dW
.. /; FM)
....-.-,:::;:;;-,,-~
The decay la\v nOw becomes: da
k"a"
dW
Us
."""'----:""....:::::... ........ .
Everything stays the same.. Plug into POLThIl\TIl. The conversion achieved is X:::: 0.266
Equas~~2. d(x) Id(w)"'-!:t:./fao
o
d(a) td (w)
1.
", ..
kd"*aA2/0s
kt",,20
£ao,,600 l
~!lE.;E.!,£
Us=:2
'"
50
o
pto:2 pt"'pto* (l"x) 'NO'" O.
We
20
0 02439·02 2<';
500
600
teD
;;
1 6
5
20
x·c,,····kc"pc·a
fao 1<;d
50
266;6
2
,
2
2 4tr'~68
·-.:10
--0 .7:5i4:
,~c
PIO-20 (C) c) Increasing Us \vill get us a higher conversion" Looking at this summary table, U. = 10 kglh and X :: 0.6
10-51
50 a loS:.5 C .024.:130: 20 600 1 Ii ;;:
~labl~
~SciaJ_val~~ Ml1XimillU._"~5!f.~ Mi:!i~'!!::1~ () 50 0
f_~,.y~l'le
::.
1
o. : l l l l l
0
0
O,SS9-i:iSS
()
CL 599686
f~o
600
600
600
600
kd
LoG
]',6
1.6
Us
1.6 10
:),0
10
10
kt
20
20
20
20
pta
:2
:2
:2
:;:
pC
:2
:2
0.e00625
0,800625
t't
-40
-l. 77917
-4.0
--1.77917
\01
.,.
x
50 111111
PIO-20 (d) For second-order decay:
For
s:::: 25 keal/rno! and E
4
:::::
10 kcalimol:
Use tlris equation in Excel to generate the following graph: Temper ature-Time Trajectory
Temperarur·c.. Time Trajectory
-
~..
soo
r------- --.------- -, - ----------1
~ %500
i~ '"
'wo
--
------
~
£,-=10 lInd E;.;2S 1600 , ...... ,.--.-.... --.,----...
~ -
t-- -------------- -----------~
"]00
" , .... "' ....... ,." .......,....... Q
100
I
.. .. ,.
400
bOD
o tim" (h)
'lime (h)
Use this equation in Excel to generate the above graph.
PIO-20 (e)
10-52
In part e, the only thing that changes from (b) is the decay law and the decay constant: 2 da krJP/a ""-'"-=-..::......:....-
dW
Us
ka = 0.2 Plugging into POLYMATH we get the following summary tables for Us:;:: 2 and 10 kglh . X:;:: 0.50 and 0.88 respectively. X will again increase as Us increases. Eq'-J,a~ions.;.
!ni tia1~1~~
d{x)/d(w)=-ra/fao
o
d(al/d(wl=~kd"'pt~2"aA2/Us
1
kd;;;.2
u5=2 fac=600 kt=20
a
~ ..~~ ~~.,.,~±~ fina:'_~ so 0 So 0 502431
0
0.502431
1.
1.
0 . 104635
0,,2
02
O.lC411J5 0.2 2
!.!'!±.ili!....yalue w
pt.o.:Z
pt"'pto" (l-x)
Us tao
:2 &00
600
sao
600
20
20
20
20
:2
:2
:;!
:2
0.9951.38
O.99Sl38
-40
··2.08253
2 ,,40
f~g.!~l_Y~: \.I~ \)
o
2.08253
!,!,=immn.:2.!~~ ~2imt.:.r::...;;:alu~ fb~l.Y.§!;b~ 50 0 SO 08']5709
(I
(I
37$'109
1
0553922
o
553922
0.2
kd
0 .. 2
0 2
02
uS
1()
10
10
€ao
600
600
600
600
kt
20
20
20
20 ,/.
2
:2
Z
2;
2
0.248581
o
-40
-2 _ 75389
.. 40
2.75389
0. 2
2
PI0-21 (a) A -7 B Elementary reaction with 1st order decay. da -=-k a dt D a = exp( -kDt)
PI0-21 (b)
10-53
243581
t The activity is never zero for first order decay. When a = exp (-kDt) there is no t such that a =O.
PI0-21 (c) Mole balance: dX= --r' A W = _ -r'_ -r' W _ AW _=_A._ dt
N Ao
CAOVO
Vo
CAO
Rate Law:
rA ' = -a [ --fA' (t -rA '(t =
=0) J
0) = klCA
Decay: a =exp( -kDt) Stoichiometry: C A = C AO (1- X ) Combine: dX W dt- = exp( -kDt )kl (1-- X)~~-
In ( -1-)
kW [ l-exp(-kD t) k Dv0
= _ 1-
1-·X
J
X=1-exp (_IsW [1-exP(--k t)JJ kDVO D
PI0-21 (d) klW -= kDVO
(0.2)(1) =2 (0.1)(1)
X = l-exp( -2[ l-exp( -1)J) = 0.97
10-54
PIO-21 (e) Decay rate increases more rapidly with temperature than does the specific reaction rate. Therefore, conversion decreases with increasing temperature.
PIO-21 (1) kjW (T) kDVO
k
D=
= 2exP[-1500(_1___1_)] =0.57 310 400
0.lexp[2000(_~ __1_)] =0.53 300 400
X =l-exp( -0.57[1-exp( -5.3)J) =0.43
PIO-22 (a) In order to get a high conversion the en.tering pressure should be as high as possible since the rate is a sewnd order function of the pressure. U should be kept low since the conversion is an indirect function of the flow rate.
PIO-22 (b) 11\e problem with such a low flow rate is that the activity will remain low
PIO-22 (c) We (an use the same eCluations that are given in example 10,7 with a few exceptions. For example the rate law, we use the one given in the problem:
I11e activity will be different because the equation given is different:
,,~:~, :;::; k naCcoke da dz
..... , ....... ,.....
kDaCcoke .... ..
::;:,....... '""- ......•.... -. .---
U
To find the concentration of coke we use stoichiometry:
c'cot(.e , == .~coke. RT We fmd that the value that gives the best conversion (X;:::: 0.337) and nses the whole reactor is U :;::; 7. See the following POLYMATH program.
10-55
:sq~aS~
Inicial ... :alu€
.:
d(a} Id(z) =·~-kd"'a""'ccoke/U
1.
d(x)idlz)=-ra/U/Cdo
.)
cao~
22-
kc.;;;;1.OO
o
;;;
o
15
1
kp=i;ne;;;:5~···5
Q
LS .a002S~34.g
00C2549';8
a
o
G 316731
cao
(} 22
0 .. 22
0.22
:) 22
eps~1.
;Cd
:00
100
leo
100
R=.OS2
k."1J::;-Z:;ne
5e: 05
5e-··CS
5,,-05
7
7
ao.:;:?
7
tlo
1
eps
pao=12
33673:
R
o
:,;-no=80
T
67)
6"1:!
673
U=:Uo*(l"e;:>s·xl
pao
12
12
pa=pao'
rho
80
Be
12 80
BO
)
9.35712
P"
L2
12
5.95425
s. 9S425
p-coke
o
}
0
) 02"288
.x:ap;:-i:ne
···0 0{)12
.~
(lox) I
(l+eps'x)
pcoke"'pao~xl (I Teps
032
082
35711
~
• xl
rapr ime=a' ( kpr ime 'p-,a ~2)
ccoke=pcoke/R/T ra:::rho""r:aprirne
e·.:-ok~
.
0.2288 51934~···O7
0.082
-·0.0072
-4 S1.934e'··O'l
o .OS~7'162
0 0547762
·3.61547e·05
12
··0 51G
PIO-22 (d) To fmd this the only change necessary was the values for the k' s because they change with temperature.
_ pr" ED (1 1 \'1' kD :;:: lOOeX ! --R6ij-i) .....
....I
The POLYlvlATH program below shows the results. The temperature is 485K and the conversion is 0 . 637 _
10-56
Equations:
Initia.l
d (x) !d(z} ;--ra!U/cao
a
d(a}/diz}~-kd*a~ccoke/U
1.
cao;.22
Uo:;;2.5 R=.082 T"'48S
pao"'12 !'ho::30 kd=lOO-e>..'P (15000/1. 987· (1/673-1/T) l kpri.me ::oSe-5*exp(30001l • .987* (l/673-·l/Tl) U:;;Uo*{1. ... eps*x}
pa=pao" (I--x) I
(l~eps"xl
pcoke"'pao*x/ (.l+·eps*x) ccoke=pcoke/R/T J:·apr.irne"'a* (-·kpri.me"'pa "2)
ra=;rho'raprime '0 "" O.
:nit:ial_'ya1:!:!.! MaximlmLy!!ue Mi~i!!!..~l.,,~ UEill value
o
15
(}
o
0 6374aS
0
0 637285
() 605597
(}
C.22
0 22 2, S
1-
o
cao
()
Uo
;J .5
2-5
2.5
eps
1
1
1
.22
o
22
15
.605697
C.082
0 082
Q
485
485
485
4S5
12 80
12
12
12
80
80
80
1.29n
1.2932
1.2932
kprirr.e
1.29 ..31 2 .. 09558.-·05
.2 09558. 05
2.09558e-05
pa
2.5 12
I 09121 L2
2.5 2.65841
2.09558.-05 4.09321 2.65841
pcoke ecok.
o o
4.670&
(}
4,6708
0.lL7445
0
0.117145
-(}. 00J01 763
-897021e-05
·0.00301763
···0.24141
-0 00711616
)cd
082
0.24141
PIO-22 (e)
10-57
082
a 9703le-O! -0 0071161&
val~
To find the temperature-·time trajectory, use equation (10-119) and add 'iNhere necessary:
C,o\
The following curve is generated from that equation.
o
() 015
01 ·rOIme, t
0 . 005
PIO-23 (a) ~
... , , - '--'"" .... _ _ ..... ,,_..., , - - _................'T
I
·-·------·l
I
~-------.~--.-
..•.,.-""..............
t------------7
,.-......",
Design equation:
Assume W=lg Cumene (A) FAD (lX)
r=FAoX ....
'"
Propene (R) FAoX
+
Benzene (S)
FAoX
The amount of cume:le hydroperoxide does not com:ribme SignL.+J.caIldy co the roml number ofmo!es.
or X =:_:!2. 1 ··Ys
10-58
X
a
.0204 1
t
0
.0165 .809 50
-
,0133 .652 100
,0107 .5145 150
,00851 .410 200
.OO563l .00311 .182 .276\
.00241
400
500
3001
1n a vs. t gives the best fit a "" c·-o.t
'Iberefore. decay is first order with decay constant of 4.27 x 10 3 (sec-I)
PIO-23 (b) C
A
:; FA"",
~AO (I-X)
U
u
F AU (1+X) Assume no AP.. ...1L "" _ll. := ...... -.~.- Vo
no
FAO
=
V .::::: ( 1+X)
vo
C A ::: £::,,!dl-~2 "" CAO l!:X) , '1,)0(1 tX) (l+X) PA
:=
CA RT =:
CAO
L~.~Xl RT (l+X)
Mass balance: F AO dX ::;: rdW
10-59
.118
F AO
= 2.0 rnol/Il"Jn :: 1/3 mol/sec
CAD
'" 0,,06
kmoVm 3 == 0.06 mol/l
R =O . 082~!L K'moi
T "" 273 + 420 :;; 693 K k ::: 3.8xlO,·3 mol/g sec ann Ct.
;:
4.27x 10--3 sec,·l
W
=:
100 kg
:=
10 kg/min:;; 1/6 kg/sec
Us
Equation (l) becomes:
2ln (l·X) -+ X =: ,CAn;:~: . !!~[e· (~t~l11
(O.06XO~082X69~~~.~.8~.~~~~!..P~J r
. .. 2m \ lX) -r X ::::: -
L~,3~~J.Q..3:.:.1.9.0.9 \. 1 Iexp \ L J 11
-·. .···{I~) {4~'27}xl0.3 3
.. x ==
PIO-24 (a)
..
6
LOO
s ..:;. W
+. CO2
:
first order, irreversible..
. . da:;:;; k<:J. an C',!1 == k.i an ill .">
··.fA
=KaC A
10-60
- -
--- -
- - --------------- ---------
-
---- -
-
---
-- -
-
-- - - - -
-
- - - - - -- -
-- - - - -
-- -
-
--------------------
For T :::::500K
o
20
X% l!1ntL\ , iI-X!
99,5
60
80
120
0.,7
0.56
0,4.5
0,38
0..29
~
178.1
2..2L7
262.6
3443
5
to
1.5
20
30
40
,.-
0.89
0,,69
0.51
0,42
0.33
111.8
1444
174.9
237.6
302.5
142
For T::::: 550 K
o
1 .,
X% ,
I
-
\
11ln 1·..... ·,1 q .. X!
8 7 .....~
.!9,,5 4W,.
I 3QQ
In(lJ(l-X)
I ~
II
+T=500K OT=550K
=1 aJ..I--'-II.
G
t
They axe Straight lines ."" n=2 For TI "" 500 K : slope:: 2.04 {t'~'h
} z:::>
[l~t]=O.Dl =('tKh
For T 2 ;: 550 K ; siope;:: 6.325 , {-c K)z ;; 0.02
}
=:>
~I
""
0.02
~=0.126S
&::..Jl.QL.: exn {-.!fLf.....L ~ _l_l} K.st 0.1265 • 8.314 \550 5001
~1 = ~1exp {- ~ (i l )} == K.i K.it
t
exp {-
88~f~4 ('s60)} = 0.02
= 1.296 x 10 3
We want to maimain Ka = constant
10-61
K
Ko exp
h"~1:a t) Ko :=
{f (f~" t)} ~ Ko (1
+
Ki t)
fE \f~" ,. l \JL 1 := v - t . IR To T I ,~
c;cp
rE( 1 IJ\'] exptR't~-- j ,- 1 t :;;;;; ,---""""-""- ,---------"""""""--""""':,'" --,
Kd
I
""",..1 480
485
490
495
500
l"""",,,
PIO-24 (b) Since the equation for the acivation is: a='""' 1 + lcd'
we cannot find a time for which a:::: 0, because it is mathematically impossible. We can, however, find a time at wbich the activity is small enough that it can be considered to be zero, The following graphs show the activation for'the two temperatures given
Activity
of
catalyst
~ 0 B
>
06
' .....·a (SaO)
::,";!Ii:",~i55~!
()
500
time
1000
1500
(days)
10-62
The graphs show that for sao K, the lifetime is about 1100 davs and for 550 K the lifetime is about 450 days. "
PIO-25 First we need to tlnd CAO '
. l!2_:::: 10 ETB
Y£TfJ
C AO
=0.1 "" !l§.tu.. = ~_~_~_03(~.1.J =0.065 RT (8.309)553
Start by guessing that the decay is fiIst order:
We were told that the reaction is zero order when the conversion is less than 0.75. This is true at any time after 2 hours. We also need to fmd the denominator as a function of conversion.
X =~:1!L~~.(~;t
CAO
CAO •.... C:
This is the graph that we get:
tn (1/cao*x» vs time
i-~[~~· o
5 10 Time (h)
15
As can be seen this graph is linear which means that the decay is first OldeL We also know that the slope will be the decay law constant so kG :: 0.2024.
10-63
PIO-26 (a) Mass balance:
Rate law:
Decay law:
.9a,. =':,~Q~ dW
Us
Energy balance:
dT
..Fa :..-. (TA"'" T ) + {"fA '( )·AH, Rx \)
-_., ....... _...".=
Stoichiometry
Evaluate the parameters:
k"'" 0.33 eXP[3777(.!-- "~,,,) 1 450 T J [' k,. :::: o.,olexP!7000( 1 .) L 450
1 \1 "T""')!J ,
Plug that into POL ThiA.TH and get the following program and answers Ug to get maximum conversion will be 17 kg/g,
10-64
Equat;h2!'~.':'
a(al/d[wy=-kd"a/:J Q
d(x) la(w) ,,··ralfao dlT)
fa (w)" (Ua r (Ta"T) ... ! -::al
~
(··Dhrl) }! (U"cps+fao*C!;>d)
fao=5 .. 42
U;1 7 1'a=323
:;).. 27
Dr.rl=···80000
Y~E~2l.,~
<;:p5=100
~~~~&... =:~.;.:i~
~'!.:;:\'1t;..'!"._va:~
~:!~ Y~1.~~
,)
so
C
v
a
cpa=40
.So
~4~
0
5.U 11
5.42 1J
f."
Er=) 777 cao".27
SC 2-::-1596
o .>S~:2:
"l'
Ed=7QOO
I2~; ...Y£t±~ ;; 211536 0 887222 6{S 6:3
He 5.42
S <12
1'1
17
321 ···3£]oao
123
32}
··S()OOC
.. 8;10000
.. 61,3
1'a
n3
eps=l
Dhrl
.. seooo
U"-"'.8
c:ps
leo
100
.:CO
kd", . Ol-exp (Ed' (1/450-1/1') )
cpa.
100 40
40
40
40
Ed.
fOCO
loon
7000
7000
E:::;
J r17
37Ti
cao
0.21
a
21
3777 027
0 .. 21
Q
B
loa
O.S 0 O!
"ca
0.,33
Q,;n
0.27
:a
,·0 069:
·0 C:Q;9:2S
k:: .. 33 'exp (Er" (1/450···1/T} ) Cd"cao" {t··x)! U .. eps·x)
l.
eps ,;;""
1.:,)4e·;.
H;:a
OS
oa
o.n
1.:"1401 4>" :~'j'lS
a
B
C Ol61349 .. 0 l.4S41
PIO-26 (b) Using the same program we can see that the maximum conversion is 0.887
PIO-26 (c) Everything is the same except the energy balance: dT
........ ., •..
-~
dW
Uaw ('fa . . . T) + hap(Ts _.- T) + (r.~ )(llH Rx )
.:.::: .......
- . -.....-...-......-
......... ....... ~
3';77
-.--.- ..
...
----
-
............... ,., .... "." ..... ............
FAOC pA
1
Choose values ofh and a!.. We fmd that Ts needs to be slightly Ingher than l' in order to get a large conversion. The maximum conversion will be different for each T5 that is \1se(i
CDPIO-A
10-65
0,0161349 ·".OISJ92S
Given: The catalytic oxidation of ethanol
CH3 CH 2 OH
+}' O2 -1- C:FI3 eHO + H:cO
Denote: A[=]CH)CHzOH. B[=]0130IO, W[=] H20 , A'S[=]ethCfhO'S Mecha111sm is belived to be the following: ,~
t-,
A-S +B-S
fA =leA [pA C§ "' CA-s CH.s!K.J
+ 23"
f:!
20-5'
-ro:= ko(P<4 (:-;, '" C~S'!Ko]
+o-S'
""'T ~
B +OH·S' +$
'TAO:::; kAo[CA'$ Co.s'· PE ({)ii-$" Cs/KAO]
~
W+S + S'
rw = kw rLCoH-$" Cli-S - Pw Cs Cs,!KwlJ
A
+ 25
(h A·S
OH·S' +H·S
It is given also that CH,s =CA,s
'''fa
i..o.;:O A
:ro,=O k.:,-
cbs' ::: Ko Po: c~, "
(-{)H-S'
or Cos,:: 'f~;p(;; c~.
Pw Cs es, .. LH-S "" -'K.;-" wltn
'r?'-""--
'0
CH-S;::; Y!'.A. .t' A
Cs
PwCS'
COrV' c;;;;; -.-- ... - - .) Kw 'iKA,P A CT;:; Cs + eH's + CA'S:::::; Cs [1 + HK;;"P;] """
('I ""
" I . . . . r . ,..--.-".--~ . -.,. e' ' ' g• + C,)-S' .;. (iJH-S':;::: (S\l+ YKo P Ch -+;:
Pw ~ Kw yKA,PA
L
,
--lAO"""
r",
1
PE c,:'.m.·s' Cs ."1' r----..,.------,,··, , Ps Pw Cs CS' k.i\O i CA.S COS' ""--;::::--"1 = kAO 'i KA. PA Ko Po:: Cg (S· - - - -..;:~ !. I\..AO .: L Kw KAO 'i KA PA
r• . 'K"K"'C C _ ?-.AO 'i . A {) ·S·S·; .,----·"tAO - "" ......... p A, ! Po"
=--_._--. FA l.
'I
K""1.::= K;.. Kw K..,o
~
kAO
.I P-a Pw .. -_ ...,,_._-_-._"'-",._-;
". K", Kw KA,cdKoi
{K;:;
iKAK:(:; Cr C r [ PA -{pc}.- . !:~w J
···r AO _ ...,...._.........................................................:"''''.....''' ...,. , ...................... - ...............
{P;'l'l
+.
2'1J{";A. PA.. . ] 1 + ·rKo-....· Pc-:-
,
Initially PB
=:;
.
-'7.
~
........,............·t.
Pw I Kwf"KA P",J
t ----.
Pw ::::; 0
10-66
(b)
With reaction 3 irreversible, '·IAO::;;; kAO CA-S (:Os', Since A·S. H·S, OH-S' are created
(and desrroyed) only at the reaction swface. A stoidriomerric relationship existS between
(a)
The same equilibrium expressions for reactions (1), (2) and (4) exist as before: (b) CoH-S' C~i-S
PwCsC s' ::;;; -,
Expressions for Cr and
•
,y>
(c)
Kw
C;. become
PwCsCs
wIth K AP.4 (, S » ··--K··"·-""".,-", IV
CT == Cs (1 + 2fKAPAJ
C~ ~ c~1+ rK~;o,;:- +fx'x] 10-67
Rate expression becomes:
···T
A
=- ...... -.. ..
CDPIO-B We can first try to come up with a rate law for this data. We can see that as P E increases the rate law also increases but slowly the amount by which it increases becomes smaller this tells us this:
We can then see a similar thing happens as PH increases so: I Fc! ""!A "" ...,..,.....,..... _ .•... _ .......
1 +KUPH
Sin,ce both reactants are adsorbed the me(hanism must be a dual site, This makes the rate law:
.~££"~JL_ ........ 1 +KEPE +KBPH
o::.,.".,.__
We can then plug this into POLYMATH. and we then get the follOWing values for k, KEf and KH • k= 14.6 Kl) =2 . 63
KH =1.76
10-68
O~90C
o.
lO"'8
--,.
t
~aa
~
i
~II
0.320
tt
f
!
c. ace
2 nOd~t;
~
'" ;(0/
1
rat"i"?;:;It;"O:e.Oh/~
t .... ·K€.')lfPE*+K~H(Ph)
~
3
~
............J?-._ ...... _...fF. . .
e
'~2
-rE'-SLdr;.;.atsl"
6
neq.,at!;\I(o 1"@$lOU.a!S ..
Sum: of
squaf9S::::
CDPIO·C (1)
NzO + S ~ N20· S
N:zO • S -+ SiO • S ·4 Si0:2 -t. 25 + N2 (Rapid) rSiOt ;:; ~ Pues PNf?
1 + KPN;;()
(3)
dF -~:::::r a dV
.
6
:<" '" 1.75797
1 "I .. 53-.... '" 2.63'17!
PQ5t':l\li:"
I!
J . ____ L.i . _. ._.
A . . __ . . _--..J............ ~."..... ""t~-.-......... !~.t...
... !.atec
vql~,,!i'"
I!
!
C .. 1SC
cc.a!~
A
'Plug Flow Reactor)
10-69
t,,"5236e~.06
k P'AO (l·X) (SB-2X)/O + EX)2
'fA
=---------:···-·--··{e- -2X}-'--1 +- KB PA~:!L-· (1 t· EX)
k P).o(1 . X}{SB .. 2X) ~. K~P:~-{~ -2X) (l'~ eX)
:; (1':; eX?
.~ 0·7 as. ; : ; 1..1.:.2:: 3 . 68 -' . E:;
('--'}~"-)(2 +- 2· 1- 2):; 0.25 11.;) + ).68 1~~------------------------.r--'
Slope =: K. == 0.7465 k'
1:300
Inr.erceut ::;:; 1 ~ k = 703
1200
!!l.rorr~"ITlin
k::;:; L49xlO- 3 NmTotr'-rr.in 11 00
K = Llx10 3 mTorr 1
A ~+---~~~--~--~---r--~--~ 300
.00
500
600
"lOa
sao
900
1000
~)(m TorT)
KB::;:; 1.1 x 10. 3 In Tou l KB
PAl
=(1.1 x 10.3)(157) -= .173
k ~o:: 1.49 x 10..3 (1.57)2::;:; 36.68 Nmin
(3·9.?: ~lli:~L_ ...._._. . _._
-r* ; : ; 36.7 Nmin
(1 +- .. 25X}2 +-.173 (3.07. 2X)(1 +ZSX)
A
FAO :::: 3.68 x 10 J gmok/min -r* ;;::; {Nrrim}:: 4.4 A
X
lO'-O r- (gmolel A
a=:; 250
(m2min)
m2/rr.in
IX ~ Ar-.....a under curve of X
..Ya.. =:; dX F · AD r I)
A
YS.
L r
4
A
10-70
x
.r~(Nmin)
.l/.r~(~)
l/.~ (m2/gmolel
...YlI...
0 0.025 0.5 0.1 0.15 020 0.25 0.30 0 . 35
73.6
0.0136 0. 0142 00149 0.0165 0.0182 0.020 00226 0.0255 0. 0288 00328 0.0378 0. 0439 00517 0.0617 0.0751 0.0936 001203 0.01617
3091 3235 3392 3739 4139 4545 5145 5784 6547 7465 8589 9983 il746
79 . 1 82.8 1783 196.9 217.1 242.3 273.2 3083 350.3 40L4 464.3
OAO OA5 0.50 -0.55 060 0.65 0.70 0.75 0.80
70.3 67.0 60.8 549 494 44.2
39.3 34.7 304 26.5 22.8 19.3 16.2 133 10.7 83 6..2
V{m3 )
FAO
0 00012 0.0012 00026 0.0029 0.0032 0 . 0036 0,,0040 0. 0045 00052 00059 0.0068 0.0080 0.0095 0.0114 0.0141
5431
14{)29
6444 7774 958 . 2 1115
17067 21262 27338 36760
0.0179 0.0236
1602
Reacror Volume per wafer:;;; 9.8 x 104 m3/wafer :. Number of wafers, n::::: V!9.&
X
10 4
A plot of conversion X. vs. number. of wafers can be obtained 80
200 --..•. - - - - - -
60
c ri
;;;
;
"2" 100
40
-0 ci
:§
....~ ';"
%
2Q
o 4"'"'"T-~--__,-...__1-..,.......,..........-+ 0 0.0
0 .. 2
0.4
O.S
O.S
1.0
10-71
Tne thickness on mest! wafers
C:l.l1
be obtained. from tb,e same plot,
Wafer#l ; X"" 0 or'A ;:;:: T3 .6 N'!nin :. After 30 min: 12Q8~.
Wafer #50: X '" 0.52
.r~ "" 20 Nmin :. After 30 tnl.'1: @ . b. Wafer #110: X:: 0,78
-r~
A
:=;
6 Nmin
., .Aite:- 30 min:
J 80 ,J.,
CDPIO-D (3/e CDPIO-E) CDPIO-E
.
(a) For simDlicitv, . ' letters were substituted for the sDecies in the
Droblem sta tement. "
The late equations for each of the three steps in the m.ecnanism is given below.
Equation.s 1 and 2. are substituted into equation. 3 because the third step (the deposition step) is assumed to be the rate l.imiting step, we want to find the rate law of the deposition, and we have data for the deposition rate.
10-72
.cuso, remember that we have the relation below:
. f v=·
1
. 1+ KcP c
Substitute everything into the deposition rate law:
We now have to check if the above rate expression agrees with
experimental observations.
..
The rate of deposition is independent of AI' and 1h. .-YES
.
tJressures of TiCI. and NFL, • At low partial , , the deposition appears to be first order in TiCl" and second order in ~n)--YES
• A.t high partial pressures of l'·Hi], th.!'! rate varies inversely with TiCl,.·YES (b) To determine the reaction rate parameters we must :earrange the rate expression to a linear £onn . f
A plot of the experimental data is shown below.
:::::i',. ,'_. .". . ".__. . ". . _._. . . ._. _ ._. 'TItanium Nitride fUm Deposition Model Veriiication
1200000 ..1
200000·
... l.2167%ii.... s: R,o\l OIl 9 91t98UE.l
f(x):;;.- L~l6l£"'l"-l:
Pal'b"2
10-73
1.
- = Y .... mtcrc:;pt :: 2.2168 x 10 l
.Jk.!'
.
k. ::::: 2.035x 1O-1l_._ ....!~~! . __'" . em-· min· mT' k.
•
.
p~
,J
.
== slope = 14 42
1\..
k~ :::; :'i . 505 X 10-5
(c)
rnT3
The experimenrai data, when plotted with the rate law deri,:ed in this problem, form a straiO'ht line .,.......... Thor'e'''''''''' . , .... .::; .. """ ttV2 ;)roposed. Iri.eC.:1arusm may O€ used to describe ti'..2 depOSition or titarl:l;,rr~ ru::ride fi.lms. _
.
...-...;
CDPIO·F Given: Ille dehydrogenation of ethyi benzt;';ne w st:;tTene:
E St +- H2 Quanur.lUve d::ua suggests re.lc:ion rare is of rhe fonn
-r;; a T11t
rJC~
Pj:;
-'~",'" ,'\-_-p-~_ :-.";::--p"'1 '!'" "L E"" ."">3 St
....
. .,-
where At, lu, A3 are consta..'1ts
that {orE) is independent of Hl suggesLS that
me reaction is i.::reversible, and that Ii2
is not adsorbed on the c;:;,ralysr Sth'iace. Also, the aoove expression for (-TE) suggests that
both E and
S~
are adsorbed on the su!face.". Try the following reaction scheme.
'E+S~·'}E.S
E "" S ·H· St .. S t 112 Sf • S ...;-)- SI t· S
-fA:::
kAlPECS· CE.sf'.r.<.A]
rs == ks fCE-s '. CSt-S PHJKsJ I'D '" kD [CSt-S - Ps,CslKDl
10-74
To insure that PH2 does not appear-in the rate expression assume that the surface reaction controls:
I~== 0
CE-S::: KA
: == 0
Cst-s
Cs PE
=C~~~f.
Then: CT Cs + CE-S + Cst-s :: Cs [ 1 + KA PE +
And ···rs =: ks
[e
E'S -
~~}
Cst-s PHll ::; ks KA Cs [FE _ PSt PH! J Ks J KAKS Ko]
~ith Ks »> 1, reaction can be considered ineversible, and its dependence on Pm drops.
kK",PE ··rs :::;; ---..-------.. . where k;: ks Cr 1;. K. P~ + EsJ..
out:
.-.
e
KD
Evaluation for k_ KA and Ko Rearranging the nue expression: P e . "'" _1-. -+ fs.
-rs
kKA
k
+_..f~.... KokKA
with PS l ---1> 0, a oiot of ~. vs . PE should be linear with slope .
"'5
A
only two points are given with Ps l ;::: 0, it is easy
1
I
C-'·~21.~· ~\ ( ··rsJl
==
k KA
{1.} and interceot (..•.L. . ). k · .kK
1.00
O:Q.!=~"-·-·
-
[0
solve analytically:
L415 x 10 3 ._+-_7.:;:..06;;;..;...·! .. 0.214 x 10 3 46.73
.J _.J
+ l (P-)l and f!:E..\ == ....L_ +.1 (PE'" k e \rsh k K." k I..
from aoove: 1..;: k
'P>:~
'P~)
{~._.jrs~L:;;: ±6~.z' . 7iliJ."Z : :; 666 7 ~~~~~~!~~ (PEh" {PEIl 0.01 1.0 . gmole
k :;;: 1.50 x 10..3. _.~?l:_
gm:;at-mm
10-75
Since
1 _.. == (PlZ - - - - "" 40 _a..Wl-g:mcat-min ana,_.-;-.. - I\ - 1. (P) E 1 "" '706 '7 .. bOO.1 ..- - : - - _ . -rs h
k KA
k
gmo!e
KA ::; 16.64 atm· l Now, taking data points 1 and 3, in which PE is constant. the value ofKo k-KA a.ndhence
KD can be dete:n:ni.ned:
(~)1 =k'k"'+~(Pdl +J~~t~-A A
Forpoim L:
'-'-I)
Forpoim3: (PE) = .. L ... +1.(PEL "IS 3 k KA k JJ
+j~~tn .... Ko k KA
Substracting, and soiving for K~'h(~' noring mat (PE); "" (PEn
:~~l . Us),]
lo.166J~·i03 "i'4;51~'lo~31 cat ....
(PEB (( 1.0 gIn lin k K ... == ··-·(Ps-;;;~·lp·~;r~· =··················i~o·::·(rO··----··--~;jf-
KD
=:
0.0075.34
.'. Rate expression is
(a)
(~vf\.V)E
0-
= 106..1 ---£.......
gmole
.. 1 grnole H..,O Inen 8) = ._.: ---~.,--- . = 0..2 :;, grnole E
YEo=···_··L.
=0.833
; P""Po =OA15
1 + 8, .'. <:: = VE {} (5 = 0. 833
. .
~ R .... PE "" C·E 1
= F:=RT ---"" ....__. = 1.)
F, RT PSt"" CstRT "" -~=u-'--
.
,
FSl( =
; XF=O . 60
; 15=2·1=1
~222.~~{.~ kg m::!)' "" 19.2 day
10.4 . 1 kg
kg ~~!. = day
FE.COX)RT CE.{}RT(1-X} YE...oP,,{l-X) ...... _ .._ ......-.... -.--...• "" ........ -.~ ............--............. ::::; •.. __..._,. .•............ 1.>0{ I +EX}
1 +EX
1+€X
"" ·1)Ji:;€.-x)"·- '1+£X
FE-o X RT ., YE.oPo X
10-76
; T=903"K
F~ eXt' t.
Rate expression for a.."lY X is: 0.025
[YE.~~X}]
~I"E ='''''--6"64" l~YE.oPJ 1-X'~1l';""=1""'32-'.7:""-'YE.-G-p--o""'X 1 + 1. - 1 ... eX j + -"," 'r'+ EX Design expression for a C'.5TR (fluidized bed) is W
=F~~F
w """ ~.5:Q_x..l..r,,,,,,,.!,,..:eXF,__ + 16.64 + p2.7 XF 1 0,,025 lYE.a Po (I,,,XF)
l-XF
j
:: _FS tt ~E,j",,,,,!,,.:tE:XE__ + 16,64 + 132.7]
- -,- daY-"·_'" W
1 day i 1 +- 0.833 (OA5) , A 132.7 (0.45)1 1 day x 0.025 iO~833 (0.415)(1~-:-45)''' 16,,64 + ---i:~45'--j x 24x~
= 7,,06xl
Cost"" 70.6 kg x Sk~' "" $T77 ::>
(b)
Plug flow reac.O!. expression for YE.O. FSr;> PEt PS~ and ·rE are the same as for the
CSTR, Tbe design equation is:
w =- __~St/._,.. !-,...._.L_., (-EX - [l+E]lnO-X]) + 10.64X + 00.5 X;: \YE.O Po
W=-O""O·~:t'x__ "I:l_p_(.€XF'[l+eJll1[l"XF])' ,-)
F \Y::..O
132.7 (-X·, in (l"X)Jq;
f
16.64XF + 132.7[XF+ III (l-XF)]}
0
w =1?:2.:.~~: ~~?,~. x.."" L"""".",,,,,, ,gm:~min 0,025 x 0.45
day
gInol
1-.833(.45) +- 1.833 In (1-A5) _ _ r" _ r _ I ."l~.",.. ""--"~08::;::;-YO-~""'-"""'''''''' 11.:l.64(O .... )}· b2"J l.o,45 + in (10.4))]1 x 24 *60 .-' l ':UA .4'), , mm
1
W
=2.71 x
1()4
Cost:: 27.1 x Sk:11 "", .5298 g
CDPIO-G Given: reduction of CO with hydrogen over Ni catalysr:
10-77
Kineric given by:
0.0183 ClL.) r ::;; - - -PH~ - "Peo ---- (gmOle -------,-.. -,-, 1 + 1,5 PH l gIn car-min
From the above expression: 1) The appearance orF2 in the denominator suggest that H2 is adsorbed on the
surface.
2) T.ile fact ll:at CO does not appear in the der.ominator suggests that it. reacr.s as a gas phase species" 3) 1be square f(x)t. dependence in the numerator suggests that Hz splits.
Therefore, suggested mechanism is: kl
III
+
S
t-
S
-'J< f"··
Ih'S
Kl
• H2'S Ii·S
_..:;
2H·S
(0-
+
(X)
CHQ..S + H·S
k, ;!.
14
... --..
" ~
CHO-S
C·S + H20 + S
ks
2Hz
C·S
".',"
,
eXit -+ S
f',
f5 '" K5 l c.c-s
~-
p>
i-h -,- Pen, CsiK sl
Remarks made in Chapter 5 of the text suggest thar reactlon 3 is rare cCEU'oiling:
CT::; C 5 -t, C:H,S+ Ce,g + CCHOS
-, -_ Cs -, 'I',1 ' Kl • p}tz -,.. vKr ---7~, Lr K: Pi'£, -t
l
"1"-
.
PCE."
1 iKl K2 J
POL PlhO
"'''''-''-:;'''''>' . , . , - - - - - - - - , - - - - - ,
Ks Pfh
pit ~ Ks
10-78
Kinetic expression becomes
Reaction is irreversible .'. let Ks _.+
QQ
Then
with K1 PH~» "K 1 Kz Pit, we get ..,r3 "".,k;-
fK11G" CT Peo· ' "Impbes that. at ml.l'l..\t;l
~_,
1 + Kl PHl active sites are occupied by the Hz mole...."Ules.
(a)
Design for a plug flow reactOr. Denote A[=] CO, B[=J Hz; then YAD. =0,25. YBO ;:: ,
~
"
F· . 0.75. as := pE:a!2.;::; 'tyJ~Q. =:: 3, XF = 0.80 AU AD
F '0' Xc
,'>.
F AO
:=
=;
F("ll
."<4
=:
2000 J1t". x ~ "" 125 lbrnol. day 16 Ib day
ill.l.bmQl.;:: 156.25l.1mml. day
0.8 day
DeSiEIl couation _.
F,o dX :; -L'~ dW or ~ }l{". =: n
f"U" dx. ~
o
___ 0.0183 FA plf 1+L5PB
-'A - - - - - . - ' . -..-.
P A :;; CART == .~ AU ~!.:.o ·X)
;:;:)1 Ai)
1+t:.X
PH. (1- X)
1+E.X
Pa "" CB RT = CAO R!J.~!!_:..,~~'!';::; fAO!!J~....:~~.~J;::; l+s:X I+eX
0.0183 [YM .'. -rA;:::: .-••• " .. "." .....
f \
~.lA{).!,O (I"X) l+eX
P.:~!:X.l1f3 YAO P'JJ~,~}r2 0.018313·ri'AO,~2J~~~Jrl2
_._l,_tX.".,,_.·
t
of
fy~o Po (l;?C)1
1 + 1.5
fd:'Ao.~2.i!-Xl1
~
I+eX
' __ ;:: _"",,,,,.,,,,,,,,,__, ,__,,,,-_.J...!:.eX _"~.'",'
EX
1 + 1.5
I + LS[-Y~~~2.E~,~ll . l+eX 1
.
)
____ ...."_. . ."l--1.±.~_JJ d.X
0.0183
1r[XAO Po (1- Xl 1312j 1 + E;;X
j
10-79
~Vf..... == FAo
0.0183 {'J(YAO pop!2
l
a~
1[L:LQ·25l?:J.312 +- 375 ri -0." 5X ]1I2 \ dX \t
0
1·· X
L I··· X .
J
J
Tt:: integral can be solved using Si.'11pson's rule
Then
f.~ f{X) d.X ::; ~(f{O) + 6i(O.2) + 2f{OA) + 4f(O . 6) + f{O.8)] 3 o
X I loX T1-O.25X 11.:Q~25X 11.q~Z;X + 3.75 (l;:q:~x:ri2 o I 1 I -.2:2 I 0.8 ," 'lf4·To.·6 I
~-lj. .-~..0.8
(0)
0.2
1
I
1
••••••
0
1
0.95 I 1.1875 0.9 ·500 0.85 T}·125 ,I ........ 0.8Q.. _ 4.000 _....
m
4.150 1.000 4.750 4.974 ~....... L090 5.420 5.250 ..._==~=·:::I':":.:2~~2'::'5·.-_-+--::6';:".4";;;3';:'0-+--=-"";'-';';;;;"';';":;' 5.875 -r-...._..!.~45=.~8;;;;.5,,:;,64.;,,+~-+;;.,;.;.;;;:;.;..! 7.950 ........1 2.000--1LS.5OO
Design of a CSTR
F eXt' Desum cauanan IS W:;:; .~.:.... .. -fA ••
' .
~
. G'"'l L-;J,'. "un \F _ 1'1171 .l. h. £.lQl 1;." •• , rv -- .I, _.oJ ~ ~,.. . X· x C.w.V £!IIlOi .. ~
fOO
-454 p'mol .;:,11 ! ".--.~~.--
... X
,\ . cav
...,---.....,.".;:. ....... __ ..,.
Ibmoi'.AX;::'O· L ... v mm
w == 3900 g CDPIO-H 10-80
(a)
.
-r~
-r;
I
/ I
0
'A
Runs 1. 3. 6, 7 . . -rts =
.
-r•
~.
Fa
Runs 4 . .5
Runs 1.2.4
t I . -. r. "" . . . . . . . -.-.-.---~ 1+ KcP c +· ....
P", ..-...-.... .. 1t· KA PA+-· ... •
,.--.;'''"'"-"".~""~
~
~
(b) Numerator: P A and Pa Denominator: P A and Pc Power of Denominator: 1 (c) Proposed Rate Law:
(d) To find the rate law parameters., reauange the rate law so that it can be plotted as a Hne with the rate law parameters as the slope and intercept of the line.
First, hold Pc constant and plot
!'~~.!!. vs . Pi\ ···r s
From the plot K Slope:: ." .....~" k
=:
5.2,
l+K P y.- Intercept =----.!:.-£. =: 3..59 k
10-81
Pa...~nett!r
Evaluation v.rith Pc=const.ant=2 atJ:n
120
25
100
20
Pa..rameter Evaluation 'W'ith P;;t:;const..ant=l atIn .-.--.. --.--.-.--------------.-..- ..........--........-
gO
'" ::::
~
60
::~1
R..... 2
.=
J
'9 999'9'~E -I
-'i=--·--.,.-·----·-..'..·-·-T-......,..........·-,....---r-....,."""",-1.
0
4
6
12
$
14
o
.
"......--1"........... /" ...
o
16 18 20
1
Pa
Second, hold P", constant and plot
2-
·'!"T-,-.,-,.-··.....'f..,~~-~·~l"'f"'-r-r :.> 4 5 " 7 8 9 10 Pc
_~,\PB. VS. Pc -.{ s
from plot below,
1 -'~'J'" . -'-.. =ntJ~.J. 51 ope = .K,-
.
k
v· ..· Imerc"'Pt :::: .1__....+K __. . .~...P_.::. ~ )- '''i)k - . -".
J
Using the four equations above we get:
"'.
to
soive for k, K A, and Ke,
mol
k :;;:: 2.60 . --·-·-----·-
gear· sec" atrn-
(e) A and C are adsorbed on the surface of the catalyst (f)
Proposed Mechanism:
c·s=c+s The irrcver'sible reaction step was assumed to be the limitL'1g step We check this mechanism and rate limiting step by rearranging and combining the rate laws for each step. If the mechanism is correct, we will obtain the rate law proposed in
pane
10-82
k5 is much smaller than kA and ·
, . :::.r A. == .:r ~ k.
1/ -- therefore ~/k
A
=
0
C
·· (~"v:;: . _ - " " -C"r - - - .A f ter 511 b stltullon, 1 + K... P A + Kc Pc
Next, substitute the above equations into the reactions step to get
Because this is the same as the rate law in part c, the rate law and limiting step assumptions have been verified.
(g) Ratio of sites of A to sites of B at 80% conversion:
Conversion at whid\ the Hnmber of sites of A equal the number of sites oi C:
X=(J76
CDPIO-I
10-83
a) To detemline the mechanism and rate-limiting step we must come up with the rate law. Looking at the rate dependence of A we See that between runs 1 and 2. P A increases fTom 1 to 1000 attn while the rate law only increases from 1 to 1.5. This tells us that as A gets larger it changes the rate law a good deal less . This tells us that A is both in the numerator and the denominator. .... f
' A
..~1 .
l+K A PA Looking at the rate dependence of B we see that between runs 1 and 3, PB incleases fmm 1 to 4.5. This teUs us that the rate law is directly related to B . ..... r~ .- PB
Looking at the rate dependence of C we see that between runs 7 and 9 Pc increases from () to 4 arm and the rate increases from 4.5 to 4.8. Also in these luns we see that P A increases from 1 to 4 atm. So one of two things is Due either Pc is both in the numerator and denominator or just in the denominator. Since C is a product it will not be in the numerator in an ineversible reactkm. "" ···r I ,................................ .
A
1+ K c P:c
So the rate law becomes
With that rate law the following mechanisrn exists:
Adsmption
A + S'-1- A S Snrface reaction
A S+ B(g»
C· S
Dissociation C ,S"'7 C+S The surface reaction is the rate· limiting step.
10-84
b) In evaluating the parameters we can also see if our rate law is a good one. Plugging into POLYMATH we can come up with the parameters. ~
r
.. aoo
r-ate llH0
4
"t.eoc
t
LoSee O.soo Model:
k 2 Ka
! !
r,
2.400
~
n
1 'r
3.200
n I I1
2
I
..,L . . 5
I
,I
II
t
!i
~
:3
~
I'
I I n I. -ltJ· . . . ".u. . . . . .'ij,. . . . ·. ' . ·r
~
II lL.
!
I
d.ata [J C"I "'" at!>d
y.all.u?
I
~
iI I,
··..If·· ..
0;
~
o QeqrEl'SS~01""I
7
a
rate=k~Cd.Pb/(l+Ka~~a~Kc.P~)
O.. GOO·H'il'ilS7
K" = 0.:;
= 2.9SSaS
We find that the !ate law is a good one.
k:::: 0'(>0045 K A =3
Kc= ..5 c) The best places to add points would be where Pc is changed, but P A and PB are not changed. d) No solution will be given.
CDPIO-J :2 C2 Hs OH? C2 H5 0 C2 Us + rhO (A) (E) (W)
A+S
+:!
A·S
A-S + A-S ~ E·S + W·S
r2
0:::;
I"4
w+S
At steady srat.e r "" - ~~6.
= kl (pA Cs- CA-slKt1 k2 [G,.$ - CE'S Cw.slKzl
f3 :::: k3 [CE-S . . PE CsJK3J
E-S :;:: E +S W·S ~!.
II
=
r == rt
=:
2r2
= k4. [Cw.s .. Pw CsJK4]
= 2r3 = 21'4
If swface reaction is controlling,
10-85
~:.== 0
=
CA-S:; Kl
:'" == 0
C;::.S :::: !'.ECs.
1<:2
== 0
I4
k4
P A Cs
~
K3
CWo:)
:::;:;!:w,. C;;i. Kl.
c·'r == c··;.· + ("·'·w·s -.,s ' C. ,{os
.
r' ·'1 C's,,i 1T' Kr 1 } ,A + .---;:-. Fe. . t· --.-.. Pw J
'
C:;
R]
K~.
=---___._c;.r.____. 1 +- Kl PA +!I{ + .1.1:. K, K4 ')k
..... r ~ 2r2, ::
I.... z p2
C~
111 +- K I PA
.
where k :::: 2k z Ct
Keq:;;
K1
PE
pwell
r,
P- Pwl
A·T klP.~ - ~-': ·····r""·"~-'-·-··"'--·--·······'·.. "···'·~· .. ·-::;;;,~ ........~~·"--:;-..::. ;::;: :::~~... . ., "' .........::..,., .....,........ ".__......._ ... "..:.~_ .. __.
- 2L l.\.l
K,-K;K:-i
.; JE
K3
+
~w 1~ K4J
I '1 .,- K··. P ,PE -+ Pw 12 ." ATK3 . K~'j
Kl
K2 K3 K4
Using points 9 and 13. PE "" 0 , Pw:: 0 ... r == __...... ~!:1._
{l + Kl P,S:
°ll =:it i; +~£ r''''''''''
A I ' fl 1 (,... K p at or "V r v'S . p-;..... produces a straight line "'1m sione 1 1,. and inteI"Ce')t __.L A • 1/ k l: k slope
={f
4.945
... k "" O.04()9
interCept =:~. =: 85.59
•. K1 := 17 ..31
Using point
H),
=:
.•
P "" ()
~=0.0399
klleck
U sing point 11. PE "" 0
=
K4
=0.0368, dose enough
Using point 6, Pw "'" 0 = K3 "'" 0.659 Finally, using point 12
= K.,q "" 0.0975
__.J~i_..~g:~?§!'~_'!'~t r=··. ....... 9._:.0_409 ) + 17.31 PA t· L517 PE ;. 25.05 PwF
.........._.
Note: Keq may aiS{) be calculated using RT In K.,q "" ···ClCio.. Interested readers are encomaged to check the goodness of fit of this rare law with the data . ---.-------------------------~--------------------------
10-86
--------.---------------
CDPIO-K CDPIO-L Rate law: -rAe;::
kCcoC ACCNaOH 2 (l+KAcCAd
Proposed Mechanism: Pd + COi--'Pd· CO -,-.-.~
Pd· CO + NaOn--Pd· CO" NaOH --'''''''-7
AC+Pd
-.~
AC·Pd
AC· Pd + Pd ,. CO· NaOH""--7C3H sCOOH + NaCl + 2Pd
Neither of the first two reactions can be limiting because they are reversible . The rate step must be irreversible because there is no subtraction function in the numerator. We will first try the third equation as the ratelimit:ing step: fA::::: kCYAC Then
lACO _
""'-·-0
kACO
,~ACO:1':!aOU_ ;:: 0
k NaOH
Combine to find C" .. " ...,_. C{ ...""."_.... "...... ,... ~_ 1+CcoKco + KCOKNa()HCCOCNAOH This is definitely not what is supposed to be on the bottom of the rate law so reaction 3 can not be rate-limiting.
Cv
;::..__
Trying reaction 4: r~ == k4CPdoACCPd
10-87
CDPIO-M a) Start \!,lith a mole bahmce:
dX
--.::;;;;
Rare law comes next: ····r'A ::::
kr a
Then the decay law:
do
.. :;;; ---k, dt " -YV'
t
::= .....
Us da
dVV Us 'We then come up with the equation
f()['
the profit:
Vvl1cre:
Then plug into POLThlATH and get the foilo\;villg program. The feed rate of solids that gives a maximum profit is 4 kg/min .
•
~:!!S!-9.£§.'
~~....Y~1:::!::
d(a) Id(w) ,;;·}Cd/Us
1
d(x)/d(w);'ra/fao
0
fao",1
kd=2 Us=4 10 ··18 :). ;;.q~~l~~ . y.~i~'1£ ~~::nu..~" ..:5!:~~ !1i:1itt.UI~.-'y'alue ~_~~.L . Y~c;~~
l
fb=fao'x P=1.601ffb~1.0~Us
a x
0
';
75
c
;;
0 73
fao l
-
2
2 1
:::-:1
··c 5
£1::
0 i5
?
4G
b) As seen above: X:::: 0.75 and a =.5
10-88
OiJ
-
"-'...
"
O. 75
"·40
SO
c) The only equation thai changes is the rate law:
.~~ --·k dt -
d
W,······W I =::_.M:1!. __ .
U ··dW dt =:: -_._-----
U da kddW . . ".. == ._.-. . . ". ,. . .dW U
-~
~.
Integrating we get this:
W::::; W MAX @ a == 1 k ==
l-,~!l~ U
a == 1- .
~£!{~Y.u.-\X_.:~J U In~~~al.'.:'5.1u~
~t;.i2r:::!;,
o
d(x)/d{w)- ·ra/fao
kr;;:;5 kd=2 Us=.8
wmax"'l
fbaif{x<11 thenlfao·x) eLse(l) 0.",1.£ (kd/Us* (w-m
P"'150*iblO"Us
o
1)
G 8
1
'·3
ra
:42
-·0
·3
142
·s
\Ve find U:= 0.8 to ma.xirnize the profit X =:: I and a:: 0 exiting the reactor.
CDPIO·N
10-89
Design Equation: ~<\ssume at t
= 0,
\V
="~~ . \O~_
For n en order kinetics .
(-J:,\ )a(t)
a(O) "" 1 and X "" I.
Now
FB(t)::::: FAoX(t);::: RON(t)
X(t) == _~O~Q '" _~:2~.~!2. 106 RON{D)
Second order kinetics and second order decay rate tIt the data very welL J
l+k .t:::: . . . . .kC~ . . . . L __ d
CAo-,C A
y == 0003x + 0.0855
Plot of -
- A____ vs t: CA
C.A(J·-
-1.lnca; lExp
Data;
i
i () t "..........". .......~"" ................... ,._"~ .. _- ....... ..,...". .... """.".. ,__....._., _ _".. ,_, ....-.J
o
100
200
300
~OO
500
t (h)
F.om the graph:
intercept slope
(b)
= 1j k =: 0.0835
~
k ::::: 11.98
= kd J k "" 0003
Activation Energies both for rate constant and decay constant can be estimated from the temperatu,re·time trajectory
CDPIO-O
10-90
Given
A
-+ R + S
Batch constant volume I"Cactor, P mCI""....ases with time NAO dX =a fa Wdt Assume a = e..(lt '0::;;
kPA
:. PA=CART
C A ;::: NAG 0··2.9.;::: .~AO (I V Va
lC
o
dX.:;:; W k
eM (i-X)
~AO RT J. ea.: dt:;::;; (W.) k RT J.' eat dt •
NAo
I-X
· ~ Le~
=:
=We..(l' k CAO (1~X) RT dt
:. NAO <.LX
f.
·;.;.t
dt wl=e
Vo '
0
0
~ =(~-)k R1
.In(l.X) == ~ (le'
a
Assume that there is very little deactivation in !he first. 10 sec
-In (i-X) =:
~.. (at) = ~t
Ct.
(for smail t)
~:;:; ~ln ( .. lJ :lln{--L-) '" 3. T7 t
l-XI
10
1-0. 037
X
10.3 sec
~=(~)kRT = k=dRT=i-~~%s-~~ioi k = 6.63 x 10 5 sec 1
10-91
At t relatively large, e-
~ -In (I-X) -=
a
~
a:= ~ (l>o~~l ==
"
3.37
~'(1
~.
10.3
-Ts;J = X
5.18 x 10'" sect
,1a(1-X) "" 0.128 (i"-e-S.1ax1 o-'1t)
" TIle assUt~rprion of a firs;: order :e2.Ction kinetics and a justified
CDPIO-P cyc1opem:ane
<
fl·
pemane
coke
Batch mole balam:."e :
AsStL."Ile q :::;;; 2
a = ..............L.". ._
10-92
order decay kinetics is
,
_I =_I_+5!t
Ifn=O
X
k-c
let
#
--1._ ;:;;: -L +.5t t ·In(l-X) lei kt'
Ifn::::l
X(%)
o
75 7 .7
20
1.33 1.414
~o. . . ,_. ;,,6 7.
_ · F
ou 112800
!
v
"'1
250 350 500
11 ..46~33
05
..,
---S--4.)8- '.~/· · ·"· ·
.__~/•. 80:,~j _
'-
i
I
42
I
361
2.381 2.778 3.333
r-' 2i ',-"-'4.545
I
30
0.721! 0.333 0.815 I OA15 0.902 II . ~;:~~--.J 1.076 O.653~1 ._-_. __ ._. I 1.245.__L_.Q;~E 1.498 I L053 1.836 I . . 1.381
2.24T·--, I
1.177
2.804
. . 2.333 --'
I I '4~023"'-'3.545 - , ···'1200··T-~. ! 6.3 _ ....L··6.TITI. ~. ·_"';'·); . .~;':;_.; .___?'6'; ;; 'O_·-_-_-+.;...-_-_··_5_·.1_3_5_:=1---,
1--';'800--
",,,l. I
+lfX
<> -llln(l-X) II1II (l-X)/X
f .
o
L.... ....~.~ ..... ~,~ .."'._.. "..... ~ ...._ ......... _ •.• ".. '"M_._ ......,
o
~
tOO
.' . -~....--.
~
ICQ
.--... 1=
From the above graph. a111ines are straight lines.. 'lberefore. q
= 2 is a good assumption.
We need to ex.amine the data to see which value of n having ~:::: constant n=:O and n "" 2 At
will have similar behavior of ~ because: At
8.= 1.._ 1
X
X
n == 1 . Av.:;;.1 f.--L--l and.1t;;::"O min
l-In( l-X}J
'.
...
~ :;;; 0.094; 0.087; 0.087; 0.084; At
0.081 ;
........
. 0.0797 lit
Therefore. ay is deo:easing gradually. It is not a constant. At
•
10--93
For n
=:;
0, n.:::2: ~r
:::: 4 x
6,
10-3 = constant.
• If n::::{) :
_
_ QX, == !<::B_~~Q,
{lX)2
at
(l,Xr "ili.." ::: •... ,. L . . . . ;- --~Q.,.,.. t .
dX
KR C AO
KR
c.-\o
n=2: Slope is negative. It is u.nr:::asonable n:::{): ~. vs.
t
is a straight line.
10-94
<> dtldX
w =0.01 kg/m] C W
't
(0.03
krlifl){0'<)1 kg)
k
= 22.-. '" ~_ ..................m... . . . :.......... .. _m 3 ... "" 20 _JL min
(b)
{m3 )2
1.5 x 10 5 kmol/min
FAO
The order of decay is q;;2 ,
k
...Q. ""
kA
(c)
4x
1(},3 =:>
k: 3 x 10 3 mini
Moving bed reac~.or: F AO =2.~&!. ; X;;: 0.80 mm
*
lia. :;: k' d1:
d
a2
In moving bed: t
.,
=if where .
Subsnrure mea equanon (1): ..
u ==
=
~~ dW' " = k.:!" '1:1' a'"
. . ~.::;kd·dW
a2
u
10-95
SUDSritllIC in~o
equation (2), vie have:
W :::::455 kg
(d)
If
u
O - k~
= ) .---::".-
ttl!.(l
w ==48 . 6 kg
CDPIO-Q
10-96
a) Mole balance:
Rate law:
Decay law:
Stoichiometry:
CA = CAtI (1" X) CD = (~O(e8 + X)
Evaluate the parameters:
d(x)/d(w)-.-'-ra)/fao d (ia) Id (w) ",·J
o 1.
kd,:6, faOi,:20 :'
~
U;:8
k';350
1.. 000
f
cao=.l
ca"'cao"(1.·x) , cb:cao * (thet.a +x)
r·a"·. ·k"ca ·CD
Wo '" o.
10-97
-, .. ,
b) The only change is in the stoichiomeuy:
POL'{;v1ATIl Initial d(xl Id(w) =a* (--ra) I£ao
o
d(a) /d(w) =··kd*co/U
1
v~~!:
kd=6 fao=20
!
~oo
U=S
i
T
K=3S0 cao::tl:"l,
C "'0..'1
1
J...
theta::: 1. alp= . Q38 ca=cao' (1"",),
(l"alp~w)'·.
5
cb=cao* (theta+x)' (l--alp'w)
~O
.. 5
ra="'k*ca*cb
Wo = 0,
wf
= 24
CDPIO-R Curoene (A)
--,
+-
Propylene (k)
Benz.ene (S)
a:s .I. . fo
(a)
We find a relationshipbetv.tcen a and C A
F X F r '" -.~W = 2W . = F"-,
(W:= 19)
.R.ill:Ll.: t
o
60
120
180
a
1
0,,594
0.491
CA
o
07.5 0.01
(LOIS
0.0243
Run 2:
PA
:::::
0.4
aL."'Il
t
0
100
200
300
400
a
1
0.833
0.733
0.583
CA
0
0.0057
0.0106
0.65 0.0148
10-98
0.0184
Plot of In a vs. C A gives a straight line passing thmugh the origin wirh slope a
(single ~te adsorbed. smface reaction controllingl
initiai
I rate are used, PR == P s = 0 and 1 » KAPA (adsolption is small at high temper.rture), then fo"" kt'KAPA·
Using data at time zao from runs 1 and 2 : k~ K.<\
=:
3.2x1O-3
Hence, overall apparent rare law is r
= k~ KA e
; k; KA "" 3.2x1O· 3
ex.::; 28.9
: : : - k:..:i amf{P A. PR, Ps) ==
-~ am P"A
Trym::::: -2 Since PAis almost constant. d~l.ring nms 1 and 2 (low conversion)
Run 1: ku PA. :;;; 5.767 x 10-3 Run 2: k
~:;::
PA ;;:; OA ill run 2 :::)
5.767 x 10.3
II ""
L29
.
Kl KA "'" 32xlO·' ~:: 11 ::::;
5.767xlO:; 1.29
t L.'1 minutes
10-99
(c) Overall conversion = 0.60
(2.8 rr.oll:Jec cu...--nene j \ 4.2 mollsec propene) L. "" 11.2 IT'.ol sec 1.4.2 IT'.o.llsec ber-.2e ne J Cu:rnt!ne "'--t Propene -+ Benzene (A)
(R)
(S)
COllIDosition at reactor oudet .. YA: 2 &. =O?S 11.2'-A
?
VR = Vs ""..::!.;;.""-."
..
-
11.2
""
0 315 "
Composition at reactor inlet. Fa :::: 7 + 3 :.= 10 mol/sec FAo ~ 1 , 3(0.25) "" 1.75
YA "'" 0.775
FRO::::: 3(03'15)
:=
Ll25
YR ;: 0.112.5
Fso
:=
LIZ5
Ys = 0 . 1115
=:;
3(0375)
Let Xf "" converS1on per pass ~rotal
flovi at :r~aC7or outlet before the recycle sc"ea.,s is:
At a.llY point along the reactor:
Assurne that rate law in (b) is srill good fOr the mO\o'ing tx:d operation (may not be rIue in practice because of the high conversion) I'::::;
• ~.. k, KA PA a k, K.,., P A=:' .. .l..."•..•.•_-_..
,
..
Moving bed react()r:
l+k.:!?A,t t """
l~;
F.,odX =rd W
lO~lOO
== ?_:15x5.7~~10-=~ (1,29+X.\O.Z9 W + _....1..7~. _._( 1.22+X.) 3.2xlO-'x2000
I-X J
3.2xlO-3
I-X
= 6.98 X 10.. 3 (1.2.2±Xj29 W + 2421 (122±X.) I-X
With X::O
>
IX
W=o
If X:: 0.542. using digital computer W :: 3.1 kg
CDPIO-S a) i'vlole balance: dX dW
a * -r'
:::::: .................• .::! .
rate law: ,
~r:4
::;: k'('',t
St.oichiometry:
Decay law:
da : : : ._kd. _. a ._............ dW Us Evaluate the parameters:
?.;:;;; (1 . . o:wt5
Po
5~ "'" (1--1OOa)05 0:::::
0.0099
rOl da kd dOO '-'Jl -~- :::: 'i~!;' JodW
!:i..;:;;; 0.023 Us
10-101
POLYMATH EguB!:ions-,. d(x)/d(w)='~B'a/fao
o
d(a) Id(wj ;-·kd*a
I
kd" 023 Yi:~X~2!~
fao:=:4
w
k=.09
x
cao::o2
a
a2.p= 0099
kc
ca=cao*(I·x)*(1-alp·w)A.5
tao
" 0
!<:
:t'a;;;-·:t',llca
:.c.':'tial val.l:.f! ~~,..:£,elue Minirnu:r:. ·,ta,l>.!.e £l!~h-y~lue l{W 100 {} 0 .75:;':51 C ,57161
0 :00251
C :)0259
D23
0 .023
0 023
C 023
~) 9
0 09
C 09
0 0099
0.,0099
C 0099
C 0099
0.04856"18
.. 0 004371.1
0 04.35678 -0 13
Cd
ra
{L1S
b) POLYMATH
J J2:)
:-
ceo
CDPIO-T Design Equation:
FAn -1~
RaleLaw:
< ;;;:k'C
Decay Law:
(for sintering)
~ a(WX·· r~) A
Stoichiometry: Combine:
dX. = _... ___ L .... ~:(l' X) dW
dX
.=.,k'
l+!'AWu Us 0
l{Ilx)~ ~:~;ln(l+ tt w J 1 1 From the problem statement a - ::::; ................. :.::: 'e.u,
1+1:< W
4
Us
Plugging in 100 kg fIx W, we can solve
C 0'1
2
CdO
fort.:! $
1
10-102
! ..
X
:=
3.43
dW
'
CDPIO-U The heat of activation is given in the problem as a function of the carbon number so we can just graph that: Heat of adsorption vs number
carbon
c: 80
~ 70
1: o
60 50 . 40
~
nI
'030 _ 20
~ 10
'r"
0
............
'H ...... _ _ """ . . . . . .
o
10
........ ,'
...........
, . , ' " ' ' . . . - - - - " . , . - - .- - _. . _ . _ . . . . . . . . . . .
40
30
20 Carbon number
To graph the activation energy we need to find its equation. Arrhenius equation.
It is the
k:;;:Ae- E/RT
Solving for E we get:
We know that as the temperature increases when n S; 15, the rate increases so k still gets larger with greater temperature so E is still positive. When the temperature increases when n > 15, the rate decreases so k decreases making E negative So we can come up with some equation with the above equation that fits this criteria and we can corne up with the following graph Activation energy vs carbon number 6
>-
....• _ .•_--........ -._.-......- ... _-_..........................._ ......................... ..
5
0'1
:;; c
.
4 3
o ---- ---.----~~
.
.. . . . . .
20-·-·-----·3"e-
10 4 1 . -------.. -----....-.------.-.... ---.. --......- -... --....- ..- -..- - -.....•-.----
Carbon number
The reason for this unusual temperature dependence is due to the fact that the higher the carbon number the less it wants to add another carbon
10-103
CDPIO-V (a)
a::: 0 at the end of the reac[Qr:
Us
:;;:?
(b)
= 1 kg/s
For Us::: 0.5 kg/s:
a
~ 1 il~ W ~ 1 {nOoN;;} ;1
(04kg')w
Catalyst Activity vs. Catalyst Weight
W (kg)
When a::: 0, the catalyst is inactive. In theory a can be negative, but in reality, once the catalyst is inactive there can be no further decrease in activity_ (c)
For a catalyst feed rate of Us =0..5 kgls:
dX:;;:: a( . . r')
Mole Balance:
F
Rate Law ;
. r,
Decay Law ;
dW
Ao
A
..... ~~.. =.~:P. =g~~.~. . = 0.4 kg! l.
dW Stoichiometry:
A
k(·' (., :;;:: '"'A'S
Us
:;;:?
a == 1·· OAW
0.5 kgls
(Assume T = To, P z Po' and 1) "" 1) Q)
C\
=C s ==(~AoCl·X)
Combine:
FAG .~= (l.'.O.4W;{kC~,,(l . ." X)"]
(k6t}(i~)l+ t~w}w ~ (~Md~~~(W-2itw'l 10··104
(l~xr!'i: (w -~ w' J From part (b) we know that the maximum catalyst weight (the point where a:::: 0) is 25 kg
We will find the conversion at this point:
(i?Xr QX~:i2'.{2.5 -$~:5) 2.5' )= 025 X=O.2 (d)
To achieve 4()% conversion:
(l~Xl =~~:{ w .. ·ittw ' ) f~~~=O{5-fJ;5') => lJ s
:=
O.667 = 1
2.5
Us
7.5 kg/s
(e)
ForU'==:~'?xj=!'¥:( W- 2(!jW') (i·~X) = (0.2)w:::: (0.2X5)= 1
=>
X =0".50
=?
CDPIO-W N~:=
Design Equation:
,
k'C'z -B
RateLaw:
- fll ;;;;
DecayLaw:
da ....• =k =0.05 dt
D
a
I
1
0
Jda =,0.05Jdt
~
a:: I,O.OSt
From tilL'> we can see that the maximum reaction time is 20 min, We will find the conversion at this point Stoichiometry :
(Assume constant volume)
ell "" ClIo(l···X)= ,~§2(1" x) V
Combine:
,~?c. =~/(l:~:~:?~)~~Q_::_~Z~ =: ,~~~.!t! (1 ""' 0.05t Xl····, X l ~
V-N 80
V2
_~_:;_ :: ~/~l!!!_ (1 . . O.05t}it ~--~ :: Y!~l!!! (t _O.025t (1" X
Y
V I " "X
"'~, =~g:g!X!gX12 ('20 . . 0,025(20'f )= 1 I-X
(1)
X=050
10-105
V
2
)
CDPIO-X a) Mole balances:
.dF . . . . _. = r A
dW
A
dF.l!.:; r ___
dFJ.:,. = r ____
dW
dW
B
C
rate 1a'N5:
stoichiometry: (
,
F..
F4 (1- aW)Q5
V
Vo
Fa
Fs (1-'aW)QS
V
Vo
= ..._ .. "" ...'-., ....... --....._,......
'-'tt
,
CB
"'" ."--,, := ...,....---..,........... "-,-,,.
decay law:
da dw
k a U,
"" ,.12"_,,
Evaluate the parameteIs:
Plugging all of tllis into POLYMATH we can change values of Us. T, and that will give us the most of product B.
Vo
CP... . o)
We find that at a tempcratllIe of 396K, a solids velodty of lOkg/s and
d (fa) fd (w) ""a
d(fb) Id (',") =,,10 Y!f,~bl(;!
d(Le) Idlw)=J::'c
;:::'~E.~~:L..y;e t~~:; ~eXit~1f!E_yalue £!~imum :!~;~~ 100
d(al/d(w)=kd'a/Us
Us"lO
fa f;;
T=396
Ec
Q
0
xci",
Os
633 oa'exp(~500011
937* (1/400·111')
J,O
11
:0
196
39G
0 .CO:iB
0 009'8
00093
0 631
0
C.066:':'49
Q
0 065::49
0.00775554
kl=, 02 'exp (lOOOOn 987* (1J400 .., 1 IT) )
kd
0 0661149
cb~fb~
(l·"alp*w)
A
,SIva
xa-;;:'-a*kl *ca rc~a'*k2'Jrcb
:cb= ··af!'
(k2~ch-kl *ca)
0 '1l19
1 . 9Ji02
396
VQ
(J..···alp""t..t) . . . 5/vo
i:
0 . 4 n,,711.93102
390
k2= OJ..*e;
Q.0913942
0 ;4t>631
10
o. 009S n 633
41p
Eb~.~.,?:~±~~ leO
3813942
.;) ,. 52'~43a
alp=9
0
633
0 6JJ
k2
OO,i55S4
o .cp:nSSS4
.00lil149 0 00715554
kl
Olj6Ln.
0,0116131
O,0176i3}
;)
Ol"l6l·.H.
1 579"; S
0.0131646
0
01Bla~S
c""
c".a
1 S'191S
0 631)42
.. 0 0278243
,c d,
.. 0 0(10620401
.. ·0 .07.::6245
0 00641366 0 02 :82";8
10-106
0 02/'S246
0991~46
0
.;}. ':;OO6:!O403 Q C01,4$'~02
"-0.0021742:3
•. 1)
O;)O~nBi$2:
b) Using the same program we can find what it takes to get the most of C possible. We find that at T:::; 396K, U,:;;;; 10, and vo:;;;; .03 we can get Imol Cis. P AO:::: 1082.4 atm
fb
~~~~ ~~_~al:.!Z Minimu,tl1 v .. l" .. !:ir..al .. ~~;,!!! 0 100 0 lOO 102629 .... 23 1 . G252ge.:n 1. l Q 0 SZ4487 a 135042"",:0
~a£±~
'"
fa
fc
()
a
1
1 1 .9370:2
1
Us
10
10
1.0
lO
l'
396
396
J96
390'
0 .. 0096
1 1.93702
<>1.;;>
0 0098 0,03
0 .. 009S 0 0)
0.0098
vo
0 03
0,03
kd
0.05511.49
o .0651149
00;;61149
0.0661149 0.00775554
k2
0.00775554
0 .. 00715554
0 00175554
kl ca
0.0116131. 33.33J3
0 0176131. J3 3333
0 .. 01'76131 1. 83'798e .. 23
ch ra
0 -0.581104
17 .. 26% .. 1 65051" .. ·24
-0. 5871C4.
-1 6505"113 .. ·14
:c
0
rb
o .5,s'1104
0 .. 13615 05971.04
'·0. C7!$781
9 56332,,-:: ·9 55332 .. ":'2
{)
0176131
4 .6379Se-23 6. 36S55e-:.n
c) To get the Time Temperature trajectory we can use the following equation to create it:
This will give us the following graph: Temperature~Time
Trajectory 1200
...
1U
.;!
.-........ -.......... - ....... -..-----.--... --..-,,-.-............ -....... .
1000 800'
lU -
d;Q. ::':::.600 ,E
......
400
__ _ __ -_._ .... ..................... .......
...-
.......
1U
I-
200-
o ...............-.. . . . . .. o
5
Time
{s}
to
d) For this we just add an energy balance We have to assume a heat capacity of the catalyst since none is given Here it is assumed to be 100 J/kgcat
Plugging this into the POLYIvlATH gives the follow'jng program. We find that the tempemture is 38SK, D, "" 10 kg/s and v 0 62 . 4 atIll.
;;::;
051 dmo/s. P AO
10-107
=
.Eq:..lationS,l, d(fa)/d('w)=:r:a
1
d (fb) /d (w) ,,:tb
o
d(fc)/d(w)=rc
o
d(a)Jd(w)~kd'a!Us
d(1') Id(w) = (:::i;l~ (-16000) .. rh T (--32000»
kd= 08*exp(15000/1. S87*(lJ4QG-I/T»
1 / (cp' (Us+'''cl+fbdc) 1
~ia::'l/;J
Us=10
t
c,,=lO()
k2=
388
~.i-;~;._·...alue ~ v.al~~ P~nimumwvaL~~ f~-,"l.~b.~£ 100 Leo G g 1:,155
Ol*exp(20000!1 987*(li400-1/T)}
o
21532&
o
o
61.1.155
t)
,215328
kl'" 02*exp(lOOOQ/l 987* (1}400·,,1/1')} alp=9,8e" 3 Vo~
40~O!,
T
382 2.:.2
C 0446169
51 cp
cb"fb*(l"alp'w)A Siva
k:::!
ta.:::::-.a,'*kl"'ca
kl
to
:,:<)
lce
10;)
o
{) .C0459207
>10
004Sn{)1 ':11J553
dlp
,;;,
0.0332906
10
10
ca;:fa* (l-,alp'*w) . . . S/vo
t;'c=a "
358
,0·098
o
O~';553-
o
.,
-Q
OC29~-!19
O,(H085Z:
0093
0098
31
5:
0 S:
3i5D}S
96{)'8
-0 Q~a':lS~
7;6.3935
:1
o :;c
:0
:3265'44
10-108
'''0,000'15307;5
0,001'73258
,·00255;.1\4
Q
:)03}'0577
0,0111477 () .. JOSS
C ,$1 0481:;.54 ,':694'1'2 -C.00:)753076
o
00)139124
~,
18S:6~-
05
Solutions for Chapter 11 - External Diffusion Effects on Heterogeneous Reactions PI I-I Individualized solution PII-2 (a)
A
2B
1,
Z 0,1'=0
WB =-2WA
cD AB dy A (1 + YA) WA = - - - - = -cD ABdin - - (1+ YA) dz dz Integrating with YA = 0 at z = ()
WA = .cD AB ·In(l + YA)
(S-z)
(1)
at z = 0 YA = YAO cDAB ( ) WA =-S-l+YAO
(2)
Taking the ratio of Equation (I) to Equation (2) to eliminate W A and solving for YA
!~(1 + YA) = In(1 + YAO)
S-z YA =1- (1+ YAO )
5 1-z/0
11-1
,,
,
"
'fA
/
EIVKD
"
", ,
,
Pll-2 (b) (g)
kc2 kcl
Tl = 300K
T2 = 350K
=(PAB2J2/3(~J1/6(U2J1/2(dplJ1/2 DABl
'\)2
U1
(11-70)
dP2
As a first approximation assume
DAB2 =~ DABl 112
AtTl=300Klll ",,0.883cP At T2 = 3.50K1l2 "" 0.380cP Assume density doesn't change that much, '\) = Il p
~=A=2.32 V2
112
1 d p1 U;-2' d P2 U2
_
_
1
-"2
rr
kl
= 4.61xlO-6 mj s
k,
~ 4.61x 10-<> m/s[2.32 J'" [~ [~
=4.65xlO- 6 mj s WA = --r; = kc2C Ab =(4.61XlO-6 mj s )(103 mOllm 3 )
-.r; = 0.00465 mollm Is 2
11-2
Pll-2 (c) (h) A 50-50 mixture of hydrazine and helium would only affect the kinematic viscosity to a small extent. Consequently the complete conversion would be achieved. Increase diameter by a factor of 5
k,2
=
kd(~:~r
=
2.9m/tr
=1.3mjs X=
1.-ex1-1.3.1~~3 0.05]
= 1-exp(-4.6) X=1-0.01=0.99 again virtually complete conversion.
Pll-2 (d) Liquid phase: e.g.. water
See margin notes on page 786 and 787 for solution,
Pll-2 (e)
~:
=
-4.,;1-;'}-4.000( 7~3
e
-
8~3J
= e-0 .59 = 0.55 Assume
~:: - (~)- o.~5 kc2 _ ( DA2 -- kcl DAl
6
)2/3 (-VI )1/ V2
_
-
( -flJ )4/6 ( -flJ )1/6 =( -flJ.)5/6 Ji2 Ji2 Ji2
k c2 . = (0.55)5/6 = 0.91 kcl
!!~~C2 U/s.
In
= (1.059)(0.91) = 0.96
1 =(0.96)(2)=1.92 1-X2
X 2 =1_e- 192 =0.85
11-3
Pll-2 (f) CAD
Assume concentration in blood is negligible (C A2
=0 at ()2). Assume quasi steady state
d(VpCAp ) ---=-WA dt A P WA =.!!ABI &.[CA --CAl ] I
W
A
= DAB2 J [CAl -0] 2
adding
[~L+~_]=C DABI A
W A DAB 2
_
CA
J
&. _2_+ __1_ DAB2 DABI CA = RCAP
WA
-
V dCAP P dt
=
ApRCAP ~ 8 - - + -2-
DABI
DAB 2
R-(~+~J!AR - DABI DAB P 2
Flow into the blood FAB
= CpA mol/time, R == [time] RV
If CpA = constant = C pAO - CpAO FAB---
R
11-4
FB
t If CAD varies
=.-1 CpA dt V R C pAO t In--=··-_· CA VR
!lCAP
CA
= CAO e-t/ VR
FB
A =--
C
81 and 82 are given in the side note.
R
FB
t Pll-3 Mol balance on oxygen:
Fo~·
0 ...
to:! ::::
0
constant liquid composition . where n is the reaction order.
. 'd AssulIlmg I eat gas Iaw app l'les: F02
P.l)" =i?:y'
• t Assummg hla.t ••..P _ . = cons tan t. ('"
by correction of 1,)0 to some reference,
v
R.T
.
h rt._ were u,,:::: oxygen uptalce rate m1...ItU
Assuming Henry's law applies (low pressures) where H is Henry's constant and Po:! is the oxygen pa.rtial pressure
11·5
Substituting into the mol balance:
Fo~
=- ['~
In u"
=:
nJn
P~
k.H"
+- in ._.". . .
C
A plot of in U o vs In
Po~
will give n as the gradient.
As the system pressure, P, given in the data is absolute: if in the reactor. the xylene is at boiling point and dissolved oxygen and oxidized xylene are at low levels, then P", =' 1 arm (open to the atmosphere) In order to deduce the correct kinet.ics of the oxidation it is necessary to find the partial oxygen pressures for rhe conditions where the rate is limited only by the reaction kinetics and not by diffusional mass transfer. Plot of stirrer speed, W, vs oxygen uptake rate, 1)", for each run will show the conditions at which diffusion is negligible.
............. 1 "
.........-"'''.~.-
Ir. can be seen that at stirrer speeds above 1200 rpm that OUR is illsensitive to W and hence the reactor is well mixed ie. no liquid diffusional limitations.
....... ~." ....
atm-
..•..." 1.0 atm. .. ··,..···.. 20 atm
:::~::]:.2 ~~. 4QO
900
1400
OUR: Oxygen Uptake Rate
W, rpm
Hence using the system pressure dam at 1600 rpm for the plot of in give n uninfluenced by diffusioIl .
Po: ;:::: P, PjI{
=:
U
o
vs In Po;: will
P, I
At 1600 rpm
·:~Ef=B~fl*~~~~= 11-6
,----_._.,-'''..... ...
_-""' •... "
,
...".,..
'"'."-'"'''''""'"'--"",-"""'-""",","-"""-"'-"","_.,""'",".,,"""''''"
y =O.9992x + 4.,6422
-1
"l.S
o
.(),,5
0.5
LN{(02)
TIle grndient, n"" .999 ::: 1, so the rate law is:
Pll-4 Diffusion in adjacent skin layers Skin
la~er
interface
P:-;~I P;>cl
P:-I~:::::: 101 kPa PH,,:::::: 0 kPn
Inner layer
Outer layer Strotum comeum
0, ::::::0.002 em .,,'--,,-,_._---.. __ ."" .• _
P:-I1:::: 0 kPa PH¢:::: 81 kPa
Epidermis
~::::::OJ)!
em
.•" _ . _ _ _ _ •••'''. _ _ •• " .•.••" ...... _''_m" '''''''''''''''''''''''''''''''''''' """_""""."""' ••••• _ _ _ _, _ .
Assnming dilute solution and constant total concentration in both layers gives: d CA
""·"-::::::KI
dz
for each diffusing component in each layer.. N~
:
Outer layer:
boundary conditions
0 , C A :::::: CAO Z :::: 01 , C,.\ : : : CAl
1.::::
c.v-C4.0 KI : : : -_.--_.-.-
01
11-7
c . == C,\O - ( C"O .- (AI).
Profile Inner layer:
:1
boundary conditions z:::; Or ,C . . . ;: CAl z=b:.,C,-\=O CAl K I == . _._._--
02 --Ot Profiie Total partial pressure profiles Outer skin layer:
7
.",
;::; 1010 -. (10 1.0 '" t008)..······:::·······t·lO·-·~· .. 0.002 0.002
::::: WIO·-lOOOz + 5000.;:: =::
Inner skin layer:
p, + Pa ::::
1010
+ 4000.;:
O' . . ~.) (b; -·o! + Pao'"
Pv-;::···~·· .. ...::~·
(O,-.~.)
(PSD ..... Pal) ._.' ....-=:..
\0 :"01
::::; 810 +- 126000(0.01··. · :::) . . . 100000(0.01 ... z) =;::
Cht.:ck:
SlO+ 26000(0.01
i\t interface (2 == 0.002 em) :
::) olllerlaver :::: inner/cryer p., +- Pa::: p, .-t. Pa
1010 + 4000:; == 810 + 26000(0.0 1····· 1018:::: lOl8
Plot these two profiles across the skin from z :::; 0 to
l ::::
11-8
OO! ern
conect!
zJ
He :
Inner layer:
boundary conditons
z
=01 ,es =Cal
z = 0:: • Co ::: eso K.I
C80 -
CST \
K z =: CBO - -( - - - - - ._.
o! --01
U51.
\ 51"01 )
0' . . .-7') 0: -"01
Profile
Outer layer:
C80'
C81
:::::: ... _-._- ... _ ..
CB :::::: CIJO'- (C80 -- CSI)( -:.-.--":::'..
boundary couditons
z:;::: 01 • C a ::: Cal z:;:::
0, e a :::: 0
Profile Total concentration profiles: Outer skin layer:
.....
CA';'
"'t..,
e8 ::::: CAD -.- (CAO'-
.....
ell).
s;Z +
., Z (8/ ~f;'
Inner skin layer:
Evaluation of C,\l and N::: :
em :
Outer layer:
IhNl [.. W,U ::: --.---CAl .-. 0].
Inner layer:
02-01'
.
Assume that flux in inner layer == flux in outer layer i.e. W AI
w"I N=?[C' .D ---AO- C"] AI ' .
-
:::
W A2
......... ....
lh-Ol
w<{
~;;; ~~,. =c.o +
Conversion of kPa to kglCIll*s2 101kPa=101OOOPa==101OOOkg/ms2:::101Okg/cms2 ,-.lkPa::::::lOkg/cms2
11-9
w
::::o _ _ _ ••
E~-_ ... -.:::: .-.- . . . . . ~} 0 _.....____ :;: 50 * 10-2...~f_.
" [~: +~£~:1 h*~~~·~%~i':1 He:
DIN, [. _ Wa ::;;:-...... (af --
Outer I:lyer :.
'"' O.
..
an's'
° ]
lrU1ef layer: Pressure Profile of skin Jayers 1050
'! E
1000
inner layer
.~
26 950
e::>
...e
900
0.
]j .., 850
interface
I-
800 0004
0002
(J
0,006
0008
001
z(cm)
...,.....
6:·· 61 .•
Hence . P~I
; :;:;
== lO08kg I em . s:
a,
0002
= lOkg I em.s:!
11·10
The maximum sum of partial pH~ssures occurs at ilie skin layer interface z
== 0002 em
PrOIa! :::: 10 18 kg/ems!
Hence the maximum sum of the partial pressures is slightly greater than the saturation partial pressure and so gas will fonn bubbles at rhe skin layer interface causing blisters.
Pll-S (a) Part (a & b)
Packed bed. mass tranfer limited, gas phase
Mol balance:
.-.. 1._4.!:!: + ra"
. a"
:=
0
where Fa: U.Ca.Ac
Ac dz 1 FaodX
.............. .•••...• _ ... _ ••• _ .••. =:;:;
Ac
ra.. . a,
dz
U . Cao.dX
..
U : const.ant superficial gas velocity
········::::ra . a.:
dz
Rate law:
Mass transfer limited boundary condition but Cas:::: 0, rapid reaction
fa" :::: kc . (Ca ... Cas) Stoichiometry •
Assume constant T. p. gas phase • ...Cao.{l······X) Ca;;::: _...... _ .........._._ . . . . (l 'fE.X)
Combining :.
(j
Cao.dX
where e:::: yaoo:::: 0.05 x 3:::: 0.15
..
......... ---: ra . a.; dz
Cao.uo..~!:..;::;.; a.;.kc. Ca
dz
.dX '. . --,,, =- a... kc.(l. . . . . _. . "........ ",X) . dz (Jo.(1 '+' O.l5X) -"~'-""'"
,,> - -
Use Thoenes & Kramers correlation for ke :
11-11
Parameter evaluation: d p
r1
=
[?v J [·~~~~lg::~·l = 1C
.
=
n'
P
~
O.238cm
..
Diffusion of cydohexane in hydrogen (assuming constant T, P) I'll'
O.OOIT175L·~~j;;+iii;J [Jab == ---,-.,..-.............-....-,----.. -.,,",--""'p[(l Va yn "1 (I Vb yny Fuller, Schettler, Giddings for binalY mixture, low pressure, non polar (Peny's handbook chem.eng,) :VIa, Me ::::: molecular masses::::: 84,2 respectively Va, Vb :::: diffusion volumes;:;:;; 122, 1.07 cmJ/mol respect.ively
4:" 0.4
Y :::'
05: +05tOj) ..... _. -" 4. _._._. _....."".. __ .. ""._ , 0 ::: l.lLi L -'7') O.. )/~
rr..dp!
Uo ",,"?'~,., = 60000
:::: 509cm / s
Ac
4
u :; ;: Uo.( I + EX)::.:: 50.9(1 + O.15X)
11-12
!l =: 0.00017 g/cm.s (Hz. 500 K , 2 atm)
Sc' =: --_Q~~:!'.. ,...0.000 19 xO.857
=:
1.044
kc
1 [, ( _ ')']112 [ 'Ji/3 =OJ'8S''' 47.3 1+' OJ)X 1.044J
a.
=:
=:
17.98(1 + O.l5X)
III
~.Q.::~.~.;~~ == 6.29cm'l
0.572 !1 =: 0.00017 g/cm.s (H 2 • 500 K , 2 atm)
+ O.l5X)0.572xO . OO019 . Re• =: 50.9(1 ....... _._...... _............ _ _......__. _ - - ::= 47.3(1 + O.l5X) 0.00017(1·· 0.4 )1.146 Sc' =: _~~~.:l.._ =: 1.044 0.000 19 xO.857
kC
=: .... 1 _.., ....
r'47.3(1+ o.])X- )'l1/z[ 1 1. .
Q44'JiIl J
0.388 .
6(1·0.4) 0572
Q, =: ._-_...._.. =:
dX
.."..... .. '=
=:
5X' liZ l ..,... 98( 1+ OJ)
.1 6.29cm
(1 ... Xl
~
t1. + O.! .5X-)ili·
POLYMATH
Pll-5 (b)
11-13
§gU~bons.:.
~Edal ~
d{xl/dlzl=2_22·(1-x)/(1~.15·xIA_5
o
Zf '"
5
~~iable :z;
o
5
o
0.99997
o o
5 0.99997
(.000
~sr;..-' x
a.BOO
Q.600
C.4CO
O.2CD
o.oeo
-+ ceO
os. 000
z !: ........... .2 625
"', (l
9S591419
:2 .5875
0,99541018
75
0.9';)081594
2 8125 ;: S'IS
O.99722ilS
.2 .9375
0.99786072
~
Q
39756518
0.993:2039 .3 .. 1.25 3.1875
25 " 3125 j
3'75
.4375
O.99
o
99g87983
G 99901579
a o
9991:3525 99924Q2
3
O.99'i33242
J. 5525
0.99941344
The results show that only :3 :3 c:c;: of [he mbe is reqllired for conversion of 99.9%, much tess rhan the fun 10 £1:
Pll-S (C)
11-14
J);.bh="$h~ kc . dp
lj)
I
U ,dp· P -11/2 [_;'(1-$).
!l
']113
[p Dab
Assuming the porosity remains the same, factors in the correlation affected by the size of the catalyst pellets are:
d!
2,1t.-···, +red.l
0,.25:
_
_
.- ....... +0.2.)(0..2))
y : : ; -_._._:1._-,_.. -. ::::; 0.286
rt.dp-
~!~
= 1.145
;; l~~?? ::~?.::?~( 1 + O~:?~~~~ ,.J!,.:_:'
dX . (i--X) -" ... ::::; 6 '8-",-"·",,--,,,_·. ,-dz -- (1 + OJ5X)II:
. (1 + OJ 5X)
POLThlATH
Problem Pll-18 parI a/b l.OoO
.".
!
L,
.!Ss.'!'..;".
c,acc
I
·. ·. x o . 6ca
0,,-100
z As can be seen from the above graph. rhe affect of reducing the dimensions of the pellets by half results in the conversion reaching 99.9 % U8 em from the entrance. This seems reasonable because reducing the size of the catalyst particles is one of the methods for increasing mass transfer and hence kc.
11-15
Pll-5 (d) rf pure cyclohexane feed were used at the same volumetric flowrar.e 60 dm3/min , then the initial bulk concentration would be greater and there would be a greater concentration gradient across the sragnant film on the pellets However the mass transfer coefficient will be reduced as the products must diffuse away fmm the surface which will be harder in higher concentrations of cydohex
(84:<2) ' _ em·.1 == 2647. 11m J I == ( ).0026)8/ (t0821x773 ," mm
• • • • •_
••••••••
)J.=? g!cm.s (cyclohexam:, 500 K, 2 atm)
Pll-5 (e) This problem gives an indication as to how changes in parameters may affect a packed bed.
Pll-6 Given,
L=O • • •
Minimum respiration rate of chipmunk, FAL = L5 /lmol 02/min Breathing rate of Chipmunk, Vo= 0.05 dm3 gas/min Diameter of hole, D = 3cm
:-.,-...,.........................
"" , ,"', ,""', :-..,"""""",.. "'............"".......,""",.. :-..,....'··s...........,"",.·.,:· ;......"'...,""',....,',.. :-......... ............... .... .. :--..... .........,",..
".
x .......,""",......
" . . ,"",. . . . ,"""',..........
","'!.,. . . . ,""',. . .
,...."""""""",...."".......""
."""··.s,,,,,,,,,,,,,,,,"'-:..
Assuming, • A represents oxygen • B represents nitrogen • Chipmunk has a constant breathing rate of 0.05 dm3 of gas/min
.....""""",. . .,···s..." ........-.. ; ,-:-.,............""""""""",-..;. ,"""',.....:...,"',....,'" :-,....."""",...",... ..""...,,
j
""' .....,""""""""",....... :..."""""""" ... ...,,, ..."""",. . . . . ." .,,'\.,."",···s ........,',................,",....:......., ............, ........,........................,............, ........, ...., ........-..;, ........, ...., ...................., ........" ........, ........,"-:.. ",
,
,
,"',
.... ....,"',....,",....,'" ....,"',....,"',............,""....;.
""""""",. :-......" ...,:.......,",........" ..
"""""",..
:-,.....
"""""""',.. :-.,....."""""",.. ..:......."""""" .. ........,""""" ...., .. :--
. . . . . . . . . . . . . . . . . . . . . . . . ." . . .
:-..................................':--...................... :............................................................. :............... ............ .... .... ..
, , , ,', """",....,"',..
: ,.............................,........".. ,. . " . . , . . . . . . . ." . . , . . . .,',. . . . . . . . ;. . . ' L = L ". . . . . . . . . ~~~~:~AL~ PT ~~~~~~_..::.._"" ~~. . . ., . ., . . . . . . ~~~~~: " "
Minimum flow rate of oxygen to the bottom,
:-..,,·..·s...:..x ........"...:...,',..
FAL
= Minimum respiration rate of chipmunk = 15 /lmo} of 02/min
11-16
Flow rate of A down the hole
= Flux of Ax Cross - sectional Area C AO -CAL
= ~B
:. FAL
L CAO _. CAL
or L=~B-
X
Ac ----(1)
xAc
FAL
~B
= 0.18 cm 2 / s = 0.18x1O-4m 2 / s
Ac= 1'lf)2 = JZ"x(0.03m)2 = 7.069x1O-4m2
4
4 6
FAL = 1.50x1O- mol /min = 2.5x1O-8 moll s
Pll-6 (a) At Pasadena, California C AO
PI = -RT X Y .A
(Ideal gas law)
where,
Py = 1.013 x 10 5 N / m 2 at Pasadena, California ( situated at sea level) R = 8.314 J / mol.K T
= 25°C = 298K (Mole fraction of oxygen in air)
YA =0.21
C AO
=
5
2
f'.! / m xO.21 ( 8.314-~-X298K) 1.013x10
mol.K
:.CAO =8.59mol/m 3 Now, Flow rate of A = (Concentration of A at the bottom) x (Volumetric intake of gas)
= CAL xVo 3 Vo = O.OS dm / FAL
. C
..
min
gas/min
8
-
FAL _ AL - - Vo
= 0. OS X 10-3 m3
2.S X 10- moll s
3.
(0.OSX1O-3
~x Imm min
60s
_ 0 03
--.
J
1/ 3 mo m
Solving for the length from (1),
L=~B
CAO -CAL
xAc
FAL
11-17
3
L = 0.lSx1O-4 m2 I SX((S.59 -0.03 0) moll m ]X7.069X1O-4 m2 2.5x1O-8 moll s :. L=4.36m
Pll-6 (b) At Boulder, Colorado Boulder, Colorado is 5430. feet above sea··level The corresponding atmospheric pressure is 0.,829 x 10-5 N/m2
=0.S29x10 5N I m 2 at Boulder, Colorado
PT
Py
C AO =-xy RT ,A
5 2 0.S29x10 N 1m
=
C
xO.21
AO
J X29SK) ( S.314 mol.K
:. C AO
= 7.03 moll m 3
Solving for the length from (1), L
= ~B ,~AO -, CAL xAc FAL
--O.O~)
3
L = 0.lSx1O-4 m 2 I sx((7·03 mol 1m ]X7.069X1O- 4 m 2 2.5x1O- moll S
:. L =3.56m
Pll-6 (C) During winter at Ann Arbor, Michigan
T
= 00 F = -17.7SoC = 255.37 K Py
C AO
C AO
='RT
xYA
= __1.013X1~5 N Im~_-XO.21 ( S.314 -J-x255.37 mol.K
K)
.. CAO = 10.02 moll m 3 Solving for the length from (1), L
=~B CAO -
CAl X
Ac
FAL
11-18
J
L=0.18xlO-4m21 sx 255 175 x ( (10.02 -0.03) mollm3 x7.069xlO-4 m2 ( 298 ) 2.5xlO-8 mol / s :. L=3.87m During winter at Boulder, Colorado
T
= 0° F = -17.78°C = 255.37 K
CAO = RT Pr xy. A CAO =
:. CAO
5
0.829x10 N I m
2
( 8.314 -~-x255.37 mol.K
xO.21
K)
=8.20 moll m 3
Solving for the length from (1), V0r CAO - CAl L -- 4B
.11 x''C
FAL
J
L = 0.18xlO-4m 2 1 sX 255 L75 x ( (8.20 -0.03) moll m 3 x7.069xlO-4 m2 ( 298 ) 2.5xlO-8 moll s :.L=3.17m
PII-6 (d) Individualized solution PII-7 (a) Given: P y = 510 nun Hg
@
35°C (from plot of In P y vs lIT)
B InPv = A - (T + C) DAB =
DAB
0 . 120 cm2/sec
Antoine Equation (from equation of Fuller, et aL)
0.00 IT' 75 ~~ + _1_ MA MB
=---P(Vl': +VV;;)2
Fuller Equation
where V A and VB are the Fuller molecular diffusion volumes which are calculated by summing the atomic contributions. This also lists some special diffusion volumes for simple molecules. Fuller diffusion volumes Atomic and structural diffusion volume increments C 15..9 F 14.7 H 2..31 CI 2LO o 6..11 Br 21.9 N 454 I 29.8
11-19
Ring -18.3 S 22,9 Diffusion volumes of simple molecules He 267 CO 18.0 Ne 5,.98 C02 26.,9 Ar 16.2 N20 35,.9 Kr 245 NH3 20,,7 Xe 32.7 H20 13.1 H2 6.12 SF6 71.3 D2 6.84 Cl2 382 N2 18..5 Br2 69,,0 02 16.3 S02 4L8 Air 19,,7
PII-7 (b) By CS z molar flow rate balance
dNA =0
- - Z=20cm
dz And
NA =K1 Fick's First Law (For NAIR = 0)
t
dXA NA =-CDAB--+XA(N A +0)
dz
- - Z=Ocm
-CDAB dXA 1-XA
CS 2
U
dz
R=0,5 cm
Equating the results of Fick's First Law and molar flow balance, then rearranging and writing in the integral fOIm
Kl1
°
J
dz = CDAB X dln(l- X A ) Xo
=.C;!?AB 1{1 ,- X2]
Kl
1- Xo
Z2
Evaluating for Z2 = 20, X2 = 0
C
= ..!..- = _,____ ....:(l....:)_at_m_ __ RT
(82.6) atm.em' (308)oK gmol.oK
= 3.95 *10-5 gmol / em 3 Xo
=
Pv -. = 510 PTotal 760
= 0.671
N = K = i~·95 * 1O~5)(Q..1221{ 1- 0.0 ] A 1 (20) 1,- 0.671 = 2.64
* 10 ··7 gmol
/ em 2.sec
PII-7 (c) 11-20
Fo, "';f:e;:) between z=0and z=20 z: = I-X, l{__ l ] l- X o
(I-X )=(I-X) [_I_I/Z2 =(1-X) (1-ZIZ 2) A OAI_X OA X B =(1- X)A
o A
=(X Bjl-Z/20)
~=~C
M
~=~C
= XAMA + XBM B
CB =XBC VA =NAICA V* = NA +NB.=7.17*1O-3
VB =NBICB =0
C
n +n V= A B =nAlp p fA =CA(VA - V*) =NA - XANA = XBNA fB = CB(VB .- V*) =-XBNA =-fA (*10 3)
(*105 ) Z
XB
XA
CB
CA
PB
PA
0 5 10 15 18 20
0.329 0.434 0.573 0.757 0.895 1
0.671 0.566 0.427 0.243 0.105 0
1.3 1.71 2.26 2.99 3.54 3.95
2.65 2.24 1.69 0.96 0.415 0
0.376 0.495 0.655 0.865 1.02 1.14
2.02 1.71 1.29 0.73 0.316 0
(*103 ) Z
VB
VA
. V
0 5 10 15 18 20
0 0 0 0 0 0
9. 9 11.8 15.6 27.5 63.5 1E+06
7.17 7.17 7.17 7.17 7.17 7.17
(*10 6) V
8.32 9.05 10 . 25 12.5 14.9 17.5
P
2.4 2.21 1.95 1.6 1.34 1.14
.J:1.1O
jA=-jB
JA=-J B
3.16 4.52 6.77 10.9 14.9 20
8.7 11.45 15.15 20 23.6 26.4
PII-7 (d)
11-21
ffis
ooA
00
0.158 0.226 0.338 0.544 0.764 1
0.842 0.774 0.661 0.456 0.236 0
60.7 56 49.4 40.5 34 28.97
Mole fraction
0. . 9 0. . 8 0..,7 0..,6 0.,,5
6
...~ g
o.A
Gl
0.,3
0.,,2 0.,,1
0.
......... ·.... ·····T····
0.
10.
5
15
20.
Z (cm)
Co ncentraiion
0
E ~ 3,5 ~ 3 0
~ 2,5 c
2
0
~ C 2!
1.5
1 c 0..5 0
u
Ill.
1
,--,--,-,~
0.
5
0.
10.
15
20.
Z (crn)
Diffusion Flux
30
CiO < 0
,...
•
25
cO
0
0 ,...
cJ,
<
Q)
III
20
• 15 IiIE
.2~ III c ra 10
. ..
11.
0
'iii
...
::J
E
5
:!::
'D
0
.................. ··T· .. N . . . . . . . . .
0
5
10
15
20
Z (em)
PII-7 (e) Evaporation Rate of CS 2 (Pliq
@
20°C = 126 gm/cc)
11-22
PII-7 (f) No solution will be given PII-7 (g) Molecular diffusion of air is taking place
NB = J B + XB(NB + N A) O=JB +XB(NA) J B =-XB(NA) = (l-XA)NA =-JA
PII-8 (a) Quasi-steady-state - no accumulation in the capillary tube or in either chamber. Volume 1: Accumulation = in - out.
dNA d(CA~) dt=-dt-=O--WAAc WA =-DABCt
+ YA (WA +WB)
:
EMCD:WA=-WB
- DAB-dCA ~ WAZoIL -- DAB CACICA2 WA -dt Ai
WA =DAB - - ( CAl -CA2 ) L
Volume 2:
d(CA~) -.- dt
=WAAc- 0
dCAI =. AcDAB dt V,L
(c -c ) Al
I
dCA2 = AcDAB dt VL
A2
(c .- C )
2
Al
A2
Subtraction gives us:
d( CAl -CA2 ) = _ AcDAB (~+~J(c dt L V, V Al I
2
-c ) A2
Now integrate:
11-23
AcDAB( - 1 _. - 1 Jt + Constant V; V2
L
Pll-8 (b) (CAl -CA2 )
= kMC
IJ t+Constant
AcDAB(1 ----
Ink+lnMC=
V; V2
L
J
AcDAB ( -1 ' - 1- t+C 2 InMC=--L
InMC =
V; V2
-DAB (O.01025)t+C 2
if we plot In(~IC) as a function of time and find the slope, that would give us -DAB
See Polymath program PIl··8··b.poL POL YMA TH Results Linear Regression Report Model: InDIC = aO + a1 *t Variable aO a1
Value 3 . 7024708 -0.0107008
95% confidence 0.0078035 9,,83E-05
General Regression including free parameter Number of observations = 9 Statistics R"2 = R"2adj= Rmsd= Variance =
0 . 9998944 0,,9998793 0.0015013 2.608E-05
DAB = 0.0107
Pll-9 Dissolution of monodisperse solid particles in exceSs solvent. Di "'" D +._.. Dil ..... [)~) = 2D*
!. .--{
Ct..!
Define conversion in tenus of volume dissolved: gives
D:: Di (t·· X) II:;
11-24
Substituting for D :
. .Dl . (l - X) lil. +._........ 1 (D':! D/t 2D*
.. ,. [D'1 (I -X ) 113]2) -== fl.. t
.
terIll 1
Surface reaction controls:
term 2
D* is large. term 2 is small cf teEm 1
Di - Di (1· X)II} ::: fI..t 1- (l .. X)
Mass transfer controls:
IIJ
fI.,(
-== --_.
Di
D* is small. term 1 is small cf term 2 1_. ( Dl·2 - [D1 . (l - X)lJ)'f).•.•. - fl.. t
2D*
.
Di... ( 1·- (1-· X) 2/3 ) _. __ 2D* Borh regimes appare?t :
= .fI._ .... tDi
Tenn 1 and term 2 apply
J
Di . . . (1- (1. X)-"13 ).' ::::"l"')"'~' fl..! [ 1 ..· (t .. X). 10] + ['--.
.2D*
•
Pll-lO (a) Mass transfer limited
A in excess, C,"" :::: C AO
T he reaction nue is equal to the mass transfer rate For small particles and negligible shear stress:
kc ""
3:pe D
11-25
;'101 balance on solids: For t mole A dissolving l mole B then . . c-'./' =.
(3," ;;;:;
kcC AO as e A, is undefined
Boundary conditions, t = 0 , D ::: Di
J
(D)dD
=:
-.~:P::~~?J de p
0,
0
. . . ". = . . . 4.De.Cw.t . . . . ". "'" 2 p ~,
~
~
Time for dissolution, te , at. D =: 0, and assuming particle density, p:::: 2000 kg/m3
tc::::
Di 2 2000 x (lOe - 5)2 ='""'""",_.. 8. De. CAD 8 x (We, to) x (2000)
_',""0"0::::;
, ~ 0.1255
ie. vinuully instama..'1eous dissolution
PII-IO (b) Surface reaction limited Mass transfer effects are not imporlant when the surface reaction rate is limiring.
~1ot
balance ()n solids:
For I mole A dissolving L mole B then
LA;'
= '" la,"
Boundary conditions, t. :::: 0 , D =: Di
C JdD ::::;.:::. .....k:.~.-:~.? Jdt D
Dr
)
I
P
0
11-26
D-Di = _ 2.kr.CAo:.!... p D::: Dj _!::!-r_~".:.?t
p Time for dissolurion,
(C ,
at D:: 0, and assuming particle density, p:: 2000 kglm3
tc == . p.pi == 2000~..QE.::::_?'! == 5xlOel2s 2.kT. C~O 2 x (toe -10) x (2000) a very long time.
....
PII-IO (C) At
t;;::
0, total moles A in tan.lc:: 0.1 x 100 == 10 mol A
Le. there is just enough mols of A to completely dissolve all of B in a well mixed tank.
Acid not in excess and dissolution is mass transfer limited (CA:I:;O: 0. and CA ;: CAo)
assume zero order in B and ,. -rAJ
kr
..
="'-0' C" 1+ -.~..
D*;::: .~:..I?e
where
kr
D*
Mol balance on solids: For 1 mole A dissolving 1 mole B then
"r.~.,";::;:
.~e -u.r·.· · · .· · ·~.·.- ] dt
D l+ .-_.
=:
"
. . lss"
where a.
2.kr.Ct = -_.j)
D*
11-27
kc
2.De = --_. D
Boundary conditions.
t
= 0 , D == Di Di······Dt
DP . . . [)J.)
:::: eLl
2D'"
As a function of radius: 2ri2
'">r2
:'...._..- 2r + at····· 2ri .....-.. _. =0 D D'
,.
r-
[··D*.a
t
..
.,]
+ D* r -+ 1.'."'-'2'-'.' "-' D'" n- rC
""
0
·-b ± Ji}':"4·7.t.·~
· ·. · . . · .iD· .-;. -.. . . .---.-..-.. -. f .1
r = ...................... _. . . . . . . . . . . . . . . . . . . . . . . -
Using the quadratic solver:
2.a
~
at D 'n. . . . . n ·2 '\I ..,. [) • ± (Of I,D ...............
r :::; ......................."'........... ................... . . .... ...... ~.................._............................
2
Time for complete dissolution, te, at r:= 0 ;
tc =
assume p == 2000 kg/m.'.. and
2kr. C, = 2x(lOe···· 18)x(OJe3) ......... _........... _.... "......_. p 2000
0:;;:;: -. __ ...........
.
D'"
gives
~{Vi ~ ·~t;;;] 2. De
2x(iOe·1O)
kr
10e ······18
::::=.._. ;;;;;............ -.::::
I
r"
tc :::: lo·~::~i9~ (lOe
-
)) +
=lOe ..... 19
_ 2xlOe8
(IOe ···5)::'1
:2. ;(:i·;io~-~::"8)J:::: 10e14s again a very long time .
PII-I0 (d) tc ::::: ._._.P . . . 2.kr.'C,
i2 [Di +e '1 2D*
To reduce te, increase C, and lor decrease Di. To increase tc, decrease C", and I Of increase Di .
PII-II (a) Irreversible, gas-phase, adiabatic. no pressure drop. packed bed. Isorhermal Mol balance :
moUgcat
$
11-28
where F Ao Rate law:
= C ..ou" = lOe-3 x lOe4::: 10 mollem'
- to\,':;; k' .C ...,
Bur C,,-, is unknown .
Assume reaction rate is mass transfer limited.
"-·f:.~,\
cm' mol "'-1 s.gecu em
k' .kC.CA
::::: ", ........ _-"." ... _ •.•
kc+k'
where k'
=0,01 cm~/s.gcat at 300 K (constant . . isothermal)
converting:
70.7 x acal;;;:: 70.7 x 60 =4242 cm3/s . gcat
0..0 lx4242xCt 4242 +0.01
······/"A. :::; •..................•.•. - ... - ...-. . - .
Stoichiometry:
gas .phase, constant pressure and temperature
C", = CAO (1- X)
and CAO = I mol/dm'
:=!
lOe·J mol/em"
POLYMATH
~~ ...Y.!1:!;!~
~S:!:2!E~.
o
d!x) fd (w) =··::a/fao
k=O.Ol fao=10 kc.. 4242
cao=O.OO1. ca"'cao"U···xl ra=· .. tk;*k;c"ca) j
Wo
= 0,
We
= 1e+05
n·29
t-*.axi.mtJ.m ,.~~
Minburn. value ~.~~~ 0
1.e+06
0
le+06 0.63212
0
0 63212
k
0 C1
0.01
0.01
0.01
tao kc cao
1.0
10
10
4.242
10 4242
4242
4242
0.001
0.001
{LOOl
Q
Cd,
0 .. 001
0.000.36783
0.Qa0367S8
ra
··9.99998e-06
0.001 ··3.6787ge-06
·'9.99993e···06
·-3.6787ge··06
yariabl~
Initial
'"x
0
valu~
001
a.aco
...•. :x
0 .. :320
0 .. 15<:]
+ ..
C.GCO
0 .. 600
..... t······ a.sob:
w
Fur a c{)(l\c:r::;ion of X :::;: 60 %. \\ :~;
:=
9:::5 kg
Scale:
!$£;Y.; ... ·······ca
PII-II (b) \ In! bal~l!KC ;lr:..i rare: b'.:I.
:.lS
in Part (a)
\vber~
11·30
E;:::: 0 ~nd To :::: 300 K
and C->,o::: lO~·-3 mollem' dF Q - W - "\.' F.tQ'P' C . (T . . . . T. )+ F X '-rT)~ L..J_ I.: -AH . .._ xx\; .._....... -_ ........_ ........ _s. . . . . _ ....__ . ._ . . . . . . .-w.... _._ . . . . .,\0 __ .............. _.-'} ::; 0
Energy bal;mcc: :
LN/:pi
dt
' ) Fio Cpi.(T -- Tio) :: [to x ::!.5 IT - 300)] + [t 0 x 75 (T. 7.50)] .........
'.
:: 1000 (T·· 30(})
.. tOOO IT-. 300} + to.X.( tOOOO) :: 0 POLY:;"'!ATH !!!.!!::!~...~~
~!~:
o
d(xl Id{",) ".·-::.al fao
1<:",o . 01 fao .. 10 k.::",4242
cao.. 0 .. 0 0 1 To=JOO
T=o(lO·x"lOOOO/lOOOl+-To
= 300 K
ca~caO'{l-xl*(To/T)
ra",-(k*kc"ca) I (k+-kc)
Wo
~ 0,
wf = 1,2e+06 value ?~~
variabb!!.
rnj.t:.l,.:!!;!;.yal!±~
Max~ .. yal.u~ MinimUll\
w
0
1.2e~()6
0
1.2 .... Q6
x
a
0.654825
0
O.654S25
k
0.01.
0.01
0.01
o.n
fao
10
10
1.0
1Q
ko:;
4242
4242
424.2
eao
~l.OOl.
0.001
0.00l.
4242 0 001
To
300
]00
.lao
3an
T
)00
365493
300
ell.
O.OOl.
O.Oell
\}
000283.331
365.483 o .000283331
··2 . S33.:le-·06
-9.39998e-06
,2. 333.3e···06
·9 .. 9939Se··05
:::a.
J.900
~b:r.! ..
0 ... <0
····x 0."80
0.320
0.160
0.000
T 1 T
t I I
t t O.COO
..... ~~._ ...
-.~·.t···.· .. ·~~-¥·~·· ... ·..·" .. "·· ..
Cl.300
+--...,.---.. .,.....~.t-.'.
0.600
O.SOO W
11-31
,~ ....... ~... ~ ........ ,. .. t····
1.200
..... ~'"'!-"--- ····'·1 1.500-
For a convc:rsion of X
= 60 % , W,at :::: 10.:0 kg
376.000T
~
3bc.aoo ..~.
!5s.:r:~
-,. r
T
J~1.coa.-.L.
I.
i ,
349.000-ii
3\~..
cco·t
T
r 296. 000 .,,-.-...... _-..••...•.-t.. -, .----~. . ... C.,CGO
~,JCO
+ .......,.. ~.-......+................... ----,. 1-........- •.--------j
C.. ~GO
O~3CO
l,2CC
L='JOO-
w
Scale! ~S;.t.!_.
.. ca
"'-''''
w
Pll-ll (C) It is possible to generalize that the addition of'temperature variation in adiabatic operation does not affect the conversion. cqncentration profIles in form, but the numerical values are sli,ghlly different. Because the reaction is exothermic, isothennal operation enhances the conversion profile along the packed bed, so that for a given bed diameter less catalyst is required. It is clear that adiabatic operation inhibits conversion as heat is not removed from the system so more catalyst is needed for the required conversion, The removal of the heat generated in the reaction allows a reduction of 85 kg of catalyst in the bed.. More detailed economics will indicate whether isothermal operation is worth it.
PI0-12 (a) Dissolution of pills Complete dissolution .
11-32
WdruJfttstUm.t.fCh
Col ;;:::: .... ----..".--.V£Ioma~h
des dt
-_.. ==
~.
"c., c". -....V,la-.......".-
W.o,,,;,
_--
...-...... gtl"'!r
Relate C" to rime: T11ree piUs.. each with different thickness OtHer layers. the inner cores dissolving at differenr times (but at the same rate) to each other, will each contri.bute to C" in the $romach.
Pill 1 D1 ::: 5 mm. Dl == 3 mm 1
Time for outer layer to dissolve tl
O.0354x(0.5 ··"OJ!) . =----....... -.----... - = 1.18 nun 8x(6:clOe- 4)xLO
Pill .2 :
.,.. ".. Ilme for outer layer to dlssolve
Pill 3 :
0.03:54x(OA"··0.3!) 8x(6:dOe·. ··4)xLO
Cl::: ._--. _ _......._._".......... _ ........ _.:::
0.. )2 - rrun .
D1 ::: 3..5 mm, Dl ::: 3 mIll
Time for outer layer to dissolve
tl:::
g:??~.~::j~.:?..:~~.:~:g~~!.::: 024 min 8.t(6xlOe ..... 4)xl.O
mol balance on drug:
and dC, k.:.S;".TC.D! 2.D~8SIfl.it [)1 . . ...."".. :;;;:: ....... "."" .. _-,--_ ....... = ... . .,. . ....... . - ........... ....._.. ~-
dt
and boundary conditions:
~
V
~,,"-
~
~
DV where W ::: mass of drug in stOmach, g V ::: volume stomach fluid, cm3 t ::: tl • C", :::: 0 t::: 12, C", == WJV = 0.5/1.200::: 4J7xl0e4 g/cmJ
c.",,=
11-33
---'-----~
((0)
D2 =4mm DI:;:;;3mm
~--
".-... -.-=,~"..,...
Outer layer:
~ (~_1C_D3p{) Mol balance on outer laver: 0 :,. 0 +- rA" ,.1C, D- ;;::----'------.--, ..dt and fA'· rate of dissolution of outer layer;;:: rate of mass transfer from pill surface W A
2.,DA8
Sh ::::: 2 , k. =""" D
. Iayer;::: mner " 1ayer ::;::-----.-_._ 500xlOe' 6 = () . 0".,)4Kg - I I em 3 1 Assume uensltv 0 f outer ... '.--.
~,: ()..3X_
"
6
Sour::;:: 1.0 kg I em)
. dD . . ". . . = .--.2.(-'r,"") .. . -,, : ; : : : -- "-2x2xDAiJ"Sout _. . . -". . . . . ,.~
-~~~~"-.~
-"~,,
p
l),p
dt Boundary conditions :
t== 0, D == I~ t= tl , D12 -
2
D::::: D!
._ ..-= 2x2DAB.S"", --"------,,t,
DII
P
where tl ;::: time for outer layer to dissolve
BC)undnc'y comli::ions :
(;;.;:0,
t
= (:, D,:
D=D, D;;;;; 0 D '.01, s,,"
. . . . -.", == --..-.. -,--.-..-.. -". . . 2, p
-~
-r!
11-34
".~-.-----"'
t:
= .. ~:g~.~:~9:.~~ __._ == L66 min 8x(6xlOe ... 4)xO.4
time for complete dissO:lmion , tr == tl + tl == 0 ..52 + 1.66 == 2 . 18 min
PIO-12 (b) blood stream
stomach
......--adsoI"ption
e s conc. in blood
,tA conc . in stomach
:@ @@ dissolutjon
stomach wall
Let rare of adsorption into bloo
Relate concentration in stomach C A • to concentlation in blood Ca : Mol balance on drug in bloodstream:
.. deB --·rA ==
dt
Scale; ~g;t!.
·-Ca
11·3.5
2.,000
--'-"~
1.6CO
.;. ..
I-~-
1.i
~--~---~---
- --
...
1
Seal e; 1.:100
~...l.-
..!$s.'t-!
/
~
·. ··Cb 0.900
J. j
o.. ~oo
T
i 0.::00
..
/
f
.1j. . --..
0,,;::00
-----t-.---.----.. -.. ~C .. :iCO
----i-----------.-~. ----.~----+-.-...----.---""~ 20.DCC
30.CCC
-+c~c~a
50,,00C
PIO-12 (C) The graph of C s against time shows how the drug concentration in the bloodstream initially varies \\oith rime and then becomes independent of t.ime as aU the drug initially in the stomach has absorbed into the blood (consumption of the drug within the bloodstream has not been modelled, unrealistic but no data). If a certain drug revel is specified, and assuming a constant size of inner core (drug) and that the pills were to be taken similtaneously, then the way to achieve this would be to use pills of differem outer layer thicknesses - to maintain an even sromach concentration and hence absorption rale over the whole period. Relate D with t.ime :.
de -= .._._~.I:::_~i~ = . . 4.
de and boundary conditions:
P
DAB
Sin
D.p
t::::: t[ ,D == Dl t::::: tz, D:= 0
Using logic to obtain the corr~ct tinting for the drug concentration profiles inside the stomach ror each pill, the toral profile is used in the relation with the concentration profile in the bloodstream.
11-36
Pll-12C , §'.gUations: d{tll.) Id( eJ =if (e>O .24) then (i.e (01.>0. 00001.) ehen( --4 "Dab-Sinl (
'Ini tial_.y!:!~ 0.:3
Ol.*rho»else{O)lels~(OJ
d (CO) Id( c) "'ka"'ea ·V/'Hbody
0
d(Cal.l/d(el=if{t>O.24)ehen(2 w Oab*Sio*3.14*DL/V}else(0}
0
d{Ca2)/d{t)=;if(t>0.52)ehen{2*Pab*SiIl"'3.l4*D2./V)else(O}
0
d(02)/d( tl =if (e>0.52) t:nen{i!: (;:)2>0. 00001) t:hen{ '4*Dab"S;in1 (
0.:3
n2·rno»else{O})else{0) d(Ca3) Id(c} ""i.f (e>1.1.8) th.en,(2"Qaj;,"Sin"3.14"DJ/V) else(O)
0
den)} Id (el ,die (t;:>l .18) then (it' (0.3>0. 00001) then ( .... 4"'Dab"'Sinl (
0.]
D3"rhollelse(O»)else(O) t).ab;O.0006 Sin.. 40Q ::'!:lo==)5 .4
ka·;O .156657
V=1200
Wbody=7S000 Ca,=Ca.1.+Ca2 ... Ca3 .... (CO·'Nbody/V)
Co '"
0,
t
e
..
co
4.5 ~!:.~1~ ~i.'Uum ...Y:!!1~ 45 a 0 ...3 0.3 2 .. 001640-05 0
Cal.
0
0.0004.3.7376
0
9. 9965e-06 2:. OO1.64e··OS 0.000417370
0
0.0004.1'1373
0
0.0,00411:373
0.3
0.3
'3.99T74e-Q6
ca:a
0
0.000411364
0
03
0,]
0.3
9 _9.9S41.e- 06
9. 99774e-01i 0.0004.1,7364 9. 9984.le· . 06
Dab
0.0006
0_0006
0 .. 0005
Sin ;;ho
400
0.0006 400 35.4
y~le t::
D1
Ca2
D2
Mini.'11um v
0
4.00
400
.35.4-
35 4.
lea
35.40.166667
V
1.200
1200
1200
1200
Wbody
75000
75000
75000
75000
Cae
0
0.000991.:1.8
0
1.0S0Ue-01S
Relate D with time:.
0.1.66667
0 . 166667
0.166661
:!!?,,;, -!:'-~~~~"-:= -.-~~.!:?~~~§~ dt
and boundary conditions:
p
t=tl, t::: tz ,
D.p
D=D1 D::: 0
Using logic to obtain the correct timing for the drug concentration prof11es inside the stomach for each pill, the total profile is used in the relation with the concentration profile in the bloodstream.
PIO-12 (d)
11-37
To maintain constant drug level by maintaining a constant stomach concentration, time needs to be allowed for the dissolution of the outer layer. for a given period of say 3 hours, a size disuibution of outer layers is needed, with thin layers for initial response and thicker layers for delayed response. This disrribution would be back ,calculated given the necessary stomach cOllcenu'ation for the required bloodstream concentration accounting for bloodstream drug consumption.. Optimization of the stomach concentration will indicate the times at which complete dissolution of the outer layer of the ends on the number of pills which can be reasonably consumed in one sitting, the period for effect and the limits of practical pill size
PII-13
----.. . . . . . . . --.. . . . . . . . . . __. _-._._. _. . . -._. . . ._. . . . . . . . _·_. _·__·__· __· . ·.· .1
The plot of the data is shown below
Particle Diameter vs Time 10-.. 9
. ;: .. '-
4> 4>
8 7
E
6
i5
5
"0
4 3
tiS
a.
2 1
0
2D
30
50
'70
60
80
00
100
110
Tlme
Initially the rate of incine:ration of the droplet in terms of diameter, is non ..linear, but apparently becomes linear after . ~ 50 time units. 'TIle linear form of the data indicates that the diameter is directly proportional to time and the rate of decrease in diameter is constant and hence not a function of diameter. TIus relationship should make it easier to estimate the reauiredincineration time for complete destruction (zero diametei). Assuming that rate of diameter decrease continues at the linear rate 1ll1til complete destruction (at time !ct and hence complete decomposition of the POlle's, the equation for the linear relation indicates 41 .~ 160 u111ts .
._--_._-----PII-14
11-38
Mol balance on layer of earth control volume:
but FA == A:..W",
dW~ ,--, = . ._.dC~ . . , _. . . .
Z -70
dz
dt
For dilute solution and constant total concentration: Gives;
Let
gives
Boundary conditions: t ::: t (present day)
0 (surface), 'V == 0
Z. :::
z::: 0 0 . t
=0 (end
'V= I
of glacial) • z > 0 ,
Gives the error function:
let
It is defined that at Tl = 1.82 , C A = 0.01
CAO
and for 1'\ > 1.82 • C", is negligibly small.
This defmes the penetration thickness, 0 (as a function of time) : Concentration Profile 1.2
..
~.-.--.'" ~
o 0.8
- _._"._,,--
«I
~ 0.0 0 0 .4
o o
5
10
15
Depth (m)
11-39
1.82
= _-;:==0=
../4.DAlJ.t
But the graph gives at, C\ = 001 C\O --» penetration thickness after time L
z::= 0,4 hence (5::::: 18 - 0,4 m , the
The time taken for 0 to reach this thickness, is the time since the difftlsion started (i.e at the end of the last glacial)
CDPII-A CDPII-B CDPII-C CDPII-D CDPII-E CDPII-F CDPII-G CDPII-H CDPII-1 CDPII-J CDPII-K CDPII-L CDPII-M
11-40
Solutions for Chapter 12 - Diffusion and Reaction in Porous Catalysts P12-1 Individualized solution P12-2 (a) (i) t=5 1
D
ex
-D
e
l-~
175
( J _T2
P Tl
(lines and angles not to scale)
P12-2 (b) (1) First Order Reaction Kinetics 2
A. = 0
W= 1
D d CA e dz 2
_
kC =0 A
'
1jf
=C A C AO'
A=~ b
1jf = A cosh..JDaA + B sinh"DaA
-dW = 0 _._ _ _-----1 A. =1 dA. symmetry:' = 0 B=--A 1jf=1
sinhrnacosh..JDa A=O
@
.~ = A..Jij-;; sinh ..JDaA + B"Da cosh A
A = 1 ..
~
=-A Tanh-vDa
@
12-1
1 = A coshvlDa A=
1 B = Tanh-v'llil. cosh viDa ' cosh viDa
cosh ..JDa A Tanh..JDa . h ~D '\ sm -yua fI. cosh ..JDa cosh ..JDa (2) Monod Kinetics \jf =
2
D d C A _ !.!max C ACc = 0 Use e dz 2 Ks + C A Quasi Steady State dCc =!.!max C ACe = 0 Analysis dz 2 Ks +C A No further solution to Monod Kinetics will be given. (3) Variable Diffusion Coefficient
clFw = Dc Wo2 1_oAc dt zWA =--De dC A.=_ DeC AO d~1 dz L dA }.=O
?Fw = D~~AO d~1
dt L Mole balance
d~:
Dc Ac/ V
dA }.=o
]
{De -------k=O dz for hindered diffusion
=
D e
2
DAB
1+ a F;/(1-- Fw)
As a first approximation, assume no variation in De with A 2
d \jf
2
kL
- -2 - - - = 0 dA DeC AO as before 2 kL <1>0=--2D eC AO
~f\jf --2<1>0=0 dA2 Solution the same as before Equation (E12-2.13)
12-2
w
°2
=W =( A
DeCAoY_~=-J=kL L A. DeC AO
The flux of 02 in does not depend upon De which is not uprising since this reaction is zero order. For the build up of material that hinders diffusion dFw_=u A kL=A Lk=kV dt C C C
Fw =kVt From a quasi steady state approximation as time goes on increases.
Fw
increases De decreases and <1>0
However, the point at which the oxygen concentration is equal to zero has to be found. We can parallel the analysis used in P12-10 switching the coordinates of A = 0 and A = 1 (see solution to P12-1O(c) in which the solution manual) we will find 1
A =--<1>0
C
~=O
5
Increasing
t
o
-------~-------~~------~~
A =1
We see that as t increases Ac decreases, that is the point at which the oxygen concentration is zero moves toward the top of the gel.
P12-2 (c)
12-3
(1) For Rl 111 =0.182 18.2% Surface reaction limited and 81.8% Diffusion limited For R2 11 = 0.856 85.6% Surface reaction limited and 14.6% diffusion limited R2 ,j-,2 (12-59) (2) Cwp = -ri\(obs)pc = 11'f'1 DeC As = (0.95)(0.9)2 = 0.77 which is less than 1 so there are no significant diffusion limitations.
1 N k SaPb
=
1 -+---=--=11 kcac Q = 0.059 So 5.9% surface reaction resistance and 94/1 % external and internal diffusion limited %R =
0.941 N
.!.. + ~aPb 11
-
0.941
0.941
6.0+ 10.96
16.96
= ------.--
kcac
% Internal diffusion reaction 0.941 x 100 x 6.0 -- = 33.3% 16.96 % External diffusion resistance = (0.941)(100) x 10.96 = 60.8% 16.96 Summary of Resistances 60.8% External Diffusion Internal Diffusion 33.3% 5.9% Surface Reactor 100.0% Increase temperature significantly. Surface reaction % resistance decreases. Increase gas velocity external resistance decreases decrease pellet size both internal and external resistances decrease. For 99.99% Conversion In 1 1 1 L2 = L 1 l n ---ln-- - / l n -- = 0.16 (1O,000) 1- X 2 1- X 2 1- Xl In.500
=0.16x~~=0.24m 6.21
P12-2 (e) 12-4
(j)
From Mears's Criterion
-LlliRx (-rA)PbRE < 0.15 hT2R g
1)
The value from the question Llli Rx = -25kcal/mol = -104.6k1 Imol h = 100BTU Ih· ft2.o F = 0.567kJ Ism2 . K E = 20kcal/mol = 83.682k1 Imol Rg = 8.3144 *1O-3 k1 ImolK From example 12.3 -rA =kSaC NO k = 4.42 * 1O-lO m 3 1m2. sec 2 Sa = 530m Ig Pb = (1- )P c = (1- 0.5)(2.8 * 10 6 ) = 1.4 * 10 6 g/m 3 R = 3*1O-3 m T=1173K At the inlet of the reactor the fraction of NO =0.02 From ideal gas law n P V RT = ~01325 * 1O~ = 10.39mol/m3 8.3144 * 1173 C NO = 0.02 * 10.39 = 0.2078mol/m 3 Substituting all value in the first equation (104.6)(( 4.42 *!.9-.~O)(.530)(0.2078»(1.46 * 10 )(3 * I_Q-3)(83.~_82) = 2.88 * 10-4 (0.567)(1173 2 )(8.3144 * 10-3 ) As the calculated result is lower than 0.15, there is not the temperature gradient. The bulk fluid temperature will be virtually the same as the temperature at the external surface of the pellet. 6
P12-2 (0 (k)
Foq = 30 use Figure 12-7.
If you draw a vertical line up from <1>1 =0.4 it should be tangent (or very, very close) to the J3 =0.4 curve. Any slight increase in temperature will cause the reaction to go to upper steady state.
P12-2 (g) (1)
For large
12-5
S'
=
a
Sa 1+ kDt
. Area 2 (1) Pore closure. ConsIder De As t -7 00 pore throat closes ~ = -7 00 , Area 1
°
-r~ --7 (2) Loss of surface area Sa. As t -700 then S~ <1>1 -7 00
-7
°
then
-7
°
Ci c -70,
11 -71, but -r~
De -70, and
=.JS:: -7 °
P12-2 (h) (m) The activation energy will be larger than that for diffusion control and hence the reaction is more temperature sensitive, If the apparent reaction order is greater than one half, then the rate of reaction will be less sensitive to concentration. If it is less than one half, the true order will be negative and the rate will increase significantly at low concentration .
P12-2 (i) In example CDRI2-1, the reactor is 5 m in diameter and 22 m high, whereas the reactor in CDR12-2 is only 2 m3 in volume. The charge is much different In CDR12-1 the charge is 100 kg/m3 and in CDRI2-2 it is only 3.9 kg/m3
P12-2 (j) No solution will be given at this time P12-2 (k) With the increase in temperature, the rate of reaction will increase. This will cause the slope of C/R j vs, 11m and, therefore, the resistance to decrease"
P12-2 (I) No solution will be given at this time"
P12-3 (a) Yes P12-3 (b) All temperatures, F H ) = 10 moJJhr" The rate of reaction changes with flow rate and increases linearly with temperature
P12-3 (C) Yes P12-3 (d) T < 367 K, F H ) = 1000 moJJhr, 5000 moJJhr. T < 362 K, Fro = 100 mol/hI.
P12-3 (e) Yes P12-3 (0 T > 367 K, Fro = 1000 moJJhr, 5000 moJJhr" 12-6
T > 362 K, FTO = 100 moVhr
P12-3 (g) actual rate of reaction -rA (at362K, Fro = 10 moll hr) - --=..:...;'-------=--=-------'--,-- ideal rate of reaction - -rA (at 362 K, Fro = 5000 mol / hr)
Q -
Q
= 0.26 = 0.37 0.70
P12-3 (h) At FTO = 5000 moVhr, there is non external diffusion limitation, so the external effectiveness factor is I.
1]=
actual rate of reaction ( at 362 K, FTO = 5000 mol / hr) . extrapolated rate of reaction ( at 362 K, Flo = 5000 mol / hr )
1]=~~=0.86 1.4
P12-3 (i) 1] =
3[¢cosh¢-1] --7= 0.86
by iterative solution
¢ =1.60
CA 1 sinh ( ¢A ) rp=--= _.CAS A sinh¢
P12-4 (a) External mass transfer limited at 400 K and dp = 0.8 cm. Alos at all Fro < 2000 moVs
P12-4 (b) Reaction rate limited at T
= 300 K and dp =0.3 cm. When T =400 K: dp = 0.8, 0..1, and 0.03 cm.
P12-4 (C) Internal diffusion limited at T = 400 K and 0.1 < dp < 0.8
P12-4 (d) 1] = -rate with d p = 0.8 = 10 = 0.625 rate with d p = 0.03 16
--_._--P12-5 Cmve A is reaction-rate limited . This is so because of the way the curve bends, implying an exponential function which is the equation form for the specific reaction rate with respect to temperature. Cmve B is inner-diffusion limited . This is because it has a dependence on temperatUIe, but that dependence is small.
12-7
Cmve C is outer-diffusion limited., This cmve has a much larger dependace on temperatme than cmve B,
P12-6 (a)
_ (1 \ sin h
Effectiveness factor:
n
cP
conditions:
., :.;:; "
(A. :::: 1) = 1 = (A. :::: 0) :: finite
cP ;:;:
Boundary
3 (A h' 1) '4'1 cos IPl - , flrst order reaction.
q:.t
At •'" -- R 1 > C-A 2' ~~:::: 7)'
::::
0.1
CAS
where CAS = 1 x 10 -3 rnpl~, I
7 X 10-4
A. = '-.-.---- = 0.7
R = 1 X 10,3 cm1 Dc= 0.1 cm2/sec
1 x 10 3
(f-}(Si~,.~.~l 1) sm h 01
A.
r
At A = ! 1. ' .
-
L sinh($l)
J
::::::::>
. ;: :; : 60 q>. 1
f
q> :.;:; __ 1 sin h (6 x 0.7)1
0.7 L
C.~ 1 x 10 -3
sin h 6
J
= -L r~4.2 - e 4:21 0.7 l eO ~ e ~ J
P12-6 (b)
" ::.: 1[4>1 cos h ~1 - 1]
- 0.80
~
2.04
<1>1
<1>12
= ~~~ SA PB De
At A.
=~ .
P12-6 (c)
-
cp :.;:; 0.1 :
Individualized solution
12-8
"
P12-7 (a) Start with a mole balance: WAI. _".WAL,..~ +. TAALlz = 0 Divide by A& and take the limit as z --> 0., dW dz
·__ ··· ..···,·,··,T
=0
A
From the flux equation:
w = -_. D .~~:i. e
d:..
Combining the two equations we get:
d[-De dCA/dz] + k =0 --_.,,-,_._dz Dividing by -De we get:
d 2 C +---=0 k dz De
_._A 1
We need boundary conditions
de
__ ._.:1"
B.C. (1):
dz
= 0 @ z =L
B"C. (2):
We can then solve for the concentration profile: dC" =~+Cl dz De Using boundary condition (1): "kL O=-+C1 De kL C1 =--
De
dC" dz
kz
kL De
-=---
De
12-9
Integrating again: dCA
=(kzDe _kLDer 1-1-.
k 2 kL C =-Z --Z+ C2 A 2De De
From boundary condition 2
CAO = C2
k 2 --Z+ kL C C =-Z AO A 2D, De P12-7 (b) 11
=
rate of reaction with diffusion
rate of reaction without diffusion
Tl
=
2D{CA - CAo) z{z - 2L) _ 2D (CA - CAO) -~---k k z(z- 2L)
P12-7 (c) Boundary conditions:
o=
k5 (~ -L)+CAO
CA = 0 atz
=L
CAO = k L2 2D
L =
L 1(2
~2DF''- = Vr~:~::~~~6::~?-::):::) :;:::
0.0640
L :;::: 0.0041 em :;::: 41
~m
P12-7 (d) The answer in pan (c) is equal to the average tail length. 11 = 1 in this problem. Ifl1 = 1, then it con1I'3dicts the assnmption of diffusion
being rate-limiting. P12-8
12-10
=
z 0 CA = CAS
I
Z
z=L
CA
= 2L
= CAS
CA = _1 C ... 10
""->
First-order irreversible reaction:
A-+B
.r~ = k CA
WA = _D~CA dz
~fole
L
balance:
.1ld
= lO-3c~
dl
2L
(d~A) + n: d r~ :: 0 ,dz
= 2 x 10-3 em
CAS = 10-3 g moleJl
D
= 0.1 crrjl/sec
tl')
At z = ~ CA = ..L CAS = CAS ( COS h['(l - ~ 10 cosh,
...L 10
=
1
cosh,
cp =2.9932 12-11
P12-8 (a) Atz = 1/2 L:
= CAS (COS h[q,{l
9A .
- ~)])
coshcp
C
= (0.001) (COS h [209932(0 05)]) cos h (2.9932)
A
C..!\. = 2.345 X 10 4 g mol/l P12-8 (b) ct>
= 2L VDkd ==
l10ld
=
tan h
~old
TInew = 0.8 =
Q a (2L) tan h (2.9932)
- . ,......99 ...~.,..
tan h Onew
==
.
Qnew
-
0.3324
= 0.8880
Onew
P12-8 (c) At z = L. CA = _C __AS-=-_ cos h 4>new
= cos gltfJ880)
= 7.038 x 10-4 g mo1ll
Thus minimum CA is now 70.38% of CAS- Therefore the suggestion is plating entire surface of the inside of the pore.
P12-8 (d) Individualized solution P12-9 (a)
12-12
Define W A > 0 in direction of increasing z. Material balance: WAAp~ - WAApIZ.4Z+-r~aApAz = 0 ..' =: kCA
-rA
;
WA
::;
dC dC -De dzA +CA V == -De dzA forsmallCA
Taking t:J.--; 0: -Ap d:A + r~ a Ap = 0
Apt-(Ded~~)- k CA aAp 2
For constant De; De ,d C A
-
dz2
Boundary conditions: z
z
=L
= 0 d
2
d CA _ 0:2 CA = 0
dz2
0
k CA a ::; 0
CA = CAS
= 0 by symmetry
where (1.2 ;;; k a
De
Assume CA = e rz then r2 y
CA
=
- (X2 ::;
0
r
= Al e -a.z + A2 eaz
Atz = L: CAS ::; Al e-aL +A2&L deAl dz .,.0 ::; 0
::;.C!.
Al + a A2 ::; 0
...
12-13
= ±a.
Therefore, concentration profile can be written as:
c = C Jeas h(az)l AS \ COS h (aL)
A
WA A;- = -De (d~A
~eL) Ap =
- Ap De CAS ex ( : :
l~) Iz.&.}
WA Ap =:= -ApDe a CAS tan h (aLl By the sign convention:
= ApDe a4-.s tanh(cxL) Tl = (~)tanh(aL) ='VfIT (De}~tanh{' Ifi'L) leaL 0; lea L 'V 0; :. 11 kCASaApL
1\
= (-v'~ t}tanh(aL)(.Jft L)
Overall effectiveness factor:
or:
•W A Ap =
n k CAO a ApL = 11 k CAS a Ap L
.n. = CAS 11
CAO
-WAAp = kcAp(CAo-CA,S) = aApDeCAs tanh(aL)
CAO = CAS [I +
a:::.
tan h (aLl]
P12-9 (b) A-7B
WAltrLI r - W AllrLI r+t.r + rAllr/).rL = 0 1d --rWAr+rA =0 r dr
12-14
EMCD therefore,
dC
WA = -De __ A dr
d
rdCA
.!. D dr = D r
dr
e
2
(d C2A -.!. dCA e dr r dr
dlf/
At A. = I, 'l' = 1 and at A. = I, -
dA
J
=0
Bessel Function Solution
(n) P12-10 (a) (0)
EMCD
dC A wA =-D--
dz
, fA =-k
In - Out + Gen = 0 WAAcl z - WAAcl z +Llz + rA~zAc = 0 dW
A ---+rA=O dz '
1\,= 0
12-15
'-_ _ _ _ el1{J = 0 ell\, J
Integrating equation (1) kL2).? \jf = + C2 2D A C As at 11,=1 \jf=1
kL2 C2 = 1 - - - 2D A C As
th. 2
_
'1'0 -
2
kL 2DaC As
P12-10 (b) (p)
Now let's find what value of A that 'If = 0 for different
<1>0
For <1>5 = 1 : 0 = 1 + 1[11,2 -.1]= 1 + 11,2 ---1
11,2=0 Therefore the concentration is zero (i.e., \jf = 0) at
111,=01 For <1>5=16 : 0=1+ [11,2-- 1]=1+1611,2- 16
11,2 = 15 = 0.938 16 Seems okay, but let's look further and calculate the concentration ratio \jf at 11,= 0 for <1>0 = 4. \jf = 1 + 16 ~0.2)2
-1]= 1 + 16[0.04 -1] = --14.9
Negative concentration. (q)
P12-10 (c)
(1)
Let's try again with
\jf = 1 + (10/
I
<1>0 =
10
~.12 -1]= 1-10 2(0.99)
\jf = 1- 99 = -99
not possible \jf will be negative for any v-a-Iu-e-o-f-·-o-gr-e-at-er-t-h--an-o-ne-.-.===_~~-_.-_~-J
P12-10 (d)
12-16
We now need to resolve the problem with the fact that there is a critical value of A, Ac, for which both 'If = 0
(s) and
d",
-=0 dA,
2
d ",
2
-dA,2 -2<1>0 =0 d", _ 2<1>5A+ C1 dA,2
'" = <1>5A2 + CIA + C2
At A = AC ' '" =0
0= <1>5At +C1AC +C 2 Subtracting
1 = <1>5 -<1>5At +C 1(I- Ac) Solving for CI _
1-<1>5~-At)
C1 --
l- Ac
Solving for C2 2
~-<1>5~-At»)
C2 =1-<1>0-------l- Ac
12-17
1.0
<1>0=2
<1>0=1
o
0.5
LO
Sketch of concentration profile for different values of <1>0
<1>0 = 1 then Ac =0 <1>0 = 2 then Ac = 0.5 That is for <1>0 =2, the concentration of A is zero half way (A =0 ..5) through the slab. 1 1- Ac = <1>0
\11= ~A2 + [<1>0'-
~(l + Ac)]+ 1- ~ - [<1>0 -~(1 + Ac)]
212 ~ 2 2 n 2 2 • = <1>0/\ + lo - <1>0 - <1>0 + <1>0 r + 1- <1>0 - 2<1>0 + 2<1>0 \II = ~A2 + 2<1>0(1- <1>0)11, + 1- 2<1>0 + ~
ForA >Ac
P12-10 (e)
P12-10 (f)
12-18
In
1.0
1 - -......
T)
0.001
'-_.....ii....-_ _ _ _;;a".,_ _ _
In
1.0
P12-10 (g) No solution will be given P12-10 (h) No solution will be given P12-10 (i) Individualized solution
P12-11 Given: second-order decomposition reaction: A -P B + 2C
= SO m6/g sec mol;
k
P = 500 kPa
dp :: O.4cm; U :: 3 mls; T
= 4.936 atm;
X
= 0.80;
= 2S()OC = 523 K;
Dc :: 2.66 x 10..a frills; Eb = 0.4;
Pb = 2 X 106 g/m3 ; Sa = 400 m2/g. CAo = L = 4,936 atm = 0.115 g mol RT (0.082 g S23 OK I
!!'flK)
Rate law:
-r;'"
k
cl
Mole balance:
D
1
d CA _ U dCA + r' p :: 0
~ ctz2
dz
1
A B
DAB d CA - U dCA - Q k SA PB dzl
dz
0
ci = 0
Neglecting axial diffusion with respect to forced axial convection. we have:
12-19
d;A = _(Qk~APBlcl -1e.. dC~ = Ink~APBll~dz CoM
CA
0
. l .. --L CA
= IOkSA PB)Z
CAO
At z = 1.;
u
\
--L = ( 1 _ I) = (-2PB k SA) L CAO
L =
In
I-X
j(
U 1 - 1) Ps k SA CAO 1 .. X " =
Internal effective factor:
~
(-Lyn n+l
~
wheren =2
(50 gSm~Ol}(4001-}{2X
cp.z = 0.2 x 10-2 m
t.z
U
106~){0.1l5 ~)(~)
2.66 x 1O..g I1J2
= 2.63 X 107 very large
11 =
(-L}112 2+ 1
3
2.63 x
107
= 9.313 x 10..a
Intcmal-diffusion limited:
n
= 11
= 9.313 x
10-8
Reactor length:
L = ( 9.313 x 10 -8) 2 X
L
3 m/s (1 _10 .8 .. 1) 106 L(SO roO ) m3 gsmol
(400 m!) (lIS g mol) g tal
= 2.80 x lO-Sm
-------------------------------P12-12 (a)
12-20
Start with the mole balance taken on a shell
Divide by -21tlM and take the limit as llr approaches zero to get:
d(lf.t,.r) dr
'i.Pc.r = 0
Next fmd the equation for equimolar counterdiffusion and plug it into· the above equation:
tt({-D_dC~)) dr
-r;PtT=(}
Next differentiate to get the following differential .equation:
We can then set the following:
.
C
rp:; .---:L
CAS
r
It=-
R
Solution: ({J
= Clo(il)
Boundary Conditions: rp=l@l=l drp =O@l=O dA.
12-21
~~ =ct>[Cl (ct>1)-C K(ct>J)] 1
2
1
11(0)= 0 Kl{O) = 00 as
dqll ~O =>Cz =0
dl ~=o
9> = C/O(Cl>A) t= C.I.{ C, =
()
10 ct>
10(<1>11)
ffJ
= l~ (<.Il)
P12-13 (a) (t)
r;
!.(r2kt ~:J+
=
(flHRx)(-rA) 0
(1)
(2)
Dividing by flHRx and using Equation (2) to substitute for -IA
dT)+~~(r2De dCA)=o (_1--J~(~r2kt --flH Rx r2 dr dr r2 dr dr
12-22
CA +
ktT De(-~HRx)
=C A
kTs DemRx
+-~-
s
T = De (-mRx) (CAs -C A)+ TS kT T=Tmax at CAs =0
P12-13 (b) No solution will be given
P12-15 (a)
CAS
Given:
A H B on the walls of a cylindrical catalyst pore.
zl = length of poisoned section -r~ = kCA
In the poisoned section:
WAxriz - WA~lz.~ +r~Pdz
For 0 S z S WA WA
Z 1,
~ Iz
=0
r;" = 0, since this area is poisoned.
- WA
m2 IZ .62:
= 0 or
-<1:
= -CDAR ~A +XA(WA +WB) 12-23
A
= 0
But WA WA
At z
=.WB , since. for each mole of A consumecL one mole of A is reacted.
= .CDAB~ and iz(CDAB~.t,) = 0
=0
At Z =
= XAS C = CAS CA = XAI = CAl
CA
t
Zl
t
. Integraang:
XAI
Atz=O,. XA = atZ=Zl,XA
=! Kl
C~ :
Z
+ K2
K2 =
C~
C~l = XAI :: C = kl %1 + ~ C = kl
=
CAl-CAS CZl
CAl" CAS Z + CAS CZl
CA XA = C -
C
CA = (CAl - CAS) (it) + CAS
The flux is WA = -C DAB dX.... = .DAB dCA (CAl - CAS)
dz
dz
P12-15 (b)
Before poisoning, 11
= tan h q,1 q,1
where 11
= q,1
= L( r
~. )112 AB
After poisoning. the differential equation and boundaIy conditions are the same for the unpoisoning region of the pore, Zl ~ Z S L if Z is replaced by L - z; and if we let CA
= CAl at z' =0 and d;A =0 at z =L - Zl.
then. for the unpoisoned section of me
catalyst pore 11 appli~s if q, is ~p1aced by :
2k"
(L - z) (r DAB
)112 -_.(1 - LZ) tPl • l.e., . 11 -
tan h
[4>1 (1 i)l
CPt (1
t)
The effectiveness factor for the unpoisoned section of catalyst pore is defined as :
WA = ll1tr{L-z)CA l 1bis can be related to the overall effectiveness factor for the entire pore by
12-24
WA ::; 11' x r L CAS = ll1trL CAl
11' CAS = 11 CAl (1 - ~)
:.
But W A = - ~;S (CAl - CAS) from pan (a)
:.
CAl = CAS _ WA Zl DAB
:.
l1'CAS =
11( 1 -
r}[CAS - ~::1]
WAxr2 = Tt'2ltrLCASk"
But
t (1 ZI)(1 21'1' Ll CAS k" Z1 \ 11 = 11 - L - ~--:rD~AB~L--';;'1
-------_. ' - P12-16
.The reaction is A -+
t - t;
a= c
1 = -
E
- CAoe!· X) where A -
(1 - 0.5X)
t ~2 = YAO 0 = -0.5 CAO
= L RT
=
8.2 atro
0.082 I atm (227 + 273 K)
gmolOK
12-25
= 0.2 gmol 1
= .l-+1.FAO
.l
2
YA
FAO = 2 FA
2 FA
(...L _1.) YA
or FA
1 _ YA F AO = .l.. _1 - 2 - YA
2
YA
= 1... FA
X
FAO
r~ = FA~X
In {-r;J
where W
.
M
2 - YA
= 4x40g =
160g
= In ko;
At =
ex; and N
= Ao+A1N D
i
i
L Mi = nAO+Al L Ni I, ~ti Ni
(2) n
n
n
= Ao
L Ni + At L Ni
1 2 3 4
F-r-n 1 2 4 6
5 6
11 20
2
(3)
i
i
i
= In CA
(1)
n
i
= 0.16 kg
= In ko + ex 1n CA
Let M = In (-r;j; A(1 .o.
= 2 .. 2YA
= 1... YA 2 ... YA
'1A
X
C~
0.21 0.33 0.40 0.57 0.70 0.81
0.88 0.80 0.75 0.60 0.46 0.32
4.285 x 10 -2 6.666 x 10 -2 8 x 10 ·2 1.143 x 10 ·1 1.403 x 10 ·1 1.619 x 10 -1
-r~
M
5.5 1.705 10 2.303 18.75 2.931 22.5 3.114 31.625 3.454 40 3.689
-3.150 -2.708 -2.526. -2.169 -1.964 -1.821
N2 9.923 7.333 6.381 4.704 3.857 3.315
-5.371 -6.236 -7.404 -0.754 -6.784 -6.717
-14.338
35.513
·39.266
N
n
2:= Equations (2) and (3) become:
17.196 = 6Ao - 14.338 Al -39.266 = -14.338 Ao -+ 35.513 Al
12-26
17.196
MN
:. Ao
= 6.36
= In leo
-?
ko = 578.25
I
At = 1.46 == I.S ::
(l
At Tl = 237 C = 510 K: kl
-r' A
= FAGWX .
where X = 2 .. 2(0.097) = 0 9490 2 - 0.097 .
(9)(0.9490) 0.16
=
-rA
C
0
A
RT
53,38
. where:
°
8.2 - 196 g mol 0.082(SI0} - . 1
= 0019 g mol
1 - 0.5(0.9490)
A
kl =
= P AO =
= 53.38 = kl cl.5
= 0.196 (1 - 0.9490)
C
. :: ..:!:6.. cks
.
1
= 2.035 x 1()4
,0.019)1.5
kl
= lco exp{f{1\--t)}
2.035 x 10' = 578.25 exp (E(llmole)
8.314
...
E = 7.55 x lOS
(-L_-L)} 500 510
J 1 gmo
P12-16 (b)
= 2naw .. 1 = 2{ 1.5) .. 1 = 2: second order Euue = 2Eapp = 15.1 x loS~
ntrue
P12-16 (c)
~ = R ... / ~ Sa Ph C,o\O = ~V
(1
X
w·2f (2.035 X 1(4)49 (2.3 X
De
0.23
12-27
X
10'"
10') 0.196
= 1.40 x 1()6
$2
(--L)I12 (-l.:){--L}l12 (1.40 3x 1()6 ) -n+ 1 $n - 2 + 1
11 -
1.75 x 10-6
P12-16 (d) To make the catalyst more effective, we should use a smaller diameter.
P12-16 (e) CA = 0.01
R To T
=
• -rA
=
527 C
leo exp {E.{..L -..l)l
k =
k
gmol 1 ;T
I
= 800 K
lOS (-L --L)) \ 8.314 500 800
= 578.25 exp 17.55 x
2.19 X 1()32
= k CA105 = 2.19 x
1()32 (0.01)1.5
= 2.19 x
Is
P12-17 (u)
2 d y ~2
n
-_. nY = 0
Multiply by 2 Y ~. 2y dy _!(dY) = 2yn2dy_ dAdA dA n dA Manipulating the L.H.S. 2 -.!(dy )2 = 2 dy d y dA dA dA dA,2
d(d )2 = nY 2dy
y dA dA
n
)2 =2<1>
yn+1 --+C 1 nn+l C d'JI Y='JIA=~ , A=O y=O --·=0 thereforeCl =0 CAO dA Taking the deri vati ve of y and evaluating at A = 1 dy . ( dA •
~YI dA A=1
=
[2~~yn+1
~- n + 1
= )..,=1
1()29 g mole
ri~
V~ + 1 12-28
The effectiveness factor is
nR2(DA dC A )
11 =
dr
r==k
kCn 4 nR3 As
3
In dimensionless form
_{~L ~
11 -
A = \IfA , differentiating gives
dy = Ad\lf + \If
dA
dA
at A=1
= d~+1
dyl
dA 1..==1 dljl =
dA
dA
~24>~ -1 n+l
For larger n
~=3~t CDPI2-A -------------, ----,
CDPI2-B (a) 3rd ed. 12-19 (a) CDPI2-B (b) 3rd ed. 12-19 (b) CDPI2-B (c) 3rd ed. 12-19 (c) CDPI2-C (a) 3rd ed. 12-20 (a) CDPI2-C (b) 3rd ed. 12-20 (b) CDPI2-C (c) 3rd ed. 12-20 (c)
12-29
CDPI2-D (a) 3 rd ed. 12-21 (a) CDPI2-D (a) 3 rd ed. 12-21 (b) CDPI2-D (a) Individualized solution CDPI2-E 2nd ed. 11-18 CDPI2-F 2nd ed. 11-19 CDPI2-G 2 nd ed.II-20 CDPI2-H 2nd ed. 11-21 CDPI2-1 2nd ed. 11-22 CDP12-J (a) 2nd ed. 12-7 (a) CDPI2-J (b) 2 nd ed. 12-7 (b) CDPI2-J (c) 2nd ed. 12-7 (c) CDPI2-J (d) 2nd ed.12-7 (d) CDPI2-J (e) 2nd ed. 12-7 (e) CDPI2-K 2nd ed.12-9 CDPI2-L (a) 2nd ed. 12-8 (a) CDPI2-L (b) 2nd ed. 12-8 (b) CDPI2-L (c) 2nd ed. 12-8 (c) CDP12-L (d) 2 nd ed. 12-8 (d) CDPI2-M (a) 3 rd ed. CDPI2-L (a) CDPI2-M (b) 3 rd ed. CDPI2-L (b) CDPI2-M (c) CDPI2-M (d)
12-30
CDP12-N 3rd ed. CDP12-M CDP12-0 CDP12-P CDP12-Q CDP12-R (a) 3rd ed. CDP12-Q (a) CDP12-R (b) 3rd ed. CDP12-Q (b) CDP12-S CDP12-T CDP12-U
12··31
Solutions for Chapter 13 - Distributions of Residence Times for Chemical Reactors PI3-I No solution will be given, P13-2 (a) The area of a triangle (h=0,,044, b=5) can approximate the area of the tail :0.11
P13-2 (b)
(3)
(4)
CSTR
LG~[CSTRtCSTR
(6) (5)
13-1
----...
I J l
AI
1
--.I -I PFR
J
PFR
r-A~
PFR
el
PFR
AJ J
P-
or PFR
Recycle
(8)
~~~ (10) (9)
~~
LFR
(11)
(12)
P13-2 (c) For a PFRICSTR Series
13-2
.
IfkSTlU
Ifksrn
P13-2 (d) X=O.75 For a PFR first order reaction.:
Da
= In(_I_) = In(4) = 1.39 l--X
where Da
= kr
For a CSTRfirst order reaction.:
Da
=
(_1_.) -1 =3
where Da
I-X
= kr
For a LFR first order reaction: Solving iteratively Hilder approximate formula with an initial value Dao (i.e . DapFR
Daex p(
~a) + Da
0.75 = ---
.
4 + Da exp (
~c.!.. ) + Da
where Da
= kr
Da=258 The ratio of the Damkohler numbers is equal to the ratios of the sizes.
V PFR
Relative sizes: -----
V CSTR
LFR = 0.46, -V- = 0.86
V CSTR
;
P13-2 (e) For a PFR, r=5J5min, first order; liquid phase, irreversible reaction with k=O.lmin- 1
X
=1- e -k1· = 0.402 13-3
For a CSTR, r=.5.1.5min, first order; liquid phase, irreversible reaction with k=O.lmin-
X
l
.
=~~=0.402 l+kr
0.385
0.402
X CSTR 0.340
Page 851, only the RTD is necessary to calculate the conversion for a first-order reaction in any type of reactor. Not good when the RTD has a long tail that is difficult to interpret or interpolate.
P13-2 (0 Decrease of 10 't' in temperature See Polymath program P13-2-f.pol
POLYMA TH Results Calculated values of the DEQ variables Variable t Xbar k
Cao
initial value
minimal value
0.0025446 0.75
0.0025446 0.75
o o
o o
X
o
o
tau
1000 500 5.0E+l0
1000 500 6.25E-08
tl
E2 E
o
o
maximal value 2.0E+04 0.6023837 0.0025446 0.75 0.9744692 1000 500 5.0E+lO 0.0023564
ODE Report (RKF45) Differential equations as entered by the user [ 1] d(Xbar)/d(t) = X*E Explicit equations as entered by the user [1 j k = .00493*exp(13300/1.9872*(1/323.15-1/313 . 15)) [2] Gao = .75 [3 J X = k*Gao*t/(1 +k*Gao*t) [4J tau = 1000 [ 5 J t1 = tau/2 [ 6 J E2 = tauI\2/2/(tA3+.00001) [7 J E = if (1
13-·4
final value
2.0E+~
0.6023837 0.0025446 0.75 0.9744692 1000 500 6.25E-08 6.25E-08
0.70~--------------,
0.56
0.42
0.28
014
O.OOU--~-·-~----------'
o
-1000
8000 t
12000 16000 20000
The decrease of lOoe in temperature has the effect ofreducing the mean conversion by 14%.
Decrease in reaction orderfrom 2nd to pseudo ]'t See Polymath program P 13-2-[:-2 .pol
POLYMA TH Results Calculated values of the DEQ variables Variable t Xbar k
Cao
initial value
o o
0 . 004 0 . 75
minimal value
o o
0 . 004 0.75
X
o
o
tau t1 E2
1000 500 5 . 0E+10
1000 500 6.25E-08
E
o
o
maximal value 2.0E+04 0 . 9391084 0 . 004 0.75
final value 2.0E+04 0.9391084 0 . 004 0.75
1
1
1000 500 5.0E+10 0 . 0024915
1000 500 6.25E-08 6 . 25E-08
ODE Report (RKF45) Differential equations as entered by the user [1] d(Xbar)/d(t) = X*E Explicit equations as entered by the user [1] k = 0 . 004 [2] Cao = . 75 [3] X = 1-exp(·k*t) [4J tau = 1000 [5] t1 = tau/2 [ 6] E2 = tauA 2/21(tA 3+ . 00001) [7] E = if (t
13-5
1.0 , - - - - - - - - - - - - - - - - ,
0.8
0.6
0.4
0.2 0.0
U--_~
o
____
4000
8000
~
t
_ _ _ __ _ '
12000
16000
20000
The decrease in reaction order from 2 nd to pseudo 1st has the effect of increasing the exit conversion by 20%. The smaller the dependency of the rate on C A means that when C A is below 1 moVdm3 then the rate of consumption of A is larger and hence resulting in a larger conversion.
Exothermic reaction in adiabatic reactor: See Polymath program P 13-2-f-3 . pol POLYMATH Results Calculated values of the DEQ variables Variable t Xbar X T
Cao tau t1
E2
initial value
o o
minimal value
o
o o o
323.15 0.75 1000 500 5.0E+10
323.15 0.75 1000 500 6.299E-·08
E
o
o
k
0.01
0.01
maximal value 2.0E+04 0.999375
final value 2.0E+04 0.999375
1
1
823.15 0.75 1000 500 5 . 0E+10 0.0022142 19 . 337202
823 . 15 0.75 1000 500 6.299E-08 6.299E-08 19.337202
ODE Report (STIFF) Differential equations as entered by the user [ 1 J d(Xbar)/d(t) = X*E [2] d(X)/d(t) = k*(1-X) Explicit equations as entered by the user [ 1] T = 323.15+500*X [2J Cao = .75 [ 3] tau = 1000 [ 4 ] t1 = tau/2 [5] E2 = tauJ\2/2/(tA3+.00001) [6] E = if (t
13-6
1.0
r--====-----------,
08
0.6
04 0.2
0. 0
U-_~
o
_ _ _ _ _ _ _ _ _ _____'
.tOOO
8000
12000
t
1600020000
The mean conversion Xbar, the integral, is estimated to be 99.9%. The reaction is adiabatic and exothermic as the temperature increases to a maximum of 1373.15 K once the batch conversion within the globules has reached 100% which occurs after only - 4 seconds . Hence, the adiabatic increase in temperature considerably increases the rate at which conversions increases with time and hence also the final value.
P13-2 (g) For a PFR, '!=40min, second order, liquid phase, irreversible reaction with k=O.OI dm3 /mol·min- I
X
= kTC~=0.76 1+ KTC Ao
For a CSTR, '!=40min, second order, liquid phase, irreversible reaction with k=O.OI dm3 /mol·min- 1 .
X
-(--)2 = kTC Ao 1·-X
--)
X = 0.58
Maximum Mixedness Model and Segregation model are given in E13-7 Maximum Mixedness :Model See Polymath program P13-2-g . pol POLYMA TH Results Calculated values of the DEQ variables Variable
z x cao k
lam ca E1 E2 F1 F2 ra E F EF
initial value 0 0 8 0.01 200 8 0 . 1635984 2 . 25E-04 5 . 6333387 0.9970002 -0.64 2 . 25E-04 0.9970002 0.075005
minimal value -0-------0 8 0 . 01 0 3 . 2490705 0.0028734 2_25E-04 0 0.381769 --0 . 64 2 . 25E-04 0 0 . 0220689
maximal value 200 0 . 5938635 8 0_01 200 8 0 . 1635984 0 . 015011 5.6333387 0 . 9970002 -0.1055646 0.028004 0.9970002 0.075005
ODE Report (RKF45) Differential equations as entered by the user
13-7
final value 200 0.5632738 8 0.01 0 3.4938093 0.028004 0 . 015011 0 0 . 381769 -0 . 122067 0 . 028004 0 0.028004
[1]
d(x)/d(z) = -(ra/cao+E/(1-F)*x)
Explicit equations as entered by the user [1] cao = 8 [2] k = .01 [ 3] lam = 200-z [4] ca = cao*(1-x) [5] E1 = 4.44658e-1 0*lam"4-1.1802e-7*lam"3+ 1.35358e-5*lam"2-.000865652*lam+.028004 [6] E2 = -2.64e-9*lam"3+ 1,,3618e-6*lam"2-.00024069*lam+.015011 [7] F1 = 4..44658e-1 0/5*lam"5-1 ,,1802e-7/4*lam"4+ 1. 35358e-5/3*lam"3-.000865652/2*lam"2+,,028004 *Iam [8] F2 = -( -9.3076ge-8*lam"3+5,,02846e-5*lam"2-.00941 *lam+.618231-1) [ 9] ra = -k*ca"2 [ 1 0 ] E = if (Iam<=70) then (E1) else (E2) [11] F = if (Iam<=70) then (F1) else (F2) [12] EF = E/(1-F) 0.60
0.48 0. 36
0.24
0.12
0.00 0
40
80
z
120
160
200
56%
P13-2 (h) Liquid phase, first order, Maximum Mixedness model 1 Rate Law: ' - fA =k1C A where kl = CAok =O.08minC A = CAo(l-X) fA _
= -k1(1-X)
CAo
_dX =
dA
rA
+ E(A).X
CAO
I--F(A)
dX
E(A)
d)'- = --k(l- X)+ 1- F(A) X dX = k(1- X)--
dz
E(A) X
1·_· F(A)
See Polymath program P13-2-h-l.pol
13-8
POLYMATH Results Calculated values of the DEQ variables Variable z
x cao k
lam ca E1 E2 F1 F2 ra E F EF
initial value 0 0 8 0 . 08 200 8 0 . 1635984 2.25E-04 5 . 6333387 0.9970002 -0.64 2.25E-04 0 . 9970002 0.075005
minimal value 0 0 8 0 . 08 0 1 . 7365447 0 . 0028731 2.25E-04 0 0.381769 -0.64 2.25E-04 0 0 . 0220691
maximal value 200 0 . 7829342 8 0 . 08 200 8 0.1635984 0.015011 5.6333387 0.9970002 -0.1389236 0.028004 0.9970002 0.075005
final value 200 0 . 7463946 8 0 . 08 0 2 . 0288435 0.028004 0.015011 0 0.381769 -0.1623075 0.028004 0 0.028004
ODE Report (RKF45) Differential equations as entered by the user [1] d(x)/d(z) = -(ra/cao+EI(1-F)*x) Explicit equations as entered by the user [1 J cao = 8 [2J k = 0 . 08 [3J lam=200-z [4 J ca = cao*(1-x) [ 5 J E1 = 4..44658e-1 O*lam"4-1.1802e-7*lamI\3+ 1.35358e·5*lamI\2-.. 000865652*lam+ . 028004 [6] E2 = -2.64e-9*lamI\3+ 1.3618e-6*lamI\2-.00024069*lam+ . 015011 ['7] Fl = 4.44658e-1 O/5*lamI\5-1 . 1802e-7/4*lam"4+1.35358e-5/3*lamI\3-.. 000865652/2*lamI\2+ . 028004*lam [ SJ F2 = -(-9 . 3076ge-8*lamI\3+5 . 02846e-5*lamI\2-.00941 *lam+ . 618231-1) [ 9 J ra = -k*ca [ 10] E = if (lam<=70) then (E1) else (E2) [11] F = if (lam<=70) then (F1) else (F2) [12] EF = E/(1-F) 0.80 1
-.--:::::::::::======::::::===
0. 64
CJ
tU8
0.32 fJ 16
O.OOl-r_ _
o
~40 _ _~---_-~ ___ J 80 z 120 160 200
At z = 200, i.e. A = a (exit), conversion X = 75 %. The decrease in reaction order from 2nd to 151 has the effect of increasing the exit conversion by 19%. Once the concentration of A drops below 1 mol/dm3 then the rate of consumption of A does not fall as rapidly (as the 2nd order reaction) and hence resulting in a larger conversion.
Liquid phase, third order, Maximum Mixedness model Rate Law: -
fA
= kC A
3
13-9
C A = CAo (l- x)
~A =-k'C Ao 2 (1-X?
Where
k'C A / =k=O.08min- 1
Ao
dX =~+ E(A) X
dA
CAo
1-F(A)
2(1_ X)3 _
dX = k'C dz
Ao
E(A) X 1- F(A)
See Polymath program P L3-2-h-2,pol POLYMA TH Results Calculated values of the DEQ variables Variable z x cao k
lam ca El E2 F1 F2 E F EF
initial value 0 0 8 0.08 200 8 0.1635984 2.25E-04 5,,6333387 0,,9970002 2.25E-04 0.9970002 0,,075005
minimal value 0 0 8 0.08 0 4.1061501 0.0028733 2.25E-04 0 0.381769 2.25E··,04 0 0.0220689
maximal value 200 0.4867311 8 0.08 200 8 0.1635984 0.015011 5.6333387 0.9970002 0.028004 0.9970002 0.075005
final value 200 0.4614308 8 0.08 0 4.3085534 0.028004 0.015011 0 0.381769 0.028004 0 0.028004
ODE Report (RKF45) Differential equations as entered by the user [1] d(x)/d(z) = ·'(-k*(1-x)1\3+E/(1-F)*x) Explicit equations as entered by the user [lJ cao = 8 [2] k = 0.08 [ 3] lam = 200-z [4] ca = cao*(1-x) [ 5] E1 = 4..44658e-1 0*lam"4-1.1802e-7*lamI\3+ 1,,35358e-5*lamI\2-.000865652*lam+.028004 [ 6 J E2 = -2,,64e-9*lamI\3+ 1,,3618e-6*lamI\2-,,00024069*lam+,,015011 [ 7 ] F1 = 4..44658e-1 O/5*lamI\5-1.1802e-7/4 *lam"4+ 1,,35358e-5/3*lamI\3-.,000865652/2*lamI\2+.028004*Iam [8 J F2 = -(-9,,3076ge-8*lamI\3+5,,02846e-5*lamI\2-.00941 *lam+,,618231-1) [9 J E = if (lam<=70) then (E1) else (E2) [10] F = if (lam<=70) then (F1) else (F2) [11] EF = E/(1-F)
At z = 200, ie., A = 0 (exit), conversion X = 46..1 %" The increase in reaction order from 2nd to 3rd has the effect of decreasing the exit conversion by 10%. Once the concentration of A drops below 1 moVdm3 then the rate falls rapidly and CA is not consumed so quickly, resulting in a smaller conversion.
Liquid phase, half order, Maximum Mixedness model Rate Law:
-fA
= k 'C A 112
13-10
CA
= C Ao (1- x)
~ C
= -k'C
Ao
-112(1_ X)1/2
Where k
008 nnn . -1 = k 'C Ao -1/2 =.
Ao
=~+ E(/L) X d/L CAo 1- F(/L)
dX
dX
dz
=kC
-1I2(1_X)1I2 _ Ao
E(/L) ,X 1- F(/L)
See Polymath program P13·2··h,J.pol POLYMATH Results Calculated values of the DEQ variables Variable z x cao k
lam ca E1 E2 F1 F2 E F EF
initial value
°°
8 0.08 200 8 0.1635984 2,,25E-04 5.6333387 0.9970002 2,25E-04 0.9970002 0,075005
minimal value
°° °
8 0,,08
0,532148 0.002873 2.25E-04
° °
0,,381769 2.25E-04 0,0220677
maximal value 200 0,,9334778 8 0,,08 200 8 0,,1635984 0.015011 5.6333387 0,,9970002 0.028004 0.9970002 0,075005
final value 200--0.9038179 8 0,,08
°
0.7694568 0,,028004 0.015011
° °
0.381769 0.028004
0.028004
ODE Report (RKF45) Differential equations as entered by the user [1] d(x)/d(z) = -(-k*(1-x)A(.5)+EI(1-F)*x) Explicit equations as entered by the user [1] cao = 8 [2] k = 0.,08 [31 lam = 200-z [4J ca=cao*(1-x) [ 5] E1 = 4,44658e-1 0*lam"4-1, 1802e-7*lamA3+ 1.,35358e-5*lamA2-,,000865652*lam+028004 [ 6 J E2 = -2.64e-9*lam A3+ 1,,3618e·,6*lamA2-,,00024069*lam+.015011 [ 'I ] F1 = 4,44658e-1 0/5*lamA5-1, 1802e-7/4 *lam"4+ 1,35358e-5/3*lamA3-000865652/2*lamA2+,,028004*Iam [8] F2 = -(-9.,3076ge,'8*lamA3+5.,02846e,·5*lamA2··.00941 *lam+ . 618231-1) [9] E = if (lam<=70) then (E1) else (E2) [10] F = if (lam<=70) then (F1) else (F2) [11] EF=E/(1-F)
13-11
1.0
~---------------,
0.8 0,6
0.4
0.2 0.0 4)'---~4-0---80---1~2-0--1-6f-)---l200 z
At z = 200, i.e. A = 0 (exit), conversion X =90 %. The decrease in reaction order from 2nd to Y2 has the effect of increasing the exit conversion by 34%. The smaller the dependency of the rate on C A means that when C A falls below 1 moVdm3 then the rate of consumption of A does not fall as rapidly ( as the 2nd order reaction) and hence resulting in a larger conversion.
P13-2 (i) Assymetric RTD: See Polymath program P IJ·2-i-l"pol
rn
POLYMA Results Calculated values of the DEO variables variable t ca cb cc cabar cbba:r ccbar cd ce cdbar cebar T
k1 k2 E1 E2 rc k3 ra re E rb Sed Sde rd
initial value 0 1 1
0 0 0 0 0 0 0 0 350 1 1 -0.004 -27.402 1
1 -2 0 -0 . 004 -1 0 0 1
minimal value 0 0.0228578 0.2840909 0 0 0 0 0 0 0 0 350 1 1 --27.414373 -27.402 0.0064937 1
-2 0 -0.0272502 -1 0 0 -0.0522659
maximal value 2.52 1 1 0.3992785 0.1513598 0.4543234 0.3570959 0 . 3178411 0.3166306 0.3029636 0.1782569 350 1 1 0.958793 0 . 9557439 1 1 -0.0293515 0.1762951 0.958793 -0 . 0807076 1.5284379 42.398031 1
ODE Report (RKF45) Differential equations as entered by the user [1] d(ca)/d(t) = ra [2 J d(cb)/d(t) = rb
13-12
final value 2.52 0 . 0228578 0.2840909 0.3992785 0.1513306 0 . 4539723 0.3566073 0 . 2612331 0.3166306 0.3026417 0.1778722 350 1 1 -27.414373 -0.0272502 0.0064937 1 -0.0293515 0.0742139 -0.0272502 -0.0807076 1. 5284379 0.8250406 "·0.0513561
[3] d(cc)/d(t) = rc [4] d(cabar)/d(t) = ca*E [5] d(cbbar)/d(t) cb*E [6] d(ccbar)/d(t) = cc*E [7] d(cd)/d(t) = rd
=
[8] d(ce)/d(t) = re [9 j d(cdbar)/d(t) = cd*E [10] d(cebar)/d(t) = ce*E
Explicit equations as entered by the user [1] T= 350 [2 J k1 = exp((5000/1 . 987)*(1/350-1/T)) [3 J k2 = exp((1 000/1.987)*(1/350-1/T)) [4] E1 = -2.104*t"4+4 . 167*tI\3-1 . 596*tI\2+0 . 353*t-0.004 [5] E2 = -2 . 104*t"4+17 . 037*tI\3-50.247*tI\2+62.964*t-27.402 [ 6 J rc = k1 *ca*cb [7] k3 = exp((9000/1 . 987)*(1/350-1/T)) [8] ra = -k1 *ca*cb-k2*ca [9] re = k3*cb*cd [ 10] E = if(k=1 . 26)then(E1 )else(E2) [ 11] rb = -k1 *ca*cb-k3*cb*cd [12] Scd = cc/(cd+.000000001) [13] Sde = cd/(ce+ . 000000000001) [14] rd = k2*ca-k3*cb*cd
If the temperature is raised, the conversion of A increases. The selectivity Sc/d increases with temperature and Sd/e decreases with increasing temperature
Bimodal RTD See Polymath program P 11-2-i-·2.pol POLYMATH Results Calculated values of the DEQ variables Variable 'z ca cb cc F cd ce cbo cao ceo cdo ceo lam T
k2 ki rc k3 E1 E2 E3 re ra rb E EF
initial value
° °0.99 0 1 1
0 1 1 0 0 0 6 350 1 1 1 1 346.34561 6737 . 4446 0 . 00911 0 -2 -1 0.00911 0 . 911
minimal value
°0.5350642 0.2660482 -0°. 0033987 0 °11 0 0 0 0 350 1 1 0 . 1424065 1 0.1019474 0.0742397 0.0061156 0 -2 --1 0 . 0061156 0 . 2083818
maximal value 6 1 1 0.2872257 0 . 99 0.2692177 0.1929233 1 1 0 0 0 6 350 1
1 1 1 346.34561 6737.4446 1.84445 0.1694414 -0 . 4084547 -0.28635 0.6288984 1.8694436
13-13
final value --6--.--
0.2660482 0.5352659 0.2745726 -0.0033987 0 . 2692177 0.1901615 1 1 0 0 0 0 350 1 1 0 . 1424065 1 0 . 20909 925.46463 1 . 84445 0 . 144103 -0.4084547 -0.2865096 0 . 20909 0 . 2083818
rd Sed Sde
1
0.1219452
1
o
1 . 0856272 17.698981
o
o
o
0.1219452 1.0198908 1. 4157321
ODE Report (RKF45) Differential equations as entered by the user [1] d(ca)/d(z) = -(-ra+(ca-cao)*EF) [2] d(cb)/d(z) = -(-rb+(cb-cbo)*EF) [3] d(cc)/d(z) = -(-rc+(cc-cco)*EF) [ 4] d(F)/d(z) = -E [5] d(cd)/d(z) = -(-rd+(cd-cdo)*EF) [6] d(ce)/d(z) = -(-re+·(ce-ceo)*EF) Explicit equations as entered by the user [1] cbo= 1 [2) cao = 1 [3) cco=O [4) cdo=O [5) ceo=O [6) lam=6-z [7) T=350 [8) k2 = exp«1000/1 . 987)*(1/350-1/T)) [9] k1 = exp«5000/1 . 987)*(1/350-1/T)) [1.0) rc=k1*ca*cb [11) k3 = exp«9000/1 . 987)*(1/350-1/T)) [1.2) E1 = 0.47219*lam"4-1.30733*lamA3+0.31723*lamA2+0.85688*lam+0.20909 [13) E2 = 3.83999*lamA6-58.16185*lamA5+366.2097*lam"4-1224 . 66963*lamA3+2289.84857*lamA22265.62125*lam+925.46463 [14] E3 = 0.0041 0*lam"4-0.07593*lamA3+0 . 52276*lamA2-1.59457*lam+ 1.84445 [ 15) re = k3*cb*cd [16] ra = -k1 *ca*cb-k2*ca [17) rb = ·k1 *ca*cb-k3*cb*cd [18] E = if(lam<=1 . 82)then(E1 )else(if(lam<=2 . 8)then(E2)else(E3)) [19) EF=EI(1-F) [20] rd = k2*ca-k3*cb*cd [21] Scd = cc/(cd+.0000000001) [22) Sde = cd/(ce+.0000000001)
If the temperature is raised, the conversion of A increases. The selectivity Sc/d increases with temperature and Sdle decreases with increasing temperature
P13-2 G) Exothermic Reaction: E=45Kj/mol See Polymath program P 13-2+ I . pol
POLYMA TH Results Calculated values of the DEQ variables Variable z x
cao T
lam ca E1 E2 F1 F2
initial value
o o
8 320 200 8 0 . 1635984 2.25E-04 5.6333387 0 . 9970002
minimal value
o o
8 320
o
0 . 2971785 0.0028744 2.25E-04
o
0.381769
maximal value 200 0.9628524 8 464.4279 200 8 0.1635984 0.015011 5.6333387 0.9970002
13-14
final value 200 0.9579239 8 463.68858
o
0.3366089 0.028004 0.015011
o
0.381769
0.01 2.25E-04 0 0.0220681 -0,8227063
0 . 01 2.25E-04 0 . 9970002 0.075005 -0.64
k
E F EF ra
1,924805 0.028004 0 . 9970002 0.075005 -0.1699892
1. 8893687 0,028004 0 0,028004 -0.2140759
ODE Report (RKF45) Differential equations as entered by the user [1] d(x)/d(z) = -(ra/cao+E/(1-F)*x) Explicit equations as entered by the user [1] cao = 8 [ 2 1 T = 320+ 150*x (3] lam = 200-z [ 4] ca = cao*(1-x) (5 J E1 = 4.44658e-1 0*lamA4-1.1802e-7*lamA3+ 1.35358e-5*lamA2-,000865652*lam+,028004 ( 6] E2 = -2,,64e-9*lam A3+ 1.3618e-6*lamA2-,,00024069*lam+.015011 (7] F1 = 4..44658e-1 0/5*lamA5-1 ,,1802e-7/4*lamA4+ 1,35358e-5/3*lamA3-,000865652/2*lamA2+.028004*lam [ 8] F2 = -(-9,,3076ge-8*lamA3+5,,02846e-5*lamA2- . 00941 *lam+,,618231-1) (9] k= .01*exp(45000/8.314*(1/320-1/T» (10] E = if (lam<=70) then (E1) else (E2) (11] F = if (lam<=70) then (F1) else (F2) (12J EF = E/(1-F) ( 1 3] ra = -k*caA2
--_._--_ _---
LO , - - - . - - - . - - - - - - - - - - - - - ,
.
0.8
0. 6
0.4
0,,2
80
40
160
120
200
Endothermic Reaction: E=45Kj/mol See Polymath program P 13·2+ 1"po)
POLYMATH Results Calculated values of the DEQ variables Variable z
initial value
minimal value
x
o o
cao
8
8
T
320 200
289.82835
8
5.5862683 0.0028739 2.25E-04
lam ca El E2 F1
0,.1635984 2 . 25E-04 5.6333387
o
o
o
o
maximal value 200 0,,3017l58 8 320 200 8
0 . 1635984 0.015011 5,6333387
13-15
final value 200 0 . 2860515 8
291. 39485
o 5.7l15879 0.028004 0,015011
o
F2 k E F EF
0.9970002 0.01 2.25E-·04 0.9970002 0.075005 -0.64
ra
0.381769 0.0017191 2 . 25E-04 0 0 . 0220679 -0.64
0.9970002 0.01 0.028004 0.9970002 0.075005 -0.0536459
0.381769 0.0019006 0.028004 0 0.028004 -0.0620023
ODE Report (RKF45) Differential equations as entered by the user [1 J d(x)/d(z) = -(ra/cao+E/(1-F)*x) Explicit equations as entered by the user [1] cao = 8 [ 2] T = 320-1 OO*x [3] lam = 200-z [4] ca = cao*(1-x) [5] E1 = 4.44658e-1 0*lam"4-1.1802e-7*lamA3+ 1. 35358e-5*lamA2-.000865652*lam+.028004 [ 6] E2 = -2.64e-9*lamA3+ 1.3618e-6*lamA2-.. 00024069*lam+ ..Q15011 [7] F1 = 4.44658e-1 0/5*lamA5-1 . 1802e-7/4*lam"4+1.35358e-5/3*lamA3-.000865652/2*lamA2+.028004*lam [ 8] F2 = -( -9.3076ge-8*lamA3+5 . 02846e-5*lamA2-.00941 *lam+.618231-1) [9] k = .01*exp(45000/8 . 314*(1/320-1rr» [10J E = if (lam<=70) then (E1) else (E2) [11] F = if (Iam<=70) then (F1) else (F2) [12] EF = E/(1-F) [ 13 1 ra = -k*caA2
0.32
GJ
0..16
0.08
0.000
40
80
z
120
160
200
P13-2 (k) Base case: ODE Report (RKF45) Differential equations as entered by the user [ 1 ] d(Ca)/d(t) = ra/vo [2 J d(Cb)/d(t) = rb/vo [ 3 J d(Cc)/d(t) = rclvo Explicit equations as entered by the user [1] vo= 10 [2] k1 = 1 [3] k2=1 [ 4] tau = 1.26 [5] ra = -k1 *Ca [ 6 ] rc = k2*Cb
13-16
[7] Cao = 1 [ 8] x = 1-Ca/Cao [9] rb = k1 *Ca-k2*Cb
PFR CA CB Cc X
K 1/K2=2 0.080 0.406 0.513 0.919
K 1/K2=1 0.284 0.357 0.359 0.716
K 1/K2=O.5 0.284 0.203 0.513 0.716
See Polymath program P13·2-k·-2. pol
POLYMATH Results NLES Solution Variable ca cb cc cao tau cbo ceo k1 k2 ra rc rb
Value 0 . 4424779 0.2466912 0.3108309 1 1..26 0 0 1 1 --0.4424779 0 . 2466912 0 . 1957867
lni Guess
fIx)
4.704E-10 -3 . 531E-10
o
1
o o
NLES Report (safenewt) Nonlinear equations
[1] f(ca) = caOHa*tau-ca = 0 [2] f(cb) '" CboHb*tau-cb = 0 [3 J f(cc) = CCOHc*tau-cc = 0
Explicit equations [1]
[2 1 [3] [4J [51 [6]
[7] [ 8] [9]
CSTR CA CB Cc
~-
cao = 1 tau = 1. 26 cbo = 0 cco = 0 k1 = 1 k2=1 ra=-k1*ca IC = k2*cb rb = k1 *ca-k2*cb
K 1/K2=1 -
0.443 0.247 0.311 0.557
-
KI/K2=2
---
0.284 0.317 0.399 0.716
K 1/K2=O.5
0.443 0.158 0.399 0.557
See Polymath program P i3-2-k-3 pol
POLYMA TH Results Calculated values of the DEQ variables
13-17
--
variable t ca cb cc cabar cbbar ccbar k1 k2 cao E1 E2 ra rc x rb E
initial value 0 1 0 0 0 0 0 1 1 1 -0.004 -27 . 402 -1 0 0 1
-0.004
minimal value 0 0.0804596 0 0 0 0 0 1 1 1
-·27.462382 -27.402 -1
0 0 -0.1353314 -0 . 0272502
maximal value 2 . 52 1 0.3678466 0.7167822 0.3050655 0 . 3350218 0.3476989 1 1 1 0.9523809 0 . 9568359 -0.0804596 0.3678269 0 . 9195404 1 0.9568359
final value 2.52 0.0804596 0.2027582 0.7167822 0.304964 0.3347693 0.3468313 1 1 1 -27 . 462382 -0.0272502 -0 . 0804596 0.2027582 0.9195404 -0.1222986 -0.0272502
ODE Report (RKF45) Differential equations as entered by the user [1] d(ca)/d(t) = ra [2] d(cb)/d(t) = rb [3] d(cc)/d(t) = rc [4] d(cabar)/d(t) = ca*E [5] d(cbbar)/d(t) = cb*E (6) d(ccbar)/d(t) = cc*E Explicit equations as entered by the user [1] k1 = 1 [2J k2=1 [3) cao = 1 [ 4] E1 = -2 . 104*tll4+4.164 *tA3-1 . 596*tA2+0.353*t·0.004 [5 J E2 = -2.1 04*tll4+ 17.037*tA3-50 . 247*tA2+62.964*t-27.402 [6) ra = -k1*ca [ "7 ] rc = k2*cb [ 8] x = (cao..ca)/cao [9] rb = k1 *ca-k2*cb [10 J E = if(k=1 . 26)then(E1 )else(E2)
Asymmetric RTD
~~gation Model CA CD Cc
--
._-
X
KJ1K2=2 0.110 0.390 0.486 0.89
KJ1K2=1 0.306 0.335 0.347 0.694
--_.
---,_._-K}1K2=O.5 0.306 0.195 0.486 0.694
See Polymath program P13-2-k-4 . pol
POLYMA TH Results Calculated values of the DEQ variables Variable t ca cb cc cabar'
initial value
o
minimal value
o
1
9.1l9E-04
o
o
o o
o o
maximal value 7 1
0 . 3678325 0.9927049 0.3879174
13-18
final value 7
9 . 119E-04 0 . 0063832 0 . 9927049 0.3879174
--
cbbar· ccbar
0 0 1 1 1 0.20909 925.46463 -1 0 0 1 0 1 .. 84445 0.20909
k1 k2
cao E1 E2
°r·a rc x
rb E4 E3 E
0 0 1 1 1 0.1017293 0.074462 -1 0 0 -0 . 1353022 0 0 . 0061369 0
0.2782572 0.3271713 1 1 1 707 . 06552 3.072E+04 -9.1l9E-04 0 . 3675057 0.9990881 1 0 1.84445 0 . 628067
0 . 2782572 0 . 3271713 1 1 1 707 . 06552 3 . 072E+04 -9.1l9E-04 0.0063832 0.9990881 -0 . 0054713 0 0.09781 0
ODE Rel10rt {RKF45} Differential equations as entered by the user [ 1 J d(ca)/d(t) = ra [2 J d(cb)/d(t) = rb [3] d(cc)/d(t) = rc [4 J d(cabar)/d(t) = ca*E [5 J d(cbbar)/d(t) = cb*E [6 J d(ccbar)/d(t) = cc*E Explicit equations as entered by the user [lJ k1 = 1 [2] k2=1 [3] cao = 1 [4] E1 = 0.47219*t"4-1 . 30733*tA3+0 . 31723*tA2+0 . 85688*t+0.20909 [ 5] E2 = 3.83999*tA6-58 . 16185*tA5+366 . 20970*t"4-1224.66963*tA3+2289.84857*tA2-2265.62125*t+925.46463 [ 6 J ra = -k1 *ca [7 J rc = k2*cb [ 8 J x = (cao-ca)/cao [9 J rb = k1 *ca-k2*cb [10] E4 = 0 [ 11 J E3 = 0 . 0041 0*t"4-0 . 07593*tA3+0 . 52276*tA2-1 . 594S7*t+ 1. 84445 r12] E = if(k=1 . 82)then(E1 )else(if(k=2 ..8)then(E2)else(if(k6)then(E3)else(E4)))
BimodalRTD .-~--
St~gregation
Model
K 1/K2=1 0.388 0.278 0.327 0.612
Kl/K2=2 0.213 0.350 0.430 0.787
Maximum Mixedness
K 1/K2=1
KI/K2=2
K 1/K2=O.5
CA CB Cc
0.306 0.335 0.347 0.694
0.110 0.390 0.486 0.89
0.306 0.195 0.486 0.694
CA CB Cc X ._----_.
-
K 1/K2=O.5 0.388 0.175 0.430 0.612
._---
Asymmetric RTD
,.------------
X
'-'--.
-.
P13-2 (l-r) No solution will be given at this time. 13-19
..
PI3-3 Equivalency Maximum Mixedness and Segregation model for first order reaction:
dC A dA
Rearranging:
-[k + 1-F(A) E(A) ]c A
Using the integration factor:
by definition
gives:
e
=
CAoE(A) 1-F(A)
k E(A) ]dA] d CAe -I[ +l-F(A) _ [k+ E(A) JdA [ CAo E(A)e I I-F(A) dA =1-F(A)
E(A )d(A) = dF(A)
-I[ I-F(A) E(A) ]dA I dF = e - I-F(A) = ed[Ln(l-F(A))) = (1- F(A))
changing the variables from')... to t in the RHS integral:
kt
~e-kA(1-. F(A)) = f[E(t )e- (1- F(t ))]dt C Ao
1- F(t)
_
Ak
CA = CAo e ( ) fE(t)e-ktdt 1-F A
(1)
r
Exit concentration is when ')...=0, F(O)=O hence eqn (1) becomes:
CA = CAo
E(t)e-ktdt
This is the same expIession as for the exit concentration for the Segregation model.
13-20
P13-4 (a) Mean Residence Time By definition
I(
E t)dt
= 1.
_ _ llr The area of the semicircle representing the E(t) is given by A 2
o T
~ J!r
Fm const,nt volumetrk flow t m
~ T ~ J!r ~ 0.8 m; n .
P13-4 (b) Variance 00
0'2
= I(t o
~
r? E(t)dt = It 2 E(t )dt - r2 0 21"
5
0
It 2 E(t )dt = It 2 ~r2 - (t -- r)2 dt = _r 4 J[cos 2(x)+ 2cos(x)+ llsin 2(x)dx = ~r4 o 0 1l 8
0'2
= 51l r 4 _r2 =_1_=0.159 8 21l
Using Polymath: See Polymath program P13-4-b pol POL,YMA TH Results Calculated values of the DEQ variables Variable t sigma tau t1 E2 E
initial value
o o
0.7980869 1.5961738
o o
minimal value
o o 0 . 7980869 1..5961738
o o
maximal value 1..596 0.15931.61 0 . 7980869 1.5961738 0 . 7980614 0.7980614
ODE Report (RKF45) Differential equations as entered by the user [ 1) d(sigma)/d(t) = (t-tau)A2*E Explicit equations as entered by the user [1] tau = (2/3 . 14)"0 . 5 [ 2 ] t1 = 2*tau [ 3 J E2 = (t*(2*tau-t»A( 112) [4] E = if (1
13-21
final value 1. 596 0.1593161 0 . 7980869 1.5961738 0.0166534 0.0166534
2
--1
and
0.20 016
1-
0.12
sigma
0.08
0.64 t
1.28
0.96
1.60
P13-4 (C) Conversion predicted by the Segregation model
x
=
J (t)E(t )dt X
o
X(t) =l_··e-kt 21"
X=l-
Je-kt(r2-(t-lljl2dt o
See Polymath program P 13-4·c . po]
POLYMA TH Results Calculated values of the DEQ variables Variable t Xbar tau t1
initial value -0-----
o 0.7980869 1. 5961738
minimal value
o o
0.7980869 1. 5961738
E
o o
o o
k
0.8
0.8
E2 X
o
o
maximal value 1. 596 0,,4447565 0.7980869 1. 5961738 0,,7980671 0.7980671 0.8 0,7210716
ODE Report (RKF45) Differential equations as entered by the user [ 1] d(Xbar)/d(t) =X*E Explicit equations as entered by the user [ 1 J tau = (2I3.14)1I().,5 [2 J t1 = 2*tau [3 J E2 = (t*(2*tau-t))A(1/2) [4 J E = if (kt1) then (E2) else (0) [5J k=.8 [6 J X::: 1-exp(-k*t)
13-22
final value 1. 596 0.4447565 0.7980869 1. 5961738 0.0166534 0.0166534 0.8 0,7210716
0.5 , - - - - - - - - - - - - - - - - - ,
0.4
OJ 0.2
OJ 0.0
L..-.......,~
0.000
_ _ _ _ _ _ _ _ _____'
0319
o 638t
0958
1277
1.596
X =44.5%
P13-4 (d) Conversion predicted by the Maximum Mixedness model
dX =.3L+ E(A) X
dA rA
CAo
1-·F(A)
= -·kCA = -kC Ao(l- X)
dX
E(A)
-dA = -"k(l- X)+ 1- F(A) X dX
E(A)
d~- = k(l- X)-l-F(A) X
See Polymath program PI J-4-d pol POLYMA TH Results Calculated values of the DEQ variables Variable z x F
k
lam tau E1 E
initial value 0 0 1 0. 8 1. 596 0.7980869 0 . 0166534 0 . 0166534
minimal value 0 0 -5 . 053E-04 0.8 0 O. 7980869 0 0
maximal value 1..596 0 . 4445289 1 0. 8 1..596 0 . 7980869 0.7980666 0.7980666
ODE Report (RKF45) Differential equations as entered by the user [1 J d(x)/d(z) = -(-k*(1-x)+E/(1-F)*x) [2 J d(F)/d(z) = -E
•Explicit equations as entered by the user k= . 8 [2] lam = 1 . 596-z [3] tau = (2/3.14)A0 . 5 [4] E1 = (tau A2-(lam-tau)A2)A0.5 [ 5] E = if (lam<=2*tau) then (E1) else (0) [1]
13-23
final value
1":5%"-0.4445289 --5.053E-04 0.8 0 0 . 7980869 0 0
0.5 , - - - - - - - - - - - - - - - , 0.4
D
0.3 0. 2
OJ 0.0
~-~--~-~--~-----'
0.000
0.319
x = 44.5%
0.638
z
0.958
1.277
1.596
as for the Segregation Model, but we knew this because for first order reactions
XMM
P13-5 (a) The cumulative distribution function F(t) is given: I
F(t)
-
0.5 -
I
20
40
The real reactor can be modelled as two parallel PFRs:
Therelative
I E(t) = { --J(t -
1"1
4
3 )+-J(t 4
1"2)
Mean Residence Time 1
tm = JtdF = (10 min * 1) + (20min*O.75) = 25 min o ~)""·························T--·-·····
t
F 13-24
Xseg =
or
P13-5 (b) Variance
=
(j2
Jrt -T)2 E(t)dt
= S{t -
!
!
tm )2[ ~o{t - Tj)+ o{t - T,}Jdt = ~ {Tj - tm )2 + {T2 - tm)2 =
o
= 75min 2 P13-5 (C)
For a PFR, second order, liquid phase, irreversible reaction with k = 0.,1 dm3 /mol min- l , t mol/dm3
X
=
k'lC Ao 1 + k'lC Ao
= 25 min and C Ao = 1..25
= 0.758
FOI a CSTR, second order, liquid phase, irreversible reaction with k = 0 ..1 dm3 /mol"min- l , t mol/dm3
X 2
(l-X )
= k'lC Ao
-t
= 25 min and C Ao = 1.25
= 0.572
X
For two parallel PFRs, tl = 10 min and tz = 30 min, Faol = 1/4Fao and Fao2 = 3/4Fao , second order, liquid phase, ineversible reaction with k = 0..1 dm3 /mol,min- l and C Ao = 1.25 mol/dm3
CAl
= C Ao -
C A2
= C Ao -
X
=
kT1C Ao
1+ kTIC Ao kT2 C Ao
CAo = 0.556mol / dm
--C Ao
1+ kT 2 C Ao
vC Ao -!VCAI -iVC A2
4
4
vC Ao
3
= 0.263mol dm 3
= 0.731
P13-5 (d) I-Conversion predicted by the Segregation Model
13-25
x=
1
X (t)E(t )dt =
o
1 0
[.!
kC Aot 8(t 1+ kC Aot 4
1"1 )
+ ~. 8(t - 1"z )]dt = 4
1 kCAo 1"1 +3 kCAoTz =0.731 4 1 + kC Ao 1"1 4 1 + kC Ao 1"Z 2-Conversion predicted by the Maximum Mixedness model
dX =~+ E(A) X dA CAo I-F(A) Z r A =-kC/ =-kC Ao (I-X? dX = -kC (1- X)Z + E(A) X dA Ao I-F(A) We need to change the variable such the integration proceeds forward:
.dX =kCAo(l-X? __E(T-~X dz I-F(T-z) See Polymath program P 13-5-d.pol
POLYMATH Results Calculated values of the DEO variables Variable --z x F cao k
lam ca t1 t2 E3 ra E2 El E EF
initial value 0 0 0.9999 1. 25 0.1 40 1..25 10 30 0 -0.15625 1..25 1.25 0 0
minimal value -0---0 -1. 081E-04 1. 25 0.1 0 0 . 3614311 10 30 0 -0.15625 1. 25 1.25 0 0
maximal value 40 0.7125177 0.9999 1.25 0.1 40 1.25 10 30 0 ·-0 . 0130632 1.25 1.25 1.25 2.8604931
ODE Report (RKF45) Differential equations as entered by the user [1] d(x)/d(z) =-(ra/cao+E/(1-F)*x) [2] d(F)/d(z) = -E Explicit equations as entered by the user [ 1 J cao = 1 . 25 [2J k =.1 [ 3 J lam = 40-z [4 J ca = cao*(1·x) [5 J t1 = 10 [6] t2=30 [7] E3 = 0 [ 8] ra = -k*caA 2 [9 J E2 = 0 . 75/(t2*2*(1-0.99)) [10] E1 =0.25/(t1*2*(1-0.99))
13-26
final value 40 0,,7061611 -1. 081E-04 1.. 25 0.1
o
0.3672986 10 30
o
-0.0134908 1.25 1.25
o o
[11] [12]
E = if ((Iam>=0 . 99*t1 )and(lam<1.01 *t1» then (E1) else( if ((lam>=0.99*t2)and(lam<1.01 *t2» then (E2) else (E3)) EF = E/(1-F)
0.731
0.706
0.731
0.758
X CSTR 0.572
P13-5 (e) Adiabatic Reaction E=10000cal/mol and T
= 325 -
500X
= 325 - 500X and the constitutive equation for k . (45000/8314*(l/325-1IT)). th MM d I k - 325 e III e mo e .
Introducing the enthalpy balance: T
The conversion is drastically reduced.
P13-5 (0 Conversion Predicted by an ideal laminar flow reactor For a LFR, second order, liquid phase, ineversible reaction with k = 0..1 dm3 imolmin- 1, mol/dm3
T
We apply the Segregation model, using Polymath:
X
=
ktC A 0
1 + ktC Ao
and E(t)
{Ofor t < 12.5min
=
625/(2t 3 )min-1 for t ~ 12.5min
See Polymath program P ].3-5-e .po]
POLYMA ~H Results Calculated values of the DEQ variables Variable t xbar cao k
tau El t1
E x
initial value 0
0 1. 25
0.1 25 3_125E+06 12,,5 0 0
minimal value 0 0 1.. 25 0.1 25 1.157E-05 12.5 0 0
maximal value 300 0.7077852 1.. 25 0.1 25 3.125E+06 12 . 5 0.0991813 0.974026
ODE Report (RKF45) Differential equations as entered by the user [ 1] d(xbar)/d(t) = x*E Explicit equations as entered by the user [ 1 ] cao = 1 . 25 [2l k = . 1 [3] tau = 25 [4 J E1 = tauJ\212/(tJ\3+0 . 0001) [ 5 ] t1 = tau/2 [ 61 E = if (1
13-27
final value 300 0" 7077852 1.. 25 0.1 25 1,,157E-05 12 . 5 1.157E-05 0.974026
= 25 min and C Ao = 1..25
We can compare with the exact analytical formula due to Denbigh .
X~ D+ -(D;}n(l + 21 Da)] ~ 0.709
with
na= kC Ao'
P13-6 (a) E(t)
Mean Residence Time 00
By definition
JE(t)dt
= 1.
The area of the triangle representing the E(t) is given by A = 2t1 0.2 = 1
2
o
tl =5min.
----7
t if t < tl Z'l tl
E(t)
2t1 )if tl 5: t 5: 2tl tl o otherwise
= - -;. (t -
P13-6 (b) Variance 00
a
2
00
= J(t-t m ?E(t)dt = Jt 2 E(t)dt-t; o
0
and
a
2
7 6
=-t
2 m
-t
2 m
t;
25 . 2 =-=--=4.167mm 6 6
See Polymath program P 13-6-b vol POLYMA TH Results
13-28
Calculated values of the DEQ variables Variable
initial value
sigma tau tl El E2
minimal value
o o
o o
5 10
5
o
o o
t
10
0.4
o
o
E
maximal value
final value
10
10
4.1666667 5 10 0.4 0.4 0,,1989341
4,,1666667 5 10 0.4
o
o
ODE Report (RKF45) Differential equations as entered by the user [1 J d(sigma)/d(t) = (t-tau)1\2*E Explicit equations as entered by the user [1] tau = 5 [2] t1 = 2*tau [3] E1 = tltaul\2 [ 4] E2 = -(t-2*tau)/tauI\2 [5] E = if (ktau) then (E1) else (if(k=t1 )then(E2)else(O))
P13-6 (C) For a PFR, second order, liquid phase, irreversible reaction with kCAo =02 min- 1,"C=5 min
X
=
ktC Ao
= 0.5
1+ktCAo For a CSTR, second order, liquid phase, ineversible leaction with kC Ao =0.2 min- 1,"C=5 min
.
X
(l-X
)2 = ktC Ao --"* X = 0.382
P13-6 (d) I-Segregation model See Polymath program Pl3-6cl·l ..pol
POLYMA TH Results Calculated values of the DEQ variables Variable t Xbar kl X tau t1 El E2 E
initial value 0 0 0.2 0 5 10 0 0.4 0
minimal value 0 0 0. 2 0 5 10 0 -0 . 2 0
maximal value 15 0,,4767547 0,,2 0 . 75 5 10 0.6 0. 4 0 . 1978698
ODE Report (RKF45) Differential equations as entered by the user [ 1. ] d(Xbar)/d(t) = X*E
13-29
final value 15 0,,4767547 0. 2 0,,75 5 10 0.6 -0 . 2 0
Explicit equations as entered by the user [1] k1 =.2 [2] X=k1*t/(1+k1*t) [3] tau = 5 [4] t1 = 2*tau [5] E1 = t/tauA 2 [6] E2 = -(t-t1 )/tau A 2 [7] E = if (t
r---------------,
0.4
0.3
0.2
0.1
12
15
X =47.7% 2-Maximum Mixedness Model
(-"..!L.
dX __ + E(T - Z) dz C Ao I-F(T-z)
xJ
rA =-kCA 2 = -kCAo 2 (1-- X )2 dX =k'(I-X?-. E(T-_~X dz 1,- F(T _. z) See Polymath program P13-6·d-2,pol POLYMA TH Results Calculated values of the DEQ variables Variable
z x F
k
lam tau El t1 E2 E
initial value 0 0 1 0.2 20 5 0.8 10 -0.4 0
minimal value 0 0 -7.513E-06 0.2 0 5 0 10 -0.4 0
maximal value 20 0.6642538 1 0.2 20 5 0.8 10 0.4 0 . 1994823
ODE Report (RKF45) 13-30
final value 20 0.4669205 -7.513E-06 0. 2 0 5 0 10 0.4 0
Differential equations as entered by the user [1] d(x)/d(z) = -(-k*(1-x)1I.2+E/(1-F)*x) [2] d(F)/d(z) = -E Explicit equations as entered by the user [1] k = 0.2 [2] lam = 20-z [3 J tau = 5 [4) E1 = lam/taull.2 [5] t1 = 2*tau [ 6] E2 = -(lam-t1 )/tauIl.2 ['7] E = if (Iam
X =46.7%
P13-6 (e) Laminar Flow Reactor For a LFR, 2nd order, liq., phase, irreversible reaction kC Ao =0.2 min- l ,T=5 min" We apply the segregation model, using Polymath:
ktC A {O for t < 2.5 min X =. 0 and E(t) = 1 + ktC Ao 25 /(2t 3 ) min -} for t ~ 2.5 min See Polymath program P 13-6-e ..pol POLYMA TH Results Calculated values of the DEO variables Variable t xbar' kcao tau El t1 E x
initial value 0 0 0.2 5 1.. 25E+05 2.5 0 0
minimal value 0 0 0. 2 5 4 . 63E-07 2. 5 0 0
maximal value 300 0 . 4506243 0. 2 5 1 . 25E+05 2.5 0.0549822 0.9836066
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(xbar)/d(t) = x*E Explicit equations as entered by the user [1] kcao = 0.2 [2] tau = 5 [3] E1 = taull.2/2/(tIl.3+0 . 0001) [ 4] t1 = tau/2 [5] E = if (1
We can compare with the exact analytical formula due to Denbigh"
13-31
final value ._-----
300 0 . 4506243 0.2 5 4 . 63E-07 2. 5 4.63E-·07 0.9836066
XMM 0.451
XcSIR
0.467
0.5
0.477
0.382
P13-7 Irreversible Liquid phase, half order; Segregation model. Mean conversion
X=
f X (t )E(t )dt = 0.1
(1)
Assume a Gaussian distribution for E(t):
E(t) =
ex p[- (t -
1
a.j271
rt]
2.a
1
=.
3.j2.71
ex p[- (t-
112
dX CAo . -= k- . - ()112 1- X dt CAo
51~]
2·3
3
and CAo=l mol/dm
The only unknown kl is estimated solving with a trial and error method Eq(1).. Using POLYMATH: kJ =0.0205 moI 1l2/dm312 . s
P13-8 (a) The E(t) is a square pulse
I
EXt)
-
OJ -
o
I
o
05
2
2j
3
t
60
Third order liquid··phase reaction: fA = kCA3 with CAo = 2mol/dm3 and k = 0.3 dm6/moe/min ( Isothermal Operation) I-Conversions Segregation Model
1 X (t) = 1- -;:;====:::::;;:
~(2kCAo 2t + 1)
X· = jX(t)E(t)dt o
=~1I 1 Jdt= IV2kC~ot + 1 .
2
~2kC~ot +1
t -=-----=-2- - = 0.53 kC Ao 1
See Polymath program P13·-8-a-l.pol
13-32
POLYMA TH Results Calculated values of the DEQ variables Variable t Xbar k
Cao t1 E2 E X
initial value 0 0 0. 3 2 1 1 0 0
minimal value 0 0 0.3 2 1 1 0 0
final value 2 0.5296583 0.3 2 1 1 1 0 . 5847726
maximal value 2 0.5296583 0. 3 2 1 1 1 0 . 5847726 0. 60
ODE Reuort {RKF45) Differential equations as entered by the user [ 1] d(Xbar)/d(t) = X*E
0.48
Explicit equations as entered by the user [lJ k=.3 [2] Cao = 2 [3] t1 = 1 [4J E2=1 [5] E = if (t>=t1) then (E2) else (0) [ 6] X = 1-1/( 1+2*k*CaoA 2*t)A( 1/2)
036
0..2-t 0.12
o.oo'---~-----,--L-~------l
2-Maximum Mixedness Model
_dX __ (~+_ E(T-z) dz - C Ao I-F(T-z)
dX
dz
0.0
OA
xJ
=kC~o(I-X)3 _ E(T-z)
X
1- F(T -z)
See Polymath program P13-8-a-2.pol POL)''MA 1]1 Res!!lts Calculated values of the DEQ variables Variable ---. z x F cao k lam ca E1 ra t2 t1 E EF
initial value 0 0 0.9999 2 0.3 2 2 1 -2.4 2 1 1 1 . 0E+04
minimal value 0 0 -9 . 999E-05 2 0.3 0 0 . 9569223 1 -2 . 4 2 1 0 0
maximal value 2 0 . 5215389 0 . 9999 2 0. 3 2 2 1 -0 . 2628762 2 1 1 1.0E+04
13-33
final value 2 0 . 5215389 -9.999E-05 2 0.3 0 0.9569223 1 -0 . 2628762 2 1 0 0
0.8
t
12
1.6
2.0
2.0
ODE Report (RKF45) Differential equations as entered by the user [1] d(x)/d(z) = -(ra/cao+E/(1-F)*x) [2] d(F)/d(z) = -E
1.6 1. 2
Explicit equations as entered by the user [1] cao = 2 [2] k= . 3 [3] lam = 2-z [4] ca = cao*(1-x) [5] E1 = 1 [ 6] ra = -k*ca"3 [7] t2=2 [8] t1 = 1 [9] E = if ((lam>=t1 )and(lam<=t2)) then (E1) else(O) [10] EF=EI(1-F)
0. 8
0.4
0.0
0.0
0. 8
0.4
z
1.2
d2X
The conversion shows an inflection point in cOIrespondence of z = 1, where start the pulse --2- = 0 .
dA
P13-8 (b) Introducing in the Segregated Model and in the MM Model :
k -- k 300' e (20000*(l I 300-11 T) See Polymath program P I3-S-b- Lpol and P13-8·b-2.pol --:::-----
X Se2 X MM
300K 0.530 0.521
310K 0.82 0.806
--
320K 0.933 0.924
330K 0.974 0.97
340K 0.989 0.987
--,------
350K 0.995-0.994
The discrepancy is greatest at 300K
P13-8 (C) Adiabatic Reaction Introducing the enthalpy balance:
T = To + (- Ml RX ) X = 305 + 40000 X LBiCPi 50 See Polymath program P 13-8-\.: pol
POLYMA TIl Results Calculated values of the DEQ variables Va:dable t Xbar
initial value
minimal value
ko
0.8948702
0.8948702
t1 E2 t2 Cao
1 1
1 1 2 2
E k
To
o o
o o
maximal value 5 0.8949502 0.8948702
o
o
1 1 2 2 1
7.6864651 305
7 . 6864651 305
7.6864651 305
2 2
13-34
final value 5 0.8949502 0.8948702 1 1 2 2
7.6864651 305
o
1.6
2.0
X
o
o
T
305
305
0 . 9430621 1059.4497
0 . 9430621 1059.4497
110U
ODE Report (RKF45) Differential equations as entered by the user [ 1] d(Xbar)/d(t) = X*E
940
Explicit equations as entered by the user [1 J ko = 0.3*exp(20000*(1/300-1/30S)) [2 J t1 = 1 [3] E2=1 [4] t2=2 [5] Cao = 2 [61 E = if((t>=t1) and (1<=t2)) then (E2) else (0) [7] k = ko*exp(20000*(1 1300-1 131 0)) [8] To = 30S [ 9] X = 1-1 I( 1+2*k*Cao"2*t)".S [101 T = To+800*X
780
Q
620 460 300
0
P13-9 (a)
(LFR, PFR, CSTR with r=lOOs) PFR
Design equation
CA
dX 2 C Ao = k CA CB dV
= CAo (1- X)
CB = C Bo (lx
V0
dX
X)
J-(-)3 =kCBo o 1-·X
2
V -
V
1 2V - - - = 2kC Bo -·+1
(1- X)2
V
Using the quadratic solution
X
1
=1---;====== = 1·- 0.168 = 0.832 1+2kC Bo
2
V
V
The conversion for a PFR X=83..2%
13-35
1
2
t
3
5
CSTR Design equation
V0
(C Ao -, C A) = -rAV
2
-rA =kCAC B
CA = CAo (1- X) C B = C Bo (1- X) CAo-CA =X C Ao X =kCBo
2
(I-XY V V
The conversion for a CSTR X=66.2 % LFR (completely segregated)
t<1:12 =1:2 l2e 1:=V/v=1000/l0=100s=1.67min E(t)=O
X
for
for
1>= 1:12
= fX(t )E(t )dt o
d~=kC2 (I-X)3
dt
Ao
WhcreX(t)=l-
~
1· ", 1+ 2kCBo t
See Polymath program PI3-9-a,.pol
POLYMA TH Results Calculated values of the DEQ variables variable t xbar cbo k
tau E1 x E
initial value 1. OE-05 0 0.0313 175 1. 67 1.394E+15 1.714E-06 0
minimal value 1. OE-05 0 0" 0313 175 1. 67 1.403E-,06 1. 714E-06 0
maximal value 100 0,,1827616 0,,0313 175 1. 67 1.394E+15 0.8315008 0,,1480215
ODE Report (RKF45) Differential equations as entered by the user [ 1] d(xbar)/d(t) = x*E Explicit equations as entered by the user [1] cbo = 0.0313 [2] k = 175 [3] tau = 1..67
13-36
final value 100 0.1827616 0,,0313 175 1. 67 1.403E-06 0.8315008 1. 403E-06
[5
E1 = tauJ\2/(2*tJ\3) x = 1-(1/(1 +2*k*cb0"'2*t))I\() . 5
[6
E = if(t>=tau/2) then (E1) else (0)
[4
The integral X
J
= X (t )E(t )dt
gives mean conversion=18%
o
P13-9 (b) (Segregation Model and Maximum Mixedness Model applying RTD of Example 13-1) Segregation model
dNA dt
---=-r V A
Batch reactor
CA
= C Ao (1-- X)
CB
= C Bo (1 -
X)
rX dX 2 .b (1- i)3 = kC Bo t
Similarly
X(t)=l--
vl1 -=-] 1
~kCB02t
=
=
f
X (t )E(t )dt and E(t) from the given data, fitted using Polymath o See Polymath program P13·9·b-regressioll..poi X
POLYMATH Results Polynomial Regression Report Model: C02
Variable a1 a2 a3 a4
= a1 *C01
+ a2*C01J\2 + a3*C01J\3 + a4*C01J\4
Value 0.0889237 -0.0157181 7.926E-04 -8.63E-06
95% confidence 0 . 0424295 0 . 0163712 0.0019617 7 . 288E-05
General Order of polynomial = 4 Regression not including free parameter Number of observations = 13
13-37
Statistics R/\2 = R/\2adj = Rmsd= Variance =
0.8653673 0 . 8204897 0.0065707 8.107E-04
See Polymath program P13-9-b-l.pol POLYMA TH Results Calculated values of the DEQ variables Variable -t
initial value
o
xbar F
cbo k x
o o
o o
o
0.0313 175
0 . 0313 175
o o
o o
E
minimal value
maximal value 14 0.4106313 1.1137842 0 . 0313 175 0.5847898 0.1566631
ODE Report (RKF45) Differential equations as entered by the user [ 1] d(xbar)/d(t) = E*x [2] d(F)/d(t) = E Explicit equations as entered by the user [lJ cbo = 0.0313 [2] k = 175 [3) x = 1-(1/(1+2*k*cboA2*t»AQ.5 [4] E = 0 . 0899237*t-0.0157181 *t/\2+0.000792*t/\3-0.00000863*tA4 0.5
r---------.------,
0.4 0..3
0.2 0.1 0.0
__ 2.8 5.6 t
'--..::;...~
0. 0
_ _ _ _J 8.4 11..2 14.0
~_._~
~
X =41%. Maximum Mixedness
~X =.!L+ E(A) dA
CAo
X
I-F(A)
Rate Law: - r A = kC A CB
2
13-38
final value 14 0.4106313 1.1137842 0.0313 175 0.5847898 0.0199021
lC = kC 80 2 (1- X)3 Ao
-dF = -E(z) dz
where z=14-A
See Polymath program P13-9-b-2.pol POLYMA TH Results Calculated values of the DEQ variables Variable z X F
Cbo k
lam Cao Ca E EF
Cb ra
initial value 0 0 0,9999 0.0313 175 14 0" 0313 0" 0313 0.0199021 199.0212 0.0313 -0,0053663
minimal value 0 0 -0.1138842 0.0313 175 0 0,,0313 0.0202322 0 0 0.0202322 -0.0053663
maximal value 14 0.3536026 0.9999 0" 0313 175 14 0,0313 0,,0313 0.156664 199.0212 0,0313 -0.0014493
final value 14 0.3536026 -0.1138842 0" 0313 175 0 0.0313 0.0202322 0 0 0.0202322 --0.0014493
ODE Report (RKF45) Differential equations as entered by the user [1] d(X)/d(z) = -(ra/Cao+E/(1-F)*X) [2] d(F)/d(z) = -E .Explicit equations as entered by the user [1] Cbo=,0313 [21 k= 175 [3] lam = 14-z [41 Cao=.0313 [5] Ca=Cao*(1-X) [61 E = O.0899237*lam-O,,0157181 *lamA 2+0,,000792*lamA 3·,000000863*lam"4 [7] EF = E/(1-F) [8] Cb=Cbo*(1-X) [9] ra = -k*Ca*CbA 2
X=35.4%
P13-9 (c) Exit time(t), internal age(a) and life expectancy;'
rE(t)dt = F(t)~ dF(t) = E(t) dt
.t
where E(t) is obtained from the polynomial fit in Part (b),
13-39
J(a) = ~[1- F(a)] V
A(A) = E(A) 1-F(A)
Intensity Function
P13-9 (d) Adiabatic reaction Segregation model
dX
dt
= kC 2(1- X )3 80
Where
k(T) = koexp[E (~.--~)] = 175exp[3000~(_1_ R To
8.314 320
T
--!)] T
T(K)=TO+(..:::::.MlrxJx =320+150X CPA
and =
J
X = X (t )E(t )dt and E(t) from the given data. o See Polymath program P 13-9··d-l ,pol
POLYMA Til Results Calculated values of the DEQ variables Variable t xbar F
x
cbo
initial value
o
minimal value
o
o o o
o o o
0_0313 320
0.0313 320
T E
o
o
k
175
175
maximal value 14 0.7585435 1.1137842 0.8973303 0 . 031.3 454.59954 0.1565966 4931. 8727
ODE Report (RKF45) Differential equations as entered by the user [ 1 ] d(xbar)/d(t) = E*x [2] d(F)/d(t) = E [3] d(x)/d(t) = k*cboA2*(1-x)A3 Explicit equations as entered by the user [1) cbo = 0,,0313 [2] T = 320+150*x [3] E = 0,,0899237*t-0,,0157181 *tA2+0,,000792*tA3-0,,00000863*tA4 [4] k = 175*exp(30000/8.314*(1/320-1/T»
X =76% Maximum Mixedness Model
13-40
final value -.:r:a-----0.7585435 1.1137842 0 . 8973303 0.0313 454.59954 0.0199021 4931. 8727
dX =~+ E(A) X dA CAo I-F(A) ~ = -kC 2(I_X)3 C 80 Ao
where
and
[E(
1 1)] =175exp[30000(-1- - 1)] k(T)=koexp - ._-R To T 8.314 320 T
T(K) = To + (-
Mlrx)x
=
320 + 1.50X
CPA
See Polymath program P 13-9-d-2 . pol
POLYMATH Results Calculated values of the DEQ variables Variable z X F
Cbo T
lam Cao Ca E EF
Cb k
ra
initial value 0 0 0 . 9999 0.0313 320 14 0.0313 0.0313 0 . 0199021 199.0212 0.0313 175 -0.0053663
minimal value 0 0 -0.1138842 0.0313 320 0 0 . 0313 0.0087977 0 0 0 . 0087977 175 -0 . 007411
maximal value 14 0.7189248 0 . 9999 0.0313 427.83873 14 0 . 0313 0 . 0313 0.1566233 199 . 0212 0 . 0313 3001 . 8809 -0 . 0020441
final value 14 0.7189248 -0.1138842 0 . 0313 427.83873 0 0.0313 0.0087977 0 0 0.0087977 3001 . 8809 -0 . 0020441
ODE Report (RKF45) Differential equations as entered by the user [1 J d(X)/d(z) = -(ra/Cao+E/(1-F)*X) [2] d(F)/d(z) = -E Explicit equations as entered by the user [1] Cbo= . 0313 [2] T = 320+ 150*X [3] lam = 14-z
[4] Cao = . 0313 [ 5] Ca = Cao*(1-X) [6] E = 0 . 0899237*lam··O . 0157181 *lamI\2+0 . 000792*lamI\3-0.00000863*lam"4 [7] EF = E/(1-F) [8] Cb=Cbo*(1-X) [9] k = 1l5*exp(30000/8 . 314*(1/320-1/T» [101 ra = -k*Ca*CbI\2
13-41
0.80.------------------.,
0.64
0.-18
5.6
z
11.2
8.4
14.0
gives X=72% If the reaction is carried out adiabatically the conversions are more than doubled .
P13-10 Irreversible, first order, long tubular reactor; constant volume, isothermal
dF
A ForaPFR - =-r
dV
A
x = l_e- kt For X = 0.865
=> kr = 2.0
For laminar flow with negligible diffusion (LFR), the mean conversion is given by: =
JX {t )E{t )dt = JX (t )E{t )dt
~
X
=
o
X{t) = 1- e-
,,/2 kt "('2
E(t) for laminar flow = - 3 where
t
2t
J(l-e
=
Therefore
X
=
,,/2
r '?2 2
J
2 =
-b
"(' e -, dt=I---dt 3 3 2t 2 ,,/2 t
-kt) "('
We can apply the approximated solution due to Hilder:
13-42
x = (4+ Da)eO SDa +Da-4 =0.782 (4+ Da)eO SDa +Da where Da=kr=2
x = 0.782 < X
PFR
= 0.85
P13-11 (a) First Moment about the mean: by definition is always equal to zero. ~
~
~
= f(t - T)E(t )dt = ftE(t )dt - T fE(t )dt = T- T= 0
ml
°=
mlCSTR
° =0 = m lLFR
m lPFR
°
P13-11 (b) Second-order liquid-phase reaction Da= rkCAo =I.O,r=2min and kCAo =05min-1• CSTR
FAo -- FA = -'A V FAo - FA = FAoX V
=
FAoX (- r A ) exit
Liquid-phase T
V
CAo -C
= - = ( .) A Vo
Second-order - r A
-'A
A = kC A2 and T =-CAo -C 2 kC A
X
-----
kC Ao (l- X)2
Solved
X
=
(1 + 2Da)-.Ji+ 4Da =0382 2Da
PFR
-dFA dV dX F Ao - - '- - rA dV X dX V =FAo 0-'11.
f--
Second-order
13-43
V = F Ao
fdX
X
--2
where
O kC A
(I-X) C A = C Ao -'---..t(1+EX)
Liquid-phase E = 0 and integrating
1 [
r= kC
X ]
I-X·
or
X
Ao
Da
= I+Da
=0.5
LFR In the ring globule of radius r
dC A - = rA
dt
ktC Ao
X =
= kC A2
(2order batch)
I+ktC Ao
E(t) =
=r
Where - rA
ofor t < 1min {4 1(2t 3 ) min -1 for t ~ 1min (E(t) LFR)
2
kC Ao 2
[._! + In(I + kC Ao
ktC AO.J]
t
Evaluate for Da=l,
t
00
= Da[I- Da
r /2
2
In(I +DaDaI2)] 12
X = 0.451
See Polymath program P 13-11-b.pol
~
_. --f..::.p.
~1~=~~5.::..;1------._-_~
.::..;FR::..:...----.
~O.5
P13-12
The criteria
a r ac 2
--
___ A A
>0
Xseg>XMM
2
13-44
The following figure shows the reaction rate as function of the concentration .
•
0.01 ,---,..,..-----,.-----.,.------y-------,
0.005
0"-----·
3
The second derivati~ is initially fitgative (X tf.6('''vl), then ptSkitive (Xseg>~M)' The flex point is for C A=8molJdm (Xseg=XMM). CA In the limit of low concentration
- rA = kC A + o(C A)2
(Fir st order) and Xseg=XMM
in the limit of high concentration
- fA
= KA
P13-13
~C . + o(~J (Reaction order=-l) and Xseg>XMM C A
A
No solution will be given
P13-14 (a) Liquid phase, Segregation Model, second order; non-ideal CSTR, adiabatic.· 00
X=
fX(t,T)E(t )dt = 0.67 o
E(t) = C(t) -00
fC(t )dt
. and
f C(t )dt = Img mini dm 3 o
o E(t)=IF (t<=I) THEN (t) ELSE ( IF (1)=2) THEN (0) ELSE (2-t)) For a batch globule: C Ao
dX = -rA dt
-
13-45
-rA = kC A
2
dX = kC (1- X)2 dt Ao Where C Ao=2 mol/dm3
k(T) = 0.5exp[~(-1 8.314 300
T =To+
Where
Cps
-!)] T
-M/ X rx
Cps +1'1CpX
=I
tJ;C pi
1'1Cp = 1I2C pb - Cpa
= Cpa + 0 = 50J 1mol· K = 10012 - 50 = 0
T=300+1S0X Iterate with Ea, using the ODE solver for values of X(t,T) and substitute these into the polynomial regression to evaluate ~
the integral X
= fX(t,T)E(t)dt =0.67 o
An activation energy Ea of 10000 J/mol gives approximately the correct mean conversion, X =0.67. Inaccuracy lies in the polynomial fit and hence the integral area. See Polymath program P IJ·14a.pol POLY~ATH Results Calculated values of the DEQ variables
variable t xbar x cao T
El E2 E Ea k
initial value 0 0 0 2 300 0 2 0 1.25E+04 0.5
minimal value 0 0 0 2 300 0 -1 0 1.25E+04 0.5
maximal value 3 0.6673438 0.9198721 2 437 . 98082 3 2 0.9933851 1.. 25E+04 2.4247011
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(xbar)/d(t) = E*x [2 j d(x)/d(t) = k*cao*«1-x)A2) Explicit equations as entered by the user [1J cao =2 [2 J T =300+150*x
13-46
final value 3 0 . 6673438 0.9198721 2 437.98082 3 -1
0 1.25E+04 2 . 4247011
(3 (4 (5
[6
[7
E1 = t E2 = 2-t E = if (t<=1) then (E1) else (if (t>=2) then (0) else (E2)) Ea = 12500 k = 0,5*exp(Ea/8,,314*((1/300)-(1fT)))
P13-14 (b) Parallel reactions, isothermal, segregation model Batch globules
dC A = rA = rAl + rA2 = -k1A C A2 -k 2C C AC B dt
dCc - = rC = -rA2 = k 2C C ACB dt
Exit concentrations
dC Aba! dt
=C
Bba dt
dC- -! dC Cba! dt
E{t) A
C B E(t )
= CCE{t)
E(t)=IF (t<=l) THEN (t) ELSE ( IF (t>=2) THEN (0) ELSE (2-t))
Selectivity
s = CBba = 2.38 !,
CCbar Iteration with k2C until S = 2..38 gives k2C = 0..3755 dm 3 /mol min See Polymath program P13··14·b.,pol
,----, ,----_._---,--------
P13·-15 Reactor: fluidised CSTR (V=lm3 ; F=lOdm 3/s, Cco=2 Kmollm3)
The system of complex reactions for the Kentucky coal n,.9 is given by
13-47
P13-15 (a) Segregation Model See Polymath program P l3-IS-a,pol
POLYMA Til Results Calculated values of the DEQ variables Variable t ca cp cc cabar cpbar ccbar co cobar k1 k2 k3 k4 k5 rc ra rp tau ro E Spo
initial value 0 0 0 2000 0 0 2000 0 0 0.012 0.046 0,,02 0.034 0.04 0 92 24 1. 667 0 0.59988 2.4E+08
minimal value 0 0 0 1383.5708 0 0 2000 0 0 0.012 0.046 0.02 0_034 0.04 -49.695094 10.804749 24 1. 667 0 3,,695E-06 0.9280349
maximal value 20 900,,42918 689.61265 2000 137.10239 46.011527 3988.5578 466.78696 9.1391124 0.012 0.046 0.02 0.034 0,04 0 92 37.542445 1. 667 36.017167 0.59988 2.4E+08
ODE Report (RKF4S) Differential equations as entered by the user [1] d(ca)/d(t) = ra [2] d(cp)/d(t) = rp [3] d(cc)/d(t) = rc [ 4] d(cabar)/d(t) = ca*E [5] d(cpbar)/d(t) = cp*E [6] d(ccbar)/d(t) = cc*E [7 j d(co)/d(t) = ro [8] d(cobar)/d(t) = co*E Explicit equations as entered by the user [lJ k1 = 0.012 [2 J k2 = 0 . 046 [3 J k3 = 0.020 [4J k4 = 0034 [5] k5 = 0.04 [ 6 J rc = -k1 *cp-k2*ca [7 J ra = k2*cc+k3*cp-k4*ca-k5*ca [ 8] rp = k1 *cc+k4 *ca-k3*cp [ 9] tau = 1.667 [ 1 0 1 ro = k5*ca [11] E=exp(-tltau)/tau
13-48
final value 20 900.42918 689.61265 1383,,5708 137.10239 46.011527 3988.5578 466.78696 9" 1391124 0.012 0.046 0.02 0.034 0.04 -49.695094 10.804749 33.425188 1. 667 36.017167 3 . 695E-06 0.9280349
[12 J Spo = rp/(ro+0,0000001)
The exiting selectivity is 0.,928
P13-15 (b) Maximum Mixedness model See Polymath program P 13-15-b,pol
POLYMA Til Results Calculated values of the DEQ variables Variable z Ca Cp Ce Co F
k1 k2 k3 k4 k5 re ra rp tau ro E Spo EF sigma Cao Cpo Ceo Coo
initial value 0 0 0 2000 0 1 0,,012 0.046 0,,02 0.034 0.04 0 92 24 1. 667 0 0,59988 2,,4E+08 5,,999E+07 3 0 0 2000 0
minimal value 0 0 0 1463,,3111 0 6.149E-06 0,,012 0.046 0,,02 0.034 0.04 -47,,836902 15,,190162 24 1.. 667 0 3.695E-06 0.9906544 3,,695E-06 3 0 0 2000 0
maximal value 20 875,,11273 631,,80972 2000 408.43732 1 0.012 0.046 0.02 0.034 0.04 0 92 37,,327161 1. 667 35.004509 0,,59988 2,,4E+08 5.999E+07 3 0 0 2000 0
ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(z) = -(-ra+(Ca-Cao)*EF) [21 d(Cp)/d(z) = -(-rp+(Cp·Cpo)*EF) [3] d(Cc)/d(z) = -(-rc+(Cc-Cco)*EF) [4] d(Co)/d(z) = -(-ro+(Co·,Coo)*EF) [ 5] d(F)/d(z) = -E Explicit equations as entered by the user [1 J k1 = 0.012 [2J k2 = 0.046 [ 3 J k3 = 0,,020 [4J k4 = 0,,034 [5J k5 = 0,04 [ 6 J rc =·k1 *Cp-k2*Ca [7] ra = k2*Cc+k3*Cp-k4*Ca-k5*Ca O[ 3 J rp = k1 *Cc+k4*Ca-k3*Cp [ 9 J tau = 1,,667 [10] ro = k5*Ca [11] E = exp(-Z/tau)/tau [12] Spo = rp/(ro+0.0000001) [13] EF=E/(1-F) [14] sigma = 3
13-49
final value 20 875,11273 631. 80972 1463.3111 408,,43732 6.149E-06 0.012 0.046 0.02 0.034 0.04 -47,,836902 15.190162 34,677371 1.. 667 35.004509 3.695E-06 0.9906544 3.695E-06 3 0 0 2000 0
[15) Cao = 0 [16) CpO = 0 [17) Ceo = 2000 [18) Coo = 0
P13-15 (C) The selectivities are reported in the following table:
I
LSifMpFR
I
XMMCSTR
I
XSffiPFR
XSffiCSTR
~. _ _ _ _ _-l.. ...;:0~.9:..::.9_ _ _ _ _-1._4;.:.;.1~7....:.4_ _ _ _ _-----l.---,0:...;.:.9...c:2,--_ _ _ _--,
P13-15 (d) Normal Distribution with T = 5min and (J = 3min
E(t) =
1 a.J21t
ex p[-
r]
(t - T 2.a
1
=
3.J2.J[
ex p[-
r]
(t - 5 2·3
Segregation Model See Polymath program P13·15-d-l.pol POLYMA TH Results Calculated values of the DEQ variables variable t ea ep ee eabar epbar eebar co eobar k1 k2 k3 k4 k5 re ra rp sigma ro tau Spo E1 E
initial ---.value 0 0 0 2000 0 0 0 0 0 0.012 0.046 0.02 0.034 0.04 0 92 24 3 0 5 2.4E+08 10 0.0331675
minimal value 0 0 0 1991.2989 0 0 0 0 0 0.012 0.046 0.02 0.034 0.04 -8.5350648 79.947816 24 3 0 5 4.1743032 10 0.0331675
maximal value 2 171.75027 52 . 879376 2000 11.116671 3.2689487 221. 39065 7.0306771 0.3200147 0.012 0.046 0.02 0.034 0.04 0 92 28 . 677508 3 6.8700107 5 2.4E+08 1.0 0.0806774
ODE Report (RKF45) Differential equations as entered by the user [1] d(ea)/d(t) = ra [2] d(ep)/d(t) = rp [3) d(ee)/d(t) = re [4] d(eabar)/d(t) =ea*E [ 5] d(epbar)/d(t) = ep*E [6) d(eebar)/d(t) = ee*E [7 J d(eo)/d(t) = ro [8] d(eobar)/d(t) = co*E
13-50
final value 2 171. 75027 52.879376 1991. 2989 11.116671 3 . 2689487 221 . 39065 7.0306771 0 . 3200147 0.012 0.046 0.02 0.034 0.04 ·-8.5350648 79.947816 28.677508 3 6.8700107 5 4.1743032 10 0.0806774
Explicit equations as entered by the user [1] k1 = 0 . 012 [2 j k2 = 0 . 046 [3] k3=0 . 020 [4] k4 = 0 . 034 [5} k5 = 0 . 04 [6] rc = -k1 *cp-k2*ca [7 J ra = k2*cc+k3*cp-k4*ca··k5*ca [8] rp = k1 *cc+k4*ca-k3*cp [9] sigma = 3 [ 10] ro = k5*ca [11] tau = 5 [12] Spo = rp/(ro+0 . 0000001) [13] E1 = 1/(tau*2*(1-0.99» [14] E = exp(-(t-tau)A2/(2*sigmaA2»/(sigma*(2*3 . 14)AO.5)
Maximum Mixedness Model See Polymath program P 13- J 5-d-2.pol ~OLYl\1A TH
Results POLYMA TH Report
0825-2005, RevS.! 233
Calculated values of the DEQ variables Variable z Ca Cp Ce Co F kl k2 k3 k4 k5 re ra rp lam r·o tau Spo sigma E Cao Cpo Ceo Coo EF
initial value 0 0 0 2000 0 1
0 . 012 0.046 0.02 0 . 034 0.04 0 92 24 2 0 5 2.4E+08 3
0 . 0296795 0 0 2000 0 2.968E+06
minimal value 0 0 0 1995.2809 0 0.9770814 0.012 0 . 046 0 . 02 0 . 034 0.04 --5 . 986657 83.563477 24 0 0 5 5.6574287 3 0.0020622 0 0 2000 0 0 . 089981
maximal value 2 120 . 77791 35.906087 2000 3.8182818 1 0 . 012 0.046 0.02 0 . 034 0.04 0 92 27.331698 2 4 . 8311165 5 2.4E+08 3 0 . 0296795 0 0 2000 0 2.968E+06
ODE Report (RKF45) Differential equations as entered by the user [1] d(Ca)/d(z) = -(-ra+(Ca-Cao)*EF) [2 j d(Cp)/d(z) = -(-rp+(Cp-Cpo)*EF) [3] d(Cc)/d(z) = +rc+(Cc-Cco)*EF) [4] d(Co)/d(z) = +ro+(Co-Coo)*EF) [5] d(F)/d(z) = -E
13-51
final value 2 120.77791 35.906087 1995.2809 3.8182818 0.9770814 0.012 0.046 0.02 0 . 034 0.04 -5.986657 83.563477 27.331698 0 4 . 8311165 5 5 . 6574287 3 0 . 0020622 0 0 2000 0 0.089981
Explicit equations as entered by the user [1] k1 = 0 . 012 (2) k2=0.046 (3) k3 = 0.020 (4) k4 = 0.034 (5) k5=0.04 [ 6 ) rc = -k1 *Cp-k2*Ca [7) ra = k2*Cc+k3*Cp-k4*Ca-k5*Ca (8) rp = k1*Cc+k4*Ca-k3*Cp (9) lam = 2-z [10] ro = k5*Ca [l1J tau = 5 [12) Spo = rp/(ro+0.0000001) [ 13] sigma = 3 [14] E = exp(-(lam-tau)/\2/(2*sigma))/(sigma*(2*3.14)"O.5) [15] Cao = 0 [16] Cpo = 0 [ 17] Cco = 2000 [18] Coo = 0 [19] EF = E/(1-F)
P13-16 Multiple parallel reactions, isothermal E(t)=0.0279693 - 0.0008527t + 1.2778e-5 t2 - 1.0661e-7
e+ 4..5747e·-1O t -7.73108e-13 t
P13-16 (a) Segregation model
dC A =--k C C 2 -·k C C dt lAB 2AB
·_3.3 k 4CCCA 2/3
13-52
4
5
dC D = CDE(t) dt
See Polymath program P13-16-a.pol Selectivities:
Exit concentrations:
CA
= 0.026mol / dm 3
SeD = 0.00196
CB
= 0.008mol / dm 3
SDE
Cc
= 6.955e -
CE
= 0.012mol/ dm 3
-
= 4.156e --7mol/ dm 3
CF
5mol / dm 3
= 3.009
SEF = 28433
P13-16 (b) Maximum Mixedness model E(t)=O . 0279693-0 . 0008527A+12778e-5A2-L0661e.. 7 A?+4.5747e-lO A4-7..73108e-13 where z=t, tp200min (extent ofE(t».
_dC A
-
dz -
r -((C -- C ) E(A) A
A
Ao
I-F(A)
J
and and so on for other species.
dF{A) = - E(A ) ._dz
gives F(A)
See Polymath program PI 3-J6-b.. pol
13-53
1.,5
Exit concentrations:
Selectivities:
CA
= 0.028mol I dm 3
SeD
= 0.0009
CB
= O.OlOmol I dm 3
SDE
= 3.004
S EF
= 1343110
Cc = 2.927e -, 5moll dm 3 CD
= 0.033mol I dm 3
CE
= O.Ollmol I dm 3
C F =7.985e-9molldm 3
P13-16 (C) Ideal CSTR -tIT
tm=t and
E(t) = _e_ T
Mol balances:
~F:1_ = F - (- r dt
Ao
FAo
A
)
= CAoVo = 0.05·10 =0.5mol I min = FBo
and so on for the other species"
CA CTo
FA FT = C Ao + CBo
=CTo--~etc ....
= 0.05+0.05 = 0.1
FT = FA +FB +Fe +FD +FE +FF See Polymath program P13-16-c"po! Exit concentrations:
Selectivities:
CA
= 0.050mol I dm 3
SeD = 0.068
CB
= 0.049mol I dm 3
SDE
= 3.342
Ce =5.63ge-5molldm 3
SEF
=815844
CD = 0.0008mol I dm 3
CE
= 0.0002mol I dm 3 13-54
CF
= 3.03e -lOmol/ dm 3
Ideal PFR
tm=T and RTD function
E{t) = 8{t - 't')
Selectivities:
Exit concentrations:
CA
= 3.98e -
CB
9mol / dm 3
SCD
= 0.004
= 0.0065mol / dm 3
S DE
= 3.368
Cc
= O.000277mol / dm 3
SEF
= 4 . 286
CD
= 0.068mol / dm 3
CE
= 0.0202mol / dm 3
CF
= 0.0047mol/ dm 3
Segregation
Maximum Mixedness
SCD = 0.00196
SCD
S DE = 3.009
S DE = 3.004
SEF
=28433
SEF
= 0.0009
=1343110
CSTR
PFR
SCD
= 0.068
SDE
=3.342
S EF = 815844
SCD
= 0.004 SDE
=3.368
S EF = 4.286
S EF For the PFR is very much smaller than for the others, because CF is not so small at the exit of the PFR in turn due to exit Cc is not so small either., The conversion of CA in the PFR is virtually complete at the exit of the PFR, hence greater Cc
P13-16 (d) See Polymath program P13-16-d.pol E(t)=IF (t<=lO) THEN (0,,01) ELSE (IF(t>=20) THEN (0) ELSE (02-0,01 t))
13-55
Segregation
Maximum Mixedness
PFR
CSTR
= 0.004
SCD
=0.004
SCD
= 0.0035
SCD
= 0.068
SDE
= 3.109
SDE
=3.106
SDE
=3.342
SDE
=3.368
S EF
= 24078
SEF
= 41503
SEF
= 815844
SEF
= 4.286
For the Segregation and Maximum Mixedness models,
S EF
SCD
is much lower than
for the CSTR but still far greater than for the PFR. The CSTR and PFR values are unchanged as they do not depend on E(t).
P13-17 Multiple parallel reactions, isothermal Asymmetric RTD: E=IF (t <=1.26) THEN (El) ELSE (E2) 4
E1=-2.104t +4" 167t3-1.596t2-0.353t.. O,004 4 E2=-2,,104t + 17 .037t3-50.247t2+62,,964t-27.402
P13-17 (a) Segregation model
dC A --= -kDlCAC B 2 -3k E2 C AC D dt
!:C B = C E{t) dt
B
dCC () --=CcEt dt
13-56
dC F (t) --=CFE
dt
See Polymath program P13-17-a.pol Exit concentrations:
Selectivities:
CA
= 0.819mol / dm 3
SCD
=0.272
CB
= 0.767moll dm 3
SDE
= 11.330
Cc
-
= 0.163moll dm 3
SEF
= 0.267
CD
= 0.600mol / dm 3
C~
= 0.053mol / dm 3
CF = 0.199moll dm 3
P13-17 (b) Maximum Mixedness model As the RTD is asymmetric we can use the same equations for E(A) as we did for E(t), with: E(A)=IF (A<=1.26) THEN (EI) ELSE (E2)
dC A
__ " "
dA -
r
~ iA
+ (C _ C ) E(A) A Ao I-F(A)
The same applies to the equations for the other species as in Part (a).. See Polymath program P t 3- t 7 -b .pol Exit concentrations:
CA
= 0.847mol / dm 3
Selectivities: SCD
= 0.281
13-57
C B = 0.824mol I dm 3
SDE
=10.7
Ce =0.162molldm 3
SEP
= 0.280
CD
= 0.576mol I dm 3
CE
= 0.054mol I dm 3
Cp
= 0.192moll dm 3
P13-17 (c) Ideal CSTR -tIT
tm=L and
E(t) = _e'['
dC ( '--=v CAo -C A) +rA A
Where r A =
dt
(-kD1CAC~ -2kE2 C AC D )
vo =v
and so on for the other species. See Polymath program P 13·17c-1.pol Exit concentrations:
Selectivities:
CA
= 1.386mol I dm 3
SeD
= 0.832
CB
= 1.774moll dm 3
SDE
= 71.100
Ce
= 0.095moi I dm 3
SEP
= 0.198
CD = 0.114moll dm 3
C E = 0.002mol / dm 3 C P = 0.D08mol I dm 3 Ideal PFR
=.
tm and RTD function
E(t) = g(t - '[')
and so on for the other species .
13-58
See Polymath program P 13-17-c-:2.pol Selectivities:
Exit concentrations:
CA
= 0.31Omol / dm 3
SCD
= 0.162
CB
= 0.338mol / dm 3
SDE
= 3.618
Cc
= 0.106mol / dm 3
SEF
= 0.497
CD
= 0.652mol / dm 3
CE
= 0.180mol / dm 3
CF
= 0.362mol / dm 3
Segregation
Maximum Mixedness
CSTR
PFR
= 0.832
= 0.162
SCD
= 0.272
SCD
= 0.281
SCD
SDE
= 11.330
SDE
= 10.7
S DE = 71.100 S DE = 3.618
SEF
= 0.237
SEF
= 0.280
SEF
= 0.198
SCD
SEF
= 0.497
S CD is significantly greater than for the others, because exit CD is lower. Similarly S DE is much greater in the CSTR than for the others, because exit CE is so low.. This is because the achievable conversion in a CSTR is not so high .
P13-18 (a) P13-18 (b) P13-18 (c) P13-18 (d) P13-19 (a) External age distribution E(t) By plotting C 105 as a function of time, the curve shown is obtained
13-59
C{t)
----------------.----.
9oor----800
700
600
500
u 400
300
200
100
o --.....-.......... -------,--........--..-.--........-....--r-----.. --.-.-....... a 10 20 30
- r '..•_..
=.-.=--:=-.::: ...... ::::::;::=""'=""""""''''l''"---~------.'''----"'--::::=...
40
50
60
70
tlmin]
To obtain the E(t) curve from the C(t) curve, we just divide C(t) by the integral
f C(t)dt = f C(t)dt + rC(t)dt + CC(t)dt + CC(t)dt
f C(t)dt
r
C(t)dt
= %(1)[1(0) + 3(622) + 3(812) + 2(831) + 3(785) + 3(720) + (650))0-5 = 4173.4 .10-5
= j(2)[(650) + 4(523) + (418))0-5 = 2106.7 .10-5
+2(136)+4(77)+ 2(44)+4(25) + 2(14)+4(8)+5)0-5 =3671.7 .10-5 JofOC(t)dt =!(5)[418+4(238) 3
roC(t)dt = !(1O )[1(5) + 1(1))0-2
5
!o
f C(t)dt = 9981.7 .10-
5
= 30 .10-5
== 0.1
We now calculate:
E(t)
= ~QL = C(t)
f C(t)dt
0.1
13-60
E(I)
009
008
0.07
006
coos
~ m:004
003
002
001
o
2
4
6
8
10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 I[minj
Using Excel we fit E(t) to a polynomial:
for 0 5: t 5: 3 E1 (t)
= -1.1675 .10-3 t 4 + 1.1355 .10-2 t 3 -
4.7492 .10-2 t 2
+ 9.9505 .10-2 t for 3 5: t 5: 20 E2 (t) E(t) = -1.7979.10-4 t
= -1· 8950 .10-6 t 4 + 8.7202 .10-5 t 3 -1.1739 .1O-3 t 2
+ 0.092343
for 20 5: t 5: 60 E3 (t)
= 1.2618 .1O-8 t 4
_
2.4995 .1O-6 t 3 + 1..8715 .1O-4 t 2
6.3512.10- 3 t + 0.083717 for t > 60
0
P13-19 (b) External age cumulative distribution F(t)
F(t)
=
!E(t)dt
Integrating the E(t), we obtain the F(t):
l3-61
-
- - - ------------------------1
12 - , - - - - - - - - - - - - -
e
1
08
~ 06
04
": __ o
~- - - - - ~- - - - - - - - -~- - - - - - - - - - ~- - - -J 20
10
40
30
50
60
\[min]
P13-19 (C) Mean residence time and variance
r
tm =
E(t)tdt
The area under the curve of a plot tE(t) as a function of t will yield tm • IE(I) 045
04
035
03
025
::
'I:l
02
015
01
005
0 10
0
20
30
40
50
Hm 0
(t-t~2E(t) -10
60
t[lIIin)
E(t)
C(t)
t
0
0
tE(t)
0
13-62
D
,--,
..... ~ ,--,
'-'
Jl ..... t m = '-' I
tm
0.4 1 2 3 4 5 6 8 10 15 20 25 30 35 40 45 50 60
)tdt =
329 622 812 831 785 720 650 523 418 238 136 77 44 25 14 8 5 1
0 . 0329 0.0622 0.0812 0.0831 0.0785 0.072 0.065 0 . 0523 0.0418 0.0238 0.0136 0.0077 0.0044 0.0025 0 . 0014 0.0008 0.0005 0.0001
-9.6 -9 -8 -7 -6 -5 -4 -2 0 5 10 15 20 25 30 35 40 50
0.01316 0.0622 0.1624 0.2493 0.314 0.36 0.39 0.4184 0.418 0.357 0.272 0.1925 0.132 0.0875 0 . 056 0.036 0.025 0
f E(t)tdt + rE(t)tdt + LO E(t)tdt + CE(t)tdt
= T = 9.88min == 10 min
We can calculate the variance by calculating the area under the Clive of a plot of:
o
10
20
30
= [(t -
tJ2 E(t)tdt
=
f (t - tJ2 E(t)dt + r(t - tJ2 E(t)dt +
C(t--tJ2 E(t)dt+ C(t-tmY E(t)dt
13-63
(t-tniE(t)
40 llmin)
(72
3.032064 5.0382 5.1968 4.0719 2 . 826 1.8 1.04 0.2092 0 0.595 1.36 1.7325 1.76 1.5625 1.26 0.98 0.8 0.25
50
60
P13-19 (d) Fraction of the material that spends between 2 and 4min in the reactor E(t) 009
008
0.07
006
::s 005 J~
;::.
~ 004 003
002
001
0
o
2
4
6
8
10 12
14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46
48 50 52 54 56 58 60
r
E(t)dt = shaded area = j[I(0.0812) + 4(0.0831) + 1(0.0785)] = 0.16
P13-19 (e) Fraction of the material that spends longer than 6min E(t)
o
2
4
6
8
10 12 14 16
18 20 22
24 26 28 30 32 34 36 36 40 tlmin)
13-64
42 44 46 48 50 52 54 56 58 60
f E(t)dt
r
E(t)dt
= shaded area =
r
E(t)dt +
CE(t)dt + fo°E(t)dt
= ~(1(0.065) + 4(0.0523) + 1(0.0418») = 0.210
fO E(t)dt = 1(1(0.0418) + 4(0.0238) + 2(0.0136) + 4(0.0077) + 2(0.0044) + 4(0.0025)
10
3 + 2(0.0014) + 4(0.0008) + 1(0.0005») = 0.367 {'°E(t)dt
= tail = 10 (0.0005 + 0.0001) = 0.003
f E(t)dt
= shaded area = 0.581
!o
2
P13-19 (f) Fraction of the material that spends less than 3min E(t)
009
008
007 006
:§" 005 '-~
.=-
~ 004 003 002 001
o
:::;: o
2
4
6
8
10 12
14 16 18 20 22
24 26
28 30
32 34 36
38 40
42 44 46
48 50
52 54
t[min)
! E(t)dt = shaded area =%[1(0) + 3(0.0622) + 3(0.0812) + 1(0.0831)] =0.192 P13-19 (g) Normalized distributions
13-65
56
58 60
Normalized RTD
e=':' E{e)=rE{t) T
09
08
07
E(8)
6
05
04
03
02
0.1
0 0
2
4
3
8 Normalized cumulative RTD
F{e)=
f E{e)de= fE{t)dt
13-66
6
7
F(e)
/'
08
----.--'-'-'
_.--.-----'
./'
06
04
02
o .'---"'-"'-"'---"'-"'-~"'----"--'"
.......... ___....___...__ .___ .__ ..._____ .________.... _. . _...._.... _______________" ________."."_____..1 "'
"'
"'
.
e P13-19 (h) Reactor Volume F= 10 dm3/min
V
= F 'T=100dm 3
P13-19 (i) Internal age distribution
I(t)=2.[l-F(t)] T
01
008
g006
004
002
o -..--.."'----..------.. o
-~-" -"'- "-"'- -~--- ---"'-~-=---==. ~==!?=~_o__o____~--------J ..----..
10
..
20
40
30
I[min]
13-67
50
60
70
P13-19
U)
Mean internal age
am =
r
I (t)tdt
=
r
I (t)tdt +
r
I (t)tdt + rI(t)tdt + CI(t)tdt =1min
P13-19 (k) Intensity function E(t)
A(t)=--1- F(t)
P13-19 (1) Mean catalyst activity Integrating the decaying rate law:
f -~~ = ! k Ddt a
1
a=--l+kDt Applying the Segregation Model with the previous RTD data:
a mean
=
r
aE(t)dt
See Polymath program P 13--1 9 "l.pol
POLYMA Tn Results Calculated values of the DEQ variables Variable t amean
initial value
o o
minimal value
o o
maximal value 70 0.5778625 0.1 0.0838847 0,,092343
final value 70 0.5778625 0.1 -2.436E+04 -21.261016
0.092343
0,,1 -2.436E+04 -21.261016
o
o
o
0.083717
1.724E-05
0.0017977
o
0.125
kd
0.1
E1 E2 E4 E3
o
o
E
o
o
0.083717 0.0838847
a
1
0.125
1
ODE Report (RKF4S) Differential equations as entered by the user [1] d(amean)/d(t) = a*E Explicit equations as entered by the user [1] kd = 0 . 1 [2] E1 -0.0011675*t"4+0,,011355*tA3-0 . 047492*tA2+0,,0995005*t [3 J E2 = -1,,8950*1 OA(-6)*t"4+8.7202*1 QA(-5)*tA3-1 . 1739*10A(-3)*tA2-1.7979*1 QA(-4)*t+0 . 092343 [4] E4 = 0 [5 J E3 = 1.2618*1 QA(-8)*t"4-2 . 4995*1 QA(·6)*tA3+1 ,,8715*1 OA(-4)*tA2-6 . 3512*1 OA(-3)*t+0.083717 [6 J E = if(t<=3)then(E1 )else(if(t<=20)then(E2)else(if(t<60)then(E3)else(E4))) [7] a = 1/(1 +kd*t)
=
P13-19 (m) Ideal PFR
13-68
2nd order, liquid phase, kC Ao =0. 1min' I, CAo=lmol/dm3 .=10 min (from P 13..1 9 (c))
X
=
k'lC Ao 1+ k'lC Ao
= 0.5
P13-19 (n) LFR Laminar Flow Reactor 2nd order, liq. phase, ineversible reaction kC Ao =0.1 min· I,.=1O min. We apply the segregation model, using Polymath:
X
=
ktC A 0
and E(t)
I+ktC Ao
={O for t < 5.00min 3 1
(l0)2/(2t )min- fort~5min
See Polymath program P13-19-n.pol POLYMAT.!I Results Calculated values of the DEQ variables Variable t xbar cao k tau E1 t1
E x
initial value 0 0 1 0. 1 10 5.0E+05 5 0 0
minimal value 0 0 1 0.1 10 1 . 852E-06 5 0 0
maximal value 300 0 . 4504221 1 0. 1 10 5 . 0E+05 5 0.2284987 0.9677419
ODE Report (RKF45) Differential equations as entered by the user [ 1 ] d(xbar)/d(t) = x*E Explicit equations as entered by the user [1] cao = 1 . 0 [2] k =.1 [3 ] tau = 10 [4] E1 = tauI\2/2/(tA3+00001) [5 ] t1 = tau/2 [6] E = if (1
We can compare with the exact analytical formula due to Denbigh .
X
~ Da[l- (~a }n(1 + 21 Da)] ~ 0.451 w;th Da=kC"",=l
P13-19 (0) Ideal CSTR 2nd order, liquid phase, kCAo =0.1 min-I, C Ao=lmol/dm3
X
-(--)2 = k'lCAo ~ X = 0.382 I-X
13-69
final value 300 0.4504221 1
0. 1 10 1 . 852E-·06 5 1.852E-06 0 . 9677419
P13-19 (p) Segregation Model 2nd order; liquid phase, kCAo =0.1 min-I, CAo =1 molldm3
x=
J (t )E(t )dt X
o
kCAot 1+ kCAot
Where X (t ) = ---"=--
See Polymath program P 13-19-p. pol
POLYMA TH Results Calculated values of the DEQ variables Variable t Xbar kCao E1 E2 E4 E3
initial value -0------
E
X
o 0.1
minimal value
o o
0.092343
0.1 -2.436E+04 -21 . 261016
0.083717
1.949E-05
o o
o o
o
maximal value 70 0.4224876 0.1 0.0836855 0.092343
o
o
o
0.083717 0.0836855 0.875
final value 70 0.4224876 0.1 -2.436E+04 -21.261016
o
0.0017977
o
0.875
ODE Report (RKF45) Differential equations as entered by the user [ 1) d(Xbar)/d(t) = E*X Explicit equations as entered by the user [IJ kCao=O.1 [2 J E1 = -O.0011675*t"4+0 . 011355*tA3-0.047 492*tA2+0.0995005*t [3] E2 = -1 . 8950*1QA(-6)*t"4+8 . l202*1 QA(-5)*tA3 . 1. 1739*1 QA(-3)*tA2··1 . 7979*10A(-4)*t+O.092343 [4] E4 = 0 [5 J E3 = 1.2618*1 QA(··8)*t"4-2.4995*1 OA(-6)*tA3+ 1.8715*1QA(-4)*tA2-6 . 3512*1 QA(-3)*t+O.083717 [6] E = if(t<=3)then(E1 )else(if(k=20)then(E2)else(if(k60)then(E3)else(E4))) [7] X = kCao*t/(1+kCao*t)
P13-19 (q) Maximum Mixedness Model 2nd order; liquid phase, kCAo =0.1 min-I, CAo =1moZldm3
Rate Law :- r A =
CA
kC A
2
= CAo (1- X)
13-70
= kC Ao 2 (1- X Ywhere k=O.l dm3 Imol min
rA
~=kC C
Ao
(1-X)2
Ao
'dF -
dz
= -
E( z)
where Z=60-A
See Polymath program P13-19-q.pol
POLYMATH Results Calculated values of the DEQ variables Variable z X F
cao lam E1 E2 E3 E4 Ca k
ra E
initial value 0 0 0 . 99 1 60 -1.284E+04 -9.8680524 2.228E-05 0 1 0.1 -0.1 0
minimal value 0 0 -0 . 010344 1 0 -1..284E+04 -9.8680524 1.806E-05 0 0.52273 0.1 -0 . 1 0
maximal value 60 0.4773052 0.99 1 60 0.0832647 0.092343 0.083717 0 1 0.1 -0.0273247 0.0832647
final value 60 0.4047103 -0.010344 1 0 0 0.092343 0.083717 0 0 . 5952897 0.1 -0.035437 0
ODE Report (RKF45) DifferentiaJ equations as entered by the user [1] d(X)/d(z) = -(ra/cao+E/(1-F)*X) [ 2] d(F)/d(z) = -E Explicit equations as entered by the user [lJ cao = 1 [2 J Jam = 60-z [ 3] E1 = -0.0011675*Jam"4+0.011355*Jam"3-0047 492*Jam"2+0 . 0995005*Jam [4] E2 = -1.8950*1 0"(-6)*Jam"4+8 . 7202*1 0"(-5)*Jam"3-1 ..1"739*1 0"(-3)*Jam"2-1..7979*1 0"(-4)*Jam+0 . 092343 [5] E3 = 1 . 2618*1 0"(-8)*Jam"4-2A995*1 0"(-6)*Jam"3+ 1. 8715*1 0"(-4)*Jam"2-6 . 3512*1 0"(-3)*Jam+0.083717 [6J E4 = 0 [ 7] Ca = cao*(1-X) [8] k = 0.1 [ 9] ra = -k*Ca"2 [101 E = if(lam<=3)then(E1 )eJse(if(Jam<=20)then(E2)eJse(if(lam<60)then(E3)eJse(E4)))
~ 1X 1 Xse I' X ~ _ _--,-.:..O•.4.;.O.::..::5_ _ _ _ _.-,-..:.O.:..:.4.::.22::" _ _ _ _ .;.:O.::..::.5_ __ MM
PFR
u
..L .
13-71
Solutions for Chapter 14 - Models for Non-ideal Reactors P14-1
Individualized solution
P14-2 (a) Approximatedformula for Segregation Model (pt reaction) 2 2
k t e = 1- kt + - - + Error. The enOl is then o((kt)\ 2 kt 3 Approximating the Error = - - . -k/
3!
kT
ffit
~
Error
rO~110·_I--,~~~~=-------------------:=~~~:O~I~~~~~7~16~6~7=_________________ _-166.7
20~--.----.----~---.----r---~
o
Ln~.... el1'orl) I.,r
-20
10
0.1
100
k'T
P14-2 (b) Parameters Dispersion Model Closed-Closed dispersion model
X = 1 - - . - - - - 4qexp(Pe,/2) (1 + qYexp(Pe,q 12) - (1-- qYexp(- Pe,q 12) Where
q = ~1+4Da/ Pe, Da
= kr = k --.I Damkohler number U
14-1
(14-26)
_
UZ
Pe , =D-
Peclet number
a
Pe,q
= ~ Pe2 + 4DaPe, 2
Where
kZ2 Rate oj consumption oj Aby reaction DaPe. = - = , Da Rate oj transport by diffusion Numerical example T
= ~ = const = 5.I5min U
= r *U = 5.I5min*0.Icm/ s = 0.309m Sc =L =1000 (Liquids region in Fig.. 14 . 11) L
DAB
= kr = 1.288
Da dt lcm
Re 10
1 dm
100
--
WOO
1m
Lldt 30.9
D/(V*d t ) 0 ..18 From Fig14.10 3.09 40 From Fig.. 105 14.11 0..309 3000 From 106 _ _ _. _ _ _ _ ....!ig.14.11
ReSc 104
----------
--
D (cm2/s) 0.018 40 30000
,---._---
Per 172 0.077 1.0310-4
Q 1.015 8.226 223.609
X ._--- _ 0.721 0.567 0.563
..
---
Is there a diameter that would maximize or minimize conversion in this range? According Fig . 14.10 there will be a radius that maximizes the conversion
P14-2 (C) (1) Vary the Damkohler number for a second-order reaction (Example 14-3(b»
For a second order reaction
Da
L
= TkC AO = - kC AO Uo
R=0.05m L=6.36m k = 0.5dm 3 /moZ.min U o = I.24m/min C AO = 0.5mol / dm 3 2 D aris = 1.05m / min Damkohler numberl Da 0.1603 0.3205
-
Conversion 0.l38 0.239 14-2
Parameter 8*VO 4*VO
0.641 1.282 2.564 5.129 10.258
2*DO DO DO/2 DO/4 DO/8
0.377 0.523 0.644 0.732 0.795
(2) Vary the Peclet and Damkohler numbers for a second-order reaction in laminar flow (Example 14-3(c»
Da = 'ikC AD
For a second order reaction
L
=-
UD
kC AD; P e = U 0. L / DAB
R=O.05m L = 6.36m k = O.5dm 3 / mol. min U D = I.24m/min
= O.5mol / dm 3 5 2 DAB = 7.6xlO- m /min -Da
CAD
-
0.1603 0.3205 0.641 1.282 2.564 5.129 f-----------10.258
Pe 8.32xIO' 4.16x105 .------;3"""---- 1-2.08xlO ----i04xl0r
--
0.52xlO' 0.26xlO' 5 0.13x10 -
Conversion 0.132 0.229 0.369 0.536 0.699 0.825 0.906 ..-
-
Parameter 8*DO 4*DO 2*DO DO DO/2 DO/4 DO/8
1.,0. 0.,9
0..6
c o "[!? 0.,5
~
§
0.4
U 0.,3 0.,2 0.,1 0.,0. DO.
5
2,DxlD
-,-- i i i 5 4 DxlD 6,DxlD 5 8 DxlD5
Pe When Peclet Number decreases less than 2 x105, the conversion is influenced significantly. Below is a FEMLAB anaylsis of the problem. (1) Vary the Damkohler number for a second-order reaction (Example 14-3(b»
14-3
-----
--
.
Da = TkC AO
For a second order reaction
L Uo
= -kCAO
R =O.05m L=6.36m k = O.5dm 3 I mol. min U o =1. 24m I min
= O.5mal I dm 3
CAO DaTi'
=1.05m
2
I min
Damkohler number/ Da 0.1603 .0.3205 0.641 1.282 2.564 5.129 10.258
Conversion 0.138 0.239 0.377 0.523 0.644 0.732 0.795
- Femlab Screen Shots [1] Domain
Ir?~i~f1 Grid I [ ] Axis equal
[2] Constants and scalar expressions - Constants
2""~=~~.= ..""--.J;xpr&~sipn . m• • • • • • • • • • •
.".. -
]1:9? 0.5
"~05"
...
. . . . -.. .-.. --.. --j:~ .............
"'.... -""..
~~,,-.~''''~
~-
. . ,. .. r..--.. ,,, ...
·t
24
" .... ,+
-
.·····f····
t ..
- Scalar expressions
14-4
Parameter 8*UO 4*UO 2*UO UO UO/2 UO/4 UO/8
COi<"'"J [c;;~J I
Apply
[3] Subdomain Settings - Physics (Mass Balance)
'Subdomain selection" '"
r Species""" ,
"
I
Library material: ['
i QllantltY
I
I® 0 1
'[J Sele()t by group
I
=-:,::='_~C~ li""'.-L-"o~a.d-~~.]ValueJExpre~sion Description
5""'"'''''' --",' ..,,'
0ts 0 isotropic
°:"v
!,[~rtifi~;i~i!i~~i;~::~]
- Initial Values
(Mass balance) cA(tO) = cAO - Boundary Conditions @ r = 0, Axial symmetry @ inlet, rBoundary
conlS!ition~
, Boundary Glonditi9n:
Ii=i~; .
I
V~lueIEHpre$sion
Descl\iplion
!2;i\d'
Concentration
I
Qu~ntity
! cAo
I @
@
No
""
.,--"
~2;£~Q'~~"':---']lhward flux
outlet, Convective flux wall, Insulation/Symmetry
14-5
Time-scaling coefficient
"]' Diffusiol) coefficient DiffusiOI) coefficient
OptiOns Draw Physics Mesh Solve PostproceSSR1g Multiphysics Help
!ill ~rl.~
. . . . i[J~.#=.~L~~. ~'gI~l? ~.~_':!I.~ • ~. .~ n .=.
!..: .. c,.c.................................... . Max: 0 523
Sur1ace: xA
0,2
0.15
-----_._-----_._---01 0.05
··005
Min: 0 110
· ,5)
(2) Vary the Peelet and Damkohler numbers for a second-order reaction in laminar flow (Example 14-3(c)) For a second orderreaction
Da = ikC AO
L
= --. kC AO; Uo
Pe = U oLI DAB
R =O.05m L= 6.36m k = O.5dm 3 I mol. min U o = 1. 24m I min
= O. 5mol I dm 3 DAB = 7.6xlO-5 m 2 I min
C AO
f---.
'---.
Da 0.1603 0.3205 0.641 1.282 2564 5.129 10.258
--.
Pe 8.32xl05 4. 16x105 2.08x105 L04x105 052x105 0.26x105 0.13x10 5
-
-_...- f---.
Conversion 0.132 0.229 0.369 0.536 - - - f-0.699 .. 0.825 0.906 --'--.
14-6
-
Parameter 8*DO 4*DO - 2*DO DO DO/2 DO/4 DO/8
---
10 0.9 08 0.7
c:
0.6
0
§ (])
> c: 0
0.5 0.4
•
()
03 0.2 0.1 0.0 00
2.0x10'
4.0x10'
6.0x10'
8.0x10'
Pe
When Peclet Number decreases less than 2 x10 5, the conversion is influenced significantly. - Femlab Screen Shots [1] Domain
[2] Constants and scalar expressions - Constants
.................................... .1
- Scalar expressions
14-7
[3] Subdomain Settings - Physics (Mass Balance)
: Vo(-DVcA+cAu)
=R, cA =concentration
rCA
Subdomain selection
Intt
Elemerd
I
Species
[Lo~d~~-'-l
Library material:
ValuelElIpression Description Cits
] Time-scaling coefficient
D isotropic
DAB
D anisotropic
R u
v
[J Select by group ~ Active in this domain
J ("---'-'--"--'-"'-'---. -.
Artificial Diffusion ..
- Initial Values
(Mass balance) cA(tO) = cAO - Boundary Conditions @ r = 0, Axial symmetry @ inlet, Boundary conditions r Boundary condition: rl~)(
ValuelExpression
Description
J Concentration
.j
] Inward flux @
outlet, Convective flux 14-8
'I Diffusion coefficient 1 Diffusion coefficient j
@
wall, illsulation/Symmetry
[4] Results ('-'u '.''''-','HI.<'Uun,
s
;}
- - .- --C.05
--------------0-,-·-·------- . 0,05
P14-2 (d) Two parameters model
T=~=lOmin va
I=2 and S=4min- 1"
fJ = .Vb_ = (~ -_!2 = 0.5 va I a= V,.=Q-fJ)_=O.013 V 'is m3 v, = (1- fJ)v a = 0.05-. nnn V, = (aT)v a = O.013m 3 T,
=y,
=0.25min
v,
14-9
-~-----'-"
-0-;---..-.-----'
CA = s
Jl+4r,kC Ao -1 kmal = 1.779-32rk m s
X = 1- CAS = 0.111 CAO Ideal CSTR
CSTR with Dead Space and Bypass (1=2.0 and S=4 min-I) X=O.lll
CSTR with Dead Space and Bypass (1=1.25 and S=0.115 min-I) X=0.51
X=0.66
P14-2 (e) Two CSTR with interchange ( 1st order reaction)
L ____________
X
.~
CAl Vo
= (fJ + ark)[p + (1- a)tk]- fJ2 (1- fJ)+ ark
See Polymath program P14--2-e.pol POL YMA TII_Re§!!lts Calculated values of the DEQ variables Variable t CTI CT2 beta alpha tau CTel CTe2 tl CTe k X
initial value 0 2000 921 0,,15 0.75 40 2000 921 -80 2000 0.03 0.5134788
minimal value 0 31_814045 164 . 15831 0.15 0.75 40 -1.275E+04 13 -80 13 0.03 0_5134788
maximal value 200 2000 1048.4628 0.15 0.75 40 2000 921 120 2000 0.03 0,,5134788
ODE Report (RKF45) 14-10
final value 200 31.814045 164.15831 0.15 0,,75 40 -1.275E+04 13 120 13 0,,03 0,,5134788
Differential equations as entered by the user [1] d(CT1 )/d(t) = (beta*CT2-(1 +beta)*CT1 )/alphaltau [2] d(CT2)/d(t) = (beta*CT1-beta*CT2)/(1-alpha)/tau Explicit equations as entered by the user [1] beta = 0.15 [2] alpha =0.75 [3] tau =40 [4] CTe1 = 2000-596*t+0 . 64*tA2-0.00146*tA3-1.047*1 QA(-5)*t"4 [5] CTe2 = 921-17.3*t+0.129*tA2-0.000438*tA3+5 . 6*1QA(-7)*t"4 [6] t1 = t-80 [7] CTe = if(k80)then(CTe1 )else(CTe2) [8] k = 0 . 03 [9] X = «beta+alpha*tau*k)*(beta+(1-alpha)*tau*k)-betaA2)/«1 +beta+alpha*tau*k)*(beta+(1-alpha)*tau*k-betaA2»
800
~~ ~J
400
o
o
40
80
t
120
160
200
Comparison experimental and predicted (13=0.15 a=0.75) concetration . For small deviations from the original value of the parameters concentration and conversion are not significantly affected.The following table show the conversion for different combinations of 13 and a..
a. 0.75 0.8
ro.s. ~. 0.5 0.5
r-o.-I0.9
_--
.
a
0.15 0.1 0.5 0.8 0.1 0.9 0.5 0.5
x ---0.51 ..
0.51 0.54 0.46 0.41 0.83 0.34 0.76
14-11
'J~
'Ji.i
0;=0.5
Ot
X
O:>::!O
""'
IJ~
~i
-1
iU
\12
'" Il"-
Q"
116
X IJ5
0..
EJ
113
o.:z
0.1
------r-
0
0.1
o.:z
03
OJ5
06
0,9
Given the interchange flowrate, the conversion is increased with the increasing of the volume of the highly agitated reactor" Given the volume of the highly agitated reactor, the conversion is increased with the increasing of the interchange.
P14-2 (f) Tubular Reactor Design The correlations between Re and Da show what flow conditions (characterised by Re) give the greatest or smallest Da and hence dispersion"
14-12
pud
To minimise dispersion a Re number of -10-20 gives the lowest value for Da. Because Re = - - , the design of the
J.l vessel could be altered (i.e. diameter) for a given fluid and flowrate,. To maximise dispersion, either a very low Re «0 . 1) or Re -2300 will give the maximum Da values . For a packed bed, the dispersion also depends on the Schmidt number as well as Re number.
P14-2 (g) Linearizing non-first-order reactions May be a good approximation if CA does not change very much with time, Le . A is in excess, in which case C Ao should not be divided by anything. Linearizing the non 1st order reactions may give significantly inaccurate results using Equation 14-27, which can be tested experimentally by recording tracer concentrations with time and using the tanks in series model for a conversion comparison.
P14-2 (h) Figures 14.3 and 14. ]b The curves in Fig. 143 represent the residence time distributions for the Tanks in Series model as function of the number of reactors. Given a CSTR of volume 1 (V=V 1), we divide the CSTR in two CSTRs (V2=1I2).. The mean residence time is unchanged (VIF) but the molecules going out of the second reactor will be delayed by the time (distribution) that occur to pass the first reactor (n=2, shift of the maximum) . In the limit of infinite division (Vn=O) , so in the single CSTRs the residence time goes to zero for all the molecules (zero variance), but their summation is the mean residence time (n=oo, PFR behaviour). The model in Fig. 14.1, assumes a "faster" (channelling) reactor and a "slower" uniform reactor.. There is an exit age distribution for the faster reactor which occurs as a distinct pulse clearly before the exit age distribution of the second reactor . The fraction of effluent which has been in the real reactor for less than time t shows a step up frum zero when flow leaves from the "faster" reactor,. This fraction is the fraction of flow in the "slower" reactor. When the flow leaves the "slower" reactor this fraction becomes one . The model in Fig. 14.1 is PFR and CSTR in paralleL The exit age distribution for the CSTR is a negative gradient curve, interrupted by the distinct exit age distribution pulse of the PFR. The CSTR will always provide a fraction of effluent which has been inside the reactor for less than time t, which increases with time. But when the effluent exits the PFR at a specified time after zero, this increased effluent is superimposed upon F(t) of the CSTR, giving a combined F(t).. Conversion
I-a
82 =--=1.5 1- fJ
v: v a=--L fJ=-1. V
v
dX
PFR mol balance: - -
dV
rA = --FAa
Rate law: - r A = kC~
14-13
Stoichiometry: liquid phase
CA
= CAo (1- X)
V = _1_[2£(1 + £)In(l- X)+ £2 X +-->-(I_+_£2-<-)X] 2
2nd order PFR:
1- X
kC Ao
V
c a
1 2
b
1 2
8 = - + - -1 = - + - -1 = 0
where
a
£=0
a
Determining
and
fJ:
a == VI ~=0 ..5 fJ
V
VI
1- ~ =Ls(I- ~ J V =1OvI _ 0 25 - VI /2.5 - 5a - 0 5 a-_ 2.5 _Vi . and fJ - - - - - - •.
Gives
10
1"1 + 1"2
1" =
Now
2
Hence
VI IVI
But
VI
=
V/5
VI
VI I VI + V2I V2
2
= 2..5vI
1 2 kC Ao
V
5 . = lll1n
+ V2 IV2 =10 -
Gives 1"1 = 2.5 min Substituting into
~=
2•.5
V2
= 7..5v 2
1"2 = 7.5min
2 [2c(l+c)ln(I-X)+c2X + (l+c )X] 1- X
Reactor 1:
2.5
=0.1
~ 2' [1 ~~,-]
1=[-~] 1-- XI CAl
-XI =0 ..5
= C AJl- XI) = 2(1-0.5) = Imall dm
3
Reactor 2:
75= __1_[ X2 ] 2 .. 0.1*2 l-X2 1 [X2] 3='0.1*22 i-x2
-X2
=0.75
C A2 = C Ao (1- X 2) = 2(1- 0.7.5) = 0.5mal I dm C . = AexlI
CAl +.EA2
2
3
= 1+0.5=075 lid 3 2 .ma m
14-14
x = CAO -CA = 2-0.75 =0.625molldm 3 2
CAo
P14-3 (a) Money for buying reactors Using the tank in series model:
X -_ 2Da + 1- .J4i)a + 1 wheI'e Da--k""Cao
. Second order reaction
Assume that 7
=7 t
.L
2Da
and that in reactors medelled as mOle than one tank, that 7
7t =-. Number of tanks
n
7
n =-
a2
rounded to the nearest integer. Reactor Maze & blue Green & white Scarlet & grey Orange & blue Pmple & white Silver & black Crimson & white _.
~(min)
2 4 3.05 2.31 5.17 2.5 2.5
:r(min) 2 4 4 4 44 2
-
n 1 1 2 3 .. 1 3 1
X 0.50 0.61 0.69 0.72 0.61 0.72 0.5
---
.--
Where Scarlet & grey: Xl = 0 ..5, C AI =0.5 -7 X 2= 0.38, CA2=0.31--t X = 0.69 Orange & blue: Xl = 0.43, C AI =0.57 ---t X2 = 0.34, C A2=0.38-7 X3 = 0.27, CA3 =0.28-7X=0.72 Using the combination of maze & blue followed by crimson & white reactor (same overall conversion either way) Xl = 0.5, CAI =O.5 -7 X 2 = 0.17, CA2=0.41-7 X=0.59 The orange & blue or silver & black reactors which both approximate to 3 tanks in series give the greatest conversion.
P14-3 (b) More money for buying reactors Try: Green & white and Maze & blue: Xl = 0 . 61, C AI =0.39 -7 X2= 0.34, C A2 =0.26-7 X= 0 . 74 Scarlet & grey and Maze & blue: Xl = 0.69, CAI =0.31-7 X 2 = 0.42, C A2=0.18-7 X= 0.82 Orange & blue and Maze & blue: Xl = 0 . 72, CAI =0.28 -7 X 2= OAO, C A2=0.17-7 X = 0 . 83 The highest conversion is now obtained from the Orange & blue reactor combined with the Maze & blue reactor..
P14-3 (C) Ann Arbor, MI East Lansing, MI Columbus, OR Urbana, IL
14-15
2
Evanston, IL West Lafayette, IN Madison, WI
P14-4 Packed bed reactor with dispersion 1st order, k1=O.0167/s, 1::=0.5, dp=O.l cm, V
= ,u =0.Olcm2 / s
L=lO cm, U= 1 cmls
P Re =
pUd p
,u
V
= 10 and Sc = - - no data concerning
DAB
DAB
From packed bed conelation for D a ' and liquid phase region of graph,
Da c
Gives - - - =
Ud p
p -_ 2*1*0.1 __ 0.4cm2/s 2 approx -7 Da ___2_U,_d_ c 0.5
UL 1 *10 Pe., =-=--=25 Da 0.4
4qexp(Pe,/ 2)
X = 1-
.---------
[(I+q)2 exp(Pe,q/ 2)]-[(I- q)2 q=
f
exp(-pe,%)]
4Da
+--
Pe,
L
Da='tK and 1: = -
U
10
=-.
1
= lOs
-7
Da=O.l67 and q=L013
X = 0.15 Conversion X=15%.
P14-5 (a) Number of tanks in series
Bo +1 2
Assuming the Peclet-Bodenstein relation: n = Where
UL
Bo = . -
Da To estimate Bo ,
diU Re = V
5*2 V 0.01 =- = 1000 and Sc = ._=- =2
0.01
From gas phase dispersion correlation chart, Gives
Da
DAB
Da ud t
=8
= 80cm 2 / s
14-16
0.005
Eo= 2*200 =5 80
n=~+1=3.5 2
P14-5 (b) Conversion Using individual reactor material balances: Reactor 1:
X
Mol balance:
rAV = -.F Ao
Rate law: - r A
= kC A
Stoichiometry:
CA
2
= CAo (1- X I)
0=0 and £=0 hence no volume
change
XI = kC Ao
2
(I-XJ2
vC Ao
=_25*0.01(I_XJ2 0.039
XI =6.366(I-X I )2 ~0.674 3 CAl = C Ao (1- XI) = 0.00326moi / dm Reactor 2:
XJ2
kCAI(lX 2 = -------- ~ X 2 = 0.507 v CA2 = CAl (1- X 2) = 0.00326(1-· 0.507) = 0.001607moi / dm 3 Reactor 3:
X3 = kC A2 (1- X 3)2 ~ X 3 = 0.387 v C A3 = C A2 (1- X 3) = 0.001607(1-- 0.387) = 0.000985moi / dm 3 Reactor 4:
C A4
kC
(1 ~ X)2 4
~ X 4 = 0.305 v = CA3 (1- X 4) = 0.000985(1-··· 0.305) =0.000685moi / dm 3
X4 =
A3
Bounds on conversion: 3 tanks
X
= C Ao -. CA3
= 0.9015
CAo
4 tanks
X
= C Ao -
C A3. = 0.9315
CAo
P14-5 (C) Change of the fluid velocity Let U=O.lcmls Re=50 and Sc=2
14-17
D
From gas phase dispersion correlation chart,
._a
udt
= 0.5
= 0.5ud t = 0.5 * 0.1 * 5 = 0.25cm 2 / s Bo = UL = 0.1 * 200 = 80
Gives
Da
Da 0.25 Bo n=-+1=41 2
The conversion is close to the one PFR 2nd order reaction: k=25dm3/(mol's) 't=l!U=2/0.00 1=2000s Da=k'tC Ao=500
X
=
Da =0.998 l+Da
Let U=100cmls Re=50000 and Sc=2 From gas phase dispersion conelation chart,
D _a
udt
= 0.21
=0.21udt = 0.21 *100* 5 = 0.25cm 2 / s Bo = UL = 100 * 200. = 190.5
gives
Da
Da
105
_ Bo 1 - 190.5 n--+ - + 1- 96
2
2
nd
The conversion is close to the one PFR 2 order reaction: k=25dm3/(mol's) 't=l!U=2/1=2s Da=k'tC Ao=05
Da X = - - - = 0.333 l+Da
P14-5 (d) Change of the superficial velocity
Re
= dtu = 0.2*4 = 80 V
0.01
From packed bed dispersion con elation chart,
DE ud p
_ a-
= 0 ..55
0.55ud p 0.55 * 4 * 0.2 D a=---= =1.1 E 0.4 Bo = UL = 4 * 200 = 727 Da 1.1 Bo 727 n = .- + 1 = - + 1 = 364.5
2
2
14.. 18
The conversion is close to the one PFR 2 nd order reaction: check k=25dm3/(mol's) .=1!U=2/1=2s Da=k'ICAo=O.5
Da X = - - - = 0.333 I+Da
P14-5 (e) Liquid instead of gas Part (a)
Re = dtu = 5 * 4 * 0.001 = 0.2 V 0.1
Sc=~-=~=2*107 DAB
5e-6
This is off the scale of the graph for liquid phase dispersion, hence Da cannot be evaluated .
P14-6 (a) Peelet numbers From Example 13.2 cr2=6.J9min2 and tm =5.15min Closed:
=_~ __ 2_(I_e-pe')---:;Pe P 2 r
(j2 2
tm
er
Per
=7414 •
Open: (J"2
2
8
Per
Per
--2 =--+---2 -~Per =11.68
tm
P14-6 (b) Space-time and dead volume T
tm =. -----.= 4.40 1+2/ Per
= T*V o = 263.8dm 3 VD =V - Vs = 156.2dm 3 Vs
156.2 %deadvolume = --- = 37.2% 420
P14-6 (C) Conversions for ]'t order isomerization Dispersion model Da=k't=O. 927
14-19
FW
Da q = 1+-- = 1221 Per
= _ X
4qexp(Pe r 12) = 0 570 1 (1 + q exp(Perq 12)- (1- q)2 exp(- Perq 12) .
Y
Tanks-in-series '['2
n = - 2 =4.35 (J
P14-6 (d) Conversions PFR and CSTR PFR: X = 1- e -k1 X=0,604
CSTR: X
= 1--1-
X=OA81
1+1k
I~~~~ __________~I_~_.~_~_-~_
-?
+X~CS1R
XPFR
I 0.604
.--~.--------------
0.481:=J .----------~
P14-7 Tubular Reactor 1st order, irreversible, pulse tracer test --7
(J'2
= 65
S2
and t m = 10 s
For a 1 order reaction, PFR: X =1- e-k~ =0.98 We need 't' and k. There being no data for diffusivity (Schmidt number) Daand hence Per cannot be obtained using tubular flow correlations. st
Calculate V= 3 m * 25 dm2 * (01 m/dm)2 = 0.,75 m3 2 3 't' = 0.75 m / (3*10 m3/s) = 25 s This is greater than tm so channeling is occuning Calculate, k
1
In-k=
1- X = 391125 s = 0.,156 S-I '['
Therefore assume closed vessel dispersion model: tm ='t' = 10 s space-time Calculate, Pe
14-20
2
(Y2
-2
= ---
Per
tm
2 --2
(l_e-per ) = 65/102 = 065
Per
Iterating -7 Per = 1.5 Calculate, X using the measured V and
4q X = 1-
Vo
giving 't = 25 s
exp( P;,) ------=
[(I+q)' exfe~*q )]+-q), ex{ -p~ *q)]
Da = 't k = 25 * 0.156 = 3.9
q=
FW Da Pet
1+ -- =
Pf.1*3.9 1+ = 3.376 1.5
4 *3.376eXpC;5) X = 1--
[(I + 3.376)'
ex{1.5 *~.376)] -[(1- 3.376)' ex{ -1.5 :3.376)]
= 0 . 88
Conversion for the real reactor assuming the closed dispersion (X=0.88) model is less than for the ideal PFR (X=0_.98)
V
'tideal
= - =25 s Vo
V=VD+Vs Vs= a *V Note, 'tRTD = 10 s = a * 't , a=10125 =0.4
./DisperSion Volume ~---------------------------7~---~
Dead Volume = (1- a) V= 0_6 V, Vs = OAV Use tm = lOs, Da = tm * k = 10*0.156 = 1.56
4*1.56 1.5
q= 1+----=2.27
14-21
XD = 1- (19.22/58.38) = 0.67
P14-8 (a) E(t):
06~~--·~------~
0.4 E(t)
0.:2
:2
4
t
Conversion T-J-S and Maximum Mixedness Model X T_I _S=05
1
r
For a fir st order reaction Xmm=X I -I-S= 1-- - - - -
(1+: k =
= JE(t)tdt=
1:=tm
2
4
2
Jo.25.t dt+ J(1-O.25.t)tdt=2min
0 0 2 (Y2
=
=
o
0
2
2
= JE(t)(t-t m)2dt = JE(t)t dt-1: 2 =-min 2 1:
3
2
n=-=6 (Y2
From the conversion it is possible to determine k at 300K:
14-22
(
k=
-1)
1
~
. = 0.367 min-I
T
n The conversion at T=31O K is given by:
k310 .
= k300 exp(
R = 1.986
E _ E ) R300K R· 310K
= 1.422 min -I
ca"-
molK
P14-8 (b) Complex Reactions
dC A dT-=-klC A -k3 C A dCB - = klC A -k 2 C B dT
!-CC =k C dr 2 B dCD.=k C dT 3 A Where k l = k2= k3=Ol min· l
dC A
dC A
"r L.
__
d/l, -
+ (C _ C ) E(/I, )
iA
-;iT = -(-kICA
A
Ao
--k3 C J+(C A
dCB _ ----(kICA -k 2 C B )+(CB
d/l,
I-F(/I,) E(/I,)
-C J1="F(/I,) A
-cBJ
E(/I,) () 1-·-F A
14-23
Changing variable: A=z-4 See Polymath program P 14-8-bpol
POLYMA TH Results Calculated values of the DEQ variables Variable z ca cb cc cd F k1 k2 k3 rc ra rb t2 rd cao cbo ceo cdc t1 lam E1 E2 E EF
initial value 0 1 0 0 0 0 . 9999999 0.1 0.1 0.1 0 -0 . 2 0.1 4 0.1 1 0 0 0 2 4 1 0 0 0
minimal value 0 0.6793055 0 0 0 2.854E-06 o. 1 0.1 0.1 0 -0 . 2 0.0537147 4 0.0679305 1 0 0 0 2 0 0 0 0 0
maximal value 4 1 0 . 142158 0 . 0181892 0.1603473 0.9999999 0.1 o. 1 0. 1 0 . 0142158 -0.1358611 0.1 4 0. 1 1 0 0 0 2 4 1 1 0 . 4965476 25 . 277605
ODE Report (RKF45) Differential equations as entered by the user [1] d(ca)/d(z) = -(-ra+(ca-cao)*EF) [2] d(cb)/d(z) = -(-rb+(cb .. cbo)*EF) [3] d(cc)/d(z) = +rc+(cc-cco)*EF) [4] d(cd)/d(z) = -(-rd+(cd-cdo)*EF) [5 J d(F)/d(z) = -E Explicit equations as entered by the user [1] k1 = 0.1 [2J k2=0 . 1 [3J k3=O . 1 [4J rc=k2*cb [5] ra = -k1 *ca-k3*ca [6] rb = k1 *ca··k2*cb [7] t2=4 [8] rd=k3*ca [9 J cao = 1 [10] cbo=O [11] cco=O [12) cdo = 0 [13) t1 =2 [14J lam=4-z [15 J E1 = D..25*lam [16 J E2 = 1-0 . 25*lam [17) E = if (lam
14-24
final value 4 0.6793055 0.142158 0 . 0181892 0.1603473 2.854E-06 0. 1 0.1 0.1 0.0142158 -0.1358611 0.0537147 4 0.0679305 1 0 0 0 2 0 0 1 0 0
0.20r-------------------, 0.16 0.12 0.08 0,04
O.OO"--~--~----~----'
0,,0
0.8
1.6 z
2.4
3.2
4.0
P14-8 (C) Use FEMLAB for thefull solution Complex reactions and Dispersion Model.
dC A dT
= -klCA -- k 3 C A
dCB - - = klC A -k2 C B
dT
dCc=k C dT 2 B
dC:!2. = k C dT 3 A
P14-9 (a) FromP13-4:
tm
=T = {2' min,
V;
0"2
= -~ = O.159min2 and k= 0 . 8 min- 1 21l
Tanks-in-series For a flrst order reaction the conversion is given by: From Example 14-1
X T-/-S
= X seg = 0.447
P14-9 (b) Closed-closed vessel dispersion model For a flrst order reaction the conversion is given:
4qexp(Pe,I_2) ____ (1 + q)2 exp(Perq 12) -- (1-- q)2 exp(- Perq 12)
X =1-.---
14-25
2 2 ( -Pe ) 0.25=----1-e ' Pe, Pe, 2 Per=683
Da = rk
= 0.638
1l=J1+ 4Da = 1+ 4*0.638 =1.885 Per
6.83
X=0.41
I
XDispersion
0.41 .
Referling to P14.2 the approximate formula for the T-I-S is not good approximation.
P14-10 (a) Combination of ideal reactors The cumulative distribution function F(t) is given:
F(t)
05
0
0
I
20
~J
40
t[min] The real reactor can be modelled as two parallel PFRs:
The relative E(t)
= {l_&(t -- 1'1)+ ~ &(t - 1'2) 4
4
P14-10 (b) Model parameters For two parallel PFRs, the parameters are L1=10 min and LF30 min,
P14-10 (C) 14--26
Conversion Fao! =1I4Fao and Fao2=3/4Fao ,second order, liquid phase, in eversible reaction with k=O ..l dm3 /molmin-! and C Ao=1.25 mol/dm3
CAl
= CAo -
krICAo CAo 1+ krIC Ao
= 0.556mol I dm
3
kr2 C Ao 3 CAo = 0.263mol dm 1+kZ"2 C Ao 1 3 vC Ao --VC A1 --VC A2 X= 4 4 = 0.731 vC Ao
C A2 = CAo -
P14-11 (a) FromP13-6 tl =
5 min
E(t)
t lif t < tl ·-2 tl E(t) = - - \ (t - 2t1 )if tl 5, t 5, 2tl tl
tm
= tland
a2
= t~ = ~5 = 4.16' 6
0.1
6
o otherwise Tanks--in-series
10
Z"2
n = -= 6 (As in 2
a
PI4.8) Second order reaction, liquid phase, kC AO=0.2min-!
C Ai -C Aout
kC Aout
2
=Z"
Solving for CAou1
C
Aout =
-1 + ~1 + 4k-xC Ain --
2kr
Six-Reactor System: T6=T/6 The Damkohler number is given by Dao= kC Ao't6=0.167
Da. I
=
--I +
F+4Da"
n
V-.l}
2
with i=1..6 In the following table the exit concentrations fOI each reactor are reported: Da! Da2 Da3 Da4 Das Da6
--
-
0.145 0.129 0.116 0.105 0.095 0.088
14-27
--
--
x = CAO -CA6 = Da o -Da6 =0.473 CAO
Dao
P14-11 (b) Peelet number The Peclet number
Ul Pe, = Da
For Closed-Closed System (no dispersion at the entrance and at the exit of the reactor P13 . 6)
~2 =~=~.--~(l-e-pe,) r2
p.2 er
Pe ,
6
Pe..=lO..875
P14-11 (C) Conversion Linearizing the 2st order according 14..2 (g): 2 ~ k C AO C - k'C , -- kC A= -rA - - AA
2
it is possible to obtain an approximated solution;
k'= k C AO = 0.lmin-1 2 Da =rk'=0.5 Pe..=10.875
(flom Part(b»
q=~l+ 4Da Per
-1-
X-
=
~*0.5·=I.088
Vl-rlO.875
4qexp(Pe, 12) (1 + q)2 exp(Pe,q 12) - (1- q)2 exp(- Pe,q 12)
X=0.382 Full solution can be obtained with FEMLAB
P14-12 (a) Remembering the physical meaning of the Peclet number:
Pe =. r
rateo! transport by convection rateo! transport by diffusion or dispersion
lim X Disp
Pe,---7=
Ul Da
= X PFR
For a first order reaction:
14-28
4qexp(Pe,/2)
X=l-·----------~--~~--~-----------
(1)
(1 +qY exp(Pe,q 12) --(l-qY exp(- Pe,q 12)
q~~1+-4D-a. Pe,
Da = 'lk For Per»1 (or for small vessel dispersion number)
~
4Da q = 1+·--Pe,
( J3 = 1+2----2(J --- 2+0---. Da Per
Da Per
Da Pe,
Introducing (2) into (1) and neglecting terms
2kr ( 4 .( 1 +--- ) e Pe
=1-e
0(~-J2 Pe,
4{1+~:)Jn
X = 1---
X
(2)
[
-kH (tt)' Pe,
pe,(!+ 2D!!.. 2Da'_)) 2
Pe
"
Pe
J
P14-12 (b) _k1:+{k1:j2
X pFR =1_e-k 1: and
X DisoPe»! = 1- e
Per
In order to achieve the same conversion:
P14-12 (C) Defining:
Pe, = UI PFR = 0.1 Da
J 2J
4q exp(per ( ___ I 1 0.99(1- e-
k
1:)
IpFR
= 1-
(1 + qYexP(pe,(_1-J q I2) - (l-qYexp(- pe,(_I_Jq I2J IpFR
IpFR
14-29
The following figure shows the solution as function of ktpFR fOl different values of lJlpFR ·
P14-12 (d) Starting from 14 . 12.1 and subtracting the conversion for a PFR:
InS.L= kr CA p1ug Pe For small deviations from the plug flow:
P14-12 (e) According P12.3 dispersion doesn't affect zero order reaction..
P14-13 From P13.J9: 2 2 T=lO min and a =74min
for 0 ~ t ~ 3 El (t)
f+ 9.9505.10-
2
= ---LI675 .1O-3 t 4 + L1355 .10-2 t 3 -
4.7492 .1O-2 t 2
t
for 3 ~ t ~ 20 E2 (t)
= -1· 8950 .10-6 t 4 + 8.7202 .10-5 t 3 -- L1739 .10-3 t 2
E(t) = - L7979 .1O-4 t + 0.092343
for 20 ~ t S 60 E3 (t)
= L2618 .10-8 t 4 -
2.4995 .1O-6 t 3 + L8715 .1O-4 t 2 -
6.3512.10- 3 t + 0.083717 for t > 60 0
P14-13 (a) 2nd order, kCAO=O 1min-!, C Ao=lmolJdm3, Segregation Model Segregation Model 00
X
= fX(t)E(t)dt o
kC Aot 1+ kCAot
Where X (t ) = --.:.=--
See Polymath program P 14-13-apoJ POLYMA TH Results
Calculated values of the DEQ variables Variable t Xbar
initial value
o o
minimal value
o o
maximal value 70 0.4224876
14-30
final value 70 0.4224876
0.1 0 0.092343 0.083717 0 0 0
kCao E1 E2 E3 E4 E X
0.1 -2.436E+04 -21.261016 L949E-05 0 0 0
0. 1 0.0836855 0 . 092343 0.083717 0 0.0836855 0.875
0.1 -2.436E+04 -21.261016 0.0017977 0 0 0.875
ODE Report (RKF45) Differential equations as entered by the user [ 1] d(Xbar)/d(t) = E*X Explicit equations as entered by the user [1] kCao = 0.1 [2] E1 [3] E2 [4] E3
=-0.0011675*t"4+0.011355*tA3-0.047492*tA2+0 . 0995005*t =-1.8950*1 QA(-6)*t"4+8.7202*1 OA(-5)*tA3-1 . 1739*1 QA(-3)*tA2-1 . 7979*1 OA(-4)*t+0.092343 = 1.2618*1 OA(-8)*t"4-2.4995*1 OA(-6)*tA3+ 1.8715*1 QA(_4)*tA2-6 . 3512*1 QA(-3)*t+0 . 083717
[5] E4 =0 [6] E = if(t<=3)then(E1 )else(if(t<=20)then(E2)else(if(t<60)then(E3)else(E4))) [7] X = kCao*t/(1 +kCao*t)
P14-13 (b) 2nd order; kCAo=O.lmin·1, CAo=lmolldm3, Maximum Mixedness Model
Rate Law: - r A
= kC A 2
C A =C Ao (1-X)
= kC Ao 2 (1- X Ywhere k=O.l dm3 /molmin
rA
~=kC (1_X)2 C
Ao
Ao
dF .--= --E(Z)
where z=6Q.··).
dz
See Polymath program P 14··13-b.pol POL YMA TH Results
Calculated values of the DEQ variables Variable --z X
F
Cao lam Ca k ra E4
initial value 0 0 0.99 1 60 1 0. 1 -0.1 0
minimal value 0 0 -0.010344 1 0 0 . 52273 0.1 -0.1 0
maximal value 60 0.4773052 0 . 99 1 60 1 0.1 -0.0273247 0
14-31
final value 6()"""--0 . 4047103 -0.010344 1 0 0.5952897 0.1 -0 . 035437 0
-1.284E+04 -9.8680524 2.228E-05
E1 E2 E3
o
E
-1 . 284E+04 -9.8680524 1.806E-05
o
0.0832647 0.092343 0.0837l7 0 . 0832647
o 0 . 092343 0.0837l7
o
ODE Report (RKF45) DifferentiaJ equations as entered by the user [1) d(X)/d(z) = -(raiCao+EI(1-F)*X) [2] d(F)/d(z) = -E Explicit equations as entered by the user [1] Cao = 1 [2] Jam = 60-z [3] Ca=Cao*(1-X)
=.
[4] k 1 [ 5 j ra = -k*CaA2 [6) E4 = 0 [7 J E1 = -0.0011675*Jam"4+0 ..Q11355*Jam A3-0.047 492*JamA2+0.0995005*Jam [8) E2 = -1.B950*1 OA(-6)*Jam"4+B.7202*1 QJ\(-5)*JamA3-1 . 1739*1 OA(-3)*JamA2-1 . 7979*1 OA(-4)*Jam+0 . 092343 [9 j E3 = 1 . 261B*1 OA(-B)*Jam"4-2A995*1 OA(-6)*Jam A3+ 1.B715*1 OA(-4)*JamA2-6.3512*10A(-3)*Jam+0 . OB3717 [1 ()) E = if(Jam<=3)then(E1 )eJse(if(Jam<=20)then(E2)eJse(if(Jam<60)then(E3)eJse(E4)}}
(Check F(t))
P14-13 (C) Tanks in series and 1st order reaction with k=O.lmin-1
For 1st order reaction is acceptable a non integer value of n for calculating the conversion
1
r
X =1----=o.527
(1+ : k P14-13 (d)
Dispersion models and 1sf order reaction with k=O.lmin-1 Peclet number open system (J"2
-
t m2
2 8 = --.. + - ~ Pe = 4.906 Pe Pe 2
FEMLAB application Peclet number closed system
2
(J"2
2 (
-=·-----I-e t'/'n Pe Pe 2 X
= 1-
q=
-pe) ~Pe=0.98 4q exp(Pe /2)
r -.--"----"-'----'-----
(1 + qy exp(Pe,q / 2) -
(l-qy exp(- Pe,q /2)
~+ 4Da
V~
Per
14-32
Da
= 'lk,
X=4.59
P14-13 (e) Dispersion models and 2nd order reaction with k=O.ldm3Imol·s and CAo=lmolldm3 Conversion Linearizing the 2 s1 order reaction rate according 14.2 (g): 2 CAO - k- r A -- kC A = - C A -- k'C A
2
it is possible to obtain an approximated solution;
k'= k
CAO
= 0.05s-1
2
Da
= 'lk,'= 30
Pe..=O.98
q=
(fromPart(b))
~+ 4Da =~1+ 4*30 =11.111 V Per 0.98 J
x = 1.-
4qexp(Pe,/2:-:-)_-:-___ (1 + qYexp(Pe,q 12) - (1- qYexp(-- Pe,q 12)
X=O . 998 Full solution can be obtained with FEMLAB.
P14-13 (I) Two parameters model (backflow) For tracer pulse input
dCI
~--=VICI2 -VICI l
dt dC 2 V2 ----= vlen dt
V OC T2
Defining:
a=
~
,fJ=}2. and
V
r=~
Vo Vo We arrive at two coupled differential equations: For tracer pulse input
ar dCI = fJ(C T2 dt
(1- a) dC 2 dt
-
ell)
= fJ(C n
- CT2 )
See Polymath program P 14-1 3-f.pol
14-33
P14-13 (g) 3
Two parameters model models and 2nd order reaction with k=O,ldm'lmol 's and CAo=lmolldm
= -ka'ZC~1 = -kC~2 (1- a}r
C AO + j3C A2 - j3CAl
j3C Al
-
j3C A2 - CA2
P14-13 (h) Table of the conversions
~X~l~I~S________I~X~M~M~______-4~X~s~eoL-________~X~_D~i~Sp~ers~io~n ____~~X~t~wo~M=~et~~s~--~
~0.527
0.405
0.422
0.5
?
P14-13 (i) Adiabatic reaction with Segregation Model 2nd order, kC Ao=O.lmin- l , CAo=lmoVdm3, Segregation Model Segregation Model
ax = kC (I-xl at AO T
= 320+ 150X
k =0.1 *exp(45000/8.314 * (1/320-lIT))
x = fX(t)E(t)dt o See Polymath program P141J-j ..pol
P14-14 (a) Product distribution for the CSTR and PFR in series T k2/kl 'tkICAO 5.0 0.2 TJ -~. 2.0--2 0.5 20 T3 0.1 200 T4 _.
Considering Anhenius equation and applying the following notation:
k'1 E, In--=--=-E
A,
RT
r1
We can write this linear system of 5 equations for the 5 unknowns (Ell' E 12 , E 13 , E 14 , AI):
k
-
-
_.
-
In.-..!L = EI2 - Ell = kl2
k
In·-1.!.. = E13 - Ell = k13
14-34
P14-14 (b-d) No solution will be given at this time . P14-15 (a) Two parameters model
Omin::;; t < 10 min C(t) = 10· (1- exp(-O.lt)) 10 min ::;; t C(t) =.5 + 10· (l-exp(-O.lt)) Omin ::;; t < 10 min 10 min ::;; t
Fz (t)
FI (t)
1 3
= 2 . (1--- exp( -O.lt)) 3
2 3
= - +-. (l-exp(-O.lt))
F(t)
t(min) The F (t) can be representative of the ideal p~ and ideal CSTR in parallel model:
14-35
node1
C{= Fractional volume 13= Fraction of flow
f3 = !3' a = ~-3'
7
= 10 min
•
7 PPR
= -Vi- = -a7 =10mIn V pPR f3
_ (I-a) _ _ ( )7 -lOmm 1- f3
7CSIR -
P14-15 (b) Conversion 2 nd order, vo=l dm 3/min, k=O ..ldm3/mol·min, CAo=L25molldm 3 Balance around node 1
fJv oC PFR + (1- f3)vOCCSTR
= VoC A
For the PFR: Second-order
= C AO .- C PFR =
X PFR
Where Da PFR
C PFR
Da PFR -- = 0556
1+ Da pFR
C AO
a
= k f3 1C Ao =1.25
= 0.556mol / dm 3
For the CSTR:
X CSTR
=
C AO
- CCSTR C AO
=
I-a
(1 + 2Da cSIR ) - ~1 + 4Da cSTR2Da cSTR
Where Da CSTR = k --1C Ao =
1- f3
CCSTR
Where
1.25
= 0.725mol / dm 3 CA = fJC PFR + (1- f3)CCSTR = 0.668mol / dm 3
X = CAO -- CA = 0.465 C AO
14-36
= 0.42
P14-15 (c) Two parameters model
omin ~ t < 10 min 10 min ~ t
F2 (t)
Fl (t)
=0
2 r n ="31 +"3' {1- exp-- LO.2(t -lO)Jf
F(t)
The F (t) can be representative of the ideal PFR and ideal CSTR in series with a bypass between them: Vb=~Vo
.--.----~
node 1
a= Fractional volume (3= Fraction of flow
1
3
P = 3' a = 4'
40.
T
= -3 rrun
TpFR -- -~ -- aT -10' rrun VO
. TCSTR -- (1( a))T -- 5 rrun
I-P
Conversion 2nd order, vo=1 dm3/min, k=O.1dm3/molmin, C Ao=1.25moVdm3 Balance around node 1
PVoCb + (1- P)vOCCSTR
= PVoCPFR + (1- P)vOCCSTR
= VOC A
For the PFR: Second-order
14-37
pFR - Da pFR - 0 556 X PFR -- CAD - C - . CAD 1+ DapFR Where Da pFR =karC Ao =1.25 C PFR = 0.556mol / dm
3
For the CSTR:
X CSTR
=
C PFR ·-CCSTR
CPFR
=
(1+2DacsTR)-~1+4DacsTR 2Da cSTR
= 0.185
I-a = k--rC PFR = 0.278 1- /3
Where
Da CSTR
CCSTR
= 0.453mol / dm 3
C -C
X=~_A =0.61
CAD Where
C A = j3C PFR + (1- /3)CCSTR
=0.487mol / dm 3
P14-16
PFRj--+--+1 :::r.
l~~J
VI,
l[
PFR
;L
(4)
(3)
CSTR
v. CS1R
t·,,.")
~
PFR
t-EJ
(6)
14-38
...
...-,
.... .1
A1
I
_-: PFR
J
,
PFR
Al
--+ I I
PFR
I I
PFR
-
I As
J
.~
or
(7)
PFR
Recycle
II
...
(8)
--1_-....J~+--~--..... z=o
Dead Zonas
z=l
(10)
(9) LFR
CSTR
LFR
(12)
(11)
P14-17 The presence of a minimum (mn2) imply the presence of a radial temperature profile that effects the reaction rate and determines the deviation from the concentration profile given by Eq . (14.51) . The enthalpy balance allows the identification of the dimensionless thermal parameters: Defining
T iJ=---~sot
C If/=._A CAO
U U =Uo
14-39
qJ=O where
013 OqJ hR
_ 013
-=0
Bi = -
Rt
= Bi(l- 13)
oqJ
ratio convection-radial conduction
Ak ratio conduction-enthalpy flux
p
~Hreac
--...:.="'--
dimensionless adiabatic heat
CpTcoolant
Da(tJ) = k(tJ)C~o'l" A pruametric study of the numerical solutions ofEq . (1451) and the enthalpy balance would give the "exact" conditions for a minimum in the concentration.
P14-17 (a) A decrease in the thermal conductivity of the reaction mixtures determines an increase of the bulk fluid temperature and can determine a minimum in concentration .
P14-17 (b) Overall heat transfer coefficient increases For a given flux this determines less deviation from the isothermal case: no minimum
P14-17 (C) Overall heat transfer coefficient decreases ." For a given flux this determines more deviation from the isothermal case and the possibility of a minimum.
P14-17 (d) The coolant flow rate increases;· The external heat transfer coefficient increases, this implies a lower wall temperatme and a resulting lower temperatme in the reacting mixtme and less deviation hom the isothermal case: no minimum
P14-17 (e) The coolant flow rate decreases: The external heat transfer coefficient increases, this implies an increase in the wall temperatme and a resulting higher temperatme in the reacting mixtme that can determine a minimum in the concentration . Below shows how to use FEMLAB for this problem. Femlab Screenshots for the baseline case (1) Domain
14-40
o
Axis equal
>-r-z limits··----·· ..· ---'- ... '
phi
~Auto
[[--OK ..... ,II (2) Constants and scalar expressions - Constants
- Scalar expressions
14-41
canceCJ
I
Apply
!
N~me
~~
~--~x;;r~~;i90- -~- -~~-~~-~~- ~-~ .-~~-~--~-
----~~---
~~-~~-------------------l
--+ti\
~?:i;~~::aY2)---~-----~-------~ ------.----. - -·-------·-·---·---·-·-1
kA-------[~A-O.:-~A)/CAO~-·---·-··-·-·-·-·-----····---.
_____ .____ y____--_--"._ ..-.-.. ---.------.-".-..".--m-'.----'-------..-------.. . . ---.'-----"1 pB . ___ .... _~~~O·X~--_- .. ______ .. _ ...____________________ ~ CC i2'cAO'xA i i-----·--i ---.-~-.".---~----~---------~~---------.-,-.---~----~----4 ~ A' exp( -ElRfTl!h..~at·(cA·cB-c~JKeq) _____.___ ~ ____________ _l Keq KeqO'exp(dHrxlR'(1 1303-1 fT)) I' ;-..,..; --------< p l:-rA)'(-dHrx) I ;~
--.~.-.---~--.--
n-~'
____ r
r---·-----~·-·I-~--·---·
~_._.
i
--.-.-.~-
. ----- -.-----.
..--..--..-.-..
~----.-~- ~----~"1
__. _____ .•___""'_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
I'
--"'u;;;.;;;;;;
b_,,_~
r
OK
l t cancel] [
(3) Subdomain settings Physics (Mass balance)
V'(-DVcA+cAu) = R, cA = concentration
Sub domain selection
r--L~-~d:~J
Library material: i
Valuell'xilression Description
[1
Cits
(~1 D isotropiC
~!.!
() 0 anisotr opic
R u [ ] Select by group
v
r;;'] Active in this domain
(Energy balance)
14-42
Time-scaling coefficient
.1Ditfusion coetficient Ditfusion coeUicient
Apply
,'Subdomain selection~-
':1
r;;;;;;i~~llnit [,,,,,,s,',t i ,-Thermal properties and heal souI'ceslsink:s-' . I
!I
i!
II i II
[LMCLJ
Library malerial:
I Quantity ,
15
I
1
ValueJl;xpression
ts
Thermal conductivity
10k (isotropic)
I
i!
I
i! -II
() k (anisotropic) c=::..::.'"-.=-..______.J Thermal conductivity
I
II
I , ! I
I!
i ii i , '
p
;:====;
U
'1,1
: i
i 'D'~~I:ct:y group
l: b
',
I 1 , J,
L!'JAoIive in thiSd!,~ain,
II
hiN o .;
r.,_A,rt,ificial,_Dlt,fu,~i;__.~..J" ~.~..:_ _ _ _•__~ ... '..
-- Initial Values
(Mass balance) cA(tO) = cAO (Energy balance) T(tO) =TO
-- Boundary Conditions @ I' = 0, Axial symmetry @ inlet, cA = cAO (for mass balance) (for energy balance) Boundary cpnditions" Boundary condition: [Heat flux -
!, ,,---
,,~-~-
ValuelExpression Description
k~;~E~!!h:l~-
jlnward heat flux Temperature
@ @
outlet, Convective flux wall, Insulation/Symmetry (for mass balance) (for energy balance)
I'-Boundary ,"" conditions '"
'
IBoundary condition:
~~~f~l!..~·:""
Qi:t~ntitY
,ValuelExpression Description
qo
[~~~f~f~r
To
Description Time"scaling c,oefficienl,'
Inward heat flux
1 Temperature
.. ,J
14·43
Heal source
P14-18 Vary the Peclet number and the reaction order in laminar flow (Example 14-3(c)) n- 1 ....l·CAO Da --UI-
-
n- 1 -'£kCAO
Uo
Pe = UoLI DAB L = 6.36m R=O.05m 3 k = O.25dm I mol. min U o = I.24mlmin C AO =O.5molldm 3 DAB
= 7.6xlO-5 m 2 I min
n -t-0.1 Attempt to evaluate noninte gral power of negative number! 0..5 Attempt to evaluate nOllinte gral power of negative number! 08
1--
-
1
15
1--
-r----
2
2.5
--- 1----
Conversion Pe 1.04xl03 I 1.04x104 I 1.04xlOs I --1.04xld' I 1.04x107 I 1.04x103 I 1.04x104 I 1.04xlOs I 1.04xlO° I --1.04x107 I j 1.04xlO 0.824 4 1.04x10 0.818 1.04xlOs 0.783 --- 1.04xlO° I 1.04x107 I 1.04x103 - 0.721 4 1.04x10 0.717 0.687 1.04x105 -1.04x106 0.653 1.04x107 0.641 --1.04xlOJ 0.526 1.04x104 0.522 1.04xlOs 0.503 1.04x106 0.481 1.04x107 0.472 j 1.04xlO 0.390 4 1.04x10 0.388 1.04xlOs 0.374 1.04x106 - 0.359 7 1.04x10 0.352 1.04x103 0.292 1.04x104 0.290 1.04xlOs 0.281
14-44
Parameter 100* DAB ---10* DAB DAB O.I*D AB O.OI*D AB 100* DAB -10* DAB DAB ------O.I*DAB -------O.OI*DAB -100* DAB 10* DAB DAB O.l*DAB -O.OI*D AB 100* DAB 10* DAB --DAB 0.1 *DAB 0.01 *DAB 100* DAB 10* DAB - -DAB 0.1 *DAB 0.01 *DAB 100* DAB 10* DAB DAB 0.1 *DAB O.OI*D AB 100* DAB 10* DAB DAB -.~-----
-
3
-
1.04xl00 1.04x107 1.04x103 1.04x104 1.04x105
0.270 0.265 0.219 0.218 0.211 0.203 0.199 0.164 0.163 0.159 0.153 0.150 0.123 0.122 0.119 0.115 0.113
1.04xl06 3.5
1.04x107 1.04x103 1.04x104 1.04xlO~
1.04xlO° 1.04x107 4
-
1.04xl03 1.04x104 1.04xlO' 1.04xlO° 1.04x107
-
---
1.0
n 0 ..8 --*-·1 . 0 1..5 ---y--- 2 . 0 -.--11-
09
-_.!!!-__•
08 0.7
c
o
.i!1
..• ·111 ................._..• __ III
06 05
~
§
O.I*DAB O.OI*DAB 100* DAB 10* DAB DAB O.I*DAB O.OI*DAB 100* DAB 10* DAB DAB O.I*DAB O.Ol*DAB 100* DAB 10* DAB DAB 0.1 *DAB O.OI*D AB
0.4
2. 5 •. ·3..0 3..5 • 4. 0
·--·-----····T------·T·-·---T_···_··__ T
<.)
.. .....
03
...........
02 0.1
00·
1if
1·-
_........
..
, '''''''I
1if
...--. -.-...
, .. ""'I
1if
1if
III
, .. ""'I
1d
, .. ""'I
1if
Pe
(1) (2)
With the increase of the reaction order n, the conversion will decrease. The conversion will also decrease when the Peclet number increases. 5 When n is less than 1.5 and Pe is 10 , the diffusion in the system becomes more important.
Below show FEMLAB screenshots useful for this problem.
Femlab Screenshots (1) Domain
14-45
DAxis equal phi limits
~Auto
rmin: rmax: zmin: zmax:
(4) Constants and scalar expressions - Constants
Li-J;me . . .Expr~55iQn ~ ,--=f:~~~ ~·-·····,-··-t~~5
C·:---"--. :.~::.·=:·.~~:~,36 _. _. __.. k
,_~,~~
... ___N'_'_. __
.....
,05 __
~t
~
_____ ,_._____ "_
36
b,25
~~==~-=f~
-il:~~:::::~:··~ .-. .~_",.
__
A_
................... _ ..•_
L~~<~~~"~W"~H'_ ~:l~.
.25 .24
~.~_
_
.8 .,,,._.
.L__".~.__ . [~---J L_ can~:~J [_AJ:~L.J
- Scalar expressions
..;~
.-.-~-
1 [_.
r
+. -,_ . " -1
. -t-- .
L.... .....____.. __ .~t~
..._,,__.
(5) Subdomain settings Physics
14-46
s;;bd~n S~ttin~ - Convection flOd Diffusi~ V.(-DVcA+cAu)
=R, cA =concentration
rSpecies'~--""'------ C"
I
Library material:
I Quantity -,'.'
[. Load...
[~=::::.~====J IQAB~::::~~::-.-
D :nisotroPic
I . [ ] Select by group
~
l_C.!~!i5i;i Diff~SiO~~J
1.1.
- Initial Values
(Mass balance) cA(tO) = cAO - Boundary Conditions @ r = 0, Axial symmetry @ inlet,
r
, I
Boundary conditiohs
i Boundary condition:
I
I Quantity I
II @ @
c;:
No
I
~!t:::
I
Valuel1:xpre$$ion pescfiption
~~~~~~'::~·_-:_~-~·:.=.]concerrtration
~~~Aq'
_. '] In~~r~flux
outlet, Convective flux wall, Insulation/Symmetry
P14-19 (a) First order reaction Input Parameters:
n =1 R=0.05m L=6.36m k = 0.251 min U o = 1. 24m I min
CAO
I
ValuelExpression Description
°ts @ D isotropic
I0 ~ Active in this domaih
•• - •.•••••' - . - - . - - - - - - - -•••• --- •••-- •• - ••.• -- ..
= 0.5mall dm 3
14-47
Time-scaling
co~fficient
Diffusion coefficient Diffusion coefficient
Conversion: xA= 0..687 @ Open-vessel Boundary: -Nin =2*UG*(I-(r/Ra)"2)*CAG xA= 0. . 726 @ Close-vessel Boundary: -Ni·n =UG*CAG Conversion: xA= 0..755 @ Open-vessel Boundary: -Nin =2*UG*(I-(r/Ra),,2)*CAG xA= 0..778 @ Close-vessel Boundary: -Ni·n =UG*CAG I Variation of Da number Damkohler numberl Da
1--. -.
_.
_.
0..449 0..898 1.795 3.59 7.18 14.36 28.72
II. Variation of Pe numbel Peclet numberlPe f--.
I.G4e3 1.G4e4 1.G4e5 l.G4e6 1.G4e7
Parameter
Conversion Closed-vessel 0..40.4 0..527 0..658 0..778 0..870. _.- 0..930. 0..964
-
--
Open··vessel 0..287 0..439 0..60.6 0..755 0..862 0..928 0..964
8*UG 4*UG 2*UG UG DG/2 UG/4 DG/8
_.Conversion Closed-vessel 0..781 0..781 0..778 -0..772 0..770.
-
OJ)en-vessel 0..781 0..776 ..- i---------0..755 ---- ' - - - - - -0..740. 0..737
-----
IlL Femlab Screen Shots (1) Domain
,;;;;";:;;;;,'1. [lAxls!i Grid t:."."...... ,.;,
[ ] Axis equal phi
lim~s
~l Auto
(2) Constants and scalal eXPlessions - Constants
14-48
Parameter
11~~*g::-DAB ----0..1 *DAB 0..0.1 *DAB
--
. . ._--_.........._-_ ..-+.._ .._.....__ ...........__...._--_ .............. __. . . . _-_. -.--..- . --....--.-
[
OK
'1 [c~ncel
1C!fPiY~
- Scalar expressions
(3) Subdomain settings - Physics
I Vo( -ElVcA+cAii) = R. cA = concentration ;
Subdomain
f;Al;~itll Element I rSpecies
["l~;d~::-J
Library material: Quantity
ValuelExp(ession
DSelect~y group
rtJ Active in this domain
Description Time-scaling CQefficient
Qts
El isotropic
Eliffu.sion coefficient
El anisotropic
Eliffll~ion coefficjent
R
Re.action rate
u
r ..velocity
v
z.:velocily
[-.-.--------:J .. ArtificiaIEiffusion .. : ....
14-49
File
Edit·
Options Draw. Physics Mesh Solve Postprocessirig. I';IulfiphysJcs Help
Dri~j~I!_~~lfII~:i.~J~:~_~_gTi !~E~.ifL~~~. .~. ~~L!____ :-_.-
-.f!l...,
Max: 4.319
Sl61'ace: Concentration cA
\~,~:
._ _ _ _ .. _ _ _ )oc'
.---.-.-.------
@
~
;;.::
g
*'
35
~:!
iii /
05
004
002
0.06
0.Q8
0,1
rSNAP f
(0.12,6)
P14-19 (b) Third order reaction with k*C 2AO= 0,,7 min-1 n - 1 = _!::-..kC n - 1 Da = rkC AO U AO
o
Pe
= UoLI DAB
Input Parameters:
n=3 R=O.05m L =6.36m k = 2.8dm 3 I mol. min U o = 1.24m I min
CAO
= O.5mal I dm 3
DAB
= 7.6xlO- m I min
5
2
I Variation of Da numbeI Damk6hler numberl Da
-1-. 1----
0.255 0.51 1.02 2.03 4.06
-_.
Conversion
-
-
Closed-vessel 0.381 -----0.466 0.560 0.655 0.741
---
14-51
Parameter Open-vessel 0.253 0.375 0.506 0.628 0.730
-- f---
8*UO 4*UO 2*UO UO UO/2
II Variation of Pe number Peelet numberJPe
• 1.04e3 1.04e4 1.04e5 1.04e6 1.04e7
Parameter
Conversion Open-vessel 0.649 0.645 0.628 0.617 0.615
Closed-vessel 0.650 0.654 0.655 0.652 0.650
-
100* DAB 10* DAB DAB O.I*DAB O.OI*DAB
(c) Half order reaction with k= 0,495 (mol/dm3)1I2rnin-l
n--1 D a = rkC AO
-
~kcn-l U AO o
Pe = UoL/ DAB Input Parameters:
n=1I2 R=O.05m
L
=6.36m
= 0.495(mal / dm 3 )1/2 min-1 U o = I.24m/min k
CAO = O.5mal / dm DAB
3
=7.6xlO-5 m 2 /min
I Variation of Da number Damkohler numberl Da
1-------
1----------"-.--.-- 0.69 1.38 2.76 5.52 11.04 22.08 44.16
II Variation of Pe number Peelet numberlPe
1-L.....-'-.
Open-vessel 0.378 0.641 0.911 1 I I I
8*DO 4*DO 2*DO DO DO/2 DO/4 DO/8
Conversion _.
1.04e3 1.04e4 1.04e5 1.04e6 1.04e7
Parameter
Conversion
Closed-vessel 0.432 0.713 0.959 1 I I I
Closed-vessel 1 1 1 I I
Parameter Open-vessel 1 1 1 0.9995 0.9994
14-52
-
100* DAB 10* DAB DAB 0.1 *DAB O.Ol*D AB
--
--
._-
(d) The radial conversion profiles for various order of reaction (1) First order
n =1 L=6.36m R=O.05m k = O.25dm 3 / mol. min U o = I.24m/min
= O.5mol / dm 3 2 5 DAB = 7.6xlO- m I min CAO
L Open-vessel:
a--
- " ! - .
[
,!
~-L---L
o
0005
001
0015
002
0025
003
Note: 1: Z= LllO; 2: z=3L11O; 3: z=Ll2; 4: z=7L110; 5: z=9L11O II. Closed-vessel:
14-53
0035
0.04
0.045
005
~
3 ."+ 4 -lr"-·5
-1--
1 2 -+-3 ,:. 4
-'if-
08
-.-9---
--JC-
5
06
04
02
-0.2
o
0005
001
0015
002
0025
003
0035
004
0.045
005
Note: 1: Z= LllO; 2: z=3L11O; 3: z=Ll2; 4: z=7L11O; 5: z=9L11O
Something is not correct with z= LIlO because of the FEMLAB! (2) Second Older
n=2 R=O.05m L= 6.36m k = l.4dm 1 / mol. min Vo = 1.24m/min
= O.5mol / dm 3 5 2 DAB = 7.6xlO- m /min C AO
1. Open-vessel:
14-.54
08
0.7
06
05
';l; 04
03
02 L...---L-.............L _ _- ' - - _ - ' - - '
---L.
o
00050010015002002500300350.040.045005
Note: 1: Z= UIO; 2: z=3UIO; 3: z=U2; 4: z=7UlO; 5: z=9UlO
II. Closed-vessel:
-->;f-
1
--8- 2
08
--+- 3 4
--*-, 5 07
06
05
04
Note: 1: Z= LIlO; 2: z=3UlO; 3: z=U2; 4: z=7UlO; 5: z=9UlO
(3) Third order
n=3 R=O.05m 14-55
L = 6.36m k = 2.8dm 3 f mol. min U 0 = 1. 24m f min CAO = O.5moll dm DAB
3
= 7.6xlO-5 m 2 fmin
L Open vessel:
-'if--
1
-_.13-... 2
-1-3 -4
-"*- 5
06
05
~04
03
Note: 1: z= LllO; 2: z=3L11O; 3: z=Ll2; 4: z=7L11O; 5: z=9L110
II. Closed- vessel
14-56
--"r-
0.7
1
-a--2 -1-3 -c, 4 -It- 5
065
0.6
0.55
0.5
~ 045
0.4
035
03
0.25
o
I
I
0005
001
0.015
002
0025
003
0035
004
0045
005
Note: 1: Z= UlO; 2: z=3UlO; 3: z=U2; 4: z=7UI0; 5: z=9UlO
P14-20 (a) Higher Peclet number Curve 1 ("closer" to the one of an ideal PFR) has a higher Peclet number, because the cumulative distribution is more concentrated around the mean residence time.
P14-20 (b) Higher dispersion coefficient Curve 2 has a higher dispersion coefficient: a higher Peclet number corresponds to a lower dispersion coefficient
P14-20 (b) Large number ofT-loS Curve 1: Increasing the number ofT-I-S corresponds to increase the Pec1et Number (i.e . Bo=2(n-l))"
P14-21 (a)
14-57
node 1
F(t) and E(t) curvesfor
a = -\1]
= O.4and
fJ = --Vb = 0.2
\1
Vo
The analytical expression for E (t) is given by:
fJc5(t) t < 7] E(t)
=
-(t-r1)t-'2
{ (l-fJ)e
t'27]
72
14-58
1
E(t) 0.5
O~~~-----L~~~====~==~
o
T./2
2
4
6
8
10
Integrating we can obtain the analytical expression for F (t): T
P t
2
1- (1- p)e +-~)f:)
T
t~-
2
1-(1- (3)/e
{3=O.2
o 'T.I_ iJ
2 514'7
4
6
14-59
8
10
P14-21 (b) Conversion 2nd order; kCAO =0.5min- 1, r=2min Balance around node 1
(1- fJ)voCCSTR + fJvOC b = voC A For the PFR: Second-order
_ CAO - C pFR -_ DapFR -- 0333 . C AO 1+ Da pFR
X pFR Where
DapFR = kC Ao (
a
1- fJ
) T = 0.5
kC PFR = 0.333 min-1 For the CSTR:
X CSTR =
C PFR -- CCSlR .-'-'---=::':'-
(1 + 2Da cSTR ) - ~1 + 4Da cSTR-
- - - = 0.268
=
C PFR I-a Where DaCSTR = k - - rC PFR 1- fJ
2DacSTR
= 0.5
kCCSTR =0.366 min-1 kC A = (1- fJ)kC cslR + jJkCb = (1- fJ)kC CSTR + jJkC Ao = 0..393 min -1
x = kCAO-kC A =0.214 kC AO In absence of bypass (P=O) the conversion would be X=0245
P14-22 Two parameters model A possible two-parameter model is the PFR with Bypass (Vb) and Dead Volume (VD)
node 1
~-----_._14-60
To evaluate the conversion we write a balance around node Ion species A:
(I-P)voC PFR +PVOCb =VOCA PFR, 2nd order, liquid phase
r A =-kCAC B Where k=L5 m3/(krnoJ.rnin) and C Ao
X PFR --
= C Bo = 2mol / dm 3
Da pFR -0857 . 1+ Da pFR a ( ) l' = 6 I-P
CAO -C PFR CAO
Where Da pFR = kC Ao
C PFR = C Ao
-
X PFRC Ao = 0.286mol / dm
3
CA = (1- p)CPFR + PCb = (1- p)C PFR + Pc Ao = 1. 143moll dm 3 X = _CAO - C A = 0.429 CAO
P14-23
0.1
E(t)
o
o
10
20
t (nun) 00
By definition: JE(t)dt
=1
o
OOJE(t )dt = 0.1 * tl = 1 ~ tl = 20 mm . o
2 2
For a triangular E(t): t1 = - E(O) The analytical expression for E(t) is given by:
14-61
For 0 5, t 5, t} E(t) = -0.1 - t + 01 . t}
For t > t} E(t)
=0
P14-23 (a) Mean residence time
tm = =fE () t tdt= o
IR --t+0.1 0.1 Jtdt= [-0.lt 0
(For a triangular E(t): t m
t1
3
3t1
2
2
2
0.1 2 +--=-6-= 0.lt1 0.lt1 667 ' +0.lt - -]11 =--t . mIn 1 2 0 3 2
= -t} ) 3
Variance =
0-
2
= JE(t)(t-tJ2 dt = o
( 0.1 0 1J 2d Jr ---t+ . t t-t
I.
o
2 _ m -
[-.0.lt
4
0.lt
3
=_~Jt3 + 0.ltI _tm2 0.1· 20 4 13
]tI=20ffiill
+-3
4tI
tl
3
2
-tm
0
3 _
6.67 2
=
22.2 min
12
Assuming closed-closed system: '[' = t m
P14-23 (b) Peclet number (j2 I'I m2- - - ' 2 - (1 ·-·e -=1=
'['2
Pe-tO
Pe r
Pe r.2
-pe) r
And Per = 0
P14-23 (C) Tanks in series
The triangular E (t) can be interpreted as an approximation for t«1' of a CSTR where the continuity impose E(O)*T=2/3 instead of 1.
P14-24 The F(t) con be representative of a CSTR in parallel with two PFRs:
14-62
no del
First order reaction r::;;;;-.--Id-ea-/-c-s-r.-'R_-_-_ --.--_l-de-a-/-/am"-in-a-r--,c_ se-g-re-g-a-tio-n--
t====
__
flow reactor
Maximum Mixedness
14-63
Disper,sion
Tanks in series
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