Brian Vicente / Max Nori / H. Scott Fogler
Soludons Manual for
Elements of Chemlcal
Reactlon
Englneerlng Fourth Edition
\_I[r. :tl{:
Prentice Hall ihtemational Series in the Physl&l and Chemlcal ~. Engíneerlng Sciences
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A Complimentary Copy from
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Solutions Manual for
Elements of Chemical Reaction Engineering Fourth Edition
Brian Vicente Max Nori H. Scott Fogler
~
PRENTICE
HALL
PTR Upper Saddle River, NJ • Boston • lndianapolis • San Francisco New York • Toronto • Montreal • London • Munich • Paris • Madrid Capetown • Sydney • Tokyo • Singapore • Mexico City
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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 Gürmen, Project Manager (2002-2004) Susan Montgomery, Project Manager (1991-1993) Department of Chemical Engineering University of Michigan Ann Arbor, MI 48109-2136 ©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.en gin.umich.edu/-cre/icm
Please report problems to
[email protected].
Performance Number Interpretation: CRE 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.= 15l41~92 Score = 1.5*(62. 7) = 94 %
Note: 75% constitutes mastery. KINETIC CHALLENGE II CzBzzAzz
Score = 2.0 * AB..C 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. TICTAC TOE zDzCzBzA
Score = 4.0 * AB.C z = random numbers
Perf. No. = 718Q318l 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. GREATRACE zzzCzABz
Score = 6.0 * AB . C z = random numbers
= 777J_8Q7_8 = 6*(07.3) = 44
Pe1f No.
Score
Note: A score of 40 is needed for rnastery of this module. Performance Number Interpretation: CRE modules
iv
**** CONFIDENTIAL **** ECOLOGY AzBCzaaD
z = random numbers a = random characters
A gives info on r/\2 value of the student's linearized plot A=Y ifr/\2 >= 0.9 A=A if 0.9 > r/\2 >= 0.8 A=X if 0 . 8 > r/\2 >= 0.7 A=F if 0 . 7 > r/\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 Cenumber of deactivated data points if at least 1 point has been deactivated C=a randomly generated letter from A to Y if O 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 wa~ used D=2 if diff erential 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.= A7213DFJ 1) A=> 0.9 > r/\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*AB.C Final flow rate = 2*DE.F
Perf. No. = 2125.181913 conversion = 2*42.J = 84.2 flow rate = 2*31.2 = 62.4
Please make a pass/fail criterion based on these values .
Performance Number Interpretation: CRE modules
V
****
CONFIDENTIAL
****
ICMs with Dos® interface Module
Format
Interpretation
Example
A=2,3,5,7: interaction done B=2,3,5,7: intro done C=2,3,5,7: review done D denotes how much they
Perf. No.= 8027435 A: W orked on interaction B: Looked at intro 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.
Performance Number Interpretation: CRE modules
vi
Solutions for Chapter1 - 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.
Pl-1.
This problem helps the student understand the course goals and objectives.
Pl-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.
Pl-3.
Helps the student understand critica} thinking and creative thinking, which are two major goals of the course.
Pl-4.
Requires the student to at least look at the wide and wonderful resources available on the CD-ROM and the Web.
Pl-5.
The ICMs have been found to be a great motivation for this material.
Pl-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 Pl-15.
Pl- 7.
Straight forward modification of Example 1-1.
Pl-8.
Helps the student review and member assumption for each design equation.
Pl-9 and Pl-10. The results of these problems will appear in later chapters. Straight forward application of chapter 1 principles. Pl-11.
Straight forward modification of the mole balance. Assigned for those who emphasize bioreaction problems .
Pl-12.
Can be assigned to just be read and not necessarily to be worked . It will give students a flavor of the top selling chernicals and top chernical companies.
1-6
Pl-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 Pl-17.
Pl-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 (f) are printed out from the web. Part (g) is usually omitted.
Pl-15.
Shows a bit of things to come in terms of reactor sizing. Can be rotated from year to year with Pl-6.
Pl-16.
Open-ended problem.
Pl-17.
1 always assign this problem so that the students will learn how to use POL YMATH/MaLab befare needing it for chemical reaction engineering problems.
Pl-18.
Parts (a) and (b) are open-ended problem.
Pl-19 and Pl-20. Help develop critica! thinking and analysis. CDPl-A
Similar to problems 3, 4, 11, and 12.
CDPl-B
Points out difference in rate per unit liquid volume and rate per reactor volume. Summary
• • •
Pl--1 Pl-2 Pl-3 Pl-4 Pl-5 Pl-6 Pl-7 Pl-8 Pl-9 Pl-10 Pl-11
Assigned AA 1
Altemates
o o
AA AA 1
1-15
s s s o
1-7
Difficulty
SF SF SF SF SF SF SF SF SF SF FSF
Time (min) 60 30 30 30 30 15 15 15 15 15 15
I I
Pl-12 Pl-13 Pl-14
•
Pl-15 Pl-16 Pl-17 Pl-18 Pl-19 Pl-20 CDPl-A CDPl-B
-Read Only
o o s AA
s o o
SF SF
30 1
FSF SF SF SF SF
30 60 15 60 30 30 15 30 30
FSF FSF FSF
AA I
Assigned • = Always assigned, AA= Always assign one from the group of altemates, O = Often, I = Infrequently, S = Seldom, G = Graduate level Alterna tes In problems that have a dot in conjunction with AA means that one of the problem, either the problem with a dot or any one of the altemates 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) FSF = Fairly straight forward (requires sorne manipulation of equations or an intermediate calculation). IC = Intermediate calculation required M = More difficult OE = Sorne parts open-ended. * Note the letter problems are found on the CD-ROM. For example A= CDPl-A. Summary Table Ch-1 Review of Definitions and Assumptions
1,5,6,7 ,8,9
Introduction to the CD-ROM
l,2,3,4
1-8
Make a calculation
6
Open-ended
8,16
Pl-1 Individualized solution. Pl-2 Individualized solution. Pl-3 Individualized solution, Pl-4 Individualized solution, Pl -5 Solution is in the decoding
algorithm given with the modules .
Pl-6 The general equation for a CSTR is:
V= FAO -FA
=r,
Here rA is the rate of a first order reaction given by: IA= .. kCA Given : CA= 0.1 CAo , k = 0.23 min", v0 = 10dm3 min", FA = 5 .O mol/hr And we know that Fa = CAvo and Fs¿ = CAovo => CAo = F Aol vo = 0.5 mol/dnr' Substituting in the above equation we get:
V= CA0v0 -CAvo
=
(0.5mol/dm3)(10dm3 /min)-O.I(0.5mq[ldm3)(10dm3 /min)
(0.23min··1)(0.l(0.5mol / dm3))
kCA V=391.3 dnr'
Pl-7 t
=
JNA
_!_
-k-NA
dNA
NAO k = 0.23min-1 dNA
From mole balance:
--dt = rA·V 1-9
Rate Jaw:
Combine:
r-0
r o
ldt=
at 'l' =O,NAo = 100 mol and 'l' = 'l', NA= (O.Ol)NAo
k1 In (NAO NA
~ t=
J
1 =-ln(lOO)
0.23
mm
t=20min
Pl-8 a) The assumptions made in deriving the design equation of a batch reactor are: Closed system: no strearns 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.
e)
The assumptions made in deriving the design equation of PFR 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, A7 B
-rA is the number of moles of A reacting (disappearing) per unit time per unit volume [=] moles/ (dm'.s) 1-10
-rx' 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. lt 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 varíes 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.
_dtu.= F-J0
-
F.+ f r.dv J
]
No accumulation and no spatial variation implies
O=F.] 0 -F.1+ fr.JdV Also r¡ = Pb ri' and W = V Pb where Pb is the bulk density of the bed,
=>
f
O= (F10 - Fj) + r/AdV)
Hence the above equation becomes
W=-1Fo-F. 1 -·r1 We can also just apply the general mole balance as
dN
--1= (Fjo -
dt
f.
F1) + r/dW)
Assurning no accumulation and no spatial variation in rate, we get the same form abo ve:
w = __F] -F. _] 0
-rí
1-11
as
- . ..\: .·i·,... ,.l ....\t•. .. '. . .. ·.. .~l. . . • .... .. Fj
',e.'1'
•
F"JO
/I"-
Pl-10 Mole balance on species j is:
F10 -F Let M ¡
dN.
V
/
+ fr.dV
=--1
dt
J
o
= molecular wt. of species j
= w ¡o
Then F¡0M ¡
N ¡Mi
= mass flow rate ofj into reactor:
= mi = mass of species j in the reactor
Multiplying the mole balance on species j by M ¡
dN
fr¡dV = M¡ ·-dto
V
F10M ¡ -F¡M 1 +MI
1
Now M ¡ is constant:
F. M. -FM. + vfM 1
¡
¡O
¡
o
fM ¡r¡dV
-.av = d0!¡NJ) = d(mJ
1
1
dt
dt
V
W¡o --
W¡
+
o
=
Pl-11 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 part e first
1-12
[In flowrate (moles/time)] penícillin + [generation rate (moles/time)]penicillin - [Out flowrate (moles/time)] penicillin
= [rate of accumulation (moles/time)]penicillin Fp,in + Gp - Fp,out =
dNp
dt
Fp,in =O
(because no penicillin inflow)
frP.dV
V
Gp= Therefore,
frP .dV -
V
dNp
F, 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, for Com Steep Liquor with Fe= O
V
=
Feo - Fe_= feo - re - re
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: -· Rank2002 1 2 3 4 5 6
Rank 1995 1
2 4 3 9
-
Chemical H2S04 N2 C2H4 02 C3H6 H2 1-13
7 8 9
6 10 -
NH3 Ch P20s C2H2Cl2
10 -+ Chemicals like H2, P205 , C2H2Cl2 has come in top 10 chemicals and C3H6 has jumped to rank 5 now then rank 9 in 1995 .
Pl-12 (b) Ranking of top 10 chemical companies in sales in year 2003 and 2002: 2003 1 2 3 4 5 6 ,_______ 7 8 9 10
2002
Company
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 source: Chemical and Enginecring News
Chemical Sales ($ million 2003) 32632 30249 20190 8371 7018 6990 6606 6545 6029 5800
--
--
may 17,2004
-+ We have Chevron Phillips which jumped to 5 rank in 2003 from 8th rank in 2002 and Air Products coming to 9th rank in 2003 from 11 th in 2003 . -+Chemical sales of each company has increased compared to year 2002 from 9%(Eastman Chemical) to28.2%(Chevron Phillips) but Huntsman Corp . has a decrease by 2 . .9%.
Pl-12 (e) Sulfuric acid is prime importance in manufacturing. It is used in sorne phase of the manufacture of nearly all industrial products .It is used in production of every other strong acid . Because of its large 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(H2P04)2) , which is formed when rock phosphate is treated with sulfuric acid. -+Used as dehydrating agent. .+Used in manufacturing of explosives, dyestuffs, other acids, parchment paper, glue, purification of petroleum and picking of metals . -+ Used to remove oxides from iron and steel befare galvanizing or electroplating. -+ Used in non-ferrous metallurgy, in production of rayan 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 5.2lx
1010 Ib/yeru
Annual Production rate of benzene for year 2002 is 1.58 x 1010 lb/year 1-14
11
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
Usa.ge
Advantage
Disadvantage
Bate 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 throughout.
l. Liquid phase 2. Gas phase 3. Liquid Solid
1 . Small sea.le pdn. 2. Used for lab experimentation.
l. High Operating cost.. 2. Variable product quality.
L Liquid phase 2. Gasliquid 3. Solid liquid
l. Used when
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. Twophase reactions possible. 4.Good Control 5. Simplicity of construction. 6.Low operating cost 7. Easy to clean l. High conversion per unit volume 2. Easy to maintain (No moving parts) 3. low operating cost 4. continuous operation l. High conversion per unit mass of catalyst 2. low operating cost 3. Continuous operation
CST R
3.
Pharmaceuticals 4. Fermentation
agitation required. 2. Series Configuration possible for different
configuration streams
PFR
One long reactor ornumber of CSTR's in series. No radial variations. Conc. changes a.long the length,
1. Large Scale Prirnarily pdn. gas Phase 2. Fast reactions 3. Homogenous reactions 4 . Heterogeneous reactions 5. Continuous pdn.
PBR
Tubular reactor that is packed with solid catalyst particles.
l. Gas
l. Used primarily
Phase (Solid Catalyst) 2.Gassolid reactions.
in the heterogeneous gas phase reaction with solid catalyst e.g Fischer tropsch synthesis.
l.
1-16
l. Lowest conversion per unit volume. 2. By passing possible with poor agitation. 3 High power Input reqd.
l. Undesired
thermal gradient, 2. Poor temperature control 3 . Shutdown and cleaning expensive. 1. Undesired · thermal gradient. 2. Poor temperature control 3. Channeling 4 . Cleaning exnensive.
Pl-14 Given
A
= 2 * 1010 ft 2
TSTP
V= 4*1013 ft3
R
= 491.69R
T = 534 . 7°R
H = 2000ft Po= latm
es = 2.04 * 10-10 lbm3ol
= 0.7302 atm ft3
ft
lbmolR
e= 4*105 cars
Fs = CO in Santa Ana wind
FA = CO emission from autos
ft3
v A = 3000-·- per car at STP hr
Pl-14 (a) Total number of lb moles gas in the system: Po V N :=-R-T
N= (
latmx(4xl013 ft3) 3
J
0.73 atm.ft_ x534.69R
= 1025 x 1011 lb mol
lbmol.R
Pl-14 (b) Molar flowrate of CO into L.A. Basin by cars. VA
FA= yAFT=YA·---
CP0
RTSTP
F
1
= ~~OOft3 hr car
x llbmol x400000 cars 359ft3
(See appendix B)
FA= 6.685 x 104 lb mol/hr
Pl-14 (e) Wind speed through corridor is v = 1.Smph W = 20 miles The volumetric flowrate in the corridor is vo = v.W.H = (15x5280)(20x5280)(2000) ft3tb1 = L673 x 1013 ft3/hr
1-17
Pl-14 (d) Molar flowrate of CO into basin from Sant Ana wind
Fs
:=
voCs
= 1.673 x 1013 ft3thr Fs = 3412 x 103lbmol/hr
x2.04
X 10-10
lbmol/fr'
Pl-14 (e) Rate of emission of CO by cars + Rate of CO by Wind - Rate of removal of CO =
=
dNco dt
--
V deco dt
Pl-14 (f) t=
o,
eco
= ecoo
Pl-14 (g) Time for concentration to reach 8 ppm.
ecoo = 2 · 04 X 10-8 lbmol eco = 2.04 X 10-8 lbmol ft
=
4ft3 3
In
1.673xl013 ft hr
t = 6..92
3
'
ft 3
4
6.7xl04 lbmol +3.4x101!bmo~-1.673x1013 ft3 x2.04x10-8!.?!mol hr hr hr ft3 6_7x104 Zbmol +3.4xl03 lbmol _l.673x1013 ft3 hr hr hr
hr
Pl-14 (h) (1)
O
t0
eco
= 2.00E-10 lbmol/ft3
VO
=
L67E+12 ft3 /hr
hrs a = 3.50E+04 lbmol/hr tr = 72
b
= 3 OOE+04lbmol/hr
1-18
»o.srxio' lbmol ft3
F,
a
+ b
= 341.23 lbmol/hr
sin( JZ";)
+
V
= 4 . 0E+13
ft3
= vdccº
F1
dt
Now solving this equation using POL YMATH we get plot between Ceo vs
See Polymath program Pl-14~h-l .pol. POLYMATH Results Calculated values of the DEO variables Variable t
e
vO a
b
F
V
initial value
o
2.0E-10 l.67E+12 3.5E+04 3.0E+04 341. 23 4. OE+13
minimal value
o
maximal value
2 . 0E-10 1 . 67E+12 3.5E+04 3.0E+04 341.23 4. OE+13
final value
72
72
2.134E-08 1.67E+l2 3.5E+04 3.0E+04 341.23 4. OE+13
1.877E-08 1.67E+12 3.5E+04 3.0E+04 341.23 4.0E+l3
ODE Report (RKF45) Differential equations as entered by the user [ 1J d(C)/d(t) = (a+b*sin(3.14*t/6)+F-vO*C)/V Explicit equations as entered by the user [ll v0=1.67*10"12 [21 a= 35000 [3] b = 30000 [ 4 l F = 341.23 [5] V=4*10"13 3.0e-8 ....------
80
(2)
tr= 48 hrs
a
+
=
Now solving this equation using POLYMATH we get plot between Ceo vs t
1-19
vdccº dt
See Polymath program Pl-14-h--2.pol. POL YMA TH Results Calculated values of the DEO variables Variable t
initial value
e
vO a b V
o
minimal value
o
2 . 0E-10 1.67E+12 3.5E+04 3.0E+04 4. OE+13
2.0E-10 1 . 67E+12 3 . 5E+04 3 . OE+04 4.0E+13
maximal value 48 1 . 904E-08 l.67E+l2 3 . 5E+04 3.0E+04 4.0E+13
final value 48 1.693E-08 l.67E+12 3.5E+04 3.0E+04 4.0E+l3
ODE Report (RKF45) Differential equations as entered by the user E-1 l d(C)/d(t) = (a+b*sin(3.14 *t/6)-vO*C)/V Explicit equations as entered by the user [1] V0=1.67*10A12 [21 a= 35000 [3) b = 30000 [41 V= 4*1QA13 2.0e-8 ~---------------~
L6e-8
JO
40
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 .
Changing b
~ The amplitude of ripples is directly proportional to 'b'. As b decreases amplitude decreases and graph becomes smooth
Changing Yo
~
As the value of Yo is increased the graph changes to a "shifted sin-curve" . And as Yo is decreased graph changes to a smooth increasing curve .
Pl-15 (a) - rA = k with k = 0..05 mol/h dnr' CSTR: The general equation is 1-20
V= FAO -FA
=r,
Here CA= O.OlCAo, v0 = 10 dnr'rmin, FA= .5 . 0 mol/hr Also we know that FA= CAvo and FAo = CAoVo, CAo = FAol ve = 0..5 mol/dnr' Substituting the values in the above equation we get,
V= CA0v0 -CAvo k
=
(0.5)10-0.01(0.5)10 0.05
7 V=99dm3 PFR: The general equation is
dFA --=rA dV
=k,NowFA=CAv0andFAo=CAovo=>
Integrating the above equation we get V CA V JdCA = JdV => V
--ºk
CAo
O
«». =-·k dV
V
=_Q_(CAO ·-CA) 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,
Pl-15 (b) - rA = kCA with k = 0.0001 s-1 CSTR: We have already derived that
V= CA0v0 -CAvo
=r,
_ v0CA0(1-0.01) kCA
k = O . OOOls-1 = 0 . 0001 x 3600 hr-1= 0.36 hr-1
7
V = (1 O_dm3 / hr )(0.5mol / dny,3 )(O.~?) (0.36hr -t )(O.O 1 * 0.5mol I dm3)
PFR: From above we already know that for a PFR
dCAvo -rA dV
-kC -
A
Integrating
e dC v _vo f-A =- JdV k CAo CA O Vo ln CAo =V k CA
Again k = 0.0001s"1 = 0.0001 x 3600 hI'1= 0..36m-1 1-21
=> v =
2750
dm3
Substituing the values in above equation we get V= 127.9 dm3
Pl-15 (e)
- rA = kCA2 with k
= 3 dm3/moLhr
CSTR:
=r, Substituting all the values we get
V= (l0dm3 / hr)(0.5mol / dm3)(0.99) (3dm3 / hr)(0.01 * 0.5mol/ dm3)2
=> V = 66000 dm3
PFR:
dCAvo --TA --kC dV
A
2
Integrating
Vo k
cf dC~ C eA
=-fdV
k
O
AO
10dm3 / hr
=>V=
=> Vo
3dm3 / mol.hr
(
1
(_!_
I_)=V
eA eAO _ _J_)
o.oic AO eAO
Pl-15 (d)
r
dN - «, -_ r V
t-
AQ
A
Constant Volume V=Vo
f
= {Ao dCA rA
.tA -
Zero order:
t=![c k
AO
-0.00lC
AO
]= ·999cAo Ü.Ü5
=9.99h
First order: 1-22
= 660 dm3
t
= _!_ k
In( J = - -in(1 0.001
CAO CA
1-) .001
= 6908s
Second order:
t=,;1[ CA1
1]
- CAD
1 1] =31[ 0.00050.5 =666h
Pl-16 lndividualized Solution Pl-17 (a)
Initial number of rabbits, x(O) = 500 Initial number of foxes, y(O) = 200 Number of days = 500
dx - = k1x-k2.xy
(1)
dt
dy
dt
= k.xy> k4y
(2)
Given,
k; = 0.02day--I k2 k3 k4
= 0.00004/(dayx foxes) = 0.0004/(dayx rabbits) = 0.04day1
See Polymath program Pl-17-a.pol. POLYMATH Results Calculated values of the DEO variables Variable t
initial value
o
500 200 0.02 4 . 0E-05 4.0E-04 0.04
X
y
kl k2 k3
k4
minimal value
o
2.9626929 1.1285722 0.02 4.0E-05 4. OE·-04 0.04
maximal value 500 519.40024 4099.517 0.02 4. OE--05 4.0E-04 0.04
ODE Report (RKF45) Differential equations as enterad by the user ¡ 1J d(x)/d(t) (k1*x)-(k2*x*y) ¡ 2 J d(y)/d(t) = (k3*x*y)-(k4 *y)
=
1-23
final value 500 4.2199691 117.62928 0.02 4.0E-05 4.0E-04 0.04
Explicit equations as entered by the user
k1 = 0. 02 J k2 = 0. 00004 [3 l k3 = 0.0004 [4} k4=0.04 [1] [2
5000.------------------
4000 • rabbíts 3000
- foxes
2000
o
o
When,
100 tfina1=
200 t
800 and
160
k3
320
300
= 0.00004/(dayxrabbits)
t
480
640
800
Plotting rabbits V s . foxes
1-24
640 320
O 378
720
1061
rabdd113
1744
2086
Pl-17 (b) POL YMATII Results
See Polymath program Pl-17-b.pol. POL YMA TH Results
NLES Solution Variable X
--
y
Value 2. 3850387 3.7970279
Ini Guess
f(x) 2 53E-ll 1.72E-12
2 2
NLES Report (safenewt) Nonhnear equanons ( 1] f(x) = xA3*y-4*yA2+3*x· 1 = O [ 2 J f(y) = 6*yJ'2-9*x*y-5 = O
Pl-18 (a) No preheatmg of the benzene feed will dmnmsh the rate of reacuon and thus lesser conversions wtll be achteved .
Pl-18 (b) An mterpolation can be done on the logarithnnc scale to find the desired values from the grven data. Now we can mterpolate to the get the cost at 6000 gallons and 15000 gallons Cost of 6000 gal reactor = 1 905 x 1 a5 $ Cost of 15000 gal reactor= 5 623 x 105 $ Coat va Volume of reactor (log • log plot)
Pl-18 (e) We are grven CA is 0.1 % of minal concentration ~ CA= o OOlCAO
1-25
.
- -
1
·---·-
1
-
.'
··-t---
-·_J_ __
l .. .::1
log volume (gallona)
..
••
Also from Example 1 J,
Substituting v0 = 10 dnr' /min and k = 0.23 min-1 we get
V= 300dm3 which is three times the volume of reactor used in Example 1-3..
Pl -18 ( d) Safety of Plant. Pl-19 Enrico Fenni Problern - no definite solution Pl -20 Enrico Fermi Problem -- no definite solution Pl-21 lndividualized solution, CDPl-A (a) How many moles of A are in the reactor initially? What is the initial concentration of A? lf 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= RT
(20a_tm)(200dm3)
(l01.33kPa)
(8.3145 kPa.dm3 J(SOO) molK
= 97_Smoles
latm
Knowing the mole fraction of A (YAo) is 75%, we multiply the total number of moles (N10) by the yA:
molesA
= NA = 0.75x97.5 = 73.1 0
The initial concentration of A (CA
e _ Ao -
0)
moles _ NA - -volume V
0
_
-
is just the moles of A divided by the volume:
73.lmoles _ 0 l Id 3 • 37mo es m 200dm3
CDPl-A(b) Time (t) for a 1 st arder reaction to consume 99% of A.
dC dt
r =--A A
Our first order rate law is:
1-26
dCA dt
mo 1 e balance: --
=kt
= 1n(
CA CAo
J,
-k
= --kCA
=>
rfdt = cfA dCA O
e
A
CAo
knowing CA=0.01 CAo and our rate constant (k=0.1 min'), we can solve
4·61 for the time of the reaction: t = __ _!_ ln(0.01) = = 46.lmin k ü.l min" 1
CDPl-A (e) Time for 2nd order reaction to consume 80% of A and final pressure (P) at T = 127 C..
rate law:
dC __!!.. = -kC1 dt
I
CA
dC
O
CAo
A
=>-kfdt= f
mole balance: :. _
e/
1
1
CA
CAo
=>-kt=--+-
We can solve for the time in terms of our rate constant (k = 0.7) and our initial concentration (CA0):
+kt
= _ __2_ +-1CAo
t=-4-=
te¿
CAo 4
( 0.7dm3 I molmin )( 0.37mol/ dm3)
=15.4min.Todetermine
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:
N8 = Ne =0.8NAo N; = (0.25)N7o+ (0.2+0.8+0.8)NA0 = 0.25(97.6) + (1.8)(73.2) = 156.lmoles N1RT P=--= V
dm3atmJ(500K) molK ---~---=32atm 200dm3
(156.lmole)(0.082
-
CDPl-B Given: Liquid phase reaction in a foam reactor, A ~B Consider a differential element, By material balance
ó V of the reactor:
FA -(FA +óFA)=-rA(l-e)óV 1-27
r· '
Where, (1-
or:
e)~ V= fraction of reactor element which is liquid. -FA =-rA(l-e)~V dF _A= r (1-e) dV A
Must relate (-rA) to FA, where, FA is the total (gas +Iiquid) 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 Chapter2 - Conversion and Reactor
Sizing Synopsis General: The overall goal of these problems is to help the student realize that if they have -r A=fi)() they can "design" or size a large number of reaction systems . lt sets the stage for the algorithm developed in Chapter4.
P2-1.
This problem will keep students thinking about writing down what they leamed every chapter.
P2-2.
This "forces" the students to determine their leaming 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 leaming.
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.
P2-5.
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 altemative to problems 8, 9, and 12.
P2-8.
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.
P2-10.
Helps the students get a feel of real reactor sizes.
P2-11.
Great motivating problem. Students remember this problem long after the course is over.
P2-12.
Altemative 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 leaming.
CDP2-C
Similar to problems 2-8, 2-9, 2-12.
CDP2-D
Similar to problems 2-8, 2-9, 2-12. Summary
Assigned
• • •
•
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
o
Altemates
Difficulty
A A
o
o
M
s
M
AA
8,9,12
AA
7,9,12
s
s
AA AA
o o o o
7,8,9 9,B,C,D 9,B,C,D 9,B,C,D 9,B,C,D
FSF FSF SF SF SF SF FSF FSF FSF FSF
Time (min) 15 30 30 75 75 60 45 4.5 45 15 1
60 .5 30 30 45
Assigned
• = Always assigned, AA = Always assign one from the group of altemates,
O= Often, 1 = Infrequently, S
= Seldom, G = Graduate level
Altemates In problems that have a dot in conjunction with AA means that one of the problems, either the problem with a dot or any one of the altemates 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
=
Fairly straight forward (requires sorne rnanipulation of equations or an interrnediate calculation). IC = Intermediate calculation required M = More difficult OE = Sorne parts open-ended. FSF
*Note the letter problems are found on the CD-ROM. For exarnple A= CDPl-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
Critica! thinking
P2-9
P2-1 Individualized solution. P2-2 Individualized solution. P2-3 Solution is in the decoding algorithm given with the modules. P2-4 (a) Example 2-1 through2-3 If flow rate FAo is cut in half. vi = v/2, F1= FAd2 and CAo will remain same. Therefore, volume of CSTR in example 2-3, V¡= F¡X_ = !!AoX -rA 2 -r A
= ]:__6.4 = 3.2 2
If the flow rate is doubled,
F2 = 2FAo and CAo will remain same, Volume ofCSTR in example 2-3, V2 = F2X/-rA = 12.8 m3
P2-4 (b) Example 2-5 2-3
Levenspiel Plot
Fao/2
45·,··---·--·--------------···---~--~--~----·----~ 4
35
.._ _ _.X overall
"'
';- 2 5
o "' u..
2 1 5
o
02
F
_¿_Q_
o
X [FAd-rA (m)
0.445
=r,
in Table 2-3 by 2 .
0.1 0.545
0.2 0.665
0.4 1.025
Reactor1
Reactor2
V1 =0.82m3 V= (F Ad-rA)X
V2
0.82
=[_~~') A
(XI)
= 3..2 m3
3.2 = (
XI
!") rA
By tria! and error we get: and X2 = 0 . 8 Overall conversion Xoverall = (l/2)X1 + (1/2)X2 = (0546+0.8)/2
X1
= 0.546
P2-4 (e) Example 2-6 Now, F AO = 0.4/2 = 0.2 mol/s, Faof.2
X1
0.55
1.61
Fao/2
08
06
Converslon
Now, F AO = 0.4/2 = 0..2 mol/s, New Table: Divide each term
0.4
X2
2-4
X2
(x2~
= O 673
0.6
1.77
0.7 2.53
0.8 4
F
New Table: Divide each term ~
o
X [FAof-rA (m)
=r:
in Table 2-3 by 2 .
0.1 0.545
0.445
0.6 1.77
0.4 1.025
0.7 2.53
0.8 4
V2 = 1.614 m3
V1 = 0.551m3
V¡= FAO
0.2 0.665
dX f--
X
0 -rA Plot FAof-rA versus conversion. Estimate outlet conversions by computing the integral of the plotted function.
Levenspiel
Plot
4.5 4 3.5
ca ....
oca 1
u..
3 25 2 1.5 1 0.5
o o
OA
0. 2
06
0.8
Conversion
ÍX::-0603 f~r V1 = 0.551m3 -~=-0.89 for V2 = L614m3 ~l conversion X0= (1/2)X1 + (l/2)X2 = (0.603+0.89)/2 = 0.746
-
P2-4 rd) Example
·-----
-·
_]
Levenspiel Plot
2-7
2.5
(1)
2
ForPFR, ~
1..5
"'
1
c5 IL
05
= 0.222m3
o
o
0.1
0.2
For first CSTR,
03
04
Conversion
2-5
0.5
0.6
0..7
FAO
X2 = 0..6, --
-rA
= 1.32m3
For second CSTR, X3
= 0..65,
= 2.0m3 ,
FAO
--
-rA
V3=
F
Ao
(X 3 -X -rA
z
)
= O.lm3
(2)
First CSTR remains unchanged ForPFR:
V=
(F }
º5 _Aº f.
02
Levenspiel
X·
TA
Plot
25
Using the Levenspiel Plot VrFR
~ 15+--~~~~~~~~-~-
= 0 . 22
-'.!:
o
{1,
1 i===::;::::;::::::;;::;~~;;;;;;::~----::::;;;;¡
ForCSTR,
o
0.1
0. 2
0.3
04
0.5
Conversion
(3) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the smaller CSTR.
...)--t( ]
.023
Conversion Xl = 0 . 20 X2 = 0. 60 X3 = 0.65
Original Reactor Volumes Vl = 0.188 (CSTR) V2 = 0 . 38 (PFR) V3 = 0.10 (CSTR_l_
ForPFR, X1 = 0..2
V,=
1( ~;: }x
Using trapezoidal rule, X0 = 0 ..1, X1 = 0.1
2-6
W.orst Arra~gement~ VI= 0.23 (PFR) V2 = 0 ..53 (CSTR) V3 = 0.10 (CSTR)
06
07
V¡=
(x~:
)[f(Xo)+f(X1)]
0
= 0·2
[l.28+0.98]m3 2 =0.23m3 ForCSTR, FAO
For X2 = 0.6, --
-rA
FAO ( X2 V2 = --
3
= l.32m ,
=r,
X1 ) = 1.32(0.6-0.2)
= 0.53
m3
For 2ºd CSTR,
2 m3 ,
FAO ForX3=0.65, --=
-rA
P2-4 (e) Example 2-8 'l
= 5hrs
v0 = ldm3/rnin = 60dm3/hr CA= 2.5 mol/dnr'X = 0.8
V
T = ·-
For CSTR,
Vo V= 300dm3 (1)
-r =CAoX =2.5x0.8molldm3hr A
T
5
= 0.4mol / dni' hr (2) V = 300dm3
(3)CA = CAo(l-X) = 0.5 mol/dm3
P2-5 X
o
0.1
0.2
0.4
FAof-rA (rrr')
0.8 9
1.0 8
L3 3
2.0 5
V= l.Om3
-
0.6
0.7
3.5 4
5.0 6
o. 8 8.
o
Levenspiel plot
P2-5 (a) Two CSTRs in series For first CSTR, V= (FAd-lAX¡) X¡ =>XI =0 . 44 For second CSTR, V= (FAJ-rAx2) (X2 - X1) =>X2=0.67
01
0.2
03
04
05 X
0.6
0.7
08
09
P2-5 (b) Two PFRs in series
V=
¡( ~;:}tx 1( ~;: }x +
By extrapolating and solving, we get X1 = 0.50 X2 = 074
P2-5 (e) Two CSTRs in parallel with the feed, F AO, divided equally between two reactors, FANEw/rAxl = 0.5FAof-rAX1 V = (0.5F Aof-rAx1) X1 Solving we get, Xout = 0 . 60
---------·-··-····
O-+-~~~~~~~~~~~~~~~-; O
0.2
0.4
X
0.6
0. 8
P2-5 (d) Two PFRs in parallel with the feed equally divided between the two reactors. FANEw/-rAxl = 0 . 5FAof-rAx1 By extrapolating and solving as part (b), we get Xout = 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) X0 = (0..60 + 0 . 74) / 2 = 0..67
P2-5 (0 A PFR followed by a CSTR, XPFR = 0.50 (using part(b)) V= (F AJ-rA-xcsJR) (XcsTR - XPFR) Solving we get, XcsTR = 0.70
P2-5 (g) A CSTR followed by a PFR, Xcs1R = 0.44 (using part(a))
2-8
1
f
XPFR
V=
F _Ml_dX -rA
XcsTR By extrapolating and solving, we get
XPFR = 0.72
P2-5 (h) A 1 m3 PFR followed by two 0.5 m3 CSTRs, ForPFR, XPFR = 0.50 (using part(b)) CSTR¡: V= (FAJ-rA-xcsTR) (XcsTR -XPFR) = 0.5 m3 XcsTR = 0.63 CSTR2: V = (F Ad-rA-xcsTR2) (XcsTR2 - XcsTR1) = 0.5 m3 Xcs1R2 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) = Vi Fao (mother). The Levenspiel Plot is drawn for the baby hippo below.
Autocatalytic Reaction
3.5 ..-,,~--------_..----<
i e
~
o
~
3 -t-~--------1--------r2.5 -t----+------------c4<------i 2 ---""'--------1--~r----=:i-1.5
;l::::~==~s-¡;;;;._.,,"'-
05
0.2
0.4
06
0.8
Conversion
v.
baby
=FaoX_=l.36*034=023 3 .. • m -r A 2
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 = 2.25 .years 2
2) If Vmax and IDao are both one half of the mother's then
mAo
G mAo-,.,)
-rAM2
-rAM2
Catalytic Reaction
and since 2
VmaxCA
-rAM2
then
KM +CA
1 -VmaxCA
-r
(
~
AM2bttby
mAo -rAM2
m
- 2
KM
+ CA
=--r .
Xi=0 . 34
1
2
1
2
baby
1
- 2
AM2
rAM2
JS= 04 Conversion
=( -rmAo_)
·-m Ao
J
o
AM2mocher
mother
mother
will be identical for both the baby and mother,
-rAM2
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 0..40
P2-6 (e) Vstomach
= 0..2 m3
From the web module we see that if a polynomial is fit to the autocatalytic reaction we get:
m
__ .!'.ill..... = 127X4 - 172..36X3 + 100.18X2 - 28354X + 4 499 -rAMI
2-10
·
fflAO
And smce Ystomach = ---
X,
-rAMl
salve V= 127X5 - 172.36X4 + 100..18X3 - 28.J54X2 + 4.499X = 0.2 m3 Xstomach = . 067. 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 e Reaction 4 5 +--------
J
----+---
i
----·-1-----1 2 1.5
1
0.5 4-----+-----------···---
o O
-----.-----r--------,-------1 04 02
x~oos1
06
O X=o 178 Conversion
08
Conversion
P2-6 (d) PFR~ CSTR PFR: Outlet conversion of PFR = 0 . 111 Autocatalytic Reaction
Catalytic Reaction 5 45 4 3. 5 ,-
.
:E ca
2
-i,
2.5
E
2
o ca
1.5
05
O X= 0.111 Conversion
o 0.4
Conversion
CSTR: 2-11
0. 6
o.e
We must sol ve V= 0.46 = (X-0111)(127X4
172.36X3 + 100.18X2
-
-
28.354X + 4.499)
X=0.42 Since the hippo gets a conversion over 30% it will survive.
P2-7 A 7 B + C
Exothermic reaction:
r(mol/dm3.min
X
)
o
1 /-r(dm3 .min/mol) 1 0.6 0.2 0.2 0.2 0.2 0.8
1 1.67 5 5 5 5 1.25 0.91
0.20 0.40 0.45 0.50 0.60 0.80 0.90
1.1
P2-7 (a) To solve this problem, first plot 1/-rA vs . X from the chart above. Second, use mole balance as given below,
CSTR:
Mole balance:
VcsrR
= FA0X = - r,
(300mol/min)(ü.4) => (5mol/ dm3 .min)
=>V CSTR = 24 drrr' PFR:
dX f--
X
VPFR
= FAO
o
Mole balance:
-r
A
0.4
= 300(area under the curve) VPFR
0.6 X
= 72 dm3
P2-7 (b) 2-12
For a feed stream that enters the reaction with a previous conversion of 0 . 40 and lea ves 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.
Reactor 0.4
OAO
0.6
X
CSTR
P2-7 (e) Vcsra = 105 dm' Mole balance:
FAOX Vcsrn = -- rA X
105dm3 = -----
300mol /min
= 0.35dm3 mini mol
Use trial and error to find maximum conversion. AtX=0.70,
1/-rA = 0.5, and X/-rA = 0.35 dm3.rnin/mol
Maximum conversion = O. 70
P2-7 (d)
PFR
From part (a) we know that X1 = 0 . 40. Use trial and error to find X2. 2-13
0.60
Rearranging,
we get
Xz -0.40 - rAlx2
At X2= 0.64,
Conversion
= ~ = 0.008 FAo
X z - 0.40 = 0.008
-rAlx
2
X¡
= 0 . 64
X
P2-7 (e)
X1
From part (a), we know that X1 = 0 . 40. Use trial and error to find X2.
Mole balance: VPFR
= 72 = FAo
f --
X2
dX
040 -
= 300
rA
f --
X2
dX
040-rA
At X2 = 0.908, V= 300 x (area under the curve) => V= 300(0..24) = 72dm3 Conversion = 0..908.
P2-7 (f) See Polymath program P2·7·-f . pol.
2-14
6.0
0.65
5. 0
052
Q
0.39
4.0 ·
[J
3.0 0.26
2.0
0.13
1.0 40
20
V
60
0. 0
100
80
o
17
33
50v
67
83
100
P2-8 (a) FsoX V=--
=r,
Fso = 1000 g/hr Ata conversion of 40% --
1
- rs
Therefore
dm'hr = 0.15--g
V= (0.15)(1000)(0.40)
= 60dm3
P2-8 (b) 1 - rs
Ata conversion of 80%, --
dm'hr = 0.8----g
Fso = 1000 g/hr Therefore
V= (0.8)(1000)(0.80)
= 640dm3
P2-8 (e)
r., f-X
VPFR
=
dX
o -rs
From the plot of 1/-rs Calculate the area under the curve such that the area is equal to V!Fso = 80 / 1000 = 0.08 X= 12% For the 80 dm3 CSTR,
F X
V= 80dm3 = --1.Q__
=r,
X/-r. = 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 1/-rs Next is a PFR with the necessary volume to achieve the 80% conversion following the CSTR. This arrangernent 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 corresponds to the minimum value of 1/-rs, first. A second CSTR with a volume sufficient to reach 80% would follow the first CSTR
=r,
=
kC5 (O.l[C50
C5
-
]+ 0.001)_
KM +Cs
1
-;: =
KM +C5 kC5 (O.l[C50 -- C5 ]+ 0.001)
Let us first consider when Cs is small C50 is a constant and if we group together the constants and simplify
since Cs <
1
r.
=
KM
KM
k1CJ + k2Cs
which is consistent with the shape of the graph when X is large (if Cs is small X is
large andas Cs grows X decreases). Now consider when Cs is large (X is small) As Cs gets larger Ce approaches O:
If -
r,
=
kC5Cc KM+ C5
then - -
As Cs grows larger, C5 >>
1
r.
KM + C5 = --'-"--kC5Cc
KM
2-16
And since Ce is becoming very small and approaching O at X = O, 1/-r5 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 mínimum.
P2-9 Irreversible gas phase reaction 2A+B ~2C
See Polymath program P2-9.po1.
P2-9 (a) PFR volume necessary to achieve 50% conversion Mole Balance
V=FAO
X2
dX
f---
500.000
X¡ (-rA)
400 000
Volume = Geometric area under the curve of (FAol-rA) vsX)
V=
G
~~ --rA
·¡
300,0CO
J
ono ·
200
x 400000x 0.5 + (100000x 0.5)
100 000
V= 150000 m3
P2-9 (b)
-- -1
soo OOú
CSTR Volume to achieve 50% conversion Mole Balance
....
400 000
V=!AoX_
r,;
(m3)
(-rA)
300
-
.
-~
ooo
?.OOOOü
V = 0.5 X 100000
100 000
V= 50000m3
o
o
o;
020~::io405
X
-·--+--1'-···-f--·+-0.60."10.aoa-10 .
P2-9 (e) Volume of second CSTR added in series to achieve 80% conversion
_!_Ao(Xz -
V:2-
500 000
Xi)_
(--rA) Vz = 500000 X (0.8 - 0.5)
(m3)
I
----l
-·-----
1
200.00Cl
V2 = 150000m3 0.1
02
os
0.4 0[5
X
2-17
0.6
0.7
• -¡---
C!.9
09
1.0
P2-9 (d) Volume of PFR added in series to first CSTR to achieve 80% conversion
1
VPFR = (~x
2
400000x 0.3) + (lOOOOOx0.3)
503.000-1.-----
V PFR = 90000m3
400.000.,
···r:;
300 000
P2-9 (e) ForCSTR, V1 = 60000 m3 (CSTR) Mole Balance
1
100 000
~· ~-- ~1
O
01
V= FAoX
·····1---!
ó.2030405
1
Oi)D"106G8·:o X
(-rA) X= 0.515 ForPFR, V2 = 60000 m3 Mole balance
V=FAO
500 000
X2
dX
-·íA
f----
300 000
x. (:--rA)
Eqn . Of 2°d line (sloping upwards)
y -100000 60000
X2
= 1.3 x106 (x -
= f (1.3 x 106 (x
0.5) X
- 0.51) + 1 OOOOO)dX
X¡
=>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= nr2h = n(45 ft/(56 ft) = 3562 ft3 = 100,865 L 2-18
Length of one segment = 23 ft Length of entire reactor= (23 ft)(12)(1 l) = 3036 ft D = 1 ft V = nr2h = 1t(0.5 ft/(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.
60
CSTR PBR
2
.4
.6 Conversion, X
6
1.0
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 detennining the area of the shaded region in the figure below.
2-19
2
A
.8 6 Convsrsion, X
1
o
The area of the rectangle is approximately 23..2 kg of catalyst.
P2-12 (e) 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.
50 :;:,
~
Ñ
40
s
C-'I·"
u.< 'r
30 20 10
.2
.6
.8
1 O
Convsrsíon, 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
.2
A
6
8
10
Conversión, 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 ti
Jf
40
'.'ªº
:J.-, ... u. 1
20 10
_L___L
'-----------L
2
.4
6
.8
__ 1.0
Oonvsrslon, X
The necessary catalyst weight is approximately 13 kg .
P2-12 (f) 2-21
08 0. 6 OA
_
0.2 (U)
..... ~~~,~
00
6.4
12.8 \V 19 . 2
25..6
32.0
P2-12 (g) For different (-rA) vs. (X) curves, reactors 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" ·····rA
6
2
.()
..... ~----r---· 0.8 1.0
"-··-- .... ·.· -·---·-7~··---·-····---- , ·-- . ····-·--T . . . ,_ ... O 0.2 0.4 (l.6
Conversíon
X
Mole balance equations far a CSTR and a PFR: CSTR: V = FAoX
=r;
PFR: V=
X
dX
f-
o- rA
Until the conversion (X) reaches 0 . .5, the reaction rate is independent of conversion and the reactor volumes will be identical. 2-22
. 1.e.
os dX VPFR = f --=-o - rA
F
os F X fdX -- AO ·-- VCSTR - rA o - rA AO
CDP2-A
For now, we will assume that conversion (X) will be less that 0.5. CSTR mole balance:
V = !AoX = VoCAoX
- rA
- rA
V
0.09m3
X=---=·
VO C
3
[
5 m x 200mo x 3 x 108
AO
- rA
s
m3
3
=3x10-13
!11 -~ mol
CDP2-A (e) This problem will be divided into two parts, as seen below: s
w~
·1A (m3-•• \ j
·-···--1
! mol J ' )
4
2
o
O
0.2
0.4
0.6
as
1.0
Conversión X
•
The PFR volume required in reaching X=0..5 (reaction rate is independent of conversion).
•
The PFR volume required to go from X=0.5 to X=0..7 (reaction rate depends on conversion),
2-23
ú1dx
V,= F.o.o (---- = v,CM 6j-· r"
* (. 10 m3•s)º1f(lOX 9
-.5 ni3 *200 mol *(10º
s
m'
= 1011rn3(
rnol
•
il.s
-2):IX
nr' •s·')*(.5X2 -2xr1 mol ,.
05
[:5(0. 7)2 --- 2(0.7) ]-['5(0.5)2 - 2(0.5)])
Finally, we add V2 to V1 and get: V101= V1
+ V2 = 2.3 xl011 m3
CDP2-A (d) What CSTR reactor volume is required if effluent from the plug-flow reactor in part (e) is fed to a CSTR to
10·•
-r:
8
./\
2
o
raise the conversion to 90 %
o
0 .. 2
(),.4
0.6
0.$
Corrvers.ion X
We notice that the new inverse of the reaction rate (1/-rA) is 7*108. We insert this new value into our CSTR mole balance equation:
VCSTR
_ FAoD.X _ VoCAOD.X _ 1 -
- rA
-
- rA
-
•4 X
1011 m
3
CDP2-A (e) If the reaction is carried 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 dt
Mole Balance: rA V - -Stoichiometry: NA
= NAO (l -
X) 2-24
Combine:
rA V
dX
= NAO
--
dt
From the stoichiometry of the reaction we know that V = Va( 1 +eX) and e is 1. We insert this into our mole balance equation and salve for time (t):
e
tfdt o
-
xI__ dX__
AO o- rA (1
+ X)
After integration, we have:
1
t =--CA0ln(l + X) -rA
Inserting the values for our variables: t = 2.02 X 1010 S That is 640 years.
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: Reaction Rate (,,-r A) ver·sus
Reactor· Volurne
(V)
4.0•10·
[,;~,i'""' ·-···-·--·--\ 2 .. 0*10·9_
....___
~
1 . 0*10-9
o
o
1.0•1011
··---···-. ····- -
..
--------------~
2 . 0•1011
3.0•1011
4 .. 0•1011
5.0•1011
V (m3)
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
o.a X
0. 6
0.4
0.2
1 . 0*1011
2.0*1011
3.0*1011
4.0*1011
5 .. 0*1011
V (m3)
The volume required for 99% conversion exceeds 4*1011 m3.
CDP2-A (g) Critique the answers to this problem. The rate of reaction for this problem is extremely small, and the flow rate is quite large. To obtain the desired conversion, it would require a reactor 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)
e
e
ºI. .".'
<.J
x,· • 0.3 O-------.:..)--x• I
o
0.1
o~
a.:i
o.• ~ o.a
0.1
o.•
0$
Convtriion. X {l)l
!aJ
For an intermediate conversion of 03, Figure below shows that a PFR yields the smallest volume, since for the PFR we use the area under the curve . A mínimum volume is also achieved by following the PFR with a CSTR. In this case the area considered would be the rectangle bounded by X =0.3 and X = 0 . 7 with a height equal to the CAof-rA value at X= 0.7, which is less than the area under the curve .
CDP2-C (b) 2-26
v
o
... 50 l/min V• I • ar~•
considercd
in p&rt ~-
+ (0.7-0.3) ""' {0.3
-
0)(10)
+
1
2
·CAO -
=,
X•0 .. 7
C0.3--0) C.SO-l.O) + {0. 7-0.3)' (15)
., 15 :r:dn So V • v. X • (SO 1/tl:ldn) (15 m.ia)
• 75::0l
• 750
=.1•
CDP2-C (e) Tho .s:ullon:
•::u
I .. (0.1--0.0)
So
V • v
o
I •
ifo Yould furtlur
cut ~ -r}.
eso)
b e achievcd.
by -c..sin¡ ouly
aae CSTRWith this
sy.st•=·
* (0.7-0)(15) • 10.5 min
X•0.7
ll=.iu
reduce
(10.5) t:lu: tot:a.l.
:tlliu - 525 l v,oli;:.:c by u.sin¡
CDP2-C (d)
2-27
a PFR at !ir.st
up to
the
to
th,ci
&::::e&
By tri1.l
of
thci roctu:i.¡lo
Ltld error
V ,. v
ForthePFR,
o
sao
tlu.t
CAO
I • (O .4-5-0)
So
,ro
up
(50
tho
specifiod.
;:(-0.4.5
is
&
:x-o.
couvarsiou.
solutiou.
• 16.55
(0.45-0)(37)
-rA
I •
to
Fo:
tb.o CS'!a,
min
7
1/:niD.} 0.6 .65
:i,.i.::,.)
832.S
•
1
I"' J°.45 o
I ..
o.os
(10
3
+ 2(20l
.- 4(15)
4(41)
o. as
•.;:..,..:,..:..
(10
3
+ 4(3Sl
2(43)
+
+ 2(20
v I • (SO 1/min)(15.72 ain) • 786
So V•
solution
i
ac
11.:i.
+ 43
~í43)
+
+ 2(50)
+so+
43)
+ 37)
H1 di!fOX'f;~Có•
1
o
Thc::c is s Ls o
-e-
+ 37}
+ 35 + 43 + 48)
.- .\(15
+ 2(<:3)
•
l..S.71.
protty
:t>o.7
T,;y I. ... o .8
csn
Foi:' t~e
I ~ (O.S-0)(33)
For
tb.e
• 26.4 =i~
PF&
o .ll
Í -
fo
• .Q..J. 3
• º3•
(10
+ 4{20)
o
+ 4(50)
+ 2.(43)
+ 4(31)
+ 2(17)
+ 43 + l.7}
+ 33}
+ 4(15) + 33) {10
+ -4(20?
I • (0.19)(30) So V .. v
+ 2(43)
50 ..... 3:Z + 1.5)
+
2.(43
• 13.7 min
t • (.50 l/ninH23.7
min) .. 1185 l Ccirn'\)
2-28
-
2.3• .9 m.i:c.
,;l
For
tlu
PFR.
...
l • 23.9
- ~ (O.S-0.79H30+33)
So V • v ! • o
.. ll.9 - 0.315
(50 1/min.) (2:3 .SS m.u)
• 23.58
mu
1179 l
•
O .ft
diffc:rcnca
o.•
o.,
CDP2-C (e)
o.o
:X CAO/-::r: A tcd..n)
o.o
-:{mi:a.)
tau (min}
10
0.1
0.3
0,.2
20
43
50
2..0
S.6
u.o
1i ·--
. ....... -., ..../f. ~ . t ' . ':'"
/ .
10:
. 7Q()
-
y
··-
43
17.2
4'
32
u.o
.
) -----
J
2-29
0.6 11
10.l
0.1
o .. s
1$
33
10 .. s
2..6 .4
CDP2-D Data taken at 101.5 kPa (10 atrn) and 227ºC (500.2 K)
;;;; XA,)~- = (o. ~~3 X1 ?]__,.. == 0.08113 zmol/drn'
Y.A.o= 0.333
e
_
0.000010
RT
Ao
--r,\ -
X
-······- c_J~-r~-- .
O s{i2.11"
....__
(0.082}(500.2)
-~
_?.:?.~~~?:~ ~?.~~?~)~f-~~-00~~?1. 0.2
04
·
0.6
16225:54 4Ú56f84 8TÍ27.68
.
·--------·---·······--·------·····
-
0.2
0.4
20000
o
X
0.6
CDP2-D (a) ·¡;PFR :;;:
30% conversion in PFR: ' (Ud.X. C Ao == 4,664.84 S =:>
f
V -- v·
O •• IA
O
1- f4 \' b"64 · . 84
f '
CDP2-D (b)
1
min J2 • - )= L -5--· ).) • - m3 s, ----6-----_------_--• -m 3/ mm
() s .
30 to 50% conversión in CSTR: 1:csTR ;;;;:
eAo'rx.-= -x- L) = 12,1692 S .... r,u
:::::::>
vcsrR
-.¡¡
CDP2-D (e)
2-30
=·(121"69''{'lminy,2. , -.-s ---_- · · m3/·.mm) == 4-·o-64 :'). ' 60s
m
3
Total Volume: = 155.5
VTorn.l
+ 405.6 == 561.1 m3
CDP2-D (d) 60% conversión in PFR:
dX
o.s
= e M, J'• ··-·-· = 70 - > 281 • 9 s
1: PFR
==>
O ···fA
VPF"'· ::::: ."
(20,281.9 s
{1..E.1.:. E.}''! . 60 S
80% conversión in PFR:
r.., is not known for X>0.60- can not do. CDP2-D (e)
50 % inCSTR: 't:::::
e.
Ao -
V= v
X
-r A
:::::
30,422.9
S
,= (30,422.9s{1¡: frm'!min)= 1014.1
O
CDP2-D (0 50 to 60% conversión in CSTR: t
= ~"º(~i . : Xi)= 8112.8 -fA2
V=
v.,=
1fo~-J2m'lmin)=
(8112.&sf
CDP2-D (g)
2-31
270.4 m3
m"
2 m3 /min)= 676.06 m3
----·-···· ·~·-· ····-·
Rate of Reactíon vs. Volume
-------.
Conversion vs. Volume O 6 r-··-------···- · ·- - ···--·-·
. · --·,
!
0.4
"'7 2 5E 06
-----
... 02
-- ~--~----- -·-· · . - ~
_J
O.OE+OO . ·-O
L. - --- - - . . - ------·· ~
t- --- ----- -1 1
O
-
(ln3)
.-.., .... ..
200
400
600
Volumc (rol
.
:.-
Critique Answers are Valid: 1,
Constant Temperature and Pressure No heat effects No pressure drop 2. Single Interpolation to X_,.== 0.15, 0.30, 0 . 45, and 0.50 allowable 3. Huge volume (the size of the LA Basin)! Raíse T? Raise P?
CDP2-E For the CSTR :
= -~?l:?.:;,(_L == F,. -------r,..
Area
0 (
=
Vt
=
Atea) l
1200 drri
A •• __,,. p:oiu::t
From the graph we can see thar X1
= 0.60
For the PFR: V2 =
·F,.,- - - {X - 1_ -Xi) ----· 0
- --TA
(
..
)
= F"° Area under curve
Area under curve == V2
= 600 dmJ
Frorn che graph we can see that X1
:::;;
0.80
2-32
800
-··---·--·-·· ·· ··--·· . ~···--· -·---~--·-· ·
CDP2-D (h)
V1
.
----·, --·-y-- __J
o . ----·- - ··--·
,---······-------·--,.-··· . 200 400 600 800
Volume
1
CDP2-F (a) Find the conversion for the CSTR and PFR connected in series. X -IA o 0.2 0.1 0.0167 0.4 0.00488 0.7 0.00286 0.9 0.00204
400 L CSTR and 100 L PFR P=lOacm C
To
P = -RT
lf(-rA)
5 59.9 204.9 349.65 490.19
Feed is 4i % A, 41 % B, and 18% I.
T=227ºC=500K
10 aun = ---·"· -· ----------·= 0.244 (0.082 L · atrn/mol · K)(SOO K)
mol/L
C Ao = 0.41 e,., = 0.41{0.244 mol/L) = 0.1 mol/L FAo = voCAo
= 1 I.Js(O. l mol/L) = 0.1
mol/s = 6 mol/min
There are two possible arrangernents of the system: L CSTR followed by the PFR 2. PFR followed by the CSTR
Case 1: CST.R ~ PFR CSTR:
V1 = F.,.<)(Area) V
400
Are a = ... -L. = ··· ·· · · = 66..67
F,,.
6
0
From the graph ~ X1 = 0.36 V2 = FAQ(Area under curve)
PFR:
Area under curve
= .-V, -. = -·100 --· = 16.667 f:_,.o
6
From the graph - X1 = 0.445
Case 1:
CSTR --> PFR
Case 2:
500-r---~----~-,.--,
PFR -·> CSTR 500,-------~-.,..-,
400
400 - ···-----·-·· -- . -
i:! 300 • ::::; 200
:::; 200
100
100
~ 300 · '
o 0 . 0 0.2
0.4
0.6
(L&
o
l.O
X
o.o
0.2 0..4 0.6 X
0.8
1.0
.- . . . --.~ . . --·~,· · -------~w•••V·--··•
'----------·· · · · ····· · ··-········-· · ····-··--······· -·"·'--· 2-33
Case 2: PFR
PFR:
--t
CSTR
Area under curve= 16.67
Frorn the graph - X1 = 0.259 Area = 66.67
CSTR:
Frorn the graph - X2
= 05 i5
CDP2-F (b) Two 400 L CSTR's in series. CSTRl:
V= F.Ao (Area) · Area
«
66.67
From the graph - X, = 0.36 CSTR2 :
=
66.67 Frorn the graph - X1 = 0.595 Area
(b) Two CSTR'.s in Sedes 500 ----~~~--....-.
(e) Two CSTR's in. Parallel 500 ,-..--------,.-,..--.
400
400
e. 300
.
i: 300
;::: 200
::; 200 ·
100
100
o~==o.o 0.2
o
0.4
0.6
0. 8
LO
0. 0
0. 2
0.4
0.6 X
X
CDP2-F (e) Two 400 L CSTR's in parallel. To each CSTR goes half of the feed Fi\~ = 6/2 = 3 mol/min V= FA (Area) 0
V
--·---·- . ·-·-·· ..e;;..;.;.._;.;;,,.;;;....;..;:;:;:......
400
Area = -- = ----- = 133.3 fAa 3 From the graph : X = 0.52
2-34
_,
0. 8
I.O
CDP2-F (d) PFR :
V = FAo ( Area under curve)
From the graph we can find the area under the curve for a conversión of 0.60:
= (0.60)(300)
= 90 2 V= (2 mollmin)(90) = 180 L
Area
. . 5º.ºr.
(d)
40ú.
.:: 300 ~ 200
---
Single
PFR
·
·
100
o
O.O
-
..
0.2
·-· ·--------
Q_4
0.8
0.6
1-D
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 tum cause a decrease in C Ao and FA.o· By looking at the design equation:
V= .S.\oX ···[A
ir is apparent that to cornpensate for the decrease in F...,, there would be and increase in X.
CDP2-F (f) Use the graph of ll·-r.~ vs. X to find values fer all volumes, (Assume a flow rate of l mol/min.) Generare the following table and graphs:
X
O
-r"
Q2
--ir:-1-· -·· · - -· · o.oT67
___o.
0.4 _,,,_ 7 0.9
V
O
3.494-
··o.o04ss- -~u:984 __..,__..,,.....,..~--------.....,,,...~-,_. 0.00286 125.878 -··0.00204
225.088
2-35
¡--···-···----···--·-···-········--·~-------··-¡
!
(O Cnnversicn
I
I
[ -
Vol u me
Í
....... -~--~
~
vs,
O 8
(0 -r., vs, Volume
l
¡
O.Ol5
0. 6
°i
0. 4
0.01
0 .. 2 1
.Q
o
o
~--· --··
CDP2-F (g) Individualized
solution
2-36
o
100
y
200
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
+--------------------------1-----J
0.005
~ :e
0 . 004
+-------------------1-
.IOI
0 . 003
o
, ....
,.
...
315
310
.. ···,--++·····-•········· .........T ......
•····--··1
320
325
330
335
T(K)
expc-- 6~~0J)
exp(- 2~~00J)
For E = 60kJ/mol
For E1 = 240kJ/mol
= 1.32X 1016
kl = 1.32 x 1016
k
E=60 kj/mol E = 240 kj/mol 6000000 3 SE-22
5000000
3E-22 -!----------------! 25E-22 +-----\~------------l
g
"'
i::. 3000000-1--------",~---------l
2E-22 -!----;-------------!
-"'
:;- 1.5E··22 -t--------+---1E-22
-t-----~\c--------
5E-23
+------~~-~
2000000+---------
Q
,T'"'"'~···"·
O 00295 0.003 0.00305 O 0031 O 00315 O 0032 0.00325
O 00295 0.003 0.00305 O 0031 O 00315 0 . 0032 O 00325
1/T(1/K)
1ff (1/K)
P3-2 (b) Example 3-2 Yes, water is already considered inert.
3-1
P3-2 (e) 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 O instead of a negative concentration. Therefore 90 % of caustic soda is possible.
P3-2 (d) Example
3-4
b
e
d
a
a
a
A +-B ~ -C +-D
So, the mínimum value of 08
1/3
= 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 = Yo. Plot:
1 -rA
0.5(1-0.14X)2 (l-X)(0.54-0.5X) 1/(-ra) vs X
180 160 140 120
f
...
::!::-
100 80 60 40 20
04
02
08
06
1 2
X
The rate of reaction decreases drastically with increase in conversion at higher conversions.
P3-2 (f) 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 . Cro = constant and inerts are varied. N204 A
f-7
-
2N02 2B
Equilibrium rate constant is given by: K e
e
z
= _ __!!:_:___ cA,e
Stoichiometry: E= y AOÓ Constant volume Batch:
= y AO (2 -1) = y AO
3-2
e - 2NAoX B -
-2CAO X
-
Vo
Plug flow reactor:
FAO(l-X)
C
v0(1
A=
+ cX) =
CAo(l-X) (1 + cX)
=
and C8
2FAOX
= 2CAOX v0(1 + eX) (1 + eX)
CAo = YAoPo = yA0(0.07176)molldm3 RT0 Combining: For constant volume batch:
2
C _ 4C2Ao X2 K e-_ ___!!_:!.___ -
CAe
.
CAo(l-X)
4CAO
For flow reactor:
=~B/ =
Kc
CA e
K e (1 - X e )(1 + eX e)
4C;ox2 CAO (1- X )(1 + ex)
4CAO
See Polymath program P3--2-f.pol. POLYMATH Results NLES Report (saf enewt) Nonlinear equations [l J f(Xeb) = Xeb - (kc*(1-Xeb)/(4*Cao))"0.5 = O [ 2 J f(Xef) = Xef - (kc*(1-Xef)*(1 +eps*Xef)/(4*Cao))"0.5 = O Explicit equations [lJ yao = 1 [2] kc=0.1 [3J Cao = 0.07174*yao [4] eps = yao 1
ai
¡; o o-~E
@
:::,
· · ···••····· ·
.
.7f~~~~~~~~~~~~~~~~~~~==~
0. 9
+-------------------------/~----'
0.8
+-------
---------------------""----!
0.6 0. 0. 5
--
0.4 ·~ 0.3 1---------------------------w 0 . 2 +-----------------------------' 0.1 +------------------------------1 0
,.
o
0. 1
·············-··········1
0. 2
,
T
,
.,
0.3
0.4
0.5
~ .• ,
1
0. 6
y inerts
3-3
,.. •... _
0. 7
"T
o.a
-
,,_,••-T•·····•·"···-·-··
0.9
-Batchl -Flow_J
Xeb
Yao
Yinert
o
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
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95 0.956
P3-2 (g) No solution
Xef
o.os
0.044
0.508 0.5217 0.537 0.5547 0.576 0.601 0.633 0.6743 0.732 0.8212 0.89 0.896
will be given
P3-2 (h) 1
1
2
2
A+-B7-C
Rate law: -
= k A CA 2 C
rA
B
and
1 dm3 kA = 25- -S
2
( mol J
~=_!!!_=-~
-1
-1/2
1/2
(dm3J
kc =ks =12.5-1 s mol
2
P3-2 (i) A+3B~2C Rate law: - TA At equilibrium,
K
e
= k A CA C B
c-,
=·--------
C
A,e
At equilibrium,
at low temperatures.
l/2c
-•A=
B,e
3/2
0, so we can suggest that - YA
=k
But at t = O, Ce = O So the rate law is not valid at t = O .
3-4
A
(CA C / 112
12 - ~:
J
Next guess: 2 _
Kc -
eC,e 2
<:c.,
ee
3 , or
=>-rA=kA(cAcB-
A
~/ CB
x;
Ce
2
2
s; CB
B -
_
2 -
o
2J
whích satisfies both the initial conditions and equilibrium rate law. Hence
-rA =kA(CACB -
s; CB J ~/
2
is the required rate Iaw,
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 Anhenius Equation For Fire flies: --- Flashes/ T(in 1/T ln(flash 2.66 K) min es/min) 294 0.00340 9 2.197 1 2.52 298 12.16 0.00335 2.498 6 303 0.00330 16.2 2.785 2.38 .....
o
2.24
Plotting ln(flashes/min) vs 1/T, we get a straight line.
See Polymath program P3-4-firetlies.pol. For Crickets: T(in K)
287.2 293.3 300
1/T
x103
3.482 3.409 3.333
chrips/ min
80 126 200
ln(chirps/ min)
4.382 4.836 5.298
-·-
Plotting ln(chirps/min) Vs 1/T, we get a straight line. ~ Both, Fireflies and Crickets data follow the Arrhenius Model. In y = A+ BIT , and have the same activation energy,
See Polymath program P3-4-crickets.po1.
3-5
P3-4 (b) For Honey1bee: T(in K)
vr
V(cm/s)
ln(V)
298
x103 3.356
0.7
-0.357
303
3.300
1.8
0 . 588
308
3.247
3
1.098
Plotting ln(V) Vs 1/T, almost straight line. ln(V) = 44.6 - l.33E4/T At T = 40°C(313K) V= 6.4cm/s At T = -5°C(268K) V = 0.005cm/s(But bee would not be alive at this temperature)
See Polymath program P3-4-bces.pol.
P3-4 (e) For ants: T(in K)
vr x1 OJ
V(cm/s)
ln(V)
283
3 . 53
0.5
-0.69
293
-303
3 . 41
2
0 . 69
3.30
3.4
1 . 22
311
3 . 21
6. 5
1.87
----··-----
------·---~·---
1.4
--
0.8
-·
Plotting ln(V) Vs 1/T, almost straight line.
0.2
See Polymath program P3-4-ants.pol.
-0.-'
-1 O'----~--~--~----~-~ So activity of bees, ants, crickets and fireflies follow 3.22E-3 328E-3 3.34E¡tf 341E-3 Anhenius model. So activity increases with an increase in temperature. Activation energies for fireflies and crickets are almost the same.
~-
Insect Cricket Firefly Ant Honeybee
1.47E-3 3.53E-3
--
Activation Energy 52150 54800 95570 141800
P3-4 (d) There is a limit to temperature for which data for any one of he insect can be extrapolate. Data which would be helpful is the maximum and the mínimum temperature that these insects can endure before death. Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperature it could be useless .
P3-5 There are two competing effects that bring about the maximum in the corrosion rate: Temperature and HCN-H2S04 concentration. The corrnsion rate increases with increasing temperature and increasing concentration of HCN-H2S04 complex. The temperature increases as we go from top to bottom of the column and consequently the rate of conosion should increase. However, the HCN concentrations (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 corrosion rate somewhere around the rniddle 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 = k¡ and at T = T2 = T 1+ 10, let k = k2 = 2k¡. Then with k = Ae·EIRT in general, k; = Ae-EtRT¡ and k2 = Ae-EtRTz, or
kz _ --e
-i(t-t)
E
or
k1
R
Therefore:
In ( kz E= R--
k1
T.1 (T.1 + io) =
J (r. (T. + 10)) 1
1
(ln2)(T¡(T¡+10)) 10
= R·---·-·
(7;-T¡) lOE
Rln2
which can be approximated by T
10E05 = ----
R In 2
P3-7 (b) Equation 3-18 is
k
= Ae
E RT
From the data, at T 1 = OºC, k1 =
E( 1 1)
Dividing gives :2· =e--¡¡
Ae··EI RT¡, and
at T 2 = lOOºC, k2 =
Tz--Ti- , or
1
3-7
Ae -et RTz
E=
[
1.99 cal ] [ 273 K ][ 373 K] mol K ln(.050) IOOK .001
= 7960 cal mol
E
A= k1 eR71 .
= 2100min-1
= 10-3 min " exp (
1.99_cal J(273K) molK
P3- 7 ( C) Individualized solution P3-8 When the components inside air bag are ignited, following reactions take place, 2NaN3 - 2Na + 3N2 . . (1) lONa + 2KN03 ---> K20 + .5Na20 + N2 . . .(2) K20 + Na20 + Si02 - alkaline silicate glass (3)
5x rxn(l)+ rxn(2)+ rxn(3) =
r:xn(4)
NaN3 + 0.2KN03 + 01Si02 7 04Na20 + l.6N2 + complex/10 Stoichiometric table: ----~--
Species -NaN3 _ KN03
Symbol A
B
Si02
e
Na20 N2
D E
(4) -
Change -NAX -0.2XNA
Final NA(l-X)
NA(}c
--OJXNA
NA( (}e - O.IX)
o o
0.4XNA l.6XNA
0.4XNA l.6XNA
Initial NA NA~B
Given weight of NaN3 = 1.50g Therefore, no . of moles of NaN3 = 2 . 3
----
NA( (}B - 02X)
--
Mw1 of NaN3 = 65
1 moles of NaN3 requires 02 mole of KN03
=> Moles ofB, KN03 = 02(2.3) = OA6 moles
Mw1 ofKN03 = 101.1 Therefore, grams ofKN03 required = OA6 x l Ol.I = 46..5 g 1 moles of Nal-l, requires 01 mole of Si02. Moles of C, Si02 = 0.1(2 .3) = 0 . 23 moles Mw1 of Si02 = 60 . 08 Therefore, grams of Si02 required = 023 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. • A void Physical damage of the bag in car. • It is stable under ordinary conditions of storage . Decomposes explosively upon heating (over 221 ° F or 105 ° 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 --
dt
= k (I - .x) and that k is a function of temperature, but not a
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 IOOºC. 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 (e)
No solution will be given
P3-10 (a) C2H4 + H2
1) C2H6 -
2) C2H4 + 1/202
-+
C2H4Ü
Rate law: -rA =
kCc2H6
Rate law: -l'A =
kCC2H4
3) (CH3)3COOC(CH3)3 - C2H6 + 2CH3COCH3 A B + 2C
c~:
2
Rate law: -rA = k[CA - C8Cc2/Kc]
4) n-C4H10
Rate law: -rA = k[
I- C4H10
+--+
5) CH3COOC2Hs + C4H90H
+
A
+--+
B
+--+
Ce H --CiC4H10 IK.:] n 4 to
CH3COOC4H9 + C2H50H
C
+
D
Rate law: -rA = k[CACB - CcCofKc]
P3-10 (b) 2A + B (1) (2) (3) (4)
C -l'A = kCACB2 -rA = kC8 -IA = k -rA = kCAcB·l
P3-10 (e) (1) C2H6 -
C2H4 + H2
3-9
(2) H2 + Br2
(3) H2 + 12
------>
------>
2HBr
2HI
P3-11 (a) Liquid phase reaction,
o
"
CHr-OH 1
CH2/- CH2 + H20 A + B CAo = 1 lbmol/fr'
--+
CHr-OH C
------>
C80 = J.47 lbmol/fr'
StoichiometricTable: Species Symbol Ethylene A oxide ---·--· Water B
Initial CAo=l lbmol/fr' Cso= 347 lbmol/ft3,
-----
Change -CAoX
Remaininz CA= CAo(l-X) = (1-X) lbmol/fr'
-CAoX
Ca= CAo(B8 -X)
CAoX
=(3.47-X) Ibmol/fr' Ce= CAoX = X lbmol/ft3
BB=3A7
c
Glycol
o
Rate law:
-rA= kCACa
Therefore,
-rA = k
At 300K
E= 12500 cal/mol,
e;º (1-X)( BB -X)= k(l-X)(347-X) X= 0 . 9, 0.1
3
3
k = O.ldm /mol.s =--x.35.315.ft
1000
llbmol.s = 0.0035.ft /lbmol.s 3
= _c_AO_X_ = (1)(0.9) --= 1000..56 S-1 -rA (0.0035)(1)2 (1-0.9)(3.47 -0.9) At 350K, k2 = k exp((E/R)(l/T-1/T2))= 0 . 0035exp((l2500/L987)(1/T-l/T2))= 0.071 dm3/mol..s Therefore, '['CSTR
CAOX
'ícsTR = --
=r,
=
(1)(0.9) 2
(0.071)(1) (1-0.9)(3.47
-0.9)
=
-1
49 . 3s
P3-11 (b) Isothermal, isobaric gas-phase pyrolysis, C2H6 C2H4 + H2 A -¿ B + C Stoichiometric table: S ecies s mbol
Enterin
A
3-10
Chan e
Leavin
o o
B
C2H4
c
H2
+FAoX +FAoX
Fa=FAoX Fe=FAoX Fr=FAo(l+X)
Frn=FAo é =YaoJ=l(l+l-1)=1 v = vo(l+ e X)
=> v = vo(l+X)
p
CAo = YAo Crn = YAo -
RT
( 1)(6atm)
=
(
= 0067 kmol/m3 = 0.067 mol/dm3
0.082 m3atm )(llOOK) K.kmol
CA=
FA = FAo(l-X) = CAo (1-X) mol/dm3 v v0(l+X) (l+X)
Ca=
FB = FAo(X) = CAO X mol/dnr' v v0(1+X) (l+X) Fe
Ce=-=
FAo(X) X 3 =CA0--mol/dm v0(l+X) (l+X)
v
Rate law:
(1-X)
=kCAo---
(l+X)
(1-X)
=0.067k~--'-
(l+X)
If the reaction is carried out in a constant volume batch reactor, =>( é = O) CA= CAo(l-X) mol/dm3 Ca= CAo X mol/dm3 Ce= CAoX mol/dnr'
P3-11 (e) Isothermal, isobaric, catalytic gas phase oxidation,
1
C2H4 + - 02 7 C2~0
2
A
1
+ -B
2
7
C
Stoichiometric table: Species C2~ 02
A B
Entering FAo Fao
Leaving FA=FAo(l-X) Fa=F Ao( 88 -X)
C2H40
e
Change -FAoX -B8FAoX
o
+FAoX
Fe=FAoX
Symbol
FAO FAO y =-----·--=. AO FTO FAO + FBO
3-11
2
3
_ _ P _ 2 CAo -YAoCro -YAo-(
RT
e - -FA - FAa(l-X)
3
_ mol 3) -0.092--3 atm.dm (533K) dm ( 6atm)
0_082
mol.K
CAO(l-x) 0.092(1-X) - --'-----'v - v0 (1 + &X) - (1-0.33X) - (1-0.33X)
A -
-
FB FAa(eB-:) e -- -
v0(l+&X)
v -
B -
0.046(1-X) - (l-0.33X)
- --'-----'-
Fe FA0X = ---'---'-0.092(X) = -- = e v v0(l+&X) (1-0.33X)
e
If the reaction follow elementary rate law
=::> -r A
_ 05 -k {0.092(1-X)}{0.046(1-X)} · (1-0.33X) (1-0.33X)
P3-11 (d) Isothermal, isobaric, catalytic gas phase reaction in a PBR C6H6 + 2H2 ~ C6H10 A + 2B ~ C Stoichiometric table: Change Species Symbol Enterinz Benzene A -FAoX FAo B H2 FBo=2FAo -2FAoX -- e o FAoX C6H10
é=y
1
.
_FA_
CA ---
V
vo(l+&X)
BB -2X)
Fc=FAoX
·-
3
P ( -1 ) = ( RT 3
FA0(l-X)
FB=F AO(
2
J=-(1-2-1)=-AO 3
CAo = CroYAo =
------··Leaving FA=FAo(l-X)
6atm ( -1 ) = 0.055mol / dm 3 3) 0.082atm.dm ( 443.2K) 3 mol.K
_ CA0(1-X) _ 0.055(1-X) -
(i-ix) (i-ix) ------
3-12
e
=
8
F8 = FA0
(88 -2x)
v (l+cX)
v
FA0X
v
v0(l+cX)
e
(l-%X)
0
_ Fe _
C---
= 0.11(1-x)
_
--
CA0X
_ 0.055X
-
-
(l-%X) (l-%X)
If the reaction follow elementary rate law. Rate law:
-rA '=kCAc; -rA
1
= 0.0007k
( 1 X)3
(1- ~X)
3
For a fluidized CSTR:
w = FAOX -r'A
w=
FAOX ( 1 X)3 0.0007k 3 (1·-~X)
k=53
k
mol kgcat min atm'
= k I exp ( E (__!__R
t;
at300K
.!.JJ = 53exp ( 80~9.Q_ (-1- -- -1-)) T 8.314 300 443
FAo = CAo* Yo v0 = 5 dm3 /min
atX= 0.8 W = 0.0024 kg of catalyst
3-13
= 1663000
mol kgcatminatm3
P3-12 A
1
+ -B
2
7
C
Stoichiometric table for the given problem will be as follows A ssummg gas p h ase Species C2H4 02 N2 C2H40
Entering FAo FBo= eBFAo F¡ = 81FA0
Symbol A B I
e
Leavína FAo(l-X) F Ao(Ba - X/2) FA081 FAoX
Change -FAoX -1/2 FAoX
-------
o
FAoX
1
-FAO 1 eB =L-=FAo 2 FAO YAo=-· =0.30, Pro
(} = l
= 0.79 F
F/0 F FAO ,
0.21
10
BO
~ (} = (} l
e = YAoP = 0.04l~ol AO
RT
dm3
C =FAº=C A
V
(1-X) AO (l+éX)
_ FB -C CB - AO v
Ce= Fe= v
0.041(1-X) 1-0.15X
1 1 (2-2·X)
.
1··- . 0.15X
CAo~ 1-0.15X
-
~.020(1-X) 1-0.15X
0.041X 1- 0.15X
P3-13 (a) Let
A=ONCB B=NH3
A+2B ~
C = Nibroanaline D = Ammonium Chloride
C+D
3-14
B
0.79 0.21
= 1.88
P3-13 (b) Specíes A B
Enterína
Change -FAoX -2 FAoX
FAo FBo= E>BFAo =6.6/1.8 FAo
e
o o
D ~--
FAoX FAoX
--·
Leavíng FAo(l-X) FB= FAo(E>B-2X) Fc=FAoX Fo=FAoX
P3-13 (e) For batch system, CA=NAN
P3-13 (d) -rA = kCACB
= NA V
FA
=.!"A TT Vo
= NAo (1- x) = eAO (1- x) ' eA = !A = FA = eAO (1- x) TT
Vo
V
F8=;
=r, BB
=~8 =~:º(B8-2X)=CA0(B8-2X), vo vo = kC~0 (1- X)( 08 -2X)
=
eAO
CBO
CAO
Vo
C8=F8 =CA0(B8-2X) Vo
= 6.6 = 3.67 1.8
= 1 . 8!'.!!lº1 3 m
[--rA-=-k(-1.8)2 (1-X)(3.67-·2X)j
P3-13 (e) 1) At X= O and T = 188ºC = 461 K
-rAo
= kCA20B8
-rAO
= 0.0202
= kACA0C80
= 0.0017
m3 . ( 1.8-kmol)( kmol) 36.63kmolrmn m m
krnol .
m 3 rrun
2) At X = O and T = 25C = 298K
3-15
11273_5!!_z m3 1 k=0.0017 exp mo -+. kmol.min [ 1.987 __c
= 2.12xl0-6
1
298)
l
m3 kmol.min
-rAo = kCAoCso= 2.52 X 10"5 kmol/m3min 3)
k ~ k0 exp [
k =0.0017-
! (;, - nJ m3 kmolmin
exp
11273-;;}¡ ( l 1.987 cal 461K molK
l
J
561K
m3 k=0.0152--kmolmin -rAo = kCAoCso -rA = 0.0152~-3 -. (1.8 kmol )(6.6 kmol) kmolmm m3 m3
~º.1
~806
L=
mmm
P3-13 (f) rA = kCAo\l-X)(08-2X) At X= 0 . 90 and T = 188C = 461K l)att=188C
- -· r,
= (o.0017 ~--.3 J(1.8 km~!-)2 (1-0.9 )(3 . 67 - 2(0.9 )) kmol.mm m
= 0.00103 kmol m3min 2)
At X = 0.90 and T = 25C = 298K
3-16
=r: =(2.12xl0-6 = l.28x10-6
m3 . kmol.mm
J(1.skm~l) (1-0.9)(3.67-2(0.9)) m 2
kmol m3min
3)
At X= 0.90 and T = 288C = 561K
=r,
=(0.0152
m3 . kmol.mm
J(1.skm~l) (1-0.9)(3.67-2(0.9)) m 2
= 0.00333--kmol m3min
P3-13 (g) v0 = 2m3/rnin
= 0..002m3/nlin
l)For CSTR at 25C -rA =
= l.28xl0-6
v = v º e Ao ( 1 - x)
kmol
mmm 3
.
- rA,X=09
1.
2)At
=-0.002m3 lminxl.8km~llm3 xO.l = 28 2Sm3 l.28x10-6 kmol_ m3min kmol 288C, -rA = 0.00333- 3 .
v = v º e Ao (1- ~) -
mmm
rA,X=0.9
V= 0.002!713 lm~~l.8kmollm3 xO.l = O.l0Sm3 0.00333 kmol m3min
P3-14 C6H1206 + a02 + bNH3
--->
c(C44HuNos6012)
+ dH20 + eC02
To calculate the yields of biomass, you must first balance the reaction equation by finding the coefficients a, b, e, 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 = 4 . 4c + e For N b = 0.86c
6 + 2a = L2c + d + 2e 12 + 3b = 7..3c + 2d
ForO ForH
Also for C, 6(2/3) = 4.4c which gives e= 0.909 Next we solve for e using the other carbon balance 6 = 4.4 (0.909) + e e=2 We can solve for b using the nitrogen balance b = 0.86c = 0.86* (0.909) b = 0.78 Next we use the hydrogen balance to solve for d 12+3b=7.3c+2d 12 + 3(0 . 78) = 7.3(0.909) + 2d d = 3.85 Finally we solve for a using the oxygen balance 6 + 2a = l.2c + d + 2e 6 + 2a = 12(0.909) + 3.85 + 2(2)
a= 1.47
P3-14 (b) Assume 1 mole of glucose (180 g) reacts: Ycts= mass of cells / mass of glucose = mass of cells / 180 g mass of cells = c*(molecular weight)
= 0 . 909 mol* (9L34g/mol)
mass of cells = 83 . .12 g Yc/s
= 83..12 g f 180 g
v., = 0.46
Y c102 = mass of cells / mass of 02
If we assume 1 mole of glucose reacted, then 1.47 moles of 02 are needed and 83.12 g of cells are produced . mass of 02 = L47 mol * (32 g/mol) mass of 02 = 47.04 g Ycto2 = 83.12 g /47.04 g Ycto2 =1.77
3-18
P3-15 (a) Isothermal gas phase reaction.
1 2
3 2
-N2+-H2 ~NH3 Mak.ing H2 as the basis of calculation:
1
2 3
H2 +-N2 ~-NH3
3
2
1
A+-B~-C 3 3
Stoichiometric table:
Species H2 N2
Svmbol A
Initial FAo
B
NH3
e
Leaving FA=FAo(l-X)
Fao=BsFAo
change -FAoX -FAoX/3
o
+2fAoX/3
Fc=(2/3 )F AoX
Fa=F Ao(
P3-15 (b)
ó=(!-i-1)=-·f e=
yA0
ó=0.5x(-!)=-i
_ CAo -0.5 (
(l6.4atm)
J 0.082 atm.dm ( 500K) 3
= 0.2 mol/dm
3
mol.K
C H2
CA (l-X) 0.2(1-X) = C = --· · = (l+eX) (1-}) = O. lmol/ dm (l-X) 0.2(X) =C =-X32 CA(l+eX) =-X32 (1-{) =0.lmol/dm 0
3
A
0
C NH3
e
P3-15 (e) = 40 dm3/mol.s (1) Por Flow system:
kN2
3-19
3
88 -X/3)
(1-X)
(1-X)
(1-f)
P3-16 (a) Liquid phase reaction --) assume constant volume Rate Law (reversible reaction):
-r A
=
k
[c e A
Stoichiometry:
B
CA =CA0(l-X),
Ce ] K e
C8 =CA0(1-X),
Ce =CA0X
To find the equilibrium conversion, set -rA = O, combine stoichiometry and the rate law, and solve for X,
CACBKc =C¿
e;º (1- x, / Kc = CAOxe
x:-(2+ CAOKC l .Jxe+l=O X,
= 0.80
To find the equílibrium concentrations, substitute the equilibríum conversion into the stiochiometric relations .
3-20
mol mol CA =CA0(1-X)=2-3 (1-0.80)=0.4-3 dm dm
mol dm
)
mol dm
C8 =CA0(1-X)=2-3 (1-0.80 =0.4-3
_ _ mol ¿ _ mol CA -CAOX -2-3 0.80-1.6-3 dm dm
P3-16 (b) Stoichiometry:
e= YAoó = (1)(3-1) = 2 NA
C =-= A V
and
Be= O
NAO ( 1- X ) ( 1 - X) =C Vo(l+ex) AO (1+2X)
C =Ne=-- 3NAOX =C 3X e V Vo (l+ex) 'AO (1+2X) Combine and solve for X.,.
Equilibrium concentrations:
Po
CAO =T.= R o
(
----;---J-
lOatm
mol
= 0.305-d3 ( 400K) 0.082dm atm m molK
_ (1-0.58) _ mol CA -0.305 ( ( )) -0.059-3 1+2 0.58 dm
c
e
=
3(0.58)(0.305) mol =0246-(1+2(0.58)) · dm'
P3-16 (e) Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction . 3-21
Stoichiometry:
e
A
=NA=NAo(l-x)=C V
ee =Ne V
V, o
AO
(1-x)
=3NAOX =3C X V, Ao o
Combine and solve for X0
KeCAO (l-Xe)
= (3CA0xe )3
xe = 0.39 Equilibrium concentrations
mol
CA =(0.30.5)(1-0.39)=0.19-3
dm mol
Ce= (0.30.5)(0.39)= 0.36--3 dm
P3-16 (d) Gas phase reaction in a constant pressure, batch reactor Rate law (reversible reaction):
-rA=+,-~:J
Stoichiometry:
e= YAoo=(1)(3-1)=2 NA cA --- V -
and
Be =0
NAO ( 1- X) -c ( 1 - X ) V0(l+&X) - AO (1+2X)
c.= Ne =-]!!Aox --=3C X e V V0(l+&X) AO (1+2X) Combine and solve for X
0:
KCCAO (1-Xe) = [3CAOxe ]3 l+2Xe 1+2Xe xe
= 0.58
Equilibrium concentrations: 3-22
e
A
=
0.305(1-0.58) mol =0.059-1+2(0.58) dm'
mol ee = 3(0.305)(0.58) =0.2461+2(0.58) dm'
P3-17 Given: Gas phase reaction A+ B 7 8C in a batch reactor fitted with a piston such that V= O.lPo
( ft3
)2
k=l.0---· lbmol2 sec -rA
=kC~C8
NAo = Nao at t = O V0 = 0.15 ft3 T = 140ºC = 600ºR = Constant
P3-17 (a) Y AO =·
NAO
=0.5
NAO +NBO
8=8-1-1=6 e= YAoJ=3 T
Now
Therefore
and ·-··
Yo
V = l0Vo2 ( 1 + eX) lOV
or
= 1, Po= 10V0,
and P
= lOV
V2 =V/(l+cX)
N
Therefore
3-23
AO
= ( YRT Aof>o ) V, o
-rA
= 5.03*10-9
[1 X]3 3 l\mol (l+3X)2 ft sec
P3-17 (b) V2 =V02(l+&X)
0.22 = 0.152 (1 +&X) X =0.259 -rA
=8.63*10-10 lbmol ft3 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
1
r
w lnW T vr --:::-",:"--~==~~~-~~,;.,..,...,,.. 6.5 13
18
1871802 2.564949 2.890372
300 310 313
0.003333 0.003226 0.003195
· · · -· · ·
-. ,-----··--· . ·-· ·---·------···
i
lnWvs1rr
3
~ e -
2 y• ·7120.4X + 25.593
1.5
R'·OSSS3
0.5
o..--·---...--
0.00315 .......... -
·---.-
0.0032
, .,N
,,~.......
, ~u•••••••••
. . ·. --.-...-·-----1
0.0(625 _,_
0.0033
1/TK
"'MUW''"''''""''''•·-··--··••••
Nu,
-,,
,N,., .. ,., ..
=""",·,, , ·,,.,_
From the graph: E= 7120
lnW =-7120 . 4*(---!··---)+25 ..593=2..95 41.5+273 W(4J.5<>C)=l9.2cm./ s
------------------------------------------CDP3-B Polanyi equation: E = C - a(-dHR) We have to calculate E for the reaction CH3• + RBr 7 CH3Br + R• Given: dHR = - 6 kcal/mol From the given data table, we get 6.8 = e - c(l 7..5) and 6.0 = C -- a(20) => C = 12.4 KJ/mol and a= 0 . 32 Using these values, and dHR = - 6 kcal/mol, we get E= 10.48 KJ/mol
CDP3-C (a)
A~B
Rate law at low temperature: - rA
= kC A
The rate law t higher temperature must: 1) Satisfy thermodynamics relationships at equilibrium, and 2) Reduce to irreversible rate law when the concentration of one or more of the reaction products is zero. Also, We know,
_ Ke -
CBe CAe
Rearranging, we get
eAe - CBe K
=0
e 3-25
.
So, lets assume rate law as
=r;
=kA(CA -
~:J
Also when C8 = O, it satisfies the given rate law. Hence the proposed rate law is correct,
CDP3-C (b) A+2B72D
-rA =kCA
Rate law at low temperature:
_
Ke -
Here,
CDe
CB
2
eAe eBe 2 -
1/2
2 2
CDe =0 K
e
=r: =kA[ CA,Cs,2 -
;>;']
But it does not satisfy the irreversible rate law at low temperatures. Hence it is not correct So, taking square root of Kc
-vte:e
CDe 112 , CAe CBe
C
=-
1\..
Ae
- rA = k A
[e
1/2
C - CDe = Ü
jKc
Be
JKc J
Ae 112C Be - CDe
Which satisfies the irreversible rate law. Hence it is the required rate law.
CDP3-C (e) A+B7C+D .
kPAPB . .: :; ___ 1 + KAPA + KBPB PcPn PcPn or PBPA -· =O PBPA x,
Irreversible rate law: -·
Weknow,
Kp
=
rA
I
=
Hence assume rate law as:
k pp
=r;
(
= .
A B
PcPnJ -·--· K
p
1 + KAPA + KBPB + KcPc + KnPn
Which satisfies both the abo ve mentioned conditions. 3-26
CDP3-D Y
e 'CO b)
C
• 0.15
NaJ.O
~ -- _ - :1'-ro v; :erro .
.
.
8 .2
.
0.082
... YNlt 3,0
NR3 .o
S
e)
P28.2 atm
->
NB3 + 4 02
C¡-0
""
um.
.
latm.l· o¡:CSOOX:)
.
O.l gm.ol/1
.gmo
C0.15) (0 .. :2 gmol/1)
• 0.03
¡mol/1
3 NO + l R:zO
Fina.l 0.15
-o.1sx
O.lS(l-X)
s
B
0.18
- 4'0.15-1.)
0 .. 18-
e
o
+
o.1sx
H2
D
o
+ t(O.l5X)
+ ·!(0 .. 1SX>
N
I
0.67
o
0 ... 67
1'
1.00
+ 1/ 4{0 .15.IJ
l + 1/4(0.lSX)
!!O
o
l
Toe al In.itio.l
3
=
Nz • 0.1:Hl-O.lS)
0.7:9(0,.8.S)
f
3
• 0.67
Initial o2 = o.as--0.67•0.18 1}
l\ • yiP .. e. • l.
1•¡.
n.; p • ~.
P.
-.!:.. .. -l.
V
RT
""" _,,,.
s.:z
ll.t.~
lli {'1+(0.lS/4lt)
11·
8 .2 u:m
:¡,
.l+0.15/4.I.
.os2
l -~ ... · n=csoou gmolºX:
3-27
o.2.
.amo1:., l
. ..
,.,
·,¡
(~0:-1-S-/4-·.-.l
i.
n.l.
A
e
t
r.as
c.1
(atm}
l.
O.lHl-X)
1-.I
1.+{0.15/4]:X
O.OJ
¡5 ( O .rsr:
O .18 -
D
P.
1-I l+[O.lS/411
(O 036 - .Q~
.I)/(1
+ ~
0.03xj 1 + o~s x
O.l5X
CO.Hll
l
.
0.67
0.0.(s/(1~
I>
O.lll/(l~
1)
V• V0 {1..-~¡1$ l)
c1
• 0.1 't\i
pi•
cin,.
(O.CslH$00}(0 .. 2)11i 'l\i
A
0 .. 1$(1-XJ
e
O.ti
-
t .(O .. UJ:)
0, .• 1.s
~i
e. 1.
1.
B
• S.2
o .. Ol . . ,,. O .. OJu -
.x
U
i (0*1SX)
1
0.67
1 .. 23 (l.-1.)
(1.-Xl ,0.-1.S
¡ ·
. X
s .. 2.co .. 11 -
f {a .. :ux.>
1,.23 X
0,.03 X
l .. t.S X O .. l.33
0 .. 1,1 ... M
o.ooax
. l.. ... 0.2 (. . 3-28
o '.1,:S
""}.
~
I)
3)
P'IO'! •
a:z -
S.ua• n
{l)
d)
o.313 1 :a.ea
4A+5B74C+6D
Mole baíance fer a PFR
dFD
sv
Rato l.aw(assutnc elemeatary): -rA
=. kCA. " cil s
Combine with equatícn 3 . 40 and 3-48
3-29
CDP3-E A -+ 2B --)
C -+· D
Symbol
Initial moles
Change
Final moles
Concennaríon
Benzoy1 chloride
A
NA.O
-N ..... oX
NAo(l-X)
NAO(l-X)/V
Atnmooia
B
NAot7a
-2,,~,WX
N Ao(S¡¡-2.X)IV
NAo9c
NAoX
NAo(Bi3·2X) NAJ/..9c+X)
NAo6o
NAoX
NA0(0D-+·X)
N Aof..0p+X)IV
Species
Benzylamide
e
Ammonium Chloride
D
N Ab(8c+X)/V
b)
, t,.-. _ -Q._.-. - N_ .X·__ - N N.NJ·. r_o·!:....
final ccnc , cf
Fir.~l cene.
e}
of
Stoichio~c:tric A
.
-:- + B 2
-->. -C2
table f or 1 f1oew iyste= n.ii:ug 1u.oni.1. u. tb.e btti s
D
+ ..... 2
3-30
Speeies
Amz:nccia Benzo chloride Benzyíamíde
Entering
Oiange
B A
Exir
Conceru:ration
FBo
-FaoX
FaoO-X)
FBo(Bc+X/2)
., e
Ammoniw:n Chlorii:ie • Table
Symbol
O
~FooX/2 Fl30(0 KX/2)
BcFno BoFBo
fBoX/2 FsoX/2
Fao(6c+~/vo ~B-O(So+'t.2)/vo
fo:r th.o fl.oY system. usin¡
ori g i1:u.l iXl the
following
1)
Mola.r flo,r
2)
e.1 ""
3)
Concentrat:ion
F.
.10
:ci.tes
considered
.c>O
fot.nd rat e
V
o
Giveu : A + B
!f B
:r&the:r 10
1::roa tla
tha.xi nu.mber
of moles.
. O
the
e.xiting
molar
flo,,,. z a t e 'by the
•
->
C t.ú:ing
pl:1.co. in
I
square: duct ..
lb molo/sec. i$
a.t equilibrium
in t.he g a.s ph a s e throughaut
Si.neo B münt&ins th.rongh.out th.o ru.etar, it
i11 difforcl\l.t
bas.is
N .. /NA
by dividing
CDP3-F
FAO .. LS
u1moni& &s the
Fno<0n+X/l)
w&ys.
/F.,,, , as oppo s ed ta
.._..ol-w::1et:ric flow
(,1.)
aAF;ao
F:ao(l-X)fvo FBo(0a.Jo/z.}/Y o
is rophcod
iu
o! B in tho liquid.
3-31
ructor
oquilib:riUJl'I v&po.r prouuro
a:, soon as 1 m.oloculo of
by a ~alcculc
the
lJ is consuzud by th.e rc1.ctio11,, Ronco, ó .. l - 1 ""' O, t "' !Ao5
(1-,.BO)PQ
YJ.•.:O f'O CAO RT
•.. . -r A (b)
At
.X •
3
k
RT
p
c.,. CB
p;
te.±] 2
k Ynot 1-yBOl
( l'""'l.)
0.5
(-C.25
atm)(.75
( • 73 O atm. f t
lb ~ole
3
~R
:tt:1n)_··-l
.
. . . .
0
X 1000 R)
{O.S) ... l.1-4
l~ole
! t so e •
2
CDP3-G a)
3 Si HQ3 + 3H2 --+ 2.Si (S) + 7HC1 + Si Hz Ch Take S ¡ HC13 as basis Si H03 + H2 ~
i Si
S¡ HC!J (g)
A
Hz (g)
B
H2HCl(g)
C
o
Si H2Cl1. (g)
D
o
si to
s
o
(s)
i
+ HQ +
FA =FAo (l-X)
F.AO
fi-io
j· SiH2Cl2
= 8'g FAO
Fa =F>.o{Sa
-x)
Fe ==1.FAOX
3
Fo =}FAoX
3-32
Assurne isothermal and constant pressure, Neglect the vapor pressure of S¡(s)
&s = 1 Stoichiomerric feed o cnly involves the changes in gas phase
= y AO O ==
E
t (t t- 1 - 1) +
L(2.) ~.L3
E=
2 3
'"
e
= !:~
Ca
= n. =
Ce
=
~A
CD=
U
u
=
e
-
o -X)
CA(~: X) ( 1 + eX) 3
J \ ··-·
...
JJ
_ o.0088 (1- X) l+X/3
AO(l+EX) -
=
Fe.= 2 . . FAoX_ -1.l
-·
Uo
(1 +- EX)
0.0088 (1 • X) l + X/3
=
2cAo 3
_x
(1 ·+ e:X)
=
o 021 •
.
X
(1 + X/3)
Fº·=.1 .. !110:!---=0.0031--~--v
3v0(1+e.Y)
The soíunon to
l+~
parts (b) and (e)
and
2 S¡ HO; ~ S¡ + 2HO + S¡ HC'.14
e)
are similar to Part (a).
CDP3-H
3.33
1-
. t
ei
F.
;-<
1
• -;-- •
So ,re
a
F.
F.
l.
l.
:.-~i }. • :;-·
1
e a a use
o
t.h.c
st.oichiometric
ta.ble
i
F,
A
FAO( r-x:
C. ' CAO{l-X.)
8
4F AO ( 1-·X.}
4C AO ( 1."-:X)
e
4FAO
D
=
Yhore
CAO
0.0658
6fte;r;
coi::..den.z;s..tion. F
t
v•v
CAQ X
FA.O
5 CA.O
g~al/l
-""v
o F
t.o
a.ad FAO
u
0.02631
FA_o (S-X) a 5 . F AO (0 .. 90)
.subst:it'O.ting
C's
fo:r
F':
:!
4CAO
X
- FbO s
x
•
•
V
Noto:
emol/s
~=X.
o 4 .. :S
at X"" 0.5 (b~ginuing o! condens.~ion} C.= e• i
l.
A
B
e
SFAtrFAoX FAo(5-X) 4FAOX
.... ---Y -·-·--···· o._1FAQC5:X)/0.9 .. Total
FAo(5-X)+FAo(5-X)/9
3-34
CAo(1-·X)(4.5)/(5-X) 4CAo(1-X)(4.5)/(5-X) 4CAoX(4.5)1(5··.X) CAo(5-X)(4.5)!(?.:,_X)J9::Q:5~AQ CAo(S·Xt/40.5
.
Stoich
Table
Assume
scoich.i.omocric
Spocies
f e ed
-~.1:.-·--
Sylllbol
CB:4
A
FA.O
Cl2
B
4FA.O
RC14
e
CCt4
Det'ore
Oi.1.~---
··F
+4FAO'S..
Fe
+F,.0I
Er
=
F AO
~~;2) ·--) ->
6 • O ---·)
Total
l?
FAO"'
o RT
FA0(1-i.)
<4FAO(l-X)
4FA0(1-:I>
4FAoA
4F A.oA
-...
O.lOFT
FAOX
o
•
CAO v0""'
O .50
O
c:onoentra.tio::i
is
0 .08205
o.3.29 -5
g:o
(0.0658)
a.t:m g!ltO l
t./l
""
g,;¡iolll
lit
X:
X
c:o:a.stant ,..
3 48 X:
o. 3 29
3~0 l
. lit
0.0658 smol/1 (0.4)
3-35
1./s
•
-
- ..---#
FTqSFA0-FA0X+O.lOFT
"" 5 FAO
.
I.c
e•
•
·-~9..;.•.;;.3,..9~...;:ª.,.;;t;,;;;t!I:..._
a-u
P.,.p......_•_P_v_r_e_m._,._i_:o._b_.1
Fr = Fr
first begins
Coude1u&tiot1.
F AO{l-X)
FD
SFAO
5 F AO
1 e .. S '1:o
-
A!ter
---
Wbea. coudea.sacion
CAO
FA•
o
Fro
c'!o
X
FB
o
D( l)
AT ""' ~p "' O
.AO
Cond.ca.sa.t:ion
P_..p_<_. _Pv ..._r_"'_m_a._i_.:a._i_n_,
-4FA
o D{ gl
Fi
0.02631
g:nol/s
CDP3-I c2a6(&)
r
+
+
23:::/i)
2oo()c .. 473 r
p ..
-)
C2H4Bt:, (;;1)
z.soo kh
:ti.
24.7
+
lFúh:(g)
acC1,
P V.- ;. e ¡1!48 - "2 = so6.5
xP1
itoich. t1.blé1
Species
Symbol
C2H6 BI2 HBr
A B C
C2ILBI2
D(g)
----·· ············· · ···
Change
In
FAo
. pA(;x
2FAo
-2FAoX
o
FAoX
o
·------- . D(l)
F'¡ After condensation Po=
Fc=2FAoX Fo =FAoX
2FAoX 0.203F'r
2FAoX
o
.
F¡ before condensation Po
Pv remaining FAú(l·X) 2FAú(l X
F' T = 3FAo· FAoX+{).203F' r
F- ' 'r
3F_. _
,-,. 0
FA(/3-X)
"" -
O .. 191
· ·
ó ;:;
--) XC "" O .60!)
o
-)
t:
,,._
o
3-36
e
T
;e:
e_
24.7
~ -
.i.o
--
~t;'O:i,
------
473 X:
0 .. 0820.S &t:x lit::
¡mol X
i
--¡;---
. .
.
..
B
'lfh.ere F
CAo(l-X)(2.39)/(3-X) 2CAo(l-X)(2.39)/(3-X) 2CAO(X)(2.39)/(3-X) CAo(0.609)
2FAoX
F Ao(3-X)/3.93 --·-··--FAo(3·X)/0.797
Total
O .106
:i
AO
g:nol/ s a ad C
,.
t
.
pr
40<
t ..U
o.:r.a, O..l Q,4
o.,41'
a.ou o.o.. o.o.u
10.
('.l.
:x
O .. 212
g:uoll l
p,11l./ •
'•
p;t1/,
p..1/•
llf 40U.-U
l1.ltllI
., J,,(ll
0.111
o
o
o.JU
O.lll
0, l.10
o.o.u
o.u•
o.uo
0.111
o.ou
o.o:n o.o.u
O.JU
0.111
o.ou
o.u,
o.ras
Q •. Jll
o.o u
l!Mll•
t.
gcal/l
Q
~ FAO(I-X)
2FAo(l-X)
C _Q__ .
~ 0.673
F;:
'o
.,. o ... ti
'r
'"°"' JJll,I,()
11-;n
el ,-111
c40u-u
!l•tJl'l,HI
t
, ..w:,<1.... 1)
o..,o,
O,Ckl.
o.ou
o.u,
o.o,.s
0,311
o.a u
o.,
0.011
O,o.4l
0,l'TO
Q,lJl
0,0lJ
\,0
o
o
º·º''
Q.lll
O.OJ,,¡
0,l•"
lF
40t
r-n
1, .Á.Ol
3-37
.u)
1.,l
'J..íJ0.1;1
e"°
o-u.
o
...c.1,,
e;,
IJ!,(:
p,ol/t
·-111
lc:J.O.( 1-XJ 1
l'I. 4li
o.:u.t Q.134
0.11,0
0.1,, tl
fH•
O.l!U
:.e A.O·. = - (Jº"'+, ;.,''· ... I) o.
11···
lC
i.,,
A.o
O .12,
tC,l~ ,,,
e
,(J =e1)
0.1,,
A.Q
lo .. '1q,i
0.11;
O.Jll .Q
O.l~
=
FA.O '"' CAO v o
........
C0.212)
gmol/1 F.
·-V
So wc can first i
ci
A
FA0(1-:U
CA0(1-:X.)
B
2.F.\O(l-l}
2C AO Cl-.X.}
D
•.. ,.....___
Total
where
FAO
t2ble,
F.
... 0.106
x
2FAO
----·F"-0 · 3
substituting
2CA.O X
!
-· CAO
.I
~ . . ...
_
3 CA.O
FAO
g:nolls
gmol/s
o
·-··-···-,. •..l..., ___
e
1/s • 0.106
1.
the stoichiometric
ttso
(0.5)
and
CAO "'"0.212
3-38
gm.ol/s
C's
for
F's
V
•
C' i
Note
V
:a
tb.a.'t
V
o
F.t
-
Fi -;s.
-
(3-X)
o 2.39
(2.39)
v
o
(3-X)
::i..t X .:: 0.609
{bcgin.n.ing
e (911011n
FCA} .;;,.. Ff8) + F (9""01/s}
+ f'(C) -e- FlD} 0.1
+ F(T)
es
01 0.1
o.a
o.9 , .o
X
CDP3-J 3-39
of
con.den.sation)
C.l.
=
C!.l.
~·~· · ·=~~l==:-t~====~~-~ -~- -.:11~
-~-:: ·:=~·:·=-tt~---------~J:6 =
._
----·-·
· _--.::.=!
1.;~F_¿: __ ·
es= ..F!2. = 1 F,..o y,..0 = 0.5
E=
Y ,. olf:::::;:. 0.5(1+ 12---3-·4)=
{
3 \
....
C..
"'º : : : y· e·
AO
ro = o .
1Pa
Pe,·T·.--.o J·i = Ü.,J
*dm·J· 8314 P ª · - ·. *973K
··.R . ·
= 6.18
mot+k
e ~ e·"º (1-.x) "
e 8
= _ 6.1s:._10.-&
1+3X
1+&
'"1.68*~--(~-~
(1-x)
J
l+3X
C __ 2.06 * 10...g X e ---.-~3X-
2.47 * 10·1 X ···· · ·1 + 3X
.·
C.=o
e· ,1,0 --
Y,...o e~ro _ - 0 -~{
Po J·• _.·R::¡~
- -~-·-···· *973K rnal r
. _ C,.. 0(1-X) _ 6.18*108(1. X) --··-··-···················-· - ---···-········· .. ·-···-··· -······-······· -} ·-t· EX 1+3X
(.. A -
e. = ·a
~~!-~~~~~:.(:_:~_={~-)
Ce -·C... .
0
l + 3X
2.06 * 10--a X • 1+3X e
.•......
·
J- . . .
-··¡,;-* lPa;¡;,_;i ·-·· - · ..... - 6.18
O·· · { ·.--- · 8314
·-·-
:::;; 2.47*107_X .•.. . . , . , ,. l +3X 3-40
]:
* 1 O -s
* 10...;g
CDP3-K (,1.)
(b)
Cg;i;¡u:i:¡ ;;;a.~
Cl:Z
A
1.0
CH4
B
CB'.2Cl2
e
o .. s o
HCl
D
o
l'otal
T
l.5
=
6
l + 2 - 1 - 2 •O;
yAO
e=
Ce)
I.iit i ll mQli:~
~DlBll
S:,stom
ó
=
is
.Qlu..
~lll.&::C:
-X
1-I
-o .sx +O
O.S{l-X)
.sx
o .. s:r:
+X
l
1.S
yAO ~ 1.0
O ¡a.s ph.r.so
u.nt.il
F0 '* 400 mmll¡
!2i..SI' (.· 760) • '4.oo 1.5
.
~ ". X"
:w
k = O .l
-rA
::14
k CA
l
l .. 5789
->
CH2 c12
l 1 ( dm· )2 -) gmolc S
3rd
~
f:, A ,,. CA.O (l-1A)
~•t -rA •
CAO (i-I.A)
1
2
3
k CAO
3
(l-XAJ ·. •
3-41
d.c11u UQt
ordor
condonie
!.t
1 at.m.
(d)
c. _
:U_!_
=,
=
A=
2 x 1012
AO
( e)
1.5
l
2k
E
.
1.- .- - --
0.0&206
3
3
CÁO (1-X.~)
(f)
dm6 .. _
s ~ol 0.2
=
2 :r. 1 O 12
E "" 7 4212
_i:1§0
=
kioo· ºC
exp
2 {-·---=_.L_-
( s.
314)
(298 .2)
}
_1.__. __ · g~ol
c:x:p
=
81 .9$
d
3 ~
c--~J..:.. gmol
J.
·-· S
3-42
1-
1
Solutions for Chapter4 - lsothermalReactorDesign P4-1 Individualized
solution,
P4-2 (a) Cooking food (effect of temperature), removing of stains with bleach (effect of bleach conc . ), dissolution of sugar in coffee or tea.
P4-2 (b) Example 4-1 There would be no error! The initial liquid phase concentration remains the same .
P4-2 (e) Example4-2 For 50% conversion, X = 0.5 and k = 0 . 31 Imin' Fe =--6.137 =12.27 lb mol / mm . X 0.5
FAo = -
V o = V Ao + VBo = 2V Ao and a)SO, V Ao V0= 24.52 ft3/min Using Mole Balance,
Also,
= VBo
V= _!oX _
= 24.52x0,5
k(l- X)
0.311(0.5)
FAO
12.27ft3
v AO = -- = ---'-C Aa min
= 78_93
=592gal This is less volume than Example 4-2 because the rate is higher.
P4-2 (d) Example 4-3 For P = 60atm, CAo = 0.0415 lbmol/fr'
C (
_ YAaPa _ Ao -
RT0
-
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) Example4-4 New D, = 3Dof4 Because the flow is turbulent
4-1
1
/J2 =A-= l)p2
0.0775= 0.1033 0.75
[)pi
2·0.103 atm ·60ft 1-
ft lOatm
1 2 1
= (-0.24)2
2/JOL ~ o , so too much pressure drop P = O and the flow stops .
Now 1 - --
r:
P4-2 (f) Example 4-5 For without pressure drop, conversion will remain same as example X= 0 . 82. With Pressure drop,
«.: = ª10 = 0.0037kg-
1
0
For decrease in diameter, Po= 2x0.01244[266 . 9x2+ 12920..8] = 334.75lb¡/ft3 = 334.75(lbrfft3)xl/144(ft2/in2)xl/147(atm/( lb¡/in2)) = 52.71 kPa/m For turbulent flow:
/J ·-· -1-
o,
a - .L => a2 = a1 !O!
and
Po
X=-
IJP1
f'o2 [)p2
kCA0w (1 aw)
1+
V0
_
2
-=
kCAaw (1 _ _qw) V0
4.6(0.949)_ 1 + 4.6(0 . 949)
2
X= 0.8136 -- virtually the same (2) Optimum diameter would be larger
a= 0.037x
52·71 25 . 8
= 0.0756kg-1
4-2
= 0.15814atm/ft
DI a 2 =a-P-=2a 1 D I p2
1-
ª2W = 1-a1W = 1-(0.037)(27.5) = -0.0175 2
Now 1-aW < O, too much pressure drop dueto higher superficial velocity.
P4-2 (g) Example 4-6 Far turbulent flow a--
1
Dp
a--
and
1
Po
a,=q(~;J~)=q[i]G)=q Therefore there is no change.
P4-2 (h) Exmple 4-7 For pressure doubled and temperature decrease Cro = 2*P JRT and T = 688K
See Polymath program P4-2--h.pol. POLYMA TH 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
o
2.26E-04
o o
minimal
o
value
l.363E-05
o
o
2.4E+04 688 0.573773 2.26E-04 0.573773 213.40078 -70.254837 2.26E-04 70.254837 3.939E-04 35 . 127 419
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
70.254837
O . 1189309
o o
o o
maximal value 1. OE--04 2.26E-04 2.124E-04 l. 062E-04 2.4E+04 688 0 . 573773 3.322E-04 0.573773 213.40078 -O .1189309 2.26E-04 70.254837 3. 939E--04 35.127419 0.9395277 0.253044 70.254837
ODE Report (RKF45) 4-3
final value l. OE-04 l. 363E-05 2.124E-04 l.062E-04 2 . 4E+04 688 0 . 573773 3.322E-04 0.0236075 213.40078 -O .1189309 2.26E-04 O .1189309 3.939E-04 0.0594654 0.9395277 0.253044 0.1189309
Differential equations as enterad by the user [ l J d(Fa)/d(V) = ra [ 2 J d(Fb)/d(V) = rb
[ 3 J d(Fe)/d(V) = re Explieit equations as entered by the user [ll E= 24000 [2J T=688 [ 3 J Cto = 2*1641/8.314/T ( 4 J Ft = Fa+Fb+Fe [ 5 J Ca = Cto*Fa/Ft [ 6 J k = 0.29*exp(E/1.987*(1/500-1/T)) [ 7 J ra = -k*Ca"2 [ 8 J Fao = 0.000226 ( 9 J rb -ra (lOJ vo = Fao/Cto [ 11 J re = -ra/2 (12] X= 1-Fa/Fao [13] Tau= V/vo (14] rateA=-ra
=
3.0e-4 ~-------------. 2.4e-4
____ ._._.. -·· ·~-
l . Se-4 12e-4 6.0e-5 O.Oe+O,___----~-~--~-~ O Oe+O 2.0e-5 t.Oe-5V 6. 0e-5
P4-2 (i)
8.0e-5
I.Oe-4
Example 4-8 lndividualized solution
P4-2 {j) Example 4-9 Using trial and error, we get maxirnurn feed rate ofB O..Olmol/dm3
= 0.0251drn3/s
to keep concentration ofB
See Polymath program P4-2-j .pol. POLYMA TH Results Calculated values of the DEO variables Variable --t ca cb
initial value
o
0.05
o
minimal
o
value
0.0063485
o
maximal value 500 0.05 0.009981
4-4
final value 500 0.0063485 0.009981
cd
o o
vOO cbO vO ca O rate
o.os o
ce k
V X
o o
0.22 0.0251 0.025 5
0.22 0.0251 0.025 5
5
o
o
equatíons 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 Explicit equations as entered by the user [1] k= .22 [2] vOO = 0.0251 [3] cbO = 0.025 [ 4] vO= 5 [5] caO = 0.05 [ 6] rate = k*ca*cb [7] 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
3.91E-05 17.55 0.5543321
l. 394E-05
o.os
o.os o
5
0.0078965 0.0078965 0.22 0.0251 0.025 5
o.os
17.55 5543321Differential
o.
0.010 0.008
Q
0.006 0.004 0.002
º·ºººo
100
200 t
300
400
500
P4-2 (k throughr) 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, Pem.
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
dC
-r =k=AeRT =---A A dt so that 4-5
Boiling Point (ºC) 99.2 94.6 88.3
Time (min) 15 17 ?
Far complete conversion (i.e . : well cooked) CA= O at time t
Therefore
e
AO
-E
= tAeRT
C. __..!!Q_
A
-E
= te RT
In ( eAO ) A
= In k = In t -
E ...!__
R T¡,
Now, plot the natural lag of the cooking time versus 1/Tb and get a linear relationship . Extrapolation to T b = 88..3ºC = 361.45 K yields t = 21 minutes.
l
21-i.
r
1
(le
1
t}
11-1
l
l.S-1
1 l
---- . ·- ----·R . . ~-. _ . . . . . . l.69
2.72
"'"~",o=·--
__ .,..__. __ . -
l.161
P4-5 (a)
4-6
C.i.a:;:;; Ca.o
;;;¡
T1
=
l 1rll,121/d;)'!.1' kt
;S dn13lttti1~ ),OOK
V;,.·::;; Vs ;;:
E.;,.
•!lll·
z;;
0:01 di':l1)Jm.oilmin 20;000 cru/mol
Using lí.be t\Jrrhem.u:s e,guatié·n •IA tk1~ CSTR ternperm.mre of 3SOK yiidds t'.bc new :íi~r::ifi,i:; reaetioc rate. , .
k
i;;
t300 ~- 2\:'fü)
2Q.Oü()r(
O,.Ot;t.é:xp I ~~87
1
l \
0
k :;;;:; S..45dtrJ.1lnI!io:l min CS't:R
I.1esI1:,";IJI El:ftt:árl,o,ü;:
X = V ( kC~0
(1-·· X )2)
10
Froin $e qtjadrati.c equgitior1: PBR
V=800 dm3
Design Eq11ation:
T=300K
dX
F.,.o,-=-rÁ
dV
___l!:_
kC~0 V
r-x
FA0
X=0.85
So, considering the above results, we will choose a CSTR.
4-7
P4-5 (b) Batch Reactor V=200dm3 NAo=N80=200 moles X=0 . 90 Assume Isothermal Design Equation:
fx dX -rAV
t=NAOlJ--
=
t
(200moles) (
t
8.4.5
dm3 . J(1 mol mol* mm dm3
)2 (2oodm1)
r-dX (1- X )2
= 1.06min
P4-5 (e) T=273K Find the specific reaction rate at the new temperature of 273 K using the Anhenius Equation.
k
= 2..54 X 10-3
k
=-r-( --
(200)(9) 3·~)
.
2.54 x 10- A 200 J
= 3.543mm = 2 . .Sdays
P4-5 (d) 1) CSTR and PFR are connected in series:
X
. = (200dm3)(007dm3 CSIR
mol.min)(lmol / dm3)2(1--X)2 lOmol / min
/
Solving the quadratic equation, XcsIR = 0 . 44 ForPFR,
dX = (0.07_dm3 / mol I min)CAo (1- X)(l - X)2 dV lOmolelmin
f-
dX . (1- X)3 0 44
= (0.07dm3
I mol/ min)(lmol / dm3)2(800dm3) lOmole / min
X =0.736 2) when CSTR and PFR are connected in parallel,
= (200dm3)(0.07
X CSTR
dm3 / mol.min)(lmol J dm3)2(1- X)2 • .5mollmm.
XcsIR = 0.56 4-8
f
ForPFR,
0
dX (1- X)2
= (0.07dm
3 /
mol.min)(lmoll dm3)2(800dm3) 5mol/ 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':13 J(60min)(24hJ mm hr day
= 14400~ol
day
If the reactants are in the same concentrations as in the flow reactors, then
V= (14400-molJ(!dm3J = 14400 dm3 day mol 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.
dX dt lR
fR
=
= -rAV = kC~0(l-X)V J\/AO
J\1AO X 2 VkCAO 1-X
=-1-~= «e¿ 1-X
J\/AO ,
.
J\1AO
and smce --
V
= CAO
l 3 dm_* 1(mol) 4_2 mol-hr dm3
º·9 =2.14hr 0.1
Assume that it takes three hours to fill, empty, and heat to the reaction temperature. tr= 3 hours t,01al = tR + tr t,0ta1 = 2.14hows + 3 hours = 5.14 hours. Therefore, we can run 4 batches in a day and the necessary reactor volume is
14400dm3 = 3600dm3 4 4-9
r 1
Referring 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.
P4-5 (f) The points of the problem are: 1) To note the significant differences in processing times at different temperatures (Le. compare part (b) and ( c)) . 2) That the reaction is so fast at 77ºC 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. lt does not matter if the reactor is red or black.
+
butano!
+ water
·-·:-
-rA=kC,..C3
elerrentary reaction
·-t,., "" k (F¡Ju) {F ¡Ju)
for liquid systerns volumetríc flow u= uo
4-10
V
- _F~P..:!_ - --r - kC2
-~º
CSTR -
,1
ri'lc
=4x
Fe
(1--xxs -X)
105 ~
, Fe:::;; 4 x 106 lb J yr , 30d J yr: operation
'
F = 4 x 106 112. ,e
V
·
yr
x .!J'.!. -l.S. ... x l lb mol 30d 24 hr
. .) l ft --. -7 .48 gal
- 1000 ga.. ~• X CS!R .
HJ.7 f,3 ""'---·----·-····
3
11 ~ _ft -~ lbmcle-hz
278 lb
""'
l-1'"3 . -1 ft.3
20_lbmoie/hr
-·- -··· · ·-
(·0.2lbmole)-(l· X) t? .,
(S-X)
X2 • éX + l.89 :: O X= O.T3
P4-6 (b) To increase conversion, use PFR, higher temperature, or use better catalyst.
P4-6 (e)
RL
e
P4-6 (d) PFR Design Equetion dX. ·- r ~"'~==~ ..
av
dX . -d.· I/t
F,\o
kC! (1-XX5·-X) ~-~ ... - - .-0
~--
F AO
V =535dmJ 4-11
P4-6 (e) 14~4 X
H'.r Ibmn] cfDBP nix:d to be ptceuccti in 10 days t ;::; N .1.0
Mok balance:
dX ft =:': r ... r 0 ....
1
t
l
5 .... X
.
= kc ,.; cs=i). h,c·S{1--x))
The above equation relates the reaction time for a batch and the conversión achieved during rhat batch. There is a trade-off berween high conversion and few batches and Iow conversión but many batches per day, What couversion will result in the smallest nurnber of reactors? N\o = 133.7 ft3
e
* CA.O
Ptoducr
N.,0X*24hr
-,;acw;,aay- -~ ;::;--- --
N,mX24 = -, ~ +-3 ·· =
:~!01:PC(:9_1!~:. = ~~}-~:!.!'~ .~9!1!?.!day
30days '278lb 480mol l dav n = --·· ···· · --·-· · · · · · · · . - --·· · ·
f
AO
= 0.2lbmol / ft:3
k~~~{}~;_f; X N,,.0X24
r···-'·'"····················--
mj
..·-···
day
·=
'
-..--·
\
l--··--··-·-· o
•· ·
---1
-
2 1: 1 O
···
'
~-------------~ -
·-- ._,
-~--· · ·v04
0.2
-(') =J
= 480-1.?!~~
(X)mol J day / reactor
25
.
~
o.s
-
1 , .•..•....
_.J
0.8
X---->
""D1e mínimum 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-48+2C
v"J-ioX -·r,¡
e -e r\ -
e= y ..¡16 = (1}(2. + l-1) = 2
(l--X)
·"'º (l+ e.X)
= ~t = (o.0JK405)=0.:;.f}
C,1o
..,.l/
(2.5mo1/min) . . · . }(1+2.)ln..L..,.1.s)
. (o.04!
= f 189.4
mh't-1
xo.3mol/run3)}
1,-
.9
drn3K5,U)
'
J
P4-7 (b) V,,, F f\oX(1+.:~J
.
CA0k(l-X)
=
$.5,,(l.X.)( \
1 )
k 10·1 .e-l\.311.ñf.,ioo
= 0.044
.
P4-7 (e) For a.= 0.00ldm·3
See Polymath program P4-7-c.pol. POLYMATH Results Calculated values of the DEO variables Variable V X
y
Co esp alfa
initial value
o o
minimal value
o o
O .1721111 0.3
i
0.3 2 0.001
2
0.001
maximal value 500 0.656431 1 0.3 2 0.001
final value 500 0.656431 O .1721111 0.3 2 0.001
4-13
- ...
-·---------------·-----------·----------
·---·------------····-----·-·-----
-
-----· -----
·------------·
-------------------·-··----·---------·--·-----
.
.
-----,-----
·-----------------------------·
e k r fo
0.3 0.044
O. 3
o.0077768 0.044
-O . 0132
O . 044 -3.422E-04 2. 5
-O . 0132
2. 5
2.5
0 . 0077768 O . 044 -3 . 422E-04 2. 5
Differential equations as entered by the user [ 1J d(x)/d(v) = -rifo ¡ 2 J d(y)/d(v) = -alfa*(1 +esp*x)/(2*y) Explicit equations as entered by the user [ll Co 0.3 (2 J esp = 2 ¡ 3 J alfa = 0.001 ¡ 4 J e= Co*(1-x)*y/(1+esp*x) [5] k = 0 . 044 [ 6 J r = -k*C [7] fo=2.5
=
0.8
0.6
N
0.4
AtV= 500, X = 0 . 66, y = Ü..17
l_yj
0.2
P4- 7 ( d)
Individualized solution
o.o ----'-----'-----------'------'
P4-7 (e) A -
O
100
B + 2C
200
300
V
400
500
)=o
Stoichiornetry :
l.+eX 4C X'
- ---~---·-· (1 + eX)""(l- X) X . ;;;::¡
-o·· .._:,_ . ,
-
= (0.90)X.,q = 0.47
:;;,:;;
.
. . . . . .... . . '
..
(2 5 mol/rnin )( 0.47)( 1 .,. 2(0.47))
-·"·""""'"'·-'''""""""''"''""""·•·•,··•·•,·•·;:•
"'""'"""""''"""""-ª""'""""''''''''"_'""'""''"''0--
.••••-•+-. ..... ,,,,,,.,,,..,':".,,•. '""."... ,'-'.••".. _~··,.
;·):(O ,.,,; to 1)4' , .. · \io ,, · ,; ')•lr . . , · \-¡;. . n.o. .,, 1) \. , · + mm A .,, mol/drrr' ,1- 0.47 > '(·_-0. ~-·.m t)..)':0..•_ .¡
1
4fc·) •
\l.'1' -
V =1300 dmJ
4-14
, 1/d
.... ,
. 0.;_;, ;:::···
..
Using these equations in Polymath we get the volume to be 290 dnr'.
P4-7 (f) A~
B+2C
Rate Law :
Kc
= -k
· r..._
1{
C"' -
=O
C3C~íL
(at equilibriurn)
= 0.025
Stoichiometry
:
C
C,.
.\
= C,."(l··· X) l+eX
= 2 and C Ao xcq =0.52 E
0
= 0.3
C
-a
=
X)!( C.,.l (l··+ eX ·X)
K _ CBC~ -( X 1(2C.,. e - C.,. - l+eX; l+eX 0
'
C,.ºX l+eX
'
C == 2C..._.,X e l+eX
)- -~- 4C~ X3 _ - (l+e:X/(1-X) 0
0
X= (0.90)XC(\= 0.47
f.FR F.
dX = -r
''" d~Vr
·rA =
ar ""
lf ~-i(
.~~
dX •
= f .. º
(l - X)-¡¡
~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 Fao Cao k
e ra
initial value
o o
0.025 2.5 O. 3 0.044 2 -0.0132
minimal value
o o
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 [lJ Kc = . 025 [2] Fao = 2.5 [3J Cao = .3 [4] k = .044 [5J e=2 [ 6 J ra = -{k*Cao/(1 +e*X))*{(1-X)-{4*Cao"2*X"3)/((1 +e*X)"2*Kc))
S::.S.IR -r;
=
ü~f~i(I
-·· ·X) ...
Ü
~~~J~~·)
V= 1300 drrr' PFR with pressure drop: Alter the Polymath equations from part (e).
See Polymath program P4-7-·.fpressure.pol. POL YMA TH Results Calculated values of tbe DEO variables Variable V X
y
Kc alfa Cao k
esp fo r
initial value
o o 1
0.025 0.001 0.3 0.044 2
2.5 -0.0132
minimal value
o o
O . 3181585 O . 025 O . 001 0.3 0.044 2
2.5 -O. 0132
maximal value 500 O. 5077714 1 0.025 O . 001 O. 3 0.044 2 2.5 -1.846E-04
ODE Report (STIFF) Differential equations as entered by the user [ 1l d{x)/d{v) = -rifo [ 2 J 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 [3J Cao = 0 . 3 [4] k = 0.044 f 5J esp = 2 [6] fo= 2.5
4-16
final value 500 O. 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*Cad'2*x"3/((1+esp*x)"2*Kc))
At V= 500 dm3 X= 0.507 and y= 0.381
P4-7 (g) Membrane reactor: A --+ B + 2C CA= CoFAIFr CB = CoFBIF1 Ce= CoFclF1 FT=FA+FB+Fc and -rx= r8 = rcf2 Using polymath, ForPFR,
See Polymath program P4-7-g.pol. POLYMA TH Results Calculated values of the DEO variables Variable V
Fa Fb Fe Kc Ft Co K kc ra
initial value
o
2.5
o o
0.025 2.5 0.3 0.044 0.08 -O. 0132
o
X
minimal value
o
1.3231889
o o
0.025 2.5 0.3 0.044 0.08 -O. 0132
o
maximal value 1040 2.5 0.3635477 2.3536223 0.025 3 . 7452437 0.3 0.044 O . 08 -3.827E-04 0.4707245
Differential equations as enterad by the user [ 1 J d(Fa)/d(v) = ra [ 2 J d(Fb)/d(v) = -ra - kc*Co*Fb/Ft [ 3 J d(Fc)/d(v) -2*ra
=
Explicit equations as enterad by the user ¡ 1 J Kc = 0.025 [ 2 J Ft Fa+ Fb+ Fe l3J Co 0.3 [4J K 0.044 [5J kc .08 l 6 J ra (K*Co)*(Fa/Ft· Co"2*Fb*Fc"2/(Kc*Ft"2)) ¡ 7 J X = 1 - Fa/2.5
= = = = =-
Solving for when X= 0.47, we get V= 1040 dnr'
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
o.o o
208
416
V
624
832
1040
P4-8 (a) The blades makes two equal volumes zones of 500gal each rather than one 'big' mixing zone of lOOOgal. Expected
Actual
- .,._ ·1 1
~].~1
.... T:,.X::.:ü 5
··-····· -····· .. ) Single Blade '"""""
Predicted
1···· 3X1 + Xf =0
2
a =[gal] 1000 gal=
X
.5
ª(1~-xyx= ª::is·= 2a a=500 gal
500:::::
38 Xi ···3X+
4-18
X,
3~t = ------·-···············-·······- = 57 2
Mensurad
X1~
,.
~ 2,
· X~--
V ~ i'J.(1··-Xf 500 gal X, 500 "'.al. :t.::l. -·-· "} ! .. º '"' l ·'-.... X)·.·u'"' . 1.·
Fer Reactor 1
;
· 12 -=. X t l. -· 2'"A1 + X"·
P4-8 (b)
-e
A CSTR is been created at the bend due to backmixing, so the effective arrangement is a PFR is in series with aCSTR.
PF~- _
[-+---
CSTRzone created due to backmixing
A-B
k = 5 min·1 Xexpected
= 0.632
v; = 5 dm3/rnin. Xactual = 0.586
V=fFAadX=vºIn[0
-rA
k
1
1- X Expected
]=1ln( 5
1
)=1.0
1- .632
Now,
4-19
r
ForPFR,
For CSTR,
v.
VP -+--1 V V
Also,
.....
.
..3
Solving 1, 2 and 3 by using polymath,
See Polymath program P4-8-b.pol. POLYMA TH Results
NLES Solution Variable Xl Ve
Value
·-~f~(_x~)~-~~-In_i_Guess
0.350949 0 . 567756 1 O . 586 0.432244
V
X2 Vp
3 . 148E-10 -3 . 297E-14
O 1
NLES Report (safenewt) Nonlinear equations lll
f(X1) = ln(1/(1-X1)) -Vp = O
[ 2 J f(Ve) = (X2-X1)/(1-X2) -·Ve= O
Explicit equations [ll
V= 1
[21 X2 = .586
(3 J Vp =V-Ve Ve= 0 . .57 dm' ; Vr= 0.43 dm";
X1 = 0.35
P4-8 (e) CAo = 2 mol/dm3 A~B
Assurning lst order reaction, For CSTR,
~
,r
= CAOX
=r,
-rA = kCA = kCAo(l-X) => rk
V
ForPFR,
= _.!!.__ == ~.4 = 0.67 1-X
= F xf AO
=>
X PFR
O
0.6
dX
kCAO (1-X)'
= 1-exp(-rk)
- rk
= xf O
dX_
1-X
=l-exp(-0 . 67) = 0.486
Now assuming 2nd order reaction, For CSTR, Now, assurning 2nd order reaction,
4-20
CAOX T=--
ForCSTR,
-rA -rA
=>
ForPFR,
=kC1=kC10(1-x)2
-r k eAa
=
T=-1-Í kC AO =>
X
2
(1-X)
= --2OA = 1. l 11
dX O ( 1-
X)
0.6
2
=-1-~
kC AO 1- X
X =1---1--=1--1-=.526 1+-rkCAO 2.111
So, while calculating PFR conversion they considered reaction to be lst 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 dnr' Mole balance:
V= FAaX
=r,
4-21
Ratelaw: Stoichrometry:
V=
FAO
koCAo'
[(!- x)' _ ;:J X
Z =-1:Ao = 2_ k0CA0
Z Now using:
x2] [ (1-x)2---Kc
= 37~5.4[0.62
=>
f (X)= O= V -(
.
_
0.4
X
O 42] 3
= 2902.2dm
3
V=(·z) z
where
V
R T0
z_) __ _!_
_
z [ (1-x)2-ic x2]
[(1-x)'
z = .!_ = exp ( E k0
--f] [J- _!_JJ
X
and
T_
Ke = Kco exp( MI~[ _T_!_ _ _!__]J R T o
Solving usmg polymath to get a table of values of X V s T.
See Polymath program P4-9 po1 POL YMA TH Results
NLE Solution variable X To T z V E
R
y
Kco Hrx
Kc
Value 0.422~453 300 305.5 2902.2 3785.4 l.5E+04 2 1.5684405 3 ·-2 5E+04 l 4169064
f(x) 3.638E-·l2
Ini Guess . 0.5
NLE Report (safenewt) Nonlínear equations
-···----------·· --------·-----~
043 042
¡ 1 J f(X) = (z/y)*X/((1-X)"2 - X"2/Kc) .. y = O
__
041
.......,....
04 -
X 0.39 0.38
4-22
------··----------·-----·--
0.37
036
¡
-------~
------~---,-------¡--·-·--r----·
~
i
Explicit equations [ll To= 300 (2) T 305.5 [ 3 J z 2902.2 (4l V= 3785.4 (5) E= 15000 [6] R= 2 [ 7 J y= exp(E/R*(1!ío-1!í)) [SJ Kco=3 ( 9 J Hrx -25000 (10] Kc = Kco*exp(Hrx/R*(1!ío-1!í))
= =
=
T(in K)
X
300
0.40
301 303 ...__
304 305
--
-
--
0.4075 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
.1
Fso - Fs + r,V = O
(Cso- Cs)Vo = I'g VYs1e = VYs1e[
P4-10 (b)
µmaxcS KM +es
ce]
Ce= YCIS [eso -es] (eso-es
)va -VYs1e
[ µMAXCS KM+Cs
ce]= o
( e; --es )va -VYCIS [ µMAXCS ]( <. -es)=
=>
KM+Cs
(30--Cs ).5-2.5x0.8x[º·5xCs ](30-Cs) .5+Cs
o
=O
Solving we get Cs = 5.0 g/dnr' or 30 g/drrr'. if Cs = Cso no reaction has occurred so the only valid answer is Cs = 0 . .5 g/dnr'.
P4-10 (e) Ce= YCJs(Cso - Cs) = 0..8(30 - .5.0)g/dm3 = 20 g/drrr' .
P4-10 (d) Vnew = Vj2 = 25 dm3/h
Using equation from above, we get C, = 1.67 g/dm' and Ce= 22 . 67 g/drrr'
P4-10 (e)
= VJ3 = 25/3 dm' Using equation from above, we get Cs = 3.0 g/dnr' and Ce= 2L6 g/dm3 Ynew
P4-10 (f) For batch reactor: Cso = 30 g/dm" Ceo= 01 g/dm" Ce= Ceo+ YCJs(Cso - Cs) V= 10dm3
See Polymath program P4-10-f.pol. POLYMA TH_Results Calculatedvalues of the DEO variables Variable t es eso Yes Km
Umax Ceo Ce
initial value
o
30 30
O. 8 5
O. 5 o. 1 O. 1
minimal value
o
0 . 0382152 30 0.8
maximal value 15
5
30 30 O. 8 5
0.5 O. 1 0.1
0.5 0.1 24.069428 4-24
final value 15 0.0382152 30
0.8 5
0.5
O. 1
24.069428
rg
rs negative_
0.0428571
-0.0535714 0.0535714
0.0912841 -0.1141052 O .1141052
5.4444349 -0.0535714 6.8055436
0.0428571 -6.8055436 0.0535714
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(Cs)/d(t) = rs Explieit equations as entered by the user [1l Cso = 30 [21 Yes=0.8 [3) Km=5 [ 4] Umax = 0.5 [5l Ceo= 0.1 [ 6 J Ce = Ceo+Yes*(Cso-Cs) [ 7 J rg = (Umax*Cs/(Krn+Cs))*Ce [Bl rs=-(1/Yes)*rg [ 9 J negative_rs = -rs 30.0r---=-===--
7.0 ~----------------,
·-------·
5.6
24.0 ·
--] l
18.0 ·
4.2
12.0
2.8
6. 0
1.4
0. 0
0.00
J.00
6.00 t
9.00
12.00
atini.rs
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
= 0.0015min-1
at 8ü°F
E= 25, 000~gmol
X = 0.90
lb MWA =MWB =58-lbmol
D1
P
= 132
psig
= 146.7
psia
T
=1
M =lOOOE:_ B
inch (I.D.)
= 260º F = 720° R 4-25
hr
L = lOft
n1 = number of tubes
1000!! Fs = hr =17.21 lbmol lb hr 58 lb mol
F AD
lbmol 17 21 = FB = · hr = 19. l lb mol X 0.9 hr
For a plug flow reactor:
V
= ni7rD/ L =F
AD
4
dX
Df.9
=r,
D
ó=l-1=0
e = AD
f
f
V= FAD Q9 dX = FAD Q9 D
At T2
-rA
D
PA =__!_ RT RT
= FAD ln ( _1_ ) kCAD (1-X) kCAD 1-0.9 dX
FADRT lnlO kP
= 260ºF = 720ºR, with k¡ = 0..0015min-1 at T1 = 80ºF = 540ºR,
k2=k1exp
1 1 -E( --( R Y¡ T2
)J =0.0015exp (25000( 1 ------1.104 540
1 )) =53.6mm__ 1 720
k2 = 53.6min--1 = 3219hr-1 19.l lb_mof )(10.73 ft3 psia )(72ü° R) V = FADRT ln 10 = _ hr lb mo~~-R-----ln kP (3219hr-1)(146.7psia) (
V= 0.72ft3 V
= !.!LTfD/ L 4
4(0.72Jt3)
4V
n =--=-r
7lD2L t
7l
(
-
J2 12 ft (10ft) 1
13.2
Therefore 14 pipes are necessary.
P4-12 A -~ B/2
4-26
10
Stoicl1iometry
l
1
s
·-·
- ,., ,.1 ·:::::; --
1/AO
=
.F~10 ,.,~·---·-Fr10 Fso
-
€
.
2
.
2
+
11.,1015
VPFH
=·
1
+ l?co
2
1/4
1x ex
,:::.:: F.40 . ()
-
··-7
.. -.
A
R.ate . T,aw__ (elementary.yeacti~>n) .• , ·- kC''2 -
·-7 A ~-· . . .4 ~
i,c·,2(
'L ' , x,'),· 2
t, ,., A, O . .,. . . . , ( 1·.· , -t· · · , X )·· ., ... ,, é ..
Combining
( fer the integration, refer A ppendix A)
from the Ideal Gas assumption,
= YAocTO X : :~ 0.8 and E = -1/4 to Eq, (4) yíelds, CAO
Substituting Eq.(5), VPFRky!oc:o
~~"------'---=
2
2(-1/ 4)(1-1/ 4)ln(l-0.8) + (-114) 0.8+
FAO
(1-1/ 4)20.8 1-0.8
=
2.9
(6)
Molar ílow rate of A cut in hall, F.~o
F~~to
- l
FÁo -J-·FBo + Feo - 3 Y:.106 = - · 1 /6
Frorn Eq. ( 4), i. ~1PFR
kC~,z ·.: AO
(1 L eJ)2 ,·VI\ , f(· l t t ')[ n. (.· 1 - X,¡) +e~ , a I 2t X + - ·--·-r·--··-.
F'AO
·
4-27
-·--·--------··-·---------···-----------·--···
--·-------------·--------·---
----------------------- ··---- -
l -·· X'
VPFRky
,2 AO
F'2
c·2TO
-
AO
i
'(·
2e 1
+
')l u·( l - ,(\.\.-',-.) t
+t
r> •)
.,'\.
( l ·· 1
2( . .:.J/6)(1 - 1/6)ln(l -- X')+ ( ---l/6}2X'
l ·'· --1
Polyrnath Non-Linear Equation Solver: X'
P4-13
= k¡ ( CA ---C9
s:
t
.tG)2 X'
=
·/_xí'
0
·
0 . 758
A <=> B , liquid phase reaction
Given: The metal catalyzed isomerization
-rA
r
+ 1+. ,._l:j:,)"_,,"\'. (l
12vt
J .
with Keq = 5.8
For a plug flow reactor with YA= LO, X1 = 0.55 Casel: an identical plug flow reactor connected in series with the original reactor.
Since YA = 1 . 0, eB
= O . Por a liquid phase reaction CA = C AO ( 1- X)
-r, =kc,.((1-x)dX
V,= FA,
f-r, =
:J
kC AOV!..
= xJ o
FAO
dX
X¡
F,,
f kCAO ((1-x)-
dX
1-(1+-1 K eq
= C AOX
- -·-
·-----------------------------·
J
(
Poi the first reactor, X¡
and CB
Jx
or
1 [ ( 1 = 1+__!_In 1- 1+ Keq
J ]
Keq
X
o
4-28
----------------·-- ····--·-----------------------------------
···--------
- --
-
-----···-
---·-
--········--------
- ---
--
--
,.
-
---
.
-
-·- -- - ·-. - ------- ----------- -·-·-··
·---·---------------------~---
·--··-··---·--
----
-
----------------· ··---··- - -
--- ---
-
·-------------·-·-
\ 1+-
-Jx
1n[1-(1 +-1
1]
Keq
x;
=-0.853ln(.355) =0.883
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__ = XJ FAo
kCAOVF FAO
º
dX
1-(1+-1-Jx «.
= 2 kCAoVi = FAO
o
!+~ 1.[i-(1+ :Jx,] Keq
.
2kCAoVi =2(0.883)=1.766 FAO 11 1.766= · 1n[1-(1+-1 1+-· 5.8 5.8
-)x
2]
X2 = 0..74 Case 2: Products from 1 st reactor are separated and pure A is fed to the second reactor,
A
x,
w 8 The analysis for the first reactor is the same as for case 1. 4-29
1 [ ( 1 J ] 1+-l_In 1- 1+
V¡ _
kC
AO FAO
Keq
«:
X¡
By perfonning a material balance on the separator, FAo,2 = FAo(l-X1) Since pure A enters both the first and second reactor CAo,2 = CAo, CBo,2 = O, 9B = O
CA= C AO ( 1 - X)
Vz
= FAo,2
dX
f--rA
x;
CB = C AOX for the second reactor.
o
X) x:
FAO ( 1-
=
dX
f (l-X)-- X
kCAo
o
«;
and since V1 = V2
kCAoV2 __ kCAoV¡ FAO
FAO
or
__ _!_11n[1--(1+-j-Jx1] 1+-K eq ~
=-
l-~ 1
l+-
1n[1-(1+-1Jx
2]
Keq
Kq
-
1-(1+ ;~ Jx, =[i-(1+ ;JxJ~' X2 -
1- 1- ( [
-
l+K.q1 J X ] 1+-1«: 1
1-\,
1
= 1-(0.356)045 =0.766
1.174
Overall conversion for this scheme: --
-
FAO - FA0,2 (
X----F
AO
X
1 - X2 ) ----
-
FAO -- FAO (
1 -· X 1 ) ( 1 - X 2 ) FAO
=0.895
P4-14 4-30
-
-1-
(
1-X1
) (
1-X2
)
Given: Orrho- to meta- and para- isomerization of xylene.
M .:»: -> p M
>0
k2
> P (neglect)
O
Pressure = 300 psig T = 750ºF V 1000 ft3 cat..
=
Assume that the reactions are irreversible and first order. Then:
=r« = k CM
+ k2CM
1
= kCM
k = k, +k2 t:=0 Check to see what type of reactor is being used. Case 1:
vO = 2500 gal
X = 0.37
hr
Case 2: V
O
=1667 ga[ hr
X =0.50
Assume plug flow reactor conditions:
FM0dX = -rMdV V-F -
V=
MO
XC
f
or
xfdX O -rM
Movo
d
=r«
X
X =vo f.
d
X·-·= vº·In(l-X) k(l-X) k
0 O CMo, k, and V should be the same for Case 1 and Case 2. Therefore,
( kV)Casel = -( v0)ca,el ln (1- Xcaiel) = -2500 ~~ ln [1-0.37] = 1155 ~~ ( kV)Case2
= -(
Vo
}ca,el ln (1- Xcase2)
= -1667 ~~ ln [1-0.50] = 1155 ~:~
The reactor appears to be plug flow since (kVkase 1 = (kV)case As a check, assume the reactor is a CSTR..
FM0X
= CM0v0X = -rMV 4-31
2
V=
CMoX
v
-rM
=
O
or kV
VoX
k(l-X)
=
voX
1-X
Again kV should be the same for both Case 1 and Case 2.
=
( kV)
O Case!
1-
Caiel
=
( kV)
X
(V )
(V )
o
Case2
X
Casel
=
2500-gal ( 0.37 ) hr
1-0 · 37
Casel
Case2
X Case2
1-- Xcase2
=
= 1468 ga
l
hr
1667-gal ( 0.50 ) hr
1---0.50
= 1667 ga
l
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
1000/t3 cat.
=1.55
gal
hr
ft' cat
Por the new plant, with v0 = 5500 gal/ hr, XF
= 046,
the required catalyst volume is:
-5500 gal_ V= -vº In (1- X F) k
= 1.155
h~ In (1-0.46) g~ hr ft cat
= 293lft3 cat
This assumes that the same hydrodynamic conditions are present in the new reactor as in the old .
P4-15
S . · : : ~-: _ : : 3~~ t ·~
A- B in a tubular reactor
J
t
----+)
~~-:-:)L:
.
... - .
:=;__]
-~
Tube dimensions: L = 40 ft, D = 0 . .75 in. ll¡= 50
4-32
500~
FAo =~= MWA V=F
AO
hr =6.86/bmol lb hr 73 lb mol
xfdX O -rA
-r = kCAo(l-X) =kC (1-X) A l+eX Ao V= FAo ͪx·= 0-rA with
e
AO
FAo Í dX 0kCA0(1-X)
= FAo In(-1-) kCAo 1-X
= __!__ = y AOP
RT
V = FAoRT kyAOP
RT
In (-1-) 1- X
or
k
=
FAoRT Vy AOP
ln (-1-) 1- X
Assume Arrhenius equation applies to the rate constant. At T1 = 600ºR, k¡ = 0.001.52 =
-E
AeR-z; -E
At T2 = 760ºR, k2 = 0.0740
= AeRTi
~ =ex{f(;,-tJ] In k2 k,
_-E(·_!_ _ _!_J- E T -T¡ 2
R
T2
E= 'I'¡T2_ln k2 R T¡-T2 k1 A= k, ro k
T¡
R T¡T2
= (660)(760)In 100
0.740 =l9 SOü°R 0.00152 '
exp[~-J RT¡
=k,
exp[-!U-:.JJ
From above we have
k
= FA0RT
80
FAoRT_ VyAOP
(-1-) In (-l ) = exp [-
VyAOP
In
1-X
1-X
k1
_!_J]
E [_!_ _ R T T¡
Dividing both sides by T gives: 4-33
FAOR
)= k; ex+f(f-t)]
ln(-1 1-X
Vy . AO P
T
mol )(10.73 psia- ft3 ( 6.86 lb hr lb mol ºR
J
ex
p
[-19500(!-
T
1 )] 660º R
-,-~~~-,--.,..-~-----'--,--~~-'-~....;_~~ln5=~--=-~~~~~~~-= (.00152-8e-c)( 3600 :; )( 6.14ft3 )(114.7psia)
T
Evaluating and simplifying gives:
0.0308º R-1
=
exp [-19500(!-
-1-)] 660º R
T
T Solving for T gives: T = 738ºR
= 278ºF
P4-16 Reversible isomerization reaction m-Xylene --> p-Xylene X, is the equilibrium conversion. -r
Rate law:
m
=
At equilibrium, -rm = O
k(C - CPk J m
e
em ek
=>
=-p
e
C
mo
(1- X e ) = CmaXe K e
K=~ e
-r m = kC AO
1-X
e
(1-_!_J X
e
P4-16 (a)
l
4-34
l
------
-
--
-
-·-·-···-·····----·--
Por batch reactor,
dX dt
Mole balance:
T=Xeln( k
= -rm V = kC AO NmO
Xe
x, --X
CAO
(1 - x,X J
)
ForPFR,
V-F -
= v0 k
fdX -rm
AO
f
;Jx f =: 1-(1+ ;Jx dX
1-(1+
dX
1
r,,. T PFR
X Xe =-eln k
X -X e
P4-16 (b) ForCSTR,
V= FmoX X TCSTR
= [ ( k 1- 1+
;e J ] X
Putting the value of K¿
P4-16 (e) .
Volume efficiency =
TPFR
rJv = --
TcsTR
Xe k
in(-
J ln(~-JJ
Xe Xe- X X e -- X = ( ) = ( xe X X k X e -X X e -X 4-35
( Xe - X ) ( X In X x, ~ X
= -
J
¡
r
n = V
l-(ft
1
X
xe Following is the plot of volume efficiency as a function of the ratio (X/X0),
See Polymath program P4-16-c.pol.
Efficiency
0.8 0.6
OA 0. 2
0. 0
-----·-0.802
e.aos OA06x:Jx1·6o4
0.010
1.000
P4-16 (d) Efficiency = V PFR / V CSTR = 1 from problem statement, which is not possible beca use conversion will not be the same for the CSTR' s in series as for the PFR.
P4-17 (a) A-YiB s = -1/2, X= 0.3, W = 1 kg, Yexit = 0..25 -rA = kC/
ForPBR,
dX
__ r,
dW
FAo
dX
--
dW
=
e =c (1-x)y 0
and
A
kC0(1-x)2 y2 v)l + EX )2
(1-X)2 y2 z~--'--(1+EX)2
and
(l+éX)
let
dy dW
kCAo
z=--
v0
= _ _!:_(1 + EX) 2y
Solving for z by trial and error in Polymath to match x and y at exit, X= 0.3 Yo= 1 and Yf= 5/20 = 0.25 we get: a= 1.043 kg" and z = 0.7 kg"
See Polymath program P4· 17-al.po1. POLYMA TH Results
4-36
Calculated values of the DEO variables Variable
initial val u e
o o
w X
y
-0.5 1.043 0.7
z
o o
maximal value 1
0.302004 1 -0.5 1.043 0.7
0 . 2521521 -0.5 1.043 0.7
1
esp alfa
minimal value
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 [ lJ esp -0.5 [ 2 l alfa = 1.043 [3J Z= .7
=
Now for CSTR:
- F0X _ X(l+cX)2 W ---------
z(1-X)2
-rA
Solving we get for W = lkg and z = 0. .7 kg' X=0.40
See Polyrnath prograrn P4 .. 17-a2.pol. POL YMA TH . . Results
NLE Solution Variable X
W esp Z
Value 0.396566 1 -0.5 0.7
f(x) -1.142E-13
Ini Guess o. 5
NLE Report (fastnewt) Nonlinear equations [ 1J f(x) = W*Z*((1-x)/{1 +esp*x))"2-x = O Explicit equations [1] W = 1 [ 2 J esp = -0.5 [3] Z 0.7
=
P4-17 (b) For turbulent flow:
G2
a= (constant)-
DP
4-37
final value 1 0.302004 O. 2521521 -0.5 1.043 0.7
= -ª1 = 0.0326
a2
=>.". Z2
and
32
= 4X0.7 = 2.8
Now solving using polymath:
See Polymath program P4-17-b.poI. POLYMATH Results Calculated values of the DEO variables minimal value
initial value
Variable w
o o
o o
X
0 . 9887079 -0.5 0.0326 2.8
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))A2 [ 2 J d(y)/d(w) = -alfa*(1 +esp*x)/(2*y) Explicit equations as entered by the user [ 1 J esp = -0 . 5 l 2 J alfa= 0.0326 [3
l Z = 2.8
So, conversion in PBR, X = 0 . 862
P4-17 ( C) Individualized solution P4-17 ( d) Individualized solution P4-18
Given a Fluidized Bed CSIR W
=
FóltX
-¡Á
o=O ·
fÁ ""'kPA
E=O,then
f\ s=J·';,0(1-X)p
No pressure drop in the CSTR PA ~
PAo(I · · · X)
\V""
p n
f\o~
kl\o(I
X)'
4-38
final value 1 0.8619056 0 . 9887079 -0.5 0 . 0326 2.8
k&
FAo unknown .. group inro a constant, use values from lst case. k
( X
FAo =.
i=x)
1
l 05 l PAoW ""'05 20(50)
k
10-3
F,\O = ~t;1~kg;;t Put PFR downstream less wasted volume
a)
vs
P4-18 (b) b) PBR: FAo ;1~- = --r,~ = kPAo(l -· X)~, ""kPAo(l ·- X)(l ··--nW)ll\since dX
=
J
dX l\0k t W ( .. )1/2 ·----·=-.-·J1 1--uW dW l X F;\O o
dW X2
X¡
!)frQk(l-X)·(·l -aW)1.12 FAO
When X
=X
1
,
\V :::: O
2 )[·l · · · ( l . .. n."W ·)3/2] I n -.l·-X -. - 1 = --kPAo( ,- ·- . · ~-··--l ······X 2.
f AO
3a
4-39
e:::: 0)
P4-18 (e) e) P == P0(I ··
2
20 atm(l · O O 18kg-1 (50kg)f12 == 63 atm
ee
P4-18 (d) For turbulent flow
=
a2 a1 (
Dpi
DP2
J J =( 2 ( ,\1 2
Ac
2
0.018kg-1
)(~)2
1
(~J = 1.5
2
0.0142kg-1
Pexit = Po(l- uW)05 = (20atm)(l-O 0142ki1(50kg))112 = 10..7 atm
ln[l-O.S]= · 1-X 2
io-·3
atmkg
20atm(
2
(
3 0.0142kg-1
)J[1-(1-0.0142kg-1(50kg)t2]
X2 = 0 . 77
P4-19 Production of phosgene in a microreactor CO + C12 --) COC12 (Gas phase reaction) A+B
-)C
See Polymath prograrn P4-l 9.pol. POLYMATH Results 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
o o
value
minimal value
o o
1 -0.5 2.0E-05 2 . OE-05 2.0E-05 2.0E-05 2.83E-07 2.83E-07
0.3649802 -0.5 2.0E-05 2.0E-05 4.32E-06 4 . 32E-06 2 . 83E-07 2. 444E-·07
70.671378 70. 671378 3.55E+05 0.004 -19.977775
9 . 1638708 9.1638708 3.55E+05 0.004 -19 . 977775
o
o
o
o
maximal value 3.5E-06 0.7839904 1 ··O. 5 2.0E-05 2.0E-05 2 . OE-05 2.0E-05 2.83E-07 4 . 714E-07 l.568E-05 70. 671378 70.671378 3.55E+05 0.004 -0.3359061 53 . 532416
final value 3.5E-06 0.7839904 0.3649802 -O . 5 2.0E-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
ODE Report (RKF45) Differenlial equations as entered by the user [ l J d(X)/d(W) = -rA/FAO r 2 J d(y)/d(W) -a*(1 +e*X)/(2*y)
=
Explicit equations as entered by the user
4-40
·---
------------~----·
-~~--~---
[ll
[2 l [4J
FAO = 2e-5 Fa= FA0*(1-X) [ 6 J vO = 2 . 83e-7 [81 Fe= FAO*X [ 1o J Cb = Fb/v [12] k = .004 [14] Ce= Fc/v
e=-.5
[3] FBO = FAO [ 5 j Fb = FBO-FAO*X [ 7 J v = v0*(1 +e*X)/y [9] Ca= Fa/v [lll a= 3.55e5
[ 13 J rA = -k*Ca*Cb
P4-19 (a) 0 . 0 .---------·------,
o.o
0.8
o.o
0.6 ·
o.o
0.4
o.o
0.2
o.o L---~--~-~--~----' O. Oe+O
7.0e- '.'
l.4e-6 W 2.Ie-6
2.8e-6
3.5e-6
o.o L---~----0.0e+O
7.0e-7
1.4e-6\V2.Ie-6
2.8e-6
P4-19 (b) The outlet conversion of the reactor is O. 784 The yield is then MW*FA *X= 99 g/mol * 2 e-5 mol/s * 0.784 = .00155 gis= 48.95 g/ year. Therefore 10,000 kg/year / 48.95 kg/ year = 204 reactors are needed.
P4-19 (e) Assuming laminar flow, a. - D/, therefore
a2 = a1 _D~1 = (3.55xl05 kg-1 )4 = 14.2xl05 kg "
DPZ
4-41
3.5e-6
2.0e-5 .----------------,
10 ~--------------,
1 6e-5
0.8
1.2e-5
Gl
w
06
8 Oe-6
OA
4.0e-6
02
o.o ....._ __
O Oe+OL---------------' O Oe+O 7 üe- 7
1.4e-6 W2 .1 e-ó
2.Se-6
3 5t>-(i
O.Oe+O
...._ 7 . 0e-7
__. L4e-6 w2 . .le-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 (0 Individualized solution P4-19 (g) Individualized solution
P4-20 (a)
Mole Balance
1fv =
'Á / F.10
Rate Law
k' = k 11 = k[!:r(
coth
$ ·-
1)]
<1>7cDp
4-42
· ----r·--·------1.0H:.:::::::::~-
Large D,..
Tl - 1 k'
cDp ~
= Tl k,
Té k=Tl
= -k'1 = k' :;: 3
Increasing Particle Síze
Then 3
3 . .. cDp
r¡----·
when Dp
= 2 mm, k'::::
0.06, r¡;::
t.k = ·036. = 0.02
3
0.02= c(2) c=75
[el) =.?.!!2r.1 For turbulent flow:
a=
2/3 P0A0p(l - (/J)
constant
=>a=--
DP
aoDPo a=-Dp1
2/30
2(0.001
atm)
:!
dm -·--'------,-------~--º - Ac(l-
a -
ªº =8.0*10- kg-l 5
P4-20 (b) See Polymath program P4-20--b.pol.
4-43
3 . 0 ~---------------,
24 1.8 1.2
0. 6
o.o~-~ o.o
O4
0. 8
16
Dp 1 2
2.0
P4-20 (e) Gas, E=O, CA =CA.,(1-x)y
-:¡ = · f;- where
a=
a{~:)
ª· = PJ;f)¡,,.~ = 2o~rd1fi¿"s"~~) xoj2~;rr 7 os and k'""'
1
kl q;2 (
l
1) J
a = 7.08 * ic 5 k/!?i:1 \ 'lJ)p)
l\1 = l, CA()= 0..207, W = O X = O y = 1.0 wt = 100
FAO
?;' =····
ew
X
ro -s
kg"'
(X
2y
where k = 3 and
=5
See POLYMATH prograrn Vary l\
See Polyrnath prograrn p4 . 20.c.pol. POLYMA TH Results Calculatedvalues of the DEO variables Variable w X y
Dp Q
Fao
initial value
o o
minimal value
o o
0 . 0075 0.5625
0.2366432 0 . 0075 0 . 5625
5
5
1
maximal value 100 0.5707526 1
0 . 0075 0 . 5625 5
4-44
final value 100 0 . 5707526 0.2366432 0.0075 0 . 5625 5
alpha Cao kprime ra
0.00944 0.207 2.9385672 -0.1259147
0.00944 0.207 2 . 9385672 -0.0012992
0.00944 0.207 2.9385672 -0.1259147
0.00944 0.207 2.9385672 -0 . 0012992
ODE Report (RKF45) 0.90
Differential equations as entered by the user [ 1 J d(X)/d(w) = -ra/Fao [ 2 J d(y)/d(w) -alpha/2/y
=
0.72
Explicit equations as entered by the user [ll Dp = .0075 [2J Q 75*Dp [3J Fao 5 [ 4 J alpha = .0000708/Dp [Sl Cao .207 [ 6 J kprime = 3*(3/QA2)*(Q*coth(Q)-1) [ 7 J ra = -kprime*(Cao*(1-X)*y)A2
=
0.54
= =
(136
0.18 0. 00
o.o
P4-20 P4-20 P4-20 P4-20 P4-20
0.1
0.2
Dp {13
04
0.5
(d) Individualized solution ( e) Individualized solution (f) Individualized solution (g) Individualized solution
(h) Individualized solution
P4-21 (a) Assume constant volume batch reactor dX Mole balance: AO = -rA
e
dt
= kC A = kC
Rate law and stoichiometry: -rA
Specific reaction rate: k ( 25º C) =
AO ( 1- X)
0.0022 weeks-1
Combine:
e
t = Ao
f kCAO ( 1-X ) = ---1k In ( 1 - X )
x
dX
O
52.2 weeks = X
-t 0.0022 weeks
1
In ( 1- X)
= 0.108
CA
= CAD ( 1-
X) but since volume and molecular weight are constant the equation can be written as:
mA
= mAo ( 1-
X)
4-45
6500/U = mAo (1-0.108)
mAo =7287/U %0U = CAO -CA *100 = 7287-6500 *100 = 12.1 %
CA
6500
P4-21 (b)
10,000,000 lbs/yr = 458 * 109 g/yr of cereal Serving size = 30g Number of servings per year = 458 * 109 / 30 =l.51 * 109 servings/yr Each serving uses an excess of787 IU = 4 . 62 * 10-4 = 1.02 * 10-6 lb Total excess per year = (l.51 * 108 servings/yr) * (L02 * 10-6 lbs/serving) = 154.11 lb/yr Total overuse cost = $100/lb * 154.J l lb/yr = $15411 / yr (trivial cost)
P4-21 (e) 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 ( 4ü° C) = 0.0048 weeks'
e
t = Ao
6 months = 26 weeks
fkCAOdX( 1-X ) = --1k In ( 1 - X )
x O
26weeks=-X
-1 ln(l-X) 0.0048 weeks 1
=0.12
= CAO ( 1 - X ) mA =mA0(l-X) CA
but since volume and molecular weight are constant the equation can be written as:
6500/U =mA0(1-0.12) mA0 = 7386/U %0U = CAO-CA *100= 7386-6500_*100=13.6%
CA
6500
P4-22 No solution necessary P4-23
CH20HCH2Cl + NaHC03
---+
(CH20H)z + NaCl + C02 4-46
FAo
= 0.1
A
+
e
B
mol/min = 6 mol/hr
+
D
+
E
P4-23 (a) dC A dt
Mole balance:
= r + Va (e - e V
AO
A
)
!Cn = r+~(-C dt V rA = -kCACB V =V0 +v0t
Rate law:
dCc=-r+vº(-C) dt V e
) B
See Polymath program P4-23-a.pol. POLYMA TH Results Calculated values of the DEQ variables Variable t Ca Cb Ce Fao Cao
k
Vo vo V r Xb Ne
initial
o o
value
minimal value
o o
0.75
l. 395E-13
o
o
6
6
l. 5
l. 5 5.1 1500 4 1500 -0.0039398
5.1 1500 4 1500
o o o
maximal value 250 0.15 0.75 0.829586 6 1. 5 5.1. 1500 4 2500
o
o o
1 2073.9649
ODE Report (RKF45) Differential equations as entered by the user ¡ 1 J d(Ca)/d(t) = r+(voN)*(Cao-Ca) r 2 J d(Cb)/d(t) = r+(voN)*(-Cb) ¡ 3 J d(Cc)/d(t) = -r+(voN)*Cc Explicit equations as entered by the user [ll Fao=6 [2] Cao = 1.5 [3 J k = 5.1 [4l Vo = 1500 [ 5 J vo = Fao/Cao [6J V= Vo+vo*t [ 7 J r = -k*Ca*Cb [8] Xb = 1-Cb*V/(0.75*1500) [9l Ne= 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
LO
O Oe+01
GJ
0. 8
GJ.
-8.0e-4
/
/
1
-1.6e-3
0.6
1l
¡
s
04
l -2Ae-·3 1
o.,
l
/ !
1
,,_,,.,,,,,,,,...-·'
-3.2e-3 ¡ .
o.o o 0.90
50
-
0 . 72
100
t
150
200
250
-4.0e-3 3000
Ca 2400
Cb
o
l
I
1l•,"*""",,""'"'"
/
¿/
/''~
50
100
t
150
200
250
GJ
Ce
0 . 54 ·
1800
0.36
1200
1118
600
0.00
o
50
100
t
150
200
250
o o
50
100
t
150
200
250
P4-23 (b) Taking average time for activities like charging, heating, cleaning = 4.5 hr So, ifthere 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 anda flow rate of 120 mol/hr, 1560 dnr' would be added to the reactor. Because the volume of the reactor is 2500 dnr', and there is already 1500 dnr' of B in the reactor, the reaction cannot not run for 19.5 hours ata flow rate of 120 mol/hr (2 mol/ruin) Also, because there is 1500 dnr' ofB ata concentration of0.75 M, there are only 1125 moles ofB to react. This means about 750 drrr' of L5 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 ata rate of 120 mol/min takes 9.375 hours, Using the Polymath code from part (a) and changing Fao to 120 and the time to 9.375 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 1 O 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 three possible times for reactions and the moles of C that are formed from the two batches. Total Moles of C
Maximum Batch Time (hr)
Formed 1723 1790 1795
10 9 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/rnin of A. (If three batches are run there will be 13.5 hours of downtime and only 10.5 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 hours).)
See the Polymath code from part (a) and vary time and flow rate,
P4-23 (e) FAo = 0.15 mol/min = 9 mol /hr vo = Fao/Ca = 9 mol/hr / 1.5 mol/dnr' = 6 dnr' /hr 1000 dnr' is needed to fill the reactor. At 6 dm3/hr it will take 166.67 hours Now solving using the code from part (a) with the changed equations:
See Polymath program P4-23-c.pol. LO
.---------------=
0.8 0.6 0.4
º·'/º
/
/
/
/
,,?
r---1 ~
o.o,...__~---~-33
/
132
66t99
165
P4-23 ( d) Individualized solution P4-24 NaOH + CH3COOC2H
5
~
CH 3COO- Na+ + C2H50H
A+B~C+D 4.49
dCA
Mole balance:
dt
= r+~(c V
dCB dt Rate law:
rA
-c A )
Aa
= r+~(-c V
)
dCc dt
B
= -r + Va (-e ) V
e
= -kC A e B
V= Va +vat
exp(
k =k0
!(: -n] 0
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 B. 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 drrr' of pure B. Assume it will take 6 hours to fill, heat, etc. the reactor. That leaves 18 hours to carry 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 1.8 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. POLYMATH Results Calculated values of the DEO variables Variable t Ca Cb Ce Cd ko
Fao Cao Vo vo T k
ra V
Ne
initial
o o
value
0.25
o o
5. 2E-05 0.04 0.2 1200 0.2 308 l.224E-04
o
1200
o
rninirnal
o o
value
0.0068364
o o
5 . 2E-05 0.04 0.2 1200
O. 2
308 l.224E-04
o
1200
o
rnaxirnal value 6.5E+04 0.1688083 0.25 0.0151725 0.0151725 5.2E-05 0.04 0.2 1200 0.2 308 l.224E-04 1.397E-06 l . 42E+04 202.92284
ODE Report (RKF45) Differential equations as entered by the user [ l l d(Ca)/d(t) = -ra+(vo/V)*(Cao-Ca) [ 2 J d(Cb )/d(t) = -ra-(vo/V)*Cb [ 3 J d(Cc)/d(t) = ra-(vo/V)*Cc [ 4 l 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 l . 42E+04 202 . 92284
[ 1l
ko == 5.2e-5 [2] Fao == .04 [3] Cao == .2 [4] Vo == 1200 [ 5 J vo == Fao/Cao [ 6 J T == 35+273 [ 7 J k == ko*exp({42810/8.3144)*(1/293-1/T)} [ 8 J ra == k*Ca*Cb [ 9 J V == Vo+vo*t [10] Ne== 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
e
Next we express the rate law: First-order reversible reaction
Transport out the sides of the reactor: RA
= kACA = kACTOFA
RB
kBCTOFB = kBCB = ----
-'-"--=--"--:.-;,.
FI
FT
Stoichiometery:
Combine and solve in Polymath code:
See Polymath program P4-25--a.po1. POLYMA TH Results Calculated values of the DEO variables 4-51
Variable
value
initial
o
V
o o
value
maximal
0.01 100 1 10 40 -10 1 0.472568
o
o
100
100
o
final
value
o
value
20 57.210025 l. 935926 61. 916043 0.01 121.06199 1 10 40 -0.542836 1 0.472568 0.6396478 100 0.4278998
20 100 9.0599877 61.916043 0.01 122.2435 1 10 40 -0.542836 1 1 2.9904791 100 0.4278998
o o
0.01 100 1 10 40 -10 1 1
X
o
57.210025
100
Fa Fb Fe Kc Ft Co K Kb ra Ka Ra Rb Fao
minimal
ODE Re(!ort {RKF45} Differential equations as entered by the user (1] d(Fa)/d(v) = ra ·· Ra [2] d(Fb)/d(v) = -ra ·· Rb [ 3] d(Fc)/d(v) = -2*ra Explicit equations as entered by the user [ll Kc = 0.01 [ 2 l Ft = Fa+ Fb+ Fe [ 31 Co = 1 (4) K=10 [5] Kb=40 [ 6 J ra = - (K*Co/Ft)*(Fa- CoA2*Fb*Fc"2/(Kc*Ft"2)) [7] Ka= 1 [ 8 J Ra = Ka*Co*Fa/Ft [ 9 J Rb = Kb*Co*Fb/Ft 3
100 ---------------
2
80
2
60
1 1
o o
a., .. ..,,..,,.,,.,.,.,,,.,.,
4
8
.... ,
~-
-.._
. . ....... ... . ...........
y
12
16
-~------------
40
- Fa
- Fb
20 20
• Fe 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. POLYMATH Results 4-52
20
Calculated values of the DEO variables Variable V
initial value
o
minimal
o
84.652698
100
Fa Fb Fe Kc Ft Ca K ra Fao
o o
o o
0.01 100 1 10 -10 100
0.01 100 1 10 -10 100
o
o
X
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
value
ODE Report (RKF45) Differential equations as entered by the user ¡ 1 l d(Fa)/d(v) = ra [ 2 l d(Fb)/d(v) = -ra [ 3 J d(Fc)/d(v) = -2*ra Explicit equations as entered by the user (11 Kc::: 0.01 [ 2 J Ft ::: Fa+ Fb+ Fe (3 J Co = 1 [4J K::: 10 [ 5 l ra ::: - (K*Co/Ft)*(Fa- Co"2*Fb*Fc"2/(Kc*Ft"2))
------------
0.50 0..45 0.40 0.35 0.30
- X_PFR
0.25
X Membrane 0.20
--------------------------i
0.15 0.10 0.05 0.00
.__
o
___¡
2
4
6
8
10
V 12
14
16
P4-25 ( e) Conversion would be greater if C were diffusing out P4-25 (d) Individualized solution 4-53
18
20
P4-26 CO + H20 <-> C02 + H2 A+B <->C +D Assuming catalyst distributed uniformly over the whole volume
dFA dW
Mole balance:
·--
=r
dF8 dW
--
dFc dW
=r
--
= -r
dFD dW
--=-r-R
Rate law:
Stoichiometry:
CD=Cro-
FT
FD FT
= FA + FB + Fe + FD
Solving in polymath:
See Polymath program P4-26.pol. POLYMA TH Results Calculated values of the DEO variables Variable -----
w
Fa Fb Fe Fd Keq Ft Cto Ca Cb Kh Ce Cd Rh k r
initial value
o
2 2
minimal value
o
0.7750721 0.7750721
o o
l. 44 4 0.4 O.2 0.2 0.1
o o o
l. 37 -0.0548
o o
1,44 3.3287437 0.4 O. 0931369 O. 0931369 0.1
o o o
l. 37 -0.0548
maximal value 100 2 2 1.2249279 0.7429617 l. 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 [ l J Keq = 1 .44 [ 2 l Ft = Fa+Fb+Fc+Fd [ 3 J Cto = 0.4
4-54
final value 100 0.7750721 0.7750721 1.2249279 0.5536716 l. 44 3.3287437 0.4 0.0931369 0.0931369 O.1 0 . 147194 0.0665322 0.0066532 l. 37 -0.002567
Hz
J Ca == Cto*Fa/Ft [ 5 J Cb == Cto*Fb/Ft [6J Kh=0.1 [ 7 J Ce Cto*Fc/Ft [ 8 l Cd = Cto*Fd/Ft [9J Rh = Kh*Cd [4
=
[10] [ 11 l
k
r
= 1.37 = -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.
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 . 4.59 2 . 0 ~-------
-----~
1.6
_,, .,.,_,,,_-----
1.2
............
0. 8
20
W 60
_
80
100
P4-27 lndividualized solution P4-28 (a) Assumc isothcrmal and e=O thcrcfore,
P=Po(l -aW).s W=99g
P4-28 (b) 4-55
o:
-TA
dW=
FAO
..!_ = (1-Wa).s Po
-r" =kCA CA-CAo(l-X) P/Po CA= CAo(l -X) (1 - 0.01 W).s Integrate from X=O to X=.9 dX dW
= kCA0(1-X)
(1-.0IW)..s
FAO
W=59.88g
FU'St 5% conversion integrate from X=O to X=.05 W=l.31 g Last 5% conversion integrate from X=.85 to X=.90 W:al0.85 g
P4-29 Individualized
solution
P4-30 (a) First order gas phase reaction,
y AO
C6H5CH(CH3h 7 C6H6 + C3H6 = 1, () = 1 :::> e = 1
p
kCAO FAo
y=-,a=--
Po
ForaPBR:
dX
a(l-X) (1 +X)
dy dW
(l+X)
dW = -=
2y
y
(l)
a
(2)
X= 0.064 and y= 0 . 123, Solving (1) and (2) by nial and error on polymath we get, a= 0.000948 (kg of catalyst)" a= 0.000101 dm-3
4-56
-----·-----·-------------·-·-----~--
--------~-------···-····-----·---------~------------------
Now solve for a fluidized bed with 8000 kg of catalyst. FAOX = --
Mole balance: W
-rA
-rA = kCA
Rate law:
1-X 1 + eX - X ( 1 + e X) __ SOOO __ X ( 1 + X) Combine: W a(l-X) 0.000101(1-X) Stoichiometry:
e =e A
AO ---
X =0.37
P4-30 (b) ForaPBR:
dX _ a(l-X)
(l+~y
dW-
(l+X)
dy
-'-----'-a
2y
dW
kCAa.
where a=
FAO
From chapter 12 we see that k will increase as Dr decreases. We also know that for turbulent flow
1
a -· -· . This means that there are competing forces on conversion when Dr is changed. DP We also know that alpha is dependant on the cross-sectional area ofthe pipe:
1
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 atea) will result in lower superficial mass velocity. The relationship is the following for turbulent flow:
G--
1
Ac
therefore,
2
and a-G
1
a - -2
.
~ If we combine both effects on alpha we get the following:
a---
1
a--
1
1
Ac A¿ A¿
So increasing pipe diameter will lower alpha and increase conversion and lower pressure drop. 4.57
-----------··
----··----------~---------
---···---------------------····---
-----
For Laminar flow:
1
a-·-Dz
p
so decreasing the particle diameter has a larger effect on alpha and will increase pressure drop resulting in a lower conversion For Laminar flow
1
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) s = 0.33(1-3) = -0..666 p AO = 0.333*10 FAo = 13.33 Mole balance: dX/dW = -ra/Fao Rate law: rA = -KP8 PA = PAo(l - X)/(1 - 0.666X) Pc=PAoX/(1- 0 . 666X) For a= O, y= l(no pressure drop)
K = 0 . 0.5
-O.OS
6
-0.16
4
-0.24!
_,.,,/
2 20
40
W 60
80
/
-040 ~-----------~---' O 20
100
40 \V 60
See Polymath program P4-31-a.pol. POL YMA TH Results Calculated values of the DEO variables Variable
w X K
Pao Pa Pb Pe Fao ra
initial value --------
º
o
0.05 3 . 33 3 . 33 6 . 66
o
13 . 33 -O . .333
minimal value
o o
O . 05 3 . 33 0.0406732 0 . 0813464
o
13 . 33 -0.333
maximal value 100 O . 995887 O . 05 3.33 3 . 33 6 . 66 9.8482838 13 . 33 -0.0040673
4-58
final value 100 O . 995887 O . 05 3 . 33 0 . 0406732 O . 0813464 9.8482838 13.33 -O . 0040673
80
100
esp
-0.666
-0.666
-0.666
-0.666
ODE Report (RKF45) Differential equations as entered by the user [ 1J d(X)/d(W) = -ra/Fao Explicit equations as entered by the user [ll K = 0.05 ( 2 J Pao 0.333*1 O ( 3 l Pa Pao*(1 - X)/(1 - 0.666*X) [4l Pb = 2*Pa ¡ 5 l Pe = Pao*X/(1 - 0.666*X) [6J Fao = 13.33 [7] ra = -K*Pb [ 8 J esp = -0.666
=
=
For first 5% conversion weight required = W1 = 2 kg For last 5% conversion weight required = W2 = 38 kg Ratio= Wzf W1 = 19 Polymath solution (4-34 a)
P4-31 (b) For a= 0 . 027 kg", Polymath code with pressure drop equation:
See Polymath program P4·31-b.pol. POLYMA TH Results Calculated values of the DEO variables Variable w X y K
Pao Pa Pb Pe Fao ra esp alfa
initial val u e
o o
1 0.05 3.33 3.33 6.66
o
13.33 -0.333 -0.666 0.027
minimal value
o o
0.1896048 0.05 3.33 0.4867002 0.9734003
o
13.33 -0 . 333 -0.666 0.027
maximal value 30 0.4711039 1 0.05 3.33 3.33 6.66 l. 0583913 13 .33 -0.04867 -0.666 0.027
final value 3()--O. 4711039 0.1896048 0.05 3.33 0.4867002 0.9734003 0.4335164 13.33 -0 . 04867 -0.666 0.027
ODE Report (STIFF) 7. 0 .----------------,
Differential equations as entered by the user [ 1J d(X)/d(w) = -ra/Fao [ 2 J d(y)/d(w) = -alfa*(1 - esp*X)/(2*y)
5.6
Explicit equations as entered by the user [ll K=0.05 ¡ 2 J Pao = 0.333*1 O [ .3 J Pa = Pao*(1 - X)*y/(1 - 0.666*X) [4J Pb = 2*Pa [ 5 J Pe= Pao*X*y/(1 - 0.666*X) [6J Fao=13.33
Pa
42
-
Pb
...
p- ...
2.8
14
4-59
O.O O
6
12
W 18
24
30
[ 7j
ra = -K*Pb
=
[ 8 j esp -0.666 [ 9 J alfa = 0.027
0.00~--------------, 0.8 -0.16
0.6
-0.24
0.4
-0.32
0.2
-0.40~-~--~-----~-~ O 6
12
W 18
24
30
[¡ Ll_J
6
12
W 18
24
30
P4-31 (e) 1) For laminar flow: Diameter of pipe = D and diameter of particle = D, Now D¡/D0 = 3/2 so G1 = 4/9G0 a= (constant)(GID/)(1/Ac) So a= ao(G/Go)(Drol Dp1/(DofD1)2 = a0(4/9)(2/3)2(2J3)2 = 0 . 00237 kg' Less pressure drop and more conversion for same weight of catalyst as in part (b). 2) For turbulent flow: p oo G2/Dr a= (constant)(G2/Dp)(l/Ac) So, a= a0(G¡/G0)2(Dpof Dp1)(DofD1)2 = ao(4/9)2(2J3)(2/3)2 = 0.0016 kg' Again less pressure drop and more conversion for the same catalyst weight. 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 =>
CA CB
4-60
----------~--·-··-
---
1.0,---=============; 0.8 0.6
o.a O.l
ºº,__ _ _._ O OOEO
v0 0X { C80-t-CA V0
( =>CA01-X => t
x2
[ = vºcAo v C .Kc(1·-X) 0
+
5.18E4
__. 1 04E5 t 1 5óE5
2.07E5
l 591:5
) =---(cAox)2 Kc
x]
80
Now solving in polymath,
See Polymath program P4-32-a.pol.
P4-32 (b) See Polymath program P4-32-b.pol. POL YMATH Results Calculated values of the DEQ variables Variable t Ca Cb Ce Cd Kc k ra va Va V X
initial value
o
7.72 10.93
o o
1.08 9.0E-05 -0.0075942
o.os
200 200
o
---
minimal value
o
O 2074331 7.6422086
o o
l. 08 9 OE-05 ·-·O. 0075942
o.os
200 200
o
maximal value 1.SE+04 7 72 10.93 3.2877914 3.2877914 1 08 9.0E-05 -i . 006E-05
final value ·-1. SE+0-4 0.2074331 9.51217 1.41783 l. 41783 1 08 9.0E-05 -a . 006E-05
200 950 0.9731304
200 950 0.9731304
o.os
o.os
l.Oe+o .....--------------. 8.0e--1
ODE Report (RKF45) D,tterential equations as entered by the user
6.0e-1
r
2:J
4 Oe-1 2.0e-1
4-61 o.oe+o o---2976 5952t 8928 11904 14880
[1
[2 [3 [4
d(Ca)/d(t) = d(Cb )/d(t) = d(Cc)/d(t) = d(Cd)/d(t) =
ra - Ca*voN ra - voN*(Cb- 10.93) -ra - vo*CcN -ra - vo*CdN
Explicit equations as entered by the user [1 J Kc = 1.08 [2 J k = 0 . 00009 [ 3 J ra = -k*(Ca*Cb - Cc*Cd/Kc) [4] vo = 0 . 05 [5] Vo = 200 [ 6 J V = Vo + vo*t [7l X= 1 - Ca/7.72 O.Oe+O r----,--,.,..-~-
2 . üe+I ....-------------,
-1.6e-3
1.6e+1
[]
-3.2e-3
L2e+l ·
-4.8e-3
80e+O
-6Ae-3
4.0e+O
-8.0e-3 '--------~---~----~--j O 2976 5952t 8928
OJ)e+O
11904 14880
L _ __::::::=~====="""'"""".J
O
2976
595.2t
8928
11904 14880
Polymath solution
P4-32 (e) Change the value of v and CAo in the Polymath program to see the changes
1.0------
0
08
P4-32 (d) As ethanol evaporates as fast as it forms: Now using part (b) remaining equations, Polymath code:
Co=O 0. 6
0.4
See Polymath program P4-32--d pol,
0. 2
POLYMATH Results Calculated values of the DEO variables Variable initial value minimal value maximal value final value t o o 7.72 0.0519348 Ca 10 . 93 6.9932872 Cb 9 . OE-05 9 . 0E-05 k .. Q.0075942 -O . 0075942 ra 0.05 O . 05 vo 200 200 Vo
o.o'----~-----~---~----' O
6000
7 . 72
10.93 9 . 0E-05 -3 . 69E-05
4-62
o.os
200
1189
2378
t 3567
6000 0 . 0519348 7.8939348 9 . 0E-05 -3 . 69E-05
o.os
200
4756
5945
V X
200
200 O
500 0.9932727
500 0.9932727
o
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(Ca)/d(t) = ra - Ca*voN [ 2 J d(Cb)/d(t) = ra - voN*(Cb- 10.93) Explicit equations as entered by the user [lJ [2J
k=0.00009
ra = -k*Ca*Cb
[3] vo = 0.05 [4] Vo = 200 [ 5 J V= Vo + vo*t [ 6 J X = 1 - Ca/7. 72 20 ,------·
16
~ 12
8
..
--
l_gJ __.
-
.
4
-8 . 0e-.30
2378
1189
t
3567
4756
5945
oo
1189
2378
t
3567
4756
5945
P4-32 (e) lndividualized solution P4-32 (f) Individualized solution P4-33 (a) Mole balance on reactor 1:
e
eAlv-rAI
cAº --vo2
eA1v-rA1V =--dNA1
AOVAO-
V
=--
dNAl
dt
.
with
1
VAO =-Vo
2
dt
Liquid phase reaction so V and v are constant, CAO CAi --· - ---- - r
2r
r
AJ
- dCAI - -·--
dt
Mole balance on reactor 2:
4-63
·---~-
-----------------
-
-----·----------------------·-------
CAl CA2 ------r
r
T
. _ -dCA2 A2 dt
Mole balance for reactor 3 is similar to reactor 2:
. - dNA3 CA2VO -CA3VO -rA3v ---dt CA2 CA3 ------r T
T
A}
_ dCA3 --dt
Rate law:
--rAi
= kCAicBi
Stoichiometry For parts a, b, and e so that
=r» = kC1;
CAi = C8¡
See Polymath program P4-33 . pol. POL YMA TH Results Calculated values of the DEO variables Variable t Cal Ca2 Ca3 k
Cao tau X
initial value
o o o o o . 025
2 10 1
minimal value
o o o o
0.025 2 10 O . 3890413
maximal value 100 O . 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 O . 3890413
d(Ca1)/d(t) = (Cao/2 -Ca1)/tau -k*Ca1A2 [ 2 J d(Ca2)/d(t) = (Ca1 - Ca2)/tau -k*Ca2A2 [ 3 l d(Ca3)/d(t) (Ca2 - Ca3)/tau -k*Ca3A2 [ll
=
Explicit equations as entered by the user [1] k = 0.025 [2] Cao=2 [3J tau= 10 [4J 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 (e) The plot was generated from the Polymath program given above. 0.90 .--------------------, 0.72
0.54
0.36 - Cal 0.18
- Ca2 - ('.a3
O 00 """"'---------~-----'-------'·-----' . O 20 40 60 t
80
100
P4-33 (d) We must reexamine the mole balance used in parts a-e. 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:
dNAI . 2 200 CAOVAO -cA!v-rAIV =--with VAO =--Vo and Vo =--
dt
2C
3
15
dN
_ ___AQ_v -C v- r V= ___Af_ Al Af ~ 3 O
Liquid phase reaction so V and v are constant.
4-65
ac;
CAO CAl ------r 21: 7:
Al
=--
dt
Mole balance on reactor 1 species B: CBOVBO -CBlv-rBlv
.
- dNBl ---
dt
and
Stoichiometry has not changed so that -rAi = --rBi and it is a liquid phase reaction with V and v constant, CEO CBl ------r 31: 7:
. Al
- dCBl --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
CAlvO -
eA2v2 -rA2v
CA2 CA3 ------r 7:
2
7:
2
.
_
A3
dNA2
=--
dt
dCA3 --dt
Mole balance for reactor 3 species B:
=z:
. - dNB3 --·rA3v
CB2V2--CB3v2 CB2 CB3 --·----yA3 T2 7:2
.
_ dCB3 --
dt
Rate law:
4-66
+ VBo = 20
See Polymath program P4-33-d.pol. POLYMA TH Results Calculated vaiues of the DEO variables Variable t Cal Ca2 Ca3 Cbl Cb2 Cb3 k
initial value
o o
minimal value
o
o o o
o o o
o
o
o
0.025
o o
0.025
maximal value 100 1.1484734 0.7281523 0.6278144 0.4843801 0.7349863 0.6390576 0.025
final value 100 1.1484734 0.7281523 0.6278144 0.4821755 0.7291677 0.6309679 0.025
Cao tau
2
2
2
2
13.333333
13.333333
tau2
10 200
13. 333333 0.3721856 10 200
13.333333 0.3721856 10 200
5
5
1
X
V
vbo
1
10 200 5
ODE Report (RKF45) Differential equations as entered by the user ¡ 1 J d(Ca 1 )/d(t) = (2*Cao/3 -Ca 1 )/tau -k*Ca 1 *Cb 1 [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 l d(Cb1 )/d(t) (1 *Cao/3-Cb1 )/tau-k*Ca1*Cb1 [ 5 J 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
=
E:xplicitequations as entered by the user [l] k=0.025 [21 Cao = 2 [ 3 l tau = 200/15 [ 4 J X = 1 - 2*Ca3/Cao [5] tau2 = 10 [6] [7J
V= 200
vbo
=5
Equilibrium conversion is 0.372. This conversion is reached at t = 85.3 minutes.
4-67
5
2.0 .-----------------, - Ca I
1.6
- Ca2 • (;a3
1.2
0.8 0.4
o.o r...,:;..=--~---'-----~--~----' O
20
40
60
80
100
P4-33 (e) Individualized solution CDP4-A CH3I + AgC104 -¿ CH3Cl04 + Agl
0.7 mol/1 CE3t "'CEO 0.5 molíl AgCl04"' V
o
= 30 dm3
rCJ.I3I
=
-k CCH3I
k "' O ,0004:2 I;;;;
CAO
c!;~104
(d:o:i3 /mol)312( sec)-l T=:29& E:
0,98 Iutegration for X
= 0.98, sel.ved nume:dcally
1 t "-··-----·"'
-;
(24.18)
0,00042 (0.5)31~ - 1.628:tlO5 s ec "' 4S. 2 hr
t -
V = V
o
{ Li qu i d pb a s e )
CDP4-B a) 4-68
d:-,.i' -··~·=t' d_t
Mole Balance :
.<\.J
. "V - F A -,-r . .,\
v.~f~.,,,c dt .... vo - e¡... v •
.. kC,,.V
.~~. . . ::: <;:,.. V.e. ~::i~.R.:t ::1 dr
V
V
Use PóLYMATtt to plot C,. vs, t -, se;;
Eguat:ions d(ca)/d(t.}
,
--ca
,.
= (cao•vo/v)
vo=70000 V"'.15000000 'kd0.0025 cf
t
¡ l
I'
•.occ
.!. '
a.scc
= o.
VS
'
.... •ce:
·· (ca• (k•v+vol /v)
cao=4 .. S
t.0
(b} (',a
l.
Kev
-
l.000 z.ecc
-l
c. coo
·~-M
·-·l---v...l,lóQ
~--.,~¿.,,..-.~-·~· e,~ '
.¡.
o.,oo
...
t
For sready-state: t
= 4.6-..!_,. = 641 hrs 1 +Tk:
CA= _cA~v. =2.930mg!dm3
kV + v
0
99% of this is 1.905 mg/dm', which is below the standard of 3.0 mg/dm) . Polymath solution(Ans CDP4-B-a) b)
...dN : . . .::..=f
Mole Balance :
dt
dN"
-·dt --··
-·F +r V ,1.,
·"
...
e
, .
== C,·,~ v ., · .: . ', v.~., - ···· !((.' ', V
Use POLYMATH to generare plot of C" and N" vs. r, Em;ar;i.;ns: d(nal /d( el "'cao•vo·~ca•vot:c-k~ca Ci:l.0-"'4.5
Ket ..~- Na
vo;-;70000
vout:=50000 k=0.0025 v=20000"t ca=na/v
to "'
l.
.,!fc:«I
ef
=
,-= '
T
Na vs t
l
'JSO a.eec
4-69
.ao. .eee
----
... .,..+.---.~----·-··:·-,t
t
!i•C.:;co:
~~:e
Ca vs t
.Ku
Ca ),000
·r 1
z.cee
l' i
LXC
O,.OQQ
.L i
L,,.... .
e.ooo
······S'····+·· ............ _,, ...... +·«·-F·>+·----+ atl..OCXJ
:n.a..ooc
•IC..'C:IXJ
Polymath solution(Ans CDP4-B-b) e)
65
6 1
Use sarne equariocs as in part (a),
57
bue
5.3 4.9
4.5 41 37
voo,7 OG(l O
3.3
v=.t.5 00 0000
29
k=O OQ25
2.5
t:f
gg
198
298
397
496
595
694
794
893
992
!.DOO
.EQlymath solution(Ans CDP4-B-c) d) This part is almost same as part(b) with minor changes: V= 15000000- lOOOOt V0= 80000 and Yout= 70000
The reason the graph looks so different from(a) is that pure water is evaporated, but water with atrazine is comingin 45
Polymath code: d(ca)/dt=cao*vo/v)(ca (kv+vout)/v) ca(0)=4.5 # vo = 80000 # V = 15000000 - lOOOO*t 0.0025 k = # 4.5 cao # vout = 70000 # t(O)=O # t(f)=lOOO
44
43 42 41 40 39 3.8 3.7 36
3.5
100
200
300
400
500
600
700
800
900
101
Polymath solution(Ans CDP4-B-d)
CDP4-C a)Find the number of moles of receptors: . 5 receptors 1 mole receptors 100011%.L . . _10 6 cells . lxlO ------···*lxlO - ·-··,-·*····--··---------.. ·····-*·-····--····... -=l.66x10 M mL cell 6.022 x 1023 receptors L .
.
1.66. xw· = 1.66 x.. lff . . . 10 ~?.,!. [, '+ .OlL .. . ··· 12moles ' . .
Design equation;
dX
N AO ---=-r dt
A
RateLaw: Stoichiometry:
Where:
e~ = CAo(l - X) Ca= CAo(eJJ-x)
Total number of moles: hence;
1.66 X 10-12 + Ix 10-.9 = 1.002 X 10-9
Combining ami solving: dX
(1--- x)(es • x)-= - .. -1:...
kC;,0dt
NAº ·
-In- · ~8=!- · · ~~;;o~
e·s -· · l
88(1 ·· x) t = 0.071 mín
N;1.o
b) Assume Cn = Cao Design Equation: Rate Law; Stoichiometry: CA= CAO(l ·
X)
CB =CB(l
4-71
Combine and solve: -~--- - _kCAO~!lQ_dt l····X NAO 1 kCAOCBOt ln ···-··· ·· == ·········· ....,.._ .. l --X NAO r;:;::().069 min A very good approximation. e)
Design Equation:
dX
NAO ----=-r dt
A
RateL1.w: Stoichiometry:
C,,. == C"0(1 • X) CB=Cso
C..,oX
Ce= Combine and solve:
N,,.0
~f
= k1CA0(l · · · X)C80 ·k,CA0X dX
dt
•••••,:M¿ ···•••••e··~,---•M"''"'"~"""'*"• .... :::·
,~· .,.,.·,.,,,,,·
k¡CAOCBO - kfCAOCJJQ -kTCAOX
f k/:so••-• X
NAO
5•
CAOdt • d• + <···•··.·-J<. X(k¡Cso + k,) o X dX
o
••>
1 k¡CBD · · · · ··-·····-··-··· · ----- In ·--------···· · ··········-······:-···· · k I C 80 +k, k¡Cso .. x(k¡Cso
_.SC,10 , .. ····-:- - --+k,) NAO
- .,} ln -· '. -~---········ -- = 500 . l+ . 1 .l····X(.l+ . i) X =05
CDP4-D Batch reaction: 2A + B -) 2C k¡ = L98 ft3/lbmoLmin, k2 = 9..2 x 103 (ft3/lbmol)2Jnrln V= 5 gal =0.67 tt3, X= 0..65 . CAo = 0.002 lbmol/ü.67ft3 = 2 . 98 x 10·3 lbmol/ft3 C80 = 0.0018 lbmol/ü.67ft3 =2.69 x 10·3 lbmol/ft3 Mole Balance: Rate law: Stoichiometry:
dNA dt
---
= -r
A
V
-rA = k1CACB + k2CAca2 CA= CAo(l-X) Ca= CAo(0a-(b/a)X)= CAo(0. 9 - 0.5X) 4-72
Combine:
eAo- dX = e Ao2 (1dt
í
X)Lk1 (0.9 -
o.sx) + k2c Ao (1-
X)(0.9 -
o.sx) 2]
Integrating between X=O to X=0.65 for t =Oto t = t gives t =24.1 min
CDP4-E
---
;. ~
Final
ly
FAF' i\!2f
V•
SO 3al.
FBF
CBF
YBF.,.
10-S
(et1;)
llqnid
P "' 500 psi¡;
··· ra "' x.
C!:!~
""
c8
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 vía absorption of H2 gas over the liquid reacting mixture. Then at 500psig,
4-73
e
.5.0s, l
ª2·f
x.
1
[1.l!f?l2b.:.1 :r [··1
'
2g
454
!bmo,le_,1 g eaoLe s
.x
• 020 81 b;.¡ol~
(3,7Jl:.ll..J
z3.l
g!'.l
VR
.1bt.n.:r_
3 .JZ lbmolc/'.1.::
~;¡¡;
l'b:::.o,le
V
o
p l:I o 2
89.3
-
________ • !!O" . TC vv -
T:" -,.
10
v :l ..2 V(l··IJC
(89.3
¡p.l/b.rH0_,..99)
;a.l. {Q ,Q2Q8 l!.,ol)
50
k = o.49
X
,.1
*""
R2,f
··--· (O.Ol)
103 ;alf(lbnole - h.::) is:
;: Ctl
2
Lct
0.0127
(
300+],4.7)
5'00+!.4.7
lbmole/gal
coave:sioD be X2 wheu p:essure ~ 300 psi¡
······-········-·;;;..;;. ;: ; e
BO
-
< 1--x > a .;.
2.
-........
"
2.300
4-74
,..5
;;a.l./!l;:.
lbmo.1 e/ ;;al
11._
l·I =
eso
2
gal>CB.49xl03 gal/lbmol-hrH0,0127 Ibmol/;;;al) 89. 3 gal/h.r
,.0127 lbgole)(S9,4g4l) gal hr
=
1.134 lbmole/hr
(3.72xl0-3) (1-0.984) 1-·1.34+3. 72
l.23d0
-s
CDP4-F NAo ~
= Vpc ( -r~}:::: Vpc kCA
Liquid Phase CA = CAo ( 1 X)=
!'l~o ( l · · X)
~~- = Pe k (1-X)
In ·-··l- = ',K 1-X t
X InJ(i.x)
t:. ·i:'~· o
5
o
10
.3
o
.36
0 ~·, Mdlim~n-.,
,.. '6
t
20
;73
.5
.69
·2..0
h.t Ci::tÜfflfUT
Ji:,
"/º
,t
1.31
s:·o · ·
.¡,.
4-75
30 .85 1.89
40
0.93 2.66
50
. 97
3.. .5
A. plot In
[[i::3q} is linear
rime so we conclude
wrt
º""
benzzzoquinoline with k'=O. L-, [
TA(T.)1
er
me reaction is ñrsr' order in 5, 6
fh ...:'
1 1
l
r,..{1.;)J - R LT, T2J
In 4;
li"' R
f-h}3 ~ 3!}1 JnA. ..31:L__··- = 6248 ºK (353{383 })
E ;;; 12,414 _;,::;i
mol
as 9ü"C"" k (363)
k (373}
k {363) == 0 . 63 k (373)
~ = ~. p_c.-z.~ = (.63) -
k1
Pc1
o o
't
X
(2) k:
1
=
1.26 ~·
1
30 0.42
20 0.82
10 0.57
50 0.99
40 0.97
CDP4-G Develop a design equation Mole balance: F A(r) - F A(r + ~r) + rA(2mMh) = O
_ FA (r) FA (r + ~r) _ (2 h) ~ -~ m ~r Rate law: - rA = k1 CA C B Assuming 8B=l and E= O Stoichiometry:
AlsoFA
eA = eAO (1-
= FA0(l-
X)
X)
and
~
dr
dr
eB = eAO ( eB -
dFA
=> --
-~ (2 m h)
_ dFA _
X)
= eAO (1-
= -FA0- dX = rA (2mh) dr
4-76
X)
dX
Combine: --
dr
2
2mhkC AD (1- X )2
= ----------'=-------
=>
FAD
X 1- X
By Integration we get: ---
X
dX
lo-
=
X)2
J (21lkC A/
)rdr
FAO
Ro
2
kllhC AD (r2 - R/) F AD
2
kllhC AD (r2 - Ro 2) FAO =>X= z 1 + kllhCAo (r2 -R/) FAO b) Now, with the pressure drop, CA= CB = CAo(l-X)y Hence, -
2
2
rA = kC AO (1 - X) Y
"'''
=>
um.
2
dX
2mhkC Ao (1- X) 2 y 2
dr
FA o
Where
2
y= (l -aW)112
.. ,.....
G
GraphTitle
rn¡¡
'"" l.Jl&J
and
,...,
W = pb1r(r2 -R/)h
1.11&4.
Using polymath, following graph is obtained:
um
,...
,~ .
Differential equations d(X)/d(r) = 2*3.1416*h*k*Cao"2/Fao*(1-X)"2*y"2*r y= (1-alfa*W)"0.5 W = density*3.1416*h*(r"2- ro"2) ro= 0.1 density = 2 Cao = 0.1 Fao = 10 k =0.6 h 0.4 alfa= 0 . 07 variable name : r initial value : 0.1
••• ,.
,/
'"
~/
/
... ,..
,
.
"''
,.
,n
lll
,~
"'
=
e) 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 CAo causes a dramatic increase of conversion. Similarly, decreasing CA0results 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 of the reactor and hence decreases the conversion. Increasing RL will increase the conversion as volume increases .
4-77
CDP4-H Liquid phase reaction A+B--..C
vcllmixed no in!lov or outtlo-.r BATCH l. Móle balance on batch reactor
2. Rate íaw -l'A""'kCACB 3. Stoichiornctry
4. Combine
4-78
Table of reaction imegrals can be found in Appendix A-10, a)
= (O.Ol
t
dm3Jmoi
~~){2~~ud;J)h~ó~9j • t = 4SO mir
l . Mole balance on CSTR
2. RateLaw
3 . Seoichíometry
liqaid phase u
=u
0
4. Combine
5. Parameter evaluation 10 moUmin (0.9)
V 4_
(0.01 d:ro:3/mol min) (2 moi!dm3f(1·0.9)4 b)
V
= 22500 (,im.3
1. Mole balanceon.PFR. FAodX= ·rA ' ·dV
if !lQ pressure or phase change (liquid phase) therefore
V=F~of ~~ 2. Ratelaw
4-79
3.
Stoíchícmetry
íiquid phasc
v
c:A~
~fM_(~-:~t~c~AoC1--Al
u.: . . . ~ ········- - (.'.1ü1 (l···X)
=- .
e)
l""X.
J. (~,}~t~XY - : J. V=:~;~\\)
V
,=
FAo
k
-·····u: ·1°.o\
~'\O
ax
1
1 ·-X)"
"''A""kfc,,._c¡¡-·~ t
KcJ
Ar equilibrium
Kc--Cce ...
=·
- CA< C¡¡.,
KC,1.0=1
X
~=4
l··2X+X2,,,,_X .. =X. KCAO 4
PFR
(.on2rr
.
AO
x_l ""kc' º 1 xf . : -_._ x_ ..._J Kc l KcCA.ol
(1-x\1I
A
í_(
1 v1
=···10-li-2X,X ·tj V ""250
f·
6 O
· &,.ix_· ···-·----·-· "J ")<:'.V X'" l ··· •• JA. +
•
V"' 250 ~-[~O)+ 4~.15) + 2t3) + 4~.45)
+
& 6J]""
250 Q,j5 fl + 4{ L 41) + 2(2.41) + 4{526) +100] V"" 250 ~[f{O) + 4ft.15) + 2~ . .3)+4ft.45) +· Q 6))
""2soW.fi + a11.41J + 2!2.41) + 4.s.26) +HXíJ 3 e
""\
'
= 1656 dm'
CDP4-I (a)
4-80
, ·r,1.=0
·=--··..A·-··<..'¡0:1-Xf C.,.d)·Xf
= .... CAoX
X2 - 225 X -t 1 = O
sx = .::!.A..= k.C re dV FAQ . "ºl
·
d) Solurion similar to pan (a) ro (e)
-uo
¡A
oo
--- roolJrrrin} ·(01 d.r:rr'/mol min)(2 rnoJJdm31
F,vÍ9R-12X)
'.)
evaluarion
v = .....
Uo
ü
c::s::::
5. P==
= 'Ua
BenzeneisA
-r A =k A (c2A - CaCc) K e
r.
forequillbrium -rA =O
4Kc{l-Xcf =Xc2 4Kc -8KcXc
+ 4KcXc 2 = X, 2
0.2X2 -2.4Xc
+ 1.2 = O
XC =0.52
CDP4-I (b) PFR V= FAo
Design Equation:
RateLaw:
-r
Stoichiometry : C Ao = _!_ =
RT
A
=k A
J -rA dX
(c2 - CBCC) A
K
e
e A= e .Ao (1 - X) S atm
CB = e Aox 2
e
e
=0.0037 lbmol
(0.13 lbmol·· ft ·atm Kiss9.6rR)· · R 3
J
4-81
ft3
= e .Aox 2
V-
FAo
V=
Combine:
kC!c,
r
(101bmoVminXImin/60s)
- (1soort3nbmol·sXo.0031)2
V= 6.76ft3
dX
[(1- X)2 - X2 ] 4Kc
rx
dX
Jo [(1-x)2-o.833X2]
Íx dX Jo 6-2X+0.167X2]
v-67j
1 ln[0.523(X-11.45Jíl) - . \. (0.617Xll.45-0.523) 11.45 X-0.523 U V =13.5ft3
CDP4-I (e) CSTR
Design Equation :
RateLaw: Stoichiometry: Combine:
v =
V
= FAox -rA
-r,
=k,(e¡- c~~c J
CA =CAo(l-X) V=
Ca=
c;x
ce=
FAoX
k,c!.[<1-xY-( !,] = 147_73ft3
(10Xo.s1X1160)
(1sooxo.oro1)'[(1-o.s1)'-( ~~~)] CDP4-I (d) Amount processed in CSTR :
(10 lbmol/minX60 minlhr X24 hr/day )= 14,400 lbmol/day
4-82
c;x
Batch t-C
- "º
1
J d.X _
CAo
-:::¡-A - kC M 2
J
dX [
(1-X)2--X2
]
4Kc
(o.161X11.~s-o.s23)i{~~~(~=~~:])
= ~,. (
t =0.30s Taking into account the time it takcs to clean the reactor and other down time assume that the total time pcr 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 lbmol/day 6nms/day = N Ao
t
V
I dX
:.
-rA
V _ N Ao
1
V
= N J dX Ao
-rA
t
(
--t-kC!o
= 2400 lbmol/run
l Jn[0.523( X-11.45 (0.167Xll.45-0.523) 11.45 X-0.523
Jíl)U
V2400 ( 1 l [0.523(X-11.45Jíl) - (4Xtsooxo.0037)2 (0.167Xll.45-0.S23) n ll.45 X-0.523 U V = 48,690 ft3
CDP4-I (e) E
= 30,202 btu/lbmol
forX =O, -rA(IOO>
-rA =kC!o
k800C!.,
ka
E( 1
1
-r,.ct4CO> = k1400C¡, = k1400 = exp R T2 - T1
.
- rA
30,202 btu/lbmol
(
-
1
1
-rA(t400> = exp 1.987 btu/lbmol-"R 125~.67"R
-r
A(100)
J )
185?.67"R
= 49
-rA(l400)
4-83
r
CDP4-J PBR ?vIB:
Combine:
dX
"""""
dW
_·····~
F,,.0
kC,..o J" o 1-- dX---X = (---·) Jo (1 ··· W a)I'Y.vdW w
.1,1
.
FA.O
MB: Combine: X_
w""' ·---:~:::;
.
kC.._0 (1 ··· X)
l 000
=
1
,
X (1-X)
· .
b) We know, s = O, so
P = P0 (l - aW)112 => 1 = 20(1-- aW)112
=> a = 9.98 X 10"4 e) For conversion to be maximum, a should be mínimum
G2
AcDP
G
Dpipe
1
And it is proportional to ---2- =
-»,
1
=
Also, we know that a is proportional to ----
Dpipe
4
2
»,
6
». 2
for turbulent flow
for laminar flow
Hence for minimum a, DP and Dpipe should be increased. d) Yes, we can increase D, and decrease Dpipe (or vice versa) according to the above equation to get the same value of u.e)
1
»:
For assumed turbulent flow,u is proportional to --
6
4-84
2 DP
ª1 -- D pipe2
Now,
ª2
Dptpel
Dpipe1 ____; ;. . . ;___ = --4 =>Dp2
.
Dp1
= kC A
= CAO (1 -
dX
Mole balance: --·
dW
. .
Combining:
k1
-1
-
1
Stoichiometry: CA
k
Dp1
2
k - - :::> k2 = k1 -DP DP2 k2 0.5 - = --= 11.4 k1 0.044
g) Rate law: - rA
___1..
6
= 0.044 cm
t) We're given,
Hence
2 D p2
and Dp1= 0.5 cm
6
Dpipe2
6
dX
--
dW
X )y
--r = __ A FAO
=·
2
1/2
k2CA0 (l-X)(l-a2W)
·-
(for
s = O)
FAO
= 11.4 and eAO k1 = 2X 10-4Integrating with limits: (X70 to X; W7 O to 1000)=> X =0.78 FAO
CDP4-K t. Mole Balance :
dX =-
rAAc , . ,., .
dz
FAo
g, Rate Law :
rA = -·k'C"A p C -(1-
<0) ,
where k', Pe, and
3
Stoichiornetry :
4-85
·---·
---------------------··
------·---~---------------------------··-----------
4.
Pressure Drop :
when neglecting the turbulent contribution
10 pressure
drop, ~ is given by 0
[{l-'o = G[.· 1~S){;··.cr .L.)" ::PT1:.J· · • ~, . PooC
p(!,
.
where all of the bracketed variables are unknown constants. , m (J =
Ac The rnass flow rate is also ami unknown constant. Thus
B
O
can be written as a con~!.ª11't~
B, over cross- sectional area, where Where B is a function of the mass flow rate, feed properties, and catalyst properties=-all of which are constant fer part (a). Note that all equations up to now apply to a tnbtfüu PBR and spherical PBR To find B and k., we must model the tubular PBR on POLYMATH by entering the above equations in addition to A<.""
.
11:.(20 dm)2 4
= 314dm2
B and leo must be arbirrarily chosen at first. Then, depending on whether POLTh1ATH high or low valúes ot y and X, one can converge on the true valúes of k, and B. In order to do so efficiently, however, one rnust make use M the following trends: (Note tnat tne rouowmg taoíe is compíetety true onív wnen e > u . )
a) From ?OLYMA..TH.:
B=843
k=0.21
x:=.61
P::::yP0=.973 * 1500=1459 ...5
4-86
Equa&.ions .:. d(y)
= . ·beta/
/d(z)
(po•y)
• (J.+eps*x)
d(xl/.4{zl=·r~·Ac/Fao poa.l.500
x and y vs z
eps=O,S Faoz950
Key
-x
R=lO
y
Cao,..0.4 k=0.;21 B"'8~3
Ac~3.1416~(R**2) Ca=Cao*y* (J. ..x:) I (l.+eps•x) beta:aB/Ac ra=··k*Ca""2 220
~ª!:.Í:™L d (y) /d(z)
=··beta/
(po•yJ
X
• (l.+eps•xJ
d(xl/d(z)"'--ra•Ac/Fao
and y
VS Z
;
po=l.500
fu:.x ---x
eps=0.5
esce
.,f.
... y
i,ao==950 R;;;25 cao"'0,4 k=0.21 l'l:843
v:24 Ac,:;3.. 1416* {R**2· · (;: ... ¡:,) '"'2)
· ..... "~-+--· ._ ... ~--.
~ooo
ca ..cao•y• ll··xl J (l+eps*x}
,i::i..
ee:
:::·a.7-k'*Ca" • 2
betaa=B/AC
zo • o.
• 48
zf
b) From polymath: Maximum flow rate = 1750 mol/s aguac.io!!!'!:. /d(zl =··'.bet:a./ (po•yJ
d(y)
• (l.+·eps•x)
d(xl /d(zJ =··ra"Ac/Fao ¡;,o.,l.500
eps=0.5
(b) Fao Ke¡_
. -x
f'ao:;1750
1750
T
c.acc
y
R.=26
=
L
c.ao=(J.. 4 1<:=0 .21 3=843 t=24 Ac"'3 .l.41.6*
ca••2 . . cz --Ll ••;;n
Ca=Cao "Y* { l·-x) ! {l.+eps •x)
ac.ccc
ra= .. ·k•·Ca*•2
z
beta=E/AC
z0
=
0,
48
4-87
·····.--1---·-··-··-"IC.;:u;Q
~:;:::,:;:;::;
'<4
e) From POLYMATR Mínimum pressure=Sü kPa E,;r..:aci~E!~:. d
/d (z) <=···be~a/ (po~yl
(y)
• (l~e¡;,s•x)
(e) Po
/d{z):-ra"Act:·ao poc:SO d(x)
epso:0
5
!l..90G
X
Fao,,,950
=
80
.i ..
y
R:=25
Cao,,,0.4. k"'0 .. 21. Bec;l34.3
!..:24. Ac=3. l.41.6" (R*•2 ... (z .... ¡.¡ '""2) CazC:ao-y• U.-·x) I (l+eps*xl :::a=o···k"'Ca * '"2
:z
beta=.B/Ac
'o = ' O,
d) Frorn POLYMATif: d(y)
/d{:d"
d(x)
/d ( z )
po=l.500
eps,,o
..beta/
(po~y)
• (l+eps•x)
(d)
= . ,r.a*Ac/2ao
Kel
s
--
Fao=950
X
y
l<.~25
Cao .. o.4. k= O .21 Bc;,843
Ac=l 14l&*(R**2 .. (z-L)**2J Ca;.Cao*y"
(1. ·X)f (1.+eps~x)
ra~···k""Ca**2 be\":a=B/Ac
zo "' o,
43
4-88
CDP4-L (a)
For a conventional PFR.:
(b)
ax
FAo-=-rA dV
···r,.,
= k(CA -C /Kc)= 1
F or the lMRCF:
kCA<>{(l-·X)-X!Kc)
dF
.....A=rv dt A ., ~!. =
··rA
Vº •..
v.,kcCs
-r, = k{C._ ···CafKc) e... =F;/V, C8 =f~N . r
V -
(FA +F FA.,
8) V . ·-········-·-· n
Use these equations in POLYMATR te generare plots of'the conversión profile. ::J,.3:CO
c. 1 ac
O.:.JOC
sri.occ
4-89
oca
._;. s:::::
o. scc
0.00:J
(e)
... fb
16 .. 000
2"1.000
30.00C
Use the same equations for rhe L\.iRCF in POLYMATH to generate the desired
pfot.
2.
•··--:::.a :::;b ··fa
. ..... __ . -+-·---.----;
······--·-+·····
0,.000
t
.
-
·.~·
.
oc
5C'.J
n.soo
'"·~ ...
;
_:;:-·:;!.;..
-- - ' --- - -----·
- ~
-------
-·-·
-;.: . :. ,.,. -- - -,- '." .... - ... - .... - . . . ,_,. ~.=-: :-: :::-: : -_ :- . :. :~ ;::::~ :::: :; :·,; :,~-~--~ ... :._.
:J. $00
~
o . coo
:;.oca
GOO
4-90
l 3. OCQ
24.J.C;O
""
;
----~-;,:.>~
(d)
By varying one parameter ata time we can see the effect of each:
Increasing the specific reaction rate causes cbanges in conversion. concentration, and molar flow rate to occur more quickly . Lowering the transpon coefficient (k,) causes an increase in both C8 and F8, whích causes a decrease in conversion . By raísing the equilibriurn constant (K,,). we cause a decrease in the molar flow rate of A and an increase in couversion, (e)
A significant increase in ternperarure for an exotherrnic reaction would drive the reaction in the reverse. This would cause a decrease in X and C8 and an increase in the CA. A significant decrease in ternperamre for an exothermic reaction would cause increase in the rate of the forward reaction. This would drive up X and C-a, while lowering CA. . An increase in temperature would drive an endothérmíc reaction forward, raísing X and C8, while lowering C,_. A decrease in ternperature would cause an increase in the reverse reaction for an endotherrnic reaction, This wonld raise CA. and lower bota X and C11.
CDP4-M No solution CDP4-N
4-91
P"A4
y FA.3
FAZ
F,U
l\5
F32 (by
F81
O (by
t i-x:
"h.erc
X is
p r o bLem
p r c b Le.n
cc nv e r-s i o n defined
with
.r e s p e c t; to
:-Al
s pe ci.f i.c a t.i.o a )
s p e c i.fLc a c i oa )
FAO
e • A1
1-y< 1--:n ,\.l so
Stoichiometric
table
ln.let.
Outl
:-Al
v0
~
:_,111 F
. Al. c.i.,
CDP4-0
V
4
et
FA,U.-!)_
~
f-
"100
Fg 100
No solution
4-92
---
-- -~-------------
CDP4-P
Al equilibrium
Initial Condition
k/~lle
CBo = Cue
(a)
=
k3Cce
C AU +
k2CB0 ·.-= k3CA0 "-?
e·-HO = -k3-~('-AO
( l)
k2
Cce
9fui
·to
tz:
CÍe + 9',, + Cce
•o L.., _o
""C1\0
(b)
Cno is a constant
dCu ···-~--- ~' -· k 1 C'. dt
J··c· O ·- '!
,,.
· AO '·
, --
h
l
"
+
k , e·-<' ,)
•
·-k1t
Problems frorn Probjot
Frorn (1) . e"·k1. l'-3 -
(e)
l''
~~;t1 dt
=
k¡C.'.
A
t
:::::>
C-A -· C. e - ·AO
k1t
For a maxima in Cn,
for a mínima in Cu
4-93
1
CDP4-Q CDP4-R CDP4-S CDP4-T
CDP4-U CDP4-V
4-94
--------~·---~-~--~-
--
-------------------
Solutions for Chapter 5 - Collection and Analy sis of Rate Data PS-1 (a) Individualized solution PS-1 (b) Individualized solution PS-1 ( C) Individualized solution PS-1 ( d) Individualized solution PS-1 (e) Individualized solution PS-1 (f) Individualized solution PS-1 (g) Individualized solution PS-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.
PS-1 (i) Example
5-2 Assuming zero arder reaction: Rate Iaw:
-,.1cA = k
I
dt
........ t(rru~·n~.)~~---º--·~--l-S_0~-1---l0_0~-1-~lS~0'---1-~2~00'---1--2~50"----1-~3~0~0~~~ CA(mol/dm ) O.OS 0.038 0.0306 0.0256 0.0222 0.0195 0.0174 zero arder reaction
lst order reaction
O 06
4.5
005 35
w
004
~
003
[
!'
{
o 002
25 2 1 5
001
05 50
150
100
t(min . )
Or, ln(l/CA) = k't + 3 t(min.) O 50 CA(mol/dm ) O.OS 0.38
100 0.0306
150 0.0256
200 0.0222
5-1
250 0.0195
300 0.0174
200
250
300
1
3.27
1
3.49
1
3.66
1
3.81
j 3.94
j 4.05
Since none of these plots are straight lines, its is not Ist order reaction or second order reaction.
PS-1 (j) Example PS-1 (k)
5-3 Because when a is set equal to 2, the best value ofk must be found.
Example 5-4
CHci O (mol/dnr') "
2
-rHci,o(mol / cm .s)xlO
7
4.0 2. 0
1.0 1..2
2.0 1.36
0.1 0..36
0.5 0.74
See Polymath program p5 . . J .. k.pol. POL YMA TH Results Nonlinear regression (L-M) Model: r = k*(Ca"alfa) Variable Ini guess k 0.1 alfa O. 5 Precision RA2 0.9812838 RA2adj 0.9750451 = 0.0341709 Rrnsd Variance = 0.0097304
Value
l. 0672503
0.4461986
95% confidence 0.0898063 0.076408
= =
Rate law:
-r"
PS-1 (1) rate law:
= l. Ictimol I cm':s
Example 5-5
rCH4
= kPt0P~
Regressing the data
-
...-----
r'(grnolf'Hz/gcat.min) 5.2e-3 13.2e-3 30e-3 4.95e-3 7.42e-3 5.25e-3
--
Pco(atm) 1 1.8 4.08 1 1 1
-
1
0.1 0.5 4
See Polymath program PS-·1-1.pol. POL YMA TH .Results Nonlinear regression (L-M)
Model: r = k*(PC0"alfa)*(PH211beta) Variable Ini guess Value k 0.1 0.0060979 alfa 1 1 . 1381455 beta 1 0.0103839 Precision RA2 = 0. 9869709
PH2 (atm) 1 1
95% confidence 6 . 449E-04 0.0850634 0 . 1951217
5-2
--
--
R"2adj = 0.9782849 Rmsd = 4.176E-04 Variance = 2.093E-06
Therefore order ofreaction = 1.14 Again regressing the above data putting
fJ = 1
POLYMATH ResuJts Nonlinear regression (L-M)
Model: r = k*(PC0"0.14)*(PH2) Variable Ini guess k 0.1 Precision R"2 = -0.8194508 R"2adj = -0.8194508 Rmsd 0.0049354 Variance = 1.754E-04
Value 0.0040792
95% confidence 0.0076284
=
Therefore, k = 0.004 (gmolCHJ(gcatmin.atm114))
PS-2 Solution is in the decoding algorithm given with the modules. PS-3 lndividualized solution PS-4 (a) The kinetics of this deoxygenation of hemoglobin in blood was studied with the aid of a tubular reactor. Hb02 "7 Hb + 02 Rate law: -lA=kC; Mole balance:
F
AO
dX =kCn (I-X)n dV
AO
dV = Acdz , where Ac is the tube cross sectional area
dX d:
= kc;OAC (1-xr FAo
In ( :: ) = In a + n In ( 1- X) therefore,
te: A
where a=-~ ,
FAo
5-3
7 A histogram plot of graphical differentiation: 0.00395 0.0039 0 . 00385 0 . 0038 eo.00375 :S 0.0037 "tl0.00365 ~ 0.0036 0.00355 0.0035 0 . 00345 0.0034
z vs . z is then produced. The values of dX/dz are evaluated using equal-area
dX/dz vs. distance
+--+--~-~+---~1---.---.~~-~-+-~-l---i O 25 5 75 10 125 15 175 20 225 25 275 30 325 35 375
z(cm)
40
Using the vaiues obtained above, a piot ofln(dXA/dz) vs. In(l-XA) is produced anda line is fit to the data ln{dX/dz) vs. ln{1-X)
In ( :; ) y= 10577x-55477
•
= In a + n In ( 1 -
where, a=
ic: A FAo
ln(1-X)
In ( :; )
= -.5 . .5477 + 1.0.577 In (1- X)
n=l
In(a) = -5 . .5 a= exp(-5 . .5) = 4.1 x 10-3 cm. Concentration of blood is 150g hemoglobin per liter of blood Molecular weight of hemoglobin = 64500 Ac=0.0196cm
CA0
= 2.3x10·-6mol / cm"
FA0 =45.7xl0-6moles/s k= FAoª
Cl0Ac
=
4}.7xl0-6moles/s (4.lx10-3cm)=4.ls-' 2.3x10- moles/cm3x0.0l96cm2
Hence rate law is,
mol =r, =4.lCA-dm •S 3-
5-4
Ao
e
X)
PS-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. POLYMA TH 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
al a2 a3 a4 as a6
Value 2.918E-14 0.0040267 -6.14E-05 7.767E-06 -5.0E-07 l. 467E-08 -l.6E-10
95% confidence
o o o o o o o
General Order of polynomial = 6 Regression including free parameter Number of observations = 7 Statistics RA2 = RA2adj = Rmsd= Variance =
1
o
1.669E-10 1.0E+99
Next we differentiate our expression of X(z) to find dX/dz and knowing that
In ( :: ) = In a + n In ( 1 ·- X ) where a= ,
ic:Ao Ae FAo
Linear regression of
In
(·:1-)
as a function of
In ( 1- X) gives us similar vaules of slope and intercept
as in the finite differences. POLYMATH Results Linear Regression Report Modal: ln(dxdz) = aO + a1*1n(1·-X) Variable
ªº al
Value -5.531947 1.2824279
95% confidence 0.0241574 0 . 3446187
General Regression including free parameter Number of observations = 7 Statistics RA2 = RA2adj Rmsd= Variance
=
n=
L28
=
0.9482059 0.9378471 0.0044015 1.899E-04
5-5
In a= -5..53, a= 0.00396
k=
FA0a = 465.7x10-6moles/s C~0Ac 2.3x10- moles/ cm3 x0.0l96cm 2(3.96xl0-3cm)=4.os-1
Hence rate law is,
- rA = 4 OC12s A º
mol 3 dm •S
PS-5 (a) Liquid phase irreversible reaction: A 7 B + e ; CAo = 2 mole/dm3
CAO -CA -kCª -
T
In ( eAO;
A
eA ) = In k + a In eA
Space time ( T )min. 15 38 100 300 1200
..
-
CA(mol/dmj) 1.5 1.25 1.0 0.75 0.5
ln(CA) 0.40546511 0.22314355
ln((CAO-CA)/T) -3.4011974 -3.9252682 -4.6051702 -5.4806389 -6.6846117
o
-0.28768207 -0.69314718
By using linear regression in polymath:
See Polymath program P5--5-a.pol.
ln((Cao-CaO/tau) vs ln{Ca)
_¡;
POLYMATH.Results Linear Regression Report
-0.6
-0.4
Model: y= aO + a1*1nCa
In ( eAO;
e = In k + a In e A )
A
Variable Value 95% confidence --------4. 6080579 O. 0162119 al 2.9998151 0.0411145
ªº .
Statistics RA2 = RA2adj Rmsd= Variance =
0.9999443 0.9999258 0003883 1.256E-04
=
s ~
i
e
-0.2
0. 2
-1 -2
,1
y= 2.9998x - 4.6081
-3 -4
......
-
;····
11
1
-7
1
··B
ln(Ca)
Hence,
a= slope e s
ln(k) = intercept = -4.6 therefore, k = 0.01 moleimin" . Rate law:
- dCA = O.OlC1mol / dm3 min
dt
PS-5 (b) Individualized solution PS-5 ( C) Individualized solution
5-6
---------------·--------------------
---
0.4
PS-6 (a) Constant voume batch reactor: Mole balance:
- dCA =kC a dt A Integrating with initial condition when t = O and CA = CAo for t
=
1
C(l-a)
AO
_ c
A
(1- a)
k
1 (2)°-a) _
=-
c:»
(1- a)
k
A
a i:-1.0
substituting for initial concentration CAo = 2
t (min.)
CA(mol/dm3)
o
2
5
1.6 1.35
9 15
1.1
22
0.87 0.70 0.53 0.35
30 ,______ 40 60
-
See Polymathprogram Pfi-ó-a.pol, POLYMATH Results Nonlinear regression (L-M) Model: t = (1/k)*((2A(1-alfa)HCaA(l-alfa)))/(1 -alta) Variable Ini guess Value 95% confidence k 0.1 0.0329798 3.628E-04 alfa 2 1.5151242 0.0433727 Precision RA2 = 0.9997773 R"2adj = 0.9997327 Rmsd 0.1007934 Variance= 0 . 0995612
=
K= 0 . 03 (moVdm3}05 .s-1 and Hence, rate law is
dC dt
P5-6 (b) Individualized P5-6 ( C) Individualized P5-6 ( d) Individualized
a= 1.5
A_=
0.03C~5mol / dm3 .s
solution solution solution
PS-7 (a) Liquid phase reaction of methanol and triphenyl in a batch rector . CH30H + (C6Hs)3CCI 7 (C6Hs)3COCH3 + HCl
A+
B
7
C
+
D
Using second set of data, when CAo = 0 . 01 moVdm3 and C80 = 0 . .1 mol/dnr'
5-7
CA(mol/dm.;)
t (h)
o
0.1 0.0847 0.0735 0.0526 0.0357
1 2 5 10 -· Rate law:
=r, =kc;c; CAo « Cao =>
For table 2 data:
1 c
Using eqn 5-21, t = -·
AD
=r, = k e; where k = kc;o I
c:» A
1
1 (0.01)(1-m) - c
k' (1- m) k' See Polymath program P5-7-a-l poi.
(1- m)
A
POLYMATH.Results Nonlinear regression (L-M) Model: t = (1/k)*((Q.1A(1-m))-(CaA(1-m)))/(1-m) Ini guess
Variable
Value
95% confidence 0.0109025 O. 0021115
1-. 815656
1 2
k m
2.0027694
Nonlinear regression settings Max # iterations = 64 Precision R"2 R"2adj Rmsd Variance
= = = =
1
0 . 9999999 3 . 268E-04 8.902E-07
Therefore, m = 2 For first set of data, equal molar feed => CA Hence, rate law becomes
= C8
=r, = kC1c; = kCi
2+n)
Observation table 2: for CAo =0.01 and C80 = 0.1 t (h)
o
0.278 1.389 2.78 8.33 .. 16.66
CA(mol/d~1.0 0.95. 0.816 0.707 ·0.50 0.37
1 c
t =-·-
k
AO
A
(l-(2+n))
k
(-1-n))
5-8
~------------
----------------~----~-
---
See Polymath program PS--7-a-2.pol. POL YMA TH Results Nonlinear regression (L-M) Model: t = (1"(-1-n)-Ca"(-1-n))/(k*(-1-n)) Variable n
Ini guess 3
k
2
Value 0.8319298 0.1695108
95% confidence O. 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:
rA
PS- 7 (b) Individualized
=0 17C2 •
A
cº·Bs
mol3 dmh
solution
PS-8 (a) At t = O, there is only (CH3)i0. At t = co, there is no(CH3)i0. Since for every mole of (CH3h0 consumed there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure. P(oo) = 3P0 931 = 3Po Po:::::310mmHg
PS-8 (b)
Constant volume reactor at T = 504 ºC = 777 K Data for the decom osition of dimeth lether in a as hase: Time O 390 777 P1(mm H ) 312 408 488
(CH3hO-> CH4 + H2 +
·~-----
1195 562
ca
YAo =1 8=3-1=2 c=óyAO =2
V = V0(
;
) ( 1-
e X) = V0
because the volume is constant.
P=P¡¡(I+cX) at t = oo, X
= XAF = 1
5-9
3155 799
·=1
1 dNA
---=-----=r
V dt
-rA = kCA
Assume
CA
= CAD ( 1CA0-
Then:
!JP ef'o dt dP dt
(i . e . 151 order)
X) (V is constant)
= kCAo (1-X)
ef'o dX 1 dP = --dt ef'o dt
Therefore:
_l
dX dt
A
P·-·R0
and X=
or
NAO dX V0 dt
= k[l-
P-f'o] ef'o
=~([l+e]f'o-P) ef'o ·
= k ( [ 1 + 8] f'o - P) dP
f-·---= fkdt [l+e]f'o-·P
P
I
R o
O
Integrating gives:
ef'o ln ( ) · [ 1 + 8 f'o -·· P
Therefore, íf a plot of ln
624 936-P
J = ln [. 3f'o2f'o- P ] = ln [ 936624- P J_ = kt
versus time is linear, the reaction is first order. From the figure below,
we can see that the plot is linear with a slope of 0.00048. Therefore the rate law is:
=r¿
= 0.00048CA
::-+----~y_=_0_.0_0_0_48_x_-_o_.0_29_0_7_~~~·
0.4
o.....,__--~--------------i -0.4
-+--~------------------<
o PS-8 ( C)
1000
2000
3000
4000
Individualized solution
5-10
PS-8 ( d) The rate constant would increase with an increase in temperature. 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.
PS-9 Photochemical decay of bromine in bright sunlight: t (min)
10 2.4 5
CA (ppm)
20 1.7 4
30 1.2 3
40 0.8 8
50 0.6 2
60 0.4 4
PS-9 (a) Mole balance: constant V
dCA =r =-kCª
dt
A
A
Differentiation T (min) ~t (min) CA (ppm) ~CA (ppm)
.L\CA L\t
(
10
20 10
2.45
PP_In) mm
30
40
10 1.74
10 1.23
50 10
0.88
0 . 62
·Ü.51
-0. 35
-0.26
-0.18
-0.071
-0. 051
-0.035
-0.026
-0.018
0 .. 01
!O
·------····-.!.. . . ,__ ··-··-_.....___.... JO 40
5-11
50
10 0.44
-0.71
o . o,
o
60 10
-J.Q
~,.o
After p lotting . an -dCA/dt ln(-dCA/dt) lnCA
1
erenuaunz b 0.082 -2.501 0.896
ecua area 0.061 -2.797 0.554 ·-
0.042 -3.170 0.207
0.030 ··3.507 -0.128
0.0215 -3.840 -0.478
0.014 -4.269 -0.821
-
Using linear regression: a= l.O In k = -3.3864 k = 0 . 0344 min-1
PS-9 (b) dN __ A·=Vr iff
A
=F
B
rA
= ·-0.0344 PJJ_m = -0.0344
Fa
=(25000gal)(o.0344
mm
m~ · at CA= 1 ppm lmm
m~ )(6omin)(-~)(3.?_8511)( lnun hr IOOOmg Igal
PS-9 ( e) Individualized solution PS-10 (a) Gas phase decomposition A7B+2C Determine the reaction order and specific reaction rate for the reaction
dCA n = -kC A dt
Assume the rate law as: --
Integrating:
t=
l ( l k(n -1) C n- 1 A
-
C
l AO
n--1
J 5-12
Ilbs_)=0.426lbs 453.6g hr
=>ln(t112) = ln (
2n-l
-lJ
(n- l)k
+ (1- n)ln(C Ao)
=> ln(t) = ao + a¡ln(CAo) Run# 1 2 3 4 5
t (min.) ---· 4.1 7.7 9.8 1.96 1.3
CAo (gmol/lt) 0.025 0.0133 0.01 0.05 0.075
See Polymath program PS-10.-a.po]. POLYMATH Results Linear Regression Report Model: lnt = aO + a1*1nCaO Variable
a:o-al
Value -~--3528748 -1.0128699
95% confidence 0.1831062 0.0492329
General Regression including free parameter Number of observations = 5 Statistics RA2 = RA2adj = Rmsd= Variance =
0.9993004 0.9990673 0.0091483 6.974E-04
3.0 --,------------~
24 1.8 1.2
0.6 0,,0 '----~---~--~-'
-4. 61
-4.20
-2,,99
-2.59
From linearization, n = 1- slope = 2.103 ;: :; 2
(2ª-1 -l)e-intercepl
k = -·
a-·-1
· =10.52 ----
lt
gmol.min
5-13
-
ln t 1.410987 2.0412203 2.2823824 0.67294447 0.26236426
lnCAo -3.6888795 -4.3199912 -4.6051702 -2.9957323 -2.5902672
- dCA =10.5C/gmol/lt.min dt
\ ~
e ee See Polymath program Pó-Iü-a.pol. POL YMATH Results Nonlinear regression (L-M) Model: t = {{2"(a-1 ))· 1 )/(k*(a .. 1 ))*{1/CaO"(a· 1 )) Variable a
2
Ini guess
k
10.52
Precision R"2 R"2adj Rrnsd Variance
Value
l. 9772287
8.9909041
95% confidence 0.093057 3.9974498
= 0 . 9986943 = 0.9982591 0.0531391 = 0.0235313
=
_!!:_CA dt
Rate law:
= 9. OC/
gmol /lt. rnin
PS-10 (b) Weknow,
k= (2ª-1 -1)·- ( --- 1 f112 (a -1) CAO a-1 Solving for k at 11 Oº C
k' =
( 2(l-l)
-1)(
2(2-1)
From these values,
1
0.025c2-1l
E
J ) = 20
l
t
gmol. rnin
(s.314
Rln k2 k
1-_-_-1~~J= (-l- __l_)
= -,(-
T,,
1
t;
º
2 J )1n mol.K 10.5
373K
5-14
383K
=
76.5 kJ/mol
PS-11 03 + wall ~ loss of 03 03 + alkene ~ products Rate law:
k¡ k2
=T.03 = __ ac; = k¡ + k2-1!!L C ·
dt
e
3
Oz
Using polymath nonlinear regression we can find the values of kl and k2 Run #
ozonerate (mol/s.dnr') Ozra
1 2 3 4 5 6
Ozone concentration (mol/s.dm') C03
le-12 le-11
0.01 0.02 0.015 0.005 0.001 0. 018
l.5e-7 3.2e-7 3.Se-7 5.0e-7 8.Se-7 4.7e-7
Butene concentration (mol/s.dnr') Cbu
le-10 le-09 le-08 le-09
See Polymath program PS-11.pol. POLYMA TH Results Nonlinear regression (L-M) Model: Ozra
=
k1 +k2*Gbu/C03 Ini guess 2.0E-07 0.1
variable kl 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, =(3.5x10-7)+(0.05) ~"" mol l dm':s 03
PS-12 Given: Plot of percent decomposition of N02 vs VfF AO
X= % Decomposition of N02 100 Assume that - rA = kC~
5-15
For a CSTR
F X
V =__M__
100
-rA
V
X
X
FA0
=r,
te;
or --=--=--
with n e O, X=k-
V FAo V
X has a linear relationship with --
as
FAO
o
shown in the figure. Therefore the reaction is zero order.
P5-13 Si02 +6HF ~ H2SiF6 +2H20
N
s
= molesof Si02=
AcPsó MW s
Ac = cross-sectional area
Ps = silicon dioxide density MWs = molecular weight of silicon dioxide = 60.0 ó = depth of Si
wpV
NF= moles of HF=----
100 MWF
w = weight percentage of HF in solution p = density of solution V = volume of solution MWF = molecular weight of HF = 20.0 Assume the rate law is -r5
dN
Mole balance: __
dt
_ AcPs do MWS dt
= r5V
s
-k( oov Jª
= /Jwª
wV MWF
1
do kMW5 -- dt =iooª !\P; - do dt
= kC;
where
(
P MWF
V
Jª Vw
a
/3 = ~MWs
-(_!!_·-)ª
100ª !\Ps MWF
V
ln(-:~)=ln/J+alnw
.5-16
~:t:: (g ~-:o,) Z
3
"'
5
In(-~~)
-16..629 2.079
In w
where (-
dJ) dt
-1432
-15.425 2.996
3.497
-13.816 3 . 689
-13.479 3.871
is in~
mm
From linear regression between slope =a= 1.775 intercept = In p = -20.462 or
In (- ~~)
p = 1.2986 *
and In w we have:
10·9
~13 ,.,......--.--------~-----------------------
In w kMWs ( p JLm V P -- 1001775 Ac:Ps MWF
,\; = (10*10-6m )(10m )(2sides )(1000wafers) = 0.2m2 MWs=60-ggmol
Ps
= 2.32 L = 2.32 * 106 mi
g3 m
(Handbook of Chemistry and Physics, 57th ed . , p.B-155)
P
=¡_.f..= 106 L ml
m3
MWF =20-ggmol V = 0.5 dm' = 0.0005 m3 p = 1.2986 * 10-9
5-17
1.775 1.2986
* 10- =
k(6o-g)
20-g-
(o.2m2)( 2.3*106
gmol
k
3 J0.775
= 3.224*10-7 ( _!!!__ gmol
:3
)(100)1115
min "
5-2.316 ( O .2) = O .107 5
Final concentration of HF =
weight fraction = 10.7% weight fraction = 20%
Initial concentrntion of HW = 0.2 (given)
dNF. dt
Mole balance for HF: --
-
pV
( 0.0005 m3)
gmol
9
dw =6k(
lOOMWF dt
= 6--dNS dt
wp
100 MWF
Jª V
where u e 1.775
_1of.1 dw -6k(·-f!_Jon5 'fdt 20 WL775 -
_1_( __1 _)107 0.775
--1-( 0.775
t
= 331
w0775
lOOMWF
=6(3 224*10-7)(
l
l
20°775
J0.775
106
t
20*100
·
20
10.7°775
o
)
= 2 389*10-4t .
min
P 5-14 (a) A+ 3B 7 C + 2D + E Observation table for differential reactor: Conc. Of Temperature(K) Aünol/drrr') ,_______ 0.10 323 0.10 333 343 -- 0.05 0.10 353 0.20 363 0.01 363 Space time for differential reactor = 2 min
Conc . Of
-
Búnol/drrr')
0.10 0.10 0.10 0.05 0.01 0.01
E;, - voCP V ------
5-18
Conc . Of C(mol/dm3) 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: = Cp = CCzH4 P
r
2
Rate law:
Where, A is Arrhenius constant B = activation energy/R x is the order of reaction wrt A y is order of reaction wrt B CA is the concentration of C2H4Br Ca is the concentration of KI Now using data for temperature 323K, 333K, and 363K, for finding the approximate value ofB because, at these temperature, the concentration of A and B are the same. Using polymath, the rough value ofB = 5500K 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 error we got B = 6500K
See Polymath program P5-14-a.pol. POLYMA TH Results Nonlinear regression (L-M)
Model: r = A*exp(-6500/T)*Ca"x*CbAy 95% confidence Variable Ini guess Value 2.928E+04 A 3.6E+05 3.649E+06 0.0032606 X 0.25 0.2508555 y 0.2 0.2963283 0.0020764 Precision = 0.9323139 R"2 R"2adj = 0.8871898 Rmsd = 3.615E-04 Variance 1.568E-06
=
Hence, by nonlinear regression using polymath A= 3.649E+06(mole/dm3)"2 6(1/s) E= 6500R = 54.015 KJ/mol X= 0.25 y= 0.30 hence,
re = 3.64E + 06e(-54º15 / RT)c125 C~30
P 5-14 (b) Individualized
solution
P5-15 (a) 5-19
mole/dm3.rnin
Model 1: Monod equation
dCC = r = µmaxcscc -8 __:;::=;.;;..__;;___;;_
K, + C,
dt
See Polymath program P5-15-a.pol. POL YMATH Results Nonlinear regression (L-M) Model: rg Variable umax Ks Precision R"2 R"2adj Rmsd Variance
=
(umax)*Cs*Cc/(Ks+Cs) Ini guess Value 1 0.3284383 1 1.694347
95% confidence 0.00686 2 . 2930643
=
0.9999439 = 0.9999327 = 0. 0038534 1.455E-04
=
PS-15 (b) Model 2: Tessier Equation
r8
=µ={!-ex{-~') ]ce
See Polymath program P5-15-b.poJ. POL YMATH Results Nonlinear regression (L-M) Model: rg
=
umax*(1·-exp(-Cs/k))*Cc
Variable umax
Ini guess 0.5 100
k
Precision R"2 R"2adj Rmsd Variance
r8
= = = =
Value 0.3258202 20.407487
95% confidence 0.0034969 5.7120407
0.9999454 0.9999345 0.0038004 1.415E-04
=0.33[1-exp( ;~; )]ce
Wdm'h
PS-15 (e) Model 3: Moser Equation
r g
=
µma.xcc I+ kC s -y
See Polymath program P5--15·c.pol. POLYMATH Results NonJinear regressfon (L-M)
5-20
Model: rg = umax*Cc/(1+k*Cs"(-y))
Variable umax
Ini guess 0.3
k y
l. 6 1
Precision R"2 R"2adj Rmsd = Variance =
= =
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
0.33Cc 1 + 162.6Cs c-2..1) ----_.c..--gldm3.h
PS-16 Thermal decomposition of isopropyl isocynate in a differential reactor .
Run 1 2
3 4
5 6
Rate (mol/s.drrr') 4.9 X lff4 1.1 X lff4 2.4 X 10·3 2.2 X 10º2 1.18 X 10·1 1.82 X W-2
Concentration (mol/dm') 0.2 0.02 --·0.05 0.08 0.1 0.06
Temperature -(K) 700 750 800 850 900 950
Rate law:
- rA -- A e (-EJRT)cnA
Where, A is Arrhenius constant E is the acti vation energy n is the order of reaction CA is the concentration of isopropyl isocynate
See Polymath program P5- l 6.pol. POL YMATH Results Nonlinear regression (L-M) Model: rA = A*exp(-E/(8.314*T))*(CA)"n Variable
Ini 100 guess _
A E n
1000 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 R"2 R"2adj Rmsd
=
0.6690419 0.4484032 = 0.0097848
=
5-21
·-
= 0.0011489
Variance
Hence, by nonlinear regression using polymath A= 10100 (mole/dm3r2 \lis) E = 58000 J/mol n = 1.7 therefore,
=r; = 10100exp ( -6976) T CA17 mole/dm3 .s CDP5-A Given the reactíon P + NH:PH --), NH:PHP where Pis Peniclllin and NfI.iOHP is hydroxylarnine acid (denoted by subscript HA)
Let A - Absorbency, then Cw,. = KA where K is sorne constan t. C, e,. . , (t.-· X) (s = O for liquid phase reaction.]
=
Ca,._
= Cp,,X = KA
Att=oo
•
A
-
r, X ·=
=~eK
K~ c.,
when X = 1 and A = A_
A
:. X·=·-· A_
Assume reactíon is irreversible:
- rJ>
= kC; = kC;_,(1 - X}"
For a batch. constant volume reactor; or or Try integral analysis first, Then
dA =K dt
or J.~A o
Assume that reaction is zero order :
= A = Kt
a plot of A vs. t should be linear if
reaction is zero order. From the plot below, it is evident that the reactíon
is not zero arder :
··········· ········---~~
06 0.5
•
•
•
•
:J_ 03
o
10
30
20
40
Time
5-22
50
Next, assume that the reactioa is first order:
dA . -. = K( A~ - A) or dt
j. A
O
dA
( A ..
. · .· = j Kdt = In (A-A) • · = -Kt t...
-· A)
.
A ..
0
· A plot of (A_ -A) vs. ton semi-Jog paper should be linear . ... ,,_,
· -· ·- -
-- . --~···-"'''"'" w
.i,---··-
Time O 10
A O
A..·A 0.685
-····"»·.
""0337 -0348
20"-
0.433
.. -· . ··-
..
•
------ .. - .. -··-·· -2C-------
-
0.252 0.190 -
S().
~
~----
··-·········-······ .
·--
-4-0-
,_
~
.
_ ..
-
.
• •
0. 1
-.. -,...·-······•· ·
-,,,
Time Ir is evident from the plot that the reactíon is not fasr order, Try second arder:
~dt = K(A_ ··A/ or
o
·-·nr
..-20 30
'"40
..
!....... · = ! · + Kt
(A_-A)
A_
A A..··A O L460 0.331'" .. --2.%7
o.
s
- _
_ .
8· 7
3.968--
+
0.539
o
•
•
2 .L-.
o
i
•
o
-·-····-····- .. ·-···· .. ···---·--.-
20Time3º
40
" •• J
so
From the plot, it is evedent that linear relationship exists between (L'A; - A) and time; Therefore the reaction is second order .
CDP5-B Determine the reaction order and specific rate constant for the isomerization reaction: A-~ B Rate law:
-rA=kCA
a
dCA
=--·-
dt
5-23
_,. -
-----:,-···--·---·-----·-----·-···-··--·--·-···--·····. ···-·---------·······CA (mol/dm ) -óC"/ó.t _-dC,,/dt
Time (min)
o
4
Q39 0..37
3
2 . 89
0.35 0.32
2.25
5
0.30 0.267
8
L45
0.25
0. .225 0 . 2[
!O
0.175
065
12
0.15 0.133
15
0.25
17.5
0.1 0.072
0.07
0.06
Plot of log --dC"/dt vs log CA shows a= 0.5
--~~ ...
k,
~
=··;if. · --0.:!f_ o 19--::WZº: __ . 65°5·· · . dm "m,tt t,,' .4
1
u=•
CDP5-C Ethane hydrolysis overa commercial nickel catalyst in a stirred contained solid reactor. H2 + C2H6 7 2CI!i
PA = CART =CA0RT(l -X) =PA0(1-X) PB =CBRT=CA0RT(BB -X)=PA0(BB -X)
X= ypFro =~= yP(l+Bn) 2FAO 2y AO 2
-r
,
A =-r
·-r,
,
FAOX
n=-W-=
Fp
2W
= kPA a pB/J
ln(-r'A) =lnk +alnPA + /JlnPB y = Ao
+ A1 Xi + A2 X 2
FTO(gmol/h)
PAo(atm)
PBo(atm)
1.7
0.5
0 . .5
1.2
0.5
0..5
0.6
0..5
05
0.3 0 . 75
0.4 0. 6
0.6 0.6
2.75 0.6 0. 4 POL YMATH Results Nonlinear regression {L-M} Model: ra:::: Variable
k*(Pa"alfa)*(PbAbeta) Ini guess
YCH4
o.o
5 0. 0 7 0.1 6 0.1 6 0.1
o.o
6
Value
Q
X
0. 0 5 0. 0 7 0.1 6
PA(atm)
P8(atm)
-rA(gmol/kg. h)
0.475
0.475
1.0625
0.465
0.465
1.05
1
0.42
0 . 42
L2
L5 1 0.6 7
0..32 0.54
0 . .52 0.54
0.6 0. 9375
0..57
0.37
2 . 0625
1
0.2 0.1
O.O
5
95%
confidence
5-24
--··---------·--
----
0.1
k
alfa beta Precision
O .1124446 0.152574 0.1668241
0.5068635 0.9828027 -a . 9669749
1 1
= = =
0.999213 0.9986883 0.0051228 = 3.149E-04
R"2 R"2adj
Rmsd variance
k = 0.5 atm gmol/hr kg; hence, the rate law is:
a= 1; /3 = -2
-r~
= 0.5 ;~
gmol I kg .hr
B
CDPS-D Since oxygen is found in excess, we assume that · -rl!O is dependent only on ~.
This gives us a rate law of the form;
-rHO
= kC:.0
From rbe uníts of the specífíc reactíon rate, ~e assume that a= 3. Now, using equaríon (5~ 18) from chapter 5, we can solve for the desíred half-Iíves,
2"·:I
-1(· ~¡-J=~~'. :~!(--t,·I= 2 -1(·+-1
k(a -1) (a)
11
CN0o
1
k(3-· 1) CNOo
2
2k.
Cw0o
1
For e_= 3000ppm:
,,,, -2\1A)'1ó..3ppl11•,nilii]( fOO() (b)
J
For ('~ 1'"
~p~)'}
ll
9.0S ~
= 1 ppm:
= 26 AXi O}p¡;;,, ·•;;;;¡;;.{(! ~m )' )= I.07 !xi o•
min
CDPS-E Given the data, postulare arate law. ··TA
-sc;
Then wrJ.tethe design equation in terms o.fthe data given, in this case volume and time. CA=NAV NA
= NAo(l-
X)
V= Yi,(l+EX)
X=Y.-~
V¡,E V-V.: l--- .. 2.. V¡,e
CA=-···-
v
5-25
Plug that into the design equation:
dC --2-=k~ dt
A
Now fínd fue derivítíve, To do thar use the graphical method. .... t ..
(l ·(V
V )fVr:Jt,)/V-f
L\t·
0
""1io'"""""'"1j,¿"
1 240 r º88 .:: .
1 t.i.1-(V-Vo)!Vr:Jt,)!V r~l/V -Vo)!Vr:Jt,)!V. ··1~t-(V Vo)Nr:Jt,)/V-
~==:: =J ::9~41 • • • ••••••••••,
60
S
••
··0.679
-·--.
-0.0I l
1+··· ~.
••••-•
8 :: · · · ::
5-26
"*-
AO•
•••••
•••:.:0:009-"""-'""""'""""••••••
e'=::::::=:: ~QQ1s.:-::::. ::..
Toe following graph is made.
Once that is done it is the namral log of thosc_ valucs is tcady to be gn.pbcd. Tbe following graph is the namral Iog of the derivitiveof thc volumc .functioa against thc natural lo·· of the volume function. .
---·-·--·· · · · · · ·------
From the graph we see that ex. is 2. We can also fiad k:
kr=:~9 =.018 5
N N - 85* 3o3.39* .2 - 020 l Ao-YAo to-. . 8.314*313 -. mo es
.:»: =.018= 9 . N;; .02 1
1
•
Toe followíng rate law is found:
-r_. =.9*C¡
Now, to determine the volume ofthe CSTR we must use the design equation aad stolchíometry:
.
V=Z.:oX -rA
C.. = CA<1(l - X) CAO
=!MJ
=
RTa
=
P.. 0 Y.. oPo = .6 * 1013.25kPa 607.95 Combining these with the rate law just determined, the fcílowína volume is found.
5-27
CDP5-F AsH1
"'
z o
1.5
p 45.3 28 . 63
6.5
129 70 50 30 18
9.0
16
(1.7)
L5 2.5
:1.0
19.4
9.5 L3
z
AsHJ "'"' 3.0
O
AsH3
=
3.0
- dI!.
ez
p
t.s
129 45
90 34.0
2.5
22
15.4
4.0
10
2.0
~ .df.
Z
p
o
129
17.4
1.5
95
15 . 0
dZ
PFR A+ B
-B"'--+-~
1
i.s
:te -i----·~,
-t--j
2.0
4
)
6
9
p M P1i.2 Task l, Rewrite the design equadon (i.e., mole balance) in terms of the measuremeat variables. Recall V = Ac Z. then
dX = -r4' d.Z
As
FA o
.5-28
Fer isothermal
pressure drop,
operarion and no
e,,..
(l
.. X)
{l.
X)
= C,.,,o { l + eX) • CA = p,,JRT
p_.
,.. = ·p AO (1 + e)()
n +e:X)P>. =
PAoO-X)
I
:(; =. . 1- P,JPAQ _
1+
ex P,JPAo>
Now we have rhe diffe:rential mole balance in terrns of the measured variables P A and
d [1 -P ,aJP.,_oJ
{ l+eh..}· P>.o
=
dZ
Postulare -rA
z.
-r,.A.:;
=
FAo
k(PA ~
-i~]
Tass: 2. Look for simplífications. A) See if volume chaage can be neglected,
e E
"'S
'"
(8 x
8 = Q_T29( . t) = 1.52
104)-o
= O therefore neglect ve lurne chango.
B) \Ve see thar fer runs 1 and 3 where PAsHi == 1.5 and 3.0 torr, respectívely, that
reos: of the Er::ln is consurned, indicaiing the equílibríum is reasonably far to the ri ght. Consequen tly, the reverse reacrion is negligible in the first pan: of rhe
reactor. i.e., ·rA
C)
\\:1:
•
C1
l3
= k PA Pa
also see that fer runs l and 3 that Bis in excess and tha: for excess B n
5-29
-~~
------~
---------
-r ..s.. = k .
_ CLPA
p~
= k
dZ
pO.
"" 30 • A ::::; 1K.' ~pa A
T ask .1. Calculare ( •
''
p,
e
A
Algorirhrn
6¡_;) and plo. vs Z ro find [ • dl()
. d.P Plot (- d.Z:i.} vs (P.-\) en log-log paper co ñnd alpha.
Task 4. TJ.ke raco of inicial rares ar P A.sH1 = 3.0 torr and PAIHJ =. L5 corr
We know y :a l oecause of thermodynamic consisteacy, Le. - rA =
o
,:r
f
k p A4
l .
Pae .. Pc~l
K ~-·~ ., . Pee ,
.
Kpl
PAe Psc:
Task 5. Evaluare k' and x, From A1H3 of 3.0 torr we see equilibriurn is rescned ar P >.e .;::.01
Pe.: :. 0.129 • O.O l == .119 Pac:
= 3.0
~ .ll9 = 2.88t
K? = {.01)(2.381) · I l9 · . = d... ,·3
mrr·t . ,.
Frern inirial rare
- ~.:. = 90 x 10-J
~-:.:
x:· C 129
torr.i (3.0 tcrr)
k' = 0.13 (torr c;nt1
5-30
CDPS-G From given data, find the rate law . Given: Oxidation of propene to acrolein Rate law: rA
= kPPªP02 b
~W=0.5g Using Polymath non linear regression, the following results were obtained: Nonlinear regression (L-M) Model: ra = k*PpAa*Po2Ab Variable Ini guess k 5 a 1 b 1 Precision RA2 = 0.9999969 RA2adj = 0 . 9999953 Rmsd 5.722E-07 Variance 4.011E-12
Value 0.006609 0.9948724 0 . 2034299
95% confidence 2.685E-05 0.0046367 0.001358
= =
CDPS-H 1)
prod
-rA =kACA
B .. h .. -+ prod
-rs =kaCa
A-··\'.~.
7
-r1 = kA CA+ k8Ca _d~.~t =k"C""(l-XA)+kaC:a.,(1-Xa) dt
C1(t)=
CA (t) t-CB(t:)
= CA0(l-XA) + Ca0(l-Xa)
C¡(t) = CA0(l- XA + E>a · · · X8) = CA0(1.75-XA -Xs) · ~~L=CA0(k;. -k.AXA +k5E>8-kaXs)
dt
--~~:1
dt
II)
l:
D
_!)
So, t(u•) C:(t)
= C .Ao[k A - kAXA + 0.75kn0a -· k11Xa)
1.m11.3
p.rod
o
10
%0
30
40
0,0H
o.01u
0.0017
O.OOH
o .0074
o
O.JU
o.531
0.700
o.i:u
5-31
'º
'°
º·º°"º
o.ooso
1.00
l.ll!
ta
rr.
I!
o
0.307
0.179
o.to
0,400
~
A.( t)
-~)
dt
,
u u. ~
If
0-11/
+l. hl0~4
•2.l:&:10 ....
•1.h10""'
.. l,lxlO.,...
•l. l:tló
1.9,.110-4
l .JlldO
?.413:r.l.O•S
7 .Q~;d.0-,
5.UhlO-S l. :n.;..10 ...,, l .• $OdO~s
1.100
l. 701
i.lU
-~·
l.37.
II •ere
true,
all
.,
... h:.ttt4
t0's
shonld
b~ the
same
· -->
•l.:h.10 ....
l.3l1
II is not trtte
CDPS-1 (a)
Experimental Plan to find the tate Iaw for rhe hydrogenation of cyclopentane on a PtlAlPi catalyst: l. Since this is a stable catalyst we don't have to worry about caralyst decay and an Integral Reactor will be used, Perform several different nms, holding C,,.0 and W constant while F,,.0 is varied from run to run.
2,.
3.. Plot XA.ótit vs . W/FAo for aU runs, 4.
(b)
Fit a curve tbrough ali poínts which passes throngh the origin. The slope ar any point is the reaction rate . Record the slope and corresponding <\0 for many different XA values. These data can be used to determine the rate law, Experimental Plan to find the rate law for the liquid-phase production of methyl
bromide from an aqueous solution of methyl amine and bromine cyanide: 1.
For a liquid-phase reaction without a catalyst, use a batch reactor.
2. While running the reaction record both C.., and C8 at equal time intervals
3. Repeat to ensure accurate data.
CDPS-1 (e) No solution CDPS-J
No solution will be given.
5-32
•0,hlCI..,.
l.203
CD PS- K
No solution will be given.
5-33
Solutions for Chapter6 - Multiple Reactions P6-l
Individualized solution
P6-2 (a) Example 6-2 ForPFR,
dCA k1 --=-
dr
k 2CA
-
-
dCx =k
k 3CA2
dr
dCB = k2CA
dCy
dT
dT
1
= k 3 cz
A
In PFR with V= 1566 dnr' we get X= LO and S81xv = 0.394 also at V = 533 dnr' SatxY is at its maximum value of 0.625 0.7 ~------------~ 03 0.2
0 . .1
o.o -0.10
313
626 tau 940
1253
1566
0. 0 O
313
626 tau 940
1253
See Polymath program P62-a.pol. POL YMATH Results Calculated values of the DEQ variables Variable tau Ca
ex
Cb Cy Cao X
kl k2 k3 Sbxy
initial
o
value
rninirnal value
0.4 1 . OE-07
-0.0439962 1 . OE·-07
l.OE-06 0.4
l. OE-06
o o
l.OE-04 0.0015 0.008
o
rnaxirnal value 1566 0.4 O . 1566001 0.1490534 0.1617665 0.4 1.1099906 1.0E-04 0.0015 0.008 0.6436717
o
o
0.4
o
1 . OE-·04 0 . 0015 0.008
o
ODE Report (RKF45) Differential equations as entered by the user [l l d(Ca)/d(tau) = -k1-k2*Ca-k3*CaA2 [ 21 d(Cx}/d(tau) = k1 [ 3 l d(Cb}/d(tau) = k2*Ca [ 4 J d(Cy)/d(tau) = k3*CaA2
6-1
final value 1566 -0.0439962 0.1566001 0.1256308 0.1617665 0.4 1.1099906 l. OE-04 0.0015 0.008 0.3946104
1566
Explicit equations as entered by the user [ll Cao = 0.4 [2 l X= 1-Ca/Cao [ 3] k1 = 0.0001 [4] k2 = 0.0015 [5] k3=0.008 [ 6 J Sbxy = Cb/(Cx+Cy)
(2) Pressure increased by a factor of 100. Now CAo = P/RT = 0.4 x 100 = 40 mol/dm3 For single CSTR, cA*
r - -v -
eAo - eA
- Vo -
- rA
-
eAo - eA ..:.=. _
___;.;;.___
- klA + k2ACA + k3Ac~
40-0.112 ------~sec 0.0001+ 0.00168+ 0.0001
r=21217sec
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 (e) Example 6-4 For k¡ = k2, we get CA = C AO exp(-k1T'
0)
6-2
Optimum yield:
dC ~
dr
= O = k1 CAO '
=> 7opt
=-
[exp(- k1 r' )- k1 r' exp(- k1 r' )]
1
k¡ X opt
and
= 1 - exp(- k1 T ') = 1- e -1 =0.632
ForaCSTR:
FAo,-FA
W= I
'[' =
CA I
-r' A
vo(CAo-CA)
=----+r' A
CAO -CA ·=-~-~ CAO -CA =r'; k1CA
= C AO ( t:' k¡ + 1) CBO -CB =-----= CB =r'» -k1CA +k2C8
T =
CB . -k1CA0(r'k1 +l)+k2C8 1 . 80.....---------------, 1.64
to find the maximum concentration of C, differentiate C8 with respect to r' and set it equal to O.
1.48
See Polymath program Pó-z-c.pol.
simplifying we get
320
Then use the quadratic formula to solve for r' . (2) Operating temperature = 325 K
P6-2 (d) Example 6-5 Kc = Ke= 0.25 r2N2
= k2N
2
2
[
t: -
CN?Cº?]
CNo - ·
6-3
340
r 360
380
400
rN2
5 1.5 = 6k1NoCNH3CNo
6-6
J
CN2Co2
2
ro2 =k2N2 [ CNo -
P6-2 (e) Example
[
2
+k2N2 CNo -
K¿
eN2K¿Co2]
2
2
-k102CN2C02
For liquid phase, Fj = V o
e
Equation (E 6-6.3):
d:;0 =v o d~;0
j
NH3 c~t
= -ic
dCNO ~--=-kCNH dr dFN02 =v dCNo 2= dV o dV
Equation (E6-6.8):
1 k 302 eN2 e:02
-
3
15 k 2 CNO -2 2N CNO 2
e
k
N2
302
- 2k2N2 C!o
e:
02
P6-2 (0 Example 6-7 For equal molar feed in hydrogen and mesitylene. CHo = YHoCro = (0.5)(0.032)lbmol/ft3=0.016 lbmol/ft3 CMo = 0.016 lbmol/fr' Using equations from example, solving in Polymath, weget Topt 0.38 hr . At z' 0.5 hr all ofthe H2 is reacted and only the decomposition ofX takes place ..
=
=
This question 0.99 0.00016 0.0042 0.0077 ---------0..38hr 1.865
Ex6-7
o.so
XH
0.0105 0.0027 0.00507 0.2hr 0.596
CH CM ~.Cx
t
-·---·
Sx,r
·---
See Polymath program P6-2-f.pol. POL YMATH Results Calculated values of the DEO variables Variable tau CH CM
ex kl k2
rlM r2T rlH r2H rlX r2X
initial value
o
O . 016 0.016
o
minimal value
o
1.64E-06 0.0041405
o
55.2 30.2 -O .1117169
55.2 30.2 -O .1117169
-O .1117169
-O .1117169 -0.0159818 2. 927E-04 -0.0159818
o
o
O .1117169
o
o
maximal value 0.43 0.016 O . 016 0.0077216 55.2 30.2 -2.927E-04 0.0159818 -2.927E-04
o
O .1117169
o
6-4
final value 0.43 l. 64E-06 0.0041405 O. 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 [ 11 d(CH)/d(tau) = rl H+r2H [ 2 l d(CM)/d(tau) = r1 M [ 3 J d(CX)/d(tau) = r1 X+r2X
0.016 0.012
Explicit equations as entered by the user [1) kl = 55.2 [2] k2 = 30.2 [ 3 J rl M = -kl *CM*(CH".5) [ 4 J r2T = k2*CX*(CH".5) [SJ r1H = r1M [ 6 J r2H = -r2T [ 7 J r1 X = -r1 M [8J r2X = -r2T
Increasing fltt decreases
P6-2 (g) Example
'opt
and
- CH - ClvI
ex
0.008 0.004 0·0000.000
0. 086
0.172tau0.258
0. 344
S xn
6-8
Using equation from example 6-8: Polymath code:
See Polymath program P6-2-g.pol. POLYMA TH.Results
NLES Solution Variable CH CM
ex
tau Kl 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-11 -L047E-11 -9. 771E-12
Ini Guess 1 . OE-04 0.013 O.OQ2
NLES Report (safenewt) Nonlinear equations [ 1J f(CH) = CH-CHo+K1 *(CM*CH".5+K2*CX*CH".5)*tau = O [2
J f(CM) = CM-CMo+K1 *CM*CH".5*tau = O
[ 3 J f(CX) = (Kl *CM*CH".5-K2*CX*CH"0.5)*tau-CX = O
0 . 020 ..-·-----
Explicit equations [ll [21
[31 [4l [5]
tau=0.5 K1 = 55 . 2 K2 = 30 . 2 CHo=0 . 016 CMo=0 . 016
0.016 ........ _..,,,,. . ...,,, __ .,_,_,fi __ ,_a_,, __
0.012 0.008
A plot using different values of t: is given. For 'l =0.5, the exit concentration are CH= 4 . 8 x10-5 lbmol/ft3 CM =0 . 0134 lbmol/ft3 Cx =0.00232 lbmol/ft3
0.004
6-5
,,..,,~-"-'"'
0.430
The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for T =0.5: YMX
Fx
Cx
= ---'-'-----
0.00232 0.016-0.0134
= ------
FMO -FM CMO -CM The overall selectivity of xylene relative to toluene is:
0.89mole · xylene · produced mole · mesitylene · reacted
= _Fx = 8.3mole · xylene · produced
S
F7
x rr
mole · toluene · produced Ex6-8 0.0089 0.0029 0.0033 0.5 0.41 0.7
CH
CM Cx '[
YMx Sxrr
·--·
This Question 4.8 X 10·5 0.0134 0.00232 ·0.5 -0.89 8.3
P6-2 (h) Example 6-9 (1)
So/U Original Problem . Membrane Reactor 2.58 PFR 0.666 Doubling the incoming flow rate of species B lowers the selectivity.
So/U P6-2 h 1.01 0.208
·-
(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
12
-
-
9 6
FB
9
FC FD FE
6
3
o o
FB
"FC
3
2
4
V
6
8
10
o o
1
2
y
3
4
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 P62 -i.pol,
P6-2 (j)
6-6
J .4e-1J
2 . 7e-1J 2.0e-13
L4e-1J
6.8e-14
- TFVIIa 140
280 t
H'VIIaX
420
560
700
At the beginning, the reactants that are used to create TF-VIIa 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-VIIa and TF- VIIaX 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 Fm Fh Fx
-
8
6
6
4
4
2
2
40
80
V
-
120
160
200
o
o
40
Fm Fh Fx
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 (1) Individualized solution P6-3 Solution is in the decoding algorithm given with the modules ( ICM problem) P6-4 (a) Assume that ali the bites will deliver the standard volume of venom. This meaos that the initial concentration increases by 5e-9 M for every bite.
6-7
After 11 bites, no amount of annvenom can keep the nwnber of free sites above 66 7% of total sites . This means that the imnal concentration of venom would be 5 Se-8 M The best result occurs when a dose of annvenom such that the minal concentration of antivenom m the body is 5.7e-8 M, will result in a mmimum of 66 48% free sites, wluch IS below the allowable mmimum.
See Polymath program P6-4-a.pol.
P6-4 (b) The victim was bitten by a harmless snake and anttvenom was mjected, This means that the mmal concentration of venom IS O. From the program below, we see that if an amount of antivenom such that the imual concentration in the blood is 7e-9 M, the patient wtll die.
See Polymath program P6-4-b.pol. POLYMA TH Results
Calculated values of the DEO variables Variable t fsv fs Cv Ca
initial value
o o
kv ksv ka kia ksa kp
kop g
-
7.0E-09 0.3344339
4 .503E-·09 0.3344339
2 OE+OS 6.0E+OS 2.0E+OS
2.0E+OB 6.0E+OB 2 OE+OS
2.0E+OS 6.0E+OS 2.0E+OS
1
1
2.0E+08 6.0E+08 2.0E+OS
5.0E-09 6.0E+OS 1.2E+09
5.0E-09 6.0E+OS 1.2E+09
5 OE-09 6 OE+OS 1 2E+09
5. OE-·09 6.0E+OS 1.2E+09
0.3
0.3 0.3
O 3
0.3
o o o
m j
0.5
o
4 503E-09
O 3
h
o
final value
7 OE-09
o
kov koa
maximal value O 5
0.6655661
o
1
eso
ºo
·---
1
o o
fsa
Cp
minimal value
-2.lE-09
0.6655661
1
o
o
o o
o
o
o
O 3
0.3
o
o
o
o
o
o
-2. lE--09
o 1
o
o o
o
·1 351E--09
o
-1. 351E-09
ODE Report (STIFF) Differential equations as entered by the user [ 1 l d(fsv)/d(t) kv * fs • Cv • ksv * fsv * Ca e 2 l d(fs)ld(t) -kv*fs*Cv • ka * fs * Ca + kia * fsa + g e 3 J d(Cv)/d{t) Cso * (·kv * fs * Cv - ksa * fsa * Cv) + h C 4 l d(Ca)/d(t) Cso*(-ka • fs •Ca+ kia * fsa) + j [ 5 ] d(fsa)ld(t) ka * fs * Ca - kia * fsa - ksa • fsa • Cv C 6 J d(Cp)/d(t) Cso * (ksv * fsv * Ca+ ksa * fsa * Cv) + m
= = = =
=
=
Explic1t equations as entered by the user [11 kv=2e8
0.92 ·-·-··-- -- -·····-----····-----·-· .. -·--·- ····--····-·-·--
[2J ksv=6e8 [3J ka=2e8 [4J kia=1 r 5 J Cso 5e-9 [6J ksa=6e8
GJ
0.84 -----~--------·----·----··--
=
(7
J kp = 1 2e9
0.76 ----- -----
0.60~---------·
o.o
6-8
~~~----·--·---·---------------···-·--····-------···-------·----·-·-·
0.1
O2
t
O3
0.4
0.5
[81 [91
kov = O koa=0.3
kop=0.3 g = ksa * fsa • Cv + ksv * fsv • Ca h = -kp * Cv * Ca - kov * Cv m = kp * Cv * Ca - kop * Cp j = -Cso * ksv • fsv * Ca - kp * Cv * Ca - koa * Ca
[10] [ 11 J [ 12 l [ 13 l [ 14 J
P6-4 (e) The latest time after being bitten that antivenom can successfully be administerd is 27 .49 minutes. See the cobra web module on the CDROM/website for a more detailed solution to this problem I
o9 :I: e: o ·,¡::;
" ~"'
J
· ···.
-
. - . ----
..
l
l
l
!
1
08
1
il,I
~t::
1.
"':!:!
·11
!i
¡...
¡;,;..
() "!
1
Antivenom Injected at t
,1
27 49 min
t
1
o, [. · ·-·· ,--,-·· ··--· ._.,,., . . , o
02
04
0.8
0.6
1.6
.1.4
Time [hour]
P6-4 ( d)
Individualized Solution
P6-5 (a) Plot of CA, CD and Cu as a function of time (t):
See Polymath program Pó-ó-a.pol. POL YMA TH Results Calculated values of the DEO variables Variable t Ca Cd Cu kl
initial value
o
maximal value 15
o o
o o
1
1
0.7995475 0 . 5302179
o
1
minimal value 0.0801802
1
1
6-9
final value 15 0.0801802 0.7995475 0.1202723 1
L8
2
l. 5 1
1
l. 5 1
1
o
o
X
100 10
100 10 l. 5
100 10
100 10 l. 5
k2
Kla K2a Cao
0.9198198
0.9198198
ODE Report (RKF45) Differential equations as entered by the user [ 1 l d(Ca)/d(t) = -(k1 *(Ca-Cd/K1 a)+k2*(Ca-Cu/K2a)) ¡ 2 J d(Cd)/d(t) = k1 *(Ca-Cd/K1 a) [ 3 J d(Cu)/d(t) k2*(Ca-Cu/K2a)
=
Explicit equations as entered by the user [1] k1 1.0 [2] k2 = 100 [3J K1a = 10 [4J K2a = 1.5 [5J Cao = 1 [ 6) X= 1-Ca/Cao
=
1.0
1.0
- Ca 0.8
- Cd
,,,,_
......
-
--
. ......_,_,.,...
0.8
0. 6
0. 6
0.4
0.4
0.2
0.2
o.o ,____----~---~---------
º To maximize C 3
0
6
9
12
o.o o
15
l~ ----· ---~-~-·-·--·-· 6
9
12
15
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.(CA= 0.53)
P6-5 (e) Equilibrium concentrations: CAe = 0.08 mol/dm3 Coe = 0.8 mol/dm3 Cue = 0.12 mol/dm3
P6-5 (d) See Polymath program P6--5·-d.po1. POL YMATH Results
NLES Solution Variable Ca
3
Value 0.0862762
f(x) -3.844E-14
Ini Guess 1
6-10
0.7843289
Cd
0.1293949 1 1 100 10 l. 5 100
Cu ca o kl k2 Kla K2a t
-2.631E-14
6.478E-14
o o
NLES Report (safenewt) Nonlinear equations [ l J f(Ca) = Ca0-t*(k1 *(Ca-Cd/K1 a)+k2*(Ca-Cu/K2a))-Ca = O [ 2 J f(Cd) = t*k1 *(Ca-Cd/Kl a)-Cd = O ( 3 J f(Cu) = t*(k2*(Ca-Cu/K2a))-Cu = O
Explicit equations (ll
[ 2] (3]
(41
[SJ [6] [7J
CaO = 1 k1 = 1 k2 = 100 K1a=10 K2a = 1.5 t= 100 X = 1-Ca/CaO
r
1 min
lOmin
lOOmin
CAexit
0.295
0.133
0.0862
C0exi1
0.2684
0.666
0.784
Cuexit
0.436
0.199
X
0.705
0.867
0.129 0.914
P6-6 (a)
(
/ )1/2 min
k1 = 0.004 ;;3 A·~B
k2 =0.3min-1 k3 =0.25
Sketch S8x, S8y and SsJXY as a functíon of CA
See Polymath program P6-6-a.pol.
1)
6-11
dm3 mol.min
32
24 16 8
o
0.0001 0.0326 0.0651<_~30.0976 0.1301 0.1626
2)
120011--·-----------, 960 720
480 240
o
0.0001 0.0326 0.0651ciP0976 0. .1301 (U626
3)
2
o
0.0001 0.0326 0.065\:atl.0976 0.1301 01626
P6-6 (b) Volume of first reactor can be found as follows
6-12
We have to maxirnize S8ix:y
•
From the graph above, maximum value of S8xv = 10 occurs at CA = 0 . 040 mol/dm So, a CSTR should be used with exit concentration CA• Also, CAo = P A/RT = 0 ..162 mol/dm3
And-rA =rx +r8 +ry =(k1CA
--
> V
* -CA ) -rA
= vo(CAo
=
1/2
3
2
+k2CA +k3CA)
* vo(CAo -CA ) . =92.4dm3 * 112 * * 2 (k1(CA ) +k2CA +k3(CA ) )
P6-6 (e) Effluent concentrations:
.
Weknow,,=9.24 mm=> r=Similarly:
Cx
*
C8
re
mol =0.007-3 dm
C8
=---::::::;,
k2CA
*
mol
Cs =0.11--3
dm * mol Cy =0.0037--3 dm
and
P6-6 (d) Conversion of A in the first reactor:
CAo-CA = CA0X-::::::;, X =0.74
P6-6 (e) A CSTR followed by a PFR should be used. Required conversión = 0.99
dV
=> For PFR, Mole balance: -
dX
-::::::;, V= 10x0.162x
FAo = --rA
o . 99
f
074(k1CA
dX 112
2.
+k2CA +k3CA )
= 92.8dm3
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:
k=A¡exp(~) ' l.98T Since we have the constants given at T = 300 K, we can sol ve for A;.
A¡ =
Ai =
.004 = 1.49e12 --20000 exp ( 1.98( 300)
J
.3
J
= 5.79e6
-10000 exp ( 1.98(300)
6-13
A,=
( ·25
-30000 exp 1.98(300)
) =1.798e21
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¡ =,¡V C¡ =r,¡ Using these equations we can make a Polymath program and by varying the temperature, we can find a maximum value for C8 at T = 306 K. At this temperature the selectivity is only 5 . 9. This may result in too much ofX and Y, but we know that the optima! temperature is not above 306 K. The optima! temperature will depend on the price ofB and the cost ofremoving 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 TH Results NLE Solution Variable Ca T R
kl k2 Cao Cb k3
tau
ex
Cy Sbxy
Value 0.0170239 306 l. 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 l f(Ca} = (Cao-Ca}/(k1 *CaA.5+k2*Ca+k3*CaA2)-1O = O
Explicit equations [1J
[2J [3] [4J [Sl
T=306 R= 1.987 k1 = 1.49e12*exp(-20000/Rff} k2 = 5790000*exp(-10000/Rff} Cao = .1
6-14
6J 7] 8] 9] 1o} 11 J
Cb = 1 O*k2*Ca k3 = 1.798e21*exp(-30000/R!T) tau= 10 Cx = tau*k1*CaA..5 Cy = tau*k3*Ca"2 Sbxy = Cb/(Cx+Cy)
P6-6 (g) Concentration is proportional to pressure in a gas-phase system Therefore:
SB I XY
p
-
¡:;;
vP+P2
which would suggest that a low pressure would be ideal . But as before the tradeoff is
lower production of B. A moderate pressure would probably be best.
P6-7 US legal limit: O 8 g/1 Sweden legal limit: 05 gil
A~B
k2
>C
Where A is alcohol in the gastrointestinal tract and B is alcohol in the blood stream
dCA =-kC dt I
A
dCB =kC -k dt I A 2 k,
= lOhr-
1
k2 --0..192-- g L hr Two tall martinis = 80 g of ethanol Body fluid = 40 L
e
40L
~~"-'----~-'--~~--.-.-·
L6
1.2
= sog =2g AO
~--
2.0
L
Now we can put the equations into Polymath
0,8
See Polymath program P6-7.poL PO.LY~'lA'I!LResults Calculated values of the DEO variables
OA 0. 0 O
Variable initial value minimal value maximal value final value t o o Ca 2 7.131E-44 Cb
O
O
kl k2
10 0.192
10 0.192
2
10 2
1 . 8901533 10 0.192
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d{Ca)/d{t) = -k1 *Ca
6-15
4
6.
10 7 131E-44 0.08 10 O . 192
8
10
¡ 2 J d(Cb)/d(t)
= -k2+k1 *Ca
Explicit equations as enterad by the user [ 1] k1 = 10 [ 2] k2 = 0.192
P6-7 (a) In the US the legal limit it 0.8 g/L. This occurs at t = 6.3 hours ..
P6-7 (b) In Sweden CB
= 0.5
gil , t
P6-7 (C) In Russia
CB
= 7 .8 hrs,
= 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 g/L . 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. Ata half an hour CA= 0.00674 g/L and CB = 0.897 g/L . The Polymath code for after the second drink is shown below.
See Polymath program Pó-Z-d.pol. POL YMATH Results Calculated values of the DEO variables Variable t Ca Cb
kl
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 l. 8069769 10 0.192
final value 10 5.394E-42 O . 08 10 0.192
ODE Report (RKF45) Differential equations as enterad by the user [ 1J d(Ca)/d(t) = -k1 *Ca [2J d(Cb}/d(t) = ··k2+k1*Ca Explicit equations as enterad 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 g/hr, Since volume is not changing the rate of change in concentration due to the incoming ethanol is 2 g/L/hr. For the first hour the differential equation for CA becomes:
6-16
= -k1 CA + 2t after that it reverts back to the original equations . dt See Polymath program P6-7-e.poL dC A
POL YMA TH Results Calculated values of the DEO variables initial value
Variable t Ca Cb kl k2
o o o
10 0.192
minimal value
o o
-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 [ 1J d(Ca)/d(t) = if(t<1}then(-k1 *Ca+2*t)else(-k1 *Ca) [2J d(Cb)/d(t) = -k2+k1*Ca Explicit equations as entered by the user [ 1 J k1 = 10 [2] k2=0.192
US: C8 never rises above 0.8 g/L so the is no time that it would be illegal. Sweden: t = 2 . 6 hours Russia: t = 5 . 2 hours
P6-7 (f)
60 g of ethanol immediately 7 CA= 1..5 g/L C8 = O 8 g/L 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 CA 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 che stomach and B be the tarzlon in the blood, Mole Balances: dC1\,_,_ = r __ dt
dC __ _ :!. dt
-
A
= rB
6-17
---~----~-----
Rate Laws :
-r..,. : :;: k1C,,. + kzCA r8
= k¡C..,.
- k~ -· k~Cs
All k valúes are given in the proble m staternent. It must be noted, however, that for C8 < O, k, must be equal to O. These equations when entered in POL Yrv!ATH generare the following results:
See Polymath program P6-8···a.po1. POLYMA TH Results Calculated values of the DEO variables Variable t Ca Cb
kl 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
final value 4 O. 3111692 0.4057018 0.15 0.6 0.2 0.1
ODE Report (RKF45) Differential equations as enterad by the user [ 11 d(Ca)/d(t) -k1 *Ca-k2*Ca [ 2 J d(Cb)/d(t) k1 *Ca-k3-k4*Cb
= =
Explicit equations as enterad by the user [ 1] k1 0.15 [2] k2 = 0.6 [3] k4 = 0 . 2 [4J k3 = if(Cb
=
P6-8 (b)
6-18
-------------~---··------·
~-~~----------~----~------
r -----------~-------~---
From the followíng graph generated using the above program in POLTuLA..TH, we can see the proper doses of the drug: . 1. First take two doses of the crug
2.
Six hours later take ene dese.
Take ene dese every four hours from then on,
3. i 3:G
:
l.050
-cc 0.7".il
0.450
1
'--
0.!3: 1'
·G.t§G
+·1 -----: --+-···--+·--- . ·T"'····-..,,..~---·+·--··· · ·+· ---·-·· ·1·
Q . DC-0 (].éQ(l
I
a.isa f-
r,
\
·•CD
l 11.I¡ , i
20 . 0CO
JLCCO
·····--···+· ..... ., -1
'7!l.nw
SG .. OCG
6-8a
¡ \ J
! !
GlóO
'(O.GfüJ
("'
;.'
-·-·--··-------·······------·----------·--··
',
\
\
\'\
J¡
I¡ \
o.2rn ~
.. w G.~CG
: --·--· -··· '. 5.C€ü
J.rn:a
_l___ ;--· · '· · · -+ ... lJ.C(l()
t,
!KOCG
+·
···+ · -··+2C . ufü
---'.
2:.m:
P6-8 (e) If one takes initially two doses of Tarzlon, it is not recomrnended to take another dose within the first six hours. Doing so will result in build up of the drug in the bloodst.ream that can cause harmful effects.
P6-8 (d) lf 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 k¡ and elimination constant k2 values. If k¡ decreases this means that the adsorption process is slow
6-19
¡
r-
and if k2 increases means that the rate of elimination of Tarzlon increases. The next graph shows the concentration profiles for k¡ = 0.10 h-1 and k2 = 0.8 h-1. Note that the maximum amount of the drug in the bloodstream is reduced by two.
O:lncentraúonProfiles -+- Ca (ll§'dm.1)
s ------------------Ch
(Illefmw)
0.6
'Ibm(h) Concentration profilefor Tarzlon in the stomach (A) and bloodstream (B). The maximum amount of Tarzlon in the bloodstream is 03 mg/dm'.
P6-9 (a) Reactor selection
A+B~D A+B--¿U rv
Sou= -=
ru
= 1 Oexp(-8000K I T)C A CB 1/2 3/2 r2A = lOOexp(--lOOOK IT)CA C8
1iA
exp(-8000K IT)CA
10exp(--8000K IT)CACB ·
lOOexp(--lOOOK IT)C/12
c/
1/2
lüexp(-lOOOK IT)C B
12
112
AtT =300K k¡ = 2.62 X 10"11 &
k2 = 3.57
At T = lOOOK k¡ = 3.35 X 10·3 &
k2 =36.78
Sotu =
9.2xl0-5C/12 CB
112
-
Hence In order to maximize S0u, use higher concentrations of A and lower concentrations of B. This can be achieved using: 1) A semibatch reactor in which Bis 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.
B~b/'J /
~¿?_~
PureA intitíally
Semibatch
Tubular reactor with slde streams
Series ol srnatl CSTAs
6-20
Also, since E0 > 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 Sou Note that the selectivity is extremely low, and the only way to increase it is to keep
(
e:e JYz
< 10-6
and
add B drop by drop.
P6-9 (b) A+ B ~ D
and
A+B~U
and
'1A
= lOOexp(- lOOOK IT)C A C B
r2A =106exp(-8000KIT)CACB
r» = l00exp(-IOOOK/T)CAC = -------exp(-lOOOK/T) 8
=-
Sou
'u
104exp(-8000K/T)
l06exp(-8000KIT)CAC8
AtT = 300K k¡ = 3..57 & k2 = 2.623 S0u = 1.14 x106 AtT = lOOOK k, = 36.78 & k2 = 3354 6 Sou = 0.103 Hence we should keep the temperature low to maximize Sou but not so Iow that the desired reaction doesn't proceed to a significant extent..
P6-9 (e) A + B -··-¿ D
and
B+ D ~ U
and
S
r2A
= 109 exp(-10,000K
IT)C8C0
_ 'iA _ 10exp(-8000K !T)CAC8 ou - r2A -109 exp(--lOOOOKIT)C8C0 =
S ou
exp(-8000/T)CA 108exp(-10000/T)C0
Therefore the reaction should be run at a low temperature to maximize S0u, 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 D~U1
and
,¡A= 4280exp(-12000K IT)CA
and
r20 = 10,100exp(-15000K IT)C0
A~U2
and
T3 A
r» SDIULU2
= ru._1_+_r_u_2
AtT = 300K k¡ = 1.18 X 10-14 &
= 26 exp( - 10800
K / T) CA
_ 4280exp(-12000K
IT)CA -10,100exp(-15000K IT)C0 10,100exp(-15000K/T)CD +26exp(-l0800KIT)CA
k2 = 1.94 X 10-lS & k3 = 6.03
X
10·15
Ifwe keep CA> 1000C0
= 1.l~xl0-
S ouno z
14CA -1.94xl0-18C0 1.94xl0-18C0 +6.03xl0-15CA
::::!2~=1.96 .603
6-21
--------------·------------------------------
--
------- ----
~---·---~-~--~
At T= lOOOK k¡ = 0.026 &
k2 = 3.1 X 10·3 & k, = 5.3 x 10·4
If we keep CA> 1000C0
S
= onnuz
0.026CA -3.lxl0-3Cv 3.lx10-3Cv +5.3xl0-4CA
=
.026 =49 .00053
Here, in order to lower U1 use low temperature and high concentration of A But low temperature and high concentration of A favours U2 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 D ~A+ B A+B~ Sv,u =-
rv ru
u
and and and
1iA = 109 exp(-lOOOOKIT)CACB
rzv = 20 exp( -2000 K IT)C T3 A
D
= 10 exp( - 3000 K / T ) e A e B 3
109 exp(-lOOOOKIT)CACB -20exp(-2000K IT)Cv 103 exp(-3000K IT)CACB
AtT =300K k¡ = 3.34 X 10"6 &
k2 = 0.025
& k, = 0.045
The desired reaction líes very far to the left and C0 is probably present at very low concentrations so that:
Sv,u =0 AtT= lOOOK k, = 45399 . .9 & k2 = 2.7 & k3 = 49.7 If we assume that CACB > 0.001C0 then,
S DIU
_ 45399.9CACB -2.7CD _ 45399 -913 49.7CACB 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 (f) A+ B-~ D
S
A....::, U
and
-r¡A =10exp(-8000K IT)CACB -r:u = 26exp(-10,800K IT)CA
U---"> A
and
-r3u =lOOOexp(-1.5,000K IT)Cu
and
_ rv 10exp(-8000K IT)CACB oto - ru - 26exp(-'10,800K IT)CA -lOOOexp(-15,000K !T)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 = O .
-rA=lüexp ( -8000) T CAC8+26exp (-10800) T CA-lOOOexp(-15000) T Cu=O CA - 10exp(-8~00)cB Cu -
+26exp(-l~OO)
lOOOexp(-1;00)
CA =-1-exp(7000)c Cu 100 T
8
+~exp(4200) 1000 T
P6-9 (g) A+B-~D
and
A+B~U1
and
--r28 = 10exp(-300K IT)CACB
D+B~U2
and
-r3D
-,¡A= 800exp(-8000K IT)C15C8 = 106 exp(-8000K IT)CDC8
(1)
800exp(-8000/T) C15C8 . 80exp(-8000/T) SDtu1 = 10exp(-300!T)CAC8 = exp(-300/T)C15 AtT = 300
s
DIU¡
= 2.098 * 10-10 0.368C15
At T = 1000
S
= DIU¡
29.43 0.7408C15
To keep this selectivity high, low concentrations of A, and high temperatures should be used.
exp ( -8000 / T) C15 eB 800C15 s otu z = 800 =-~ 106 exp(-8000/T) CDC8 106C0 To keep this selectivity high, high concentrations of A and low concentrations ofD 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 ata low temperature to maxirnized So/UJ in a membrane reactor that allows only D to diffuse out (2) SD/U1U2
= __ (
800 exp ( -8000 / T) C15
)__
6
(
e
B
·--)
lüexp -300/T CAC8 + 10 exp -8000/T CDC8
800 exp ( -8000 I T) C15 S D/UlU2 = 10 exp ( -300/T ) CA +10 6 ----'-(---)exp --8000/T At T = 300
6-23
CD
s
=
DIUIUZ
cº
2.09 * 10-9 5 A ::::O 3.67 A + 2.62 * 10-6 D
e
e
At T = 1000 and very low concentrations of D
=
S DIUIUZ
0.268C15 7.408CA+335CD
.03617 C15
If temperature is the only parameter that can be varied, then the highest temperature 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) rD
-,u -
exp(-7000KIT)C/2 lOC/2
dFA
--=r
dV
dFB
-·-=r,
dV
A
dFu
--=r.
dV
+R
B
B
dFD
--=r
dV
u
D
R = FAO B V: T
A+B~D
1iA = --10exp(-8000K IT)CACB
c/'
r2A = -lOOexp(-lOOOK /T)C/2 TA
2
= rB = 1iA + r2A FA
CA =Cro-F¡
mol
CTO =0.4--3
dm
s
FAo = Crovo 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.pol. POL YMA TH Results Calculated values. of the DEO variables Variable V
Fa Fb Fd Fu Cto
initial value
o
4
o o
o
minimal value
o
maximal value 10
o o o
4.5141833 3.034E-06 3.4858137
0.5141833
Ft
0.4 600 4
600 4
Ca
0.4
0.0241566
T
Cb
o
4
O.4
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 O . 0241566
o o o o o o
rla
rd r2a ra rb ru Vt Fao Rb Sdu
o
-4.575E-07
o
4 . 575E-07
o
.4461944 5 4 0.8 5.423E+05
o o o o
-0.4461944 -0.4461948 -0.4461948 5 4 0.8 2.894E-07
5 4 0.8 5.423E+05
-8.297E-08 8.297E-08 -0.2867066 -0.2867066 -0.2867066 0 .. 2867066 5 4 0.8 2.894E-07
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(Fa)/d(V) = ra ¡ 2 J d(Fb )/d(V) = rb+Rb ¡ 3 J d(Fd)/d(V) = rd [ 4 J d(Fu)/d(V) = ru Explicit equations as entered by the user [ ll Cto = ..4 [21 T=600 ¡ 3 J Ft = Fa+Fb+Fd+Fu [ 4 J Cb = Cto*Fb/Ft ¡ 5 J Ca= Cto*Fa/Ft [ 6 l r1 a = -10*exp(-8000/T)*Ca*Cb [7 l rd = -r1a [ 8 J r2a = -1 OO*exp(-1 OOO/T)*CaA. 5*CbA1 . 5 [ 9 J ra = r1 a+r2a ¡10 J rb ra [ 11 J ru = -r2a [12] Vt=5 [13J Fao=4 ¡ 14 J Rb = Fao/Vt [ 15 J Sdu = exp(-7000/T)*CaA.5/(10*CbA.5+.00000000001)
=
5
6.0e+5
4
4.8e+5
3
3.6e+5
2
2.4e+5
1
l.2e+5
o
o
2
4
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 (1) No solution will be given
6-25
·--·-----------· o ·-··-··--2 . --,8 4 V 6
10
P6-10 (a) Species A:
dC
...........t.== r
dt
.\
-(~ = k1C,\
Specics B:
Species C:
dC ~=r. dt e re= k.. Ca
Plugging imo POL YM.A..TH: · gets the following. Variable
:yiltial. value
t.
o i
k.l= , 4
ca cb ce
k:2= . 01
lcl
raac ..·kl*ca
k2
0.4 0 . 01
rC=l<:2*cb
ra re rb
Equations:.. d(ca)/d(t)=ra d(cb) /d( t) =z·b d(cc)
.
/d( e) =re
rc,,)c¡, •ca .. ·k2 *cb
~a "'
O,
c.,...
"
J.00
e o
.s
-o
0.4
0.4
0.4
0.01
0.01
0.64
o . 6~
P6-10 (b)
For CSTR, -r = 0.5h First calculate k¡ and k2 :
1 --;¡: l)J = 0.4exp ( 10,000( R 373
6-26
6. 7898e-18
final_value 100 6. 789oe··18
O .. 9963
o
... c.
o
.0.60)7
o
key
Minimum value
o o
. 2.nsne
64
-Ca -Cb
k,
Maximum_yalue 100 1.6 l.4556
Ol.4556
1a
0.9963
-o , 64
º"ºl 7.l592e-18
o
O.Ó060l7
-o. 0132417
-2.
0.006037
k =0.0lex (20,000(_l __
p
2
R
373
]_)J T
See Polymath program Pó-Iü-b.pol. POL YMA:J'H Results NLES Solution
T
kl ra k2 rb re
Ini Guess l. 5
f (x)
Value 1 . 5268515 0.0319385 0.04121 0.5 760 0.0958161 -0.146297 2 . 580579 O. 0638771 0.0824199
Variable Ca Cb Ce tau
-6.446E-13 7.28E-14
o o
o
NLES Report (safenewt) Nonlinear equations
[ 1 l f(Ca) = tau*(-ra)-(1 . 6-Ca) = O (21 f(Cb) = tau*(rb)-Cb = O [ 3 J f(Cc) = tau*(rc)-Cc = O
Explicit equations [ll [21 [3 l [41 rsJ ¡6l [7J
tau= .5 T = 760 k1 72*exp(-10000/(1..987*T)) ra e -kt=Oa k2 1457152*exp(-20000/(1.987*T)) rb k1 *Ca-k2*Cb re k2*Cb
= = = =
Cb vs, temperature
3.50E·02 3.00E-02 +----------2.SOE-02
f
ll
2.00E02 +-------------1------
1
{l 1 . SOE-02
1 OOE-02 5.00E-03 +----------·--+--------
-----r··----··
O.OOE+OO O
200
400
600
800
1000
temperature (K)
Therefore, C8 is maximum at T=760K.
6-27
1200
P6-10 (e) (e) ?art is similar to part b except for two rate laws:
" k 1 *(' 'A r¡¡ =k 1 *C'A -k ··l *C B -k,*C 8 c.
rA-
-k
-t1
*(' -s
Using those rate laws inPODCMATH produce the following:
_ __
... Eauations:
lni tialvalue
d(ca) /d(t) .::ra
1.6
........-,.....,.,
o o
d(cb)/d(t}=rb d(ccl /d(t:l =:c·c
klr=8 .33e·S
2.Jc;:
--·
\ . scc
-·
klf=.0001 k2=2. 78e-6 rc=k2 .. cb
! .. 2CC
ra=k1r"'cb··k1f"ca rb=klf*ca--klrwcb-·k2*cb ci:
==
350000 o • .11cc
e.oco
Variable
o
Final value .,..,..,_......._ _
350000
350000
l. 6
l. 6
0.436316
0.416316
o
0.833237
o
0.5159
O. 647"784
o
O. 647784
o
ca cb ce
valu':::
.....
a. 33e-os
8 . . 33e-OS
0.0001.
0.0001
2.78e·06
2. 78e-·06
2.3164e·-06
o
1. 4342e··06
· 6. 57168e··07 0.00016
-0.00016
-6. S7168e--07
k.lr
8 33e···OS
S .33e--05
klf
0.0001 2.78e06
0.0001 2.78e·06
k2
o
re ra
·O. 00016 0.00016
rb
P6-10 (d) Th is
----~ _
Maximum va Lua Mínímuzn va.Lue ..,._. .,,....
faj.tiah
t
is similar to part d except for one rate law: r, = k, *Ca--· k_2 * Ce
Using that in POL Th1AT:H produces the following:
6-28
-r.
20632e··06
--7 .77034e-07
Equa CÍS:!:!.: d(ca) /d ( ::) =z-a d{cb)/d(t.)=rb d(cc)/d(t);rc
2
ccc
T
i
.5.:JC
.L
i
klr=8 .. 33e-·5
klf= 0001 k.2f=2. 78e···6 k2r=L39e-6
ra,,,kl:::·•cb·· k.l f
=
C."CC
o.eco
¡,:
c.oco
,,. . . ,.. __ ,.~ ....... ,··-···-·. ·--~-----~---~ .... --~-·-t t $OC z.scc J.. 2:lC ... ,::
a. soc
350000
Variabl.e t
ca cb
ce klr klf
o
150000
o
350000
1 6
1 5
0.490306
O 490306
o o
O 833769
O
0.58J662
0.526032 8 ne-05 O 0001 2 i8e·06 l 39e·06
O
0.526032
8.33e-OS
8.33e-05
0.0001
0.0001
2.JBe-06
2.78e-06
l. 39e 06
1.. 39e· . 06
.4 USBe··O'I
-4.11573e-07 8.91395e-07
3 . .33e··05 0.0001 2.78e·06
k2f
k2r za
l. 39e·Oé
re
o
2 .. 25569e·06
-0.00016 O
O 00016
O 00016
· L l24.93e
··O
0001.6
06
··4. 79824e··O
7
P6-10 (e) When kl>lOO and k2<0.1 the coucentration of B immediately shoots up to L6 and then slowly comes back down, whiíe CA drops off immediately and falls to zero, This is because the fírst reaction is so fast and the second reaction is slower with no reverse reacrions, When k2 = l then the concenrrarion of B spikes again and rernains high, while very little of C is formed. This is because afrer B is formed it will not got to C because tbe reverse reaction is faster. When 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 reacrion.
(e)
P6-11 (a) Intermediates (primary K-phthalates) are formed from the dissociation of K-benzoate with a CdCh catalyst reacted with K-terephthalate in an autocatalytic reaction step:
CAo
.r.. RT
A~R~S R+S~
Series 2S
Autocatalytic
IlOkP: =0.02molldm3 (s.314 kPa.dm 683K) mol.K
J(
6-29
Maximum in R occurs at t = 880 sec. See PolymathprogramP6- l I-a. pol.
POLYMATH Results
Variable t A R
initial value
o
0.02 O
minimal value
o
0.003958
s
o
kl k2 k3
0.00108 0.00119 0.00159
o o
O . 00108 O .00119
0.00159
maximal value 1500 0.02 0.0069892 O . 0100382 0.00108 O. 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 [ 1J d(A)/d(t) = -k1 *A [2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S) [ 3 J d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [1] k1 = 1.0Be-3 [2] k2 = 1..19e-3 [3J k3 = 1 . 59e-3
P6-11 (b) 1) T= 703 K CAo = 0 . 019 mol/dm"
k; = k;
ex{i{~ - -;. )J
)exp( (l.987cal/ (4260~cal I mol) __ (_._l l -)J = 2_64x10_ mol.K) 683K 703K
k; = (l.OSxl0_3 s_1
3
Similarly,
k~ =3.3xl0-3s-1
and
k~ =3.lxl0-3dm3 /mol.s
Maxima in R occurs at around t =320 sec, See PolymathprogramP6- l 1 · b Lpol
POLYMATH Results Calculatedvalues of the DEO variables Variable t A R
s
kl k2 k3
initial value
o
O .019 O
o
O 00264 0.0033 0.0031
ODE Report (RKF45)
minimal value
o
3 . 622E-04
o
o
O . 00264 0.0033 0.0031
maximal value 1500 O . 019 O . 0062169 0.0174625 0.00264 O . 0033 O . 0031
Differential equations as entered by the user [ 1J d(A)/d(t) = -k1 *A [ 2 J d(R)/d(t) = (k1 * A)-(k2*R)-(k3*R*S) [ 3 J d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [ 1 l k1 = 2.64e-3 r2 J k2 = 3.3e-3
6-30
final value 1500 3 . 622E-04 8 . 856E-04 O . 0174625 0.00264 0.0033 0.0031
s_1
¡ 3 J k3
= 3.1
e-3
2) T = 663 K CAo = 0 . 19 mol/dm3
Is= (l.OSxl0_ s_ )exp( 3
1
(-1
(42600cal /mol) (l.987cal/ mol.K)
1 _)) =
683K 663K
0.42x10_3 s_1
= 0.4xl0-3 s-1 k~ = 0.78xl0-3 dm' / mol.s k~
See Polymath program P6-ll-b2.pol. POLYMATH Results Calculated values of the DEO variables Variable
initial value
o
t
A R
minimal value
o
s
0.019 O
o
2.849E-04 O
kl k2 k3
4. 2E-04 4.0E-04 7.8E-04
4. 2E-04 4.0E-04 7.8E-04
o
maximal value 10000 0.019 0.0071414 O . 016889 4.2E-04 4.0E-04 7.8E-04
final value 10000 . 2.849E-04 0.0012573 0.016889 4.2E-04 4.0E-04 7. 8E-04
ODE Report (RKF45) Differential equations as entered by the user l 1l d(A)/d(t) = -k1 * A l 2 J d(R)/d(t) = (k1 *A)-(k2*R)-(k3*R*S) [ 3 l d(S)/d(t) = (k2*R)-(k3*R*S) Explicit equations as entered by the user [ l J k1 = 0.42e-3 [2 J k2 0.4e-3 [ 3 l k3 = 0.78e-3 lndependent variable variable name : t initial value : O final value : 1 0000
=
0.020 .---------·-·-----~
0016
0.012
0008
0. 004
2000
4000 l
6000
8000
10000
Maxima in R occurs around t = 2500 sec.
P6-11 (e) Use the Polymath program from part (a) and change the limits of integration to O to 1200. We get: CAexit = 0.0055 mol/dm3 CRexít = 0.0066 mol/dm3
6-31
CSexit = 0 . 0078 mol/dnr'
P6-12 (a) rlA
r,~ r,
·'"'
P6-12 (b)
= -k\A C, = ··(.07)(.1)
Ita
dm • s
r,0 = ······""' - == --,-(.3)(.51)\1) --=
3
. _ ··rlA _ (.7)( . 1) . mol -·-·-·------·-=Ü.023 .. -····
= ·--0 . 01-=EfL .
3
•
IDC?.1._
drrr' •s
"0
·' ·
..,.._
drn' • s
P6-12 (d)
_ -riA _ (.7)(.l) r,c .... -··- - ---
mol
= 0.023 ---, r
3 3 d.nr•s _ -2 * r10 _ ·-2(.3)(.5 lf (.1) _ rzc ---- ------------·0.00:>2
3
3
·-(.2)(.049)(51)
r3E :
dm3
r.8 = O * r. ., = O _ rn~1--
= o * rJE = o-E!.º.L.. drrr' •s
=
3
r,n :::;: o* ro = O-
mol
---0 00?6-·-····· . -· · - dm' •s
P6-12 (e)
Ijc
3
mol ·-·-
dn/•s
= ··0.005- ~!.:?.1 •.
dm" •s
P6-12 (e)
_ ;;;: Ü···--· mol··'it : : : Q * líA dm3 •s
'íE :::::
. Ü ~- r;0
•
mol = Ú······:::----·
dm' •s
r,E = (.2)(.049)(.51) = 0.005· _ mol
drri3 «s
P6-12 (0 ÍA ::
P6-12 (g)
v- .FAo
··Ü.07 -Q.0026 = -0.0726-_Eli!_ . dm''•s
--F - A_
·-fA
- vo(C~:\!L--CAl Ii_:
= 0.023-·
--r;..
0.0052 ·· 0.005:::: 0.0128-_E.'l_?]_ __
100(3 .. -0.1) 0.0726
dnl•s
= ----·------·---·-- ··-
ro :::: 0.0078 ···· 0.033 == O. 00 l J . . . !:t~!.... drn •s
P6-12 (h) Mole balance:
Rate law:
C AO -- e A =(-·TA
)zCC = (re )zCn = (ro )z-
r,
= -[
k ACA +ik 1
20
CAC¿
6-32
J
= 4000dm 3
•S
Solving in polymath: Sam = ralro
= 247
Sa1c = 1.88
See Polymath program P6-12-h.pol. POLYMATH Results NLES Solution Variable Ca Cb Ce Cd Ce kd ka rb ra ke re rd re tau Cao
Value 0.0068715 0.9620058 0.5097027 0.0038925 O . 2380808 3 7 0.0160334 ·-O. 049885 5 2 O . 008495 6.488E-05 0 . 003968 60 3
f (x)
-2.904E-10 -l.332E-15 -·1. 67E-08 -2.391E-08 l.728E-08
Ini Guess 3
o o o o
NLES Report (safenewt) Nonlinear equations [ l l f(Ca) = Cao-Ca+ra*tau = O [2 J f(Cb) = Cb - rb*tau = O [ 3 J f(Cc) = Cc-rc*tau = O [ 4 J f(Cd) = Cd-rd*tau = O [51 f(Ce) = Ce - re*tau = O
Explicit equations [1] kd = 3 [2 J [3J [4J [SJ
[6J [7
J
[8J [9 J l lOJ
ka= 7 rb = ka*Ca/3 ra = -(ka*Ca+kd/3*Ca*CcA2) ke 2 re= ka*Ca/3 - 2/3*kd*Ca*CcA2 - ke*Cd*Cc rd = kd*Ca*CcA2 - 4/3*ke*Cd*Cc re = ke*Cd*Cc tau= 60 Cao = 3
=
P6-12 (i) For PFR and gas phase: Mole balance:
dFA
--
dV
= rA
ar¿
dFB
--=r
dV
B
--·=r
dV
6-33
dFD
e
--=r
dV
D
Rate law:
Stoichiometry:
r; =FA +FB «r; +Fn +FE dy dV
-a
Fr
---2y FTO
Plot of CB and Ce are overlapping.
See Polymath program P6-l2-ipol. POLYMATH Results Calculated values of the DEO variables Variable V
initial value
o
minimal value
o
Fa Fb Fe Fd Fe
20
y
1
Ft Cto Ce ka kd
20
0.9964621 13.330407
o
o
7
7 3 2
ke
Ca rb ra
Cd
Fto re rd re alfa
X
o o
o
o O. 2
3 2
0.2 0.4666667
-1. 4
o
20 0 . 4666667
o
o
1.0E-04
o
9 . 147E-04
o o o o
maximal value 100 20 6. 66381 71 6 .. 6442656 0 . 0201258 0.0043322 1
20 0.2 O . 0993605
O. 2
7 3 2
367E-05 3 . 191E-05
0.2 O 4666667 -9.586E-05 3 . OE-04 20 0.4666667 8 . 653E-04 5.908E-05 l.OE-04 O . 9999543
l.
-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 J d(Fa)/d(V) = ra [2] d(Fb)/d(V) = rb [3J d(Fc)/d(V)=rc ¡ 4] d( Fd)/d(V) = rd [5] d(Fe)/d(V) = re
6-34
final value 100 9.147E-04 6 . 6638171 6 . 6442167 O . 0171261 O .. 0043322 O . 9964621 13. 330407 0.2 O 0993325 7 3 2
367E-05 3 . 191E-05 -9.586E-05 2.56E-04 20 -1 . 923E-05 -6.742E-05 5 . 087E-·05 1.0E-04 0.9999543 l.
[6]
d(y)/d(V):::: -alfa*Ft/(2*y*Fto)
Explicit equations as entered by the user fll Ft:::: Fa+Fb+Fc+Fd+Fe [2l Cto=0.2 [3] Ce:::: Cto*Fc/Ft*y [4] ka=7 [5] kd e 3 [61 ke = 2 [7J Ca= Cto*Fa/Ft*y [8] rb = ka*Ca/3 [9] ra:::: -(ka*Ca+kd/3*Ca*Cc"2) [10] Cd:::: Cto*Fd/Ft*y [11 J Fto = 0.2*100 [12] re= ka*Cá/3 - 2/3*kd*Ca*Cc"2 ·· ke*Cd*Cc [13] rd = kd*Ca*Cc"2 - 4/3*ke*Cd*Cc [14] re= ke*Cd*Cc [15] alfa= 0.0001 [16] X:::: 1-Fa/20
P6-12 G) Changes in equation from part (i): dFe dV =re-Re
Re
=
kdiffusecC
k diffuse -2 . -l nnn
See Polymath program Pó-I'Z-j.pol.
6-35
20
0. 020
16
0.016
[:]
.
0 . 012
)
12
~
8
0 . 008
4
0.004
~~- ...~··--'"'"''_......_...._ ...... 20
40
V
60
80
100
40
20
V
60
80
100
P6-13 (a) m-xylene A m-xylene
A
benzene + methane B + M --> p-xylene --> p
-->
-->
See Polymath program P6 J3 . . a.pol. POL YMA TH Results Calculated values of the DEO variables Variable V Fa Fb Fp Fm
Fi Ft
kl
k2 Cto Ca rl Cb r2 Cp Spb tau
y
X
initial value
o
75
o o o
25 100 0.22
o. 7l o.os
minimal
o
value
6.1072984
o o o
25 100 0.22
o . 7l o.os
0.0375 -·0.00825
0.0026257 --0.00825
-0.026625
-0.026625
o
o o o o o
maximal value 6000 75 16. 297198 52.595503 16.297198 25 116.2972 0.22
o. 71 o.os
0.0375 -5.777E-04 0.0070067 -0.0018643 0.0226125 3.2272267 3 0.7634409 0.9185694
o
o o o o o
final value 6000 6.1072984 16.297198 52.595503 16. 297198 25 116.2972 0.22
o. 71 o.os
0.0026257 -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 [ l J d(Fa)/d(V) r1 +r2 [ 2 J d(Fb)/d(V) -r1 [ 3 J d(Fp)/d(V) = -r2
= =
Explicit equations as entered by the user [ll Fm = Fb [2J Fi = 25 [ 3 J Ft = Fa+Fb+Fp+Fm+Fi (4] k1 = .22
6-36
-------~·-----
-
----····------- ·------ -----------
- -- ------- --- ------~----~-
k2=.71 [6J Cto = .05 [ 7 J Ca= Cto*Fa/Ft [8J r1 = -k1*Ca [ 9 J Cb = Cto*Fb/Ft [ 1 o l r2 = -k2*Ca ¡ 11 l Cp = Cto*Fp/Ft ¡ 12 l Spb = Cp/(Cb+ . 0000001) [ 13 J tau = V/2000 ¡ 14 l Y = Fp/(75 . 00000001-Fa) [ 15 J X= (75-Fa)/75 [5]
a r = 2.8 is necessary to achieve 90% conversion 1 O
4.0
0.8
3.2
Q
0. 6
2.4
04
L6
0.2
0. 8
1200
2400 V
3600
4800
r=:
0. 0 O
6000
1200
2400 V
3600
4800
P6-13 (b) C~SIR Mole Balances :
Ej,. -=F, ,"+r.V )._ 0
f,1,,
.F. s -· · .\, -rl.l. r? ; ;: :k z eA
Rate Laws .
Using these equations and Polymath we find the optimal temperature is 1194 K. The maximum concentration of p-xylene is 0.013 mol/drrr'
See Polymath program P6-13·b.pol. fOLYMATH .Resul!§
NLES Solution Variable Ca Cb Cm Cp Cao tau klo El R
k2o E2
Value 0.012197 0.0122301 0.0122301 0.0130729 0 . 0375 0.5 0.22 2.0E+04 1..987
o.
f(x) 7 .15E-ll 5.284E-12 5.284E-12 -3 . 069E-11
Ini Guess 0.0375
o
o o
71
10000
6-37
6000
Yp
0.5166557 1194 2.0054175 0.0244601 0.0244601 2.143628 0.0261459 -0.050606
T
kl rb rm k2 rp ra
NLES Report (safenewt) Nonlinear equations [ 1 J f(Ca) = Ca-Cao-ra*tau = O [ 2 J f(Cb) = Cb-rb*tau = O [ 3 J f(Cm) = Cm-rm*tau = O [4]
f(Cp)
= Cp-rp*tau = O
Explicit equations ¡ 1 J Cao = .0375 (2 J tau= .5 k1o = .22 E1 = 20000 R = 1.987 k2o .71 [ 7 J E2 = 10000 [8J Yp = Cp/(.03750000001-Ca) [9J T= 1194 [10] k1 k1o*exp((E1/R)*(1/946-1/T)) [lll rb k1*Ca [12] rm e ktrOa [ 13 J k2 = k2o*exp((E2/R)*(1/946-1/T)) (14) rp = k2*Ca [15] ra = -k1*Ca-k2*Ca (3J
[4J (51 [61
=
= =
P6-14(a) 50dm3 PFH Mole balance:
.dC,1 . . ,. .;"' = .r,. .,1. . ,
dCe TB dV. - ·vo
dV dCc
re
dCD
d(:¡
'IB
dOF
vo
dV
Vo
·dv· -· "'.'uo dV
=
1· /)
dV rp
:;::: -._..,._-,,..
6-38
Rate laws: tr: - ::~,
rA
=t'
FB
··2ro1
re
r'1J1
ro
TJ)J
rp
re:,
....
E2
Tf'?,
+ rE2
'2.r¡,3
···
27'&2 f
rv1
e ("'2 k 01-'A.,B
rz;;,2
k¡;;2CACD
rr«
1, (Y2 ri.:Ji':J .,EJ -'C
rr»
e
The equation for the conversión of A is : CAo. ··· ---· C\1 C'A:o
l( -
,.
-
See Polymath program P6-14--a.pol. POLYMATH Rcsults No Titlc 08-01-2005, RevS.1.233 Calculated valucs of the DEQ variables Variable t Fa Fb Fe Fd Fe Ff vo Cao Cbo Ft Cto kdl ke2 kf3 Ce Cd Cb Ca rdl re2 rf3 re rf rd ra rb re
initial value
o
15 20
minimal valu e ------
o
0.2090606 1.3440833
o o o o
o o o o
l.. 5
l.. 5
10
10
2 35 3.5 O . 25 0.1 5
2 15.582463 3.5 0.25 0.1 5
2
O . 3018965 O . 0469574 0.0010699
o o
l. 5 l. 5
o o o o
l. 5 --1 . 5 -3 l.. 5
maximal value 50 15 20 l. 9655663 7 . 2554436 2.5920934 4.6265981
10
1.5 2 35 3.5 0.25 0.1 5 0.228291 l. 4503322 2 l. 5 l.. 5 0.1004639 0.3632767 0.1004639 0.3632767 l. 5 -O . 0215011 -0.0116593 l. 5
o o
o o o o
-0.042376 -1.. 5 -3 -O . 0962952
final value 50 0.2090606 l. 3440833 0.3535564 6.4570707 2.5920934 4.6265981 10 l.. 5 2 15.582463 3.5 0.25 0.1 5 0.0794128 l.. 4503322 0.3018965 0.0469574 0.0010699 0.0068104 0.0095194 0.0068104 0.0095194 -0.0030314 -O . 0215011 -O. 0116593 -0.0111585
ODE Report(RKF45) Differential equations as entered by the user
6-39
i··
J J [3J [4J [ 5J [6J [l
[2
d(Fa)/d(t) = ra d(Fb)/d(t) = rb d(Fc)/d(t) = re d(Fd)/d(t) = rd d(Fe)/d(t) = re d(Ff)/d(t) = rf
Explicit equations as entered by the user [l] VO= 10 [2J Cao = 1.5 [3J Cbo = 2 [ 4 J Ft = Fa+Fb+Fc+Fd+Fe+Ff [ 5 J Cto = Cao+Cbo [6] kd1 = 0.25 [7J ke2 = .1 [8] kf3=5 [ 9 J Ce = Cto*Fc/Ft [ 1o l Cd = Cto*Fd/Ft [ 11J Cb = Cto*Fb/Ft ¡ 12 l Ca= Cto*Fa/Ft [ 13 J rd1 = kd1 *Ca*CbA2 [ 14 l re2 = ke2*Ca*Cd [ 15 J rf3 = kf3*Cb*CcA2 [16] re= re2 [17] rf=rf3 [18] rd = rd1-2*re2+rf3 [ 19 J ra = -rd1-3*re2 [20] rb=-2*rd1-rf3 [ 21 l re = rd1 +re2-2*rf3 20 • Fa - Fh
16
- Fe - Fd
12
- Fe
8 4
o
o
10
20
V 30
40
50
P6-14 (b) {b) Determine the effluent concentration and conversion frorn a 50dm:3 CS'I'R. l'VIole Balance: -rAV
ro V re V 1·vV
6-40
FE
rEV rpll
Fp 1
== 110' v combinina" rate law and mole balance,
-
>
f(CA) J(Cn)
f(Cc) f"' e-o ,)
---Ce+ rar --Cv + rvr
f(CE) f (Cp)
--C'E
>
\
+ TFJT "-C:p + 'fpT
Polymath code, f (ca),.,ca-caO··ra*tau
f(cb)•cb-cbO-rb*tau f(cc)=rc*tau-cc f (cd),==rd*tau-cd f(ce)=tau*re-ce f(cf)"'tau*rf tau=V/vo V=50
-cf
vo=10 ca0=1 5 cb0=2.0 rd1=kd1*ca*cb''2 re2=ke2*ca*cd rf3=kf3* cb= cc ··2 kd l=O . 25
ke2=0.
1
kf3=5 r a= rb=:
rd1··3*re2 2*rd1 ·rf3
rc=rd1.+re2 ·2*rf3 rd=rd1~2•re2+rf3 re=re2 rf=rf3 ca(0)=1
cb(0)=1 cc(0)"'1 cd(O)=l ce(O)=l
cf(O)=l (Ans) CA= 0.61, CB = O 79, Ce , 011, C» = 0.45, CE
(U4.,Cp
O 25
P6-14 (e) (e) V{)= 110dm'1 Semi-Batch mador l} A is fod to B,
(1) A is fed to B, (2) B is fed to A
6-41
dC.4.
dt dCn
dl' dCo dt
dCD
Cv
Vo·v·
1'1)
dt
CE
dCs dt
re ..... llo-·v
dCF
rp
dt
....
Cp
vo --· V
(CBo .---CB) Cso
){
---····
aud modifyíng corresponding polymath cede. d(ca)/d(t)=ra·vo/V*ca d(cb)/d(t)=xb+vo/V*(cbO·cb) ca(0)=1.5
cb(O) .. O x=(caO·ca)/caO Diffot enees (1) Because C8o is higher than CAo (Le ;JO% higher molar ílow rate], case (2) reachee X"" 1 in earlíer time. (2) With the same reason, case (2) produces D and F more (3) With the sanie reason, Ce (in case 2) increases more drestlcallyíexcessively] than CA (in case l)
(e) case l concentration vs. time 1.00
,:,.
un
'"" 1AO
GláJ>hliHe
\
[ ' .:] o ~t, ;: ,;! '" + ce !";
<:;(>
..
!
f
! .. ;!J
1 '\
6-42
Case
conversión
vs. lime
f llO O 90
0.!;0 [l.'10
case {'l)
OJO l)2(1.
0.10 ....., ... _.i.-..,
.,:..
-e.oo
50JDO t
Case 2 concentrarion vs. time 1W
6-43
....,
..._
&000
70.00
.._
13.000
QOOD
......
lDO 00
Case 2 conversión vs. time LOO 090
oso O 70 0.00 050
case (2) 040 0 ..30
020 O 11! 0.00
,_ __
....._
0.00
10.00
..._
_._ __
20.00
__,c._.,,·-···· 40 . 00
........ 50.00
.., 60.00
701)0
80.00
00.00
10000
P6-14 (d) As 08 increases the outlet concentration of species D and F increase, while the outlet concentrations of species A, C, and E decrease . When 08 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 kdl ke2 kf3 Ce Cd Cb Ca rdl re2 rf3 re rf
initial value
o
20 20
o o o o
minimal value
o
18.946536 18 . 145647
o o o o
100 40 0.4 0.25 0.1 5
100 38.931546 0.4 0.25 0.1 5
O.2 O.2 O . 002
0.1864364 0.1946651 0.0016916
o o o o o o
o o
o o o o
maximal value 500 20 20 0.9342961 0.8454829 0.0445942 0.0149897 100 40 O.4 0.25 O. 1 5 0.0095994 0.0086869 0.2 0.2 0.002 l. 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 O . 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
0.002 -0.002 -0.004 0.002 1
rd ra rb re Sed Sef
o
0 . 0014393 -0.0021989 -O. 003469 0.0016889 1.1734311 1.9686327
0.002 -O . 002 -0.003469 0.002 1.1734311 83.266916
0.0014393 -0.0021989 -0.004 0.0016889 1
o
ODE Report (RKF45) Differential equations as entered by the user [ l J d(Fa)/d(V) = ra [ 2 J d(Fb)/d(V) = rb [ 3 J d(Fe)/d(V) = re [ 4 J d(Fd)/d(V) = rd [ 5 J d(Fe)/d(V) = re [ 6 J d(Ff)/d(V) = rf Explieit equations as entered by the user [1] vo 100 [ 2 J Ft = Fa+Fb+Fe+Fd+Fe+Ff (3 J Cto = .4 [4) kd1 0.25 [5 J ke2 = .1 [6] kf3 5 [ 7 J Ce = Cto*Fe/Ft [ 8 J Cd = Cto*Fd/Ft [ 9] Cb = Cto*Fb/Ft (10] Ca=Cto*Fa/Ft f 11 l rd1 = kd1 *Ca*Cb"2 [ 12 l re2 = ke2*Ca*Cd [ 13 l rf3 kf3*Cb*Ce"2 [ 14] re= re2 [15] rf=rf3 [ 16 l rd = rd 1-2*re2+rf3 [ 1 7 J ra = -rd1-3*re2 [ 18 l rb = -2*rd1-rf3 [ 19] re= rd1+re2-2*rf3 [ 2 o l Sed = re/(rd+ . 0000000001) [ 21 l Set re/(rf+.00000000001)
=
= =
=
=
1.20 0.96
---~--·-~----
90 ~-------72
0.72
54
0.48
36
0.24
18
0 . 00
o
o
100
200 V 300
P6-14 (f) The only change from part (e) is:
400
d;- =
dF
500
rD - kcDC D
6-45
...
O
100
~-
._.,-.,-.,..
.. ~
200 V 300
.
400
500
See Polymath program Pó-l-t-f.pol.
------ -
1.15~-------------,
......
0.92
100
0.69
75
046
50
0.23
25 100
200 V
300
o o
500
400
100
200 V 300
400
500
P6-14 (g) dF dV
The only change from part (e) is: __.!L
= r.
where V r = 500 dnr' and F80 = 20 mol/min
B
-
F VT
_QQ_
See Polymath program P6--14-g.pol. 1.2 ,-------------·
700 .--------
1.0
560
0. 7
420
0. 5
280
0.2
140
O.O"----·
o
100
200 V
300
400
500
o L---=---·---········ ··-·-·"""'"'' '''"""-··· . · · ·-·"'" o 100 200 V
P6-15 (a)
6-46
300
400
500
(a) En ter the given prngrarn inro POLY?vlATH Equations for the concentrations muse be added.
The following maximums can be seen in the graph given (More exact valúes can be found in che corresponding table in POLYi'.VlATH) CCm.u
== 0 . 0434 and
CDtrcu;::;:
0.0033
Eguati9ns :.
Init,;al_va,lue
d(fc)/d(v)=kl•(fa/ft}*(fb/ft)**(l/2}-k3•(fc/ft}+k4•{fw/ft
O
} .. ( fd/ft) d ( fal
/d{v)
* ( fb/ft) " ( 112) · k2 * ( fa/f tl *"'2
= ·kl* ( fa/ft)
d{fbl/d(vl=-kl/2*(fa/ft)~(fb/ft)••(1/2)
4.91
d ( fd) /d (vl =k2 /2" ( faJ ft.J "'2 --k4 • ( fd/ ft:) * ( fw/f t l
o o
d(fe)/d{v}::ak3*(fc/ft)
o
d(fw}
d
(fg)
/d(v)
"'k3* (fc/ftl ··k4* {fw/ftl
/d{v} ;k4* (fw/:et.)
* (fd/ft)
o
* (fd/ft)
kl.;;;0. 04
ft=fa·,·fb+fc+fd+fetfw+fg ~-k2=0.007 k3=0
014
k4.=0
45
C,.6C'.J
vo=lOO c t.oe O . 147 ca,,,cto"(fa/ft)
C 1·0C
o .. 2ca
cb::cto*(fb/ftJ cc=cto*(fc/ft)
c .. oca
cd=cto"(fd/ftl ce=cto*(fe/ft.) cw=cto•(fw/ft} cg"'cto• (fg/ft)
"o
9. 83
= O,
1000
O GDO
P6-15 (b)
6-47
(b) Overall
yield of HCOOH:
Selectivity of HCHO
to
CO :
s- AE -- FF,--~~E
S = fo
Selectivity of HCOOCH3 to CH_;OH: Selectivity of .HC:OOH to HCOOCH3
eº2
DG
F
G
:
Ca =e-· A
Add these equations to the previous program and use it to generate the desired plots .
o . aco O SCJ
o zcc
: .. :J:;C
lr;02
P6-15 (e)
6-48
~,,sas.
(e)
Modify
the original
by adding y to each of the Also add che following equation:
POLYJ\-IATH prograrn
concentrarion terms
!t-~(!~) a=ü.002 Fro =cJ5 The graphs of concentration down the reactor are very similar to those generated in part (a). The only rnajor difference is thar with the change in pressure, the máximum reactor volurne is significanrly smaller: E~a!::.ion~_: d { fe l /d (v) =k1 * ( fa/ft:)
* ( fb/ft) "* ( 1/2)
w/ft)*(fd/ftl*Y d(fa) /d(vl '=-kl* (fa/ft) d(fb}
/d(v}=
·kl./2*
(fa/ft)
*
**
(fb/ft)
* (fb/ft)
/d(v)
d(fd)
/d{v)aak2/2"
(.fa/ft)
*Y+·k4* ( f
!nitialya3::ue O
{1/.2) •y·k.2* {fa/ft) ""*2*y
10
**
5
=k3* (fc/ft} *y--k4* Cfw/ft)"
d(fw)
* ( fe/ ft)
"y·k3
"*2•y-k4*
(1/2) *y
(fd/ft) *y
(fw/ft)
*
(fd/ft)
O *y
O
d(fe)/d(v):k3*(fc/ft)*y
O
d(fg)/d(v)=k4*(fw/ft)*(fd/ft)*y
O
d(y)/d(v)=-0.002/2/y•(ft/fto)
1
k2=0.007 kl=0.04
Cencentraííen profile with premue
dia:nge
ft,,,fa+fb+Ec+fd+fe+fw+fg
k3=0. 014 k4;;;Q.45
vo=l.00 ft.0""15 ca=fa/vo cb=fb/vo
cc=fc/vo
cd==fd/vo ce=fe/vo
o . coo
cg=fg/vo
Cnneentratíen profile with preHure change
"o "' o,
:. 6GO
o.coc
o.coo
P6-15 (d) 6.49
•
k,, ""' ;;: .
''
'"
{. c.c I»l ex111·. =. 1 -
l
¡t
--· ·- · ). \ R\ T T,, j
Subsritute rhis equadon in for ali of the k vruues- Vary T and find out what ternperature maxímizes the yield of C. The best temperature at . .vhich to run the reactor ¡5 523 K or 250"C.
P6-16 (a) (a)
Mole Balances:
dFc ....... -::::r dV
dFo ·-·-= rO dV
e
re = -·k1 Ce -- k2Cc
Rate Laws :
rA
=k1Cc +k1Cp ··-k.;CA
·ksCA
eo = ero (·F·{F. ). )
Stoichiometry :
1'
FI =F.C +F· P +F A +FO Use these equations in POLYMATH to generate a plot of the flow rates vs,
't ..
\
f:c
. sccc
. \
\
\
fp fa fo
,.-"
, '-.. ~
,...,.
. ~:c.::.:.:.
e.eco
....
•'
. .,,;- '"' "' ··-
1(1000
-- .... _ ~~-" H
: ,_,_
~__
20.:J.:C
'"''
.:=::;===--·-----._ . . "'
P6-16 (b) (b) For a CSTR Mole Balances:
Fe
= Feo + re V r, =k1Cc ··-k3Cp+k4CA
Rate Laws :
ro
6-50
= ksCA
Stoichiornetry : Combine:
í(Cc)=O=Cc
Ce" +(k1 +k1
-x,c, +k,C\)t
f(Cp)=0=Cp-(k1Cc f(C;J=O=CA f(C0 )=O= C0
·(kiC~c ....
Tcr
k3Cp--k,CA
-k/'.:,.)i-
k5C.J
Use these equations in POL YMATH to generare values for the flow rares at different values of 1: Use these values to generate the desired curve.
Iaj,5:ial
Ec,--ua t: i.ofis : f (ce) e.e e .. eco+ {kl+k2)
f (cp)
-"'CP··
(kl'"cc-kJ
•cp+k4*cal
=t
o
au
f (ca} =ca· (k1 '"cc-·k3 =cp -k4·•ca···kS*ca) f
va Lue
2
•cc"t.au
o
• tau
o
(c o ) ,;,co···kS"ca*tatl
cco=2 kl-=0 .. 12 Sol ut L or i
k2"'0.046
-f
k:3,:(). 02 U.038:3701 0 . 211711
ce
J<.A;;,;() .O:J4
c:p ee ca
voo=J.O k5=0 . 04
CUJ0633865 0.07506'.38 2 C.t2
C:Cd
Va:3000
k l k2 k3 k4
fcm:vo•cc fp=vo•cp
fa;vo•ca
0 . 046
c. 02
CL034 10
VO
foc=vo"co
c.
k5
u
tat.1=V/vo
()4
3COU
lp
D. JSJ.70! 2 .. l 17 t 1
fo
O.C6T3865 C).760638
fe
300
(b) Flow rates vs. 1: 2.CO iso iso Hú 120 l(XJ
. fC ....... fp: -FA
80 60
FO
4()
20
o
o
P6-16 ( C)
50
!CO
150 200 ,; (mh!)
250
300
Individualized solution
6-51
350
<
··8.S45e .. ,7 .. L078e··16 ····2 .. '.30Se··l5
··8.5!8e
·18
P6-17
Individualized solution
P6-18 (a) Blood coagulation living example See Polymath program P6-18.pol OLYMA TH Results Example CD Solved Problems - Blood Coagulation
08-25-2005,
Rev5.l 233
Calculatedvalues of the DEQ variables Variable initial value t o TF 2. 5E-11 VII l.OE-08 TFVII o VIIa 1. OE·-10 TFVI!a o Xa o IIa o X l.6E-07 TFVIIaX o TFVIIaXa o IX 9.0E-08 TFVIIaIX o IXa o II l. 4E-06 VIII 7.0E-10 VIIIa o IXaVIIIa o IXaVIIIaX o VIIIalL o VIIIa2 o V 2.0E-08 Va o xava o XaVaII o mIIa o TFPI 2.5E-09 XaTFPI o TFVIIaXaT o ATIII 3.4E-06 XaATIII o mIIaATIII o IXaATIII o TFVIIIaAT o IIaATIII o kl 3.2E+06 k2 0.0031 k3 2 . 3E+07 k4 0.0031 k5 4.4E+05 k6 l. 3E+07 k7 2.3E+04 k8 2.5E+07 k9 l. 05 klO 6 kll 2.2E+07 kl2 19 k13 l.OE+07 kl4 2.4
minimal value
o
8.24E-14 3.513E-10
o
l. OE-10
o o o
l.426E-07
o o
8.994E-08
o o
-3.41E-24 -2.024E-28
o o o o o
-r . 55E-52
o o
-3.938E-26 -8 . 77E-25 2.094E-09
o o
2.00lE-06
o o o o o
3.2E+06 0.0031 2 . 3E+07 0.0031 4 . 4E+05 l. 3E+07 2.3E+04 2.5E+07 l. 05 6 2.2E+07 19 1.0E+07 2.4
maximal value 700 2. 5E-11 l. OE-08 2.027E-ll 9.724E-09 3.361E-13 l.481E-09 2.487E-07 l. 6E-07 1. 869E-13 5.673E-l4 9.0E-08 7.2E-14 3.579E-ll l. 4E-06 7.0E-10 5.352E-10 2.988E-12 5. 372E-12 6.585E-10 6.585E-10 2.0E-08 1.943E-08 l.492E-08 2. 281K-10 3.788E-07 2.5E-09 3.867E-10 l. 881E-11 3.4E-06 6.073E-10 8 . 247E-07 1. 301E-ll 8.354E-14 5.734E-07 3.2E+06 0.0031 2.3E+07 0.0031 4.4E+05 l. 3Et07 2.3E+04 2.5E+07 l. 05 6 2.2E+07 19 l.OE+07 2.4
6-52
·---------------·
final value 700 8.24E-14 3.513E-10 5. 71E-12 9.724E-09 l.665E-13 l.481E-09 l.846E-09 l.426E-07 8.423E-14 2.608E-14 8.994E-08 3.568E-14 3. 579E-11 -a . 05E-25 -l. 026E-38 3. 366E-11 2.873E-12 4.995E-12 6.585E-10 6.585E-10 2.793E-90 5.077E-09 l.492E-08 -6.977E-27 l. 663E-25 2.094E-09 3.867E-10 l.881E-11 2.00lE-06 6.073E-10 8.247E-07 l.301E-11 8.354E-14 5.734E-07 3.2E+06 0.0031 2.3E+07 0.0031 4.4E+05 l.3E+07 2. 3E+04 2.5E+07 l. 05 6 2.2E+07 19 l.OE+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 rl r2 r3 r4 r5 r6 r7 r8 r9 rlO rll r12 r13 r14 r15 r16 r17 r18 r19 r20 r21 r22 r23 r24 r25 r26 r27 r28 r29 r30 r31 r32 r33 r34 r35 r36 r37 r38 r39
l. 8 7500 2.0E+07 1.0E+07 O . 005 l . OE+08 0.001 8.2 0.006 2.2E+04 0.001 2.0E+07 4.0E+08 O,, 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
l. 8 7500 2.0E+07 1.0E+07 O . 005 1.0E+08 0.001 8.2 0.006 2.2E+04 0.001 2.0E+07 4.0E+08 O.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
5.75E-14
1 . 735E-14
o
o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o
l. 8 7500 2.0E+07 1.0E+07 O . 005 l. OE+08 0.001 8.2 0.006 2.2E+04 O . 001 2.0E+07 4.0E+08 0.2 l. OE+08 103 63.5 l.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 6.283E-14 5.75E-14 1.042E-15 9.923E-16 1.444E-11 4. 073E-11 l. 318E-12 l.962E-13 l.121E-12 5.381E-15 1.078E-12 3. 024E-13 L 728E-13 l. 296E-13 2.883E-13 l.159E-11 4.674E-14 1.494E-14 4. 406E-11 5 . 372E-15 4.405E-11 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 l. 372E-13 4.492E-14 2.065E-15 3.19E-15 4.387E-12 7. 708E-09
o
o o o o o o o o o o o o
-1. 6E-29
-9. 831E-28
o o o o o o o o
-2.108E-52
o o
-3.763E-24 -4 . 056E-24 -2.5E-24 -l.733E-25
o o o o o o
-l.248E-26
l. 8 7500 2.0E+07 1.0E+07 O . 005 1 . 0E+08 0.001 8.2 0.006 2.2E+04 0.001 2.0E+07 4.0E+08 0.2 l.OE+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 l. 782E-14 l. 855E-14 5.189E-16 2.696E-17 6 . 953E-12 l.602E-14 5. 971E-13 8. 896E-14 5. 083E-13 5.381E-15 4.982E-13 l.506E-13 8.609E-14 6. 456E-14 4 . 511E-30 -4.231E-40 l. 217E-14 l. 439E-14 4.106E-11 5.006E-15 4.105E-11 2.044E-13 9.527E-15 5.006E-15 1.152E-91 2 . 999E-09 2.974E-09 6 . 12E-25 l.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
6-53
·----------
--------·---
--------------
r40 r41 r42
Total
o
o
o
o
o o o o
3 . 502E-14 4.224E-09 1. 78E-16 5.749E-07
3.502E-14 2.704E-11 7 . 705E-17 l.903E-09
ODE Report (STIFF) Differential equations as entered by the user [ 1 J d(TF)/d(t) = r2-r1-r3+r4 [2 l d(Vll)/d(t) = r2-r1-r6-r7-r5 [ 3 J d(TFVll)/d(t) = r1 -r2 [ 4 J d(Vlla)/d(t) = -r3+r4+r5+r6+r7
[ 5 J d(TFVlla)/d(t) = r3-r4+r9-r8-r11+r12-r13+r14-r42-r37 +r15 [ 6 J d(Xa)/d(t) = r11+r12+r22-r27 +r28-r33+r34-r38 [ 7 J d(lla)/d(t) = r16+r32-r41 [ 8 J d(X)/d(t) = -r8+r9-r20+r21 +r25 [ 9 J d(TFVI laX)/d(t) = r8-r9-r1O - TFVIIa [ 1 o J d(TFVllaXa)/d(t) = r1O+r11-r12-r35+r36 TFVUaX [ 11J d(IX)/d(t) = r14-r13 2.72E-1J - IFV!laXa f 12 J d(TFVI lalX)/d(t) = r13-r14-r15 *_IFV IX [ 13 J d(IXa)/d(t) = r15-r18+r19+r25-r40 2 . 04E-1J [ 14 l d(ll)/d(t) = r30-r29-r16 [ 15 J d(Vll 1)/d(t) = -r17 [ 16 J d(Vllla)/d(t) = r17-r18+r19-r23+r24 1.J6E--13 [ 17 J d(IXaVllla)/d(t) = -r20+r21+r22+r18-r19 [ 18 l d(IXaVlllaX)/d(t) = r20-r21-r22-r25 r 19 l d(Vllla1 L)/d(t) = r23-r24+r25 6.80E-14 [ 2 o J d(Vllla2)/d(t) = r23+r25-r24 [ 21 J d(V)/d(t) = -r26 O.OOEO O . OO [22 J d(Va)/d(t) = r26-r27+r28 138.93 277.8~ [ 2 3 J d(XaVa)/d(t) = r27-r28-r29+r30+r31 [24] d(XaVall)/d(t) = r29-r30-r31 [ 2 5 J d(mlla)/d(t) = r31-r32-r39 [ 2 6 l d(TFPl)/d(t) = r34-r33-r35+r36 [27] d(XaTFPl)/d(t) = r33·-r34-r37 [ 2 8 l d(TFVllaXaTFPl)/d(t) = r35-r36+r37 [ 2 9 J d(ATll 1)/d(t) = -r38-r39-r40-r41-r42 [30] d(XaATlll)/d(t) = r38 r 31 l d(mllaATlll)/d(t) = r39 [32 l d(IXaATlll)/d(t) = r40 [ 3 3 J d(TFVlllaATll 1)/d(t) = r42 [34] d(llaATlll)/d(t) = r41
Explicit equations as entered by the user [11 k1 = 3.2e6 [ 2 J k2 = 3.1 e-3 [31 k3=2.3e7 [4] k4 = 3 . 1e-3 ¡ 5 J k5 = 4.4e5 [6l k6 = 1.3e7 [7l k7 = 2.3e4 [8l k8=2.5e7 [9] k9 = 1.05 [10] k10=6 [lll k11 =2 . 2e7 [12] k12 = 19 [13] k13=1.0e7 [14] k14 = 2 . 4 [ 15] k15 = 1.8 [16] k16=7 . 5e3
6-54
416.79
555.72
694.64
(17]
[18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43] [44] [45] [46] [47] [48] [49] [50] [51] [52] [ 53] [54] [55] [56] [57] (58] [59] [60] [61] [62] [6 3] [64] [65] [66] [67] [68] [69] [70] [71) [72] [7 3] [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 = 9e5 k34 = 3.6e-4 k35 = 3.2e8 k36 = 1 . 1e-4 k37 = 5e7 k38 = 1.5e3 k39 = 7..1e3 k40 = 4.9e2 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 = k?*lla*VII rB = k8*TFVlla*X r9 = k9*TFVllaX r10 = k10*TFVllaX r11 = k11 *TFVlla*Xa r12 = k12*TFVllaXa r13 = k13*TFVlla*IX r14 = k14*TFVllalX r15 = k15*TFVllalX r16 = k16*Xa*II r17 = k17*11a*VIII r18 = k18*1Xa*Vllla r19 = k19*1XaVllla r20 = k20*1XaVllla*X r21 = k21 *IXaVlllaX r22 = k22*1XaVlllaX r23 = k23*Vllla r24 = k24*Vllla1L*Vllla2 r25 = k25*1XaVlllaX r26 = k26*11a*V r27 = k27*Xa*Va r28 = k28*XaVa r29 = k29*XaVa*II r30 = k30*XaVall r31 = k31*XaVall r32 = k32*mlla*XaVa r33 = k33*Xa*TFPI r34 = k34*XaTFPI
6-55
J r35 = k35*TFVllaXa*TFPI r36 = k36*TFVllaXaTFPI r37 = k37*TFVlla*XaTFPI r38 = k38*Xa *ATIII [81] r39 = k39*mlla*ATIII [ 82 J r40 = k40*1Xa*ATIII [83] r41=k41*lla*ATIII [ 84 l r42 = k42*TFVlla* ATIII [85] Total= lla+1.2*mlla [ 77
[78] [791 [ 8o l
P6-18 (b) No solution
will be given
P6-19 (1)
C2H4 + 1/202 E + 1/20 -
F10 = 0.82F10
(2)
C2H40
C2H4 + 302 - 2C02 + 2H20 E + 30 - 2U1 + 2U2
D
= 0.007626
P6-19 (a) Selectivity of D o ver C02
S= FD Fui
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 _º _____
minimal value
5 . 58E-04 O. 001116
l. 752E--10 4. 066E--·05
l. OE-07
1 . OE--07
o
o
o
o o
0.007626 0.0093001 6.5 4.33 2 O .1.199987 0.2399974 0.15 0.088
0 . 007626 0.0091804 6.5 4.33 2 3.817E-08 0.008858 0.15 0.088
-0.0024829 -0.0029803
-0.0024829 -0.0029803
o o
o
o
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 O . 2399974 0.15 0.088 0.9999997 O . 4101512 -3.692E-10 -8.136E-10
Differential equations as enterad by the user [ 1 J d(Fe)/d(W) = r1 e+r2e [ 2 l d(Fo)/d(W) = 1 /2*r1 e + 3*r2e [ 3 l d(Fd)/d(W) = -r1 e [ 4 J d(Fu1)/d(W) = -2*r2e r 5 J d(Fu2)/d(W) = -2*r2e Explicit equations as enterad by the user [ 1 J Finert = 0.007626 [ 2 J Ft = Fe+Fo+Fd+Fu1 +Fu2+Finert [3J K1 = 6.5
6-56
final value 2 l. 752E-10 4 . 066E-05 2 . 395E-04 6. 372E-04 6. 371E-04 O . 007626 0.0091804 6.5 4.33 2 3.817E-08 0.008858 0.15 0.088 0.9999997 0.3758225 -·3. 692E-10 --8. 13 6E-·10
=
[4J K2 4 . 33 [5] Pto = 2 [ 6 J Pe = Pto*Fe/Ft [ 7 J Po = {Pto*Fo/Ft) [8] k1 = 0.15 [ 9] k2 = 0.088 [ 10 J X= 1 - Fe/0 . 000558 [lll S = Fd/Fu1 [ 121 r1 e= -k1 *Pe*Po"0.58/(1 +K1 *Pe)A2 [ 13 J r2e = -k2*Pe*Po"0.3/(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 from part (a):
dF0 =-r 1 1e +3r2E+R0 --
2
dW
Ro= 0.12x0.0093
and
F0 (o ) == O
= 0.001116 mol 2
W
kg.s
From Polymath program: X = O. 71 S = 0 . 04 (mol of ethylene oxide)/(mole of carbon dioxide)
See Polymath program P6- l 9-b..pol
P6-19 (e) Changes in equation from part (a):
= rE
dFE
1
dW
+r2E+RE
and
FE (o)= O
R = 0.06 x 0.0093 = 0.000558 mol E W 2 kg.S From Polymath program: X= 0. 96 S
= OAl(mol
of ethylene oxide)/(mole of carbon dioxide)
See Polymath program P6-19-c .poi
P6-19 (d) No solution will be given. P6-20 Solved on web Go to http://www.wits.ac.za/fac/engineering/procmat/ARHomepage/frame.htm
P6-21 (a) Isothermal gas phase reaction in a membrane reactor packed with catalyst.
·
A~7B+C
1i.c
=k« [ CA ---CBCc] K1c
A7D 2C+D 7 2E
6-57
.
.
~------·--------------·-···--·--·. ·----------·---------------·····--------------·---·--------------·--------------------------
C AO
= __!__ = RT
= Ü.6mo[ / dm3
24.6atm (0.082dm3atm/mol.K)(500K)
See Polymath program Pó-Z l-a.pol POL YMATH Results Calculated values of the DEO variables Variable initial value minimal value
o
w
Fa Fb Fe Fd Fe
10 O
0.349438 O
O O
O O
1 0.4 0.2 10 0.6
O . 3404952
0.4 0.2 9.7581913 0.6
0.6
0.0073158
O
y
k2d Kle Ft Cto Cb Ca Cd Ce kb kle r2d k3e rle ra r3e rd rb re re Ce alfa Fto
o
O
O O 1
O
1
0.4 O.2 13.220737 0.6 0.1403618 0.6 0.1019635 0.2117037
O
O O
1 2
1
2 0.24 5 1.2 -1.44
2 0 . 0029263 5 0 . 0051635 -1 . 44
0.24 1.2 1.2
O .. 0051635
O
maximal value 100 10 3 . 2375418 4.9873025 2.7304877 1.3722476
0 . 24 5 l. 2
-0.0080898 0 . 0216828 0.24 1.2 1.2 0 . 0216828 0.0398819 0.008 10
O
0.0014457
-0.0042625
O O
O O
0.008 10
0.008 10
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 0.0571654 0.1017844 1 2
0.0029263 5 0.0051635 ··O. 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 ¡ 1l d(Fa)/d(W) = ra [2J d(Fb)/d(W) = rb·(kb*Cb) [ 3 J d(Fe)/d(W) = re [ 4 J d(Fd)/d(W) = rd ¡ 5 J d(Fe)/d(W) = re [ 6 J d(y)/d(W) = -alfa*Ft/(2*Fto*y) Explieit equations as entered by the user [1] k2d = 0 . 4 [21 K1e=0 . 2 [ 3 J Ft = Fa+Fb+Fe+Fd+Fe [ 4 J Cto = 0.6 [ 5 J Cb = Cto*(Fb/Ft)*y [ 6 J Ca = Cto*(Fa/Ft)*y [ 7 J Cd = Cto*(Fd/Ft)*y [ 8 J Ce = Cto*(Fe/Ft)*y [9] kb=1 [lOJ k1e=2 [11] r2d = k2d*Ca [12) k3e 5 [ 13 J r1 e = k1 e*(Ca-(Cb*Ce/K1e)) r 14 J ra = -r1e-r2d
- Fa .. Fb .. Fe
8
- Fd
6
4
2
=
o
6-58
-
_..
O
.
-====--~-·----~----' 20
40
\V 60
80
100
[ 15 l [ 16 l ¡ 17 J
r3e = k3e*(CcA2)*Cd rd = r2d-(r3e/2) rb = r1c (18] rc e rtc-rae [ 19 l re= r3e [ 2 o J Ce Cto*(Fe/Ft)*y ¡ 21 l alfa = 0.008 [22] 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 ( C) lndi vidualized 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 (ac) Observation 1: -r1 and -r2 decreased Observation 2: 'tp¡ increased Observation 3: -rP2 decreased Now, i-1
t"p1
=(
= _!_ ln( µ~ J and i-2 = _!_ ln(Po* J e
e
µ1
21l *2
µl +
K
-[(1U
_ k0 * e--, µ12 k2
'
=(
)112 and t"p2
2Ku) ±
2
µ2
21l µ2
*2
+
.
K
U
)112
.Jf= 8Ku ]
112
From observation 2 and 3, we get Decreasing µ1 • and increasing µ2 • => Ku = (ku/ki) will increase Also, from observation 1, e increased => kofk:2 should be increased Now, P '7 A k = ko (1) A7 B k=ku (2) A+2B 7 3B k = k¡ (3) B7C k=k2 (4) Hence the reaction (1) and (2) are more temperature sensitive than reaction (4)
P6-22 ( d) lndividualized solution
6-59
.
-----------------------~---··----·-------·------------------------
--------
P6-23
Individualized solution
CDP6-24
•
-uo Ca + I'B W
=O
·Uo Ce + re W
=O
··Uo Co + ro W
=O
r,1.,.
= ·k1 CA• k3 e,,,,
I'B
=k1 CA ··k2 CB
re= k2 Ca
l
s. aa
KE.Y: 1
-
ca
2 - cb 3 •· ce 4 ··
cd
CDP6-25 (a) PFR:
6-60
Mole balance:
dC:,,.t ---·-· = dT
r
dCH ........ . ;:.. . =
rH
dT
dC dr
.......... !, .. -··r
-
M
X
dC!l.1<, . - . . . . . .t dr Me
dC, ...... .,_. =
dT
dC8
···-·n..:;..:
dr
fT
rB
Rate laws
C,{O = 0.021
CMO = 00105
·r = .5
Plugging those into POLYMATH gets the following:
6-61
CSTR:
Mole balance:
FHo···FH =(·-rrn-·r;H····r,H)V FMO- FM;:: -··
r1:,..1v
r, =(r1x+rzx)V FT = {r:1+ r;r) V f:-.1. =(r1M~+rz'.',1o+r,1.~i.)V F6
= r111
RateLaw:
Stoichiometry:
= voCH
FH
F..,,1 = voCM
= voCx
fx
Fr = voCr
F:. .
i.
=
=
VoCMc
F6 v0C3 Combíníng all of these:
(··e •ñ()
= vo(Cuo ···· C11)
= "'íi[(<\m
H ::;;;; ·.(k·,¡ ('ºJe· H '"M
-
CM)····
+ k. , :: (~"ti05C :<
e, -· C1] k C'º·Sc·· H T ) t'
tr, . :;
. C'ºs<,·'Mt 5c· k· c•J ..s<, ) ., (.,·x ;:: [k c·º ·'H M ·- ! ""H · X "
C.,"MO
e
.... -:,..1 ::::: i<¡ "!! ··!
e· ·r = [k'·1e-x c~o.s
H · •
CM< =
k., ('-u05("'--x )""~
C,;o - CH
C8 = (CMo -CM) .... Cx ·-CT The follcwing i:s the POLTht-".TH program and rhe concenrrations.
me summary
table showing ali of
:gq,.,¡acions:. f (ch) :ch -crio» (kl "ch~. 5 •c:n··k2~ch". S •cx+k3 .. ce •ch~. s) "cau
f { cm) =cm . ·cmo·-+· kl '!lf'cn- ~ s-...c:r.
f()
kl=55.2
0.0077651 S 0.00301658 o.00;31 ;:,457 0.00286611
k2.:30.
0.021
f(cc)a(k2•ch-.S•cx cho=.021
kJ•ct•ch-.Si•cau-cc ex
2
ld=!.l.2
k1 k2
taw=.Sl
\d
55.2
30.2
11.2
0.5!
cmo"",0105
cmo
crnee cho . -ch
cm;;,
cb"' (cmo-ccrn} --cx··cc
6--62
0 . 0105 c.o i 32348 0.00144254
2.ss:::ie--1s l. 463€;'
14
2.79e··l4 L495e
·I 4
The comersion of Hydrogen ami Mesitylene are then:
x.
== C,m __ H
e'HO
cH= o.021--ooon=0.63 o · 0'.::.1 1
q_~5~?_ :·:.9:9.22Q.
XM = 5~~!.!L:-· (~M.= 2 = .71 ' C:,.10 O.Ol05 CH= 0 . 0078 lb mol/fr' CM= O 0()30 lb mol/fr' Cx: = 0.00.32 lb mol/ft" C1= 0.0029 lb mol/ft' C= = O O 13 lb mol/fr' C8=0.0014 lb mol/fr'
CDP6-25 (b)
When 0M is reduced to 1.5, it now takes a T of 0.24 h to achieve a maximum of xylene. Increasing 0M to 10 now requires a T of only 0.08 h
CDP6-25 (e)
To find out the reactor schemes needed, use the attainable region to get these graphs
Using a .PFR would m:ax.imize If we used a of mesirylene ro hydrogen W, the.o we would only have to have a t of .08 hours, So our voulme of che reactor
woutd _ only 3Kü8 ft\. So úur entedng eoneentraions would be JJW5 lb molift.3 of Hydrogen and . 105 lb molt'ft' Mt:SÍ[ylene:.
CDP6-25 (d) First fmd rhe propomonalíry constants of Arrhenius
k:: :.;
= =l
IL2
=
;;;;8.28x 1ri Once that ¡5 by
and error come up w(th rhe
6-63
raee consranrs using
Initial value
~----···~--
§~at:iOi:;:! . .:..
O. 0105
d(cm) /d ( t) =rm ct{chl /d(cl
ezh
d(cxl /d(t)
=rx
d{cme)
/d(t)
0.021
o
o
-rme
o
d(ct) /d t c) ;::rt d(cb} /d ( e) =rb
o
Scale: 1U2 2.:::
rm=-55 . 2 .. cm.. ch? . S
~'L.
zb=Ll, . 2 .. et *ch
......
A.
S
cm
rx=-rm-30.2*cx*ch~.s
ch
rt.=30 .. 2*cx"ch".
eme
ex
5-rb
rh=rm-30.2"'cx•ch~.s-rb
L~C
: . :ice
... - ~.,
......
---
-··
.
:.:!.::
Scale:
Kfil;.
w3
s.:::
..
et c:b
: .. :::.;
......
O,JCO
Variabl.e
I:ü.t:ial.
t;
o o.
cm
ch ex
Y~l~
Ó.!.05
O. 021.
e.ne
o o
cz
o
cb
o
=m
·-O . 083992
=b
o
:::x
0.083992
:'t:.
o
Max!~.
o
Y!':~:!!!: ninimum. ..value
5l
O. Ol.OS
o o
.021. 00506614,
O. 0171.815 0.00477233 0.00J...30452 ·0.002J8715 O. 00)5212'7
:::·h
··O. 083992
0,083992 o 01.46868 .. Q.0126384
rme
O .08J992
0.081992
Final
. . value
o
O .. 51
0.000699836 o .00381847
0 .. 000699836 O . 003818,4 7
o o o
0.00372332
O.Ol71.8lS . O. 004772:D
o ··0 .. 083992
o ··O. 00523484
O. 00.130452 ··O .002J8'7lS 0.00330258 -0.004561.2 o 00)64547
o ·O. 083992
·O. 0126384
0.0126384
0.0126384
The máximum concennation of xylene occurs at t
..
;
= 0 . 19 h.
6-64
--------------------------------------------~-----·--·-----------------------
- - ------ ----
---
---
------
---------
---------
-- ----
CDP6-26 (a) Sean with the mole balances: ['.
.df, ....- = r
_,,,,, ...... ::: f,
----'-'-·:::. r
dF,
1 ----'-;;;;;;
dV
dF, dV
dV
u
9
dFH dV
'
H
Then the rare laws: 1ú = . . . k 1 ('5('' ·ií
1H ;;;:
1'"' 'e··, kI ere ,ii -11 + k 2~'1{ H) 'i'
Finally .the stoichiometry: Putting ali of rhose together ami pm program and aaswers d(cll)/d(V) = rl/vo
ÍJ
into POL YMA"I1I and get the following # cll(0)=0.137
d(c9)/d(V) = (-r2+r3)/vo
# c9(0)=0
d(clO)/d(V) = (-rl+r2)/vo
# clO(O)=O
d(c8)/d(V)
(--r3+r4) /vo
# c8(0)=0
d(c7)/d(V)
( -r4+r5) /vo
# c7(0)=0
d(c6)/d(V}
-r5/vo
# c6(0)=0
d(ch) /d(V)
(rl+r2+r3+r4+r5)/vo
# ch(0)=0.389
S9o c9/(clO+c8+c7+c6+0.0000001) S87 c8/(c7+0.0000001) # S89 c8/(c9+0.0000001) # S910= c9/(c10+0.0000001)# k5*10 kl k5*17.6 # k2 k5*2.7 k3 k5*4.4 # k4 2.1 vo 1 # k5 rl -kl*chAO.S*cll # r2 -k2*chA0.5*c10 # r3 -k3*chA0.5*c9 # r4 -k4*chA0.5*c8 # r5 -k5*chA0.5*c7 # X= 1-cll/cllo #ello= 0.1 V(O)=O # V(f)=0.8
#
6-65
006 0 . 05 O 05 0 . 04 0.03 0.03 0.02 0.02 0 . 01 0.01 0.00 0.00
0.08
0.16
0.24
0 . 32
040
0.48
0.56
0.64
0.72
O
V
• The ratio of hydrogen to pentamethylbenzene is 2.83 and the volume is 0.8 m3. Polymath solution
CDP6-26 (b) The polymath program is the same as the first, we see that the value of cl l(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 S89• To maximize S87 it follows that the volume would be smaller because earlier the reaction ends the less C7 is formed. 67.0 60 . 3 53. 6 46.9 40.2 33 . 5 26.8 20 1
13.4 6.7 O.O
0.00
0 . 67
1.35
2.02
2 . .70
3.37 V
CDP6-26 (e) Individualized solution CDP6-27
6-66
4.05
4 . .72
5 . 40
6.07
6..74
,/di.UC
1
8
O.S v,,,10,00 fhó=lOI.JOO fb=4320 fbo:..;72ú0 Kl.=.0264
k2::. O"l T=40J Poo:;;;1400
RzB.309 l?b= I?ho"' . 6 vo=2 .. 41'\il:7/?oo
Ph=fh•8.309*403ivc < fb<:.,- fb} PC'*tc*3.309"T/vo
y= fe/
:é::>"' · · {kl *Kl "l?b"Ph''. ch= (;::b)
rc~--{.tbJ
S) / ( l +.Kl .. J?b)
-k2:"K2'"l?c"Ph/
(l+K2•?c)
=k2"'K2"Pc*Ph/(l+K2"'b'C)
r· ca
Value 1 9,,5, 2 ·S.08·i 26 0.852181
re
8. 23208
Ua,-· 1 ab fh
re a
u U'.O
ib
- oo kl Kl k2 K2
Pbo
63.5619 8232.08 €l62. 18:
Pho
5~826EJ····l3 i.8•17(';' 12 2 8•!2@ l 4
y·
lUDU l 4 '.320 7200
-9. 93644
oooo
2.7 0 .. 0264 D. 07 CJ. o "i 4U3 1400 8 .. 30S
Pb
84U
vo Pt.
7214. J L2.J64l
o
2.85836 L6Ul.3l
The highest yield occurs at pressures: Pno
= 1400 kPa
Pao = 19452 kPa
6-67
CDP6-28 (a)
Mole Balances: dF -~=r¡,,. +ri,,_ dV dFD
··-=r. dV
2A
Rate Laws : -ru
= k (c . . c!--CcfKt) 1
-ru. : k2 (e
A
CD - CECs!K.2)
-t3c =kaCc
Use these equarions in. POLThIATil. Vru:y P, and T0 to find the optímal conditions. W e determine these to be:
T.., =315.8 K
Po = 160 atm ,:;:;a.=Cr.:.o" ( fa/: t:) • (':o /'l' l
dlfe)/d{v)•·r2a d ( fo) /d
(V)
d( fa) /d(v)
'"2 *rla
6 ::'2a· ::;-3<:;:
"r1a,-z-2a
cd=Ct:o • ( fd/ f;,:¡ "{To/T)
1.4
ce=C~o~(fe/fcJ•(To/T)
5
d(!c)/d(vJ•-::::la•:Jc
O
d( !g) id(v) :.-;-::::Je
O
d ( fd) Id
(v)
4
To=31.S
a
e;r2a
CbJ#.Cto .... { fb/ ft). (To/T) CC•Cto~{fc/fc)•(TO/T)
:::3c"'··k3 "ce * (<.;a1""cb""2,.,,.c-c/L
rl..a==-·kl
::::·Z.a"-k2"'
"o ·~ o ,
T"'TO
kt". 933 "e:,c;; ¡ ( 2. S • (31400/1. 537 ~ ( ::./33()-r /'t') ) ) ) (O. 00::.98 *'.t') "'2"'..ixp (30520/ L 987"' (l/'!'··1./298)) k2=. 6J6wexpi1SOOO/l. 987 .. ( U 300-lJT) l K2=.l03943 *ex.p ( 9834/ l. . 987""{ 1 /T-l/298.))
Ktz13J.667"
k3 :::; . :244 *é,xp (l. 51t28956/ l. 98'7 * ( 1./325-l/'l'}}
6-68
(ca."'cci-ce/K2)
G
200
o
6
5
J.A
9. 99892 lí'.3038
6
5
0.184792
o o
0.682081
o
0~907053 1. . 1375,4
fd
4
4-
O 0010827
O. 00441.766
ft.
30
30
2.7. 4837
2'? ·• 4933
J.15.8
31.5 8
V
te
9 . 99558
14
·ro
31.5.8
315
?o
160
15.:'.l
150
8
Ci 1:34792
1.13754
T
.ns.a
ns a
J 1.5 S
:ns . 11
kl
0.00428571
0.00428571
O .. 00428571
O 00<1,28S7l
2791 12
2791 12
2791,.12
2. aa.1.22
2 881.22
2.88122
Kl k2 X2
2. 831.22 4ú76t.5
40161 S
4076L5
o _0343899
O .. OJ4J.899
0.0343899
6 17855
ó . 17866
6 .17366
l. 23573 2.83337
1 23$7.3
O . OH5214
ó J.7866 O. 0415214
k.J Ct.o
3. 48136 n 15'.3258
2.88337
ce
o
O . 20249.8-
o
cd
0~823-821
0 . 823821
0.000236552
O 000992.ól.5
ce
l..2357]
2 24735
.L 23S73
2.24.593
rJc
rla
--~rj ..
{)0215647
··O .00696389
·O. 00527054
·O - ú440299
-o
-2.93305 V
zo
J:J
(b)
4 00ü4le: ·05 Me::h.anol Synthe$:i.$ ,.., .• ·-·-
'}
1)02:564?
o O. •H82044.5
40
O 68570232
60 80 100
e
82334782
O, S'Hl4271 O . 90829749
l.80
0.89276$42 O .$554.1262 0.$0470055 O 7466007J
200
O .. 6S5.3895
Use the same POLYMATH programas above and vary the ratio ofentering reactants The optima! ratio would be:
fs
hydrogen gas,
11
carbón monoxide, and
!
carbón dioxide
These results are similar to those in part (a) in that the optima} volume is still 100 dm', and the concentration profile is very similar in shape, '111e primary
difference is that the Fe valúes are more than doubled,
6-69
Me::ha.nol syn:hesis eguations : d1Fa)/d(Vl=~la+r2a Q(Fb)/d(Vl=2•rla-~2a-rJc .:l(Fc)/d(V)=-rla+~3c d(Fe)/d(V)=-r2a. ,:!(Fg) /d.(V} =-::le ~(Fd)/d.(V)=::2a.
:tnir..1.al value 8
16
o
6
o o
1'0=315.8
Po=lóO T=To J.CCO
T
t 1
~-
-re
2~
t.lDI
!
'°'
(l.C(X;
O.t>;Q
l
f
z.zn
. . · .--,--..-. -- , . :
10.:c~
ea cee
uc.ccc
u
--+---, ~.oc;:
2:x;.::
kl=.933•exp((2.S•(3l400/l.987•(l/330-l/T)))) Kl=l31667•(0.00198•T)~2•exp(30620/l.987•(1/T-l/298}) k2=0.636•ex;>(l8000/l.987•(l/JOO-l/T}) lt2=-l0.3 943 •ex;> (9834/1. 997• {l/T-1/298 l l
kJ=0.244~ex;,c1.s•2a9ss11.9a1•c11l2s-11T>> Cto=?o/(.OS2•To) Ca=C~o·(Fa/Ft)*(To/T) C~=Cto•(Fb/Ft)•(To/T) lc~Cto•(?c/ft)•(To/T) Cd=Cto•(Fd/:~)•(To/T) Ce=Cto•(Fe/ft)•(To/T) rla=-kl•(ca•cbA2-Cc/:Cl.l
r2a=-k2•(ca•cd-ce•Cb/K2)
llaChM!.ol. ~i.,
.,
~~~ial
V&r!.al:>le
...
value
o
200 8
8 i.s
r.,
kAxiDD val ...
!i
nu
J.Y.tu.miUIII v.al'-le o
1 u~as 7 .i3H9
o
2 OSl.'9
2
'
S .. 9'9"1
r11
s o
N
Q
JO
21 IHS
4 00756
a oaos,ou,
200
! Ul.05 7 118711
a
fe Ye
Fl..n&l val. ...
9 .H,n
o
4 QC75'
o
O 0005903)9
r::
n
To
l!S I
l.5
"°
lf,')
.t60
1'
llS i
ll.5.1
us.,
itl.
o Q04ll5H
Q.004215TI.
O.OOUU.71.
O 00'2157%.
lit~
21'%. l2
279%..12
nn
27'1..12
!el
2 au22
2 981J2
2.,11.22
JO
·~i,1
lcl
Ql4lU9 , l78i6
t. &.\"15~ l.29528
o
C;.o
'-'"'C:l>
2~ Has
us • 160
JU.a 12
2 11122
407U.5
10751..5
Q 03438'9
, nu,
O Ol4lU9 6 ?.7866
a 01uu,
•. 6f,7i4
O. 547917
O 54'7'H7
J lt521
2 22&H
1.
-e
-o
·O
··O.Q75a71
~.01.LUOB
•0.0?'511
1.2H7J
6-70
ltO
~ UlU
C:•
.-i....·
•
o oaounu
O!
.-J1:
,o,n s
ll'S
o o
o o
C:c
S
e
O "7241'9
, ::rass
2 21,sa
O 579213
ns7l
oiou
O 000%.H~a. ! S93'5!.
-o.ou,.n •O . QUUOS
:Met:hanol V
----~···-
o
.
Synthesis Fe
o
'.20
1.2H8568
40
2.0129151
60
2. 4406221
ªº100
2 6366715
120
2. 6292203
1~0 150
2 . 51.93559
2.680932
2 376518
l80
2.2171669
200
2.0519865
CDP6-29 No solution will be given
CDP6-A a) .A+ B -~ C + D
C-+B-"' E+D First, find t'. To do this use the original design equation fer a CSTR: V=F~ Then since FAo
= CA v
0 0
···fA
and r
«
V/v0 then rhe design equation becomes:
. - CAOX
1: - ...... , . -· . ···-·
·-rA Using the rate law and sroichiometry we find: -rA = k· 1 C'Aú (1-X) . Combining ali these and solving for 't when X= .3, CAo
= .l and k, = .412
CAo~ .•.. = _:_l_~} __ = l.04h k1CAO(l-X) .412 * .1 * .7 Once thar has been calcnlated, redo all the mole balances: eAO --· C\ = -r...r t: =
CBo·· · ··Ca=-rat Ce= rcr C0 = r0t
6-71
--------------
-
-------------------
-
-------------------
-- -------
---
Then do the rate laws:
= k1C"
-rA
= k1CA + k2Cc
.... rs
fe = k¡f:A - k2Cc r0
= kC
=
1
A
+ k2Cc
rE k2Cc Combining and rearranging into a function:
/(CA)= o= C; -e"º+ k.C; r
f( C8) =O.= C8 -C80 + (k1CA + k2Cc)'t" f(Cc)= O= (k1CA -· k2Cc}r-Cc f(C0) =O= {k1CA + k2Cc)'t" -· C0 f(CE) =0
= ki~c't"·· ··CE
Plug those into POL Y1fA TH: Soluuoi
6-F Em¡.ations= f(ca}:ca-cao+kl•ca•cau f ( cb} =cb . -cbc .. (kl "'ca+k2 =cc) •cau f(ccJ;(kl"ca-k2•ccJ*cau-cc f(cd}=[k1*ca•k2*cc}*cau-cd f{ca)=k2*cc•cau-ce ca.o=.l
Uar1ab!e
Fe Ce
c.b
0.106382
ce
0.0283726 0.0316i8i 0.0016229! 0.1 0.112 !.Oi O.H
cao Id tau cbo
k2
C~ :::: .028 need Feo= 10 .
=
0.0100015
ce
kl=.412 ta.u=l.04. cbo=.14 k2=.055
v0
ca
cd
= 1.~-= 357gall .028
V= n,0 = 1.04
ro
Uah.ie
3.36le-I5 ·t.02Je-! l 9. 112~-13 . 9. iO:e-· i 3 I.S?Se·! I
o.res
'
h
* 357 = 311.3gal
b)No solution will be given.
CDP6-B A
Xt ---?
8
k.J;,t D »:
~e
Bao:h. reactor; yAO = l
6-72
---- --
-----
-- -
---- - ---
--- -
-- -------- -------
-
---
- -- -- ------------------·---·----------------·-------------
k1 = 0.01. ~c·l •
(a)
NAo«=-rA dt CAo
a;-=
t
= L5 ruin= 90 s
V
= k1 C,,.,o ( i -X}
k1 CA-
{¡:1-Xt·) = kt t __j__ = et,t In
l·X1
e,.,, = CAO ( 1-Xl) = CA.o e·k¡t . c6. = e··Í:¡t = e·{0.01}(90) = 0.41 CAO (b)
A
-, B
B
"""7
e
re
= k:2 Ca
X2
B
"""?
D
ro =k3 Cs
X3
·ta= -rs1
rs1 = -rA :::: k1
+rc+ rn
--~~ = -k1 CAo
::<1~8-=--0.002[1
-
X1
= - k1 CA+ k2 Ca+ k:3 Ca
e·k,t +
;·dCa. = -{0.01)(0.2) dt
CA
{k2 + k3) Ca
e·-0.0lt
+ (0.003 + 0.002) CB
e·-0°ttJ +0.005 Ca
U síng a Runge- Kntta Gill numerical solurion, we find that. for t
= 2 min = 120 s. CB =
0.136 gmol/dm3.. (e)
!!~=,e=
ki Ca
;
!!~·= k2 Ca
; Ce~ k2
J.'c.dt
From the solurion in part (b), we have valúes of C:s at intervals as srnall as 5 sec, so we can use Simpsoa's rule to obtaía Ce: t
= 1 min = 6-0 sec. Ce=
0.003
{lf-)co + 4 (0.0449)
+ 2(0.01s2J + 4<0.102) + o.11ss]
Ce = 0.0129 gmoifdm3
6-73
e= 2 roin = 120 sec
Ce=
0.003
(f)(o + 4 (0.0782) + 2{0.1183} + 4(0.134) + o.135.S]
Ce = 0.0366 gmoL'dm3 {d)
0. .20. G.18 016 'J.14
0.12 'J'flOllai. dm..
0.1 O O.OS
0.06 0.04 •102
Timot (i)
The concentrntion of B is highest at t = 100 sec, where its value is 0 ..1363 gmol/dm3
teJ
Balance on A: FAo- FA +rAV =0 U CAo -
U
CA ~ k1 CA V
CAo - CA • k1 CA 1: = O
Balance on B:
= o·
FBo - FB + rs V = O -u
Ca + (k¡
Ca) V= O
CA ·• k1 Ca - k3
·Ca + k1 CA t - k2 C.a -r: - k3 CB 't = O
CAo - CA (1 + k1 1:) :..: O
CB (1 + k2 t + k3 "t) ;:-;. k1
CA = l ---CAo O+ k¡ -t)
CA= Ca
't
CA
(1 + k2_t+ k3 't) k1
t
6-74
·--------·----------·---
---
----
---------
------
-
------
.fA_ = __ l_. za Ca (1 + k:2 't + i.:1 .ac) CAo
~= CAO
l + k1 't
CAO k:1 't
k 't
(l
+ kt t) (l + k:2 't + k3 't)
't(s)
10
so
CBJ'CAó
0.08
0.26
Ca.m~ oecurs ar. t
100 0.33
200 0.33
150 0.34
300 0.3
:500 0..23
= 150 A
CDP6-C First, fínd the values for k,
= 2989er-ios11(().H9!it·100»
k1
k z = 9466e{-rs w(.0019a,•1oon
= L2 ::::
k.J = l l 127e("·LS.06/(0019ll7•700)) Isobutylene == r Methacroleín = .M
In thís problem
.18
::::;: _22
CO,=D
ce/== e
Oxygen== O Toe mole balances for these species are as follows: F10 .... Ff (--·r't¡ -· r2r - r31)V
=
F.'>{=
r¡M V
F0 = rw V Fe==
r30V
Oxygen is in exeess so we will assume that F., = Fo0· Toe rate Iaws far rhese reactions are as folíows: ·· · ··Ii_r = --1¡0 = r1M = kLC1C00 -4r.,. l -r21 = ··2.sº·: : 4 rm k2C1Coo
=
6-75
---------------------------
-
--- -- - - - -
-
-- -- ----
-
---- ---- - -- ---
-
----
-- -
-
- - -- - --
--------- -------
-----·---~----------
----
-------
-··r = JI
4rJO = . !.r -17 4 3C
= k l Ct.C OO
Combine all of fuese and come up wirh the following: C10 -C1 = (k1C1C00 + k2C1C00 + k3C1C00)'r
C00
-
C0 = (k1C1C00 + ~¡-k2CrCoo + }} k3C1C00 )'í
C.\,f = k1C1Coc/r C0 =4 * k2(\C00-r Ce = 4 * k3C1C00
t'
Before plugging into POL YMA TH, evaluate the parameters and derive equaríoas for conversion.
e . = .J?. . . = -.... .?
= 0.034 RT .082 * 700 Cm= Y AOCr .1 * .034 = 0.()034 CCQ =0.031 1
=
~;!l..,- ,el e, =e . . '.M.
X1 =
X
u
C
'to
X =-~:'.!L 21
4. C·¡o
Xlt =X1
······Xt1 ······X21
Then plug into POL Ylv1ATH to get the following: Egua
I,ütial
ti,~.:..
f (cil
:=ci· .. ci.o+
(kl. "ci. "coo+k.2 "ci. *coo+k.J
*ci •coa)
O .. 001
"tau
f (cml =kl *c:i. •coo"t.:au···cm
O ,.001
f (cdl =4"k2 "ci"'coo •c.au.··cd f (ce) =4 •ci •coa" ta.u-ce f {col =co···coo+- (lc1.•ci."coo+2S/
0.004. 4 .. k2"'ci."coo+17
cio=.0034 kl.=l.
2
coo=.0:31. k,2;; .. 1,8
k3::.22 t:;.au=lO xl.=cm/ci.o ;x:2c:=. 25*cd/ci.o x={cio·ci.)
/ci
x.3=x··Xl .... x:z
6-76
/2 "'k:3 •ci
... cool
•cau
0.004 O .• 01
.value
.Sot ut
1
en
'º
Ualue
e: l
0.00227273
cm
0.00084-5"155 O.D00507273 0.00281818
Cd
ce: c;o
0.0034 1.2
C:00
o. 18 c. 2:2
k:2
10 D.2
xl
x2
0.0372995 fL.496 0.210037
X
x3
X1
z
-l.775e-12 1..0112-12 -1. 381 e-12 3.. 835e·i3
-?)---· X 11 -0, -
0.496
X21
= 0.037
X11 == 0.21
CDP6-D A-··tB C-:,,D+E Fr == 1 O lbmol/sec
«c.
~-:::::TA:::::
F.4 =CAVO
e10 = :. "o
Q:.:?..
0.73*900
= 7 .61 * 1 o-~
= 13140
V= 1005ft3 See POL YMi1.]1I solution below
c¿guati9ns:
r::.itiad va.l;i+s::
d(fa.)/d(V);,r.a d ¡ fe) /d(V) =zc d(fb)/d{V)"' ra d(fd)/d(V):rc
5 5
o
o o
d(fe)/d(V)<=«rc k.1=10
k2=.03
6-77
---
....
---------------------·-----
-- --- ------
~
--
-- - ------- ---
---- ---- -------------------
ft=:fa~·fc +·fh+fd+ fe fao=:5 cto""S/(900*.73) fto;lQ :rc=-k2 X"' { fao·~fa)
J fao
avo=fto/cto ca=cto*
( fa/ ft}
:ra=-kl*ca
yariabl~
!n:J:3;.ial. y~ue, O
V
Ma;ximum,,~valu.e 105
o
fk"1a.l value
105
fe
5 5
5 5
2,49621 L65
l.BS
fb
O
2.5037.9
fd
O O 10 0.03
3.15 3.15 10 0.03 13 15
o o o
2.50379 3.15 3.15 10 O 03 13.15
fa
fe kl
xa
10 5
::ao cto
10 0.03 10
5
2.4.9621
5
O. 00161035
O . 00761035
ft:.o
10
10
re
··O 03
O
· O 03 0.500758
o
(l.
avo
1314
1314
131.4
1314
ca
D.00380518
0.00JBDS18
0.001.44464
O 001A44'54
0.00761035 10 -0.03
CDP6-E a) Usíng the equation for the equilibrium constants:
W e can come up with the eqnations for CA, Co, and Cx.
c:A ...... cCc:JL. Ke1C1J
CD
= .!!l~Ac_»_ Ce
C\ = ~:.:tf.CC~f!.. Cy
The test can be found with stoichiometry .
Cy =Cx
e11 :;,: c»o -·CD······ e, Ce
= CAO eA -
Cx
6-78
0.00761035 10
~0.03 500758
POLYlvlATH 6--s a
6··5
~qµat:ions :. f (ca) "'Ca---cc•cd/ ne. •cb) E (cdl c;cd--K1 •ca•cb/cc
Uar,able ca ~
l:{cx) =cx--·K2""cc •cb/cy
h ,.··
Ualwe Cl.0306182
J.[J49e-
D.714386
!.084e·· t8 ·· L l05e-· l 6
o. l 09652
ex
K2aal
Kl
c.y=cx
i()
---~-·-"-··-·---"-·-··"···-·
cd
E<:1=4
"'
4
K2
cbo"'LS
c.ao e L .. 5 cc:=cao--·ca-c:x Jea.o
X"(C$;0'-•C3.)
Cl.J
0.714°386
cbo
L5 L5
ce
cb"'cbo-cd--cy
0.754995 0.979588 0.675961
X
cb
e, =0.68
e, =0.0306
e, =ü.71
(\ =0.71
W e also find that X = 0 . .98 b) With the new equation we must find the new equilibrium equatíons,
e, = K,i~,1.Cx
e, = .!::lts;:~~~<\
C
'D -
K,1C,1.CB - ...
Ce -
The rest are the same except for: Cx. =Cr -···Cz
CA= CAo
CD
flczJ=ci:-k3"ca"cx :E(cyJ=cy,.k2•cc"cb/cx f. ( cdl =ocd~kt•ca "cb/cc k;J ,,,5
cx;c·y~ . . cz k2:,;.:l" 51 cao=l.
S
c!:lo=LS
ca=cao--cd--cz:
cb:::cbo-cd-cy
.c, Ua,,-
Val
cz
o . 890725
cy cd
L07384 CL 2681 O l
ex
CJ. CJ831 135
caoeoo T sdz
k3 ee k2 cb kl X
ce
!.5 1.5 300
Cl.2706 l l 1.08389 48.4255 o. 24 l J 74 o. 480271 o. r 5806 8.26888 0.839217 L 17571 1'
L
-~
-r ·•
6-79
,,
j.
.... I"" ......
·"l:;J:::,
10
.OS2e--13 Ll44e·l2 2 .607c,-15
19
Wefind that e, =0.65
e, =0.48
e, =0.91
CD =0.37
CB =0.22
Ce =0.76
S0z =0.57
Syz = 1.41
e, =0.27
X =0.68
s.; =2.8
e) When the temperature is raised from 300 K to 500 K. Scx goes down, Soz goes
up and Syz goes up.
d) First find the proportionality constants from the Arrhenius equation. 0.002 ==
A,e·il0000/19SViOO)
A1 =38603 _ 4. -( ;CXXJO/l 987 • )00)
O . 06 - , ie A 2 --
-'j . '24 , X
1013
0.3 = A,e··P.000011 ssr-soo)
= 2.16
A3
X
1021
Once those are known, come up with equations for the equations in terms of space-time 't.
r - _.... C,rn .. . -CA . _ . k, + k2CA + k:.;C~ tk; +-r:k:C~ +· rk.JC~ - C,.10 + CA . . ..........
e'1
j
,
,
=O
.. -···············-·············"""''····. ')
~,····--···,
.
-·-{ 1.k2 + l) + ~( 1:k2 + 1 )- ·· 4(rk, )( tk, - CAo) = -··-·--·--- . ---- _ _ 2rk.,
··C3 .. -r = ......
--Cr ~' -· k·~(,;
1: :::: ·----~········ ... -c-
---·k,
e, = k; r
C8 = k-.,.1CA e, = k3i-C1 Use EXCEL solver in order to find the temperature that maxirnizes C6 T Ca 312.5336841
0.016513937
ex
k1
0.003919233
0.039192325
k2 0.230891512
Cb 0.038129279
k3 2.260443602
Cy
0.006164458
CDP6-F (a) Mole balances:
6-80
-------~---~~----
- ------------------
-
------
--
-
------- ------ ----- -- --
--
---- ---
-
----
--
-------
-----------
---
------ ----
---- ------
-------
dFA/dV = -r0-ru
dFofdV = ro dFu/dV =ru Rate laws:
= k1C; ru = k2CA rv
k¡ = 15 ft3/lbmoLs k2= 0.015
s'
Stoichiometry: CA= CT0(FA/FT), F1 =FA+ Fo+ Fu Cost = 60F8 - 15Fc -lOFAo Using these equations in polymath to find the necessary volume to maximize Cost: V= 1425 ft3 POLYMATH Results Calculated values of the DEQ variables Variable initial value minimal value o V o fu o o fa 0.06705 0 . 0018652 fd o o kl 15 15 k2 0.015 0.015 fao 0.06705 0.06705 ft 0.06705 0.06705 0.00447 cao 0.00447 ca 0.00447 l.243E-04 -0.6705 Cost -0.6705 rd 2.997E-04 2. 319E-07 ru 6.705E-05 l.865E-06
maximal value 2000 0.0237312 0.06705 0.0414537 15 O . 015 0.06705 0.06705 0.00447 0.00447 l. 46684 2.997E-04 6. 705E--05
final value 2000 0.0237312 0.0018652 0.0414537 15 0.015 0.06705 0.06705 0.00447 1.243E-04 1.4607519 2°.319E-07 l.865E-06
ODE Report (RKF45) Differential equations as entered by the user ( l J d(fu)/d(V) = ru ( 2 J d(fa)/d(V) = -ru-rd l 3 J d(fd)/d(V) = rd Explicit equations as entered by the user (1] k1=15 [2] k2=0 . 015 [ 3 J fao = 0.06705 ( 4 l ft fu+fa+fd ( 5 J cao = 0 . 00447 l 6 J ca = cao*(fa/ft) (71 Cost = (60*fd-15*fu-10*fao) [ 8 J rd = k1 *(ca"2) l 9 J ru = k2*ca lndependent variable variable name : V initial value : O final value : 2000
=
CDP6-F (b) Mole balances: FA = F Ao + rAV
Fo = ro V
Rate laws: rA = -r0
k,
-
ru
= k;o exp
Cost = vo(60F8
--
rD
Fu = ruV
= k1C;
ru
= k2CA
1 ) ( -¡tE.)(1T - 919.67
15Fc --lOFAo)
Use these equations in Polymath and vary T from 860 ° R to 1160
° R and find maximum value of cost,
6-81
--------·--------------·------
----------·--------------
------
--
-------- ---
-
---
-
-
-- -------
-------
-------------
-
Cost steadily rises with temperature and reaches a maximum at 970
.fOLYMATH Results NLES Solution Variable ca cd cu cao El R T k2o E2 klo vo V tau cost kl k2 rd ru ra
Value 0.0023213 0.0016225 5.262E-04 0.00447 10000 1.987 970 0.015 2.0E+04 15 15 400 26.666667 0.6713603 11.292187 0.0085009 6.085E-05 l.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 [ l J f(ca) = ca-cao-ra*tau = O [ 2 J f(cd) = cd-rd*tau = O [ 3 J f(cu) = cu-nrtau = O
Explicit equations l1 l cao = 0.00447 [2 l E1 = 10000
[3 J R = 1.987 [41 T= 970 l5 J k2o = 0.015 [ 6 l E2 = 20000 [7 J k1o = 15 [8]
V0=15
[9J V= 400 [ 1 o J tau = V/vo [llJ cost = vo*(60*cd-15*cu-10*cao) [12J k1 = k1o*exp((E1/R)*(1/T-1/919.67)) l 13 J k2 k2o*exp((E2/R)*(1/T-1/919.67)) [ 14 J rd = k1 *(caA2) [15] ru = k2*ca [ 16 J ra = -k1 *(caA2)-k2*ca
=
CDP6-G3
6-82
° R(510.33 ° F) .
ac, ···;·-·:;:;:
Mole Balances :
r ...
dt
dt
:;:;: ft
dC .......... : . . = rf dt
RateLaws:
C'E<>
Stoichiorne try :
= ('-4k J.Y • mol/dm um '-
--10 rno11·d·· ;;;:: )4' · m3
frorn Henry' s Law :
C\,, = 5. 9 MPa (o .058 kmol/m' Use these equations in the following POLYMATH program Ce, C,, and CF as a funcrion of time
o o o
í
d(ce)/d(t:l=:::e /d(t:)c::rf
w~1.o kl;Q.000468 Ka=:22 . 76 k2;Q.000227 k4;;() 00147 k3"'0
graphs of C,., c•.
540
=rb
d cc ) /d(t:.) =zc d(cí:)
to generate
O 3422
d(ca)/d(t)""ra d(cb) /d(tl
· MPa) = O .3422 mol/dm '
00282
rl=w*kl*ca*cb/(l+Ka•ca) r2=w*k2*ca*cc/ll+Ka•ca) r4-=w .. k4 *ca*cei
( l+Ka "cal
20
6-83
YE:~ c.
o
20
o
20
ca
O 3422
0.3422
9 1.B5l7e-20
9 185'.Ue-·20
e~
540
540
539658
539 65$
o
0.205675
o o
O 116398
ce
o
0.311641
ce
o
O 136398
o
6 .. 37928e·OS
10
10
10
O 000468
10 O 000468
0.0004.68
0 . 000468
Ka
22 .76
22.,76
22.76
22.76
l<.2
0 .. 000227
0.000221
0 . 00022'7
k4.
0.001.47
0.0014'7
0.00147
k3
0.00282
0.00282
0.00282
0.000227 0.00147 0.00282
rJ
o
0.00878821
o
rl
0.0984025
0.0984.025
2.31981e
cf kl
6 . 37921:le-·05
0.00580002 . 19
2.31981.e-19 4 .28838e··23
x:·4
o
4.96363e
rb
··tL0984025
--2 . J l.9Sle-l9
re
o
0.00878695
o o -o o
rf
o
2 .. 07178e··05
o
2 .270Sle-·22
.. 2 .32208e
... 0.091H025
··2.32208e· 19
o
J::'2
ra
L62le-OS
O, 0984025
0.0984025
~. •;: ~.;.
ca ce ce
:.n: :U•!
a, ,,::
t
.. o6
.. 19
0 .. 098i,HJ2S
Ca, Ce, Ce
V$
l .84167e··22 0984025
·2 31981.e·•l9 0.00560002
-o , 00834598
-o. 00580002
Cb VS
t
t
l l
T
:l.J.&O:
1 & ooc
............ 'i
2iLJOC
Cf vs . . , .
1z.,.::cc
CDP6-H
6-84
t
Thc equilibrium
constant fer the reacuon
tr OL -;.:~. cs--OL
can be estimarcd as a function of rcmperarure frorn rhe mole fraction data below lOOºC.
T
323
375
Yes
.8 .2
.25
Kp
.__.)
.33
.75
-, -
!n Kr
..
lOOO!f
.3.09
LJ9
-t.r l 2.63
}
.: _~39 :f.:_l:I_~¿ x 1000::: -683 ,.09 --- 2.68 6.H
=< ••
= ~HE_ R
L3 57 .sal, roo!
CDP6-I 2, 9-17 CDP6-J CDP6-K 2, 9-13 CDP6-L Given: Líquid feed to CSTR
Ga.:seou:s D
LJ.twd Peed
-----,.
CMJ .. 0.4 tm.clll
e ro .... o .4 r;m.clll "jo - .2 llmm . Liq uid P:rod ua
V~ 1201
6-85
With the follcwing reactíon sequence
kt
= k ¡ C.,. -r2 = k2 CA
C
A -·~
k2
A ........,,. B k3 B -,. C k,¿
B +Di
F-,.
k:1
= 0 . 0 I min-1
k:2
= ü.02 min•1
·r3 = k3 Ca
k;3
= 0.07 min·1
-n = k... C}
k,.
= 0.50 Vgmol • rnin
. r¡
k2 k3 {a) Sínce C is an ~ prcduct, formed through intermediare B by either A-+ B ......,.. Cor k.&. k3 k.1 F -+ B + D i -.;, C. or din:ctly from A ._,...,. C. the máximum ccncemrarioa of C occurs
Cm.u= C11.o + C¡:o., (wíth e= O).
when ali A and F have been converted to C: (b)
Borh A and F are only deecmpcsed by the above seheme:
Balance on A: ~ C.-.o • Uo CA "',C.rA) V= (~r1 • r2} V ={k1 + k2) CA V
c...,o :.C, = {k.1 + k2) y
uo
CA
or CA =
C~-l + {k1 : kzJ -r
""(kt + kz},:
=
0.4gmol / 1 . . 1 + (0.0l + o.02) min:' (21J~-}
= ._0.4 gmol / l
H· (0.03) (60)
e,..= 0.143 gmoi 11 Balance on F:
uo C¡:,J • Vo C¡;
= (-rF) Y_.,,,~ .
wnere
.
C~ V V
1: ::: --
Uo
""
t:.r\
v..;·
•
mm
Solving we get: CF = O.Igmol/I BalanceforB:
O-v0C8
= (-r8)V = (k3C8
-k2CA +k4C;)V
(k2C A+ k4Cj )r (O.Olmin-·1x0.143gmal /l + 0.51.moZ-1 min-·1 O.lgmol /l2)60min C8 = 1 + k3r -= 1+0.07min-160min CB
= 0.0907 gmol f l
BalanceforC:
Ce
= (--r~)V = (-k3C8 -k1CA)V + k1CA)T = (0.07min-1x0.0907gmal/l
O-v0Cc
= (k3C8
+O.Olmin-1x0.143gmal/l)60min
Ce= 0.467 gmol/ l Mole fraction of C =
CA+C8+Ce+CF
=
0.467 0.143 + 0.09 + 0.467 + 0.1
6-86
-~~z. =
=0
0.8
0.583
CDP6-M 4A + SB -) 4C + 6D
-,¡A
= k!ACAC;
2A +l.SB-) E+ 3D
-r2A
= k2ACACB
2C + B-) 2F
-'"3B
4A + 6C -) SE + 6D
-r4c
= k3BC~CB = k4cCcC1'3
Rate laws:
-rA = r1A + r2A + (2/3)r4c -re = -rIA + 2r3n + r4c -rE = -:0.5r2A - (5/6)r4c
-rn = l.25r1A + 0.75r2A +r3n -rD = -l.5r1A - l.5r2A - f4c -rp = -2r38
Using these equations inpolymath to find the exiting molar flow rates.
POLYMA TH Results NLES Report (safenewt) Nonlinear equations f(ca) = vo*ca-fao-ra*W = O f(cb) = vo*cb-fbo-rb*W = O [3] f(cc) = vo*cc-rc*W = O [4] f(cd) = vo*cd-rd*W = O [5] f(ce) = vo*ce-re*W = O (6] f(cf) = vo*cf-rf*W = O [l) [2]
Explicit equations [1] [2)
[ 3] [4] [5) (6)
[7 J [8]
[9] [10] [11] [12] (13)
(14) [15] (16]
(17) (18) [19) [20] (21] [22) [23) [24) [25)
= 10 tao= 10 W=3 fbo = 10 rho = 0 . 0012 k1 = 5 k2 = 2 k3 = 10 k4=5 fa= vo*ca fb = vo*cb fe= vo*cc fd = vo*cd fe= vo*ce ff = vo*cf rl = rho*k1 *ca*{cb"2) r2 = rho*k2*ca*cb r3 = rho*k3*cb*(cc"2) r4 = rho*k4*cc*(ca"(2/3)) rf = 2*r3 re = 0.5*r2+(5/6)*r4 ra = -r1 +r2-(2/3)*r4 rb = -1 . 25*r1-0. 75*r2-r3 re = r1 -2*r3-r4 rd = 1.5*r1 + 1.5*r2+r4 VO
NLES Solutiou Variable ca cb ce cd ce cf va
fao w fbo rho kl k2 k3 k4 fa fb fe fd fe ff rl r2 r3 r4 rf re ra rb re rd
Value 0.998927 0.997227 0.0017849 0.0037612 3. 613E-04 1. 285E-12 10
f(x) -1. 05 6E-15 1. 08 2E-15 -1. 42 2E-16 -1.041E-16 -1. 73 5E-18 1.254E-13
!ni Guess 1 1
o o o o
10
3 10
0.0012 5 2 10
5 9.98927 9. 9722697 O. 0178487 o. 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
···-·-----------------------·---------------
-------- -------··--·------------------------------
---···---··--·····---·---------···---·--···---
--
---·····----·--·-------- ------------------ - --------·--·--·----------------------------·-------------------
Y AE
=
FE FAo -FA
y BF
=
FF
(b)
y AC =
FBo -F8
= 3.6xlo-3 = 0.336 10-9.989
- 2.29xlo-7
10-9.973
= 8.25x10-6
Fe - 0.0178 = 1.663 FAo - 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 IJI will remain same if CS2 concentration changes. If concentration of M increases the slope of line will decrease.
P7-1 (b) Example 7 -2 For t = O to t = 0..35 sec, PSSH is not valid as steady state not reached . And at low temperature PSSH results show greatest disparity. See Polymath prograrn P7-I-b.pol. POL YMA TH Results Calculated values of the DEQ variables Variable t el e2 e6 e4 e7 e3 es es ePS ePl k5
initial value
T
kl k2 k4 k3
o
0.1
mí.n.i.ma L value
o
2.109E-04
o o o o o o o o
0.1 3.98E+09 1000 0.0014964 2.283E+06 9 . 53E+08 5.71E+04
o o o o o o o o
2.166E-04 3.98E+09 1000 0.0014964 2.283E+06 9.53E+08 5 . 71E+04
maximal value 12 0.1 l.311E-09 3.602E-09 2.665E-07 0.0979179 0.0012475 0.0979179 6.237E-04 0.0979123 O.1 3.98E+09 1000 0.0014964 2.283E+06 9.53E+08 5.71E+04
final value 12 2.109E-04 1. 311E-09 3.602E-09 1.276E-08 0 . 0979179 0.0012475 0.0979179 6.237E-04 O . 0979123 2 . 166E-04 3.98E+09 1000 0.0014964 2 . 283E+06 9.53E+08 5.71E+04
ODE Report (STIFF) Differential equations as entered by the user f 1J d(C1 )/d(t) = -k1 *C1-k2*C1 *C2-k4*C1 *C6 [ 2 J d(C2)/d(t) = 2*k1 *C1-k2*C1 *C2 f 3 J d(C6)/d(t) = k3*C4-k4 *C6*C 1 [ 4 J d(C4)/d(t) = k2*C1 *C2-k3*C4+k4*C6*C1 ·· k5*C4A2 [ 5 J d(C7)/d(t) k4*C1 *C6 [ 6 J d(C3)/d(t) = k2*C1 *C2 ¡ 7 J d(C5)/d(t) k3*C4 r e 1 d(CS)/d(t) = o.s*k5*C4A2 [ 9 J d(CP5)/d(t) = k3*(2*k1/k5)/\Q.5*CP1A0.5 f lo J d(CP1 )/d(t) -k1 *CP1-2*k1 *CP1(k3*(2*k1 /k5)/\Q.5)*(CP 1 /\Q.5)
3 2e-9 2 .4e-9
=
L6e-9
=
8 Oe-10
=
· · · ··-·-·· ·· ·"'·----·--
O.Oe+O ,.__-~-O.O 24
P7-1 (e) Example 7-3 The inhibitor shows competitive inhibition.
See Polyrnath program P7-·l·c.poL
7-1
-------~---
-~-------
- -
-·-
-------·---------·-·-·-----------
-----------------· ·---
----···-····----------------------
- ----
-----
4.8 t
7..2
9.6
12 O
0.40.---------------,
0.32 0.24
- r_inhibitor r
0.16 O.OS
o.oo~--------~--~---'
5.0e-3 6.0e-3 7.0e-l:ur\De-3 9.0e-3 LOe-2
P7-1 (d) Example7-4 l)Now
Curea
= 0.00lmol/dm3 and t = 10 min = 600 sec. t
ln(-1-) +
= _.!_y_
1-X
VMAX
_ 600 sec -
_Cu,ea X VMAX
0.0266mol / dm" ln ( -- 1 ) +--'--(o.OOlmol/ dm3...,___ )x 0.000266mol / s.dm' 1- X 0.000266mol / s.dm'
Solving, we get X= 0 . 9974. 2) For CSTR, t = t = 461.7sec
<:«
r=-----
-r~,ea
-·-· ru,ea
·- ru,ea = ~
=
VMAXCu,ea K C M + urea
Cu,eaX T
0.000266mol / s.dnt' xO.lmol / dm3(1-X) _ (o.lmol/ dm" )x . - --·---·--0.0266mol / dm" + O.lmol / dm" (1- X) 461.7sec
Solving, we get X = 0 . 675
See Polymath program P? . l-c.pol. POLYMATH Results NLE Solution Variable X
Value 0.6751896
f(x) Ini Guess -8.062E-10 0.5
NLE Report (safenewt) Nonlinear equations [ 1 J f(X) = 0.0000266*(1-X)/(0.0266+0.1 3) ForPFR,
T=
Cfo
<:
C""'
-
*(1-X))-0.1 *X/461. 7 = O
and
Cu,ea
r
7-2
= Cu,eaO (1- X)
r=-- kM
=>
1 n( -- 1
1- X
VMAx
<:»
) +---
VMAX
Same as batch reactor, but t replaced by r
P7-1 (e) Example
X= 0. 8
7-5
y
·=-!1C5=-(238.7-245)=218 SIP tlCP 5.03-2.14
YPIS
.
/ g g
1 1 =--=--=0.46g/g
2.18
Ys,P Ys,c+P
1
1
C+PIS
0.075+0.46
=y
= 1.87g I g
Yes there is disparity as substrate is also used in maintenance .
P7-1 ( f) Example
7-6 1) if we go for 24 hrs, fermentation will stop at 132 hrs as Cp =
e,' .
See Polymath program P7-1-f-l.po1. POL Yl\!IATH Results Calculated values of the DEO variables Variable t Ce
initial
o
value
minimal
o
value
es
1
1
250
39.292786 0.01 12.5 5.6
m
0.01 12.5 5.6 l. 7 O . 03 0.33 0.03 0.33 0.3277712
o
Cp rd Ysc Ypc Ks
umax rsm kobs rg
o
l. 7
maximal value 13.2 16.558613 250 92 . 981376 0.16559 12.5 5.6
final value 13.2 16.550651 39.292786 92.981376 0.1655065 12.5 5.6
0.03 0.33 0.4967701 0.33 2.1455962
0.03 O . 33 0.4965195 0.0039386 0.0624825
l. 7
0.03 0.33 0.03 0.0039386 0.0624825
l. 7
ODE Report (RKF45) Differential equations as entered by the user [ 1l d(Cc)/d(t) = rg-rd [ 2 J d(Cs)/d(t) = Ysc*(-rg)-rsm l 3 l d(Cp)/d(t) = rg*Ypc Explicit equations as entered by the user [ 1 J rd = Cc*0.01 [2 l Ysc 1/0.08 [3J Ypc = 5.6 [4J Ks = 1.7 [5J m = 0.03 [ 6 l umax = 0.33 [7) rsm m*Cc [ 8 J kobs (umax*(1-Cp/93)"0.52) [ 9 l rg = kobs*Cc*Cs/(Ks+Cs)
=
= =
7-3
-·-·--·--·---
-----------------------------·--·-·--·
-------·------···-----------------··-·
---·-····---··---------····----·-·····-------·----··--------···--·-
----------
------·-·-----·---··---------
---------------------·-·----------
2)Semi-Batch reactor:
See Polymath program P7-1-·f-2.po1. POL YMATH Results Calculated values of the DEO variables Variable t Ce es Cp rd Ysc Ypc Ks
m
umax rsm kobs rg eso vo Vo V
initial value
o
1.0E-04 l.. OE-04
o
l. OE-06
12.5 5.6 l. 7 0.03 0.33 3.0E-06 O . 33 1. 941E--09 5 0.5 1 1
minimal value
o
1.0E-04 l. OE-04
o
l. OE-06
12 . 5 5. 6 l. 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 l. 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 [ l J d(Cc)/d(t) = rg-rd ¡ 2 J d(Cs)/d(t) = vo*CsoN + Ysc*(-rg)-rsm [ 3 J d(Cp)/d(t) = rg*Ypc Explicit equations as entered by the user [ 1 J rd = Cc*0.01 [2 J Ysc = 1/0.08 [3J Ypc = 5.6 (4J Ks=1 . 7 [5J m = 0 . 03 r 61 umax = 0.33 (71 rsm = m*Cc [ 8 J kobs = (umax*(1-Cp/93)"0.52) [ 9 J rg = kobs*Cc*Cs/(Ks+Cs) [10] Cso = 5 [11] vo = 0. 5 [12 l Vo = 1 r 13 J V = Vo+vo*t
7-4
final value 24 0.0474697 12.206266 0.2748298 4.747E-04 12.5 5.6 l. 7
0.03 0.33 0.0014241 0.3294925 0.0137289 5
0.5 1
13
0.30
15
0 . 24
12
..
018
0 . 12
9
6
)
3
0.06 000
o.o
9. 6
4.8
t
14.4
19.2
o
24. 0
O.O
4.8
9,6 t
19.2
144
24.0
3)Changes from part(2)
See Polymath program P7-L-f-3.poJ. POL YMA TH Results Calculated values of the DEQ variables Variable t Ce Cs Cp rd Ysc Ypc Ks
m
umax rsm kobs Ki Cso VO
Vo V rg
initial value
o
l.OE-04 1.0E-04
o
1.0E-06 12.5 5.6 l. 7 0.03 0.33 3.0E-06 0.33 0.7 5 0.5 1 1 1.941E-09
minimal value
o
l. OE-04
1.0E-04
o
l.OE-06 12 . 5 5.6 l. 7 0.03 0.33 3.0E-06 O . 3299991 0.7 5 0.5 1 1 1.941E-09
maximal value 24 l. 514E-04 12.823709 4.669E-04 l. 514E-06 12.5 5.6 l.. 7 0.03 O . 33 4.541E-06 0.33 0.7 5 0.5 1 13 8.215E-06
final value 24 1.514E-04 12.823709 4.669E-04 1.514E-06 12.5 5.6 l.. 7 0.03 0.33 4.541E-06 0.3299991 0.7 5 0.5 1 13 2.568E-06
ODE Report (RKF45) 15
Differential equations as entered by the user [l l d(Cc)/d(t) = rg-rd [ 2 J d(Cs)/d(t) = vo*CsoN + Ysc*( -rg)-rsm [ 3 J d(Cp)/d(t) = rg*Ypc
12 9
Explicit equations as entered by the user [ 1 J rd = Cc*0 . 01 [2) Ysc=1/0.08 [3 l Ypc 5.6 [4J Ks=1.7 [5J m 0 . 03 [ 6 J umax = 0 . 33 [7 J rsm = m*Cc
6
=
=
3
o o.o
4. 8
9.6 t
144
19 . 2
24 . 0
7-5
- --
-
-----------
- ----
--
- - --- --- -- - ---------------------------- -------·----·-··----------·-·
-------··--------·--··------------ --··-·--------------···---·-- ----·--------·· · ··-----
--------····----------··-····----------·--···-
-----·---·-----·-----------------
. -~----··---
[8J
kobs:;:;: (umax*(1-Cp/93)A().52) [9J Ki 0.7 [lOJ Cso = 5 [11) vo = 0.5 [12] Vo=1 [ 13 J V= Vo+vo*t [ 14J rg = kobs*Cc*Cs/(Ks+Cs+Cs"2/Ki)
=
1.60e-4
5.0e-4
1.48e-4
4.0e-4
l.36e-4 ·
3.0e-4
1.24e-4
2 . 0e-4
1.12e-4
LOe-4
-·--~--'-
1.00e-4
o.o
9.6 t
4.8
14.4
19.2
6
8
24.0
O.Oe+o
GJ
1<---~-~----~
o.o
4. 8
4)After 9.67 hrs, Cs = O. 300 .--------· 240 180
- C~c C.s
120
Ci
60
o o
2
4
t
10
P7-1 (g) Individualized solution P7-1 (h) Individualized solution P7-1 (í) Individualized solution P7-1 (j) Individualized solution P7-1 (k) Individualized solution P7-1 (1) Individualized solution P7-1 (m) Individualized solution P7 -1 ( n) Individualized solution P7-2
Solution is in the decoding algorithm given with the modules.
7-6
9.6 t
14.4
19.2
24. 0
P7-3 Buming: "'·-~
1 \, CO·+OH·
substimti.ng
for{O:) .
71- (H~OJ2k,{O-J+k.{H·ÍÍO,l
(OH··J ·~· ::~:: ...... ,~...... :::.« }·Jt\
("J 1(--
--·=~~:. . :~~--=1: .•
.
(01f}.- 4k1.(Cn}C:~fH·){C~J i'..:;3 {
-..:Jj
-reo ""·reo,"" k.3{Cú}{OH·} -reo= 4k1(02) + 3~(H)(02) -rH
= k3(CO)(OH) - ~(H)(02) - ks(HCI)(H) - k6(Cl)(H)
-rCJ.
= O = ks(HCI)(H)
-- k6(Cl)(H)
7-7
CL _ k5 (HCl,l • k6
r¡¡ =4k1 (02)+ 3k.i(H )(O:!)· kt(H·}(02)- 2ks(HCl)(H·) r11, = 4k¡ (0:?) + 2[k.: {0:d · ks {HCl)] rH
=a+
(H·}
b {H ··)
Ccnsranr Volume
_ .· ·-b, e·-b, dCR · ·· e··, · *b * e . h, -ae dt
H
Using the imegraring factor d (e-oicH.) -· ..- -·---·- - a e··bc dt CH, = .. t
=O
ff
CH.= O
+ K¡ ebt therefore K1 ==
l:~::t1~~l: JJ k,: (Oz)
> ks {HCl)
CH·
!f
t
k5 (HCt)
t
, then "b" is posirive
)
>
kJ (02} , then "b" is negarive
.,,, , , , .,
e.
b
7-8
de>." ·
;·J.
r· 4k¡ {02) . + 3k,qH 1 , - } (02)]
=reo=··
0
L
p( CH )= 3k,, (O/f:11 p = 3k4 (oi) a= 4fs (02)
=-a··p(C¡+)· = .. a .. p l.Jé..· . ;,Dl ~ ..... o :
t
=o
Ceo
i) ¡
··CO
-----·-·
•
l
1
1 l :
¡
·-· · -· · - .. .....
= c.-0.
-
•
= Ceo o
r···.· - -· ---·· · ··- ·,Ceo.
~
-
J
ar .. p 1L eºr + prj b-' ........
·
.....,..._._ ......._,_
.
Raaction Pathway with HCt..
Reaction Pathway w/o HCL
[• ?;_:J· · · ······ · · · · · ··[~;=~
-.
[~E]
[lHJ
_
·-I
lLJ P7-3 ( d) Individualized solution P7-3 (e) lndividualized solution P7-4 The reaction sequence is:
r1
= k!
CH3CHO--~--·,-Cfl3
r,-
= b c . . ('' c(-1' .
·+CHO·
CHi ·+CHiCHO-~CH3 ·+CO+CH4 CHO·
scn cno-.»-,
,,_-.(-'H· •
· ---~ .. ....;¡,' C'--1
.3
CH3 • ·+2CO + H2
H·6
7-9
CA.e -
,.... ...
•. •J
= k1 CcHo- CAc r4 = k.i c.J., ~~ .. -.11•
r3
• C-CHO·i 2 : = eAC (k ! + k2 C--CH;·+ KJ Active imen:nediates: ··rc:h
(1)
CH3 -, CHO·
= -r 1 + r2 • r2 ·· r3 + 1. r = O 2 4
}
== ·k¡C:_,.c + kiC~;,cCc.11). - k,CCH0.C,ic · · · lczC;,cCcH, ·· ::= ••. k ¡(,AC --rcHo- = -r1 + r3
or
=
r>:
'--{.;HO-
=
+- k:2 eACe·'CH;
k:
e·
.... '3 'CHO
+2 k· 4 c2CH,
(-~·'.1C+21 k4 e:
'CH;·
(2)
·k1 CAc + k3 CcHO· CAc:::: O
k1 = -k3
(3)
Substimdng (3) in (2) gives:
k-, .n With Ykl ~~ > > l or CAc > > l . ·rAc et (CAcr' - as suggested by equation (7-2). CH3CHO ,.
.
•
--· .•
C2Hs
P7-4 ( d) Individualized solution P7-4 (e) Individualized solution P7-5 (a) 7-10
Gas phase reaction
with third order kinetics andan apparent negative activation energy. Consider the following mechanism, in which N03 is an active intermediate,
r¡ =k1[N0][01J-k1[NO,] r2
~ k¡[NO][ 02] k1IN0,]
=
NO
k2[NO]fN03]
Is[ NO][NO,]
5I:Y(,,)!ed.
r No31:,:; '-I
=:;
k
1
+Js[NO]
k!NO][o.J+ (k 1 · · ~
1
Ir?11
k1[NOJ) .~tU.'v'.e.k1[NO] ·· k Ir
_ ··k¡J.; [ NOJ'[ 02] ····· k/s[ NOj2 [O,] . k_1 +J.;f NO]..... . _ 2k1k2f NOJ2[ C\] k ·
k_1 1
+ kzf NO]
>> k2[NOJ
····r"tm
. z¡. 02 ] = · · · · 2k¡f<.., k···"'··[NO] ..
-i
For the overall activation energy to be negarive,
-··(E1 + E2)+ E~1 == E0v < O =::::;,
E. ..t « (E l +E)2 .
As long as all energies are positive values
P7-5 (b)
k3
Cl+CO~COCZ ks
COCZ + Cl2 --"7 COCZ2 + Cl
7-11
=O= k3 ( Cl)( CO)-k4 ( COCZ)-k5 ( COCZ)( Cl2)
Tcoa
k3 ( Cl) ( CO) ( C OC! ) = ----'-"'--'-----'-
k 4 + k5 ( Cl2)
rcoct2
= k5 ( COCZ ) ( Cl2 ) =
k1k3 ( CO) ( Cl )( Cl2)
(
k4 + k5 Cl2
)
rc1 =O= k1 ( Cl2)-k2 ( cz)2-k3 ( Cl)( CO)+ k4 ( COCZ)+ k5 ( COCZ)( Cl2) add rcoc, to re,
rct + rcoct = O+ O = k1 ( Cl2 )- k2 ( Cl)
2
( Cl)2 = 5-( Cl2) k2
(el)= Tcoa 2
5-(czi) k2
=
k1k3 ( CO) 5-( Cli)05 ( Cl2) k2 k¿ + k5 ( Cli)
)f
k.k; !5__ ( CO)( Cl2 k2 · = _ _.:: _ k4 + k5 ( C/2)
k¿ >> k5 ( Cl2) Tcoci;
k¡2k3 ( CO)( Cl2 )i2 = --
k2k4
P7-5 (C) lndividualized solution P7-5 (d) lndividualized solution P7-6 (a) K1
N02 + hv ~
NO + O
02 +O+M
~
K2
03 +M
K3
03 +NO
-~ N02 + 02
UsingPSSH,
P7-6 (b)
7-12
-------------------------
------------···--------------------
--
--
---
----------------
---
-
--
--------------------------------
--------·-·--·--
CJ,. M and O, aonear in the denorninaror .....
..)
•
~
-
..
..
._
~
active scecies sucsestec ~o . . -> ......... ;..5
•
._
..
~.....
o
,.
A.pplying rule 3 of rabie 7-1 to 02 and O,: ÜJ
+ O --', 2(}2
-o,
02+0
03 and M appear in the numerator. Applying rule 1 of Table 7 · l to O¡: 0,-7 02 + O If the second ami rhird equations me cornbined, and M is added to each side of the equatíon:
uio
.
.!L,O,. . +O ·+,H
3t~·-3~-<>
(\+O·
.SL¡,2()2
A rnechanism is proposed which satisfies all the rules of thumb:
r01
= ·· · r1
~>
= r¡ -· 'r. -· i; == k1 [M][03]
+
r2 · ····
r3 = - k1
[Al)[<.J.,]
1"
fs[01JO · ][1\tí]
k;[O -)[03]::::
kzÍD;a}O ·][M]- k:i[O
O
·10 l= O 3
[O] ;;:;--k~[M][C'~_, . _ k2[02.IM]+Jc,[03)
o,
2:~.L~:l.. í M} [ 12 b ' . . . '1 2k, k-, re" ::: . •.... :_,___ - -------- -· - - ·· wrn1 K == ---~---·
[0:i] (M}
+
~}f O,}
P7-6 ( d) Individualized P7 -6 (e) Indi vidualized
and
k2
solution solution
P7-7(a)
7-13
------------------------------------ ---- ----- -
--
-------
-
----
- - -----
~
--
- --·---------------------------·-----
-
------·-·--------------·-----------·------·----·-------------
LOWTEMPERATURES • NO ANTIOXIDANT
di•]
dt
=
r¡
= 2ko [I2] - k¡ [I•][RH]
(A)
= -k¡(I•][RH]-kp1(R~·J[RH]
e(~
(B)
c(R:·] = kp,(R·][Oi]- k,i[ROze][RH]-k,[R~·]2
(C)
e(:·] = lq[RHJ[I•]-kp [R•][(h] + kp [R~·](RH]
(D)
1
1
PSSA.... {A)=
O
e{ RADICALS]
dt
ri
= O
= O
[I•]
=
e{:·] = O = ki [RH](I•]r
2ko[h]
k¡ (RH] kp1
[R•][O:i] + kp1[RQi•][RH]
2ko[Ii] k¡(RH]
<(~<;·]
= 0 = kp: [R•][Oil-
kp:z(R02·HRH] • k! [R02·j~
Substitute for[R•]:
2ko [12] - k, [ROi· F = O [ROi-] = . /2ko [Il]
'V
k,
Now substitutc the expressions for me radicals into <(:~] thc expression for the dcgradation of thc oil.
<(RHJ = -k·I (2ko [h]l [RH] • k dt k¡ [RH]J P::
[2kok,[fiJll] r.[RHl•
J.
[I•]
=
ª
2ko [Iil -[2lc2iko]r[Iil1 [RH] 7-14
---
--------- - ---
--- -- ----- ---
-
---
-- -------- ------
--- ---------
--
ODE to be solved- low temperarures, no antioxidant:
[
·
- c{RH] = -,1,,_ "I? 1 + 2~1 [ko] dt
-AQ
L -·
kt
11nr-LI ]112 [RHl 2
J
..
P7-7(b) Low temperatures with anti-oxidant
<{:·] = k¡(RH](J.]-kp [R•][Oi)+kp (R0 ·J[RH] 1
dÍA•] d[
2
2
(same)
= kA1 [AH][R02•]- kA2[A•](ROr]
c(RH] = • k·l [I•l..[RHl- - kp..• [RO,·] íRH] dt -ApplyPSSH:
d[I•]
ÍI•] = 2ko [Ii] .. k¡ [RH]
-=O dr
d(R•l = o _____;.. dt
d(R2 = O 2·]
2ko[!i] - kt[ROi-]2 - 2(kAt [AH)[RQi•])
:. - k1 [R0:?•)2- 2kA1 [AH] (RO:z•] + 2ko [12] = O kt [R(h•f+ {2kA1 [AH])[ROr]- 2ko [li]
=
O
[ROie] = • 2kAt [AH]± ../(2kAt [AH])2 + 8kt ko [fi] 2kt
Now (FINALL Y!)~ let's substitute into d(:~] :
7-15
Quadradtic in [R(h•]
[ROi•] MUST be positivc
c{RH] = • 2ko [12] _ kp [RH] - 2kAt [AH) + ~{2kA 1 [AH))2 + 8kt leo (li) l ?k dt t
P7-7(c)
If the radicals are formed at a constant rate, then the differential equation for the concentration of the radicals becomes:
d [I·] dt =k
0
and
-k; [I•][RH]
=O
[I•] =-~--0
k, [RH]
The substitution in the differential equation for R· also changes. Now the equation is:
d [ R •] = k, [ l •] [ RH] - k Pi [ R •] [ 02] + k r: [ R02 •] [ RH] = O
dt
.
.
.
.
and solving and substituting grves: [ R•] =
k0
+ kn
[ R02•][ RH] [ ]
kPl 02
Now we have to look at the balance for ROi· .
f [ROz•] dt
= kP1 [ R•](02]-kp [R02• ]( RH]-k, [ R02•
r
]2 = O
and if we substitute in our expression for [R] we get
O= k¿ - k, [ R02 •
[RO,·]~
Jt
which we can sol ve for [R02·).
Now we are ready to look at the equation for the motor oil.
d [ :~·]
= -k; [ l•][RH]-kp2 [R02•][ RH]
and making the necessary substitutions, the rate law for the degradation of the motor oil is:
d[RH•] dt=
rRH
~~ = -·k0 -kp2~'s [ RH]
P7-7(d)
With antioxidants
Without antioxidants
7-16
-----~----~-------------------------·--------·--·--·-------··--·-------------·-·--·-------·--·-·--------·-·--··--·-------··-·-·-------------·----
. ---------···-······-----·------·-----------·--------------·····------·---·------·--·--··-------------------·-···-------·--·-----------·--·---------------------------·
·rJ vi "'º J._¡
t t'-,,
1>
¡
\.-
H 1.
- · 7,,
~!~- .. - >·r
()a
R
f
:i •
,\¡
,n .::u:~·t IV&
P7- 7 (e) Individualized solution P7 · 7 (f) Individualized solution P7- 7 (g) Indi vidualized solution
P7-8 (a) Given: Illness mechanism
H-+ I
r¡
= k1
[H]
k2
I + H --, '.2.I 1-+ H
--t
(heairhy person ccnaacts
disease from ill cerson)
k:3
I
rz::.ki[H)[IJ
(Healthy persor; gers Hl)
D
(HI person gers well) (Ill person di-es)
7-17
ro= k, [I) Applying pseudo steady srate hyporhesis to 1: r¡ = r1 + rz - r3 ·- ct=.:c O
or k 1 [IrtJ• 1
[ 1r 1 [ + k2 H_ 1.1.1 ·• k3 • t
·3·
- k.t •. IJ1 ""'O or l• 11i, (
J
.-. ro
_ -
k ¡ [HJ
:--·----·· -a . -··---.-:-""'
{k:; + k-1) •
kzlHJ
= .. k1 . •~. [H] k3 + k..t - k2[HJ
P7-8 (b) k-1 + k.t: at [H] = --rhe death rate becomes infinire, k2
P7-8 (e) Ir is enlightening to calculare rn:
=
rH
·r1-r2+r3
= ·k1[H]-k2[H][IJ+k3[f]
·k¡ k.tfHJ
rH = ·····-----·---:-7··
(k3 + ~ - k2tHJ}
= -k¡[H]··{k2[H]-·k;}(I]
= · ro -c 0
This expression srates rhat for every person who 1$ ill and dies, a healthy person becomes ill
For
the population as a whole, "illness is bue a 'way srarion' on the road to 'dearh'." Note further rhar I}{
< O, and therefore the population will die off eventually. Inirially, the death rate will be slow,
unril [H]
·i-
{k3 + ~} / k2 . I11e rnodel neglecrs the possibility of birth, In practice, it would appear
to be useful in describing epidemic-like diseases, which occur ever a short time so that the birth can be neglected.
rate
P7-8 (d) See Polymath program P7-8-d.pol. 1.0e+9
l.OOE9
8. 0e+8
8.00E8
6.,0e+8
6.00E8
fi"========-~-·--. ---,
CJ I
4.0e+8
400E8
2.0e+8 ·
2.00E8
0,.0e+O
)
·--·-·-----~· -----·....
0.0e+O 2.0e+9
4.0e+f
6.0e+9 8.0e+9 LOe+lO
O.OOEO L-""""'===---~· -------~--O . OOEO 2.00E9 4.00Ei9 6.00E9 800E9 l.OOElO
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 start dying.
7-18
P7-8 (e) Individualized solution
P7-8 (f)
Individualized solution
P7-9 (a) Starting with the design equation for a batch reactor dC -···-·f.
:=: Tp
dt
kC
TI:'==
~c~\1 ·f!) e
C== (c~0
ac
r
· ·d/ =klCso Crll--i~.J
1
in te grating
(
.,,
··-
i.s"')
e
! .
"p ....
ln'l.933---··········· C p · ··IA
·- -
·-
,~
.. J
=-= 7.14kt
+e .
.. _.,
Finding k, b40
y= 0.002x + U697 l
O
~
_c.
20
40
,.....................
.•
60
80
time
1--·----··:-· ···
·
.•..
1.. .
- ..
f
l
ti
q
·---==--=~r-- ~.-_ -·. ·---- . · --·- - _ ..f
I_
Using the data at 40°F and 45°F the following ~raphs are made.
_
.
t
Finding k lO---·-·---·-· ····· ·- ······ ··-·-·····--·------·-·-·-···· · · ··--~---~
o
o·o
y= O.f.:02x + l. l697
• •- •-- • • • ...... ,;
o
20
40
•
_
60
timé
7-19
__,
,
80
=
k,o 2.8*10-~ el{):::::: 1.1697 k~5 = s.26*10 ... C45 = 1.1883 Toe activation energy then is:
from mese:
e -1.s) ln .93 3-=-a... . ·-- . ::::: 7 .14kt (
e, ---1.4
+e
E:::::: - Rln(k1fki)
1f T2 --· l/1;
= 108120 Bt1:,_ mol
P7-9 (b)
~{~(:¡};7- . - 5~))
k3s = k4510"(2
= 1.8 * 10
4
=
Using interpolation, C l.18086 Using the same equation used on the graphs we can. solve for t when C, t = 49 days.
P7-9 (e)
'(i~IR{ 5{6· ·-· ·:s·t¡))
kn = k45 l01
= 1.1
= .0023
e= 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) lndividualized solution P7-9 (f) Individualized solution P7-10 (a) E+S kl >E•S E•S kz >E+S E•S_E_~P+S P+S·-~E·S rE·S = k, ( E)( s)-kz (E·S)-k3 ( E·S) + k4 ( P)(S) since Sis not consumed: rE.s
= k1 ( E)[( ST )-(
S7
= S + E•S
or
S
= S7 -
E•S
E•S)]-k2 (E·S)-k3 ( E•S) + k4 ( P)[(
S7 )-( E•S)]
(E·S)= k1(E)(Sr)_+k4(P)(S1) k1 (E) + k2 + k3 + k 4 ( P) -rE
=k1(E)(S)-k2(E•S)
-r
_-k ( E )( S )[ 1E
l
I
k1(E)+k4(P) k1(E)+k2+k3+k4(P)
] - k1k2(E)(S1 )+k2k4(P)(S7) k¡(E)+k2+k3+k4(P)
7-20
P7-l0 (b) E·S k2 E+S~E·S k4 E·S k3 E•P E·P~P+E rE,P =0=k3(E·S)-k4(E•P)-k5(E•P) ( E·P)
E+S E·S
= k3 ( E•S) k4 -+-ks
rE,S
= k¡ ( E)( s)-k2 (E·S)-k3 ( E·S)
ET =E+ E•S + E•P
since E is not consumed:
E= E7 - E•S -(
+ t,( E·P)
k3
~+~
J E·S
or
E= ET - E•S - E•P
E= E1 - E·S
(1 + ~+~ k J 3
Insert this into the equation for rEs and sol ve for the concentration of the intermediate:
(E·S)=
k1(S)(E7)
[ r p
1+
k3 k¿ + k5
]k1 (s)+k2 +k3 - k4k4k3 + k5
=k (E·P)=kls(E·S) 5 k4 +k 5 k¡kls ( s)( ET) [ k3 + k4 + k5] k1 ( S) + k2k4 + k3k5
r - ----'----'--'---'-~P -
P7-l0 (e) No solution P7-l0 (d)
will be given
k3
E•S1 + S2 µ E•S1S2 k4 ks
E•S1S2 --'7P-t- E =O=k1(E)(S1)-k2(E•S)-k3(E·S)(S2)+k4(E•S1S2)
(1)
rE,s,
(2)
rE,S,Sz
= Ü = k3 ( E·S) ( s2 )-· k4 ( E·S1S2 )-k5 ( E•S¡S2)
If we add these two rates we get: (3) rE,S,
+ rE,S,S2
= Ü = k¡ ( E)( s)-k2 ( e-ss-», ( E·S1S2)
Plug this into equation 3 and we get:
7-21
P7-10 (e) k¡
E+Sµ(E·S) ki
1
k3
( E·S)1 µ( E·S)2 + P¡ k4 k5
(E•S)2 ~Pz +E =O=k1(E)(s)-k2(E·S)1-k3(E·S)1 +k4(E·S)2(P¡) (2) ~E·S)2 = Ü = k3 ( E·S)¡ -k4 ( E·S\ ( P)¡ -k5 ( E·S)2 k3(E•S)1 (E•S)2 =k 4 (P.)+k 1 5 (1)
~E-s)1
Add (1) and (2) to get
~E·S)¡
+ ~E·S)2
= Ü = k¡
(E)(s)-k2
k1(E)(S)
7-22
(E·S\ -k5 (E·S)2
k1
k
k5k3
2
(E•S) = 2
r =k P2
5
(E)(S)
+ k5 + k, ( P¡)
k3k1(E)(S)
k2k4
(
P¡) + kis + ksk3
(E·S) = 2
k1k3k5(E)(S) k2k4 ( P¡) + k2k5 + k5k3
( E7) = ( E)+( E·S)1 +( E·S)2
P7-10 (f) k3
k,
E+SµE•S-~P k2 k4
E+PµE•P k5 ko
E·S + PµE•S•P k, ks
E•P+SµE·S·P k9
r¡,
= k3 ( E·S)
c1)
~e,s)
=o=
(2)
~E·P)
=0=k4(E)(P)-k5(E·P)-k8(E·P)(S)+k9(E•S·P)
C3)
~E·S·P)
(2) + (3):
k1 ( E)(s)-k2 ( E·S)-k3 ( E·S) + k6 ( E·S)( P) + k, ( E·S·P)
=o=
~E·P)
k6 ( E·S)( P)-k1 ( E·S·P) + k8 ( E·P)( s)-k9
(
E·S·P)
+ ~E·S·P) =O= k4 ( E)( P)-k5 ( E•P) + k6 ( E·S)( P)-k7 ( E•S•P)
now add (1) to this:
7-23
---------------------------------
-----------------·--
--
.
-
.
.
.
- -----·-------·--------------·-·-------------------·----------·-----------
+ '(E,P) + '(E·S·P) =o= k1 ( E)( s)-k2 ( E·S)-k3 ( E·S) + k4 ( E)( P)-k5 ( E·P) (E•S) = k1 (E)(S)+k4 (E)(P)-k5 (E•P) k2 +k3 '(E,s)
(E•P)=
k4(E)(P)+k9(E·S·P) k5 + ks ( s)
(E•S•P)
= k6 (E•S)(P)+k8 (E•P)(S) k; +k9
k4 (E)(P)+k9 (k6 (E•S)(P)+k8 (E·P)(S)J k1 + k¿ __ _...::....
( E•P) = k5 (E•P)
= k.k¿ (E)(P)+k9k4
( E•P)
= k1k4 ( El_( P)
-t
ks (s)
(E)(P)+k9k6 (E•S)(P)+k8 (E•P)(S) k5 + ks ( s)
+k9k4 ( E)( P) +k9k6 (E•S)( P)_ k5
( E·S) = k¡ ( E)( S) + k, _( E)( p~ -k,_( k7k4 ( E)( P) + k.,k. ( ~: ( P) + k.,k, ( E•S)( P) k2
J
+k3
(E•S) = k1 (E)(S)+k4 (E)(P)-k7k4 (E)(l~)+k9k4 (E)(P) k2 +k3 -k6k9 (P) rp
= k3 ( E•S) = k3 ( E) k1 ( S) + k4 ( P)-· k.k¿
( P) + k9~4 ( P) k2 + k3 - kl9 ( p)
(Er )=(E)+(E•S)+(E•P)+(E•S•P) (Er) = (E)+ k1 (E)(S)+k4 ( E)( P)-k7k4
(E)( P)+k9k4 (E)(P) k2 +k3 ·-k6k9 (P)
k..k¿ ( E)( P)
+ k9k4 ( E)(
P)
+ k9k
+--
.
+ k4 ( E)( ~
(k1 ( E)( S) + k4 ( E)(P)-k7k4
k 6
+
(k1 ( E)(S)
6
_
P)--·· k1k4 ( E)( P) + k9k4 ( E)( P)J( P)
k2 +k3 --k6~9 (P) ·---
( E)( P) + k9k4 ( E)( P)J k2 +k3 -k6k9(P) (P)
-
7-24
-
kl +k9 ali 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 (E1), (P), and (S) can be made and plugged back into the equation for rp.
P7-10 (g) k¡
k3
E+S µE•S ,-:,,P k2 k4
E+PµE•P k5 ~E·S)
= 0 = k1 ( E)( S)-k2 ( E•S)-k3 ( E•S)
(E·S)= k1(S)(E) k2 +k3 ~E·P) =0=k4(E)(P)-k5(E•P) (E·P)= k4(E)(P) k5 rp = k3 ( E •S )- k¿ (E)( P) + k5 (E• P) rP
= k1k;(!~(E) 2
k1k3
(
S)( E)
-+k3 (ET )=(E)+(E·S)+(E·P) rP
=
__ k4(E)(P)+k4(E)(P)
3
k2
(ET )=(E)+ k1(S)(E)+ k2 + k,
k4(E)(P) k5
(Er)=(E)[l+-k1(S) + k4(P)] k2 + k3 k5 klk3 ( S) (ET) r = --·----,=----'---'---'---~
P
r P
(k2 +k1)[1+
k1. (S) + k4(P)] kz + k3 ks k3 (S)( E7) - ~-----'----'-----(k2+k1 +(s)+ k4(k2+k3)(P)J
kl
k¡k5
7-25
-----
-
----------
------- ---- -
- --
-
--···--·------------------·-·------
-- -- - ----
--- - -
-
P7-10 (h) No solution will be given P7-10 (í) 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:
.L dN.s = .QCs = rs = .:Y max ~1. V de d K ..
t
rn
ar
+ Cs
t
=
O
,
e·S =
Csa
Reaua.nging and incegraring:
l f~·s +
tldCs
es.
cs..
= ·
J
r
Jo
v;.ax de
or L In Csn = _ Csn.:~.s. t Cs Km e · A . , plot. or t
.L
·¡
e
e
11
+
= Km In
e Cs. e;;+
Cs<> "" -Vmll <.
Ym,u
x,
Cso . Cso - c.. . . with I -e•5 vs -·. t' ~. should be linear .... . s,ope . - j · -·Km _
-·-· _ft_
...lr
Cso
·¡ n __Q Cs
Cs
Cso ··
e,
Csa..: Cs_
--L-.. O 10
.0200 .01775
1 .0 1.1268
20
.01193
.0158
.00225
1.2654
.000225
50
.01179
.0106
.0042
.00021
roo
l.8867
0!2"70
.0050
.0094
.000188
4.0000
.01386
.0150
.0001.5
..
·-··-- ....·--- .......
t
.0000
00140 0.0135
1.1n t
e~
0.0130
Cs
0.0125 00120 0.0115
r-:i···-r1-···r1 15
1s
11
18
19
20
1 21
1--1T-
22 23
24
(C{so) - Cfs))Jt x 1 ES
7-26
--
--------
t
-------------~-----
25
_
013· ()]1 ¡· ·-----------;·J min·l '! --·•---"'---· F rorn t he grapn,· slope "' ···-~-(17.5 . 20.6) x 10 5 ,g mol l min 'J
Atl.Jn Cso. e Cs V . Km
__.l:llll.
Y;:·
=
=
Km
=
JLH.
~--Cs
min
C Cs
.0310 g mol/ l
C
l. In ~º- + ~ e
!J?.:< . ~-°-~g!:1~!.
""
C
..-:.. -2.
Kmt
1-min
=
~ . .ül :> rrun
l
= . Ol8ó5
= (.013 + :5.64 x 10 ·3) rnin
~
(
L75 x 10 4 g mol/l-rnin + ---::;··--- ------·-·
-1
O, 10
min
s mol/l
1
. · l)(
gmol\ g mol Vm:u. ::::: .018fo x mm .. 0310 --i"-1 == ~.'-. 78 X· iO 4 -· l-rnin
P7-11 (b) V m= a. [EJ. If the enzyme concéntration is
increased by a factor of three,
.3(- . 78 * 10.4 gmol Vmax _ - .. _ ::>. . .
-igmol ~;·;~;~;-
l_ .1 ..
1-; ~~- )-
c
km In -~1- + C5
Gw
....
C so ;;:; · -1734
"
.)_4
1 .)4
* 1<
.
y
l
* 10 4 .. 2;'!:l!.·.---" * 20 min = · -34. 7 * 10 L ~n
4
then
ff..'.11_:?l
L
This equation shouid be solved by trial and error. Rearranging,
.:
..
,
.02 exn
( ··S :::::
'
Assume a Cs • calculare a new ene from Assume Cs
o.o 15 . 0.@077 o.@®~7
0.0091
0.()€)~3 0 . 0®,2
rl l
~:Q!47- ~ºt1 .03 lO
J
che RHS ef the absve equaríon, New Cs % difference ·-·1)~.00:,:~7:=7---~,.,..4~. o;, _.,..g _
ze.s
@.@@97 @.oo, 1
0.0093
®.® 2.2
o.omn
o
0.0092
1.1
P7-11 (e) Individualized solution P7-11 ( d) Individualized solution P7-12 (a)
7-27
...
-
.----------------
--~
.-------
Given the reacrion sequence;
E+S =E•S
The plat of -r, vs -~ is shown below,
the mecbanism for the above reacdcn is:
. kCs ~-. ; For C"-s << l: -rs"" k C Et in qualitative agreemem: wíth the graph at Cs + K2 CJ
-r,. = -
l + K1
es. fur es >> l: -r, = izÍ
low values ef
-
l
. This is aíse in
agreemenr
with the graph.
Cl.Oii .
i•
o
At?1-~:~!.
zo
e;
n
=
oCs
se
4Q
~IM•rl
. ~ . Cs.E,t~.1.t>Z)<:2_~:,} ,,. . . _.~_.§dL::.;K1 C¡} ... l + K¡ C\ + K2 Ci
(1
+ K1 C, + K2 C~)2
(
l
+
"" O
K1 C, + K2 C})2
2k E1 ( l__ - K5 C;}fK L : . . 2K2 S'.sl. ( l + K1 e, + K2 C~f d2(-·r_t') •
1k
- :'>
clC~
E r ~K C K · K2 C3]
+ K1 Cs + K2
¡
'il
CHJ
7 1 < O :. (- r,) goes rhrough a maxirnum. 'I K.2. This observanon also agrees with che above graph 2k Et [. K1 ·· 3 vKi" t
+
P7-12 (b)
7-28
=
uo
For a CSTR operatin g with V;:::; 1000 l :
32 1/min V
J ~ODQ = 313 rnin, J
Uo
= 50 m rnoles/l
with CAo
. r. ,
.
r'\
= 50'""l--·~-:::. ·
.
,)
"'r
•
.)
,)
.. ~.
chis (linear) eauacion is plotted on the accompanying ¡,
r,J v s Cs curve frorn the rare of reacrion ar
graph, the equarion inrersects the {
""o.•s4m~ ., min >I .
es =
Stabiliry of points: assume that a perrurbation
Cso -
Cs occms
where the overbar designares
the steady-stare condition . Material balance for any time:
or ~• dCs ---· dt
=
rs:
-i-
}
"\
Cso ·· Cs
ar steady stare O = rs t + Cso + Cs I ÍS1!1ª a Tavlor ,,
.
::::-
<;;..,. ~
Combining
d6 _ r-a/=(rs --r,)r+Bso
abou:
5~,.;,,.S .>v~"'1Si('[1 ~-...... ~}/.¡, ..;.., ,1
.........
....
~
{
"';
,.,,~
·e- . ·'::},
to linearize re · ?s: ~
1 ··d ¡s ·11 8s \2 Cs fc5
rs .. fs ·1
.... des V---~· dt
1
li es + 6so
a: ,
··Bs
...
J
-
.
and substituting
= o . 8s
=
o
,l
For this solution of the above equnrion 10 be stable:
(!Jit' -
r .
l <0
or
\
( -~-r-2. \ r ¡_ \.o~ e·-s 1c5
< t
••
or(~- _r~~ ·.· ~ C d
Lr .
>
-1
·S 1c5
At C,. = 2.J m moles/l, ·r, = 0.154 m moles/ ! -rnin. the sloce of che reacrion rare curve is posirive Therefore, this operating point is srable. Fer rhe other rwo poirus. the srabiliry rnay be examíned by estimaring the derivarive gr;ipbically: At ( .,.. 5
z;
0,
.
.
.4 m moles/l ,
(º-··-:--!_ ( rs) \
d(s
,i('. ··~'):
,,,n ,1 . ;¡ '"'t
úCs
7-29
= · Q,JJ.Q. 2 min. · 1
:. the point Cs
= 9.4.
At Cs e: 2.1 m rnoles/l ,
··rs "'(U32 is unsrable .
¡~.aL.:~!.\
Cs fes= 2.1
a
~L::~1. Di.Cs
040672 .. -. O.O 18 . 24. 44
= . Q~2.5.
min -1
20
= 5.1 m rnoles/l
. -rs "" 0.154 m.ill.Q!~~ appears to be stable. bue more accurate calculation l= min .rablisl .. 1. d ennmve1v. •.... o - ······- 1s necessar.; ro esra 1sn mis conctustort o Cs , •
At
Cs
,
f( a(-rs)).
.
X= eso -es=
eso
.
50-2.1 =.958 50
P7-12 (e)
=r. =---1 + K e 1
+ K2es
2
1
If E¡ 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=0..2
P7-12 ( d) Individualized solution P7-12 (e) Individualized solution P7-13 (a) Data on Bakers Yeast at 23.4 ºC
~\:~
Po,.
2(}
na
suífanilamldc
····o.o
__ L.
mg
si.düi.n ilamidc/ml adccd to mcdium
··--o:ó"-·· ..... -
Q0i
no sulíanilarnidc
Ü.0----·-········-
-L
Q.:>z
10 mg sulfanilamide/ml
····--·- ·---·········~to.médium .0575 .0391 .03246 .02747
O 5
LO
l.5
?___ '.} -
7-30
.
1
r;· - v:ll s ~-m
..
} v:;l l;( - VKm(1' m:TPo,
+·
+V
1 ~;l; -
l
Qc>t
1- will have a slope of-!:Sm ... andan inrercept of Un
Plot of nl~.-º versus
·p·
Ymax:
O;i
'
0.06
•
+ Wlth
( lltl tn
(tt
05
LS
U)
l/1)(.o) (mm Ha) Frorn che graph, slope r.
Vm.u ·
=
=
ffJQj,t' ,!}Ji
=
.0165 I ntercept = O.O 19
52.63 -· .!111 OL .• nr mg cells
Km = 0.0165
Vrnax =
= 0.8684 mmHg
((L0165)(52.63)
P7-13 (b) E +· S ,:::::, E • S
Now, with comperiuve innibirion:
f+E.QI•É E·S~P+E Rare law becornes: r;1
= __
In this case, the slope is :
V :max $
S +K Km\.l
- .
l t + K¡J ) m l +
·+-}
K1 ---·~.. ,-··---·· 'v rnax
or
I< m
(1 +-.KL}
J_ = · ··-···-.. fp
V rnax
¡·S -!\ ·-·-
l.. ¡ ) +
. ·1 ( h e .inrercepr is. t.h e sarne. wnue
max
. case ( a ) as .in
7-31
--·-·-----·---·-··---------·--
-- --
-- -
--·~-------------~---
For the case of uncomperitive
inhibition:
E+ S
= E. S
E·S+lc:::d•E•S E·Se;-,P+E
Rate law becomes:
1
+..L
In this case. che slope is che same, bur the inrercept is
·,;¡
And fer the case of non-competitive inhibition:
E+l=E.•l
K¡
rnax
E•I+S~I·E·S E + S e::>l:.: • S I·E•S(:;:>E•S+l E•S<::::>P+E
l"p
= {S ~
Vmax S Km){l +
l +-L
J;}
l
r;·
or
=
K1
V m-;;
l +l
+--!;1-(1-) V mu S
In this case both the slope and intercept change, Ptoning rhe data of p~ in mml:ig versus
o;;
with sulfanilamide on the sarne plot as was plotted the data for the case with no sulfanilamide, it is seen tha; the slopes are different, but the intercept is che same, Therefore the inhibition is
comparanve,
P7-13 ( C) Individualized solution P7-13 ( d) Individualized solution P7-14 For No Inhibition, using regression,
1-= + a1(!)S aO
Equation model: -
-r
For Maltose,
ªº
s
= 0 . 008
Equation model: -
al= 0..0266
1-= + al(!)
=r,
aO
aO= 0.0098
S
al= 0.33
For u-dextran,
7-32
Equation model:
1-= + a1(!) aO
-
-r
S
s
aO = 0.008
al
= O 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
rp =k(EHS) (EHS) = KM (EH)(S) rP = kK M (EH)( S) (EH;)= K2 ( H+ )( EH) (EH) = K1 ( H +) (E-) ( E- ) = j__EH) K1 (H+) ( E1 ) = (E-)+ (EH) + (EH; ) (EH) (Er)=-K1(H+) (EH)=·
+(EH)+K2
(Er) !+
(
+
H )(EH)
-·-
K,(~') +K,(H')
Now plug the value of (EH) into rp rP
= kKM
(EH) ( S) =
kK (E)K 1 .
M
l+K1
1
(H+)(s)
---2
(H+ )+ K K (H+) 1
2
At very low concentrations of H+ (high pH) rP approaches O and at very high concentrations of H+ (Jow pH) also approaches O. Only at moderate concentrations ofH+ (and therefore pH) is the rate much greater than zero . This explains the shape of the figure .
Fp
P7 -15 (a) Indi vidualized solution P7-15 (b) Individualized solution
P7-16 (a) For batch reaction,
7-33
----
------- ----
--~--------
----··----·--··--·-··-----------
-··----------·---------·---
- -- - - --- - - - --
-
- --
----
-
--- ---
----·-------------------·--
-----
---------- -------------
dC5
rs --
--=rs
dt
&
See Polymath program P7-16-a.pol. POLYMA TH Results Calculated values of the DEO variables Variable t es eso
initial value
minimal value
20 20 0.25 0.5 0.1 0.1 1 -0.0987654 0.0493827
6. 301E-11 20 0.25 0.5 0.1 0.1 1 -8 . 0781496 1.273E-09
o
Km
Yes Ceo Ce umax rs re
o
maximal value 10 20 20 0.25 0.5 0.1 10.1 1 -2. 546E-09 4.0390748
final value 10 6.301E-11 20 0.25 0.5 0.1 10.1 1
-2.546E-09 1.273E-09
ODE Report (RKF45) Differential equations as entered by the user ¡ 1 J d(Cs)/d(t) = rs Explieit equations as entered by the user [ ll Cso = 20 [ 2 l Km = 0 . 25 [3] Yes= 0.5 [4l Ceo=0.1 [ 5 J Ce= Ceo+Yes*(Cso-Cs) [6J urnax e 1 [ 7 J rs = -umax*Cs*Ce/(Km+Cs) [ 8 J re= -Yes*rs
o.o
20
--1.8
16
[:]
-3.6 -5.4
12 8 4
-7.2 · -9.0
o
2
4
6
8
o
10
P7-16 (b) For logistic growth law:
dCc dt
---=r
g
7-34
o
2
4
6
8
10
See Polymath program P7-16-·b.pol. POL YMA TH Results Calculated values of the DEO variables Variable t Ce umax Coo rg
initial value
o
minimal value
o
0.1 1 1 0.09
0.1 1 1 0.0080739
maximal value 7 0.9918599 1 1 0_2499857
final value 7 0.9918599 1 1 0_0080739
ODE Report (RKF45) Differential equations as entered by the user [ l l d(Cc)/d(t) = rg Explicit equations as entered by the user [ll umax « 1 [2] Coo = 1 [ 3 J rg = umax*(1-Cc/Coo)*Cc O.JO~-------------~
1.00 0.82 0.64
012
0.46
0.28 0.10--~-----~------' 0. 0 1.4
Forpart(a), C=
2.8
4.2
t
5.6
OJ}O.__-------~------'
o.o
70
14
2.8 t
4.2
5.6
7.0
=Yc1sCs0 +Ceo =(0.5)(20)g!dm3 +O.lg/dm3 =10.lgldm3
P7-16 (e) ForCSTR,
= -Y
r g
r
ct s s
= YCISµmaxCsCc
Dilutionrateatwhichwashoutoccurs=
K
M
+e
S
YctsµmaxCso - O.Sxlhr-lx20g/dm3 =0.494hr-1 KM+ Cso 0.25g I dm' + 20g / dm'
P7-16 (d) D maxprod -Y - CISµmax Dmaxprod
[1 - ~M 5 x lhr - ]KM + Cso -Ü.
= 0.44hr-1
7-35
-1 X [
1-
f;
0.25g/dm ·-3----- 3 0.25g I dm + 20g / dm3
]
Ce
=(YC/s(Cso + KM )·J(YC/SµmaxcSO YC/Sµmax -D
KM +eso
-DJ
C = (0.5(20+0.25)g / dm3 J(0.5xlhr-1 x20g / dm3 -0.44hr-1J = 9.0Sg I dm' e 0.5xlhr-1 -0.44hr-1 20.25g / dm" DKM =1.83g/ dm 3 YCtsAnax -D C; = Yp¡cCe = 0.15x9.08g / dm' = I.362g I dm3 C8 =
_
r8
_ µmaxCsCe _ lxl.83g / dm3 x9.08g I dm3 = 7 99 Id 3 h • g m . r KM + C8 0.25g / dm3 + 1.83g / dm 3
P7-16 (e) If rd= k.tCc
m
= Ce Vo = (rg
- rd )v
D = (rg - rd) = (YCISµmaxcS ~ -kd Ce KM +Cs
Divide by CeV,
= . ( D + k d )~_M__
Cs
Ye,sµmax -- D - rs = Ys,erg
Now
~
Ce= Ye,sD(Cso -Cs) D+kd For dilution rate at which wash out occur, ~ Cso=Cs
eso--
~
Ce = O
(D+kd )KM .. Ye,sµmax -D
DMAX = Ye,sµmaxCso -e,«; = ~:5xihr-1 X 20g I dm3 -0.02hr-1 x0.25g I dm3 = 0.493hr_1 Cso +KM 20g/dm3 +0.25g/dm3 There is not much change in Dilution rate while considering cell death to one where cell death is neglected.
DC
=Yc,sD2((!so-Cs) D+kd d(DCe) For D max prod , =O dD
Now
C =-(D+kd)KM s Yc;sµmax -D
e
D max
prod
= 0.446
hr"
P7-16 (f) Now -rm=mCc
DCe=rg m = Ccv0
&
D(C80-C8)=-rs-rm
= (rg )v = uc; V 7-36
D = µ = (Yc,sµmaxCs) KM +Cs
Divide by CeV, Solving
DKM
=>
Cs =----'~Yc,sµmax -D
- ''s = Ys,c'c
Now
Ce
=
=>
[D(Cso -Cs )] Yc,sD+m
For dilution rate at which wash out occur,
Ce= O
Cso = Cs,
eso --
DKM YetsAnax -D
DMAX _ Yc,sµmaxC50 _ O.Sxlh.r-1 x20g I dm3 = 0.494hr-1 Cso +KM 20gldm3 +0.25g/dm3 d(DCc)
Now
For D max .prod
Dmax
prod
'
---;¡¡y-.- = O
= 0.4763 hrº1
P7-16 (g) Individualized solution P7 -16 (h) lndi vidualized solution P7-17 Tessier Equation,
rg. -_ µmax (l -e
-C5 I k
)e e
(a) For batch reaction,
ac, =:=>r». dt Ce
rg _ - µmax (l -e
=r, =Ys,crc,
= C co + Yu s ( C so
-C tk)c 5
e
- Cs )
See Polymath program P7-17-a.po1.. POL YMATH Results Calculated values of the DEO variables Variable t es Ceo Yes eso Ce k
umax Yse rg rs RateS
initial
o
value
20
O. 1
0.5 20 0.1 8 1 2
0.0917915 -0.183583 0.183583
minimal value _º _
rnaximal value 7
0.0852675 O. 1 0.5 20 0.1
20
8 1
8 1 2
O. 1 O. 5
20 10.057366
2
0.0917915 -7.7126957 0.183583
3.8563479 -0.183583 7.7126957
7-37
final value 7
0.0852675 0.1 0.5 20 10.057366
8 1 2
0.1066265 -0.213253 0.213253
ODE Report {RKF45) Differential equations as entered by the user [ 1 l d(Cs)/d(t) = rs Explieit equations as enterad by the user ¡ lJ Ceo= 0.1 [21 Yes= 0.5 [3J Cso = 20 (4J Ce= Ceo+Yes*(Cso-Cs) [5] k= 8 [6J urnax e 1 [7J Yse = 2 [Bl rg=umax*(1-exp(-Cs/k))*Ce [ 9 J rs = -Yse*rg [ 1 o J RateS = -rs
16
6.4
12
4.8
8
3.2
4
1.6
o (e)
0. 0
1.4
2.8 t
DCc=rg m
5.6
42
ro... RateS
7.0
µ = r.: (1- e-Cs I k)
D(Cs0-Cs)=rs
= CcVo = (rg )v = µCe V
Divide by Ce V,
D = µ = Anax
µ:-J
(1-· e-Cslk)
e, =-k1n(1Now
Por dilution rate at which wash out occur,
(d)
=>
Cso = Cs
=>
DMAX
= =k
Ce = O
ln(1-_Q_J µmax
= s.; (1-e-Csolk )= lhr-1(1-e-20gldm' l8gldm• )= 0.918hr-l
DCc =DYc,s(C,0
-C,)
DCc =DYc,s[ C,0 +kin(i-
C, =-kln(l-
µ:JJ 7-38
µ:J
Now
For D max
prod
,
d(DCe) dD ::: O
D max .prod = 0.628 hr-1
P7-17 (a) Indi vidualized solution P7-17 (b) Individualized solution P7-18 (a) rg = µCe
µ::: µmax Cs KM +Cs
ForCSTR,
DCe ::: rg
D(C50 -C5)::: -r5
-
C5 :::Cs0(1-X):::10g/dm3(1-0.9):::lg/dm3
r5
:::
Y5,erg
D:::
vrv
Ce =Ye15(C50 -Cs)=0.5(10-l)gldm3 =4.5gldm3 DC
e
= r = µmax CsCe K M +C S
s
~4.Sg dm' I
V
= 0.8hr_,
xlg / dm3 x4.5g / dm' (4+I)g!dm3
V ::: 6250dm3
P7-18 (b) Flow of cells out = Flow of cells in
Fe =voCe Cell Balance: Fe + r8V - Fe
dC e=r __ dt
dC
= V--c
dt
g
µmaxcsce r = --"==---"-----=g KM +Cs This would result in the Cell concentration growing exponentially. This is not realistic as at sorne 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 (e) TwoCSTR's For 1 st CSTR, V = 5000dm3 ,
DCe
=r
g
7-39
=r,
r g-
=Ys,crs
ee
AnaxCS +C
K
M
S
See Polymath program P7-18-c--lcstr.pol. POLYMA TH Results NLES Solution Variable Ce Cs urnax Km
Csoo eso Yse rg rs V vo D X
Ceo
Value 4.3333333 l. 3333333 0.8 4 10 10 2 0.8666667 -1.7333333 5000 1000 0.2 0 . 8666667 4 . 33
f (x)
9 . 878E-12 l.976E-11
Ini Guess 4
5
NLES Report (safenewt) Nonlinear equations [ 1 J f(Cc) = D*(Cc)-rg = O [ 2 J f(Cs) = D*(Cso-Cs)+rs = O Explicit equations [ll umax= 0.8 Km=4 Csoo = 10 [4l Cso = 10 [SJ Ysc = 2 [ 6 J rg = umax*Cs*Cc/(Km+Cs) [7) rs=-Ysc*rg [8J V= 5000 [9) vo = 1000 [10) D=voN [11) X= 1-Cs/Csoo [ 12 l Ceo = 4.33 [2J [3J
X= 0.867 Cp1 = Yp1cCc1 =0.866 g/dnr'
Cc1 = 4.33 g/dm3 Cs1 = 1.33 g/drrr' For2ºªCSTR,
D(Cc2 -Cc1)= r,
See Polymath program P7-18--c-2cstr..pol. POL YMATH Results NLES Solution Variable Ce es urnax Km
eso o Csl
Value 4.9334151 0 . 1261699
f (x) 3 . 004E-10 6.00BE-10
Ini Guess 4 5
O .. 8
4 10
1.333
7-40
2 0.120683 -O. 241366 5000 1000 O.2 0.987383 4.33
Ysc rg rs V
vo D X
Ccl
NLES Report (safenewt) Nonlinear equations [ 1 l f(Cc) = D*(Cc-Cc1)-rg = O [ 2 l f(Cs) = D*(Cs1-Cs)+rs = O
Explicit equations [ 1 l umax = 0.8 [2l Km=4 [3 l Csoo = 10 [4J Cs1=1.333 [51 Ysc=2 [ 6 l rg = umax*Cs*Cc/(Km+Cs)
rs = -Ysc*rg V= 5000 vo = 1000 [10] O= vo/V [ 11l X = 1-Cs/Csoo ¡ 12 l Cc1 = 4 . 33
(7]
[8] [9]
X= 0.987 Cr1 = Yr1cCc1 =0 . 9866 g/dnr'
Cc2 = 4.933 g/dm" Cs2 = L26 g/dm"
P7-18 (d) For washout dilution rate, Ce = O
=
D max
DMAXPROD
= µmax
f"I{~--] [ 1-v¡¿;¡ e~~
= O.S?hr_1
J!.:_8hr-1 xIOg / dnt' 4g/dm3 +10gldm3
µmaxCso KM + Cso
= 0 . 8hr
Production rate = Ccv0(24hr) = 4 . 85
-1[
r·-·4gldm3
P7-18 (e) dC c_=r __ dt
dC5 dt
---=r g
= 0.37hr
g / dm3 x1000dm3/lnx24hr = 116472.56g/day
For batch reactor, V= 500dm3,
·--]
1--V4g / dm3 +"íoi1 dm3.
s
Ceo= 0.5 g/dm3 Cso = 10g/dm3 ro = _l!max º
See Polymath program P7-18 e.pol. POL Y1\1ATH Results
7-41
e e S
K M +C S
e
-1
Cakulated values of the DEO variables Variable t Ce
es
Km
Ysc umax rg rs
initial value
o
minimal value
o
0.5 10
0 .. 5
4 2
4 2
0.8 O .2857143 -0.5714286
maximal value 6
4 2
4 2
0.8 0.1486135 -2.8064061
6
5.4291422 0.14171.55
5.4291422 10
0.1417155
final value
0.8 0.1486135 -O. 2972271
0.8 l. 403203 -O. 2972271
ODE Report (RKF45) Differential equations as enterad by the user ¡ 11 d(Cc)/d(t) = rg [2 J d(Cs)/d(t) = rs Explicit equations as enterad by the user (11 Km=4 (2) Ysc=2 [ 3 ] umax = o.a [ 4 J rg = umax*Cs*Cc/(Km+Cs) [ 5 J rs = -Ysc*rg
For t = 6hrs, Ce= 5.43g/dm3. So we will have 3 cycle of (6+2)hrs each in 2 batch reactors ofV = 500dm3• Product rate = Ce x no. of cycle x no. of reactors x V = 5 .43 g/dm3 x 3 x 2 x 500dm3 = 16290g/day.
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 =(aµ+ f3)Cc] µmax = 0.5 h-I, f3 = OIg/g.h,
Ks = 2.0 g/L, Yx/s = 0.2 g/g,
a= 0.2 g/g, Yp1s = 0.3 g/g
Cs,o= 50 g/L
Solution:
Cs,o
= 50
/L 0.1 Ce
es.ce
A steady state material balance on the Bioreactor (including the recycle device) gives:
Accumulation =
outlet +
inlet
Cells:
o
=
Substrate:
o
= D *Cso
Product:
o
=
o o
- D*O.l * Ce +
generation µ*Ce
(7.19.1)
-D
* Cs
-µ*Cc/Yx/s -- rp/Yp1s(7.19.2)
-D
* Cp
+
(7.19.3)
7- 43
~--------------------------·---------------------------------------------·---
From equation (7 .19 .1) :
C,
=
0.1 * D* Ks µ max -0.1 * D
. D * (Cs.o - Cs) From equation (7.19.2): Ce= ( µ aµ +. f3J .
--
Yx,s
+
(7.19.4)
D * (Cs.o -Cs) = ( 0.1 * D a* 0.1 * D + f3J'
YP,s
Yx,s
+------'-
YP,s
(7.19.5) Rate of production
rp = ( a µ + füCc =
(a*O.l*D+/J)*D*(C
s.o
- O.l*D*Ks) O • 1 *D µmax -
( 9.1Yx,s* D + a * 0.1Yp¡s* D + /JJ
(7.19.6)
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 = 3.765 hr". -----,---------------------------------,
Recycle CSTR - Fogler 7-19 80 70 60 50
------------,
40
-
30
--Cell Conc
20 10 0
--·
Substrate conc
--· Rate of Production
J~---::::::::=--::::::::::.... .......~~====:::::;=====:====:::=======--~o
1
2
3
Dilution Rate, 1/h
4
'--·------------------------7-4.3a
5
P7-20 (a) X1 + S -·-> More X¡ + P1 X2 + X1 -> More X2 + P2 ForCSTR,
se, _ ( eso
-- D dt
-
dC
~=D(-c dt
Xi
)
es - Ys IX
ac,
.
1
(
·--'dt
rgX
1
=D-·C
X1
) +r
gX1
-Y
X1 I X21
rgX2
)+rgXz
rgx1 =µ¡Cx 1 See Polymath program P7-20-a.pol. POLYMA TH Results Calculated values of the DEO variables Variable t
initial value O
minimal value O
rnaximal value 1
final value 1
7-44
·--.
.
-
...
·
. .
----
---·----
.---
. ·-
.
.
.
.----~-----
es
exl ex2 Kml Km2 ulmax u2max rgxl rgx2 Yxls Ysxl Yx2xl Yxlx2 eso
D
10
25 7 10 10 0.5 0 . 11 6.25 0.55 0.14 7.1428571 0.5 2 250 0.04
l. 2366496 25 7 10 10 0.5 0.11 1 . 4008218 0.55 0.14 7 .1428571 O. 5 2 250 0.04
10 25.791753 7.2791882 10 10 0.5 0.11 6.25 0.5748833 0.14 7 .1428571 O. 5 2 250 0.04
1.2366496 25.456756 7.2791882 10 10 O. 5 0.11 1.4008218 0.5748833 0.14 7.1428571 0.5 2 250 0.04
ODE Report (RKF45) Differential equations as enterad by the user [ 1 l d(Cs)/d(t) = D*(Cso-Cs)-Ysx1 *rgx1 [ 2 J d(Cx1)/d(t) = D*(,·Cx1 )+rgx1-Yx1x2*rgx2 [ 3 J d(Cx2)/d(t) = D*(-Cx2)+rgx2 Explicit equations as enterad by the user fll Km1=10 [21 Km2 = 10 [3 J u1max = 0.5 [ 4 J u2max = 0.11 [ 5 J rgx1 = u1 max*Cs*Cx1/(Km1+Cs) [ 6 l rgx2 = u2max*Cx1 *Cx2/(Km2+Cx1) [7] Yx1s = 0.14 [8] Ysx1=1/Yx1s [ 9 J Yx2x1 = 0.5 [ 10 J Yx1x2 = 1/Yx2x1 f 11 J Cso = 250 [121 D = 0.04
8.2
c---1
[_~J
6.4 4. 6 2.8
LO
o.o
0.2
04
t
0.6
0.8
LO
7.30 ....-----
7.24
GJ
718 25.32
712
25.16
7.06
25.00'---~-·
o.o
0.2
--------~--~ O 4 t o.s
0.8
1 O
7.00 --~·--~O.O
7-45
0.2
0.4 t
0.6
0.8
LO
0580 ~-------------~
7.0 58
~\
4.6
\[;;]
0.574 0.568 ·
0562 ·
3.4
22
~
1.0 O O
0. 2
....._. 0.4
t
0.556 "_·~·
0.6
P7-20 (b) When we increase D, Cs
0.8
,., .....
l.O
0 . 550-----~--~--~--O.O 0.2 0.4 t
0.6
0. 8
increases, Cx1 decreases, and Cx2 has very Iittle decrease.
P7-20 (e) When Cso decreased, Cs and Cx1 both decreases, Cx2 has no noticeable change . When Cs¡ increased, Cx1 increases, Cx2 has no noticeable change for Iarge t P7-20 ( d) lndividualized solution P7-20 (e) lndividualized solution P7-21 (a and b) Run
No Y east Extraer Yeast Extraer
#1
#2
The percent volume of the growth prodnct H2S collecred above rhe broth was reponed as a function of time: celI
+ nutrient
... ,. more cells + producr
(a and b) Cx
= or
Cxo&(1··i....)
Cx
=
Cxo eme·µ.~ ..
curve fit exponenrial curve:
7-46
..,-------···---
LO
where A
= Cxo e iJ.
tiq
Run 1 (between 15, 20, 30 hrs)
Run 2 (points 10, 15, 20 hrs)
µmu
= 0.2125 hr :'
µmu
A=
7492.6
A
t1og
=
t1ag
= 6.0 hr
§2~t.) ==
ln (~{4
=
=
0..3124 hr"
55717
0.21252 ilag
=
5.1 hr
P7-21 (e) Srationary tstationary between
Time 45 to 55 length of time 10 hr
25 ro 15 20hr
P7-21 (d) Production starts at the end ofthe exponential (for both runs) P7-21 (e) dCc. = D (Ceo · C'--c} + µCe = O dt
D{C'..co ·Ce)+ µCe -DCc+µf:C
Ce =
=
O
=O
O or D
=
µ wash out occurs when D > llmu
P7 -21 (f) Individualized solution P7-21 (g) Individualized solution P7-22 (a)
se- = O ( Cr- .... C',,r_·) + r_g --""'" dt .-..... e,: ...... -
~lma:..
= 1.5 hr
1
Cs;;.: == Ceo = 0.5 g/dm3 K;
=
50 g/dm1
Ks == 1 gjdm3
O = - DCc .. + r.,"'
D = 0.75
Cso = .30
)'CJS
= 0.08
7-47
e, Ce \l e-·S (i. . ·k Cs)1 f1
µrna.
O = ·· DC',e
+ ---·-·····---------(
\ KS
·1·
-t-
l
f - ··-----------·-· -·-
1 Ks: 1
1,D 1
=
}l:r.ax
Cs
'1'
D
Cs ( 1-·+-~-)).
\ , K¡i., ······•4···-.. ---L------······-~- . ·-·-· - .. ,
........... 1.20 __
, ¡' Ce ¡ l + (.. s \.1 ,.' -2..! 8 )
¿, ......... - •.
\l Cs -··-··-
=
Cso -
YSiX
Ce
·~-~-----·------···--·····-.....----
I
by definition
!2.5 Ce :::;
YC/S
(Cso. Cs}
pick Cs , calculare D and (A":
¿¡;~ !tíl
Cso == 30.0 g/dm-3 D = (
¡;
Ce "" (30.0 · · Cs)(0.08)
For D = 0 . 876hr-1 (Cs = 2..5g/dm3),production rate is maximum
P7-22 (e) ··········· es
Ce
35
z.so 1 oº 2.00 1
oº
1.50 1
oº
30 25 20 1 5
s.oo o.on
1 () 1 0·1 1
5
o
oº
-5.00 1 O
1
.o
2
O
0.2
0.4
D
0.6
1 2
0.8
For D = 0.27hr-1 , Ce= O if Ceo= 05 g/dnr', For D = O.J14hr-1 , Ce= O if Ceo= O g/dm", And for maximum production rate, D = 0.876hr-1
P7-22 (d) For batch reactor,
7-48
o (/}
dCc
--=r
dt
g
See Polymath program P7-22-d.poL POLYMA TH Results Calculated values of the DEO variables Variable t Ce Cs Ki umax Ks rg Yes Yse rateS
initial value
o
minimal value
o
maximal value
final value
2
2
0.5 30 50
0.5 2.0lE-07 50
2.9 30 50
l. 5 1
l. 5 1
2.9 2. OlE-07 50
0.4591837 0.08 12.5 5.7397959
8.744E-07 0.08 12 . 5 l.093E-05
2 . 9019168 0.08 12.5 36.27396
8.744E-07 0.08 12.5 1. 093E··05
l. 5 1
1.5 1
ODE Report (RKF45) Differential equations as entered by the user [ 1 l d(Cc)/d(t) = rg [ 2 l d(Cs)/d(t) = -Ysc*rg Explicit equations as entered by the user [ll Ki=50 [2] umax e 1.5 [3] Ks= 1 [4] rg = umax*Cs*Cc/(Ks+Cs*(1+Cs/Ki)) [Sl Ycs=0.08 [6] Ysc e t/Ycs [ 7 J rateS = Ysc*rg
40 24
32
18
24
12
16
6
8
b=============:::'S=1 04
o e.e
0. 8 t
1.2
1.6
2. 0
o
P7-22 (e) For semi-batch reactor,
dCc dt
=r g
- voCc_ V
µmaxCsCc = --·--=:::;;__::_-=.._
r
Ks+Cs(I+Cs/ See Polymath program P7--22-e.pol. g
K1)
7-49
o.o
··-···-·~..,-04
0.8 t
1.2
L6
2.0
POL YMA TH Results Calculated values of the DEO variables initial
Variable t Ce Cs Ki vo Vo
o
value
V
Csin umax Ks rg Yes Yse rateS
l. 5 1
l. 5 1
1
1
2.2464622 0.08 12.5 28.080778 530.94351
0.2329022 0.08 12.5 2. 9112771
0.487013 0.08 12.5 6.0876623 5
vce
o
0.2329971 O . 8327919 50 50 10 10 30 l. 5
0.5 2 50 50 10 10 30
final value 4.5 2.2593341 0.8327919 50 50 10 235 30 l. 5
maximal value 4.5 2.2593341 24.016878 50 50 10 235 30
minimal value
5
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*CcN [2] d(Cs)/d(t) = -Ysc*rg+vo*(Csin-Cs)N 30
Explicit equations as entered by the user [ l] 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] Yes= 0 . 08 [10] Ysc = 1/Ycs (llJ rateS Ysc*rg (12] VCc = V*Cc
w
18 12 6
=
o
30
[:J
24
r.;.
24
119
O.O
600
---------·
480
1-
'
18
360
12
240
6 . ..
o o.o
__ ._.~ ~··-"'. , .,., .
L8
3. 6
2.7
t
4.5
vcc]
120
....
- -~·-----·. 0. 9 L8
--
2.7
3. 6
4.5
o
o.o
0.9
1.8
t
2.7
3.6
4..5
7-50
-------------------
-
-----
--
-
- -------------------------------------------
-----
.
---------------------------------
P7-22 (f)
Individualized solution
P7-22 (g) Individualized solution P7-23 (a) Fit the data to the equation: D
= c;sµ_.
Ks+Cs Using POLYMATH, fínd the values for S-·29
J. 1
J_
l./'
.------
,,.,., . ,."'·-"'' /
-----
µnw: and
Ks·
They are 198 and 0.97 respectively.
--·-------···=--.
1 ¡/
c.. S·JC
'Y
··ti •.•. _ . l .. ooc
.· -··· 3.. 000
. ~·-. . .... . .
'5.0CG
.
-~--··Y-:
7 .. 0C:J
----··:t í..S s.coo ¡¡,QQC
P7-23 (b) Using this equation, solve for Y c,'S:
With rhe given information Ycis:::: 0.099, therefore, Y5,.c is equal to the inverse of that, 10J)75.
P7-24 No solution will 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 will be given. CDP7-B
7-51
---
-·--------
----·--··
---
Given the following reaction schcme:
i · + OCl · ·-, OI ·
+ C! ·
with the following rate law: Active intermediares
assumed to be HOCl and HOI
From table I, che first rule of thumb suggests:
k¡ OH·+· HOCI ""' OC! . + H2.0. -r,
k¡
= k¡
CoH C¡¡oc¡ • k
.¡
Coa·
CH;0
k2 OH.. + Hor
OI .
<,·-',
H20 ,
+
'"T2 "" k2 (~)H · CHO! ...
k ·2 Cor C¡.¡p
k ·2 Toe rhírd rule of thumb suggesrs that the reverse reacríons occur, A chain propagation srep,
involving the ccnversion of orre intermediare into the other mighr tie in both the reactíons above, k3
I + HOC! · , HOI + Cl
,
r3
This step makes the overall reacrion sequence r · + OCl ro, :;:;:: r2 :;:;: · k.2 cor CH ¡O ... ruo,
= ··r2
k3CrCHoc1
+ fj ""ki,'orr
= kzCoH
Then;
(',{)H ·
·roI == ~
k.iC:01 Cnp C H(X.l
r
CH(X.1
:! k;,_C¡
k1 CoH
=
-i-
= KC. .....eL ..COC'
=
·rw
.. k · l c(Xl
CrhO
k3 Cí·
where K
is cíose; with k1 >> k3
«», = ----k 1
An alternative approach assumes that reaction l
... .
. .
Ihen ·-
= k,(., re .
rol
.,
-,.
+ h C¡ CH
, . , k .i Ccx::1 CH-0 k .. ¡ c(XT CH,:0; t.e., CHCx.1 "" ;----·----. ......... · K¡ CoH·· + k3 Ct
C.:'zx;r.<;i:1;o
_.!__
01
k.zCor CHiO ·~ k3(,\ Cuoct zz: O
C:1101
CHO!
+ k3 C¡ . ]
· · .:; OI · + Cl · possible:
kz CHO CHO{
. rmx.1 ::::: -r¡ • l3 "" k I CoH ·
or [k .· 1
= k1 C¡ · CHoc1
·1·k .3 k
L
'. . HOCI ;;;:
·'1
quickly attains equilibrium, then:
,
k,
eotr
These two approaches me basically equivalent
CDP7-C
7-52
(a) Assumptions: • Transfer rate frorn bubble bulk to fluid interface is not rate limiting, i.e., C¡ is the equilibrium oxygen concenrranon • System is.ar pseudo sready-state with regard to rhe particle size, i.e., parricle growth is slow compared to oxygen rransfer, • Rate of oxygen consumption is directly proporrional to the cell growth rote:
ro1 = y~
=
re
where yo1 yield of cells on oxygen
{i~~~~J
This irnplies thae any oxygen utílizaríon requíred to maintain the cells is neglígible, and ihere is no signiñcant merabclic produce being synrhesraed.
Oxygen balance:
wbere:
k¡ ah = overaíl mass transfer resistance frorn the bubble to the bulk,
ac k y0¡ Rearrange
= surface area per gram of cells = mass rransfer of cells on 02 = yield of cells on 02
Oi balances:
Ro, ::; --·-·· k1 ah
C'¡
Y9i1% = Tlk C'.c
-
e,¡,
(1)
e
(3)
-s
Add equations 1, 2 and 3
i~ = k/ab
+
J-c (kc1ac+ ~~)
(4)
(1) When oxygen consumption by the cells is slow, the process is reacnon rate límited .
Thus r¡ ··.) 1 and:
CL Ro.
= .. L . . + .101 _ k1
ab
<=<: k
(2) When oxygen consumption is much faster than mass transfer, the mass transfer · · · .::1actor. · C; =,mes r h• e ¡·umnng --'-·· . .,..· -· ····· l ··-Rei k¡ kc oc Ce
i..---
ªº
7-53
(b) To increase the growrh rare, you could: • Increase k1 ab by increasing fermentor agítation, • Increase the concenrraricn of cells (since this is an autocatalytic reaction) • Increase clump surface area and the effecriveness factor by decreasing the partícle size (also by increasing ferrnenror ágitarion) . (e) Re ""' yo1 R<'h From equation (4). have: 1n. ~e:, ~ 1(; + (k~.' 1111 C; "" YOi L... L "'l ªb . ..,. ""\,. 1 ,
'{k;:;"
c
ac:
( 10:ic; [¡{~~
ªº
Ce · · CcQ
-~k¡
+
{
+ ~.;:
+
~t]
dCc "
1 . ;Y0,) Ce k(-;;;-¡;.,. 1k In Ce~
+ Yt~] -1 nk j
L
di
.~
'"" Yoi C¡t
(d) Assumprions: .. There is a constant nurnber of particles
• Each pellet is roughly spherical and has constant density, We do not k:now which resistances
are controlling, so we know there are no reactíon
Iimitaríons, but may be either interna! or externa! diffusion Iimitations.
~~ =
y0z
C¡ [¡}ª~+ ¿c·(~:1~; + ~~)j .
¡
As panicle growth increases, kc, ac and T\ will change as functions of the particle diameter. Thus need to find particle diamerer as a funcrion of Ce
where n = number concernrarion of particles ( 111)
Ce ""' npc V e Ye= zi;. d:,
·
d3
P
=
6 P §9;;.,,
Pe :::: the densiry of the particles (gil)
rcn pe
=
Ve
particle volume (l)
Internal diffusional resistance can be rnodeled as: ¡,.
:::::
T)I\.
a, do•
External diffusional resisiance with
or without snear is: l,,5$b.$2
ac
Re where
ab
(:t:1
:o::.
c.¡
1
!
.
+ ·-'·--·. Ce \ª1
nn Pe
and
C·e!,',- + a., C./. ~ h'3)] ~ ,.
a;
=
o:2
.J
rrn Pe
Dropping the primes and simplifying:
qi )
7-54
Cc1
¿t
{c:¿13 · c¿&3) + 2~1.(c¿i13• c¿,¡f} =
0 + 3c,1
y01 C¡
t
(e) From pan (e), we have:
For a vigorously
stirred fermentor,
assurne that fluid shear is sufficienrly
high,
that transpon to the edge of the floc is neg!igiblem:
kcac
-·} o
Toe mass transfer resiszance frorn air bubble to bulk liquid depends on the fermentar design, aír flow rate, agitation tate and a number of other factors, For a 1 O l laborazory scale fermentar, k¡ ab was found to be 150 mM..Qk;. during the growth phase ofthe >-
l hratm
fermeruaticn . C2) Dividing through by Henry's low constant:
k:1 ab
= {150 m.MQ!t)(0.88 m1n.L. . }(;:;JJ1r. . ) l hr arm
= 3.67 Effecaveaesa factor:
X
mMole
:,600 s
10•2 S ·1 Microbial growrh en multíple subsrrates (here oxygen and glucose)
is typically rnodelled usíng Monod type kinetics:
Re = By represenríng
µC;;
the reactíon as first-order with respect to oxygen, we are essentially
assuming a low oxygen concentrauon,
relative to rhe intrinsic rate parameter;
7-55
Ko:
_ _¡__ kcac
--t
o
The mass rransfer resistance from air bubble to bulk Iiquid depends on the ferrnentor design, air flow rate, agitanen rare anda nurnber of other factors. For a 10 l laboratory scaíe fermenror, k¡ ab was found to be 150 mM~ during the growth phase of the 1 hr atm fermentarion.(2) M
Dividing through by Henry's low constant:
= [iso ml hrMole) (ü.S8 .. .mmL){-.1.Ji..r.) aun rn Mole 3600 s .: 3.67
X
Effectiveness factor.
10·2 S
-i
Microbial growth on múltiple substrates (here oxygen and glucose)
is typically rnodelled using Monod rype kinetics: µ ( hr · l)
=
Re
=.
CN--· } ( µrn ( :--·..::::··
--=.1..-Co.
.Ko
KN -r CN
Cn
)
t
µ(~:
By represenring the reaction as first-order wirh respect to cxygen, we are essentially assuming a low oxygen concentradon, relative to the intrinsic rate parameter, K0: For a fírst-order reacrion. the effecriveness factor is:
i]
r¡ "" · ~ ( cosh
qt
TI
=
0.45
Reacnon rate constant; 8 . .3
µmax
k
= . Ka
=
10 S
;:·;··;·;~·::i§i. X
S .. J
=
l Finally, assume an inicial cell concentrarion of 0 . 25 g cell/l, the cell concentration equaríon now becomes: ¡·,
, - g ce 11 ( Ce · · 0.25) ;:;;_____
3.67
X
10
-2
s
-l
j
+
1
! {();S·¡'{o:~ L
-¡
"' cell 5 s,....::::.:_ crQ
• ·· \
¡
~-OJ-:-~:,,In •·1¡ g
4(:'c
.: ;:
2S .
27.25 (Ce · 0.25) , 128 In 4Cc = 0.012
t
Clearly, rnass transfer from the gas to rhe liquid phase and internal diffusion play i.mponant roles in deterrnining the cell growth rate
7-.56
Cell mas vs time. Start at 0.25 gil t (hr)
Ce (g/1)
O 1 2 3
0.25 .,._ L13 2.37 3.74
4
5.17
5 6 7
Ce ( í!./1)
t ihr)
. 8 9 10 11
1 Ll5 12.67 14.21 15.74 17.28 0.62 1..72
12
6.64 8 . 13 9.63
0.5 LS
From the graph it can be seen that growth stans out exponenrially and becornes linear as the fermentation becomes lirnited by gas-liquid mass transfer, Sensitivity analysis: TI1e gas-líquíd mass rransfer ccefficíent is related to the agitadon rate to the 0.95 powe.0): k1 ao et N0.95 What is the effect of lncreasing the agitatíon by 50%? k¡ a2
=
k1 a¡ (1.5°·95)
::=
(3.67
X
1Q ·2}s ·l(1.50.95}
=
{5.39
X
10
(see graph)
Since cell growth has an exponentíal portion, another way to increase the growth rate would be
to
increase the innocuous size, C<:o. What happens when Ceo is quadrupled?
=
C'-co
LO g/l
(see graph)
From the results shown in the graph, a relarively small increase in the agitazion rate leads to a significant íncrease in rhe cell growth rare, while an increase in irmocaíum size rneans thar the fermentarion reaches a gas-liquid transfer-limited state more quickly, but the growrh rate remains the same,
cell mass (g/1)
Cell Mass vs Time lora STR P( Ierrnentatíon
18
-
16
...
14
......
12
.
I
¡ I
I J
I
\<:¡ a " 3 . 67 X )O -l S l
10 .. ,
:~
I j
J.
!
4
1
2 ......
Ceo = o . 25 g/i $:¡ a = 5. 39 X !O Ceo " O 25 g.,1
/
k¡
a
=
J 6.7
X
Ceo "' 1.0 g/1
'/
"T
·-i 2
4
-r- r·~1-r .. "T" 6
8
10
12
14
t (br)
7-57
10
·l S -:
zs
l
References: (1)
James E. Bailey and David F Ollis, Biochemica!
EngjneerinP'
Fnndamemals,
(NY: McGraw-Hilf. 1977), Chaprers 7 and 8.
(2) GF. Payne, PhD Thesis, University of Michigan, (1983). (3) D.LC. Wang, et al., F~rmenratiQ.Q_and_Enz,,rmeTechnolo2'v, Sons, 1979),
(NY: John Wiley &
Chapter 9
CDP7-D No solution
will be given,
CDP7-E Since the denitrification follows Michaelis-Menron kinetics, fírsr determine Vmax and Km from Lineweaver Burk plot. Inirial
[ N O-z] ( prn) _____ E .... -.
r--;--1---·-
Time for 50.6º
tN02]
·--- - -
reduc1ion.Jm02~{!::om/mjn) -----
25 50
.04 .02
35 38
75
.013
44
100
.01 O
200 (given)
.357 .658 852 LOOO l.515
50 66 (given)
.005
1/r
_
2.800 1520
1.173 LOOO 0.660
initial [N02 'l / 2 rime fer 50% reducrion
where the rate of reaction is found by tne ratio: 3.0
Rare of reacdon
··1
2 . 5 ··, 1/r {~~~)
2 ..0 1 .5
LO
l
0. 5 "T" 0 . 00
- - - ---·-···---· 1 - ---
0.01
0. .02
,·--·L-~ [NOd
-T-0.03
·---¡ 0.04
npm .. ¡
..
7-58
-·--------·--·~--------------------------------------------------------·----------------
------··--------------------------------------------------------·--~----------------------
Lineweaver Burk: From the plor, V max
1--- · = 2 . 73 EE~ intercept mm
-
:. Km = 165 ppm Next, need conversion as a function of time. Design equarion: RateLaw:
··Is
Stoichiomerry:
Pinally, Vm,u;
Nso~
==
O.!
vmax
-rs V
s
=Km+ s S
=
So (1 · X)
=
So X - Km In ( l ·· X) K.now V rr.ax , Km , t = 68 hrs = 4080 min t
ol.} ·(46 .mol JL.) {!QQQ 1!:~} { LPP~) = g , mg/1
So "" 0.25 ·(rr1 • l Iteration
to
11,500 ppm
find me conversión obtained after 68 hrs gives:
The [N02J level is: ( l ·· 0.930) (l l.500 ppm)
=
X
=
O. 930
805 ppm
Since t, Km, and So are fixed by the system, the change rnust be made in Vmax· X
Desired conversion: Desired Yr;,u
=
=
1 · _j()O __
l 1,500
= . 9565
11,,sooJ.9565) • _165In(1 _ ·_.9565) 4080
=
2.823 EP~!: mm
Since Vmax o. [Ei] , increasing the concemrarion of whole cells in the emulsión will increase
Vmu :
(2..8.ll) 50 - ~[_~el~--273 rnl emulsión
== 51.7 ·- rng c:lls -- . ml emulsión
Therefore, increasing the cell loading to 52 .... mg~ell_s - would results in a level of (N02] < mi emulsión 500 ppm after 68 hours .
CDP7-F No solution will be given.
7-59
CDP7-G Mole balances in a CSTR DC50
·-
-DCP -·
C =
-~
LJC5
+ rs
rl',;;::
O
.,
(
:::=
O
¡
L
D
the rate law as given is: V
C ' ..:.e. )
r ;;:: ·-· JU!1,.~::á . l 1 ..
r
e''p )
,,. , C 1 l\.s + -s \
J
Plugging those into POLYMATH and using different values of D and Ce, come up with the following: Using a volurne of 200 drrr' and a cell concentration of 50, get the best production of me L malic acid, Ec¡.M:.iot:~s ;_
.,,
ftcsJ=cs-cso+rp/D cso=2
Solwt!on i()
CC'-'SG
k.·.n== 04.8 cps t.e rv L. 59 rp;::ro·,,,1max~c.s1
o. :35 i 657
es
v.na.xo=76
cp
L 64834
eso
2
ce
5C
vrnax km
76
u {;<:,.rn"'·cs:}
* {1.
cp/cpstar)
•,~e
3.222e.>·· 15 <3. 2770-· 15
1.59 82.·'>171
CDP7-H Michaelis-Menton Kinetics: ""[
s
--
vro-,/'s_ Km +Cs
Km :::: 6.2
X
10-·2 ml/ml = 6.2% E ==40 mg
7-60
-·-----------------------------------------
~----
...
.
--------------~---~-----------------
Mole Balance: .. dC.s. dt CS'i!,
V m;,,Cs% K.n +C5\!,
===
= C:80.., (1 ·-· X)
lA--0.2 x=-···-""'0.86
::::::>
L4
Where C:5~ is the percent of fish oil. For rnost oils:
., density 0.93 C1 =··---=··-·-=·. MW 300
o'
xl
e, = CrCs<» =>
ºJ2K 14
3
J •(':"'"' dt:::::: j · 1
1
··s%
= 3. X 10 jlffiO 11m 1
ml
5.6xl0.c3
V
+e
m(. ~ --~{iCs%:::::: 1.4)
mol/
O
'T
O
3.
· · dt
5.6x 10-·3
.
=> ~In ( O.Z +(1.4 -0.2)=--3--·-t =>
t=·
6.2 ln('l)+ 1·2 rnm== · '7106 · 118 4 hr · _ ·.mm=·. •·s
5.6xl0 3/3
CDP7-1 No solution will be given. CDP7-J {a)
No solution will be given,
(b)
X
n
l = ----· 1-- p
For X0 == 5: X,=10:
X"
= 20:
P = 0. 80 p=0.90
p = O 95
7-61
.
Use these equations
to generare the desired graphs:
y¡ vs, j for X0 = 5, 10, and 20 O 25
t
0 . 20
Ii
-·-:·:-·:-:-. . -
¡:
{ 1
0.15 ;¡...J
t•¡' :
j
Xn"S -
1
-Xu.;;10 ·Xn=20
----··-··--------···--··-·---·
¡ 1
t
¡
O. !O¡',
1 \\ 1 ;
O.OS·~
o.oo
L----~~--;:, .=. o 20
~.··~""."'¡¡¡;;.;;;;::.
40
j
=--60
.
80
100
P-J = M o ',fl P )z PJ 1 = M(l - p) o • 2p9 Use the above equation to generare a graph of P10 vs, p:
o (e)
wj
p
04
0.2
0.6
= + j(l - p )3 (j . - 1 )pj
0.8 2
Use rhese equations to generare graphs ofyj and YJ vs,
o lo . .,
for p :.:: 0.80, 0.90, and 0 . 95
.,
'' '
-
p "'0.90
- ·p"'o95_
002 ,., O 00 -002 0.04 -ü06 -.O 08 ......__,.
i 20
60
40
0.07
¡ ~
0 . 06
f
,.,-------------------.. --.. ----··--------·-----·---- ..·-------·
....,
¡
t! +
7-62
1
i· /
p -- --- - p
,. ·
·---¡.
l
= 0.90
_p=0.95
\
/ ·'\-· /
V; . .
O 00
..... ,,,
,
\,
¡
\
.
<, <, -,
\ \, - '"" '---..
<,
.
¡
• - .•
"· . , -
20
40
i
60
jl
- - .
-
80
100
k..-------------~- ........,-...:~ .... -.
o
j
Í \\
¡ 0.05 -fi 0,04 0 . 03 0.02
80
.
t
O 01 -. ,._,
vs. j.
w J vs, j fe r P :.:: O.SO, 0.90, 0.95
'
008 0.06 0.04
wí
(d) M = µ lv1.r 11
-
11
.
"
=
. •...
Mw=µwMM
MM_
1-p
·
l+p
=MM-1····-p
D=µ"'. =(l+.lJl(l-p)=l+p µn 1-p) We must find the valué of p.
mt~(!t:tnex{=t~)-1];-Q.067 Nf =2.803
I2
=I
exp(-k t)
20
0
p == {J = ----·
= (O.OOl)exp[(-l.4
X
10.3)(14,400)] = 2.794x 10·9
-· · ---· . . . -. -~PM_···-· ···- .,. ;.
· ... · · · ·
kpM + kmM + k.C + k$S + ~2ktkJ(Ii)
= 0.99991
This can then be used to calculare the desired valúes: 104 = --·--·· · · -- = l. lJ)- ~ x. 1. O l ·-0.99991
-·- . .
NI
a
6
(e) Mole fraction of polystyrene of chain length 10 (y 10).
I2 = 110exp(-··k t)
~t' M ; M,exp[( 0
8
r(
ex~
:t' )-1 )]
p = f3 = ····-· ···----~E~'! r··=· =-::-=· = k p M + k m. M + k.S + "\j12k.k S , O· f(I 2 )
Y0=(l-p)p1H
n=lO
Use the above equations to plot Y« vs.t.:
7-63
y
10
O'' 04 · t ---·--+ ·-"·--------·--
vs, t ,.,.,,,
O 035
1
0.0.3
1 '
0.025
;.o 02
1¡
0.015 0.01
¡
0.005
¡
¡
..::=:::::s-------. . . J 2
5
3
t (hr)
CDP7-K Reaction R.+¡.
5L.,P,
J
¡
(a) and (b)
L;j·= ( ~,,;:· .
XN ==
''L.,¡r
"\:' '], ¡
--r
e-r-
J
,L..ir.
=r+Y'fk --·R. r L., ~ p M(R. · J
)+1k a +k'
J-l
\:
J
)R."J -'Rk L
1'
MR.¡+k a CR.J+kSR..s +· k.fR. )r, J
M '·
J
Let "'\:"""'R ,L., .1 == R' , ... r,_. == r:r k
XN
XN (e)
=
·_.'i
+ R. (k m i'vf + kS + k.,.!) ->
•
t,~~-fttp::~,s·_!c,!J'J"- I
= rt . ..-'~ "
.,1
+
lm·+ kk,~1 p
p'
+-:·~í·] 1 p
'
~[~-+ !: •{i+{'if ~ (1rrft <:+ :;!+k:tl' Frórn the above derivation we know that
Neglecting the solvent
term and rearranging yields:
Substituting in for -rM and r¡ and simplifying:
7-64
-------
---------------------------------------·---------·----------
--------
-
_ 2k,.J(I2)/?v1(-rM)
1
.
krn
k1)
""""" ·····~- + ·- + -···k!(2ki(I2).1k.1) k? kPM
=······ - ' ""'' ' ~., X¡,¡
l _kJ·rM)
. .....,.,..
XN
·~
''1··_
.
k;(Iv1)
. km , k0I
...,.... ...
k,
-.-..,.,..._
---·
kpM
To determine rate Iaw parameters experimentally from a CSTR both the final XN
(d)
value and the final concentrations of M and I must be recorded. These data can be used in the above equation to find valúes for the parameters.
An increase in ternperaiure would cause an increase in ali three primary steps of
{e)
free-radical polymerization (initiation, propagarion, and termination). the overall rate law:
l2k0(l,
= k?M1í·---·-···=M kt
r .
By looking at
)f
·
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 polymerization.
CDP7-L a)
PFR:
d]Vf
---dr = r:M r-:--·--·-
J2k0L,f Ii-.f = -k?M~~;; ... Plug those into POLYMATH
r
l:
o L..
:::; . . . k
to gel this graph,
§~ati2!:;!,:,
g!i:. t.i~l: Y§l-11:ie
d(m)
3
/d(tau);rm
d(i)/d(cau}=:·i.
n Ol
kp=lG ko,.le-3
ceo
!,E,t~!!;1....v.3l.~
M.Axi~ . .:t!l.Ifft
O
199·93
~"Í;'fi4l..VA~U# 7j9
1 -,·n;n
.J 01
s . ~~: ::... -,a
s .o.;;.:·:e·H
l~ O 0-01
lO 0.001
tt• :) :~:
10 0 .. JOl
s"'= .... !1'1
se+•~n
s-e ... cn
·•1e·•05
··S 0~1.'11,e.···'U.
"'1.J4¡¡:_~ .. i)$
··i 99:lS8·e .. ·23
sc ... ::r; "':,« . <)$ .. ~ 'i.:.1.-S4e·-,0;5
o .o;
kc:"'5e7
:rm.. ,,k;p*m•sqrt(2*ko~i""f/ktl t:.au.0 sa n ,
va:.u~
:.t .17l29:
J
f.,,5 zi.w- .. ko*i.
~:.n.~::iu-m...
o.s
7-65
e.s
es
o.s
·S.·:.Hi!."1l.e···4:
~2 9-g:Se1: l:3
J
CSTR:
I:w --- I2-= ···r1: *"",. · . YM"ATH tb i 11 are The rate laws are the sarne so agam usmg POL · . e o ow inzo ograpas . .M 0 --M:;::: -····r 1 * t "
.
·
generated. lnltiator Concentratlon vs spaee Tlme CSTR
Monomer vs space time es TR ¿ q.
!
e
3 ......
z
··-··········-----------------·---·
'i
• • • • • • • • •
•
...•..••.
,
•..
.
••
'
:i:. o . . ;:,·t ¿ O 009
•
2.a :2 7
,:2 O .006 a.001
!
u,
'.2:-•
.g
g
2 3 2 l
~ ~-:~~; r. o . ,.
~ 46 :1.S
! E
C. o ces 0.005
r. o: +.!.. . .:!-.~
B.
=.:;;
~ z~ 50000
J.
OOOOC-
1 SO
'ZCiOD(IO
f
.0, .. 002
G.001
r
~ .. . .. ..+-·---·-··----+
seccc
T.au 1J'::1
1
ceceo Tau.,
1
sccco
zococc
z sccee
H:,.
2.5
o . oc 1. :mo:::3 t.au
e
Se Ll,
ioz.01 kp.=:1 (} ko"-'le-·<3
i=.5 kt:.=5e}
b)
For two CSTRs, the design equations change just a bit ,¡··
M.o ··M¡ ;;;;;é··rM, *·:-2
M
'
t.f' ::;::: ' . L -
"':
*r
2
I )¡ -
•.•
I,_._, :;:::
* T2.
The rate laws are the same with the exception that instead of just lz or M, I2l' 122, M1, or M, are used depending on which reactor they carne frorn ami the following graphs are generated
7-66
zoo
en., csrn"" Two CSTRs .l
,
.
O.. !)t
1
•'•
i'.'i!i
II
1
·::
1
z .••
···- .. ······-~·~·., .. ~ .. ·------·-···
..... .,
~-00:9 .;
iI•
.. [
a--,-----·-·-.
••
•
...... .. . . . .
!,H)OIXI
,,,.,, l:OCOQQ
...
'.·
r
"00• [ 00(),l
•
0,01
,.
• ,
i ~QQQ:
'.•
t< .:.,
O.v'C;i: ~
4:QQ(XXJ
o
...,
,; ----·· T.-..
lh
1:$0ó!k1
e) Making k0 bigger causes L2 to decrease rapidly and M does not gez formed as much, Increasing kt> causes M to decrease slightly, but noc by that mue h. ú1c1easing ~. causes M not to decrease by very much stayíng very close to 3.
CDP7-M No solution will be given CDP7-N l +.o/1 --5-~R
+ ,vi
R. I
a)
--11
Balance on I
=
=> I
-
ca
l
!:.e. :,. R .
r1
= k/vfl
.
l+ik,M
-
= k¡MI + kpM'.¿Rj
--t~I
-
As¿Ri
=I
0
--
I
j;,!
··rM
.. , ...•.. .., !
O .?OS L
•
··~-··········-···-···
t
0.00·1
º ºº'
.
r
o aca
.,
= k , MI+
k p M(I O -- l)
7--67
--------
..
Balance on M: _ lvf0 - ¡y{ M¿ - M r ··------···· - -------------····«-·---. ---·-·-· · r;,1 k.Ml + k/d( 10 ······ !)
!) ""M0 -
:::.::>
rk/vll + rk/v!( 10
=:>
rk,.Nlf,1 ·t'k/ol'4(rk,1W) ...... ·--+ ··-·- .--·-lVI¡·-}1 l +·rkM l+rkiW ¡
-
•
¡
:::::::;. rk,10A1 + -r2 k ¡}JO l1,i2 == ( M O :::.::>
,k,
M
(1 + rk,., ~ )M~ 1
M){ 1 + rk/d)
-
+ ( 1 + 1:k,10
-· 1
==>Al= . :(1 + tk.I¿ ·· rk)v/0)
.
--------· . ·-----·-- .
A10 = O
'!k,i'YfO )AJ
··-······--·=·---~
,,
+ ,,/(l+rk)0~·
·~----··:..;·"'"'""············"'"---······--;-·----·····················--·----------
,k,1YJ0)" + 4 * rk,(1+ rk/0 )M0 2-rk,(1 + tk 10) .. 1
..
<
p
,r
.
,
b) ·-rR, = +k MI+ k r·· MR l Balance on R1 1
Sirnilarly,
=
R,
1
\
j
e)
l
1 +k lvf·t ,
(_
R=
'\
R2 ¡r ············tk.1!../l¡f-- . - • p
)
t.
'¡]
!K.
1
l+tk.A f k p 1\ 1 + rk p iW !
.
1
)
Initiation Rate constant •1<:¡ < < ~ propagation rate constant
Hence, nearly no change in the concentration of Initiator (I). Mo lo ki
kp
1
t 0015
?OC()
7-68
~ ~ ~t
!----~·i·"'l"I!'-·-_·······_·-···_·······_-··_·-···_···--_·--_····-·_··-·=====-·
.lllo __
···_-···---=~-~ ---1 1
0.6 · 0.5 · 0 ~U. oi?S
1
:,i!03I
. . _·····-·-- ----.... ~M/Mo
0 .2 . . 0.1 · .
oi o
1,.
--··---···------'
, · · -· · · · · ., ,. ; ,., . .,., _......... . . - · · : · · -· · · · · · · · 0.5
i
1
1,
:. . .1.5
1
tau d)
7-69
,. . 2
,':,,
¡
.2
As2..,J x ¡=l
x(l+x)
= -(-l:=~~-)T
CDP7-0 a)With the reaction self catalyzed the mole balance and rate law becomes:
dlCOOfll
-- ·-L-:-~
at
. . . , = k[COOH]'
We can then get [COOH] as a function of time. The following graph shows both the given values of p and the calculated value as a function of time where
p=
lCOOl(h:[CC~OHJ p vs time
0.8
•
O6
0.4 0.2
o
500
HXXJ
l5(XJ
nme
It appears to follow thís aoove 500 min,
7-70
20(XJ
b)
The new mole balance and rate law is:
- d[C~t(?!!J = k[COOH]['OHJ[ll"] [OH]= [COOH] COOH;~coo- +H+
= ~q!!. }irJ
K
[COOH]
eq
[coo·· ]= [n+ ]
[H+ 1 = K~q[COOH]
.. .:!l.~99.{!J = kJCOOH] 12 dt . Solving for [COOH] as a function of time gives the following graph:
p vs time
- ~--;··0. 8 . .•• • 0.6 .
o.
'··- ·. '.----
. . . ...............................
-
.
0.4 0 .. 2 ...
o • o
500
1500
1000 time
2000
It foliows the data above 200 rnin, e) This mechanism can be made to :fit either rate law, depending on whether UA dissociates before or after the first reaction.
CDP7-P CDP7-Q
7-71
1
I
+M
R
.!ii.__""7
l
+ i'VI __ -5?.._ __ 7 R J+! -
R, J
.u
~
------, - k0Ml -- k M }kl ----o
dt
dl
.. -- ·"·-------
p
di --= dR J_
dt
I) __ ¡ k,., .
Mdt k
=> I
!J.. 1
ae
·--
k
--
k/v!dt
dii=]:
kp
p
ke :=;
=l
O
- -~ e •,
=k MI + k MR o
=>
p
d\__:_~j de ·'
¡}
\
I
:=::
,'
kol,L' _\2___If'i e(!_ i-~'o 'J'-
1 _
K
k 1' ( 1 -- ___Q_ -,
k !')
'
k,, _ k Jo [. ·,:,ª => Ri - -----. _ ._..',. w•• e , -- e k kp l - k')_ .
J
(
. . R' __ =;,
J .
.
:-,-~¡r~-
~¡¡
~Q·· J'_· k,
k .' _1-- _ _Q PI\ , k;, k 1
---. ! .JLQ
e -e
.
··----8 o e o
k -¡,-º
r
k,. 1 l
Y1
;o //
\
e i;'
k
(
l.
p
k.,, l·o
\.'
.
kuft>e
j I
l1-
6
·i
1 -- ._ _ _ . _ ,
l
_._.,.. .
- . 2 --1! . k,/o._ - . T 8 e _ - .. -----
(
k¡, l -
~Q_
''\.
.kp.,,
l -
,_
')
'9ie·,)
;-_-,1
e
r·._--,~::j_~Q\~JJ=l . . . ~ ~-"r---;J:-·:¡: ¡
-,
I
kp
. l
(i)!! 1 --
-_º_ -
kp
Forj>1
·
CDP7-R
7-72
~-------····------------··--
---
-------------------,-~-------~--
---
R.J +lvf·---~-·~R.
J+l
- _I r :::: ...lA ...... k0A.fl
e'= k11i'vl'r: e'= ~í!:L.::: . !.l k0l
I'f
k/o - - . k08' +s,
= 1'tf .. Q .....__ M...... k/1110
lt1 =
- -1vf ---0
kiJor + 1
e' = . .~i~!?.!. . k/0-r
+1
, __ R, . k01"'vfl ·-· kPAfR,
r =,
=
RI 1:
k&Lte', . . ••.
k¡,,(1 + 8')
= --· , , ,_, _ Ri....... . . .
R2
k/dR1 R18'
·····
k/v!R2
k0k/08'2
k,i08'2
=1::;.z,-; = k)i~e}·{i)i:;;~); c1·.~-e}~~ª;·;::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 acet/g 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 cells/liter.
Additional parameters: µmax
= 1.2 hr-1, KG = 1.0 g/1 , Ko = 1 mg/1, Yxts = 0.5 g/g, Ypts = 0.3 g/g, q01x = 1000 mg/g
Oxygen mass transfer rate kja = 500 hr-1, 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 fedbatch reacator . (a) list ways you can work this problem incorrectly . (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 = O 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 = O SATG = G/(KG+G) SATO = 0/(KO+O) SATP = 1.0 - (P/10) mux = mumax * SATG*SATO*SATP*X
d/dt(X) = mux d/dt(G) = - mux/Yxs d/dt(O) = kLa*(7.5 - O) - 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 critica! 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 critica! values of 0.5 mg/1 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 critica! 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
t: 43 steps in o seconds 18
o
o
0.5 ·
1
1.5
2
2.5
3
4
TIME
In the second part of the batch culture, acetic acid is getting produced and glucose is still above its set point of 1 O g/1, 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 acid/g g cell mass and integrated from the end of first part of batch culture.
R0n .t 2ti steps¡n o s~conct~
.~..:--,;;,;,,....:--,...,,,........,....:--'T""""..:--......+-.....;.;....;,.,,.....;.;.....:--+-.....;.;..."""T".....;.;........;.;...f-,,-;;....,...,.... 45
From the simulation results, we see that glucose concentration reaches the predetermined value of 1 O gil (for tuming 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 g/1. 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 INIT INIT INIT INIT
G =10.0364 X= 41.8957 O= 0.1978 P = 2.85834 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 0.1
=
SATG = G/(KG+G) SATO = 0/(KO+O) SATP = 1.0 - (P/10) mux = mumax
* SATG*SATO*SATP*X
d/dt(X) = mux - X*vin/V d/dt(G) = - mux/Yxs +(vin/V)*(500 - G) d/dt(O) = kLa*(7.5 - O) - q*mux d/dt(P) = alpha*mux - P*vin/V d/dt(V) = vin The constant value for vin, 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 vin values are tabulated below:
vin
time, hrs
Volume
X cell concent
0.05 0. 1
407.8
100
87..53
208
100
87.58
0.2
107.7
100
87.6
0. 3
75 . 3
100
87.5
0.4
57.4
100
87.3
7-78
0.5
47.7
0. 6
41
0.7
36.2
87.5
100 100 100
87.6 87 . 5
lt 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 lt 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 PS-1 Individualized
solution
PS-2 (a) Example 8-1 ForCSTR
_FA0X _
X
-rA
u0k(l-X)
V--------rk X=--=---
1+-rk
-rAe-E/RT l+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
f(X)= X--
-rAe-E/RT -E/RT l+Ae
f(T)= To - LlHRxX_ CpA Enter X, A, E, R, CpA, To and ~HRx to find -r and from that you can find V.
PS-2 (b) Example 8-2 Helium would have no effect on calculation
%Error=
-LlCp (T- TR) -[Mlix +LlCp(T-TR)]
1270 = ---
23,210
xl 00 = 5.47%
PS-2 (e) Example 8-3 (a)
V= 0.8 m
3
See Polymath program PS-2--c.pol.
8-1
--
-·-----------··---·--·---------·-
-·-·--·-·····-------------·----- ---------
----·-·---··---------·----
------·-·---------·----·-----~----~
POLYMA TH Results Calculated values of the DEO variables variable
initial value
o
V
o
X
T
Kc k Xe ra rate
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
Cao Fao
minimal value
maximal value 0.8 0.5403882 9.3 146.7 363.39881 2.8783812 38.191248 0.7421605 -79.470998 110 .4184
final value O. 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 enterad by the user [ l J d(X)/d(V) = -ra/Fao Explicit equations as enterad by the user [ll Cao = 9.3 [2 l Fao = . 9*163 [ 3 l T = 340+43.3*X [ 4 l Kc = 3.03*exp(-830.3*((T-333)/(T*333))) [5] k = 31..1*exp(7906*(T-360)/(T*360)) [61 Xe = Kc/(1+Kc) [71 ra=-k*Cao*(1-(1+1/Kc)*X) [ 8 l rate = -ra
PFR T
330 0.26
X
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
04 0.3 02
330
384
438 T
492
546
600
CSTR has the same trend.
PS-2 (d) Example 8-4 Counter-Current: Guess Ta at V= O 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, Rev5.1.233
Calculated values of the DEO variables initial value
Variable V
o o
X
310 330.7 9.3 14.67 3.6518653 .0.9004084 0.7850325 -8.3737978 -6900 5000 159 8.3737978 50 75
T Ta Cao Fao Kc k
Xe ra dHrx Ua Cpo rate m Cpc
minimal value
o o
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) = -ra/Fao r 2 J d(T)/d(V) = ((ra*dHrx)-Ua*(T-Ta))/Cpo/Fao [ 3 J d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc Explicit equations as entered by the user [ll Cao = 9.3 [2J Fao = .9*163*.1 [ 3 J Kc = 3.03*exp(-830.3*((T-333)/(T*330))) [ 4 J 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 [SJ Ua = 5000 [9J Cpo = 159 [ 1 o J rate = -ra [lll m = 50 [12J Cpc 75
=
PS-2 (e) Example 8-5 At V= O, Ta= 995.15 and gives a counter current entering temperature of 1250 K. See Polymath program P8-2--e.po1. POLYMATH_Results Calculated values of the DEO variables Variable V X
T Ta Fao Cpa delCp Cao To
initial
o o
1035 995.15 0.0376 163 -9 18.8 1035
value
minimal value
o o
972. 39417 986.00676 0.0376 163 -9 18.8 1035
maximal value O. 00;1. 0.3512403 1035 1249.999 0.0376 163 -9 18.8 1035
final value 0.:001 0.3512403 1034.4748 1249 . 999 0.0376 163 -9 18.8 1035
8-3
-·--------------
--- -~---- --------
--------
-----
7 . 414E+04 -67 . 304 l . 65E+04 O .111 34 . 5
7 . 414E+04 -67.304 1.65E+04 O .111 34.5
dHrx ra Ua me Cpc
7 . 47E+04 -6.3363798 1.65E+04 O .111 34.5
ODE Report (RKF45) Differential equations as enterad by the user [ l J d(X)/d(V) = -ra/Fao [ 2 J d(T)/d(V) = (Ua*(Ta-T)+ra*dHrx)/(Fao*(Cpa+X*delCp)) [ 3 J d(Ta)/d(V) = -Ua*(T-Ta)/mc/Cpc Explicit equations as enterad by the user [ll Fao = .0376 [2l Cpa = 163 ¡ 3 J delCp = -9 [4l Cao = 18.8 [5] To= 1035 ¡ 6 J dHrx = 80770+de1Cp*(T-298) [ 7 J ra = -Cao*3.58*exp(34222*(1!To-1/T))*(1-X)*(To/T)/(1 +X) [ 8 J Ua = 16500 [9] me= .111 [ 1 o J Cpc = 34.5
(b)
P8-2 (f) Example 8-6
Energy balance will remain the same
XE8 = 2 x10-3 (T-300) for 2A-72B K
cz
e¡ (1-xf
X ' lst Order
'
= e
.Jf
\ \
2nd Order
\
\
(e)
xz
=------ª- = ---
\
''
' '-....
P8-2 (g) Example 8-7 8-4
7.414E+04 -31.792345 1.65E+04 O .111 34.5
Both X, and
XEB
will change. The slope of energy balance will decrease by a factor of 3.
X
T
Also X, will be more temperature sensitive K = K exp AHRx e
e
R
(J_T _ _!_JT 1
The dotted line in the plot below shows an increase in -AHRx
\ \ \ \
X
\ \
''
' ' ....
T
PS-2 (h) Example 8-8
(d) (1) (2)
CAo will decrease but this will have no effect 't
will decrease 't'=
(3)
401.1 ft3 466.1 ft3
/s
In the energy balance the slope of the energy balance of X vs . T will be greater
8-.5
í:E>iCPc = 35 +(18.65X18)+4 X (1.67X19.5)= 35 + 335.7 + 130.2 =501
I
/
,-
BTU kmolºR
XM
---.
XM, T
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
PS-2 (i) Example 8-9 Change Cp = 29 and -L.\H = 38700 POL YMA TH Results NLES Solution Variable X T
tau
A
E
R
k
Value 0.7109354 593.6885 0 . 1229 1.696E+l3 3.24E+04 l. 987 20.01167
f (x)
Ini Guess 0.367 564
2 . 444E-11 l. 2E-09
NLES Report (safenewt) Nonlinear equations [ lJ f(X) = X-(397.3*(T-535)+92.9*(T-545)}/(38700+7*(T-528)) = O [ 2 J f(T) = X-tau*k/(1 +tau*k) = O Explicit equations tau= 0.1229 A::16.96*1QA12 [3J E= 32400 [4] R = 1.987 [ s J k = A*exp(-E/(R*T)) [lJ
(2]
Vary the heat exchanger area to find the effect on conversion.
PS-2 (j) a= 1.05 dm3 See Polymath program PS--2--j.pol. POLYMATH Results Calculated values of the DEO variables
8-7
------·----------------
-----------------------------·-·-
--·-··---·-···-----··--··------···-···-··---
-- ----- ------
-
- -- - - -- -- -- --- -- -
- --- ----- - --- - - ----------
---- -- ------
-
-
---- - --- -----------------------
------------------
Variable V
Fa Fb Fe T
initial
o
value
minirnal
kla k2a Cto Ft To Ca Cb Ce rla r2a Fto alpha
value
100
2.738E-06
423 1 482.8247 553.05566 0.1 100 423 0.1
423 0.3120454 482.8247 553.05566 O. 1 77.521631 423 2.069E-09
o o
y
o
o o
o o
o o
-48.28247 -5.5305566 100 l. 05
-·373. 39077 -848.11153 100 l. 05
maximal
value
1 100 55.04326 22.478369 812.19122 1 4.484E+04 l.48E+07 0.1 100 423 0.1 0.0415941 0.016986 -5.019E-05 -l.591E-ll 100 l. 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-ll 100 l. 05
ODE Re)!ort (RKF45} Differential equations as entered by the user ¡ l J d(Fa)/d(V) r1 a+r2a ¡ 2 l d(Fb)/d(V) = -r1 a ¡ 3 l d(Fc)/d(V) = -r2a/2 [ 4 l d(T)/d(V) (4000*(373-T)+(-r1 a)*20000+(-r2a)*60000)/(90*Fa+90*Fb+ 180*Fc) [ 5 J d(y)/d(V) = -alpha/2/y*(FVFto)
=
=
Explicit equations as entered by the user [li k1a = 10*exp(4000*(1/300-1/T)) [2 l k2a = 0.09*exp(9000*(1/300-1/T)) [3 l Cto = 0.1 [ 4 l Ft = Fa+Fb+Fc [5J To= 423 [ 6 J Ca= Cto*(Fa/Ft)*(To/T) [ 7 J Cb = Cto*(Fb/Ft)*(To/T) [ 8 J Ce= Cto*(Fc/Ft)*(To/T) [9] r1a=·k1a*Ca [ 1 o J r2a = -k2a*Ca"2 [lll Fto = 100 [12] alpha = 1.05
(e)
PS-2 (k) Example 8-11
VaryUA
UA= 70,000
J/m3 •s•K
only the lower steady state exists at T
UA= 60,000 only three steady states exist T intersection on the graph.
UA= 700
=
= 318 K S8c = 0.05
J/m3 •s•K
= 318, 380 (about)
J/m3 •s•K
and 408 (about) depending how you read the
=
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 S8c remains low at 0.05, meaning that the reaction has neared completion to form species C therefore reactor is too large. 8-8
----·------·---------------··-------------·-------------------------·-----------------------
--------·-·-·---·--------------------------------- ---·-·-··---------------····-----
-
----------------------··--------------------------------------
T0 = 27 5, very little effect.
Vary 't 't = 0.001 only the lower steady state at T = 316 about and other off scale SBc z = 0.0001 only are steady state at T = 316 and others off scale SBc = 0.05 -r = 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
a--dp Po
ª = ª dp:P 2
1
0:
1
dpPo ( 1
1 )
= (1/2) (2) = l
No effect for turbulent flow if both dp and P changed at the same time.
PS-2 (m) Example TS-3 m, = 200 gis
See Polymath program PS-2-rn.pol. POLYMATH Results Calculated values of the DEQ variables Variable
w
Ta y T X
alpha To Uarho Me Cpme Hr Fao thetaI CpI CpA thetaB CpB Cto Ea Ke ka ya o xe Cao sumep Ca Cb Ce ra
initial
o
value
minimal value
o
320 1 330
320 0.3044056 330
o
o
2.0E-04 350 0.5 200 18 -2.0E+04 5 1 40 20 1 20 0.3 2.5E+04 66.01082 0.046809 0.3333333 0 . 8024634 0.1 80 0.1060606 0.1060606
2.0E-04 350 0.5 200 18 -2 . 0E+04 5 1 40 20 1 20 0.3 2.5E+04 O . 8247864 0.046809 0.3333333 0 . 3122841 0.1 80 0.0137198 0.0137198
-5.265E-04
-O . 0143957
o
o
maximal value 4500 334.77131 1 385 . 31436 0.5645069 2.0E-04 350 0.5 200 18 -2.0É+04 5 1 40 20 1 20 0.3 2.5E+04 66.01082 11.205249 0.3333333 O . 8024634 0.1 80 0.1060606 0.1060606 0.0724316 -1.745E-05
8-9
.
---------------------------------
---
--
--
-
- - - ------- -----
-
final value 4500 334.77131 0.3044056 338.18498 0.5645069 2.0E-04 350 0.5 200 18 -2 . OE+04 5 1 40 20 1 20 0.3 2.5E+04 31.551036 O .1177827 0.3333333 0.7374305 0.1 80 O. 0137198 O. 0137198 0.0355685 -1.745E-05
= 0.05
ODE Report (RKF45) Differential equations as enterad by the user [ U d(Ta)/d(W) = Uarho*(T-Ta)/(Me*Cpme) [2 l d(y)/d(W) = -alpha/2*(Tffo)/y [ 3 l d(T)/d(W) = (Uarho*(Ta-T)+(-ra)*(-Hr))/(Fao*sumep) [ 4 J d(X)/d(W) = -ra/Fao Explieit equations as enterad by the user ¡ 1 l alpha = . 0002 (2) To=350 [ 3 J Uarho = 0.5 [4] Me= 200 [5] Cpme = 18 [ 6 J Hr = -20000 [7J Fao=5 ¡ a J thetal = 1 [9J Cpl = 40 [lOJ CpA = 20 [lll thetaB = 1 [12] CpB=20 [13J Cto = 0.3 [14] Ea= 25000 [ 15 J Ke = 1000*(exp(Hr/1.987*(1/303··1ff))) [16] ka= .004*exp(Ea/1.987*(1/310-1ff)) [ 17 J yao = 1/(1 +thetaB+thetal) [ 18 l xe = Ke"().5/(2+Ke"().5) [19] Cao = yao*Cto [ 2 o J sumep = (thetal*Cpl+CpA+thetaB*CpB) [21] Ca= Cao*(1-X)*y*To!f [22] Cb = Cao*(1-X)*y*To!f [ 23 l Ce= Cao*2*X*y*Toff [ 2 4 l ra = -ka* ( Ca*Cb-CeA2/Ke)
PS-2 (n) (1) The concentration of Anear 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. l. Parameters in simulation on the tubular reactor from Example 8-12 (First Order reaction):
A+ B ~ C
Reaction:
A- propylene oxide; B- water; C- propylene glycol (1) operating parameters Reactants • Peed rate of A FAo = 0.1 mol/s FA0M • InletflowrateofA VAo =
PA
A
=
o.rxsa.rxro' 830
8-10
=7x10
_6
3
mis
• • •
• •
• •
InletflowrateofB
v80 =2.5x2xvA0
=35xl0-6
m3/s
Inlet total flowrate v0 = 2v AO + v 80 = 14xl0-6 + 35 xl0-6 = 49x10-6 Inlet concentration of A CA0 = FAo = O.l v0. 49x10-
6
= 2040.8 mol/rrr'
. F v p 35xl0-6 xlOOO 3 Inlet concentratíon of B C80 = __!!_(}_ = ~- = = 39682.5 mol/m v0 M 8v0 l8x10-3 x49x10-6 Inlet temperature of the reactant T0 = 312K
Coolant flowrate, m¡ = 0.01 kg/s Inlet temperature of the coolant, Ta0 = 298 K
•
(2) properties of reactants • Heat of reaction, !iHRx, dHrx = -525676+286098+ 154911.6=-84666.4 J/mol • Activation energy, E= 75362 J/mol • Pre-exponential factor, A= 16.96x1012 /3600 lis • Specific reaction rate k¿ = 1.28/3600 lis @300K • Reaction rate k • • • • • •
rr?/s
= k¿ ex p[ E (_!__ _!_ )] R ~
""7
T
or k
=A
exp[- !!__] RT
Gas constant, R = 8.314 J/mol·K Rate law - rA = kCA Thermal conductivity of the reaction mixture, ke = 0.559 W/mK Average density of the reaction mixture, p, rho = 1000 kg/m' Heat capacity of the reaction mixture, Cp = 4180 J/kg·K Diffusivity of all species, Diff = 10·9 m2/s
(3) properties of coolant • Overall heat transfer coefficient, U, = 1300 J/m2·s·K • Heat capacity of the coolant, Cp1 = 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
(3) Subdomain Settings - Physics ( mass balance)
8-12
-------,-----
.
------------------------------------------------------------
---------------------
(Energy balance)
¡ ·Subdomain selectio~
1--~~j¡
11
.
'¡¡
!
1
!
11
l!
JI
1
I¡
ll
·1
11 11
l!I!
: 1
],
I
¡1 ·[1
I
T!
1
.· .
tJ Sel~
j. 0
I! ;;¡;(
c. Q
~-i by group
... :
A~i~: in this don,si11. '.· [
(Cooling Jacket) rEguation
I V•í: l,
F
- Initial values (Mass balance) cA(tO) = cAO (Energy Balance) T(tO) = TO (Cooling Jacket) Ta(tO) = TaO
(Source Term) F = Taz-2*pi*Ra*Uk*(T-Ta)/(CpJ*mJ)
- Boundary Conditions @ r = O, 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) Physics
-Mesh SOlve Poslprocsssilg
!
il'lh ¡[1-1.6. &; &. =fl :::¡ = ~ 1 ~
¡w·-
---·-·-
.. --·-·
·---·--
-----.
~ltlphysics
He/p
_
1 fil .e jJ >t> 1 il i2 aa o ef9>1 , -.
------,.-···--·-·---·-----·-····,
.- ...
..Z.-J
llJ !
i::il
~l
*l ••• I
~-1
0.6
0.4
0.2
¡
0.2
8-14
--- ..
_---
.
-·-.··---·
,--·.··· ..
_--
<.
.·, '
Second order reaction [l] Domain
[2] Constants & Scalar expressions
( 1) Constants
~
r
-f:~ - - - ~
D
~--- - r!~~- . . ---- - --. -- - --- - -
1
(2) Scalar expressions
[~:~-f~{ªi~). ~ =~~~=:~~==~=:===:=-t,~ f~.·. . . .
f L
-~q
J:~:~;~-~IRfT)"rhoCat•(cA .•cB~cC!l\eq) . . . . . . . ... . •. . . . . . . •. . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... . . f<89º*exp(dHrxiR•(113_03-1 ff)) __ -··--·------·--·· ·--··-·P!,-,)•~-dHrx2_ ---·-·· . ····· . ·····-·· ----··
r-'
-··----·--------····-·······--·-·--1· .
1 - -~-·-.----~--------~.,.
· ---··-;
i
1 · .:
1 ¡
. , ., ,. . .,. ,.,.,. . . . , .,., . ,. . . . . ,.,
8-15
-·--------------------
-----------------------·------------
-------------·-·····-----·---------·-------
-----
- -- ----- ------- - -- -----
-----
-
------- -------
--------- ---------
[3] Subdomain Settings (1) Physics
(Mass balance)
1
;'Subdomain selection -- ·-- ---
! jll
l i
('!
.
I
1111 111 HI 111111111111111 ,~;
l
il¡ ¡
i
Quantity
6ts
i l
11
!-l jij
R
: L,.. ! 1'.
! ¡
- ~-------
~ :,.:.,-
D Select by group ~~~¡~-ªin t~i; d~-~~i~
u
!
V
J
(Energy balance)
h1N[),¡
~Á~~~ 1P:ttúS'~O~ain
1,:
_[~cj~¡~~Á
(Cooling Jacket)
(Source Term) F = Taz-2*pi*Ra*Uk*(T-Ta)/(CpJ*mJ) 8-16
(2) Initial values (Mass balance) cA(tO)
= cAO (Energy Balance) T(tO) = TO (Cooling Jacket) Ta(tO) = TaO (3) Boundary Values @ r = O, Axial syrnmetry @ 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
8--17
PS-2 (o) Individualized solution
PS-3 Solution is in the decoding algorithm available in the beginning of this manual. PS-4 Find the reactor temperature at steady state (prior to shutdown) Let M = mass of the ~N03 in reactor. FAO = lbs/hr of ~N03 fed to the reactor. Mass balance: FAo -FA0(1- X)= -rAV = kM X= k(T)M
FAO Energy Balance: -FAO LB¡[HiO -H¡(T)]+ FA0XMIRxn(T)+ FA0(1-X)MivA = O Where H¡(T) is the enthalpy of i at the temperature of the reaction, MI VA is the heat of vaporization of A, and MI Rxn (T) is the heat of reaction at the outlet temperature.
8-18
NH4NO(l) ~ 2H20(g)+N20(g) A~2B+C
The last term FA0(1- X)WvA accounts for the unreacted NRiN03, which exits as vapor rather than the liquid. Now, we can make sorne substitutions .17 B 88 =-, A =I,Bc =O .83 Heat capacity of A is given, and the enthalpy change for water 200º F (l) ~ 500º F (g) is also given. So, after dividing by FAO, we obtain cpA (T-200)+ BB[H 8(500º F) + e; (T-500)]+ W Rxnx + (1-X)WVA = O The previous equation assumes that the heat capacities are constant over a reasonable temperature range. the phase change NH4NO(aq) ~ NH4NO(l)is isenthalpic. In addition, we must account for the effect of the temperature dependence of MI Rxn
MI Rxn(T) = MI R.xn(TR) + (T-TR)[80
44
Cp, -cpA] = MI Rxn(Tr)+
•
a -rnsc,
Let MI Rxn = HB (500) - HB (200), we have CPA(T-200)+B8(Mlw +Cps(T-500)+ X[Mf
Rxn
+~C/T-500)]+(1-X)MlvA = O
or,T[CpA +BBCp, ]+[-CpA (200)+BB(Mlw -sooc.u- X[Ml
Rx11
+ ~Cp(T-500)]+ (1-- X)MlvA
Numerical Substitution with K(560) 5.03 and k(510) .53
=
=
(J_ _ _!_) Y¡ -=> E= -ln(k )(J_ _ _!_)- = 44499 ln(k2) =_E k,
RT2
R
k1
2
1
T2T¡
~ = k, exp(-)E = 4.51424xl019 RT¡
Btu cpA =.38-;Cp, =.466 lb.R
(P = 1 atm over 450 - 500°F Himmelblau) Cpc
= .2521· Btu
lb.R
(Himmelblau, App E, over 230-265°C)
8-19
--------------
---------·-------------
--- -- -
- - -- -- --- - ----- -----
=o
8 = .:!2_ = .205 B .83 Mi
w
= 1034 Btu .03 lb
= 257.3 lb NH4N03 hr 44·02 36·03 ACP = (.2521)+ (.466)-.38=-.0316 Btu 80.05 80.05 lb.R MIRxn = cpA(410-T) Substituting all these into the mass and energy balances: FAO
)M
4 . .5lx1019 ex ( -44499 p T+459.67 Mass balance: X = ------'----~257.3 Energy balance: O= .48T +88.21 + X(-320.20-.03T)+ (l-X)(l.55.80 ·-.38T) Assume X = .96 and M=500. Then, from mass balance, T = 510°F
PS-5 A+B~2C -· -·
..
Fío ( lb -h~ole) -
e--
T¡o(F)
-·
-·
C
( Pío
--·----
Btu
lb moleº F
MW ,(
)
___!1!_) lb mol
P,.(;)
A
B
10
10
80
80
51
44
128 ------63
= 20 000 R
'
----47 . .5
94
222
67.2
65
--
-
Btu
lb mol A'
Energy balance with work term included is:
8-20
-·-
O.O
-- f.o------
-------
Mi
e --
Q=UA(Ts-T)
Substituting into energy balance, UA(Ts -T)-Ws -FAOAflRXAF = FAo[ cpA +Cps][T-To] ~UA(Ts -T)·-Ws -FAOAflR ={FAo[ cpA +CpBJ+uA}[T-To]
+ UA(T5 - T) - W5 - FA0Afl R
T_
T,
-
o
-W s
FAo[ cpA +Cps]+UA
= 63525 Btu hr
:. T =199ºF
PS-6 A+B~C Since the feed is equimolar, CAo = C80 = .1 mols/dnr' CA= CAo(l-X) Cs = Cso(l-X)
Adiabatic: T = T0 + ¡[-_Afl R (T;i)I
8
-HA
'°'~ 0¡ci- = cpA + B Cp 8
8
= -41000-(-15000-(-20000)= -6000 cal/mol A
= 15 + 1s = 30--cal mol K
T =300+~Q.~9X = 300+200X
30 -rA =kC1o(l~X)2 =.01k(l-X)2
8-21
í -----------------------·-----------------------
-------------
-
----------
------
-------- --------
-----
------
------
---------
For the PFR, FAo= CAoVo = (.1)(2) = .2 mols/dm3
See Polymath program P8-6-a.pol. Calculated values of DEQ variables t 1variab1el1nitia1 ;;~ r-¡:;-~ ------T -----· --, ·· ......... --------
111x .
¡
o
1~un1~a1 ;~1;e·¡,.,,~~.~~1\r;,;~¡i1~~!!~¡~~:J
~=~~
10
·····-------~--------
---------------------
----r··
-----·-- -----
·--
T"
-
·
--
·
··rcú -- . . . ~ i:-ª~--:-¡~~~ f¡3ºL ~ - -r2--:::_-: ~~~~] ;5 ¡T 1300. 00. 70. 470. _ ~: ·:l~~~~: :~. · · - .l º:·ai·_-_ ~:~=----- -~. _14~iso37s _- :---:~=-1·4~iso3?s·-~J V
131cao... .. .
~ +~ - -~l?Jr~
O 0.1
. J~9.gg9~
'O
1
j~0:00~8~~1.
-
io.ss [c.ss I ----------------- - - -- - -------------r-------------------------1 08.2917 1308.2917 j 0.1 ····--- ---- . -. lo.i--------- -- J
¡-0.0001
.
1
t9·999~~~-~J
Differential equations d(V)/d(X) = -FaO / ra Explicit equations CaO = .1 FaO
= .2
T = 300 + 200 * X k = .01*exp((lOOOO / 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/dm3Is V=!AoX = .lx2x.85 =175 dm" -rA 9.71xl0·4 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.
PS-6 (b) X[-MlR] T=T+-·-o
¿B¡Cll
For boiling temp of 550 k, 550 =To+ 200 To= 350K 8-22
PS-6 (e) 0.90
500
0 . 72
460
Q
0.54 0.36
380
0.18 ·
340
0.00
o
62
123
185
V
EiJ
420
246
308
300
o
62
123 V 185
246
308
PS-6 (d) V
. CSTR
=> X
=
FAOX -rA
= VcSTR ( -rA)
FAO For V = 500 dnr', F Ao=.2 -rA =kC10(l-X)2 =.Olk(l-X)2 T =300+200X Now use Polymath to solve the non-linear equations. See Polymath program P86-ct. . l.pol. Calculated values of NLE variables ·--1· -· . .. i ! ¡variable !Value jf(x) ¡Initial Guess,
r·-r -------··· ..-r- _, --- ····-. cr ·-- . - -- .. ·-
rr·· ·-·····¡···
1 ¡T r- .• , ¡
-----·
.-
r---··
·r
--T
-
¡484.4136 ¡O
-··· - · -· · !480.
!?!~ !°.'.?2_206~1J-2.041E-O?J°.:? J" ·r , , ... ¡variable Jvalue I 1
.
¡
J
j
iilk . . ,6~072856 ·j
¡·r·¡2 [ra """' .:
'----·T·---·-·
.···
.
...1
¡0.0003688 !
, ,
, !
~ .. ,
,.~--
.- .. ,"
~
Nonlinearequations f(T) = 300 + 200 * X - T
ti f(X) = 500 - .2 *X/
ra
=
O
=O
Explicit equations k = .01 * exp(10000 / 1.98 * (1 / 300 - 1 / T))
8-23
---------
----
-- ----------
-~-~
-----------------
--
-
---·----·-----------------------------------------~-~--
= 0.01 * k * (1 - X)
ra
A
2
Hence, X= .922 and T = 484.41 K For the conversion in two CSTR's of 250 dnr' each, For the first CSTR, using the earlier program and V = 250 dnr', Calculatedvalues of NLE variables LJY~r,i~-~l~.[!~~ue. jf(x) 11IT··· . , .. , r-·h·
.
IIniti~~:~~-~!~J
!476.482 1.137E-13 _,., . ,,T-------····· ...•.• ,
_
480. " .. -------
J~.:-~~.?-~~.L=?:~~~~~o9J~.:~
J~J~
,
--- _
1 j
1
r Tv~-~i~b1;fv~1~;·, - . , t·---·--·~-- ,.--,,,.,, .. ,.,.•.••..••............. --~<-·······----·--···············"·~<,···········J
!;J~
. !?:1~??7~ I
.;. ... ,s......................
···----··································
¡2¡ra
¡0.0007059¡
.. ··········--·)
Nonlinear equations
= f(X) = f(T)
300
+ 200
250 ·· .2
* X-T = O
*X/
ra
=
O
Explicit equations k = .01 * exp(lOOOO / 1.98 ra
= 0.01 * k * (l
- X)
A
* (1 /
300 - l / T))
2
T = 476.48 ad X = .8824 Hence, in the second reactor, V = FAo(X -X1). CSTR
-rA
V => X =-csrn (-r )+ X F A 1 AO
T
= Tout,CSTRI
+ 200(X
-· X¡)
See Polymath program P8-fr. d-Z . po1. values of r·Calculated 'º('''"'"'' -.~ ,_,.· ··-s.-r=· .. -.. 'v·,
j
J.
·
¡variable!Value
!i•J±:••-·-·-~·-· · ·
E I~
.....
·.,.
NLE variables -"--··--··---~~··r·~~,.,.--- .... f(x) !Initial Guessl
· · · ¡JO.~E>~36~~J~ i~3.~!~.~-- .¡.º1 '.~S9E-g?J •· ;~~ª:········-· ·- · · · · .·· ~:~~~~
·_· · · -· · ·]
8-24
1--··¡-·-·-
--M-~--·-·••·•~·••••••'
.,· .• -~-·••••----··"
J3 JX1
¡o.8824
·····-········--······················-·····--
•• •
•••-····-·j
!
....... ...J
Nonlinear equations
+ 200 * (X - Xl) - T = O
f(T)
=
476.48
f(X)
=
250 - .2
* (X - Xl)
/ ra
=O
Explicit equations
= .01 * exp(lOOOO/ 1.98 * (1 / 300 -· 1 / T)) ra = 0.01 * k * (1 - X) /\ 2 ,__ , Xl = .8824 !il k
Hence, final X= .9694
PS-6 (e) Individualized solution PS-6 (0 Individualized solution PS-7 (a) For reversible reaction, the rate law becomes +rA = k
X=~ e
(c
A
J
CB - K Ce e
l+K e
T=300+200X
-.!.))
k = k(300)exp(E (-1--
R 300
T
1 1 )] K = K (450)exp [ MI Rxn ( ----e e R 450 T
Stoichiometry: Ce= CAoX CA= CAO(l-X) CB
= CAO(l-X)
See Polymath program P87· a.pol, Calculated values of DEQ variables
jr · r¡vari;ble mal value I M~xhnal ~alue f Final value I . - lr·Initial . . value . ... 1i Mini .. f - . . .. . . - T . - . .. - -, ¡O
l¡V
¡O
f:iíx - - --· ·¡o . 1
!._; ....
y~""",S'·•····~>
.,.;.
... ,...
"""······'
:10. --· IO.ooso8oÍ5
110. ·11a.oososo6
·110 '•
••w•••,•••···········'S,•V
,., .. , · ··•••••••--·•·'·"•
•·•
l.V"'"""'
1
1 '
8-25
----
-
------------
----
-- -
-
·-·-····-···---·--
--
-
---
------ ---
-- --
-
- --------·-···--------···--··-·· -···--
- ---
- -----
-
--
-- ---
----
----
------- ------------
---------------~---------
f 3h
·
·
hoo. · · · · · · ····· !ioo.
· ·· ·
t··t··········---··-·····¡·······- -··· · ·········--······ · ¡························ !41FaO ¡5 ¡Kc
0.2 ¡o.3567399
r-·r
¡0.01
1 j-·-4---~,---~•••--•••--•••A•r•--•••-•••-""'-""''''"'""'"'•"'··•--.s··•
r·T·······--······· ··--·r···
¡6 ¡k
---·-···················r
¡7 ,cao
r··T·-········ ······-··
J.~lr.~ -
¡~J~~ . .
10.2 J0.3567399
¡0.01
T
....•.... ····•···•·······
. i~g:g~~1.
··
,.--•·, .. ,·.,w·•--••••"'-'"'°'''"'··•·'••••-•••-•
•·•••••~r-··•·-••~=••••••-'•"••••-••••••-•••'•,•O•,•O··•••-·-••MeaA¡•-·•···••··•·"''"'•-'••"•"'•-••'"'·"S-••••--'S•A•"••AA<.o>O,~
·····-···········--·-- . r······--··········-···--
··T10.1 ·······················
····¡3oi.-o161 ·Tioioi6i···1 ··r······· · · ········ ··-· ·-··· ········ ····· · · · · · -·········· ······· í ¡0.2 . 0.2 J¡ o.36898_?7-_J 10.3689877
··········••
¡0.1
·-·
···,.······· r··
···················· .
¡0.0105787
······
'T" ·········
·········-r· ··-···-··········· · ····•···
. .. .• .
- .. .
¡0.010578! _
T .. · · ····.... . . • . . . . ···.
.........•
T¡0.1 · · ··· · · · · · · · ·· ·
t~:º?.~1.g3.~·-·· · · l-~:~g~1.
J~:??293~1..
¡~.2629391
¡
•
0.2695332
¡0.1
. J
.
·r·········· ··• •·· ······ ··· · ·
[~·?.~~-1.g3-~_J
]~·?6?53.3.?
j
Differential equations d(X)/ d(V) = -ra / FaO Explicit equations T = 300 + 200 * X FaO
=
.2
= 10 * exp(6000 J 2 * (1 J 450 - 1 J T)) k = .01 * exp((lOOOO / 2) * (1 / 300 - 1 / T)) CaO = .1 ra = -k * ccao 2) * ((1 - X) 2 ·- X/ Kc) Xe = Kc / (1 + Kc) Kc
A
A
302.0
O.JO~-----------.
3016
0.24
301.2
0.18
300. 8
0. .12
300.4
0.06
.....
300.00
2
4
V 6
8
10
{LOO
--
_.
~. -
ITJ
---··-·----
.
o
2
4
V 6
PS-7 (b) When heat exchanger is added, the energy balance can be written as dT _ Ua(Tª ·-T)+(-rA)[-AfIRxn(T)] dV -
FAo(¿Bicpi +~Cp)
So with ~CP= O, ¿B¡Cp; =30, Af!Rxn= -6000 cal/mol dT
Ua(Tª -T)+(-rA)[6000]
=---------
FA0(30) Where Ua = 20 cal/m3/s/K, Ta = 450 K dV
8-26
8
10
See Polymath program PS-7-b.pol.
Differential equations d(X)/d(V)
= -ra
d(T)/d(V)
= (UA * (Ta - T) + (,-ra) * 6000)
/ FaO / 30 / FaO
Explicit equations
= .1 FaO = .2 Kc = 10 * exp(6000 / 2 * (1 / 450 - 1 / T)) k = .01 * exp((10000 / 2) * (1 / 300 - 1 / T)) ra = -k * (CaO 2) * ((1 - X) 2 - X J Kc) I Xe = Kc / ( 1 + Kc) UA = 10 CaO
A
liTa
A
= 450
!IA = 10
8-27
-----------------
470
1.0
466
0.8
[iJ
462
0. 6
458
0.4
454
0.2
4500
2
4
6
V
10
8
4
2
6
V
8
PS-7 (e) For a co-current heat exchanger,
-T)[!-ex{ ni~:,J]
Q-nicC,c(T.,
Cpc = lcal/g/K, Ta1=450 K, m = soL sec See Polymath program P87·c.pol. Calculated val.ues of t>EQ va.ria bles
¡:·_-¡ v;·;¡·;11,~TI~-iti~,-~~·~~l.~-!~·~~(y~,.~~J•~.~~·~·~1:·~;,~;··rt=in~i····~~·1·~~J 11
V
¡2
J~ . .
; ' f ······
_,
} !T.
i4 Icao ¡·j-··· ···r·
O
·····T
. .
f _
.
jo
lº _ .
. .. -r-
.
J450. 0.1
1.
!13 A
)i4l~~···
[isJ§··
110:·· - l ... , J5o.
t.:
l°.
J 5.~:5192 ¡0.1
d(T)/d(V)
.
~~!!:!~ª
1
.
.¡, ....
10.
. l~°-·
10.
50.
1-139.4201
......... L.
= -ra / FaO = (Q + (-ra) * 6000)
j ···j
. ....
o
45g'.
l-22.83309
/ 30 / FaO 8-28
1
l !
'"'1
l
9]
10. ·-·,· · · 5· · -o· · · .· · · · ··· - · "'1
Ll50.
'
j
t ~w~~r · Ji:¡ttttº .. - f2.1s1765
. . Tia.···· · · · · · !10.
'1
li6~37812
·11.
3.734633
Differential equations d(X)/d(V)
T
1.
J2.ssG106
110. 1450.
J4?~--~s2s 10.1
º
-1··•<
r
10. ··· . . . . 0.575571
li2.46524 ' . "' (...
1.
..
i
r10. '. . -1°-·_57~571
•.
.
i:.
0
· · · · · · · ·····
..
¡45g: 0.1
l! :: l¡t 2.586706
········· . . . . . . .
·j
1
·¡
10
Explicit equations
=
CaO
1ª-1
.1
FaO = .2 Ke
=
10 * exp(6000 / 2 * (1 / 450 - 1 / T))
Cpe = 1
!)
k
=
ra
.01 * exp((lOOOO/ 2) * (1 / 300 - 1 / T))
= -k * (CaO
Xe = Ke / ( 1
A
2)
* ((1 - X)
A
2 - X/ Ke)
+ Ke)
UA = 10 Ta= 450
fij]A
=
10
me= 50 Q =me* Cpe * (Ta - T) * (1 - exp(-UA /me/ Cpe))
~ 470
LO
466
0.8
[iJ
462
ITJ e
0.6
458 0. 2
454 4,;o
- o
2
4
8
V 6
10
2
4
V
6
8
Next increase the coolant flow rate and run the same program to compare results.
PS-7 (d) For counter-current flow, Q=rhcCpc(Tª1 -T)[l+exp(-
-J]
. UA «e:
See Polymath program P8-7-d.poI.
¡
¡
Calculated values of DEQ variables
1.va~i_é3 ~IE!_ ~-".itié31··"-~luE!l-~-ini111_a_1_ yé3l_~E!lM~~-~ 111as. "-é3lue·1-~~"..a1. "-ªl~e-J
[!Jf-·- . f:;.· ·.-~f 0
¡s~-~ .J~~:!~~ :&.~~~f 2···
8-29
J
10
Differential equations ~ d(X)/ d(V)
I
d(T)/d(V)
= -ra /
FaO
= (Q + (-ra)
* 6000) / 30 / FaO
Explicitequations CaO
il
FaO Ke
= .1 = .2
=
10 * exp(6000 / 2 * (1 / 450 - l / T))
ecpe =
1
fii k = .01 * exp((10000 / 2) * (1 / 300 - 1 / T)) II ra = -k * (CaO 2) * ((1 - X) 2 - X/ Ke) II Xe = Ke / (1 + Ke) A
-~ UA
IITa
=
A
10
= 450
[~IA=10
frj me=
50
mi Q =me*
Cpe * (Ta -T)
* (1 + exp(-UA /me/
Cpe))
8--30
---------
---
-
-
-- ---------··---------------
452.0 .----------------,
1.0
45L6
08
4512
e.s
450.8
0.4
---
450.4 450.00
2
4
V 6
02 00
10
8
o
2
4
V
6
8
10
PS-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.
PS-7 (f) lf the reaction is irreversible but endothermic, we have -rA = k 0(1- X)2 = .Olk(l-X)2 as obtained in the earlier problem.
C1
MI Rxn = 6000cal / mol For co-current flow of coolant,
Q = rhcCpc (Tª1 -r)[l-exp(-_;!!-~JJ
«e:
See Polymath program P8-7-f-co.pol. we use 8- 7f cocurrent.pol Calculated values of DEQ variables
r··-··-,····-·-····--· ··,---····-····---·,---···----··--········· ¡ ¡variabte!Initial valuelMinimal
!i'JY-- - ·.w· · K-. -Tli
- - t~sº~
f¡6H-,C:p~d~ - JirL·-
'.7 ¡s
. .
·rlk- . .
T
¡~~ liaJ+~ - . . · · 14so. ·1 ····· ····
------Tª-- -
····-
t~-39.9908
r
1~}
r1: ·- -
······
··---!10···-· -
··· ¡ ; value¡Final value]
---··· · ·
--
t1~::-~?
896
'!
--+---
i~_~J~~-- . J5.º·
t~:1i. i
· - ··Tia·:·... }+so. · · · . ·T·14so. L5.~·
.
T."
· 12::is2492·· 1
. Tia:
···1· · •
J t1ll{:!r'--- ~
l!?·
-tr1:·i+-- · ,:
.
·12:ss6706 ·r2:oo's972'' - fa.586706 1~a-.02ss1571 · FD.02ss6n l-o.oos3s73
¡g··-TuA ·-· · ·-··· 110·. · ¡·
.
· r·
value¡Maximal
.. J~9:..
'" 8-31
1
Fo:oos3s73 J ha. - · 1 l4so. - -- ·- J¡ .15.P· J r····
f12
íQ
l..
•.,---··. · ···············--······· ···----· ·· · · ··-·"-·······
··--·-10···------------·ro···-·--------······-·lgo.71779 ,.
,._,
-------134.54357 ,.
-.,.,
--··-····--··
I
··· ···---- -·-······-.]
Differential equations
I I
= -ra / FaO = (Q + (-ra) * (-6000))
d(X)/d(V) d{T)/d(V)
/ 30 / FaO
Explicit equations leao = .1
IJ FaO = .2 !il Cpc = .,,,·~"#
1
I k = .01 * exp((lOOOO/ 2) * (1 / 300 - 1 / T)) [!il ra = -k * (CaO 2) * ((1 - X) 2) I UA = 10 A
A
!!Ta= 450
imc = 50
t~ Q =me* 450
Cpc * {Ta -T) * (1 - exp(-UA/me/ Cpc)) 0.60
-------------
0.48 0.36 438
0. 24
434
0.12
430 ~-~---------~
O
2
4
V 6
8
10
Q
O 00
-
-·
o
2
4
V
6
8
10
For counter-current flow, Q=rhcCpc(Ta1 --T}[l+exp(-
.UA )] mcCpc
See Polymath program P8--7--f-counter.po1. POLYMATH program 8- 7f countercurrent. pol Calculated values of DEQ variables
:il~riab~e;~:tialv~:~l:::;-:uelfí~: ~val~1~;;~::~1 8-32
---
-------------------------------------··--·---·-··--·-·--··-------------------------·-------------------------------
--------------------------------·-------
!4 ¡Q¡a··-·---·Tcú - ------··· · ¡o.i [s jFaO
ío.2 ·
1 ··--·· (
. ra.1 - -·
!0.2
-·-T········ ··-
. fo.2 . .
r···
.,
- "f
¡6 [Cpc·········--······--r····1·1· . ········ ·····--·-· ·r·¡·l· ·. ,-········-¡·· J7 [k ¡2.586706 j2.4901 ,·-•••••
("'H-··•··~••-'"'-
!8 Ira
r-~·M1-·-- - ---~~--
"O'oMh"-··r-~-··•··•~• •-' _.._._~-- ..••''-"'-·'~
l-0.0258671
-·--·----~---·-·M-------~- . . . ·---·
¡9 ¡U........... A . .............. ,,..110. . ,. . ··-"'l"····--·· 1450. r.jlO· · " r··Ta --· · · ·" . '"'". ,_.
.
. - .. , ~---
[so,
¡11¡me
••-~')
-
.
ío.2
.--···-•••""'·•-,-···'
···•,-••··•••
-----·"""'''°' ''"··•·•-·••,·•
1 ,
. ,M;
-0.0049788 1 --------1--------·-----------···-------------·---'0---~';·-,···-·····M·--·-----------"--- -- ¡-----------~ ---------------1 110. 110. . ¡10. ¡ T ··r·1 ¡450. . r-·450. · . .,.. '"" , · · - ·----· ··¡~¡4· 50. .,,. . , _. ..1 [so, [so, ¡so. 1 Jo . . . . I__i39.7022 . . . _J3.~.09601 - .• .__
=--'>·" - .
l-0.0049788
¡2.565639
-··r·• ·•·--·---- . ·--··-·~••"'"'"' •"·
! ¡
11. T
12.586706
•• •• •.. ,·,•,••-••······••••-'•H•••-•
.
··-·r····-- ······--·--·-··
¡,l.· · ·
¡-0.0258671
··"'""""'"'""1
f;?Jc~ _ J~--
To:1-···--·· · ··¡
., ..,, . ._ ...• ,, .. ~, ..• ,.. , •
..
J
Differential equations d(X)/d(V) ~ d(T)/d(V)
= -ra /
FaO
= (Q + (-ra)
* (-6000)) / 30 / FaO
Explicit equations cao = .1
I
FaO = .2
~~"
¡;cpe=l k = .01 * exp((lOOOO / 2) * (1 / 300 ·· 1 / T))
íS ra = -k * (Caü fsí ~ UA =
/\ 2)
* ((1 - X) /\
2)
10
Ta= 450 me= 50 ~ Q =me* Cpc * (Ta -T) * (1
+ exp(-UA /me/ Cpe))
450.0
0.60
----
449 . 6 449.2
OAS 0.36
[J
448 . 8
0 . 24 0.12
448.4 448.0
o
-
2
4
V
6
8
10
0 . 00
o
PS-7 (g) For a runaway reaction, the following must be true: RT,2 T - T >-'
e
E
8-33
2
4
V
6
8
10
and T =To+!(['ª= 300+3*450 =412.5 e l+K 1+3 So if we plug this value into the original equation we get: L9S7 T2 -T 1 +450 > O
10000
r
Tr= 499 K
PS-8 (a) A+B~C Species Balance:
rA
dX -=--
dW
v0
I
FAO
= 20dm3 / s
Po= lOatm Stoichiometry: CA =CAo( l-X
l+ex
).!_,where é=l TO
=>CA= CAo(l-X).!_ l+X T0 Rate Law is:
_!_)]
-rA = kCA, with k = 0.133exp[E(-1- _ R 450 T E=31400 Af-IRxn
=-20,000] /mol
Energy Balance: T =To+
X[-_MIR(To)l _
¿B¡CP; + X!iCP
se, =15+24-40=0 T = 450+
20,000 X = 450+ 500X 40
See Polymath program P8-8-·a.pol. Calculated values of DEQ variables
rTv~;iabl~ Initi~t;~·,~; fMi~¡;;,;,~1~~ f M~;¡·~al value l Final val.ue
lfi~ .. f = =-tF- · · =fü\3;:=-=:R~~~--:; ·¡
8-34
--
----
------~----·----------------·-----
---
------·-----------------------------------·----
Differential equations
d(W)/d(X)
= vO * (1
+ X)
* T/
k / (1 - X)/ TO
Explicit equations
T = 450 + 500 vO
*X
= 20
TO= 450 k
= .133 * exp(31400
/ 8.314 * (1 / TO - 1 / T))
0.8
900
0.6
800
0. 5
700
0. 3
600
0. 2
500
o.o o
9
18
W 27
36
45
400
GJ
o
PS-8 (b) Species Balance for CSTR: FAOX WCSTR
=--,-
·-rA
T
= 450+500X = 450+500(.8) = 850K
k
= .133exp[31400
WCSTR
8.314
(-1450__._1_)] = 6.9 850
= 39.42kg
PS-8 (e) Individualized solution PS-8 (d) For pressure drop, an extra equation is added
8-35
9
18
w 27
36
45
ª(!_Jt; _
dP =dW 2
__!:1_(1+&X)
(P/11,)
1·-X) T P CA= CAo ( 1 + X To 11, See Polymath program P8-8---d.pol. Using POLYMATH program CRE_8_8d.pol For a=.019 Calculated values of DEQ variables
[dt'=bl=t==•111!•~e.1ti~l\l~uel~ttl~111alu=!tf 1 ~~lueJ
!¡4JT"" I ~ . . · ·.· · ·. ]4sa:· ii.013E-t-06 lsT~o · · · 'io:
[i[k-
: s I Po
·
'~~~~3
·¡1:oúE-t-06
r9·1~·1ph~ · · ¡0.019
li.002E+06
Fisci.'. . l:20: · 1~~~;3
.
· f 20:·
lJi.oúE:-t-06 ::~332 ·
i.oiss-os 0.019
W:6:;;:ri~ tir~?~::6üi. ::rit.
. ··raso......... . - .
0.019
. -•- ¡0:019
Differential equations d(X)/d(W) = k J vO * (1 - X)/ (1 +X)* TO J T * P / PO d(P)/d(W)
= -alpha J 2 * (T J TO) * PO
A
2
. . . _j
.904332 1 1.013E+06
J P * (1 + X)
Explicit equations T = 450 + 500 * W
vo = 20 TO= 450 k = .133 * exp(3l400 / 8.314 * (1 / TO - 1 / T)) PO= 1013250 alpha = .019
8-36
0060
900
0 . 048
800
Q
0.036 0.024
600
0. 012
500
0.001)
O 00
0..16
0.32 W 0.48
GJ
700
0.64
400
080
0.00
0.16
0.32 W 0.48
0.64
0.80
PS-9 (a) We use the same equations as problem P8-8, except that the energy balance changes as: ( T -T ) +(-r , )(-MI ) -Ua p a A Rxn dT dW
=-'----
Where -MI
FAocpA Rxn
= 20,000 J/mol, Ta=323 K, CpA=40 J/mol/K
See Polymath program P8--9-a.pol. Calculated values of DEQ variables r-·lvariablejlnitial val~Minimalvaluel M~ximal~aluelFinalvaluej
r r · ···r· ···· ·· r·-·-· · T · . 1 ¡W ¡O ¡50. 1o r2 ·1x .. · · · ··· · la· · · · . .... · · ·· ra· . . __ · · ·· · · · · · · ra:111G1a1 ··
[3 Ti . . .. _,,4so. r . ... !· . j4 ¡TO 1450.
¡ .
lisi.'1sss i .
!450.
f4so 1450.
. .. . ~.. .
1·· ·
¡SO.
·········
·¡
¡
lo:iii°i5isi· ·, l3si.18ss ··1 T . . ... . .
1
¡450. , Js Fa·· · · · r2a.·· 1~º· · · · · ·- · · i20.· . . . f 20. · ..... · ·; [6 ll<· · · . . "ío:133 . fo.0292331 ·· · 10.133 . ·· ra.029233i i f1 'jlí~rh~ fo.os . .... 'jo.os" · · . . [o.oe . fo.os" . . . ·i 1s.lr~·- . ··129i···· ·· figi··· ·- · · -·r29i · 1293:···· . ··· 1 Ji jPO·· · · ... ·· ¡iol:3E_;()6 . . , 1.013E+06 . f1.oi3E~()6 .... ·rl.013E;()6 ¡
1iakia··· · · · r210:s2á3·· · · fúa-:s2s3·---·-· fa16.s2s3- · · ·T21ci.s2si··1 r11¡c¡· · · ·ráa~s2s3 · · 1242.3648 · · . í210:s2s1 · · · · · · ·1242.36~i,i · : 11.2J~~:·~-- · }36:02017 · r~36.02.?il" 1~?·ºªso~4- · · · :F?·~~-~ª~~ .
J
Differential equations
d(X)/d(W) d(T)/d(W)
= k * (1 - X)/ (1 +X)* TO/ T / vO = (Uarho * (Ta - T) + rA * 20000) / vO / CAO / 40
8-37
Explicit equations JITO=
450
livo =
20
k,.,.-,1
11 k = 0.133 ~ ,lii1 Uarho
* exp(31400 / 8.314 * (1 / TO - 1 / T))
= 0.08
[~Ta= 293
ij
I I I
PO
=
1013250
CAO = PO/ 8.314 / TO CA = CAO * (1 - X) / (1 rA = -k
+ X) *TO/ T
* CA 0.20 0.16 0.12 0.08 0.04 ·
10
20
W 30
40
50
0.00
o
10
20
W 30
40
50
If !!_A was increased by a factor of 3000, we use the same program with the new value. The profiles
A
are in the graphs below.
450
0.20
434
0.16
[J
418 · 402
0 . 08
386
0.04
370
o
10
20
w
30
40
Q
0.12 ·
50
0.00
o
10
PS-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 ºC
0.20 016 012 0. 08
408
0 . 04
50
º·ººo
10
20
W 30
40
50
countercurrent: d~
--= dW
«e;
Tat= 50 ºC guess and check Tao until Ta = 323 K at W = 50 Tao= 438.8 K 450---------------
0.20
436
0.16
422
0..12
408
0.08
394
0. 04
380 '------~--10
º
20
W 30
40
50
O 00
o
10
PS-9 (e)
For a fluidized CSTR with W = 80 kg, UA = 500 J/s/K,
8-39
20
W 30
40
50
w
-CAO Species balance: X MB= __!_!:___, r = ..:..p-"'b__ l+rk FAo
UA (T-Tª)
+C
F
pA
(T-T)
o
Energy balance: X EB= """"p-"b'-___,_A=º-------Ml
Rxn
XEB =XMB
Solving, X= .95, T = 323 K
PS-9 (d) For a reversible reaction, we have all the previous equations, but the rate law is modified as: ·-rA = k1CA -k,C8Cc C =C =C _}!__To B e AO l+X T
Plugging the equation for k., and solving using POLYMATH program, we get the plots. Only the co-current program and plots are shown. See Polymath program P8-9-d.po1. Calculated values of DEQ variables
¡···1v;·;i;b1e'Ji~it1at;aí~~l M1~¡;;,;a1~elMaxima1 va1~eTFi~a1 ;;·,~; l
[~_:]Y'(····· · · · · · · i§··· · · · · · · ·
? )~ .
J? . . j3 ¡ T !450. ,. ..... · r············r······ 14 ITa ¡323.
Js !-ro ,·r·
l6 ¡va
.. ··
·l4s·a:
. . 1º 1420.7523 j323. "T"
..
t
!20.
120.
t rn::ho 11:3 . li~42131 !9]Po ...... h:~13E+O? li.oÚE:+06
liif~~(--liiü:üii112jTkr ¡·
··n .. •· .. ·.',., .. -.
·······r __ . ,_,,,,,., •' .......-~...- ·' .,, ·-,h·,,,,.
.. - . . .
r¡0.2.
(i!l'¿~ . ···t6
.
-•w·-
J?·??7~~~ ...J 1420.7523 J! 142.0.7565 1
•·-¡""··""······
t~3 .
120.
..
]4so:
·¡
r
J
120.
1
t!42i3i I
f
'li:cú3E+06
·_ 11:oi3E+06 ]
li~rn~ · ¡!;~: m li~~1¿ii-l
'''f'"'••
· · l¡o.076962 .. v• -"~
. . . [6 .
li~F~- · · · · :F36~62aú·
w·.º57593 !450. r· 1426.1627
f4so: · · · ·14so.·
.. .. ...
("''""·· '• '"1'"'•••'"'"''··'•''
· · "·'Js~_:· ·· ···::-:]
·1ª·························· . ····· laó
_
,
, •. ,
·r-36:02017
.2
,
.,
l..¡o.076962. =· ·-;:
,
·l
'"'""..;;
1
.Tt-º:????j?~ i::;;::~ I-~.§ª6?4_ 1i::;;j:i?~]'· ·j
Differential equations d(X)/d(W) = -rA /va/ CAO
8-40
d(T)/d(W) d(Ta)/d(W)
= (Uarho =
* (Ta - T)
+ rA * 20000) / vO / CAO / 40
Uarho * (T - Ta)/ .2 / 5000
+ sign = cocurrent, -ve sign = countercurrent in RHS ot eqn
Explicitequations TO= 450
= 20 k = 0.133
vO
* exp(31400 / 8.314 * (1 / TO - 1 / T))
= .08 * 3000
Uarho
PO= 1013250 CAO = PO/ 8.314 / TO
= CAO * (1 - X) / (1 + X) * TO / T kr = 0.2 * exp(51400 / 8.314 * (1 / TO ce = CAO * X/ (1 + X) *TO/ T CA
(!
CB
= CAO * X /
(1
1 / T))
+ X) * TO / T
rA = -(k * CA - kr * CB * CC) 450 ..--------------,
0060
444
0.048
438
O 036
432
l)J}24
426
0.012
16
32
vV 48
64
80
0.000
o
16
32
\V 48
64
80
PS-9 (e) Individualized solution PS-10 (a) A---?B+C CA =Cr-
FA FT
er =!LF
A
c1 = C, +c1 Fr =FA +F1
¡ 1
8-41
1
~
e
AOI
= cAº +crn
BI + l
PS-10 (b) Mole balance: dX = -rA dV FAO Rate law: -rA = kCA . h. S toíc iometry:
eA -- eAoi 1+ 1-- x To XT 8
e= YAoó ó=l+l-1=1 y = FAº·= FAO __ 1_ . AO Fro FAo + Fw 1 + O¡ 1
&=--·
l+B.1
T=
-XMIRX
+(CPA +O¡Cp;)I'o
CPA +O¡Cp; Enter these equations into Polymath See Polymath program PS.-10-b.pol. POLYMAT.H Results Calculated values of the DEO variables variable
V X
Cao Cio theta Fao Caol e To dHrx Cpa Cpi T k
ra
initial value
minimal value
0.0221729 0.0221729 100 10 4.391.E-04 0.009901 llOO 8 . OE+04 170 200 1100 25.256686 ···O . 01.10894
O . 0221729 0.0221729 100 1.0 4.391E-04 0.009901 1100 8.0E+04 170 200 1098.3458 24.100568 -0.0110894
o o
o o
maximal value 500 0.417064 O. 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 [ 1 J 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 24.100568 -O . 0061524
Explicit equations as entered by the user [ 1 l Cao = 2/.082/1100 [2j
Cio = Cao
[ 3 J theta = 100 [4] Fao = 10
[5J [6J [7] [8J [9J
Cao1 = (Cao+Cio)/(theta+ 1) e= 1/(1 +theta) To= 1100 dHrx = 80000 Cpa = 170 [lOJ Cpi = 200 [ 11 l T = (X*(-dHrx)+(Cpa+theta*Cpi)*To)/(Cpa+theta*Cpi) ¡ 12 J k = exp(34.34-34222/T) [ 13 l ra = -k*Cao1 *(1-X)*To/(1+e*X)/T Conversion
LO
..,..,.,
0. 8
-
0.6 0.4
_,.,,.
0.2
o.o o
100
~200
V
Tempera ture
----
1060
No inert Modera te Laree
- No inert - Modérate
1020 980
,,.........
-- -----~~--
940
300
400
500
900
....
o
100
200 V
300
400
500
1.00
[i] OAO 0 . 20
ºººo
J
6
Ihela
12
15
PS-10 (e)
There is a maximum ate= 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.
PS-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--10-d.pol. 8-43
Temperature
Conversión .......-------,
1.0n------=
1600.------~--------, 1500
0.8 - No inerts Modérate - J.,.uue
0.6
1300
0.2
1200
o.o w:::;...-~--~-----·-0
100
200 V 300
- Moderare - Laroe
1400
0.4
400
500
40
so
- No ínerts
1100 .----------·:· O 100
···-· · ----. ·-····_· --~ 200V 300 400
500
1.00..-----~---------, 0.92 0. 84 0.76 0.68
.~--~-~-·---·---· 10
20
JO Thela
The maximum conversion occurs at low values of theta (0 < 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.
PS-10 (e)
We need to alter the equations from part (e) such that -rA = kC1 and CAo = 1 A plot of conversion versus theta shows a maximum at about e = .5. See Polymath program PS l.O-e.pol.
8-44
0 . 90~-------------,
O 82
-
2
4
Ihet3
8
10
PS-10 (0 W e need to alter the eq uations from part (e) such that 1-X
=r, = k [ C, -
C;: e ]
T
° . Now weed need expressions for Cs and Ce. From l+EX T stoichiometry we can see that Cs Ce. In terms of CAo we find that: We already know that CA= CA0
=
C =C =C B
e
.
Ao
X
'Fa
l+EX T
We also need an equation for Kc:
te; = K0
exp[!}Jf
Rx
R
ººº~(-
(_!_ _ __!__]] = 2exp[8 '.I'i
T
1- _ __!__)]
8.314 1100 T
When we enter these inot Polymath we find that the maximumn conversion is achieved at approximately 0 = 8. See Polymath program PS-10-f.pol. 1.00.-----------------,
..--------·---.J
3
6
Ihet3
12
15
PS-10 (g) See Polymath program P8-10-g.pol.
8-45
Fb
10
8 - Part (d) Pnt (e)
6
4
2
o o
100
200
V
300
400
500
PS-11 (a) Start with the complete energy balance:
dE . . -LE¡F; 1 in-LE¡F; 1 out -=Q-Ws dt The following simplifications can be made: lt is steady state. In part a, there is no heat taken away or added There is no shaft work That leaves us with o= -LE¡F; lin -LE¡F; lout Evaluating energy terms: In: H AOFAo + H soFso + H coFco Out: HA(FA +RAV)+H8(F8 +R8V)+Hc(Fc +RcV) Simplifying, HA(FA +RAV)+H8(F8 +R8V)+Hc(Fc +RcV)-HAoFAo -H80F80 -HcoFco = O If only C is diffusing out of the reactor we get: RAFA +HBFB - n¿ (Fs +RsV)-HAoFAo-HaoFso -HcoFco =0 Now we evaluate F¡ FA = FAO ·- FAOX FB = FBO + FAOX Fe= Feo+ FAoX -Re V Inserting these into our equation gives: HAFAo -HAFAoX +HsFao +HBFAoX +HcFco +HcFAoX -HAoFAo -HBoFBo-HcoFco =0 and note that FBo = Feo = O H AFAo - H AFA0X + H 8FA0X + H cFA0X - H AoFAo = O and combining and substituting terms gives: FAo(HA--HAo)+FAoXMI.Rx =0
8-46
FAOCPA(T-T0)+FA0XWRx
=Ü
Differentiating with respect to V with LiCp = O
dT dX + FAO -( ¿V[ Rx ( T)) = Ü dV dV dT (rA)[ WRx(T)]
FA0C p
-
-.-=
dV
I.B¡Cp
Combine that with the mole balance and rate law:
=r
dFA dV
AdV
dFB
= -r
AdV
dFc
= -r
A
-k
e
ce
-rA =kCA
e
A
=C
FA To TO F T 1
e =e
FB To roFrT
B
e =e
Fe To roFTT
e
= 10
Forkc
See Polymath program PS-11 a.pol.
f¡
Calculated valúes of DEQ variables ·····r··-·· ··· ·····-········
············ - ·· · - .... ·T······ ··"· ··--· · · · • ·
l
I
·· · ··
···
¡variable I Initial value ¡ Mini mal value Maximal val u
¡i ¡v···
r ·· · ··· ·
·¡
e¡ Final value I
· · · · ·¡a·· · · · · · · · Jo· · · · · · · · · ···· · · · · · Tsa:· -- · · · · ·- · ······· Tsa·. ····· · · · ·- · ·1
r2··1F~·· · · · ··ro··· · ········· · ·1a· r.r. ,. ·., .
¡3
••O
•
,a.a,
¡Fb
-r"" ,_.
'
¡O
¡¡ F~~ · · · · fs:42
¡s· r,: · · · · -_·· 1j76-lf<. [dhrx
f4sa:·
.42 · · J-2.0E+04
T
¡a· F< · · · · · To.iii .. · · · · · fg ·rk~
jio¡ó~
. · ·110.
f ii f ca
f~!li -
· ··-····10:0012968 .. ···ro:aoos2Gi··1 'T ·<· · r "'• ,., ,, , IO j0.1978837 ¡0.1978837 . . J ls.222ii6 · · ¡ s.42 lS.222116 1 f4so. ·· · · · · ·Ts48:94ia·· · ··rs4s.941a·· 1 ff~i- · · · · · 15.42129?_ · · -1~·i_~-ª~~~~ · 1 ¡-2.0E+04 ¡-2.0E+04 l-2.0E+04 ·¡a:i:ii · · . . . · · 10.6036997.. ··10.6036997 ·
' '"'' 'r•
•••o•,,•••···•···".·
.. ,,.··,,,""""""''".''••'•····
•••',••'O•·•·-A, ... ,·, .....
, ,• ~,,'.,._,,,_,_,'
íisl~~·· · · ·
T2.iiooú· · · ·12.610831 · · fa.710027 Fo.3604336. ··r:o.3604336 · ··r-0.0026249 . .. . :J¿o:_ •.• . .•.
·.· · · · · · ··:·.· _· ::·.·_-]4a::···-·· · ·:··:·:.· :·······
I d(Fc)/d(V) = -ra - kc * Ce fl d(Fb)/d(V) = -ra 8-47
-·······--
··- .. ·--
.
¡
'.i 1
:=1~:~!ti: ==t~tf*q
Differential equations
-- --
~. , . ._,,.,
1
t~·OOÓ69ÓS j~.OÓ02635_
r1.G·l~P~·· ····· · . •. J4a.· ·- ·
- -- --
'°"""ª"'"''
... · · .... f10. ·- -- . · · · · · Tia: -· · · .. · · Tia·.······· · ., ·_ . · ·¡2.noo27 · '[2~710027 · · · li.710027 · · lz.710027 . T2.710027 . í2.610831 ¡2.710027 12:Gioá:31
¡i~dcb
·--
,.,, ... ,,
f 2.6ios3i
}ó:00262491 ....
·.-.i4~--~· ····· •· · ··········1
1
:
d(Fa)/d(V) = ra d(T)/d(V) = ra * dHrx / Cpa
Explicit equations Ft = Fa + Fb + Fe dHrx = -20000 k = .133 * exp((31400 / 8.314) * (1 / 450 - 1 / T)) ke
=
10
Cto = 100 / .082 / 450 Ca = Cto * Fa / Ft Ke = .01 * exp((dHrx / 8.314) * (1 / 300 - 1 / T)) Ce=Cto*Fe/Ft Cb = Cto * Fa / Ft ra = -k * (Ca - Cb * Ce/ Ke) Cpa = 40 6.0 .---·--------------, 4. 8
3.6 2.4 1.2
10
20
V 30
40
50
vary kc to see how the concentration profiles change.
PS-11 (b) Now, the hear balance equation needs to be modified. dT = Ua(Ta -T)+ FAo(rA)[ M!Rx (r)] dV
FAO"i:,(l¡Cp
See Polymath program PS-· 11-b.pol.
8-48
6.0 ~-----------~ 4. 8
3. 6 24 1.2
o.o o
10
20
V
40
30
50
P8-12(a) To find the necessary heat removal, we start with the isothermal case of the energy balance
Because there is no shaft work Ws = O. 11CP =60-25-35=0 And for isothermal operation T = TO If we simplify the energy balance using this information we get:
or Q = FA0X !1H ex = CA0vX !1H 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: 11Cp
= 60-25-35
=O
Q = O and Ws = O -X
su = ¿ B¡Cp; (T-T RX
0)
Because feed is equal molar in A and B, 88 - X !1H RX f1H RX
= (e + e PA
PB )(
=1
T -To)
(25+35)(350-300) = ----·--·------0.4
kJ mol
= - 7500-
Now go back to the isothermal case: · Q=CA 0vXl1HRX=
( 1000m mol)( m3 ) (0.2)( -7500mol kJ ) 3 .5min
8-49
.
kJ
Q = -750000-.
mm
P8-12(b) We start with the energy balance for the second CSTR (already simplified): UA
Ta - T )-( X 2 - X¡) su RX
-(
= ( C PA + C PB) ( T - To)
FAO
This equation has two unknowns (T and X2) and so we need another equation. Now wee need the mole balance for the second reactor FAO (X2 -X¡) FAO ( X2 -X¡)
v2 =
= kCAo 2 ( l-X2 )2
--rA2
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 kas 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. = 1
FAOXI
kCAO 2 (1-X 1 )2
k=
X
vX
FA0
VC10(1-x)2 -
VC~:(l-Xf
Solving for k at 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
( k E(l 1) we get E/R = 2664.
In - 2) =-- ---k1 R T., T2
Which meaos k(T) ~ k1 exp [
!(:. - ; )]
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-b.pol. POLYMA TH Results
NLES Solution Variable T X2 UA
Value 327.68712 0.4214731
4
f {x)
-2 .274E-13 -6.666E-12
Ini Guess 340 0.4
8-.50
350 0.5 0.2 -7500 300 1
Ta vo Xl dHrx To V k
3.309E-04 1000 500 -110.73657
Cao Fao ra
NLES Report (safenewt) Nonlinear equations [ll f(T) = (UA)*(Ta-T)/Fao-(X2-X1)*dHrx-60*(T-To) = O ¡ 2 J f(X2) V-Fao*(X2-X1 )/(-ra) = O
=
Explicit equations
r n UA = 4
[2] Ta= 350 [3 J vo = 0. 5
=
X1 0.2 dHrJ<=-7500 [6J To= 300 [7J V= 1 [ 8 J k .00015625*exp(2663.8*(1/300-1/T)) [ 9 J Cao = 1000 [10] Fao = Cao*vo [ 11 J ra = -k*CaoA2*(1-X2)A2 [4J
[5J
=
P8-12(c) Now we need the differential form of the energy balance dT
Ua(~
dV =
uatt, -T)-rAMIRx
-T)-rAMIRx
FA0¿/l¡Cp¡
FAo ( CPA + Cp8)
=
we also need the mole balance far a PFR. Far this case it simplifies to: dCA dCB --=--=-r dt dt A with =r: = kCACB and we can use the same equation for kas in part (b). When we put these equations into Polymath we get an outlet conversion of X = 0.33 See Polymath program P8-l 2-c.pol. POLYMATH Results Calculated values of the DEO variables Variable ---V T X
Cao Ua Ta dHrx Cb
initial value
o
300 0.2 1000 10 300 -7500 800
minimal value
o
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
0.5 500 25 35 l.563E-04 800 -100
V
Fao Cpa Cpb k
Ca ra
0.5 500 25 35 l . 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 [ l J d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb)) [2 l d(X)/d(V) = -ra/Fao Explicit equations as entered by the user [lJ Cao = 1000 [2J Ua = 10 [3l Ta=300 [4l dHrx = -7500 [Sl Cb = Cao*(1-X) [6] V=0.5 [7 l Fao = Cao*v [8l Cpa = 25 [9l Cpb 35 [ 10 J k 0.00015625*exp(2664*(1/300-1/T)) [ 11l Ca Cao*(1-X) [ 12 J ra = ··k*Ca*Cb
=
= =
P8-12(d) In this case we need to replace the rate law we used in part (e)
-r A
=k
[e
A
CB - K Ce ] e
We also need an equation to calculate K, at different temperatures. Ke
= K ci exp[Af!..Rx.(_!_--~JJ R T.T 1
be careful of the units when entering Kc1 into Polymath. Also note that the initial temperature is diff erent than in part (e) We get an outlet conversion of X= 0.48 See Polymath program P8-12-d.pol. POLYMA IH Results Calculated values of the DEO variables Variable V T X R
Ua Ta dHrx Cao
initial value
rninirnal value
350 0.2 0.0083144 10 300 -7500 1000
314.93211 0.2 0.0083144 10 300 -7500 1000
o
o
rnaxirnal 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
V Fao Cpa Cpb k
Ca Cb Ce Kc ra
O. 5 500 25 35 5.556E-04 800 800 200 0.002 -300 . 01803
o. 5
O.5 500 25 35 2.381E-04 519 . 53058 519.53058 200 0.002 -335.38132
500 25 35 5.556E-04 800 800 480.46942 8.63E+l21 -64 . 253241
0.5 500 25 35 2 . 381E-04 519.53058 519.53058 480.46942 8.63E+l21 -64 . 253241
ODE Report (RKF45) Differential equations as entered by the user [ l] d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb)) [ 2 J d(X)/d(V) = -ra/Fao Explicit equations as entered by the user [ 1 J R = 8.3144/1000 [21 Ua = 10 [31 Ta= 300 [4J dHrx = -7500 [ 5 J Cao = 1000 [6] V= 0.5 [7 J Fao = Cao*v [BJ Cpa = 25 [9J Cpb = 35 [ 10 J k = 0.00015625*exp(2664*(1/300-1/T)) [lll Ca= Cao*(1-X) [12] Cb=Cao*(1-X) [ 13 l Ce = Cao*X [ 14 J Kc = .002*exp((dHrx/R)*(1/350-1/T)) [ 15 J ra = -k*(Ca*Cb-Cc/Kc)
PS-12(e) Individualized solution P8-12(t) For the gas phase the only the stoichiometry changes.
CA= CAo(/:c: )(;) ande= YAoó = 0.5(1-1-1)
= -0.5
From Polymath we see the exiting conversion is X = 0.365 See Polymath program P8-12 f.pol. POLYMATH Results Calculated values of the DEO variables Variable V T X
To Ua Ta
initial value
o
300 0.2 300 10 300
minimal value
o
279 .3717 0.2 300 10 300
maximal value 1 300 0.3650575 300 10 300
8-53
final value 1
279 . 3717 0.3650575 300 10 300
-7500 1000 0.5 500 25 35 l.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 l.563E-04 -0.5 888.88889 888.88889 -56.423752
-7500 1000 0.5 500 25 35 8.lllE-05 -0.5 834.06666 834.06666 -123.45679
ODE Report (RKF45) Differential equations as entered by the user [ l J d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb)) [ 2 J d(X)/d(V) = -ra/Fao Explicit equations as enterad by the user [1] To= 300 [21 Ua = 10 [31 Ta= 300 ¡ 4 l dHrx = -7500 [51 Cao = 1000 [6) VO 0.5 ¡ 7 J Fao = Cao*vo [81 Cpa = 25 [91 Cpb = 35 [10] k = 0.00015625*exp(2664*(1/300-1/T)) [11] e =-.5 [ 12 l Ca= Cao*(1-X)*To/(1+e*X)/T [13 J Cb = Ca [ 14 J ra = -k*Ca*Cb
=
PS-13 K = CcCD =
CACB
e
~x
= e
T=T0
x;
(1-· x, )2
ji(;
I+jif; w Rx
=
300- ( --Joooo) (25+ 25)
x
= 300+ soox
See Polymath program PS. 13.pol.
~-·-1--
, .
Calculated values of NLE variables -r::-·-----·---1 .· ~-- ~ -¡t ¡variableValue f(x) iinitial Guess J !1 l~ • 0.9997644 3.518E-11 i?·~ (O < ~~ ~ r.; .
r·
l
--
-
- ·-. .
;.JJ
8-54
-7500 1000 0.5 500 25 35 8. lllE-05 -0.5 834.06666 834.06666 -56.423752
Nonlinear equations f(Xe)
= Xe - (1 - Xe) *
Kc
A
0.5
=
O
Explicitequations
= 300 Kc = 500000 * exp(-30000 /
T
1.987 * (1 / 323 - 1 / T))
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
0.4
480
0.26
500
0.1529
520
0 . 091
540
0.035
560
0..035
8-55
Xe 1 --r-~~--¡¡¡¡;;;¡;;~-----~-------------·-·---~---···---~----~1
e
.
0.9
o.a
o ·¡¡; 0.7 ~ 0.6
s::
8
§.
0.5 0.4
!!:!.. 0.3 CI)
X 0.2
0.1
o 400
350
300
450
500
550
Temp (K)
PS-14 For first reactor,
K
X = •1 e 1-X
or X
· el
K = e el I+K
e
For second reactor K
e
= 882+xe2 1-X
= Kc -882
orX
e2
e2
l+K e
For 3rd reactor
K = 883+xe3 orX = Kc -883 e 1-X e3 e3 I+K e 1 st reactor: in first reaction X, = 0.3 So, FB = FA01(.3) 2nct reactor: Moles of A entering the 2nd reactor: FA02 882
=
.2FA01
1.7FAO
-FA02'i:-CP¡B¡
= .12 (T-To) + FA02x
(-MIR)= Ü
( e; +88CpB )(T-T o)
X=-----
-M!R
Slope is now negative 3rd reactor:
8-56
= 2FA01
- FA01(.3) = l.7FA01
x., = 0.3 eB = (.2FAOl)+.3FA02 = FAoJ.2+.3xl.8) FA03 = FAO! + FA02 (1- X eJ = FAOl + 1.8FA01 (1- X eJ = 1 + 1.8(1-
Say
FA03 = 2.26FAOI Feed to the reactor 3: (2FA01) + .3FA02 = FA01 (.2 + .3xl.8)
= 0.7FA01
=
.74 2.26 Feed Temperature to the reactor 2 is (520+450)/2 = 485 K Feed Temperature to reactor 3 is 480 K Xfinal = .4 Moles of B = .2FA01 + .3FAo2 + .4FAo3 = FA01(.2 + .54 + (.4)(2.26)) X = FB/3FAOl = .54
883
.3)FA01
,, ,............ -
= 1.64 FA01
Xe.
----~¡ Xo.:,.
/
·'i ,'\.
PS-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)/2 = 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'CP;B¡ (T-To )+ FAOjx
(e
pA
(-MIR)= o
+ 88CpB) ( T -TO)
X=---------
-MIR
and 98 for reactor 1 = O. For reactor 2, 98 > O. This means that the slope of the conversion line from the energy balance is larger for reactor 2 than reactor 1. And similarly 98 for reactor 3 > 98 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. )(e.,
k'l.
~3
X
PS-15 (a)
Substrate 7 More cells + Product
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.
_ FA0X _ Fs0X r8 V - -- -and since -rs = then, -~
r8 =µC
-~
~/!
e
e
andµ=µ(T)--··-s_ K s +C s
;º) ~ºº)
0.0038*Texp( 21.6where µ(T) =
Jlimax
l+exp(153-
6
48
if we combine these equations we get:
V= ~,sFsoX µ(T)CcCs K, +Cs
8-58
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 =
e
PO (
T - To )
The heat removed by the heat exchanger = UA(T-T.)/Fso
UA
R(T) = CPs (T-·'J'o)+-¡-·(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= Ll.
See Polymath program P8-15-a.pol. Differential equations as entered by the user [ 1 J d(T)/d(t) = 1
Adiabatic Case
[3]
----]
20000 ~-----------~
Explicit equations as entered by the user [ll rnurnax » .5 [2l Yes= .8
10000
V0=1
[4] Ta= 290 [ 5 J mu= mumax*(..0038*T*exp(21.6-6700/T))/(1 +exp(153-
48000/T)) [6] Ks=5 [7J V= 6 [8] Cso = 100 [ 9 J Fso = vo*Cso ¡101 Cps = 74
o
"
.
-10000
-
G(t)
RJi,
-20000
-30000 L.--~------~----------' 294 299 303 T 308
312
8-59
-- -
-------·--------·----··-·-------
-
---------·
·--·-·--------
-·-· ·--
-----------------
--
---------------·----------------
·---------·
-----------··---·---
317
J dH = -20000 J UA = 0*300 [ 13 J kappa= UA/(Cps*Fso) [ 11 [12
Non-Adiabatic Case
[14J To= 280 [15] X= 1-((Fso*Ks)/((mu*V*(CsoA2))-Fso*Cso)) [l6J Gt = X*(-dH)*Ycs [ 17 J Rt = Cps*(T-To)+UA*(T-Ta)/Fso
20000 ~--------------, 14000
lndependent variable variable name : t initial value : O final value : 30
8000 2000 -4000 -10000 "------~------~----' 2 4 299 303 308 312 1
PS-15 (b)
317
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 file 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,
R(T)
= mcCpc
(T-Tª
{1-exp(
\
-U~J)
mcCPc
+ Cps (T-T0)
Now we set R(T) equal to the maximum value of G(T) which is 15600 J/h
G(T)
J)
= 15600Fso = mee PC (T -T" { 1- ex{;,,~~~c + FsoCPS (T -T,)
And now plug in the known values . Assume the maximum coolant flow rate and that will give the minimum heat exchange area.
1s600I.(100L) = g hr (6ooooJ;)[74·/i)310K-290K
J+exp
-VA
-·--
_!_) ( 60000 .1..)[14 hr gK
8--60
+(100!)[74 /K)310K-288K)
1560000 ~ hr
= (88800000
~) hr
-UA
1 - exp
+ ( 162800 :, )
( 4440000 -1-) hrK
UA=704l5-1hrK A=l.4 m3
PS-15 (e) 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.
PS-15 (d) Increasing TO enough will eliminate the Iower 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. lncreasing T ª will lower the exit concentration of cells.
PS-16 (a) G(T) =D..HRX X
=
.ss: r = 6.6xl0- exp[-1!..(-1- _ _!_)] l+rk R 350 T 3
R(T)
= CP0 (1 + K)(T-Tc)
cp0 ='LB;Cpi =BAcpA K=
Te=
+e1cp1
=20+30=50
= 8000 =2 cpoFAO (50)(80) UA
ToFAOcpO +UATª UA+CpOFAO
=
KT +T,o
ª
I+K
=350
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.
PS-16 (b) First, we must plot G(T) and R(T) for many different T0's on the same plot. From this we must generate data that we use to plot T, vs T0•
8-61
1
1
¡e ¡_
G('l) \S. R(1)
10000
'J; w. T.
42.5
8000 400
;
6000
C)
2000
375
E• 4000
lSO 325
o. ZTS
325
375
T(K)
300
425
200
250
300
lSO 400 T. (JQ
450
,00
550
PS-16 (e)
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 wíll provide for higher conversions.
PS-16 (d) For a temperature of 375 K, the conversion is .968
PS-16 (e) The inlet extinction temperature is 375 K.
PS-17 The energy balance for a CSTR: -FA0X/J.HRx(T)
UA = FA0[~B¡Cp;(T-T¡¡)+-(T-.Y 0)] FAO
FA0X =-rAV Cps is independent of T.
e
pO (
T - To ) = epi
G(T)=(-/J.HRx)(-~V) FAO R(T) = FAOcpS +UATr -UAT Differentiating: dR(T) = UA dT dG(T) _ -1:,.HRx d(-rA) dT FAO dT d(-r) E where A = --2( -rA) dt RT Setting these two equations equal to each other and manipulating, 8-62
r
In 1.421 = [E(323-313)] 1.127 R 323*313
E=19474 FAoCps + (T -T) > RT,2, andit will berunawayreaction
UA
E At A< 2 . .5 m2 it will become a runaway. r
a
PS-18 (a) Mol Balance :
~ = FAoX =
=r,
voCAoX k[CA-C8!KJ
(:. } [ ( 1- X) - X I K,] X [I + i-k(l + 1/ Ke)] X=
= rk
Tk l+rk(l+l/ Ke)
G(T) = -!:ú/ RxX k=lmin·-1
= 80000X
r=lOmin K, =100
10 =.901 1 + 10(1.01) G(400) = 72080 cal/ mol X=·
PS-18 (b) UA R(T)
= 3600 10*40
=9
= CpA (1 + K)(T-Tc) = 400(T-Tc)
T = 1'o +KT,, =310 e
l+K
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
------~-------~-------
-
-----~---------------
-
--------~~
-----
-.------
PS-18 (e) 3 lOK and 418.5 K are locally stable steady-state points
PS-18 (d) R(418.5) = 400(418.5-310)=43400 G(418.5} = 43400 = 80000 * x 43400 X= SQOOO = .54
PS-18 (e)
-
The plot below shows Ta varied. Tos175
~~
_... .. ,······
.....
......
.....
...
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
...
200
_
.
, ,._ 150
-
...
·---'---~---·-·-------
PS-18 (f)
-
This plot shows what happens if the heat exchanger quits. Toe upper steady-state temperature now becomes 431.5 K.
T(IC)
PS-18 (g) At the maxinrnm 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) = G(T). See Polymath program PS-18-g. pol.
8-65
UA
K'=---
FAOcpA
R(T) = CpA (1 + K)(T - Te) = 73520
where T
= T0 + «tª
l+K If we plug in the values and salve for UA, we get: UA = 7421 cal/min/K e
PS-18 (h) Individualized solution PS-18 (í) The adiabatic blowout flow rate occurs at V T = 0.0041s =-
T
= 0.0041s
vo
v0 = TV V0
= 0.0041 * 10
dm3 =0.041-. mm
See Polymath program PS-18--i.pol. 6000.-------------,
/
4600
/.l\.!
i;,(¡
3200
"
1800
f
RH)
--
/
/
,#/
400
30
60
t
90
/
I
\
1
\
\
'\
120 150
PS-18 (j) Lowing TO or Ta or increasing UA will help keep the reaction running at the lower steady state.
PS-19 Given the first arder, irreversible, liquid phase reaction: A-~B
UA = l.Ocal / min/ºC
8-66
C
Pure A Peed= O.Sg mol/ mm 6.C p
= 6.C pA -
6.C pB
º•
=C
Po
=2cal/g-mol/ºC
MIR= -200
=O
cal gmolA
= const
FAoX Vo Design Ean V --------
=r,
k(l---X)
Rate Law : - rA = kC A Sunchiometry CA == CAO (1- X) Energy Balance -UA(T-T,1)- FA0X MIR
= FA0"i,(J;Cp;CT-T0)
Símplífyíng,
.
rk
X=--·-=---
l+rk
o=s ==F
l l+-1
rk == F40(--Af!R)_
X(--Afl) AO
R
l+-
l
rk
This is the curve plotted in the problern statemcnt
From the equation for heat generation curve, wc
get
= --·
Qg - = - Qz = Qz F40(--Af!R) 5*200 100 The equatron for hcat removal curve is QR = FA0C p A (T -T0) + UA(T -T4) == .5 * 2 * (T ---T0) + l O* (T -T1) X
QR
== 2T-- 1~
-100
Plot this along with the heat generat1on curve far various To This
1~ shown
¡ r:~:~:···-Qg··--·--1 1
1- - ··-To=1501
1
¡- · -
i· · · · · To=160¡
·To=1701 i ¡- · · ·-To=180 ¡
l I ______ J
-To=1901
.-
-To=200
l. - - To=2101
'-----·---·-~
140
120
160
180
220
200
T {C) "
'
,.
.
.....
-
.. --·-"'"''~---··--
.... •-Ho,oO,,,.,--,OoO, .. -·--·---·----·-
The mtersecuon between Qg and QR can be used to prepare the igrution - extinction curve shown m figure P8-22-'.2 The valúes fo: T~ as a fnnction of inlet temperatme, To, are tabulated below.
8-67
m the figure
TO ( degree C) 150 157 160 170 190 202 210
T s ( degree C) 132 13.5, 172 137, 168,176 142.5, 167, 181 154.5, 165, 193 . .5 162, 199 204.5
PS-19 (a) To obtain high conversion, the reactor must operate at or beyond point 14 in figure. So the mínimum inlet temperature for high conversion is To 2 202 ºC.
PS-19 (b) The temperature of the fluid in the reactor corresponding to temperature in part 1 is Ts > 199 ºC (point 14 in figure).
PS-19 (e) The inlet temperature of the fluid is To = 202 + 5 = 207 ºC. This will be somewhere between points 14 and 1.5 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 ºC. From figure, Ts = 19.5 . .5 ºC. From equation (.5), Qs = 2*196.5-19-100 = 95cal/mol
QR
= Qg = 96 cal/ mol
So X= Qg =0.96 ' 100
PS-19 (d)
Extinction temperature is the temperature corresponding to point 3 in the figure= 157 ºc
PS-19 (e) Individualized solution
PS-20 (a) The following are the explanations for the unexpected conversion Case 1: Broken preheater or ineffective catalyst Case 2: The equilibrium conversion was reached due to a problem Case 3: Broken preheater or ineffective catalyst Case 4: The equilibrium conversion was reached due to a problem Case 5: lneffective catalyst Case 6: The equilibrium conversion was reached dueto a problem
PS-20 (b) 8-68
and temperature profiles with the heat exchanger with the heat exchanger with the heat exchanger
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 preheater or ineffective heat exchanger Case 4: Ineffective catalyst
PS-21 Below is the FEMLAB solution. l. Parameters in simulation on the tubular reactor in Problem 8-6: Reaction: A+ B ~ C (1) operating parameters
Reactants • Inlet concentration of A CAo = 100 mol/m' •
Inlet concentration of B C80 = 100 mol/m3
•
Inlet total flow rate v0
•
Inlet temperature of the reactant T0 = 300K
= 2xl0-3
m3/s
(2) properties of reactants
• Heat ofreaction, 1'1HRx, dHrx = -41+20+15=-6 kcal/mol=-25100 J/mol • Activation energy, E= 41840 J/mol • Specific reaction rate k¿ = O.Olxl0-3 m3/mol.s @300K • Reaction rate k = k¿ exp[ E (_!_ - __!_ )] R T0 T • Gas constant, R = 8 . 314 J/mol·K • Rate law - rA = kCACB Assumption: • Thermal conductivity of the reaction mixture, ke = 0.68 W/mK
(needed in the rnass balance and the energy balance)
• Average density of the reaction mixture, p, rho = 1000 kg/nr'
(needed in the energy balance)
• Heat capacity of the reaction mixture, Cp = 4200 J/kg-K (needed in the energy balance)
• Diffusivity of all species, Diff = 10-9 m2/s 2. Size of the Tubular Reactor (1) Volume of reactor sized by a PFR (2) From FEMLAB • Reactor radius, Ra = 0.1 m • Reactor length, L = 10.0 m
=
0.317 m'
8-69
---------~~-------
---- -- --- ~- - --
---
-----
- ---
-------
3. Femlab Screen Shots (1) Domain
(2) Constants and Scalar Expressions - Constants
- Scalar Expressions
(3) Subdomain Settings - Physics
8-70
(Mass balance) V•(-DVcA+cAU)
= R, cA = concenlration
Library material: Quaotity
5ts D isolropic D anisotropic
R u V
(Energy balance)
6ts (.') k (isotropic)
O k (anisotropic)
- Initial Values
(Mass balance) cA(tO) = cAO (Energy Balance) T(tO) = TO
- Boundary Conditions @ @ @ @
r = O, Axial symmetry inlet, cA = cAO (far mass balance) T = TO (for energy balance) outlet, Convective flux wall, Insulation/Symmetry (for mass balance) Thermal Insulation (for energy balance)
(4) Results
(Concentration, cA)
8-71
File
Edlt
Ó~ riiI [!]
Optlons Oraw Physlcs Mesh salve Posq::,-ocessing Multiphyslcs Help
~ii iii;í; j !íftJ¿, ,i&,j ~ :ÍI ~ 10
Afü
gj@!fe' ¡a_¡i 1bj_,ici'l c,Q D Corfour: Concentration,cA ------------------~==~ ---------------------__ -Xi:),,.:ti
/
3
2
.
-01
0.1
0.2
lííi /
4
02
8-72
'--------------------
_
----------------------------------
...
_
PS-22 (a) Liquid Phase : A+ B ~ C dX
-r
dV
FAO
-=-A
-rA =kCACB k
=.Ol*exp[-!G-
3~)]
CA =CA0(1-X)=C8 The energy Balance: dT Ua(Tª -T)+(-rA)(-MlR)
=
dV-
a=-
FA0 [
c:A +
cPB
J
4
D
AssumeD=4 U =~X
m2Ks
leal
4.1841
x lm2 =.0120
l0dm2
cal
dm2 Ks
~ee Polymath program PS-22-a.pol.
8-73
---------
--- --------~
----- ~--~--
-------------
- ---
----
------
-----
----------
Differential equations
d(T)/d(V) =(U* a* (Ta -T) + (-ra) * (-Dhrl)) / (fao *(epa+ cpb)) d(X)/d(V) = -ra / fao Explicit equations Ta= 300 R
=
1.988
E= 10000
cao
=
.1 ca = cao * (1 - X) cb = cao * ( 1 - X) k
=
* exp(-E / R * (1 / T-
.01
1 / 300))
ra = -k * ca * cb cpb = 15 epa= 15 fao = .2 Dhrl = -6000 a= 1 U= .0120 500
1.0
460
0.8
EiJ
420
0.6
380
0.4
340
0. 2
300
o
200
400 V 600
800
1000
200
PS-22 (b)
8-74
400 V 600
800
1000
Gas Phase: A
µ
B+C
dX = -rA dW
FAO
= k¡CA
-rA
-krCBCC
C =C =C ~-Tº B e Ao l+ X T k =.133*exp[- E(_!_ __ l )] R T 450
--!)J,Er
k: = .2exp[E, (-1 ' R 450
T
= 51.4kl /mol
The energy Balance: dT Ua(~ -T)+(-rA)(-MlR) ~-=~-=~~~~~--'-~ dW
Mi
R
= -40-
50 + 70=-20 kJ/mol
U=5 See Polymath program P8-22-b.pol. Calculated values of DEQ variables
[¡v;_n~b·~v,;1~,;~1;;e1~¡.;¡,;;¡;alueJ M;;;¡mal ,,~¡.;;l Final ~~;..;jJ
i1--lvV
w
· lº rI--·ffº· · t:66:-· ¡? . lx
Is ¡k •· 1~0··
r6·
fa.133
lío. --
r7· T1c;· · · fo.2
líi JU;rt,;
'Ti~
·rs.
t--·
. - -11!~t ~ 9.~~- ·
· ro.06s1696
15?:.... ..,,¡5-~·--·· ·, ¡o-os6oass . . . _ f.?:_~s6oa5-_5- j
..
·t!rt9.~~
t:66~
2·
To.133
·120: · - -- · · · · ·
·-r2·a: ....
· ro.0622149
··-J
¡o.o6si696 ·
c.
1
120:-- ..... - --1
••
· '0.0622149 J
ro.2.....
is:
1 I
Js:
Js:
¡~92.8948_ . ·
¡;f¡:~: . - nf~tf I
1.
f32i - -bii . . l:i2i ..... ' fióí PO. 'li.013E+06 f1.ol3E~Ü6 ' fi.oi~3E+o6 f i:oi3E+C)6 · [ii¡cAO , . . · · r.304.6819 . T304-.6819·· - ,. f304:·68Í9 . ·r3o4:i5f319·· ¡ 19
iHf
t.323 .. --··
f
lt6819
)i4lcB·
lo
-lo
,.isf~~··:· · · · · · · ,.· ·t~40.52,269 _· ·-·.r·~.~?-52~~.9.
..
f 17.40323
117.40.32.3-
_· ·¡~o:2446624
· ·1.=~:24~~~24··1
Differential equations
d(X)/d(W)
=
-rA / vO / CAO
8-75
·¡
= (Uarho
d(T)/d(W)
* (Ta - T)
+ rA * 20000) J vO / CAO J 40
Explicit equations TO= 400
= 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 k
Ta= 323 PO= 1013250
= PO/ 8.314 / TO CA = CAO * (1 - X)/ (1 + X) *TO/ ce = CAO * X/ (1 + X) *TO/ T CB = CAO * X/ (1 + X) * TO/ T rA = -(k * CA - kr * CB * CC) CAO
T
0.060
400
0. 048
394
Q
0. 036
388
O..o24
382
0.012
376
0. 0000
......._.__......._
10
20
W 30
40
50
370
.._·-- ... (l
10
20
W 30
---·--·
40
50
PS-23 First note that ~Cp = O for both reactions. This means that AfIRx(T)= ~HRx Now start with the differential energy balance for a PFR; dT _ Ua(Ta-T)+¿r¡1(-MlRxiJ)
_ Ua(Ta-T)+1i.c(-MlRxic)+rw(-·MIRxw)
dV -
-
¿F¡Cp¡
¿FjCPJ
If we evaluate this differential equation at its maximum we get dT --- = O and therefore, Ua(i: -T) -- 'i.cMI Rxic -- rwMI RxZD = O dV
We can then salve for rrc from this information.
8-76
O
for both reactions.
¡},j/ RxlC
'ic =
10(325-500)-0.4
( 0.2) ( 0.5) ( 5000)
-50000
= 0.039
'ic = 0.039 = k1cCACB = k1c (0.1)(0.2)
k1c
= 1.95
k.c,,oo, ~ k.c¡,oo¡ exp[ ln ( k1c(soo) E =
R
J
k!C(400)
( 4~0
! ( ~O - 5~0)]
= 7628
5~0)
E=l5158 cal ' mol K
PS-24 (a) See the additional homework problems in chapter 8 at http://www.engin.umich.edu/-cre for the full solution. Iso1hennal: T = 321 K l, Q = 40 mol/L
RMAt\ = 1
.. .. ···-·- · ···
·-
.. .
···- -- .
.
.
6 5
2
O+"-~--.-~~-.-----..-~~-.-~--l
o
-. 2M1B
02 ·---···-·2M2B
0.4
X
06
-Me0H
0.8 ---·T.AME
PS-24 (b) See the additional homework problems in chapter 8 at http://www.engin.umich.edu/-cre for the equations to enter into Polymath. Then vary to see its effect.
8-77
PS-24 (e) See the additional homework problems in chapter 8 at http://www.engin.umich.edu/-cre far full solution Non-Isothennal: To= 353 K, Tw = 298 K RMIL\ = 1.0,
-
$o
Q = 200 LIÍnin
4
!. 3 u
2
o
02
-2MIB
0.6
0.4 X
----2M2B
-MeOH
0.8
··-TAME
PS-24 ( d) No solution will be given PS-25 Mole balance: dFA -=r dW A RateLaws:
dFB dW
-=r
B
rA = --r2B + 'iA + 'iA re= -r3A
-r¡A
= klCA
-r28 = k8C8 ·-fjA
= k3Cc
Stoichiometry:
e =e A
C B
FA'I'o 1
F¡T
=C
Fn'I'o
T
FT T
Energy balance: dT = Ua(I;,-T)+(-,¡A)(-AfIR1A)+(-rR2n)+(-1jA)(-AfIR3A)
dW
FACpA + F8CP8 + FcCpc 8-78
16(500-T) + (-r¡A)1800 + (-rR2s)1800 -·dT- = -------=""-------'=----~--
+ (-1:iA)llOO
lOO(FA + FB + Fe)
dW
k, =0.5exp(2(1-320/T)] k =~
K
2
e
k3 = 0.005exp[ 4.6(1-460/T)]
s; =10exp(4.8(430/T-1.5)] See Polymath program PS .. 25.pol.
¡r··-·-·r¡···········Variábt~ }r.Initial · · ·· . . . . . ·;~.u~·,.,-¡~~al-~~i~~ 'T'' ' Maximal V~-,~~ rfi~~,;~,u;IJ Calculated values of DEQ variables
¡O 'fi_·
¡1 ¡W
Ei
Ífb . ·-
!
1
/3 F~ . · · ¡1: •
~-6 · F - -· .
1
1
1
Ua lr; ..
To
1
1
. ,...
~!tri ¡1¡r,<3 · ·
j
~
Js f 011~i~ -- Fisoa. Fisoo. · rg . Joti~3~ · Fiioo~ · · f-lloo. l 10l~p~ . . ·1100. Tioo~f iil~pb . . ¡ 100. . . .. ·· fiao: !
· ·········T···
·-· ·· --10:2Cl44i34_----Jo.2~44ij;il ?3~'. . . . . . . 330. . ..... +4;6.. .3..º6.~ . ... --+~-!-~:3-~.6.-~- _1 ¡ 16. 16. ¡ 16. , 16. f soo. ., . . ' lsoo. J 500: f 5()(). l
f
[1'
IO ¡100. ¡100. . ·- . 1 o.926124i ·- Ti.368476.. ·-· ,,éi.9261241· 1 · f 0:6296429 · · · · ¡ 1. · · · · · · · · · · · · lo~s69462s· ·1
¡o . -
~¡-~
1
r····
.
r
[ioo. . . .. .
· r1soo.
risóo: . . . . -1~1100. 'Jioo.- .....
flioo. 'T100. · 1100: · T
.
..
Tiaa·:·- · · · · 1 .. ..
.
t6:f~t566
•
•
í ·l
1 J . •
¡¿:i~12401 ]O.s312401 . ·--l6~1~S66j !o.óoosiis ' lo.0008165 · To.oa.30853 · 1a:003óss3· i Jisfct J · Ti:·---······ -----Ti.- -·Ti.--·-····t r1G11t ··· J;i i2. · 12. · - ]
K íi71T~ . - ····r33a:··
r iia:
¡11i9¡1<2 -s.1~~- _ :·_: . lfª .8-~º~~·. ·: . . )i:0_~23.3.2 io.1367403
r¡2.~1¡~J~. . : · ¡[x.i.. . . cb
0.1367403
1
. J~:-~6~~~~9
133(). liici, ._J3.~~.~?~? . _ .... .... F.:.a.~233i · ~ ¡o.74756 10.74756
....
·:w:~~~~ª~~ ·J
F· . . .
i
¡o.7341242 ¡1.253213 0.7341242 ¡22Fi~· · · }o.s3i2401 1-o.5748799 · · Fo.362406··· F6..'s473402·1 !iiF3~ ·· ·-·- · Fo·.-6008165 F>:002138 T-o.ooo7s94·· · · . Fa:óo2i:i64. i f24f~¿ fo:066s'16s"' ' '. '. lo.ÓoÓ7594 ·ra.'ocái3'8" ' ·rr·c):ooii~i64-·i~ ···r·. . . ' "' . r ""' j25jr2b J-0.1367403 ¡-0.5770243 J-0.1367403 !-0.5488019 l 1F26Fb·-· · . -- ·¡-···ra:3944998 ---Fo.as22701 · . .·.rra·:3944998 . .... . · . .. Fci: aai4Gi,· · ·.~1 r· . . · r· . . . J~?J~~ . J=0.3953164 t?. :3.??3.1.~~ J?·~?1.?.?~1 '=~:~?~~6.~?J 1
1
"T'
.
..
8-79
1
Differential equations
d(fb)/d(w) = rb d(fa)/d(w) = ra d(fc)/d(w) = re d(T)/d(w) = (Ua * (Ta -T) + (-rla) * (-Dhrla) + (-r2b) * (Dhrla) * (-r3a) * (-Dhr3a)) /(fa* epa+ fb * cpb + fe * cpc) Explicit equations Ua = 16 Ta= 500 Dhrla = -1800 Dhr3a = -1100 epa= 100 cpb = 100 cpc = 100 kl = .5 * exp(2 * (1 - 320 / T)) k3 = .005 * exp(4.6 * (1 - (460 / T))) et= 2 ft = 2 To= 330 Kc = 10 * exp(4.8 * (430 / T- 1.5))
= kl / Kc ca = et * fa / ft *To/ T cb = et * fb / ft * To / T k2
rla = -kl * ca r3a = -k3 * ca re= -r3a r2b = -1<2 * cb rb = -rta + r2b ra = -r2b + rla + r3a
PS-25 (a)
As seen in the above table, the lowest concentration of o-xylene (A)= .568 mol/dm3
PS-25 (b) The maximum concentration of m-xylene (B) = L253 mol/dm3 PS-25 (e) The maximum concentration of o-xylene = 1 mol/dm3 PS-25 (d) The same equations are used except that FBo = 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
PS-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.
PS-25 (f) Individualized solution
PS-26 (a) AHB+C A-+D+E A+C-+F+G We want the exiting flow rates B, D and F Start with the mole balance in PFR:
dF dV dFE --rE dV _A=ra
dFB dV
-·=r
B
dFF =r dV F
dFc dV dFG --=r. dV
--=r.
e
dFD dV
--=r
G
Rate Laws:
rD
= r28
rE
= r28
rF
= ':ir
10925)(PA- P.~e P ) r¡, =p(l-<6)exp ( -0.08359--Tr28 = p(l-~)exp
25000)(PA) ( 13.2392- T
r31 =p(l-~)exp
( 0.2961-
11000) T (PAPc)
8-81
D
Fe Pe =-Pro F1 ~=~+~+~+~+~+~+~+~ F, = steamratio x .0034
Energy Balance: dT -( ,¡,W RIA+ rwí1ll R2A + '3111H R3A) -= dV FA *299+ F8 *283+ Fe *30+ FD *201 + FE *90+ FF *249+ F0 *68+ F, *40 KP1 =exp(b1 +
i
+b3In(T)+[(b4T+b5)T+b6]T)
See Polymath program P8-26.pol. ForTO
= 800K
I ila_rial>J1tlilJI !alueltimal val_uet::xi~~VIJl~eJ::alvalu_e I ,2Tt• · · Jo.CJ0344 · w~a249Í; · Jó.003.¡.¡ · · . ¡0.002496 1 Calculated values of DEQ variables
/1
1: . .
¡::-
.
1V1:1f~~rr · Ttt--
Ta· · · · ·
¡a-·Jig · · fo··· jg.lT
-t
¡iilci::~i~ 1 -······ -ir=1i:cit:!~r lt::itrs:1 Ta·· · · · · · · ---· T3.saáÉ~as···· · · risssE~os 1~:~~:::::
·- · ·· ·1a· · - · ···fissBÉ~-os .. )BoO: . ···¡76úi7 ···¡aoo~ -••
li.isÉ;os. ·1·1:as2E+as··
jia.H·I~ jl.18E+05 . 1.isE~OS r11· 1•H2~•· · · •·• · · · r1•:os~E-r•os·• · · •1•i:os2E-ros
JisssÉ=ai [765:úl
·1
·¡
''''Ji.i8É-r05 ·j ·1.os2E-+as·· · 1
!
lrnt~ : fitf} tf¡j}E~ !fitE~ii4~~: l~t}tf~o J f;Q4 .
4 • •. .
í14Tphi ·Ja'4·-· · · · · · ·····ra.4·· · · · · -· · 1 Ís[ KI . ¡ 0.0459123¡0.0196554-
fi6¡;~-
· · · · 114.s·
··--·.1;¡:s
¡ii. ii.. · ··ro:a49ss
1.
¡ill]ft: ]i9]Pa
·········· ·1
4.
· - ·Ta.4·· · · ·
l. Ü.0459123
··· ·fa.4
.0196554 j
· -- µ;¡:s- ·-·- ·- ¡i<1.s _J o.a498a · · ·¡o:049s11 · · ··· ·¡o.M9as··--·J . . . -1_.
O.
o.05332 ]cúíSil:i · · ; 0.0542282 · -To-:0542282. 1 IO.is48387·· · Tó.iio4652 "' lo'.is48387 . · · ·ro.iio4652 · l ¡ióTPb. · Jo · · · ·• · · · · ·· ro·· · · ·· · ·· · ····- - - - ---·· - 1a.Q391iss·-- ¡a:03911ss J,
li I1~-i;:i]:t:~]t~_~J~:füt~ : ·lm:rl 8-82
-------------·---·---------~-----~-
---
-----·-----------------·
- -----------------------------
--------------------------------·---·-·--·---------·----
r24F;
t~~ _ ~ __ _ _
···
¡i·:991E=a6
r~;¡;_t ¡27!rg
¡o
1 ,_ . ---··T··--·-·M·--·-M·H----"t''-··~-- ~
¡28jrls
,·--··1-----·--···--·
¡29 ¡ rb
jo
..
----····-·····~-¡------·· -····-··----~--~·-·-·····-··-·····-·¡·-------·--
J0.0002138
l2.481E-05
¡0.0002138
j2.481E-05
.. - .,
fjoF~--
. -. .--"·--~·-----
· Ts.iGE-07
--------·-""·-···--·--- .. ~----
· 10:"oo62ú8·· · ·-¡2:a66E=os-
J: : ~;~:~_.J:: ~:;:~: 1 12.991E-06
.
¡4.196E-06 -------------·--------
. 1s:16E-07
.. ·~·-··· .
¡0.0002138
j2.481E··05
¡0.0002138
!2.481E-os
-ro.Ooo213s
¡
--·1
---·-···--- ----·--·-···-···-·---·+····------·--·"
· ¡2:·a-66E=-as···¡
. .•. . t-0.·.0.°-.0.2 ~~? l. =.O.. º?O.! 16 !. . . • . . .1=?: ?~.~.~-°-~·· ·- ·
1~~1r.~·-· · · ·
¡4.151E-06
_T
1
1
.
I
.. 1=?: ~48~-~º? J
Differential equations
i
d(fa)/d(v) = ra d(fb)/d(v) = rb d(fe)/d(v) = re d(fd)/d(v)
=
rd
d(fe)/d(v) = re d(ff)/d(v) = rf d(fg)/d(v) = rg d(T)/d(v) = -(rls * Hla + r2b * H2a + r3t * H3a) / (fa * 68 + fi * 40)
* 299 + fb * 273 + fe * 30 + fd * 201 + fe * 90 + ff
Explicit equations
Hla = 118000 H2a = 105200 H3a = -53900 p
= 2137
phi= .4
= exp(-17.34 - 1.302e4 / T + 5.051 * ln(T) + ((-2.314e-10 * T + 1.302e-6) * T + -0.004931) * T) sr = 14.5 KI
fi
= sr *
.00344
ft=fu+fb+k+fd+~+ff+~+fi
* 2.4 Pb = fb / ft * 2.4 !ljll Pe = fe / ft * 2.4 r2b = p * (1 - phi) * exp(13.2392 . . 25000 / T) * Pa Pa = fa / ft
l'il rd ~F';'
=
li]re=
r2b r2b
r3t = p rf
=
* (1 - phi) * exp(.2961 ·-
11000 / T) * Pa
* Pe
r3t
iifii rg = r3t
rls = p * (1 - phi) * exp(-0.08539 - 10925 / T) * (Pa - Pb * Pe/ KI) 8-83
--~--~--~-~-~-~-
---
-
-- -- - - --------------------------
= rls
rb
re= rls - r3t
=
ra
-rls - r2b - r3t
= 0.0008974 Fbenzene = l.078E-05 Ftoluene = 3.588E-05 Fstyrene
= 19.2
Ss1BT
PS-26 (b) To= 930K
= 0.0019349 Fbenzene = 0.0002164 Fstyrene
Ftoluene
= 0.0002034
= 4.6
Ss1BT
PS-26 (e) To= llOOK = 0.0016543 Fbenzene = 0.0016067 Ftoluene = 0.0001275 Fstyrene
Ss1BT
= 0.95
PS-26 (d) Plotting the production of styrene as a fünction of To gives the following graph. The temperature that is ideal is 995K
f\
i•rb(S~)vs.
To1
º·ººn r------==-----.,/ ft
0.0022, c.0021 G.~02 0.0011
C.0018 1 C.0017
j
0.0011 T 0.001$ T
0.001, -· ----100
1¡ 100
.\ ......--~--J
'ººº
1100
1200
8-84
PS-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 .,.. sr
• J:b (~•) Q . 002
-------------
+
0.COtt
00011 • !
O 0017 O 0011-
0.0015 . . 10
.,j
20
10
(0
PS-26
When we add a heat exchanger to the reactor, the energy balance becomes: dT
=
Ua(Ta -!)-(r¡,fVIR1A +rwfVIR2A +r3rfVIR3J
FA *299+ F8 *283+ Fe *30+ F0 *201 + FE *90-t FF *249+ Fa *68+ F1 *40
dV
With Ta= 1000 K Ua 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. 4.l)e-3
300
3 2e-3
240
2Ae-3 1.6e-·3 8 Oe-4
-
180
fb fe fd fe t'f
120 60
4
V
6
8
10
o o
2
PS-26 (g) Individualized solution PS-26 (h) Individualized solution
PS-27 (a) 8-85
4
v
6
8
10
Adiabatic exothermic, adiabatic endothermic, exothennic 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 (e) 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 E8-10.l, 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-28 (a) 2A-···-·~B a) Design equation:
:!X= dV
-rd. F,.,0
Rate law:
Stoichiometry: Energy balance:
'Áx"
dT. _ .ua(T.4_-T)+( dV FA0cP"
Aff RX.)
e.. dV
.s(700····T)+(231+.012(T-298)X--rA) 5*.122
Plugging those into POL YMA TH gets the following program and the following graphs. Toe conversion achíeved is 0.36.
8-86
See Polymath program P8-28-a.pol. POLYMA TH Results Calculated values of the DEO variables initial
Variable
o
V T X
value
Ua
Ta dHr Cpa k
675
5 5
5 5
value
o
final value 10 704.76882 0.3580882
maximal value 10 715 . 55597 0.3580882 5 5
5 5
700 -235.524 0.1222 0.0734336
700 -236.01067 0.1222 0.0734336
700 -235.524 O . 1222 0.3658299
700 -235 . 88123 0.1222 0.243008
-0.0734336
0.6419118 -0.3335975
-0.0734336
O. 6419118 -O .1001316
1 1
Cao Ca ra
o
675
o
Fao
minimal
1
1 1
1
ODE Report (RKF45) Differential equations as entered by the user [ lJ d(T)/d(V) = (Ua*(Ta-T)+(·ra*(-dHr)))/(Fao*Cpa) [ 2 ] d(X)/d(V) = -ra/F ao Explicit equations as entered by the user [ll Fao=5 [2] Ua = 5 [3J Ta= 700 [4] dHr = -231-0.012*(T-298) ¡ 5 J Cpa = .1222 (6] k= 1.48e11*exp(-19124/T) [7J Cao = 1 [ 8 J Ca = Cao*(1-X) [ 9 J ra = -k*Ca"2 720.-------------------,
0.40~--------------~
700 690 680
2
4
V 6
8
10
2
PS-28 (b)
8-87
4
V 6
8
10
Using the same POL YMA TH program we were able to change the entering temperature and come up with this graph.
í1
Conversion vs temperature
1.2 ,------------------.
)(
__ }
1
50.a
f0.6
!o.4
80.2
O+----.----,.-----,...---.,.....--~
o
200
400 600 Temperature, K
800
1000
PS-28 (e) Again using the same POL YMATH program, we can vary the ambient temperarure until the reaction runs away. As the following summary table will show the maximwn temperature is 708 K. POLYMA TH Results Calculated values of the DEO variables Variable V
initial value
o
T X
675
Fao Ua Ta dHr Cpa
5
o
k
Cao Ca ra
minimal value
o
675
o
5 5
5 708.2 -235.524 0.1222 0 . 0734336
708.2 --236 . 5196 0.1222 0.0734336
--0. 0734336
O. 5137156 -1.078707
1 1
maximal value 10 758.02032 0.4862844 5 5 708.2 -235.524 0.1222 1.6320169 1 1
1
-0.0734336
final value 10 712.2666 0.4862844 5 5 708.2 -235 . 9712 0.1222 0 . 3233503 1
O. 5137156 -0.0853333
PS-28 (d) When the reaction becomes adiabatic the energy balance will then become:
(-rA X"-MI ) T-To +-··---····R .
cpA
However, the heat of reaction is a function of temperature. This is a circular reference, so we need to find Tas a function of just X. 1~CpA -231X -3.576X
T--------- . ·-----. · -
C¡,1. ·-0.012X Plugging that into POLYMATH gets the following program and graphs.
8-88
See Polymath program P8-28-d.pol. POL YMA TH Results Calculated values of the DEO variables initial value
Variable V
o o
X
k
ra
o o
5 5 708.2 675 0.1222 618.33416 1 0.9695163 -235 . 524 0.0054738 -0.0734336
5 5 708.2 675 0.1222 675 1 1 ·-235.524 0.0734336 -0.0734336
Fao Ua Ta To Cpa T Cao Ca dHr
minimal value
maximal value 10 0.0304837 5 5 708.2 675 0.1222 675 1 1 -234 . 84401 0.0734336 -0.0051452
final value 10 0.0304837 5 5 708.2 675 0.1222 618.33416 1 0.9695163 -234.84401 0.0054738 -0.0051452
ODE Report (RKF45) Differential equations as enterad by the user [ 1. J d(X)/d(V) = -ra/Fao ' Explicit equations as entered by the user [ll Fao = 5 [2] Ua = 5 [31 Ta=708.2 [4] To=675 [ 5 J Cpa = .1222 [ 6] T = (To*Cpa-231 *X·3.576*X)/(Cpa-0.012*X) [ 7] Cao = 1 [8] Ca= Cao*(1·X) [ 9 J dHr = -231-0 012*(T-298) [ 10 J k = 1.48e11*exp(-19124/T) [ 11] ra = -k*Ca"2 0 . 040 ,-----,-------------, 0 . .(132
666
0.024
652
0 . 016
638
11
L2_J
624
0.008
2
4
V
6
8
10
610
o
2
4
V 6
8
PS-28 (e) When it becomes reversible with inerts tbe two equations that change are the rate law and the energy balance.
-,. =f!-<;; J
8-89
10
We can use those equations in POL YMATH and the following graph is made:
Conversion vs. temperature, reversible 1 ,-~~~~~~~~~~~~--. ~ 0.8 ! 0.6 ~ 0.4
__
8
j
02
O+-~~~~~~..-~~~~~---f
o
500 Temperature
1000
There is no maximum. because the reaction is a nmaway at a certain temperature and the conversion goes to close to one at that point.
PS-28 (f) No
solution will be given
PS-29 A+B ~ 2C
Toe elemcnmy, reversible, gas phase teaelion
= fBo = 20 moVs = 1200 mol/min. Po = 580.5 kmol = 5.74 atm.
Feed: FAO
To= 77°C = 3SO°K YAo=Yso=O.S CAo = y AO Po =
RTo
Rate
= 0.1 gmo1Jl
(0.5) (S.74 aun.)
io.082 1 aan }(3SO°K) \ gm:>lºK
conswu: k = k exp(¡({ -{)] 1
= 0.035 exp [7~3~ (2j3 - +)]
i -+}]
: 0.035 exp [8419.5 (2 3
(1)
8-90
Equilibrium constant: MIR {TR)
= 2Hc - HA • Hs = 2(45,000)
+ 40,000 + 30.000
= -20,000 J/mole
= 2'1,c - e;,. . - e;,. = 2(20) - 25 - Is = o
ac;,
So AHR (O = constant = -20,000 J/mole
(2)
exp[~'{i;-t}] Kc = 25,ooo exp [-i~:i~ {2Js -})] Kc = 25,000 exp (-2405.6 (~ -})] Kc=Kc1
(3)
Kc can be calc:ulatcd by equation (1) and (3) ifT is given.
k and
Kc=__ff_ CACB
CA=CAO(l-X)f
wbcref=Th T
C¡¡ = CAO(l - X) f Cc=2CAOXf
Substitute CA, Ce, ami Ce into Kc
Kc =
4C¡,q xi f 2
Cio (1 • X)2 f 2
Calculate
,¡i;
=
4Xl (1 • X)2
(4) (5)
as a function oftemperature from equation (3)~ substitute in equation (5)
to get Xe,g. as a function ofT. Energy balance for adiabatic conditiorr
-X AliR 9A
= :E 8¡ e,{T • To)
= 9B =
1 •
(6)
0c = O • To = 3SOK
Substiture : + 20,000X = (25 + 15) (T - 350) or X
= T = 500X + 350
= 0.002 (T - 350)
(7)
Equations (3) = (4):
{..J._-
4X2 = 25000 exp [-2405 6 1 Jíl (1 - X)2 • 298 350+ soox ~
8-91
Xeq = 0.8667 0.85Xeq :: 0.7366
= 0.7
(a) Plug flow reactor design cquation:
law: -rA = k[CACe
Ratc
·rA
(8)
-~l
r. (2X}1l = k c2AO f'.\(1 • X)2 - Kc ]
(9)
(10)
To evaluare the integral, we necd to evaluare f(X) as a funcrion ofX. This is done in the table below:
I!K
Kc
(cqn. 7)
o
350
0.1
k
f!Xl
(eqn. 3)
(eqn. 1)
(eqn. 9)
30.96
400
7534.8 3191.1
0.2 0.3
450
1635.9
500
958.5
6491.5 4.216 X 1()4
0.0323 2.58 X 10-3 3.98 X lQ-4
0.4
550
0.5 0.6
600 650
618.9 429.9
1.949 x IOS 6.978 x lOS
0.7
700
315.8 192.8
2.054 X 1(.)6 1.155 X 1Q7
X
626 .. 1
1.08
X
10-s 10-s 10-s 10-s
4.98
X
lQ-6
9.89 3.53
X X
1.70 X
The integral can be evaluatcd using Simpson's rule for the first six segments and
trapeziodal rule for the 1ast segment.
I = ['
f(X) dX 5
'11- [0.0323 + 4(2.58 X 10-3) + 2(3.98 X 104) + 4(9.89 X JQ·S)
+ 2(3.53 X 10-5) + 4(1.70 X 1Q·S) + 1.08 X 1Q·5] + I • 1.5 X ¡Q·l
VpfR
= ( 1.2
x 10S)( 1.5 x 10-3) = 180 L 8-92
Of [t.08 X tQ·S + 4.98 X 10-6]
· Exothermic, adiabatic
T X
distance down the reactor
(e) CSTR. design equation: V=
V_ FAO XA -rA
FAoX
k cio f [(1-Xf-4/t] 1
X
kf2 [(l
(1 -
· =
·Xf-4:;J
xf. !X: Kc
ISOOL x (O.l)2(mo!.f l
1200:::i::
=
- O 01"5 - . -
so...X... f2k
l-2X + x2. 4X2 = fillX.
Kc
(i· ic:)x2-{2+ Let b1 ~
f2k
f~ºk)x+ 1 =O
= 1 - -.!... and bi = 2 + ...fill.
X=
Kc
f2k
(11)
b2±.J~ - 4b1
2b2 Nonadiabaric energy balance:
-10~x2
m2 K 1200 mi. min
x
(12)
lkW lhr 3.6 x 106 J lOOOW x fJOmin x lKW.hr (T- 290) + 20000X
-T + 290 + 20.000X
= 40(T-
350)
= 40T - 14,000
T = 348.5 + 487.8 X
(13)
8-93
So thc proccdure to calculare X is as follows: 1. Choose incremcnts in T and calculate X as a function of T from equation ( 13). This is the valuc givcn by energy balance.
2. Choose increments in T and calculate b¡ and b2 from equation (11). 1 or X < O. This is the value of X gi.vcn by material balance. 4. Plot X vs. T given by equation (13) and (11) on the samc grapb. The intcrSeClion'g_ives the conversion in the reactor. A typical graph looks like the following: -, ,
X
Opetating paint
The aaual calculation gives: X • O which is the conversion in CSTR.
d) The same equations can be used except that ~ = 20000 and T0 = SSOK. The following graph shows the equilibrium. conversion for this case.
Conversion vs temperature
:o.a· . :!0.6 ! • ~0.4
go.2 o "'-~~~~~~~..p:;;;.....~~~~~~-;..~~~~ o
50
100
150
Temperature
200
250 ·
The following POL YMATH program gives the PFR volume necessary to get a
conversion of .65.
See Polymath program PS--29.pol. POL YMA TH Results Calculated values of the DEO variables variable V X
Fao T
initial value
o o
1200 550
minimal value
o o
1200 225.22683
maximal value l. OE+08 0.6495463 1200 550
8-94
final value 1 . 0E+08 0.6495463 1200 225.22683
O . 064 550 1 . 949E+05 1.01E+06 1 0.064 0.064
Cao To k
Kc f
Ca Cb Ce ra
o
-798.17344
O . 064 550 5.047E-05 1841. 4832 1 0.0547713 0.0547713
o
-798.17344
0 . 064 550 l.949E+05 1.01E+06 2 . 4419826 O . 064 O . 064 0.2030311 -1.503E-07
O . 064 550 5.047E-05 1841.4832 2.4419826 0.0547713 0.0547713 O. 2030311 -l.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 [ll Fao = 1200 [ 2 l T = -500*X+550 [ 3 J Cao = .064 (41 To= 550 [ 5 l k = . 035*exp(8419 . 5*(1/273-1/T)) (6] Kc = 25000*exp(2405.6*(1/298-1/T)) (7) f = To/T [ 8 J Ca = Cao*(1-X)*f l 9 l Cb = Cao*(1-X)*f [ 1o J Ce = 2*Cao*X*f r 11 J ra = -k*(Ca*Cb-Cc"2/Kc)
Toe CSTR conversion can be found similarly to the equilibrium conversion. Toe following graphwas made to find the conversion. Conversion vs temperature
i~::rr--------/1
F:_ 230
280
2~ 330
380
Temperature
The graph shows a conversion of .39 ata temperature of 351.8K e) lf the reaction is left in a large enough volume it will "runaway'. lf any of the quantities get bígger, then it will run away even faster. f) The ambient temperature around the CSTR has little effect on the conversion in theCSTR.
PS-30 (a)
P8-6 is adiabatic so the radial reactor has no effect on it, PS-30 (b)
8-95
dX
-r
r:
-=__:i.
dW
-rA -kCA
1-·X T, CA =C.·---·-ºv ".40 l +X T.,
dy -a ---=>!!'--. dW 2y dT
dW
_ U(r);(T, -T)+ (-rA)(-iiliR)
-
.........
·-~--....
·--
·-··
Since ~ and p are unknown, we will assume that they are both equal to one. We will also assume that r varíes as W5 varies. so:
U(r)(~~ -T)+(-,:~)(-L\HR) -dT _- -.....---~-:...-..:.:~-._..:.~
F"0cp,\
dw
T) U= U(ro) ( w..s ;. )··5 ( To
5
See Polymath program PS-JO-b.pol. POLYMATH Results Calculated values of the DEO variables Variable
w
y X T Fao alpha Ta dHrl Wo To E R
Cao Xl yl
u k
initial value
o
1
o
minimal value
o
0.8062258
o
450 5 O,, 007 300 -·2. OE+04 O . 01 450 3.14E+04 8 . 314 0.2498645
450 5 0.007 300 -2.0E+04 0.01 450 3.l4E+04 8.314 0.2498645
1000 5 . 0E+04 l
806 . 22577 0.9244844 1
o
o
maximal value 50 1 0.7845538 756.08452 5 0.007 300 -2.0E+04 O . Ol 450 3.14E+04 8.314 0.2498645 784.55376 1000 5 . 0E+04 29.893648
8·96
final value 50 0.8062258 O. 7845538 756.08452 5 0.007 300 -2.0E+04 0.01 450 3.l4E+04 8.314 0.2498645 784.55376 806 . 22577 0.9244844 29.893648
Ca Ce
Cb Kc ra
0.2498645
0 . 0320395
0.2498645 1635 . 8981 -0.0624323
O . 0320395 187. 88715 -o . 1234947
o
0.2498645
o
0.2333455 0 . 2498645 1635.8981 -0.0220235
0.0320395
0.2333455 0.0320395 187.88715 -0.0220235
ODE Report(RKF45) Differential equations as entered by the user [ l J d(y)/d(W) = -alpha/(2*y) [ 2 J d(X}/d(W) = -ra/Fao [3) d(T)/d(W) = (U*(Ta-T)+(-ra)*(-dHr1))/(Fao*40) Explicit equations as entered by the user [ll Fao = 5 r 2 J alpha = .007 [3] Ta= 300 [ 4 J dHr1 = -20000 [5J Wo=.01 [6] To=450 [7J E= 31400 [SJ R = 8.314 [ 9 J Cao = 9 . 22/(.082*450) [ 10 J X1 = 1000*X [ 11) y1 = 1000*y ¡ 12 J U = 5*(WoA.5/(WA.5+.000000001 ))A.5*(T/To)A.5*yA . 5*(1 +X)A.5 [ 13 J k = exp((E/R)*(1/450-1/T)) [ 14] Ca= Cao*(1-X)*To/T [15] Ce= 2*Cao*X*To!T [16] Cb = Ca [ 17 J Kc = 25000*exp((dHr1/R)*(1/298-1/T)) r 18 J ra = -k*(Ca*Cb-CcA2/Kc)
o.so.---------------~
800
0 . 64
720
0.48
640
0.32
560 480
20
W 30
40
50
4000
PS-30 (e)
8-97
10
20
W 30
"º
50
W = p V = p,cr2h
= dW =
2:n:prhdr
Material balance:
Taking the limitas tJ.r ~ O == ( rA) 21tprh
~
= ~A
FA = FAo (1-XA) so:
= -FAo ~"
dXA -2ltrhp (rA) dr = FAO
(1)
or; dXA = • ..!A.. • dW FAO
(2)
Assuming pressure drop is negligiblc. The rate cquation is:
-rA = k (~)1/2 [P0i -{ Ps~ )2] Pso, KpPs0i The stoichiomeny :
S(h + l.(h- ~ SO) 2
A+l.B=C 2
Ó = -0.5, YAO
Ps0¡ = Psau
= 0.11 :. E = YAO~ 1 - XA
. Pso, 1 + e XA '
Toen: ·rA ... k(t)(l
= -0.055 ; 0c = 0 ;
= Ps0i.,o
XA
. P0i 1 + E XA '
-XA)l/2[Pseu(º·91-0.5XA)-
XA
=
0n =
w=
0.91
(0.91-0.SXA)
Pse>,,o :..----...;..;. 1 + e XA
Xi l
l-0.055XA (1-XJ Ki]
(3)
Equation (3) is true for XA < 0.05 for XA ~ O.OS - fA
- rA
= k(t) (4.35) 't0.22 ( 0,91-Q,025 } • 0.0(}25 l 1-0.05:,(0.05) ( l-0.05)2KiJ = k{t) [0.848 -
Q.0~~0~l
(4)
8-98
1-
---------------------------
-
-··----------------------------------------·--------------····-------------------·-----
--------------·---------------·---···------------
-···-·---------------·-·--·-·
Enerzv -J balance:
F AO (l:0¡ Cp¡ + X .6.Cp) nr - FAo (1:0¡ Cp¡ + .6.Cp X) Tlr+Ar
U (r) (T-TA) = O
+ rA ~W(AHR)-(2)(.M)
or: F AO (!9¡ Cp¡ + X ~Cp) Tlr - FAO (!9¡ Cp; + ACp X) Tlr-1-Ar + rA 2r.:rhp (.1HR} !J.r-(2}(2xr Ar) U(r} (T-TA) = 0
.
Taking the Iimít as Ar -+ O: FAo (.I:9¡
e, + X .ó.C;,) ~
= -U(r) 41tr (T-TA) + (-.rA){-MiR) 2mhp
Rearranging:
dl:. _ -U(r} 4m(T-TA)
-U(r)2h.(T- TA)+ (-rA){-MlR) _ FAo (!0¡ Cp¡ + X ACp}
.4I.. =--P dW
•
(5)
FAo (!0¡ Cp, + X ~Cp)
dr or:
+(-rA){-aHa} 2milp
(6)
Assumc that: U(r) = U(r0){.32..\ll2 ; u= 'U°rrº (1 +eX)J,. ~}
~o
'\fo)1
Thcrcfore: U(r) = U(r0} {!f-)1
12 (1 - 0.055XA)112
From examplc 8-10, we have:
Kp k
= exp [42R~ 1
- 11.24] (Kp in aun·112 • T in °R)
('!)
= exp [-176.¡.008 - 110.1 ln T + 912.84]
(8)
aHRrn
= -42,411 - c1.s63>CT- 1260) +(1.36 x 10·3)(T2 -
-(2.459 ·
X
10·1)(I3 -
12603) wherc AHR in lpmole Bru
3
12602) (9)
8-99
--
------------------------~
l:9¡ Cp¡ = 57.23 + 0.014T - 1.788 x 10-6 T2
(10)
Sincc cquations ( 1) to ( 1 O) must be sol ved together as two pairs of coupled differcntial equanons, they must be solved on a computer, employing numerical methods such as
Runge-Kuna, Toe results follow:
i
T 12IIO ICI» T
IOO
r t
~¡_/
(-a.) 600
....
t
O
4
------
1
~
_; '
1
12
16
2D
24
'9
:SZ
..;
40
44 49
U
S6
l. '. .
~:
60
y (\t,·
16IIO
T
14'111 12IIID
-
lllllD
,·u r
1111)
. .:·'
....... ...
-- -~~·--------------~-¡L,O
o.e.
••
....
4CD
:ID
o
¡,;;;_,~--------~--------~-----------zs n u ,. ....'°o o ' • ,: ,, :m. 2,
... --
., (,~,/
'O
...
41
i
T
/
1000
/
I
'°°
16
to
--------f
-·- ---
(.l.:
400
:a,
o
I I
: t)
4
1
12
,,
:O
%4
:S
1%
16
.,, c..-..¡
.0
oM
41
'2
]71
PS-30 (d) 8-100
'6
60
)1
i
dX _ -21rrhp(-r.4.)
............_,
dr =t'
A
-
-
=
k(c e - _c~.K'? J-
--··"-·"'···-·
r:o ..
A
8
e
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 O. 5
o
350 373 -2.0E+04 1200 25 15 0.5 0.5 350 7.0E+04 8.314 O. 1 33.3 30.955933 0.1 -3.0955933
minimal 0.5
o
value
350 373 -2.0E+04 1200 25 15 O.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 O. 1 33.3 3.398E+04 0.1 2.09E-06
ODE Report (RKF4S) Differential equations as entered by the user [ 1 J 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 [ll Ta== 373 [2] dHrx == -20000 [.3 J Fao = 1200 [4J Cpa == 25 [5] Cpb = 15
8-101
final value 1000 l
373 373 -2.0E+04 1200 25 15 O.5 0.5 350 7 . OE+04 8.314 0.1 0.769425 136.44189 -3.125E-17 4.264E-15
h = .5 [7 J ro= .5 (8) To= 350 (9) E= 70000 (10] R=8.314 (11] Cao = .1 ¡ 121 U= 33.3*(ro/r)''.5*(Tffo)A.5 ¡ 13 J k = .035*exp((E/R)*(1/273-1!f)) [14] Ca= Cao*(1-X)/(1+X)*(Toff) [15] ra=-k*Ca (6]
PS-31 (a)
Mole balances: V= F.4.o -~ -rA
V=Fª=Fc rs
re
rate laws: ~rA
rs
= k,,CA
= lciCA -ki.Cs
re =k.iCs
Stoichiometty: C.=F;I_ t
Vo
To
F8 =lOFc .5
= .05-~
0.5 f';. = .025
llFc = .025 Fe =.023
F8 =.023
From this we can use two of the mole balances to solve for T
8-102
---
--~--------------------
~~-
----------------~
-- ----
-
-- --~----
------
~
------
~o-~
ki~
=-
1
E;
k.iFs
0.1 Are-E' KI - Aie-E' KI T=269ºF PS-31 (b) Knowing the temperature we can then solve for the Volume:
PS-31 (e) We then need the energy balance:
~Ocpa(T-Ta)+ v[(.6HR1)(,¡A)+MfRi1is)] = Ü
UA(i: -T)-
Solve for A ami we get: A= 399 ft2
PS-31 (d) In order to get multiple steady-states, the kappa, tau and feed temperarure had to be changed. K 0.1, t = 0.0005 and T0 would be changed around.
=
This first graph is G(T), R(T) vs T at TO = 2000 ºF 20000 - ··-··"
. . . ,_.
,
___,
15000 10000 · 5000
-5000 -·10000 ....
500 ··--·-··-
1000
15' O
------
As can be seen there are three steady-states.
8-103
Ts vs To 1200 ·---~~~~~~~~--,
1000 800
· •
~600
,
400 ~~· 200 .
o
• • •• • • • • •
+--~~~~--~~~__,
o
10000
5000
To
PS-32 (a) Energy balance
a.; Ua/ p,,(T.-T)+(-rAX-AHR(~)] dW
~o(LB;C,.+XAc,)
4T _ Ua/ P.(i;, -T)+(-rAX-MIR(TR)]
J:..0C,..
dW
dX
Mole balance: dW Pressure drop: dy dW
-r = -4.
~o
=~Iic1+x) 2y T
Rate law: Stoichiometry: CA = CA0(1-
XXI_}
l+X
Evaluating the parameters:
7;,
k = ex]E(_!__
1lR
450
.!.)l = ex]3776.7J' .i, _ .!.)l T J 1l \450 T J
Plugging these equations into POL YMA1H we get the following plots. See Polymath program P8-32--a.pol. OL YMATH Results Calculated values of the DEO variables 8-104
iariable
w
initial
o
value
minimal
o
T
u a To Ta dHrx Fao Cpa k
Cao Ca ra
5 0.007 450 300 -2.0E+04 5 40 0 . 0232094 0.25 0.0825702 -0.25
5 0.007 450 300 -2.0E+04 5 40 1 0.25 0.25 -0.25
value
50 1 450 0 . 175758 5 0.007 450 300 -2.0E+04 5 40 1 0.25 0.25 -0.0019164
o
o
X
maximal
0.6823861 310.69106
1 450
y
value
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) Dífferential equations as entered by the user [ 1 J d(y)/d(W) -a/(2*y)*(To!T)*(1 +X) [ 2 J d(T)/d(W) = (U*(Ta-T)-ra*dHrx)/Fao/Cpa [ 3 J d(X)/d(W) = -ra/Fao
=
Explicit equations as enterad by the user [ll U=5 [2 J a= .007 [3 J To= 450 [4) Ta= 300 [ 5 J dHrx = -20000 [6J Fao = 5 [7J Cpa = 40 [SJ k = exp(3776.76*(1/450-1!T)) [9J Cao = .25 [ 1 o J Ca= Cao*((1-X)/(1 +X))*(T/450)*y [lll ra e -k'Ca 1.0
---=---------------,
500
0. 8
460
0.6
420
OA
380
0. 2
____
'
20
W 30
340
.
40
50
300
o
10
20
W 30
-40
50
PS-32 (b) From the Polymath summary table, it is apparent that the maximum value for -rA occurs at the beginning of the reactor.
PS-32 (e) 8-105
r
The maximum value for the temperature also occurs at the beginning of the reactor.
PS-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.
PS-33 Mole balances V= vo(CAo-CA) -rA
Rate laws: -rA =k1CA +k2Cs -rs =k1CA +k2Cs Energy balance: -FAo[<;,A(T-TAo) +Cpa(T-Teo)]+
ru =k2Cs V(rA1X.MiR1(TR)+ACp1(T-TR))+
V(rA2XAHR2(TR)+ACp2(T-TR))
Evaluating the parameters: T=400K k1
=loooexp(-2~)=6.73
ACp1 =50-20-30=0
FAo 60 vo =-=-=6000CAo .01
Simplifying:
k2 =200oexp(-30T00)=1.11
ACP2 =40-30-20=-10
dm3
C80 =CAo
min
CA= 't ..rA +CAo
Ce= 't •re +Ceo
Co = -r .. r0
Cu = t"' ru
171000 V=--------20190 • eA.~ 6660 • e~
We can plug those into POLYMATII and fínd the exit concentrations of U and D and find the volume of the CSTR. See Polymath program P8 33.pol. POLYMA TH Results NLES Solution
8-106
Variable Ca Cb Cd Cu V
Cao Cbo vo kl k2 tau rd ru ra rb
Value
0.0016782 0.0016782 0 . 0071436 0.0011782 3794.94 O . 01 0.01 6000 6.73 1.11 0.63249 O. 0112944 0.0018628 -O . 0131572 -O . 0131572
f {x)
-L 472E-16 -1.472E-16 -1.154E-16 -2 . 017E-17 L 018E-09
Ini Guess
0 . 0017 0 . 0017 0.0072 0 . 0012 3794
NLES Report (safenewt) Nonlinear equations [ 1 J f(Ca) = tau*ra+Cao-Ca = O [ 2 J f(Cb) = tau*rb+Cbo-Cb = O [ 3 l f(Cd) = Cd-tau*rd = O [ 4 J f(Cu) = Cu-tau'ru = O [ 5 J f(V) = 171000/(20190*Ca+6660*Cb )-V = O
Explicit equations [ll
Cao = .01
[2
l Cbo = . 01
[3 J vo = 6000
k1 = 6..73 [5] k2=1.11 [ 6 J tau = Vivo [7J rd=k1*Ca [ 8 J ru = k2*Cb [ 9 J ra = -k1 *Ca-k2*Cb [4J
[10]
rb e ra
PS-33 (a) Cu= .0012 C0= .0072
PS-33 (b)
V=3794dm3 PS-33 (e) Individualized solution
8-107
Solutions for Chapter 9 - Unsteady State Nonisothermal Reactor Design P9-1 lndividualized solution P9-2 (a) Example 9-1 The new TO of20 ºF (497 ºR) gives a new LiHRn 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 explosíon occurred without cooling failnre. Isothermal operation throughout (T = 175ºC) Máximum cooling rate:
Qr = UAl448---298] = 21300 BTU/min
= 142 * 150
Maximum Qs at t == O (máximum concentration and reaction rate) · MI Q g =k!!Ao!!_ao.*V(··· V2
Ri
)
= 0.0001167[2:~.~}~]2.34 * 105 = 15914.2 For all t:
. 5119 . BTU/min
Qi < Q,
No explosion
'í
9-1
1 ~------
----------------------~-----------·----
To show Lilac no explosion occurs with cooling shut down for 10 mi. after 12 hrs. Isothermal
t=[
operation for 12 hrs.
(at T
= 175ºC)
;AJ( eB ~. 2} :SBú··~~;) 0
Ja -- 2x · = 1 276 88(1-·-·x)
3. 64 -· 2.:c = 1276 X=
* J.64(1- X)
0.38
Qg ar t =- 12 hrs .
---/º (8.
N (1- x)N
Qt ::: k ..
AO -··-·····
= 0..000 116
-2r)
!!·----, -( --t:,}f fa)
,¡_~:.º~~Q_:=·=1~At:=-~~
:3.~2}34 * l 0
6
= 77844 BTU/min. Adiabatic operation for 10 mm,
VA= O
After l O minutes x
= 0..385,
=
T
=
184ºC
Qg 10000 B TUrmin, When we restare the coclínz flow rate
=
Q,l,,,., 21,300 BTÜ/rnin. Ternperature will drop to 175ºC No.e~Iosion
P9-2 (e)
Example 9-3
Decreasing the electric heating rate (Tedot in polymath program from example 9-3) by a factor of 10 gives a conversion of9.72 % 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 ofthe coolant rate to 1000 kg/s gives a Tmax 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 codeas seen in Example 9-4, we were able to change rhe various
parameters. The two grapas we have show T0:::.: 70 and 120ºC, T¡ = 160 and 40ºC, ami Cai
= . 1 and . 2.
Each set of parameters hada Temperature time
trajectory anda temperature-concentration phase plane, These are the four graphs,
9-2
~:,!CQT !9'5.000! KEY:
-T
Startup of a CSTR To=70,
·~i
·~-t
¡::Js.ooo+
n..coo _.,
_
.........-------·+-~~~-.;~·~------+t~~~----lt 2. 'IOQ
0.000
TT \
Startup
of
a, 200
'I, 000,
a CSTQ To=70.,
Ca1:::. 1
..¡... T
KEY:
-Ca
C.090
Q.0<10
o.oca
!
l t
t -e-
~----+·-·-------·--,--··1~QtJO
l::l~.000
t1'.!i.COO
..
l!ffi,000
--; l~.tlC:
T 2'2tl, (.."\JU=·
. }
.r--
!1 <, ~····
toc.cocl
-
KEY:
-T
>~OMt
Startuc of a CSTR To=120~ ···.··.··-006~··-·+··-··.
f
J
,=.,·+-y. ·/ 60,COO
r
ze.eco
J-. . . . . c. ceo
~---+,-,·,.,,=· ,..=.·· -·-- _.,=...,-,+~"""'""'"~·-· 1100 1. ,a::i
o.
'""'*"---~,,., _.,._.,~-~--""""""' ;v~,ao
9-,3
::i.
wc
-. ooo
O~
;i,oo
CLt60
a CSTR
~....-
1
To=120
T1;4Q Ca1=.2
t· .L
T
P9-2 (f) Example 9-6
Using the code from 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 temperature. These are the graphs of those:
Conversion vs To 0.,9
e
.. ,. . .....,.._
_ -:·. --·-
______
~
- -·-
,
0.8 0.7
o 0.6
f
(l)
0.5
> 0. 4
e:
8
0.3
0.2 0.1
o
_._ . ,.... 66
68
.,.
. .,........-------,,-·······-·"'········
70
72
74
To
9-4
...............
76
"""""r
78
T
..
82
-··----·-··-
---
1 '
-.-
..
··· ----------------
..
- ------------·--·---------
X vs coolant flow
-, .
-·--·------------
·-----
0.9 0.8 · e 0.7
o 0.6 iii ....CI.) 0.5
~ 0.4 o
o
0.3 · 0.2 · 0.1
o
·--···.-------..
o
2000
4000
6000
10000
8000
Coolant ftow, moVh
T vs. coolant flow
----·-··
180
..
··--- .. ··
160 · u, 140 0> 120 ·
...
ZCl:J .... 8_
100 80 ·
E CI)
60 ..
¡,...
40· 20
o ..-
-
o
-2000
_
-r
- ·--------
4000
--
8000
6000
Coolant flow moVh . ~ . - .._ .,. .. .
---------·-----·---·- -
--~
,---
· ~·-·-
10000
-··-----· --.
P9-2 (g) Example 9-7 The temperature trajectory changes significantly . Instead of a maximum in temperature, there is now a mínimum. The concentration profiles also change. C8 no longer goes through a maximum and Ce
second reaction
P9-2 (h) Example
RE9-1
Using the code from Example RE9-1 • we can determine the value of k. for which the reactorwill fall to the Iower steady-state and when it becornes unstable. The following two graphs show those points when ~ .2 and 24 respectively. The third graph shows what happens when T0 = 65. It becomes unstable ata much lower temperature.
=.
9-.5
.
t 1 .....
~eo...!...
KEY:
~T
l
l¡ '
-~---~-..+----,,..
-;
... coo
l.óOO
:no.COJ-
i
I rrt €?qr al
Control
-~
t
P9-2 (i) Example RE9-2 Using the code frorn Example RE9-2 , we can change the valúes of k, and t1 and find valúes that produce the Iowest oscillatíons and the quickest return to steadystate and gettína ~ = 150 and t"1 .I. The following graph shows the result.
=
PI Con1~oller
kc=150
t
P9-2 (j)
No solution will be given
P9-3 9-6
tau=.I
Find time to explosión For the unsteady-szare, we assume the following: (i) The inlet is closed, but the ontíet is not closed (ii) Operation is adiabaríc, and PV terms are negligible so that H.,,,. U
Equarion 8-61 reduces to: (-6.liR) (-rAV} = NA('-p"
f
(-~HR) (kNA) ~ Nt\Cp" fil:. .
4I_
dt
dt
= AHR:i:n k(T) <;. .
k = .53 exp { 44499 {9jo •
+n / 60 min,
See Polymath program 1'9-3.pol POLYMA TH Results Calculated values of the DEO variables Variable t
initial value _º ____ 970 -336 O. 38 0.53
T
dH Cpa
k
minimal value
o
970 -336 0.38 0.53
maximal value 2.9 1.119E+23 -336 0.38 4.438E+l9
ODE Report (RKF45) Differential equations as entered by the user [ 1 l d(T)/d(t) = -dH*k/Cpa Explicit equations as entered by the user [ll dH =-336 [2J Cpa = .38 [ 3 J k = .53*exp(44498*(1/970-1/T))
P9-4 Mole balance:
FA0
······
F,., + Vr
A
= cl_.dt ~'Y_tl
F.ao - F.B + VrB ;; 4.!'!.ªdt f~0
•··
Fe + Vr,
= 4_l!s. dr
9-7
final -~---
value
l .119E+23 -336 0.38 4.438E+l9
so F80 and F C1J
= O.
=
FA O. There is no B or C entering the reactor The amount of B and C leaving is equal to the reaction rate so
There is no A leaving the reactor so
F8 == Vr8 and Fe= Vr Simplifying: 0•
F + 'Vr AO
= ·dNA ·· ·dt·-'-
A
~~!L-o dt
dNc
---
:=:
dt
o
Rate law:
' Stoichiornerry:
e ::;; !'V!.1. A
Simplifying:
Energy balance:
Evalúate the parameters:
ln(:-ji)= !(sfm· . 4~) E== 3467
{t1r:{4lm
= · 19 ex1 Mí( 400) = - 8kj I mol=
,k
1\)J llli(800)
= snt:
200):::, Af-I(T)
Assume a heat capacity for A of20 J/molK and assume there is no A already in the reactor. W e can plug these equations into POL YMA TH and get this answer See Polymath program P9-4 .pol POLYMATH Results No Title 08-Il-2005, Rev5.l.233 Calculated values of the DEO variables Variable t Na T
Fao
dHr
Cpa
initial value
o
1.0E-09 400 100 -8000 20
minimal value
o
l. OE-09 400 100 -8000 20
maximal value 100 326.10947 801. 376 100 -8000 20
final value 100 312 . 51175 800 100 -8000 20
9-8 1
.1
400 0.19 -1. 9E-10
To k
Vra
400 0.3202745 -l.9E-10
400 0.19 -101.76832
400 0.319988 -100
ODE Report (STIFF) Ditferential equations as enterad by the user [l J d(Na)/d(t) = Fao+Vra [ 2 J d(T)/d(t) = ((Vra*dHr)-Fao*Cpa*(T-To))/(Na*Cpa) Explicit equations as enterad by the user [ll Fao = 100 [2 l dHr = -8000 (3J Cpa = 20 [41 To= 400 [5J k = .19*exp(3467/8.314*(1/400-1/T)) (6) Vra = -k*Na
NA= 326.125 When the flow is tumed off, F ..,0 = O and we get this graph of T and N...._. 2000
1.0e-9
1600
8 . 0e-10
Q
1200
6.0e-10
800
4.0e-10
400
2.0e-10
o o
2
4
t
6
10
8
O.Oe+o O
2
4
t
8
6
10
P9-5 (a) Mole balance : dNa
~-
dt
= ra.V
dNb --- = r«. V+ Fbo dt
dNc
· --=-rri.V
dt
= k(T).Ca.Cb
Rate law :
-ra
Stoichiometry :
Ca= _1:!_a V
Nb Cb = --··· V
IF (t<50) THEN (V= V0 + u.t) EISE (V:::V0 + 50) dm3/min
9-9
Ne
Ce=·-···V
where Vo = 50 dm", u= l
Combining :
. Na Nb --ra :::: k(T).. ·---· k(T) :::: ko.
exp[..R!(.LTo _. ~J\]. T .
V
V
= 10000
where E
cal/mol
R = 1.987 cal/mol.K ,
ko = 0.01 dm3/mol.min To= 300 K Energy balance : dT .·
Q · ···· Ws · · · ·
d~- - ----
where
L Fio.Cpi.t'T ... Tio) + e Af!rx(T)).(··ra. V)
.. .
. L Ni. Cpi
.
.
-
Q = Ws = O
.6.Hrx(f) = 1.\Hrx(freí) + ACp(T-Tref) ACp = !:__.Cp--~. Cpb-« Cpa = 30 · ·· 15--15
a
a
.6Hrx(273)
= [(-41)-(-20}{
15)].103
=
=O
6000 cal/mol
.6.Hrx(f) = -6000 cal/mol
=
I.Fio.Cpi.(f~Tio) f1)o.(15).(T~323) INi.Cpi = NaCpa + Nb.Cpb + Nc.Cpc == 15Na + 15Nb + 30Nc See Polymath program P9-5-a.pol POLYMATHResults 08-11-2005,
Rev5 . l .233
Calculated values of the DEQ variables Variable --t Na
Nb
Ne
initial value
o
500
o o
T X
298
Fbo Nao Cbo
10 500 10 0.0089352 1 50
k
vo V ra
o
o
minimal value
o
0.1396707
o o
298
o o
500 10 0 . 0089352
o
50
-o . 3365379
maximal value 120 500 42 . 43357 499.86033 510.44411 O. 9997207 10 500 10 10.085112 1 100
o
final value 120 0.1396707 0.1397745 499.86033 510.44411 0.9997207
o
500 10 10. 085112
o
100 -r . 969E--05
ODE Report (RKF45)
9-10
L 1
Differential equations as entered by the user [ 1 l d(Na)/d(t) = ra*V
l [3J [4l [5J [2
d(Nb)/d(t) = ra*V+Fbo d(Nc)/d(t) = -ra*V d(T)/d(t) = ((6000*(-ra*V))-(Fbo*15*(T-323)))/(15*Na+ 15*Nb+30*Nc) d(X)/d(t) = -ra*V/Nao
Explicit equations as entered by the user [ 1J Fbo = if(t<50)then(1 O)else(O) [2J Nao= 500 [3J Cbo = 10 [ 4 l k = .01 *exp((10000/1.987)*(1/300-1/T)) ¡ 5 J vo = Fbo/Cbo [ 6 J V= if(t<50)then(50+(vo*t))else(100) [ 7 J ra = -k*Na*Nb/(V"2) 600~----
1.0
520
0.8
440
0.6
G
360
Q
0.2
280 24
48
t
72
24
120
96
48
t
72
96
120
P9-5 (b) This is the same as part (a) except the energy balance .
Energy balance : dT _ Q-- Ws -
-d¡·where Q = UA(Ta · · T)
.
-
¿Fio. (pi.(1'··· Tio) + (-·Afirx(T)). (=ra. V) ¿Ni.Cpi
UA = 100 , Ta= 323
See Polymath program P9-5-b poi
9-11
. ·.···
___ N____ N"-·--"N
1.0
500
440
0. 8
Q
0.6
GJ
360 320
OA
260 ·
36
72
t
108
144
160
200
o
72
36
t
144
108
180
P9-5 (e) This is the same as part (b) except the reaction is now reversible.
-ra = kl(T),Ca.(:b - k2(r).Cc
Rate law :
Combining :
=ra =
kl(T).(-~·.·1-}·k2(D.(
k2(T) = ko
:e)
exp[i{};-J )] where E= 16000,
R = 1.987.
ko = 10 , To = 300
See Polymath program P9--5-c poi 500--------------
0.9 · --·'60"""'
......... ----
440 0.5
380
OA
320
0.2
260
36
72
t
108
144
180
200
o
36
72
t
108
P9-6 (a) Mol balance : Rate law :
dNa -·- ·= raV dt · ra
= k.Ca
But Ca.V= Na
!_f!a =-kNa dt
9-12
144
180
Na
In-·-= Nao Na= Nao.e -kt
+k.t
where Nao = 0,5 x 50
= 25
Energy balance :
-.dT : ; :; Q-Ws. . dt
Í:.,Fio.Cpi(T-Tio)+ (--Mlrx)(·-ra.V) . ··--"···--···--" _ " ··-· -·· Í:Ni.Cpi
-o
constant T
Q=Ws=O
LFi.Cpi(T -Tio) = (-AHrx)(-ra.V) Fco.Cpc.(T - Tio) = (-óllrx)(k.Na)
= (.-AHrx)(k.Nao.ekt) (-Mfrx).(kNao.e
b)
Feo = ····-···-'--- · · · · · · · · · · · · · ····-Cpc.(T -Tio)
. . 25000_x 0.000l2x25 x exp(-0.()0012 x 2 x36~} 0.5. (l 00 - 80)
= 3.16 lb/s
P9-6 (b)
Av = 1 OOOBtu /lb Nao= 25 lb/mol
Vo :::. 300 ft3 of which 250
fr is solvent
dT dt
·-=0 Fs.
Av = = (-Lllirx)(-ra. V)
Fs
(-··Aflrx). (kNao.e->.t) exp(--0.00012 x 2 x 3600) = --· . _ --¡: - 25000 . . x 0.()0()12X - 25..x1000 . .
Fs = 0.0316 lb/s
P9-7 9-13
Batch problem.
dX
dX
eao--·=·-ra
Nao-r-- = --ra V dt
Mol balance :
dt
Rate law : · Ca:,;; Caot l-X) = 0 . 1(1-X)
Stoichíometry :
Cb = Cao(L25-X) = 0.1(1.25-X.) Ce = Cao(O+X)
kl(D =
ko.exp[-!{./~· _. ·})]
= O IX E= 100000 J/moL, R = 8.3 l4
where
J/moLK,
ka= 0.002 /s , To= 373 K
E;-;;.: 150000 J/mol., R = 8.314 Jzmol.K, ko = 0.00003 /s, To = 373 K Energy balance :
st
. dt. ~.
(··AHrx(n)(··ra. ¿Ni.Cpi
V)
where }:.Ni.Cpi =Nao(í:0íCpi + ACpX) L"':.Cp = -~-. Cpc ----~ . Cpb -· Cpa ;:;;: 40 - 2:5 · · · 25 = · ···1 O a a ,·, a: ("" . 7 171.. ,pi
--'
·
Cbo = Cao ······-· . e·,pa + ··--·. Cao
Cao
, b Cp ·
. +--Ceo . -- .. Cpc
= 25 + o 125101 25 + O AHrx(f)
10..(f--298)
Nao V=·············
Cao
.4! = [~-~-~~~~!-~. -~98)]( dt
::::: 56.25
= Aflrx(Tref) + ACp(T-Tref)
= . . 40000
subs
Cao
:ra·r~;)_
Nao(56.25 -·· · · 1 OX)
9-14
r --ra)
[40000 + lO.(T -· 298)).l·· . ........ . .. . = -~-·..Ql.. _._ dt (56.25 lQX) dT
cancelling
See Polymath program P9-7 .poI
POLYMATH Results Calculated values of the DEO variables initial value
Variable t
o
minimal value
o o
o
X T
373 0.002
kl Ca Cb
373 0.002 0.0749517 0.0999517 3.0E-05
O. 1
0.125 3.0E-05
k2
o
o
Ce ra
-2.236E-04
-·O .1344598
final value 10 0.2504829 562.91802 106 .13612 0.0749517 0.0999517 366.75083 0.0250483 l.3E-07
maximal value 10 0.2504829 562. 91803 106.13627 0.1 0.125 366.75159 0.0250483 8.644E-06
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(X)/d(t) = -ra/.1 [ 2 J d(T)/d(t) = ((40000+(1 O*(T-298)))*(-ra)*(1/.1 ))/(56.25-(10*X)) Explicit equations as entered by the user [ l J k1 = .002*exp((100000/8.314)*(1/373-1ff)) [2 l Ca= .1*(1-X) [ 3 J Cb = .1 *(1.25-X) [ 4 l k2 = .00003*exp((150000/8.314)*(1/373-1ff)) (5] Ce= .1*X [ 6 J ra = -((k1 *(Ca".5)*(Cb".5))-(k2*Cc)) 0.15~------------~ 540
0.12 --·- . ·-··-----
480
0.09
t------\
.. ---
. . _, _
---
- Ca 0.06
420 360
{_\:
· · ·-···· -·-··,· -"
2
Check answer :
- Ch
0.03
4
6
Ce
8
10
0.00
~-···
o
2
- kl
·..fca.m-
k2
0.02504 1..06 --·· · = --- .J0.07495 X .J0.09995 3.67
= 0.289
9-15
4
t
6
8
10
P9-8 (a) Use Polymath to salve the differential equations developed from the unsteady state heat and mass balances. See Polymath program P9-8-a.pol POLYMATH Results 08-11-2005. No Title
Rev5 .1.233
Calculated values of the DEO variables Variable t Ce es T
Iprirne Km
rnulrnax Yes mu Q
Cps Hrxn V rho rg Cpe rs
initial
o
value
0.1 300 278 0.0866522 5 o. 5 O. 8 0.0426159
o
74 -2.0E+04 25 1000 0.0042616 74 -0.005327
minimal
0.1 43.080524 278 6.499E-04 5 0.5 0.8 2.912E-04
rnaxirnal value 300 205.63558 300 333.55016 0.3593146 5 0.5 0.8 0.1725264
74 -2.0E+04 25 1000 0.0042616 74 -30 . 908876
74 -2 . 0E+04 25 1000 24. 727101 74 -0.005327
o
o
value
o
ODE Report (STIFF) Differential equations as entered by the user [ 1J d(Cc)/d(t) = rg [ 2 J d(Cs)/d(t) = -rg/Ycs [ 3 J d(T)/d(t) = (Q+(-Hrxn)*(rg))/(rho*Cps) Explícit equations as entered by the user [ l J lprime = (0.0038*T*exp(21.6-6700/T))/(1 +exp(153-48000fT)) [2] Km= 5.0 [3 l rnutrnax e 0.5 [4] Ycs=0.8 [ 5 J mu= rnu1 rnax*lprime*(Cs/(Km+Cs)) [6] 0=0 [7J Cps = 74 [ 8 J Hrxn = -20000 [9J V= 25 [ 1o l rho = 1000 [lll rg = mu*Cc [12J Cpc=74 [ 13 J 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
o
74 -2.0E+04 25 1000 0.0598648 74 -0.074831
--··
326
312 ·
180
298
120
284
60
60
120 t
180
240
300
o o
----·60
120 t
180
240
300
0.40~-----------------, 32
0. 32 0.24
G
24
0.16
16
0.081,-----
8
60
120 t
180
240
300
o .... , O
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
290
300 To 310
320
330
P9-8 (e) Ce can be maximized with respect to T0 (inlet temp), Ta (coolant/heating temperature), and heat exchanger area. Therefore, if we are to find the optima! heat exchanger area the inlet temperature and coolant/heating temperature needs to be specified. If we take TO = 31O and Ta = 290 we find that the optima! heat exchanger area is an infinite amount of area . As A increases Ce increase wíthout a maxímum in 24 hours. Ce= 118.
9-17
~----------------------------------
--------------------~------
-----------------------------------
P9-9 First order liquid phase, CSTR First solve the steady state problem for the heat exchange area A for normal operation
T= 358 K. ·•·
Mol balance :
Ce
·· ··• r;1
'X
l;n(f})--······
E:::;;
1
1
R.T2
R.Tl
zn(}}) ·-··· í , __ .
=·-·-·"f'"_ ·····.
·······---·--·-··
,,
·•···· ········--··-·,·-·----
8.314(323) 8.314(313) 94852
k
= JI mol K
~?~(- l)l_=l07A /min J
= 1Jexp[?1 .
1
8314
313
T
Steady state solution :
v,, = f:...~. = 2~ CAo
1.·=
CAo
V V
90*2
= 500dm /min 3
) . = 200 -···=l.4/mm 500
= 2 M = 2*90 = 180 g/dm3
CA= 180 107.4 (0.4) CA C';t
= l+
180 (04*107.4)
rA = 1074 * 4 1
dCs V ::::: ···-··· = O dt
=
; : ;,
idm3
440 g/min.dnr'
9-18
forT:;;:.; 358 K
Ca= - ( - 440) OA = 175. 9 g!dm3
Energy balance :
.dT . . ._ -
Q·-Ws- LFio.Cpi(T···Tio)+(-AHr.x)(- . ra.V)
.
dt
,____
LNi.Cpi
-
= (120*60)A(273 ·• 1) F Ao..CpA·(T - TAo) = 90000*2*(1' - 313) Q == UA(Ta-T)
A = -180000.(358·-313)+(250)(440x200) _._. -..·-·-7200(273 - 358)
---··· - o
J/min
J/min . 2 = 22.7m
Use the unsteady state equations to determine what heat exchange area A will give a
runaway reactíon, where
Í: Ni. Cpi = Cp
50¡
Psol V
= 2 X 900 X 200 = 360000
J/K
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
o
180
o
313 0.4 22.696 1..1 200 -198 O.9
o
minimal value
o
4.1007565
o
313 0.4 22.696 1..1 200 -439 . 75367 0.0205038
o
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 l d(Ca)/d(t) = ((180-Ca)/tau}tra [ 2 l 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 [li tau= .4 [2 l A= 22.696 [3] k=1 . 1*exp(11409*(1/313-1/T)) [4J V= 200 [ 5 J ra = -k*Ca [61 Na= Ca/V [71 Nb = CbN
9-19
360 ..-----------~
340 330 320 310'---~-~--~--~-_____J O 2 4 t 6 8 10
A sensitivity analysis for different valúes of A show that a new steady state is reached ar every new A but ar an increasingly higher T. The required A depends on the definítion of reactor runaway.
P9-10 (a) X
= ¿ e¡Cp¡
[(T-To )1(-Ml
(1)
Rx )]
P9-10 (b)
e =e A
Ao
(1- x)
(2)
Insert (1) in (2)
(3)
For X= 1, Tr= T expression (1) becomes
- Mi
Rx
="°'e Cp. · (T -T ) <=:=! T L. , , o t
-T
o
= ¿e;Cp¡ - Mi Rx
Insert the inverse of (4) in (3)
9-20
(4)
(5) Remember that C80 is not constant here. Similar to CA, use equations (1) and (4)
¿0;CP; (T-T0) _ (T-T0) (T-T0) _ CB =CBo-·CAO. (-MIRJ ·(T-To)=CBO-CAO. (rl -ro)-CBO .eB. (rl -ro)-CAO. (rl -ro)-
e
AO
[e,
-e, ·T,)]
·TI -T+(T, T -T l
o
For developing -rA use
(6)
e
B
=C
AO [
eB . T1
-T + (T º T -T f
en . Tº-)]
(7)
o
(8) Insert (7) and (8) in (9)
P9-10 (e) The unsteady energy balance for a batch reactor is showed in equation 9-11.
dT
----;¡¡·=
Q+Ws +(-MIR>J(-rA
·V)
¿N;CP;
Adiabatic operation ( Q =0), neglecting
.
W, and assuming ~Cp=O gives the following using the expression for -rA:
9.. 21
P9-10 (d) Puta= 1, p = 1, 08 = 3 in the equation
from part (e)
P9-10 (e) Use the reaction constant expression from the great Swedish chemist Arrhenius and develop expression from 9-IOd.
P9-10 (0 First use a plot of T vs . t to get TO (329 K) and T r ( 439 K).. Checking the concentration of species B gives 08=3. Make a table showing t, T, dT/dt, left hand side of equation P9-10..7 and 1/T.. Plot left hand side of equation P9-10.7 vs. 1/T in Polymath and use linear regression to get E from the slope and k, 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
--·---·-------------------------·--·-
--·---~-----·--------------------
----------------------------·-- -----------
---
---~------ ---·-·------------·
-----·----------------------------------
-----·---------------·-----------------
aO
= 17.534 = 1n(
k1 ·CAo 2 T ( j -T,) o
J+~
= 1n(k1 ·6.7)~
R · 'Fo '---v-'
12100
k,
= 12100 ·exp11s34 = 1.298·1011 6.7
neglect
P9-10 (g) Follow the procedure from example 9-3 gives heat of reaction,
X=l and T=Tf gives -IV/Rx
= ¿eicp¡ ·(T-To)=(CPA +ea ·CPBXrl -To)
= (189.7 + 3 · 75.4
)(439-329K)
J
mol·K
= -45.75kJ
/ mol
P9-11 (a) . dNa
Mol balance :
····-·-:::::: dt
dX
raV
Nao---·- = ··ra V dt
dNb ---·- == 2.ra. V+· Fbo dt
d.Nc
... Rate law :
dt
.
= -rav
- ra::::: kCa.Cb2 k( T) = 0.0005 ex
. Ca
Stoichiornetrv : .;
p(-~1.987 9.~~
Na_ V
> ....
Cb
= fil:.. V
Ce=
Energy balance :
.rr
d; . ==
Q- Ws.. .
¿ .Fio.. Cpi.(T-Tio) V) .. :t ,Ni. Cpi+ . ( -,···6.Hrx(T)).(-.-ra. .... . , __ , . _,, . ·--
\\ here Q == UA.(Ta - T) V./s
=O
= 250.(390
- T)
i\Hrx(T) :::;; -55000
I:Fio.Cpi
==
Fbo.Cpb = Cbo. u0.Cpb = 4*uº *20 = 80uº 9-23
---
------·--·-----~~------~-------~-------~----------
/:!.E V
= NaCpa
>':-..;-i_Cpi
•
-
...
+ Nb Cnb + Nc.Cpc .
.j.
=::
35Na + 20Nb + 75Nc
-· 325) + (-55000). (-ra. V)
dT 250. (290 - T) - 80.1.k(T -- = ~ . . . . -.-· -...
35. :"la+ 20.Nb + 75.
,.ve
See Polymath program P9-l l-a.pol POLYMA TH Results Calculated values of the DEO variables variable t Na
initial value
o
o o o
o o o
Ne X T
283 . 60209 l. 5 2.3E-04 10 6 2.202E-04
300 l. 5 5.0E-04 10 6 5
vb k V
Fbo Ca Cb Ce ra
o
0.3324469
50
Nb
minimal value
o o o
o o
-0.0028611
maximal value 1000 50 5900.6649 49.667553 O. 9933511 300 l. 5 5.0E-04 1510 6 5 3.9077251 0 . 0718765
o
final value 1000 0.3324469 5900.6649 49.667553 O. 9933511 298.76383 l. 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)H80*vb*(T·325))+(-55000*(-ra*V)))/(35*Na+20*Nb+75*Nc)
=
Explicit equations as entered by the user (ll vb=1.5 [2 l k = .0005*exp((8000/1 . 987)*(1/300-1/T)) [ 3 l V = 1 O+(vb*t) [4] Fbo = 4*vb [5) Ca= Na/V [6] Cb = NbN [7l Ce= NcN [ 8 l ra = -k*Ca *CbA2 5
300
4
296 292
288 284
O O
200
·400
600
800
1000
280 ~-~-----~--'----'
O
200
9-24
400
t
600
800
1000
0.8 0.6 0.4 . 0.2
200
400 t
Table ! Iterating wrt
--
Uo
dm '/min ,
K
1000
r. - ~- __ _. - ·- - . ·-·. _ ···-· _ · .
for X=0.8 and T<403K and daily Ne+ 120 mo!C
_._¡_. . T. (X=0.8) ..
~º
800
600
t
'
(X:O 8)
Ne
min
(X=O 8)
Daily Ne
.. Ne(24*60/t+30)
mol
A flowrate of 15 mol Brrnin would produce 120 mol C/ day , with X "' 80 % and T< I 30°C at ali times
P9-11 (b) lf the rnax. coolant rate falls ro 200 mol/min, then it may not be prudent to assume that ihe coolant leaves at the entering ambient temperature, Ta. It should be assumed that the coolant temperarure varíes spatially along the heat exchanger pipes and the required terrn for the heat exchange would be:
Q ==
m: Cp<=001 (Tal
- Ta2)
where Tal Ta2
u:
= ambient T coolant entering = ambient T coolant leaving
Ta2 = 1'-(T-Tal).exp(· __ 4 )·. rrkCpco11 The reduced flowrare and hence heat exchange, may increase the reactor temperature to approaching 130°C, the upper limit, at conversions approaching 80 %, and so more caution is required. Toe incorporation of temperature control would be prudent,
P9-12 CSTR startup Need sready-state valúes at To= 75°F
9-25
Mol balance :
dNa dt = F ao ···· Fa + ra .V
dCa
(Cao-Ca).Vo
dt
V
-··-=
-
dCb -=·· dt
(Cbo - Cb)v.,
dCc
(Ceo·- Cc)u,
dt
V
----
··· ·
V
+ra
+ra
ra
dCM (CMo - Cw)tk. --·~ ·---· dt V .....
Fio
Cio=--
Cao = Faq. = 0.18157 lb mol/ ft3
·v.,,
Va
= 2.2696
• Fbo Cbo = ..... U::,
-, 3
lb mol/ fr
Cco=O
.
Ewo
....
Cuo = ·--·····- ;;:; 0.22696 lb mo113ft v., 1 V = ····--·-'· x 500 7.484 Fao
Fbo
F.t10
80
pao
pbo
f}MO
ü.932
1000 100
vo = ··'". ·-- + - -···- ·+ ·-·----··- : -· ---- + -Rate law :
- ra = k.Ca
3.45
+ --- .:::: 440 6 fr I br 1.54
·
(Cb in excess)
k = 16.96el
2.~{-¡_ g'ji:ff4&)1) 9
Energy balance ; dT
dt -
Q-Ws·- LFio.Cpi(T·-Tio) +(-·-ólirx)(-ra.V)
.. . ..
í:_Ni.C;i
9-26
....
Ws=O
Q = Inc.Cpcoo1.(Tal - Ta2)
= 1000 X 18 x (60-Ta2)
1:Fio.Cpi.(T ~ Tio) = [Fao.Cpa + Fbo.Cpb + FMo.CPM ].(T - 75) AHrx
= ~36000 Btu/lb mol
í:,..Ni.Cpi = Na.Cpa + Nb.Cpb + Nc.Cpc + NM.CPM
= 53Ca V+
18Cb.V + 46Cc.V + 19.SCM.V
Ta2 = T- (T -~ Tal).exJ - · · · · UA · J, t'~ m::Cpcrx,1
= T--(T. ..
60) . .
1600°.-). exp( l ()()(h:18
Ta2 = T-0.4llll(T--
dt
~
l8000(60-(T-0.4111 l(T-60)-22750(T-75)+ (36000)(-ra.V) ·--!<"'.'!'"~"· ..--·35Ca.V + 18Cb.V + 46Cc.V + 19.5CJ1.V
Initial conditions : To
If
60)
= 75, T = 138.5 °F.
Ca= 0.03780 • Cb - 3.3062 • Ce= 0.0144. CM= 0.2269 lb mol/ ft3
To drops from 75 to 70 °F
P9-12 (a) P-control only :
manipulated variable
= lllc
controlled variable = T ID<:= lll.;o
where Illco = 1 OOOlb mol/h
+ kc.(T- Tsp)
Tsp = 138.5 °F kc= 10
See Polymath program P9-12 a . pol POLYMA TH Results Calculated values of the DEO variables Variable t
initial value O
minimal value O
maximal value 4
9-27
final value 4
Ca Cb Ce Cm T I
Fa O TO V
Tsp UA
Tal ke k FbO FmO meo ra NCp ThetaCp vO Ca O CbO Cm O tau X
me Ta2 Q
0.03789 2.12 O . 143 0.2265 138.53
o
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 O . 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 O . 6775218 872 . 17351 102.00022 6.594E+05
0.0584442 2.1423214 0.143 0.226519 138.53
o
80 70 66 . 809193 138 . 5 1.6E+04 60 10 24.990212 1000 100 1000 -O. 7913918 3385.3399 284.375 441 . 46403 O . 1812152 2.2651902 0.226519 0.1513355 0.7909116 1000.3 106.23674 8.325E+05
ODE Rellort {RKF45} Differential equations as enterad by the user [1] d(Ca)/d(t) = 1/tau*(CaO-Ca)+ra [2] d(Cb)/d(t) = 1/tau*(CbO-Cb)+ra [.3] d(Cc)/d(t) = 1/tau*(O-Cc)-ra [4] d(Cm)/d(t) = 1/tau*(CmO-Cm) [5] d(T)/d(t) = (-Q-FaO*ThetaCp*(T-T0)+(·36000)*ra*V)/NCp [6] d(l)/d(t) = T-Tsp Explicit equations as entered by the user [ll FaO = 80 [2J TO= 70 [ 3] V= (1/7484)*500 [4J Tsp = 138.5 [5] UA = 16000 [óJ Ta1 =60 [7J kc=10 [ 8] k = 16.96e12*exp(-32400/1.987/(T +460)) [9] FbO = 1000 [10] FmO = 100 r 11 J meo = 1000 (12 J ra = -k*Ca [ 13 J NCp = Ca*V*35+Cb*V*18+Cc*V*46+Cm*V*19 . 5 (14] ThetaCp = 35+Fb0/Fa0*18+Fm0/Fa0*19.5 [ 15 J vO = Fa0/0.923+Fb0/3.45+Fm0/1.54 [16] CaO = FaO/vO (17] CbO = FbO/vO [ 18 l CmO = FmO/vO (19] tau= V/vO [ 2 o l X = (CaO-Ca)/CaO [21] mc=mcO+kc*(T-Tsp) [22 J 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 O .1513355 0.703827 899.92991 102.98479 6.963E+05
Q = mc*18*(Ta2-Ta1)
[23]
140
1100
136
1040
GJ
132 128
920
124
860
120
o.o
0. 8
1.6
t
3.2
2.4
4.0
800
P9-12 (b) manípulated variable
!-control only :
Q
980
o.o
0.8
1.6 t
3.2
2.4
4.0
= m,
=
controlled variable T kc mc = 111..c.i, + - I
wbere
1J
m:o = 1000 lb mollh kc
d1 dt
-=
= 10. 't1 = 1
Tsp = 138_5 °f
(T-Tsp)
See Polymath program P9·· 12-b.pol
---
140 134
940
128
880
122
820
116
760
110
o.o
0. 8
P9-12 (e)
1.6
t
2 .4
3. 2
4. 0
700
[mJ
0. 0
0.8
L6 t
2.4
3.2
4.0
rnanipulated variable == me
PI-control only :
controlied variable = T m:
=meo+
kc kc(T - Tsp) + +-I
where rn.,
't,r
= 1000 lb mol/h
kc == 10,
dl
1:1
;l
Tsp = 138 . .5 °F
······-·=(T-1sp) dt See Polymath program P9-12-c.pol 140 .--------------.
1100
136
1020
132
940
128 ·
860
124
780
120
o.o
08
L6 t
3.2
24
4.0
700
O.O
0. 8
L6 t
2.4
3. 2
P9-13 dCa
Mol balance :
-=--· dt
(Cao ···· Ca)
·---+ra 'Z
Cao = 0.1 kmol/m' Cao ;;:;: 0.1 mol/dm3
dCb
( Cbo -· Cb)
-- . · - -- ·- . . · ···- . ·+ ra dt -i dCc
dt Rate law:
(O·- Ce)
·=----· t
Cho=Cao
,; = 50 s
+ra
· · ra = k.Ca.Cb . . f'" l 0000(· l k(T);;:;;;0.0lexp ··----· ---- l )]" . L 1.987 .· 300 T
9-30
4.0
dT
Energy balance :
V = Uo
• 't
dt =
= 2 X 50 = 100 dnr'
Fao;;;;Cao .
'\Jo=
0.1 x 2 = 0.2 mol/s = Fbo (equimolar feed)
IFio.Cpi.(r - Tio) Mírx(T)
Q-W.S"··· "'iFio.Cpi(T-Tio)+(·-Mlrx)(-ra. V) . LNLCpi .... . . -
= 2 x 0.2 x (l5)(f
- 300) = 6(T - 300)
= tilirx(To) + ACp.(T-To)
= - 41000- (- 20000) - (·· 15000) = -· 6000 cal/mol
but .6.Cp = O
f!1' = j_(T__::-300) + (6000),(-ra. V) dt
15.Ca. v+ 15.Cb. V +30.Cc. V
No control: See Polymath program P9- l 3--a.pol
POL YMA TH Results Calculated values of the DEO variables Variable t Ca Cb Ce T
V k
ra
initial value
o
O . 001 O . 001
o
300 100 0.01 --1 . OE-08
minimal value
o
l.027E-04 l.027E-04
o
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 l.893E+05 -0.001998
ODE Report (RKF45) Differential equations as enterad by the user ¡ 1 J d(Ca)/d(t) = ((.1-Ca)/50)+ra ¡ 2 l d(Cb)/d(t) = ((.1-Cb)/50)+ra ¡ 3 J d(Cc)/d(t) = (-Cc/50)-ra [ 4 J d(T)/d(t) = ((6*(T·300))+(6000*(-ra*V)))/(15*Ca*V+ 15*Cb*V+30*Cc*V) Explicit equations as enterad by the user [ll V= 100 [2 l k = .01*exp((10000/1.987)*(1/300-1rr)) [ 3 J ra = -k*Ca*Cb
9-31
300000
0.10 0.08
240000 ·
.
Ca
.
.
Ch
180000
Ce 120000 ·
60000
OJJ2
º·ººo
80
160
t
240
320
400
o o
80
160 t
240
320
The results withont control, indicare a runaway reaction, as T continúes to increase after the concentrarions have approached their steady-state values Control aspects : assume that the operaring T sbould not exceed 550K, the boiling point of the Iiquid, Without given data for heat exchange : manipulated variable = Tío (inlet feed T) conrrolled variable = T
Try
P-control
Tío= IF (T <550)'I1iEN (300) EL.SE (Tioo + kc.f'I' - Tsp)) where Tsp ~
= 550 K. Tioo = 300
K, kc = -10
Tío rnanipulation not feasible for T control for any kc (requires Tio 2000 K) »
See Polymath program P9-· l 3-b.pol 800 ----------------,
700 600 500
400 300
'----=-----~--~-
o
Try
80
160 t
240
320
400
manipulated variable= u0 (inlet volumetric flowrate) controlled variable = T
9-32
400
P-control
u0 == IF (T<550) TIIEN (2) El.SE (u00 + kc.(T --Tsp)) where Tsp = 550 K, u00
= 2 dm",
kc
= 10
See Polymath program P9-13-c.pol 600 ~------------, 540 480 420 360
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, Ce=:: 0.0144, CM= 0.2269 lb mol J ft3
T drops from 138.5 to 133.5 °F As in P9- l 2B except : If
dCa
Fao - Ca. vo
-----=- -- -----· + ra dt V Vo
Fao · + --1000 ·r 100 0.932 3.45 1.54
= - .. ,
- (T -· 0.411 ll(T · ··-60)-(35Fao + 19950).(T,., -- _75) + (36000)(--ra. V) -dT = --·18000(60 ""--·--·•• •· ·-A·-·-·-··--·--•-·• ·•·•· · · .. --·•·• • •••·••--·••"-•_ . __ .,. _ _, .. ,_ dt
35Ca. V +18Cb . V+ 46Cc..1'+ 19.SC.l-l. V
Lcontrol only :
manipulated variable= Fao
controlled variable = T . kc
Fao == Faoo+-1 1,¡
where Faoo == 80 lb mol/h,
say kc:;: -0.2,
-r, = D~I
9-33
,-
d.[
-
dt
Tsp = 138.5 °F
:::: (T- Tsp)
See Polymath program P9-14-a poi POL YMA TH Results
Calculated values of the DEO variables Variable t
Ca Cb Ce Cm
T I
Fa O o
TO
V Tsp UA
Tal ke k
FbO FmO
meo ra NCp Fa O ThetaCp vO CbO CmO tau Ca O 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. 324Et·05 0.7909116
minimal val u e
o
-27.922973 2.12 -r , 7469378 0.2265 17.999228 -307.34656 80 70 66.809193 138 5 l.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 1.6É+04 60 0.2 24.990212 1000 100 1000 0.72.3285 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 -O. 2979677 2 3835154 938. 55771 34.305432 -4.3418+05 12.667586
ODE ReW!rt íSTIFF) Differential equations as enterad by tne user [ 1 J d(Ca)/d(t) 1/tau*(CaO-Ca)+ra (2 J d(Cb)/d(t) = 1/tau*(CbO-Cb)+ra [ 3 l d(Cc)/d(t) = 1/tau*(O-Cc)-ra ¡ 4 l d(Cm)/d(t) = 1/tau*(CmO-Cm) ¡ 5 J d(T)/d(t) = (-0-FaO*ThetaCp*(T-T0)+(-36000)*ra*V)/NCp ¡ 6 l d(l)/d(t) = T-Tsp 150 ---------·-----·--
=
144 138 132 126 ,__
120
9-14
o.o
__.
08
1.6
t
2.4
3.2
-1.0
(10] (11] [12]
[13] [14] [15]
[1.6] [17] [18] [19] [20] [21] [22]
[23] [24]
FmO = 100 meo= 1000 ra = -k*Ca NCp = Ca*V*35+Cb*V*18+Ce*V*46+Cm*V*19.5 FaO = FaOo+(ke/.1)*1 ThetaCp = 35+Fb0/Fa0*18+Fm0/Fa0*19.5 vO = Fa0/0.923+Fb0/3.45+Fm0/1. 54 CbO = FbO/vO CmO = FmO/vO tau= V/vO CaO = FaO/vO me= meO+ke*I Ta2 = T-(T-Ta1)*exp(-UA/(18*me)) Q = me*18*(Ta2-Ta1) X = (CaO-Ca)/CaO
90 88 86
84 · 82 80
P9-14 (b)
o.o
0.8
1.6
t
4.0
3.2
2.4
T.1 = 55ºF
L-conrrcl only :
manipulated variable
e
T0
controlled variable :::; T To :;:; :
r». + -~~¡
whereToo » 75 °F, say kc:::: ""0.2,
di
't'¡
= 0.1
Tsp = 138.5 °F
dr = (T- Tsp)
See Polymath program P9-14-b.pol 150
85
144
82
138
79
132
76
126
73
1200.0
0. 8
1.6
P9-14 (e) No solution
t
2.4
3. 2
4.0
70
0. 0
0. 8
1.6
t
2.4
3.2
4.0
will be given
P9-15 9-35
----
---------~·--------------------------
------~----·---------·-·-
nitrous oxide B
amrnonium nitrate A
Feed , mo
= mao
+rnco e 0..83 mo + 0.17 mo
stearn
e
(meo
= 17% Iiquid water in feed)
d!via ------ - =mao ··O+m. V
Mol balance :
Ma = lb A, rnao = lb/h A
dt
dMb
---··--- =0-mb···ra.V dt mb = mao.(Bb
leaving :
+; x)
== mao ; 80 = O, b/a = 1
(X= 1, i.e. ali of A entering reacts, but there is some initially inside) di14c
-·-··--· = meo - me -- 2. ra. V dt
leaving : me ::::: mao,
(oc+ l- x) = meo +Zmao ; 0(.4;(), e/a = 2
assuming all water entering with the ammonium nítrate leaves as steam.
Rate law :
-ra, V
E- ..
= k.Ca. V = k.Ma
;i~~.
R.T2
R.Tl
Ln
:) ( o-.,,) 5.03.
l 1.987(560+460) = 88500 Btu /lb mol
where
Ws=O
Q = UA.(Ta -T) 9-36
J_
l 1.987(510+460)
Imio.Cpi.(T ~ Tio) == mao.Cpa(T -Tio) = 0.83 X 310 X 0.38 X (T- 200) mco.(Hi -Hio) = mco.[Hg(T) * H1 (Tioj] 0.17 x 310 x [(1202 + 0.47(T- 500))- 168]
=
assumes all liquid water in feed leaves as steam,
Hg (T) = Hg (Tref) + Cpc.fT - Tref) = 1202 + OA7.(T ~ 500) H1 (200) -== 168 Btu /lb .i:Mi.Cpi
= 0.38 Ma + Cpb.Mb
+ 0.47 Me
From ChemCAD m physical propertíes systems:
=
Cpb (NiO, 516°:F) 1065 J/kg.K. (JQ~S
Btu). (0454
:J (~-)
= 044 Btu/lb N20. K
P9-15 (a)
ar t::: O Ca= 500 lb, T = 516 °F
-T)- ··· 97..77(1'. 200) - 52.7((1202 + 0.47(T --500)) · · 168]+ 336.(-raV) .dT . . ., = 10000.(515 ·•· · _. ···-·-·dt 0.38Ma +044.Mb + OA7Mc »
.,......................_
,u,
- .. -~.-~
_
-
-..
..
.
See Polymath program P9-15-a.pol POLYMA TH Results Calculated values of the DEO variables Variable --t Ma
Mb
Me T
meo mao k
rav
rob
me
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 [ll d(Ma)/d(t) = 310*.83+raV [ 2 l d(Mb)/d(t) = -mb-ra V [ 3 J d(Mc)/d(t) = mco-mc-2*raV [ 4 l d(T)/d(t) = ((10000*(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 [ll meo= 310* . 17 [2 J mao = 310*.83 [ 3 J k = .53*exp(44540*(1/970·1/(T +460)))
9-37
[4]
(5] (6)
raV = -k*Ma mb = mao me = mco+2*mao
518
500
517
460
GJ
516
420
516
380 ·
515
340 ·
'il4
- o
4
2
t
6
8
10
300
o
2
4
t
6
8
10
P9-15 (b) UA = 10000 Btu /h.ft2 Pvcontrol only :
manipulated variable ;;;:, Ta controlled variable =T
Ta== Tao + kc.(T - Tsp) where Tsp == 516 ºF, kc = -5, Tao= 975 °R = 515 °F See Polymath program P9 l 5-b. poi 516.20.---
500 460 420 380
515 72 515.600
34{1
-----·--·---·- . 2
4
t
6
8
10
300
o
P9-15 (e)
9-38
2
4
t
6
8
10
UA = 10000 Btulh.ft2
manipulated variable = Ta controlled variable = T
PI-control:
Ta= Tao+kc.(T·~Tsp)+ k<:.__¡ 1J
dl
-· ·
dt
where Tsp = 516
=(T-Tsp)
°F, kc = . 5, 't1 = l. Tao= 975 °R = 515 °F
See Polymath program P9-15-c.pol 516.20~------------,
500
516.12
460
516.04
420
515.96
380
515.88
340
515.800
2
4
8
6
t
10
300
o
---~---· 2
4
t
6
---~
8
P9-15 (d)
U A ;;:;: 10000 Btu /h.ft2 PI-control loop 1 :
manipulated variable= Ta controlled variable = T kcl t:Jl
Ta= Tao+kcl.(T-Tsp)+-·· ·-· ./1 dll dt
-=(T-Tsp) where Tsp Pl-control loop 2 :
= 516 °F, kc l = -5, 1:11
= l. Tao= 975 ºR = 515 °F
manipulated variable e mao controlled variable = M (= Ma + Mb + Me) 9-39
10
kc2 .
+ kc2. (¡\ti - M.sp) +-·· ·-·. 12
mao == maoo
t/2
dI2 dt
-·-· = (lvf where Msp
::=
lvf sp)
500 lb, kc2 = 25,
't'r
2
= 1,
maoo ::::; 31 O lb/h
See Polymath program P9-15-d.pol 516.20 .--------------,
500
516.12
460
516. 04
420
515.96
380
51588
340
515.80'--~-~--~-~8----' O
2
4
t
10
6
300
o
2
4
t
6
P9-16 (a) Plot R(T) vs G(T):
R(T) == UA.(T ·- 7;,)
+f)V0Cp(T
····
Ta)
G(T) = -,.MIR;: VkC~ Evaluare the parameters in those equations:
k =2*7.08* lOnexp(-30000/1.987/T) MI RX
;;;;;
-30000BTU / lbmol
o = 50lb I .ft3 lIA = 150
* 250
= 37.500
See Polymath program P9- l 6-a poi POL YMA TH Results Calculated values of the DEO variables Variable t T
dHr
V
Cao VO
UA
initial
o
value
520 -3.0E+04 48 0.5 400 3.75E+04
minimal value
o
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
Ta rho To E k
0.3478362 0.12 0.479966 0.75 2.404E+05 -5.25E+05
tau Ca Cp Gt Rt
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
530 50 530 3.0E+04
164.49672 0.12 0.0241085 0.75 5.711E+06 6.825E+06
164.49672 0 . 12 0.479966 0.75 5.711E+06 6.825E+06
ODE Report(RKF45) 7.00E6 --------------~
Differential equations as entered by the user [ l J d(T)/d(t) = 1
5.40E6
Explicit equations as entered by the user [ l J dHr = -30000 [2] V=48 [3J Cao = .5 [4) vo = 400 [5] UA = 250*150 [6J Ta=530 [7] rho = 50 [8] To=530 [9] E= 30000 [lOJ k=2*7.08e11*exp(-E/1.987/T) [11] tau= Vivo [ 12 J Ca = Cao/(1 +tau*k) [l3J Cp = .75 [14) Gt = -dHr*V*k*Ca [15) Rt = UA*(T·Ta)+rho*vo*Cp*(T-To)
3.80E6 2.20E6 6.00E5 -1.00E6'----~--~-~---~-...., 520 548 576 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.lºR
., -- O • 4.?5 ("A -
T
C• A 0.319 . CA._= 0.068
T = 57L3ºR T 628.6ºR
=
=
The extinction temperature is TO= 506ºR. At this point R(T) is tangent to G(T) at the upper steady state.
P9-16 (e) Using the unsteady-state equations for a CSTR we get the following Mole balance: d~:i- = .~o{CAO ··<:A)._ kCA dt V
Energy balance:
See Polymath program P9- l ó-c. .pol 9-41
632
660
i,~l~~M~illl,;t~B~-P9"rt:~~+:}~i!'lt,
:t'.~f''!íI['tmM~.ll~¡~·l;~j~;~}l~;;.~~1
;itf¡~,f[·';,i~> {,t,ii > · . ;'' -' ,t~A'.'l
Calculated values of DEQ variables ---·-··-· 1····-----· ·····- ---- - --~--------------·-· -···· !···---·- ·-- _, 1-1·----... ·-y·-·-----~·-·~··---~---·-·-1---·---~-------------- ,, ·~- i ¡ ¡variable Initial value I Mini mal value Maximal value Final valueJ
!t
I
r---···,,~-----
f-·!1 ·- r
¡2 ¡ea
---t·········'·· -- ----.
!O -\"
·
¡13 j,T
¡o.04847
_....
_···1200.· -
1
,530.
f
fj- TA~- . . --Íi416É·¡12 . . .
[B jR-,9 tu r-·· J .--
¡250.
1'110¡Ar·--
111Jv0
¡400.e
<'=r•··ao.-,~
..
r13F/.... . - --- ·í4s. ¡12¡Ta
¡530.
"""
14jE
r!isfk . ··r--· 16,tau [i7lrA
_ ..
1
~
¡0.4253139
-----c---·-·-------- . ---- ¡· j11
r¡5. 47.0711 ··------ --
. -_ j260.-_
1530. Ji.4i6E;Ú-
- f ioo:·
¡0.5
]
1530.
j
1530. - ll.416E+
¡
"'"' 1--- - · ·-·
j400. ·•·------¡-~·· •v·• • '·,-.••,-· '
---·>•,Y'-'-~·~•·•-·--·-
·<""
.,..
•
j3.0E+04 Ti.is5967-
··· r_..,
r
figr N~o- -·-·124: L..
·¡'24:·--··
1is~ -· · · -· - I;,s.' .
f iolcp~ .
, .
.
•..•
-..-. •••• ,.... v,r1400. ~·,·•, v,~ . .
"'"'"t·'·"
~
!3,0E+04 -·- r15.24i79
- ----------¡
,0.12 ,0.12 F3.49i943. --·-1-3.971954 rioE+04 l~i6E~04- ,
1
· ¡1250. .
--
-
.,,-•a.•'•••
, ..
.1
--~J
1 J
·¡
,- .. ·,Ta,.· · ·
4
l3.0E+04
1
10.12
1
Ji:463353 ·1i T -- -- -----
-···
,0.12 1-0.4433316 f ~ioE+04 · ···--¡24:· .. ----
r-0.6223844·¡ ·7!:_j:ci'E".;Ó4
124:·
hs. -
--11s. · · ·-
. , ..!
,
.., . - .
¡530. ¡530. ¡530. · · --- -· 14a: -· · - - - . - -·14a:· · --- - - · · -·- ·l¡s~- . - --
l3.0E+04 ·-¡st:21669
[i8jd~IH
1
., 1250. .
!400.
•''
•·•-0••"·"·•·-<>·•••··•-.a
T···~---..
1
'
--
----------
·
~j6.
. -T1638.1791 ¡O.S
. T.1250. .
T...
r--··,., . r . . ,., ., , . ~. _ . . ,.,
1
,,1··----···---·····-"·----···"*----·-·-----·-
12 ------lDiicsÉ~l.2· 987 . . µ.987 . .• . . . /ús1... . .. '¡1-987 . j j ,., 150. . . .. --- -r¡ 150. . ----.. --¡- l 150. " ------¡--·--·- ··--1-150. . -·-·1
1
r"·· ·· · r~ ,
.-
¡0.4290427
·-· , --
1542.6046 r........ .. ... j0.5
- -· ¡200.0.5 - · ·
1s+F~o 6 1TO
. ---
¡
--- · ·-········"·····-- -·------··
16. · ---+
. --- .. ----------···
T!628. ,
4 ¡CaO
I
-------··-----·····r-·--------
.. - ---~···· .. ···---- .. ·-·-----··---·--
jo · · --·--·-t····--- - - ---- -
¡0.0681
,----r-- . ·--- . ·--·-- -v-r-_ 1
"-·-~-·------·--·····-'",-·
·--1
l
.!
j
1
J
Differential equations
d(Ca)/d(t)
=
rA + CaO / tau - Ca/ tau
d(T)/d(t) = (U * A * (Ta - T) - FaO
* Cpa * (T -- TO)
+ (rA
* V) * delH) /
(NaO
Explicit equations
CaO = 0.5 FaO
=
200
TO= 530 Ar= 1.416 * 10
R
A
12
=
1.987 U= 150 A= 250 vo = 400 Ta= 530 V= 48
9-42
----------·------
--
-- ---
- -- - ---- ~--
- - -- -- ---------
---·---·--------·--------
* Cpa)
E= 30000 k
= Ar * exp( -E /
R / T)
tau= V/ vO
= -k * ca delH = -30000 rA
*V Cpa = 37.5 * 2 Nao= CaO
0.5
(
0.4 0.3 0.2 0.1
~
o.o o. o
1.2
2.4 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 T0 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
0.40
740
0.30 ·
680
1111\j,..,..,...,..,.... 0.20
620
0.10
560
0. 00
o.o
1.2
2.4 t
3. 6
4.8
6.0
500
o.o
1.2
._._
2.4
t
3.6
48
6. 0
Ifwe 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 mínimum in CA is reached. 9-43
0.5
0.5
04
OA
GJ
03
0.3
0. 2
0. 2
0.1
(U
o.o 500
560
620 T
P9-16 (O lndividualized
680
740
solution
P9-16 (g) The following Polymath program gives the linear analysis of the problem A= 1.175, B = 6.03, J = 1600, -e= . .12 See Polymath program P9-16-g.pol
9-44
j2a·1 rA----· --r-3.491943 ····-r-3.971954 j :0.4433316·····-·--}0.6223844 .. . ····,····-··-- ······--··1-·---···-···--·---·-·--r·-··"·-·----------····-··~·--··-. ·····-···· -- ··-·--·······T····---_j 1 ldelH ¡ 3.0E+04 j-3.0E+04 l 3.0E+04 ¡ 3.0E+04 j 121
-124. _ --+4·__ _ _ . • 124. _ _ __
[22JNaO ¡23~J 124 e
1800. !6.
·-- ---··i··· ·· ··
r··-----·· ·-··-····- ··
t·· -:r ····-- -
1800. -
·····
¡6.
1800.··-····· ·····- · · ·· - -···--··-· 6. 16.
···· ···---------·· ····¡'··· ·····
---·· ···-··--···----······--·····-··--1·-··-····---···-···-··--········· ···-T
¡25¡8
j6.03
!?~I~- .. .11:~?~··.
··--·----···--·····--··---·--·
¡6.03
. ..J!:.~?.~- ·
.• _ ..... · . 1 800. J; 1r24
¡6.03
..
J~·-~?~--.
· ···
1··· ---------------·
j6.03
•.... l~:~?~.. .
J.
j
J
Differential equations
!Id(y)/d(t) I d(x)/d(t)
= (-J * (1 - A)* X+ (B - C) * y) = (-A * x - B * y/ J) / tau
[ffl d(Ca)/d(t)
= rA + CaO / tau - Ca/ tau
ti d(T)/d(t) = (U * Area * (Ta - T) - FaO * Cpa * (T - TO) + (rA * V) * delH) / (Nao * Cpa) Explicit equations
llcpa = 37.5 ¡ite~i>~ cao = o.5
*2
11 FaO = 200 TO= 530
\IJ Ar=
1.416 * 10 /\ 12
t~ R = 1.987 l)l U= 150
DArea = 250 llvo = 400 !ITa = 530 l:~\!~ili ~ V= 48
[1,J E=
30000
ilJ k = Ar * exp(-E / R / T) l!ltau =V/ vO
!B] rA =
-k
t~~1'~
* ca
li!i! delH
= -30000 IINao = cao *V
!ti!
J = -delH/Cpa / CaO
jl"!I¡,•
llL,1 e= 6
Rs ~""I
~;¡Ji A
= 6.o3
=
1.175
P9-16 (0 Only the lower steady state plot of xl and yl will be shown.
9-45
0.50 .---------------,
0.50 .-------------,
0.38
0 . 38
0 . 26
0.26
014
0.14
0.02
(U)2 . . ........ .
-0.10 ~---~--~-~--
-O.JO~-~------~-~ O.Oe+O 1.6esJ
__
----·---- --..·-··-- -..........
.....
.
o.o
12
24
t
3.6
48
6.0
3.2e-?,1
4.Se-3
6.4e-3 8.0e-3
P9-17 Batch reactor - series reaction Mol balance on A :
dCa
---·-= raI dt
= kl.Ca
-ral Mol balance on B :
Ca= Cao.expi-kl .t)
.dCb . ...... = rbl + rb2 dr
rbl
= · · ral = kl.Ca
dCb
and
·-;¡:¡·· = kl.Cao.exp(
Mol balance on C ·;
Energy balance :
Tb2 =k2.Cb
kl.t) · · k2Cb
dCc dt - -rb2
Y)+ (·6.Hrxb2)(·rb2.V) _LNi.Cpi
dT .~ Q+(lillrxal)(-·ral
dt
9-46
Q = UA.(f a- T) LNi.Cpi = Ca. V.Cpa + Cb.V.Cpb +Cc.V.Cpc dT dt
UA.(330-T)+55000(-raLV)+ 71500(-rb2.V) 200.V.(Ca + Cb+ Ce)
P9-17 (a) dT di
UA=O
. ...... ......
V[55000(--·ral.) + 71500(-rb2.)J - ---· ----·-------·- ..··-···... 200.V.(Ca+ Cb+ Ce)
See Polymath program P9- l 7-a. pol
POLYMATH Results Calculated values of the DEO variables Variable t Ca Ce Cb T UA V
kl k2 ral rb2 rbl
initial value
o
0.3
o o
283
o
10 1.1172964 4.081E-09 -0.3351889
o
0.3351889
minimal value
o
l. 34E-65
o
-2.02E-44 283
o
10 1.1172964 4. 081E-09 -35.016552 -27.963241 -6.345E-60
maximal value --0.2 0.3 0.3 0.2895784 915.5
o
10 2.141E+05 1.041E+06 6.345E-60 2.103E-38 35.016552
final value 0.2 l. 34E-65 0.3 -1.864E-65 915.5
o
10 2.141E+05 1.041E+06 6.345E-60 -3.974E-59 -6.345E-60
ODE Report (STIFF) Differential equations as entered by the user [ l l d(Ca)/d(t) = rat [ 2 J d(Cc)/d(t) = -rb2 [3] d(Cb)/d(t) = rb1+rb2 [ 4 l 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= O [2] V=10 [ 3 J k1 = 3.03*exp((9900/1.987)*(1/300-1/T)) [ 4 J k2 = 4.58*exp((27000/1.987)*(1/500-1/T)) (5] ra1 =-k1*Ca [ 6 J rb2 = -k2*Cb [ 7 J rb1 = -rat
9--47
1000
840
0.22 - Ca
0.14
Cb
006
520
360
-0 . 02 -0.10
[iJ
680
- Ce
L..--~--~------~ 0 . 04 O 08 t 0.12 016
0.00
O 20
200 ~-~--~------~ 0 . 00 0 . 04 O 08 t 0 . .12 0.16
0 . 20
P9-17 (b) UA - 10000 J/min.K
V= 10dm3
0.22
840 -
0.14
Ca
680
- Ce -
Cb
-0.02
360
-0 . 10
0.00
0.04
0 . 08 t
0.12
0 . 16
UA = 40000 J/nün.K
020
200
0 . 04
008 t
012
(U6
0 . 20
V= 10 dm3
O.JO
900
0.22
-
0.14
-
Ca Ce
620
Ch
480
-0 . 02
340
0.00
0 . 04
0 . 08 t
0.12
UA::;.: 100000 J/min.K
r.
760
(J.06
-O.JO
000
0.16
O 20
200
O 00
V= 10dm3
9-48
0.04
008 t
0.12
0.16
0.20
0.30 ~----,.---------
500 ~------------,
0.24
440
0.18 0.12
320
0.06 ·
260
0·00
0.00
0.04
0.12
0 . 08 t
0.16
0.20
200 O. . 00
o.os
0.04
t
0.12
0.16
0 . 20
At UA = 10 0000 J/min.K the 2nd reaction for C is totallv suppressed
P9-17 (e) UA
= 40000
J/rni.n.K
To =320 K
0.40 ~------------
900 ..--,--·------------,
0 . 30 \
780 • Ca
. e,
0.20
•
660
Cb
540
0.10 ll.00
1) ------------1
420
1'-11
-O.lO0.00
0.04
o.os
t
0.12
0.16
0.20
300
0.00
0.04
0.08 t
0.12
P9-18 Semíbatch with parallel reactícns Mol balance ;
dNa
-·d; · = Fao +· (rl + r2). V
~- = Fbo+ (rl+r2). V
dNd . -· -·-==-rLV
dNu
dt. == -r2.V
dt
Rate laws :
-rl:;:;;kLCa
-r2
9-49
= k2.Cb
0. .16
0 . 20
Na Ca=----
Stoichiomeuy:
V
•
Nu
Nd
Cd=-·V
Cu=--
V= Vo + oo.t+ ub.t
assume Vo=O
VQ::;;;-
Fbo th::::; -···-·· . --
Fao-
Cbo
Cao Cao
= 5 mo.lldm3
Cbo = 4 mol/dm'' Fao
Fbo
where W == -········- + ·· ····· Cao Cbo
V= uo.t But Fao
V
= Fbo (equirnolar feed)
uo;:;:::0.45.Fao V = 0.45Fao.t
Energy balance :
dT
Q · · · Ws · ·
L n« Cpi.(T-Tio)
dt- - ···- ·-· ····--· . . . . . . . .
LNi.
+ (-AHrx(T)).{-ra. V)
Cpi
where Q=Ws=O Af!rx(T) = .ó.Hrx(Iref) + tlCp(T-Tref) ACpl = :f... Cpd . · .?...Cpb-Cpa a a
= 50-20--30 = O
L'lHrx 1 (T) •::z ··· 3000 cal/mol A ACp2 = .1!.. Cpu · -~-. Cpb ·· Cpa = 40 - 20 - 30 a
Afirx2(T)
a
= · · 5000 ·
= -· 1 O
1 Off - 300)
í.:.Fio. CpL(f·Tio) = 2CtFa0.(T-Tao) + 30.Fbo.(T-Tbo)
:ENi.Cpi = NaCpa + Nb.Cpb + Nd. Cpd + Nu.Cpu dT dt
Faotl' _ 1~o)+ 30.fbo.~T-~:!.~~?.t~_?..~>.9.::.L:!LY1 . ~J:S?OO +_IO(T :-_300)](-r2. V} = -·20 -· · ~·.. --· ~· · ·-.· · · 20. Na+ 30 Nb +: 50. l·ld + 40. Nu
Selectivity
s- rl---r2
P9-18 (a) 9-50
The selectivity is proportional to CA/CB and therefore a small concentration of B in the reactor is preferable. To maintain a low concentration of B it would be beneficia} to run a sernibatch reactor where B is slowly added to A.
P9-18 (b) Let the rate Iaws be :
-r2 =k2.Ca
- rl = kl.C'b
Now it would be best to slowly add A to B in a semibatch reactor.
Let the rate laws be :
-rl -rt
= kl.CaCb
- r2;;;;; k2.CaCb - r2;:;;; k2.Ca
= kLCa
-rl =kLCb
-r2=k2.Cb
For these rate law combinations a sernibatch reactor will not improve selectivity.
Let the rate Iaws be :
· · rl = kl.Ca.Cb ·• rl =kl.Cb
-r2 =k:2.Ca - r2 k:2.Ca.Cb
=
Now it would be best to slowly add A to Bina sernibatch reactor to maintain high concentrations ofB and low concentrations of A.
Let the rate laws be :
- rl
- rl
= kl.Ca
- r2 = k:2.Ca.Cb -r2=k2.0>
= kl.Ca.Cb
Now it would be best to slowly add B to A in a sernibatch reactor to maintain high concentrations of A and low concentrations of B..
P9-18 (e) AHrxl ¡¡,..··-.. •·•"'''''"''''"''"'-•'"'"'''" .. "''w,
cal/mol A ..~"•--"-··-•""''"''"'"
3000 •••·--·•-•"w_,_.,.,,.--~'"'°"--
.. "'ª"'•""
Afirx2(300KJ cal/mol A ,.,...... ~-.
•·""'--•·--·"~---•"'"
·"''"·~·•,.,-
. -----·~®-•. •••-•• . . . . . . . . . . . __ !OOQQ_.______
~--------
,.._,
__
.....
l 000 3000 3000
""'"""'"M'"'
.
5000 .. ,..,--.-~--
_,,,,o-•·"•"·--
.. ,..,.,,,. .. ,.,,.,,,,.., . .,
,~,-
~-··"
·-"-"""""'
,
,_
5000 5000 ·,.,=·--·· .. ,, 1000 iooco
,.,.. --•···--
.. --,·
5000 --·-··· _ _
Temp,
K
362 ,.
Selectívity
,. ...... ,,_.
_. 399 --
__
7.9··..,-······--. -----· 6.1
484
323 354 375
s
,__;;.,,,
¡.....
-1
__ .
3.9 __ 11.0 8.5 --·--·7.2 __
If AHrx2 is constant, then if !\Hrxl increases, T increases and S decreases. If AHrxl is constant , then if Affrx2(300K) íncreases, T increases and S decreases. S is dependent on r l and r2. which depend on kl and k.2, which depend on the temperature. The greater the temperature in the reactor rhe smaller the ratio between kl and k2. hence reduced S.
P9-19 Plotting the data oft vs. T gives the initial and final temperature ofthe reaction, T0 = 52 . .5 ºC and Tt=1668 ºC.
9-51
200 160 120 80
40
o o
s
16
t
40
32
24
Recalling equation (8-29) with .1Cp = O, X= 1 and T = T r gives:
Mi R x
= -18
cal
moi-K
(166.8-52.S)K
= -2057.4
cal
mol
Make a table of t (min), T (°C), dT/dt, 1/T (K1). ln(dT/dt) and plot ln(dT/dt) vs 1/K (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-I 9. pol
Model: !
lnTdot = ao + al *T_inverse_kelvin
..
...
.
¡variablejvalue
¡ rn.88162
r- ··-·· -
¡ao
Ía1
··r
,.. .
. ...••....
..... 195°/o confidence
T.......
¡2.425175
.. .
· · ·r-6224·:af6J~so:s96 . .
Statistics
1~~-2.-~:.·_Tª··-~7is_~~.~•1
RA2adj ¡o.9720758 i1 1Rmsd í0.05979881
1 ¡····
,
!varia~i~Jo.::a,~ 13783 j Remember that Ts is the self-heating rate, which occurs after the onset temperature is reached.
lnTs = --6224 +18.88 T E
6224 =-'R
9-52
E=6224K·l.987
cal
cal
=-12367mol·K mol
P9-20 No solution will be given CDP9-A CDP9-B CDP9-C
Adiabatic, batch, reversible, Design equation :
dX. N,i.o---· = ·--n V dt
Rate law : Stoichiomerry : Cornbining :
Parameter evaluation : E=
ki Ln - .. kz
0211
R.Ti R.T1
1· ... _
,_
,
1
In- ------- ---1-0324
~......... ..
8..314(340)
· •· = 8498J I mol 1. . ... 8314(300) .. - . .
.ó.Cp = 2 Cp8 ~ Cp.,.. = 20 -- 12 = 8 Jzmol.K Mm:
= .ó.Hrx (300)
+ ó.Cp(T - 300) = - 75000 + 8 (T -- 300)
9-53
NAo == 0.6667
* 900 * 254 ==
272 127 mol A
N10 = 0..3333 * 900 *454 = 136 064 mol l
= NAo /V= 272127 / (50 *28.12) = 192 mol/ Cps = 10¡Cp¡ = 8ACPA + 8s Cps + 61Cp1
CAo
dnr'
= (1)(12) +O+ (136064 /272027)*(15) = 19.5 Jzrnol
l-
Afirx(300).X] (-~ 75000).X T - To+---·----.~- 300+·-· --····· Cp« + LlCp.X 195+ 8.X
Energy balance :
~
POLYMATH tr-x: - (4•caó• txA~U
d(x} /d( el :zkl~ (
o
/Ke} l
To,.,300 Cá0=<.192 T"'To• (7$000 .. x/ {19 .S+ {8":x:}))
kl~, 217"exp ( 102'2 • ( l{:14.0«l/T)) Dh-.r-x=-75000+ C 8'" (1'=300)) Kecz70000 • e:x:p { Ol:tr:x/
a.
31,4 .. { 1JJ O 0-1./1' l )
9-a !ni:::iaJ
Variable e:.
ca.o
V<.\J_ue
Max,:..num"_:::alu~
l:tinimum Vjelue
Fina.l
D
2
o
o , 044.9,57.3
.300 192
300
o
2
O
o.
aco
JOO 192
192
0449573
valu~
192
T
s cc
aoo
0.145345
D.49785
0.145345
0.49785
Dhr-.<
-75000
-73641.B
-75000
-73641.8
Ke
700{)0
70000
l. 62532
l. 62532
469.781
,s:i.n1
3 ··3
"".::::
j
ccc
1
.•..
1
- l .1
!
'
1
1
! J
.;.¡ V --: .. .:ce -----· -·-"·•-·h•---··----~j
c.coc
-···-·=-0>=····-·
,,.
• . _----···-4·,.,
: 2::c
9-54
+ .. ~•--·
,-.-
~
.
B-·a
'\8{,LC0'::),...
1.
¡;e~,
t
... ·r
1,ccc::i .
·-·-··--··----------~""""'"'"'-......
.,.. ...--··-...····-·-----------····--··----··· f
l.
~cc,t .1 +
1
J
,.,.oc,t J J
I
¡¡
J20~,occ-+f
1
2sc.oco-;.j~~·
••·••--t ·
Equilíbríum conversión , Xe
90 % of Xe = 0.9
.. ~ .. ·--•·'""'""'""'""'•'f"'"'""•".•-•-·•··~•· ..
o.~oo
o . cco
o.. .ioc
-·•u+....... c . ,_.,._......,- . ,.,. .. i·"'• ...... :.200 i..soo
...... .-.,._.,,,,,,¡
zcnc
= 4.5 %
* 4.5 = 4.05
% conversión
Time for this conversión "" 0.11
min = 6.6 s
Check: (:'Ao(I · · Xe)
at equílibrium :
4.C..w2X2
= ····--··--·Ke
(4.Cw2).Xe2 +(Ke.Ci0).Xe-Ke.<.7Ao using the solution to a quadratic : where at T
Xe -
-
_
-Ke.Cw±- ÍKe 2 ~cAi--16.Ke.(~i · ,, . ~- · ---
Xe - ·---
= Te = 470 K,
..• (1.625 * 192) + )(1.625)2 -
=O
8.C.w·
Ke = 1.625 mol/ dni3
09'2) ···Úi* (1.6 ;·(192)3.] 2
- _ -···-·---·-·-·--·· . ··· 8 *(192)2
From POLYM.A..''Til , Xe ... 0.045
CDP9-D
9-55
0.045
CSTR startup CSTR starrup, gas phase, isobaric - no pressure drop
Part (a)
dFa
___ _; = (Fao - Fa)+ ra . .V
Mol balance :
dt
dFb ---· - = (O - Fb) ··· 2.ra.V dt
Fa
ur = na + ub
Ca=-···
VT
Fr
vr = 1110---·
T
..
Fro To
Fr =Fa+ Fb
Fro == Fao + Fbo V = 1>-r:t
Energy balance :
st
dt -
't
Q··· Í.,Fio.C'pi.(T ·· Tio) +(··Aflrx)(--ra.V) .
.
L,Ni.Cpi
UA ·-· · = 51 / kgcat.s.K pcat UA
= 500 s
. ...
but kg catalyst
= 5 x 50 = 250 J I s.K
ACp=O
40.Ca.V +20 . Cb.V
9-56
= 50 kg
No control:
Ta = 300 K, Tío ::: 450 K., Fao = 5 mol/s !nitial value
E:9.::ta t. ion s : d{faJ/d(cl~lfao-fal•[ra·v1
le-OS
d(::bl/d(t)=l-fa!-(2"'ra"V) é.•,T) /d{t.} :;;( ;250• (Ta-T} }-t,W*fao ... (T-Tio) l
+
(20000* (-::::a"V} l) I ( (
40'"Ca"V) H20"Cb•VJ l ':'a=300
Tio=4.50 '!'o;:,450
cau=SOO Tau=SOO k-;a:;exp < tnAoo, a. 314 ¡ •t !J.J 450 i - t 11t1 lI cao::a0.25
C::a,..fa/v Cb=fb/v V=v"t:au
ra~-k"Ca
9-57
O 450
o fa
le-03
o
30 Q_Ol06221
O le-OS
)Ó
O 44S.6J4
299.098
0.00969749
450
299.098 451_548
:fa.o Ta
s
5
5
s
)00
300
T:..o
450
esa
300 450
fe
le-08
299.107
le-08
JOO 450 i99.l07
To
450
450
450
450
soo
500
500
SO-O
sao
sao
soa
sao
l
1.0292
0.921054
l.02512
0.25
0.25
o . .25
.$
5 20
Ta.u k
a.2s fto
5
s
VO
20
20
20
1200.55
4e-08
V
o.zs :ra
451. 548
1200.SS 8.07756e~06
0.251181
8.07756e-06
o
0.2500')8
O
o. 2491.JS
2e-0.S
600273
2e-05
$00273
-0.25
-8.J1.J39e-06
-0.25
-a
.31339e· . 06
--.52.acc-, 1
Part (b)
Mol balance ;
dCa
(Cao --Ca)
··· · · · · · ··· = --· · -· · · ····-----·dt 1:
Cao e 0.1 krnol/ui3 = 0.1
+ ra
9-58
---·--------------------------------
mol/dm"
P9-IO cont'd dCb (Cbo- Cb) --· == ·--·--+ ra dt
Cbo == Cao
-¡;
dCc
(O··Cc)
-c=50s
-· - -·-·-- --· ··• ra dt •
- ra = k. Ca.Cb
Rate law:
k(T) ==
dT
Energy balance :
V = u, . 't = 2
O.Olexp[-.~ ~-(-L . ... 1.987 300
Q-· Ws ······
LFio. Cpi(T ··· Tio) + (--Mlrx)(··
dr =
X
!.).]
T
ra. V)
LNi.Cpi
50 = 100 dm3
Fao = Cao . 1>0 = 0.1 x 2 = 0.2 mol/s == Fbo (equimolar feed)
= 2 x 0.2 x (15)(1' - 300) = 6(f - 300)
IFio.Cpi.(T-Tio) AHrx(f)
= tllirx(To) + L\Cp.(T -To)
but L\Cp
= - 41000 - (- 20000) - (- 15000)
=O
= - 6000 cal/mol dT -=dt
6(T - 300) __+.. ( 6000). ( +ra. V).. 15.Ca.v+ 15.Cb. V+30.Cc. V
~gµation:s: d(ca} ld(t)=(
Initi.Al valu.e (0.l.-c<\) /50)
+t;>a
0.001
d!cb)/d{t)•{{0,1-cbJ/SO}•ra
0.001
d(cc)/d(t);C{O-cc)/50}•ra
O·
dfT)/d{t)=C(,•tT-JOD)J•!6000~t-ra*V))}/l(lS•ca•v)+(lS*cb• 300 V)+{JO"'CC*V}) V<:lOO
l<:=-0. 01 *exp { (10000/ l. 98'7} • ( ( l/JOOl •· { 1/T))
)
ra .. -k·•c:a' .. ,cb
t
o
400
o
400
ca
0.001.
0.0732615
0.000l0274
0.00010274
cb
0.001
O. 0001021'4
0.00010274
ce
o
0.0732515 0.099864
o
0.099864
T
300
251380
257380
V
100
100
300 100
k
0.01
1892130
0.01
-le-OS
-le-08
-o , 0110 071
9-59
100
rasaao
-o. 00199'796
J
Sc::4\e,: KtY:. -T
10
.•
s2,;,oo
T
z. 100
+
t.5CO
o.. seo
í
T1
T ,.Tr
1. .. .,--' ,/
0.300 ,., __ 300 .. ' -
•
J
/
-_ ·_~------="'-==::. ·------------------- ------
---=--------~-.-_-_-_-_--_ _-_-_-_-
.. r_,
_-_-_---·---
..,...
e.eco
··-·
-
- ::o:!:
:í
/'s
...... t-·--··--··+···-··· i
sc.cco
-:
!_(Y: ..
/
~·
se.eco
,.
· ···++-"""" -····-··-·--··+··· . -··-·· . ·<-·-·...-t e.eco azo.oco 100 oc
2 ...
----·· -·····-··
f
;/
j
1'1 f. ·------~. . ----==-'"'--
1_.•.1 /
b
c . co::
¡
l
sc.ccc
J2c . :cc
i sc.ccc
9-60
---
-:-----------
--.----~----
----------.
---
--
..
~-··.--_-
:·-··-----
,,
.
.
_-
.
--·-:------
.
-----~
Solutions for Chapter 10 - Catalysis and Catalytic Reactors Pl0-1 Individualized solution Pl0-2 (a) Example 10-1 (1) Pentane isomerization nP
iP
Pt
Assume that Pt is the catalyst used. Maximum f = 5 molecules/site/sec Minimum f = 3e-3 molecukes/site/sec Maximum rate:
=r- =
/D(-1-J MWP,
%Pt 100
=r» =5(0.5)(-1-)-1-=I.28*10195 100
mol s gcat
4
Minimum: -rp =3*10-3(0.5)(-1-)-1-=7.69*10-7 mol 195 100 s gcat (2)
so
2
1 +--0 2
2
~
so
3
No turnover frequency is given so this rate law cannot be determined by this method (3) H2 +C2H4
-~
C2H6
=JD(~J:~~
Assume Cobalt is the catalyst.
=r»,
Maximum: f = 100 molecules/site/sec
=r»
2
=100(0.5)(-1-)-1-=0.00849 mol 58.9 100 s gcat
Minumum: f = 0.01 -rH 2
1 ) 1
=0.01 ( 0.5 ) ( ··-·-=8.49*10 58.9 100
_7 mol
s gcat
Pl0-2 (b) Example 10-2 10-1
(1) Cv =
1
C1 1 + KrPi- + K8P8 K 8 = 1.39 K7 = 1.038
Pro= Y1o"Pro1al = 0.3*40 = 12 For 60% conversion Py = Pro (1-X) = 12*0.4 = 4.8atm P8 = Pr0X = 12 *0.6 = 7.2atm Cv = l . =-1-=0.063 CT 1+(1.038)(4.8)+(1.39)(7.2) 15.99 6.3% of the sites are vacant (2) X= 0.8
Cr.s = _CvKrI'iCT C7 CT•S -= C7
=
KrI'il+KrI'r +K8P8
(1.038)(1)(1-0.8) 0.2076 = =009 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,I'i1 KBPB KTPy ---=-+--+---r1
k
k
k
Pl0-2 (e) Example 10-3 Increasing the pressure will increase the rate law. P.2 -rA·-· __ T_ - Py l+KrI'r
lf 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 Elü-3.1 we see that when X= 0.6, W = 5800 kg.
Pl0-2 (d) Example 10-4 With the new data, model (a) best fits the data (a)
POL YMATH Results Nonlinear regression (L-M) Model: ReactionRate
=
k*Pe*Ph/(1 +Kea*Pea+Ke*Pe)
10-2
Ini guess 3 0.1
Variable k
Kea Ke
2
Value 3.5798145 0.1176376 2.3630934
95% confidence 0.0026691 0.0014744 0.0024526
Precision = 0.9969101 0.9960273 = 0.0259656 Rmsd Variance = 0.0096316 RA2 RA2adj
=
(b)
POL YMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/(1+Ke*Pe) Ini guess 3 2
Variable k
Ke Precision RA2 RA2adj
Value 2.9497646 1. 9118572
95% confidence 0.0058793 0.0054165
=
0 . 9735965 0.9702961 Rmsd = 0.0759032 Variance = 0.0720163
=
(e)
POL YMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*Pe*Ph/((1 +Ke*Pe)"·2) k
Ini guess 3
Ke
2
Variable
Value 1.9496445 0.3508639
95% confidence 0.319098 0 . 0756992
Precision = 0.9620735 0.9573327 = 0.0909706 Rmsd Vari.ance = 0.1034455 RA2 RA2adj
=
(d)
POL YMATH Results Nonlinear regression (L-M) Model: ReactionRate = k*PeAa*PhAb Variable _k _
Ini guess 3 1 1
a b
Precision RA2 RA2adj
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 notas 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 RA2 values. (e)
POL YMA TH Results Nonlinear regression (L-M) Model: ReactionRate
=
k*Pe*Ph/((1 +Ka*Pea+Ke*Pe)"2)
Variable
Ini guess
k
3 1 1
Ka Ke Precision RA2 R"2adj Rmsd Variance
Value 2.113121 0.0245 0 . 3713644
95% confidence O . 2375775 O. 030918 0 . 0489399
=
0 . 9787138 = 0.9726321 = 0 . 0681519 = 0.0663527
(f)
POL YMATH Results Nonlinear regression (L-M) Model: ReactionRate
k*Pe*Ph/(1 +Ka'Pea)
Ini guess 3
Variable k Ka Precision R"2 R"2adj Rmsd Variance
=
1
Value 44 .117481 101. 99791
95% confidence 7.1763989 16 . 763192
= -0. 343853
= -0 . 5118346
= 0.5415086 3 . 6653942
=
Pl0-2 (e) Example 10-5 (l)X =1-
1 ktk
(l+kdt) d As t approaches infinity, X approaches 1.
10-4
larga kll~
X
t (2) Second order reaction with first order decay. dX -=-rA-
, W
dt
NAO
-r A '=ak'C . A2 a= exp[-ki] I
_dX = Wk C2
dt
N
AO
AO
(1- X)2 exp[-·k t] d
dX =k(1-x)2 exp[-ki]
dt X
k
l ·-X = ,;--(1-exp[-kdt]) d
X as t ~. -, mfini iruty -·--
1--x
X
= -k
kd
=_}{d 1+ k/
/kd (3) First order reaction with first order decay dX Wk'
dt- =-¡¡-CAo (1-X)exp[-kdt] AO
dX
-
dt
= k(l-X)exp[-kdt]
tn(i!x )=:, (t-exp[-k,tl) t
"7 infinity X = 1-exp[-:,
]
Pl0-2 (f) 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 CA= 0.607 If tau= O.Olthe minimum CA= 0.5088
Pl0-2 (g) Example
10-7
(1) If the solids and reactants are fed from opposite ends,
da
kda
dW
ti,
-·=-
atW=We,a=l C 1
= kdWe
us
This gives the same expression for conversion as in the example. (2) Second arder decay 1
a=----I+ kdW Us
.s.: kdFAo 1-X
kC10Us ln(l+ k,1WJ U5
1.24 = (0.6)(0.075}2 (0.72)(30)
u, ln(l
+ (0.72)220?0J
u,
Salve for Us by trial and error ora non-linear equation solver.
U,
= 0.902
(3) If e= 2 dX F --AO dW
= kC 2
(1.,...x)2
-----·a
AO ( 1
+ eX )2
10-6
·---
-·····-····--
- -----
---------------·--·-
····-------------------·--
-
---·-·-·--·------------·----------·-·---
-----~--~----
( 2e ( 1 + e) In ( 1 - X ) + &2 X + l+e)2 X = kC2Ao Us ( 1 - exp [-k d We 1-X kdFAo U5
]J
9X
12In(l-X)+4X
+--=1.24 1-X
X= 0.372
Pl0-2 (h) Example 10-8
Uo = 0.25
0.6
0.6
0.4
0.4
0.2
0.2
ººo
2
4
8
6
z
10
Uo =2.5
1.0
o.o o
2
0.8
0.6
06
0.4
0.4
z
6
Uo=25
LO
0.8
4
0.2
,___-~-----~----·· _____] o.o O 2 4 6 8 10 z
Pl0-2 (i) For EA = 10 and Ect = 35, for first order decay we rearrange Eq 10-120 to:
J
In (1- kdofEd = Ed ( 1 -· 1 EA R T T0 1 1 =- Ed ( ·----
R
t;
T
J J
10-7
8
10
1
R -In Ed
1
1
Yo
T
= ---
1~ kdotEd EA
400
Yo
T=
=--------,---~~~ l+0.07948ln(
Pl0-2 (j)
1 ) 1-0.00286t
Individualized solution
Pl0-3 Solution
is in the decoding algorithm given with the modules
Pl0-4
w. s + I • SHTBA • s + s TBA•SHTBA + S
Pl0-4 (a) Surface Rxn Límited
eI•S =KC'C l ·1 . V
r,WJ
·-·=0 kA
(~ 'T8A•s
e e
= . .. ., 1BA K .,- -y-D
: : : K'. IRA C.
'mA
C
-v ,
. smce
K--l ;:; ;: K-nu D
10-8
r 1
········-··-------·-···-------------------·-··--··-·--
---
------·-··-----···--
····--
----------·-··------···-·-·-------------··---·---·----·-·-------------
-·-··-·-··-······-·--------------
. ···--·····--···----------·······-····----····--······-·-------------- . ·-------·· -
- --
·········-----·······-----
-
-·-·······-··---···---··-----------------·-·-·····-·---·-·
-- - -- -----·-·-··--··--··--·---·······--------·-·····--··---··--·--·---···---···-----··
Pl0-4 (b) Adsorption of isobutene limited
'Alll = kAi[cI -
e:~]
C,, •
Pl0-4 (e) Eley Rideal Kinetics
· CTBA•S1. [ eW eI•S -···-K--. s ~
rS =k S •.
Pl0-4 (d)
10-9
--------------------·--------------··----·--------······-------------·-·
------·-·------------------··------------·------------···------------
-··-----------·--------------·--.---------·
.
·----
--·---·-··-----------·-----------------·-----·------------··-----
---·-·-------------·---------------·
W•S2+ l•S1f4TBA+S2+S1 r
s
=k
S
e e .
[
W•S2 I•Sl
cy1=.vz]
-~l.BA
K5
Cn == CVl + CI•Sl
en == CV2 + CW•S2 cl•Sl "'1S el CVJ '
CTI
= CVl ( l+rs e¡)
cw.s2""Kwcwcv2. Cn=Cvi(l+K...,,.Cw)
Pl0-4 (e) Individualized solution Pl0-4 (f) Indivídualized solution Pl0-5 (a)
Ci\
1\ + C2H4 ~ H+E~A Eley Rideal E•S~-:#A+S
E+ S
H
E • S , ':E•S = KE PE Cv
rs = ks[ CE•S l?H] CT=Cv+CE•S
,.1.....,. ·~ =ks ~
r
P p
ci_.1+~i;_
1
Pl0-5 (b) Indívídualízed solution Pl0-5 (e) 02 +2S µ 20·S
~+2Sµ2A•S
C3H6 + O·S ~ C,H50H ·S C3H60H •S µ C,H50H + S
B+A•S ~ C·S C•S ~ C+S
=r, = rs = k3PBCA,s
¡
i
r
2 C1.s] rAD =kA [ PA,CvKA
10-10
rAD =0
kA CA·S = Cv~KAPA
=r« = rs = k3P8Cv~KAPA Tc.s
= kv [
<: - P~:v] = kv [ <:
-KcPcCv]
rc.s =0
kD Cc.s
= KcPcCv
e; = Cv + CA·S + Cc.s = e, [1 +~KAPA -rB
= rs =
+ KcPc J
k3 CT PB .[if;i; l+~KAPA +KcPc
Pl0-6 (a) A (butanol)
¡:t B (butene) + C (water)
Possible mcéhan:ism:
r., = K -~·
A·S,t·S
C·S
(·.pe
AA·
.4
s
····---~.-A·
K.
s.,)
M
¡:.± B•S+C·S
rs "" ks (CA.s Cs - Ca.s Ces Ks)
~
roe = k1x {Ces- Pe Cs/Koc)
C+S
Assume surface reacrion eontrolling;
, _ P¡:¡ Cs _ p K (' (··B·S --K. ··- · B AB ··S DB
e-c-s == -K··Pe. -Cs = Pe K' oc
e
AC -S
10-11
j _~ ks K,,/-s-.2(.p .
' ( .. ,t -- P¡;PcC; -r; - rs - ksl PAK"-l's -- K K K \
S
;: _.......
K
eo
h.
.
w ere
K - K". K v K eq -
s
AAn.oc
ks KAA Ct (PA ~ ?¡¡ Pc/K.,q} ..
.
......... --·- . ·--·---·
·-·
{ 1 + P A K.A.A + PB K.A,B + Pe KAcl2
If PBo
= O and Peo = O
.; = ~2 K.Mf.i.~M.,,,, A.O
DC
!ziPc jl
CT = Cs + CA.S + Css + Ces "" Cs (1 + KAA P A + KAB PB + KAc Pe)
site balance: : • "·rA
DB
A --
{l
+ P AO KA.4.}2
, then l + k2
1:.1 Peo
-·
Pio + k3 pAO
This is consistent with the observation,
Pl0-6 (b) From the figure, Poim number
2
3
4
5
6
-.:AO (J~_¿u... ¡~} hr- !b car
o
0.275
0.5
0 . 77
0.77
0.5
PAo (atm)
O
4.5
27
54
112
229
405
6 . 45
8 . 14
l2.06
21.4
10-12
DE
Ar Iarge P..1.0: -r~0 = ~!l'..i.o k2 P~o Ar small P "º:
.;AO
= k1
0.06! + 5.34xl04
P..,o
k2
--·-;
P;,,,o + k3 P AO
k3
= 7.05xlQ·2
Using point 4:
k3
= 3.19xIQ·2
Using poinr 5:
k3""' L05xl0··2
Using point 3:
2:5.0·-:: -
= (229) · ·
; using point 6 : ~
so
~ k3
(O.S) = 114 ·· ·
·
= 0061 = k2 :::i: 5.34x104
PAo; using point 2: k1
-r,,.o = --·l
=~ {--1-) k2 P AO
k3
= Q~··· PL. 5.34x104 -r
PA
AO
AO
= 3.19xI0·2
(The
reason for tite differenr
values of k, is from reading the graph)
}
y .. 4.3103 + 0.073937x R= 0.99583
20.0 -
"'
.
"'::::: is.o: ""< .., 7
:::;:
" e,
i
10.0·-~
i g/
5.0
lntercepb1!k0·5
=
4.31
Slope == KA/ko.E
=
0.074
k: 0.054
;¡,("·
"-'
KA::: 0.32
.;
.¡
0.0--·t·,-,
~,
o
...•
30
--,-,-','
100
i
•..•
p 150
-,
·.··.s.··~\
200
250
a
Pl0-6 (e) Find the percent of vacant sites
Pb and Pe
= O so that reduces
to:
1 1 . -----·--- -:::: 0.41 1+ KAAPAo 1+0.01596 * 90
%vacant = -·· · · · ----·--·
Find the percent of sites occupied by A and B. No B will have occupied any sites at X = O. So: or A. •.•. :::;::. /,i.i\:::;; ·-·····KAA. · . -·-· P..:....
1 + K.-'\AP A
0.01595~90 : 1 ·t-0.01595 * 90
-:::: O .. "'9 ::)
Pl0-6 (d) lndividualized solution Pl0-6 (e) lndividualized solution Pl0-7 10-13
ME
DME + H20
·--+
. . . ....... __
---
100
200
300
Toe rate of formarion ofDME is greater initially. This is a resulr of more vacanr sir.es beíng initially available for react.ion beeause water is not adsorbed on the sises, As time goes on
the equilibrium concentrarion of warer sítes is reacbed, Water is strongly adsorbed oa this caralyst, Probable Mechanism
2iYfE • S -..:, W • S + D:/víE + S
w.s ;.:
W+S
Assume Surface Reaction Controls
rs =ks C~B CME.S
-= KME PME
Cw.s ""Kw
CV
Pw· Cv
Pl0-8 Gíven: Kineric rare expression for i:.he reduction of NO over a salid caralyst; Pz..c = parcial pressue NO Pe
= parti.al
pressure CO
Assume that overall reaction is of the form
NO +CO . .....,.tN2 +CC}z
Pl0-8 (a) 10-14
It is seen that neither N2 or
COi appear in the denominaror, This infers that neither is
adsorbed on the catalyst, On the other hand, it can be infered that both NO and CO are adsorbed on the surface. The squared denominator suggests a dual
síre surface reactíon of
the adsorbates ofNO and CO. Therefore the following mechaaism is proposed,
P... ; ;: P.110
kA,,""1'
Pe= Peo
NO(g) + S ~ NO· S ka{
··ro,¡
CO(g) + S ~ CO • S
ks NO • S + CO • S --,.
= k<:N [Pe e, - Cco.s/Kc.N]
1 N2 + C(h + 2S
With the surface reaction controlling
Cco-s = kc:N Pe Cs
= Cs + CNo.s + Cco,s = Cs (1 + KAN PN + K'.cN Pe] and therefore reaction is .. rs = ks C¡..¡o,s Cco,s = ks () PN Pe KAN KcN
Then Cr
or -rs =
!5s..~~ ..Ko!. !'1t?s: . gL
[1 + KA."'I Pi..+ N.:N Pcf
with
et Km
k.1 = ks K1 :z:Km
KcN
K1=KcN
. . rs-
k1 Pe PN" - -~~---
_.
[ l + Kt PN + K2 PcJ2
Pl0-8 (b) Assume that Pe>>
flN. Then Pe changes ver¡ litde during the course of the reacdon
and remaíns constaru. A maxirnum in (,rs) then occurs, for a fixed value or PN ar;
~:rs) dPc
= ·-.
k1 PN ··---- _. 2 k1 !:5:.PN K, _ = 0 [l + K1 PN + K2 PcJ2 (l + K1 PN + K2 Pc]3
10-15
Toe rate of reaction will increase with an íncrease in Pe until che above value is reached,
, afrer whích it will decrease. Ir appears thar there is an excess pressure which will minimíze reactor volume .. Operanng ar excess pressure greater
man this
valúe will deerease (-rs), and
hatee increase V. This analysis is exact if the catalytic reactor is a CSTR. lf che reactor is
treated as PFR. the crirical value of
Pe is oníy
approximare, but me general observazion is
qnalitarively che same, This
analysis funher assumes mat me excess CO can be eliminared easily and economically
downsream frem the NO converter;
Pl0-8 (e) The conditions for which the rate law and mechanism are consistent are the following. The CO S surface reaction must be the rate limiting, P cJP No rnust be small, The mechanism must be a dual site rnechanism (which it is).
Pl0-9
Methyl eihyl ketone (MEK) is an importan; industrial solvent that can be produced from the dehydrogenation
of hutan-z-ol
(Bu) over a zinc oxide catalyst
Bu --,MEK+ H2 A )B+C The following data giving the reaction rate for MEK were obtained in a differential reactor at 490ºC. 1
5
1\iu: (atm) Pu, (atm¡ cat}
o
o
2 l
0.044
0 . 040
0.069
()
o
()
0060
o
10
0.043
0.059
Pl0-9 (a) Suggest arate law that is consistent with the experimental data From data sets 2 and 5 Data P8" (atm) f~w¡;¡.:{atm)
O
P11, (atm}
O
(mol/h x
cat)
0 . 040
0.043
we can say that an increase in Bu partial pressure slightly increases the reaction rate,
10-16
From data sets 1 and 5
Data Set -· -·· ---·-· P11u (aim)
(atm)
P1v1eK
l
5
2 5
2 O
-------- . · ··---- .. ,- ---···-·----~---··-----
···--------·--···-·--·-····--
o o ~:?.~~i_ . . -·-· .?:?~3.
Pu, (atm)
__ rME(((rr_~ollh x 2_cal)
we can see that the MEK partial pressure has little if no effect 011 the rate law. From data sets 4 and 6 Dcpf §.~t ... _ P8,, (aim)
. _ =-·-
4
_
1 1
P.weK ( atm] P112 (atm¡
. . . ,:;m;: (mollhxg
c,a) . .
....... .. , ..
~
-
6.-.
1 O
1
10
0.060
0.059
..
Is seems that the partial pressure of H2 has no effect on the reaction rate.
If MEK and H2 are weakly adsorbed (or not adsorbed at ali) we can proposed initially the following
,
-- r~ I
= ro, =
k1P1 ---L
l+k2PA
.
But, from the complete data set ·-·---·-·••"'"""''"'"'"''"'"'"'''"'''''"'''""""''''-•'••••--•••••••••••"'"'"'
'""""""""'""""""""''"""''''"-'""""'''''"""•"""''"'"'"""••--••••••••..,...••,•"""'"''"
"'••••••••••••••-••*•"'""-""""""""""-••••,o-·-••••••-••-•••-••••••••••••-••u•"••-·••••••••-
-.1·-------·-·--· - -~---··
_!__··--··-···· . ········· 2 ·-····----·_3_ 2 0.1 0.5 0.044 0.040 0069
•• l_)_at_a_S_et••• ••»••»»»• cat)
1 0.060
We can see that the reaction rate goes through a maximum o.os om
r 1 r
0.06 _j
!
,!::
]
0.05
004
1003
...
-
·j 1
J
0.02
001
o o 1
l. -
¡-1
¡
t L
1
¡
1
j
1
1
¡
1
05
1.5
2
Puu (arrn) '"
---------··-···
.•..•.
10-17
l
1
l
2 0.043
- -~--- . ···
1 0.059
Pl0-9 (b) Suggcst a reaction mechanism and rate-Iirniting stcp consistent with the rate law: One possible mechauism is the following one (1) A +S H A··S (2) A S
t
S
B·S +C·S
H
(3) B S H B +S
(4)C·SHC+S lf the limiting step is #2 (irreversible surface reaction) and the others are at Pseudo Steady State (#L, 3and 4)
· ·· r;~
= k2C,15Cv
e
s-; ····-···
J
A site's balance will yield
Therefore,
Solving for
Cv
e,,== J+PK,1
,.. l
Pn
/~.
K,
K1
-t---+--'-·
Substítuting the expressions for Cv and CAs into the equation for -r' A
r:i
== k.z(~ASC\,
= k~K,Pt\C~~
== ··(
k K.
·---,~
l+ P K ,1
1
'º
¡P,/.:1~---··::;·
!!c_ j-
+ 1:!L +
KJ
K.• 4 )
which for the case of weak adsorption of MEK and H2 reduces to
. ,l
·' -_ .. k-------· 2KJ',1Cf -······-·-·· t1 (1 + P K 1 )2 ¡\.
Pl0-9 (e) Individualized solution
10-18
----:----_
·--e---·---------,-.-. -·---------e-~.--------~
-·- .---------,---
.
.
-----------
Pl0-9 (d) First we need to calculate the rate constants involved in the equation for -r' part (a). We can rearrange the equation to give the following
JE-·Jr:+1(P,
which is a linear equation with slope equal to
YJTi·
and intercept equal to
y.¡r; .
A
in
Shown below is the linear
regression we did using the problem data
g 7
6<
5
'72
:)1 4
s ~
3
· (P¡t¡/í~;,)l'(J5 = 2.7298Puu R1~0.9997
.f·
1 3362,
+·····-··------ ··-··1---:7'~--.
2
o
o
0.5
2
2. .5
Pu" (atm)
Thus from the slope and intercept data mol
k, =0.56·
--·-·· h· gcat-atm
l
and
k2 = 2.04-· ·-· atm
Thus, 056P\ r , - r_r - ···----'"···-----··--··' .... --A B (1 + 2.04PA )2 The design equation for the PBR is
F AO --~-· = ·-r' dW A From stoichiometry (gas phase)
e, = c¡\({/;~-\-)%{From the reaction E= l + 1
P,1
=
l . Assuming isobaric and isothermal operatíon ami using ideal gas law
PAo(h})
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 Pl0-9-d.pol. POL YMA TH Results Calculated values of the DEO variables Variable
w
initial value
minimal value
o o
o o
X
1 10 0.0042521 560 2.04 600 -68.5622 2.3403948
1 10 10 560 2.04 600 -12.228142 12.228142
e Pao Pa kl k2 Fao ra rate
final value 23 0.9991499 1 10 0.0042521 560 2.04 600 -2.3403948 2.3403948
maximal value 23 0.9991499 1 10 10 560 2.04 600 -2.3403948 68 . 5622
ODE Report (RKF45) Differential equations as entered by the user [ 1J d(X)/d(W) = -ra/Fao Explicit equations as entered by the user [ll e= 1 [2l Pao = 10 [ 3 J Pa = Pao*(1-X)/(1 +e*X) [4] k1 = 560 [5] k2 = 2.04 [6] Fao = 600 [ 7 J ra = -k1 *Pa/((1 +(k2*Pa))A2) [8] rate = -ra
. d., t h at F\w lt shou Id 1 )C mennone
= .l)(.
mol . ;;: ;.oü ()..mol·· mm hr
---
1.0
0. 8
70 56
GJ
0.6 0.4
28
0.2
14
,o.o
o.o
4.6
Q
42
9. 2 \V 13.8
Pl0-9 (e) Individualized Pl0-9 (f)
184
23.0
o o.o
solution
10-20
4.6
9.2
\V 13.8
18.4
23.0
Now consíder the change in pressure: Stoichiometry :
fe
1-X -· ···RT PA =C A RT=C Ao ·--. p ( }+i¡{ o
º'ª( 1 + X ) . dy . . - .. = ·-···-
Pressure :
dW
2y
Use these new equations in the Polymath program from part (d). See Polymath program Pl0-9-f.pol. POLYMATH Results Calculated values of the DEO 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 J d(X)/d(W) = -ra/Fao [2 l d(y)/d(W) = -alpha*(1 +X)/2/y Explicit equations as entered by the user [ll e= 1 [2J Pao=10 [ 3 J Pa = y*Pao*(1-X)/(1 +e*X) [4] k1 = 560 [5] k2 = 2 . 04 [6J Fao=600 [ 7 J ra = -k1 *Pal( ( 1 +(k2*Pa) )"2) [ s J rate = -ra [ 9 J 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
/\ o/\ '
70 ,------------::---, 0.8
·
56
0. 6
42
0.4
28
0.2
14 ., •.. ,,~,·····"'
0·0 oi...o~-4~.. -6------9.-2-,,-, ~1J-.-s-~1-s.-.~-_,2J .o
o
.,,..-
\
\,.
W 13.8
18.4
1 ; 2 and 4, O< f) -c 1 ; and 2 and 5. -1 <)'
From
o.o
4.6
9.2
23.0
Pl0-10 (a) Iso-octene + Hydrcgen=-e iso-octane
A
+·
B
.....;.
C
Discriminarion of models:
Assume -r A ""' k G ~ ('~ For runs 2 and 3. O<
o:<
Perry's handbook, 5th ed., p. 4-8; the reactíon is probably surface reactíon rate controllíng. Mechanism I (H. Alvord):
+2 B.2 + 2S f'.! A+S
A·S+2B·S¡! C·S Hence,
A·S 2B • S C·S+2S
•~ C+S • Pc!Ke,J · -. r A ,,. --- .... _,,.. k f•P·-A ...PB . ,-.,-,,,, [1 +-KAPA+Kf
P'fr+KcPcP
. .,
Mechanism II (S. L Mullick): A+S ~ A·S· B+S
~B·S
A· S +B • S ~!: C • S + S C·S~C+S Hence,
. _
··IA - ···-··-.
. lc[PA
Pa- PcfK«.J - ......•..
···--··-·-·-·-·-·
,
- ..
[1 + KA PA + KsPB + Kc PcF
10-22
·---
----------·-----·-·-------------
From runs 2, 9
11. 12. P A
= Pa = Pe = P, a plot of -rA vs. P shows a parabollic
behavior, therefore we will drop the second terrn in the denominator for easy linearizarion. The readers can calculare ~
value by Gibbs free energy change in this equation (up to 650
K. the reverse reaction is negligible) The linearized regression rnodel is:
Using given 12 data peines to soíve íor these four unknowns: y
= 3.0 + l.42 P,.. + 0.97 Pa +:
1.42 Pe
Toe final results are:
= · -··· · ·
··fA
{l
-t
A P-B ·--0.1113 ·--··-- . -p. ;......-_._ . 0.475 P A+ 0.322 PB --;- 0 . 414 Pcf
Toe comparison of the percentage error berween the model and the experimental data are: Run
PB
l
l 1
1
2 3
1
3 l
4
5
6
l
10 1 2.
7
8 9
10 11
PA
0.2
12
0.1
5
3
l l 10 1
1 2
0.2
0.1 5
Pe
o 1 1 1
3
o o
10 2
0.2 0.1 5
r (exp.)
(~A!ll.)0 .5
ü.0362 0.0239 0.0390 O . .o351 0.0114 0.0534 0.0310 0.0033 0.0380 0.0032 0.0008 0.0566
r
ex¡:, ..
(calc.)
% error
:5.26 6.47 8.77 9.25 9 . .37 13.69
0.0345 0.0227 0.0410
-4.8 -4.7 +5.0 -4.8
17.96 10.41
10.26 3.54 3.54 21.02
0.0334
0.0120 0 . 0505 0.0302 0.00315 0.0380 0.00288 0.00089 0.0599
+5 . 7
-5.4 -2.6 -4.5
o
-9.8
+10.7 ..... :..L2 -
lsuml =59.2
avg,
= 4.9
Pl0-10 (b) Discussion: Toe readers
may check the valídity of mechanism L To reduce the
accumularion of error in calcularions, the readers should have used all data points and.
solved. ali unknowns simultaneocsly. To get the maximum informatíon of compíex kinetics of a reaction from the least nms, it is advantageous to do planned experíments snch as factorial design . c:,N. G . Hunter. and A C. Atk.inson.
Chemjca! En~neering, p. 159, June
6. 1966). A paper discussing chernical reaction rate equations from experimental data is in: C. H. \Vue Jr., Surnrner Computer Simularion Conference, Proceedings, 1975, Part l. p . 368. 10-23
Pl0-10 (e) c¡A
= ·-····
f1
0.2223 C~.o (RT)2 { 1-Xf / ( 1-0.5X)2
-·
-
-
·-;
+ CAOo ~T {(0.475 + 0..322 (1-X)} + 0.414X}1~
1- . .:,X
l
·
O,Jl 13 {RT CAor (l·.XF /{1-0.5Xr
-rA :;;;;:;--~---··---~-
.
Í1 :
+ C.:..o RT tO 797 . O 383 1
1-0 .5X
•
'
•
x.,¡2 •
.
1
PAo zz CAo RT = L5 atm FA.o · = ,,.... 5 mol. min = 1 50 hr h
CSTR:
w - . , _,
.., 150 X Ü.8-
_
_.
. \2 0.11L3(1.5}2(1 ·08}2/\1 ·0 . .5x0 .. 8¡
r 1 + ... . .LÍ .•_ .... (0.797 • 0.383 ¡ l · · 0.5 X 0.8
X
Q . 8f
-
l
,
J
W = 21.380 g:::: 2L4 kg
w=
1.soJ 0
X
0.1113 {1.5}2(1-Xr,/L0.5 Xf l 1 + - ........1 .l __ {0.797 . 0.383
r
L
1 - 0.5 X
TI
X)I~ 'j
10-24
W= . - ... .J~Q
0.1113
X
2.25
lx
G (X}dX
0
. 1.5 (0.797. - 0.383 X)J2 (1 [ l + -···-· ---------· G(X). 1 -Q~.X __
where
• 0.5 Xf
.
(1-X)2
~,
G{X)
X
o
~1
4.82
0 . .1 0.2 0.3
5.38
OA
8.66
05
11.00
0.6 0.7
15.03 23.15
0. 8
44 . 62
:1
6.13 ~
7.16
o
o
oz c..s
a..1
Using Simpson s rule. area under the W
curve
tz:
1O
= · ...... J.~º---·-·· X 10 = 6()()() g :::: 6 kg 0.1113
X
2..25
Pl0-10 (d) Consider the differential secuon berween L and L In - Out + Generation
r_J::4,L)A1 PcÜ · · · t/J)dL
r: FAo
= Accumuladon
= ··dlf
= N)L)
But: A1v(L)CJL)
+ dL
f·= TA pc(l ··
1·
A1v{L)C)L)]
= F~0(I
X)
$) rA {Pi}
Relati.ng the expression r A to X and L :
"" lf0 (l·X) 2
= }Fo (l·X)
Fe
= FAO X =
i Fo X
pA.,.pA..,1:.:X.Pr 2·X
F.,1
1 = -(2······ X) 2
p -- FA-
=:»;
1 -2
X.
Pc::=..X-Pr 2-X
LetP=P.r,
10-25
x
0.4
"
a.:r
c.,.
o:r
u
From the Ergun equation, we have:
AU the things on RltS are constant, except far p
p~p.(~h~:d
dP
o:
• dL
Z-X 2
!P o\11) \p J 0
'P ge Dp
L Ler L" =--·-
:
p·
L ~ •o ·= 2!. p~
·~
L-qc:.
where
~:- . . . L~O~)_ Po Po~ Dp
I
J:L.. .
3 t¡1
l\_lj q)
1 ~ 50
1 l
P Po
=
) •
· . men:
-t-r-:
( 1-4> )J:. + 1.7 5 G Dp
dP '>-X -. = =d.L 2P"
R>o•
o
l 1so
L75 a)
gr--<1>) µ_ ..
$=04
= 1,L in) (....Lf!.J = .L, fr \16 12 inJ 192
Dp
G
( 150 gmol_) (2 + 112} hr . gmol 4).:>.6 g .
AT
Tlr-{~1.\2 ft2
G=l~
o
ft2 hr .
-
4
12
J
.íbm.
T
p ;;; (3 ann) (14-?. psi'} p0
..1.--{---:lh-,-.-·)
= {FAo mA + f-Bo mB) =
1 arm
(1441 ft2in:}= 6348.7 -ft2 ~
Cr fü = (51.la..) J.:... lbmol RT
1-
-
-
-J:~-l~o¡)_{~--~~t--~·
¡o.7301 fr3atm . . \ lbmolºR, 1\[{200+273.15)(1.8}JºR
10-26
.
0.4
IÍ) ;::
Dp = (l
16
in} LLfL} = _1 ft \12in 192
&e= 32.114
G
lb:m....JL
lbr sec2
( 150 gmol) {2 + 112) , . .
= {FAo IIlA + Fso II1B) = . . . .
hr
AT G=~
p0
.l.-··{·--dL-) 15.,.6 g
gmol
T 1L{il..2.52J fiZ 4
12
.J.bm..
T
f{Z, hr
= (3 atm)(l¡:~i)(t~f~~,) = 6348.7 ~·
__
Po z. Cr
m ={l~in~i"lfi
(57 ..J.b'"7)(3 aun)
= (0.7301 ..ft3 snm~pmlL(:~ + 273.15){~.8).1 ºR lbmolºR,
p0
= 0.2750 l!.mt ft3
µ (3 ann. 200ºC}
= 9.4719 x 10·3 cp {t C.8, =
X 1()"3 + 8.6211 = µ = {9.4719 -. ---·---·-·· . 2 X
µ=o n. • ),Jo
~· He)
}
-· 8.6211 x 10·3 cp (ali Cs) X
10·3)
· ·
.
cp
PPROP
•
6. 719 7
X
· 4
lbm
l 0- --.· ·---ft sec cp
JfillQ_.z~. hr
02188
~n.
ft nr
(--1 hr •.•yz 3600 seer (6348.7 lbr) (ü.2750.ll2m.\ (82.174 JtlmfL}. (· 1 fr.,\ (0.4t (35 ft]
(1.546.9. Jbn:L) { 1 T
ftl
, X
fr2
hr
ftl }
. {
j l.50(1 - 0.4} 0.02188
1------· ·---. 1 -· - ft {l} L 192
• 0.4)
lbt sec2 192 /
·· ~r le) . P.~.. · · +· 1 75 1754609\ -·~
. \Ti..,· ft~ hr
10-27
where
K ¡>; ( l. -X\2 º .,_xJ -· '
= ;:---·--··"
RA
p*2
__ _.. . -
l I + (KA Po+ KB Po){Px-) p" + Kc P., fi\) P"J
R4 ·- ·-·"'·· .
p2
1.·
..
o¡
K~_12(1- X)2 _
-_!_.....,.... + ozp•z -
tro
+ K-s )(, ,...I __ !X ) -•· K( -~--y 1J · e\ \_¿·--2 ,~/J
(K
·A
KP2(1
f?._A .:.:; ·-
----·---------·-····-·--
X2)
0..... . .
l-2~~~ + (KA + K )i¡¡(l--
'...........
X)+
3
Lec a. Then:
•
LA. T p~ (1--,) K?; m---·-~ ·· ; F,...o
dX =a:-··· .. Jl X_i2
l p• f2-X
cl.L
v• .
+ ¡;.l
. •.-
.,
.
J ,
.
KcPoXj
K~ ={KA+ Ks}P0 _
K• X_,.,12
·1·
J
For a given T. we can solve the rwo ODFs to get X(L .. ) . We need to guess T unril X= 0.8. For the rate constants, we use K = 0 . 1118 -·· .. ~:::'.?! . ._. __ gcar, .. hr • a....~· Kc
= 0.414
ami
me solution of 6.10 (a) ;
KA"" 0.475
arm! ; KB = 0.322
1
,,,. 3 -- 2 .. 391 l\.l • (O .. 4·"', I.; .... O,.,,..,,,, ...)..:..-, , ~ ~ 1 A O ' 4·--7-.::, .... · () ·~,.,."""} ~,)-··l.. - • 149 K .2-- (O .........
a•
. "·y·,t)p\·"" . · 1 .. ·t-) K" ·•on!
= I,..r.
_. =
T (35 fi:)(3 . 14/4) \'Q,22.12• 7.'j2 frZ
F AO x
{o ... 1.,
1 ' .,
a:""' 85.958
gmol \ ·-:-··-· ·· ·;·l.. gcat nr aun- /
....
(2.6 . .JL.))
- _ ~ {·') 5 gmol) (fil mjn) -... rnin 1 hr
r·.,~ ann. 1
\2
f't'n)
(1() 4l ~ .. ~
n
T
10-28
atrn'1
f
O
= (~
l Q-4 ft:.ht.. ) (378.{)9 + .l.J..2.QI) jbn..
X
T
11:m ·
ft2 hr
T
¡3: = _ Q 2~691 _ 8 ~;55 A FORTRAN program is written to solve the equations. The results show that any numbers of tubes sufl:icient 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 different L, and it only changes the dimensionless parameters. With L = 20, the problem is still unsolvable. For L
= 10 ft
, T = 2.16 tubes
= _fp. - = 0.46 in
Note: Using the modified program with
Dp=rgin L
= 10 ft
, µ = 9.05 x 10'3 cp A= 2.15 in1
==>X= 0.80 • P" ""0.7537 Using l 1/2 in schedule 80 pipe (!.D.= l.5 in) • A= 1.76715 in2 and rhe length to get
X ""0.8 is ímpossible (Pr < O). With 2 in schedule 80 pipe {I.D. = 1.939 in) , A= 2.952877 in2 This gives L "" 6.67 ft . X l 1/2 in schedule 40
= 0 . 80 and P = 0.9173 = A = 2.03580
in2
L = l0 . 94857 , X= 0 . 8 and P = 0.6903
Pl0-11 (a)
10-29
Assume a rate .. limiting step: start with surface reaction A·S+s--·-B S+H S B
s·--s+s ... ----··-
H s-·-H+S ~·-~-, Assurne a rate-litrniting step: start with surface reaction
Come up with ways to find P A•S' P s-s- and P H•S
.:.~o.= O kAD
o
~BD.:::
kBD
-~t.i!2. = kHD
o
Find the expression for C,
Combine ali of those to get the following rate law suggestíon . r. _ ~ s
-.........
k,K;.P,/~ 2 .. .
_
-
(l+KAPA+K8P¡¡+KuPa)"
Checkinz to see if it fits, we see that for high P A' increases in P A cause decreases in the rate. \ve see that if P 8 or P H increase the rate will ge down, which is consistent with the rate law.
Pl0-11 (b) Now using POLYMATH's non-linear equation regression we can find the valúes for the parameters. We find that k = 0.00137 Kª = 4.76 KB 0..259 Kc:::::: 0.424 In the problem it is given that K,..,, is 1 or 2 orders of magnitude greater than K8 and ~: which is true so this is a good answer.
=
10-30
-f.000 1''11'10°
3.200 2."'fOO
1.600
o.acc o.ooo i'lodet: lo; l(.¡1
T
o ;;¡li'<;" "'l>!i'1. ddt a
I
tI ~.. ,.'_._llttJ l
!•I 1
~
2
=
3
"I
s
6
7
a
s
ocat;;~i.-t~d v at . . u?
10
r a=i<. •P.a.,.c l ~Ka•Pa ..Kb•Pb ..l(hll:Ptü-2
0, 00137022
"'
o
º"
Kb _. O.259382
'+.7609'\o
7 pos1 u ve
r·e>S1
t
dual s,
3 n-.,qan
v e- ,. eos1 d•Jal s..
Sum al
squa.-~s
"' S. 27033EP- t 5
Pl0-11 (e) The estimates of the rate law parameters were given to simplify the search techniques to make sure that it converged on a false mínimum. In real life, one should make a number of guesses of the rate lw parameters and they should include a large range of possibilities
Pl0-12 (a) Assume that the second reaction is the rate--limíting step.
Using PSSH, we know that
10-31
Perform a site balance:
Combíning all of these we find: k¡CsiH4 C\
kCsm4
1 + KCsiH4
1 + KCsiH.¡
t, - --------- - ------
s
This rate law is consistent with the data. As the concentration gets larger, the rate change gets smaller which is consistent with the rate law as gíven,
Pl0-12 (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.
Pl0-13 Assume the rate law is of the form
=
rD ep
At high temperatures K rDep
-1.-
kPirz~o 1 + KP V71PO
as T Í and therefore KP::.rIPo « 1
= kPV~IPO
rDep
--=k PirIPo
Run 1
_p..:028 = 11.2 (0.05)2
Run 2 0.4S (0.2)2 7 Run 5 ·2 (0.8)2
= 11.28 =
11.25
At low temperature and low pressure rDep
= kPirf PO
r
-.!l!!!_
Pir1Po
=k
Run 1 0.004 = 0.4 (0.1)2 10-32
= 0.375
Run 2 O.Ol5
(0.2)2 These fit the low pressure data At high pressure KPv~IPo » 1 r
Dep
kP~lPO k =KP.2 K
=
VTIPO
This fits the high pressure data At PvTIPo = 1.5, r = 0.095 and at PvTIPO = 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 ( ~ )-
! (:. ;} ! ( \~T¡) E(
l (11.2) 473-393 n 0.4 = R ( 473)(393)
J
E
-=7738 R
E= 15375 ca!_ mol
Pl0-14 r110¡, :;::
ks.fi.s
1 = fv + ÍJ.s
~1~: ,_
= K¡,
= fv( •
l + K,P¡}
Pt = KpPtm>/Pp1
_ ks Pr Kr 1 + P1 K¡
• u(h - --· -·-· ...
10-33
r.Ti01
--k --.
.¡ SJ J-•S
h-s = P¡K1fv
/i.s = l Pr
Pu=
P-hn,
Kp
:1:¡f
_ ks PI K1 ·1 -+· Pr Kr
rn0,. -
kP¡,.1p I
r¡;o, --
Low
High
P-rm,:
r;
kP}.¡rp
1:;r~~í?·l·l;:)i = ¡>;;·;:P:;;;K Rxn is second arder
Since
1 >> K Pfm(Pp¡
Pm:P:
i << K Pfr¡p/Pp1
Reaction is zero order High Temperarure Kr vr::ry small such thar
Pl0-15 (a) Using Polymath non-linear regression few can find the parameters for all models: See Polymath program Pl015.pol. (1)
POL YMA TH Results Nonlinear regression (L-M) Model: rT = k*PM"a*PH2"b Variable
Ini guess
k
1 O. 1
b
0.1
a
Precision
Value 1.1481487 0.1843053 -O . 0308691
95% confidence 0.1078106 0.0873668 O . 1311507
= = =
0.7852809 0 . 7375655 Rmsd 0 . 0372861 Variance = 0 . 0222441 RA2 RA2adj
a= 0.184 ~ = -0.031 k = 1.148 (2)
POL YMATH Results Nonlinear regression (L-M)
10-34
Model:
rT = k*PM/(1+KM*PM) Ini guess
Variable 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
k = 12.26 KM= 9.025 (3)
OL YMATH Results Nonlinear regression (L-M) Model: rT = k*PM*PH2/((1+KM*PM)"2) Variable
Ini guess
k KM
1 2
Value 8.4090333 2.8306038
95% confidence 18.516752 4.2577098
Precision R"2 = -4.3638352 R"2adj = -4.9002187 = 0.1863588 Rmsd Variance = 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
1 2 2
KM
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 R"2adj Rmsd Variance
= -3.2021716
= -4.1359875 = =
0.1649487 0.4353294
k = 102 KM = 83.6
Km= 67.21
Pl0-15 (b)
10-35
We can see from the precision results from the Polymath regressions that rate law (2) best describes the data.
Pl0-16 Using Polymath non-linear regression few can find the parameters for all models: See Polymath program Pl0-16.pol. (l) POL YMATH Results Nonlinear regression (L-M) Model: r
=
k*KNO*PNO*PH2/(1 +KNO*PNO+KH2*PH2) Ini guess
Variable
k--
1 1 1
KNO
KH2
Precision R"2 R"2adj Rmsd Variance
Value 0.0030965 57.237884 101.9967
95% confidence 3.702E-05 l. 0353031 2.2870513
= 0.9709596 = 0.9645062 = 5.265E-07 = 4.436E-12
k = 0.0031
KNo = 57.23
Ktt2
= 102
(2)
• POL YMA TH Results Nonlinear regression (L-M) Model: r
=
Variable k
KNO KH2
k*KNO*PNO*KH2/(1 +KNO*PNO+KH2*PH2) Ini guess 0.1 10 1
Value -4:-713E-06 -108.42354 l.046E-05
95% confidence 1.297E-05 4.9334604 2 . 878E-05
Nonlinear regression settings Max # iterations = 300 · Precision R"2 R"2adj Rmsd Variance
= -9.6842898 = -12.058576
= 1.01E-05 = 1.632E-09
k = -0.00000471
KNo
= -108.4
KH2
= 0. 00001046
(3)
POL YMA TH Results Nonlinear regression (L- M)
10-36
Model: r = k*KNO*PNO*KH2*PH2/((1 +KNO*PNO+KH2*PH2)"2)
Variable k
Ini guess 0.1 10
KNO
KH2
1
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 = 0.9767486 Rmsd = 4.262E-07 Variance = 2.906E-12 R"2 R"2adj
k = 0.000519
KNo = 13.19
Ktt2
= 18.49
The third rate law best describes the data.
Pl0-17 (a) Mole balance:
dX~" '"' =. (--ri)a --• --····--·..,,.,,•,w
dW
FAO
Rate law:
Decay law . : _da__ = __ k¿-C,.,_ dW u, Stoichiometry:
{1 "X) e - ·C·-.,,º·(l+EX) -···· ·-···--· 'A~
Bvaluate the parameters:
e=.8 POLYMA'Ill §.SJ:!ation:!¿_ d(x) /d (w) =a* ( ··ra) /fao
!nitial
o
a
d(a)/d(w)=-kd*a*ca/Us k.d=>9
fao=4000000
10-37
valué
Us=250000
¡
k=90
.ooc
(a)
Hovi:n.g--,Bed Reaccor
Kf;t:,:
-·- a
cao~~B
Ca
8 ca=cao* (l··x) f (l+eps*xl eps=
-
o.aoc
X
ra:-k*ca 50000
Pl0-17 (b) Mole balance:
/
dC,,,4= dt
"C +ra ·t kJ 14
z
Everything else is the saine and we need to know ,. V
r= · · · =0.004 1.\, 5 CS':t'R'a in
Egua!:l,2ns_: d(a)/d(tl~-kd•a•ca d (ca) /d ( t) "'ca4/t.au· ( d(ca4) /d(t)"ca.3/tau-· d(call /d(tl "'cao/tau· d(ca2) /d{t) =cal/tau· d (ca3 l /d(t) "'Ca2/tau· kd,z9
Series
( (l+ya.4 l I CHca/ctol l tt,au•a *k) *ca/tau ( ( (l+ya3) I (1+ca4/cto) l +tan*.a*k) *ca4/t { ( (l.,yao) / {l+cal/ctol) ttau*a"'k) •ca1/t ( ( (l+yal) I (Hca2/cto)) +tau*a'"k) •ca2/t ( ( (l+ya2) / {1+ca3 /cto)) +ta\1•a*k) •ca.3/t
10-38
!n:j,!:ia). value 1 O a O 8 O. 8
o. a o. 8
tau=O . 004
1 •. cco
cto=l k=45
. .. a
cao=0.8 ya4=ca4/cto yala=cal/cto ya2==ca.2/cto ya3,,,ca3/cto
---
{b)
5 CSTR's
ü1 Series
IJ,lJOC
Ca
~
• :::-- ~~<~~
X .
f("''··----~
0.600
x= (cao .. ca) /cao
yao==cao/cto to = o. tf
_
O .. {lQO
-;-~+- . · ·••+-·-·--f----·•·._
, .. C.CCC
tLJ.00
C,.200
... · ·-----:t--··~··-~"" · -~· Q,.;JQQ
0,'W'O
·--1
l},.$G(J
0.5
Pl0-17 (e) The only change from part (a) is the decay law:
<}_a_= -k aC dt
d
A
w .. w
t - -···"""'--·
lis
dt=~~'.~. Us da dW
kdCAa
----=-· U5
Integrating: ln a =
-~-tL~'.:i~ + k Us
W=Ww,x ® a= 1
k - _kdCA~l,fAX,
Us
a=
exp(!l?~{i; -~~~2.)
Eg',!,at,;,on~,;. d(x) /d(wl =a* k=SO
Ini;!al __ y~
o
( . ,ra} /fao (e)
fao=4000000
Moving .. ·Bed
kd=::9 Os==250000
..miax=SOOOO cao"' . 8 eps=.8 ca=cao•(l.-x)/(1.+eps•x) ra=-·k*ca a=exp{kd*ca/Us*Cw-'Nn!aX)) w0 O, wf "'50000
~
a
...
Ca
... -
X
=
Pl0-17 (d) 10-39
Reactor
(councexcur cenc )
To find the Time-Temperarure
Trajectory we need to use the following equation
for firs t -order decay . t
=
E f k~-oc:E:t 1
Í E.-~d " l.
"ll
exP¡_-;-l. T . .. 300J J. ..·¡
,
1
Since no initial ternperarure was given, we assumed one of 300K. Th.is is the graph of that equation Ternper ature-Tlme 1000 .,111 -800 !· ~ ---500 E f; 400 · ·
tra Je et o r y
... !
: _ :, 2oi • ~--·-:- . --·.:_= .
o
.
O .1 Time
J.·······-··-··-i
_ J
·-·
0.2 (h)
0.3'
Pl0-17 (e) The two energy of activations are switched and this is the new graph rnade.
Tempe:rature-Time
Tr aj e c t o r y :,
1000 .•
.
~ ,..-.,800
t
~ºº
o..~ºº E,... 200
41 1-
_
...,_
o . ---o
· ·-
]
..... ....
.
The graph looks the same just the time is rnuch smaller,
Pl0-18 (a)
1
10-40
1
f
For all of the parts, the mole balances and rate laws are the same, They are:
se,
---·=r dW
r,,
A
V
ac, O
=--·kaC,,,
dW = rsvo
ra =kaCA
Find the equation needed for a.
da
·· . ·-·=kdaCA
dt
a= exp(--kdC
At)
a=l when t=O Assuming valúes for vo, k, and k(l come up with the following graphs according to the cases described. Equations !. d(cb) /d(w) =rb*vo d(ca)/d(wl=ra*vo vo=l.O a=l kd=l t=20 k=l
Ini tial __ya! ue
o l. (al ··-·
Ca
Cb
_ .. a
O,"-'C
rb=k .. a ... ca 50
.U.X -~
,::b a
¡.-,.
:3.fY
.......
t•lOOO
-~- - - ·- . . .. ·-·- ----...
i::a=·-k*a•ca
-
Case I@
(~;:),
., ce
-a
10-41
-
.
Pl0-18 (b) Find the new equation far a: ·- !f..C!_
dt
=k
d
a2
e
i\
1
a ::::: ·····-··········-······ . . l+k¿C,1t Using same values come up with these graphs: Initialvalu!::
~~i.~.E ..ó
o
d(cb) /d{w) =:::b•vo
1
d(ca)/d(w)=:::a•vo
vo=lO
{b}
kd=.001 t=lOOO
Ca
k=l. a=l/
-·-
(1.+kd~ca*t)
a
ra"···x•a•ca :c·b=k·"a:t"ca l
_,...,...!
Ca
Cb ;a
10-42
Ca:se
I
@.
t::: 100{}
1.000
¡s¡;y Ca
Cb
-
a.aoc
a
Q,óOO
\
/T
(b )
Case
t \ -· ' / ·~
:
\
.'
\"
II:i'.'.
@
_
t.=lG·
!
+ ·/x· . -, \
::: k=-~- ,~.,: 1'-
. ._ · •
\
¡
o.ecc
e.sec
1.200
i.see
:c.:Ooo
a.coc»
Pl0-18 (e) Find the new equation for a: ___ da= k aC ·
dt
d
B
a= exp(-kdC 11t) The following graphs are made: Eouations:
"'-"------
Ini tia l. val!f§-
d (ca)/ d (w) =l:'a *vo
1
d(cb) /d(w) :.rb*Vo
o
vo=lO kd=-001.
(el
t:;;;1000 k=l a=exp ( -.J,d*c:b•t)
rb.:k*a"ca
KlU
......
Ca
.
Ch a.
-
.ria=··k*a*ca
10-43
Case t I t•lOOO
Case
1:
@
t.:::lúC
~-EX Ca
e.seo
Cb
a
s.oco
l.QOC
{e}
.KEY: Ca " C:b .a.
c.acc
d. 2GD
Pl0-18 (d) Making a into a differential equation we come np with this:
......4.!!:.. == kdCAa dt t;;::;
w
U,
____ da_::::: kJC,.a dW U,
10-44
Eguat.ions:
~tia!
d(ca)/d(wl=ra*vo
l
valu~
d(cb)íd(w]=rb~vo d(al/dtwl=·kd
1
.. a~ca/Us
vo=l.O
kd=.001
1 .. :;;00
Us=lO UY ····•·
Ca
Cb ... . _ a
:rb=k•a•ca :::a=··k*a"ca
T\
Case
{d}
rp. -t\
,·
I
... , .•..••...• ·... ,rttt-
.
·+· ~ i
\.'
l /\
1
U\ j :'
0.000
{d) 1.000
Cáse
T.L ~.•.
e.e-o:,
KEY
"'-. . ..
.. .... -.,t........... · · ··------:::::::::,--....::·::;=---'+; a.. .:1':io o. -.co c.600 C;,.•QCC
II
~.-·~ ~-
~~~~- -~~~-e:..-~~~~_. "",......
"'
~-~-~-·-·~~
Ca Cb
a
O.liOO
¡
1
¿
ii
t/
\
\
\
<,
r '..¡. . _.. ,-,~~-----·
(l,000
L.000
KlU'.
Ca Cb
0.200
id)
o.seo
-+-- ....... c.eec
c.see
Case III
~~--·····=--·---- . . . _. _
e.see
a
0.600
0.200
0--000
l
. . ·------ ....
___
f
-·-. .
-r
J. .
c .. :ooo
Pl0-18 (e)
10-45
.............
.... _,_~
1._00(]
.. ...... -4 1.0:-:,
Everything from part (d) is the sarne except for the decay law .
-~~ =r-ko C dt
a
·A
~ i~ttAX -·
t ~····"·····
u,
dt=> da dW
....,. ... *"~
'IV
dW .....
u,
k¿C,a
:;.:=: ·~
u,
·-
Integrating:
W=WMAX@
k
e= .. .
a= 1 kd(\
w:,tA,'{
u,
a "" exp[~~CA
t7;;· .!~it.. x)-]
d{caJíd(w)"'ra~vo d(cb)/d(w)=rb~vo
o
vo= l.ü
kd=l
Case
k=.001
~y_
wmax;;lQ
C.:;:
a=exp(kd*ca/Us'"
a
(w· wmax)
J
Cb
a..JJC;'.:;
1:·bmik*á.*ca
"'o "'
O,
"'f
10
10-46
(e) r.cco
e.scc
'.'.""'."" a
4
0.4Co
m¡:_ .. . ..
0.100
.)
ir
lt
.
~t
0...20t
-000(:
......... ;
t \ -·· .
é.000
K~Y ...
Cb
-· a
(é}
·•.OOO
,$,.OCO
11.000
Case !!!
t
e.seo
Ca.
2,00C
T ...
t.occ -
H
l/
~
Ca ..• Cb
Case
0.600
...,,,
c.soo
0.00!)
l
-.,.-·
... -.U:ufl's'·\I'. ---~ ·-·"-"-~-~--
0.000
·r--·~· - . ~ . vv ~~:: ... :-~: .:
~,,ooo
:2-G·OC
~ .• cec
.: . : . .: . . . . .
._. oca +·--~~:~
s.
~é .. occ
P10-19(a) da =-k
dt
W-Ut - s
V
_!!!__
a= 1 _ kvW
k W
U .5 then W=-5=-=2.5kg kv .2
lf 0=1--v-
U8
Us
dX dW
= -rA =~(-rA(O))=
dX dW
=
f
FAO
(1-
dX
(1-x)2
FAO
J
1-
akC10(1-x)2 FAO
W kC AO ( X )2 U8 v0
-~V
= kCAo Vo
Ji- kvW dW u,
Activity is zero for W > 2.5 kg, so the catalyst weight only goes to the effective weight.
10-47
---·-----
·-----·
- ········· ·--··--·
X -----
1-X
_kCAo[ W ---kvW/]_1(0.2)[ 2.
e
V0
2Us
1
.5- 0.2(2. .5)2]--0.25 2*0.5
X =0.2
P10-19(b)
o
2.5
5
P10-19(c)
Por infinite catalyst loading a = l. dX
= kC AO ( 1- X )2
dW
v0
_JI_ = kC AO 1-X X=0.5
v0
w=1
P10-19(d)
.s.: 1-X
kCAº[w-kvW2] "e _ 2U,
0.4 =0.2[5- 0.2*25] 1-0.4 2U, . Us =l.5 kg s
P10-19(e) a=l-kvW Us 0=1- kDW_ Us
10-48
=kvW=0.2*5=1·kg
Us
s
P10-19(f) a= O means there is no reaction is taking place. Activity can never be less than O.
P10-19(g) U=-Us . da = kv dW ti,
when W
= w; , a = 1
kW a=-D-+C Us
l _kDw; ---+ Us
C
a= kDWe +1- kDw; Us
Us
Nowfind We. O =-kDWe_+ 1---kDw; ti,
u,
We
= Us [kv~---l]=.:~[]*5
u,
kv
we =2.5 dX·=(l-
kDw;
dW
u,
___.J!:_ = ~~AO 1-X
.2
Vo
v0
a
+wJkCA0(1-x)2
\Ve
"'
v0
r( 1- kDw; + w 1,_:,w
Jl
x _ kCAo [(w;
1-X
.5
-l]
_
u,
f
-w.)(l- kvw; J+·w; --w.2] U¿ 2 _ 2
(1-
___.J!:_ = 0.2[( 5 _ 2.5)
1-X X --=1.875 1-X X =0.65
J
0.2 *~-) + 25 - 6.25 0.5 2
P10-19(h)
$ = 160FAOX -IOUS
10-49
\Vt
kDW
a=l---·
Us
dX =ka=k(l-~ºW) Us
dW
X =kW_ kkDWZ
2Us To maximize profit, a maximum in profit is reached and so we set the differential of profit equal to
O. · d$ =0=160F dX -10 dU s AO dU s dX - kkDW2
dUs -
2UJ
160F kkDW2 = 10 AO
zu:s
8FAokkvW2
= uJ
u, = ~8FA kk W 0
0
2
= ~8(2)(.2)(.2)(25)
U s =4 kg. mm
P10-19(i) No solution will be given
Pl0-20 (a)
Start with the mole balance for a batch reactor:
dX.,,, dt
··r;w NTO
Rate law:
Decay law: da , , ·- -·-- == "- adt d Sroichíornerry:
Plugging those into POLYMATH gets the following program and the following graph
10-50
Equatfons :. dlx)/dlC)•-rt•w/nto
o
d(a} /d{t}
l
~·· kd-j:a"2
w=S kt=20
¡
1 :::ic:;
kd"'i...6
v=l Ra:.082
r
+T //
:).&OC
T=735
pt.=pto• ( 1 rt:·-kc*pt•a nto=pt.O
/'
-:
.,,.~--~--
tí
··X)
TI !/ Tl
.. V/ (R*T)
k-·-····---~-----------· -
,_ .. ,
O .. Ct'lO
·.+-----· a.. ccc
' ,,..._..
Pl0-20 (b) For the rnoving-bed reactor the mole balance now becomes:
dX ··11 ....-.-.-. -~ dW
FAO
The decay law now becomes:
da - k,,a 2 -"' . dW Us
···-·----
Everything stays the same . Plug into POLTh!.t\T.H. The conversión achieved is X= 0.266
Eguas~~.:.
Ini7ial
d(xí /d(wl=-tt/fao
o
d{a)/d(w);;·kd--aA2/0s
1.
value
ktp20 fao"600 l
6
~!E.11!.E.1.~
"'
tJs-=2
50
pt,.pto* :i:·t
w0
ll··x) ...... kt; .. pc•a
=
O,
wf
fao
0 . 2SS1.6
2C
20
O .0243-SH!2 .2(;
500
i'iOO
tCD
kd
50
50
O 266;6
pto"'2
6
2
l
6
5
2
--0
7!.Si4!
45768 -ic
Pl0-20 (e) e) Increasing U5 will ge; usa higher conversion, Looking at this summary table, U; = 10 kg/h and X 0.6
=
10-51
2C 600 l ¡j 2
2 -·-40
C .024.HC:
!'.!!iah1~
!riicial_v.tltM
o
,.i
o
:ao
600
~'lirn.'1.ll'tl
l
O. 1.11111
so
a X
l-m::
o.
Sft%sa 600 r,6 10 20
l. 6
10 20
o
fit:%1 . .Vª±*'
'.'!~:l:~
so o llllll.
o L6
O 599ij.1.18 600 1.6
20
20
600
io
ló
2
2
2 ~l.. 7791.7
c.aeosas
o.aoos:as
·40
··L7'7917
Pl0-20 (d) For second-order decay:
Far
= 1 O kcal/rnol:
E, ;: :; 25 kcal/mol and E,
Temperature-Tlme Trajectory
Ternperature-Time Trajectory ~
1900
~ :,..: 1.500
...:, E
~
·- --
-·----
- - ' - ~ ~ . .. ,
----
--· -- -- -·-··· ---
------ -~ ·-·--
•
···1
:
200
,,,,,,•,••••••••H,,HH
JO()
·ª··
·-------
i
.
·---·----
--
.
·-·h---......
IWO
OH, ,.,,,,,,,,,_,,,,,.,.,,,,,,,,.»Mo•••-••oooOhO,O•OOH<"•O•••••••••'••••••
¡;;,..is
.,H•·---·-·------••••••••••••••"•·-•••--•-••a,;.,.,.
. ..., ---
¡ ¡.- - -----·--·-···----· · -·----1
¡...
E,-=10 and
1600
..
-400
600
o
1b;ne (h)
Use this equation in Excel to generate the above graph,
Pl0-20 (e)
10-52
time (h)
_
·-- - - --·· -·~ -
-
- - -
-- .... -·H-- .. -~~
-
In part e, the only thing that changes from (b) is the decay law and the decay
constanr:
da = _.........;..._ kdP/a2 .. _,.__ dW u, k,¡ = 0.2 Plugging into POL YMATH we get the following surnmary tables for U5;; 2 and 10
kg/h . X= 0.50 and 0.88 respectively, X will again increase as U5 increases. Eg:-J.aEions.;,
!ni tial
d{x)/d(w)=-ra/fao d(a)/d(w)=-kd"ptA2"aA2/Us
1
~1~!!
o
kda-.2
Us=2 fao=600
I~i.cial ....v~
pt.0"'2 pt=pto'" ( L-oc)
final_v~lue 0.502431 0.104635
600 20
O 0.1.04,;Js 0.2 2 600 20
2
O. 9951.38
0.99Sl38 ..:2. 08253
so O 502431
X
kt=20
&:lime..~.... ~±~
Maximum ...~~
o
w
a kd Us
i.
1
0.2
O 2
tao
600
2 20
50
z
2.oa2s:i
.,4()
lll
l.lie
o o
kd éao kt
0 .. 2 10 600 20
!'{aximwn .:2±lt.~ MipJ:tni+m 50
0.8,5709 1 O 2 10 600
20
2
;;¡
2
2
"'·º
yalue l"inal
,tj\lue
O
so
o
O 375709
0 . 553922
O 553922
1) 2
íl.2
1
o
600 20
600
20 2
2 75.389
0.248581
O 248581
-40
·2,75389
Pl0-21 (a) A 7 B Elementary reaction with 1 st order decay. da -=-ka dt D a= exp(-k0t)
Pl0-21 (b)
10-53
o
·40
so
0.2 2
600 20
1.0
a
t The activity is never zero for first order decay. When a= exp (-k0t) there is no t such that a= O.
Pl0-21 (e) Mole balance: dX
=r' A W =--A-= +r' W __ =r' A_W
-=
dt
CAoVo
NAO
CAo Vo
Rate Law: rA '=-a[-rA '(t =0)] -rA '(t
=o)= klCA
Decay: a= exp(-k0t)
Stoichiometry:
eA = eAo (1 - x )
Combine: dX
W
dt
V0
- = exp(-k0t )k1 (1-X)-
ln(--1)= k,W [1-exp(-k0t)] 1-·X
k D vO
t)JJ
X =1-exp(- k¡W [1-exp(-k0
kovo
Pl0-21 (d) k1W k0v0
--
-
(0.2)(1) -2 (0.1)(1) -
X= l-exp(-2[ l-exp(-1)])
= 0.97
10-54
Pl0-21 (e) Decay rate increases more rapidly with temperature than does the specific reaction rate. Therefore, conversion decreases with increasing temperature.
Pl0-21 (f)
k W (T) = 2exp[-1500(-1---1-)] = 0.57
1 kDVO
310 400
k 0= O.lexp[2000(-1---1-)] 300 400 X= 1-exp(-0.57[1-exp(-5.3)])
= 0.53 = 0.43
Pl0-22 (a) In order to get a high conversion the enteríng pressure should be as high as possible since the rate is a second order function of the pressure. U should be kept low since the conversion is an indirect function of the flow rate.
Pl0-22 (b) 111e problem with such a low flow rate is that the activity will remain low
Pl0-22 (e) We can use the same equations that are given in example 10-7 with a few
exceptions. For example the rate law, we use the one given in the problem:
111e activity will be different because the equatíon giveri is different: da
·· ·dt· = k naCcoke , da_. ·= kuaCcoke ·. · · · -· ·--- .
dz
U
To find the concentration of coke we use stoichiometry:
ecoxe. = !'coke RT We fínd that the value that gives the best conversión (X= 0.337) and uses the whole reactor is U= 7. See the following POLYMATH program,
10-55
~qud.t:i()~s
.. :
Inicial
d(a}
=·--kd*a .... c.co k e z'U
1.
/d(z)
o
(x)/dlz)=-ra/U/cao
22
cao"'
Inir ..i.<11. "value-
.kC;;:;100
o
kprüne=S~····S
l
Uo=:7
eps~1.
0.000254948
ca o
e 22
0 .. 22
0 .. 22
O 22
kd
100
me
lCO
5-e 05
Se---05
100 Se-05
7
7
T:::;673
Vo
7
;cho"'80 IJ=:[Jo*(l+e::;:,s~xl pa=pao•
{l·x)
pcoke"'pao~xt
I (l+eps•x) (.1.Teps •x)
rapri.me=a • ( ·· kpzí.me •pa~2) ccoke,,pcoke/R/T :ra:::rho""'ra:prime
O }l67E
1
ép5
pao"'l2
.aoo2s;94.g
O 316731
k.c"'fl:::-i::ne
=
o
15
o
?.ca .082
z.0
.. :alue
032
a .osz
R
O 082
T
673
5.·13
673
·pa,o
082 673
12
i.z
12
12
r.ho
ªº
ac 9: .. 3571.2
80 --¡
BO 9 .. 3571.2
-pa
12
o
12
5.95425
S .. 9S42S
pcoke
O
) :02288
.xapri:ne .<:•.::-oke
-··O O-Oí'2
·-.i;.
o
O-.
}
02283
5l9·34e· .. 07
0547 762
·3.61547e-05
O,
-·O., 00-12 ··O
-4 5193"1e:··07 o .. os;T1&2
57G
Pl0-22 (d) To find this the only change necessary was the vaíues for the k' s because
they change with temperature,
l \'1 ··R 6i3 ·· i)
. -r··ED ( 1
k0 ~ lOOex¡\
;,..,
.
1
.
....1
The POLYrv1ATH program below shows the results. The temperature is 485K and the conversion is 0.637.
10-.56
Equat;ions: d(x)/d(z};-ra/U/cao d(a)/d(z)~-kd•a~ccoke/U cao=.22 Uo:a2.5
Initial value o 1
R;.082 T:;,;485
pao::12 :·ho,,,80 kd=lOO•e.i...-p
{ 15000 /L
987• ( 1/673 • l/T)
kpri.me=5e-S*exp{3000
/1. :987* (1/673
l ··1./Tl)
tJ=.Uo• { 1.+eps"xi pa::pao .. (l··X)
/ (l~eps":x;)
pcoke:cpao*x/ (.l+·eps*x) ccoke=pcoke/R/T raprimea::a* (-kprime .. pa~2} ra""rho•raprime
Y!!~ i
:tnir.:ial _~al~
Maximwn... ~!ue
Mi.r;¡j.mum . . value
Fin.al valu,e
O
15
O
as
x
o
o 631:;rns
o
o. 637285
a
l.
0.22
O 605597 C.22
O .. 605697
cao
l O 22 2-5
2.5
2.5
l
l
1
!
R
O 082
T
4SS 12
0.082 485
,as
criss
Uo
o . 22 0.082
O 082
12
12
12
80
80
80
80
kd
1.29.32
1.2932
1..2932
l.2932
ic¡,rime
2 . 09SS8e··OS
2.5
pa
12
2.09558e-05 2.5 2.65841 O
2 . 09558e--OS
o
2 09S5Be OS 4 09321 12 4..6708
pcoke
o
4. 09321
2.65841 4. 6708 G.117445
ccoke
o
·O. 00301763
.a nn;..e-05
:-a
···O ,24141
-o . 007:7616
Pl0-22 (e)
10-57
To find the temperature-time
trajectory, use equation (10-119)
r
l-·expl EA_-·_r!A+Ed t_ ::= --················e,
.. ,_
,_
(
~-- 4;Jj
,,
c·--···--·--
k ct,-0(1, · · n +E· a.(EA )C~,k~ ~u ~
The following curve is generated from that equation,
1400 _ .
Time-Temperature trajectory , __ . . ... . . . . . . . . . ,
,,
, . , ... . ·
1200 ~:1000
ECil aoo
600
~ 400
¡...
200
Pl0-23 (a)
~ ... . _,, ·- ..···--·· ..... - ... ·-
·-.--1---------1
Design equarion:
r=f~
Assume W=lg
r=FAoX
Cumene (A) FAo (1 X)
....... q_ ..
t·-----..¿
<;
-~- .... -_.......,,
1
w Propene (R)
FAoX
+
Benzene (S)
FAoX
The amcunt of curnene hydroperoxide does not contribnte significaruly to the total numcer óf moles, or X= •.. Ys. 1 · · Ys
10-58
,e
and add C,0
where necessary:
X a
.0204 1
.0165
o
t
.0133
.~U':I
.652
50
100
.0107 .5245 150
.00857 .420
200
.005631 .00371 •.00241 .118 ,2761 .182 500 400 3001·
ln a vs. t gives the best fit a== c-o.t d.= 4.27 x lQ-3 (sec-')
Therefore, decay is first order with decay constam of 4.27 x 103 (sec+)
Pl0-23 (b) CA= FA=
u
(l-X}
~AO
1.)
. . n Assume no AP. •. ·· J.L _ -- ...,,._ Uo
CA=
no
AO ( 1 +X) = ..F. ··-· --- ~ u-= ( 1 +X) u FAo
0
~Aril-~2 = CAofrX) (r-x)
Uo{l+X)
• . PA = CA RT
{1-X)
= CAo ff+xj' RT
Mass balance : F AO dX == rdW r = e.at k P A t
= . Y:L üs
10-59
F AO
= 20 rnol/min = 1/3 mol/sec
CAo "'0 . 06 kmoiím3 = 0.06 mol/l
R == 0 . 082 T k a.
atrn:"l.
Krmol
= 273 + 420 = 693 K = 3.8xl0--3 mol/g sec ann = 4.27xl0·.3 sec··l
W ""lOOkg Us
= 10 kg/rnin = 1/6 kg/sec
F.quation (1) becomes;
:.:: -L4 x 103 X:
LOO
Pl0-24 (a) S · .,. W + C02 : firsr order, irreversible .. .- an c·,m . . da. d¡ -- k·d 'A
... rA
Assume n=Z
10
=KaCA a
= ·--··L_. l+kcit
fr\J = Tf~~-i·
__ _..1, In
:
= "di- .. an
h:xJ
-
"tk· + ~-
t
10-60
--·--------------------------------·------
-------------·--------- -·-··-------
------···------·--------------··----------·-----------------------·--·--------------------------------·--------------------------
For T : : 500 K
o
20
40
60
80
120
0.. 7
0 . .56
0.4.5
0.38
0..29
90, . 5
1424
178.l
1.21~7
262.6
3443
o
5
10
1.5
20
30
40
1 ., 4~-
0.89
0 . 69
057
0.42
__ ..,. 87 ~
0.33
111.8
144. 4
174.9
237.6
302 . .5
t(min)
X% Ll.J il-Xl
1!1n
'
·
Fer T:::: 550 K t(rni.'1)
.,
X%
l/ln ¡ ¡ q .. x1 '
l
\
49.5
1
3QQ ;..
1
ln(l/(1-X))
\
+T=500K
=1
0T=550K
HlQ
a---o They are Straight lines : . n=2 For T1 = 500 K : slope = 2.04 {-t~·h
[l~t]=0.01
} ::::::>
=('tKh
Far T 2 = 550 K : {~lKo~ "":: : .. } • u: 0602325
._
_,,
~
~I
.::::
0.02
= 0.1265
& = ..JlOL = exn {__§_f...L ~ _J_\} K.si
• 8.314 \550 5001
0.1265
E.!= 84,344 v ·
""<11
=
1c~ ·
•"'
(l)lf = v... ·
exn• {- E.i R T1
K.i1= L296
X
"
! :" mot
exp.
·344 I\ _ 84 8.314 (--L·}\¡ 500 = 0.02
10 3
We want to maintain Ka = constaat
10-61
K
x, exp
h--¡l t) =
{f (t--t)}
e:
x, (1
1
'"""
i E( 1
lj\.]
¡· · 1
expi_ ,R. tI'o -- T
t
t
+ K¡ t)
= ""' · t
exn f E.\[ .l . . . ·- _l\J\ -1 . \R To T
K.,
= . ---·· · - . . ···-· •"·· · --·· ··:. . --KJ
exp /!~~coo f_.L .. jJ}-1 = . .... ..! ... 8,.:iL± ~:!filL_I!_ _ L296
TrK) · 1 480 485
490 495
r:11tl
107 exp { ·
X
.
· t(mm)
1
O
1
44.3
!
87.3 l30.4 -·
i
::ff
·
120
J
1 ~·. ~ ·¡¡ 60 ·
-300- -·rn-f74_9
40·
20
·l
o-------·---···········-··~·········--·----
485
480
..·,-·--·-·--·
,
.
490
495
500
Pl0-24 (b) Since the equarion for the acivation is: a=····
I+kct,
we cannot find a time for which a= O, because it is mathemancally ímpossible. We can, however, find a time ar which the activirv is small enough that it can be considered to be zero. Toe following graphs show the activation for'tb.e
tWO
Actlvlty '
~
O 8
>
O 5
-(.) ~
O 4 0.2
o
..
+
. ···~-
.. DC'
... ••••••;,,,.,.,,.
500
time
.. ~,.,_
temperatures
of ,;,
-.,.
given
cataly st ,
.. ,
1000
-
,,
__
,
.
1500
(day s)
10-62
The graphs show that for SOO K, the lifetime the lifetime is about 450 days
is about 1100 davs and for 550 K
,
Pl0-25 First we need
to
find CAo·
J:lL-10 ETB
Yero= 0.1
C
PYere
AO
RT
= 3* ltr(O. l)_ (8.309)553
0.065
Start by gnessing that the decay is first orden
We were told that the reaction is zero order when tbe conversión is less than 0.75. This is true at any time after 2 hours, We also need to ñnd the denominator as a function of conversión.
X= ~,1Q_::_<:,t CAO
So we graph rhis:
This is the graph thar we get: ln
(1/cao*x)) vs time
i-![~~0
5 lime
10 (h)
15
As can be seen this graph islinear which means that the decay is fírst order. We also know that the slope will be the decay Iaw constant so kct = 0.2024.
10-63
'
Pl0-26 (a) Mass balance:
Rate law: ··-r;,._
= kaCA
Decay law:
_9ª.. = dW
··_kDa
us
Energy balance: Fa _:··--(TA · · · T ) + ¡\rA) dT .............. - .;.. ..;:: _ _ --
·(
. , ·AHRx J ..
Stoichiornetry
eA = e""·º -1-\--::_: ·-i-·
t.1:1.
Evalúate the parameters: k
= 0.33exp[3T77(-4}() r
k- """OJHexpl 7000( .> L
·-})J
l 450
1 \! ··T:--··)1_J
Plug that into POLThiATH and get the following program and answers máximum conversión will be 17 kg/s.
10-64
l\ to get
E~at~~.L d
(a.Ud
(w)' =-kd•afU
d(xl /d(w) :--ra/fao d('J)J /d (w)
= (tra
r
Q
(Ta··TJ +(-::a)•
( ··Dhrl)}
! (U•cps+fao•c9a)
450
fao=S.42 U=17
,ra::323
10···27
Oh.rl=-80000
YE;.4bl~
cps=lOC
¡~~~~4l
... v~:...:e
M.3x:.!,.¡:it,;.,"':.
o
a
cpa=40 Ed=7000 Er=3777
cao=<,27 eps=l
ºª"' ·ª kd: .. OPexp(Ed•
(1/450-1/T))
k= .. 33•ex¡;,(Er• ( l/ 450···1/T))
ca .. cao*{l-x)/(l+eps•x) . ra= . . k·~,...ca
va.¡".;~
50
::f:::!i~cm.~~
ra~ . va.:.:::!
o
e o.
271.53'6
s
42
o
2: 1595
x
o
o.as,:,:
T
450
:i,S.ó.13
450
5..42 1"7
fa.o
5.42
5 .42.
17
ll
Ta Dhrl
nJ
323 ···3!JOOO
n3 100
aco
"ººººº
·80000
epa
40
4C
~o
40
Zd
70-00
7000
7000
sz cao
j r77
i'COO 3777
J777
37""!7
e, :27
o
O . :Z1
0.21
o.s
o.e
eps ~a
21
l
o.a
k'3
c. o¡
"ce
o. n
l .. U4C! 4 .19·;:a
o .. oi o 33
o. 27
O .27
O Ol61349
:a
··O
·O
0891
c:a::92s
•O, 1,S4l
Uaw (Ta······ T) + hap(Ts -·- T) + (r_¡ )(~'\HRx) . - ·-· -···""*'-"'"' "-·----··-····-
,
.-
.
FAoCpA
We also need an energy balance on the catalyst particles .
~~1 _ . . . ha p{°I~ ::.. :~) dW
l\Cps
Choose values of h and a., We find that T5 needs to be slightly higher than T in order to get a large conversion. The máximum conversión will be different for each T5 that is used,
CDPlO-A
10-65
---·--···-··-·---··-·-·--
··80000
lCO
Everythíng is the same except the energy balance:
aw
323
lbó
Pl0-26 (e)
- -~· . · -
1·1
C''.;IS
Pl0-26 (b)
dT
88722.2 6{5 613
o
Using the same program we can see that the maximum conversion is 0.887
,· ,
se
o. a 2..!.!401 ~ . 19·11s O. Clol.349
-o.
0133928
Given: The catalytic oxidaaon of ethanol CH3
Oh
OH + }" 02 ~ c:H3 CHO + H20
Denote:A[=]CH.3CH20H.
B[=J0130IO,
W[=]H20,
A·S[=JCÜ3CH20·S
Mechanism is belived to be the following: -~
A·S +H·S
[PA C! .. CA·S CH-s!K.J
A
+ 2S
o.
+ 2S"
¡:! 20-S'
-ro= ko(PÜ1. C.';,··
A·S
+O-S'
-···+ ~
B +OH•S'+ S
·rAo = kAo(CA·S Co.s· • Ps ~ñH" Cs/KAo]
~
W+S + S'
rw = kw ÍLCoH·S
+-·
OH•S' +H·S
TA= kA
With reaction (·rAo) controlling
PwCs· CoH·-· ""'----···---
"
CT
= Cs + Cu-s
Kw 'iKAPA +
CA·S
= Cs [1 + HK~Pj]
K-eq == KA Kw KAo fKi:;
Pc,;· . !:(w]
iK:,;:
kAo Kú Cr Cr [ pA-f ···r AO - ...•. . .•..... " ·············•··· · ' ············•·········· ··························--····· . , · · · ~ {p;_ 0l l + 2 f'J{'";A P l + ,¡-Ko-:-···· Pe-;- t --· __ Pw
·
·
;J ....
·
r..
•·•··
j
KwfKA PA J
Initially P3 "" Pw == O
10-66
c:t,..S'!Ko]
C¡.¡.5 -
Pw Cs Cs·IKwlJ
With reaction 3 irreversible, .. rAo = kAo CA·S C-0s·- Since A·S, H·S, OH·S' are created
(b)
(and destroyed) only ar the reaction surface. A stoichiomerric relationship exists berween
(a)
Toe same equilibriumexpressionsfor reactions (1), (2) and (4) existas befare: (b} CoH-S'
.
C¡.¡.s
~.,
s
with KA P,4C >>
CT=Cs(l
PwCsCs· Kw
= ·-
(e)
··PwCsCs --¡~ ·-·
+ 2l'KAI5A]
e; =c,.[1+pt;P~+~]
10-67
------------···------------·------·---
-
··-·---·----------------------
-------
---- --
- --- - -
---
-
- -----
-
---- -- ------
-----
-----
Rate expzession becomes:
CDPlO-B We can first try to come up with arate law for this data. We can see that as PE increases the rate law also increases but slowly the arnount by which it increases becomes smaller this tells us this:
'
PE
---rA .... ··---··---~--------
l+KEPE
We can then see a similar thing happens as PH increases so:
.,
- - rA ••
Pu
i':t:KnP;~·
Since both reactants are adsorbed the rnechanism must be a dual site This makes the rate law:
, rA
=
k.PEPH 1 + KE.PE + KHPH
yYe can then plug thís into POLYMATH and we then get the foilowing values 'fer k, Kv and KH.
k'::::14.6 KE:::
2.63
KH= L76
10-68
0,90C 0 .. 6.,C
lO···B
--··
t
[3C.a!c.:.-!atea v.ali.i~
o ..... ac 0.320
rtodet;
rat~;:;it;.xO:e•Oh_,·<
t ... -Ke)lfPe+KtucPh)
,: "
l"'l .. 53"'1"'1 )(11 => 2, 63.,7 l l
pos1r.1ve
'(h
-rEt-SLdl;..ats,. 6 neq.at.t\l@,
-'
.·~2
!.?5797
i"@$lOu.als..
Surr.: cf
squa,·95
CDPlO-C (1)
NzO + S -~ N20 • S SiH2 Cl2 + 2N20 • S •..., Sí(h + 2N2 + 2.I!Cl + 2S
N20 • S + SiO • S ·4 Si0:2 + 2S + N2 (Rapid) rsiOi ;: ~Pues. PNp_ 1 + KPN.O
(3)
dF -11.:::::r - a áV
A
, Plug Flow Reactor)
10-69
::::
I .. ~·S236e~·os
k Plo (1 .. X) (eB-2X}/(l + EX}' rA =
1 .,...· K' B pAa;;----· (eB-2X'} (l +· EX)
- (1
eB.
+
k ~o(l . X}{6B··2X) cX)2 + KB P xo (~ ·2X) { l + €X}
= 1.L. 3 = ., o. ·7 3 . 68 ...
E:::(··-·:?~-·-) (2 + 2 • 1 - 2) = 0.25 11.;; + ,.68 Slope
1300
Inrerceor •
1200
m. .Torr2--miri A
k
11 00
=
9CO-i-
_,.... __ .....,..__ _,..__ -...f
"ºº
300
500
600
700
800
900
1 000
KB = 1.1 x 10.J m Torr' Ka P Al= ( 1.1 x 10·3) (157}:;:: .173
=
L49
36.7 hmin
x
10"3 (157)2
= 36.68 Á/min
. . (3.Q.?.:~Hl:~L- ..... - . . ·-·(1 + . 25X}2 +· .173 (3.07 • 2X)(1 + 25X)
3.68 x 10.J gmole/min -rA ={Álmm}=4.4x
10-ór:(~i}
a= 250 m2/rr.in
:ª. M
=
1x ~ ~ .·o
rA
k'
= l. k = 703
L49xI0·3 Á/mTon2--rnin
K.::: Llx1Q·3 m'Torr
1000
k Pio
= K. = o. 7465
Area
under curve of X vs, ~-
rA
10-70
1
X
··r:(Á/min)
·l/-r:(~}
1/·r: {m2/gmo!e)
...Ya.
';\m3)
o
736
0. 0136 0 . 0142 00149 0 . 0165 0.0182 0.020 0 . 0226 0 . 0255 0 . 0288 0..0328 0.0378 0 . 0439 0.0517 00617 0.0751 0.0936 O..ot203 0 . .01617
3091 3235 3392 3739 4139 4545 5145 5784 6547 7465 8589 9983 i!746 14029 17067 21262
79.1 82.8 1783 196.9 217 . 1 242.3 273.2 308.3 350.3 4014 464.3 543.1 644.4 77"7 4 958.2
36760
!602
0 . 0012 0.0012 00026 0. 0029 0.0032 0.0036 0.0040 0 . 0045 00052 00059 0.0068 0.0080 0.0095 0.0114 00141 0 . 0179 0.0236
0.025 0.5 0. 1 0 . 15 0.20 0.25
0..30
0.35 0 . 40 0 . 45 0.50 > 0 ..55 060 0.65 0.70 0 . 75 0 . 80
70.3 67.0 60 . 8 54.9 49.4 44.2 39.3 34 .. 7 30.4 26.5 22.8 19.3 16.2
133 10.7 8.J 6..2
'
Reactor Volurne per waíer :. Number of wafers, n A plot of conversion
«
FAo
o
1215
27338
9.8 x 104 m3/wafer
= V/9.8 x 10-4
X. vs, number, of wafers can be obtained 80
200 ------·-----
60
,::
i
G>
l
o
100
40
:s i
5.
••';-
ci
:z
20
o.o
0.2
OA 0.6 Conversion,x
o.a
1.0
10-71
The rhickness en rhese wafers ca.'l be obtained from the same plot, Wafer s l
: X= O
-r: =
73.6 Álmin
:. After 30 min:
2208A
Wafer #50: X= 0.52 -rA ""'' 20 Áfrnin :. After 30 min: fl'Q .A. Wafer #110: X"" 0.78 _/A =6 Á/rnin .. After 30 rnin: _l 80
A
CDPlO-D (3/e CDPlO-E)
CDPlO-E (a) For simplicirv, letters were substituted nroblern sta ternent.
.
....
~
B=NH1 ¡, V
=
Cl (v"NH. 3 \)7. •
T' 1 l ., ~
s
fer the suecies in the li,
1v1 ) (,-·= 1··1c·1.. ,,,·1J2
The rate equations Ior each of the three steps in the mechanism is given below.
· · re= kcf Pcfv '
···
t: ·1
1 ·-~·=:::-~ f0 = KcPcfv
~.
Equations 1 and 2. are substituted into equation 3 because the third step (the deposition step) is assuzned to be the rate Iimiting step, we want to find the rate law of the deposition, and we have data for the deposition rate.
10-72
•Also, remember that we have the relation beiow:
1 fv=·------
1+ KcPc
Substitute everything into the deposiríon rate law:
We now have to check if the above rate expression agrees with experimental observations,
.
The rate of deposition is independent of Ar and :Eh. ·-YES
•
At low partial pressures of TrCl, and NH., the deposinon appears to be first order in TiCl, and second order in ..
.
.t
)
!'íH1 --YES
•
At high partial pressures of ?-,ffi3, the rate varíes inversely
with TiCl , . · ·YES (b) To determine the reaction rate parameters we must r earrange the rate expressíon to a linear fonn.'
A plot of the experimental data is shown below. Títanium Nitride Film Deposition Model V erlfü:ation
:::::l··········-·····""••••"••·•-- ·········"········--·---·--1200000--1
~
1000000·
¡t
800000
"-
"
600000
g
.;00000-
;:;< E
200000-
l.2167%¡..._5'. Ra'\la119.97t98UE·l
ffx) :;;.- L "4tl6ZE-,._l'":t: +
Pal'b"2
10-73
l '1-kl = v~ ··
k.
.
mtcrceo,
= 2.035x
~
= 2 . 2168x 10
l
10-u --.·--~~~!···-3 crrr rrun- mT >
.
k.
_:._ = slope = 14 .42
.,fk. .
(ci
The experimental data, when plotted with the rare law derived in. this problern, forrn 'a straizht " line .. Thereíore .. . . -, the proposed mecnanisrn rnav be used to describe ti:e deposition of titaruurn nitride filrns .
.
. .
CDPlO-F
Given: ·1:1e dehydrogenarion of ethyl benzene to styrene:
e, Hrn -+
Cs Hg + H2
E St + H2 Quanrirarive data suggests reaction rare is of che form -rz a -1·.··-;\
..,.. ,-..2
A1 Pi=
p:_-;..~;_---p· ··- where Al, e · .-..3
St
,A.2,
AJ are constants
111e facr rhar (·r E) is independent of H2 suggests that the reaction is irreversible, and that H2 is not adsorbed on rhe caralyst surface, Also, the above expression for ( -aj suggests that both E. and S.c are adsorbed on the surface :. Trv the followinz reacticn scneme. E+S·HE•S 1
-rA
E • S f·+ St • S + H2 St • S {·) St +· S
·rs
= kA [PECs • CE.sf.t<.A] = ks [CÉ·S . . Cs-s PHJKsJ
rn "" ko [Cs,·s - Ps,Cs/Ko]
10-74
------------·---·-----------
--------------------~
--
- - ---- ----
-- -----
--
-
-----
To insure that PH1 does not appear in the rare expression
assume that the surface reactíon
controls:
f~=º
Ca-s
:; = o
= c~¿~r.
Csc·s
Toen: Ct And -rs
= KA e, Pa
e, + Ca-s + Cst·S = Cs [ 1 + KA PE + ~~]
= ks (cE·s - Cs¡K·s PH1}:::: ks KA e, (PE s
, e ' ""-··· ----=-·--
Pst PH,] lC5 ·TK AÍp I E··-· -;;-·"
..1r ··rs
Keo P ·-'-·-·
1 + KA PE + .::.~!.
,
' ·. wnere
K· eq
Psi
J
PH, KAKs Ko]
= KA K s K o
Ko
with Ks >>> l, reactíon can be considered irreversible, and íts dependence on Pm drops..
kK,.i..PE ·rs - ---·-----~ · , where 1 + K,..., P~ + fs-1. . t Kn
out:
k = ks Cr
Evaluation for k, KA and Ko Rearranging
the rare expression:
Pe.= -rs
i
k KA
. + Pg_ + _ k
PSL ...
Ko k KA
with Pst -....:, O, a olor of ~ vs, PE should be linear with slope
·
·,s
(1-k ) and intercem• {--1. . . ). -kKA
only two points are given with Pst a:: O. ir is easy to solve analytically:
. 2,.
Poinr
PE (atm)
1
1.00
l_.4_1_5_x_l_0_3
7__
0.01
0.214 X
46.73
(-aj gmol/min-gmcat
···-·----·-'-------.!:E\ = k L._+ l(P-}i and (~.\ = L-+ ( -rsJ1 KA k e rs r: k KA ---
PEJ'·rs
6_:_7·· · j
0_
1Q3
l.(PEh
k
10-75
Since
• 1 and -·-:-··-·· == (P;; -1 \ - 1. (PE ) 1 ""
k~
~h
arm-gmcat-min = 40 --·--:---· ~~
706 ---· . ·7 ·· 000.1
k
KA= 16.64 anrr! Now, taking data poinrs 1 and 3, in which PE is constanr, the value of Kn k-KA and hence
Ko can be determined;
= k-k-··A +
Por poinr L: (~)1
(PE)
Forpoim3:
-rs
!(PE)1 + J-~t't-
=--L . . +1.(PE'- +J~1th_ .... k KA
3
k
,
[(,J~t · \_;s),]
(PE);, (Ps:}t{Pstl1
-- 1--·· - 53173
KD k KA
Ko
Ko k KA
~
LO
[o:·1661~"io:3 1.41}~·10~3] gm car .. rnin LO ·· 00
-
.
gmol
pn_c:u.:rnin gmole
= 0. 007534
.", Rate expression
is
P-c/40 . . . . 16 64 p . +· J:'sLw. . " 0 . 007534
··rF """" --
1
(a)
JJ
R;;{¡¿;, noring that (PEh ""{PE):3
Subsrracting, and solving for
K~'tK.~""'
A
•"-!)
:::!
0.25 P;; (. e:u:role . . ~... . !".H·n·· l + 16.64 Pr: + 1327 Ps1 """"" ...... ... ...
~1':.~,.
"¡•
"
O"
(iv{\.V)E = 106.. 1 - - "' -
gmoíe
.. 1 zrnole H,,,Q Then 8¡ = ---: ---~-----:::, grnoíe E
YEo=·-J-
=0833
l + 81
'. <:': = VE o ó =
· ·
~ .. . PE - LER I
7
O. 833
F:=RT ----------
l.)
Psl ·::: CsiRT"""
·
;
·
'
= 0. 2 P=P0=0.415 Fsl(=
u
FE.O:1 X)RT
-
X¡:,=0 . 60
i?:~day. .!S~ (l104 kg m::!) . 1 kg)
= """"" ·--··-- . .;:: l)o{l+EX}
!::.s.&1:. - FE.o X
;
CE.oRT(l--X} · -···· l+e:X
;
8=2-1=1
;
"" 19.2 kg ~:?!. = F- oXF day e., -
RT -- YE.o.Po X - l +EX
1Jo( l+e..X)
10-76
YE..oPo{ l X) -----· l+EX
.
T=903"K
Rate expression
·I"E -
X is:
for any
[YE..r~,;..+~.;;,;;~.,,.-X_}.....] __
_.,., . 64 YE.oP o{ 1-X}l 132.7 YE.O Po X 1 + 1 6. - 1 + e:X j +-··· 'i'+ EX
l~
0.025
Design expression for a C".STR (fluidized bed) is W = F~~F
W = ~_S::Q_X...f..f ......!...: EXF ·-- + l6.64 + l32.7 XFJ 0.025 LYE.O P0 {l·-XF) l·XF j
+_EXF
_ Fst, Xr .. J
... 1. --+ 0.025 XF tYE.o P0 {1-X¡:)
_ 19.?.xl03 gmol
- -·- day-.. -·--
1 day
X
í
16.64 + 132.7 XF.l 1-X¡: j 1
+
0.833 (0.45}
.
o:025 t()~833 ( Ü.415) ( l~-:-45) .,.
16"64
A
132.7 (0.45)]
l dav
+ ---i:~45·--¡ X 24x6Q mi~
W = 7.06xl04 g Cosr = 70.6 kg x Sk1;:· = $777 :;,
(b)
Plug fiow reactor. expression fer YE.o, Fsr;. PE, Ps~ and ·I"E are the same as forthe
CSTR. The design equazion is:
W=
--~5''--
. , ¡_ ..._.l___ (-eX • [l+E} ln[l-X]) + 16.64X + 132.7
00-5 X;: \YE.o P0
W=-0 .. :_.l_p-(·eXp ·[l+eJln[l--X¡:]) · 0-~~}'x··Í _;:, F \Y::..O o W=
l?}:: 103 ~~?.~ . x day
L Ü . 025
X
0.45
j·.833(.45)+L833ln(l--.45).
¡"'----·'[ri833)0.415) W
= 2.71
Cost == 27.1 x S.k:11 g
X
+
(-X·· In (1-X)Jll;
f
16.64X¡:: + L32.7[X¡: + ln{l·X¡:)]}
gm car min gmol
1~ . . . )JJI X ...24 *60mi~
.1 .• .,-,0 S l (l 1_64(QA-, o, . )J .. s.s: . l l. .4 + n 04-11\
104
= S298
CDPlO-G Given: reducríon of CO with hydrogen over Ni cataíysr:
10-77
Kineric gíven by:
_
1 -
0.0183' -- Pfü Peo (gmole·!",........... Cfu) .. • 1 .,... l .J PH:t . gm car-rmn h•
.
·-
--
From the above expression: 1) The appearance of Ph2 in the denominator suggest that H2 is adsorbed on the
surface,
2) The fact that CO does not appear in the deaorninator snggests that ir. reacts as a gas phase species. 3) The square root dependence in the numerator suggests rhat H2 splits. Therefore, suggested mechanisrn is:
'
H1
+
s
H2·S
+-
s
H·S
+ C'O
CHO-S + H·S
C•S
2H2
k1 .-:, ~---·
k2 --~ fok· .)
;::.
Ih·S --1:
2H·S
l
:o:
k 2 [e'H •Se"''S 2
--· ("'
2 'H•S
! K :z;.J
CH().S
~
C·S + H20 + S
--~ ~-ks
' ·T 5 = k 5 !•e ('-<~-s
CH4 + S
-··"1"
~
p'H2 . . pCH.
eSi l\.5J f'T
i
Remarks made in Chapter 5 of the text suggest that reacnon 3 is rate ccnzrciling:
/1=º K¡ e-<· ·S :: ·--·Pea, --~-· • e-S .
Ks Pfi:
CcHo-s - Q;§PH::0Cs __ . Pc1-LPH::0.q ... - ..... Pc,L PH,0 Cs_~
K, C¡1-5
K, Ks
PA
C,Vi
2
Ka Ks ~
CT = Cs +- CH,S +· Cc-s + c:.~HOS -,.
Lr
= C.s., '1· ' 1
l
• •
-r-
--·'··;-
K1 Pni + YK1 K2 PH,
. -
PCH..
-t ·:- ..
.
Pea, P¡{ . l ,---· :=-___, P~ Ki x, iK1 K2J .n
-·.'í·· .,. . . .
Ks Ptti
10-78
P?i;
Kinetic éxpression becomes
. _ ··IJ -
k3 .rv--··K-·· l .l'I..J z Cr f. PH3 Peol Ptt, so.
l
'1
. --. ···-·
-
. L
-
....lj·
. -
Ppi01
_
. PH, JIJ. -~ J.. + K1 + Y!\.¡ !\.2 P!-i + PCH.. ;, YPH, _:_.,, + Pcn, ----·--·2 Ks
•
Reacrion is irreversible r. let Ks -rj -_ .
PCH..___::¡_ PH o
K1 -·-K2 K-3 Kt .. Ks ·
1
k3
--+
{Ki"K2 CT P~2 Peo• • .. •
00
Mf;[l + K1 Pi-h + YK1 K2 PH:] with K1 PHz >>
Ks .
Toen
._ -
.
Ka
k3
iKi Kí . .....Cr Pi
._
Peo . .
-
1 + K1 PH1+ vK1 K2 P¡.f.
YK1 K2 Plf¡. we get
k; {K'j""IG" CT Pre Peo . . . . . _,.¡_ ~th . ,r3 ""·· · · ~----..r:!.l--. This. expression irnplies that at moderare pressnres, most or e 1 + K1 Pfü active sítes are occupíed by the H2 molecnles, Design for a plug flow reactor, Denote A[=] CO,~ B I=J H2; then yAO. F . • •
(a)
0.75.
ªª = FE:eQ. = 'ty_-ªQ..,:;: 3, XF = 0.80 AO
,.,.•
.
= 0.25. YBO = ...
AO
FAO X¡:= Fc:p"' = 2COO F AO
.,
.
:y ¡m~ x
= 125
11:~
= 12..5J.bmai. = 156.25 llim.ul. 0.8 day
day
Desizn eonarion F ,0 dX =
•
...,
= -r,
·"'
ex
dW or .E.. . = f.o."" FAO • rA o
__ 0.0183 PA P.jf ·-· l+L5PB
"lA----
pA;;; CA
RT = -~AO
~!_( 1-X) l+E.X
= y A-O PB .(1-X) l+E..X
p8 = CB RT- CAo RTfe:a_- _ 3X} = CAo RT{3< 3X)_ = 3 YAo_P0 (l ..X) l+s:X
0.0183 :. -rA -.-
L._t::X
1 +
\
l+e:X
[YAO p;~!:Xllf3 YAO Po (1.~)rZ .
_.
f
!+EX
· .. J-.+ EX
i.s f Y~0 P º (1;:Xl] ~
1+€:X
,. -::::
0.0183
.
1 + 1.5 f¿:A0_~2.(l·XJ] ) ----- -··-··-. l--1±.~L.\ dX
0.0183
fr[·
¿'AO Po (l-X)l3!2j l + 1;;X J
10-79
r
,r3·r¿'A0.~2J!~~Jf'2
µ~· . i.s [. Y ~[~2.E~.~Jl
__. ,_.L __ J
+
.
l+eX
1
Tl;e integral can be solved using Sirnpson's rule
Let f(X)
Toen
" ,,
f..,
~X) dX s
tt-[
I
1-0 ·-?5X L:Q.~ 1 1-X . l. X 1 o 1 1 l l 0.2 1 0.8 0.95 1 1.1875 0.4 1 o.6 ,
-01;-ro::n- o.o.9H-{soo ss · 2J2s
cf.8-¡-o.:rt- o.so . T 4.ooo í(X;rl.X
1 (\ .., sx 1 n '} ,x1112 ;¡!..::1l~~ + 3.751!--=1!.d:.,¿_ . 1-X 1-X J r
,r
J.
~O) + tú( 0.2) + 2~0.4) + 4~ O. 6) + ~ 0.8)]
X
~1
·-
= o.;z.[4 7 50
11:Q.2.5.X + 3.75 1S
(l:Q,2,5x.y12
-
-
~-875
1 1
+ 21.679 + 12.860
-~.:~
, .9)o T + 34.257
z.ooo
w
f(X)
1-X
4.750 1.000 . 4.974 1.090 ~ .... 1.225 5.250 ...- .... . ....... -~·-·
4.750 5.420 6.430
4
8.564 is.soo
4 l. . . ....
l
2
Wf(X)
4.750 21.679
12..860
34.257
15.500
+ 15.500] = 5.936
j
• w- ----··
_2.2.2Q
···-·-·gmcar·_rr>inx.U6Iil!JmoLx454gmol .,,., l . lb . gmo day iomot
0.0183A(025xlOfi-
X
Jday 24x60niin
= 2334 g
W
(b) Design of a CSTR · · . enuanon . is... W :: ~~-'F cXi= Desizn y
•
.
-~
!25 lbmolidayx08
W =···----··---··--·--···"-.:--;0018.., .1"' .. (Ü'" 10\!f.
j
I
J
· -- X •
,
r¡1--{.25L8))H2. L 1-.8 ,
l¡-----··,
W = i 2371 l!m.19.i x gm cat • min x da. y gmoi
.
_
_
-,·{l ))(.2.;){10)
--·-:-. ... ·¡!E:;_:_':__.:= 1-.8 ! J zmol -
:_~7 g~l x ·-L~~I---lbrnol
{J . {.25X,8J\Itz1emcnt·mi·
· ~.,4...x60 min .
W = 3900 g
CDPlO-H 10-80
(a)
.
=r,
í
o
'A
Runs 4 . .5
Runs L.2.4
Runs l.3. 6. 7 , P¡,. ···-es ~ - -- - -----~ - - -. 1 t· KA P A+- .....
(b) Numerator: Denomínator:
t
P A and PB P A and Pe
Power of Denomínator: (e)
1
Proposed Rate Law:
(d) To find the rata law parameters, reauange the rate law so thar it can be plotted as a line with rhe r ate law parameters as the slope and íntercept of the line.
rb .... ¡5
= IS:. P .. : Kr . P + i k " k ~ k e-
First, hold Pe constant and plot I'..a~!t vs. P
¡\
""ís
From the plot Slope
K = .... ~ . = 5.2, k
l
··-· r, ·- ........ ---·-·· ~ l+KcPc-+· ....
l+K,..P y Intercept = --·-· ...,..¡;_. = 3..59 +
k
10-81
p.ararne eer Evalu.ation with
Pa..rameter Evaluation
Pc=const.ant=2 atm 120---------- ---···························-·· .. ···········-··--·-
Paeecorrsaa n teej,
,
25
100
"'
·-·····--·--------------------··-··········-······· '
20
80 O::::
\'oi'ith
atm
.,
15
""-
60
,;?
~
~ fb:} = S. t':17:22.5.···0.. x
.. 3.58Th4E.-O
R""2,,,, 9 999'1~E ·l
O
1.
4
6
8
U
H
o
16 18 20
2.
3
Pa
Second. hold P,;_ constant and plot -~"'PB. --r s
vs.
Pe
Frorn plot below, --~-J"' . . -'-. = l . OlJ-J. Sl OPe = . K,.
k
,v ·· Intercept "'·
+K P = ---l ------, , .. k. . l .....::. -~ ,- . .,,_
1 )-
Using the four equations above to solve for k, KA, and Kc, we get: mol k ... '2.60----------- . . . gcat • sec • aun..,.
KA ... l3. 52 ann".
K::-
= 4 1665 ami .,
{el A and C are adsorbed en the surface of rhe catalyst. (f)
Proposed Mechanism:
C·S=C+S The irreversible reaction step was assumed to be the liminng step We check this mechanism and rate lirniting step by
rearranging anrl combining the rate laws for each step. H the mecnanism is correct, we will obtain the rate law proposed in pan c.
10-82
4
5 6 Pe
8
9
10
=rc = ketr Cc-s. . ... ..
·¡ PcC K . z. e .
-r
.
.
-r('.
k, is much smaller than kA and kc, therefore, ---' = ____,,.:,_ = O le"
and
. . ~ After substirution, Cv
C'-c-s
= -~.r._Cv K.
= K e P e· C-v- where ·
e
kc
x, = .L K. e
= ---··-·-Cr-·-·-·. -
l + K" P; + KcP e
Next, substitute the above equations into the reactions step to get
Because this is the same as the rate law in parte, the rate law and límiting step assumptions have been verified, (g)
Ratio of sites of A to sites of B at 80% conversión:
Conversion at which t.he number of sites of A equal the number of si tes oí C:
e
.s
..:::.f.:l.
= ·•·K.O-·X)
c(:·S
.e,
,,,,_ ...... _
KcX
-
(1J.s2ic1 ·X). , ----·----.. ··(41665)(X)
-
l
X =076
CDPl0-1
10-83
a) To determine the rnechanism and rate-Iimiting step we must come up with the rate Iaw. Look.ing at the rate dependence of A we see that between runs 1 and 2, P Aincreases frorn 1 to 1000 atm while the rate law only increases from 1 to 1 .5. This tells us that as A gets Iarger it changes the rate law a good deal less . This tells us that A is both in the numerator and the denominator.
··· ·rA
,
-
. PA_ l+KAPA
Looking at the rate dependence ofB we see that between runs 1 and 3, PB increases from 1 to 4.5. This tells us that the rate law is directly related to B. ··--r~ ·-
PB
Looking at the rate dependence of C we see that berween runs 7 and 9 Pe increases from O to 4 atm and the rate increases from 4.5 to 4.8. Also in these runs we see that P A increases from 1 to 4 atrn. So ene of two things is tme either Pe is both in the numerator and denominator orjust in the denominaror. Since C is a product it will not be in the numerator in an irreversible reaction . --··r
•
I
l+ KcPc
A
.
So rhe rate law becornes l
·' -A
kP.{PB ~---··-···-·•
.. · -··.
1+ K A pA +K e pC
Wíth that rate law the following mechanism exists:
Adsorption A+S ·--?A··S Surface reaction
A
S+
B(g) ~ C · S
Dissociation C ·S ···3' C+S The surface reaction is the rate-Iimiting step.
10-84
b)
In evaluating the parameters we can also see if our rate law is a good one. Plugging into POLYMA.TH we can come up with the parameters. 4.$00 r-atell'10~ -t.COC
3.200 2.400 L.SCO
o.soo
r
n
l
T'
f_
íl
!
r
n
1
1
1
¡
1,
········~···
2
.
3
!l ·······•·•···· t'·· . - ..J . . 5 1
I
1
íl
11
. . . . un . _. . . ij
íl
1
4
,1
lI
o Regressior, ddta
íl
[J Cal c:.;l at ed v.all.1€'
1
1 1
11
.!J .
1
.... 171 ......
6
7
1
.......
a
Model: r~te=k•~a•Pb/(l+Ka~~a~kc•P~) k 2 0.. 00044'ilSS7 Ki:: = 0.5 Ka = 2.sssas
We find that the rate law is a good one,
k
= 0. 00045
KA=3
Kc= ..5 e) The best places to add poinrs would be where Pe is changed, but PA and P8 are not changed, d) No solution will be given.
CDPlO-J 2 C2 Hs OH .,, C2 Hs O C2 Hs + fhO (A) A+ S
+:::
(W)
(E)
r1
A·S
r2 ""' k2 [ G..s - CE·S Cw.sfKzl r.3 = k3 [CE-s PE Cs/K3]
A•S + A·S +:! E·S + W•S E·S
..-•• :¡.
W·S -·'t t·-·
s w +s E+
At steady state r
= . !:!~t. =
= k1 (PA Cs - CA-s!Ki1
I4
r = r1
= ~ [Cw-s
= 2r2 = 2r3 = 2.r
4
lf surface reaction is controlling,
10-85
- Pw Cs!K.4]
=
r4
~
O
Cw
--
s
: : !=w
C1.
K4
e
t:
.
P + :_!f_ pl 1· 1 + K, P,,. + ---!!-· ' . R.3 Ki.
es¡
1
Cc=C -r = Cs + c.{•S + (~.s
:::
CT --,-- - . 1+KP +!!L+~I A K3 K..,
== _
·'S
ci
"Jk [ K2 2· . ·T ···2 1· l p..._.
:. r - 7r2 - ·-::· e_____
!1
.. ---
PE Pw Cfl
PE Pw]2 + K 1 p.A -r-. -+-.-
l·
1 k• [•, Pr2'",' -PE~ Pw .... - -:,. L - .. - l _ Í.l -. K. P . P:- + Pwi,·2
- ·-K·--·.,.. .... -N.•. -. 1 1'.-1__J_ - ---
K3
K4J
l ,
A
!
+
K3 K~]
where k :::: 2k2 Ct Ki
K'.eq = K1 K2 K1 K4 Using poims 9 and 13. P1,. "" O • Pw r
== - ..... _ kPt
yI· = .L ..L ,f f r
Yk PA
!Si v'f
vs, _pl produces a straight line with slope
A plor of
slope
_
K1 PAF
(1 + +
=O
/f =
=1
A
4.945
ee
~L k
k ""0.0409
m ............ K, = g )• ..)-9 ....... ._~p t =-:¡{' Using point 10, P
¡{I" and intercept 1 k ·..
K1
= 17 . 31
~-=0.0399
O
i:,'J:¡eck U sing poiru 11. PE
=O
~ K4
K3 = 0.659
Using poinr 6, Pw =O=Finaily, using point 12 ~ r
=
= 0.0368, clase enough
K,:q = 0.0975
-0.0409 (Pl -
· 10.256 Pw) ---- -PE - ----------
·-·
(.1 + 17.31 PA +· L517 PE+ 25.05 Pw]2
Note: Ka¡ rnay also be calculated using RT In Keq = -6G
0•
Interested readers are
encouraged to check the goodness of fü of this rare law with the data ..
10-86 1
í
CDPlO-K CDPlO-L Rate law: -rAc
= kCcoC. ACCNaOH 2 (l+KAcCAc)
Proposed Mechanism: Pd + co+---·pd · co
-·---~
Pd · CO + Naorr--Pd .,
--
· CO · NaOH
AC+Pd--1- AC·Pd AC· Pd + Pd · CO · NaOH .... --7C3H5COOH+ NaCl + 2Pd
Neither of the first two reactíons 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 rate limiting step: Then
= kCJ\c
rA
.kAco _._º X:'\CO _
rACO·NaOH_
-0
kNaOH
CPdCO·NaOH
= KcoKNacmCvCNaoHCco
Combine to find C... Cv
= ··· - -
__ Cr
-..................... ···---··
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. Trying reaction 4: !4;::
ki.~Pd•ACCPd.C:O•NaOH
cannot be found so therefore this rate law is also not rate-limiting so none of the mechanisms reactions agree with the rate law. Crd•Ac
10-87
CDPlO-M
a) Stan with a mole balance: cfX
-r¡
dW
F,...0
--··=··_·~-··'··
Rare law comes next: ····r'A =kar Then the decay law: da ;;;:---k, dt "
iv
t::;:: -·- .. -
v.
· --da ---· · = -·-· - -k,,~ -·
dVV U5 \Ve then come up with the equation for the profit:
P -:;: ; 160 * ( F8) -- 10 * ( U5 ) W11ere:
FB -F AO *X Then plug into POLY1v1ATH and gel the following prograrn The feed rate of solids that gives a máximum profit is 4 kg/min .
•
~!S!9E.§..,.
rni tia:!c...YA±~ 1
d(al/d{w);;;-J
o
fao"'l
kd=2 lis=4 10····18 :l.
ra.=-·krtl¡"a fb=.fao•x !?=l.GO~fb·
lO
C.$
~us
O 75
x
e s
o . 75
tao J
2 l ··C
S
!b
-,.,J .
5
C . 75
p
·-4C
80
so
b) As seen above: X= 0.75 anda =.5
10-88
-----··-------------
--------·-·--·---------·-----
---
..
.
...
----------
.
--
--- -··---------·---------------·-----------------·-
------·-----·--·------·-------------------
----------------------~-
e) The only equation that changes is the rate law:
tj!!_ - -·k dt -
l : ,
d
w
.. w u
.::.M.:;;,A:;,X__
dt = -~~~~~
u
kJdW__da _ = ·---~dW
U
Integrating we get this: k¡}W
a-=---+k
u
W
= WMAX@
a= l
k-1-·
k"W
a=l-
kd(i~1ux ·W)
u
u
Initial_value
Eauat:i_2B!:!:. d(x)/d(w)•·ra/fao
o
-: 'kr;::;5 · fao;::;l kd=2
Us=.8 wmaxeL fb,;:if {x-e L) t hen ( fao*xl e Ls e (l) a=if (kd/Us
P=lSO•fb
* (wmax-w)
l t.hen (l··kd/Us* (wmax wl l e Lse (O)
lO*Us
ra·!::j . . a~kx: W-
t
!~;.. tial . yalye
= l
o
M~x~~~- val;,;.':! ~ ~ .. v.d li.;e
5 fao 2 C.B
,:, . 5
fb l'
··8
·S
:-a
··O
-5
We find U= O 8 to rnaximize the profit. X= I ami a= O exiting the reactor.
CDPlO-N
10-89
142 .;
(a) CAo ... CA
ü aít)c=-2..
W=--F\0X _ )a(t)
Design Equation :
(--r~
\,
\F
f\0:::;;
==>
For nrh order kinetics.
u
·~
kC;\
106
Now F8(t)
= FA X(r) = RON(t) 0
X{t) == ~01:_(t)
ros
"".!::.2~J!2. RON{O)
~econd order kinetics and second order decay rate fit the data very well. ) ·da k a-t'(.=-·
2
. .. l 1 5
Plot of --
e}
(.Ao ·-
CA
.
vs.
y= 0003x
+ 00855
t :
+ o From the graph :
intercept slope
(b)
Exp Data
-1.1near t Exp Data:
o
·············-···--···-·········•··-···~·--·~··-··-·-·--~-·--~
= I I k == 0.0835
= kd I k = 0.003
tOO
200
------ .... ~ -·------·----~-~~·
.
300 t (h)
iOO
1
¡
500
= 11.98
=::;,
k
=>
kd == 0 . .00025
Activation Energies both for rate constant and decay constant can be estimated from the temperamre time trajectory
CDPl0-0
10-90
Given
A --+ R
+S
Batch consranr volurne reactor, P increases with time
NAo dX
=ar
0
Wdt
Assurne
'º::;: kPA :. PA=CART CA= NAO
O ··2.Q. = -~,\O ( I. :s;J_ = CAo ( 1 ·X)
V
:. NAo d..X
V0
= We-0., k CAo (1-X)
ex = w k NAo RT J. f. . 1-X l(
~AO
o
=~
o
Le=
dt where
•
ea.:
RT dr
dt = {W·) k RT Yo,
f.' e(tt
dr
o
~ =(~·)k R1
~ ( 1 e=)' -ln ( 1-X) == ;¡ Assume that there is very little deacrivation in the first l O sec -ln{l-X)= ~ .. (at)= ~t (forsmall t) a.
~=l. in (-·lJ = .L, In {-·····-L-) = 3 ..Tl x 10·3 sec 1-Xl
t
~·(~)kRT k
10
=
= 6.63 x
1-0.037
k=(~jRT"i-;~~".:~iOí 10s sec!
10-91
At t relaríveíy Iarge, e--01 = O ~ . - :: -ln ( 1-X) et
CI
~ = ---= --3.37 · X. . 10·3. . = 5.18
~
x 10-~ sec··•
1n(-2x") 1n(r.0_511) ·-ln (I-X) = 0.728 (i··-e·:S.1ax1o-'lt) u
o
,.
10
0.037·····-
10
0.071
40
....o.T30
--·30-,...
0.102
..._ 60-·
0.1&0
--80 ..... ] lG'0····-··1
0.223
150
... . . .. -200 . ·-· . __ }00
..
. 0379
o.
1. 1
1
6
0.099
1
0.127
0.177 0.219 CJ.2-=-55,,._· _,
-r· 0.325 0.375
...
OA37
. 1
1
..,_.o.4'ir· !
fsci·-· ·-t
.... ~ooo
o
-o..,,.03,,_6 ......-
-·T
0.2)9-1 0.330
,
g~~~l : -%1i-8·1 0.s1, ···-¡ o.s1s_._j
TI1e assumprion of a firsr order reaction kinerics anda firsr order decay kinetics is jusrified .
CDPlO-P cyclopentane
< <;
Batch mole balance :
d/.l
n - pentane cake
. . --dt~=
-rAW=
dCA.
kA .• ,.
k,J:;w
. d/. =--~-¡;·e Aw da
. . dt : : kat.ll e,; : ; :; kdaq,
WÚh
'
1 + kdt
10-92
e
A
constant
,
Ifn=O:
l..::_I
X
k:t
+5it k't ,
If n
=1
__¡__
_.Q_
~~--
---- -
. .. 800
1200
kt'
l.33 l.414 l.493 1.653
1
l 1
42
1
-
1.811
f24s-·
2.053
L498
1
--r·1
'
t............ •-•>A••<•~ •••••
o
~
!
1.053 1.381 1.777 2.333 3.545
·3·_¡35¡
1
1 1
l~(l-)q
• (1--X)/X
.. •--·••••-O•OH••-·--·•••••H•,0••00••"•0--
u,o
~~~~~--+1
5.620
1
_Q.81~~ ..
1
2.241
:
~/
:~
0.653
!
~-,- -»:". , ~~'.,. ...
~_.,,,,,
\
º·22~_j
-º
~/~/
:I
U.415
-:»
!
o
1
0.333
1 1 1 1 1
l.836 .•.
4.545 22 16.3 . .........1 --6.135 -
"',.). 1
,r '\
0.902 1.076
2.381
-x·-·
1
0.815
36 1 --· 2.778 3.333 30•...•. __...__ 1
Tl
¡1:x¡
-lnO-Xl 0.721
l(
1
48.7
1 1
. .. _1..___
.l.
1 1
55.2
1 1 ..... .....1 1
) 500
.5t t 1
70.7 67 60.5 •.
40
•.
Ici
75
20
120 180 250 3·0
_j_ +
X(%)
1
t
T
=;
·ln(l-X)
.-·-----.
1coa
,ca
UlD
From the above graph, all lines are straigh; Iines, Therefore, q = 2 is a good assumption. :
'
We need to examine the data to see which value of n having ~ J
will llave similar behavior
of Ay ~use:
At
•
= constant, n = O and n = 2
li.t.
1-X=l..-l
X
X
n == l ; .1.y ;;:;: L\ .
Áy
llt
[·--ln(-L--1 1-X}J
and
.
Át
= 20 mín
""0.094; 0.087; 0.087 ; 0.084; .... ; 0.081 ; ··-;
0.0797
Therefore. tiy is decreasing gradually. lt is nota constant, At
10 .. 93
For n
e
O, n=2: ~;::. = 4 x lQ-3 = constant, ót
, If n::::D : .
9-~i = KR ( ··· · ·...l-:···· ') l
\
i + kd t
4X KR í l \ . -- -1 ·--····-·J cu
CA.o \1 + k~ r)
If n=2 : - ec. -~· = KR ar T'
(
1 -----1....• , 1
..
\
¡ cz
A
" -r- K¿ t ¡
... J. __ dX = KR CAo. {1X)2 dt 1 +kdt {l·· X'f . OL = dX
1-.
KR CAo
+ -~d-
-··· t
KR CAo
(1 X}2 .&!L dX
n=2 : Slope is negative, It is unreasonable n={):
c&-· vs,
t
is a straight líne,
10-94
O dt/dX
a
W
= 0.01 kgím'3 e w
't - ~-·
FAO
(b)
(0.03
= ,_
m
ni .. -20
'
1.5 x 10 5 kmol/mín
k
g mín
{m3)2
The order of deeay is q""2 , k .. Ji "" 4 X
kA
(e)
krrf.L) (0.01 kg)
tn·3
=:>
k: 3
X
103 rnin
¡
Movingbed reactor: FAO = 2.~&!. ; X = 0.80 *
mm
iia. = k, al dl:.
d
In moving bed :
t
= if· where u == r:J::·
dt=J.u dW
.
.
.
.
Substinne mto equanon (l):
~~
k:i "
.....~ . . =·-u·· · a-aW
. ~.::;kd.dw a2
u
10-95
a - ···--.
¡
----
•.l ·1-·-·--K.é w u
Substirute into equarion (2), we nave:
fk-~ol,x ex
,. .
==
,ó
JÍ" dW -~
'+·ki~'U. 1
!:_~Q X = 1L L., ( r + K.é V.,:) k:\ .
W
K,_:
=
• \.
u_ · Í exn ( X
k¿ \
• \
u
f..t:Q_k:1) . U kA
l.l/\
W == 455 kg (d)
u
u =
~ k:g O . ) ·--';"-·· tmn
W
=48.6 kg
CDPlO-Q
10-96
a) Mole balance:
Rate law:
Decaylaw: . da. _ -- k,1 (' dWU5 'B Stoichiometry:
CA
= CA(¡(l ·· X)
CB = CAo(es +X) Evalúate the parameters:
e¿
88=-·--
(rn
o.or.... -0.1 0.1
rn;.,cia_l.,.va-1.ue
o l 1 .. 000
k=350 c ao= cL
ca=cao* í 1. ·x) \ cb=cao * ( thet.a +x) ra=:·.. ·k"ca•cb
"'o"' o.
10-97
b) The only change is in the stoichiornetry:
C, = C,.0(1-X)(l-
C8 = C,.0(811 + X)(l
aH')º' -r-
aW)05
POLYMA1Tl ~~d(Xl /d(w) =a• ( -za) I f ao
Inltiál
d(a)/d(w)=-kd*cb/U
1
V~h!.§:
o
kd=6 fao:20
!..
U=S k=3SO cao« .. 1
r------ . .
:::oc
e.oc:
TT
c . 600
. ...¡...
T
t.heta= l. alp=,,038 ca=cao·• (l -x) • {l··al¡;,*w) '·, 5
cb=c ao> (theta ..x) • { l··alp*w)
~o . 5
ra=·--k*ca*cb wO = O,
"'"f
=
24
CDPlO-R Cumene (A) a;J
--t
+ Benzene
Propylene
(S)
(k)
..
ro
(a)
We find a relaríonship berween a and CA {W=lg)
.Run 1: (
o
a
l
CA
60 120 0.75 0.594 0.01 0.018 p A~ 0.4 arrn
o
Run 2:
180
0.491 0.0243
t
O
100
200
3CX)
400
a
l
0.833
0.733
0.583
O
0.0057
0.0106
0.65 0.0148
eA
10-98
0.0184
Plot of ln a vs, CA gives a straight line passing through the origin with slope
a
( single sire adsorbed. surface reaction controllíng] I initial rateare used, Pa == Ps = O and 1 » KAPA (adsorption is small at high temperature), then ro"" ki ·KAP,..... Us:ing data at time zero from runs 1 and 2: k~ KA= 3.2x10-3 Hence, overall apparent rare law is r
= k; KA e.a.e... P A
;
k; KA ""3.2x10·3 a.= 28.9
: = - k:¿ amf {PA, PR, Ps} = -k:¿ am P"A Trym =-2 Since P A is almost constant .. duríng runs l and 2 (low conversion)
Run I : k:¿
P\
= 5.767
x 10-J
Run 2: k¿ PA = l.769 x 10·3 PA == i in run 1 => k:¿ == 5.767 x I0-3 PA ""'0.4 in run 2 => n"" L29 e
k1 KA= 32xl0·3 k¿ = 5.7ó7xl0:; 11::::;
1.29
t in minutes
10-99
= 0.60
(e) Overall conversión
{2.8 scoüsec curnene j \ 4.2 mol/sec propene > L.= 11.2 mol sec l,4.2 ll'.OJlsec ber.zeneJ Cumene ··-t Propene -t- B enzene (A) (R) (S)
Comnosíríon ar reactor outlet •
YA'=··2·8
11.2
=0"5 . ·-· A ")
VR "" Vs :::;,. .::?....... ""
' .
'
11.2
0 . 375
Composition ar reactor inlet, F0
= 7 + 3 = 1 O mol/sec
FAo ""' l + 3(0.25) = 7 .75 FRo = 3(0375):: 1.125
YR = 0.11""
Fso"' 3(0.375) "" l.125
Ys = 0 . 1125
Let X¡ :;;;; conversión per pass Total flow ar reactor outlet before the recycle stream is:
At any poiru along rhe reactor:
Assume that rate law in (b) is still good for the moving bed operarion (rnay not be true in pracnce because of the high conversión) • --
r e a k, K.,o,PA '··.
k, KA PA =-I+k.:!PAt .. ·-····--···
Moving bed reactor:
F;,,odX
=
t
=
tj~
rd W
10-100
= ?.:75x5.7~~10-=~(1.29+X.\º·29W + _... 1..7~---·-( l.29+X.) 3.2xI0·3x2000
= 6.98
X 1Q'·3
1-X J
3.2x10·3
1-X
(l..22±Xj29 W + 2421 (122.±X.) 1-X
ix
With X=O , W=O If X= 0.542 • using digital competer W
= 3.1 kg
CDPlO-S a) Mole balance: dX dW
= a* -r'
1..
rate Iaw:
-,:4' ; k'C ·. 1! Sroichiometry:
Decay law: ·-· da = ,-k,1 .~ a dW U5 Evalúate rhe parameters:
t
= (l
¿
= (1-- lOOo:)05
Po
5
o:W)°s
o:= 0.0099
rº l da kd ·-·11 -;- = "i~~-
J/W ~100
!-.tL = 0.023 Us
10-101
POLYMATH Egua~ions.,__
Ir:it:ial_value
d(x) /d (w) ea---ra•a/fao
o
d(a) /d (w) a.-·kd•a
l
kd= .. 023 fao=4
11}0
k= .09
X
cao=2
a
alp= 0099
kd
ca,,,cao* (1.-·x)
G
O _75;:51
C.02J
0 .. 023
o o
k
e.
O., 09
e 09
eco
¿.
alp
O. 0099
O.C09S
e
f ao
.. {l···alp•w) ~. 5
0·9
ra
·O OOH1il
·O. 18
T
a. uec
o
..:!;"y~ .
.. ,e
·:JJW
I
ü.:60
c.ccc
CDPlO-T FAo
Rate Law :
r;;:: k'CA
DecayLaw:
coa
-1~ = a(WX-r;)
Design Equation :
(for sinrering)
Stoichiornetry : dX
Combine:
1n(i-\<)= From the problem statement a_. ,
exn
k'
!:·~:1{1+ü!w) 1
=···-···= 1 ,.·
1:,
"'u,
Plugging in 100 kg for W, we can sol ve for
l
W
ti!
·--·
4
s
!2...::0.03
Us
ln(--.1__; ')· = 1- X
-~({.:.~92=~- ln (1 + (O 03 X100 )) = l .232
(3.nXO.(;_,)
:::::}
o. 0'1 2
ca
,b)POLYlVlATH
:002s:i {)2J
100 C.757161 0.:00259 e 021
1-X
X= 0.708
10-102
· ·=343
dW
0099
0 009,
0.0435678
0.04S5ó7S
---0 .is
··O
004371.1
CDPlO-U The heat of actívatíon is given in the problem as a functiori of the carbón number so we can just graph that:
r·----------
------- .
.... ------ ----
Heat of adsorption
i
vs
------------·-·¡
carbon
, 1
!
number í:
- -----·--¡
80 · 70
.g
eo 60
-
~ 50 "'40
o
30
-
';"' o
··.········------
o
10 Carbon
To graph the activation Arrhenius equation.
.......-30
20
40
number
energy we need to find its equation.
Ir 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 temperatura 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 sorne equation with the above equation that fits thís criteria and we can come up with the following graph Activation
>,:n
.
;
e:
energy vs number
carbon
5 4 . 3
e: o 2
.; > :¡;; u
<(
1
o
---------~----·································.··················· 10
20
-~
--·-····--·--·--···-·----·-----··---------···-Carbon
4
number
The reason for this unusual temperatura dependence is due to the fact that the higher the carbon number the less it wants to add a.nother carbón 10-103
CDPlO-V (a)
a= O at the end of the reactor: da
---·--·-dW
"..
j - da = 1
1-0 = (b)
For Us
a=l
k0
u, k w R. dW Uso
J
(2{t'}
==>
l·· a =-~2-W
us
==>
kg)
Us
= 1 kg/s
= 0.5 kg/s:
-f¡~ W = 1 ( Jsii; )w = 1 -(0.4 kg ')w Catalyst Activity vs. Catalyst Weight
to: ---·- ----'-05
l
w (kg)
When a= O, the catalyst is inactive. In theory a can be negative, but in realiry, once the catalyst is inactive there can be no further decrease in activity.
(e)
For a catalyst feed rate of Us = 0 ..5 kg/s:
xi= a(·r:)
Mole Balance :
fAo
Rate Law :
-r~ =kCACa
Decay Law : Stoichiometry :
.. ---~~-- = -~:P . = ~-?.S. -1. = O .4 kg 1 dW
Us
0.) kg/s
(Assume T = T P ""' P and u "" u 0,
CA =CB =CA (l· 0
0,
:::::,
O)
X)
Combíne: FAo
~= (1, .. o.4w)Íkc~o(1 .
-xy]
(!t}í~jl+·· iJ~w}w 10--104
a = 1 · OAW
J
(1?'x)- ki: ( w-2~,w'
Frorn part (b) we know that the máximum catalyst weight (the point where a= O) is 25 kg We will find the conversión at this point:
Ci~xi = i1XJ1:f2'.{ z.s
-$is) )= o 2.5'
25
X=0.2 (d)
To achieve 40% conversión:
(1~xj=~r:{W ··fitw')
r~t~ = o { 5-fJ;5') lJ5
O . 667
=>
=1
2.5
= 7.5 kg/s
(e)
ForU, ==:
z¡·?xr!'f:{W · 2¡!)W')
(i"!X) = (o.2)w = (o.2X5)= 1
=>
x =O . .so
==:?
CDPlO-W N
Design Equation :
Bo
~:=··r:aw dt
Rate Law :
• tgI ;;;; k'('2'B
DecayLaw:
da , , ··= k D = 0.05 dt a
I
Jda= ·0.05Jdt
~
a=l·0.05t
o From this we can see that the máximum reaction time is 20 min We will find the conversion ar this point l
Stoicbiometry :
(Assume constant volurne) CB
Combine:
= CBo (1 · · X)=
.~~2 (1 . X) V
.~?c.== ~'(1:~:~?.t)~!,Q.::~.L~--~:~.!1.(¡ . . Q.05tX1 ·X)2 ~
v-N
· ~ -:;· = !-:'~.!!!!. (1 · · 0.05t)It
(1 .. X)'
V
V2
Bo
=>
... ~. = 1g:9!X!?t)2 (w . o.02s(20 'f )= 1 1-X
(1)
X =0..50
10-105
--~
1 •X
= .~:!~.!!!!. (t "' 0.025t V
2)
CDPlO-X a) Mole balances: dF
--..!!.-::::
c!W
'ª
rate laws: 1~-=
ak2C,1
stoichiometry: CA
5 == . .F. . 6.:::: F(l-aW)° .. '1. •• -·-··· --V
Vo
. _ F8 _ F~(I ·· aW)05 CB - ... _ - -------···-·-·· · · · V
V
o
decay law: da
-·· ---·
dw
- k·--a~ 0
..
U,
...
Evaluare the parameters:
Plugging all of this into POL YMA 11f we can change valúes of l\, T, and v (P ,._0) that will give us the most of product B,
0
•
We find tha; at a ternperarnre of 396K, a solids velocity of lOkg/s anda gas volumetric velocity of O 633 dm3ts that corrcsponds to a partía! prcssure of 51 3 aun we can the máximum yield for B. Eguat.ions: d ( fa) /d (w) "ra d(fb) /d
(w)
=z:'b
d(fc)/d(w)'-'rC 1.00
d(al/d(w)=kd~a/Us
.O
üs=lO
f::,
T-=396
fe
o
Os
10 396 0.0098 o 633
10
o.o65U.49 o. OC:77S5S, 0 .. 01761.31 l 57973
al.;:>=9. VO=
o. s24::3a o.~71.'171 l.93'i'02
Se-·3
633
kd=.ó$•exp(1SOOO/l
987•(1/400··1/'I'))
k2= Ol*e;q:,{20000/1.
987* (1/400··1/'rl)
alp
kl=. 02 •exp( 10000í1. 9S7• ( 1/400-1/T))
kd
o
ca,,,f.a• (1--·alp~w)
k2
o 00 I i555~
kl
ó.0176131
ca cb
1 5'19"18
A
5/vo
cb=fb•(l-alp*W)A.5/vo :·a-:.:--a*kl
*ca
rc=a"*k2'"'cb ·:c.b=···aff {k2*ch-k1."'ca)
0661!<1
-o . 02,a2,a O .0::!'?'8248
10-106
o o
lGO OQl:394:2
o . co·~a O. 6:3.l
O. 630·42
···O . 000620403
o
0064.lJ66
Q·.
0278246
O .:;,4f,6)1
l.93702 10 l.96
196
0 .. 09llS4Z
O·, 0098
!.O O .0093
o .,S3l o 066!H9
o .saa
O, 00775554
o. 00775554
0.0176131 O. 01.8184.6
-o ,o:08240 ··O .00217423
O . 066l!4:S O OlHUl 01618,~
o
O . 09~;:n 546 ··O .·~00610403
O .COli:$'~02 .. o. ~ooa·7a;:;.2:.
b) Using the same prograrn we can find what it takes to get the rnost of C possible.
We find that ar T 1082.4 atm.
= 396K, U,= 10, and v
~~!E-f.! Ma.ximum
V_e:::i~
= .03 we can get
0
lmol C/s. P AO =
value fir..al .. ~!~
val,::~ Mínimum
"'fa
o
100
o
100
l
l
r 02629e···23
l.C2529e-23
fb fe
o
o 524487
l . 35042e··.:O
o
l
l
a
l
l ... 9370i
l
Us
10
10
l.O
10
'.I'
396
396
:!96
396
al.;;,
O 0098
0.0098
"º
o
0.0098 O 03
Q.0096
0 .. 03
kd
0.0661149
O .0651149
O .0661149
0.0661149
k2 kl ea
O. 00775554
0.00715554 O 0175131.
0.00775554
O. 0077555(
O 0176131
O . 017613!.
J3. JjJJ
t. 8}798e·--23
o
4.83'19$e-23
ch ra
-o
o . 01·1s131 33 3323
0)
17.26~6 .581104
re
o
r:O
0.587104
··l
1.93702
0.03
6 .. )655Se-l0
'º. 587104.
65057e--·24
-1
o
0 .. 13615
o.sano.
-c
óSOS'le .. ·24
9 S6332e-:2 -9 . S~3J2e- .. ::z
.Cr7lS7S2
e) To get the Time-Temperature trajectory we can use the following equation to
creare ir:
t
=
1-·.
expr·--"'---~----{}---- - - - . ···4~o)] -------· kd0(Ed!E..,,)
This will give us the following graph:
Temperature-Tim-e Trajectory
1 ~m · E
400
1-
200
<11
o
J
i
. o
5 Time (s)
10
d) For this we jnst add an energy balance. We have to assurne a heat capacity of rhe catalyst since none is given. Here it is assumed to be 100 J/kgcat dT
(r¡){AfIR>-)+(,;)(Afiiu)
dW = Cp/Js + cp(FA +Fa+ Fe) Plugging this into the POLYMATH gives the following prograrn. We find that the temperature is 388K, U, 62.4 arrn,
= l O kg/s and v = 051 drrr'zs. O
10-107
P Ao
=
;Eq,..1ations . .:. d(fa}/d(wl=ra
!nitial
. value
l.
d ( fb) /d (w) =rb
o
d(fc)/d(w)=rc
o
d(a)/d(W)=kd•a/Us
l 388
d(T) /d(w) = (ra• (-16000) +rb• (··32000)) / (e¡::,• (Us+::'a+fb..fc) J kd:: .. 08:*exp ( 15000/l
.. 987*' (1/ 4-0G-··l
/T))
~ia~le
;n_h:;:Od.l _•.,alue
Us=lO
~.axi!!:Ufflv~l!:.,.~
P..i.-:d.cnum.~~
10-0
r..k:_~1_~ 1CO
O ,17.)516
cp=lOO
k2" Ol'exp ( 20000/1. 987* (1/400-1/T)} kl=< 02*exp ( 10000/L
987* ( 1/ 400-1/T))
fe a
D .. 51.U.56
o
0.611156
O 2.15326
o
D. 21$328
lt04Cl
alp=9 . 8e
T
asa
:)63
)S.l.46~
JBZ.242
voe .. 51
kd
e o.;.;c,169
0. C•Uii2&9
C.0319"75.3
ro
10
LO 10-0 .. OC29Hl9 0,ól08521
0 .. 0332906 10
o . . ,111,i17
0.0098
O .0099
ca=fa*
(1--alp•w)
A
cb=fb*
(1-·alp•w¡
A.,
.•
$/ve Stvo
cp k.2
ta=-·a*kl•ca
kl
t:·c=a*k2•Cb
<.ÜP \/0
:,e
lCC
.-004sn-01 ,'.)1)553 .0-096 O 31 900"73
D. 004.59207
013553 .0038
s;
51
02:S~· 114
o .0:2.6S·i.44
10-108
···O. 00·0"153076
O .V027125a
'$1
o
96()'?8 O 7SJ935
-c e
100
.Oi!SllS'i
O .:0941'2 ··O
0265'74-<:
-o . aoo'153076
o .oo:ni,224
l, l85.l6{~~ 05
Solutions for Chapter 11 - External Diffusion Effects on Heterogeneous Reactions Pl 1-1 Individualized
solution
Pll-2 (a) A
2B
1,
Z ó, y=O
WB =-2WA
w
A
= cDAB dyA =-cD dln (l+yA) (1 + YA) dz AB dz
Integrating with y A
=O
WA = cDAB.ln(l+ YA)
(S-z)
at
Z
=Ü
at z = 8 (1)
y A = y AO
cDAB WA =8- (l+YAO )
(2)
Taking the ratio of Equation (1) to Equation (2) to eliminate WA and solving for yA ln(l + y A)_ ln(l + y Ao)
8-z YA =1- (1+ YAO )
8
1-z/'6
11-1
YAO
' ' -,
YA
/
ElV[CD
''
''
O
Pll-2 (b) (g)
kc2 kcl
''
i.o
T2 = 350K
T1 =300K
=(D AB2)2/3(~)I/6(U2)I/2(dp¡)l/2
DABI 'U2 Ul As a first approximation assume
DAB2 DABI
dp2
(11-70)
=~
µ2
then
kc2 = ~ 5/6( dp¡ )l/2( U2 Jl/2 ( 'U2 ) kc1 dp2 u1 At T1
= 300K
µ1 == 0.883cP
At T2 = 350K µ2 == 0.380cP Assume density doesn't change that much, u=µ
o
~ = .!:i = 2.32
v.
µ2 U2 1 dp1 -=-,-=U¡ 2 dp2 k,
1 2
= 4.6lxl0---{j m/ s
"'~ 4.6lx!O_.
m/s [2.32 t
[H' [~f
= 4.65xl0-6 m/ s WA = --r; = kc2CAb = ( 4.6lxlff--{; m/ s )(103 mol/m3)
--r; = 0.00465 mol/m
2/
s
11-2
Pll-2 (e) (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 -k,i(1;J2 - 2.9m/tr = l.3rn/s
J
X= 1-exi{-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) U2 = e-4,ooo(_!_ _ .l_Je -4,ooo(_l T1
U¡
T2
1_)
773 873
= e-0·59 = 0.55 Assume DA2 DAI
-(·/i¡ Jµ2
_l
0.55
k c2 _ ( DA2 J2/3 ( V¡ Jl/6 _ ( Jii J4/6 ( Jii Ji/6 ( Jii J5/6 ---- - = -· kcl
DAI
V2
µ2
µ2
µ2
kc2· = ( 0.55)516 = 0.91 kcl
ul~c2
U2k1
= (1.059)(0.91) = 0.96
ln-11-X2
= (0.96)(2)
= 1.92
X2 =1-e-192 =0.85
11-3
Pll-2 (f) CAO
Assume concentration in blood is negligible (C A2
d(VPCAp) ----=-WA dt
w
A
A
= O at 82).
p
[c -c Al ] · A
=.!!ABI
Ó,
1
w = A
DAB2
Ó,
[cAl -o]
2
adding W A
[_!_L+~-]=C
WA -
DAB2
DABI
A
CA Ó, . _2_+ __1_ J,
DAB2
DABI
CA =HCAP V dCAP dt
p
=
ApHCAP Ó¡ 82 --+-DABI
DAB2
R-(~+_!i_J!A -
DABI
DAB2
p
H
Flow into the blood FAB
= CpA mol/time RV
If CpA = constant FAB
, R =[time]
= CPAO
_ CPAO _
R
11-4
Assume quasi steady state
t If CAD varíes !CAP =--1 CPA dt V R
In CPAo
=--t-
CA VR CA= CAo e-1/vR
CA
FB = -
R
81 and 82 are given in the side note.
FB
t Pll-3 Mol balance on oxygen :
Fo2 • O +
ro2
=O
constant liquid composítíon. where n is the reaction order.
. id r: Assuming l e·a! gas l aw appues : F02 t p Assurnín• zº tnat ···· R.-T····
. = constant . . .•
('. ,
:::::
P.»; ir"
W h. ere
·~1, rate mí u, = oxygen uptaxe . ...1~Ul
by correction of 1.lo to sorne reference,
then Fo, = C. u, Assurning Henry's law applies (low pressures)
C0i
= H.Pai
where H is Henry' s constant and Po2 is the oxygen partial pressure
11-5
Substituting
into che mol balance :
C u, = k Co-"
n k: . H" P o,.-•
e
U= o
ln
= k.(H.Pozt
Uo .:::
i.n:
n.ln Po, + ln ·-···"'
-
e
A ploc of ln u, vs ln Po: will give nas che gradient. As che systern pressure, P, given in the data is absolute : if in the reactor. the xylene is ac boiling point and dissolved oxygen and oxidized xylene are at low levels, then P, 1 atm (open to rhe atmosphere)
=
In arder to deduce the correct kinetics of the oxidation it is necessary to find the pardal oxygen pressures for the condirions where the rate is limited only by che reaction kinetics and not by diffusional mass transfer, Plot of stirrer speed, W, vs oxygen uptak:e rate, \).,, for each run conditions at which diffusion is negligible.
will show the
Ir can be seen rhat at stirrer speeds above 1200 rprn that OUR is insensitive to W and hence the reactor is well mixed i.e. no Iiquid diffusional lirnitations.
.. ,,- ·-········. ..... ···+·· t 2 atm-
, .......
--.-
··• ..-1.6 ...... -2.0
.:::..*·· +o 400
900
atrn atm
e!_m.
1400
OUR: Oxygen Uptake Rate
W, rpm
Hence using the systern pressure data at 1600 rprn for the plot of In u, vs ln Po:! will give n uninfluenced by diffusion . Po:!
= P ···
Px
=P
· l
- ~- _;.!i}-t~~~+~ ~ . i rr~=: 1 t : ~ ,r~~ [=J . - · At l600 rpm
··-t-··· · · · }~~- ·
- · ,-· . ·-···--··/o· · · ·..
11-6
---···-M9·--- . - ·-
~---·--·-~--··· ··-·-·-··~-··---·"'·----···--·---·--
y
·1.5
·2
= 0.9992X + 4 .. 6422
o
-0..5
0.5
LN{P02)
Toe gradient, n
= .999 = 1, so the tate law is:
Pll-4 Diffusion in adjacent skin layers Skin laver interface P:-11
P:"'1 P:,i~ ::: l O! kPa PHc
Inner layer
Outer layer Srroturn corneum
= o kPa
ó, =0.002 cm
z=O -···· -·•· -·
PN~
Epidermis
~=O.O! cm
_
-------···--······--··-···· •""'"''-"""""•
-------...
Assuming dilute solutiou and constant total concentration in both layen; gives:
-·dC,1 ···· = K, dz
for each diffusing component in each layer, N~ :
Outer layer :
boundary conditions
z =O,
CA= C...0
z = ó, , c. . = CAJ
11-7
= O kPa
Pi;~= 81 kPa
C., = Cw - ( Cw -- C.A1 ).
Profile
Inner Iayer :
boundary condítions
.
Profile
CA :::: ··-., . ··- ..
¡
1
z = or , c. . . = CAI z :::; &:!. ' CA :::; o
. (82-··z)
(,11 .. -··· ···;:-· -·· ....... ·-·..
\Oz --
01.
Total partial pressure profiles
P,, + Pe = Pso ··· ( PAo - PM).. !:. + Pe: .3:. . 31 s.
Outer skin layer :
= 1010 ·-1000 = Inner skin layer : P,
+ Pe
= Pv
1010
z + 5000.z
+ 4000.z
0'·....·····'..::~. ) + Peo ( -;::···:b:--81
(Pso ·····
(º'-"')
Pa,) ·--.:.......::. \0:··01
= 8lO+126000(0.01 ..... :::) · · l 00000(0.0 ""' S l O + 26000( Check:
o.o l
l ··· z)
2:) outerlaver = inneriayer
At interface (z == 0.002. cm) :
P, + Pe
>
P, ·+· Pe
1010 + 4000 ::
= 81.0 + 26000(0.0
1018.::::: 1018
Plot these two pro files across the skin frorn
z
= O to t. :::: O O l cm
11-8
correct!
l . . . z)
He :
Inner layer :
boundary conditons
. K1
=
Ceo +Ce: = ······--·-·· · · · -· · · · · ·
K:
81 --81
Cso·· e» \ = Cso= ( ~-----. -· U52
\ 81··&1 )
º' ..::.7'). ( ó:--81
Ce = Ceo ·- ( Ceo - Ca,) _:
Pro file Outer layer :
z = Ó1 , C8 = Ca1 z = O:: • Ca Cao
boundary conditons
z = Ót , Ca = Ca, z =O, Ca= O Ce! K1=--
K::=0
ó1
e, =
Pro file
Csr(t·)
Total concentration profiles : Outer skin layer :
Inner skin layer : Evaluation of C\l and Cm : N::: :
Outer layer :
Inner layer :
W,u
. -· o]. = --···--- r·e,,., D2N2
02-61.
Assume that flux in inner layer
· D1.v1 -W..\--- = [C' ·
AO-
IV,h:;;
.
= flux in outer
layer Le. W Al = WA2
e· .] Al
+
·
~~;i,]
íh-01
= C,o
Conversión of k:Pa to kg/cm*s2
10lkPa=l01000Pa=l01000kglms2=1010kg/cms2 r. lkPa=lOkg/cms2
11-9
w -
E_Ao -···
·=·-·-···---~-10--····----=50*10-2
. [o~,+~é:J r~~i~~·~%~i':1 He :
an•s'
Outer Iayer:
D,H,· ¡· ] Wa = -······ " .. Ca1 -- O O,
Inner íayer :
Ws
D:Hc¡· = ·---------. lh-·61.
Ceo
Ca,]
Pressure Profile of skin Jayern 1050 e------····--,-·-------·-··-····---·--····---···-·-----·--·----,···--··· --1
'!°E
¡ é e
.•
1000
innerlayer
outer !ayer
950
::,
e
900
o. ]j 850 o
interface
1-
800
o
0.004
0002
0.006
0008
001
z(cm)
Hence
=
P)I
~"Vi D2,vz
189 el Oe ··• 2
L5xl0e ·· ,-
-
0.008 = l008kg I cm. p¡¡¡ -
VlHc
Wa
5x10e-2 ·······----1 Oe . ... ::,
15,
0002
ts.__
...
... - ·········- .... ---
= lOkg I cm.s'
11-10
The máximum sum of partía! pressures occurs ar the skin layer interface z ::: 0 . 002 cm
PTo1..i:::
1018 kg/crn.s '
P== 101 kPa= 1010kg/cm.s:!
1018kg/cm.s2
Hence the maximum sum of the partía! pressures is slightly greater than the saturation partial pressure and so gas will form bubbles at the skin layer interface causing blisters.
Pll-5 (a) Packed bed, mass tranfer limited, gas phase
Part (a & b)
where Fa= U.Ca.Ac
Mol balance :
-"""'l"" ··FaodX · ·-"·-···-··" = ra.. . ac Ac
d:
U.Cao.dX ..................... . . . .... = ra .. .a.
U
dz
= constant superficial
gas velocity
Mass transfer limited boundary condition
Rate law :
ra" = kc . (Ca · · Cas) Stoichiometry
but Cas =O, rapid reaction
Assurne constant T. P, gas phase ,
Cao.(I · X)
Ca - ···-···--- . . -- . ··· (!+E.X)
U Cao.dX
..
. . . . . . ·-=ra dz
Cornbining :
Cao.Uo .:
-·dX - -· d:
1!:..: dz
where
yao.S == 0.05 x 3
E=
.a,;
as . kc Ca
kc.(l- X) -. ~ .. Uo.(l + 0.15X)
--a-·-
c-,
,.....,,
;,
Use Thoenes & Kramers correlarion for kc :
.
( 1 = [".µ.Ú·-ñ~b- \1 ··$/y
kc.dp
1
U.dp . p
]112[
$).7.
11-11
µ p.Dab
]":l
= O. 15
Pararneter evaluation : d
P
=[.?.v J =[·i~~~L9: :~-1 = 0.238cm n: Jt
'
Dab
O.OOIT1·75
r
P
.
~
..
Diffusion of cyclohexane in hydrogen (assuming constant T, P) i
l
-·ti'
L-~~¡;, + -¡;¡¡;J -
P[(l Va yn + (I Vb r
r
= . . -------e------•··Ú·----------- . ·---··--
3
Fuller, Schettler, Giddings for binary mixture, low pressure, non polar (Perry' s handbook chem.eng. ) Ma , Me= molecular masses Va, Vb = diffusion volumes
= 84, 2 respectively
= 122, 1.07 cm'rrnol
d2 -s-tc.d.l
. . . ..... . . . . . . . . =
respectively
o.s' + o.s.o.s: , ···-">---··-······-----------·-'"""'
0572
= Ll 46
4
U= Uo.( l +EX)= 50.9(1 + 0.15X)
11-12
µ = 0.00017 g/crn.s (H2, 500 K, 2 atm)
Se'- -
9.ooon. ..- -1.044
0.000 l 9x0.857
kc
1 l47.3 . .
ae
= ~.Q.:::.~·;~¿ = 6.29cm·
1
0.572
11 = 0.00017 g/cm.s (H2. 500 K, 2 atm) . 50.9(1 + 0.15X)0.572x0 . 00019 Re = _ -- . ····------·· ·. 0.00017(1--0.4)1.146
Sc'=-~00017_
0.000 l 9x0.857
kc= . ··-··1
. = 47.3(1 + 0.15X)
-1.044
, ( -··)·¡112[. L 044-vn .. ·S( l+OJ·) . sx· 112 . [47.31+0.bX J =lt.9 1
0388 ·
6(1 ··0.4)
a, = ---· ·· ·- ... 0572
·
. .. = 6.29cm
1
POLYi\•1AlH
Pll-5 (b)
11-13
§g\lacbon:s;.
fni,tial
d{xJ/d(z)=2.22•(1-x)/ll•,15*xlA.5
o
¡:f
=
:':!~iable
5
o
:z;
va.lue
5
o
0.99997
o o
5 0 .. 9999'7
t.000
KEY:_
0.800
··-· X
Q.600
C . "'tCO
0.2CQ
o.oco
z z
:z.
.
2 S25 2.6875 .t. T5
O 9%91419
2 8125 2 8'15
O. 99 72288
2 9375
0.99786072
O 99541018
0.9':ló84SSM 0.99156518 O. '.:l.9312039
0.99SJ4853 3 .. 1.25
O 99854898
3 .1815 25
0.9937251 O 9SS8798J O 99301579
3 3125 J 375
o
9991JS2S
O 9992402
.4375
3 l 5525
0.99'133242 O. 99941344
The
show that only 3 3 cm of rhe tube is required for conversion of 99 .9%,
rnuch less rhan the fon ?O rn
Pll-5 (e)
11-14
Assuming the porosity remains the same, factors in the correlation affected by the size of the catalyst pellets are:
= [ !··
dp
Vp
.]'''
.
O.., - , .., -1•n = [_~L=--~~g::?. = 0.2S6cm
,(
:2.n:. --~- +x.d.!
y - ----·
kc
= · .!
6( l . 0.4) ---·--·---
= 11.. . :::,9cm
0.286
~L~ = !~:?,9.t:2539(1 ,fa dX
dz
+ O l 5X)'12
(l-·X)
· - 6.28-----
= l.145
~
/
!
e.seo 0.600
0.-100
J¡ -~--I
//_
. ····--·········-"'""""""""
,11
J
T
(l- XJ__
POLThlATH
T .L
·····X
_
. (l +OJSX)
(l+O.l5Xf·
1.000
_!:$.!;;.'!'.E ...
·- --
0 . -86
-1
50 ..9 ..
·-· - -- 2 -·- ..,··-
[2368(1 + OJ 5X)J'![t.044]'n == 25.39(1+ 0.15X)"~ ·
0194
a,=
4 . . •
tc.dp:
o.zs:
__ ·::~.". +0.25(0.2:5)
/ r
¡. ¡ ¡I
:: L. 0.000
-+--
o.->oc
-+----+----+o.aoo i.zco z
1..;00
--~ 2. . 000
As can be seen from the above graph, the affect of reducing the dimensions of the pellets by half results in the conversion reaching 99.9 % Ll8 cm from the entrance. This seerns reasorrable 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) If pure cyclohexane feed were used at the sarne volumetric flowrare 60 dm3/rrún , then the initial bulk concentration would be greater and there would be a greater concentration gradient across the sragnanr film on the pellets However the mass transfer coefficient will be reduced as che produces must diffuse away frorn the surface which ,.,..¡11 be harder in higher concentrations of cyclohexane. 'lhe equanons used m Part (a) do nor incorpórate Cao, the initial cyclohexane concentration (Cao is cancelled out), so the only affect this change would have is to alter the physical properties of the bulk f1ow : P
P (84:c2) = · RlY/1H. ·---R-·. ·-1.:--· - = ·····-· · ·-- : = 2 . 647 gm / l = C).00265g Q.082lxTT3
, / cm·
µ = ? g/cm.s ( cyclohexane , 500 K , 2 atrn)
Pll-5 (e) This problem g1ves un indication as to how changes in pararneters may affect a packed bed.
Pll-6 Given, • • •
L=O
Mínimum respiration rate of chipmunk, FAL = L5 µmol 02/min Breathing rate of Chipmunk, v0= 0.05 dm3 gas/min Diameter of hole, D = 3cm
,,,,,,,,,,,,,,, ,,,,,,,, . .,,,,,,, ..
:,,,,,,,,,,,,,,,,,,,,,,,, -. ,,,,,,,,,,,, ,,,,,,,,,,,,,,,,,,,, ....,,,,,,,....
.
.......
:
'''''''''''''''''''''''''''''''''''''''''''"'>, ;. .··,,,,,,,,,,,,,,,,,,,,,"
.vv,,,,,,,,,,,,,,,,,,,-.. .;
.·,,,,,,,,,,,,,,,,,,,,," '''''''''''''''"''''''-......;_
.,,,,,,,,,,,, ,,,,,,,,, ,,,,,,,,,,,,,,,,,,,,, .... ,,,,,,,,,,,,,,,,,,,,,, :-. ,,,,,,,,,,,,,,,,,,,,'-; .,,,,,,s,,,,,,,,,,,,,,-. . . .,,,,,,,,,,,,,,,,,,,,,~ .....
Assuming, • A represents oxygen • B represents nitrogen • Chipmunk has a constant breathing rate of 0.05 dm3 of gas/min
.,,,,,, . . ,.,,,,,,,, . x,,,,,-. .;
. .
l
,,,,,,,,,,,,,,, ..
¡ ,,,,,,,,,,,,,,, . i ,,,,,,,,,,,,,,, 1
..
,,,,,,,,,,,,,....., ..
·'''''''''''''''''''''"
i
.,,,....,,,,,,,,,,,,,,,,,-. .:.
,,,,,,,,,, ,,,, ':\,.,,,,,,,,,,,,,,. ,,,,,,,,,,,,,,,
..
,,,,,,,,,,,,,,,
..
•,,,,,,,,,,,,,,,,,,,,,...::.
i
.,,,,...,,,,,,,,,,,,,,,,,-...
, ~'. L
.:~~~~:e.AL..
11-16
,,,,,,,,,,,,,,,.. Ai i ""''''''''"'''''
.,,,,,,,,,,,,,,,,,,,,,, ,,,,,,,........,,,,................,,,,,
Minimum flow rate of oxygen to the bottom, FAL
.
1.
, , , , , , , ,.
¡ '''''''''''''''" .... .
.,,,,,,,,,,,,,,,,,,,,,~
.
= Mínimum respiration rate of chipmunk = 15 µmol of 02/min
j
:
,,, ....,,,,,,,,,,,. ,,,,,,,,,,,,,,, . ,,,,,,,,,,,,,,, . ,,,,,,....,,....,,,,,. ,,,).,,,,,,,,,,, . ,,,,,,,,,,,,,,,. ,,,,,,,,,,,,,,, ..
Pr ~~~~~~
,,,,,,,,,,,.. ,,,, .
= L :~~~~:
---=-...:;.."'"',,,,,,,,,,,,,,,.
Flow rate of A down the hole = Flux of Ax Cross - sectional Area CAO -CAL
= ~B
:. FAL
L
or L=~8-
CAO -· CAL
X Ac
----(1)
XAc
FAL
= 0.18 cm" / s = 0.18xl0-4 m2 / s
~8
Ac=
= JZ"X(0.03m)2 = 7.069x10-4m2
J'l[)2
4
4
FAL= l.50xl0-6mol /min = 2.5xl0-8mol/ s
Pl 1-6 (a) At Pasadena, California P1 CAO = -RT x .y A
(Ideal gas law)
where,
P¡ = 1.013x 105 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
1.013X105J.!./ m2
CAO =
X0.21
( 8.314-_{-x298K) mol.K =8.59mol/m3
:.CAo
Now, Flow rate of A= (Concentration of A at the bottom) x (Volumetric intake of gas) FAL
= CAL
XVo
v0 = 0.05 dm' / min = O. 05 x 10·3 m3 gas/min
. cAL ---- FAL -
••
2.5 X 10-smol/ s 3
(0.05x10-3 .!!!:_X lmin min 60s Solving for the length from (1), Vo
L=~B
CAO -CAL
J
-
--
o • 03 mol / m 3
X,\.
FAL
11-17
I L = 0.18xl0-4m2 / sx( (8.59 -0.030) mol m3 Jx7.069xl0-4m2 2.5x10-8mol / s :.L=4.36m
Pll-6 (b) At Boulder,
Colorado
Boulder, Colorado is 5430 feet above sea-Ievel: The corresponding atmospheric pressure is 0 . 829 x 10-5 N/m2 .
P7
= 0.829xl05 N / m2
at Boulder, Colorado
p
C AO =RT 1xy. CAo
=
A
0.829x105 N /m2
x0.21
J x298K) ( 8.314 mol.K :. CAo
= 7.03 mol/
m3
Solving for the length from (1), L
= f!A eAo -· eAL x A F
AB
AL
L = 0.18xl0-4m2
/
e
sx((7.03--0.03) mol I m3 Jx7.069xl0-4m2 2.5 x 10-smol/ s
:. L =3.56m
Pll-6 (e) During winter at Ann Arbor, Michigan
T
= 0° F = -17.78°C = 255.37 K
eAO =·PRT ¡ xy . A l.013x10s N / m2
CAo -
1-x255.37 ( 8.314 mol.K CA0
= 10.02 mol/
K)
x0.21
m3
Solving for the length from (1), L=~s
CAo -CA1
xAc
FAL
11-18
J
L=0.18xl0-4m2 / sx 255 175 x ( (l0.02 -0.03) mollm 3 x7.069xl0-4m2 ( 298 ) 2.5x10-8mol / s :. L=3.87m During winter at Boulder, Colorado
T = 0° F =-17.78°C = 25.5.37 K
e
AO
Pr
=
xy
RT .
0.829x10s NI m2
CAo =
( :. CAo
A
8.314 _ _!__x255.37 mol.K
K)
x0.21
= 8.20 mol/ m3
Solving for the length from (1 ), - Cu L - 4B
e
e
AO -
.11 X,-'C
Al
FAL
L = O.l8x10_4m2
J
s»: 255 L75 x ( (8.20 -0.03) mol I m 3 x?.069x10_4m2 ( 298 ) 2.5xl0-8mol Is
/
:.L=3.17m
Pll-6 ( d) Individualized
solution
Pll-7 (a) Given: P, = 510 mm Hg
@
35 ºC (from plot of In P, vs 1/T)
B lnPv = A - ( T + C) DAB =
O. 120 cm2/sec
0.00!T' DAB
75
Antoine Equation (from equation ofFuller, et al.)
~_! + _1_
= ---P(VV:
+ ~ )2 MA
MB
Fuller Equation
where V A and V8 are the Fuller molecular diffusion volumes which are calculated by summing the atomic contributions. This also lists sorne special diffusion volumes for simple molecules. Fuller diffusion volumes Atomic and structural diffusion volume increments
C
15 . .9
F
14.7
H O N
2..31 6..11
Cl Br 1
2LO 21.9 29.8
454
11-19
Ring -18.3 S 22.9 Diffusion volumes of simple molecules He 2 . 67 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 38.2 N2 18.5 Br2 69 . 0 02 16.3 S02 41.8
Air
19.7
Pll-7 (b) By CS2 molar flow rate balance --
dNA =O
Z=20cm
dz
And
NA =K¡ Fick's First Law ( For NAIR = O)
t
NA =-CDAB dXA +XA(NA +O) --
u
CS2
Z=Ocm
-CDAB dXA 1-XA d:
dz
R=0.5 cm
Equating the results of Fick's First Law and molar flow balance, then rearranging and writing in the integral form
K1
J d: = CDA x,XJ dln(l-XA)
z-
8
O
K¡ =-CDAB 1{l·-X2]
z2
1-X0
Evaluating for z2 = 20, X2 = O
e = __!.__ = -·---RT
_(l_)_at_m _ (82.6) atm.cm3 (308)ºK gmol.ºK
= 3.95 * 10-s gmol / cm" X0 = Pv -· = .510 = 0.671 Pro,al 760 N =K =Q.95*10~)(0.1221{ l
A
= 2.64
(20)
* 10 ·-7
1-0.0 1-0.671
gmol / cm
2
J
.sec
Pll-7 (e) 11-20
Fo,
"'r:e;:) between = O z
,: =
1-X,
and z =
20
1{--l ] 1-X0
(1-X )=(1-X) A
XB = (1-X\
O
[-1-I'Z2 Al-X
=(1-X)
(l-Z/Z2)
OA
O A
= (XBjl-Z/20)
~=~C M = XAMA + XBMB CB =XBC VA =NA/CA
~=~C V8 =N8/C8 =0
v· = NA +Nn·=7.17*10-1 e
n +n V= A n =nA/p p ]A= CA (VA -V*)= NA -XANA = XBNA JB = CB(VB ·-V*) =-XBNA =-]A (*105)
z
Xs
XA
Cs
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
1.3 1.71 2.26 2.99 3.54 3.95
o
o
(*103)
z
o
5 10 15 18 20
Vs
o o
o o
o
o
VA 9. 9
11.8 15.6 27.5 63.5 1E+06
. V
7.17 7.17 7.17 7.17 7.17 7.17
(*103)
CA
2.65 2.24 1.69 0.96 0.415
o
PB
PA
0.376 2.02 1.71 0.495 0.655 1.29 0.865 0.73 1.02 0.316 1.14 o
p
2.4 2.21 1.95 1.6 1.34 1.14
(* 1 06) .J:1.Q'.) V
8.32 9.05 10.25 12.5 14.9 17.5
jA=-js
JA=-Js
3.16 4.52 6.77 10.9 14.9 20
8.7 11.45 15.15 20 23.6 26.4
Pll-7 (d)
11-21
O)n
ffiA
O)
0.158 0.226 0.338 0.544 0.764 1
0.842 0.774 0.661 0.456 0.236
60.7 56 49.4 40.5 34 28.97
o
Mole fraction
1 0. 9 0.8 0.7 0. 6 0.5 0.4 0.3 0.2 0.1
6
...~ G)
~
o
o
10
5
15
20
15
20
Z (cm)
Co ncentration
o E
4 . :5 4
~ 3.:5 3 ~ 2.:5 e o 2 ~ 1 . :5 1 fj
:ro e
e 0.:5 o o o
o
10
5
2 (crn)
Diffusion Flux
30
¡;¡, < o ,...
25
.
có o G)
<
O
III
r cJ»
20
8
15 uí ..2 ui LL. 111 e ra 10 o ·¡¡; E 5 :::,
-:s-
o
-.-.
-
lli!!I
-
..................
o
5
NT"'N,NN
10
.,..••
15
20
Z (cm)
Pll-7 (e) Evaporation Rate of CS2 (p1¡q
@
20 ºC = 1..26 gm/cc)
11-22
(2.0 * 10-5)(
A
Q=~t= Puq
1l )(3600
4 1.26
* 24)
= 1.075 cm3 (@ 2oºc day
Pl 1- 7 (f) No solution will be given Pll-7 (g) Molecular diffusion of air is taking place
NB = JB + XB(NB + NA) O=JB +XB(NA) JB =-XB(NA)=(I-XA)NA =-]A
Pll-8 (a) Quasi-steady-state - no accumulation in the capillary tube or in either chamber. Volume 1: Accurnulation = in - out.
dNA d(CAV¡) dt=d-t -=0--WAAc WA =-DABC,: + YA (wA +wn) EMCD: wA= -WB
deA ~ WA=- DAB-dt
wAzoIL =-DAB eA e
WA =--DAB ( CAi
)
L
1cA2 Al
-CA2
Volume 2:
d(CAV¡) dt =WA,\.-0 dCA1 =· AcDAB dt V,L
(e
Al
l
dCA2 = AcDAB dt VL
(e
2
Al
-c ) A2
-c .) A2
Subtraction gives us:
d ( eAl - CA2) = _ ,\.D AB (_!_ + __!__) ( e _ e ) dt
L
V,
l
V
2
Al
A2
Now integrate:
11-23
!\;DAB (_!_ _ _!_Jt+Constant
V¡
L
V2
Pll-8 (b) (CAi
-CA2)=kMC
Ink + lnAfC lnAfC
J
=
!\;DAB ( ---1 1 t+Constant L V¡ V2
= - !\;DAB L
(_!__J_Jt+C2
V¡
V2
lnAfC =-DAB (ü.01025)t+C2 ifwe plot ln(~IC) as a function oftime and find the slope, that would give us-DAB
See Polymath program Ptl-Svb.pol. POL YMA TH Results Linear Regression Report Model: lnDIC = ao + a1 *t Variable
ªº al
95% confidence 0.0078035 9.83E-05
Value 3.7024708 -0.0107008
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 salid particles in excess solveru,
Di ..... D + __ .!._·-·{Df 2D*
Define conversión
· · D1)
= cu
in terrns of volume dissolved :
gives
O= Di (l .. Xt;
11-24
Substituting for D :
l (Di't 2D*
Di. · Di. (1. - X) in- +
[•D.1 ( l ·
term l Surface reaction controls :
-
X ) in]2) -==a. t
term 2
D* is large. term 2 is srnall cf term 1 Di - Di ( l ·· X)113 = et.t l - (. I ·· X)
Mass transfer controls :
C/. . t
l/J
-== ---· Di
D* is small, term 1 is smaíl cf term 2 l . -2 ·-·-·-·(Di
2D*
·rD1. (l
- X)u;·f') --
·
Di 2D *
.
-· ····{1 ··- (l ·· X) 2,3 ) Borh regirnes appare?t ;
a.t
= C/..t ·-·-Di
Term l and term 2 apply
[1 ··- (1
. "l. + [--··--(l-(1 Di
·· X)
.2D*
J =. a..t
· X)-"'t3 )_
··l···)··.·1:·
Pll-10 (a) Mass rransfer limired
A in excess , CA ::::: CAo
The reaction rare is equal to the rnass transfer rare For small partícles ami negligible shear stress :
kc =
11-25
3:.Pe D
Mol balance
on solids :
For l mole A dissolving
.dD ... ... = dt
S.oundary conditions,
t
= · · ra," = kc C,,.o
r.,/'
l mole B then · ·
•ií"2.·-· ·(·--·r.-1.!')-J ·-·p·-··-· ..
c.oe.c_._ »
== •· ·
1
p. D
.J
as C"-' is undefined
.
= O , D = Di
j (D)dD = --~.:P.::0"':.?.j dt
o,
p
o
4 . De . Cso.t
2
D
=
=-······-··
p
·11,2
.~ 8.De. lD,. -·-··--;-. í
r·
.
!
Cso.t
.,
Time for dissolution, te , ar D = O, and assuming particle density, p p. Di2 ce== ··---·-· 8 De. CAo
= 2000 kg/m3
(lOe - 5)2 = ·-···- 2000·-··X·-·····--· - -- . . . .. = o.rzs,~
8 x (lOe ···· lü) x (2000) i.e. virrually ínstaruaneous dissolution
Pll-10 (b) Surface reacrion lirnited Mass transfer effects are not important when the surface reaction rate is limiting.
Mol balance on solids : For 1 mole A dissolving 1 mole B then
Boundary condirions,
t.=
O, D
rA.; ·
= Di
D
') k e --_r._.,10
Di
p
J dD =
= . ra,"
J dt I
o
11-26
D-Di-
_ 2.kr.C.o.t p
D = tn : '3.kr. (¿~o.t p Time for dissolurion, te. at D
= O, and assurning particle density, p = 2000 kg/m'
2000>e(l0:.::::..?1. ... =5xl0e12s te= _p._Di = 2.kr. Co 2 x (!Oe - 10) X (2000) a very long time.
Pll-10 (e) At
t:;::
O. total moles A in tan.le= 0.1 x 100
= 10 mol
A
i.e. there is just enough mols of A to cornpletely dissolve ali of B in a well míxed tank, Acíd not in excess and dissolution is mass transfer limited (CA:s:t: O and CA';!:: CAo) W Ar:;:: kc(CA - CA.)
W.-1.,·
= kc. (c.
,A
.,
kr
= - rx," ::: k,. CA$
-C,u) =-·rAJ
+r»» = . · ·-¡f"'· l+-·-· D*
.•
e~
;,,
assume zero order in B
= -kc.kr --.C,i.
and
D*
= 3_._De
Ice +kr
where
kr
Mol balance on solids : For l mole A dissolving
l mole B rhen ··
rA,:·
!e dt = -u.f.. . ~··--] L+·--D \
= · rss''
where a.
2.k,.C, p
= --·-
[)*
11-27
2.De kc=--D
Boundary
conditions,
t
= O , D = Di
=
Di= D + ... --1 . ... (Di2 · · · D1) 2D* As a function of radius :
2r2
2ri2
o: = O
-- -· ·- 2r + ca D
·· 2ri -·· -----
tr"'.n . ·-· ri:·'] = o
>*.CLt r:' + D*-r+ [''I·-.;· ---
L
=
r
Using the quadratic solver:
Time for complete dissolution,
-
--b ± J¡}::. 4. a. e _ 2.a
te, at r
=O:
re = {{ Di +
assume p
= 2000
{)j{;]
2 kr. C1 ··· 18)x(0Je3) = ---·-·· · ·· · = 2x(l0e ··--·----·--····-···· --
· kg/nr' and ri
p
.
2. De
D"' == ····-· kr
gives
t
(J.
te =
I
2000
= 2x(
= lOe -19
lOe • 10)
······- = .2x10e8
IOe ··· · 18
r··
lo~·:=·i·9· ~ ( IOe
_
. ))
(1 Oe -· 5)2
1
+ i:;(2-;L0~~::·8)] = l Oe 14s again a very long time .
Pll-10 (d) te=
-··p[Di 2.kr:c,.
+ .J?i_~- -, 2D*
To reduce te, increase C.\ and I or decrease Di. To increase te, decrease C,,, and I or increase Di .
Pll-11 (a) Irreversible, gas-phase, adiabatic. no pressure drop, packed bed. Isotherrnal Mol balance :
mol/gcat s
11-28
where F,,.0
= C ..o u" = lOe-3 x
Rate law :
10e4
= lO mol/cm'
Bue Ca., is unknown . Assume reaction rate is rnass transfer limited. W,,.= kc (CA - C,J = k' . CA,
e; = . ~:.:. e~. kc+k'
cm;
. -.,·.,. . ~- = '"k' .kc.Cs . _. kc+k'
where k'
kc =
s.gcat cm
= 0.01 cm'zs.gcar at 300
De dplOO(u.dp) ···-:;--
111
converting :
mol . ··- l
70.7 x
{oe-2
K (constant · · isothermal)
= --0:-1--100
ac,11 = 70.7
02-J
(10xOJ\uJ
x 60
---0_
... = 70.lcm Is
= 4242 cm3/s . gcat
. . -,.,(¡ = 001x4242xCt - - . -···-·. 4242 +0.01
Stoichiornetry :
gas phase, constant pressure and temperature
where
é=YAoó=0.5(1-1)=0
C.,. = C,,.o(l • X)
and CAo = 1 mol/dnr' = lOe-3 mol/cm"
POLYMATH
~t:ioo.s, d!x) /d
(w)
~~.L.~11!~ o
=··ra/fao
k,;0.01
fao=lO kc .. 4242
cao=O 001. ca"'cao*U
. ·xl
ra= .. · (k*kc•cal
wo
=
O,
wf
I {k+kc]
=
le•06
11--29
Minin:ium val u e ?ina¡_va~
yaríabl~
:rnit:ial
"'
o
J.e+06
o
0.63212
o o
O. 01
0.01
o
10
10
1.0
10
4242
4242
4242
4242
0.001
0.001
{l.001
o 001
o
O .. 001
0.000.36788
··3 . 67879e-06
--9. 9999Se···06
X
value
o Cl
k
tao ltc ca.o ca ra
001
···9. 99998e-Oti
Maximwn
val1,.:,e
1.e•06
o
63212 01
o
00036788 -3 .67879e-·06
c.soo 0.6~0
0.320
O.ISO
e.eco
•••• .. "A••··-+ t.000"
0.000
w
Fura conversión
--·····
of X= 60 % . \\ ;~,
= 9::..5
kg
Sc:ale: KEY:_ ···-Ca
Pll-11 (b)
A. (
r, = o o 1 ex Pr_' =~!.9g ( . . !_.
l_ L%7\ 300
-··
.L. f] T )_
where
Sroichiornetry
11-30
E
= O and To :::: 300 K
and C.,.0
dF
· ·-·•·"·· _-
Energy balance :
L, Fin.Cpi.(T--
Q- WS - _'-, · T.w.. )+ F ,\O. X t.: L.,¡ F C p,. (T · \ AfJ · KX \fT)c\ '}
- -.. . . . .
dt
Tio)
.
10
~ -.,··--LN/;pi
,
= [ ro x 25 .r ~ 300)] + [ to x 75 (T =
= lOe·-3 mol/crn
- .. --·- :::; o
. . 750)]
t 000 (T .. 30&)
·· lOOO íT·· 300) + LO.X.( 10000)
=O
POLY:,..!ATH Init:.itl
Equaci~L d(x}/d{~)=-~a/fao k:0 . 01 fao=lO kc;c4.242 cao ..O .. OO.?.
o
_va.lue
To=-300
'l';;;:(lO•x•lOOOO/lOOOl+To
C'a:cao•{:t.-xl ra=- (k•kc•ca)
wo "'
O,
= 300 K
*(To/'t') I (k+ke)
"'f ., l,2e+06
t:o.ic.i..al. .Y~~
v'a.ri.~le
o o
w X
0.01
le fao
Minimwn val u e
Mal<~--.'.!~
o o o .ot
300
JOO
300 o 00028l331 -9.99998e-06
O. 0002833 3l
lec
cao
0.001.
To
JOO
T
365.483
ca
)00 0.001
:::a
· 9 .. 99:998e-05
-2. 8331e-·OS
KEY:.
O.MO
·-··X
o.~ao 0.320
0.160
Q.000
l. 2e•06
0.654825 O .01
10 4242
10 4242
J.900
~~-
1.2e+06 0.6$4825 0.01 10 4242 0.00l ]00
o
O .001.
001.
10
42~2 o 001 365.483 ;2. 83J3e··06
T l T T
I 1
T
r
T. o.coa
· ····-·••··-·~··t,·~· .. ·----·-·•"···" 0,300
_.,_
. .,...._.t-·····,·~-·--·..+0.900
o• .soo w
11-31
+ ~,-
, 1 .. 200
-,---
4 ¡.500•
¡'
For a
(Of1Vc'.[SÍOll
of X
= 60 9'c • w_,, =
l020 kg
376.ººº.T
r
360 .. COO·l-
¡
T 3.,1.cco-L
l. i
329.. 000·t
!
:lt2 ..
C00t·l t
296.000··-t--····-···-·--+····----·······
a. 3CO
0.CGO
·+·
_
C .. 5QC
+
o.seo
_
-t-···--·~---l L2CC
?..!50C·•
w 1 ..
Scal e:
103
KE.Y;_
a .. :i
o.ase
0.720
ca 0,560
o.•CO
O ... Z
T
t'
Ti
rT 1
1
T1
t .¡
1
. ............. ..,.,,•·----0·---··•""''
o.coo
o.aoc
, "'···i- ..... 0,600
·---+·-···-· ··-+'""·h,,----~··· 0.900
. .. , ... ~-------+-·-·--····~-~----·-·-? 1.200
!.500•
w
Pll-11 (e) It ís possible to generalizo that the addition of temperature variation in adiabatic operation
operation enhances the conversión profile along the packed bed, so that for a given bed diamerer less catalyst is required. It is clear that adiabatíc operation inhibits conversión as heat is not removed from the system so morecatalyst is needed for the required conversión. The removal of the hear generated in the reaction allows a rednction of 85 kg of catalyst in the bed. More detailed econornics will indicare whether isothennal operation is worth it,
Pl0-12 (a) Dissotution of pills Complete dissoíurion
11-32
CA -
.,
W..intT,tt.slunu4cJt.
_
Vilomach
___ §.:!"'~. g;,,..J;,.min
dCs_ == k». C,1_ V"'ª"'"'·' dt W1,,,..i, Relate
C,. to time :
Three pills .. each with different thickness outer Iayers, the inner cores dissolving ar. different times (but ar the sarne rate) to each other. will each contribute to C,. in the stornach.
Pill 1
D:: = 5 mm. Di
= 3 mm
Time for outer layer to dissolve Pill .2 :
. .
. .
D::=4mm,D1
. .
.
.
8:c(6xl0e - 4)xl.O
=S'rnrn
.
Time tor outer layer to dissolve Pill 3 :
. ..). )~4 .X\ro .)-1 -· o . 3:?. ) . Oº ··- -1.18mm
t1 - ··----··--····-----·
ll
D'!::: 3.5 mm, Di
t::::: t1
--
= ?.:?.?.?.~:~.1~·.!2~::gt}.
dt
k:.S,,..rr..D!
T.{
• , . . .,
••
V
t t
= t1
d'\-V:ln,Nin.:m,muc.:lr
IC. 1)¡· :: ·-·-
·····-··""""""
dt
r.," = k.:.S¡,,
2.D.~sSm.rr.D1
0 ••• ~ ·--·---
and boundary conditions :
.
= 0.24min
8:r(6x10e ·· 4):d.0
and
se, -
_
8x(6xl0e ----·4)xLO
mol balance on drug :
.. ~-
0.31)
= 3 mm
Time for outer layer to dissolve i1
Ac
1
0.0354x(OA
= ------·---.,----·-· = 0)2mrn
,
<
_
,,,,...,...~ •• --,
,H'~---··
D.V where \V
= rnass of druz in srornach, g V = volume sto~ach fluid , cr~3
CA= O
= t1, CA= C.....= WJV
= 0.5/1200
11-33
=4'..17xl0e-4 g/cm3
D2 =4 mm D1:;::;;3mm
Outer layer :
~o- = ·-·--··-· ( ¿ · -·-·······-·p .1t.D3
Mol balance on outer laver :
-
O:,. O +o.'·
and rA' • rute of dissolution of outer Iayer
k:
Sh=2,
.11: .. e
0)
dt
rate of rnass transfer from pill surface W A
= 2· ·.· ·DAB · · · ··
. . . . Assurne densitv. of outer layer . = muer ··. ·.· layer . .
D
500x10e ·
6
,,_
6
Sou[ = LO kg I cm3
·dD • ·•" '"'" =
2.(-·rA'')
•w_, ..,-~~••D·~··-·
::::::=:: - ·
2x2xDAI,Sour ·"n•••·.,.,.-"
p t = O, D = I>.? t= t1, D::::: Di dt
Boundary conditions :
In.1 where
t1
= time for outer
· · · = 2x2
2
••·-•~"''""•~•·-·••••
D.p
DAa.S,,,,, ·····-··· . ti p
layer to dissolve 032)
= --0.0354x(OA · · ···-·· ·1
t¡
Inner
D11
-
8x( 6xl0e · · · 4 )xl.O
__ . -- = O:i2rnm
Cüt'e . t t
""º, = i1
D= ,
•" " " " "' = 2
inner core
to
D ""O ·••·•-••••on•••~
.
=-·-·-·-··· ···:- -· = 00:,:,4kg l cm3 11:.(0.:~)3
...
p
..'.?!CAA
•
t:
dissolve
11-34
t:
= . . ~:9~.~.:~9:.~~---- = l.66 min 8x(6xl0e ····· 4)x0.4
time for complete dissolution , t-r ==
t1
+ t1 == 0 . .52 + 1.66 == 2 . 18 rnin
Pl0-12 (b) blood strearn
Ce conc, in blood
stomach
, (:A cene, in stornach
@ @@ dissolution
stornach wall Let rate of adsorption into bloodstream
Relateconcentration in stomach CA. to concentration in blood Ca: Mol balance on drug in bloodstream :
..
+r« x:
dCs dt
.4.~::~. = ks. CA dt
Se ale;
!$g;YL -·Ca
11-3.5
, , ,.,,.,...-·~-----~~--, . .sCO
Se ale;
KE'!'..ó .. ··---Cb
1.:z 0 0
•
1
.Í.1 i _¡_
0.900
L
o .. ~oo
Tf
o.:;cc
' J _ji o.eco
¡ / ¡ Í ----; -re.eco
··
; ---
20 .. eco
· t··----ao.cco
;
-- .
-o .. eco
---i so .. coc
Pl0-12 (e) The graph of C8 against rime shows how the drug concentration in the bloodstream initially varíes wíth time and then becornes independent of time as all' the drug initiaíly in the sromach has absorbed into the blood (consurnption of the dmg within the bloodstream has not been modelled, unrealistic but no data}. tr a certain drug íevel is specified, and assuming a constant size of inner core (drug) and that the pills were to be raken similtaneously, then the way to achieve this would be to use pills of different outer layer thicknesses - to mainrain an even srornach concentration and hence absorption rate over the whole period. Relate D with time : . dl?_ - . .. !:.:.~~-?_;~- _ . 4 D,.s,S-.,. dt D.p p ami boundary conditions : t= ti, D=D1 t
= t2• D = O
Using logic to obtain the correct tim.i.ng for the drug concenrration profiles inside the stornach for each pill, rhe total profile is used in the relación with the concenrratíon profile in the bloodstream,
11-36
P11-12C
, s,guations: d{tll.)
/d( t:) =if {e>O .24) then
!nitial
(i.f (Dl>O. 0000:t) t:hen(-·4 *Dah•Si.n/
(
__ya.l~
O. 3
01.•rho))else{O))else(Ol d (Ch} /d( t) :ka*Ca d(Cal)
"'V/"Wbody
O
/d (el =if { t>O .24) t:hen(2•0ab*Sio*3
.. l4*DL/V} else
(O}
O
=aif (t>O. 52) t;.hen{:2"0ab*Sin"'3 .1.4"02./V) ebe(O) d(02)/d(t)=if(t:>0.S2Jt:hen{iE(02>0.0000l)t:hen{·4*Dab•Sin/( o2•rhol)else{O})else{O) d (Ca3l /d(t:l =i.f ( t:>l.18) tl::.en.(2 *Oal:,"Sin•3 .. 14 "D3/V) else {O) d(!l:l} /d(tl =>if {c>l .18) then(if (0.3>0. 00001) then( ... 4•Qab*Sin/ oJ•z:hol J else(Ol )else(O)
O
d{Ca.2) /d(t)
0.3 O (
o. 3
t)ab;0.0006 Sin .. 400 ::·!?.o-"'35.4 lca.o0.166657
V=l200 Wbody=75000
Ca,;,Ca.l.+Ca2+Ca3·· .. (c.b"''..body/V)
eo = o,
tf.,
45 :tnitial
o
o.l
Maxi."tlum Y:!!!~ 45 o . .J
Mini,"ltum value O 9.996Se . ·06
Fina! _!<1..l~ 45 9.996Se-0'6
Ch
O
Z.00164e-OS
Cal.
o
2.001S4e-05 0.000417370 O .•. 000411:373 9. 99774e-06 0.00041~364 9.99841e-D6 0.0006 400 35 4 0.166667
Variable t::
Dl
va.lue
o
0.000417376 O. 00041'1313
02 Cal
0.3
0.3
o
0.000417364
O O O 9. 99T74e-06 O
03
O.] 0.0006
a .166667
0.3 O 0006 400 35 4 O.l.66667
9.99841e-06 0.000& 400 35.4 O 156667
l.200 75000
1200 75000
1200
1200
75000
75000
o
O 000991.:l.8
o
1.09044e-06
Ca.2
Dab Sin ;;ho ka
Ca
'ºº
35.4
and boundary ccnditions :
t=t1,
D=D1
t:::tz,D=O Using logic to obtain the correct tíming for the drug concentratíon profiles insíde the stornach for each pill, the total profile is used in the relation with rhe coneentration profile in the bloodstream,
Pl0-12 (d)
11-37
To maintain constant drug level by rnaintaining 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 distribution of onter Iayers is needed, with thín layers for initial response and thicker layers far delayed response. This disrribution would be back ,calculated given the necessary stomach concentration for the required bloodstream concentrarion accounting for bloodstream drug consumption. Optimization of the stomach concentration will indicare the rimes at which complete dissolution of the outer layer of the 'next' pill is required to maintain this level. TI1e range of pill sizes depends on the number of pills which can be reasonably consurned in ene sitting, the period for effect and the limits of practical pill size.
Pll-13
----1
The plot of the data is shown below ..........
---- ...
~-·-···-
. - -.------
·---·---------··--··--··--·-----
Particle Oiameter vs Time 1 O -. . . --
~
,-·-··· · -
- ·-· · -·-· ·-----·-··-··· -·
---·-··--·-··--""···· •""'"-····· ·· · ·
. ·--------· · · - ··· ---------·· ---- -
9
....CI
8 7
-¡; E
6
Q
5
:s
4
til
o• a.
3 2· 1
o
.·
20
30
. .
-- . --
50
40
~----
-····-····
-
-·-···-
~- ,
-
70
60
- •" · , -----·· .. -···:
90
. •.
r-
100
b
.
110
Tlme
---··----········-·· -· · · · · -· -··· -- --'--------------
Initially the rate of incinerarion of the droplet in terms of diameter, is non-linear, but apparently becomes linear after ,~ 50 time units. TI1e linear form of the data Indicares that the diameter ís directly proportional to time and the rate of decrease in diameter is constant and hence not a function of diameter, This relationship should make it easier to estímate the reouired incineration time for complete destruction (zero diameter), Assuming that rare of diameter decrease continues at the linear rate until complete desnuction (at time to and hence complete decomposition of the POHC's, the equation for the linear relation indicares 41 ·~ 160 units.
Pll-14
11-38
Mol balance on layer of earth control volume:
ac,
F~!. -·- F~I, - _.,=.,A.. l\z. "--~-
dt
A . W~L - ,., . . . Ac. W1I:
z --~
o
dC,1 dt
=
Ae, óz
dW~
ac,
dr
dt
For dilute solution and constant total concentraríon :
W,1
dC,1 = ····DAa.--· d;
Gives : C,1
Let
'V=···C.o
gives
Boundary condirions : t t (present day)
=
z == O (surface), 'I' = O
=
z t
= O (end
Gives the error function : It is defined that ar T\
'V= I
00.
of glacial) • z >O,
'I'
1.j1
.,
= erf .•
= 1
let
-
. -J4-.D,1st
o (as a function of time) :
Concentration Profile
- ---·---
1.2
.
1 '
....-,.----
0.8
~ 06 O 0.4 0..2.
o
·--·-····-·-O
J4.DA8.t
= l.82 , CA= O.O 1 CAo and for n > 1.82 , CA is negligibly small,
This defines the penétratíon thickness,
0
z ll=-·===
.
.. 5
·····' 10
_..
..
-
,.
.
15
Oepth (m)
11-39
1.s2
= -·-==s= .J4.D,,.a.t
But the graph gives at, CA= 0.01 C,.0 --,. penetration rhickness after time t.
z::: OA hence 8 = 18 -· 0.4 rn , the
o
The rime taken for to reach this thickness, is the time since the diffusion started (i.e, at the end of the last glacial)
CDPll-A CDPll-B CDPll-C CDPll-D CDPll-E CDPll-F CDPll-G CDPll-H CDPll-1 CDPll-J CDPll-K CDPll-L CDPll-M
11-40
Solutions for Chapter 12 - Diffusion and Reaction in Porous Catalysts P12-1 Individualized
solution
P12-2 (a) (i)
t =5
1
5
D
ex
-D 1-~ e
( J
l.75
_T2
p Ti
(lines and angles not to scale)
P12-2 (b) (1) First Order Reaction Kinetics
D d2CA -jc e 2 dz
-di.v = O dí\. Symmetry:
___, 7i.. = 1
A
=0
,'lf
= CA A='!:_ C , b AO
'lf = A cosh..Ji>aA + B sinh..Ji>aA
·~ = A ..JDa sinh ..Ji>aA + B..Ji>a cosh A
=0@ A=l
B = --A sinh..Ji>a = -A Tanh..Ji>a cosh..Ji>a 'lf=l@ A=O
12-1
1 = A coshvfDa A=
1
cosh
\lf =
coshvfDa cosh
vfDa ' A
vfDa
(2)
B = TanhvfDa_ cosh vfDa
- TanhvfDa
cosh vfDa Monod Kinetics
. h ~D '\ sm -v ua 11,
D d2CA _µmaxCACc =0 Use e dz2 Ks + CA Quasi Steady State dCc = µmaxCACc =0 Analysis dz2 Ks +CA No further solution to Monod Kinetics will be given. (3) Variable Diffusion Coefficient
dFw dt
= 'De Wozl Z--oAc
W A= - De dCA_ =- DeCAo d\lfl dz L dA ;..=O
_?Fw
= DeC AO d\lfl uc Ac/V dt L d).. ;..=O Mole balance {De dCAJ ___ d_z __ k=O dz for hindered diffusion
=
D e
DAB
1 + a2F;/(1 ·-Fw)
As a first approximation, assume no variation in De with A d2\lf kL2 -----=O
dA2
Solution the same as before Equation (El2-2.13)
12-2
W
02
=W
=( DeCAo'f_~=._J=kL L A_ DeCAo
A
The flux of 02 in
dt
e
e
e
Fw =kVt From a quasi steady state approximation as time goes on increases.
Fw
increases De decreases and o
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 = O and A = 1 (see solution to P12-10(c) in which the solution manual) we will find 1
"A =--e
o
~=0.5
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 (e)
12-3
(1) For R1 111 = 0.182 81.8% Diffusion limited
18.2% Surface reaction limited and
For R2 11 = 0.856 85.6% Surface reaction limited and 14.6% diffusion limited
(2)
CWP
=
-rA(obs)pcR
DeCAs
2
,¡.Z
= ll'f'l
(12-59)
= (0.95)(0.9)2 = 0.77 which is less than 1 so there are no significant diffusion limitations.
1 =--~1 kuSaPb Q
=
-+--11 kcac 0.059 So 5.9% surface reaction resistance and 94/1 % external and internal diffusion limited
%R =
0.9.,,41 = 0.941 __ 0.941 _!__ + ~aPb 6.0+ 10.96 16.96 11 kcac
6·0 -- = 33.3% 16.96 10·96 % External diffusion resistance = (0.941)(100)x = 60.8% 16.96 % lnternal diffusion reaction 0.941 x 100 x
Summary of Resistances Externa! Diffusion 60.8% 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 intemal and external resistances decrease. For 99.99% Conversion 1 1 1 ln(lO,000) L2 =L1ln In ln--=0.16-~-~ 1-X2 1-X2 1-X1 ln500
¡
=0.16x
9·2 =0.24m 6.21
P12-2 (e) 12-4
(j)
From Mears's Criterion
-LlliRx (-rÁ)PbRE < 0.15 hT2R g
1)
The value from the question LlliRx =-25kcal/mol = -104.6kJ /mol h = lOOBTU /h · ft2 · ºF = 0.567kJ /sm2 · K E = 20kcal/mol = 83.682kJ /mol Rg = 8.3144 *10-3kJ /molK From example 12.3 -rÁ =kSaCNo k = 4.42 * 10-10m3 /m2 · sec Sa =530m2 /g Pb = (1- )Pe= (1-0.5)(2.8 * 106) = 1.4 * 106 g/m3 R = 3*10-3m T=1173K At the inlet of the reactor the fraction of NO =0.02 From ideal gas law n p
V
RT
= 1.01325 * 105 = 10.39mol/m3 8.3144 * 1173 CNo = 0.02 * 10.39 = 0.2078mol/m3 Substituting all value in the first equation (104.6)((4.42 * 10-10)(530)(0.2078))(1.46
* 106)(3 * 10-3)(83.682)
= 2.88 * 10-4
(0.567)(11732)(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 externa! surface of the pellet.
P12-2 (f) (k) Por y= 30 use Figure 12-7.
If you draw a vertical line up from 1 = 0.4 it should be tangent (or very, very close) to the í3 = 0.4 curve. Any slight increase in temperature will cause the reaction to go to upper steady state.
P12-2 (g) (l)
a= __ 1 1 + k0t
,
$¡ ~ Rl'DPce~ª-
For large
12-5
S' =
ª
Sa
1 + k0t
. Area2 (1) Pore closure. Consider De As r-» oo pore throat doses ~ = Area 1 <1>1 ---)
00
-fÁ ---) o
(2) Loss of surface area Sa. As t ---) oo then
s; ---) O then
---)
00,
crc---) O, De---) O, and
---) O 11 ---) 1, but -rÁ =
.Js: ---) O
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 CDR12-1, 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-l the charge is 100 kg/m3 and in CDR12-2 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¡ vs . 1/m and, therefore, the resistance to decrease .
P12-2 (1) No solution
will be given at this time .
P12-3 (a) Yes P12-3 (b) All temperatures, Fro = 10 mol/hr . The rate of reaction changes with flow rate and increases linearly with temperature
P12-3 (e) Yes P12-3 (d)
=
T < 367 K, Fro 1000 mol/hr, 5000 mol/hr . T < 362 K, Fro = 100 mol/hr,
P12-3 (e) Yes P12-3 (f) T > 367 K, Fro = 1000 mol/hr, 5000 mol/hr. 12-6
T > 362 K, Fro = 100 mol/hr .
P12-3 (g) at362K, F10 = lOmol/ hr) -=--,- ideal rate of reaction - =r; ( at 362 K, F70 = 5000 mol I hr)
Q - actualrateof reaction - _ -rA
(
_c.:_.:...._
Q
= 0.26 = 0.37 0.70
P12-3 (h) At Fm = 5000 mol/hr, there is non external diffusion limitation, so the extemal effectiveness factor is L
r¡ =
actual rate of reaction ( at 362 K, FTo = 5000 mol/ hr) . extrapolated rate of reaction ( at 362 K, FTo = 5000 mol I hr)
r¡=~=0.86 1.4
P12-3 (i)
3[ {bcosh {b-1]
r¡=---¡--=0.86
{Ó = 1.60 1 sinh ( {ÓA) A sinh
by iterative solution
CA CAs
rp=-=
é
P12-4 (a) External mass transfer limited at 400 K and dp = 0.8 cm. Alos at all Fro
P12-4 (b) Reaction rate limited at T
= 300 K and dp = 0.3 cm. When T = 400
P12-4 (e) lntemal diffusion limited at T = 400 K and 0.1 < dp
< 2000 mol/s
K: dp
= 0 . 8, 0.1,
and 0.03 cm.
< 0. 8
P12-4 (d) 17
= _ rate with dP = 0.8 = 10 = 0_625 rate with d; = 0.03 16
P12-5 Curve 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 . Curve B is inner-diffusion limited. This is because it has a dependence on temperature, but that dependence is small.
12-7
---
- -------
-------- --- -------------
---
-
- ·---------------------------------------------------
Curve C is outer-diffusion
limited . This curve has a much larger dependace on temperature than curve B.
P12-6 (a)
Toen
(l.\ sin
h cb r A )) sin h ct,1
Effectiveness factor.
(91 cos h $1 -
= O)
conditions:
Tl = · \·
r) =
Boundary
1
= finite
1} first orderreacri.on.
~t
1 .....- R 2 ' -,,_. = ?' CA = 0.1 CAs
At
7x104
11. = ·-·------
1
X
10
= 0.7
.3
1
=
1 x l O ·3 cml De = 0.1 cm2/sec
(f-)(si~sm~_!I. A.) h 9 A.
At ÍI.
R
where CAS = 1 x 10-3 mole
1
=},
=
0.1
: 0..1
:::: 2
f sin h (4>1 J 2)1 L
0.7
1 x 1 O -3
6.0
J
..L ¡sin h (6 x0.7}1
q> =
CA.
sin h {
!
r l
sin h 6
= _J__ s;4.2 -
O. 7
e -4.21
j
e6 - e '"6 J
CA = 2.36 x 1 O 4 g mol l
P12-6 (b) Tl = --;-[
i2
=
At A =
k ~~ SA PB
1·
De
(!) =
0.I
P12-6 (e) lndividualized
$1 = 6 (see part (a))
solution
12-8
~~
------------
----
----
----------------
--
---- - --
----
----
---
-
---
-
----
- ------
------
P12-7 (a) Start with a mole balance:
WAlz __ WAL ... i: + rAAdZ = O Divide by A& and take the limitas z --> O.. dW ·-· · · r =0 dz A
From the flux equation: W=--·D dC4. e d; Combining the two equations we get:
d[-Pe!.~'A /dz] + k = Ü dz
Dividing by -D, we get:
d2c dz'
k De
---A+--·=Ü
W e need boundary conditions
B.C. (1):
dC
----·1·· = O@z
dz
=L
B.C. (2): We can then solve for the concentration profile: dCA =~+e, dz De Using boundary condition (1):
.
kL
0=-+C¡ De C,=--
kL
De dCA - kz dz De
-----
kL
De 12-9
Integrating agaín: dCA
=(kzD, - kLD,r\,/ ..
C =-Z k A 2D,
2
kL --z+ D,
e2
From boundary conclition 2 C =-zk A 2De
2
kL --z+ De
CAo = C2
eAO
P12-7 (b)
rate of reacrion with diffusion 11 -- rate of reaction without diffusion 2D(CA -CAo)
z (z - 2L) n = ~_,.;....--~--k
2D (CA - CAo) k z(z- 2L)
P12-7 (e) Boundary conditions:
0 =
k5 (~ - L}
CAo =
CA= O at z
=
L
+ CAo
k.L.:.. 2D
L =
~2D;,o· _
1..1rz
=
~::;~-:6:a:,~·:;¡~::!)
0.0640
L;::: 0.0041 cm=
41 µm
P12-7 (d) The answer in pan (e) is equal to the average tail lcngth. 11 = 1 in this problem. If 11 = 1 , then it cono:adicts thc assumption of diffusion
bcing ratc-limiting.
P12-8
12-10
z
=o
z=L
cA
= 2L
z
I
CA = CAS
=
CA
CAS
=10 1 c ~.... A -,. B
First-erder irreversible reaction: -r~ = k CA WA = -DdCA
Mole balance: ··
! d d:A) + 2(
dz
1t
d r¡ = O
L = 10 ·3 cm, 2L = 2 x 10 -3 cm CAS = 1 O -3 g mole/1
D
=
0.1 an2/sec
Atz = L, CA= ..LcAS
lO
.L -
= CAS
cos (
h[+{1-L\l) LJJ cosh,
1
10 - cos b cp
q, = 2.9932 12-11
--··--------------··----
--------------------·----
--
---------------------····--
---·----·---------···--------
---------------------
--------··-·-------------·---·-
P12-8 (a) At z = 1/2 L:
= CAS
9A
(cos h (q,{ 1 - ~))) cos h e
·
e =
{O.OOI)(cos h [2.9932(0.5)]) COS h
A
CA = 2.345 x 104
P12-8 (b) et>
=
2L
Tlold =
VDkd
==
tan h
-
Qa
= 0.8
T'lnew
=
s mol/1
(2L)
tan h
$old
(2.9932)
.,
(2.9932)
... ., 99 -· ~-
tan h Ónew
-
=
0.3324 0.8880
Onew
P12-8 (e) Al
z
z
L. CA
=
CAS h C?new
COS
= cos e('lfisso} =
7.038 x 10 4 g molll
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 > O in direction of íncreasing z, Material balance: WA Ap ~ - WA Ap lz.a + -r~ a Ap Az = 0
. WA =. -D, ctcA . . v -rA.. = kCA; dz +CA,
=- -n e acA dz . Az--, " . o : -Ap dWA Taking dz. + rA a A p ; o
ror snmalleA .e
H
Apf (ned~4.)-k CA aAp ::: o For constant De ; De d2 CA. - k CA a = O dz2 Bounda.rycondítions: z = L
z d2C.A
dz2
~
- a. ..
CA
=
O
= 0
d~A
cz
CA :: CAS
=
O by symmetry
where a.2 ;;;
Assume CA == erz. then r2 - a2
=
k ·nª e O
r
CA = A 1 e -cz + A2 eCU At z = L: CAS =:: Ar e-CIL+Az&L d~.;, 1,..0
= O = -a.A1 + a.A2::
0
. .. From above:
A1 =
CAs
e-al+
eaL
12-13
=
±a
Therefore, concentration profile can be written as:
e = e Jcos h(az)l A
W A A;- =
WA Ap
AS
\cos h {cxL)
-o, (d~A ~..) Ap
= - Ap De CAS
ex(: : l~) lz.1.)
= ·ApDe a CAS tan h {aL}
By the sign convention:
= ApDe a4..5 tanh(cxL) Tl = (~)tanh(aL) = '\/fITfi; (ºe}.ltanh{. m:L} kaL ka L 'V rS; :. 11 kCASaApL
TI= (~t}tanh(aL)(-v1tL) Ovcra.ll effectivencss factor:
or:
•W A Ap =
n k CAo a ApL = 11 k CAS a Ap L
U. = CAs T1
CAo
-WAAP = kcAp(CAo-CAS)
CAo = CAS [ I +
ª:::·
= cxApDcCAs tanh(aL)
tan h (aL)]
P12-9 (b) A7B WAJtrLj r - WA1lrLj r+Isr . + rAJlr/),,.rL = O 1 d --rW r dr
A r+r A
=0
12-14
EMCD therefore,
drdCA
.!. D dr = D r
e
dr
dC = -De __ A dr
WA
e
(d
2CA .: dCA dr' r dr
J
rA =-kCA 1
d
--rW r dr
A r+r A
=0
D [ª2cA .: acAJ-kc e dr" r dr
A
=O
1/f= CA_,
A=!_
CAO
R
d21/f 1 dl/f ----------·=O d,12 A d,1 d21/f 1 dl/f -----Dl/f=O d,12 A d,1
kR21/f De
ª
AtA. = 1, 'JI= 1 and at/1.
= 1,
dl/f
-
d,1
=O
Bessel Function Solution
(n) P12-10 (a) dCA
(o) EMCD WA =-D--
dz
, rA =-k
In - Out+ Gen = O W A Aclz - W A Aclz+Llz + rA~zAc = O dWA ---+rA=O
dz
'
12-15
Integrating equation (1)
kL2A2
\JI=
+ C2
2DACAs at A=l \Jf=l
kL2 C2=1---2DACAs
,¡,.2 _ '1'0-
2
kL 2DaCAs
P12-10 (b) (p) Now let's find what value of A that
\ji=
O for different 0.
Por <1>5=1: 0=1+1[A2-1]=1+A2--1 )...2 =0 Therefore the concentration is zero (i.e., \JI = O) at !A=O!
Por <1>5=16: 0=1+ [)..Z--1]=1+16,.2-16 15 = 0.938 16 Seems okay, but let's look further and calculate the concentration ratio \JI at A= O for o= 4. )...2
=
\JI= 1 + 16 ~0.2)2-1
]= 1 + 16(0.04 -1] =--14.9
Negative concentration. (q)
P12-10 (e)
(r)
Let's try again with
\JI= 1 + (10)2 ~.12 -1
o = lO
]= 1-102(0.99)
\JI= 1-99 = -99 not possible \JI will be negative for any value of o greater than one.
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 'JI= O
(s)
d\lf
and-=0
dA d2\lf 2 ·dA2 -2o =0 d\lf - 25A + C1 dA2
\lf = <1>511.2 + C1A + C2
At A = Ac , \lf = O
O=511.t +C1Ac +C2 Subtracting 1 = 5-5At +C1(l-Ac) Solving for C1
_ l-5~-11.t)
C¡--
1-Ac
Solving for C2
2 ~-<1>5~-11.t )) C2 =1-o-------1-Ac
12-17
LO o=l
o=l
o Sketch o
LO
0.5
of concentration
= 1 then
Ac
profile
for different
of o
=O
o = 2 then Ac = 0.5 That is for o = 2, the concentration 1-Ac =-
values
of A is zero half way (A
1
o
\11 = <1>5A2 + [o ·- <1>5(1 + Ac)]+
l-<1>5 - [o -<1>5(1 + Ac)]
212 r, • = o/\, + Lo -
11
o - o + o r + 1- o - 2<1>o + 2<1>o 2
2
\11 = <1>5A2 + 2<1>0(1ForA
2
2
o)A + 1-2<1>0 + <1>5
>Ac
P12-10 (e)
P12-10 (f)
12-18
= 0 . .5) through
the slab.
ln
P12-10 (g) No solution will be given P12-10 (h) No solution will be given P12-1 O (i) Individualized solution ,
P12-11 Givcn: second-order dccomposition rcacti.on: A -+ B + 2C k = SO m6/g sec mol; dp P = 500 kPa
=
=
0.4cm; U = 3 mis; T = 2SO°C = 523ºK;
4.936 ann : X
=
0.80; De = 2.66 x 10 .a rril/s; Eb = 0.4;
Pb = 2 x 106 g/m3 ; Sa = 400 m2/g. CAo = L
RT Rate law:
4,936 atm
=
(0.082g
-r;.. .
!!'flid 523 ºK
= 0.115 g mol
l
k C¡
Mole balance: D
~
dl CA - U dCA + r. p ctz2
dz
A B
=O
2 cA - u dCA - a k sA PB ci dz
DAB d
dzl
=
o
Ncglecting axial diffusion with rcspect to forccd axial convcction. we havc:
12-19
,
_
.J_ _ _L
CA CAo At z = L:
=
¡ak Ps)z SA U
....L = ( 1 CAo
1-X
L = /
·
.
- 1) = (· 2PB k S A) L U
U
\n Ps k SA CAo 11 =
Intemal cffective factor:
'2 =
TI -
- 1)
1•X
(-L) 112 n+ 1
..1. ·
whcre n
q)n
=2
2 ){400 m!)(2 x 1<>6 ..!...)(0.115 ( 50 g smmol g m2
0.2x 10-2 m
'2 =
)( 1
2.66
X
10 .g
n¡2
g
mol)(lili) 1 ml
2.63 x 107 very large
(_L)l/2
- 2+ l
3
2.63 X 107
= 9.313 x 10 ..g
Q = 11 = 9 .313
Intcmal-diffusion limited:
X
l Q -S
Reactor length:
3 mis ( 1 -10.8 - 1)
L =
.!..(so ro' }(400m!)(11s gmol) ml gsmol g m3
( 9.313 x 10-&)2 x 106
L = 2.80 x 10-sm
P12-12 (a)
12-20
Start with the mole balance taken on a shell
Divide by -21t1M and take the limitas l1r approaches zero to get:
r.
d(~,.r) di'
'i.Pc.
r= O
Next fmd the equation for equimolar counterdiffusion
equation:
and plug it into· the above
ii({-v. dC~)) dr
-rlPJ=O
Next differentiate to get the following differential ,equation:
We can then set the following:
. e
CAS
ít=-
r R
Solution:
Boundary Conditions:
dA.
12-21
dtp
!
= [C/1 (l )-C2.K1 (<1>1 )]
dl
(0)= O K .(o)=oo
/1
1
as
d.,,I ~º=>e,=º dl ~ .. 0
= C/0(<1>2)
= C.J (<1,)=>
tp »
0
C,
= I, ()
I0(l) 10(4>)
Dividing by 8HRx and using Equation (2) to substitute for -rA _1 __ J__!_(~r2kt dT)+__!_~(r2De dCA)=o ( --8H Rx r2 dr dr r2 dr dr
ktT C1 =----+C2 De(-8HRx) r C1 = O because T & C must be finite. ktT CA+ =C2 De(--~HRx) r = R cA = cAs and T = Ts CA+
12-22
P12-13 (b) No solution
-
will be given
P12-15 (a)
CAS
z=O Givcn:
z1
L
1
A H B on the walls of a cylindrical catalyst pore.
z1 = length of poisoncd section -r~ = kCA In the poisoned section: WA
xr lz - WA ~
For O S z S z 1
w A 1tt""., 1:t WA
lz.~ +r;P dZ = O
=
,
r;
O , since this area is poisoned .
-
w A m-" 1Z•6Z
= O or -dW dz A =
= -CDAB dXA +XA(WA + WB) dz 12-23
o
But W A = -Wa, since, for each mole of A consumed, one mole of A is rcacted.
W A == · CDAB ~.\
and
!:(CDAB ~"}
= 0
or d2XA = O dz2
= o • cA = xAS e = cAS At z = z1 • CA = XA1 = CA1
At z
. Integranng:
XA1 =· K1
Atz=O. XA
= C~:
at z
Z
+ K2
K1 = C~
e 1 = z1, xA = xA1 = ~ e =
X A=
k1 z1 +
~
e
=
k1
=
cA1-CAS
e~
CAt • CM z+CAS
CA _ e-
C
Cz1
CA= (CA1-CAS)(fi")+CAS The flux is W A = -C DAB dXA = -DAB dCA (CA1 - CAS) dz
dz
P12-15 (b)
Before poisoning, ll
=
tan
h q,1 where 11
q,1
=
cl>t = L { ~ • )112 r
AB
After poisoning, the differcnrial equation and boundary conditions are the samc for the unpoisoning region of the pore, z1 ~ z ~ L if z is replaccd by L - z; and if we let CA
=
CA1 at z' = O and d;A = O at z = L- z1. then, for the unpoisoncd scction ofthe
cataíyst pore Tt applics if <, is replaced by :
2k"
(L- z) ( rDAB
)l/2 _- (1-L)4>1 z_ · .1.e .• 11
-
tan
h[t1 (t -t)] ci>1 (1-t)
Toe cffectivencss factor for the unpoisoned section of catalyst pore is defined as : WA = tptr{L-z)CA1 This can be relatcd to the overall effectiveness factor for thc entire pore by
12-24
WA = 11' x r L CAS = ll
11' CAS = 11 CA1 ( 1 - ~)
r,
WA = -
But
~,;8 (CA1 - CAS) from pan (a}
r,
CA1 = CAS - W A Zt DAB
r,
1l1CAS =
ll{t -y-}[cAS-
~~1]
WAx r2 = Tt'2lt r L CAS k"
But 1l
1t r L CA1
, =
1'1
(t -!l)ll · 211' LrDABL 2 CAS k" Zt \ L J
P12-16
_ The teaetion is A
6 =
e
A
t-
1 =-
t;
-+} A_2 E
=
yAO O
=
-0.5
CAo(l - X) where CAo = .L = 8.2 atm = 0.2 gmol ª ( 1 - 0.5X) RT 0.082 l atm (227+ 273 K) 1
gmolºK
12-25
= .L+!.FAo
_L YA
2
FAo = FA
2 FA
2(..L-.l) YA 2
1 _
FA
YA
= 2 - 2yA
X = 1 .. FA = 1 - YA FAO 2 • YA r~ = FA,!,.X
ln(-r;J
=
YA
or FAO = _¡_ _ l - 2 - YA
2 - YA
wherc W = 4 x40g = 160 g
= 0.16 kg
ln ko +
LctM = ln(-r;J; Aa = lnko; A1 = cx;andN
. ..
M
= Ao+A1N
= lnC,..
(1)
n
D
i
i
L M¡ = nAo+At L N¡
(2)
n
n
n
i
i
i
L ~ti N¡ = Ao L Ni + At I
N/
e~
i
Frn
YA
X
l
1
0.21
0.88
4.285
X
2
2
0.33
0.80
6.666
X
3
4
0.40
0.75
4
6
0.57
0.60
1.143
-r~
M
N
N2
MN
10 ·2
5.5
1.705
-3.150
9.923
-5.371
10 ·2
10
-2.708
7.333
-6.236
18.75
2.303 2.931
-7.404
22.5
3.114
31.625
3.454 3.689
-2.526. 6.381 -2.169 4.704 -1.964 3.857 -1.821
3.315
-6.717
-14.338
35.513
-39.266
8 X 10 •2 X
(3)
10
·l
5
11
0.70
0.46
1.403 X 10 ·l
6
20
0.81
0.32
1.619 X lQ ·l
40
n
I=
Equations (2) and (3) become: 17.196 = 6Ao - 14.338 A1 -39.266 = -14.338 Ao + 35.513 A1
12-26
17.196
-ó.754 -6.784
=
A.o
r,
6.36 = In ko -+ leo
= l.S = a.
=
578.25
1
At = 1.46
.
At T1 = 237C = 510K: k1 = ~
cl_·s
-r ·
A
=
FAO X where X = 2 • 2(0.097) = O 9490 W 2- 0.097 .
-r. = {9)(0.9490) 0.16
A
CAo
e
= PAo = RT
k
8.2
=
0.082(510)
1
ct.S A
..
where.
o• 196 gmol 1
= 0.196 ( 1 - 0.9490) = 0 019 g mol 1 • 0.5(0.9490) • _J
A
53,38
k1 =
= 53 ·38 --
(0.019)1.5
= 2.035
X
104
koexp{f{fo--t)}
k1 =
(_1_--L)}
2.035 x 10' = 578.25 exp (E(J/mole) 8.314 500 510
:.
E = 1.55 x 10S
P12-16 (b)
naue
=
Euue
2napp -
=
1
1
gmol
= ~ 1.S) • l =
2Eapp = 15.1
X
2: second order
lCP ~
{1 X lQ·lf (2.035 X
104)49 (2.3
0.23 X 104
12-27
X
10') 0.196
$2 = 1.40 X }()6 n
·1
=
{
? )112 f..J...) f 3X 1()6 ) n:T \4>n = { 2.;' 1 }1/2 \1.40
= 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
gmol ;T 1
= s21 e = sao K
k
=
leo exp
k
=
{E.R (l. l.)I = 578.25 exP 17 ·558.314 x lOS (-L. -L)) To TI 500 800
2.19 X
}()32
- 2.19 X
1032 (0.Ql)l.S = 2.19 x ¡()29 g mole Is
P12-17 (u)
d2y __
el).}
ti.
'l'n
y n =0
Multiply by 2 y : dy ~(dy)=tl\2 nzdy 2 y dA dA d11. 'l'nY dA Manipulating the L.H.S. ~(dy)2 d11. dA
= 2 dy d2y
dA dA2
d (dy) dA dA 2 = nYn 2dy dy)2 Yn+l 1 (·-dA =2<1> n --+C n+l
·y=\j/A=~e
d\Jf , A=Ü y =Ü --=O therefore C¡ =0 CAo dA Taking the derivative of y and evaluating at A =1
dyl _
dA .t.=I
2<1>~Yn+I n+ 1
.
Á=l
-tl
n+ 1
12-28
L 1
The effectiveness factor is
1tR2(DAdCA) l'} = dr r=k kCnAs 41tR3 3 In dimensionless form
_\~L q>~
l'} -
A= \j/A, differentiating gives dy = A d\Jf dA dA at A=l
+ \JI
dyl = d\j/ + 1 dA A.=l dA
d'lf; ~2~~-1 dA
n+l
'
,¡;3g.~
For larger cJ>n
CDP12-A CDP12-B (a) 3rd ed. 12-19 (a) CDP12-B (b) 3rd ed. 12-19 (b) CDP12-B (e) 3rd ed. 12-19 (e) CDP12-C (a) 3rd ed. 12-20 (a) CDP12-C (b) 3rd ed. 12-20 (b) CDP12-C (e) 3rd ed. 12-20 (e)
12-29
CDP12-D (a) 3rd ed. 12-21 (a) CDP12-D (a) 3rd ed. 12-21 (b) CDP12-D (a} Individualized solution
CDP12-E 2ºd ed. 11-18 CDP12-F 2ºd ed. 11-19 CDP12-G 2ºd ed. 11-20 CDP12-H 2ºd ed. 11-21 CDP12-I 2ºd ed. 11-22 CDP12-J CDP12-J CDP12-J CDP12-J CDP12-J
(a) 2ºd ed. 12-7 (a) (b) 2ºd ed. 12-7 (b) (e) 2°d ed. 12-7 (e) (d) 2ºd ed. 12-7 (d) (e) 2°d ed. 12-7 (e)
CDP12-K 2ºd ed.12-9 CDP12-L CDP12-L CDP12-L CDP12-L
(a) 2ºd ed. 12-8 (a) (b) 2ºd ed. 12-8 (b) (e) 2ºd ed. 12-8 (e) (d) 2"d ed. 12-8 (d)
CDP12-M CDP12-M CDP12-M CDP12-M
(a) 3rd ed. CDP12-L (a) (b) 3rd ed. CDP12-L (b) (e) (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) 3rded. CDP12-Q (b) CDP12-S CDP12-T CDP12-U
12--31
Solutions for Chapter 13 - Distributions of Residence Times for Chemical Reactors P13-1 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
,·I
P13-2 (b) --+
v¡,
PFR
PFR
~
(3)
~
(4)
CSTR
r¿FR
J • [
CSTR
(6) (5)
13-1
CSTR
1-
PFR
.------.i
1---.-,
1
PFR
r,_A1_--1 i----...
PFR
or PFR
Recycle
(8)
~~~. (10)
(9) PFR
-[
'-==-1--· L::J (12)
(11)
P13-2 (e) For a PFRICSTR Series
E(t) = {:-(,:,: ::. T,
13-2
fl3-2 (d) X=0.75
For a PFRfirst arder reaction:
Da
= ln(-1-) = In (4) = 1.39 where 1-X
Da = kt
Fo, a CSTR.first order reaction:
1-·)-1= 3
Da= (-
1-X
where
Da= kt
For a LFR first order reaction: Solving iteratively Hilder approximate formula with an initial value Daº (i.e . DaPFR
Daexp( ~a)+ 4 + Da exp( ~a)+ Da Da
0.75 = ---
where
Da= k.t
Da=2.58 The ratio of the Damkohler numbers is equal to the ratios of the sizes. VPFR
Relative sizes: -----
VcsTR
VLFR
= 0.46, ---
VcsTR
= 0.86
P13-2 (e) For a PFR, r=5.15min, first order, liquid phase, irreversible reaction with k=O.lmin-1
.
X= 1-e-ki' = 0.402
13-3
¡-
For a CSTR, r=5.15min,first
X=_!:_'!__= l+kr
order, liquid phase, irreversible reaction with k=O.lmin-1•
0.402
0.385
0.402
XcsTR 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 (f) Decrease of 1 O 't' in temperature See Polymath program PI 3-2-f.pol POLYMA TH Results Calculated values of the DEO variables Variable t Xbar k
cao
X
tau tl
E2 E
initial value
minimal value
0.0025446 0.75
0.0025446 0.75
o o
o
1000 500 5.0E+lO
o
o o o
1000 500 6.25E-08
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 J 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] Cao = .75 [ 3 J X = k*Cao*t/(1 +k*Cao*t) [4J tau= 1000 [ s J t1 = tau/2 [ 6 J E2 = tauA2/2/(tA3+.00001) ¡ 7 J E = if (t-ett) then (O) else (E2)
13-.4
final value 2.0E+~ 0.6023837 0.0025446 0.75 0.9744692 1000 500 6.25E-08 6.25E-08
tl70~--------------,
0.56 042 0.28 014 0·00o
4000
sooot
12000
16000
20000
The decrease of lOºC in temperature has the effect ofreducing the mean conversion by 14%. Decrease in reaction orderfrom 2"d to pseudo l" See Polymath program P 13-2-f-2.pol POLYMA TH Results Calculated values of the DEO variables Variable t Xbar k
Cao X
tau tl E2 E
initial
o o
value
minimal value
ºo
·---
0.004 0.75
0.004 0.75
1000 500 5.0E+lO
1000 500 6.25E-08
o
o
o
o
maximal value 2.0E+04 0.9391084 0.004 0.75 1
1000 500 5.0E+lO 0.0024915
final value 2.0E+04 0.9391084 0.004 0.75 1
1000 500 6.25E-08 6 . 25E-08
ODE Report (RKF45) Dífferential equations as entered by the user [ l J d(Xbar)/d(t) = X*E Explícit equatíons as entered by the user [ l] k = 0 . 004 [2J Cao=.75 [3] X= 1-exp(-k*t) [4] tau= 1000 [ 5] t1 = tau/2 [ 6 J E2 = tauA2/2/(tA3+.00001) [ 7 J E = if (kt1) tnen (O) else (E2)
f-
13-5
1 1
¡
1.0 .----------------,
0. 8
0.6
0.4
0.2
o.o--~--~--~-~--~ O
4000
8000t
12000
16000
20000
The decrease in reaction order from 2nd to pseudo 181 has the effect of increasing the exit conversion by 20%. The smaller the dependency of the rate on CA means that when CA is below 1 mol/dnr' then the rate of consumption of A is larger and hence resulting in a larger conversion. Exothermic reaction in adiabatic reactor: See Polymath program Pl3-2-f-3.pol POLYMATH Results Calculated values of the DEO variables Variable t Xbar X T
Cao tau t1
E2
E k
initial value
o o
o
minimal value
o o o
323.15 0.75 1000 500 5.0E+lO
323.15 0.75 1000 500 6. 299E-·08
0.01
0.01
o
o
maximal value 2.0E+04 0.999375
final value 2.0E+04 0.999375
823.15 0.75 1000 500 5 . 0E+l.0 0.0022142 19 . 337202
823 . 15 0.75 1000 500 6.299E-08 6.299E-08 19. 337202
1
ODE Report (STIFF) Differential equations as entered by the user [ 1J d(Xbar)/d(t) = X*E [2] d(X)/d(t) = k*(1-X) Explicit equations as entered by the user [ 1] T = 323.1 S+SOO*X (21 Cao = .75 [ 3 J tau = 1 000 r 4 J t1 = tau/2 [ 5 J E2 = tauA2/2/(tA3+.00001) ¡ 6 J E = if (t
13-6
1
r--==----------,
LO 0.8
0.6 0.4
0.2 0 . 0 --~--------~-~ o 4000 sooot 12000 16000 20000 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., t=40min, second order, liquid phase, irreversible reaction with k=0.01 dm3 /mol-min" .
X=
krC~=0.76 1+ KrCAo
For a CSTR, t=40min, second order, liquid phase, irreversible reaction with k=0.01 dnr' /mol-min' .
X (
l·-X
)
2
=
krCAo -) X= 0.58
Maximum Mixedness Model and Segregation modelare given in El3-7 Maximum Mixedness Model See Polymath program Pl3-2--g.pol POL YMA TH Results Calculated values of the DEO variables Variable z X
cao
k
lam ca El E2 Fl F2 ra E F EF
initial value
o o
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
o o
8 0.01
o
3 . 2490705 0.0028734 2.25E-04
o
0.381769 -·O . 64 2.25E-04
o
0.0220689
maximal value 200 0.5938635 8 0.01 200 8 0 . 1635984 O. 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
o
3.4938093 0.028004 O. 015011
o
0.381769 -0.122067 0.028004
o
0.028004
1-:
[1
l d(x)/d(z) = -(ra/cao+E/(1-F)*x)
Explicit equations as entered by the user [ll cao = 8 [2 J k = .01 [ 3 J lam = 200-z [ 4 J ca= cao*(1-x) [ 5 J E1 = 4.44658e-1 O*lam"4-1.1802e-7*1am"3+1.35358e-5*1am"2-.000865652*1am+.028004 [ 6 J E2 = -2.64e-9*1am"3+1 . 3618e-6*lam"2-.00024069*1am+.015011 [ 7 J F1 = 4.44658e-10/5*1am"5-1.1802e-7/4*1am"4+1.35358e-5/3*1am"3- .000865652/2*lam"2+.028004*lam [ 8 J F2 = -(-9.30769e-8*1am"3+5. 02846e-5*1am"2-.00941*lam+.618231-1) [ 9 J ra = -k*ca"2 [ 1 o J E = if (lam-ee70) then (E1) else (E2) [11] F = if (lam<=70) then (F1) else (F2) [12 l EF = E/(1-F) 0.60 0.48
[]
0. 36 0.24 0.12 O . 00o
40
80
120
z
160
200
1-[-:6=M~=:------.--_+-._I-X_6""'1_ %:~=------R;.,__~6_p;_:-
-l};..__:_i~_;-;;-·
5
P13-2 (h) Liquid phase, first order, Maximum Mixedness model RateLaw: ·-rA =k1CA where k¡ =CA0k=0.08min·1
cA = cA (1-x) 0
rA CAo
= -k1(1-X)
E(,1) ·X dA CAo 1--F(A) dX E(,1) di= --k(l- X)+ l-F(,1) _dX =~+
dX = k(l-
d:
X
X)-- E(,1) X l·-F(,1)
See Polymath program PI 3-2-h-l .pol
13-8
]
POLYMATH Results Calculated values of the DEO variables Variable z X
cao k
lam ca El E2 Fl F2 ra E F EF
initial value
o o
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
o o
8 0.08
o
1.7365447 0.0028731 2.25E-04
o
0.381769 -0.64 2.25E-04
o
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
o
2.0288435 0.028004 0.015011
o
0.381769 -0.1623075 0.028004
o
0.028004
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 [ ll cao = 8 [2] k=0.08 [3J
tam e zco-z
[4J [5J [6J [7J [8J [9J [ 10 J [ 11 J [ 12 l
ca= cao*(1-x) E1 = 4.44658e-1 O*lam"4-1.1802e-7*1am"3+1.35358e-5*1am"2-.000865652*1am+ . 028004 E2 = -2.64e-9*1am"3+ 1.3618e-6*1am"2-.00024069*1am+.015011 F1 = 4.44658e-10/5*1am"5-1 . 1802e- 7/4*lam"4+ 1 .35358e-5/3*1am"3- .000865652/2*1am"2+ .028004 *lam F2 = -(-9.30769e-8*1am"3+5.02846e-5*1am"2-.00941 *lam+.618231-1) ra = -k*ca E= if (lam<=70) then (E1) else (E2) F = if (lam<=70) then (F1) else (F2) EF = E/(1-F)
0. 64 (US
Q
0.32 0.16 0.00'--~--~--~-~-~--o 40 80 z 120 160
200
At z = 200, i.e. A= O (exit), conversion X= 75 %. The decrease in reaction order from 2ºd to 1 st has the effect of increasing the exit conversion by 19%. Once the concentration of A drops below 1 mol/dnr' then the rate of consumption of A does not fall as rapidly (as the znct order reaction) and hence resulting in a larger conversion.
Liquid phase, third order, Maximum Mixedness model Rate Law: - rA = kC / 13-9
1
I· 1
cA
= cA0(1-x)
~A =-k'CA/(1-X)3
Where
k'CA/ =k=0.08min-1
Ao
dX =Í+ dA CAo
E(J) X 1-F(J)
dX = k'C 2(1- X)3 dz Ao
E(J) X
1-F(J)
See Polymath program Pl 3-2-h-2.pol POLYMA TH Results Calculated values of the DEQ variables Variable z
X
ca o
initial value
o o
8 0.08 200 8 0.1635984 2.25E-04 5 . 6333387 0 . 9970002 2.25E-04 0.9970002 0 . 075005
k
lam ca
El E2 Fl F2 E F EF
minimal value
o o
8 0.08
o
4.1061501 0.0028733 2.25E-04
o
0.381769 2 .25E···04
o
0.0220689
maximal value 200 0.4867311 8 0.08 200 8 0.1635984 O. 015011 5.6333387 0.9970002 0.028004 0.9970002 0.075005
final value 200 0.4614308 8 0.08
o
4.3085534 0.028004 O. 015011
o
0.381769 0.028004
o
0.028004
ODE Report (RKF45) Differential equations as enterad by the user r 11 d(x)/d(z) = ··(-k*(1-x)"3+E/(1-F)*x) Explicit equations as entered by the user [lJ cao = 8 [2] k = 0.08 [ 3 J lam = 200-z [ 4 l ca= cao*(1-x) [ 5 J E1 = 4.44658e-1 O*lam"4-1.1802e-7*1am"3+1.35358e-5*1am"2-.000865652*1am+.028004 [ 6 J E2 = -2.64e-9*1am"3+ 1.3618e-6*1am"2-.00024069*1am+.015011 [ 7 J F1 = 4.44658e-10/5*1am"5-1.1802e-7/4 *lam"4+ 1 . 35358e-5/3*1am"3-. 000865652/2*1am"2+.028004 *lam [ s J F2 = -(-9 . 30769e-8*1am"3+5.02846e-5*1am"2-.00941*lam+ . 618231-1) [9J E= if (lam<=70) then (E1) else (E2) [ 10 J F = if (lam<=70) then (F1) else (F2) [11] EF = E/(1-F)
At z = 200, Le. A= O (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 mol/dnr' then the rate falls rapidly and CA is not consumed so quickly, resulting in a smaller conversion.
Liquid phase, half order, Maximum Mixedness model RateLaw:
-rA
= k'CA112
13-10
cA = cA (1-x) 0
ÍC
Ao
= -k'C Ao -112(1-X)112
dX =~+ dA CAo dX =kC d:
Where
k
. -i = k 'C Ao -vz = O . 08 nun
E(;i,) X 1- F(A) -1,2(l-X)112 _
Ao
E(;i,) X
1- F(;i,)
See Polymath program Pl3-2-h-J.pol
POLYMATH Results Calculated values of the DEO variables Variable z X
cao
k
lam ca El E2 Fl F2 E F EF
initial
o o
value
8 0.08 200 8 0.1635984 2.25E-04 5.6333387 0.9970002 2 . 25E-04 0.9970002 0.075005
minimal value
o o
8 0.08
o
0 . 532148 0.002873 2.25E-04
o
0 . 381769 2.25E-04
o
O . 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
o
0.7694568 0.028004 O. 015011
o
0.381769 0.028004
o
0.028004
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(x)/d(z) = -(-k*(1-x)"(.5)+E/(1-F)*x) Explicit equations as entered by the user [ll cao = 8 [2] k = 0. 08 [3] lam = 200-z [ 4 l ca= cao*(1-x) [ 5 J E1 = 4.44658e-1 O*lam"4-1.1802e- 7*1am"3+ 1 . 35358e-5*1am"2-.000865652*1am+ 028004 [ 6 J E2 = -2.64e-9*1am"3+1.3618e-6*1am"2-.00024069*1am+.015011 [ 7 J F1 = 4..44658e-10/5*1am"5-1. 1802e-7/4 *lam"4+ 1 .35358e-5/3*1am"3-.000865652/2*1am"2+.028004 *lam [ 8 J F2 = -(-9.30769e-8*1am"3+5.02846e-5*1am"2-.00941 *lam+ . 618231-1) [ 9 J E= if (lam<=70) then (E1) else (E2) [ 1 o J F = if (lam<=70) then (F1) else (F2) [ 11 J EF = E/(1-F)
11
13-11
1.0 ~---------------,
0.8 0.6 04 0.2
o.o f),_---4-0--~80
1~2-0-~l-64-) _ _,200 z
At z = 200, i.e. 'A= O (exit), conversion X= 90 %. The decrease in reaction order from 2ºd to Y2 has the effect of increasing the exit conversion by 34%. The smaller the dependency of the rate on CA means that when CA falls below 1 mol/dnr' then the rate of consumption of A
P13-2 (i) Assymetric RTD: See Polymath program PlJ-2-i-Lpol
rn
POLYMA Results Calculated values of the DEO variables Variable t ca cb ce e abar cbbar cebar cd ce cdbar cebar T
kl k2 El E2 re k3 ra re E rb Sed Sde rd
initial value
o l l
o o o o o o o o
350 1 1 -0.004 -27.402
minimal value
o
0.0228578 0.2840909
o o o o o o o o
1 -2
350 1 1 --27. 414373 -27.402 0.0064937 1 -2
-0 . 004 -1
-0.0272502 -1
1
o
o o
1
o
o o
-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 J 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 O . 4539723 0.3566073 0 . 2612331 0.3166306 0.3026417 O .1778722 350 1 1 -27.414373 -0.0272502 0.0064937 1 -0.0293515 0.0742139 -0.0272502 -0.0807076 l. 5284379 0.8250406 . . o. 0513561
d(ee)/d(t) = re d(eabar)/d(t) = ea*E d(ebbar)/d(t) = eb*E d(eebar)/d(t) = ee*E d(ed)/d(t) = rd [ B l d(ee)/d(t) = re [ 9 l d(edbar)/d(t) = ed*E [10] d(eebar)/d(t) = ee*E [3J [4l [ 5J [6J [7 J
Explieit equations as enterad by the user [lJ T= 350 [ 2 J k1 = exp((S000/1 . 987)*(1/350-1/T)) [ 3 J k2 = exp((1000/1.987)*(1/350-1/T)) [ 4 J E1 = -2.104*t"4+4.167*t''3-1.596*t"2+0.353*t-0.004 [ 5 J E2 -2.104*t"4+ 17.037*t"3-50.247*t"2+62.964*t-27.402 [ 6 J re = k1 *ea*eb [7] k3 = exp((9000/1.987)*(1/350-1/T)) [ 8 J ra = -k1 *ea*eb-k2*ea [ 9 J re = k3*eb*ed [ 1o J E = if(k=1.26)then(E1 )else(E2) [ 11 l rb = -k1*ea*eb-k3*eb*ed [ 12 l Sed= ee/(ed+.000000001) [ 13J Sde = ed/(ee+.000000000001) [ 14] rd == k2*ea.~k3*cb*cd
=
If the temperature is raised, the conversion of A increases. The selectivity decreases with increasing temperature
Scid
increases with temperature and Sctie
Bimodal RTD See Polymath program P l 3-2-i--2 .pol POLYMA TH Results Calculated values ofthe DEO variables variable
'z
ca cb ce F cd ce cbo ca o ceo cdo ceo lam T k2 ki re k3 El E2 É3 re ra rb E EF
initial
o
value
1 1
o 0.99
o o 1 1
o o o 6 350 1 1 1 1 346.34561 6737.4446 O. 00911
o
-2 -1 O. 00911 o. 911
minimal
o
value
O . 2660482 0.5350642
o
-0.0033987
o o
1 1
o o o o 350 1 1 0.1424065 1 0.1019474 0.0742397 0.0061156
o
-2 -·1 0.0061156 0.2083818
maximal value 6 1 1 0 . 2872257 0.99 0.2692177 0.1929233 1 1
o o o
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
o o o o
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
o
o
o
o
1
1 . 0856272 17.698981
0.1219452 1.0198908 l. 4157321
ODE Report (RKF45) Differential equations as enterad by the user [ 1 J d(ca)/d(z) = -(-ra+(ca-cao)*EF) [ 2 J d(cb)/d(z) = -(-rb+(cb-cbo)*EF) [ 3 J d(cc)/d(z) = -(-rc+(cc-cco)*EF) [ 4 J d(F)/d(z) = -E [ 5 J d(cd)/d(z) = -(-rd+(cd-cdo)*EF) [ 6 J d(ce)/d(z) = -(-re+(ce-ceo)*EF) Explicit equations as enterad by the user
[ll cbo « 1 [2J cao = 1
[3]
CCO=Ü
cdo e O [5] ceo=O [6] tam e s-z [7l T=350 [ 8] k2 = exp((1000/1.987)*(1/350-1/T)) [ 9 J k1 = exp((S000/1. 987)*(1/350-1/T)) [lOJ rc=k1*ca*cb [ 11 J k3 = exp((9000/1.987)*(1/350-1/T)) [ 12 J E1 = 0.47219*1am"4-1.30733*1amA3+0.31723*1amA2+0.85688*1am+0.20909 [ 13 J E2 = 3.83999*1amA6-58.16185*1amA5+366.2097*1am"4-1224 . 66963*1amA3+2289.84857*1amA22265.62125*1am+925.46463 [ 14 J E3 = 0.0041 O*lam"4-0.07593*1amA3+0.52276*1amA2-1.59457*1am+ 1.84445 [ 15 J re = k3*cb*cd ¡ 16 l ra = -k1 *ca*cb-k2*ca [ 17 J rb = ·k1 *ca*cb-k3*cb*cd [ 18 J E = if(lam<=1.82)then(E1 )else(if(lam<=2.8)then(E2)else(E3)) [19) EF=E/(1-F) [ 2 o l rd = k2*ca-k3*cb*cd [21] Sed= cc/(cd+.0000000001) [ 2 2 J Sde = cd/(ce+.0000000001) [4]
If the temperature is raised, the conversion of A increases. The selectivity Scid increases with temperature and Sd/e decreases with increasing temperature
P13-2
G)
Exothermic Reaction: E=45Kj/mol See Polymath program PB-2,J l .pol POLYMA TH Results Calculated values of the DEO variables Variable z X
ca o
T
lam
ca
El E2 Fl 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 O. 015011 5.6333387 0.9970002
13-14
final value 200 0.9579239 8 463.68858
o
0.3366089 0.028004 O. 015011
o
0.381769
O . 01 2.25E-04 0.9970002 0.075005 -0.64
k
E F EF ra
0.01 2.25E-04
o
0.0220681 -O . 8227063
l. 924805
0.028004 0.9970002 0.075005 -0.1699892
l. 8893687
O . 028004
o
0 . 028004 -0.2140759
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 [ll cao = 8 [ 2 J T = 320+ 150*x [3J lam=200-z [ 4 J ca = cao*(1-x) [ 5 J E1 = 4.44658e-1 O*lamA.4-1.1802e-7*1am"3+1.35358e-5*1am"2-.000865652*1am+.028004 ¡ 6 J E2 = -2.64e-9*1am"3+ 1.3618e-6*1am"2-.00024069*1am+.015011 [ 7 J F1 = 4.44658e-10/5*1am"5-1 . 1802e-7/4*1amA.4+ 1.35358e-5/3*1am"3- .000865652/2*1am"2+.028004 *lam [ 8 J F2 = -(-9.30769e-8*1am"3+5.02846e-5*1am"2-.00941 *lam+.618231-1) [9] k= .01*exp(45000/8.314*(1/320-1/T)) [ 10 J E= if (lam<=70) then (E1) else (E2) [ 11 J F = if (lam<= 70) then (F 1) else (F2) [12] EF = E/(1-F) [ 13 J ra = -k*ca"2
LO r:
0.8
Q
0.6 0.4 0. 2 O.O O
80
40
120
160
200
EndotherrnicReaction: E=45Kj/mol See Polymath program P 13-2-j- l. poi
POL YMA TH Results Calculated values of the DEO variables Variable z X
ca o T lam
ca El E2 Fl
initial
o
value
o 8
320 200 8 O . 1635984
2.25E-04 5.6333387
minimal value
o
o
8
289. 82835
o
5.5862683 0.0028739 2.25E-04
o
maximal value 200 0.3017158 8 320 200 8
0.1635984 O. 015011 5.6333387
13-15
final value 200 0.2860515 8 291.39485
o
5. 7115879 0.028004 O . 015011
o
0.9970002
F2 k
0.01 2.25E-04 0.9970002 0.075005 -0.64
E F EF ra
0.381769
0.0017191 2.25E-04
o
0 . 0220679 -0.64
0.9970002
0.01 0.028004 0.9970002 0.075005 -0.0536459
0.381769
0.0019006 0.028004
o
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) Explieit equations as entered by the user [ll eao = 8 [ 2 l T = 320-1OO*x [3 l lam = 200-z [4] ca e caor(t-x) [ 5 J El = 4.44658e-1 O*lam"4-1.1802e-7*1amA3+1.35358e-5*1amA2-.000865652*1am+.028004 [ 6 J E2 = -2.64e-9*1amA3+1.3618e-6*1amA2-.00024069*1am+.015011 [ 7 J Fl = 4.44658e-10/5*1amA5-1.1802e-7/4*1am"4+1.35358e-5/3*1amA3-.000865652/2*1amA2+.028004*1am [ 8 J F2 = -(-9.30769e-8*1amA3+5.02846e-5*1amA2.00941*lam+.618231-1) [ 9 J k = .01 *exp(45000/8.314*(1/320-1/T)) (10] E= if (lam<=70) then (El) else (E2) [ 11 J F = if (larn-ee70) then (Fl) else (F2) [ 12 J EF = E/(1-F) [ 13J ra = -k*eaA2
040.------
·---
0.32
0 . 08
ºººo
40
80
z
120
160
200
P13-2 (k) Base case:
ODE Report (RKF45) Differential equations as enterad by the user [ l J d(Ca)/d(t) = ra/vo ( 2 J d(Cb)/d(t) = rb/vo ( 3 J d(Ce)/d(t) = relvo Explieit equations as entered by the user [1] VO= 10 [ 2] k1 = 1 (3) k2=1 [ 4 J tau = 1 .26 [ 5 J ra = -kl *Ca [ 6 J re = k2*Cb
13-16
¡ 7J Cao = 1 [8] x = 1-Ca/Cao [ 9] rb = k1 *Ca-k2*Cb
PFR CA CB Ce X
K1/K2=2 0.080 0.406 0.513 0.919
K1IK2=l 0.284 0.357 0.359 0.716
K1IK2=0.5 0.284 0.203 0.513 0.716
See Polymath program Pl3-2-k-·2.pol POLYMATH Results
NLES Solution Variable ca cb ce ca o tau cbo ceo kl k2 ra re rb
Value 0.4424779 0.2466912 0.3108309 1 1..26
f (x)
4.704E-10 -3.531E-10
o
Ini Guess 1
o o
o o
1 1 -0.4424779 0.2466912 0.1957867
NLES Report (safenewt) Nonlinear equations [ 1 J f(ca) = cao+ra*tau-ca = O [ 2 J f(cb) = cbo+rb*tau-cb = O [ 3 J f(cc) = cco+rc*tau-cc = O
Explicit equations [ll cao = 1 ¡ 2 l tau = 1 .26 (3 J cbo = O [41 ceo= O [ 5] k1 = 1 [6] k2=1 [ 7 J ra = -k1 *ca [ 8 J re = k2*cb f 9 J rb = k1 *ca-k2*cb
CSTR CA Cii Ce X
K1IK2=l 0.443 0.247 0.311 0.557
K1/K2=2 0.284 0.317 0.399 0.716
See Polymath program P 13-2- k--3 .pol POLYMA TH Results Calculated values of the DEO variables
13-17
K11K2=0.5 0.443 0.158 0.399 0.557
Variable t
ca cb ce e abar cbbar cebar
kl k2
ca o
El E2
ra re
initial value
o
E
o
value
0.0804596
1
o o o o o
o o o o o
1 1 1 -27.462382 -27.402 -1
1 1 1 -0.004 -27.402 -1
o o
o o
X
rb
minimal
-0.1353314 -O . 0272502
1.
-0.004
maximal value 2.52 1 0.3678466 O. 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 O. 7167822 0.304964 0.3347693 0.3468313 1 1 1 -27.462382 -O. 0272502 -0.0804596 0.2027582 0.9195404 -0.1222986 -O. 0272502
ODE Report (RKF45) Differential equations as enterad by the user [ 1 l d(ea)/d(t) = ra (21 d(eb)/d(t) = rb [ 3 l d(ee)/d(t) = re [ 4 l d(eabar)/d(t) = ea*E [ 5 l d(ebbar)/d(t) = eb*E [ 6 l d(eebar)/d(t) = ee*E Explieit equations as enterad by the user [l] k1 = 1 [2] k2=1 [3 l eao = 1 ¡ 4 J E1 = -2.104*t"4+4.164*t"3-1.596*t"2+0.353*t-·0.004 [ 5] E2 = -2.104 *t"4+ 17.037*t"3-50.24 7*t"2+62.964 *t-27 .402 [6] ra = -k1*ea [ 7] re= k2*eb [ 8 l x = (eao-ea)/eao [ 9 J rb = k1 *ea-k2*eb [lo J E = if(k=1.26)then(E1 )else(E2)
Asymmetric RTD Segregatíon Model CA CB
Ce X
K1IK2=2 0.110 0.390 0.486 0.89
K1IK2=l 0.306 0.335 0.347 0.694
K1IK2=0.5 0.306 0.195 0.486 0.694
See Polymath program PI.3-2-k-4 . pol POLYMA TH Results Calculated values of the DEO variables Variable t
ca cb ce e abar
initial value
o
1
o
o o
minimal value maximal value o 7 9.119E-04
o
o o
1
0 . 3678325 0.9927049 0.3879174
13-18
final value 7
9 .119E-04 0.0063832 0.9927049 0.3879174
cbbar cebar kl k2 ca o El E2 •ra re
o o
o o
1 1 1 0 . 20909 925.46463 -1
o o
X
1 1 1 O .1017293 0.074462 -1
o o
-0.1353022
1
rb E4 E3 E
0.2782572 O. 3271713 1 1 1 707.06552 3.072E+04 -9.119E-04 0.3675057 0.9990881 1
o
1.84445 0.20909
o
o
1.84445 0.628067
0.0061369
o
O . 2782572 O. 3271713 1 1 1 707 . 06552 3.072E+04 -9.119E-04 0.0063832 0.9990881 -0.0054713
o
0.09781
o
ODE Re[!ort {RKF45} Differential equations as entered by the user [ 1 J d(ea)/d(t) = ra [ 2 J d(eb)/d(t) = rb [ 3 l d(cc)/d(t) = re [ 4 J d(cabar)/d(t) = ca*E [ s J d(cbbar)/d(t) = eb*E [ 6 J d(ecbar)/d(t) = cc*E Explieit equations as entered by the user [ 1 J k1 = 1 [2] k2=1 [ 3 J eao = 1 [ 4] E1 = 0.47219*t"4-1.30733*tA3+0.31723*tA2+0 85688*1+0.20909 [ 5 J E2 = 3.83999*tA6-58.16185*tA5+366.20970*t"4-1224.66963*tA3+2289.84857*tA2-2265.62125*t+925.46463 [6] ra = -k1*ca r 7 J re = k2*eb r s J x = ( cao-ca)/cao r 9 J rb = k1 *ca-k2*eb [10] E4 = O [ 11] E3 = 0 . 0041O*t"4-0.07593*tA3+0.52276*tA2-1.59457*t+ 1 . 84445 r 12 l E= if(k=1.82)then(E1 )else(if(k=2 . 8)then(E2)else(if(t<6)then(E3)else(E4)))
BimodalRTD Segregatíon Model
CA CB Ce
X
Ki/K2=1 0.388 0.278 0.327 0.612
Ki/K2=2 0.213 0.350 0.430 0.787
Ki/K2=0.5 0.388 0.175 0.430 0.612
Ki/K2=1
Ki/K2=2
Ki/K2=0.5
0.306 0.335 0.347 0.694
0.110 0.390 0.486 0.89
0.30ó 0.195 0.486 0.694
Asymmetric RTD Maximum Mixedness
CA ,...CB Ce X
----
P13-2 (1-r) No solution will be given at this time. 13-19
..
P13-3 Equivalency Maximum Mixedness and Segregation model for first order reaction: Maximum Mixedness Model:
dCA
dA
E(A)
= kC A + 1- F().) (CA -
C Ao
)
Rearranging:
d CA e [ C Ao Using the integration factor:
by definition
E().
J[k +1-F(,t) E(,t) "
E(A )e
=-
d).
-
J
[k+
E(,t) ]dA l-F(,t)
1- F().)
)d().) = dF().)
gives: changing the variables from ')... to t in the RHS integral:
_.S_e-u (1- F(). )) = CAo Ak
CA= CAo
e ( ) 1-F A
r
r[
E(t )e-kt (1- F(t )) 1-F(t)
l,t J'
f E(t)e-k dt
(1)
1
Exit concentration is when l=O, F(O)=O hence eqn (1) becomes:
CA= CAo
E(t)e-k1dt
This is the same expression as for the exit concentration for the Segregation model..
13-20
P13-4 (a) Mean Residence Time By definition
T
=
l·
fE(t)dt
2
= 1. The area of the semicircle representing the E(t) is given by A= .1Z';
o
Fo, constant volumetric flow
tm
=T=
l
=1
and
= 0.8 min .
P13-4 (b) Variance 00
a2
= f (t - r)2 E(t)dt = ft2 E(t o
ft E(t)dt = ft ~r 2
2
o
5.1Z' 8
=-T
4
-T
2
5
o
2
o
2
r2
o 2-.
ee
O'
)dt -
-(t --r)2 dt = -r4 J[cos2 (x )+ 2cos(x)+ 1 ]sin 2(x)dx = ~r4 8
n
1 =-=0.159
2.1Z'
Using Polymath: See Polymath program Pl3-4-b pol POL YMA TH Results Calculated values of the DEO variables Variable t sigma tau tl E2 E
initial value
minimal value
0.7980869 1.5961738
0.7980869 1.5961738
o o
o
o
o o
o o
maximal value l. 596 0.15931.61 0.7980869 1.5961738 0.7980614 0.7980614
final value 1.596 0.1593161 0.7980869 1.5961738 0.01.66534 0.0166534
ODE Report (RKF45) Differential equations as entered by the user [ 1J d(sigma)/d(t) = (t-tau)A2*E Explicit equalions as entered by the user [11 tau=(2/3.14)AQ.5 [ 21 t1 = 2*tau [ 3 l E2 = (t*(2*tau-t))A(1/2) [ 4] E = if (t
1. 1
13-21
0 . 20~--------------,
0.16 ·
1-
0.12
sigma
o.os 0.04 0·00,J."'-oo-~o-.J-2-~0.-64-, -t·-o-.9-6-~1-.2-s _ __.1.60
P13-4 (e) Conversion predicted by the Segregation model
J
X = X (t)E(t )dt o
X(t) =I-··e-kt 21"
X = 1- Je-kt
(z-2 -
(t -
i-)2 ]'2 dt
o See Polymath program P 13-4-c.pol POLYMA TH Results Calculated values of the DEO variables Variable t Xbar tau t1
initial _º
o
0.7980869 l. 596173 8
o o
E2
E k
0.8
o
X
_
value
rninirnal value
o
o 0.7980869 l. 5961738
o o
0.8
o
rnaxirnal value l. 596 0 . 4447565 0.7980869 l. 5961738 0 . 7980671 0.7980671 0.8 0.7210716
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 J tau = (2/3.14)11().5 [ 2 J t1 = 2*tau [ 3 J E2 = (t*(2*tau-t))A(1/2) ¡ 4 J E = if (t
r6J
k= .8
X = 1-exp(-k*t)
13-22
final value l. 596 0.4447565 0.7980869 l. 5961738 0.0166534 0.0166534 0.8 0.7210716
0.5 ~---------------, 0.4 .
03 · 0. 2 0 . .1
o.o '--"""""'::.......-----~-~----' 0.000
0319
O 638t
0 . 958
1277
1596
X =44.5%
P13-4 (d) Conversion predicted by the Maximum Mixedness model
dX dA
=Í+.
CAo
E(A) X
1-· F(A)
rA =-kCA =-kCA)l-X) dX
dJ
= -·k(l- X)+ E(A)
dX d~- =
1-F(A)
X
E(A) k(l-X)-~F(JJX
See Polymath program P 13-4-d. pol
POLYMA TH Results Calculated values of the DEO variables variable z X
F k
lam tau El E
initial value
o o
1 0.8
l. 596
0.7980869 0.0166534 0.0166534
minimal
o o
value
-5.053E-04 0.8
o
O . 7980869
o o
maximal value l. 596 0.4445289 1 0.8 l. 596 O . 7980869 0.7980666 0.7980666
final value
1.596--
0.4445289 -5.053E-04 0.8
o
0.7980869
o o
ODE Report (RKF45) Differential equations as entered by the user [ 1J 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 lam = 1 . 596-z [3J tau=(2/3.14)"0.5 [ 4 l E1 = (tau112-(lam-tau)112)"0.5 [ 5 J E = if (lam<=2*tau) then (E1) else (O) [l]
[2]
1.
13-23
0.5
0.4
D
0.3
0.2 0. .1
O.O 0.000
X
0.319
= 44.5%
0.638z
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 cwnulative distribution function F(t) is given: 1
F(t)
-
05 -
1
1
20
40
PFR
The real reactor can be modelled as two parallel PFRs: The relative
1 3 E(t) = {-J(t - r1 )+-J(t 4
4
PFR -1"2)
Mean Residence Time 1
tm = ftdF = (lümin*l) + (20min*0.75) = 25min o
t
F 13-24
Xseg
=
or
ft-t5(t-r1)+-t5(t-r2) [1 3 J =-r1+-r2 1 3 =25mm . 4 4 4 4
00
t
m
=ftE(t)dt= o
P13-5 (b) Variance
f
a2 = J(t-r)2 E(t)dt = (t -tm o
)2[: t5(t - r1 )+ ! t5(t ·-rJ]dt
!
=: (r1 -tm)2 + (r2 -tm
)2 =
= 75min2
P13-5 (e) For a PFR, second order, liquid phase, irreversible reaction with k = 0.1 dm3 /mol-min", r = 25 min and CAo = 1 ..25 mol/dnr'
X=
= 0.758
kiC Ao l+kiCAo
For a CSTR, second order, liquid phase, irreversible reaction with k = 0.1 dnr' /mol-min", mol/dm3
X -(1-- x-)-2 =
kiC Ao -
X
t
= 25 min and CAo = 1.25
= 0.572
For two parallel PFRs, t1 = 10 min and t2 = 30 min, Fa01 = 1/4Fa0 and Fao2 = 3/4Fao , second order, Iiquid phase, irreversible reaction with k = 0..1 dnr' /mol-min" and CAo= 1.25 mol/dnr'
CAI
= C Ao -
C A2
= C Ao -
X=
Ve Ao -
kr1CAo 1 + kr.C Ao kr2CA0 l+kr2CA0
1 4
- V
e Al
vCAo
3 4
C Ao
= 0.556mol I dm
-C Ao
= 0.263mo/
- -- V
dm
3
3
e A2 =0.731
P13-5 (d) 1-Conversion predicted by the Segregation Model
13-25
X= jx(t)E(t)dt=
0
0
1
4
1
kCAorl 1 + kC Ao 'Z"1
+l 4
kCAot [!8(t-r1)+i8(t-r2)]dt= 1 + kC Aot 4 4
kCAoT2 =0.731 1 + kC Ao r 2
2-Conversion predícted by the Maximum Mixedness model
dX =Í+ dA CAo
E(A) X 1-F(A)
rA =-kc/
=-kCA/(1-x)2
dX =-kC (1-x)2 + E(A) X dA Ao 1-F(A) We need to change the variable such the integration proceeds forward: dX =kCA
dz
º
(1-x)2-
E(T-z)
1- F(T-
z)
X
See Polymath program P 13-5-d poi POLYMA TH Results Calculated values of the DEO variables Variable z X
F cao k lam ca tl t2 E3 ra E2 El E EF
initial value
o o
0.9999 l. 25 0.1 40 1.25 10 30
o
-0.15625 l..25 1.25
o o
_º____
minimal value
o
-l.081E-04 l. 25 0.1
o
0.3614311 10 30
o
-0.15625 l. 25 1.25
o o
maximal value 40 0.7125177 0.9999 1.25 0.1 40 1.25 10 30
o
··O . 0130632 1.25 1.25 1.25 2.8604931
ODE Report (RKF45) Differential equations as entered by the user [ l J d(x)/d(z) = -(ra/cao+E/(1-F)*x) [ 2 J d(F)/d(z) = -E Explicit equations as entered by the user ¡ 1 J cao = 1 .25 [2)
k
= .1
[ 3 J lam = 40-z [ 4 J ca= cao*(1-x) [ 5] t1 = 1 O [6] 12 = 30 [7J E3 = O ¡ 8 J ra = -k*caA2 [ 9 J E2 = 0.75/(!2*2*(1-0.99)) [101 E1 =0.25/(!1*2*(1-0.99))
13-26
final value 40 O. 7061611 -1. 081E-04 l. 25 0.1
o
0.3672986 10 30
o
-O. 0134908 1.25 1.25
o o
[ 11
[121
J E= if ((lam>=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) X seo
0.706
0.731
0.731
Pl3-5 (e) Adiabatic Reaction E==lOOOOcal/mol and T Introducing the enthalpy balance: T
k _- k 325 . e (45000/8..314*(1/325-IIT)).
0.758
XcsTR 0.572
= 325-500X
= 325 - 500X
and the constitutive equation for th MM d I m e mo e .
The conversion is drastically reduced.
P13-5 (f) Conversion Predicted by an ideal laminar flow reactor For a LFR, second order, liquid phase, irreversible reaction with k = 0.1 dm3 /mol-min", mol/dm3
t
We apply the Segregation model, using Polymath:
X
= ne,
and E(t)
0
1 + ktC Ao
= {Ofort<12.5min · 625/(2t3)min-1
far t
e 12.5rnin
See Polymath program P 1.3-5-e.pol POLYMA TH Results Calculated values of the DEO variables Variable t xbar cao k
tau El t1
E X
initial value
o o
l. 25 0.1 25 3.125E+06 12.5
o o
minimal value
o o
l. 25 0.1 25 1.157E-05 12.5
o o
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 [ 1J d(xbar)/d(t) = x*E Explicit equations as entered by the user I 1 J cao = 1 . 25 · [2] k = .1 [3 J tau= 25 [ 4 J E 1 = tauA2/2/(tA3+0. 0001) [5] t1 = tau/2 ¡ 6 J E= if (kt1) then (O) else (E1) [ 7 J x = k*cao*V(1+k*cao*t)
13-27
final value 300 0.7077852 1.25 0.1 25 1.157E-05 12.5 l.157E-05 0.974026
= 25 min and CAo = 1..25
We can compare with the exact analytical formula dueto Denbigh .
X - Da[1-( ~a }n(l + 2/ Da)J-0.709
withDae kC,.,s
P13-6 (a) E(t)
Mean Residence Time 00
_ 2t¡ 0.2 _ l JE( t )dt = 1 . The area of the triangle representing the E(t) is given by A -
By definition
2
o
t1 =5min.
~
if t < 11
-21
l
t¡
E(t) =
-~(1
-211
)if
s 1 s 2t1
t1
11
O otherwise oo
tm
=
11
211L2 \
2
2,1
ftE(1 '}it = f ;d1 - f4dt + f21 dt = 11 1
O
O 1
t1
t¡
t¡
t¡
.
P13-6 (b) Variance
f
00
a2
f
00
= (t -lm )2 E(t)dt = o
t2 E(1 )dt
o anda
2
7 6
=-t
2 m
-t
-t; 2 m
2
t 25 . =_!!!_=-=4.167mm 6 6
2
See Polymath program P13-6-b.pol POLYMA TH Results 13-28
r
Calculated
values of the DEO variables
Variable t sigma tau tl El E2
initial
minimal value
value
o o
o o
5
5
5
10
10
10 0.4 0.4 0.1989341
o o
o
0.4
o
o
E
maximal value 10 4.1666667
final value 10 4.1666667 5 10 0.4
o o
ODE Report (RKF4S) Differential equations as entered by the user ¡ 1J d(sigma)/d(t) = (t-tau)112*E Explicit equations as entered by the user (1 J tau= 5 [ 2 J t1 = 2*tau [ 3 J E1 = t/tau112 [ 4 J E2 = -(t-2*tau)/tau112 [ 5 J E= if (ktau) then (E1) else (if(k=t1 )then(E2)else(O))
o2=4 . l 67min2
Pl3-6 (e) For a PFR, second order, liquid phase, irreversible reaction with kCAo =0..2 min·',i-=5 min
X
=
k'lCAo
1 + k'fCAo
= 0.5
For a CSTR, second order, liquid phase, irreversible reaction with kCAo =0.2 min-1,i-=5 min
X
---=k'lCA --'>X =0382
(1- x)2
º
.
P13-6 (d) 1-Segregation model See Polymath program PU-6-d-Lpol
POLYMA TH Results Calculated values of the DEO variables Variable t Xbar kl X
tau tl El E2 E
initial
o o
0.2
o
value
minimal value
o o
0.2
o
5 10
5 10
0.4
-O . 2
o o
o
o
maximal value 15 0.4767547 0.2 O . 75 5 10 0.6 O.4 0.1978698
ODE Report (RKF4S) Differential equations as entered by the user [ 1J d(Xbar)/d(t) = X*E
13-29
final value 15 0.4767547 0.2 O . 75 5 10 0.6 -0.2
o
Explicit equations as entered by the user
k1 = .2 X=k1*t/(1+k1*t) tau= 5 t1 = 2*tau E1 = t/tauA2 E2 = -(t-t1)/tauA2 E = if (t-etau) then (E1) else(if(t<=t1 )then(E2)else(O))
[1]
[2]
[3 l [4l [5l [6J [7J
0. 5 .----------------, 0.4 0.3 0. 2 0.1
6
3
12
9
15
X =47.7% 2-Maximum Mixedness Model kCA0=k'
--(Í+
dX dz rA
CAo
E(T-z) 1-F(T-z)
xJ
= -kCA 2 = -kCAo 2 (1--x )2
dX =k'(1-x)2dz
E(T-z) X I-F(T--z)
See Polymath program Pl3-6-d-2.pol POL YMA TH Results Calculated values of the DEO variables Variable
z
X
F
k lam tau El tl E2 E
initial value
o o
1 0.2 20 5 0.8 10 -0.4
o
minimal value
o o
-7.513E-06 0.2
o
5
o
10 -0.4
o
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
o
5
o
10 0.4
o
Differential equations as entered by the user [ l l d(x)/d(z) = -(-k*(1-x)"2+E/(1-F)*x) [ 2 J d(F)/d(z) = -E
Explicit equations as entered by the user [1] k = 0.2 [2] lam = 20-z ¡ 3] tau= 5 ¡ 4 l E1 = lam/tau"2 [ 5 J t1 = 2*tau [ 6 J E2 = -(lam-t1 )/tau"2 [ 7 J E = if (larn-etau)then (E1) else(if (larn-eett) then(E2}else (O}}
X =46.7%
P13-6 (e) Laminar Flow Reactor For a LFR, 2nd order, liq . phase, irreversible reaction kCAo =0.2 min-1,T=5 min. We apply the segregation model, using Polymath:
X=·
ktC A 0
1 + ktC Ao
and
E(t)=
{º
for t < 2.5 min
25 /(2t3) min-1 for t ~ 2.5 min
See Polymath program PI 3-6-e . pol
POLYMA TH Results Calculated values of the DEO variables Variable t xbar kcao tau El tl E X
initial value
o o
0.2 5 1.25E+05 2.5
o o
minimal value
o o
0.2 5 4.63E-07 2.5
o o
maximal value 300 0.4506243 0.2 5 l. 25E+05 2.5 0.0549822 0.9836066
ODE Report (RKF45) Differential equations as entered by the user [ 1 l d(xbar}/d(t) = x*E Explicit equations as entered by the user [1] kcao = 0.2 (2] tau= 5 [ 3 J E1 = tau"2/2/(t"3+0.0001) [ 4 l t1 = tau/2 [ 5 J E= if (kt1) then (O) else (E1) [ 61 x = kcao*t/(1 +kcao*t) We can compare with the exact analytical formula dueto Denbigh .
X = Da[1-( ~a }n(l + 2/ Da) J = 0.451
with Da= kC,01
13-31
final value 300 0.4506243 0.2 5 4.63E-07 2.5 4.63E-07 0.9836066
XLFR
XMM
x,
0.451
0.467
0.477
XcS1R 0.382
0.5
P13-7 Irreversible Liquid phase, half arder; Segregation model..
X =
Mean conversion
f X (t )E(t )dt = 0.1
r] =
Assume a Gaussian distribution for E(t):
E(t)
=
1
a.fii
CAo -dX = k--·dt
exp[
112
(t - i2.a
1
(1)
exp[- (t2-· 35!~]
3,Jz.;i
(1- X )112 and CA =l mol/dm3 0
CAo
The only unknown kl is estimated solving with a trial and error method Eq(l). Using POLYMATH: k, = 0.0205 mol112/dm3n . s
P13-8 (a) The E(t) is a square pulse
E(t)
0.5
o
1
-
1
...
-
1
o
1
0 . .5
1
2.5
3
t
60
Third arder liquid-phase reaction: rA = kCA3 with CAo = 2mol/dm3 and k = 0.3 dm6/moi2/rnin ( Isothermal Operation) 1-Conversions Segregation Model
1 X (t) = 1- -;::;=======.: 2t
~(2kC Ao + X= jx(l)E(t)dt o
=
1)
'[¡ _
/l
2
l ]dt= t )2kC~0t + 1 1
)2kC~0t + 1 2
kCAo
See Polymath program Pl3-8-a-l.po1 13-32
=0 . .53
POL YMA TH Results Calculated values of the DEO variables Variable t Xbar
initial value
o o
minimal value
o o
0.3 2 1 1
0.3 2 1 1
k
Cao tl E2 E X
o o
o o
rnaximal value 2 0.5296583 0.3 2 1 1 1 O . 5847726
final value 2 0.5296583 0.3 2 1 1 1 0.5847726
0. 60
ODE ReRort {RKF45} Differential equations as entered by the user [ l J d{Xbar)/d(t) = X*E
0.48
Explicit equations as entered by the user [1) k= .3 [2] Cao = 2 [ 3 J t1 = 1 [4] E2=1 [ 5 J E = if (t>=t1) then {E2) else {O) [ 6 J X = 1-1 /( 1 +2*k*CaoA2*t)A( 1 /2)
0.36 0. 24 012 0.00'------------'-~--~----' o.o 0.4 0.8 t
2-Maximum Mixedness Model
--(Í+
_dX d: rA
CAo
E(T-z) 1-F(T-z)
xJ
= -kC A 3 = -kC Ao 3 (1-X )3
dX =kC~º(1-x)3d:
E(T-z) 1-F(T-z)
X
See Polymath program Pl 3-8-a-2 pol
POL YMA TH Results Calculated values of the DEO variables Variable z X
F cao
k
lam ca El ra t2 tl E EF
initial value
o o
0.9999 2 0.3 2 2 1 -2.4 2 1 1
1.0E+04
minimal value
o o
-9.999E-05 2 0.3
o
0.9569223 1 -2.4 2 1
o o
rnaxirnal 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
o
0.9569223 1 -0.2628762 2 1
o o
12
1.6
2.0
2.0
ODE Report (RKF45) Differential equations as enterad by the user [ 1 l d(x)/d(z) = -(ra/cao+E/(1-F)*x) [ 2 l d(F)/d(z) = -E
1.6 12
Explicit equations as enterad by the user [ll cao = 2 [2] [3J [4)
[ 5J
[6J [7)
[ 8] [9J [10)
k
= .3
0. 8
lam = 2-z ca e cao'(t-x) E1 = 1 ra = -k*ca"3 t2=2 t1 = 1 E = if ((lam>=t1 )and(lam<=t2)) then (E1) else(O) EF=E/(1-F)
0.4 -
o.o o.o
0.8
0.4
z
d2X
1.2
The conversion shows an inflection point in correspondence of z = 1, where start the pulse --2- = O •
dA
P13-8 (b) Introducing in the Segregated Model and in the MM Model :
k _- k 300 • e (20000*(1/300-1/T))
See Polymath program P 13-8-b- l.pol and Pl3-8-b-2.pol 300K 0.530 0.521
x., XMM
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 (e) Adiabatic Reaction Introducing the enthalpy balance:
T
= To + (- Mi RX ) X = 305 + 40000 ¿B¡Cp¡
X
50
See Polymath program P 13-8-c. pol POLYMA TH Results Calculated values of the DEO variables Variable t Xbar ko
initial value
o o
0.8948702
tl E2
1 1
t2 Cao E
k To
minimal value
o o
0.8948702
maximal value 5
0.8949502 0.8948702
final value 5
0.8949502 0.8948702
2
1 1 2 2
1 1 2 2
7.6864651 305
o
1 1 2 2 1
7.6864651 305
7.6864651 305
7.6864651 305
2
o
13.34
o
1.6
2.0
o
o
X T
0 . 9430621 1059.4497
305
305
0.9430621 1059.4497 1100
ODE Rel!ort {RKF45} Dífferential equations as entered by the user [1] d(Xbar)/d(t) = X*E
940
Explícit equatíons as entered by the user [1] ko = 0.3*exp(20000*(1/300-1/305)) [2] t1 = 1 [ 3} E2 = 1 [4] t2 = 2 [5] Cao=2 E= if((t>=t1) and (k=t2)) then (E2) else (O) [6J [7] k = ko*exp{20000*{1/300-1/310)) [8] To= 305 [9] X= 1-1/(1+2*k*Cao"2*t)A.5 [10] T = To+800*X
780
Q
620
460
300
o
P13-9 (a) 3'd order, k=175 dm6/(mol2min), C80=0.0313 dm3/min
(LFR, PFR, CSTR with r=lOOs) PFR Design equation
V
oc
dX
Ao -
dV
cA = cA)1-x)
=kc
A
c
2 B
Ca= CB0(1-x) X
dX
f-(1-·X -)
3
2 V =kCBo -
v
0 X
dX =kC f (1-X)3 Bo
2
1
V V
O
2
V
---=2kC80 -·+1
(1-X)2
V
Using the quadratic solution X
1
= 1--,::===== 1+2kC80
2V
= 1- 0.168= 0.832
V
The conversion for a PFR X=83.2% 13-35
1
2
t
3
4
5
CSTR
v)C Ao -· eA)= -rA V
Design equation
-rA =kCAC8
2
C, = cAJ1-x)
C8 = C8)1-x) CAo-CA CAo
=X
X= kC8/(1- X)3 V V
The conversion for a CSTR X=66.2 % LFR (completely segregated)
E(t)=0
for
t<-r/2 /2t3 -r=V/v=1000110=100s=l.67min
for
='r2
t>= -r/2
f
X= X(t )E(t )dt o
dK_ = kC,¡ (1- X)3 dt
o
Where X (t) = 1 -
1 2
1+ 2kC80 t
See Polymath program Pl3-9-a . poJ POLYMA TH Results Calculated values of the DEO variables Variable t xbar cbo k
tau El X
E
initial value
minimal value
1.0E-05
l.OE-05
O .0313 175 l. 67 1.394E+15 l.714E-06
0_0313 175 l. 67 1. 403E-·06 l. 714E-06
o
o
o
o
maximal value
final value
100 0.1827616 0.0313 175 l. 67 1.394E+15 0.8315008 0 . 1480215
100 0.1827616 0.0313 175 l. 67 l.403E-06 0.8315008 l. 403E-06
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 fll cbo = 0.0313 (2] k = 175 f 3 J tau = 1.67
13-36
[4 [5
E1 = tauA2/(2*1''3)
X= 1-(1/(1+2*k*cb=tau/2) then (E1) else (O)
[6
The integral
X
=
fX (t )E(t )dt
gives mean conversion=18%
o
P13-9(b) (Segregation Model and Maximum Mixedness Model applying RTD ofExample 13-1) Segregation model
dNA dt
---=-r
A
V
Batch reactor
cA = cA)1-x) C8
= C8)1-X)
fx dX 2 Ji (1-X )3 = kCBo t
Similarly
X
=
fX (t )E(t )dt
=
and E(t) from the given data, fitted using Polymath
o
See Polymath program P 13 .9. b-regression. pol
POLYMA TH Results Polynomial Regression Report Model: C02 = a1*C01 + a2*C01A2 + a3*C01A3 + a4*C01A4 Variable al a2 a3 a4
Value 0.0889237 -0.0157181 7.926E-04 -8.63E-06
95% confidence O . 0424295 0.0163712 0.0019617 7.288E-05
General Order of polynomial = 4 Regression not including free parameter Number of observations = 13
13-37 r-
1
Statistics R"2= R"2adj = Rmsd= Variance
0.8653673 0.8204897 0.0065707 8.107E-04
=
See Polymath program 1'13-9-b- Lpol POLYMA TH Results Calculated values of the DEO variables Variable t xbar
initial value
o o o
F
k
o
0.0313 175
o o
X
E
o
o
0.0313 175
cbo
rninirnal value
o o
maxirnal value 14 0.4106313 1.1137842 0.0313 175 0.5847898 0.1566631
ODE Report (RKF45) Differential equations as entered by the user ¡ l J d(xbar)/d(t) = E*x [ 2 J d(F)/d(t) = E Explicit equations as enterad by the user [lJ cbo = 0.0313 [2] k = 175 [ 3] X= 1-(1/(1+2*k*cboA2*t})AQ.5 [ 4 l E = 0.0899237*t-0.0157181 *t"2+0.000792*t"3-0.00000863*t"4 0.5 .----------·----~ 0.4 . 0. 3
0.2 0.1
o.o
L.-...::;..~--~----------...J
0. 0
2.8
5.6 t
8.4
11.2
14.0
X =41%. Maximum Mixedness
RateLaw :-
rA = kCACB
2
13-38
final value 14 O. 4106313 1.1137842 0.0313 175 0.5847898 0.0199021
Í C
= kC
2 Bo
(1- X )3
Ao
dF -= -E(z)
where z=14-A
dz
See Polymath program Pl3-9-b-2.pol POLYMA TH Results Calculated values of the DEO variables Variable z X F
Cbo k
lam Cao Ca E
EF Cb ra
initial value
o o
O . 9999 0.0313 175 14 0.0313 0.0313 0.0199021 199.0212 0.0313 -0.0053663
minimal value
o o
-O .1138842
0.0313 175
o
0.0313 0.0202322
o o
0.0202322 -0.0053663
maximal value 14 0.3536026 0.9999 O. 0313 175 14 0 . 0313 O . 0313 0.156664 199.0212 O . 0313 -0.0014493
final value 14 0.3536026 -o .1138842 O. 0313 175
o
0.0313 0.0202322
o o
0.0202322 -0.0014493
ODE Report (RKF45) Differential equations as entered by the user [ l l d(X)/d(z) = -(ra/Cao+E/(1-F)*X) [ 2 J d(F)/d(z) = -E .Explicit equations as entered by the user [ll Cbo = .0313 [21 k= 175 [ 3 J lam = 14-z [41 Cao=.0313 [ 5 J Ca= Cao*(1-X) [ 61 E = 0.0899237*1am-0.0157181 *lamA2+0.000792*1amA3-0 . 00000863*1am"4 [ 7 J EF = E/(1-F) [8J Cb=Cbo*(1-X) [ 9 J ra = -k*Ca*CbA2
X=35.4%
P13-9 (e) Exit time(t), intemal age(a) and life expectancvñ
( E(t)dt = F(t)~ dF(t) = E(t) Ji dt where E(t) is obtained from the polynomial fit in Part (b).
13-39
I(a) = ~[1- F(a)] V
Intensity Function
P13-9 (d) Adiabatic reaction Segregation model
dX dt
= kC
Bo
z (1- X )3
ººº~(-
Where
k(T) = koexp[E (___!___ ·--~)] R To T
and
T(K)=To+(..:::::.MlrxJx =320+150X
_!_)]
= 175exp[3 1- __ 8.314 320 T
CpA
J
ee
X = X (t )E(t )dt and E(t) from the given data. o See Polymath program Pl3-9-·d- Lpol POL YMA TH Results Calculated values of the DEO variables Variable t xbar F X
cbo
T E k
initial value
o
o o o
minimal value
o
o o o
0.0313 320
0.0313 320
175
175
o
o
maximal value 14 0.7585435 1.1137842 0.8973303 O . 0313 454.59954 0.1565966 4931. 8727
ODE Report (RKF45) Differential equations as entered by the user [ 1] d(xbar)/d(t) = E*x [ 2 J d(F)/d(t) = E ¡ 3 l d(x)/d(t) = k*cboA2*(1-x)A3 Explicit equations as entered by the user [lJ cbo = 0.0313 [2 J T = 320+150*x [ 3 J E = 0.0899237*t-0.0157181 *tA2+0.000792*tA3-0.00000863*t"4 [ 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
Í C
Ao
where
and
= -kC Bo 2(1-X)3 1 k(T)=koexp [-E( --. R To
T(K) =To+
1 )] =175exp [30000( ---1 T 8.314 320
(-Aflrx)x CpA
1 )] T
= 320 + 150X
See Polymath program P 13-9-d-2 . pol
POL YMATH Results Calculated values of the DEO variables Variable z X
initial
o o
value
0.9999 0.0313 320 14 O. 0313 0.0313 0 . 0199021 199.0212 0.0313 175 ~0.0053663
F
Cbo T
lam Cao Ca E EF
Cb k
ra
minimal value
o o
-O .1138842
0.0313 320
o
0.0313 0.0087977
o o
0.0087977 175 -o , 007411
maximal value 14 0.7189248 0.9999 0.0313 427.83873 14 0.0313 0.0313 0.1566233 199.0212 O .. 0313 3001. 8809 -0.0020441
final value 14 O. 7189248 -0.1138842 0.0313 427.83873
o
0.0313 0.0087977
o o
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 l d(F)/d(z) = -E Explicit equations as entered by the user [ll Cbo = .0313 [2] T 320+150*X ¡ 3 l lam = 14-z (4) Cao = .0313
=
¡5J í6J l 7] ¡8J r9J [ 10
Ca = Cao*(1-X) E = 0.0899237*1am-0.0157181 *lam"2+0 . 000792*1am"3-0.00000863*1am"4 EF = E/(1-F) Cb = Cbo*(1-X) k = 175*exp(30000/8.314*(1/320-1/T)) J ra = -k*Ca*Cb"2
13-41
0.80..----------------,
0.64
0.48
0.32
0.16
o.oo u·"". o--""'"2-.8--5~.-6-z ~8-.4--~11-.2-----'14.0 gives X=72% If the reaction is carried out adiabatically the conversions are more than doubled.
Pl3-10 Irreversible, first arder, long tubular reactor, constant volume, isothermal
dFA
ForaPFR --=-r
dV
A
X= I-e-k1 For X
= 0.865 => k t = 2.0
For laminar flow with negligible diffusion (LFR), the mean conversion is given by: ee
~
O
r/2
X = JX (t )E(t )dt = JX (t )E(t )dt X (t) = 1 - e -kt
'('2
E(t) for laminar flow =-3 where
2t
~
Therefore
X=
J (I-e
r/2
t
'l
'?. -
2
2
-k1 )
T' 3dt=I-- T' -
2t
2
2
-kt
J -e 3dt
ee
r/2
t
We can apply the approximated solution dueto Hilder:
13-42
=0.782
X= (4+Da)e05Da +Da-4
(4+Da)e050ª +Da where Da=kr=2
X
= 0.782 < X
PFR
= 0.85
P13-ll(a) First Moment about the mean: by definition is always egua! to zero. 00
00
00
m1 = J(t- r)E(t )dt = JtE(t )dt-T JE(t )dt = T-T = 0 o
o
m1CSTR :::: mlPFR
o
= m1LFR
=0
P13-11 (b) Second-order liquid-phase reaction Da= rkCA0=1.0,r=2min
and kCA0=0 . 5min-1.
CSTR
FAo -- FA = -r A V FAo -FA= FAox V=
FAox (- rAtxit
Liquid-phase T
.
.
= -V = CAo( -CA)
Second-order -
rA
V0
= kC A2
-rA
and
T
=
CA 0 -CA
X
2
kCA
Sol ved
(1 + 2Da)-.Ji+Wa
X=
=0..3SZ
2Da
PFR
-dFA dV
dX
FAo-=-rA · dV
Second-order
13-43
V = FAo Í-d_X_2 where CA = C Ao -'(I_-_x~) o kCA (I+éX) Liquid-phase
O and integrating
E=
1 [ X r= kCAo 1-X
J or X= I+Da Da =0.5
LFR In the ring globule of radius r
dCA -= rA dt
X=
Where -
»c:
rA
= kC A2
(Zorder batch)
I+ktCAo o for t < lmin
E(t) =
{4 /(2t3)
min -I for
t ~
1 min
(E(t) LFR)
r dt fXE(t)dt =-kCAo f ( )t 2 1 + ktC Ao
X=
2
oo
r2 =-kCAo 2
oo
2
T
112
2
= r2kC
2
Ao
[·-! t
+ kC Ao ln(l + ktC Ao·J]
Evaluate for Da= l ,
t
t
12
r
A 2
>O
)2
]
t
P 13-11-b.pol
-· --+-!-~-~----·
-+-l~-~-5-1
The criteria
a ac
12
(
Da ln(l + Da/2)] 2 Da/ 2
P13-12
2 -___
T
1
[
X = 0.451
See Polymath program
tE
= Da[I-
00
f 2+ (1 +kCktCAo Ao )---kCtAo
oo
Xseg>XMM
A
13-44
-_~
dt =
The following figure shows the reaction rate as function of the concentration .
•
0.01
0.005
º..______ The second derivatiée is initially mgative (X a-~M). (Xseg=XMM). CA In the lirnit of low concentration
- rA
= kC A + o( CA )2
(First order) and Xseg=XMM
in the limit of high concentration
- rA
=
!
KA CA
+
a(~J CA
Pl 3-13 No solution
(Reaction order=-1) and X,eg>XMM
will be given
P13-14 (a) Liquid phase, Segregation Model, second arder; non-ideal CSTR, adiabatic:
fX(t,T)E(t)dt
= 0.67
= = C(t)
f C(t )dt = lmg mini dm3
ce
X=
o
E(t)
fc(t )dt
and
o
o
E(t)=IF (t<=l) THEN (t) ELSE ( IF (t>=2) THEN (O) ELSE (2-t)) For a batch globule:
cAo -dX = -rA dt
13-45
The flex point is for CA=8mol/dm3
-rA = kCA
dX = kC
dt
2
Ao
(1- X )2
Where CA =2 mol/dm3 0
k(T) = 0.5exp[__É!:_(-1_..!..)] 8.314 300
T =To+
Where
-MI
rx
T
X
Cps +!l.CpX
C ps =
¿ tl¡Cpi = C pa +O= 50] I mol·
K
!l.Cp = 112C pb - C pa = 100 I 2 - 50 = O T=300+150X Iterate with Ea, using the ODE solver for values of X(t,T) and substitute these into the polynorrúal regression to evaluate ce
theintegral
fx(t,T)E(t)dt
X=
=0.67
o An activation energy Ea of 10000 J/mol gives approximately the correct mean conversion, X =0.67 . Inaccuracy lies in the polynorrúal fit and hence the integral area. See Polymath program PlJ-14--a . pol POLY~ATH Results Calculated values of the DEQ variables Variable t xbar X
cao
T
El E2 E Ea
k
initial value
o o o
2 300
o
2
o
l.25E+04 0.5
minimal value
o o o
2 300
o
-1
o
l.25E+04 0.5
maximal value 3 0.6673438 O. 9198721 2 437.98082 3 2 0.9933851 l.25E+04 2.4247011
ODE Report (RKF45) Differential equations as entered by the user [ 1J d(xbar)/d(t) ::: E*x [2 J d(x)/d(t) = k*cao*((1-x)"2) Explicit equations as entered by the user [ll cao = 2 [2 J T = 300+150*x
13-46
final value 3 0.6673438 0.9198721 2 437.98082 3 -1
o
l.25E+04 2 . 4247011
(3
(4
¡s [6 [7
E1 = t E2 = 2-t E= if (t<=1) then (E1) else (if (t>=2) then (O) else (E2)) Ea= 12500 k = 0.5*exp(Ea/8.314*((1/300)-(1ff)))
P13-14 (b) Parallel reactions, isothermal, segregation model Batch globules
dCA -= rA = rAI + rA2 = -klAC A 2 - k 2cCACB dt
dC8 _ _ . _ -- rs - -rAI + rA2 - klACA dt
2
k - 2cCACB
dCC = rC = -r A2 = k 2C eA eB --
dt
Exit concentrations
dC Abar dt
=e
E(t) A
dC Bbar = e E(t) dt B dCcbar dt
= CcE(t)
E(t)=IF (t<=l) THEN (t) ELSE ( IF (t>=2) THEN (O) ELSE (2-t))
Selectivity
S
=
CBbar. = 2.38 Ceba,
Iteration with k2c until S = 238 gives k2c = 0.3755 dm3 /mol min See Polymath program PB--14-b.pol
P13-15 Reactor: fluidised CSTR (V=lm3; F=10dm3/s, Ccº=2 Kmol/rrr') r
= 1000dm3s/10dm3
= lOOs
The system of complex reactions for the Kentucky coal n . 9 is given by
13-47
r
P13-15 (a) Segregation Model See Polymath program P 13- J 5-a . pol
POL YMA TH Results Calculated values of the DEO variables Variable t ca cp ce e abar cpbar cebar co cobar kl k2 k3 k4 k5 re ra rp tau ro E Spo
initial value
o o o
2000
o o
2000
o o
0.012 0.046 0 . 02 0.034 0.04
o
92 24
l. 667
o
0.59988 2.4E+08
minimal value
o o o
1383.5708
o o
2000
o o
0.012 0.046 0.02 0.034 0.04 -49.695094 10.804749 24 l. 667
o
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
o
92 37.542445 l. 667 36.017167 0.59988 2.4E+08
ODE Report (RKF45) Differential equations as entered by the user [ 11 d(ca)/d(t) = ra [ 2 J d(cp)/d(t) = rp [ 3 l d(cc)/d(t) = re [ 4 J d(cabar)/d(t) = ca*E ¡ 5 J d(cpbar)/d(t) = cp*E [ 6 J d(ccbar)/d(t) = cc*E [ 7 J d(co)/d(t) = ro [ 8 J d(cobar)/d(t) = co*E Explicit equations as entered by the user [ 1 J k1 = 0.012 [ 2 J k2 = 0.046 [ 3 J k3 = 0.020 [4] k4 = 0 . 034 [5] k5=0.04 [ 6 J re = -k1 *cp-k2*ca (7 J ra = k2*cc+k3*cp-k4*ca-k5*ca [ S J rp = k1 *cc+k4 *ca-k3*cp [9] tau= 1.667 [lOJ ro= k5*ca [llJ E=exp(-t/tau)/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 l. 667 36. 017167 3.695E-06 0.9280349
[ 121
Spo = rp/(ro+0 . 0000001)
The exiting selectivity is 0 . 928
P13-15 (b) Maximum Mixedness model See Polymath program Pl3-15-b poi POLYMA TH Results Calculated values of the DEQ variables Variable z Ca Cp Ce Co F kl k2 k3 k4 k5 re ra rp tau ro E Spo EF sigma Cao Cpo Ceo Coo
initial value
o o o
2000
o
1 0.012 0.046 0 . 02 0.034 0.04
o
92 24
l. 667
o
0.59988 2.4E+08 5.999E+07 3
o o
2000
o
minimal value
o o o
1463.3111
o
6.149E-06 0.012 0.046 0.02 0.034 0.04 -47.836902 15.190162 24 l.. 667
o
3.695E-06 0.9906544 3.695E-06 3
o o
2000
o
maximal value 20 875.11273 631 . 80972 2000 408.43732 1 0.012 0.046 0.02 0.034 0.04
o
92 37.327161 l. 667 35.004509 0 . 59988 2.4E+08 5.999E+07 3
o o
2000
o
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 l.. 667 35.004509 3.695E-06 0.9906544 3.695E-06 3
o o
2000
o
ODE Report (RKF45) Differential equations as entered by the user [ l J d(Ca)/d(z) = -(-ra+(Ca-Cao)*EF) [ 21 d(Cp)/d(z) = -(-rp+(Cp-Cpo)*EF) [ 3 J d(Cc)/d(z) = -(-rc+(Cc-Cco)*EF) [ 4] d(Co)/d(z) = -(-ro+(Co-Coo)*EF) [ 5 J d(F)/d(z) = -E Explicit equations as entered by the user [lJ k1 = 0.012 [2) k2 = 0.046 [3 J k3 = 0.020 [4] k4 = 0.034 [5] k5 = 0.04 [ 6 J re = -k1 *Cp-k2*Ca [ 7 J ra = k2*Cc+k3*Cp-k4 *Ca-k5*Ca -r 8 l rp = k1 *Cc+k4*Ca-k3*Cp ( 9 J tau = 1.667 [ 1 o J ro = k5*Ca [ 11 J E= exp(-z/tau)/tau [ 12 J Spo = rp/(ro+0.0000001) [13) EF=E/(1-F) [ 14] sigma= 3
13-49
1
L
1
[15) Cao = O [16J Cpo = O [17) Ceo= 2000 [18) Coo = O
P13-15 (e) The selectivities are reported in the following table:
I XMMcsTR
jsifMPFR ~--------'--·
o_._99
I
XsmPFR ___.__4_.1_7_4
I
Xsmcs1R .,_ __ o._92
P13-15 (d)
exp[- (t2.a - r] = 1 exp[- (t - 5r] 3..{i;i 2·3
Normal Distribution with T = Smin and a = 3min
E(t) =
1
a.fii
T
Segregation Model See Polymath program Pl3-15-d-l.pol POLYMA TH Results Calculated values of the DEQ variables Variable t ca cp ce e abar cpbar cebar co cobar kl k2 k3 k4 k5 re ra rp sigma ro tau Spo El E
initial value
o o o
2000
o o o o o
0.012 0.046 0.02 0.034 0.04
o
92 24 3
o
5 2.4E+08 10 0.0331675
minimal value
o o o
1991.2989
o o o o o
0.012 0.046 0.02 0.034 0.04 -8.5350648 79.947816 24 3
o
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
o
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 l d(ea)/d(t) = ra [2 J d(ep)/d(t) = rp [ 3 J d(ee)/d(t) = re [ 4 J d(eabar)/d(t) = ea*E [ 5 J d(epbar)/d(t) = ep*E [ 6 J 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
k1 = 0 . 012 k2 = 0.046 k3=0.020 k4 = 0.034 k5 = 0.04 [ 6 J re = -k1 *cp-k2*ca [ 7 J ra = k2*cc+k3*cp-k4*ca-·k5*ca [ 8 J rp = k1 *cc+k4*ca-k3*cp [9J sigma= 3 ¡ 1 o J ro = k5*ca [lll tau= 5 ¡ 12 l Spo = rp/(ro+0 . 0000001) [ 13 l E1 = 1/(tau*2*(1-0.99)) [ 14 l E= exp(-(t-tau)"2/(2*sigma"2))/(sigma*(2*3.14)"0.5) [1]
[2 l [3] [4] [5}
Maximum Mixedness Model See Polymath program P l 3- J 5-d-2.pol roLYJ\!A TH Results POLYMA TH Report
0825-2005,
Rcv5 .1.233
Calculated valúes of the DEQ variables Variable z Ca Cp Ce Co F kl k2 k3 k4 k5 re ra rp lam ro tau Spo sigma E Cao Cpo Ceo Coo EF
initial value
o o o
2000
o
l 0.012 0.046 0.02 0 . 034 0.04
o
92 24 2
minimal value
o o o
1995.2809
o
0.9770814 O . 012 O . 046 0 . 02 0.034 0.04 ·-5 . 986657 83.563477 24
maximal value 2 120 . 77791 35.906087 2000 3.8182818 1 0 . 012 0.046 0.02 O . 034 0.04
o
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 . 0296795
5 5.6574287 3 0.0020622
92 27.331698 2 4 . 8311165 5 2.4E+08 3 0 . 0296795
2000
2000
2000
2000
2.968E+06
0 . 089981
2.968E+06
0.089981
o
5 2.4E+08 3
o o o
o o
o o o
o o o
ODE Report (RKF45) Differential equations as entered by the user [ l l d(Ca)/d(z) = -(-ra+(Ca-Cao)*EF) [ 2 J d(Cp)/d(z) = -(-rp+(Cp-Cpo)*EF) [ 3 J d(Cc)/d(z) = +rc+(Cc-Cco)*EF) [ 4 l d(Co)/d(z) = -(,-ro+(Co-Coo)*EF) [ 5 l d(F)/d(z) = -E
13-51
o
4 . 8311165 5 5 . 6574287 3 0.0020622
o o o
Explieit equations as entered by the user [ll k1 = 0.012
[2] [3] [4] [5] ¡6J ¡7J [8] [9l [10]
[lll [12 l ¡ 13 J [ 14 l [15] [16]
[ 17 J [18] [19]
k2=0.046 k3 = 0.020 k4 = 0.034 k5=0.04 re = -k1 *Cp-k2*Ca ra = k2*Ce+k3*Cp-k4*Ca-k5*Ca rp = k1*Ce+k4*Ca-k3*Cp lam e z-z ro= k5*Ca tau= 5 Spo = rp/(ro+0.0000001) sigma = 3 E= exp(-(lam-tau)"2/(2*sigma))/(sigma*(2*3.14)"0.5) Cao = O Cpo = O Ceo = 2000 Coo = O EF = E/(1-F)
P13-16 Multiple parallel reactions, isothermal
E(t)=0.0279693 - 0.0008527t + l.2778e-5 t2 - l.0661e-7
t3 + 4..5747e--10 t4-
7.73108e-13 t5
K1=5 . 0 m6/kmol2min, k2=2.0 m3/kmol.min, k3=10 m6/kmoi2min,k4=5.0 m2/kmol213min
P13-16 (a) Segregation model
de A = - k 1 -dt
eA eB 2 - k2 e A e B -- 2 k 4 e C eA 2'3 3
13-52
1
t
dCD
dt
= CDE(t)
See Polymath program PIJ-16-a.pol Selectivities:
Exit concentrations:
CA
= 0.026mo[ f dm3
ScD = 0.00196
C8
= 0.008mol / dm3
SDE
Ce
= 6.9.55e-
SEF = 28433
5mol / dm3
= 3.009
CD= 0.036mol I dm3
CE= O.OI2mol/ dm3 -
CF
= 4.I.56e--7mol
/ dm
3
P13-16 (b) Maximum Mixedness model E( t)=0.0279693-0.0008527A+ 12778e-51..2- L066 le-·7
1? +4.57 4 7 e-10 A4- 773 lOSe-13 1c5
where z=t, tp200min (extent ofE(t)).
_deA _ r d: - A
-((e _ e ) I-F(A) E(A) J A
Ao
and and so on for other species.
dF(A) = -E(A) ·-dz
gives F(1c)
See Polymath program PI3-16-b.pol
13-53
Selectivities:
Exit concentrations:
CA
= 0.028mol / dm'
s.; = 0.0009
C8
= O.OlOmol/ dm3
SDE = 3.004
Ce = 2.927e-· 5mol / dni' CD
= 0.033mol / dm'
CE
= 0.01 lmol / dm'
S EF
= 1343110
CF = 7.985e-9mol / dm"
P13-16 (e) Ideal CSTR -t
tm=r and
E(t) = _e -
tr
T
Mol balances:
fFJ_ = F - (- r ) dt Ao A FAo
= CA v = 0.05 ·10 = 0.5mol /min = FBo 0
0
and so on for the other species .
FA CA =Cro--~etc FT CTo = CAo + CBo
....
= 0.05+0.05 = 0.1
F7 = FA +FB +Fe +Fv +FE +Fp See Polymath program PlJ-16-c . pol Exit concentrations:
Selectivities:
CA
= 0.050mol / dm3
SeD = 0.068
CB
= 0.049mol / dm3
SDE
= 3.342
Ce
= 5.639e-5mol / dm3
SEF
= 815844
CD = 0.0008mol / dm3
CE
= 0.0002mol / dm3 13-54
CF
= 3.03e-l0mol/
dm3
Ideal PFR tm=r and RTD function
E(t) = 8(t - 't')
Exit concentrations:
Selectivities:
CA = 3.98e -9mof / dm3
ScD
= 0.004
CB
= 0.0065mo[ / dm3
S DE
= 3.368
Ce
= 0.000277mol / dm3
SEF
= 4 . 286
CD
= 0.068mof / dm3
CE
= 0.0202mof / dm3
CF
= 0.0047mol/
Segregation
dm" Maximum Mixedness
CSTR
PFR
SCD
= 0.00196
ScD
= 0.0009
ScD
= 0.068
S DE
= 3.009
S DE
= 3.004
SDE
=3.342
SDE
=3.368
SEF
=28433
SEF
=1343110
S EF
= 815844
S EF
= 4.286
SCD
= 0.004
S EF For the PFR is very much smaller than for the others, because Cp is not so small at the exit of the PFR in turn due to exit Ce is not so small either . The conversion of CA in the PFR is virtually complete at the exit of the PFR, hence greater Ce
P13-16 (d) See Polymath program P13-16-d.pol E(t)=IF (t<=lO) THEN (0.01) ELSE ( IF(t>=20) THEN (O) ELSE ( 02-0 . 01 t)) 13-55
Segregation
CSTR
Maximum Mixedness
PFR
ScD = 0.004
ScD =0.004
ScD = 0.0035
ScD = 0.068
SDE =3.109
SDE =3.106
SDE = 3.342
SEF = 24078
SEF = 41503
SEF = 815844 SEF = 4.286
SDE = 3.368
For the Segregation and Maximum Mixedness models, S EF 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).
Pl3-17 Mu/tiple parallel reactions, isothermal Asymmetric RTD: E=IF (t <=1.26) THEN (El) ELSE (E2) El=-2.104t4+4.167t3-l.596t2 -0.353t--0.004 E2=-2.104t4+ 17 .037t3-50.247t2+62.964t-27.402
P13-17(a) Segregation model
dCn = C8E(t) dt
ac; --=CcEt dt
()
13-56
------,-----,--~---~-------------------------~
--~----~-~-----~-------------------
dCp --=CpE
dt
()t
See Polymath program P 13-17-a.pol Selectivities:
Exit concentrations:
CA
= 0.819mo[ / dm3
Scv =0.272
C8 =0.767mol/dm3 -
Ce= 0.163mol/ dm CD
3
SDE
= 11.330
SEF
= 0.267
= 0.600mo[ / dm3
e~ = 0.053mol I dm3
e, = 0.199mol/ dm3 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<=l.26) TREN (El) ELSE (E2)
dCA dA -
- """
~
r + (e iA
A
e ) Ao
E(A)
1-F(A)
The same applies to the equations for the other species as in Part (a), See Polymath program Pl 3-17-b.pol Exit concentrations:
CA
= 0.847ma[ / dm3
Selectivities:
Scv
= 0.281
13-57
C B = 0.824mo[ f dm3
SDE
=10.7
Ce =0.162mol/dm3
SEF
= 0.280
CD
= 0.576mo[ f dm3
CE
= 0.054mo[ f dm3
e, = 0.192mol/
dm3
Pl3-17(c) Ideal CSTR -t
tm=r and
E(t) = _e -
tt
'['
dCA ·--=v
( CAo -CA ) +rA
dt
Where
rA
= (-kDICAC~
-2kE2cAcD)
vo=v
and so on for the other species. See Polymath program Pl3-l 7 c-1.pol Exit concentrations:
Selectivities:
CA
= l.386mo[ f dm3
Sev
= 0.832
CB
= l.774mo[ f dm3
SDE
= 71.100
Ce
= 0.095mol / dm3
SEF
= 0.198
Cv = 0.114mol/ dm3 CE
=
0.002mo[ f dm3
Cp = 0.008mo[ f dm3 Ideal PFR tm=• and RTD function
E(t) = t5(t - T)
and so on for the other species.
13-58
See Polymath program P 13-17-c-2 poi Exit concentrations:
Selectivities:
CA = 0.3 lOmo[ f dm3
ScD
= 0.162
C B = 0.338mo[ f dm3
SDE
= 3.618
Ce = 0.106mol / dm3
SEF
= 0.497
CD = 0.652mo[ f dm3
CE = 0.180mo[ f dm3 C F = 0.362mo[ f dm3 Segregation
Maximum Mixedness
CSTR
PFR
ScD = 0.272
ScD
= 0.281
ScD
SDE = 11.330
SDE
= 10.7
S DE = 71.100
SEF
= 0.280
SEF
SEF
= 0.237
= 0.832
ScD = 0.162
= 0.198
S DE = 3.618 SEF
= 0.497
S CD is significantly greater than for the others, because exit C0 is Jower. Similarly S DE is much greater in the CSTR than for the others, because exit CE is so Jow. This is because the achievable conversion in a CSTR is not so high .
------------------
P13-19 (a) Externa/ age distribution E(t) By plotting C 105 as a function oftime, the curve shown is obtained
13-59
------------~--
C{t)
900~---------· 800 700
600 500
u 400
300
200 100
o --·-···-· ·-· ····------,.-·- . -. ,--··-·--···-··-·-·-···-r-----···--·-·-·-····--,---- =- -=·--==-:::... .. ::: . . :::::;::==------,.-------···----··--o 10 20 30 40 50 60
70
t[mln]
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
f r
+
r
C(t)dt +
J:º C(t)dt + fo°c(t)dt
C(t)dt = %(1)[1(0) + 3(622) + 3(812) + 2(831) + 3(785) + 3(720) + (650))0-5 = 4173.4 .10-5 C(t)dt = j(2)[(650)
+ 4(523) + (418))10-5
JotºC(t)dt = _!_3 (5)[418+4(238)
= 2106..7 .10-5
+ 2(136) + 4(77) + 2(44) +4(25) + 2(14)+ 4(8) + 5 )o-s
(>O C(t)dt = _!_ (10 )[1(5) + 1(1) )0-5 = 30 · 10-5
lo
2
f C(t)dt = 9981.7 · 10-
5
= 0.1
We now calculate:
E(t) = ~QL_ = C(t) C(t)dt 0.1
I
13-60
= 3671.7 .10-5
E(I)
009
006 'i: 0.05
!
~004 0.03 002 0.01
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 t[min]
Using Excel we fit E(t) to a polynomial:
for O 5: t 5: 3 E1 (t)
= -1.1675
-10-3 t4
+ 1.1355 -10-2 t3
-
4.7492 -10-2 t2
+ 9.9505 -10-2 t for 3 5: t 5: 20 Ei(t) E(t) =
= -1 · 8950 -10-6 i' + 8.7202
-10-5 t3 -1.1739 -10-3 t2
-1.7979 -10-4 t + 0.092343 for 20 5: t 5: 60 E3 (t)
= 1.2618 -10-8t4
-2.4995 -10-6t3 + 1.8715 -10-4t2
-
6.3512 -10-3 t + 0.083717 for t > 60 O
P13-19 (b) External age cumulative distribution F(t) F(t)
=
!
E(t)dt
Integrating the E(t), we obtain the F(t):
13-61
--·-·---·----~------·-----~-
---
---~------~----
· · ·-· ·- - -- -------------1
12 ~----------·-
e
1
o.a
~ 0.6
0.4
02
o
· · - - - - - - - --~------~--------~- - - - - ------~--- - --- - - - ~-------j
o
10
20
40
30
50
60
t(mln)
P13-19 (e) Mean residence time and variance
f E(t)tdt
tm =
The area under the curve of a plot tE(t) as a function of t will yield tm. tE(t) 045
04
0.35
tm='t'= 1 Omin 0.3
e
025 ·
'!:!
02
015
0.1
005
o
o
10
20
30
40
50
t-tm
(t-tni}2E(t)
60
tlmin)
o
C(t)
o
E(t)
tE(t)
o
o
13-62
-10
D
0.4 1 2 3 4 5 6 8 10
15 20 25 30 35 40 45 50 60
,--, ....,
.._,
~,--,
.Jl .!. t m : .._, tm
)tdt
0.0329 0.0622 0.0812 0.0831 0.0785 0.072 0.065 0.0523 0.0418
0.01316 0.0622 0.1624 0.2493 0.314 0.36 0.39 0.4184 0.418
-9.6 -9 -8 -7 -6 -5 -4 -2
3.032064 5.0382 5.1968 4.0719 2.826 1.8 1.04 0.2092
0.0238 0.0136 0.0077 0.0044 0.0025 0.0014 0.0008 0.0005 0.0001
0.357 0.272 0.1925 0.132 0.0875 0.056 0.036 0.025
5 10 15 20 25 30 35 40 50
0.595 1.36 1.7325 1.76 1.5625 1.26 0.98 0.8 0.25
329 622 812 831 785 720 650 523 418
=
= T = 9.88min
r
238 136 77 44 25 14 8 5 1
E(t)tdt +
= lümin
r
E(t)tdt +
o
o
foº E(t)tdt + 1E(t)tdt
We can calculate the variance by calculating the area under the curve of a plot of:
....
·--·-·-·-··--··-·-·-···
CY2
r
= [ (t-tJ2 E(t)tdt = (t-tJ2 E(t)dt +
J:º (t -tJ E(t)dt + fJt 2
r
---
..............
"""·······-·····----···--·-·······
tfmin)
(t -tJ2 E(t)dt +
- tm )2 E(t)dt
13-63
----------
,_
(t-tm/E(t)
o
o? = 73.8lmin2
= 74min
2
P13-19 (d) Fraction ofthe material that spends between 2 and 4min in the reactor E(t) 0.09
000
0.07
0.06
'.s' 005 ,:: _.,,
;:::. ~ 004
0.03
002
0.01
o
r
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
52
58 60
E(t)dt = shaded area = i[l(0.0812) + 4(0.0831) + 1(0.0785)]= 0.16
P13-19 (e) Fraction ofthe 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
t[minl
13-64
38 40
42
44
46
48
50
54
56
f E(t)dt
r
E(t)dt
= shaded area =
r
E(t)dt +
e
E(t)dt +
Ía E(t)dt 0
= ~ (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)
lo
3
+ 2(0.0014) + 4(0.0008) + 1(0.0005)) = 0.367 f'ºE(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 O 07
006 '.§' 005
.=-~ 004 "~
O.D3
002
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 56 58 60
t[min)
!E(t)dt = shaded area =
%[1(0) + 3(0.0622) + 3(0.0812) + 1(0.0831)]
P13-19 (g) Normalized distributions
13-65
= 0.192
Normalized RTD
8=.!_ t
E(8)=iE(t)
09
08
07
6
E(8)
0.5
04
0.3
0.2
0.1
o
o
2
3
4
e Normalized cumulative RTD
F(8)=
! E(e)de=
{E(t)dt
13-66
-·-------------~--
6
7
F(8)
_/"''
./·
08
----·
----·-·
-·--·-----,
06
04
0.2
o .. ·---·-·-·---·-·-~·----·--··
··········---····-,,----·-·--·---·--· ··---.------·---·----····-,-···-····- ······------------,----··---·--·-····----·-·.
¡
e P13-19 (h) Reactor Volume F= 10 dm3/min V= F ·T=100dm3
P13-19 (i) Internal age distribution
1 I(t)=-[1-F(t)] T
0.12
01
0.08
g 006
004
0.02
o
----------------------~-·-----·-------------~----------~-=--=---~=Q==---~--------_J
O
10
20
30
40
t[min]
13-67
50
60
70
U)
P13-19
r
Mean internal age
a;
=
=
l(t)tdt
f
l(t)tdt +
r
l(t)tdt + J:°1(t)tdt + cl(t)tdt
= lmin
P13-19 (k) Jntensityfunction A(t)=-E(t) 1- F(t)
P13-19 (1) Mean catalyst activity Integrating the decaying rate law:
f - d~a = Í k dt 0
1
a=--I+k0t Applying the Segregation Model with the previous RTD data:
amean
=
r
aE(t)dt
See Polymath program P 13--19-1. poi POLYMA TH Results Calculated values of the DEO variables Variable t amean kd El E2 E4 E3
initial value
E a
o o
0.1
minimal
o o
value
0.092343
0.1 -2.436E+04 -21.261016
0.083717
l.724E-05
1
0.125
o
o
o
o
o
maximal value 70 0.5778625 0.1 0.0838847 0.092343
o
0.083717 0.0838847 1
final value 70 0.5778625 0.1 -2.436E+04 -21. 261016
o
0.0017977
o
0.125
ODE Report (RKF45) Differential equations as entered by the user ¡ 1 J d(amean)/d(t) = a*E Explicit equations as entered by the user [l) kd = 0 . 1 [ 2 J E1 = -0.0011675*t"4+0.011355*tA3-0.047 492*tA2+0.0995005*t [ 3] E2 = -1.8950*1 QA(-6)*t"4+8.7202*1 QA(-5)*tA3-1 . 1739*1 QA(-3)*tA2-1.7979*1 QA(-4)*t+0 . 092343 [4] E4 = O [ 5 J E3 = 1.2618*1 QA(-8)*t"4-2 . 4995*1 QA(·6)*tA3+1.8715*1 QA(-4)*tA2·6 . 3512*1 QA(-3)*t+0.083717 [ 6 J E= if(k=3)then(E1 )else(if(k=20)then(E2)else(if(t<60)then(E3)else(E4))) [ 7 J a= 1/(1 +kd*t)
Pl3-19 (m) ldealPFR 13-68
2nd order, liquid phase, kCAo =0.lmin-1, CA0=1mol/dm3
-r=lü min (fromP13.19 (c))
X
=
k'lC Ao 1 + k'lC Ao
= 0.5
Pl3-19 (n) LFR
Laminar Flow Reactor 2nd order, liq. phase, irreversible reaction kCAo =0.1 min-1,-r=lü min. We apply the segregation model, using Polymath: X
=
ktC A 0 and E(t) l+ktCAo
=
{º
for t < 5.00mii1
(10)2/(2t3)min-1
fort~5min
See Polymath program P 13-19-n .pol POLYMAT.!I Results Calculated values of the DEO variables Variable t xbar cao
initial value
k
tau El tl E
o o
minimal value
1 O. 1 10 5.0E+05 5
1 0.1. 10 1 . 852E-06 5
o o
X
o o
o o
maximal value 300 O . 4504221 1 O. 1 10 5 . 0E+05 5 0.2284987 0.9677419
ODE Report (RKF45) Differential equations as entered by the user [ 1J d(xbar)/d(t) = x*E Explicit equations as entered by the user [ll cao = 1 . 0 [2] k = .1 [3 l tau= 10 [ 4 J E1 = tauA2/2/(tA3+0. 0001) [ 5 J t1 = tau/2 [ 6 J E= if (t
We can compare with the exact analytical formula dueto Denbigh .
X= Da[i-(~a )in(l + 2/ Da)J = 0.451
with Da=kC""<=l
Pl3-19 (o) Ideal CSTR 2nd order, liquid phase, kCAo =0.1 min", CA =1mol/dm3 X 0
-( --)2
1-X
= i«:Ao -
X
= 0.382
13-69
final value 300 0.4504221 1
o. 1 10 l . 852E-06 5
1.852E-06 0 . 9677419
P13-19 (p) Segregation Model 2nd 01de1; liquid phase, kCAo =0.1 min", CA =l mol/dm3 0
X= f X(t)E(t)dt o kCAi 1 + kC Ai
Where X () t = ----
See Polymath program P 13-19-p.pol POLYMA TH Results Calculated values of the DEO variables Variable t Xbar kCao El E2 E4 E3
initial value
E X
o
o
minimal value
o
o
0.1
0.092343
0.1 -2.436E+04 -21.261016
0.083717
l.949E-05
o
o
o o
o
maximal value 70 0.4224876 0.1 0.0836855 0.092343
final value 70 0.4224876 0.1 -2.436E+04 -21.261016
0.083717 0.0836855 0.875
0.0017977
o o
o
o o
0.875
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(Xbar)/d(t) = E*X Explicit equations as entered by the user [ll kCao=0.1 [ 2] E1 = -0.0011675*t"4+0.011355*tA3-0.047 [3
J
[4l
E2
= -1.8950*1QA(-6)*t"4+8.7202*1
492*tA2+0.0995005*t QA(-5)*tA3-1.1739*1 QA(-3)*tA2-1.7979*1QA(-4)*t+0.092343
E4 = O E3 = 1.2618*1 QA(··8)*t"4-2.4995*1 QA(-6)*tA3+ 1.8715*1QA(-4)*tA2-6.3512*1 QA(-3)*t+0.083717 [ 6 l E= if(l<=3)then(E1 )else(if(1<=20)then(E2)else(if(1<60)then(E3)else(E4))) [ 7 J X = kCao*V(1+kCao*t) [5J
P13-19 (q) Maximum Mixedness Model 2nd arder; liquid phase, kCAo =0.1 min", CA =lmol/dm3 0
2
RateLaw :- rA = kCA
cA = cA)1- x)
13-70
= kCAo 2 (1- X )2
rA
Í=kC C
Ao
Ao
1
dF
-
dz
=
where k=O. l dm3 /mol min
(1-x)2
-E( z)
where z=6Ü-A
See Polymath program P13-l 9-q.pol
POLYMATH Results Calculated values of the DEO variables Variable z X F
cao lam El E2 E3 E4 Ca k
ra E
initial value
o o
minimal value
o o
O . 99 1 60 -1.284E+04 -9.8680524 2.228E-05
-L284E+04 -9.8680524 1.806E-05
1 0.1 -0.1
0.52273 0.1 -O . 1
o
o
-O . 010344 1
o
o
o
maximal value 60 0.4773052 0.99 1 60 0.0832647 0.092343 0.083717
o
final value 60 0.4047103 -0.010344 1
o o
0.092343 0.083717
o
O . 5952 897 0.1 -0.035437
1 0.1 -0.0273247 0.0832647
o
ODE Report (RKF45) Differential equations as entered by the user [ 1] d(X)/d(z) = -(ra/cao+E/(1-F)*X) [ 2 J d(F)/d(z) = -E Explicit equations as entered by the user [ll cao = 1 [2 J lam = 60-z [ 3 J E1 = -0 . 0011675*1am"4+0.011355*1am"3-0047492*lam"2+0.0995005*1am [ 4 J E2 = -1.8950*10"(-6)*1am"4+8 . 7202*10"(-5)*1am"3-1. 1739*10"(-3)*1am"2-1..7979*10"(-4)*1am+0. 092343 [ 5 J E3 = 1.2618*10"(-8)*1am"4-2.4995*10"(-6)*1am"3+1.8715*10"(-4)*1am"2-6.3512*10"(-3)*1am+0.083717 [61 E4 = O [ 7 J Ca = cao*(1-X) [8] k = 0.1 [ 9 J ra = -k*Ca"2 [ 1o J E = if(lam<=3)then(E1 )else(if(lam<=20)then(E2)else(if(lam<60)then(E3)else(E4)))
~---'-'-~_.:
__~_5
_,_, _~_42_g 2
___.l_~_.;_FR
13-71
_
I Xcsrn 0.382
J
Solutions for Chapter 14 - Models for Non-ideal Reactors Pl 4-1 Individualized solution P14-2 (a)
Approximatedformula for Segregation Model ( 1" reaction] e-kt
k2t2 = 1- kt + -- + Error . The error is then o((kt)\ 2
Approximating the
kt3 Error=--· 3!
i-[-:_; __ -_-_---~------------------~--~----_-_-_-_+.._l-=_1-r._}_:_~-~-7_1_6-=_6-7_-_-_-_-_-~---_-_-_-_-_-_-_-_-_J
o
LuQenor[)
-20
10
0.1
100
k1 P14-2 (b) Parameters Dispersion Model Closed-Closed dispersion model
X =l--·---4qexp(Pe,!2) (1 + q )2 exp(Pe,q / 2)-(1-- q )2 exp(- Pe,q / 2) Where
q = ~r-1 +-4D_a_/_P_e_, Da
= kr = k--l
u
Damkñhler number
14-1
(14-26)
Ul
Pe,
Peclet number
Pe,q =~Pe22 +4DaPe, Where
kl2 DaPe =-= r Dª
Rateof consumptionof Aby reaction Rate of transport by diffusion
Numerical example '!'
= !::.._ = const = 5.15min
u
= t *U= 5.15min*O.lcm/ s = 0.309m µ Se = -- = 1000 (Liquids region in Fig. 14 . 11) L
DAB Da = kt = 1.288 d, lcm
Re 10
ReSc 104
L/d1 30.9
1 dm
100
105
3.09
lm
1000
106
0.309
. _____
--Per 172 0.077 1.03 10-4
Q 1.015 8.226 223.609
D/(U*d1) 0.18 From Figl4.10 40FromFig. 14.11 3000From fig. 14.11
D (cm2/s) 0.018 40 30000
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 (e) (1) Vary the Damkohler number for a second-order reaction (Example 14-3(b)) For a second order reaction
L
Da= 'TkCAo =-kCAo Uo
R =0.05m L=6.36m k = 0.5dm3 /mol.min U0 = l.24m/min CAO D aris
= 0.5molf
dm'
= l.05m 2 / min Damkohler number/ Da 0.1603 0.3205
-
Conversion 0.138 0.239 14-2
Parameter 8*UO 4*UO
0.641 1.282
0.377 0.523
2*UO
2.564 5.129 10.258
0.644 0.732 0.795
U0/2 U0/4 U0/8
uo
(2) Vary the Peclet and Damkohler numbers for a second-order reaction in laminar flow (Example 14-3(c))
Da
For a second arder reaction
L
= -rkCAD = -
kC AD ; Pe
= UD LID AB
UD
R=0.05m L = 6.36m k = 0.5dm3 / mol.min UD= l.24m/min
= 0.5mo[ / dnt' DA8 = 7.6xl0-5m2 /min -Da CAD
..._
0.1603 0.3205 0.641 1.282 2.564 5.129 10.258
-
-
Pe 8.32xlü' _______4.16x10 -;:3'"" 5 ___ 2.08x10 ----l.04x10r
___________ --
f--•
0.52x10' 0.26x10' 0.13x10 5 -
Conversion 0.132 0.229 0.369 0.536 0.699 0.825 0.906 ---
Parameter 8*UO 4*UO 2*UO
uo
-
1.0 0.9 08·
0.6
e: o "[!? 0.5
~ §
o
04 0.3 0.2
0.1
-,---·-,----,---..--r--
O.O OO
2.0x105
4.0x105
6.0x105
8 Ox105
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
U0/2 U0/4 U0/8
-----
For a second order reaction
.
L
Da= rkCAo =-kCAo Uo
R =0.05m L = 6.36m k = 0.5dm3 / mol.ruin U0 = l.24m/ruin CAO
= Ü.5mo[
D aris
= I.05m
f dm' 2
/
ruin
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 [l] Domain
[2] Constants and scalar expressions - Constants
- Scalar expressions
14-4
Parameter 8*UO 4*UO 2*UO
uo
U0/2 U0/4 U0/8
f'--"---------··--+'-lu_o•_k•_cA_O....,"(_n-1)
---·-·-··········------------·-····
~-------------!--··-·-·---------·---------·------------c-------------t-------------------···--·-------
r---------··-
-··---------------·----·--------·-·----
[3] Subdomain Settings - Physics (Mass Balance)
- Initial Values
(Mass balance) cA(tO) = cAO
- Boundary Conditions @ r = O, Axial symmetry @
@ @
outlet, Convective flux wall, Insulation/Symmetry
14-5
[4] Results
m11r11m1eml!D!!III1tlll
(Concentration,_ c!Al)t·~1·
~He Edir Op\jons . Draw • F'hysk:s . Mesh
S?lyE, •.
Pos1proc~ssk1g. Multiphysic~ Help
.•
,(+) J3
oíl
t.··.·~;;¡ ~Ti~ lRil!~J¿4il ~.·~ .,;,, !liil@l L:J.
¡:;~
_
1·~~- .. xA~
n
... .····•··. . • . : • > . •. .• .· ' i ·. @f.eiij _ ,>:'. > . · _.· ·.· , . . . ·.·
.>Uf ........
.¡¡¡. /
3
(0.1, 5)
(2) Vary the Peclet and Damkohler numbers for a second-order reaction in laminar flow (Example 14-J(c)) For a second order reaction
L Da = tkC AO = --· kC AO ; Pe = U O L I D AB
Uo
R =0.05m L= 6.36m k = 0.5dm3 / mol.min U0 = l.24m/min C AO
= Ü.5mo[
DAB
=7.6xl0-5m2
~-
Da 0.1603 0.3205 0.641 1.282 2.564 5.129 10.258
f dm3 /min Pe 8.32x10) 4.16x103 2.08xl05 l.04x105 0.52xl05 0.26x105 0.13x105
Conversión
0.132 0.229 0.369 0.536 0.699 0.825 0.906
.
14-6
Parameter 8*UO 4*UO 2*UO
uo
U0/2 U0/4 U0/8
···.
~.x: 0:523
1.0 0.9 0.8 0.7
e: o
f? Q)
0.6
0. 5
>
e: 0.4 o
o
0.3 0.2 0. 1 O.O O.O
2.ox10•
4.0x105
6.0x105
a . ox10•
Pe
When Peclet Number decreases less than 2 x105, the conversion is influenced significantly. - Femlab Screen Shots [1] Domain
[2) Constants and scalar expressions - Constants
- Scalar expressions
14-7
[3] Subdomain Settings - Physics (Mass Balance)
- Initial Values
(Mass balance) cA(tO) = cAO - Boundary Conditions @ r = O, Axial symmetry @ inlet, riúUndary COnd~i~ns r
¡
Bq1.1qdary c:i,;¡qdfücn; ·•·. jf!ux _ . .
I Qµjnt¡ty 1¡ :; @
.
· . ~~f~f /1;1'~~~~,,0~ i.;~.,Cfiptiro
,t;~-=-==J:::;r
.... outlet, Convective flux
. ·····
.
14-8
@
wall, Insulation/Symmetry
[4] Results (Concentration,
9AiiJ)lft.. llífl!IWffl!
c
P14-2 (d) Two parameters model
T=~=lümin Vº
I=2 and S=4min-1.
/3 = _vb_ = (~ -}l = 0.5 Vo
I
V, (1- /3) a=-·=----=
V
V
rS
0.013 m3
= (l-/3)v º = 0.05-.nnn s
Vs = (ai-)v o = 0.013m3 T
s
=
y, v,
= 0.25min
14-9
e
=
A,
X=
J1+ 4-rskC Ao -1 kmol =1.779-2'f s k m'
1- CA, CAo
= 0.111 CSTR with Dead Space and Bypass (1=1.25 and S=0.115 min") X=0.51
Ideal CSTR
X=0.66
CSTR with Dead Space and Bypass (1=2.0 and S=4 min") X=0.111
P14-2 (e) Two CSTR with interchange ( 1 st arder reaction)
a=-
Vi V
X= (/J+atk)[,8+(1-a)tk]-/32 (1- /J)+ atk See Polymath program Pl4--2-e . pol POLYMATH Results Calculated values of the DEO variables Variable t CTl CT2 beta alpha tau CTel CTe2 tl CTe k X
initial value
o
2000 921 0.15 0.75 40 2000 921 -80 2000 0.03 0.5134788
minimal value
o
31.814045 164 . 15831 0.15 0.75 40 -l.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 -l.275E+04 13 120 13 0.03 0.5134788
Differential equations as enterad by the user [ 1 l d(CT1)/d(t) = (beta*CT2-(1+beta)*CT1 )/alpha/tau [ 2 J d(CT2)/d(t) = (beta*CT1-beta*CT2)/(1-alpha)/tau
Explicit equations as enterad by the user [ll beta= 0.15 [2l alpha = 0.75 [3] tau= 40 ¡ 4 l CTe1 = 2000-59 6*t+0.64*tA2-0.00146*tA3-1.047*1QA(-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 J CTe = if(k80)then(CTe1 )else(CTe2) [8] k = 0.03 ¡ 9 l X = ((beta+alpha*tau*k)*(beta+(1-alpha)*tau*k)-betaA2)/((1 +beta+alpha*tau*k)*(beta+(1-alpha)*tau*k-betaA2))
40
80
t
120
160
200
Comparison experimental and predicted (f3=0 . .l.5 a=0.75) concetration . For small deviations from the original value of the parameters concent:ration and conversion are not significantly affected.The following table show the conversion for different combinations of f3 and o;
a 0.75 0.8 ,....__ 0.5 0.1 0.5 0.5 ,____ 0.1 0.9 1---
B
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
lJ!4
ua
cF,=Ü.5
:Ji
X'
¡1-,,
u~ (14
U.j
uz o;
Oll
---··---·-·--·-----··--·--···---·-·· ···----···--·--····--------------
Cl?
X
Cl6
115
Cl•
Cl3
02
a·1
o
0,1
03
0.5
06
D.'r
0.9
Given the interchange flowrate, the conversión 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 D. and hence dispersion.
14-12
pud
To minimise dispersion a Re number of -10-20 gives the lowest value for Da. Because Re= --
µ
, the design of the
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
P14-2 (h) Figures 14.3 and 14.lb The curves in Fig.143 represent the residence time distributions for the Tanks in Series modelas function of the number of reactors. Given a CSTR of volume 1 (V=V1), we divide the CSTR in two CSTRs (V2=1/2). 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==, PFR behaviour). The model in Fig. 14.1, assumes a "faster" (channelling) reactor anda "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 from zero when flow leaves from the "fas ter" 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.J is PFR and CSTR in parallel, The exit age distribution for the CSTR is a negative gradient curve, intenupted 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
I-fJ
a
V, V = _.1_ fJ = __!_
V
V
dX
PFR mol balance: --
dV Rate law: - rA = kC1
= - --r, FAo
14-13
e A = eAo (1- X)
Stoichiometry: liquid phase 2ºd
V= -1-[2c(l
orderPFR:
v
+ c)ln(l-X)+
&2
kCA/
X +~(l_+_E2~)x] 1-X
e b 1 1 8 = - + - -1 = - + - -1 = O a a 2 2
where
c=O Determining
a and fJ :
a== V1 ~=0 . .5 V vi
fJ
1- ~ =Ls(I- ~ J V =10v1 V1 / 2.5 _ 5a _ 0•.5 a-_ 2.5 _vi -_ 0 . 25 and fJ _-------
Gives
10 vi
Now t'
=
'Z"1
+ 'Z"z 2
=
V/ 5
vi/ vi + v2 / Vz
V1 tv, + V2 lv2 But V1 = 2 . .5v1 Gives 1"1 = 2.5min
2
2 . .5
5 . = mm
= 10
Hence
.........
V2 = 7 . .5v2 1"2
= 7.5min
Substituting into
r. V
1 kCAo
2
[2&(l+&)ln(1-x)+&2X + {l+&z)x] 1-X
Reactor 1:
2•5
= 0.1 ~ 2'
1=[-~] 1-X1 CAi
[1 ~~.-] -x1 =0 . .5
=CA0(l-X1)=2(1-0
75= 1 [ X2 .. 0.1*22 l-X2
l
1 [ i-x2 X2 3=·0.1*22
-xz
Reactor 2:
l
. .5)=1mol/dm3
=0.75
eA2 = eAo (1- X 2) = 2(1- 0.7.5) = 0.5mol / dm' e Aexzt.
=
cAI +_EA2 2
= l+0.5=075.t smo lid m 3 2 14-14
X= CAo -CA= 2-0.75 =0.625mol/dm3 CAo 2
P14-3 (a) Money for buying reactors Using the tank in series model:
· S econ d or d er reaction
Assume that
r = r,
X -- 2Da + l - .J4Da + l 2Da
where
Da--k,,.Cao .•
and that in reactors medelled as more than one tank, that
r, r=- . n
Number of tanks
rounded to the nearest integer. Reactor Maze & blue Green & white Scarlet & grey Orange & blue Purple & white Silver & black Crimson & white
~(min) 2 4 3.05 2.31 5.17 2.5 2.5
:r(min) 2 4 4 4 4 4
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: X1 = 0..5, CA1=0.5 ~ X2 = 0.38, CA2=0.31 ~ X= 0.69 Orange & blue : X1 = 0.43, CA1=0.57 --t X2 = 0.34, CA2=0.38~ X3 = 0.27, CA3=0.28~X=0.72 Using the combination of maze & blue followed by crimson & white reactor (same overall conversion either way) X1 = 0 . .5, CA1=0.5 ~ X2= 0.17, CA2=0.41~ 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 moneyfor buying reactors Try: Green & white and Maze & blue: X1 = 0.61, CA1=0.39 ~ X2= 0.34, CA2=0.26~ X= 0 . 74 Scarlet & grey and Maze & blue: X1 = 0.69, CA1=0.31 ~ X2= 0.42, CA2=0.18~ X= 0.82 Orange & blue and Maze & blue: X1 = 0.72, CA1=0.28 ~ X2= OAO, CA2=0.17~ X= 0 . 83 The highest conversion is now obtained from the Orange & blue reactor combined with the Maze & blue reactor.
P14-3 (e) Ann Arbor, MI East Lansing, MI Columbus, OH Urbana, IL
14-15
r2
n=a2
Evanston, IL West Lafayette, IN Madison, WI
P14-4 Packed bed reactor with dispersion lst order, k1=00167/s, e=0.5, dp=O.l cm,
V=µ
= O.Olcm2 / s L=lO cm, U= 1 cm/s p plld ; V Re = = 10 and Se = -no data concerning D AB µ DAB From packed bed correlation for Da , and liquid phase region of graph,
Da& = 2approx ~ Gives -UdP
D ª -- _2_u_d_P-- 2 *1 *0.1 -- 0.4cm2 Is
e
0.5
Pe . = UL = .!.._* 10 = 25 ' Dª 0.4 4qexp(Pe,/
2)
X=l-----------------~ + q)2 exp(Pe,q/
[(1
q=
q)2 exp(-Per%)]
2)]-[(1-
r::..
4Da. V.._' Pe,
Da='tK and
1:
L
=-
U
10
= -· = lOs
1
~
Da=0.167 and q=l.013
X= 0.15 Conversion X=l5%.
P14-5 (a) Number of tanks in series
Bo
Assuming the Peclet-Bodenstein relation: n = Where
Bo =-
+1
UL Dª
ToestimateBo, Re=-
du v
5*2 =--=lOOOandSc 0.01
From gas phase dispersion correlation chart, Gives
2
Da = 80cm2
/
=--
0.01 =--· - = 2 0.005
V
D AB
Da = 8 ud,
s
14-16
---
----
-- - ------~------------
---
--- -----
--- - ---
-- - -
--- ------------·-----·-·-----
--
---
-----
-
---
---------
----
Bo= 2*200 =5 80 n=1+1=3.5 2
P14-5 (b) Conversion Using individual reactor material balances: Reactor 1:
rAV = - ·-
X
Mol balance:
FAo
rA
Rate law: -
= kCA
2
e A = e Ao (1- X 1)
Stoichiometry:
O=Ü and c=Ü hence no volume
change
x,
= kcA/(1-
<,
xi)2
= _25 *0.01 (l-
xi)2
0.039
X1 =6.366(1-X1)2 ~0.674 C Al
= e Ao (1- XI)= 0.00326mo[ / dm'
Reactor 2:
kCA1(1-xJ2
X 2 = -------
e
A2
~ X 2 = 0.507
V
= eAl (1- X 2) = 0.00326(1-- 0.507) = 0.001607mol / dm3
Reactor 3:
X3 = kC A2 (1- X 3 )2
~
X 3 = 0.387
V
eA3
= eA2 (1- X 3) = 0.001607(1--0.387)
= 0.000985mol
/ dm"
Reactor 4:
X4 =
kC
e =e A4
AJ
(1 -:-- X 4 )2
~ X4 = 0.305
V
A3
(1- X 4) = 0.000985(1-··· 0.305) = 0.000685mol / dm'
Bounds on conversion: 3 tanks
X
=
e
Ao --
e
A3
= 0.9015
e
A3.
= 0.9315
CAo
4 tanks
X=
e
Ao CAo
P14-5 (e) Change of thefl.uid velocity Let U=O.lcm/s Re=50 and Sc=2
14-17
D
--ª ud.
From gas phase dispersion correlation chart,
= 0.5
D ª = 0.5ud1 = 0.5 * 0.1 * 5 = 0.25cm2 / s
Gives
Bo = UL = 0.1 * 200 = 80
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=2/0.001=2000s Da=k'tCAo=500
X=
Da =0.998 l+Da
Let U=lOOcm/s Re=.50000 and Sc=2 From gas phase dispersion correlation chart,
D
-ª
ud,
= 0.21
Da= 0.21ud1 = 0.21 *100* 5 = 0.25cm2 / s
gives
Bo = UL = 100 * 200_ = 190_5
Dª
_ Bo
10.5 l _ 190.5
---+ 2
n--+
2
_ 1 - 96
The conversion is close to the one PFR 2nd order reaction: k=25dm3/(mol·s) t=l!U=2/1=2s Da=k'tCAo=05
Da l+Da
X = ---
= 0.333
P14-5 (d) Change ofthe superficial velocity
Re= d1u = 0.2*4 = 80 V
Ü.Ül
From packed bed dispersion correlation chart,
DE -.-ª - = 0.55 udP
0.55 * 4 * 0.2 =1.1 0.4 Bo = UL = 4 * 200 = 727 Dª 1.1 Bo 727 n =·-+1 =-+1 = 364 . .5 2 2 0 . .5.5ud P Dª=---= E
14--18
The conversion is clase to the one PFR 2nd arder reaction: check k=25dm3/(mol·s)
1:=l/U=2/1=2s Da=k'ICAo=0.5
Da = --= 0.333
X
l+Da
P14-5 (e) Liquid instead of gas
Part (a)
Re= d,u = 5 * 4 * 0.001 = 0.2 V 0.1 Se=~-= DAB
__!_QQ_ = 2 *107 5e-6
This is off the scale of the graph for liquid phase dispersion, hence Da cannot be evaluated .
P14-6 (a) Peclet numbers
From Example 13.2 cr2=6.19min2 and tm=5.15min Closed:
ª =-3--p e, Pe,2-(1-e-Pe,)---'tPe 2
tm2
2
r
=7414 '
Open:
2 8 =--+---2---'tPe, =11.68
0"2 --2
tm
t»,
Pe,
P14-6 (b) Space-time and dead volume T
tm = ··-----·= 4.40
1+2/ Pe,
= T*v = 263.8dm3 Vv = V - Vs = l56.2dm3 V,
0
156.2 = --= 37.2%
%deadvolume
420
P14-6 (e) Conversions for
1'1 arder isomerization
Dispersion model Da=kt=0 . 927
14-19
= 1221
q= .
X =1-
4qexp(Pe,/2)
(1 + q )2 exp(Pe,q !2)-(l-q)2
=0.570 exp(- Pe,q 12)
Tanks-in-series '['2
n =-2=4.35
a
P14-6 (d) Conversions PFR and CSTR PFR:
CSTR:
X = 1- e:" X
= 1- --
X=0.604
1
l+tk
X=OA8 I
I
I 0.604 XPFR
Xcs1R 0.481
P14-7 Tubular Reactor l st order, irreversible, pulse tracer test ~
a 2 = 65
2
S and
tm = 1 O s
For a 1 st order reaction, PFR: X =1- é~ = 0.98 We need 't and k. There being no data for diffusivity (Schmídt number) Daand hence Pe, cannot be obtained using tubular flow correlations. Calcula te V= 3 m * 25 dm2 * (0..1 m/dm)2 = 0 . 75 m3 't = 0.75 m3 / (3* 102 m3/s) = 25 s This is greater than tm so channeling is occurring Calculate, k
k=
1 ln-l-X
=391/25s=0.156s-1
T
Therefore assume closed vessel dispersion model: tm = 't = 10 s space-time Calculate, Pe
14-20
f
CY2 -2
= --
2
Pe,
tm
-
--2
2
(1-e-Per)= 65/102 = 065
Pe,
Iterating ~ Pee = 1.5 Calculate, X using the measured V and
v0 giving 't = 25 s
4qexp( P;,)
[{l+q)2 exf\*q )J+-q)' ex{-P~ *q )]
X=l-
Da=
-----
't
k = 25 * 0.156 = 3.9
4*3.376exp(\5)
ex{ 1.S * ~376) ]-[{t-3.376) ex{-1.5 :3376)J
X= 1-·
[ (1 + 3.376)'
= 0 . 88
2
Conversion for the real reactor assuming the closed dispersion (X=0.88) model is less than for the ideal PFR (X=0.98) 'tideal
V
= - = 25 S Vo
V=V0+Vs v;» a *V Note, '°CRTD = 10 S = C( * 't , CC=l0/25 =0.4
./Dispersion Volume
.------------------;¡----·~
Dead Volume = (1- a) V= 0 . 6 V, Vs = OAV Use tm = lOs, Da= tm * k = 10*0.156 = L56 q=
4*1.56 1.5
1+----=2.27
14-21
4*2.27exp
X0= 1- (19..22/58.38)
( 21.5)
= 0.67
P14-8 (a) E(t):
06
0.4 E(t)
oa o
o
4
:l t
Conversion T-1-S and Ma.ximum Mixedness Model XT+s=0.5 For a first order reaction Xmm=X1_1_s=
1- ----
1
( 1 +~k ee
2
4
O
2
r
1:=tm = fE(t)tdt= fü.25·t2dt+ f(I-0.25·t)tdt=2min O
o?
=
=
=
o
O
fE(t)(t-tm)2dt = f E(t)t2dt-r2
2
= -min2 3
1:2
n=-=6
ª2
From the conversion it is possible to determine k at 300K:
14-22
--------
---~----
---
----~--
~---~--
--
-
---
( 1 -1)
k= ~
T
·=0.367min-1
n The conversion at T=310 K is given by:
exp( R300K E E ) = 1.422minR·310K
k310 = k300
1
R=l.986~
molK
P14-8 (b) Complex Reactions
dCA dr·=-k1CA dCB
-k3CA
--
= k¡CA -k2CB
dCc
=k
dr
dr
C 2
B
deº -k e 3 A dr
Where k1= k2= k3=0.l
mín'
Where CA0=1mol/dm3 and CBo= Ceo= Cn0=Ü.
14-23
Changing variable: A=z-4 See Polymath program P 14-8-b poi POL YMA TH Results Calculated values of the DEO variables Variable z ca cb ce cd F
kl k2
k3 re ra rb t2 rd ca o cbo ceo cdo
initial
o
value
1
o
minimal value
o
0.6793055
o
o o
0.9999999 0.1 0.1 0.1
2.854E-06 0.1 0.1 0.1
o
o
-0.2 0.1
o
o
-0.2
maximal value 4 1
0.142158 0.0181892 0.1603473 0.9999999 0.1 0.1 O. 1
0.0142158 -0.1358611 0.1
4
O. 0537147 4
0.1
0.0679305
0.1
o
o o o
o
tl
2
2
lam El E2
1
1
o o
4
o o o
E
EF
1
o
o o o o
4 1
o o
4
0.6793055 0.142158 0.0181892 0.1603473 2.854E-06 0.1 0.1 0.1 0.0142158 -O .1358611 O. 0537147 4
0.0679305 1
o
o
o
2
2
4 1 1
1
0.4965476 25.277605
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(ca)/d(z) = -(-ra+(ca-cao)*EF) [ 2 l d(cb)/d(z) = -(-rb+(cb-cbo)*EF) [ 3 J d(cc)/d(z) = -(-rc+(cc-cco)*EF) [ 4 J d(cd)/d(z) = -(-rd+(cd-cdo)*EF) [ 5 J d(F)/d(z) = -E Explicit equations as entered by the user [l) k1 = 0.1 [2) k2 = 0.1 [3] k3=0.1 [4] re= k2*cb [ 5 J ra = -k1 *ca-k3*ca [ 6 J rb = k1 *ca··k2*cb [7] t2=4 [SJ rd = k3*ca [9 J cao = 1 [lOJ cbo=O [11]
final value
CC0=0
cdo = o [13] t1 =2 [14] lam=4-z [15J E1 = o . 25*1am [ 16 J E2 = 1-0.25*1am [ 17] E= if (larn-ett) then (E1) else ( if (lam-cetz) then (E2) else (O)) [18] EF=E/(1-F) ¡121
14-24
o o
o o
0.20....---------------,
0.16
0.12
0.08
0 . 04
0.8
1.6
z
2.4
3.2
4.0
P14-8 (e) Use FEMLAB for thefull solution Complex reactions and Dispersion Model.
dCA
dr = -kiCA --k3CA
dCB
k¡CA -k2CB
--=
dr
dCc·=k C
dr
acn dr
-k -
2
a
3
eA
P14-9 (a) FromP13-4:
tm = T =
/2 min,
'{;
a? = -1- = 0.159min2 21l
and k= 0.8 min"
Tanks-in-series For a first order reaction the conversion is given by: From Example 14-1
X T-1-S
= X seg = 0.447
P14-9 (b) Closed-closed vessel dispersion model For a first order reaction the conversion is given:
_1_ X -
4qexp(Pe, 12)
(1 + q )2 exp(Pe,.q / 2)--(1-- q )2 exp(- Pe,.q I 2)
14-25
l + 4 * 0.638 = 1.88.5 6.83 X=0.41
x_T=-I=-S~~~~~~~~-~~~-t-X=m=spe=rs1="ºº'----~--~~~~~-?
_~0.447
=:J
_ 0.41 .
Referring to P14.2 the approximate formula for the T-1-S is not good approximation.
P14-10 (a) Combination of ideal reactors The cumulative distribution function F(t) is given: 1
F(t)
os -
-
1
1 ºo~~~-'--~---'-~----~ 20
40
t[min] The real reactor can be modelled as two parallel PFRs:
S vo
V1=a\T
L
L
The relative
E(t)
1 = {-J(t
4
3 -- r1) + -J(t - r2)
4
P14-10 (b) Model parameters Far two parallel PFRs, the parameters are -r1=10 min and -r2=30 min,
P14-10 (e) 14-26
Conversion
Fao1 =114Faoand Fao2=3/4Fao ,second order, liquid phase, irreversible reaction with k=O.l dnr' /mol-min" and CA0=1.25
mol/drrr'
CAl
= CAo
CA2
= CAo
-
kr.C Ao 1 + kr.C Ao
CAo
= 0.556mo[ / dm 3
CAo
= 0.263mo / dm3
kt:2 e
Ao
1 + kr2C Ao
1 3 vCAo --vCAl --vCA2 4 4 X=
<,
= 0.731
P14-11 (a) FromP13-6 t1 = 5min
E(t)
if < f¡
t
l t
-2
t¡
E(t)= --\
t¡
tm
(t-2t1)ijt15:t5:2t1
O otherwise
= t1and
ª2 = t~ = j5 6
6
= 4.16'
Tanks-in-series '!'2
n=-=6
(Asín
ª2
Pl4.8) Second order reaction, liquid phase, kCA0=0.2rnin-1
-c.: = t: «.:
CAi
2
Solving for CAout
-1 + .jI+4.kzC
eAº"' =------2k'l'
Ain
Slx-Reactor System: T6=r/6 The Damkohler number is given by Da0= kCA0't6=0.167
-1 + ~1+ 4Dau-i> Da.=----'----
'
0.1
2
with i=l..6 In the following table the exit concentrations for each reactor are reported: Da1 Da2 Da3 Da4 Da5 Da6
-
0.145 0.129 0.116 0.105 0.095 0.088 14-27
t1
10
X= ~Ao -CA6
=
= 0.473
Dao -Da6 Dao
CAo
P14-11 (b) Peclet number The Peclet number
Pe,
= -Ul Dª
For Closed-Closed System (no dispersion at the entrance and at the exit of the reactor Pl3.6)
Pe,=10.875
P14-11 (e) Conversion Linearizing the 2•1 arder according 14..2 (g):
=r, = kC; = k CAo CA= 2
k'CA
it is possible to obtain an approximated solution;
k'=k CAo = Ill min " 2 Da= ik'= 0.5 Pe,=10.875
q=~l+
4Da
Pe,
=
~--4 1+
(from Part(b))
*0·5 = 1.088
10.875
4qexp(Pe, 12) (1 + q )2 exp(Pe,q / 2)-(1-q )2 exp(-Pe,q 12)
X =1 X=0.382
Full solution can be obtained with FEMLAB
P14-12 (a) Remembering the physical meaning of the Peclet number:
Pe. =. r
rateof transportbyconvection rateof transportby diffusionordispersion
Ul Dª
lim X Disp = X PFR
Pe,~oo
For a first arder reaction: 14-28
1
4qexp(Pe, 12)
X=l---~----------~·-------
(1)
(1+q)2exp(Pe,q/2)--(1-q)2 exp(-Perq/2)
q ~ ~!+ 4Da. Pe, Da= rk Far Pe,»>I (or far small vessel dispersion number)
Da Pe,
=1-t-2--2
q=
(
Da Pe,
-
J
2 +o-.( Da J3
(2)
Pe,
Introducing(2) into (1) and neglecting terms
a(_J-. J Pe,
2
4-(1+ ~:}[';'] X= 1--2kT
4 . ( 1 +-Pe X
[
= 1 -e
-kH
(ix)' Pe,
)
Pe,(!+ 2D!!. _ 2Da'-)) ( Pe Pe e 2
r
r
J
P14-12 (b) -k1:+ (k1:)2 XPFR
=l-e-k1:
and X DisoPe>>I = 1 - e
Pe,
In arder to achieve the same conversion:
Pl4-12 (e) Defining:
Pe, = UlPFR = 0.1 Dª
4qexp(Pe,(-1J12J ZPFR
0.99(1- e-k1:) = 1-
(1 + q)2 exp(Pe,(-1-Jq/2)-(1-q)2 exp(-Pe,(-1-Jq 12J ZPFR
ZPFR
14-29
4,rPFRk
q= 1+--Pe,
The following figure shows the solution as function of k'tPFR for different values of 1/lpfR·
P14-12 (d) Starting from 14.12.1 and subtracting the conversion for a PFR:
ln~..L=
kr
C
Pe
Aplug
For small deviations from the plug flow:
P14-12 (e) According Pl2.3 dispersion doesn't affect zero order reaction.
P14-13 From PB..19: 't=IO min and cr2=74min2
far
O<;:;.
t
<;:;.
3 E1 (t) = ·-·L 1675 -10-3t4 + 1.1355 -10-2t3 -4.7492 -10-2t2
+ 9.9505 -10-2 t far 3 <;:;. t
<;:;.
20 E2 (t)
E(t) = -1.7979 -10-4t
= -1 · 8950 -10-6 t4 + 8.7202 -10-5 t3
-1.1739 -10-3 t?
+ 0.092343
far 20 ~ t ~ 60 E3 (t) = 1.2618 -10-8t4 -2.4995 -10-6t3 + 1.8715 -10-4t2 6.3512 -10-3 t + 0.083717 far t > 60 O
P14-13 (a) 2nd order, kCAo=O.lmin·1, CAo=lmol/dm3, Segregation Model Segregation Model
fX (t )E(t )dt
00
X
=
o
kC Aot 1 + kC AJ
Where X () t = ----
See Polymath program Pl4-13-apol POLYMA TH Results Calculated values of the DEO 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
kCao El
o
0.092343 0.083717
E2 E3 E4 E
o o o
X
0.1
O. 1
o o o
o
-2.436E+04 -21.261016 l. 949E-05
0.0836855 0.092343 0.083717 0.0836855 0.875
O. 1
-2.436E+04 -21.261016 0.0017977
o o
0.875
ODE Report (RKF45) Differential equations as entered by the user [ 1 J d(Xbar)/d(t) = E*X Explicit equations as entered by the usar [ll kCao = 0.1
= =
[ 2 J E1 -0.0011675*t"4+0.011355*tA3-0.047 492*tA2+0.0995005*t [ 3] E2 -1.8950*1 Q/\(-6)*t"4+8.7202*1 OA(-5)*tA3-1.1739*1 Q/\(-3)*tA2-1. 7979*1 OA(-4)*t+0.092343 [ 4] E3 1.2618*1QA(-8)*t"4-2.4995*10A(-6)*tA3+ 1.8715*1 Q/\(-4)*tA2-6 . 3512*1 Q/\(-3)*t+0.083717 [5J E4 O [ 6 J E= if(k=3)then(E1 )else(if(k=20)then(E2)else(if(k60)then(E3)else(E4)))
= =
[7J X
= kCao*V(1 +kCao*t)
P14-13 (b) 2nd arder, kCAo=O.lmin·1, CAo=lmol/dm3, Maximum Mixedness Model
= kC A 2 = C Ao (1- X)
Rate Law : -
CA rA
rA
= kC Ao 2 (1- X )2
Í=kC C Ao
where k=O. l dm3 /mol min
Ao (1-x)2
dF
·-- = --E(z)
where z=60-·A
dz
See Polymath program Pl4-13-b.pol POLYMA TH Results Calculated values of the DEO variables Variable z X
F
Cao lam Ca k ra E4
initial value
o o
0.99 1 60 1 0.1 -0.1
o
minimal
o o
value
-0.010344 1
o
0.52273 0.1 -0.1
o
maximal value 60 0.4773052 0.99 1 60 1 0.1 -0.0273247
o
14-31
final value 60 0.4047103 -0.010344 1
o
0.5952897 0.1 -0.035437
o
-1.284E+04 -9.8680524 2.228E-05
El E2 E3 E
o
0.0832647 0.092343 0.083717 0.0832647
-1 . 284E+04 -9.8680524 1.806E-05
o
o 0 . 092343 0.083717
o
ODE Report(RKF45) Differential equations as enterad by the user [ll d(X)/d(z) = -(ra/Cao+E/(1-F}*X) [ 2 J d(F)/d(z) = -E Explicit equations as entered by the user [ll Cao = 1 [2 J lam = 60-z [ 3 J Ca = Cao*(1-X}
k = .1 [ 5 J ra = -k*CaA2 [6J E4 = O [4]
[7J
E1 = -0.0011675*1amA4+0.011355*1amA3-0.04
7 492*1amA2+0.0995005*1am
[ 8 J E2 = -1.8950*10A(-6)*1amA4+8. 7202*1 QJ\(-5)*1am"3-1.1739*10A(-3}*1amA2-1 . 7979*10A(-4)*1am+0.092343 [ 9 J E3 = 1.2618*1 QA(-B)*lam/\4-2.4995*1 QA(-6)*1amA3+ 1.8715*1 QA(-4)*1amA2-6.3512*1QA(-3)*1am+0.083717
[ 1o J
E = if(lam<=3}then(E1
)else(if(lam<=20)then(E2)else(if(lam<60)then(E3)else(E4)}}
(Check F(t))
P14-13 (e) Tanks in series and lst arder reaction with k=O.Jmin-1 T2
n =-2 =1.35 a
Far lst arder reaction is acceptable a non integer value of nfor calculating the conversion
P14-13 (d) Dispersion models and lst arder reaction with k=O.lmin-1 Peclet number open system
a?
t;
-
2
8
Pe
Pe2
= --- + -
~ Pe= 4.906
FEMLAB application Peclet number closed systern
2
t;
X
Pe
= 1_
Pe2
(1 + q )2
(l -e -Pe) ~Pe=0.98 __4q_e_xp_(P_e_,I 2_) _ exp(Pe,q / 2)-(1-q )2 exp(- Pe,q / 2)
q~~I+ 4Da Pe,
14-32
Da= 'lk X=4.59
P14-13 (e) Dispersion models and 2nd arder reaction with k=O.ldm3/mol·s and CAo=lmol!dm3 Conversion Linearizing the 2•1 order reaction rate according 14.2 (g): -
- ic: -= k --cAo e - k'C
rA -
A
A -
2
A
it is possible to obtain an approximated solution;
k'= k CAo = 0.05s-1 2 Da= 'lk'= 30 Pe,=0.98
(fromPart(b))
q= ~+ 4Da =~l+ 4*30 =ll.lll Pe, 0.98
f
X =l·-
4qexp(Pe,/2~) _ (1 + q)2 exp(Pe,q /2)-(1-q)2 exp(--Pe,.q/2)
X=0.998 Full solution can be obtained with FEMLAB.
P14-13 (f) Two parameters model (backflow) For tracer pulse input
dC1
V¡--=
dt dC2
V2 ----· dt
v1C72 -v1Cn
= v1Cn
-v0Cn
Defining:
a= V¡ ,fJ=}:i . and r=~ V v0 v0 We arrive at two coupled differential equations: For tracer pulse input
ar dC¡ dt
= /J(CT2 -crJ
(1-a) dC2 = /J(Cn dt
-Cr2)
See Polymath program P 14- I 3-f.pol
14-33
P14-13 (g) Two parameters model models and 2nd arder reaction with k=0.Jdm3/mol·s and CAo=lmol!dm3
e
AO
+ /JC A2
/JC At
/JC A2
-
/JC Al =-kaiC1¡
-
e
-
= -kC12 (1- a}r
A2
P14-13 (h) Table o 'the conversions X1
l S
0.527
XMM
x.,
X Dis
0.405
0.422
0.5
ersion
Xtwo arameters
?
P14-13 (i) Adiabatic reaction with Segregation Model 2nd arder, kCAo=O.lmin-1, CAo=lmol/dm3, Segregation Model
Segregation Model
ax = kc O(
AO
(1 - x )2
T = 320 + 1.SOX
k
= 0.1 *exp(45000/8.314
* (1/320-1/T))
X= Jx(t)E(t)dt o See Polymath program 1'14· 13 .. i . .pol
P14-14 (a) Product distribution for the CSTR and PFR in series T kz/k1 't'k1CA0
,-1:J.
5.0 2.0 0.5 0.1
T2 T3
~---·
0.2 2 20 200
--
Considering Arrhenius equation and applying the following notation:
k,T E,. ln--=--=-E
Ar
-
RT
rt
We can write this linear system of 5 equations for the 5 unknowns ( E11,
1 n-k11 -- E 12
k12
-
E12, E13, E14, A1 ):
E 11 --
1 n·-k-11 -E-· n- E11 --
k¡3
14-34
I_ 1 !
P14-14 (b-d)
No solution will be given at this time.
P14-15 (a) Two parameters model Omin s; t < lOmin C(t) = 10· (1-exp(-0.lt))
Iümin s t
C(t)
= 5 + 10 · (1-exp(-O.lt))
Omin s; t < lOmin Iümin s; t
F2 (t)
F1 (t) 1
2
3
3
=
2 · (1--- exp(-0.lt)) 3
= -+- · (1-exp(-0.lt))
F(t)
t(min) The F (t) can be representative of the ideal P~ and ideal CSTR in parallel model:
14-35
V1=aV
nodel
a.= Fractional volume 13= Fraction of flow
fJ = .!..3' a = .!..3' •
1"PFR
r
= 10 min
a = 10 m1n . = --V1 = -"[' VPFR fJ
't'csrn
(1- a) . = (1-/J ) 1" = lümm
P14-15 (b) Conversion 2nd order, v0=l dm3/min, k=Oldm3/mol·min, CAo=L25mol/dm3 Balance around node 1
fJv ºC PFR + (1- fJ)voCcsrn
= voC A
ForthePFR: Second-order
C AO - C PFR = DaPFR __ = 0 . .5.56 CAo l+DaPFR a Where DaPFR = k fJ iC Ao = 1.2.5 X PFR
=
CPFR = 0.556mo[ f dm3 For the CSTR:
Xcsrn = Where
CAO ~ Ccsrn
DacsTR
CAo
=
1-a
= k--iC 1-/J
= 0.72.Smol/ dm3 Where eA = fJC PFR + (1-
(1 + 2Dacsrn )- ~1 + 4Dacsrn2DacsTR Ao
= 0.42
= 1.25
CcsrR
fJ)CcsTR = 0.668mol / dm3
L
x = eAo :- eA = 0.465 CAO
14-36 1
¡¡
P14-15 (e) Two parameters model
Omin s t < lümin
IOmin $ t
F1 (t) = O
F2 (t) = _!_ + ~ · {1-exp-[0.2(t-10)D 3 3
F(t)
The F (t) can be representative of the ideal PFR and ideal CSTR in series with a bypass between them: vb=~Vo
nodel
a.= Fractional volume 13= Fraction of flow
fJ = 'CPFR
1
a=
3,
3 , 4
i-
= 40 nun.
3
-- -V¡ -- ar --10 nun ·
V
0
-
(l - ª))r-- 5 nun · 1- /J
'ícsrn - (
Conversion 2°d order, v0=l dm3/min, k=0.ldm3/mol. min, CAo=l.25moVdm3 Balance around node 1
/Jvocb
+(1-/J)voCcsrn
=/JvoCPFR +(l-/J)vocCSTR
=voCA
ForthePFR: Second-order
14-37
X PFR = C AO - C PFR =
DaPFR = 0.556 l+DaPFR kcac Ao = 1.25
CAo
=
Where
DaPFR
CPFR
= 0.556mol f dm3
For the CSTR:
Xcsrn =
C PFR - Ccsrn C PFR
=
(1 + 2DacsTR )- ~1 + 4DacsTR
2DacsTR
= 0.185
1-a
= k---rCPFR = 0.278 1- /3
Where
DacsTR
CcsTR
= 0.453mol I dm3
X= CAo-CA =0.61 CAo Where
e A = /3C PFR + (1- /J)Ccsrn = 0.487mol / dm3
P14-16
(3)
(4)··· ,,
CS1R
~a
v. CS1R
(5)
~
PFR
...
(6)
14-38 ~-
1
A1
PFR
A1
PFR PFR
"" PFR
CSTR
~ 1
(7)
J
PFR
...
As
A.
or
t
PFR
...
1
Recycle 1
(8)
Oaad Zones
(9) PFR
LFR
CSTR
LFR
(12)
(11)
P14-17 The presence of a minimum (run2) 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 r
rp=R
A=~ L
fJ=-
T T¡sot
C U lf/=-A U=-
e
AO
u,
14-39
q., = o
aiJ =O
aq.,
hR
where Bi = -
1
A'k
"t
R CP
q.,=1:
- aiJ = Bi(l- tJ) aq.,
ratio convection-radial conduction
ratio conduction-enthalpy flux
8H,eac
----
dimensionless adiabatic heat
CpTcoolant
Da(tJ) = k(tJ)C~0T A parametric study of the numerical solutions of Eq .( 14 ..51) and the enthalpy balance would give the "exact" conditions for a mínimum 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 mínimum in concentration .
P14-17 (b) Overall heat transfer coefficient increases :·
For a given flux this determines less deviation from the isothermal case: no mínimum
P14-17 (e) Overall heat transfer coefficient decreases . ·
For a given flux this determines more deviation from the isothermal case and the possibility of a mínimum.
P14-17 (d) The coolant flow rate increases:
The external heat transfer coefficient increases, this implies a lower wall temperature and a resulting lower temperature in the reacting mixture and Jess deviation from the isothermal case: no mínimum
P14-17 (e) The coolant flow rate decreases:
The externa! heat transfer coefficient increases, this implies an increase in the wall temperature and a .resulting higher temperature in the reacting mixture that can determine a mínimum in the concentration. Below shows how to use FEMLAB for this problem. Femlab Screenshots for the baseline case (1) Domain
14-40
[:) AXiii
equal
'r-í: 1¡n1~s - .-
--
. (min: .·{§_.1
. t }}rtax:. fo.2
;:!~t1
] =:].
(2) Constants and scalar expressions • Constants
• Scalar expressions
14-41
[N1:1me . io ~z ·-!xA
;-----'cB
v', Ex~ressfooT
,c.,,, "'
wO/(pi•RaA2)
< . _ .,·.
>Y;;;:./,,
~~)"2)______________ (cAO-cA)/cAO k:BO-cAo•xA
V , .2< · ,}, +J¡) . ~-.1 .
-·------·-------
·
ti,~]·
!'"º'l ,1
--·--··-----·-----------·---------~·---·---
~-- , ~A --- . . . --. --------·------------------------·--·r~--·-·I' A LA •exp( -E/RfT)*rhoCat•( cA'cB-cCIKeq) --------·-----·-----------·· ,Keq
){eqO•exp(dHrxlR'(1
/303-1 ff))
'G (-r A)'(-dHrx) · r--·----,~ ,·------·,·----------·----·--· -··---------·------------·· -·~------- .. - ........
¡
; ~~
-------------------------¡,,,,,;
-r,
· ... · ·------..,.,,.--""""-""'"""""""""""""" ..... ~-----
(3) Subdomain settings Physics (Mass balance)
Lil:>rary maferlal: Quantity ó,s
@ D isdtropic
O D eu,isotropic R u
(Energy balance)
14-42
.......... •:::;:;:::;;¡
1! -subdomain selection-~-~:i ¡gj_( j
iI
ji
iI Ii
j
r;;;;;sic;i
1 1
[i
1
I
! r
o
lnil
,,;;, ,,s,-,t
k (isotropic) k ( anisotropic)
P .CP Q
- Initial Values
(Mass balance) cA(tO) = cAO (Energy balance) T(tO) = TO - Boundary Conditions @ r = O, Axial symmetry @ inlet, cA = cAO (for mass balance) ( for energy balance) ,·Boun f
- ••.. ~ ..
;~iÍ
Boundary condition: iHeat flux !.'ó·---
ion Description
Qmmtíty
@ @
~
qo
[~ho"Cp"TO"uz ·. -_· J lnward heat 11ux
To
[!~ , __ .
:-:J
Temperature
outlet, Convective flux wall, Insulation/Symmetry (for mass balance) (for ener y balance) nditions·· Boundary conditi,on: QqJntíty
qo
To
flux
Valliel1:xpression Description
·-----·-··-·--
. ...
,
lnward heat flux Temperature
14-43
P14-18 Vary the Peclet number and the reaction arder in laminar flow (Example 14-3(c))
Da -- rkcn-i -!:_kcn-i AO - ij AO o
Pe= U0LI DAB
L = 6.36m R=0.05m k = 0.25dm3 / mol.min U0 = l.24m/min CAO
= 0.5mo[ f dm'
DA8 = 7.6xl0-5 m2 /min n 0.1 Attempt to evaluate nonintegral power of negative number!
- e--·
0. 5 Attempt to evaluate nonintegral power of negative
number! 0.8
-
1
,---.
L5
·-
2
2.5
·--
-
Pe l.04xl01 l.04xl04 l.04xl05 l.04xl0° l.04xl07 l.04xl03 l.04xl04 l.04xl03 l.04xl06 l.04xl07 l.04xl0j l.04xl04 l.04xlü"~ l.04xl06 l.04xl07 l.04xl0j l.04xl04 l.04xl03 l.04xl06 l.04xl0' l.04xl03 l.04xl04 l.04xl05 l.04xl06 l.04xl07 l.04xl0j l.04xl04 l.04xl05 l.04x10° l.04xl0' l.04xl03 l.04xl04 l.04x105
Conversion / / /
/ / / / / / / 0.824 0.818 0.783 / / 0.721 0.717 0.687 0.653 0.641 0.526 0.522 0.503 0.481 0.472 0.390 0.388 0.374 0.359 0.352 0.292 0.290 0.281
-
14-44
Parameter 100* DAB -·-10* DAB DAB O.l*DAB O.Ol*DAB 100* DAB 10* DAB DAB --·0.1 *DAB 0.01 *DAB 100* DAB 10* DAB DAB O.l*DAB O.Ol*DAB 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 0.0l*DAB 100* DAB 10* DAB DAB
--
-
3
3.5
4
l.04xl0° l.04xl0' l.04xl03 l.04xl04 l.04xl05 l.04xl0° l.04xl07 l.04xl03 l.04xl04 l.04xl05 l.04xl06 l.04xl07 l.04xl03 l.04xl04 l.04xl0' l.04xl0° l.04xl07
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
O.l*DAB O.Ol*DAB 100* DAB 10* DAB DAB O.l*DAB O.Ol*DAB 100* DAB 10* DAB DAB 0.1 *DAB O.Ol*DAB 100* DAB 10* DAB DAB 0.1 *DAB O.Ol*DAB
1.0
n
0.8 --•-1.0 ---11-
0. 9 O8 0.7
e:
o
o
--·- -----·-
..
-.
0. 6
º1!1 05 g;?
§
---·----.
0.4
.A
... k,
T
..•....
·-----
---
·.<\
. . ....
J¡,
1..5
.............-2. 0 ~ 2.5
---
•
3.5 4.0
---·T·-·---T-···-··--T
0. 3 0.2
O .1
·-·· -····
-··
e - ·-··-···- e ---
·1··-
··-·--··
.....
.... .. ·--- ---- •
Pe
(1) With the increase of the reaction order n, the conversion will decrease, The conversion will also decrease when the Peclet number increases. (2) When nis less than 1.5 and Pe is 105, the diffusion in the system becomes more important. Below show FEMLAB screenshots useful for this problem. Femlab Screenshots (1) Domain
14-45
(4) Constants and scalar expressions - Constants
- Scalar expressions
(5) Subdomain settings
Physics
14-46
- Initial Values
(Mass balance) cA(tO) = cAO
- Boundary Conditions @ r = O, Axial symmetry @
@ @
outlet, Convective flux wall, Insulation/Symmetry
P14-19 (a) First order reaction Input Parameters:
n =I R=0.05m L=6.36m
k =0.25/min U0 = l.24m/min C AO
= Ü.5mo[ / dm' 14-47
Conversion: xA= 0.687 @ Open-vessel Boundary: -Ni·n =2*UO*(l-(r/Ra)"2)*CAO xA= 0 . 726@ Close-vessel Boundary: -Ni·n =UO*CAO Conversion: xA= 0.755@ Open-vessel Boundary: -Nin =2*UO*(l-(r/Ra)"2)*CAO xA= 0 . 778@ Close-vessel Boundary: -Ní·n =UO*CAO l. Variation of Da number
-·
0.449 0.898 1.795 3.59 7.18 14.36 28.72
Closed-vessel 0.404 0.527 0.658 0.778 0.870 0.930 0.964
11. Variation of Pe number Peclet number/Pe
-
l.04e3 l.04e4 l.04e5 l.04e6 l.04e7
Parameter
Conversion
Damkohler number/ Da
-·
Oeen-vessel 0.287 0.439 0.606 0.755 0.862 0.928 0.964
Open- vessel 0.781 0.776 0.755 0.740 -- -------0.737 --~· "
IIL Femlab Screen Shots (1) Domain
(2) Constants and scalar expressions - Constants
14-48
uo
U0/2 U0/4 U0/8
-·--·
Conversion Closed-vessel 0.781 0.781 0.778 0.772 0.770
8*UO 4*UO 2*UO
Parameter lOO*DAB ·10* DAB DAB 0.1 *DAB 0.01 *DAB
--
FI-·--·-···---···-···---·-·---·----··----.. , . ,. __
,.,
(3) Subdomain settings - Physics
14-49
---·--······-·-
.
- Initial Values
(Mass balance) cA(tO) = cAO - Boundary Conditions @ r = O, Axial symmetry @ inlet, Flux N0 = 2*uO*(l-(r/Ra)"2)*cAO : Open-Vessel Boundary Flux N0 = uO*cAO : Close-Vessel Boundary @ outlet, Convective flux @ wall, Insulation/Symmetry (4) Results
(Close-vessel Boundary)
14-50
P14-19 (b) Third arder reaction with k*C2 Ao= 0.7 min-1
Da -rkcn-i - ~kcn-i AO - ij AO o
Pe= U0LI
DAB
Input Parameters:
n=3 R=0.05m L = 6.36m k = 2.8dm3 /mol.min U0 = l.24m/min CAo
= 0.5mol / dm'
DAB
= 7.6xl0-5m2 /min
l. Variation of Da number Damkohler number/ Da 0.255 0.51 1.02 2.03 4.06
..
Conversion
Closed-vessel 0.381 0.466 0.560 0.655 0.741
Parameter Oeen-vessel 0.253 0.375 0.506 0.628 0.730
14-51
8*UO 4*UO 2*UO
uo
U0/2
0.809 0.865
0.813 0.867.
8.12 16.24
II Variation of Pe number Peclet number/Pe
•
Parameter
Conversion
l.04e3 l.04e4 l.04e5 l.04e6 l.04e7
-
U0/4 U0/8
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 0.1 *DAB O.Ol*DAB
(e) Half order reaction with k= 0.495 (mol/dm3)112min-1
D = rkcn--I = ~kc"-I a AO U AO o
Pe= U0LI DA8 Input Parameters:
n =112 R=0.05m L=6.36m k = 0.495(mol / dm' )112 min -i U0
= l.24m/min
C AO = Ü.5mo[ / dm'
DA8 =7.6xl0-5m2
lmin
l. V ariation of Da number Damkohler number/ Da
¡.-------------·
0.69 1.38 2.76 5.52 11.04 22.08 44.16
------
-
Conversion Closed-vessel 0.432 0.713 0.959 1 / / /
II. V ariation of Pe number Peclet number/Pe
>---
l.04e3 l.04e4 l.04e5 l.04e6 l.04e7
Parameter Ooen-vessel 0.378 0.641 0.911 1 / / /
Conversion Closed-vessel 1 1 1 / /
uo
U0/2 U0/4 U0/8
Parameter Ooen-vessel 1 1 1 0.9995 0.9994
14-52
8*UO 4*UO 2*UO
100* DAB 10* DAB DAB 0.1 *DAB 0.01 *DAB
--
(d) The radial conversion profiles far various arder of reaction (1) First arder
n =1
L=6.36m R=0.05m k = 0.25dm3 I mol.ruin U0 = 1.24m/min
= 0.5mo[ / dm' DAs = 7.6xl0-5m2 /min C AO
l. Open-vessel: --1 ··-e-2 -1-3 .. ..,,. .. 4
08 07
...¡¡-
0. 6
0.5
~04
03
. -~<,r//-- ~
02
0.1
,,/:
~--·-·:-,,._,. __;__,,,_.:..--.'--;----·J.:.;.,-•~..-~>"""'
:
..l..___L.~-L~-'-~---'-~---'-~--'-~--1...~--1....---'-~-'-~ O
0.005
O 01
0.015
0.02
0.025
0.03
Note: 1: z= L/10; 2: z=3LIIO; 3: z=L/2; 4: z=7L/10; 5: z=9L/10 11. Closed-vessel:
14-53
0.035
0.04
0.045
O 05
5
1 2 -+-3 ..,:, 4 -
08
---9--
---5
0.6
0.4
02
\ .
'\ .
~\ .
\'
\
. \
-0 . 2
O
O 005
Oü1
O 015
O 02
0.025
O 03
0.035
O 04
0.045
O 05
Note: 1: z= L/10; 2: z=3L/10; 3: z=L/2; 4: z=7L/10; 5: z=9L/10
Something is not correctwith z= L/10 because of the FEMLAB! (2) Second arder
n=2 R=0.05m L= 6.36m k = l.4dm3 /mol.min U0 = l.24m/min CAo
= 0.5mol / dm'
DA8 = 7.6xl0-5 m2 /min I. Open-vessel:
14-54
--1
0.8
. :
0.7
,·,?
05
0.4
~-:
Á:..,..,.~.......:
...·,-··
//
:
:
~
: -~-·-·:---,,;,-... , .
~·:
.
:
:
:
: y
-~~//. .
:
.,?'"
-e-- 2 -+--3 -,¡-
4 5
-,;¡-
1
•3
L-1--- :-~---r=<~,-~ : //~~~-:
06
';l;
~-
:
,0"'.....~-·---i:t·~
.
.,.,..)/· '.
·,a"': /"
,
.
03
"•
0.2 J......__..L-..........J
___L_
O
O 005
O 01
0.015
0.02
O 025
O 03
O 035
__
0.04
_.__ _
0.045
_,__,
O 05
Note: 1: z= L/10; 2: z=3L/10; 3: z=L/2; 4: z=7L/10; 5: z=9L/10 II. Closed-vessel:
--s-2 --+-3
08
4
--·5 07
0. 6
';l; 0.5
04
"t
O
O 005
0.01
O 015
0.02
O 025
0.03
Note: 1: z= L/10; 2: z=3L/10; 3: z=L/2; 4: z=7L/10; 5: z=9L/10 (3) Third arder
n=3 R=0.05m 14-55
0.035
0.04
0.045
O 05
L = 6.36m
= 2.Sdm3 / mol.ruin U O = l.24m I min k
CAO = Ü.5mol/ dm'
DA8
= 7.6xl0-5m2
/min
l. Open vessel:
_,,,_. 1 ···5-··· 2
-t-3 4
--+-- 5
06
05
io4
__...,.._.,...ff ....
ra----- -
-·--:-·El-" -~
.a- _
L__ a<
0.3
02
L-L--L--L---'-----'---..L------'----'-----'--___¡_
O
O 005
O 01
0.015
0.02
O 025
O 03
Note: 1: z= L/10; 2: z=3L/10; 3: z=L/2; 4: z=7Lll0; 5: z=9L/10
11. Closed- vessel
14-56
O 035
0.04
_
O 045
___¡__¡
O 05
--1
0.7
---2 -+-3 ,) 4
O 65
-lf-5 0.6
0.55
0.5
~ 0.45
0.4
0.35
0.3
0.25 O
O 005
O 01
0.015
O 02
0 . 025
0 . 03
0 . 035
O 04
O 045
O 05
Note: 1: z= L/10; 2: z=3L/10; 3: z=L/2; 4: z=7L/10; 5: z=9L/10
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 of T-1-S Curve 1: Increasing the number ofT-1-S corresponds to increase the Peclet Number (i.e . Bo=2(n-l)).
P14-21 (a)
14-57
nodel
F(t) and E(t) curvesfor
a= -\11
\1
vb = 0.2 = 0.4and /3 = --Vo
The analytical expression for E (t) is given by:
/Jo(t) E(t)
=
{ (1-/J(
t < r1 -(t-r1)tr2
t'2T1 t'z
14-58
1 E(t) 0.5
o
4
1
O ,/2
6
8
10
lntegrating we can obtain the analytical expression for F (t):
fJ F(t)
t
= 1-(1-
T
<-
2
p)e-(, ;){3;)
i
e=T 2
1 1-(1- {3)/e
0.5 {3=0.2
o ,,,_' )
,.,
~ 5/41
4
6
14-59
8
10
P14-21 (b) Conversion 2"d arder; kCA0=0.5min·1, t=Zmin Balance around node l
(1- p')v ocCSTR + /JvoCb
= VoC A
Forthe PFR: Second-order
XPFR
=
CAo-CPFR CAo
Where DaPFR
= kC Ao
= DaPFR
=0.333
l+DaPFR (
a
1-/3
) T = 0.5
kCPFR =0.333min-1 For the CSTR:
- C PFR - Ccsrn (1 + 2DacsTR )- ~1 + 4Dacsrn X CSTR C PFR 2DacsTR l-a Where DacSTR = k-iC PFR = 0.5 1- /3
= 0.268
kCcsTR = 0.366 min -i kC A = (1- /J)kCcsrn + /JkCb = (1- /J)kCcsrn + /JkC Ao = 0.393min-1 X= kCAO -kCA =0.214
ic ¿
In absence of bypass (~=0) the conversion would be X=0.245
P14-22 Two parameters model A possible two-parameter model is the PFR with Bypass (vb) and Dead Volume (V0)
nodel
14-60
To evaluate the conversion we write a balance around node lon species A:
(1-p)vocPFR PFR,
r,
2ºd arder,
+PvoCb =voCA liquid phase
=-kCAC8
Where k=L5 m3/(kmol·min) and
X PFR
=
C AO - C PFR
=
CAo
Where
DaPFR
ePFR
= eAo
e
Ao
=
DaPFR
e
Bo
= 2mol / dm3
= 0.857
l+DaPFR
a
= «; (1-p)T= 6 X PFRC Ao = 0.286mol / dm3
-
= (1- P)C PFR + /JCb = (1- P)C PFR + /JC Ao = l.143mol/ x = eAo - eA = o.429 CA
CAO
P14-23 0.1
E(t)
o
o
10
20
t (nli:n) 00
By definition: JE(t)dt o
OOJE( t )dt
o
=1
= O.l*tl = 1 ~ 2
. t 1 = 20 mm
2
For a triangular E(t): t1 = -E(O) The analytical expression for E(t) is given by:
14-61
--·---·----------
.
--------~--------
-
-
.
.
...
- ------- -·------------·······---·---------·~------------·-
---------------·---------
dm3
0.1 For05't5't1 E(t)=--t+
O
t¡
.1
For t > t1 E(t) = O
P14-23 (a) Mean residence time
J ()
1
00
tm = Et tdt= 0
R
0.1 --t+0.1
0
) tdt= [-O.lt3 +--O.lt2]11 3tl 2
t¡
0
0.1 12 +-2-=-6-= o.u,' O.lt12 6 . 67 mm . =--t 3
t¡
=- )
(For a triangular E(t): t m
3
Variance
a2 = J E(t )(t - tm )2 dt = 00
o
iJ:r(--t+0.1
O 1J zd z -[-O.lt4 O.lt3]11=2omin z . t t-tm +-=t; 4t1 3 .0
t1
O
= - ~J t31 + O. lt/ - t2 4
3
m
0.1 · 203 - 6.672 = 22.2 min 12
Assuming closed-closed system:
r
= tm
P14-23 (b) Peclet number
a2 -=l= r2
¡·1m---·-2
Pe--tO
Pe r
2 (l ·-·e Pe r 2
-Pe r )
AndPe,=0
P14-23 (e) Tanks in series r2
n=-=1
ª2
The triangular E (t) can be interpreted asan approximation for t<<-r of a CSTR where the continuity impose E(O)*-r=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 ldealPFR
Ideal CSTR
.. ldeal laminar tlow reactor
Segregation
Maximum Mixedness
14-63
Dispersion
Tanks in series
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