Elements of Chemical Reactor Engineering

~CD-ROM
~INClUDED
Brian Vicente / Max Nori / H. Scott Fogler
Soludons Manual for
Elements
of Chemical
Reaction
Engineering
Fourth Edition ~
Prentice Hallitaternational Series
in the Phys!ptl and Chemical
E.ngineering Sciences
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Solutions Manual for
Elements of Chemical
Reaction Engineering
Fourth Edition
Brian Vicente
Max Nori
H. Scott Fogler
~
PRENTICE
HALL
PTR
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ISBN 0-13-·186383-5
Text printed in the United States at OPM in Laflin, Pennsylvania.
First printing, November 2005
______________...________ .________ :
**** CONFIDENTIAL ****
UNIVERSITY OF MICHIGAN
INTERACTIVE COMPUTER MODULES FOR CHEMICAL ENGINEERING
CHEMICAL REACTION ENGINEERING MODULES
H. Scott Fogler, Project Director
M . Nihat GUIllen, Project Manager (2002-2004)
Susan Montgomery, Project Manager (1991-1993)
Department of Chemical Engineering
University of Michigan
Ann Arbor, MI48109-2136
©2005
Regents of the University of Michigan
- All Rights Reserved -
INTERPRETATION OF PERFORMANCE NUMBERS
Students should record their Performance Number for each program, along with the
name of the program, and tum it in to the instructor. The Performance Number for
each program is decoded as described in the following pages.
The official site for the distribution of the modules is
http://www.engin.umich.edu/-cre/icm
Please report problems to [email protected].
PerfOimance NumbeI InteIpIetation: eRE modules
iii
**** CONFIDENTIAL ****
ICMs with Windows® interface
Module
Format
Example
Interpretation
KINETIC CHALLENGE I
CzBzzAzz
Score = 1.5 * AB.C
z =random numbers
Perf. No. =15.641~92
Score = 1.5 *( 62.7) =94 %
Note: 75% constitutes mastery.
KINETIC CHALLENGE II
CzBzzAzz
Score = 2.0 * ARC
z = random numbers
Note: 75% constitutes mastery.
Perf.. No. =Q3176.167
Score = 2.0*(47.0) = 94 %
A even: Killer and victim
correctly identified
A odd: Killer and victim
not identified
z = random numbers
Perf. No. = 50132
Score: No credit
MURDER MYSTERY
zzAzz
Note: An even number for the middle digit constitutes mastery.
TIC TAC TOE
zDzCzBzA
Score =4.0 * AB.C
z = random numbers
Perf. No. = 718Q3281
Score =4*(15 . 0) =60
configuration 7 completed
Configurations
Note: Student receives 20 points for every square answered correctly.
A score of 60 is needed for mastery of this module.
GREAT RACE
zzzCzABz
Score = 6.0 * AB . C
z =random numbers
Perf No. = 777J.8Q18
Score = 6*(07.3) = 44
Note: A score of 40 is needed for mastery of this module.
PeIformance NumbeI Interpletation: eRE modules
iv
**** CONFIDENTIAL ****
ECOLOGY
AzBCzaaD
z = random numbers
a =random characters
A gives info on rl\2 value of the student's linearized plot
A=Y if rl\2 >= 0.9
A=A if 0.9 > rl\2 >= 0.8
A=X if 0..8> rl\2 >= 0.7
A=F if 0..7 > rl\2
A=Q if Wetland Analysis/Simulator portion has not been completed
B gives info on alpha
B=l to 4 => student's alpha < (simulator's alpha ± 0.5)
B=5 to 9 => student's alpha> (simulator's alpha ± 0.5)
B=X if Wetland Analysis/Simulator portion has not been completed
C indicates number of data points deactivated during analysis
C=number of deactivated data points if at least 1 point has been deactivated
C=a randomly generated letter from A to Y if 0 points deactivated
C=Z if Wetland Analysis/Simulator portion has not been completed
D gives info on solution method used by student
D=l if polynomial regression was used
D=2 if differential formulas were used
D=3 if graphical differentiation was used
D=4 to 9 if Wetland Analysis/Simulator portion has not been completed
Perf No . = A7213DF2
1) A => 0.9 > rl\2 >= 0 . 8
2) 2 => student's alpha < (simulator's
alpha ± 0 . 5)
3) 1 => one data point was
deactivated
4) 2 => differential formulas were
used
STAGING
zCBzAFzED
z = random numbers
Final conversion = 2*ARC
Final flow rate = 2*DE.F
Perf. No.. =
2125.:!8~913
conversion = 2*42 ..1 = 84.2
flow rate = 2*31.2 = 62.4
Please make a passlfiil criterion based on these values .
Performance Number Interpretation: eRE modules
v
**** CONFIDENTIAL ****
ICMs with Dos® interface
Module
Format
Interpretation
Example
A=2,3,5,7: interaction done
B=2,3,5,7: intra done
C=2,3,5,7: review done
D denotes how much they
did in the interaction:
Perf. No. = 8027435
A: Worked on interaction
B: Looked at intra
C: Looked at review
D: found parameter values,
didn't find mechanism
HETCAT
zzABzCD
D<2
Not done
2 < D:5 4 Dependences
4 < D:5 6 Parameter values
6<D
Mechanism
z = random numbers
Note: Performance number given only if student goes through the
interaction portion of the module
HEATFXl
zzAzz
A even: score> 85 %
z = random numbers
Perf. No. = 53~07
Score> 85 %
Note: Student told they have achieved mastery if their score is greater than
85%
HEATFX2
zzzAzz
A even: completed interaction
z = random numbers
Perf. No. = 407.2.82
Interaction not completed
Note: Performance number given only if student goes through the
interaction portion of the module.
PerfclImance NumbeI InteIpretation: eRE modules
vi
Solutions for Chapter 1 - Mole Balances
Synopsis
General: The goal of these problems are to reinforce the definitions and provide an understanding of the
mole balances of the different types of reactors . It lays the foundation for step 1 of the algorithm in Chapter
4.,
PI-I.
This problem helps the student understand the course goals and objectives.
PI-2.
Part (d) gives hints on how to solve problems when they get stuck. Encourages
students to get in the habit of writing down what they learned from each chapter.
It also gives tips on problem solving.
PI-3.
Helps the student understand critical thinking and creative thinking, which are
two major goals of the course.
PI-4.
Requires the student to at least look at the wide and wonderful resources
available on the CD-ROM and the Web.
PI-S.
The ICMs have been found to be a great motivation for this material.
PI-6.
Uses Example 1-1 to calculate a CSTR volume. It is straight forward and gives
the student an idea of things to come in terms of sizing reactors in chapter 4. An
alternative to PI-IS.
PI-7.
Straight forward modification of Example 1-1.
PI-S.
Helps the student review and member assumption for each design equation.
PI-9 and PI-IO. The results of these problems will appear in later chapters. Straight
forward application of chapter 1 principles.
PI-II.
Straight forward modification of the mole balance. Assigned for those who
emphasize bioreaction problems.,
PI-12.
Can be assigned to just be read and not necessarily to be worked, It will give
students a flavor of the top selling chemicals and top chemical companies.
1·6
PI-13.
Will be useful when the table is completed and the students can refer back to it
in later chapters. Answers to this problem can be found on Professor Susan
Montgomery's equipment module on the CD-ROM. See PI-17.
PI-14.
Many students like this straight forward problem because they see how CRE
principles can be applied to an everyday example. It is often assigned as an inclass problem where parts (a) through (0 are printed out from the web. Part (g)
is usually omitted.
PI-IS.
Shows a bit of things to come in terms of reactor sizing. Can be rotated from
year to year with PI-6.
PI-16.
Open-ended problem.
PI-17.
I always assign this problem so that the students will learn how to use
POLYMATHIMaLab before needing it for chemical reaction engineering
problems.
PI-IS.
Parts (a) and (b) are open-ended problem.
PI-19 and PI-20. Help develop critical thinking and analysis.
CDPI-A
Similar to problems 3, 4, 11, and 12.
CDPI-B
Points out difference in rate per unit liquid volume and rate per reactor
volume.
Summary
•
•
•
PI--l
PI-2
PI-3
PI-4
PI-5
PI-6
PI-7
PI-8
PI-9
Pl-IO
Pl-11
Assigned
AA
I
0
0
AA
AA
I
S
S
S
0
Alternates
Difficulty
1-15
SF
SF
SF
SF
SF
SF
SF
SF
SF
SF
FSF
1-7
Time (min)
60
30
30
30
30
15
15
15
15
15
15
•
PI-12
PI-13
PI-14
PI-IS
PI-16
PI-17
PI-18
PI-19
PI-20
CDPI-A
CDPI-B
I
I
0
0
S
AA
S
0
0
AA
I
SF
SF
FSF
SF
SF
SF
SF
- Read Only
FSF
FSF
FSF
30
1
30
60
15
60
30
30
15
30
30
Assigned
• =Always assigned, AA =Always assign one from the group of alternates,
o = Often, I = Infrequently, S = Seldom, G = Graduate level
Alternates
In problems that have a dot in conjunction with AA means that one of the
problem, either the problem with a dot or anyone of the alternates are always
assigned.
Time
Approximate time in minutes it would take a B/B+ student to solve the problem.
Difficulty
SF =Straight forward reinforcement of principles (plug and chug)
FSF = Fairly straight forward (requires some manipulation of equations or an
intermediate calculation).
IC =Intermediate calculation required
M =More difficult
OE = Some parts open-ended.
*Note the letter problems are found on the CD-ROM. For example A == CDPI-A.
Summary Table Ch-l
---------------,
Review of Definitions and Assumptions
1,5,6,7,8,9
Introduction to the CD-ROM
1,2,3,4
1-8
Make a calculation
6
Open-ended
8,16
PI-I Individualized solution.
PI-2 Individualized solution.
PI-3 Individualized solution.
PI-4 Individualized solution.
PI-5 Solution is in the decoding algorithm given with the modules .
PI-6
The general equation for a CSTR is:
V
= F AO
--FA
- rA
Here rA is the rate of a first order reaction given by:
fA=·- kC A
Given: C A= O.lCAO , k = 0.23 min-I, Vo = 10dm3 min-I, FA = 5.0 moVhr
And we know that FA = CAVo andFAo = CAOVo
3
=> C AO = F Aol Vo = 0.5 moVdm
Substituting in the above equation we get:
v = CAOVO -CAVO = (0.5mol/dm 3 )(lOdm 3 lmin)-0.1(0.5mq!ldm 3 )(lOdm 3 I min)
(0.23 min--I)(O. 1(0.5mol I dm 3 »
kC A
V =391.3 dm3
PI-7
t
= INA
-~-kN
dNA
A
NAO
k = 0.23 min-l
dNA
From mole balance:
-=IA' V
dt
1-9
Rate law:
Combine:
at T:::: 0, N Ao = 100 mol and T:::: T, NA = (O ..01)NAo
1 (N AO J
-+ t= "kIn
NA
1
:::: -In(lOO) min
0.23
t = 20 min
Pl-8
a)
The assumptions made in deriving the design equation of a batch reactor are:
Closed system: no streams carrying mass enter or leave the system.
Well mixed, no spatial variation in system properties
Constant Volume or constant pressure.
b)
The assumptions made in deriving the design equation of CSTR, are:
Steady state.
No spatial variation in concentration, temperature, or reaction rate throughout the vesseL
c)
The assumptions made in deriving the design equation ofPFR are:
Steady state .
No radial variation in properties of the system.
d)
The assumptions made in deriving the design equation of PBR are:
Steady state .
No radial variation in properties of the system.
e) For a reaction,
A-7 B
-rA is the number of moles of A reacting (disappearing) per unit time per unit volume
(dm3 s)
1-10
[=J moles/
··rA' is the rate of disappearance of species A per unit mass (or area) of catalyst [=] moles/ (time.
mass of catalyst)..
rA' is the rate of formation (generation) of species A per unit mass (or area) of catalyst [=] moles/
(time. mass catalyst)..
-rA is an intensive property, that is, it is a function of concentration, temperature, pressure, and the
type of catalyst (if any), and is defined at any point (location) within the system. It is independent
of amount On the other hand, an extensive property is obtained by summing up the properties of
individual subsystems within the total system; in this sense, -rA is independent of the 'extent' of the
system.
P 1-9
Rate of homogenous reaction rA is defined as the mole of A formed per unit volume of the reactor per
second . It is an Intensive property and the concentration, temperature and hence the rate varies with spatial
coordinates.
rA on the other hand is defined as g mol of A reacted per gm. of the catalyst per second.. Here mass of
catalyst is the basis as this is what is important in catalyst reactions and not the reactor volume.
Applying general mole balance we get:
dN.
dt
--.L=F· o -F. + fr.dV
J
]
J
No accumulation and no spatial variation implies
o= F·Jo -
F.] + fr J.dV
Also Ij = Pb rj' and W = VPb where Pb is the bulk density of the bed.
=>
f
0 = (FjO - F) + r;CPb dV )
Hence the above equation becomes
Fo-F
W=_l
__,_l
--rj
We can also just apply the general mole balance as
-dN j = (FjO dt
Fj ) + f'r/dW)
Assuming no accumulation and no spatial variation in rate, we get the same form
above:
FO-F
W =._1_._,_1
-rj
1-11
as
Fj
,/ i' F JO
PI-IO
Mole balance on species j is:
v
Po -P + fr.dV
I
I
o
I
dN.
=__
1
dt
Let M j = molecular wt. of species j
= W jO
Then FjoM j
N jM j
= mass flow rate of j into reactor:
= m j = mass cif species j in the reactor
Multiplying the mole balance on species j by M
v
FjOM j
FjM I +M I frjdV = M j
-
o
Now M
I
dN
1
._ _
dt
is constant:
F M. -PM. + vfM rdV
10
I
I
I
I
o
I I
= d0! j N j ) = d(m1l
dt
dt
v
w jO -- Wj
+
fM jrjdV =
o
PI-II
Applying mole balance to Penicillin: Penicillin is produced in the cells stationary state (See Chapter 7), so
there is no cell growth and the nutrients are used in making product
Let's do pari c first
1-12
[In flowrate (moles/time)] penicillin + [generation rate (molesltime)]penicillin - [Out flowrate (molesltime)]
penicillin =
[rate of accumulation (moles/time)]penicillin
Fp,in + Gp - Fp,out =
dNp
dt
,., .,(because no penicillin inflow)
Fp,in = 0 """ ''''', ...
v
Gp =
fr~.dV
Therefore,
frp .dV -
dNp
v
Fp,out =
dt
Assuming steady state for the rate of production of penicillin in the cells stationary state,
dNp
--=0
dt
And no variations
Or,
Similarly, fOI Corn Steep Liquor with Fe = 0
Assume RNA concentration does not change in the stationary state and no RNA is generated or destroyed.
Pl-12 (a)
Ranking of 10 most produced chemicals in 1995 and 2002 are listed in table below:
.--'
Rank 2002
1
2
3
Rank 1995
I
2
4
4
5
9
6
-
3
Chemical
H 2SO4
N2
C2H 4
O2
C3H 6
H2
1-13
6
NH3
10
8
Ch
9
P20S
10
C2H 2Clz
-+ Cherrucals like H2 , P 20 S , C2H2Cl 2 has come III top 10 cherrucals and C3H6 has jumped to rank
5 now then rank 9 in 1995 .
7
Pl-12 (b)
Ranking of top 10 chemical companies in sales in yeru 2003 and 2002:
2003
I----
1
2
3
4
5
+-.
8
9
10
Chemical Sales
Company
2002
($ million 2003)
1
2
3
4
8
5
6
7
11
9
-
Dow Chemical
DuPont
ExxonMobil
General Electric
Chevron Phillips
Huntsman Corp.
PPG Industries
Equistar Chemicals
Air Products
Eastman Chemicals
32632
30249
20190
8371
7018
6990
6606
6545
6029
5800
----
--
SOllIce: Chemical and Engineering News may 17,2004
-+ We have Chevron Phillips whichjumped to 5 rank in 2003 from 8th rank in 2002 and Air
Products coming to 9th rank in 2003 flom 11 th in 2001.
-+Chemical sales of each company has increased compared to year 2002 from 9%(Eastman
Chemical) t028.2%(Chevron Phillips) but Huntsman Corp . has a decrease by 2 . 9%.
Pl-12 (C)
Sulfuric acid is prime importance in manufacturing. It is used in some phase of the manufacture of neruly all
industrial products It is used in production of every other strong acid. Because of its lru·ge number of uses,
it's the most produced chemicaL Sulfuric acid uses are:
-+ It is consumed in production of fertilizers such as ammonium sulphate (NH4hS04 and
superphosphate (Ca(H2P0 4)2) , which is formed when rock phosphate is treated with sulfUric acid .
-+Used as dehydrating agent.
-+Used in manufacturing of explosives, dyestuffs, other acids, pruchment paper, glue, purification
of petroleum and picking of metals.
-+ Used to remove oxides from iron and steel before galvanizing or electroplating.
-+ Used in non-ferrous metallurgy, in production of rayon and film.
-+as laboratory reagent and etchant and in storage batteries
-+ It is also general purpose food additive.
Pl-12 (d) Annual Production rate of ethylene for year 2002 is 5.21x 1010 lb/yeru
Annual Production rate of benzene for yeru 2002 is 1.58 x 1010 lb/year
1-14
Annual Production rate of ethylene oxide for year 2002 is 7.6 x109 lb/year
Pl-12 (e) Because the basic raw material 'coal and petroleum' for organic chemicals is very limited and
their production is not increasing as production of raw material for inorganic chemicals.
1-15
Pl-13
Type
Characteristics
Phases
Usage
Advantage
Disadvantage
Batc
h
All the reactants
fed into the
reactor. During
reaction nothing
is added or
removed . Easy
heating or
cooling.
Continuous flow
of reactants and
products .
Uniform
composition
tluoughout
1. Liquid
phase
2" Gas
phase
3" Liquid
Solid
1, Small scale
pdn.
2. Used for lab
experimentation.
3.
Pharmaceuticals
4. Felmentation
1. High
Operating
cost
2. Variable
product
quality.
L Liquid
phase
2. Gasliquid
3. Solid -,
liquid
L Used when
agitation required.
2. Series
Configuration
possible for
different
configuration
streams
PFR
One long reactor
or number of
CSTR's in
series .
No radial
variations .
Conc. changes
along the length,
1.
Primarily
gas Phase
1. Large Scale
pdn .
2 . Fast reactions
3. Homogenous
reactions
4, Heterogeneous
reactions
5" Continuous
pdn,
PBR
Tubular reactor
that is packed
with solid
catalyst
particles"
L Gas
Phase
(Solid
Catalyst)
2.Gas -solid
reactions.
1. Used plimarily
in the
heterogeneous gas
phase reaction
with solid catalyst
e . g Fischer
tropsch synthesis.
L High
Conversion per
unit volume.
2. Flexibility of
using for
multiple
reactions"
3. Easy to clean
1. Continuous
Operation"
2.Good
Temperature
Control
3 . Two phase
reactions
possible.
4.Good Control
5. Simplicity of
construction.
6. Low
operating cost
7. Easy to clean
1. High
conversion per
unit volume
2. Easy to
maintain (No
moving parts)
3 . low
operating cost
4. continuous
operation
1. High
conversion per
unit mass of
catalyst
2" low
operating cost
3., Continuous
operation
CST
R
""
_.
1·16
1. Lowest
conversion
per unit
volume.
2. By passing
possible with
poor
agitation.
3 High power
Input reqd.
1. Undesired
thelmal
gradient
2. Poor
temperature
control
3" Shutdown
and cleaning
expensive"
-
L Undesired
thelmal
gradient
2" Poor
temperature
control
3.
Channeling
4" Cleaning
.~~pensiv£:. __
Pl-14
Given
A
= 2 *1010 ft 2
TSTP
= 491.69R
T = 534 . 7°R
Po= latm
cS = 2.04 *10-10 lbm301
=0.7302 atm ft3
R
H = 2000ft
ft
lbmolR
C = 4*105 cars
FS = CO in Santa Ana wind
VA
FA = CO emission from autos
ft3
= 3000-·- per car at STP
hr
Pl-14 (a)
Total number of Ib moles gas in the system:
POV
N :=-R-T
latmx(4xlO 13 ft3)
N= (
3
J
= L025
X
1011 lb mol
0.73 atm·ft_ x534.69R
lbmol.R
Pl-14 (b)
Molar flowrate of CO into L.A. Basin by cars .
F F A -YA T -YA
VA CPo
·Rr
STP
= ~~00ft3
F
T
FA
hr car
= 6 . 685
X
X llbmol
x400000 cars
359 ft3
(See appendix B)
104 lb mollhr
Pl-14 (C)
Wind speed through corridor is v = l.5mph
W = 20 miles
The volumetric flowrate in the conidor is
Vo = v.W.H = (l5x5280)(20x5280)(2000) fe/ill
l3
= L673 x lO fe/ill
1-17
Pl-14 (d)
Molar flowrate of CO into basin from Sant Ana wind
FS:= vOCS
= 1.673 x 1013 ft 3fhr
Fs = 3412 x 103lbmolflu
x2.04xlO- 1O Ibmolfft3
Pl-14 (e)
Rate of emission of CO by cars + Rate of CO by Wind - Rate of removal of CO
=
V dC co
dt
(V=constant, Nco
Pl-14 (0
t= 0 ,
Ceo
= Ccoo
Pl-14 (g)
Time for concentration to reach 8 ppm,
C = 2.04 X 10-8 ~bmol C, = 3..:94 X 10-8 ~~mol
coo
ft3 ' co
4
ft3
From (f),
_ V In (FA +Fs--vo.CcooJ
t--VO
FA + Fs --VoCca
=
4ft 3
--In
1.673xlO13
t=
ft'
hr
6.92 hr
Pl-14 (h)
(1)
Ceo = 2.00E-1O IbmoUft3
tl = 72lus
a = 3.50E+041bmoUln
v0
b
to
0
L67E+ 12 fe flu
= 3,OOE+04 IbmoUln
1-18
dN
=-
co
dt
= Cco V )
Fs
a + b
=
v = 4.0E+13
341.231bmol/hr
Sin( ~
JZ"
)
ft 3
dCco
= v
+ Fs
dt
Now solving this equation using POLYMATH we get plot between Ceo vs
See Polymath program Pl-14···h·-l.pol.
POLYMATH Results
Calculated values of the DEQ variahles
Var'iable
t
C
vO
a
b
F
V
initial value
0
2.0E-10
1.67E+12
3.5E+04
3 . 0E+04
341..23
4.0E+13
minimal value
maximal value
0
2 . 0E-10
1 . 67E+12
3.5E+04
3 . 0E+04
341 . 23
4 . 0E+13
final value
72
72
2.134E-08
1 . 67E+12
3.5E+04
3 . 0E+04
341..23
4.0E+13
1.877E-08
1.67E+12
3.5E+04
3.0E+04
341.23
4.0E+13
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(C)/d(t) = (a+b*sin(3 . 14*t/6)+F-vO*C)/V
Explicit equations as entered by the user
[1] vO=1 . 67*10A12
[2] a = 35000
[3] b = 30000
[4] F = 341 . 23
[5] V = 4*10A13
3.0e-8,---------------
24e··8
80
(2)
tf= 48lus
F,=O
a + bSin(
JZ"~)
Now solving this equation using POLYMATH we get plot between Ceo vs t
1·19
v dCco
dt
See Polymath program P 1-14-h--2.pol.
POL YMA TH Results
Calculated values of the DEQ variables
Variable
t
C
vO
a
b
V
initial value
0
2 . 0E-I0
1 . 67E+12
3 . 5E+04
3 . 0E+04
4.0E+13
minimal value
0
2 . 0E-I0
1 . 67E+12
3 . 5E+04
3 . 0E+04
4 . 0E+13
maximal value
48
1 . 904E-08
1 . 67E+12
3 . 5E+04
3 . 0E+04
4 . 0E+13
final value
48
1 . 693E-08
1.67E+12
3 . 5E+04
3.0E+04
4.0E+13
ODE Report (RKF45)
Differential equations as entered by the user
t 1] d(C)/d(t) = (a+b*sin(3 . 14*t/6)-vO*C)/V
Explicit equations as entered by the user
[1] vO = 1.67*10A12
[2] a = 35000
[3] b = 30000
[4] V = 4*10A13
2 Oe-8
r-------- .----------,
50
(3)
Changing a
-+ Increasing 'a' reduces the amplitude of ripples in graph
It reduces the effect of the
sine function by adding to the baseline.
-+ The amplitude of ripples is directly proportional to 'b' .
Changing b
As b decreases amplitude decreases and graph becomes smooth.
Changing Vo
-+ As the value of Vo is increased the graph changes to a "shifted
sin-curve" And as Vo is decreased graph changes to a smooth
increasing curve .
------------------ - - - - - - - - - - -
PI-IS (a)
- rA =
k with k = OD5 mol/h dm3
CSTR: The general equation is
1-20
v = FAO -FA
-rA
3
Here C A= D"DIC Ao , Vo = 10 dm /min, FA = 5 . 0. mollhr
3
Also we know that FA = CAVO and F Ao = CAOVo, CAO = FAoI Vo = 0..5 mol/dm
Substituting the values in the above equation we get,
V =
CAOVO -CAvo
= (0.5)10-0.01(0.5)10
k
0.05
-7 V=99dm3
PFR: The general equation is
dFA
dCAv O
--=r
=-·k
A =k,NowFA=CAvoandFAo=CAovo=>
dV
dV
Integrating the above equation we get
V
CA
k
V
V
-~ JdC A = JdV
=>
0.
C AO
V =--.!l(CAG ,,- C A )
k
Hence V = 99 dm3
Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of
concentration.
PI-IS (b)
- rA = kC A with k = 0.,,0.0.0.1 s"!
CSTR:
We have already derived that
V
= CAOVO -CAvo
- rA
_ vo C Ao (1-0.01)
kC A
k = D. DDDls"! = 0. . 0.0.0.1 x 360.0. hr"!= 0..36 hI'''!
-7
3
V = (10"dm / hr )(0.5mal / dny3 )(O.~?)
(0.36hr -1 )(0.01 * 0.5mal / dm 3 )
PFR:
From above we already know that for a PFR
dCAv O -rA
-- kCA
dV
Integrating
dC
f _A
=- JdV
V
C
_VG
k
Vo
k
CAO
CA
0.
In C AG =V
CA
Again k = D.DDDls"! = 0..0.0.0.1 x 360.0. hI != 0..36 ru-!
o
1-21
=> V
=2750 dm3
Substituing the values in above equation we get
V = 127.9 dm3
PI-IS (c)
- fA
= kC A2 with k = 3 dm3/moLhr
CSTR:
-rA
Substituting all the values we get
V = (lOdm
(3dm
3
3
/
/
3
hr)(O.5mol / dm )(O.99)
hr)(O.Ol * O.5mol / dm
3
=> V = 66000 dm3
)2
PFR:
dCAv O _- r -kC
2
A
A dV
Integrating
CA
V
--~
k
f
dC
_A
C/o
CAD
=> V
vII
V
=- fdV =>~-(----)=V
=
k CA
3
10dm I hr (
1
3
3dm lmol.hr O.OlCAO
CAO
_~_)
CAO
PI-IS (d)
t-
t
. dN
-- r V
A
AO
WA
Constant Volume V=Vo
t=
fAO
dC A
1A -
rA
Zero order:
t
.999C Ao
= -1 [CAO -O.OOlCAO ] = - - = 9.99h
k
0.05
FiIst Older:
1-22
= 660 dm3
=~ln(CADJ
t
CA
k
=_I_ln(_I_)
=
0.001
.001
6908s
Second order:
t
=k1[1
C
A -
1]
CAD
1
="31[ 0.0005
-
1]
0.5 = 666 h
PI-16 Individualized Solution
PI·17 (a)
Initial number of rabbits, x(O) =500
Initial number of foxes, y(O) = 200
Number of days = 500
dx
dt = k1x-k2 ·xy .................................... (1)
dy
'dt
= k3 XY _. k4 y
.............. .
................ (2)
Given,
kl = 0.02day--1
= 0.00004/(dayx foxes)
k3 = 0.0004/(dayx rabbits)
k4 = 0.04day-1
k2
See Polymath program P 1 . 17 -a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
x
Y
kl
k2
k3
k4
initial value
0
SOO
200
0 . 02
4 . 0E-OS
4.0E-04
0 . 04
minimal value
0
2.9626929
1.128S722
0 . 02
4.0E-OS
4.0E-04
0 . 04
maximal value
SOO
S19.40024
4099.S17
0 . 02
4.0E--OS
4.0E-04
0 . 04
ODE Report (RKF45)
Differential equations as entered by the user
[ 1 J d(x)/d(t) (k1 *x)-(k2*x*y)
[ 2 J d(y)/d(t) = (k3*x*y)-(k4 *y)
=
1-23
final value
SOO
4.2199691
117.62928
0.02
4.0E-OS
4.0E-04
0.04
Explicit equations as entered by the user
[1] k1 = 0 . 02
[2] k2 = 0 . 00004
[.3] k3 = 0.0004
[4] k4=0 . 04
5000.------------------
4000
3000
2000
o
o
200
When, tfinal= 800 and
160
kj
t
300
400
500
= O.00004/(dayxrabbits)
320 t
480
640
800
Plotting rabbits Vs . foxes
1-24
160lli-----------------
o 378
720
1061
rabddPl
1744
2086
PI·I7 (b)
POLYMATII Results
See Polymath program Pl-17-b.pol.
POLYMA TH Results
NLES Solution
Var~able
Value
Ini Guess
f(x)
X------- 2.3850387 --'253E-iI"
3.7970279
y
2
2
1.72E-12
NLES Report (safenewt)
Nonlinear equations
( 1] f(x) = xA3*y-4*yA2+3*x-1 = 0
[ 2 1 f(y)
=6*yJ\2-9*x*y-5 =0
PI-IS (a)
No preheatmg of the benzene feed wtli dlmmlsh the rate of reactIon and thus lesser converSIOns
achIeved .
WIll
PI-IS (b)
An lDterpolation can be done on the logarithnnc scale to find the desued values from the gIven data.
Now we can lDterpolate to the get the cost at 6000 gallons and 15000 gallons
Cost of 6000 gal reactor = 1 905 x lOS $
Cost of 15000 gal reactor = 5 623 X 105 $
Coat va Volume of reactor (log - log plot)
PI-IS (c)
We are given C A IS 0 . 1% of minal concentration
~ CA =OOOlCAO
..
.
1
--~-
--
.,
1-25
"--t---
-
--~-"
1011 volume (1I8110na)
..
••
be
Also from Example 1.3,
Substituting vo= 10 dm3/min and
k = 0.23 min-! we get
v = 300dm 3
which is three times the volume of reactor used in Example 1-3.
PI-18 (d) Safety of Plant
PI-19 Enrico Fermi Problem - no definite solution
PI-20 Enrico Fermi Problem -- no definite solution
PI-21 Individualized solution .
CDPI-A (a)
How many moles of A are in the reactor initially? What is the initial concentration of A?
If we assume ideal gas behavior, then calculating the moles of A initially present in the reactor is quite
simple. We insert our variables into the ideal gas equation:
n=
!.V =
(20~tm)(200dm3)
(lOL3~~a) = 97.5moles
3
(8.3145 kPa.dm )(500)
molK
RT
latm
Knowing the mole fraction of A (y Ao) is 75%, we multiply the total number of moles (N1o) by the YA:
molesA = N Ao
= 0.75x97.5 = 73.1
The initial concentration of A (C Ao) is just the moles of A divided by the volume:
CAo
=
moles
volume
= N Ao =73 .1mol~~ = 0.37 moles / dm 3
V
200dm
CDPI-A(b)
Time (t) for a 1st order reaction to consume 99% of A.
dC
dt
r =_-A
A
OUf
first order rate law is:
-rA =kCA
1-26
mole balance: dCA =--kC
dt
A
-kt =
In(
-k t
fdt
=>
= CfA dCA
CAO
o
C
A
CA ) , knowing CA=O.OI C Ao and our rate constant (k=O.1 min-I), we can solve
CAO
for the time of the reaction: t
= -..!.In(O.OI) =
k
4.61 1
O.lmin-
= 46. 1min
CDPI-A (c)
Time for 2nd order reaction to consume 80% of A and final pressure (P) at T = 127 C.
rate law:
1
1
=>-kt=--+CA CAo
dC
mole balance: :. _2. = --kC~
dt
We can solve for the time in terms of our rate constant (k = 0.7) and our initial
concentration (C Ao):
-kt = _~ +_1_
CAo CAO
t=_4_=
4
----=15.4min.Todetermine
kCAo (0.7dm 3 / mol min)( 0.37moll dm 3 )
the pressure of the reactor following this reaction, we will again use the ideal gas law.
First, we determine the number of moles in the reactor:
NB =Nc =0.8NAo
NT = (0.25)N yo + (0.2+0.8+0.8)N Ao
= 0.25(97.6) + (1.8)(73.2) = 156.1moles
3
(156. Imole) (0.082 dm atm] (500K)
N 1RT
molK
p =_
_ =
_
.___
---''---__ = 32atm
V
200dm 3
CDPI-B
Given:
Liquid phase reaction in a foam reactor, A ~B
Consider a differential element, ~ V of the reactor:
By material balance
FA -(FA +~FA)=-rA(l-e)~V
1-27
Where, (1- e)~ V = fraction of reactor element which is liquid.
or:
-FA
=-rA(1-e)~V
dF
=r
_A
dV
(I-e)
A
Must relate (- rA) to FA '
where, FA is the total (gas +liquid) molar flow rate of A.
-rA=rate of reaction (g mol A per cubic cm. of liquid per sec . ); e
flow rate of A (g mol/sec . ); V = volume of reactor
1-28
= volume fraction of gas; FA = molar
Solutions for Chapter 2 - Conversion and Reactor
Sizing
Synopsis
General: The overall goal of these problems is to help the student realize that if they have -rA=j{X) they
can "design" or size a large number of reaction systems. It sets the stage for the algorithm developed in
Chapter 4"
P2-l.
This problem will keep students thinking about writing down what they learned
every chapter.
P2-2.
This "forces" the students to determine their learning style so they can better use
the resources in the text and on the CDROM and the web.
P2-3.
ICMs have been found to motivate the students learning.
P2-4.
Introduces one of the new concepts of the 4th edition whereby the students
"play" with the example problems before going on to other solutions.
P2-S.
This is a reasonably challenging problem that reinforces Levenspiels plots.
P2-6.
Novel application of Levenspiel plots from an article by Professor Alice Gast at
Massachusetts Institute of Technology in CEE.
P2-7.
Straight forward problem alternative to problems 8, 9, and 12.
P2-S.
To be used in those courses emphasizing bio reaction engineering.
P2-9.
The answer gives ridiculously large reactor volume. The point is to encourage
the student to question their numerical answers.
P2-l0.
Helps the students get a feel of real reactor sizes.
P2-ll.
Great motivating problem. Students remember this problem long after the
course is over.
P2-l2.
Alternative problem to P2-7 and P2-9.
CDP2-A
Similar to 2-9
2-1
CDP2-B
Good problem to get groups started working together (e.g. cooperative
learning.
CDP2-C
Similar to problems 2-8, 2-9, 2-12.
CDP2-D
Similar to problems 2-8,2-9,2-12.
Summary
•
•
•
•
P2-1
P2-2
P2-3
P2-4
P2-.5
P2-6
P2-7
P2-8
P2·-9
P2-10
P2-11
P2-12
CDP2-A
CDP2-B
CDP2-C
CDP2-D
Assigned
0
A
A
0
0
S
AA
S
AA
S
AA
AA
0
0
0
0
Alternates
8,9,12
7,9,12
7,8,9
9,B,C,D
9,B,C,D
9,B,C,D
9,B,C,D
Difficulty
M
M
FSF
FSF
SF
SF
SF
SF
FSF
FSF
FSF
FSF
Time (min)
15
30
30
7.5
75
60
4.5
45
45
15
1
60
5
30
30
45
Assigned
• = Always assigned, AA = Always assign one from the group of alternates,
o =Often, I =Infrequently, S =Seldom, G =Graduate level
Alternates
In problems that have a dot in conjunction with AA means that one of the
problems, either the problem with a dot or anyone of the alternates are always
assigned.
Approximate time in minutes it would take a B/B+ student to solve the problem.
Difficulty
SF =Straight forward reinforcement of principles (plug and chug)
2-2
FSF
= Fairly
straight forward (requires some manipulation of equations or an
intermediate calculation).
IC = Intermediate calculation required
M = More difficult
OE =Some parts open-ended.
*Note the letter problems are found on the CD-ROM. For example A == CDPI-A.
Summary Table Ch-2
Straight forward
1,2,3,4,10
Fairly straight forward
7,9,12
More difficult
5,6,8
Open-ended
6
Comprehensive
4,5,6,7,8,9,12
Critical thinking
P2-9
-
P2-l Individualized solution .
P2-2 Individualized solution.
P2-3 Solution is in the decoding algorithm given with the modules .
P2-4 (a) Example 2-1 through 2-3
If flow rate FAD is cut in half.
VI = v/2, F I = FAd2 and CAD will remain same .
Therefore, volume of CSTR in example 2-3,
F;.X 1 FAOX 1
VI =----=---=-6.4=3.2
- rA
2 -r A
2
If the flow rate is doubled,
F2 = 2FAD and CAD will remain same,
Volume ofCSTR in example 2-3,
V 2 = F2X1-rA = 12 . 8 m3
P2-4 (b) Example 2-5
2-3
--
Levenspiel Plot
Faol2
45-.------------------------------------4 ~;==;==;==;===~
35~~~
J--O+X overall
'"';- 25
o
'"
LL
2
15
05
o+-~~r=~~~~~T=~-=~--~
Faol2
o
02
04
06
08
Conversion
Now, F AO = 0.,4/2 = 0...2 molls,
F
New Table: Divide each term ~ in Table 2-3 by 2.
cx:= ---.--/
-- 'A
0.-·10.·-1- . - - -0..2--'--,
,
0-.4- . - - ,0.-.6
0..445 TI545
0..665
1.0.25
1.77
~-rAJ(m3)
Reactor 2
3
V 2 = 1.2 m
Reactor 1
VI = 0.. 82m3
V = (FAol-IA)X
0.82 =
(_~~o J (Xl)
A
XI
By trial and error we get:
and
X 2 = 0. . 8
Overall conversion XOver.1I = (l/2)XI + (l/2)X2 = (0..546+0.._8)/2 = 0..673
XI
= 0..546
P2-4 (C) Example 2-6
Now, F AO = 0..4/2 = 0..2 molls,
Fao/2
X1
....--+X overall
Faol2
2-4
_Jj)4J.~J
_ ,/ 0.---,,-.7
~.53
k.-J
F
New Table: Divide each term ~ in Table 2-3 by 2 .
-rA
0.1
0.6
1.77
0.4
1.025
0.2
0.665
3
VI = 0.551m
V; = FAO
x dX
I--
o -rA
Plot FAof-rA versus conversion. Estimate outlet conversions by computing the integral of the plotted
function.
Levenspiel Plot
45
4
3.5
...ca•
"0
ca
LL.
3
25
2
1.5
1
0.5
0
0. 2
0
06
0.4
0.8
Conversion
[ XI = 0603 f~r VI = 0.551m
"---X;-=-O. 89 for Vz = L614m3
Overall conversion Xo= (l/2)XI + (1I2)Xz = (0 . 603+0.89)/2 = 0.746
3
-
._----
-.~
_J
Levenspiel Plot
P2-4 (d) Example 2-7
25
(1)
2
ForPFR,
~ 15
c5
'"
IL
1
05
= 0.222m
3
0
0
0.1
02
For first CSTR,
03
04
Convelsion
2-5
05
06
07
F
X 2 = 0..6, _AO.
-rA
_
3
-1.32m ,
For second CSTR,
F
3
X3 = 0.65, -AO
- = 2.0m ,
-rA
V 3-
FAO (X 3 -X 2 )
.
-rA
= 01. m3
(2)
First CSTR remains unchanged
ForPFR:
V=
05(F } X'
f.
_AO
02
rA
Using the Levenspiel Plot
Levenspiel Plot
25
2+-----------------~---
*
15+----------------------
VPPR = 0..22
"0
.f
ForCSTR,
1
o5
i=====~=iO;;;;;:~------~
+-------~----
o +---,---'-'-'--o
01
02
0.3
04
0.5
06
07
Conversion
(3) The worst arrangement is to put the PFR
first, followed by the larger CSTR and finally the smaller CSTR.
'[J_O~23J)-IIT53 ~
-r)- (
Conversion
Original Reactor Volumes
~
Worst Ana"itgement_,,___
Xl = 0..20
VI = 0.188 (CSTR)
VI = 0..23 (PFR)
X2 = 0..60
V2 = 0.,38 (PFR)
V2 = 0.53 (CSTR)
LX:.::.::..3_=-=0..:.:.6:.::.5_ _ _ _ _ _ _ _V.3 = 0.10 (CSTR) _____ y3 = 0.10 (CSTR)
ForPFR,
Xl = 0..2
V. =
f( ~~J1X
Using trapezoidal rule,
Xo = 0,1, Xl = 0.1
2-6
~ = (X~:D)[f(XD)+f(Xl)J
= 0.2 [1.28+0.98]m 3
2
=0.23m 3
ForCSTR,
FAD
FAD
V2 = -(
X2 - )
Xl = 1.32(0.6- 0.2) = 0.53 ill3
For X 2 = 0 . 6, - - = 1.32m ,
3
-rA
-rA
nd
For 2 CSTR,
FAD
-rA
3
For X3 = 0.65, - - = 2m ,
P2-4 (e) Example 2-8
T
Vo
= 5hrs
= 1dm3/min = 60dm3/hr CA = 2.5 mol/dm3X = 0.8
For CSTR,
T
V
= .-
Vo
V= 300dm3
(1)
--r A= CAOX = 2.5xO.8 moll dm 3 hr
T
5
3
= OAmol / dm hr
(2) V = 300dm3
(3)C A = CAO(1-X)
= 0.5 mol/dm3
P2-5
m3 )
0
0.1
0..2
0.8
9
1.0
8
13
3
-
OA
0.6
0..7
O.
2. 0
5
3.5
5.0
6
8
8..
0 ._.
4
Levenspiel plot
_
..............- .... _.. ... _._......_..- .... _"
P2-5 (a) Two CSTRs in series
For first CSTR,
V = (FAd-rAXl) XI
=>Xl =OA4
For second CSTR,
V = (FAd-rAX2) (X2 ·- XI)
=> X2 = 0.67
~5~----------·-------------------~L----.~
g
~4~--------------------
01
02
03
05
04
x
06
07
08
09
P2-5 (b)
Two PFRs in series
4
v~ 1( ~~:}x +)( ~~:}X
By extrapolating and solving, we get
Xl = 0..50.
X 2 = 0.74
P2-5 (C)
3
----f--------~
+------------
F"o
-rA 2
(m')
1+-----------~~----------------~
.....-
Two CSTRs in parallel with the feed, F AO,
divided equally between two reactors, FANEW/IAXI = o.5FAd-IAXI
V = (o..5FAd-rAXI) Xl
Solving we get, X out =0.,,60.
----_ ..
--_.._....
..
o +------,-----.------.------.----~
0,,8
o
0,2
0,4
0.6
1
X
P2-5 (d)
Two PFRs in parallel with the feed equally divided between the two reactors,
FANEW/-rAXI = o.5FAOI-rAXI
By extrapolating and solving as part (b), we get
X out = 0..74
P2-5 (e)
A CSTR and a PFR are in parallel with flow equally divided
Since the flow is divided equally between the two reactors, the overall conversion is the average of the
CSTR conversion (part C) and the PFR conversion (part D)
Xo = (0.,60. + 0.,74) 12 = 0.,67
P2-5 (0
A PFR followed by a CSTR,
X PFR = 0...50.
(using palt(b»
V = (FAd-rA-XCSTR) (XcSTR - X PFR )
Solving we get, X CSTR = 0.,70.
P2-5 (g)
A CSTR followed by a PFR,
X CS1R = 0.,44 (using part(a»
2-8
V
=
X pFR
f
F
-.MLdX
X CSTR -rA
XPFR = 0. . 72
By extrapolating and solving, we get
P2-5 (h)
A 1 m3 PFR followed by two 0.5 m3 CSTRs,
ForPFR,
XPFR = 0..50.
(using part(b»
3
CSTR 1: V = (FAd-rA-XCSTR) (X CSTR - XPFR) = 0..5 m
XCSTR = 0..63
3
CSTR2: V = (FAd-rA-XCSTR2) (XCSTR2 - XCSTRI) = 0..5 m
X CSTR2 =0..72
P2-6 (a) Individualized Solution
P2-6 (b)
1) In order to find the age of the baby hippo, we need to know the volume of the stomach.
The metabolic rate,-rA, is the same for mother and baby, so if the baby hippo eats one half of what the
mother eats then Fao (baby) = Y2 Fao (mother).
The Levenspiel Plot is drawn for the baby hippo below .
Autocatalytic Reaction
5
-
.•....................................... __ ..._•....•.•. .....••.•............................,
4~------------------~--4
35~r----------------~----4
i
e
-.l.
o
~
3 -l-\---------------+------1-'-------:-:--::-~
2 5 t-~~--------------j~----I
2t\-~~---------~~-.I_~r~--~-~
1.5 t=~==~S;;;;;;;;o.."","'--~_r-----;
05
02
04
06
08
Conversion
2-9
Since the volume of the stomach is proportional to the age of the baby hippo, and the volume of the baby's
stomach is half of an adult, then the baby hippo is half the age of a full grown hippo.
Age =
4 ..5 years
2
= 2 .25. years
2) IfVmax and mao are both one halfofthe mother's then
Catalytic Reaction
and since
2
then
1
-vrnaxCA
..=2:....-.._ _
-rAM2
KM
bllby
+C
A
mAo
(
-rAM2
J
baby
1
=- - r
.
2
o
AM2mothef
1
'-m
2 Ao
----1
- 2 rAM2
Conversion
mAo
=
JS= 0.4
(
-
rAM
~
J
mother
mother
m
-rAM2
~ will be identical for both the baby and mother
Assuming that like the stomach the intestine volume is proportional to age then the volume of the intestine
would be 0.75 m3 and the final conversion would be 040
P2-6 (C)
VSlomach = 0.2 m
3
From the web module we see that if a polynomial is fit to the autocatalytic reaction we get:
m
- - AO
= 127X4 _ 172.J6X3 + 100.l8X2 - 28354X + 4.499
-rAMI
2·10
·
r.n AO
And smce Vstomach = - - - X,
5
solve V= 127X
Xstomach = . 067.
-
-rAMI
4
2
172.36X + 1OG.18X3 - 28..354X + 4 . 499X = 0.2 m
3
For the intestine:
The Levenspiel plot for the intestine is shown below. The outlet conversion is 0 . 178
Since the hippo needs 30% conversion to survive but only achieves 17.8%, the hippo cannot survive .
Autocatalytic Reaction
Catalyti c Reaction
5
45
+--------
4+\---35~--------------+_
i
----+---·l
3-··
n:t
!!: 25
fil
E 2
-f-..'~-----------j'-------i
2
1.5
1-·······----
I
t - - - - - - - . - - - -.. - - -
05
o
-----,-----r-------.,..------/
o
X~0067
02
04
06
o
08
Conversion
X=0178
Con v er·si on
P2-6 (d)
PFR~ CSTR
PFR:
Outlet conversion of PFR = 0 . 111
Autocatalytic Reaction
Catalytic Readion
5
45
4
3.5
,...
..
:E
ca
2
-.!. 2.5
0
ca
E
2
1.5
05
o
X= 0.111
Conversion
0
0
0.2
0.4
Conversion
CSTR:
2-11
0.6
08
We must solve
v = 0..46 = (X-QJ 1l)(127X4 -
172.36X3 + 100. J 8X2 - 28,354X + 4.499)
X=O,42
Since the hippo gets a conversion over 30.% it will survive"
P2-7
Exothermic reaction: A ~ B + C
3
r(mol/dm ,min
X
1/-r(dm 3.min/mol)
1
0.,,6
0..2
0.,,2
0..2
0..2
0. . 8
1.1
)
0.
0. . 20.
0..40.
0..45
0..50.
0..60.
0..80.
0..90.
1
1.67
5
5
5
5
1 . 25
0..91
P2-7 (a)
To solve this problem, first plot 1/-rA vs" X hom the chart above. Second, use mole balance as given below,
CSTR:
Mole balance:
= FAo
V
CSTR
!
= (300mollmin)(0.4) =>
3
(Smol / dm .min)
- 'A
=>V CSTR = 24 dm 3
PFR:
x dX
VpFR = F AO
Mole balance:
f--r
0
A
0.4
= 3QQ(area under the curve)
0.6
X
V PFR = 72 dm3
P2-7 (b)
2-12
For a feed stream that enters the reaction with a previous conversion of DAD and leaves at any conversion
up to 0. . 60., the volumes of the PFR and CSTR will be identical because of the rate is constant over this
conversion range.
0040
CSTR
0.60
P2-' (C)
3
V CSTR = 105 dm
Mole balance:
FAOX
VCSlR = - -rA
x
._. fA
105dm 3
- - - . = O.35dm 3 mini mol
300mollmin
Use trial and error to find maximum conversion..
At X= 0..70.,
1/-fA = 0..5, and X/-rA = 0..35 dm3 . minlmol
Maximum conversion = 0..70.
.
P2-' (d)
From part (a) we know that Xl = DAn.
".
1
Xl
PFR
II
1
Use trial and error to find X 2 .
X2
CSTR
2-13
--
.....
-
Reananging, we get
X 2 - 0040
- rAlx2
At X 2 = 0.64,
= ~ = 0.008
FAD
X 2 -OAO
- rA
I
= 0.008
X
2
Conversion = 0.64
P2-7 (e)
From part (a), we know that Xl = OAO. Use trial and error to find X2
Mole balance: VPFR
= 72 = FAO
dX
f --.r = 300 f-r
X2
dX
040 -
A
X2
040 -
At X 2 = 0 . 908, V = 300 x (area under the curve)
=> V = 300(024) = 72dm3
Conversion = 0..908 .
P2-7 (£)
See Polymath program P2·7fpoL
2-14
A
-lIrA
0.65
6.0
5. 0
052
Q
0.39·
GJ
·to
3.0
0.26
2.0
0.13
1.0
0.00
0
20
40
V
60
0. 0
100
80
0
17
33
50
V
67
83
100
P2-8 (a)
FsoX
V=---rs
Fso = 1000 glhr
1
At a conversion of 40% - -
-rs
Therefore
V
3
dm hr
= 0.15··--g
= (0.15)(1000)(0.40) = 60dm 3
P2-8 (b)
1
_. rs
At a conversion of 80%, - -
3
dm hr
= 0.8----g
Fso = 1000 glhr
Therefore
V
= (0.8)(1000)(0.80) = 640dm 3
P2-8 (C)
x dX
VPFR = Fso
f--
o -rs
From the plot of 1/-rs Calculate the area under the curve such that the area is equal to VIPso = 80/ 1000 =
0..08
X= 12%
For the 80 dm3 CSTR,
F X
V = 80 dm 3 =~­
-f
s
XI-rs = 0.08.. From guess and check we get X = 55%
2-15
P2-8 (d)
To achieve 80% conversion with a CSTR followed by a CSTR, the optimum arrangement is to have a
CSTR with a volume to achieve a conversion of about 45%, or the conversion that corresponds to the
minimum value of 1I-rs Next is a PFR with the necessary volume to achieve the 80% conversion following
the CSTR. This arrangement has the smallest reactor volume to achieve 80% conversion
For two CSTR's in series, the optimum arrangement would still include a CSTR with the volume to achieve
a conversion of about 45%, or the conversion that cOllesponds to the minimum value of 1I-rs, first. A
second CSTR with a volume sufficient to reach 80% would follow the first CSTR
_ kC s (O.l[C so - C s ]+ 0.001)
-r
l
KM
+ Cs
Let us first consider when Cs is small
Cso is a constant and if we group together the constants and simplify
since Cs < KM
1
-
KM
klC~ + k 2 Cs
r,
which is consistent with the shape of the graph when X is large (if Cs is small X is
large and as Cs grows X decreases),
Now consider when Cs is large (X is small)
As Cs gets larger Cc appwaches 0:
kCsCc
1
then - KM + Cs
r,
As Cs grows large!, Cs » KM
=
If - r
l
KM + C s
= ---''-'---
kCsC c
2-16
And since Cc is becoming very small and approaching 0 at X = 0, 1/-1s should be increasing with Cs (or
decreasing X)., This is what is observed at small values of X_ At intermediate levels of Cs and X, these
driving forces are competing and why the curve of 1/-rs has a minimum.
P2-9
Irreversible gas phase reaction
2A+B ~2C
See Polymath program P2"-9.poJ.
P2-9 (a)
PFR volume necessary to achieve 50% conversion
Mole Balance
V=FAO
X2
dX
Xl
(-rA )
f---
Volume = Geometric area under the curve of (FAoI-rA)
vsX)
V
V
= GX400000XO.5) + (100000 X0.5)
= 150000 m3
x
P2-9 (b)
CSTR Volume to achieve 50% conversion
Mole Balance
V=!AO~_
(-rA )
V = 0.5 X 100000
V = 50000m3
-I- -+"o.s
0 <1 10
x
P2-9 (c)
Volume of second CSTR added in series to achieve 80%
conversion
__l!.AO(X2- Xl).
V2(-rA )
V2
= 500000 X (0.8"- 0.5)
V2 = 150000m3
x
2-17
P2-9 (d)
Volume of PFR added in series to first CSTR to achieve
80% conversion
1
VPFR = (-x 400000 x 0.3) + (100000 X 0.3)
2
VPFR = 90000m3
P2-9 (e)
ForCSTR,
VI = 60000 m3 (CSTR)
Mole Balance
V=
x
FAOX
(-rA )
x == 0.515
ForPFR,
V2 = 60000 m3
Mole balance
V=FAO
X2
dX
Xl
(-rA )
f---
nd
Eqn . Of 2 line (sloping upwards)
y -100000 =1.3 x10 6 (x - 0.5)
X2
60000 = f (l.3x10 6 (x -- 0.51) + 100000)dX
x
Xl
=> X2 = 0 ..746
P2-9 (0
Real rates would not give that shape. The reactor volumes are absurdly large .
P2-10
Problem 2-10 involves estimating the volume of three reactors from a picture., The door on the side of the
building was used as a reference., It was assumed to be 8 ft high.
The following estimates were made:
h = 56ft
d = 9 ft
V = nr 2h = 11:(45 fti(56 ft) = 3562 ft' = 100,865 L
2·18
Length of one segment = 23 ft
Length of entire reactor = (23 ft)(12)(II) = 3036 ft
D=lft
V = m 2h = n(O.5 ft)2(3036 ft) = 2384 ft3 = 67,507 L
Answers will vary slightly for each individuaL
P2-11 No solution necessary.
P2-12 (a)
The smallest amount of catalyst necessary to achieve 80 % conversion in a CSTR and PBR connected in
series and containing equal amounts of catalyst can be calculated from the figure below.
CSTR
PBR
Conversion~
X
The lightly shaded area on the left denotes the CSTR while the darker shaded area denotes the PBR This
figure shows that the smallest amount of catalyst is used when the CSTR is upstream of the PBR
See Polymath program P2-12.pol.
P2-12 (b)
Calculate the necessary amount of catalyst to reach 80 % conversion using a single CSTR by determining
the area of the shaded region in the figure below .
2-19
60
2
,4
6
Conversion, X
8
10
The area of the rectangle is approximately 232 kg of catalyst.
P2-12 (C)
The CSTR catalyst weight necessary to achieve 40 % conversion can be obtained by calculating the area of
the shaded rectangle shown in the figure below.
60
10
_____ ~ __-LI____~_____
4
10
6
8
~
2
Conversion, X
The area of the rectangle is approximately 7,6 kg of catalyst
P2-12 (d)
The catalyst weight necessary to achieve 80 % conversion in a PBR is found by calculating the area of the
shaded region in the figure below,
2-20
6
Conversion, X
The necessary catalyst weight is approximately 22 kg .
P2-12 (e)
The amount of catalyst necessary to achieve 40 % conversion in a single PBR can be found from calculating
the area of the shaded region in the graph below.
60
f
1;;
40
If
<;J,-1...~30
u.
I
20
10
2
.4
.8
6
Conversion. X
1.0
The necessary catalyst weight is approximately 13 kg.
P2-12 (1)
2-21
1.0
~
~
0. 8
0.6
04
0.2
0.0
0. 0
6.-1
12..8 \V 19. 2
25 . 6
32.0
P2-12 (g)
For different (-rA) vs . (X) curves, reactOIs should be arranged so that the smallest amount of catalyst is
needed to give the maximum conversion.. One useful heuristic is that for curves with a negative slope, it is
generally better to use a CSTR Similarly, when a curve has a positive slope, it is generally better to use a
PBR
CDP2-A(a)
Over what range of conversions are the plug-flow reactor and CSTR volumes identical?
We first plot the inverse of the reaction rate versus conversion.
10"
r
8
A
/
/'
/
/
/
o ,. -.........-..-.---.-,.-.. . -.. --.-------,--.. .--.--.-----,-.. .-.-.. ----..-.. . .,.----.---,-.---..
o
0.2
OA
,)_6
0 .. 8
1.0
Conversion X
Mole balance equations for a CSTR and a PFR:
CSTR: V = FAOX
-- r A
PFR: V=
x dX
S-
0-- r A
Until the conversion eX) reaches 0.5, the reaction rate is independent of
conversion and the reactor volumes will be identical.
2-22
. vPFR = 05f -dX- = F-AO- 05fdX --
1.e.
o - rA
-
rA
0
F X
AO
-
rA
- VCSTR
'-
CDP2-A(b)
What conversion will be achieved in a CSTR that has a volume of 90 L?
For now, we will assume that conversion (X) will be less that 0.5.
CSTR mole balance:
V = !AO X = voC AO X
- rA
- rA
V
X =---=.
V C
3
o AO 5 m
- rA
S
0.09 m 3
=3xlO-13
l
3
X 200 mo X 3 X 108 !'1 .~
m3
mol
CDP2-A (c)
This problem will be divided into two parts, as seen below:
8
l()~
-'A
(m3 •• '\
,-
, ---··--1
! mot 1
)
0.2
{I.4
0,6
1.0
Conven,ion X
•
The PFR volume required in reaching X=O.5 (reaction rate is independent of conversion).
•
The PFR volume required to go from X=O..5 to X=o.,7 (reaction rate depends on convelsion).,
2·23
lO
= 8*10 m
3
Finally, we add V 2to VI and get:
Vtot =VI + V2
=2.3 xlO 11 m 3
CDP2-A (d)
What CSTR reactor volume is required if effluent from the plug-flow reactor in part (c) is fed to a CSTR to
10'
r
l'
1113
1/
/\
~,~
ruol
f
2
I)
C;onversion X
raise the conversion to 90 %
We notice that the new inverse of the reaction rate (I/-rA) is 7*108. We insert this new
value into our CSTR mole balance equation:
CDP2-A (e)
If the reaction is canied out in a constant-pressure batch reactor in which pure A is fed to
the reactor, what length of time is necessary to achieve 40% conversion?
Since there is no flow into or out of the system, mole balance can be written as:
_dNA
Mole Balance: rA V
- --
Stoichiometry: N A
= N AD (1 -- X)
dt
2-24
dX
Combine: r AV = N AO - -
dt
=
From the stoichiometry of the reaction we know that V Vo( 1+eX) and e is 1. We insert
this into our mole balance equation and solve for time (t):
Vo
dX
N AO
dt
-rA -(1+X)=-tSdt - C XS __d_X__
o
-
AO 0 -
rA (1 + X)
After integration, we have:
1
t =--CAoln(l + X)
- rA
Inserting the values for our variables:
t = 2.02 X 1010 s
That is 640 year'S.
CDP2·A (f)
Plot the rate of reaction and conversion as a function of PFR volume.
The following graph plots the reaction rate (-rA) versus the PFR volume:
Ii' ~..
~"~"".'"
"''''''''''' " '" ""''''''''''''''' '"'''''' '
il
Reaction Rate (--rA) ver'sus Reactor' Volume (V)
4 . 0*10-
'I
II-'A 3O'10_9"--"--------~
11[~o:s)
2 .. 0*10-9_
------
-
.'-..._.•. ......_,"'_.","",.
1 . 0*10-9
o
o
- - - - t - - -___----,>----_,--_,
1 . 0*10"
2 . 0*10"
3 . 0*10"
4 .. 0*10"
5 . 0*1011
V (m 3 )
Below is a plot of conversion versus the PFR volume. Notice how the relation is
linear until the conversion exceeds 50%.
2·25
Conversion (X) versus Reactor Volume (V)
1 .. 0
0 .. 8
x
0 .. 6
0 .. 4
0. 2
0--1<----_--;,----..-,-----t,----o,
o
1 . 0*10"
2.0*10"
3.0*10"
4 .. 0*10"
5 .. 0*1011
V (m 3 )
The volume required for 99% conversion exceeds 4*1011 m 3 .
CDP2-A (g)
Critique the answers to this problem.
The rate of Ieaction for this problem is extremely small, and the flow rate is quite large. To obtain the
desired conversion, it would require a reactm of geological proportions (a CSTR or PFR approximately the
size of the Los Angeles Basin), or as we saw in the case of the batch reactor, a very long time .
CDP2-B Individualized solution
CDP2-C (a)
o
0.1 O~ a.3 0.4 ~ e.G 0.1 0.1 o.~
COfivtrll01l. X
!a.}
For an intermediate conversion of 03, Figure below shows that a PFR yields the smallest volume, since fOI
the PFR we use the area under the curve . A minimum volume is also achieved by following the PFR with a
CSTR In this case the area consideIed would be the rectangle bounded by X =0.3 and X = 0 . 7 with a height
equal to the CAr/-rA value at X = 0.7, which is less than the area under the curve .
CDP2-C (b)
2-26
so t/min
v ... Vo I. 'Where
v o ....
,0 .3
I
,.
-C
J_
+ <0.1-0.3) -
o
AO
-%A
*" <O.l - 0)(10)
-+
1
2'
(0
X-O.1
.3-<D (,50-10) + <0 .. 1-0.3)
(15)
"" IS :::c.in
So V .. v .. 1:
III
(SO l.J:ad.n) (15 lJdn) .. 75,0 1
III
750
=.t.o
CDP2-C (C)
The $~allo$t ~:O& caa be achieved by asi~, ouly Qua CSTR with this systom.
C"I'I
I ... (O.1--{).Ol....u::.;.
* (0.7-0)(15) - 10.5
-r).
mi~
'1-0.7
So
V - v
o
I .. (SO) Ihniu (10.5) lAiu -
525 1
We Yo~ld further roduco the tot~l vQl~e by usin& a PFR at first up to the
CDP2-C (d)
2-27
to t11(1
of tlu rcctan.;lc up to tho specified c;ouvcr:Jion.
&::C&
I '" (O.45-0)
V ~ v
So
o
eM
-r
(0.45-0)(37) - 16.55
~in
A :X ..O.1
I . (50 I/min}{16.65 min.)
832.5 1
u
For thePFR,
o. 4S
I ..
fo
4(48) ... 2(43) + 31)
..
So V
..9.....Sli
3
(10 ... 4(1.5 ... 3S ... 43 ... 48) ... 2(20 + 43 ... 50 + 43) ... ]1) ... :1.5. n
v I -
p
o
There is also
(50 I/min} {15.12 miu) - 186 1
~
solution at au X>O.1
Tr:r I. .. O.S
For the PFR
+ 4( 15) .... 33)
""
For
.!hl.
3
tl:a
UO ... 4'\ 20 +- SO +- 3Z .... 1.3) ....
~H43 .... 43 .... 1.1} ... 33) ...
CSU
I • (O.19)(30) - 13.7 min
So V ...
'V
c
t ... (SO
1/;:dn)(23.7 min) ... 11S! 1 Cdm~)
2·28
Z3.9
For t:.h.a PFR.
I - 23.9 -
~
... (O.S-Q.79)(30+33) - 23.9 - 0.315 - 23.58 mi:
So V .. v I '" (50 I/mill)(23.SS mal -1179 1
o
o.~
diffc:rcu:u:o
CDP2-C (e)
x
CA0/-:r A{J:).inJ
-;(minJ
0.0
0.1
0,,2
0.3
0.4
0.5
0 .. 6
0.1
0.8
10
20
43
SO
43
32
11
15
33
0.0
l.O
S .. t)
15.0
17.2-
16.0
10,.2
10 .. 5
2,6.4
3CI
1$
20
o
0.1
02
0.4
X
v
"= . - , .
7'00 1.
50 If::r.i.a. "" 14 m.in.
2-29
0.7
CDP2-D
Data taken at WI.;; kPa (10 atrn) and 227"C (500.2 K)
y/\O
(0.333XlO) ==.
00811"-)'d-'
c~ ; ; ; .y,,(}p
__ ...-;;:;; ---------,J gInO I m
=0333
Ao
RT
.. -...
100000
~
(0,082X500.2)
~"'--"""----'-
.
-l
~,-- -------.~-~-.
!
60000
I
40000
... -···········-·······----1
I
I
·-----·--···------·-·-··-~i
zoooC)
t
() -~---.- .... -.- ......- ... --... ---.~..-----..-.-.- ....-•... -.---.-...
o
0.2
x
l
--.-.
0,6
0.4
CDP2-D (a)
30% conversion in PFR:
,
-C pfR ;:;:
C AO
0.3
cLX
5--....... :;: ;: 4,664 . 84
s
V --
=?
f
~64 .84 s,
V o "£ -- (4
\' 0
0'-' fA
1
min }2 m '/mm
.)
····6······.:·-.·--
, Os
- - m3
= L5 ).)
CDP2-D (b)
30 to 50% conversion in CSTR:
c
(X,·-·X)
1:CS'IR ;;;:; .... _Au, .. : ............... !...:
.... IAl
:;::;: 12,169.2 s
=?
"1
Y
CSIR ::::::
CDP2-D (C)
2-30
1
(12 ,169
min ]2 In 3/..)
- 64 m 3
. _')S{' -.......
IIlln =40 3.
60s
f
Total Volume:
VTouu == 155.5+405.6 == 561.1 m
J
CDP2-D (d)
60% conversion in PFR:
1:PFR ::-:.
C At>
J."~X.:;;;; 20.281.9 s
o ""fA
80% conversion in PFR:
fA
is not known for X>O.60 - can not do.
CDP2-D (e)
50 % in CSTR:
't' :::;
X
C Ao ~- :::; 30,422.9 s
- fA.
V:::; v 0"' ;;;; (30,422.9 s {
\
I
~~:~ Y.2 m>Imin)-;;;:: 1014.1 m 3
DOS
J
CDP2-D (1)
50 to 60% conversion in CSTR:
C.\o(X) -Xl) :::::: 8112.8
.
t -;;;:: --.:...--.....:::."".....- fAl
V = v.' = (8112.8S{To':-}2 m'/min)= 270.4 m'
CDP2-D (g)
2-31
Rate of Reaction vs. VolUl:oe
IOE,05 ,"-""'.'--"---
5 OE
'--1
I
061 .._, ---, --
2 SE 06
O.OEtOO
06
I
I ... --- "--, -,--
7 5E 06
?
.,
Conversion vs. Volurne
0.4
0.2
~
-, - -
L__,_._"~ ____ ,_~_,,,_J
o
400
200
600
~--~-
0,
800
o
200
3
Volnme(m )
400
Volume (m
600
3
)
CDP2-D (h)
Critique
Answers are Valid:
1. Constant Temperature and Pressure
No heat effects
No pressure drop
2. Single interpolation to X .. :;;; 0.15, 030,0,45, and 0.50 allowable
3. tiuge volume (the size of the LA Basin)! Raise T? Raise P?
CDP2-E
For the CSTR :
VI
F X
=: , . ",::'l£, . .L ::;;;;
.,['....
F...o ( Area)
]
Alea == VI == 1200 drri
A ----
From the graph we can see thaI Xl :: 0.,60
4000
For the PFR:
Fio
XI) == FAo (Area und.er curve)
V7. :::: .F"o(Xl ---,"._,
orA
''''''rA
We
2000
1000
J
Area under curve == V, == 600 dm
From the graph
3000
can see that X z ::;;;; 0. . 80
2-32
poiu:::t
800
CDP2-F (a)
Find the conversion for the CSTR and PFR connected in series
·,rA
X
0
0.2
0.1
0.0167
0.00488
0.4
0.7
0.00286
0.9
0.00204
400 L CSTR and 100 L PFR
lI(-rA)
5
59.9
204.9
349.65
490.19
Feed is 41 % A, 41 % B, and 18% L
T :::;:; 2270 C = 500 K
P ;:: 10 attn
10 atm
= ..-P ;;;:;.--..' .,...----.-......
,-----...--",.,.".,--- = 0_244 mollL.
C
RT
' To
(0.082 I.. . atmfmol "K)( 500 K)
CA. ;;:: OAlC,o;:: 0.41(0.244 maUL)::::: 0.1 mol/I..
FAo
:;;; t)OC AO :::::
1 Us(O.l moUL)::::: 0.1 molls::: 6 mol/min
There are tv.'o possible arrangements of the system:
1. CSTR followed by the PFR
2. PFR followed by the CSTR
Case 1: CSTR -.,. PFR
CSTR:
VI = F. . )Area)
V
400
Area =: """,l". = '.... , ,:;;:;: 66.67
FA.,
6
From the graph
PFR :
V1
:;;:;:
~
XI = 0.36
FA., (Area under curve)
V,
100
Area under curve:;: ,-- ;;: -""".,.,:;:; 16.667
F:. .., 6
From the graph .. X~ = 0.445
-
....... '.' ..~' ...... '.~-'- . ~-..-.,...-"."..,.,..-~--........
Case 1:
CSTR ••> PFR
Case 2;
500r-----~-------.,,~_,
•
..
-~ ~-
PFR --> CSTR
500~-------------r~
400
~
.
400
,.: :3 00 ,+.••,..."...",...
.
300
;;::; 200
::::; lQO +." .."-",,.,
100
100
o
0,0 0.2
o
0.4
0.6
X
<.,.------,--."'~""."""
Q,.8
1.0
0.0 0.2 0.4
0.6 0.8
x:
. "". ."""" -"""'' ' -' ' ' ' ' ' -' ' ' ' -'
2-33
1.0
Case 2: PFR --+ CSTR
PFR:
Area under curve:: 1667
From the graph - Xl ::;; 0.259
CSTR. :
/\rea::: 66.67
From the grapb . X 2
::;;
0.515
CDP2-F (b)
Two 400 L CSTR's in series.
CSTRl:
V;;:;: FAQ(A.rea)
mea;;:;; 66.67
From the graph - XI ::: 0.36
CSTR2 :
Area = 66.67
From the graph . X z = 0.595
(b) Two CSTR's in Series
(c) Two CSTR's in Parallel
500 1"'----------:Jr-.
400
500·~·--------------r~
400
. 30G
. 300
~
~
::: 200
::.; 200
100
100
o
o
0,0
0.2
0.4
0.6
0.8
LO
0.0
X
Two 400 L CSTR's in parallel.
To each CSTR goes half of the feed
:::;:;
6/2 ;:::: 3 moUmin
V::: FAo(Area)
V
400
Area ::: .._,-- :: ------ = 13 3...3
FA"
3
From the graph:
0.6
X
CDP2-F (C)
FAo
0 . 2 0.4
X = 0.52
2-34
0 .. 8
1.0
CDP2-F (d)
PFR : V ::: FAo (Area under curve)
From the graph we can find the area under the curve for a conversion of 0.60:
Area == (0 . 60)(300) :.::: 90
2
V = (2 mol1min)(90) = 180 L
~--~.-
-------
..
...................
-.-,-----..
".,.--~
- - -.......-.-.. .. .. ...........
............
~
,
"
(d) Single .PFR
500 T-----------.---.---.--~._,
400
.
.::: 300
-
200
1010
o
0.0
0.2
0.6
OA
0.8
1..0
X
CDP2-F (e)
Pressure reduced by a factor of 10.
A decrease in pressure would cause a decrease in the overall concentration which
would in turn cause a decrease in CAo and F.-\,o" By looking at the design equation:
v::: !.",,,X
"'[A
it is apparent that to compensate for the decrease in F.Aa there would be and increase
in X.
CDP2-F (1)
Use the graph of 1/.,(... VS. X to find values for all volumes. (Assume a flow rate of
1 mol/min.) Generate the following table and graphs:
...... "
X
-
0
0.1
... ,,0.4
- - ; : ,........
"
i----
0. 7
0. 9
-r"
0.2
"·"'0.'01'67
V
0
-;-3 . 494
··0.0048·S-- -"4i984
=+. .
0.00286
0.00204
125.878
225. 088
2-35
1"""'"""" .. """'-'----,.,-'"'-'-'-"'t
!
~
Conversion
(f)
i
""'---~-~~---'--"
-
"'S.
-------,
!
''flolume
(0-1.""
!
_",~ _ _ _ _ _ •
'IS.
Volume
DO! -,--.---'-------~
0&
0.015
0.,6
.; 0,01
0. 4
o
.,.,,_,_.,.,_...,__,-..-J
IL-_, _____0 ,__ , ,_,~
100 V 200
__ "'
CDP2-F (g)
0005
I
0,2.
300
____
o ,..___ ,_..,._..
Q
,J
Individualized solution
2-36
~
100
__,___..J
V
ZOO
300
Solutions for Chapter 3 - Rate Law and Stoichiometry
P3-1 Individualized solution.
P3-2 (a) Example 3-1
0 . 008
0 . 007 - t - - - - - - - - - - - - 0 . 006 - t - - - - - - - - - - - - - - - - - - - - - - - # - - - - j
0 . 005
~
~ 0 . 004 + - - - - - - - - - - - - - - - - - - F
.iOi
0 . 003
0 . 002
+-----------~~-------
0 . 001 - t - - - - - - - - - =.......~---
o
.•....
. T
310
315
.....,.-++.... _................,
"
320
325
330
335
T(K)
For E = 60kJ/moi
k
For E1 = 240kJ/moi
= 1.32 X 1016 expC-- 6~~OJ )
kl = 1.32 X 1016 exp( -
2~~OOJ)
E=60 kj/mol
E = 240 kj/mol
6000000
35E-22
3E-22
5000000
t---+------------i
4000000
25E-22 t---~._---------___j
g
2E-22
:;; 1 5E'22
1E-22
. . ...............................-_ •.•
"'2.
t----\--------------I
t------\-----
t----~-=--
3000000 t - - - - - - - - " . . : : : - : - - - - - - - - j
-'"
2000000 t - - - - - -
t-------'b------1000000
5E-23 t--------"'c--~
t---.----------~-__l
o
o 00295
0003
0 00305 0 0031
000315 00032 000325
000295 0003
0.00305 00031 000315 0.0032 000325
1fT (11K)
1fT (11K)
P3-2 (b) Example 3-2 Yes, water is alJeady considered inert.
3-1
P3-2 (c) Example 3-3
The solution to the example at a conversion of 20% would remain unchanged. For 90 % conversions of the
caustic soda, the final concentration of glyceryl sterate is 0 instead of a negative concentration. Therefore
90 % of caustic soda is possible.
bed
P3-2 (d) Example 3-4 A + - B ~ -c + - D
a
So, the minimum value of @B
a
a
113
= b/a = -- = 0.33
1
P3- 2 (e) Example 3-5
For the concentration of N2 to be constant, the volume of reactor must be constant. V = Vo.
Plot:
1
0.5(1-0.14x)2
-rA
(I-X)(0.54--0.5X)
1/(-ra) vs X
180
160
140
120
f"100
::!:...
80
60
40
20
02
04
08
06
12
X
The rate of reaction decreases drastically with increase in conversion at higher conversions.
P3-2 (0 Example 3-6
For a given conversion, concentration of B is lower in flow reactor than a constant volume batch reactor.
Therefore the reverse reaction decreases .
C ro = constant and inerts are varied.
N 2 0 4 *"72N0 2
A
~
2B
C
2
Equilibrium rate constant is given by: K c =.~
CA,e
Stoichiometry: £ =
YA0 6 = YAO (2 -1) = YAO
Constant volume Batch:
3-2
CB -- 2N AO X --2CAO X
Vo
Plug flow reactor:
= _~AO(l- X) = C Ao(l- X)
C
v
A
o
(1 + eX)
and
B
(l + EX)
Y;o;0o = YAo(O.07176)mol! dm
CAO =
C
=
2FAO X
v o(l + eX)
= 2CAO X
(1 + eX)
3
Combining: For constant volume batch:
C 2
4C 2 X2
-~Ao
.
K cCAe
C AO (1 - X)
4CAO
For flow reactor:
C
4C 2
2
=_._~=
K
C
K C (1 - X e)(1 + eX e)
X2
Ao
CAO (1 - X)(1 + eX)
CAe
4C AO
See Polymath program P3··2-LpoL
POLYMA TH Results
NLES Report (safenewt)
Nonlinear equations
[1] f(Xeb) = Xeb - (kc*(1-Xeb)/(4*Cao»I\() . 5 = 0
[2] f(Xef) = Xef - (kc*(1-Xef)*(1 +eps*Xef)/(4*Cao»A0 . 5 = 0
Explicit equations
yao = 1
kc = 0.1
[3] Cao 0 . 07174*yao
[4] eps yao
[l)
[2]
=
=
1 .....•.••..................
<:
0.9
+-------------------------:.7-~---i
0. 8
+------------------~--------......,-.mIIIJIIl''----------i
.§~ ~;~~-~:i
-~~
-;-;:;~~~~~~_=--
-Batchl
:::J
-FIOW.J
Qi 0 . 6
oE
____
0.7
~
0. 5
~ 04
'3
Jr
0. 3 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0. 2 + - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1
0. 1 - 1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - 1
o
...
o
·······1· .. ·........ ".. ," ... 1 .... · ........ , ... ·.. T.... · ................... ·.. T"·
0. 1
0. 2
0.3
04
.. ... " .. ' ..... "1' ........... " ....... ,•• 'r~ ........ ".' ..' ".. ·.. ·T·~ .. ' ..·.. · " .... n~ ..·T
........ '1.'''...
0. 6
0. 5
y inerts
3-3
0. 7
0. 8
0.,9
Yinert
Xef
Xeb
Yao
0.44
0.458
0.4777
0.5
0.525
0.556
0.5944
0.6435
0.71
0.8112
0.887
0.893
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.05
0.044
0
0.1
0.2
0.3
0.4
- 0.5
0.6
0.7
0.8
0.9
0.95
0.956
0.508
0.5217
0.537
0.5547
0.576
0.601
0.633
0.6743
0.732
0.8212
0.89
0.896
P3-2 (g) No solution will be given
P3-2 (h)
1
1
2
2
A+-B-7-C
Rate law: - r A =
kAC A2C S and
dm
2
3
1
kA = 25- - s ( mol J
~=.~.= rc
-1
-112
112
(dm 3
J2
kc =kB =12.5-1 s mol
P3-2 (i)
A+3B -7 2C
Rate law: -
rA
= k A C A CB
at low temperatures.
At equilibrium,
C
K = ..---- Ce'
C
C I12C
A,e
3/2
B,e
At "Iuil;b,;um, -, A = 0 , so we can sugge" th,t -
rA
= k A ( CA 1/2 CB 112 -
But at t = 0, Cc = 0
So the rate law is not valid at t = 0 .
3-4
~~ J
Next guess:
P3-3 Solution is in the decoding algorithm available separately from the author.
P3-4 (a)
Note: This problem can have many solutions as data fitting can be done in many ways.
Using Arrhenius Equation
For Fire flies:
--- r;:::-_.
T(in
K)
294
298
303
-
11T
0 . 00340
1
0.00335
6
0.00330
0
Flashes/
min
9
In(flash
es/min)
2.197
12.16
2.498
16.2
2 . 785
2.66
2.52
2.38
2.24
Plotting In(flashes/min) vs lIT,
we get a straight line .
See Polymath program P3-4-firetlics.pol.
210L---~·----~----~--~
3.30E-:3 .332E-3 3.34E
h 3.36E-3
1
3.38E-3 3.40E-3
For Crickets:
11T
~2
x103
3.482
chrips/
min
80
In(chirPsi
min)
4 . 382
293 . 3
3.409
126
4.836
300
3.333
200
5.298
T(in K)
5.3
5.1
-49
Plotting In(chups/min) Vs lIT,
we get a straight line .
-+ Both, Fireflies and Crickets data
follow the Arrhenius Model.
In y = A + BIT , and have the same activation energy.
·~.5
See Polymath program P3-4-crickcts.pol.
4..3
3-5
'---~--~--~--
3.HE-3 1.36E-3 3.39Eih 1.42E-3 3.45E-3 3.48E-.3
P3-4 (b)
F or H oney!bee:
T(in K)
1rr
x10 3
3.356
298
3.,300
3,,247
303
rsoa-
20 , - - - - - - - - - - - - - - - - ,
V(cm/s)
In(V)
0.7
-0.357
1.4
1.8
3
0 . 588
1.098
o. s
Plotting In(V) Vs liT, almost straight lme.
In(V) = 44.,6 - 1.33E4/T
At T = 40°C(313K)
V = 6Acmls
At T = -5°C(268K)
V = 0,005cmls(But bee
would not be alive at this temperatme)
See Polymath program P3-4-bces.pol.
0.2
-0.4
-1.0 L-_ _ _ _ _ _ __ _ ___'
3.25E-3 ,'.2'7E-3 3.29E
HIE-, 3.3.3E-3 3.36E-3
~
~
1lr
~
~
P3-4 (c)
For ants'
1rr x1Q"
V(cm/s)
In(V)
3,53
0,,5
-0,,69
293
3,41
303-
3,,30
2
3,4
0 . 69
1,22
6,,5
1.,87
T(in K)
-,---
283
---,,-,------'-
3,,21
311
~--
'------
20
r---------------,
1.4
--
o.S
'------"
0.2
Plotting In(V) Vs liT,
almost straight line,
See Polymath program P3-4-ants.pol.
-10L--~-----So activity of bees, ants, crickets and fireflies follow
3.22E-J .uSE-J J,J,tE1tr 3.41E-J
Arrhenius modeL So activity increases with an
increase in temperatme, Activation energies fOl fireflies and crickets are almost the same,
---:-c-
Insect
Cricket
Firefly
Ant
Honeybee
3.47E-33.53E-3
--
Activation Energy
52150
54800
--95570
141800
P3-4 (d)
There is a limit to temperature for which data for anyone of he insect can be extrapolate, Data which would
be helpful is the maximum and the minimum temperatme that these insects can endme before death.,
Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperatme it could be
useless.,
P3-5
There are two competing effects that bring about the maximum in the corrosion rate: Temperatme and
HCN-H2S04 concentration. The corrosion rate increases with increasing temperatme and increasing
concentration of HCN-H2S04 complex, The temperatme increases as we go from top to bottom of the
column and consequently the rate of conosion should increase" However, the HCN concentr'ations (and the
3-6
HCN-H2S04 complex) decrease as we go from top to bottom of the column. There is virtually no HCN in
the bottom of the column, These two opposing factors results in the maximum of the cOIrosion rate
somewhere around the middle of the column.,
P3-6 Antidote did not dissolve from glass at low temperatures.
P3-7 (a)
If a reaction rate doubles for an increase in lOoC, at T = T 1 let k = kl and at T = T 2 = T 1+ 10, let k = k2 =
2k • Then with k = Ae-EIRT in general, kl = Ae-E1R1i and k2 = Ae-E'RT2 , or
l
k2-_e -~(*-*)
kl
E
or
R
Therefore:
In (k2 J(T. (T. +10))
E = R--
(In 2)(7; (7; +10))
"
10
kIll
= R·---,-·
(7;-7;)
T. (T. + 10) = lOE
1
1
RIn2
which can be approximated by T
!OED 5
= ---RIn2
P3-7 (b)
E
Equation 3-18 is
Dividing gives
k = Ae
RT
E( 1 I)
~2, = e -Ii Tz""-"1;"
, or
1
3-7
1.99 cal ] [ 273 K][ 373 K]
E=
mol K
In(·050) = 7960 cal
lOOK
.001
mol
[
E
A = k1e RT, = 10-3 min -I exp
.
P3-7 (c)
= 2lO0min- 1
Cal
1.99_
](273K)
(
molK
Individualized solution
P3-8
When the components inside air bag are ignited, following reactions take place,
2NaN3 --7 2Na + 3N2 ... ·.. . , '
, ............. ' .. '
,,(1)
, ,(2)
lONa + 2KN03 --> K20 + 5Na20 + N2 ,
K20 + Na20 + Si02 --7 alkaline silicate glass"
",.,.,(3)
5x rxn(l) + rxn(2) + rxn(3) = rxn(4)
NaN3 + 0.2KN03 + 01Si02 -7 04Na20 + 1.6N2 + complex/l0., ... , ,(4)
Stoichiometric table:
-
r - - ' - - - ,--_.
Species Symbol
NaN3 _
A
KN0 3
B
Si02
C
Na20
N2
D
E
Initial
NA
NABe
Change
Final
---NAX
NA(1-X)
-O.2XNA N A( BB - 02X)
-··OIXNA N A ( Be - 01X)
0
0
O.4XNA
1.6XNA
NA~B
Given weight of NaN 3 = 150g
Therefore, no., of moles of NaN3 = 2 . 3
O.4XNA
1.6XNA
M wt of NaN3 = 65
1 moles of NaN3 requires 02 mole of KN0 3
=> Moles ofB, KN0 3 = 02(2..3) = 0.46 moles
Mwt ofKN03 = lOLl
Therefore, grams ofKN03 required = 046 x lOLl = 46..5 g
1 moles ofNaN3 requires 01 mole of Si02,
Moles of C, Si02 = 0.1(2.3) = 023 moles
M wt of Si02 = 60,08
Therefore, grams of Si02 required = 0,23 x 60,,08 = 13,8 g
Following proposals are given to handle all the un-detonated air bags in cars piling up in the junkyards:
• Store cars in cool, dry, ventilated areas"
• Avoid Physical damage of the bag in cal.
•
It is stable under ordinary conditions of storage" Decomposes explosively upon heating (over
221 0 For 105 0 C), shock, concussion, or friction,
3-8
•
Conditions to avoid: heat, flames, ignition sources and incompatibles.
P3-9 (a)
dX
From the web module we know that - - = k (1 - .x) and that k is a function of temperature, but not a
dt
linear function. Therefore doubling the temperature will not necessarily double the reaction rate, and
therefore halve the cooking time .
P3-9 (b)
When you boil the potato in water, the heat transfer coefficient is much larger, but the temperature can only
be 100°C.
When you bake the potato, the heat transfer coefficient is smaller, but the temperature can be more than
double that of boiling water.
P3-9 (C) No solution will be given
P3-10 (a)
1) C2H6
~
C2H 4 + H2
Rate law: -fA = kCC2H6
Rate law: -fA = kC C2H4 C~:2
2) C2H4 + 11202 -+ C2H40
3) (CH3)3COOCCCH3)3 ~ C2H 6 + 2CH3COCH3
~
A
4) n-C4HIO
+
B
Rate law: -fA = k[C A- CBCc2IKcJ
Rate law: orA = k[ CnC4 H 10 --CiC4 H IKe]
1- C4H lO
+-+
2C
lO
5) CH3COOC 2H s + C4H 90H +-+ CH3COOC4H9 + C2H sOH
+
A
B
+-+
C
+
D
Rate law: -fA = k[CACB - CcCnlKcJ
P3-10 (b)
2A
(1)
(2)
+ B-- C
orA
orA
orA
orA
(3)
(4)
= kC ACB 2
= kC B
= k
= kCACB - 1
P3-10 (C)
(1) C 2H 6
~
C2H 4 + H2
3-9
(2) H2 + Br2
(3) H2 + 12
---->
---->
2HBr
2HI
P3-11 (a)
Liquid phase reaction,
o
/"-
CH2-·-OH
I
CH2 - CH2 + H2 0·~ CHT-OH
A
+B--~
C
3
C AO = 1 Ibmol/ft
C BO = 3.47 lbmol/fe
Stoichiometric Table·
Species
Symbol
Ethylene
A
oxide
-B
Water
Initial
C Ao=1Ibmol/ft3
C BO = 347lbmol/ft1,
----
Change
-CAOX
Remaining
C A= CAO(l-X)
= (l·X) Ibmol/ft3
-CAOX
CB= C AO ( BB -X)
CAOX
=(3.47-X) lbmol/fe
Cc = CAOX
= X Ibmol/ft3
....-
BB=347
f--
Glycol
'----
C
0
----
- - - - _.. _-----
Rate law:
-rA = kCAC B
Therefore,
-rA = k C~o (1-X)( BB -X) = k(l-X)(347-X)
At 300K
E = 12500 cal/mol,
X = 0 ..9,
0.1
3
3
Ilbmol.s = 0.0035ft Ilbmol.s
1000
CAOX
(1)(0.9)
-1
=--- =
- - 2 - - - - - - - - - = 1000..56 s
-rA
(0.0035)(1) (1- 0.9)(3.47 - 0.9)
3
k = O..ldm ImoLs =--x35.315ft
'rCSTR
At 350K,
k2 = k exp«E/R)(lIT-lIT 2»= O.0035exp«125001l.987)(11T-IIT2»= 0.071 dm3/moLs
Therefore,
'rCSTR
CAOX
-rA
= -- =
(1)(0.9)
(0.071)(1) (1- 0.9)(3.47 - 0.9)
2
P3-11 (b)
Isothermal, isobaric gas-phase pyrolysis,
CZH 6
C2H 4 + H2
A
~
B + C
Stoichiometric table:
3-10
_-1
-
49.3s
0
0
FTO=FAO
B
C
C 2H 4
H2
+FAOX
+FAoX
FB=FAoX
Fc=FAOX
F~FAO(1+X)
e =Yaob= 1(1+1-1)= 1
v = vo(1+ eX)
=> v = vo(l+X)
CAO = YAO CTQ = YAO -
P
RT
(I)(6atm)
=
(
C =
A
0.082 m3atm
FA
v
K.kmol
3
= 0..067 kmol/m3 = 0..067 mol/dm
J(1100K)
= FAO(l- X) = C
(1- X)
AO (1 + X)
v0 (l + X)
mol/dm3
FAO(X)
X
3
=CAO - - - mol/dm
vo(l+X)
(I+X)
FB
v
CB = - = -
Fe
v
Cc = -
=
FAO(X)
vo(l+X)
=CAO . -X- - mol/dm3
(I+X)
Rate law:
-fA
(I-X)
(I-X)
(I+X)
(I+X)
=kC AO - - - =o..o.67k---
= kCA
If the reaction is carried out in a constant volume batch reactor, =>( e = 0.)
3
C A= C AO (1-X) mol/dm3 C B = CAO X mol/dm
C c = CAOX mol/dm3
P3-11 (C)
Isothermal, isobaric, catalytic gas phase oxidation,
1
C2H4 + - O2 ~ C2~0
2
A
1
2
+ -B
~
C
Stoichiometric table:
Species
--- .-::::------,--"------
Entering
F AO
F Bo
Change
-FAOX
Leaving
FA=FAO(l-X)
O2
Symbol
A
B
-BBFAOX
FB=FAO ( BB -X)
C 2H 4O
C
0.
+FAOX
Fc=FAoX
C2~
3-11
_
_
P _2
( 6atm )
_
mol
CAO - YAOCro - YAO-- (
3)
-0.092--3
RT 3 0.082 atm.dm (533K)
dm
mol.K
FA
C - -
A-
V -
FB
C - -
B-
FAo(1-X) CAO (1-X) 0.092(1-X)
- --~---.:..
Vo (1 + eX) - (1-0.33X) - (1-0.33X)
-
V
FAo(eB-~) - --'-----'0.046(1-X)
-
vo(1+eX) - (1-0.33X)
-
Fe
C = -- =
e
FAoX
0.092(X)
= ---'---"vo(1+eX) (1-0.33X)
V
If the reaction follow elementary rate law
_ {0.092(1- X)}{0.046(1- X)}O.5
=> -r -k
.
A
(1-0.33X)
(1-0.33X)
P3-11 (d)
Isothermal, isobaric, catalytic gas phase reaction in a PBR
C6H6 + 2H2 ~ C6HlO
A + 2B ~ C
Stoichiometric table:
Change
Species
Symbol Entering
FAO
Benzene A
-FAOX
B
FBO=2FAO -2FAOX
H2
C
C6H lO
1
-- --
0
FAOX
(1)
P
RT
-
3
CA
=(
Fc=FAOX
-,-
3)
vo(1+eX)
-1)
6atm
(
3
= 0.055mol / dm
0.082 atm.dm (443.2K) 3
mol.K
_ FA _ FAo(1-X) _ CAo (1-X) _ 0.055(1-X)
------
---
V
FB=FAO( BB -2X)
2
e=y, AO J=-(1-2-1)=-,3
3
CAO = CroYAO =
--------
Leaving
FA=FAO(1-X)
(1-~X)
(1-~X)
3-12
= FB = FAO (BB -2X) = O.l1(I-X)
C
v
B
C
V
= Fe =
c
V
(1-~X)
o (1+£X)
FAoX __ = CAOX
=
0.055X _
(1-~X) (1-~X)
vo (1+£X)
If the reaction follow elementary rate law.
Rate law:
·-rA'= kCAC;
(1
X)3
-rA'= 0.0007k----3
(1- ~ X)
For a fluidized CSTR:
W
= FAoX.
-rA '
W
=
FAoX
_
(1- X)3
0.0007k--"-3
(1-~X )
mol
" - - at300K
kgcat min atm 3
k=53 -
!.J)
k = kj exp(E(l.. __
R ~ T
=
53exp(80~9.Q_(_I_""_-I-_)J=
1663000
mol
8.314 300 443
kgcatminatm
FAO = CAO * Vo
3
Vo = 5 dm /min
atX= 0.8
W = 0.0024 kg of catalyst
3-13
3
P3-12
1
+ -B
A
2
-7
C
Stoichiometric table for the given problem will be as follows
A ssummg gas plhase
Species
Symbol
C2H 4
O2
N2
C2H 4O
A
B
I
C
8B
-
1
2 FAO
1
e
=---=-
FAO
Entering
FAO
FBo = eBFAO
FI = elFAO
0
2
I
Leavin~
Change
-FAOX
-112 FAOX
FAoCI-X)
F AOce B- x/2)
FAOel
FAOX
-----_.
FAOX
=_FIO F
= 0.79 F
FAD'
0.21
10
BD
=?
e =e
I
F
YAO = -.:iQ.. = 0.30,
FTO
C
AO
= YAO P = 0.041!!1013
RT
F
C
=_AO
A
dm
=C
V
'AO
(1-X)
(1 + EX)
0.041(1- X)
1-- 0.15X
(!--~-X)
_ FB _
2 2
0.020(1---- X)
CB ---CAO
---------V
1-----0.15X
1-0.15X
C =J:c = CAO~
0.041X
c
V
1- 0.15X 1- 0.15X
P3-13 (a)
Let
C = Nibroanaline
D = Ammonium Chloride
A=ONCB
B=NH3
A+2B
~
C+D
3-14
B
0.79
0.21
= 1.88
P3-13 (b)
Species
A
Entering
FAO
F BO = 0 BFAO
=6.6/1.8 F AO
0
0
--
B
c--:::------C
D
---
Change
-FAOX
-2 FAOX
FAOX
FAOX
---
Leaving
FAO(1-X)
FB=
FAO C0B-2X)
Fc=FAOX
Fo=FAOX
P3-13 (C)
For batch system,
CA=NAN
P3-13 (d)
-rA = kCACB
FA = _~A = _~A = ~:O (1- X) = CAO (1- X) , CA = !A = FA = CAO (1- X)
Vo
Vo
V Vo
FB
=-; =-v,B =- T:O(BB-2X)=CAO(BB-2X),
Vo
N
N
N
o
F
CB =-1L=CAo (BB- 2X )
Vo
-rA = kC~o (1- X)( BB -2X)
BB = _~BO_ = 6.6 = 3.67
CAO 1.8
CAO = 1.8i'!!lol
3
m
FrA =k(1.8r(1-X)(3.67--~~)]
P3-13 (e)
1) At X = 0 and T = 188°C = 461 K
kmOl)( 6.6kmOl)
3
2
-rAO = kCAoB
B = kACAOCBO = 0.0017
E
-rAO
-
= 0.0202
m . ( 1.8--kmolnun
m3
krnol
3
.
In nun
-
2) At X = 0 and T = 25C = 298K
3-15
-
m3
k~ ko exp( !(:o + nJ
m3
k =0.0017
[ 1.987
kmol.min
.~ (461
112735!il
1
1
mo - + -
exp
1
298)
mol.k
= 2.12xlO-6
-lAO
m3
kmol.min
= kCAOC BO = 2.52 X W
3)
k
5
3
kmoJ/m min
t- nJ
~koexp[ !(
11273~ ( 1
m3
k =O.0017---exp
kmolmin
1.987 cal
461K
1
561K
J
molK
m
3
k=0.0152--kmolmin
-rAO = kCAOCBO
-rA
=0.0152-~~-. (1.8 kmol)(6.6
kmol)
3
3
kmolrmn
m
m
kmol
-rA = 0.1806-3- . m rmn
I' - - - - - - - - P3-13 (f)
lA
= kC AO\1-X)(e B-2X)
At X = 0.90 and T = 188C = 461K
l)att=188C
- _. r
A
= (0.0017
--.!!!~-J(1.8
km~!_)2
(1- 0.9 )(3 . 67 - 2(0.9))
kmol.rmn
m
= 0.00103 kmol
3
m min
2)
At X = 0.90 and T = 25C = 298K
3-16
- rA = (2.12XlO-6
-m~J(1.8
km~1)2
(1-0.9)(3.67 - 2(0.9))
kmol.rmn
m
= 1.28 X10-6 kmol
m 3 min
3)
At X =0..90 and T
- rA
=288C = 561K
= (0.0152
km~1)2 (1-- 0.9 )(3.67 - 2(0.9))
m3. J(1.8
kmol.rmn
m
=0.00333 __ kmol
3
m min
P3-13 (g)
3
3
Vo = 2m /min = 0..002m /min
1)Fof CSTR at 25C -fA =
= 1.28 X10-6
kmol
3
.
mmm
- r A : X =09
.9.0.02m3 /m~nX1.8km~!/m3 xO.l
= 281.25m 3
1.28xlO- 6 _ kmol_
m3 min
kmol
2)At 288C, -rA = 0.00333- 3 .
m rmn
v = v o CAo (I-· ~)
- rA :X =O.9
V = 0.002!11 /m~?<1.8kmollm3_~JJ...:..! = 0.108m 3
3
0.00333 kmol
m 3 min
P3-14
C 6H 120 6 + a02 + bNH3
---+
C(C44H7 3N0860L2) + dH20 + eC0 2
To calculate the yields of biomass, you must first balance the reaction equation by finding the coefficients a,
b, c, d, and e.. This can be done with mass balances on each element involved in the reaction. Once all the
coefficients are found, you can then calculate the yield coefficients by simply assuming the reaction
proceeds to completion and calculating the ending mass of the cells.
P3-14 (a)
3-17
Apply mass balance
For C
6 = 44c + e
For N
b = o..86c
Also for C, 6(2/3)
6 + 2a = L2c + d + 2e
12 + 3b = 7.,3c + 2d
ForO
ForH
= 4,4c which gives c = 0.909
Next we solve for e using the other carbon balance
6 = 44 (0.,,90.9) + e
e=2
We can solve for b using the nitrogen balance
b = o..86c = 0..86* (0.,90.9)
b = 0.78
Next we use the hydrogen balance to solve for d
12 + 3b = 7..3c + 2d
12 + 3(0..78) = 7.3(0.,,90.9) + 2d
d =3.85
Finally we solve for a using the oxygen balance
6 + 2a = 1.2c + d + 2e
6 + 2a = L2(o..9o.9) + 3.85 + 2(2)
= 1.47
a
P3-14 (b)
Assume 1 mole of glucose (180. g) reacts:
Y e/s= mass of cells / mass of glucose = mass of cells /180. g
mass of cells = c*(molecular weight) == 0.,90.9 mol* (9L34g1mol)
mass of cells = 83,.12 g
Y e1s
Y els
= 81.12 g /180. g
= 0.46
Y e/02 = mass of cells / mass of O2
If we assume 1 mole of glucose reacted, then 1.47 moles of O2 are needed and 83 . 12 g of cells are
produced"
mass of O2 = L47 mol * (32 glmol)
mass of O2 = 47,0.4 g
Y e/02 = 83.12 g /47,0.4 g
Y e/o2 =1.77
3-18
P3-15 (a)
Isothermal gas phase reaction.
1
3
--N +-H2
---7NH
3
2
2
2
Making Hz as the basis of calculation:
1
3
2
3
H2 +-Nz -'?- NH 3
1
3
2
3
A+-B-'?-C
Stoichiometric table'
Species
H2
N2
Symbol
A
Initial
F AO
B
' - - - - - 1----
NH3
C
Leaving
FA=FAO(1-X)
FBO=BBFAO
change
-FAoX
-F AoXl3
0
+2FAoXl3
Fc=(2/3)FAOX
FB=FAO( 0B -Xl3)
--
P3-15 (b)
6=(~ -i-1)=--~
e= YAo 6 =0.5X( _
CAO
-
~)=-i
(16.4atm)
0.5 (
J
0.082 atm.dm~ (500K)
3
= 0..2 mol/dm
mol.K
C
H2
=C =
A
CAO
(~_~~_Xl = 0.2(1- X) = O.lmol/ dm 3
(l+eX)
(1-})
2 CAo (l-X)
C
3
= C =--x
NH3
C
(l+eX)
2
O.2(X)
3 (1-{)
=-x
=O.lmol/ dm
P3-15 (C)
kNZ = 40 dm3/mol.s
(1) For Flow system:
3-19
3
(1- X)
(I-X)
(1--})
P3-16 (a)
Liquid phase reaction -? assume constant volume
Rate Law (reversible reaction):
-rA =k[CA CB
_
Stoichiometry:
-~]
K
c
CA =CAO (1-X), CB =CAO (1-X), Cc =CAOX
To find the equilibrium conversion, set -rA = 0, combine stoichiometry and the rate law, and solve for Xe
CACBKc =Cc
C~o (1- Xe
r
Kc
X: -(2+ CAoKC
1
= CAOX e
JX e +1=0
Xe =0.80
To find the equiliblium concentrations, substitute the equilibrium conversion into the stiochiometric
relations.,
3-20
mol
mol
CA = CAO (I-X) = 2 -3 (1-0.80) =0.4-3
dm
dm
mol
)
mol
CB =CAo (I-X)=2-3 (1-0.80 =0.4-3
dm
dm
CA = CAOX = 2 mo! *0.80 = 1.6 mo!.
dm
dm
P3-16 (b)
Stoichiometry:
e= YAo t5=(1)(3-1)=2 and Be =0
NAN AO (I- X)
(1- X)
C ---C
A - V - Vo(1+eX) - AO (1-t-2X)
C = Ne = __.3NAO X = C
3X
e
V
Vo (1 +eX) 'AO (1+2X)
Combine and solve for x.,.
K C
C
.Q..-- Xe) =[c 3Xe ]3
AO (1+ 2X e)
AO (1+ 2X e)
Kc (1- Xe)(1 + 2Xe)2 = 27C~oX:
J
27C~o X 3 +3X +1=0
- 4+--K
e
e
(
e
Xe = 0 ..58
Equilibrium concentrations:
Po
lOatm
mol
C =--=--------------=0.305--AO RTo (400K) 0.082 m 3 atm
m3
molK
(d
CA
_
-
C =
e
J
d
(1-0 ..58) _
mol
0.305 (
(
)) - 0.0.59--3
1+2 0.58
dm
3(0 ..58)(0.30.5)
mol
=0246-·3
(1+2{0.58))
.
dm
P3-16 (C)
Same reaction, rate law, and initial concentration as part (b) gas phase, batch reaction.,
3-21
Stoichiometry:
C =NA =NAO (1-X)=C (I-X)
A V
V
AO
o
C =Nc =3NAO X =3C X
c
V
AO
V
o
Combine and solve for Xe
KCCAO (1- Xe) = (3C Ao X e )3
Xe
=0.39
Equilibrium concentrations
CA
mol
=(0.30.5)(1-0.39) =0.193
Ce
=(0.30.5)(0.39) =0.36--3
dm
mol
dm
P3-16 (d)
Gas phase reaction in a constant pressure, batch reactor
Rate law (reversible reaction):
-rA =k[CA ._
C~]
K
e
Stoichiometry:
£=
YAoO=(I)(3-1)=2 and Be =0
NAN AO (1- X)
(1 - X)
C ---C
A - V - Vo(1+£X) - AO (1+2X)
C
c
= Nc =_l!'AO X --=3C
V
Vo (1+£X)
X
AO (1+2X)
Combine and solve for Xe:
KCCAO (1- Xe)
1+ 2X e
Xe
= [3CAO X e ]3
1+ 2X e
= 0.58
Equilibrium concentrations:
3-22
C =
0.305(1·-0.58)
1+2(0.58)
A
C =
3(0.305)(0.58)
c
1+2(0.58)
mol
=0.059--3
dm
mol
=0.246-3
dm
P3-17
Given: Gas phase reaction A + B -7 8C in a batch reactor fitted with a piston such that
V = O.IPo
(ft3f
k=l.O---·
lbmol 2 sec
-rA =kC~CB
N AO = N Bo at t = 0
Vo = 0.15 ft 3
T = 140°C = 600 0 R = Constant
P3-17 (a)
=
N AO
=0.5
N AO +NBO
8=8-1-1=6
e = YAo8=3
YAO
T
and .-.- = 1,
Now
1'0
Therefore
v = lOVo2 (1 + eX)
lOV
Po = lOVo ' and P = lOV
or
N
Therefore
3-23
AO
= ( YRT
AOPo ) v:0
= 5.03*10-9
A
[1
-r
-
X]3
3
Ib mol
3
(1+3X)2 ft sec
P3-17 (b)
V2 =Vo2(1+eX)
0.22 = 0.15 2 (1 + eX)
X
= 0.259
-rA = 8.63*10- 10 }b mol
ft 3 sec
P3-18 No solution will be given.
P3-19 No solution will be given.
P3-20 No solution will be given .
--_..._ - - - - - - - - -
CDP3-A
3-24
w
6.5
InW
1871802
13
18
2.564949
2.890372
r
1fT'
300
310
0.003333
313
0. 003195
..........-.--..
---.--.--.-..i~w;;1rr
3.5 , . . - - -..- .._ ...-._.
0.003226
. ··-·---················__·· . ···1
•......_ .... _ I
3
2 ..5
~
e
-
2
1.S
I
y '" -7120.4x + 2:5.593
Fr.O.S8$3
0.5
0+-,----.-o..(XX:I1S
' - - . - -.."'.. _ _.. , - . , - - - - - 1
0.0002
0..0032S
1trK
~." ....... ~,- .. " •••• , .. "" ..... ,
"".,.
~ ...... -~~ ................... - , - - " -
"......... ".".~ .... -"--"-,,~ .......... ,.... _ ............... ,....7"" .............. -. __ ........ .
From the graph:
E= 7120
In W = --7120.4
W(4L5<>C)
*(_._
. -!. ,,,.--) 1- 25.593 == 2.95
41.5+273
=19.2cm I s
__._--------
---------------------------_.-_.
CDP3·B
Polanyi equation: E = C - a(-.1HR)
We have to calculate E for the reaction
CH3 0 + RBI' -7 CH3Br + Ro
Given: dHR = - 6 kcaUmol
From the given data table, we get
6.8 = C - a(17.5)
and
6,.0 = C -- a(20)
=> C = 12.4 KJ/mol and a = 0..32
Using these values, and .1HR = - 6 kcaUmol, we get E = 10.48 KJ/mol
CDP3-C (a)
A-7B
Rate law at low temperature: -
rA = kC A
The rate law t higher temperature must:
1) Satis(y thermodynamics relationships at equilibrium, and
2) Reduce to irleversible rate law when the concentration of one or more of the reaction products is zero.
Also, We know,
KC
__CBe
-
CAe
Reauanging, we get
C
Ae
- C Be =0
K
c
3-25
So, lets assume rate law as
- rA
=kA(C ~:J
A -
Also when CB = 0, it satisfies the given rate law, Hence the proposed rate law is conect.
CDP3-C (b)
A+2B72D
- rA =kC A
Rate law at low temperature:
1/2
CB
Here,
But it does not satisfy the irreversible late law at low temperatures"
Hence it is not COlrect
So, taking square root of Kc
fj{
-V 1\' C = -
C De
C
112'
CAe
C Be
112 C
Ae
(c
- rA = k A
_ CDe = 0
~Kc
Be
De
Ae 1I2CBe - ~CK C
J
Which satisfies the irreversible rate law,
Hence it is the required rate law.,
CDP3-C (C)
A+B-7C+D
.
, k P A PB
=
...::::--1 + KAPA + KBPB
PCPD
PCPD
Kp =
or PBPA -·
=0
Kp
PBPA
IrreversIble rate law: -, r A
We know,
Hence assume rate law as:
PCPDJ
k PP -'--.
K
( A B
p
- rA = .
."----1 + KAPA + KBPB + KcPc + KDPD
Which satisfies both the above mentioned conditions,
3-26
CDP3-D
y
a)
- 0.15
A$$umi~1
C
b)
NlIJ • O
-
eo
CNH
ideal gas law
~Q
...
Va
3 ,.0
~
8.2 Itra
' .. 0.2 ,mol/l
· OKCSOOK)
0.082 latln
. ,:no 1
.. -
H:ro
.. y~
r_~ ~ (0.15)(0.2 ,mol/1) .. 0.03 gmol/1
........3.0 '"""I ..
c)
Compound.
S:r=bol
~!p. ...
J.
;)
Il1it.i .. l
Chan,,,
0.15
-O.lS:.t
5
Find
0.15(1-X)
°l
B
0.18
- 4<0.151:)
0 .. :18- f<0.15X)
!!O
c
0
+(l.15X
0.151:
H2 O
D
3
... t'
0 .. lSI:)
N
1
I
°0.67
3
+ t'0.151:)
0
0 .. 67
Totill
l'
1.00
+ 1/4(0 .. 15,I,)
1
In.itiol N2 ... 0.19(1-0.15) "'" 0.1.9(0.8S). 0.67
Initial 02 = O.a5-0.61~O_lg
lJ
~i
Pi ... YiP ...
ttr
~i
P -
8.2
;J.t::l
(1+(0.151411; )
3-27
+ I/4( O .. lSl'.)
i.
P.
Catm)
l.
A
0.15(1-1.)
5
4'
0.18 -
'-------,
C.
l.
'''''-'''''''--
I-I
0.03 1+(0.15/411
1.23 1+{O.15/4]X
(0 036 - .Q~ X) /(1 + .Q...ll I)
(O.lSI.)
4
c
4
OJ1311+ O~5 X
O.I5X
1.
::t CO.HI)
1
a.lll!(l~ 1}
0.67
v ...
Ci
d
Vo (l..,O+B
I)
, ....
0 .. 1 1\1
Pi .... CiI::I ... <O.OllHjOO)(O.2)l1
1
'Ill 1
S .,
0.1.1 - -:; (O .. Ul:)
0.15 X
~
I
8.2 ~:L
Ci
o .:l.,${ 1;-1:)
II
Ill!
i
Pi
1 .. 23 (I-X)
0 .. 03 Ct.-X)
Q
....... 6
n
"".I
_
0.03
o. '15, :x
"
1 .. 1(0 .. 11 ....
:x
(0.151:)
0.67
1.%3 X
1"","
0 .. 1.33
'0 .. 1.99 + 0.0011
... 0 .. 2 (l ... 0 ~1.5 .x)
3-28
:r5 {Q .. UX)
X
P'tO't'" 1.2 - 0 ... 313 I .. taI.
3)
Suu; u
d)
(l)
4A+5B-1'4C+6D
Mole balance for a PFR
elF
•
A
dV
dF
"... b . . =
dV
dF,c
......
dV
-r,
= -rc
dF v
dV
Combine with equation 3,,40 and 3-48
3-29
CDP3-E
A -+ 2B --)
C
-+-
D
a.)
Species
Symbol Initi..a1 moles
Ben:royl chloride
A
A.:rom.ooia
B
Be!".zy~e
C
Ammonium C1l1oride
D
b)
CAO
4!'!-
2.
D
2 gma-i/l
AO IV "'"
N
A ... n _)
-
NAG
NAoSs
N Ao6 c
NA o6 D
D
+""'"
2 :2
3-30
(llange
Final moles
Conceruration
--·NAoX
NAo(l-X)
N Ai)( 1-X).fV
-2.'""lAQX
NAO{~·2X)
NAoX
NAoX
NAo(8 c +X)
NAQ(8 D -+-X)
NAoC&g-1XW
N Ab(6c.--+X)!V
N A o(6D+XW
Species
Ammooia
Benzo chloride
Symbol
B
Entering
Olange
Exit
Fao
-FsoX
A
SAFBO
~FooXI2
Fao{1-X)
F}3O(S KXI2)
... C
Conceruration
F:ao(l-X)/vo
F:ao(9 a .JlY'z,}/v0
9cFBO
FBO(ec+XI2)
FBo(ac+~/vo
0
eoFBO
FMXI2
Fsr:XJ2
fBO<9 n+X/2)
~B<I(&o~/vo
Benzylamide
Ammorrlum Chloride
•
1)
:Moh.r floy :l:1.tcs considered rll.ther tha.n numb..,r of moles.
2)
e.1 ""
3)
Concentration found by dividing the e~i1:ing molar floy rate by the
F. IF.", , as opposed to N. INA
10.;:>0
~olnmetric
10
flo. rate
'IT
.0
•
"
CDP3-F
Given. : A + B -) C taking placo. in .. sqU&xo: duct ..
T
F AO .. loS Ib molel sec.
(~)
If B is at equilibrium in the gAS phase throughout the reActor
Since 13 m&int&ins its oquilibriUll'l v&por prouure
throughout the re&ctor.
it
is replaced by
&
&$
soon as 1 moloculo of B is consumed by the re&ctioll.J
lIIalccuh of B in tha liquid.
3-31
Ronco, 0 ... 1 - 1 • 0, ~ '" TAOQ
. .. ..
(1:1)
.At 1: ,..
0.5
[- ( .25 a'tm) ( .75
.em f~3
( .730
.
.~
11:1 lnole OR
3.'t1ll) ..........
-1
2
% lOOOOR)
1 'b.mo 1 e
{O.S]" .114'; 3
•~t: ~An~
.W~_
CDP3-G
a)
3 5i HC13 .+ 3Hz ~ 15i (S) + 7HCl + 5i H2 Ch
Take Si HC13 as basis
Spede~
Symtx:!l
Emering
ClJani.'C
Si HO] (g)
Hz Cg)
H2HCl (g)
A
-FACiX
B
FAC
f=Bo =&s FAo
C
0
+2FAoX
Si lhC1 z (g)
D
0
Si (s)
S
0
+}FAoX
2.FA oX·
~F}.D<X
3
3-32
L&mdDr;
FA
n
= FAO (I-X)
=FAo(ea .. X}
Fe :::2FAOX
3
Po =lFAoX
3
Assume isothermal and constant pressure. Neglect the vapor pressure of Si(S}
as = 1 Stoichiometric feed
o only involves the changes in gas phase
E
= YAO 0 =:; .1 (1. -+- L - 1 - 1)
233
e=}tt)=t
CAO - Y 0 Po . - 1 (
2,Q
)
.
A R Tn - 2 (O_OR2HI127\
""
c" = £~,
:= CAO
\)
:=
Cc
= Fc.
1,)
tl
{l- ifL
(1 + EX)
g. ::::
Ca
"" _.
IJ
= 0.0088 (1- X)
C {Ss-=-X} ::::
A
(1 + EX)
== 7 .,.' F AoX _.
3 \)0 (1 +. EX)
I \ .
1'1- X/3
O.~88 (1·X)
1 + X/3
= 2 CAO __ ,_,X____
3
(1
CD == F D .::::}
l!~9.~ -=0.OO31 ....- ~-.-.
v
3vo(1+c:X)
1+~
...
1he SOlUllon to parts (b) and (c)
and
e)
a...""e similar to Part (a).
CDP3-H
3-33
'1-
£X)
::: 0 021
•
X
(1 + XJ3)
$0 vc can use the stoichiome~ric tAble.
i
FAO(l-X)
CAO{l-X)
B
4F AO(l--X)
4C
4F AO .l.
D
_ F 00
5
titer
~--
A
C
whor~
C.,.
F,
... ...
CAO
=
:x
FAC
3ubstit~ting C's for F's
AO
4C
{l"-X)
AO
eAQ
:!
X
5 CAO
0.0658 g~ol/l aDd FAD - 0.02631 Bruol/s
cQ~gell~a~ion
F
t
V"'v o -""'V
F
Q
t.o
Not~:
at X - 0.5 (beginning of
condensation) C.; c·
i
1.
A
B
C
5FAirF A (}X
FAO(5-X)
4FAOX
..... __"_--1?. ___. . . . _........._. 9:.1FAO(5·~?f)!Q. 9.
Total
FAO(5-X)+FAO(5-X)/9
3-34
CAo(l-X)(4.5)/(5-X)
4C A0(1-X)(4.5)/(5-X)
4C,,,oX,(4.5)J(S. -X)
C Ao (5-X)(4_5)/(5-X)/9=O.5C
._-...,._..- .... .. A{) .
C A o(5Xt/40.5
__ _-_
"
....
Pv
95
.--::w-P
9S0
u
0.10
Scoich Tl.ble
Spocies
DoCore Co ndel1s;1I.l:.i on
Pp < P y r4111&ining
After COl1dens&tiol1
C1:l&ngo
FAOU-IJ
___ -._,----_ -.
In
Sy:ubol
........ ,.. ....
........
....
CB:4
A
FAe
,·,F X
,Ac
FA - FActI-X)
C1
2
13.
4F AO
-4FAcf.
Fa
liC1 4
C
+4FAOl:
Fe
eel.",
.0
0
D(g)
DO)
+FAOI
...
-
Fn
AO (1-X)
4F
4FAOA
Pp - Py rem&il1il11
AO (1-lJ
4F
4FAQ1O.10FT
.. FAOX
0
---"
. ....
"'.","',,"-_
Fr ...
SF AO
Fro
When conde •• atioD first begins
Fr
=
5 FAa
.
'r
0.50
1)
AI' "'" AP "" .0
""..,.)
p
-
.0
--} e ,.. 0
Total concentration is consta,llt
o
" ,RL'
-""
o
C.0820S atm lit X 348 K
gmol K
3-35
,----.
..
FT-SFAO-FAOX+O.10FT
,
F
(S'-X)
AO
FT ...
.0.90
CDP3-I
T+
473 K
200~C ~
tl
10E
:;:: "v::
P
506,S;:; 0203
2500
•
)toich t~bl/!l
In
Species
'" .....,..... __ ~ •... "~ '''_~""'''''''''
C2H6
BI2,
HBr
C2F14B1 l
A
B
C
D(g)
FAll
2FAO
o
o
o
Fi before
F'i After condensation PD :::::
condensation PD
Pv remaining
<
P
remaining
v
.. _.... -.-' ...... _ ....... "....... " -...... •. _....
w'"....... __,.......
...-, .. _.,.". ........,_.......
,,,FA(;X
FA : .:.: FAO( 1"X)
FAO( IX)
2FAO( IX
-2FAoX Fu ;:::: 2FAO( I-X)
Change
_~~
2FAoX
FAOX
Fc == 2FA()..'X
FD
Bafar. coadeas&tioa
---) e '" 0
3-36
=FAaX
,4. _ _ _ . _..... _ .. _ . o K "
2FAoX
O.203F'r
P
'Clt
_._0
a-I
CAO •
o .. 0820S
40
1
3' <=-rIO
0.673
:d
••
'A
t
prAll.
.
t
24. • 7 . . '/... ttti
-
= -
'JOn~u
3
&t=
.
lit
~
l:
1=01 I:.
0.673 gcol/l
413 I
,mol/l,. 0,.212 ,moll1
'.
pAll.
l '~ll.-X)
'0
...11.
IM1I.
a-lJ 1
11 J.&ll
';41
lPA.O
<::.40 u -%)
Fe
&IN".
C
4
'r
0.H16
.0.:1%
0
0
0.11.1
0.111
G.1
a,au
Of .Ur.)
0.041
Q.OU
Il.ua
<LUG
0,4
13.1>44
0.111
0.o"
0.041
0.311
0.U1
o.'¢~
0.041
O.Ul
(I.U'
Q.Q'"
o.JU
0.01l
t
, .AQ(1.-11
l).fO!!)
a.IML
t.'
toO
IF>\I.'l'thIl
o-n
.u:n
IFClIl;l'l
l'>\I.'l'l
F.w-r:Ti
'J.J::IJ.1;1
a.ou
O.t.1f
0 • .,(.5
O.ltl
O.OU
0.011
0.641
0.110
a.aH
<f.Ul
0,037
1)
0
Q.lll
O.OH
Q.l"
\)
3-37
CAl)
O-t) \I
lCjD(l-Xl
~."1'.4
Q.llt
"'PC
.10
A,Q
Q.134
0.:110
0.1"
O.lll
1"';-1211: \l' J
ll ... X)
1e
"
Ml
d.l"
ttl~
i).1J'
,t"
e
U-l)
(Q ..
iq71
J..Q
0.1,:$ ,
O.Hl
F AO
=
v
CAO
=<
Q
F.
C
:0=
l.
-
(0.212) gmal/1 (0.5) lIs
F.
1-
V
'If
0
0.106 gmal/s
F.:l
1.
:::t
n
(l+f:l.)
... v
0
So we ca~ first use t~e stoichia~etric t~ble. SUbstituting C's for F's
i
....
F.
C __ .~
_._:..L
(l.-:n
C el-I.)
AO
._.__...•,.•..l...._ _
A
F
B
2F AO (1-·,X)
c
2F AO
-
D
Total
AO
.!
___:.10 1_
3 F
2C
AO
:lC
(1-I.}
AO
-- ehQ
X
X
.--
.3 CAO
AO
3-38
v .. v
F 'C '
o F
to
F.
C
1
i
""
". V
-::Ill
--1.
F.1. (2.392...
v
C3-IJ
2.
"Y
o
,.,. C(A)
.... e(8)
C(9710111)
.. C(Cl
C(O)
0.
OQ
0.1
Q.l
as
1),4
os
04 01 04 09
"L
j --.--.
OJ
I.Q
____~. .
-~
+
FCA}
<>- F(6)
F ((1no1lS) OJ.
,.. F(C)
<lo F{D)
""'Fm
0.1
O.O~~-::--
____-=::::~~
Qj
Q.4
V
04 U
1.0
X
CDP3-J
3-39
,13-X)
o 2.39
Fso
as =,,--=1
F AO
YAo :::;; 0 . 5
E
== Y ADO =0.5(1+ 12,.-3--4)= 3
_
C
C.~o(I
A --
- X) _ 6.18*IO·-&(1-X)
l+&Y- -~'-'i+3X
= 1.68 *~-a(~_ ~~
C
)
1+3X
8
C = 2.0~:'10-a X
c
• 1 +- 3X
.
2.47 * 10-7 X
C =
-.D
1+3X
CAO
= Y .... O(::1'O
=
o_i'It)=
o.'){.'.···.---.·P;;-~~;3.'.... ".J=
--l
*
8:314··_··········
mol*k
_. _ C.-IO (1-, x) _ 6.18" 1{)8 (1 ' x)
C. A. - -"--"'-"'" ................ -. - . _....................... _ ...... _ ......"........ -'-'
l+EiX
•
('1J
1+3X
J
1.68 * 10-$(1 . . . ~~:.
= --_.. _.-........................ . . . .:"..........................1+3X
2.06'"
lO·-a__ ....x-C c = ...............
- .....................
• 1·+ 3X
c... = .2.47*107X. . . . . . . . . . . . . . . . . . _.-.. . . . . . . ..
'0
1+3X
3·40
97.3K
6.18"" lO-lt
CDP3-K
CA)
C QUllZ Q :t\l;;:a. t
~D:tIQl
el2
A
bilid m!.lh~
1.0
CH4
B
0 • .5
c
0
Rel
D
0
l'ot31
T
1.5
CIUel2
(b)
0 "" 1
C
'"
-I-
YAO
2 - 1 I) '"
Q:r:~'.
~ 7(0)
A
~A
:m
:lI!
1-1
O.S(l-X)
-0.5:1
.5.1
0.51.
+.1
X
1.S
0
c
-400 m::dl,
.. <400
k C 1
A
CAO (l-l.AJ
~;:I. .. l..
:2
-:1
+{)
..... X" =- 1.5189 - } C1I el docs
1
z
-r
~
2 '" 0 • YAO ... 1.0
s,.S1:O': is lao,s ph"'$c Ulltil F
(c)
Chu,U
C
(l-ll)
AO
3-41
U,Q
t
cgn.d~n.:u
!. t
1 a 'bI ..
(d)
c.AO 3E....!....""
1.5
.1.
O~og206
= 0.02724 (1-O.6) ~ 0.01090
,mol
3
W:zt
-r
A
Klmpl
~3
(f)
A e::t:.p
k
-
A
"" 2
.:t
.10
(-
.$
Lz
Rr"
~2.
d<e;6
s
gm.ol
2.
at:
gJ:llol
(--
-E
E.
(
373.2
3-42
2S"'C
Solutions for Chapter 4 - Isothermal Reactor Design
P4-1 Individualized solution,
P4-2 (a)
Cooking food (effect of temperature), removing of stains with bleach (effect of bleach cone,,), dissolution of
sugar in coffee or tea.
P4-2 (b) Example 4-1
There would be no enur! The initial liquid phase concentration remains the same"
P4-2 (C)
Example 4-2
For 50% conversion, X = 0.5 and k = 0.311min- 1
.
F AO =Fe
- =6.137
- - = 12 .27 lb rna If mm
X
Also,
0.5
FAO
VAO
=C=
12.27 ft3
min
AO
Vo = VAo + VBo = 2VAo and also, VAo = VBo
Vo= 24.52 fe/min
Using Mole Balance,
~oX _ = 24.52xO,5
k(l- X) 0.311(0 ..5)
= 592gal
V
=
=
78.93
This is less volume than Example 4-2 because the rate is higher..
P4-2 (d) Example 4-3
For P = 60atm,
C AO = 0.0415 lbmol/fe
(C
= YAO Po =
AO
RTo
60
)
0.73x1980
Using equation E-4-3,,6, for X = 0.8
We see that the only thing that changes is CAO and it increases by a factor of 10, therby decreasing the
volume by a factor of 10"
VOO
1
P
P4-2 (e) Example 4-4
New Dp = 3DoI4
Because the flow is turbulent
4-1
1)p2
fJ2
=A-
1
=0.0775- =0.lO33
0.75
1)PI
I
2.0.lO3 atm . 60ft
1ft
10atm
2fJo L
Po
Now 1 - - - ~
0
,so too much pressure drop P =
2
I
= (-0.24)2
°
and the flow stops,
(0
P4-2
Example 4-5
For without pressure drop, conversion will remain same as example X = 0,,82.
With Pressure drop,
anew
= a o = 0.0037 kg- I
lO
For decrease in diameter,
Po = 2xO.01244[266.,9x2+ 12920.,8]
= 33475lbtlft3 = 334.75(lbr/ft')xlI144(ft2/in2)xll147(atml( Ibtlin2»
= 52.71 kPalm
For turbulent flow:
fJ ,_.'
_1_ and
a-.L => a 2 = aI to!
Poz
Po
1)p
X -
kCAOW
va
(1 aw)
2
1)pI
Dp2
_
4,6(0.,949)
-~~C:W(l_.9'W)- - + 4~6(0.,949)
1
Vo
2
X = 0,,8136·- virtually the same
(2) Optimum diameter would be larger
a = 0.037x 52.71 = 0.0756kg- 1
25,,8
4-2
= 0. 15814atmlft
Dl
a2 =a
-P-=2a1
1D
p2
aW = 1-a W = 1-(0.037)(27.5) = -0.0175
1
1 __
2 2
Now l-uW < 0, too much pressure drop due to higher superficial velocity.
P4-2 (g) Example 4-6
For turbulent flow
1
a-Dp
a,
and
1
a·---
Po
~~( ~:-)( ;':)~~[ilG)~~
Therefore there is no change.
P4-2 (h) Exmple 4-7
For pressure doubled and temperature decrease
ero = 2*PJRT and T == 688K
See Polymath program P4-2--h.poJ.
POLYMA T!l Results
Calculated values of the DEQ variables
Variable
V
Fa
Fb
Fe
E
T
Cto
Ft
Ca
k
ra
Fao
rb
vo
re
X
Tau
rateA
initial value
---0
2.26E-·04
0
0
2 . 4E+04
688
0.573773
2.26E-04
0.573773
213.40078
-70 . 254837
2 . 26E-04
70 . 254837
3 . 939E-04
35 . 127419
0
0
70.254837
minimal value
0
1.363E-·05
0
0
2.4E+04
688
0.573773
2.26E·-04
0.0236075
213 . 40078
--70 . 254837
2.26E-04
0 . 1189309
3 . 939E-04
0 . 0594654
0
0
0 . 1189309
maximal value
-1-. 0E--~2.26E-04
2.124E-04
1.062E-04
2.4E+04
688
0 . 573773
3.322E-04
0 . 573773
213.40078
-·0 . 1189309
2.26E-·04
70 . 254837
3.939E-·04
35.127419
0.9395277
0 . 253044
70.254837
ODE Report (RKF45)
4-3
final value
1. OE-04
1. 363E-05
2 . 124E-04
1.062E-04
2 . 4E+04
688
0 . 573773
3.322E-04
0.0236075
213.40078
-0.1189309
2.26E-04
0.1189309
3.939E-·04
0.0594654
0.9395277
0 . 253044
0.1189309
Differential equations as entered by the user
[1 J d(Fa)/d(V) = ra
[2] d(Fb)/d(V) = rb
[3 J d(Fc)/d(V) = rc
Explicit equations as entered by the user
[1] E = 24000
[2] T=688
[3J Cto = 2*1641/8.314rr
[4] Ft = Fa+Fb+Fc
[ 5] Ca = Cto*FalFt
[6 J k = 0.29*exp(EI1.,987*(1/500-1rr))
[7] ra = -k*Ca"2
[ 8] Fao = 0,,000226
[9 J rb -ra
[10] vo = Fao/Cto
[11] rc -ral2
[12] X= 1-FalFao
[13] Tau Vivo
[14] rateA=-ra
=
=
=
3,Oe-4..-------
.2 .4e-4
O.OHO
OOHO
2.0<,-5l.Oe-5
v
............. .., .........
_--".......... .
60e-5
S,Oe-5
""
1.0e-4
P4-2 (i) Example 4-8 Individualized solution,
P4-2 (j) Example 4-9
Using trial and error, we get maximum feed rate ofB
Q.,Olmol/dm3 ,
= O,,0251dm3/s to keep concentration ofB
See Polymath program P42-j.,pol.
POLYMA TH Results
Calculated values of the DEQ variables
Variable
-t
ca
cb
initial value
o
minimal value
o
0,,05
0,0063485
o
o
maximal value
500
0.05
0.009981
4-4
final value
500
0.0063485
0,,009981
0
0
0
0
0 . 22
0.22
k
0.0251
vOO
0.0251
0.025
0.025
cbO
5
5
vO
0.05
0.05
caO
0
rate
0
5
5
v
0
0
x
equations as entered by the user
[1] d(ca)/d(t) :::: -k*ca*cb-vOO*ca/v
[2] d(cb)/d(t) :::: -k*ca*cb+vOO*(cbO-cb)/v
[ 3] d(cc)/d(t) :::: k*ca*cb-vOO*cc/v
[4] d(cd)/d(t) :::: k*ca*cb-vOO*cd/v
0 . 0078965
0 . 0078965
0.22
0.0251
0 . 025
5
0.05
3.91E-05
17.55
0 . 5543321
cc
cd
Explicit equations as entered by the user
[1] k:::: . 22
[2J vOO:::: 0.0251
[3J cbO:::: 0 . 025
[4J vO:::: 5
[5J caO:::: 0.05
[6 J rate:::: k*ca*cb
[7J v:::: vO+vOO*t
[8] x:::: (caO*vO-ca*v)/(caO*vO)
If the concentration of A is tripled the
maximum feed rate becomes
0..064 dm3/s
0.0078965
0.0078965
0.22
0 . 0251
0.025
5
0.05
1. 394E-05
17.55
o . 5543321 Differential
0.010
0.008
GJ
0.006
0.004
0.002
0.000
0
100
200 t
300
400
500
P4-2 (k through r) Individualized solution.
P4-3 Solution is in the decoding algorithm given with the modules.
P4-4
We have to find the time required to cook spaghetti in Cuzco, Peru.
Location
Ann Arbor
Boulder
Cuzco
Elevation (km)
0.21
-- 1.63
3.416
--
Pressure (mm Hg)
739
625
493
~-
Assume reaction is zero order with respect to spaghetti conversion:
-E
de
-r =k=Ae RT = ___
A
A
dt
so that
4-5
Boiling Point (0C)
.. 99.2
94.6
88.3
Time (min)
15
17 ..?
For complete conversion (i.e . : well cooked) CA = 0 at time 1.
Therefore
-E
CAO =tAe RT
C
-E
·~=teRT
A
In ( CAO )
A
= In k = In t _ E ~
R~
Now, plot the natural log of the cooking time versus liTb and get a linear relationship . Extrapolation to T b =
88 ..3°C = 36145 K yields t = 21 minutes.
I
2l-L
I
I
I
Ull t}
11~1
I
lS-t
I
I
2.69
::'.72
2.767
P4-5 (a)
4-6
e Aa :;:;:; coo;;;; 2: !il;mtld:ft~1>
kl!l!i,
VA;;;; VB =;S drt1:'ltttif~
E. . . ;;;; ZO..COO cillmol
0;01 <!:t:n;;hnoilmin
TI=:lOOK
'LTsi'll.g the
t\lirheniu:s
eqUi.lillc·l} .!i. tll¢
CSTR rernpeli'il'i.WRof 3,SOK y~~!ds, tbe: new sp!l!tifi(:
reactio'll rate.
CSTR tleslf:,";o Btruatilll1::
.".
"
\' y\ - r,<,,}('r::
;"',1' <!r AOJ
rt r'lI
_~=I,
X = V ( kC~o (1_. X)2)
10
From the :quadratic ~qu~tion:
PBR
V=800 dm3
T=300K
Design Equa.tion:'
dX
F"o'-=-rA
dV
~___ kC~oV
1-- X
FAD
X=O,,85
So, considering the above results, we will choose a CSTR.
4-7
P4-5 (b)
Batch Reactor V=200dm' NAO=NBO=200 moles X=0.90 Assume Isothermal
Design Equation:
t=NA O
rX dX
-lJ -rAV
r
t
=
t
dm'. J(lmal)2(200dm3)
mal * nun
dm 3
= 1.06 min
(200males)
(
_dX
8.4.5
(l-XY
P4-5 (c)
T=273K
Find the specific Ieaction rate at the new temperature of 273 K using the Anhenius Equation.
k
= 2..54 X 10-3
(200)(9)
.
k = -r-(- - ' - - 3'~) = 3.543 nun
2.,.54 x 10- A200)
= 2..5days
P4-5 (d)
1) CSTR and PFR rue connected in seIies:
X
. = (200dm 3 )(007dm' / mal.min)(1mall dm 3 )2(1-- X)2
CSJR
lOmal / min
Solving the quadratic equation, X CSIR = 0 . 44
ForPFR,
dX = (0.07dm
1.
dX
044 (1- X)3
3
/
mall min)C,40(1- X)(l- X)2 dV
lOmale/min
= (0.07dm 3/mallmin)(lmall dm 3)2(800dm 3 )
lOmale/ min
X =0.736
2) when CSTR and PFR rue connected in parallel,
= (200dm 3)(0.07dm 3/ mal,min)(lmal / dm3)\1- X)2
X
CSTR
•
.5mal/nun.
X CSIR = 0.56
4-8
ForPFR,
1
dX = (0.07dm 3 / mol.min)(1moll dm 3 )2(800dm 3 )
o (1- X)2
5moll min
XPFR = 0.92
Hence, final conversion X =
0.56+0.92
=0.74
2
P4-5 (e)
To process the same amount of species A, the batch reactor must handle
2M
(5d~~J(60min)(24hJ = 14400~lOl
hr
mm
day
day
If the reactants are in the same concentrations as in the flow reactors, then
v = (14400-mo.lJ(~dm3J = 14400 dm
day
mol
3
day
So the batch reactor must be able to process 14400 dm3 every 24 hours"
Now we find the time required to reach 90% conversion" Assume the reaction temperature is 300K.
tR
=
N AO
X
.
N Ao
2 - - , and smce - - = CAO
VkCAO I-X
V
1
X
1
0.9
tR =-_._--=
3
--=2.14hr
kCAO I-X 4.2 dm_*I(mol) 0.1
mol·hr
dm 3
Assume that it takes three hours to fill, empty, and heat to the reaction temperature.
tf= 3 hours
ttotal = 2. 14hours + 3 hours = 5.14 hours.,
Therefore, we can run 4 batches in a day and the necessary reactor volume is
14400dri = 3600dm3
4
4-9
Refening to Table 1-1 and noting that 3600 dm3 is about 1000 gallons, we see that the price would be
approximately $85,000 for the reactor.
P4-S (f)
The points of the problem are:
1) To note the significant differences in processing times at different temperatures
(i . e.
compare part (b) and (c)).
2) That the reaction is so fast at 77°C that a batch reactor is not appropriate. One minute to react
and 180 to fill and empty.
3) Not to be confused by irrelevant information. It does not matter if the reactor is red or black.
+
'rA=kC",CB
k (F Ai,,) (F NU)
":A ""
butanoi
+-
..."
eiernentary reaction
for liquid systems
volumetric flow u =uo
4-10
water
V
F~oX
Fe
- --.------ = -.--.-----.--2
CSTR -
- ..
kC
rA
£7\: =4
A
_~o
(l---XXS--X)
106~.
6
, Fe = 4. x 10 Ib / yr , 30d / yr operation
F := 4. x 106 It!.. x ..!:x! _1.£.... x 1 Ib mol
.C
yr
30d 24 hr
278 Ib
V
•.
CSIR
-lOC'fl
~1
I'I.J ga.
=20.0 !t'~1
1ft"
1 ..j.)
~_",-3
x. ------::::
J u.
7.48 gal
"
133.7 h 3 = . ~ ___ .._._ ..... 20.1bmole/hr_ . ._._ ._._
J
12-;--..fE.--_.-;-··...
lomol<:::·lll
Xi. • 6X,· 1.89
=:
2
(.Q:~lbmOle) (1 ·X) (5-X)
f?
0
X"" 0.33
P4-6 (b)
To increase conversion, use PFR, higher temperature, or use better catalyst.
P4-6 (C)
ME
Rl.
s
c:
E
P4-6 (d)
PFR Design Equation
d)(
dV
dX
F Ac
=:
IcC!o (1- X XS"'_· X )
W-= ___
~_
, _-
.. ""." •• " .• "...... ~" .." . - , _ _ _.~
dV
V = 535dm 3
4-11
P4-6 (e)
MOle baiance.;
The above equation relates the reaction time for a batch and the conversion
achieved during that hatch.. There is a trade···off bct\vcen high conversion and few
batches and low conversion but many batches per day_ What conversion will
result in the smallest nurllber of reactors?
N AO
:::;;;
133.7 ft;'
* C AO
C AO
:=:
__ .!2.(?:!!!:.c..~.__ . . _'!:lQ ;y*-_~~~C = .!!~AOX ~~. = . . . . . .
reactor * day
t
.
+3
t
bau:h
r=
O.2lbmoll ft>
'Y~QJ!:!:_~_. ____ = f{X )
1
(
<; ...
X \
'A:i":~~; In ,.~.... '5X
J1- 3
J~I'?l:pr.5:c!~:!. =2~}~:!?~19J,!q! = 480.~q!.
day
30days 278lb
day
480moll
f(X)moll dayJ reactor
n :::;:.-._--...
o
0.4
0..2
U.O
0.8
x---->
The minimum occurs at X
= 0.82 and corresponds to 4.192 or 5 reactors
P4-6 (f) Individualized solution
P4-6 (g) Individualized solution
P4-6 (h) Individualized solution
P4-7 (a)
4-12
Elementary gas phase reaction,
A-4B+2C
c
",c (1-X}
...0 (1+
r\
£X)
P,IQ
10
mol
CAO '" RT ::: (0.0$2){400) '" O.3'dml
V;;:967
P4-7 (b)
v", FAc X(l+!!:l
CAJ(l-X)
P4-7 (c)
For a = O. OOldm-3
See Polymath program P4,. 7 . c.pol.
POLYMA TH Results
Calculated values of the DEQ variables
Variable
v
x
initial value
0
0
Y
1.
Co
0.3
2
0_001
esp
alfa
minimal value
0
0
0_1721111
0_3
2
0.001
maximal value
500
0 . 656431
1
0.3
2
0_001
4-13
final value
500
0_656431
0 . 1721111
0_3
2
0_001
C
k
0.3
0.044
-0" 0132
2.5
r
fo
0. 3
0 . 044
-3.422E-04
0.0077768
0.044
-0 . 0132
2.5
0 . 0077768
0 . 044
-3 . 422E-04
2. 5
2.5
Differential equations as entered by the user
[1 J d(x)/d(v) = -r/fo
[2 J d(y)/d(v) = -alfa*(1 +esp*x)/(2*y)
Explicit equations as entered by the user
[1] Co 0 . 3
[2] esp = 2
[3] alfa = 0 . 001
[4] C Co*(1-x)*y/(1 +esp*x)
[5] k = 0 . 044
[6] r = -k*C
[7] fo=2 . 5
1.0 . . . . . . : : : - - - - - - - - - - - - - - - - - - ,
=
0.8
=
0.6
~.
LrJ
0.-:1
AtV= 500,
x = 0..66, Y = 0..17
0.2
P4-7 (d)
Individualized solution
0.0
o
P4-7 (e)
A
If--_ _-'--_ _--'-_ _- '_ _ _-'--_ _--I
~
B + 2C
Law:
100
200
v
300
400
)::: 0
X
Stoichiometry:
1.+ eX
esc:!
Kc =---~.=
C"
s : : ; 2 and C Ao
;;;;;
0 ..3
X "1:1 --O-?
.31_
X:::; (O,,90)X~ :::: 0,.47
4-14
500
3
Using these equations in Polymath we get the volume to be 290 dm
,
P4-7 (0
~
A
B+2C
Rate Law:
Kc == 0,,025
C ::::C"'Q(L:.~l
.. \
l+eX
Stoichiometry:
CBC~ _ (_C,-oX 1(2~~oX)!(
K -
C ...
C -
-
heX) l+eX
C
,-:a
=C",,,2S.
1+- ex'
Cc=2C...o X
l+eX
__C".,(l··X)
. !,7 ex );; ,(1+eX}2(1-X)
__ ,,_4C~oX3..__
e::; 2 and CA,o::: 0.3
Xeq::: 0.52
X::: (O.90)Xeq::: 0.47
,E.EB.
,...
t"
M
orA
dX
r
dV'"
- - ::: -
0r
dV F
-,,,..,,.' ;;;; """.ia.
dX -r. .
=(i~~~)((1- Xl - (1:~fd
Using Polymath to solve the differential equation gives a volume of 290 dm3
See Polymath program P4-7-f.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
-X
V
Kc
Faa
Caa
k
e
ra
initial value
0
0
0.025
2,,5
0.3
0.044
2
-0,0132
minimal value
0
0
0,,025
2.5
0.,3
0.044
2
-0,,0132
maximal value
0.47
290,,23883
0.025
2 "5
0,,3
0.044
2
··9" 391E--04
ODE Report (RKF45)
4·15
final value
0,,47
290.23883
0,,025
2,,5
0.3
0.044
2
-9.391E-04
Differential equations as entered by the user
[1 J d{V)/d{X) = Fao/{-ra)
Explicit equations as entered by the user
[1] KC= . 025
[2] Fao = 2 . 5
[3] Gao= . 3
[4] k = . 044
[5] e=2
[6] ra = -(k*Gao/{1 +e*X))*{{1-X)-{4*Gao/\2*X/\3)/{{1 +e*X)/\2*Kc))
V:::: 1300 dm)
PFR with pressure drop: Alter the Polymath equations flom part (c)..
See Polymath program P4 . T. fpressure.pol.
POLYMA TH Results
Calculated values of the DEQ variables
Variable
v
x
Y
Kc
alfa
Cao
k
esp
fo
r
initial value
0
0
----"---
1
0 . 025
0 . 001
0.3
0 . 044
2
2.5
-0.0132
minimal value
0
0
0 . 3181585
0.025
0 . 001
0.3
0 . 044
2
2.5
-0.0132
maximal value
500
0.5077714
1
0.025
0 . 001
0. 3
0 . 044
2
2. 5
-1.846E-04
ODE Report (STIFF)
Differential equations as entered by the user
[1] d{x)/d{v) = -r/fo
[2] d{y)/d{v) = -alfa*{1 +esp*x)/{2*y)
Explicit equations as entered by the user
[lJ Kc =.025
[2 J alfa = 0 . 001
[3] Gao = 0 . 3
[4J k=0.044
[5J esp = 2
[6J fo = 2 . 5
4-16
final
value
---500
0 . 5077714
0.3181585
0.025
0.001
0. 3
0.044
2
2. 5
-1.846E-04
[7]
r = -(k*Cao/(1 +esp*x»*(4*Cao"2*xI\3/«1+esp*x)1\2*Kc»
At V = 500 dm3 X = 0.507 and y = 0.381
P4-7 (g)
Membrane reactor: A -+ B + 2C
Cc = CoFCIFI
CA= CoFAlFr CB = CoFBIF1
FT = FA + FB + Fe
and orA = rB = rel2
Using polymath,
ForPFR,
See Polymath program P4·7···g.pol.
POLYMA TH Results
Calculated values of the DEQ variables
Variable
v
Fa
Fb
Fc
Kc
Ft
Co
K
kc
ra
X
initial value
0
2.5
0
0
0.025
2.5
0.3
0.044
0.08
-0.0132
0
minimal value
0
1. 3231889
0
0
0.025
2. 5
0.3
0 . 044
0.08
-0.0132
0
maximal value
1040
2.5
0.3635477
2.3536223
0.025
3 . 7452437
0.3
0.044
0 . 08
-3.827E·-04
0.4707245
Differential equations as entered by the user
[ 1 J d(Fa)/d(v) = ra
[2] d(Fb)/d(v) =ora - kc*Co*Fb/Ft
[3 J d(Fc)/d(v) -2*ra
=
Explicit equations as entered by the user
[1] Kc =0 . 025
[2] Ft = Fa+ Fb+ Fc
[3J Co =0.3
[4J K = 0.044
[5J kc = . 08
[6 J ra = - (K*Co)*(Fa/Ft· CoI\2*Fb*FcI\2/(Kc*FtI\2»
['7 J X = 1 - Fa/2 . 5
Solving for when X = 0 . 47, we get V = 1040 dm3
4-17
final value
1040
1.3231889
0.0684325
2.3536223
0.025
3.7452437
0.3
0.044
0.08
-3.827E-04
0 . 4707245
3.0
2.4
-
Fa
Fh
1.8
1.2
0.6
0.0 0
208
416 v
832
624
10-10
P4-8 (a)
The blades makes two equal volumes zones of 500gal each rather than one 'big' mixing zone of IOOOgal.
Si71.g1e B12c:'e
Predicted
'V =
F
.~.
AO kll
\., _.
X)-
X
;:;; a ~-'-"' __ 7
(1··
X).
rnal
[aj
=:;SL. :.-.----..-:"
CAok
-2:
~~~;;,;l ~~l)
1···· 3X1 ·+
Xl
xf =0
=_ .... -::;:::. .38
2
Reac:or 2
a = [gal]
X ··?. = a . -.......
.5 :::::. 2a
1000 gal = a---..
(l··-Xt
.25
1-· 2X-:> + X~:;:;:X, ··38
""
:..,
a=500 gal
.,
./;.,.
.
X1--3X + 1.38=0
4-18
2
Measured
500 gal == 5U{10 ~~!;~~L
.!.-. X.·I·~
... '"
. -~ £.."
P4-8 (b)
A CSTR is been created at the bend due to backmixing, so the effective arrangement is a PFR is in series
with aCSTR.
-c PF~ 8-]-~__
----------~-~.--
CSTRzone
created due to
backmixing
A--+B
3
k = 5 min- 1
Vo = 5 dm /min .
x.,xpected = 0.632 Xactual = 0.586
V = xfFAOdX = va In [ _
1
]
o -rA
k
1- X Expected
=1 In (
5
1 ) =1.0
1- .632
Now,
4-19
=
Vp
ForPFR,
Vc
For CSTR,
Also,
In (_1
J.
I-Xl
. ..1
= FAO(Xactual -
Xl)
-rA
Vp
Vc_
V
V
= (Xactual - Xl)
(1-· Xactual)
............. .2
........................... ..3
-+--}".
Solving I, 2 and 3 by using polymath,
See Polymath program P4-8-b.pol.
POLYMA TH Results
NLES Solution
Variable
Xl
Vc
Value
0 . 350949
0.,567756
V
1
X2
Vp
0 . 586
0 . 432244
f(x)
lni Guess
--~3~,~1~4~8E~--cl~0~--0~·
-3 . 297E-14
1
NLES Report (safenewt)
Nonlinear equations
[ll f(X1) = In(1/(1-X1))-Vp = 0
[21 f(Vc)
=(X2-X1 )/(1-X2) -, Vc =0
Explicit equations
l11 V = 1
[21 X2 = .586
[3] Vp = V - Vc
Vc = 0..57 dm3 ; Vp = 0.43 dm3
;
Xl = 0.,35
P4-8 (c)
CAO = 2 mol/dm3
A~B
Assuming 1st order reaction,
For CSTR,
'r"=
CAOX
~
-rA
-rA = kC A = kCAO(l-X)
=> rk
ForPFR,
V
= -~ == ~.4 = 0.,67
I-X
=F
AO
=> X PFR
0.6
X
f
o kC
dX
_ rk
(I-X)'
AO
= Xf~_
0
I-X
=1- exp(-rk) =1-exp( -0 . 67) = 0.486
Now assuming 2nd order reaction,
For CSTR,
Now, assuming 2nd Older reaction,
4-20
ForCSTR,
CAOX
1:=--
-rA
-rA =
kC~=kC~o (1- X)2
=> 1:kCAO =
ForPFR,
X 2 = O.~ =1.111
(I-X)
0.6
1:=-1-1 dX 2
kCAO 0 (1- X)
=_1_~
kCAO 1- X
=> X =1- _ _
1_.=1-_1_=.526
1+1:kCAO
2.111
So, while calculating PFR conversion they considered reaction to be 1st order. But actually it is a second
order reaction.
P4-8 (d)
A graph between conversion and particle size is as follows: Originally we are at point A in graph, when
particle size is decreased by 15%, we move to point B, which have same conversion as particle size at A.
But when we decrease the particle size by 20%, we reach at point C, so a decrease in conversion is noticed"
Also when we increase the particle size from position A, we reach at point D, again there is a decrease in
the conversion,
1
x
A
n
dp
P4-9
A<=>B
Kco (300K)= 3.0 V = lOOOgal = 3785.4 dm3
Mole balance:
V =!.AO X
- rA
4-21
Rate law:
StOIchIometry:
Now using:
v =(-Z)
Z
where
X
[(I-X)'
=>
_of]
K
C
R To
(Z) __-.!___
Z
[
X2]
(1-X)2 -'K
c
z=.!. = exp(E[J_ . _!])
ko
f (X) = 0 =V _
and
T.
= K co exp(MI~[_!._!..])
R
T
T
o
Solving usmg polymath to get a table of values of X Vs T..
See Polymath program P4-9 po]
POLYMA TH Results
NLE Solution
variable
X
To
T
z
V
E
R
Y
Kco
Hrx
Kc
Value
0.422~453
f(x)
3.638E··12
Ini Guess
. O.S
300
305.5
2902.2
3785.4
1.SE+04
2
1.5684405
3
·-2 SE+04
1 4169064
NLE Report (safenewt)
043
Nonlinear equations
( 1 J f(X) = (z/y)*x/«1-X)"2 - X"21Kc) .. V = 0
042+---
- ..... - -..' •.' -..
---------.-.---~
,
--------,
i
---+--.-------~
041
x
04 - - 0.39
038
4-22
- - - - - - -..- - - - -.. -----.-----
0.37
o36
-··---~--~I------_.,_-·-·__r_-~--
Explicit equations
[1] To = 300
[2] T =305.5
[ 3] z =2902.2
[4] V =3785..4
[5] E = 15000
[6] R=2
Y = exp(E/R*(1rro-1rr))
[8] Kco=3
[ 9 ] Hrx = -25000
[10] Kc = Kco*exp(HrxlR*(1rro-1rr))
[7]
----
T(in K)
X
300
0.40
301
0.4075
303
304
f--
305
--
0.4182
---
-._--
--
0.4213
0.4228
305.5
0.4229
305.9
0.4227
307
0.421
310
0..4072
315
0.3635
~.
--
_._------- _._-----_._We get maximum X = 0.4229 at T = 305.5 K.
P4-10 (a)
For substrate:
4-23
Fso - Fs + rs V = 0
(Cso - Cs)vo = rg VY S1C =
VYSIC[ ,urn.xCS Cc ]
KM +CS
P4-10 (b)
= YC/s [C so -
Cc
Cs ]
(C so - Cs )Vo - VYS1C [,uMAXCS Cc ]
KM+Cs
=
°
(C so ---Cs )Vo - VYCIS [,uMAXCS ]( Cso - Cs ) =
KM+Cs
=>
°
(30--C s )5_25XO.Sx[O.5XCs ](30-Cs ) = 0
5+Cs
Solving we get Cs = 5.0 g1dm or 30 g/dm if C s = C so no reaction has occurred so the only valid answer is
C s = 0.5 g1dm3•
3
3
P4-10 (C)
Cc = Y ClS(C SO - C s)
= 0..8(30 - 5.0)g1dm3 = 20 g1dm3
.
P4-10 (d)
3
V new = vJ2 = 25 dm /h
3
3
Using equation trom above, we get C s = 167 g/dm and Cc = 22 . 67 g1dm
P4-10 (e)
3
V new = V J3 = 25/3 dm
Using equation from above, we get C s = 1.0 g1dm3 and Cc = 216 g/dm 3
P4-10 (f)
For batch reactor:
C so = 30 g/dm3 Ceo = OJ g/dm 3
Cc = Cco + Y ClS(C SO - C s)
V = lOdm3
See Polymath program P4-1O-f.pol.
POLYMA T!lRe§ults
Calculated values of the DEQ variables
Variable
t
Cs
Cso
Yes
KIn
Umax
Ceo
Ce
initial value
o
minimal value
o
maximal value
15
30
30
final value
15
0.0382152
30
0.8
30
30
0 . 0382152
30
0. 8
5
o. 5
0. 8
5
0.8
5
0. 5
0.5
0. 1
0. 1
0. 1
0.1
o. 1
0.1
24.069428
24.069428
4-24
5
0.5
rg
rs
negative_
0.0428571
-0.0535714
0.0535714
0.0912841
-0.1141052
0.1141052
5.4444349
-0.0535714
6.8055436
0.0428571
-6.8055436
0.0535714
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cs)/d(t) = rs
Explicit equations as entered by the user
[1] Cso = 30
[2] Ycs=0.8
[3] Km=5
[4] Umax = 0.5
[5] Cco =0.1
[6 J Cc = Cco+Ycs*(Cso-Cs)
[7] rg = (Umax*Cs/(Km+Cs»*Cc
[8J rs=-(1IYcs)*rg
[ 9 ] negative_rs = ·'rs
30.0r---==:=::;:;;::::::--·-------1
7.0 . . . . - - - - - - - - - - - - - - - - - - ,
24.0 .
5.6
-]
c::
"
18.0 .
4.2
12.0
2.8
6,,0 .
1.4
0,,0
0.0
0.00
3.00
6.00 t
9.00
12,00
15,,00
tU
"~!O
'-="""'"'~==
0,00
3.00
__ ____ ___.!J
~
~
6.00 t
9,00
1200
15,00
P4-10 (g)
Graphs should look the same as part (f) since reactor volume is not in the design equations for a constant
volume batch reactor.
P4-11
Gaseous reactant in a tubular reactor: A
~
B
-rA =kCA
k = 0.0015 min-l at 80 P
0
E
= 25 , OOO~gmol
X
=0.90
M
B
=1000~
hr
lb
MWA =MWB =58--lbmol
Dt =1 inch (I.D.)
L = 10ft
P = 132 psig = 146.7 psia
T = 260° F = 720° R
nt = number of tubes
4-25
1721lbmol
F = FB = '
hr = 19 .1 1b mol
AO
X
0.9
hr
1000!!
FB =
hr =17.21Ibmol
58 lb
hr
lbmol
For a plug flow reactor:
V
2
= nt 1lDt L = F °f·9 dX
AO
4
° -fA
6=1-1=0
c
= PA
AO
(E(
1-1
k2 =k1 exp - R ~ T2
RT
=~
RT
)J =0.0015exp (25000(
.
- - - -1- - -1))
- =53.6mm
540 720
1.104
k2 = 53.6min-- 1 = 3219hr-1
19.11bmol)(10.73 it3 psia )(720 R)
V = FAORT In 10 = .
hr
lb mo~~_R_____ In 10
1
kP
(3219hr- )(146.7psia)
0
(
V
= 0.72ft'
V
= !!L,!Dt L
2
4
4V
4(0.72/t
n =--=-_.
t
1£D2L
t
1£
3
)
(112ftJ2 (10ft)
13.2
Therefore 14 pipes are necessary..
P4-12
A
--4
BI2
4-26
-1
Stoichiometry
1
(5
2
!/AO
1
2
1
FAU
FAO +Pno+
,--
IE
11 AO()
1
2
-
1/4
_v .. v
.1
X
V p I;' R
-
FAO
o
dX
---7'11
Rate Law (elementary
reaction)
.M._. ____ .................
....
_~
.....
~._
.• .,.¥ .. _" ...
~._
..••.. "._._..
¥ ............... " . _ . _
_.,.~,
_ _ . ••
(for the int.egratitJll, refer Appendix A)
from the Idea.l (jus assumption,
= YAoCm
0.8 (),nd E ~1/4 to Eq. (4)
CAO
Substituting
Eq.(5)~
X
yields~
VPFRky~oC~O = 2(--1 / 4) (1-1/ 4) In(1- 0.8) + (-1/ 4) 2 0.8 + (1-1/ 4)20.8 = 2.9, ... ".. .. ,(6)
FAO
1-0.8
--'-'-'~"""---':..:;:.,-
1\1oh\1' flow 1a to of A cut iu haH,
F.~1.0
-
.F~~to
1!
~'1A_O
c'
lihHll
IF'
2,., .-'10
F~~. :~""FBU"':f~ Fc;;'; ~
-
Y~1UJ =1/6
Eq. (4),
FAO
i{
2E/(1 -+ t')ln(l
4-27
1
:3
VPFR ky ,2AD C'2TO_
F'2AD
1" I/G)2 X'
2( --'-1/6)(1 - 1/6)ln(l -- X') + ( l/fi) 2 X'
Polymath ='fol1-Liw:cll' Eqnation Solver, X f
I
XI
:.::c
(}.758
P4-13
Given: The metal catalyzed isomerization
CB-rA = kl ( CA - Keq
A q B , liquid phase reaction
J.
WIth Keq = 5.8
For a plug flow reactor with YA = LO, Xl = 0.55
Casel: an identical plug flow reactor connected in series with the original reactor.
Since YA = 10,
e = 0 . For a liquid phase reaction e A = e AO (1- X) and e = e AOX
B
B
-r, = kCAo((l- X)-
i::J
For the first reactor,
~ =F
AO
Xl
dX
o
-fA
dX
Xl
f-. "= F f kC ((1- X)-~.----"-J"
AO
0
AO
or
K
eq
4-28
\
In[I-(1 +_I_JX 1 ] = -0.853In(.355) =0.883
1+Keq
Keq
Take advantage of the fact that two PFR' s in series is the same as one PFR with the volume of the two
combined.
~CAOVF._ = X
F AO
dX
Jl-(l+_l_Jx
0
o
Keq
kCAOVF = 2 kCAOl-i
FAO
FAO
=
1+~1+++:JX2]
Keq
2
.
kCAO
l-i =2(0.883)=1.766
FAO
1.766 =.
.)X
11 In[I-(I+_1
1+--.
5.8
5.8
2]
X2 = 0..74
Case 2: Products from 1st reactor are separated and pure A is fed to the second reactor,
The analysis for the fIrst reactor is the same as for case 1.
4-29
kC AO
F
l+_l_ln 1- 1+ Keq1 JXI ]
v; -
1
AO
[(
Keq
By performing a material balance on the separator, F AO ,2 = FAO(l-X I )
Since pure A enters both the first and second reactor CAO ,2 = C AO , CBO,2 = 0, 9B =
C A=CAO (1 - X)
X2
V2
= FAo,2
dX
0
CB =C AOX for the second reactor.
f- =
FAO (1- X)
Xl
kCAO
0
o -rA
dX
f (l-X)-X
Keq
and since VI = V2
V
kC AO 2 _ _kCAOV;
FAO
FAO
or
---.!-1
In[l--(l +-}-Jx
1+eq
l]
K
~
=-
1- ~l In[l-(l +_1
JX2]
1+Keq
K
~
1-(1+;~ Jx, +-(1+ ;Jx,t'
X
1- 1- (
[
_
2 -
1 J ]l-~l
l+K.
Xl
q
1+_1_
1
= 1- (0.356)045 = 0.766
1.174
Keq
Overall conversion for this scheme:
.- _ FAO - FAO ,2 (1 - X 2) _ FAO -- FAO (1 _. Xl) (1 - X 2) _ (
)(
)
X - - - - F- - - -1- I-Xl I-X2
AO
FAO
X =0.895
P4-14
4-30
Given: Ortho- to meta- and para- isomerization of xylene.
M
M
>P
>0
kj
k2
o
>P(neglect)
Pressure = 300 psig
T = 750°F
V = 1000 cat.
fe
Assume that the reactions are irreversible and rust order.
Then:
--rM = kjCM + k 2CM = kCM
k = kl +k2
£=0
Check to see what type of reactor is being used .
Case 1:
Vo
= 2500 gal
hr
X = 0.37
Case 2:
V
o
=1667 gal
hr
X =0.50
Assume plug flow reactor conditions:
FMOdX =--rMdV or
dX
J-o
x
V=FMO
V=
-rM
Xc
f
"0
d
x
MOVO X =vo f-'~.-'= VO• In (I-X)
-rM
0 k(I-X)
k
C MO , k, and V should be the same for Case 1 and Case 2.
Therefore,
(kV)casel = -( Vo )casel In (1- X casej ) = -2500 ~:IIn [1-0.37] = 1155 ~~
(kV)case2 = -( vO)caSel in (1- Xcase2) = -1667 ~~ In [1-0.50] = 1155 ~~~
The reactor appears to be plug flow since (kVkase 1 = (kV)Case 2
As a check, assume the reactor is a CSTR
FMOX
= CMOVOX =-rMV
4-31
vX
kV=_oI-X
or
-rM
Again kV should be the same for both Case I and Case 2.
( kV)
( kV)
=(
V )
X
0 Casel
Casel
=
I-X Casel
Casel
=
(V )
0 Case2
X
Case2
1-· X Case 2
Case2
gal ( 0.37 )
25001
hr
= 1468 ga
1-037
hr
•
gal ( 0.50 )
16671
=
hr
= 1667 ga
1---050
hr
•
kV is not the same for Case 1 and Case 2 using the CSTR assumption, therefore the reactor must be
modeled as a plug flow reactor .
kV =1155 gal
hr
1155 gal
k=
hr _=1.55 gal
1000 ft3 cat.
hr It 3 cat
For the new plant, with vo =5500 gal / hr, XF = 046, the required catalyst volume is:
-5500 gal_
h
-v
V = _ 0 In (1- X F) =
~
In (1--0.46) = 2931ft3 cat
k
1.155--.-K~
hr ft cat
This assumes that the same hydrodynamic conditions are present in the new reactor as in the old .
P4-15
A-4 B in a tubular reactor
Tube dimensions: L = 40 ft, D = 0 ..75 in.
n t =50
2
V
= nt 1[D
4
2
(50)n- 0.75
L=
(
4
12
)
40 = 6. 14ft3
4-32
500~
F =~=
AO
MWA 73
hr =6.861bmol
lb
hr
lbmol
x dX
V=FAofo -rA
-r = kCAo (I-X).=kC (I-X)
A
1+8X
AO
V = FAO SdX.= FAO S
dX.
o-rA
okCAo (I-·X)
.
P
YAOP
wIth CAO = - = - RT
RT
V = FAORT
kyAOP
In (_1_)
or
k
=
FAO In(_I_)
kCAO
I-X
I
n
(_1_)
VYAOP
= FAOR~
1- X
1- X
Assume An'henius equation applies to the rate constant.
-E
o
At T J = 600 R, kJ = 0.00152
= Ae
R1
;
-E
At T2 = 760 oR, k2 = 0 . 0740
= Ae RT2
~ =ex+~If(:' -t)]
In ~ =
-:(k- n= !T~;;
E = TrT2 ._ In k2 =(660)(760) In 0.740 =19,500oR
R Tr - T2 kl
100
0.00152
A = kl exp
'0
[~-]
RTr
k=k\ex
p[ -
!u -n]
From above we have
k
= .FAoRT
Vy. AO P
so
In (_1_)
l--X
FAoR~P In (_1)= kl exp [- E (~-~J]
VYAO
I-X
R T
Tr
Dividing both sides by T gives:
4-33
(
6.S61bmal)(10.73 psiaft3 )
hr
lb maze R
-In 5 =
sec)(
6.14ft3 )(114.7 psia)
( .00152-)(3600
sec
hr
Evaluating and simplifying gives:
exp [-19500 (! - _1_)]
T 660 R
T
0
0.030So R-1 =
Solving for T gives:
P4-16
Reversible isomerization reaction
m-Xylene -> p-Xylene
Xe is the equilibrium conversion.,
Rate law:
-rm
= k(Cm _ Ck p )
e
At equilibrium,
C
p
C=m
ke
-rm = 0 =>
K=~e
l--X e
1+ _1_ = (1 + 1- Xe ) = Xe + 1- Xe
Ke
-rm
Xe
Xe
= _I_
Xe
= kCAD (1-~-)
X
e
P4-16 (a)
4-34
ex [-19500(!- 1 )]
P
T 660 R
0
T
For batch reactor,
dX
Mole balance:
= -rmV = kCAO
dt
T=X e1n (
k
N mO
Xe
Xe --X
CAO
(1- XeXJ
J
ForPFR,
V =FAO
= va
k
fdX
f
dX
1-(1+ :JX
1
TpFR
-rm
=-;;
f
dX
( 1J
1- 1+- X
Ke
= Xe
T
k
PFR
In--~
X-X
e
P4-16 (b)
ForCSTR,
V= Fma X
-rm
X
T CSlR
=
~k[;::-1--(-:-1' +-K-1e-J-X-=]
Putting the value of Ke,
TCSIR
X( XeXe-X J
=k
P4-16 (C)
Xe
In(.
Xe . )
In(-~-)
X --X
vOlumeefficieney~qV=T::~ ~(; )=(~)=('x }{X,~£)
T,
k
k
X -X
X e --X
4-35
X e -X
X -X
X
1]
v
=
1-(~)
1
X
Xe In
X
1--·
Xe
Xe
Following is the plot of volume efficiency as a function of the ratio (x/Xe),
See Polymath program P4-16-c.pol.
1.0 , ; : : : - - - - - - - - - - - - - - ,
Efficiency
0.8
0.6
04
0.2
0.0
0.010
- - - - - "- OA06v.\.1.604 0.802 1.000
0. 208
P4-16 (d)
Efficiency = VPFR I VCSTR = 1 from problem statement, which is not possible because conversion will not be
the same for the CSTR' s in series as for the PFR
-------_.
P4-17 (a)
A--7Y2B
E
= -1/2, X = 0.3, W = 1 kg, Yexit = 0..25
ForPBR,
C =Co (l-X)y
A
(1+ eX)
and
dX
rA
dW
FAO
dX
(1-x)2l
dW
(1+eX)2
--=z
kC o (l-
XY y2
let
vJl+eXY
and
kC Ao
Z =---Vo
dy =-!:'.(l+eX)
dW
2y
Solving for z by trial and error in Polymath to match x and y at exit,
Yo = 1
and
Yf = 5/20 = 0..25
X = 0.3
we get:
a = 1.043 kg- l and z = 0.7 kg'l
See Polymath program P4-I7-al.pol.
POLYMA TH Results
4-36
Calculated values of the DEQ variables
variable
initial value
0
0
W
x
1
Y
esp
alfa
-0.5
1.043
0.7
Z
minimal value
0
0
0 . 2521521
-0 . 5
1.043
0.7
maximal value
1
0.302004
1
-0.5
1.043
0.7
Differential equations as entered by the user
[1] d(x)/d(W) = Z*«1-x)*y/(1+esp*x»)A2
[2) d(y)/d(W) =·alfa*(1+esp*x)/(2*y)
Explicit equations as entered by the user
[1] esp = -0.5
[2] alfa = 1.043
[3] Z=.7
Now for CSTR:
_ FoX _ X(l+cXY
W -----_._--T
z(l-XY
A
Solving we get for W = lkg and z = 0..7 kg·'
X=OAO
See Polymath program P4·17-a2.pol.
POLYMA TH . . Results
NLE Solution
Variable
x
W
esp
Z
Value
0.396566
1
-0 . 5
0.7
fix)
-1.142E-13
Ini Guess
0.5
NLE Report (fastnewt)
Nonlinear equations
[1 J f(x) = W*Z*«1-x)/(1 +esp*x)}A2-x = 0
Explicit equations
[1] W = 1
[ 2 J esp = -0 . 5
[3J Z = 0.7
P4-17 (b)
For turbulent flow:
G2
a=(constant)Dp
4-37
final value
1
0.302004
0 . 2521521
-0.5
1.043
0. 7
= _a = 0.0326
j
a
=>:. Z2
and
32
2
= 4xO.7 = 2.8
Now solving using polymath:
See Polymath program P4-17-b.poI.
POLYMATH Results
Calculated values of the DEQ variables
Variable
w
x
minimal value
0
0
0,9887079
-0.5
0 . 0326
2. 8
initial value
0
0
1
Y
esp
alfa
-0 . 5
0.0326
2 "8
z
maximal value
1
0.8619056
1
-0.5
0.0326
2.8
ODE Report (STIFF)
Differential equations as entered by the user
[1 J d(x)/d(w) = Z*((1-x)*y/(1 +esp*x))"2
[2 J d(y)/d(w) = -alfa*(1 +esp*x)/(2*y)
Explicit equations as entered by the user
[ 1] esp = -0,,5
[2] alfa = 0,,0326
[3]
Z = 2.8
So, convelsion in PBR, X = 0,,862
P4-17 (C) Individualized solution
P4-17 (d) Individualized solution
P4-18
Given a Fluidized Bed ('SIR
w=
F[~&X
'fA
0= 0'
E =0,
thcll PA
=
p
J'\o(t- X) p
n
No pressure drop ill the CSTR
PA == l\o(I . . . X)
'vV =f"AO)l:
kPAO(IX) ,
4-38
final value
1
0.8619056
0 . 9887079
-0.5
0 . 0326
2.8
k & F A(} unknmvl1 .. group into a constant, use values from 1st case,
~o C~x,) PA:\V ~j~20(150)
=
~o ~t;1~~~)~~t
==
=
Put PFR downstream" less \Nasted volume
a)
'IS
P4-18 (b)
b) PBR;
.,~~
dW
=
tAo:i~" == --r,~ = kPJ\o(l·- X)~)::;: kPAO(1-· X)(l ,,--nW)1l2(since e =0)
P~Qk (I _ X)(l- aW)li2
FAO
PAOk rW (
. )112
l·-uW
dW
F;\0 ()
X z dX
JxII·····
X
·----"=--'JI
When X :;:: XI , \V.:::: 0
In } .- ~L
.1 ",., X 2
=_~_I~.Ao. (~_?_ )[ I .m (I ' ', u W) 3/ 2 ]
1"",0. 5 1
F\o
3a
In [ l"::x-~. ""'atmkg 20dtm
lO3
'."
2
3("0.oi8)""
[
'.'
1~
1-,,(I-.O.018kg )Okg)
k(yb
X2
=0.756
4-39
3{2]
P4-18 (c)
C)
P == PO(!··' (lVV')
I·)
ito
(
;=
1
.')t(Z
20 aIm J" 0018kg- (50kg) , = 63 alm
P4-18 (d)
For turbulent flow
(~J2 = 0.0142kg-l
l Dpz J2 (Ac.lAc2 J2 =(0.018kg- l )(,,~)2
1
1.5
a 2 = a (D Pl
P exit = P o(1- UW)05 = (20atm)(1-0.0142kg·!(50kg»1!2 = 10 ..7 atm
3
In[1-0.5]= 10-.
1-X2
atmkg
)][1-(1-0.0142kg - l (50kg
l
3 0.0142kg-
20atm( (
2
)t2]
X 2 = 0.,77
P4-19
Production of phosgene in a microreactor .
CO + Cl 2 -> COCl2 (Gas phase reaction)
A+B ->C
See Polymath program P4·19.pol.
.POLX,MA I!lResults
Calculated values of the DEQ variables
Variable
W
X
Y
e
FAO
FBO
Fa
Fb
vO
v
Fe
Ca
Cb
a
k
rA
Ce
initial value
0
0
1
-0.5
2"OE-05
2,OE-05
2.0E-05
2"OE-05
2,,83E-07
2,,83E-07
0
70,,671378
70,,671378
3.55E+05
0,,004
-19.977775
0
minimal value
0
0
0.3649802
-0,,5
2"OE-05
2"OE-05
4.32E-06
4,,32E-06
2,83E-07
2" 444E-·07
0
9,1638708
9.1638708
3,,55E+05
0,,004
-19,,977775
0
maximal value
3,,5E-06
0,,7839904
1
··0" 5
2.0E-05
2.0E-05
2,OE-05
2"OE-05
2,,83E-07
4,,714E-07
1,,568E-05
70.671378
70,,671378
3.55E+05
0.004
-0.3359061
53,532416
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = -rA/FAQ
r2 J d(y)/d(W) = -a*(1 +e*X)/(2*y)
Explicit equations as entered by the user
4-40
final value
3.5E-06
0.7839904
0,,3649802
-0.,5
2"OE-05
2.0E-05
4,32E-06
4 ,. 32E--06
2,.83E-07
4,. 714E-07
1" 568E--05
9,,1638708
9.1638708
3,,55E+05
0.004
-0.3359061
33,,259571
[1]
e
=-.5
=
FAO = 2e-5
Fa = FAO*(1-X)
[6] vO = 2 . 83e-7
[8] Fe FAO*X
[ 10] Cb Fblv
[2]
[4]
[3] FBO FAO
[5] Fb = FBO-FAO*X
[7] V vO*(1 +e*X)/y
[9] Ca Fa/v
[11] a = 3.55e5
[13] rA = -k*Ca*Cb
=
=
=
=
[12]
[14]
k=.004
Ce
= Fe/v
P4-19 (a)
0 . 0 ..- - - - - - - - . - - - - - - - - - ,
1.0
0.0
0.8
0.0
0.6
[j
Fa.
- Fb
- Fe
0.0
0.4
0.2
0.0
0.0
...--------------_=____,
IL--_ _ _ _ _
OOe+O
7.0('·· '7
~
__
~
_ _ _ ____l
lAe-6 W 2.1e-6
2.8e-6
0.0
.3.5l'-6
L -_ _ _ _ __
O.Oe+O
7.0l'-7
1.4e-6\V2.1e-6
2.8e-6
P4-19 (b)
The outlet conversion of the reactor is 0 . 784
The yield is then MW*FA *X = 99 g/mol * 2 e-5 molls * 0 . 784 = .001.55 g/s = 48 . 9.5 g/ year.
Therefore 10,000 kg/year / 48 . 95 kg/ year = 204 reactors are needed.
P4-19 (C)
Assuming laminar flow, a - D p ·2, therefore
a z = a1 .D~1 = (3 ..55x105kg -1) 4 = 14.2x105kg- 1
DP2
4-41
3.5e-6
-5.---------------,
2.0 ...
16...-5
10 . - - - - - - - - - - - - - - - - ,
0.8
lj
1.2\'-5
80\'-6
"F:l
•. lih
0.6 .
- Fe
OA
~."
W
':'()e-':'
OOHO
l..4e-6 W2.1e-6
2.8e-6
0. 01'+0
_
_
'
_
~
_
-
0.0
3.5('-(;
L
O.OHOIL-----~-----·----l
I
02
_ _ _ _ _ _ _ _ _ __ _ l
':'.Oe-':'
1.41'-6 ",,2 ..1e-6
2.8e-6
3 . 5e-6
P4-19 (d)
A lower conversion is reached due to equilibrium, Also, the reverse reaction begins to overtake the forward
reaction near' the exit of the reactor"
P4-19 (e) Individualized solution
P4-19 (f) Individualized solution
P4-19 (g) Individualized solution
P4-20 (a)
Mole Balance
_4~
=z lA' IF ...
dW
. . .o
Rate Law
k k[!:r(
k' :.:;;; 11 ;;;;
t!>
com t!> ""'1)J
<l> ':' c Dp
4-42
k'
'''''-'''''~'--'''''-''''-~'---- i1 - 1 k' = Tl k, k = -11
1.0·H;:::::::::::;1..-.._
Largen..
cD p
~
Increasing Particle Size
Then
3 . ""
11 == 3 :;;:: -_
~-,
cDp
<l>
when Dp::: 2 mm, k' ::: 0.06, 11:::;:
{,= :~6,:::;; 0.02
3
0.02=: c(2)
,:=75
[~_~J~
For turbulent flow:
2{3
"
PoAop(l- fjJ)
a =-
constant
=>a=-----Dp
a
aoDpo
Dpl
=--"-----.:..:..
P4-20 (b)
See Polymath program P420"b.pol.
4-43
k'
= -1 = k" = 3
3.0
2.4
Q
1.8
1.2
0.6
0.0
0.0
04
0.8
16
Dp 12
2.0
P4-20 (C)
Gas, £=0, C A =C"Q(l-X)y
~~. iy' whele a~a{~;)
Ct,
~ pJi~)p,.~ =2oru;;;6---Jl%:~"s'i:~!ki:S2d;;;'r 708 x 10' kg'
and
?Y
;:;;;
dW
a
2)'
See Polymath program P4 . 20·e.pol.
POLYMA TH Results
Calculated values of the DEQ variables
Variable
w
X
Y
Dp
Q
Fao
initial value
0
0
1
0 . 0075
0.5625
5
minimal value
0
0
O. 2366432
0 . 0075
0 . 5625
5
maximal value
100
0 . 5707526
4-44
1
0 . 0075
0 . 5625
5
final value
100
0 . 5707526
0 . 2366432
0 . 0075
0 . 5625
5
alpha
Cao
kprime
0 . 00944
0.207
2.9385672
-0.1259147
ra
0 . 00944
0.207
2 . 9385672
-0.1259147
0 . 00944
0.207
2 . 9385672
-0.0012992
0.00944
0 . 207
2.9385672
-0 . 0012992
ODE Report (RKF45)
0.90
Differential equations as entered by the user
[1] d(X)/d(w) = -ra/Fao
[2] d(y)/d(w) = -alpha/21y
0.72
Explicit equations as entered by the user
[lJ Dp = . 0075
[2J Q 75*Dp
[3J Fao 5
[ 4 J alpha = .0000708/Dp
[5J Cao .207
[ 6 J kprime = 3*(3/QA2)*(Q*coth(Q)-1)
[7 J ra = -kprime*(Cao*(1·X)*y)A2
=
0.54
=
=
036
0.J8
0.00
0.0
0.1
0.2
Dp 03
0.4
0.5
P4-20 (d) Individualized solution
P4-20 (e) Individualized solution
P4-20 (1) Individualized solution
P4-20 (g) Individualized solution
P4-20 (h) Individualized solution
P4-21 (a)
Assume constant volume batch reactor
Mole balance:
dX
CAD -
dt
= -rA
Rate law and stoichiometry: - rA
= kCA = kCAD (1- X)
Specific reaction rate: k ( 25° C) = 0.0022 weeks- l
Combine:
dX
--1
f----·= --In (1- X)
kC (1- X)
k
x
t=
CAD
D
AD
--}
52.2 weeks = 1 In (1- X)
0.0022 weeksX =0.108
CA = CAD (1-·· X) but since volume and molecular weight are constant the equation can be written as:
m A = mAD (1- X)
4-45
6500IU
mAO
= mAO (1-0.108)
= 7287IU
%OU = CAO -CA *100 = 7287 -6500 *100 = 12.1 %
CA
6500
P4-21 (b)
10,000,000 lbs/yr = 458 * 109 glyr of cereal
Serving size = 30g
Number of servings per year = 458 * 109 /30 =1.51 * 109 servings/)'1
Each serving uses an excess of787 IU = 4 . 62 * 10-4 = 1.02 * 1O-6 1b
Total excess per year = (L51 * 108 servings/yr) * (102 * 1O-6 Ibs/serving) = 154111b/yr
Total overuse cost = $100/lb * 154.1 Ilb/yr = $15411 /)'1. (trivial cost)
P4-21 (C)
If the nutrients are too expensive, it could be more economical to store the cereal at lower temperatures
where nutrients degrade more slowly, therefore lowering the amount of overuse . The cost of this storage
could prove to be the more expensive alternative. A cost analysis needs to be done to determine which
situation would be optimaL
P4-21 (d)
k ( 40° C) = 0.0048 weeks -)
6 months = 26 weeks
-1
fo kCAOdX(1- X )= -In
(1- X )
k
x
t = CAO
-1
26weeks =-)In(1- X)
0.0048 weeksX =0.12
CA = CAO (1 - X) but since volume and molecular weight are constant the equation can be written as:
m A =mAo(1-X)
6500IU =mAo(1-0.12)
mAO
= 7386IU
%OU = CAO -CA *100= 7386-6500_*100=l3.6%
CA
6500
P4-22 No solution necessary
P4-23
CH20HCH2CI + NaHC0 3
-7
(CH2 0Hh + NaCI + CO 2
4-46
A
FAa
+
c
B
= 0.1 mol/min = 6 mol/hI'
+ D
+ E
P4-23 (a)
Mole balance:
dC A
dt
= r + Vo (C
V
!:!CB = r
dt
- C )
AO
+ V0
V
r A = -kCACB
Rate law:
A
(_
C )
dCc
dt
B
= -r + Vo (-- C
V
V =Vo +Vot
See Polymath program P4-23-a.pol.
POLYMA ~H Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
Cc
Faa
Caa
k
Va
va
V
r
Xb
Nc
initial value
0
0
0.75
0
6
1.5
5.1
1500
4
1500
0
0
0
minimal value
0
0
1. 395E-13
0
6
1.5
5.1
1500
4
1500
-0.0039398
0
0
maximal value
250
0.15
0.75
0.829586
6
1.5
5.1
1500
4
2500
0
1
2073.9649
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = r+(voN)*(Cao-Ca)
[2] d(Cb)/d(t) = r+(voN)*(-Cb)
[3 J d(Cc)/d(t) = -r+(voN)*Cc
Explicit equations as entered by the user
[1] Fao =6
[2J Cao = 1..5
[3 J k = 5.1
[4] Vo = 1500
[5] vo = Fao/Cao
[6] V=Vo+vo*t
[7] r = -k*Ca*Cb
[8] Xb = 1-Cb*V/(O.75*1500)
[9] Nc = Cc*V
4-47
final value
250
0.15
1.395E-13
0.829586
6
1.5
5.1
1500
4
2500
-·1.067E-13
1
2073.9649
)
C
LO
O,Oe+O"-j-----------~>
....-----------~
D
0. 8
-8.0e-4
0.6
-1.6e-3
04
-2 Ae-·31
/I
II
II
l
I
!
".
/
/
/
"P,""";'#'/
-3.2e-3 \
1
/
~/~
1" ¢~,#,,r
1
_-I.Oe_3l.!s:::..,"-~--~--~--- __- - : '
,
0.0
0
50
100 t
150
200
,/.i'
o
250
50
100 t
150
200
250
3000.---------------,
0.90....------·
- Ca
2400
0.72 '
054'
036
600
1118
o,ool.::------------~~;...
o
50
100 t
150
200
250
o I)
50
100 t
150
200
P4-23 (b)
Taking average time for activities like charging, heating, cleaning = 4..5 hr
So, if there is one batch per day, the time for reaction = 24-4.5 hr = 19.5 hr
Since at the temperature at which we are operating the reactor no side reactions occur, the quickest way to
run the reaction will be at the highest flow rate of A (2 mol/min or 120 mol/hour)
But at 19.5 hours and a flow rate of 120 mol/hr, 1560 dm3 would be added to the reactor.. Because the
volume of the reactor is 2500 dm3 , and there is already 1500 dm3 of B in the reactor, the reaction cannot not
run for 19.5 hours at a flow rate of 120 mol/hr (2 mol/min)
Also, because there is 1500 dm3 ofB at a concentration of 0.75 M, there are only 1125 moles ofB to react.
This means about 750 dm3 of 15 M A is all that can be reacted. More A may be added to the reactor to
keep the reaction rate high as the concentration of B drops, but adding twice the necessary volume would be
a waste of time and material (on top of being physically impossible !)..
To add 1125 moles of A at a rate of 120 mol/min takes 9375 hours. Using the Polymath code from part (a)
and changing Fao to 120 and the time to 9375 gives 1107 moles of C. Allowing the reaction to go for 9.5
hours results in 1115 moles of C and a reaction time of 10 hours gives 1124 moles of C.
Now consider multiple batches per day.. If two batches are run, then there will be 9 hours of downtime,
meaning that the time for the reaction will be 15 hours - split between the two batches, with a maximum
4-48
batch time of 10 hours., The following table shows the tluee possible times for reactions and the moles of C
that are formed from the two batches.
Maximum Batch Time
(hr)
10
9
Total Moles of C
Formed
1723
1790
1795
..-
8
So the best setup will be to run 2 batches per day. One batch will run for 8 hours and the second batch for 7
hours. Both will be run with a flow rate of 2 mol/min of A.
(If tlu'ee batches are run there will be 135 hows of downtime and only 105 hours for the reaction. A
maximum of 1257 moles of C can be formed if the time is split evenly for each batch (3.5 hows).)
See the Polymath code from part (a) and vary time and flow rate.
P4-23 (C)
F AO = 0.15 mol/min = 9 mol flu
vo = FaofCa = 9 molJlu f 1.5 mol/dm3 = 6 dm3flu
3
1000 dm3 is needed to fill the reactor. At 6 dm flu it will take 166.67 hows
Now solving using the code from part (a) with the changed equations:
See Polymath program P4-23-e.pol.
1.0 , - - - - - - - . - - - - ,
0.8
0.6
0.4
66
t
99
132
165
P4-23 (d) Individualized solution
P4-24
NaOH + CH3 COOC2 H s ------t CH 3 COO- Na+ + C2 H sOH
A+B~C+D
4·49
Mole balance:
dec =-r+ va
dt
V
(-c)
C
Rate law:
V=Va+Va t
k =koex p (
!Uo -nJ
To produce 200 moles of D, 200 moles of A and 200 moles of B are needed. Because the concentration of
A must be kept low, it makes sense to add A slowly to a large amount of R Therefore, we will start with
pure B in the reactor. To get 200 moles of B, we need to fill the reactor with at least 800 dm3 of pure R
Assume it will take 6 hours to fill, heat, etc . the reactor. That leaves 18 hours to cany out the reaction. We
will need to add 1000 dm3 of A to get 200 moles in the reactor. We need to check to make sure the reactor
can handle this volume if only 1 batch per day is to be used . Since we add 1800 dm3 or 18 m3 and the
reactor has a volume of 4.42 m3 we can safely carry out a single batch per day and achieve the necessary
output of ethanol.
Now vary the initial amount of B in the reactor, the flow rate of A, and the temperature to find a solution
that satisfies all the constraints . The program below shows one possible solution .
See Polymath program P4--24.pol.
POLY1VIA TH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
Cc
Cd
ko
Fao
Cao
Vo
vo
initial value
o
o
0 . 25
minimal value
o
o
0 . 0068364
o
o
o
o
5.2E-05
0.04
5 . 2E-05
0 . 04
0.2
1200
0. 2
1200
0.2
308
1.224E-04
ra
o
308
1..224E-04
V
Nc
1200
1200
o
o
T
k
maximal value
6 . 5E+04
0.1688083
0 . 25
0 . 0151725
0 . 0151725
5.2E-05
0 . 04
0.2
1200
0. 2
308
1 . 224E-04
1.397E--06
1 . 42E+04
202 . 92284
0.2
o
ODE Report (RKF45)
Differential equations as entered by the user
[11 d(Ca)/d(t) = -ra+(vo/V)*(Cao-Ca)
[2] d(Cb)/d(t) = -ra-(vo/V)*Cb
[3] d(Cc)/d(t) = ra-(vo/V)*Cc
[ 4] d(Cd)/d(t) = ra-(voN)*Cd
Explicit equations as entered by the user
4-50
final value
6.5E+04
0 . 1688083
0 . 0068364
0.0142903
0 . 0142903
5.2E-05
0 . 04
0. 2
1200
0.2
308
1 . 224E-04
1.. 412E-07
1 . 42E+04
202 . 92284
[ 1] ko == 5 . 2e-5
[2] Fao == .04
[3] Cao ==.2
[4] Vo == 1200
[5] vo == Fao/Cao
[ 6] T == 35+273
[7] k == ko*exp«42810/8.3144)*(1/293-1fT))
[8] ra == k*Ca*Cb
[9] V == Vo+vo*t
[10] Nc == Cc*V
P4-25 (a)
A ~ B + 2C
To plot the flow rates down the reactor we need the differential mole balance for the three species, noting
that BOTH A and B diffuse through the membrane
dFA =r -·R
dV
A
A
dFB =r -R
dV
B
B
dFc
--=r.
dV
c
Next we express the rate law:
First-order reversible reaction
Transport out the sides of the reactor:
kACroF
A
RA =kAC A = -.!.!......!..!!.-!.!..
FI
kBCroF
Fr
B
RB = kB CB = --"'--'-""--':::..
Stoichiometery:
Combine and solve in Polymath code:
See Polymath program P4-25-·a.pol.
POLYMA TH Results
Calculated values of the DEQ variables
4-51
Variable
v
Fa
Fb
Fe
Ke
Ft
Co
K
Kb
ra
Ka
Ra
Rb
Fao
initial value
0
100
0
0
0.01
100
1
10
40
-10
1
1
0
100
0
X
final value
20
57.210025
1. 935926
61.. 916043
0.01
121.06199
1
10
40
-0.542836
1
0 . 472568
0.6396478
100
0.4278998
maximal value
20
100
9.0599877
61.916043
0.01
122.2435
1
10
40
-0.542836
1
1
2.9904791
100
0.4278998
minimal value
0
57 . 210025
0
0
0.01
100
1
10
40
-10
1
0 . 472568
0
100
0
ODE Rel!ort {RKF4S1
Differential equations as entered by the user
[1 ] d(Fa)/d(v)
ra .. Ra
[2] d(Fb)/d(v) ··ra .. Rb
[3 ] d(Fe)/d(v) -2*ra
=
=
=
Explicit equations as entered by the user
[1] Ke = 0..Q1
[2 J Ft = Fa+ Fb+ Fe
[3 ] Co = 1
[4]
[5]
[6]
[7 ]
[8 ]
[9]
3
K= 10
Kb = 40
ra = - (K*Co/Ft)*(Fa- CoI\2*Fb*FeI\2/(Ke*FtI\2))
Ka = 1
Ra = Ka*Co*Fa/Ft
Rb = Kb*Co*Fb/Ft
r-...,....,---------.-----
100
2
80
2
60
1
40
~---
1
0
0
..j
8
~.~
16
20
v
20
..., ..........--.-~-..........
(I
12
o
4
8
y
12
16
P4-25 (b)
The setup is the same as in part (a) except there is no transport out the sides of the reactor .
See Polymath program P4-25-b.pol.
POL YMAIH Results
4-52
20
Calculated values of the DEQ variables
Variable
v
Fa
Fb
Fc
Kc
Ft
Co
K
ra
Fao
X
initial value
0
100
0
0
0.01
100
1
10
-10
100
0
final value
20
84.652698
15.347302
30.694604
0.01
130 . 6946
1
10
-3.598E--09
100
0.153473
maximal value
20
100
15.347302
30 . 694604
0.01
130.6946
1
10
-3.598E-09
100
0.153473
minimal value
0
84.652698
0
0
0.01
100
1
10
-10
100
0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(v)::: ra
[2] d(Fb)/d(v) ::: -ra
[3] d(Fc)/d(v) = -2*ra
Explicit equations as entered by the user
[1] Kc = 0.01
[2] Ft = Fa+ Fb+ Fc
[3] Co=1
[4] K=10
[5] ra = - (K*Co/Ft)*(Fa- CoA2*Fb*Fc"'2/(Kc*FtA2»
-------------------
0.50
0 . 45
0.40
0.35
0.30
[ --]
- X PFR
0.25
... X l\fembmn.
0.20
-.---..----.-
............
0 . 15
~------
............-.--.-...
--.. .....--_........_---_...
""
0.10
0 . 05
0 . 00
t -_ _ _ _ _ _ _ _ .. ________
0
2
4
6
8
~
10
___
V 12
14
16
P4-25 (C) Conversion would be greater ifC were diffusing out
P4-25 (d) Individualized solution
4-53
18
20
P4-26
co
+ H20 ...... CO2 + H2
A+B ...... C
+D
Assuming catalyst distributed uniformly over the whole volume
dF
dF
Mole balance:
A =r
._-
Rate law:
r = rA
= rB=C
-.y'
dW
RH2
=
KH2
dF
B =r
--
dF
dW
e = -r
--
dW
- -D= - r - R
dW
= -1D = -k
[c
H2
C _ CeC
AB
K D]
eq
CD
Stoichiometry:
FD
CD=C ro FT
FT
= FA + FB
+ Fe + FD
Solving in polymath:
See Polymath program P4-26.pol.
POLYMA TH Results
Calculated values of the DEQ variables
variable
---W
Fa
Fb
Fe
Fd
Keq
Ft
Cto
Ca
Cb
Kh
Ce
Cd
Rh
k
r
initial value
0
2
2
0
0
1,,44
4
0,,4
0,,2
0.2
0,,1
0
°°
1.. 37
-0,,0548
minimal value
maximal value
0
0,,7750721
0,,7750721
0
0
1.44
3.3287437
0.4
0.0931369
0.0931369
0.1
0
0
0
1.. 37
-0,,0548
100
2
2
1,,2249279
0.7429617
1,,44
4
0.4
0.2
0.2
0.1
0.147194
0,,0796999
0,,00797
1..37
-0,,002567
ODE Report (RKF45)
Differential equations as entered by the user
[ 1 J d{Fa)/d{W) = r
[2 J d{Fb)/d{W) = r
[ 3 J d{Fc)/d{W) = -r
[ 4 J d{Fd)/d{W) =- r -Rh
Explicit equations as entered by the user
[1] Keq = 1..44
[ 2] Ft = Fa+Fb+Fc+Fd
[3] Cto = 0.4
4-54
final value
100
0.7750721
0,7750721
1.2249279
0,5536716
1. 44
3,,3287437
0,,4
0" 0931369
0,,0931369
°"
1
0,,147194
0.0665322
0.0066532
1..37
-0,,002567
[4] Ca = Cto*FalFt
[ 5] Cb =Cto*Fb/Ft
[6] Kh=O,,1
[ 7] Cc = Cto*Fc/Ft
[ 8] Cd = Cto*Fd/Ft
[9] Rh = Kh*Cd
[10] k= 1 . 37
[11] r -k*(Ca*Cb-Cc*Cd/Keq)
=
For 85% conversion, W
= weight of catalyst = 430 kg
In a PFR no hydrogen escapes and the equilibrium conversion is reached.
= CeCD = C~OX2
K
eq
CACB
=_
C~o(1 __ X)2
X_2_ = 1.44
(l_X)2
solve this for X,
X= 5454
This is the maximum conversion that can be achieved in a normal PFR.
If feed rate is doubled, then the initial values of Fa and Fb are doubled. This results in a conversion of 459
2 . 0 r;;----;::====;_
1.6
_..,,,.,---_.....
1.2
--
...
0.8
04
20
·40
W 60
80
100
P4-27 Individualized solution
P4-28 (a)
Assume isothermal and £=0
therefore,
P=Po(l-aWl5
1=10 (1-.01 glW).5
W=99g
P4-28 (b)
4-55
dX -rA
dW= FAO
.!..=(l-Wa)-'
Po
-rA =kCA
CA ==CAo(l-X) PlPo
CA =CAo(1 -X) (1 - 0.01 W).5
Integrate from X=O to X=.9
ax = kCAO (1-X) (l-.01W).5
dW
FAo
W=59.88g
Fust 5% conversion integrate from X=O to X=.OS
W=1.31 g
Last 5% conversion integrate from X==.85 to X=.90
W:alO.85 g
P4-29 Individualized solution
P4-30 (a)
First order gas phase reaction,
C6HjCH(CH3h -7 C6~ + C3~
YAO = 1, g = 1 ~ [; = 1
P
kC
Y = - , a = -AO
-
Po
FAO
For aPBR:
dX
dW
a(1-X)
= (1 +
iT
dy
(I+X)
dW
2y
-=
y ........................ (1)
a" ....................... (2)
X = 0.064 and y = 0 . 123,
Solving (1) and (2) by trial and enOl on polymath we get,
a = 0.000948 (kg of catalystr1
a = 0.000101 dm- 3
4-56
Now solve for a fluidized bed with 8000 kg of catalyst.
FAOX
Mole balance: W = - -rA
Rate law: -fA = kCA
I-X
I+cX
Stoichiometry: CA = CAO - - Combine: W
= X (I + eX) = 8000 = ___. X (I + X)
a(1-X)
O.OOOlOI(I-X)
X =0.37
P4-30 (b)
ForaPBR:
dX _ a(1-X)
dW - (I+XfY
dy _ (I+X)
---- - ' - " - - a
dW
2y
where
a = kCAO
FAO
From chapter 12 we see that k will increase as Dp decreases_ We also know that for turbulent flow
I
a-· - .. This means that there are competing fOIces on conversion when Dp is changed .
Dp
We also know that alpha is dependant on the cross-sectional area of the pipe:
I
a - ._--
Ac
But alpha is also a function of superficial mass velocity (G). If the entering mass flow rate is held constant,
then increasing pipe diameter (or cross-sectional area) will result in lower superficial mass velocity . The
relationship is the following for turbulent flow:
I
2
G - - and a-G
Ac
I
therefOIe, a _. -'2 .
~
If we combine both effects on alpha we get the following:
I
I
a--Ac~
I
a-~
So increasing pipe diameter will lower alpha and increase conversion and lower pressure drop .
4·57
For Laminar flow:
1
a-·D2
p
so decreasing the particle diameter has a larger effect on alpha and will increase pressure drop resulting in a
lower conversion
1
For Laminar flow a·- G and so a .-. -2 .
A;
This means increasing pipe diameter will have the same trends for pressure drop and conversion but will
result in smaller changes.
P4-30 (c) Individualized Solution
P4-31 (a)
K = 0. . 0.5
= 0..33(1-3) =-0..666
P AO = 0..333*10 F Ao = 13.33
Mole balance:
dXldW = -ralFao
rA = -KPB
Rate law:
P A= P AO(l - X)/(l - o..666X)
Pc = P AoXl(l - o. . 666X)
For a = 0., Y= l(no pressure dlOp)
I>
10..---------------#_-
0. 00
8
-0.08
6
-0.16
.-----------.--$-.__---:=
1
4
-0.24
2
-032/
o I)
-040
20
·to
W 60
80
/
L -_ _
o
100
~
_____
20
40
~_.
\V 60
See Polymath program P4-31-a . pol.
POL):''l~'IA!!1
Results
Calculated values of the DEQ variables
Variable
initial value
-------
X
o
K
0.05
3 . 33
3 . 33
6 . 66
0 . 05
3 . 33
0.0406732
0 . 0813464
Pao
Pa
Pb
Pc
Fao
ra
o
minimal value
o
o
W
o
13 . 33
-0 ...333
o
13 . 33
-0.333
maximal value
100
0.995887
0 . 05
3.33
3 . 33
6 . 66
9.8482838
13 . 33
-0 . 0040673
4·58
final value
100
0.995887
0 . 05
3 . 33
0 . 0406732
0.0813464
9.8482838
13.33
-0 . 0040673
_ _ _....J
80
100
esp
-0.666
-0 . 666
-0 . 666
-0 . 666
ODE Report (RKF45)
Differential equations as entered by the user
[1 j d(X)/d(W) = -ra/Fao
Explicit equations as entered by the user
[1] K = 0.05
[2] Pao = 0 . 333*10
[3] Pa = Pao*(1 - X)/(1 - 0 . 666*X)
[4] Pb = 2*Pa
[5] Pc = Pao*XI(1 - 0 . 666*X)
[6] Fao = 13 . 33
[7] ra = -K*Pb
[8] esp = -0.666
For first 5% conversion weight required = WI = 2 kg
For last 5% conversion weight required = W 2 = 38 kg
Ratio = Wi WI = 19
Polymath solution (4-34 a)
P4-31 (b)
For a = 0 . 027 kg· l ,
Polymath code with pressure drop equation:
See Polymath program P4·31-b.pol.
POLYMA TH Results
Calculated values of the DEQ variables
Variable
w
X
Y
K
Pao
Pa
Pb
Pc
Pao
ra
esp
alfa
initial value
0
0
1
0.05
3 . 33
3.33
6 . 66
0
13.33
-0.333
-0.666
0 . 027
minimal value
0
0
0.1896048
0.05
3 . 33
0.4867002
0.9734003
0
13 . 33
-0 . 333
-0.666
0.027
maximal value
30
0.4711039
1
0.05
3.33
3.33
6 . 66
1.0583913
13.33
-0.04867
-0.666
0.027
final value
~----
0.4711039
0.1896048
0.05
3.33
0.4867002
0.9734003
0.4335164
13 . 33
-0 . 04867
-0 . 666
0.027
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(X)/d(w) = ·ra/Fao
[2] d(y)/d(w) = -alfa*(1 - esp*X)/(2*y)
Explicit equations as entered by the user
[1] K=0 . 05
[2] Pao = 0.333*10
[3] Pa = Pao*(1 - X)*y/(1 - 0 . 666*X)
[4] Pb = 2*Pa
[5 J Pc = Pao*X*y/(1 - 0.666*X)
[6] Fao=13.33
7.0
5.6
42
2.8
14
4-59 0.0 OC""""-~---'---~--~---'30
6
12
W 18
24
ra = -K*Pb
esp -0.666
alfa = 0,,027
[7
=
[8
[9
0.00,---------------,
1.0 . . . . . , . . . - - - - - - . - - - - - - - - ,
0.8
0.6
0.4
0.2
-040~-~--~--~-----~
o
6
12
W 18
2-t
0.0
L -_ _ _ _ _
o
30
6
~
_ _ _ _ _ _ _- - '
12
W 18
2·4
30
P4-31 (C)
1) For laminar flow:
Diameter of pipe = D and diameter of particle = Dp
Now DIlDo = 3/2 so G I = 4/9Go
a = (constant)(GID/)(lIAc)
So
a = ao(G/Go)(Dpof Dpl)2(DofDI)2
= ao(4/9)(2/3)2(2/3)2 = 0 . 00237 kg· 1
Less pressure drop and more conversion for same weight of catalyst as in part (b).
2) For turbulent flow:
~
G2IDp
a = (constant)(G2IDp)(lIAc)
So, a = ao(G I/Go)2(Dpof D pl )(DofDl)2
= ao(4/9)2(2/3)(2/3)2 = 0.0016 kg·!
Again less pressure drop and more conversion for the same catalyst weight
eX)
It is better to have a larger diameter pipe and a shorter reactor, assuming the flow remains the same as
through the smaller pipe.
P4-32 (a)
At equilibrium, r = 0 =>
cAeB
4-60
1.O,-:::::==========;
0.8
0.6
04
0.2
OO~~~------~----------~
0081:0
=>
CAO (l-X(CBO ~t-C
V
AO
o
AO
=> t= VOC [X2
VOC BO ,Kc(l'-X)
+
1 0..1:5 t 1 56E5
5.181:4
2.07E5
2591:5
xJ= (CAOXY
K
C
x]
Now solving In polymath.
See Polymath program P4-32-,a.pol.
P4-32 (b)
See Polymath program P4-32-b.pol.
POLYMATH Results
Calculated values of the DEQ variables
Var~able
t
Ca
Cb
Cc
Cd
Kc
k
ra
va
Va
V
X
initial value
0
7.72
10.93
0
0
1.08
9.0E-Os
-0.0075942
0.05
200
200
0
minimal value
0
0 2074331
7.6422086
0
0
1. 08
9 OE-05
--0.0075942
0.05
200
200
0
---
maximal value
1.SE+04
772
10.93
3.2877914
3.2877914
1 08
9.0E-OS
··1.006E-Os
0.05
200
950
0.9731304
final value
'-L5E+O-40.2074331
9.51217
1.41783
1. 41783
1 08
9.0E-Os
-·1.006E-OS
0.05
200
950
0.9731304
1.0e+o
,
8.0e--l
ODE Report (RKF4S)
Differential equations as entered by the user
6.0e-l
..,.'
. /',;',."
..
.,.~ ~,,,"~ ~~
.........
-.--
~-"
..
~ ... ~- . . .
¥ ...
~~
[oJ
--_
..
I
f
".
!
-,
..
I
I
40e--l
2.0e--l
4-61
O.Oe+o 0
2976
5952t
8928
11904 14880
[1
[2
[3
[4
d(Ca)/d(t) = ra - Ca*voN
d(Cb )/d(t) = ra - voN*(Cb- 10 . 93)
d(Cc)/d(t) = ora - vo*CcN
d(Cd)/d(t) = ora - vo*CdN
Explicit equations as entered by the user
[1 J Kc = 1.08
[2] k = 0,00009
[3] ra = -k*(Ca*Cb - Cc*Cd/Kc)
[4] vo = 0,05
[5] Vo = 200
[ 6] V =Vo + vo*t
[7] X = 1 - Ca/7.72
2(le+l . . . . - - - - - - - - - - - - . ,
-1.6e-3
.'
c;:J<
• ell
1.6e+1
D
-3.2e-3
.~-n
L2e+l
I\'-_--------l
-4.8e-3
8.Ue+O
·6 ...:/e-3 '
'.O'ffi~
-8.0e-3
Lo
2976
5952
t
8928
t),Ue+O
1190-1 1,4880
o
2976
5952
t
8928
11904 1-1880
Polymath solution
P4-32 (c)
Change the value of Vo and CAO in the Polymath
program to see the changes
1 II , - - - - - - - .
08
P4-32 (d)
As ethanol evaporates as fast as it forms:
Now using part (b) remaining equations,
Polymath code:
CD=O
11,6
04
See Polymath program P4·32··d.po!.
POLYMATH Results
Calculated values of the DEQ variables
11 . 2
Variable initial value minimal value
maximal value final value
t
0
o
7.72
0.0519348
Ca
10.,93
6.9932872
Cb
0.0
k
ra
va
Va
9.,OE-05
--0" 0075942
0,,05
200
9,OE-05
-0,0075942
0,,05
200
'----~---------~----'
I)
6000
7 . 72
10,,93
9"OE-05
-3,,69E-05
0.05
200
4-62
1189
2378
t
3567
6000
0,,0519348
7.8939348
9"OE-05
-3 .. 69E-05
0.05
200
4756
5945
V
X
200
200
0
500
0_9932727
500
0_9932727
o
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra - Ca*voN
[2] d(Cb)/d(t)
ra - voN*(Cb- 10.93)
=
Explicit equations as entered by the user
=
[1 J k 0.00009
[2] ra = -k*Ca*Cb
[3] vo = 0.05
[4J Vo = 200
[5] V
[6] X
=Vo + vo*t
= 1 - Cal7. 72
------.-------
20
O.Oe+O~-----
16
12
8
""'--.".,,~.....
--
-... ..-....
"
.........................--.'.........,._..... .,..
4
-8.0e-.3 0
{I
-~---~::-::--~-------
1189
2378
t .3567
4756
5945
o
P4-32 (e) Individualized solution
P4-32 (0 Individualized solution
P4-33 (a)
Mole balance on reactor 1:
- dNAl
CAlv-rAlV - - dt
C
dN
-.AQ. v -C v - r V=--..A!...
2 0
Ai
Al
dt
CAOV AO -
.
with
1
2
_
V AO - - Vo
Liquid phase reaction so V and v are constant
CAO
CAl
2r
r
--- - ---- - r
Al
_ dCAl
- -----
dt
Mole balance on reactor 2:
4-63
--~-:----....::::::::::::::;:====~
1189
2378 t
3567
____--!
4756
5945
CAl
C A2
, _ dCA2
------r
--
T
T
dt
A2
Mole balance for reactor 3 is similar to reactor 2:
, _ dN
C A2V O - CA3VO -rA3V - - -A3dt
C
C
_ dC
A3
- A2
- - - -A3- r - - -
T
T
A3
dt
Rate law:
--rAi
=kCAiCBi
Stoichiometry
For parts a, b, and c C Ai = CBi
so that -rAi
= kC~i
See Polymath program P4-33"pol.
POL YMA TH Results
Calculated values of the DEQ variables
Variable
t
Cal
Ca2
Ca3
k
Cao
tau
X
initial value
0
0
0
0
0" 025
2
10
1
minimal value
0
0
0
0
0,,025
2
10
0,3890413
maximal value
100
0,8284264
0,,7043757
0.6109587
0.025
2
10
1
ODE Report (RKF45)
Differential equations as entered by the user
4-64
final value
100
0.8284264
0.7043757
0.6109587
0.025
2
10
0" 3890413
d(Ca1)/d(t) = (Cao/2 -Ca1)/tau -k*Ca1A2
[2] d(Ca2)/d(t) = (Ca1 - Ca2)/tau -k*Ca2A2
[3] d(Ca3)/d(t) (Ca2 - Ca3)/tau -k*Ca3A2
[1]
=
Explicit equations as entered by the user
[1] k = 0,,025
[2] Cao=2
[3] tau 10
[4] X = 1 - 2*Ca3/Cao
=
From Polymath, the steady state conversion of A is approximately 0.39
P4-33 (b)
99% of the steady state concentration of A (the concentration of A leaving the third reactor) is:
(0.99)(0.611) = 0.605
This occurs at t =
P4-33 (C)
The plot was generated from the Polymath program given above.
0.90 r - - - - - - - - - - - - - - - - - - ,
0.72
0.54
- Cal
0.18
- Ca2
- (:a3
o00 ~:....--........---~-----'------'.------I
.
0
20
40
t
60
80
100
P4-33 (d)
We must reexamine the mole balance used in parts a-c. The flow rates have changed and so the mole
balance on species A will change slightly. Because species B is added to two different reactors we will also
need a mole balance for species B .
Mole balance on reactor 1 species A:
CAOVAO -CAlv-rAIV
dNAI
.
= - - - wIth
dt
dN AI
-_·v -C v-r V=---
VAO
2
200
3
15
=--Vo and Vo = - -
2CAO
3
0
Al
M
~
Liquid phase reaction so V and v are constant
4-65
CAO
- - -CAl
- - r =dCAI
-27:
7:
Al
dt
Mole balance on reactor 1 species B:
,
_ dN BI
CBOV BO -CBIV-yBIV - - - and
dt
Stoichiometry has not changed so that -rAi = --lBi and it is a liquid phase reaction with V and v constant
CBO
C
, _ dC BI
- - - - -BI- Y - - 37:
7:
Al
dt
Mole balance on reactor 2 species A:
We are adding more of the feed of species B into this reactor such that V2 = Vo + VBO = 20
CA2V 2 -
CAlVO -
dNA2
rA2 V
=--
dt
Mole balance for reactor 3 species A:
-rA3V
C A2 V 2 - CA3 V 2
C
C
7:
2
3
, _ dCA3
A2- - -A3- Y
--
7:
2
dNA
=dt
A3
---
dt
Mole balance for reactor 3 species B:
C
,
_ dN B3
V --C V --,Y V B2 2
B3 2
A3
dt
C B2
C
, _ dCB3
-· - - -B3- Y
-A3
7:2
7:2
dt
Rate law:
4-66
See Polymath program P4-J3-d.po1.
POLYMA TH Results
Calculated vaiues of the DEQ variables
Variable
t
Cal
Ca2
Ca3
Cb1
Cb2
Cb3
k
Cao
tau
X
tau2
V
vbo
initial value
o
o
o
o
o
minimal value
o
o
o
o
o
o
o
o
o
0 . 025
2
13.333333
1
10
200
5
0.025
2
13.333333
0.3721856
10
200
5
maximal value
100
1.1484734
0.7281523
0 . 6278144
0.4843801
0.7349863
0.6390576
0.025
2
13.333333
1
10
200
5
ODE Report (RKF45)
Differential equations as entered by the user
[1 J d(Ca1 )/d(t) = (2*Cao/3 -Ca1 )/tau -k*Ca1 *Cb1
[2 J d(Ca2)/d(t) = Ca1/tau - Ca2/tau2 -k*Ca2*Cb2
[3 J d(Ca3)/d(t) = (Ca2 - Ca3)/tau2 -k*Ca3*Cb3
[4] d(Cb1 )/d(t) = (1 *Cao/3-Cb1 )/tau-k*Ca1 *Cb1
[5] d(Cb2)/d(t) = Cb1/tau+Cao*vboN-Cb2/tau2-k*Ca2*Cb2
[6 J d(Cb3)/d(t) = (Cb2-Cb3)/tau2-k*Ca3*Cb3
Explicit equations as entered by the user
[1] k=0.025
[2J Cao =2
[ 3 1 tau = 200/15
[ 4 J X = 1 - 2*Ca3/Cao
[5] tau2 = 10
[6] V = 200
[7] vbo = 5
Equilibrium conversion is 0.372 .
This conversion is reached at t = 85.3 minutes .
4-67
final value
100
1.1484734
0 . 7281523
0.6278144
0_4821755
0.7291677
0.6309679
0.025
2
13.333333
0.3721856
10
200
5
2.0 . - - - - - - - - - - - - - - - - - - ,
- Cd
1.6
- Cal
1.2
0.8
---,_."._" ......"" . ".,,---1
0.4
0.0
"""""=--~---'-----~--~----'
o
P4-33 (e)
20
40
60
80
100
Individualized solution
CDP4-A
CH3I + AgCl04
0.7 molll CE31
0.5 moli1 AgCI0
V '" 30 dm
o
;=
4
~
CH3Cl04 + AgI
CBO
~
CAD
3
3/2
C
tCH 1 ~ -k C;CH IAgCI0
3
4
S
k ~ 0.00042 (dm 3!mol)3!2(sec)-1 T=298 K
I "" 0.93
Integration for! "" 0.98, solved lll!lllerically
1 .
t =
~---.- ••.• ~
(24.18)
0.00042 (O.5)3fb
t "
v = Vo
CDP4-B
a)
4-68
1.628:t10
5
sec '" 4S.:2 hr
d=".
-"-'=1:"'
dt
<\..1
:VIol.! Babnce :
-
F ...-r
' "v
:~
,\
V~~~.
= C vdt'"
C v·· kC., V
.
~
';0...
0
.~s:.~ :::5: A. V.e,. _ S::{~.2.. :~~1
dt
V
V
Use POLYMATH: to plot C" vs. t
.... sc:;:
(b) C,a lIS t
1.,
Eqgal::ions '.
d(ca) Jd (t.) '" (cao·vo/v)·· (ca"" (k"v+vo) Iv)
,
,
cao=4 .. S
vo=70000
v",15000000
k"O.0025
t.O '" O.
~
I:: f
1000
':.&.aoC
I
.
~""~"'''"''''''''- ..f-.-.~",.""~- .. ~""",,,,,,,,,,,,,,,._,, .• ~ . ........... .,._.
:3..'%00
c,~
f:I.':X
o"coo
t
For steady-state:
t ::.:: 4.6 -...:.......
1 + 1:'k:
::::: 641 !:us
CA ;:;; .CADV. ::::2.930mgldm3
kV+v~
99% of this is 1.905 mg/dm \. which l5 below rhe standard of 3.0 mg/dm) .
Polymath solution(Ans CDP4-B-a)
b)
dN
dt
Mole Bal::Ulce :
Use POLYMATH
...: ....i..:=
[0
F - F -+ r V
.-I"
."
gener.ate plot of C ... and N"
Em;"- £.~2.!1:!:.
.~Ii¢l
d(nal/d( cl .. cao"vo·~ca"VOt;::;;-k~cil
cao",4.5
Ke.!
yo,,,,70000
...
'IS. L
T
- I
,.=
.
~
.._~ Na
vouc=50000
11:=0.0025
"=20000"1::
ca=na/v
t
Na vs t
f '" '150'
t
4-69
Ca vs t
.Ku
Ca
),000
•• /XX)
I,:)',)(l
'T
I
1,
i
.1
i
Oc,_
L---'
D..COO
----·---+---··---"-,-';O,,-,,,··,-"-+----+-I
lfPl,Q(:)Q
:n.a...:;oo
"iC...'C:OO
'tu.coo
Polymath solutionCAns CDP4·B·b)
c)
60
61
tjsc same equaIio[;s as in pan: (a),
07
but
0,3
4,9
40
d (ca 1 fd
(t:)
= (C;;to~".ro/v) -
lea· (}C"_'.-vo) Iv)
41
37
vo •• 7 OfJ-() 0
3,3
v'" ,:,5 ,,0 noO()
29
20
}<=0_0025
1:
f
.EQlymath solutionCAns CDP4·B·c)
d) This part is almost same as part(b) with minor changes:
V = 15000000 - 10000t
Vo= 80000 and Vout = 70000
The reason the graph looks so different from(a) is that pure water is evaporated, but water with atrazine is
coming in,
45
Polymath code:
d(ca)/dt=cao*vo/v)(ca (kv+vout) Iv)
ca(O)=4.5
#
vo = 80000
#
V = 15000000 - 10000*t
0.0025
k =
#
4.5
cao
#
vout = 70000
#
t(O)=O
#
t(f)=1000
44
43
42
41
40
39
3,8
3.7
36
3,5
100
200
300
400
500
600
700
800
900
101
Polymath solutionCAns CDP4-B-d)
CDP4-C
a)Find the number of moles of receptors:
10 cells
;5 receptors
1 mole receptors
l000mL
-10
I x 10 "'----" * 1 x 1O-m,--,,_- *,'''---------------''""- '*' '--'-'--'-"-.- == 1.66 x 10 M
mL
cell
6.022 x 1023 receptors
L.
1_66.',.
XW'lO .~£!. + .OlL:: 1.66 x 1O- 12 moies
L'
"
.
Design equation:
dX
NAO ---::-r
dt
A
Rate Law:
Stoichiometry:
CA = CAo(l- X)
Cs =CAO(aB-X)
Where:
Total number of moles;
1.56 x 10-12 + 1 X 10-9 == 1.002 X 10-9
hence;
Combining and solv41g:
dX
k<:;'odt
·(i.-:x)(e~:~:X)' ==
o '
ii:
-_.}:"--, In . _.~!!::L._ = ~5;;o~
8'/1 ··,·1 aB{l" X)
N AO
t=O.071mm
b)
Assume en == egO
Design Equation:
ax
N,1.0 dt-- == "rA
Rate Law:
Stoichiometry:
CA
=C;\O(1····· X)
CB = eRG
4-71
Combine and solve:
1-· X
N Ao
1. --. :::: . kCAOCBOt
In .---.
.-.. .----.. -.1 -- X
N A.O
t==O.069 min
A very good approximation .
Design Equation:
c)
dX
N ·---=:.-r
AO dt
A
RateL1.w:
Stoichiometry:
CA == C"o(l- X)
Co == CSO
Cc =:' C"oX
Combine and solve:
N AO~;:::: kjCAo(l··· X)C so
·····kJ~·AOX
dX
kjCAOCSO
dt
kfCAOCBO -krCIlOX
x
dX
5
D
k/'BO--X(kjCslJ"f k,)
0
N AO
CAodt
f-:---·-·---·~"·:--·--·-·-:;;: S. .
X'-
kfC
1.
_ .sC
no -- -..................-:-. - .---AO
-.. . . . . ---.. -.-- -........--- In ...-----...... ........... _. .-...--....
kfC SO +k, kjClJ()--X(kjC so +k,)
N Ao
...L. --.ln -.--. -. 1+ ,1
::~
,1···· XC.l
-.
+ . 1)
= 500
X =0.5
CDP4-D
Batch reaction: 2A + B -) 2C
k j = L98 feJIbmoLmin, k2 = 9..2 X 103 (feJIbmolilmin
V = 5 gal =0. . 67 fe, x = 0..65
.
CAO = 0. . 0.0.2 IbmollO.67ft3 = 2 . 98 x 10-3 Ibmol/ft3
CBO = 0..0.0.18 IbmollO.67fe =2.69 x 1O-3 lbmollft3
Mole Balance:
Rate law:
Stoichiometry:
dNA
dt
---
= -rAV
2
-rA = kjCAC B+ k2C ACB
CA = CAO (1-X)
CB = CAo (8 B-(b/a)X) = C Ao(D . 9 - D5X)
4-72
Combine:
dX
CAD -
dt
=CAD 2(1- X)lkr l (0.9 -
0.5X) + k 2 C AD (1- X)(0.9 - 0.5X)
Integrating between X=O to X=O.65 for t = 0 to t = t gives
t =24.1 min
CDP4-E
Ha..--_ _-..
e.
i o - - - -.........;>
A,.
Finally
F
AF
, FH2f
FBF
CBF (ete)
YBF "" 10- 5
"'r
a
=
k C!:!A Cn
"'
Liquid feed is a mixture of A and B YAo = 1 - YSo= 0.999
Because of the low concentration of B in the feed, such properties as SG, the specific
gravity, Mw, the molecular weight, and the solubility of H2 are essentially that of
component A. assume that any H2 depleted via reaction is instantaneously replaced via
absorption of H2 gas over the liquid reacting mixture. Then at 500psig,
4-73
2]
·~>
l
1
;:(.
[~2b':'1 :r [ __ ..Ll~.le._.) .x (.L...1l!.ll..J
2&
l
454 g:tloles
g!'.l
.UfL.:U!L.aI.
ll.L..1.~
.020 81 b::.t91~
zal
3 .72 Ibmo1eJ~.::
1 b::::.o~le
v
o
constant ..
P Jl
~
2.
or
0
r.'
,··5
10 Ibmolel;;a.l
• !:to' 1.'0
----V
.;l
U:.ing ec;,nll.tio:;. for
3.
CSTR
(89_3 zallhrHO.:.21L.......__ _
50 .. d
(Q .QZQ8 11,r'9U (0.01)
,11.1
k
= 0.49
x 10 3 ;;alf(l~ole - hr)
is:
:::
C
a2
(
0.G127 IbwoleJgal
Let
4-74
300+14.7)
500+14 * 1
314.7 1""'01.,
{.020l0 l'Si'4:71
Ed
~
1-1
=
2
F
(SO gal)(S.49x10 3 gal/lbmol-hrHO.0127 Ibmol/g-all
89.3 gal/h.r
= en
B2 ,F
<';2.F
= (.0127 Iba o1e )(S9.4 g al) = 1.134 Ibmole/hr
v
gal
0
hr
3
(3.72x10- ) (1-0.984) = 1.23xlO-S
1-'1.34+3.72
CDP4-F
NAo
~ = VpC (.rJ "" VPC kC A
Uqtlid Phase C A
"'" CAO
OX) "'" ~~. (IX)
QX=
Pc k {I Xl
de
• ,
t
X
In--L -.
(I-X)
o
o
o
5
.3
.36
10
20
.73
1.31
.5
.69
4-75
30
.85
1.89
40
0.93
2.66
50
.97
3.:5
A. plot In [[i:~} i:; lir:ear v.n rime so we concllmc me reaction is filsi order in 5. 6
benzz:roqumoline with k'=O.o,,¥ th'''':'
f '" .___. ,. ".ru
......_ '" 6248 oK
.30.__
(353{383 ))
E ;;; 12,414 .;.:;;i
mol
k (363)
k (36.3)
=:
0,63 k (373)
~ =~. £'.:2.,5.. "" (.63) (2) k:
-
kl
Pel
:: 126 k'
1
1
In --L..."" k't
i-X
o
o
't
X
30
0.42
20
0.82
10
057
40
50
0.97
0,,99
CDP4-G
Develop a design equation
Mole balance: FACr) - FACr + L1r) + rAC2nrMh) = 0
_ FA (r) FA (r + ~r) _
~
-~
~r
Rate law:- r A
=
(2
m h)
_ dFA
~
-~
dr
(2
m h)
k1C ACB
Assuming 8B=1 and E = 0
Stoichiometry:
Also FA
CA
= CAO (1- X) and CB = CAO (eB -
= F AO (1- X)
dFA
=> - -
dr
dX
= -FAO -
4-76
dr
X) = CAO (1- X)
= rA (2mh)
J(21lkC
=
2
AO
)rdr
FAO
Ro
x
By Integration we get: - - -
I-X
b) Now, with the pressure drop,
C A= CB = CAO(l-X)y
Hence,
-rA =kCA0 20-X)2y2
dX _ 2mhkCAO
dr
Where
2
0_ -
y = (1- aW)1I2
2
X) Y
"",,----_._-----------,
..,......
=>
1~"
2
G"phTItl.
tJ2&3
and
W = PbJZ"(r 2 - RO 2 )h
Using polymath, following graph is obtained:
Differential equations
d(X)/d(r) =2*3 . 1416*h*k*Ca0"2/Fao*(1-X)1\2*yI\2*r
y = (1-alfa*W)"O . 5
W = density*3_1416*h*(rI\2- 1'0"2)
ro 0,,1
density = 2
Cao = 0.1
Fao 10
k=0,,6
h=O.4
alfa = 0 . 07
variable name: r
initial value: 0.1
oro.
'--""'---_~---'-_~
~
~
~
~
~
_ _ _ _.
~
,
~
~
m
=
=
c) Increasing the value of k increases conversion while decreasing it decreases the conversion" Increasing
FAO will decrease the conversion and decreasing it will increase the conversion. Increasing C AO causes a
dramatic increase of conversion.. Similarly, decreasing CAoresults in a large decrease in conversion.
Increasing the height will only slightly increase the conversion and decreasing the height causes a real small
decrease in the conversion. Increasing Ro decreases the volume ofthe reactor and hence decreases the
conversion. Increasing RL will increase the conversion as volume increases .
4-77
CDP4-H
Liquid phase rexrion
A+B--rC
vellmixed
no inflov or outtloll
BATCH
L Mole balance on batch reactor
2. Ratelaw
-1'A "'" Ie eACa
3. SroichiCll:!l!'!try
liquid phase
V "" V" (batCh) (if flow u
=u",]
4. C.ommne
4-78
Table of reaction imegrals can be found in Appendix A-lO.
a)
I. Niok baIa...,ce on CSTR
2. RateLaw
3. Stoichiometry
liquid phase
u ::::: U o
4, Combine:
5. Parameter evaluation
10 moUmin (0.9)
V=: - - - - - - - .
..._--
(o,m d:a:J!l/mol min) (2 lllOvd,m3f(1-0.9)2
b)
V"", 22500 4m3
1. Mole bala.'1ce onPFR
FAO JiX.:::::
'dV
-fA
if!lQ pressure OJ.' phase change (liquid phase) therefore
r
V=F~o ~~
2. Ratelaw
4-79
5, P = = evaluation
C:A
~
:;=
fM ~l·:~~., = (~AD Cl--A)
U
___ :-QQ.rooVrrrin}· ___
(01 cio:r'Jmol min)(2 rnolJdm'J
F,,,,,ieR_hX\
:: __:...~.:__.,.....1t., ....~_;;:; (:AD (I···X)
C::s ::::
e)
"rA=kfcACS··~
t
KcJ
Arequilibrium
Kc - _Cce ....
(A
V""rAO
I
;j.~: .
li·X)'-
10
=
- CAe CJl"
""
x
~=4
_X,.= X.
KCN)
X2-2.25X-t 1=0
PFR
.dX.",.::!A.""kC rC (1-X'I1.X.j""kC2 0 [(lXf·,:;. X ..J
dV FAG
AOl AO
I
Kc
A \
KcCAol
f6 .-. . ."...")-''''X",.,",....X'_.
£.,i
o 1·· • • JA,
+, •
V", 250 ~.[~O).,. 4~, 15) -+ l~ ,3) + 4~,45) .. i\ 6}]::::
150
Q'j 5[11 4{ 141)t 2(241).· 4{5.26) +100J
V", 250 ~[r{O) + 4~.,15) + 2~,.3)+4ft.45)+ ~6)J
.:;: 250 llir
;~11.411 +-li2Al)
+ 4{5 26) +100J
3 c1.,
"'\
•
'" 1656 c:n'
CDP4-I (a)
4-80
,·,,,,=0
,.,.CAolL.= __ ..,..x._.._
CAd)-Xf
C'idl-xf
KCAo=l~.
1 .,. 2X +- X2
V", 250
.
d) Solution si::::ni1:at· to part (a) ro (e)
-UO
'J
·····,i·f1g·
v = .....
DO
4
Benzene is A
-rA =k A (C A2
for equilibrium - rA
=0
C
CBC
K )
c
_
:.
4Kc:{1-Xc'f =Xc
2
2
4Kc - SKcXc: + 4KcXc = Xc
2
0.2X2 - 2.4Xc: + 1.2 = 0
Xc: =0.52
CDP4-I (b)
PFR
Design Equation:
V = FAD
I -rA
dX
-rA =k A (C 2A
RateLaw:
_
C
CBC
K )
C
Stoichiometry :
C Ao
=~ =
RT
C
,,= CAD (1 - X)
CB
=C 2 X
Ao
C
c
5 atm
= 0.OO371bmol
(0.73
·atm K1859.67-R)·
ft3
lbmol·- R )
fe
4-81
=C ADX
2
Combine:
v=
FAo
kC!c,
r[O-Xy _ X
dX
2
]
4K,
V_
rX
(IOlbmoUminXlminl60s)
dX
- (ISOOfellbmol.sXO.OO37Y Jo [(I-XY -O.S33X 1 ]
v - 6 76ft 3rX
-.
dX
Jo &-2X+O.167X z]
1
In[O.523(X-11.45)~)
- . \. (O.617X11.45-0.523) 11.45 X-O.523 U
v- 671
v =13.5fe
CDP4-I (c)
CSTR
Design Equation:
RateLaw:
Stoichiometry:
Combine:
V=
V = FAOX
-rA
-r.
=k.(
C! _
CA =CAo(l-X)
V=
C~~c )
Ca
FAo
=C~X
X
k.C~[(l-X)'-(
(lOXO.51X1l60) .
(1800XO.oo37)'[(l-0.51)' -(
Cc
=C~X
!:J
4~~))]
= 147.73ft 3
CDP4-1 (d)
Amount processed in CSTR ;
(10 IbmoVminX60 minlhr X24 hrlday)= 14,400 IbmoVday
4-82
Batch
t-C
-
Ao
Jd.X - kCC J
-::;-A
Ao
2
[
Ao
d.X
2
X ]
(l-xf-4Kc
1
In[0.s23(X-l1.45)~)
- kC Ao (O.167Xl1.45-0.523) 11.45 x -0.523 U
t __
I (
=
t 0.30s
Taking into account the time it takes to clean the reactor and other down time assume that
the total time per run is 4 hours. Assuming that the reactor can be used twenty-four hours
a day there can be 6 runs per day.
14,400 Ibmollday =2400 Ibmollrun
6 runs/day
= N Ao
V
f -rA
dX
V - N Ao
1 (
t
- t kC!o
:.
J
V = N Ao dX
t
-rA
In)
1
In[0.523 (X -11.45
(0.167Xl1.45-0.523) 11.45 X-O.S23 U
2400
(
1
1 [O.523(X-ll.45)~)
- (4X1800XO.0037Y (O.l67XU.45-0.523) n 11.45 X-O.S23 U
v-
V = 48,690fe
CDP4-1 (e)
E
=30,202 btullbmol
for X =0,
-rA =kC!o
- rA(800)
k800C!o
k800
- rA(l400) = k'400 C!o = k l400
E( 1
=exp R
.
1)
T2 - Tl
-
- rA (800) = ex 30,202 btu/lbmol (
1
_
1
)
-tA(I400)
P 1.987btullbmol·-R 1259..6rR 185?67·R
-r
A(800)
= 49
- r A (1400)
4-83
CDP4-J
PBR
?vfB:
Combine:
ME:
w ::::;~~Q.>!:­
-fA
Combine:
X
W r; .•..f'...o
---. .... _ ...... _ .
kC llo (1 . . . . X)
1000::;:: ---
1
X
(I- X)
X =.18
b) We know, s = 0, so
P = Po (1- aW)l!2
=>
1 = 20(1-- aW)l!2
4
=> a = 9.98 x 10.
c) For conversion to be maximum, a should be minimum
2
G
Also, we know that a is proportional to - - - - =
AcDp
G
And it is proportional to - - - 2 - =
AcDp
1
6
2 for turbulent flow
D pipe Dp
1
4
2 for laminar flow
D pipe Dp
Hence for minimum a, Dp and D pipe should be increased"
d) Yes, we can
increase Dp and decrease D pipe (or vice versa) according to the above equation to get the same value of a.e)
For assumed turbulent flow,a is proportional to .-
1
6
D pipe Dp
4-84
2
Now,
6
al
2
_ Dpipe2 Dp2
-
6
a2
-1
2-
D pipe! D pI
Dpipel
---''---=--4
and Dp1 = 0.5 cm
6
Dpipe2
=>D p2 = 0.044 em
.
1
t) We're gIven, k ~ Dp
k2
Hence -
kl
Dpl
=> k2 = kl - Dp2
0.5
= - - - = 11.4
0.044
rA
g) Rate law: -
Stoichiometry:
= kCA
CA = CAO (1 -, X) Y
dX ·-r
Mole balance: -
...
dW
dX
Combining:--
=_-A
FAO
=
k C 2(1_ X)(l-a W)1I2
2
k2 _
-
- 11.4 and
2
AO
dW
,- (forE = 0)
FAO
CAOkl
= 2 X 10
--4
..
..
Integratmg WIth hrruts: (X -70 to X; W -7 0 to 1000) => X
FAO
kl
CDP4-K
1.
Mole Balance:
dz
2,
Rate Law :
lA
FM
;;:;-·k'C:\pc (1--
<p)
wherek',pc' and <p are unknown constants 111CY will be groupedimo one unknown
constant, ko '
3
Stoichiometry:
4-85
=0.78
Pressure Drop:
when neglecting the turbulent contribution to pressure drop, ~ 0 is given by
= G[- l~S)~~_, ,p}~~-l
r<
Po
'
(T
'[)"
~3
p<i' ,
where all of the bracketed variables are unknown const£l.nts,
,Pooe
'
(J
=
ill
Ac
The ma.ss flow rate is also and unkrlOwn constant Thus Po can be written as a con~t,~t~
E, over cross·· sectional area.
where
Where E is a fUllction of the mass flow rate, feed properties, and catalyst PIOpclties-'--<llI
of which are constant f6r part (a). Note that all equations up to now apply to a tnbulal
PBR arld spherical PER To find Band k., we must model the tubular PER on
POLYMATH by entering the above equations in addition to
(70'om )2 :::: -'14d
J
mZ
A (,,;;;;; It'_
-
4
B and leo must be arbitrarily chosen al first Then, depending on wheLher POLTh1ATH
gives high or Im·v values of '/ and X, one Cffi1 converge on the true values of}~ and B. In
order to do so efficiently. hm.vevcr, \Jue must rnakc usc of the foHm'iing trends: (Note
that tile tollO\vmg taNe IS completely tnj(~ on.1 YwIlen c > (J.)
a) from POL YMATIi:
B==S43
k==O.21
x=.67
P=yPo=·973* 1500=1459..5
4-86
£qua&.ions .:.
dey) /d(z) ",···beta! (po"y)' (~+eps*x)
d( x1 1.4fzl=ra,·Ac/Fao
po,"2500
x and y vs z
'eps;O ,.5
!.QCQ
l:'ao"'950
Re10
Cao"'O.4
k=O.:21
B",8~3
~~
AC;=3 .1416" (R H 2)
Ca;Cao"Y*
(l
....--....._..
"
.......... ~~~~.....-" .... ".
"',
·x) I (l.+eps·x)
beta;B/Ac
ra=··k~Ca·"2
220
~!:12E..L
x and y vs z
d (y) Id(z) ""'oeta! (po"y) .. (l+eps·x)
d(x} Id(z) =·'·ra-}l.c!Fao
po=1500
;
fu:.l
-x
eps=0.5
l:ao==950
~iIOQ
i
....
... y
R==25
Cao"'O.4.
k=0.21
ll=84.3
L=24
Ac",3 .. 14lli*{R""2··(z"'L) ''''2)
Ca"'Cao"y· (l"'xJ 1 (1 + eps"x)
:::'a=-k"Ca .. " 2
beta.="B/AC
Zo :
0,
zf
= 48
b) From polymath: Maximum flow rate = 1750 molls
aqua cio!!!!:,
dey} Id(zl= .. oota/ (po·y) .. n,·eps·x)
(b)
d(xl/d(z) ",··ra"Ac/Fao
Fao ... 1750
",0=,,1.500
T;
ep8=0.5
a..DC::'
'''ao=>:1750
"",25
cao==(j ..
4
1<;=0.21
3=843
L"24
Ac==3.1.41.6 w (a""2 .. · (z .. L) .";;l)
Ca""Cao"y· (l'-x)! (I+eps"x)
~:;.:eo
ra='"k-Ca*-,2
beta"B/Ac
Zo '" o.
43
4-87
.......-1"...-.-.•• --.•.
"C.;3CO
."
c) From POLYMATH
Nllnimum pressurc""gO kPa
Eg:,.:aci~!:!~:.
d
(y)
Id (zj ",···beta! (po ~y)
• (l~·eps·x)
(c)
d (Xl Idl z) ::~ra -Ac/l"'ao
Po;; 80
po"'80
e?so:O 5
x:
Fao",SSO
y
....."""
R:=26
',.
Cao",O . 4..
k'=O .. 21.
13"'843
L::24
he""3 .1.41.6* (R"'·2··· (z;· ..·Ll·".2)
Ca::Cao"Y" U ..~x) I fl+eps*xl
:::30= . ·'k"'Ca '" '"2
1:
bee ta=:S
, I AC
d) Flom POLYMATI1:
d(y)
Id{:d " ..bet.a/ (po~y)· (l+eps~x)
Cd)
d(x) Id (z)" .... );.a*Ac/23.0
Kex
pO=I.5CO
eps"O S
-x
:;;'ao=950
y
...
1<.=25
Cao .. O.4
k=0 . 21
13",843
he=3 141&*CR··2 .. (z-L).'Z)
Ca"'Cao*y· (1. ·xl f (l~eps ~x)
;::'a~""k""Ca1/("*2
De';:a=B/Ac
43
4-88
~".,..~
..•".
CDP4-L
(a)
C
Kc::: ._l!
C'"
C. . "" C"'o(l- X) , c.s == C",oX
Kc =_.C . . oX.._.::;: .~
C",.(l .. X) 1·.. · X
X"l "" _~s..". '" _..2:~._.
1-+ Kc 1 -+ 0.5
"" 0.33
For a conventional PFR:
(b)
dX
FAo--;::;:-rA
dV
"'r" =k(C .... -Ca/K,}=kCAQ{(l-.X) -XlKc )
For the IMRCF:
dF
-"'::::::rv
dt
"'.,
~.!!. == "J:'"v. '"
v"kcCs
··rA := k{C" ... CafKc)
CA
:::
en =F~N
(1'. . + FrJ
._._-....•.
__ .
FAN ,
V, =:: v
•
FA"
Use these equations in POLYMATH to generate plots of the conversion profile.
::1 .. 3::0
8. lac
0. :20
8.:.JOC
::leo
4-89
",
l::::.CC
!,
-===
............
o .. ooc
0.000
(c)
-~.-+
...
18.,::)00
'''-'--1'-'''-'-'-'~'''-"-'1
2"1.000
3CLOGC
Use the same equations for the IMRCF in POLY¥ATIi to generate the desired
plot.
2.
t::c
... fb
tL9GG
..;..
_.....
1 s~ ceo
4-90
:3~J, 8(J::~
(d)
By varying one parameter at a time we can see the effect of each:
Increasing the specific reaction rate causes changes in conversion. concentration.
and molar flow rate [0 occur more quickly .
Lowering the transpon coefficient (k,) causes an increase in both C a and Fa. which
causes a decrease in conversion .
By raising the equilibrium constant (K.,). we cause a decrease in the molar flow rate
of A and an increase in conversion.,
(e)
A signitlcant increase in temperature for an exothermic reac1.ion would drive the
reaction in the reverse This would cause a decrease in X and C s and an increase in
the CA'
A significant dect-ease in temperarun.': for an exothermic reaction would cause
increase in the rate of the forward reaction. This would drive up X and Cll• while
lowering C....
An increase in temperature would drive an endothermic reaction forward, raising X
and Ca. While lowering C"
A decrease in temperature would cause an increase in the reverse reaction for an
endothermic reaction . This would raise CA' and lower both X and C II .
CDP4-M No solution
CDP4-N
4-91
~A4
Y F A3
F
F A2
F
,.1,.0
.- y 1',.1,.3
FAl (1·-X) "nere X is con.version defined with respect to • .'1.1.
A1
Also
Stoichiometric :able
v
2
t"Al
CAl
v.
1
v
lDO
1
_~~l_
F!~1
t.A1
100
CDP4-0 No solution
.-----------------------------------_.----------------
4·92
CDP4-P
A
(n)
At equilibtimll
Initial Condition
ki~lle = k"CCe
e no "" CUe
k 2C nO
C ;\(Ii
.'"
k 3C AO
(1)
(b)
= --k[C"
elt
=> C A == C AO e
klt
dec
,cit = +k;2C' llO ... k3 C(,
is it constant
U::::. 1"1 C-AO
I:~ ··k1t
,.
Problems fwn, I'robjot
From (I)
de
ell
>0
C:::;
for a minima in ell
4-93
vtoto " L~ocle 0,'
t
'''0
CCe ""Cc\O
+ CCe
CDP4-Q
CDP4-R
CDP4-S
CDP4-T
CDP4-U
CDP4-V
4-94
Solutions for Chapter 5 - Collection and Analysis of
Rate Data
P5-1
P5-1
P5-1
P5-1
P5-1
P5-1
P5-1
(a) Individualized solution
(b) Individualized solution
(C) Individualized solution
(d) Individualized solution
(e) Individualized solution
(0 Individualized solution
(g) Individualized solution
P5-1 (h)
Example 5-1
The graphical method requires estimations of the area under and above curves on a plot as well as in
reading the intersection of lines on the plot. This can lead to small inaccuracies in each data point
The Finite differences method uses mathematical estimates to calculate the rate. It is only possible to use
this when the time interval of each data point is uniform,
The graphical method uses polynomial regression to approximate CA as a function of time. The derivative
of the polynomial is then used to calculate the rate,
P5-1 (i) Example 5-2
Assuming zero order reaction:
Rate law:
_.4C
A
dt
= k'
~~~__~+-____~__~_______~~__~~__-+~~___+~3~OO~~-J
.-L..:='::'~--L.::0:.:.:::'O 17r=J
zero order reaction
1st order reaction
4.5
35
~
~
25
2
15
05
50
dC
Rate law: ___
A =k'C
dt
100
150
t(min,,)
t(mln . )
A
5-1
200
250
300
TI27 1 3.49
1 3.66
13.94
1 3.81
14.05
Since none of these plots are straight lines, its is not 1st order reaction or second order reaction.,
P5-1 (j) Example 5-3 Because when a is set equal to 2, the best value of k must be found.
P5-1 (k) Example 5-4 -C
(moVdm'i)
HC10
:
2
-rHel,O(mol / em .s)xlO
7
4.0
2.,0
1.0
L2
2.0
1.36
0.5
0.,74
0.1
0..36
<--.
See Polymath program P5···1··k.poL
POLYMA TH Results
Nonlinear regression (L-M)
Model: r = k*(Callalfa)
Variable
k
alfa
Precision
R"2
R"2adj
lni guess
0.1
0.5
Value
1.0672503
0.4461986
95% confidence
0.0898063
0 . 076408
= 0,,9812838
= 0.9750451
= 0.0341709
Rmsd
Variance = 0 . 0097304
Rate law: _r
P5-1 (I)
n
= 1.1C~:;:mol/ em 2 .s
Example 5-5
rate law: rCH4
= kP:!oP~
Regressing the data
~oiCHJgcat.min)
5.2e-3
13.2e-3
30e-3
4.95e-3
7.42e-3
5.25e.. 3
Pco(atm)
P H2 (atm)
-
1
1.8
4.08
1
1
--
--
See Polymath program PS .. I-Lpo1.
POLYMA Tn..&§.!!lts
Nonlinear regression (L-M)
Model: r = k*(PCO"alfa)*(PH2 I1beta)
Variable
k
alfa
beta
Precision
R"2
lni guess
0. 1
1
1
Value
0 . 0060979
1 . 1381455
0.0103839
95% confidence
6.,449E-04
0.0850634
0,1951217
= 0.,9869709
.5-2
I
1
I
0.1
0.5
4
---
-
-
R"2adj
= 0.9782849
Rrnsd
= 4.17BE-04
Variance = 2 . 093E-OB
Therefore order of reaction = 1.14
Again regressing the above data putting
fJ =1
POLYMATH Resu1ts
Nonlinear regression (L-M)
Model: r = k*(PCOAO.14)*(PH2)
Variable
Ini guess
Value
o.:r-'---
k
Precision
95% confidence
0-:.0040792
0.0076284
R"2
= -0.8194508
R"2adj
=-0.8194508
Rrnsd
= 0.0049354
Variance = 1.754E-04
Therefore, k = 0..0.0.4 (gmoICHJ(gcat.min . atm I14»
P5-2 Solution is in the decoding algorithm given with the modules.
P5-3 Individualized solution
P5-4 (a)
The kinetics of this deoxygenation of hemoglobin in blood was studied with the aid of a tubular reactor.
Hb02 -7 Hb + O2
-IA=ke~
Rate law:
FAO dX =ke~o(1-Xr
Mole balance:
dV
dV = Acdz , where Ac is the tube cross sectional area
= ke~O~.L (1 - X
dX
dz
FAo
r
In( :~)=lna+nln(l--X)
therefore,
ken A
where, a = __ ....&Q._L
FAo
••
~
1
0.
Position (cm)
rA~'(clIl)
!
..........
'-'
,.
!.5
,......... ..
IConversion ofHb02 (XA)
I 0..0.0.0.0.
(l-XA)
lioOOD
.ax
"Ii xl Ii z-(cm: i )
r
<
Electrode Position
I
I
I
2
3
I
.5
10
,1
.5
.5
IT
I 0..0.382 I
r
f
0..0.193
0..0.568
r 0..9618', 0..9432
'f 0. . 0.193" r 0..0.189 I
10.980.7
0. . 0.0.386
I
I 0..0.0.378
5-3
0..0.0372
,
'r
.5
20.
5
f····
I 0..0.748
I
I,
'r
6 ....
!"
I
7
3D
5
Electrode Position
dXldz'(cm- l )
A hi~~t.ogr~~ plot ~TLi XIii ~ vs . z is then p~oduced Th~values of dXldz are ~~~iuated using~qual-area
graphical differentiation:
0 . 00395
0.0039
0 ..00385
0 . 0038
00375
g 0 . 0037
dX/dz vs. distance
EO
~0.00365
~ 0 . 0036
0.00355
0.,0035
0.,00345
0.,0034
+--+--~+--~+--~t--~f--~r--l~--+~~f.----l
o
25
5
75
10
125 15
175 20
225
25 275
30 325
35 375 40
z(cm)
Using the values obtained above, a plot ofln(dXA/dz) vs. In(l-XA) is produced and a line
is fit to the data
.
In(dXldz) V$, In(1.Xj
-
~~~~~~~~~~~~~~~
In (
y = 1 0577x- 5 5477 •
~~ ) = In a + n In (1 - X )
where a =
,
kCn A
Ao c
F
Ao
In(1-X)
In(
~~) = -S.S477 + l.OS77In (1- X)
n=l
In(a) = -5.S
a = exp(-5.S)
= 4 . 1 x 10- 3 em.
Concentration of blood is 150g hemoglobin per liter of blood
Molecular weight of hemoglobin = 64500
Ac=D., 0 196cm
= 2.3x1W6 mol I em 3
FAo = 45.7 X 10-6 moles I s
CAO
k
=
FAoa
C1of\:
=
6
4S.7x1O- moles Is . ____ ( 4.1x1O-3 em)
2.3 X 10-6 moles I em 3 xO.0196em 2
Hence rate law is,
-rA =4.1CA
mol
3
dm ·s
5-4
= 4.1s-1
P5-4 (b)
First we fit a polynomial to the data. Using Polymath we use regression to find an expression for X(z)
See Polymath program P5--4-b.pol.
POLYMATH Results
Polynomial Regression Report
Model: X = aO + a1*z + a2*z"2 + a3*z"3 + a4*z"4 + a5*z"5 + a6*z"6
Variable
aO
a1
a2
a3
a4
a5
a6
Value
2.918E-14
0.0040267
-6.14E-05
7.767E-06
-5.0E-07
1. 467E-08
-1.6E-10
95% confidence
0
0
0
0
0
0
0
General
Order of polynomial = 6
Regression including free parameter
Number of observations = 7
Statistics
R"2 =
Rl\2adj=
Rmsd=
Variance =
1
o
1.669E-10
1. 0E+99
Next we differentiate our expression of X(z) to find dXldz and knowing that
In ( :~) = In a + n In (1·- X)
where a =
,
ken A
Ao
r
F
Ao
Linear regression of
In (.:;-)
as a function of
In (1- X)
as in the finite differences .
POLYMATH Results
Linear Regression Report
Model: In(dxdz) = aO + a1*ln(1·X)
Variable
aO
a1
Value
-~. 531947
1..2824279
95% confidence
0 . 0241574
0 . 3446187
General
Regression including free parameter
Number of observations = 7
Statistics
R"2 =
R"2adj=
Rmsd=
Variance =
n= 128
0.9482059
0.9378471
0.0044015
1.899E-04
5-5
gives us similar vaules of slope and intercept
In a = -5.53, a = 0.00396
4;.7XlO-6m~les/s
k= FAOa =
C~oAc
2
(3.96xlO-3 em)= 4.0s- 1
2.3xlO- moles / em xO.0196em
Hence rate law is,
_rA = 4 •0 CAl 28
mol
3
dm ·s
P5-5 (a)
Liquid phase irreversible reaction:
A -7 B -+- C; c AO = 2 mole/dm3
CAO-CA = kc a
A
T
In ( CAO ; CA)
_. Space
=In k + a In CA
j
time ( T )min.
~.
38
100
300
1200
-
CA(molldm
1.5
1.25
1.0
0.75
0.5
)
In(CA)
0.40546511
0.22314355
0
-0.28768207
-0.69314718
In«CAO-CA)/T)
-3.4011974
-3.9252682
--4.6051702
-5.4806389
-6.6846117
By using linear regression in polymath:
See Polymath program P5-S·a.pol.
In((Cao-CaO/tau) vs In{Ca)
POL YMA TH, Results
Linear Regression Report
Model: y = aO + a1*lnCa
-q.s
-0.6
-0.4
-1
-2
In ( CAO ;- CA)
Variable
= In k + a In CA
y = 2 . 9998x - 4 . 6081
-4
Value
95% confidence
0 . 0162119
2.9998151
0.0411145
":"a=O=-=-=--'--'~--47.-:6:-::0-=8-=-0-=5:79
a1
Statistics
RI\2 =
RI\2adj =
Rmsd=
Variance =
. . . . .,. ~. ,_._"'.-. . . '. .-.'.'. ::s.
...............
0 . 9999443
0.9999258
0 . 003883
1.256E··04
··8
In(Ca)
a= slope::3
In(k) = intercept = -4.6
therefore, k = 0.01 mole- 2min- 1 .
dC
dt
-6
-7
Hence,
Rate law:
-3
3
A
--= O.OlCAmol / dm
3·
mm
P5-5 (b) Individualized solution
P5-5 (C) Individualized solution
5-6
....................... ; ....
PS-6 (a)
Constant voume batch reactor:
Mole balance:
A-7B+C
_ dCA =kC a
dt
A
Integrating with initial condition when t = 0 and CA = C AO for a =t 1.0
1 C(l-a) - C(1-a)
1 (2)(1-a) _ C(l-a)
t =
AO
A
=_
A
............. ,' ... ,' "substituting for initial concentration C AO = 2
k
k
(1- a)
(1- a)
t (min.)
0
5
r---"
9
IS
r---.
22
30
--
~-
60
CA(m~Vdm3)
2
1.6
1.35
1.1
0.87
0.70
0.53
0.35
See Polymath program P56··a.pol.
POLYMATH Results
Nonlinear regression (L-M)
Model: t = (1/k)*((2A(1-alfa».. (CaA(1"alfa)))/(1·alfa)
Variable
Ini guess
Value
95% confidence
k
0.1
0.0329798
3.628E-04
alfa
2
1.5151242
0.0433727
Precision
RA2 = 0.999'7773
RA2adj = 0.9997327
Rmsd
0" 1007934
Variance = 0,,0995612
=
K= 0..03 (moVdm3yo5"s·1 and
Hence, rate law is
dC A
dt
a = 1.5
-._-
= O.03CA15 mol / dm 3 .s
PS-6 (b) Individualized solution
PS-6 (C) Individualized solution
PS-6 (d) Individualized solution
PS-7 (a)
Liquid phase reaction of methanol and triphenyl in a batch rector .
CH30H + (C6Hs)3CCI -7 (C6Hs)3COCH3 +
HCI
A
+
B
-7
C
+
D
Using second set of data, when CAO = 0 . 01 mol/dm3 and CBO = 0.1 mol/dm3
5-7
------
CA (molJdm3)
0.1
0.0847
0.0735
0.0526
0.0357
t (h)
0
1
2
5
10
-'
Rate law:
-rA =kC:C~
«
For table 2 data:
C AO
Using eqn 5-21, t
= _.
1
k
CBO
=>
C~l;m)
-
-Y'A
= kI
= kC;o
-
(1- m)
I
I
1 (O.01)(l-m) _ C(l-m)
cCl-m)
A
C; where k
A
k
(1 - m)
I
See Polymath program P5··7-a-l.poL
POL XMA l]!.!l.esnlts
Nonlinear regression (L-M)
Model: t = (1/k)*((O.1 !\(1-m))··(Ca!\(l·m)))/(1·m)
Ini guess
Variable
k
1
m
2
- -Value
----
95% confidence
0.0109025
0 . 0021115
L 815656
2.0027694
Nonlinear regression settings
Max # iterations = 64
Precision
R"2
R"2adj
Rmsd
Var iance
=
=
=
=
1
0 . 9999999
3268E-04
8 . 902E-07
Therefore, m = 2
For first set of data, equal molar feed => CA = CB
Hence, rate law becomes -rA = kC~C; = kC~2+n)
Observation table 2: for
C AO
=0.01 and CBO = 0..1
C A (molJdm3) 1.0
'(J.278 . 0.95.
r-.---0.816
1.389
0.707
2.78
-0.50
8.33 -0.37
16.66
t (h)
0
t
1 C(l-C2+n)) = ___
AO
k
C(l-C2+n)) _ _
1 (O.l)C-l-n) A
(1-(2+n))
k
C C- 1- n )
A
(-I-n))
5-8
See Polymath program P5-7-a--2.pol.
POL YMA TH Results
Nonlinear regression (L-M)
Model: t
= (1"(-·1-·n)··Ca"(--1-·n))/(k*(--1-n))
Ini guess
3
2
Variable
n
k
Value
0.8319298
0.1695108
95% confidence
0.0913065
0.0092096
Nonlinear regression settings
Max # iterations 64
=
Precision
R"2
R"2adj
=
Rmsd
Variance =
= 0.9999078
0.9998848
= 0,,0233151
0,,0048923
Therefore, n = 0,,8
Hence rate law is:
__ r = 0 • 17 CA2 CBO.8 mol
3
A
dmh
P5-7 (b) Individualized solution
P5-8 (a)
At t = 0, theIe is only (CH3hO. At t = 00, there is no(CH3hO. Since for every mole of (CH3hO consumed
there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure"
P(oo) = 3Po
931 = 3Po
Po :::o31OmmHg
P5-8 (b)
Constant volume reactor at T
=.504°C =777 K
hase:
"] 1195
562
(CH 3 h O ~ CH 4 + H2 + CO
YAO
=1
£5=3-1=2
e=£5y. AO =2
V
=Vo( ; ) (1- eX) =Va
because the volume is constant.
P=Po(1+eX)
at t = 00, X = X AF = 1
5-9
1 dNA
N AO dX
---=-----=r
V dt
Vo dt
A
st
Assume -rA = kCA (i . e . 1 order)
CA = CAO (1- X) (V is constant)
dX
=kCAO (1- X)
dt
po_oR
Then: CAO and X
=
0
cPo
dX
1 dP
Therefore: =- - dt cPo dt
_1 :!P = k[l- P-Po]
cPo dt
or dP
dt
cPo
=~([l+e]Pa -p)
cPo·
= k ([1 + e] Po - p)
dP
f-----= fkdt
[l+e]Po--P
P
t
R
0
o
Integrating gives:
In [ ( cPo
) - ]
1+ e Po --- P
Therefore, if a plot of In
= In [2Po] = In [624 ] = kt
3Pa - P
936 - P _
624
vel sus time is linear, the reaction is first order. From the figure below,
936-P
we can see that the plot is linear with a slope of 0.00048.
Therefore the rate law is:
---"y_=_0_.0_0_0_48_X_-_O_.O_29_0_7_~~/
j
~
1.6
1.2
+ -__
...
0.8
+---------7"'''---------------1
0.4
0+£-------------------1
-~4
I
o
P5-8 (c)
1000
2000
3000
Individualized solution
5-10
4000
P5-8 (d) The rate constant would increase with an increase in temperatme, This would result in the
pressure increasing faster and less time would be need to reach the end of the reaction, The opposite is true
fro colder temperatures,
P5-9
Photochemical decay of bromine in bright sunlight:
t (min)
CA (ppm)
20
1.7
4
10
2..4
5
30
1.2
3
40
0.8
8
50
0.6
2
60
0,,4
4
P5-9 (a)
Mole balance: constant V
dCA =r =-kC a
dt
A
A
Differentiation
T (min)
~t (min)
CA (ppm)
~CA (ppm)
10
20
10
2,45
.L\CA ( pp.m )
L\t mIll
30
40
10
1.74
50
10
10
0.88
1.23
0,,62
·0.51
-0..35
-0.26
-0.18
-0,071
-0,,051
-0,,035
-0.026
-0.018
.:,:,::~
0.,1)4
0 .. 01
.,
10
.,..,---,_._.
)0
40
5·11
50
.._'••"__-.1.-.;),
50
10
0..44
-071
0,,06
.t
60
10
-}.a
-4.0
3.£
After piI
' an
oUmg
-dCA/dt
In( -dCA/dt)
InCA
1
erentIatmg by equa area
0.061
0.082
-2.501
-2.797
0.896
0.554 .-
0.042
-3.170
0.207
0.030
··3.507
-0.128
1.0
0.014
-4.269
-0.821
0.0215
-3.840
··0.478
-
Using linear regression: a = 1.0
in k = -3.3864
k = 0 ..0344 min- l
P5-9 (b)
!!NA=Vr =F
lit
A
B
rA =
FB =
·-0.0344 p~m
-0.0344 lmm
m~ .
mm
(25000 gal) (0.0344 lmm
m~ J(60 min_)(_lL'J(3.?~51lJ( IlbS_J
hr
1000mg
19a1
453.6g
=
at CA= 1 ppm
P5-9 (C) Individualized solution
P5-10 (a)
Gas phase decomposition
A7B+2C
Determine the reaction order and specific reaction rate for the reaction
dC A
dt
-kCA
Assume the rate law as: - - =
Integrating:
t = __
1- [
k(n-l) C
1n
A
-1
-
n
C
1n
AO
--1
J
5-12
=
0.426
lbs
hr
n 1
-lJ
=>In(tl/2) = In 2 ( (n -l)k + (1- n)In(CAO)
Run #
CAO (gmolllt)
0.025
0.0133
0.01
0.05
0.075
1
_.
2
3
4
5
t (min.)
- ~.1
7.7
9.8
1.96
1.3
See Polymath program P5··10··a.pol.
POLYMATH Results
Linear Regression Report
Model: Int = aO + a1*lnCaO
a:o--
Var'iable
Value
-2:"'3528748
a1
-1.0128699
95% confidence
0.1831062
0.0492329
General
Regression including free parameter
Number of observations = 5
Statistics
RJ\2 =
RJ\2adj =
Rmsd=
Variance =
0.9993004
0 . 9990673
0 . 0091483
6 . 974E-04
3.0 - - . - - - - - - - - - - - - - .
24
1.8
1.2
0.6
0. 0
L - ._ _ _ _ _
-4 . 61
-4.20
~
__
~_.
-3 . 80 In(dojO
-2 . 99
-2.59
From linearization, n = 1- slope = 2 ..103;:::;; 2
(2 a - 1 _l)e-interLePI
it
k=-·
. =10 ..52 - - - a··-l
gmol.min
5-13
-
In t
1.410987
2.0412203
2.2823824
0.67294447
0.26236426
InC Ao
-3.6888795
-4.3199912
-4.6051702
-2.9957323
-2.5902672
_ dCA =1O.5C/gmolllt.rnin
dt
.c'"
i:.
\1 ~
C aO
See Polymath program P5-· lO··a,pol.
POLYMA TH Results
Nonlinear regression (L-M)
Model: t = ((2J\(a·1 ))··1 )/(k*(a··1 ))*(1/CaOJ\(a·· 1))
variable
a
Ini guess
2
10.52
k
Value
1. 9772287
8.9909041
95% confidence
0.093057
3 . 9974498
Precision
R"2
= 0 . 9986943
R A 2adj
0.9982591
Rrnsd
0 . 0531391
Variance = 0.0235313
=
=
dCA =
-----
Rate law:
dt
9 .OC 2 gmol 11'
t.mm
A
PS-IO (b)
We know,
( 2 a--l
-
1) (
1
k = t1l2 (a·- i) C AO a-I
J
Solving for k at 110° C
k'= (2(2-1) -1)( __1_)=20
it
2(2-1) 0.025(2-1)
gmol.rnin
Rln k2
From these values, E =
kl
U-;J
(8.314
=
J
mol.K
)In. 20
10.5 = 76 ..5 kJ/mol
(37~K - 38~K)
5-14
~,
ij ~
P5-11
0 3 + wall ~ loss of 0 3
0 3 + alkene ~ products
Rate law:
-~
0
3
dCo3 =kJ
= __
.
kJ
k2
C
Coz
+k2~
dt
Using polymath nonlinear regression we can find the values of kl and k2
ozone rate
(mol/s.dm3)
Ozra
Run
#
Ozone concentration
(mol/s.dm3)
C03
le-12
le-ll
0.01
0.02
0.015
0 . 005
0.001
O..oIS
1.5e-7
3.2e-7
3.5e-7
5.0e-7
S.Se-7
4.7e-7
1
2
3
4
5
6
Butene concentration
(mol/s.dm3)
Cbu
le-lO
le-09
Ie-OS
le-09
See Polymath program PS· 1 l.pol.
POLYMA TH Results
Nonlinear regression (L-M)
Model: Ozra
= k1 +k2*Cbu/C03
lni guess
2.0E-07
0.1
Variable
k"l-k2
Value
3 . 546E-07
0.0528758
95% confidence
4.872E-ll
1.193E-·05
Nonlinear regression settings
Max # iterations 300
=
Precision
R"2
R"2adj
Rrnsd
Variance ==
= 0.7572693
= 0.6965866
= 4.531E-08
1 .848E-14
Rate law:
-ro,
= (3.5xlO-7 )+(O.05) ~BU mol/dm 3 .s
03
P5-12
Given: Plot of percent
decomposition of N0 2 vs VIFAO
X= % Decomposition of N0 2
100
Assume that -rA
= kC~
5-15
ForaCSTR
or
F X
V =~
-rA
x x
v
--=--=-FAD
kC~
-rA
V
withn=O, X =k-FAD
V
X has a linear relationship with - - as
FAD
shown in the figure .
Therefore the reaction is zero order .
P5-13
Si02 + 6HF ~ H 2 SiF6 + 2H 2 0
N s = moles of Si0 2 = AcPsD
MWs
Ac = cross···sectional area
Ps = silicon dioxide density
MWs = molecular weight of silicon dioxide = 60.0
o= depth of Si
N F = molesofHF=
wpV
lOOMWF
w = weight percentage of HF in solution
P = density of solution
V = volume of solution
MWF = molecular weight ofHF = 20 . 0
Assume the rate law is
dN
Mole balance: _ _S
dt
-rs = kC;
= rs V
_ l\;Ps do _ k(
wV
MWs dt
lOOV MWF
__ do
dt
]a V
= _ kMWs __ (~]a Vw
lOOa !\Ps
MWF
- do = fJwa where fJ = ~MWs
dt
a
_(--.£
__ ]a V
MW
lOOa !\Ps
F
In(- ~~) = InfJ + aln w
5-16
In(- ~~)
-16,629
In w
-15.425
2,,996
2.079
where (-
dJ)
dt
is in
-1432
3.497
-13.816
-13.479
3.871
3689
~
rom
FlOm linear regression between
In (- ~~) and In w we have:
slope = a = 1.775
intercept = In ~ = -20,462 or ~ = 1.2986 * 10-9
~13 .,..------:-""'-----------------,
-:g
-Il,}
"a
"C
2.5
..14
3.5
3
y ;: 1.7746x - 2Q.461
-15
t
c: ..16
·17
In w
kMWs
(p
fJ = 1001775 AcpsMW:;-
Jl775
V
Ac = (10 *10-6 m) (lOm)( 2 sides )( 1000 wafers) = 0.2 m 2
g
MWs =60-gmol
Ps
= 2.32 L = 2.32 *106 -.1]
ml
m
(Handbook of Chemistry and Physics, 57 th ed, p"B-155)
pzl-~=106L3
ml
m
g
MWF =20--
gmol
= 0.5 dm 3 = 0.0005 m 3
fJ = 1.2986 *10-9
V
5-17
1.775
106~
1.2986 *10-
9
= _--.!!L
g
20--
gmol
3
k =3.224*10-'7 ~
(
gmol
Final concentration of HF
J0775
=
min-I
5-2.316
(0.2) = 0.107
5
weight fraction = 10.7%
weight fraction = 20%
Initial concentration of HW = 0.2 (given)
Mole balance for
_
=
pV
.dw
100MWF dt
f
_lO 7_
dw
WI
775
20
dN
dN
dt
dt
F
HF: - = 6--s
6k( 100wpMW Ja V
F
=6k('_~J0775 tfdt
100MW
°
F
_1_( __1_) 10.7
0.775
W0
775
where u = 1.775
= 6(3.224*10-7 ) (
20
__~_,( 1 _1_) = 2
0.775 10.7°775
t=331 min
20°775
o775
6
10 J . t
20*100
389*1O-4t
.
P 5-14 (a)
A + 3B -7 C + 2D + E
Observation table for differential reactor
Cone., Of
Temperature(K)
A(molJdm3)
rm-
SQ,--
c--.
Cone, Of
B(molJdm3)
0.10
0.10
0.10
0.10
333
343
-,- 0.05
0.10
353
,-- ~~--,0.01
0.20
363
0.01
0.01
363
Space time for differential reactor = 2 min
V= F;, = VoCp
--rp
-rp
5-18
-
Cone., Of
C(molJdm3)
0.002
0.006
0.008
0.02
0.02
0.01
--
Rate
(mol/dm3 .min)
0.001
0.003
0.004
,-0.01
0.01
-,0.005
--
r,
p
= Cp = CCzH4
r
2
Rate law:
rc =AJ-BIT)C:C~
Cc =Ae(-BIT)CC
2
A B
Where, A is Arrhenius constant
B = activation energy/R
x is the order of reaction WIt A
Y is order of reaction WIt B
CA is the concentration of C2H 4Br
CB is the concentration of KI
Now using data for temperature 323K, 333K, and 363K, for finding the approximate value of B because, at
these temperature, the concentration of A and B are the same. Using polymath, the rough value of B =
55o.o.K
While using polymath for solving the rate law apart from guessing the initial values of n, m, and A , we
change the value of B in the model to get the optimum solution . So after trial and enOf we got B = 65o.o.K
See Polymath program P5-14-a.pol.
POLYMAI!!. Results
Nonlinear regression (L-M)
Model: r = A*exp(··6500/T)*GaA x*CbA y
variable
Ini guess
Value
A---3 . 6E+05
3 . 649E+06
x
0.25
0.2508555
Y
0.2
0 . 2963283
Precision
R"2
= 0,,9323139
R"2adj
= 0.8871898
Rmsd
= 3 . 615E-04
Variance
1.568E-06
95% confidence
2.928E+04
0.0032606
0.0020764
=
Hence, by nonlinear regression using polymath
A = 3649E+o.6(mole/dm3 2 6(1/s)
E = 65o.o.R = 54,,0.15 KJ/mol
x = 0..25
y= 0..30.
hence,
r
rc = 3.64E + 06e(-54015 / RT)C~25 C~30 mole/dm3.min
P 5-14 (b) Individualized solution
P5-15 (a)
5-19
Modell: Monod equation
dCc = r = JlmaxCsCc
--
----'=='---"--~
dt
Ks + C s
g
See Polymath program P5-15··a.pol.
POL YMA TH Results
Nonlinear regression (L-M)
=
Model: rg
(umax)*Cs*Cc/(Ks+Cs)
variable
lni guess
Value
umax
1
0 . 3284383
Ks
1
1.694347
Precision
R"2
= 0.9999439
R"2adj
= 0.9999327
Rmsd
= 0.0038534
Variance
1A55E-04
95% confidence
0 . 00686
2.2930643
=
P5-15 (b)
Model 2: Tessier Equation
Tg
=tL={I-ex
p( - ;
J]Cc
See Polymath program P5···15-b.pol.
POLYMATH Results
Nonlinear regression (L-M)
Model: rg = umax*(1··exp(-Cs/k))*Cc
Variable
umax
lni guess
0.5
100
k
Precision
R"2
RA2adj
Rmsd
Variance
Tg
=
=
=
=
Value
0.3258202
20.407487
95% confidence
0.0034969
5.7120407
0 . 9999454
0 . 9999345
0 . 0038004
1.415E-04
=0.33[I-ex{ ;~~ J]CC
Wdm'h
P5-15 (C)
Model 3: Moser Equation
=
r
g
f.1rruJ.x Cc
1+ kC --Y
S
See Polymath program P5··1Sc.pol.
POLYMATH Results
Nonlinear regression (L-M)
5-20
Model: rg = umax*Cc/(1+k*CsJ\(-y»
Variable
umax
k
y
Precision
RA2
RA2adj
Rmsd
Variance
Ini guess
0.3
1.6
1
Value
0 . 3265614
162,,599
2,,0892232
95% confidence
6.984E-04
34.273983
0.0461489
= 0 . 9999447
= 0 . 999917
= 0.0038269
= 1.794E-04
O.33C
1 + 162.6Cs (-2.1)
c
---...:::.....,--gldm3 .h
P5-16
Thermal decomposition of isopropyl isocynate in a differential reactor .
Run
1
2
3
4
5
6
--r------
Rate
(mol/s.dm3)
4.9 x 10-4
1.1 X 10-4
2.4 x 10-32.2 X 10-2
1.18 x 10- 1
1.82 x 10-2
Rate law:
--rA -- A e(-EIRT)CAn
Where, A is Arrhenius constant
E is the activation energy
n is the order of reaction
C A is the concentration of isopropyl isocynate
See Polymath program P5···16 . pol..
POLYMATH Results
Nonlinear regression (L-M)
Model: rA = A*exp(-E/(8 . 314*T»*(CA)An
var'iable
A
E
Ini guess
100--1000
n
1
Value
1.01E+04
5.805E+04
1.7305416
95% confidence
327.35758
237.32096
0 . 0134196
Nonlinear regression settings
Max # iterations = 64
Precision
RA2
RA2adj
= 0 . 6690419
= 0.4484032
Rmsd
:::: 0 . 0097848
--
Concentration Temperatwe
(mol/dm3)
-(K)
0.2
700
0.02
750
1-------0.05
800
0.08
850
0.1
900
-0.06
950
5-21
Variance = 0.,0011489
Hence, by nonlinear regression using polymath
A = 10100 (mole/dm3 2 6(1/S)
E = 58000 J/mol
n = 1.7
therefore,
r
-YA =lOlOOexp ( -6:76)C~7
mole/dm3 s
CDP5-A
Given the reaction P + NH;PH -? NHzOHP
where P is Penicillin and NH10HP is hydroxylamine acid (denoted by subscript HA)
Let A """ Absorbency, then CHA, ::; KA where K is some constant.
Cp
"""
C HA
Cp., (1,,,· X) (s = 0 for liquid phase reaction.)
=Cp"X "'" KA
Att=-
A
•
KA
.', X= ~- when X
C v"
=1 and
A = A..
C
-
=,.......l:2.
K
- rp' := kC;
Assume reaction is irreversible:
:=
1 dN
kC;" (1- X)*'
-~
For a batch. constant volume reactor:
Vdt
dC?
=-,= To
dt
l."
or
,~~e.=Cl''''~''''''
or
'~'=k(~~rl(A_'A)"=K(A_.'Ar, WhereK-k(~:r-l
dt
dt
dA"",.-kC;o(l-X)"=,-kC;"(l-A/A_)"
A_ dt
Try integral analysis first. Assume that reaction is zero Older:
Then
,~~;;;;;:
K
dt
or
rdA = A = Kt
Jo
a plot of A vs. t should be linear if
reaction is zero order. FlOm the plot below, it is evident that the reaction
is not zero order;
06
05
4)
(,,)
c
4-
0.4 '
«I
...
..Q
(:)
03"
.-
II)
..Q
<
::L_o
10
20
30
40
Time
5-22
50
Next, assume that the reaction is first order;
dA
- :;:: K( A_ - A) or
dt
J(A. ..ciA_. A) :;: JKdt ::: In (A. -A-A)
== - Kt
...
A
t
0
0
. A plot of (A_ - A)
VS. t
on semi -log paper should be linear.
-.. ---...-..-, •.•....-..-.-. ..'-•......
..............
.~:- .....w... .......-2C--.. - ..... 30-............. --.40-.............
*--_.. _. . . . . . _................_..................._. . . . . . . . . . . . . . . . . . . . . . . . --..
••••• _ •• _ ...... _ _ .. _ _
-
~._
.. -.. ....."- ..
Time
A
0
0
10
t-20"-
A.,.-A
-0.348
0.433
0.495
0.539
0.561
0.685
0.252
0.190
0.146
0.124
0
30
40
50
"'"
.
0.685
"'om
~
-
.
•• _ .................. _ . . . . . . . _ _ _ _ _ _ •• _
•• _ .
.......• -,....
~
or _... •.!. ..-.-
(A_-A)
o
--1'0·
·'-20
o
1.460
"-'2.967
. . 68--
0.33'1"
0..
_,
A_
A.. .. A
1
.. A plot of-----vs. t should be linear
(A_-A)
s ........__. -.. -.. _...........................................
8
•
7
30
•
'"40 0.539
50
0.56 f
_ _
,-
0.1 ................- .. - .....- ,-. ,..................................................- ..,..... .."..........................._"' ......-
: : : ...!_. +- Kt
A
~
-.-.--~.-.-
..
It is evident from the plot that the reaction is not first order. Try second order:
dt
•• _ ••••••••• _ . . . . . . . . .
.
Time
?A "" K(A_ .. Al
"W'
-8])65
2
o
o
10
20
Time 30
4U
50
From the plot, it is evedent that linear relationship exists between (VAx, - A) and time;
Therefore the reaction is second order .
CDP5-B
Determine the reaction order and specific rate constant for the isomerization reaction:
A7B
Rate law:
a
dC A
-rA=kCA =-_.dt
5-23
..--.:r;-...-..- . -..----.--..-.-......--..-.....- .....- ...-.--.--.---....... . -f,.CA/f,.t
.-dCA/dt........____ .._
Iime_(min) .._ _ _. C A (mol/dm)
o
4
0 . 39
0.37
2.89
3
035
0.32
225
5
0..30
0 . 267
1.45
8
0.25
0.225
to
0 . 21
0\75
12
065
0.15
0.133
0.25
15
0.1
0.072
17.5
0.07
0 . 06
Plot of log --dCA/dt vs log C A shows a. "" 0.5
de.,
kA
=···~iitM.;:; O~~S019·d;:11~1:
CDP5-C
Ethane hydrolysis over a commercial nickel catalyst in a stirred contained solid reactor.
H2 + C2H6 -7 2CH4
PA = CART =CAoRTCl- X) =PAoCl- X)
PB =CBRT=CAORTCBB -X)=PAoCBB -X)
X =!SFTO =~= Yp Cl+BB)
2FAO
2YAO
2
I
I
FAOX
Fp
-r A =-r B=- W--'--2W =
.- r'A
Y pFAO
2W
= kPA a pB P
InC-r'A) = Ink + aInPA + fJInPB
y = Ao
+ Al Xl + A2 X 2
FTo(gmollh)
PAO(atm)
PBO(atm)
1.7
0 ..5
05
1.2
05
05
0.6
0.5
0.5
0.3
0.75
04
0. 6
06
0.6
2.75
0.6
0.4
POLYMATH Results
Nonlinear regression {L.M}
YCH4
0.0
5
0.0
7
0.1
6
01
6
0.1
0.0
6
X
0. 0
5
0. 0
7
01
6
0.2
0.1
0.0
5
Q
PA(atm)
PB(atm)
-rA(gmollkg . h)
0.475
0.475
1.0625
0.465
0465
L05
1
042
042
L2
L5
1
0.6
7
032
0.54
0..52
0 ..54
0.6
0 . 9375
057
0.37
20625
1
Model: ra = k*(PaAalfa)*(PbAbeta)
Variable
lni guess
- -Value
--_.
95% confidence
5-24
0.1
k
alfa
beta
Precision
0 . 1124446
0.152574
0 . 1668241
0.5068635
0.9828027
-1. 9669749
1
1
:::: 0.999213
:::: 0.9986883
R"2
R"2adj
RInsd
:::: 0.0051228
Variance :::: 3.149E-04
k = 0.5 atm gmollhr kg;
hence, the rate law is:
a == 1;
P == -2
-r~ = 0.5 ~ gmol / kg .hr
B
CDP5-D
Since oxygen is found in excess, we assume that.-r·1iO is dependent only on ~.
This gives us a rate law of the form:
- rHO
=kC:.o
.
we
From the units of the specific reaction [ate,
assume that ex == 3. Now, using
equation (5-18) from chapter S, we can solve for the desired balf-.Jives.
(a)
For ~ == 3000ppm:
t",
(b)
~ 2\iA"iO}""".'I~(3000~)' )= 119.05_
For C"'No<>
t",
=1 ppm:
=2\1.4,,10. 3pp;;;:>J;;;i~{(1~y L071x 10' mID
)=
CDP5-E
Given the data.. postulate a rate law.
=kc:.
·rA
Then write the design equation in tenns of the data given, in this case volume and time.
CA=NAV
NA =N"o(l-X)
V=
Yo(l + EX)
V-· V.
X=·· . ···. . _J1.
Vue
V-v.
=__ . Yoe
1-·_··· ........2..
C
"
V
.5-2.5
Plug that into the design equation:
de
--2..=kC:
dt
A
5-26
The following grapb is made.
Once that is dorie it is
ready to be graphed. The
following graph is the natUral log the derivitive ofthc volunle function against
the natUral 10 . of the volume function.
From the graph we see that ex. is 2. We can also find k:
9
k == :9 =.018
J
5
N
.. 0
303.39*.2 020 1
=YAO NTO =. 85* 8.314*313
=. rna es
k'
.018
k =. -N-a - I = M!, •.9
.v...
.40
The following rate law is found:
2
-r:A - .9*cA
Now, to decennine the volume of the CSTR. we must use the design equation and
stoichiometry:
,
V;: F,..oX.
-rA
CA
:;:
CAo(l- X)
C"o =!MJ
RTa
~o ;: Y"ofQ =.6 * 1013.25kPa == 607.95
Combining these with the rate law just determined. the followinl! volume is found.
5-27
CDP5-F
Asr't, '" U
"riP.
dZ
Z
P
0
129
1.5
2,5
70
50
45.3
28,,63
19.4
:l.O
30
9.5
6.5
18
L3
9.0
16
(1.7)
_de.
AsHi "'" 3.0
AsHi = 3.0
Z
P
dZ
0
129
1..5
45
90
34.0
2.5
4.0
22
15.4
10
2.0
_df.
Z
P
dZ
0
129
1.5
95
17.4
15 . 0
PFR
A ... B ::C
-8
I
1.5 2.0
I
)
6
9
,
I
4
PM PAl.
Task I, Re\\l'I'ite the design equation (i.e.. mole balance) in. te:rms of me measurement
variables. Recall V ::: Ac Z. then
ax. = :r,,&
dZ
FAo
5-28
For iSOthe."!I1ai operation and no pressure drop.
X)
rr:.. eX)'
(l ..
CA =
CAO
CA
'=
P,JRT
(1. X)
= PA 0(1 +"e;C)
PA
(I + EX)
PA = PAO (1 - X)
Now we have the differential. mole balance
in terms of the measured variables PA and
Z.
Posrulate
-tA '=
k[PA ~ -~~j
Task 2. Look for simplifications.
A) Sc:e if volume change can be neglected.
a therefore neglect volume change.
E ::
forr..:ns 1 and 3 where PAsH} "" 1.5 and 3.0 torr. respectively. [har
most of tht E!;In is consumed. indicating the equilibrium is reasonably fJr to
the ri giu. Conse:quen ely. the reverse reaction is negligible in the fU"St pan of the
B) \Ve see
L1:U
reactor. i.e.
C)
\'v:: ::lise
see i..l-tat for nlnS 1 and 3 that B is in excess and that far excess B
II
5-29
-r,
• ,,,,\
:::::
k
,
R
pi-' DO.. 80 .. A
=
lK' pet
~ r\
Algorlcnm
Task -' Co1=ul>te (0
~;) Jnd plot vs Z to find (0 ~)
. dP
Plod ~ dZ-'\) vs (PA) on log-log paper to find alph3..
Task 4.
T~e
Task 5.
Ev:uua~e k" wd K?
r:ldo of inidal races at PAsH]
=:
3.0 torr and PA1l:1J
=:
L5 corr
From AjHl of 3,,0 corr we see equilibrium is reached at P Ae :=:.0 {
PC<:::;:;
0.129 ~ 0.01 ::::: .119
3.0 •. t 19 = 2.SS1
K ==
.! i2
: : : J...I] torr.. oL
"?
(.01)(2.381)
P:ac
-
~...
<=
:: 90 :< 10-)
tg :;.: x;: (.119 torr.l (3.0 ~orr)
k: :::; 0.23 ({or. c~tt
5-30
CDP5-G
From given data, find the rate law"
Given: Oxidation of propene to acrolein
Rate law:
rA = kPpapo
b
2
~W=O.5g
Using Polymath non linear regression, the following results were obtained:
Nonlinear regression (L-M)
Model: ra = k*PpA a*P02A b
Variable
k
a
b
lni guess
5
95% confidence
2.685E-05
0.0046367
0.001358
Value
0.006609
0.9948724
0,,2034299
1
1
Precision
RA2
= 0.9999969
RA2adj
= 0 . 9999953
Rmsd
5.722E-07
Variance
4.011E-12
=
=
CDP5-H
1)
A ..- ..!~-7 prod
B
.,,-.!~..~
- f1
= kAC ... +kaCa
-fA
prod
=kAC A
- fa = kaCll
_ d~·~I =kAC",,(l-X ... )+kBC:ao(l-X a )
dt
C1(t) = C A (t)t-Ca(t:):= C ...o(l-X ... ) t CaQ(l-X a)
C1(t) = CAo(l- X ... + 9
g ......
XB)
=C
Ao
(1.75 - X ... - Xs)
.~~L=CAO(kA -kAX", +k a9 a -k aXB)
dt
_.~~:l
= C.Ao (k A dt
II)
tn
n -) prod
Crr(t)
crt}
kal/.3
Xt t. • • 14
de II
-
... Cn(t)
So,
ten.)
~ ~
kAX", + O.75kaea •. kaXa)
2
2:
d;- '" -r II '" knC n '" kn eno (l-ln)
"" eno
7;
(I-In)
0
0
10
lO
30
40
.50
'0
0.01,.4
o.ou~
0.0011
0.00'"
0.0074
11 ,00i40
o.ooso
O.31J
I).SlI
0.700
0 ••1$
1.00
1.lU
0
5-31
If
n. 10
0
~
Alt}
-~)
<il:
cio (1-10,2
If II.
-:LhlO
~l.hJ.O-4
.. l.hlO- 4 • Z.h10'"
1.9hlO
LJ7S
Ln
~.
-4
0.400
0.301
0.179
-1.ldO
+1.h10-4
-4
0.571
0.411
-ldO
"1.1:tlO-4
~.
.
-1.4110
-4
-
·l.ltlO
Q.6.43
-ldO
-4
+l.ltlO·'
..0,1110 ....
1.3UdO
9.Uhl.O- 3 7.06dO-'
5.41b10-'
l.'UtlO-5
1.'Odo~$
1.60.1
1.700
l.lU
3.311
l.lO]
'~4
1.701
If I! were true .. all kO's should 0(:: the same .._-) II is not true:
rime (s)
.------,---------------~
CDP5-I
(a)
Experimental Plan to tind the rate law for the hydrogenation of cyclopentane on a
PtIAll)l catalyst:
1. Since this is a stable catalyst we don't have to worry about catalyst decay and an
Integral React(l" will be used.
2, Perform several different nms, holding CAO and W constant while
from run to run.
3., Plot X A•OUI
V$,
FAG
is valied
WIF;>.o for all IUllS,
4. Fit a curve through all points which passes through the origin,. The slope at any
point is the reaction late Record the slope and COII'cspouding CAl} for many
different X A values. 111cse data can be used to determine the rate law.,
(b)
Experimental Plan to find the rate law for the liquid-phase production of methyl
bromide from an aqueous solution ofmerhyl amine and bromine cyanide:
L Fot a liquid-phase reaction without a catalyst. use a batch reactor.
2. Vlhile running the reaction record both CA and en at equal time intervals
3.. Repeat to ensure a.ccurate data.
CDP5-I (c) No solution
CDP5-J No solution will be given.
5-32
CDP5-KNo solution will be given.
5-33
Solutions for Chapter 6 - Multiple Reactions
P6-1 Individualized solution
P6-2 (a) Example 6-2
ForPFR,
dC A
- dr
= -kJ -
dCB
dr
=k
kzC A
Z
-
dClL - k
J
dr
k3 C A
dCy
dr
C
z
A
=k
CZ
3
A
3
In PFR with V = 1566 dm we get X = LO and SB/XY = 0.394
also at V = 533 dm 3 SBIXY is at its maximum value of 0 . 625
0.04 . - - - - - - - - - - - - . - - - - - - ,
0'.7r-----
o.J
0. . 6
0..2
0..4
0.1
0.",3
0.,,0.
0.,1
-0..10.
313
626 tau 940.
1253
1566
0.",,0. . - - - " - - - "-----~----'
1253
1566
0.
313
626 tau 940.
See Polymath program P6--2·a.poJ.
POL YMATH Result"
Calculated values of the DEQ variables
Variable
tau
Ca
Cx
Cb
Cy
Cao
X
k1
k2
k3
Sbxy
initial value
0
0.4
1.. OE-07
0
1.0E-06
0,,4
0
1.0E-04
0 . 0015
0.,008
0
minimal value
0
-0 . 0439962
1 . OE·-07
0
1. OE-06
0.4
0
1,OE-·04
0,0015
0,,008
0
maximal value
1566
0,,4
0,1566001
0.1490534
0 . 1617665
0..4
1.1099906
1 . OE-04
0.0015
0,,008
0,,6436717
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(tau) = -k1-k2*Ca-k3*CaI\2
[2] d(Cx)/d(tau) = k1
[3] d(Cb )/d(tau) = k2*Ca
[4] d(Cy)/d(tau) k3*Ca1\2
=
6-1
final value
1566
-0 . 0439962
0 . 1566001
0 . 1256308
0,,1617665
0.4
1 . 1099906
1,OE-04
0 . 0015
0,008
0.3946104
Explicit equations as entered by the user
[1] Cao = 0 . 4
[2] X = 1-Ca/Cao
[3] k1 = 0.0001
[4] k2 = 0 . 0015
[5] k3=0.008
[ 6] Sbxy = Cb/(Cx+Cy)
(2) Pressure increased by a factor of 100.
Now CAO = PIRT = 0.4 x 100 = 40 mol/dm3
For single CSTR, C A ' does not change but
r - -v - CAO
-
Vo
CAO - CA
- ----'=----=..:...- rA
- klA + k 2A C A + k3AC~
T=21217sec
-
CA
40-0.112
---------sec
0.0001 + 0.00168 + 0.0001
P6-2 (b) Example 6-3
(a) CSTR: intense agitation is needed, good temperature control.
(b) PFR: High conversion attainable, temperature control is hard - non-exothermic reactions, selectivity not
an issue
(c) Batch: High conversion required, expensive products
(d) and (e) Semibatch: Highly exothermic reactions, selectivity i.e. to keep a reactant concentration low, to
control the conversion of a reactant.
(t) and (g) Tubular with side streams: selectivity Le . to keep a reactant concentration high, to achieve higher
conversion of a reactant
(h) Series of CSTR's: To keep a reactant concentration high, easier temperature control than single CSTR.
(i) PFR with recycle: Low conversion to reuse reactants, gas reactants
(j) CSTR with recycle: Low conversions are achieved to reuse reactants, temperature control, liquid
reactants
(k) Membrane Reactor: yield i.e. series reactions that eliminate a desired product
(1) Reactive Distillation: when one product is volatile and the other is not
P6-2 ( c) Example 6-4
For kl = k2' we get
C A =CAoexp(-klT')
6-2
and
X opt
= 1- exp·( k(l' ') = 1- e -1 =0.632
ForaCSTR:
W= FAo-FA = vO(CAO-CA)
-r' A
r
-r' A
, CAO -CA CAO -CA
=
= ---""'---'-"-r'A
kjCA
CA = CAO (r'kj +1)
C
CB
, C -CB
r = BO
= - - -B- - =
.
-r'B
-kjCA+k2 CB -kjCAO(r'kj +1)+k2 CB
1. 8 0 . . - - - - - - - - - - - - - - - . .
1.64
to find the maximum concentration of C, differentiate
CB with respect to .' and set it equal to 0 .
148
See Polymath program P6·2··c.pol.
simplifying we get
320
Then use the quadratic formula to solve for .'.
(2) Operating temperature = 325 K
6-3
340
r
360
380
400
Equation (E6-6.8):
P6-2 (0 Example 6·7
For equal molar feed in hydrogen and mesitylene.
3
CHO = YHOC ro = (0.5)(0.032)lbmollft =0.016Ibmollfe
3
CMO = 0.016lbmollft
Using equations from example, solving in Polymath,
we get
T opt = 0.38 hr., At1' = 0.5 hr all of the H2 is reacted and only the decomposition oiX takes place . ,
XH
CH
.. CM
C !-,
- - ! .x
l'
----_.
Ex 6-7
0.50
0.0105
0.0027
0.00507
0.2hr
0 . 596
--
This question
0.99
-0.00016
-0.0042
0.0077
--_._---0..38hr
._-1.865
SXIT
See Polymath program P6·-2-f.pol.
POL YMATH Results
Calculated values of the DEQ variables
Variable
tau
CH
CM
CX
k1
k2
rIM
r2T
rlH
initial value
-----
minimal value
o
o
0 . 016
0.016
1.64E-06
0.0041405
o
55.2
30.2
-0.1117169
o
-0.1117169
r2H
o
rlX
r2X
0_1117169
o
o
55.2
30 . 2
-0.1117169
o
-·0.1117169
-0 . 0159818
2.927E-04
-0.0159818
maximal value
0.43
0.016
0 . 016
0.0077216
55.2
30 . 2
-2.927E-04
0.0159818
-2.927E-04
o
0.1117169
o
6-4
final value
0.43--1. 64E-06
0.0041405
0.0077207
55.2
30.2
-2.927E-04
2.986E-04
-2.927E-04
-2.986E-04
2.927E-04
-2.986E-04
ODE Report (RKF45)
0.020,.-----------------,
Differential equations as entered by the user
[1] d(CH)/d(tau) = r1 H+r2H
[2] d(CM)/d(tau) = r1 M
[3] d(CX)/d(tau) = r1 X+r2X
0.016
0.012
Explicit equations as entered by the user
[1] k1 = 55,.2
[2] k2 = 30.2
[3] r1 M = -k1*CM*(CHI\.5)
[4] r2T = k2*CX*(CHI\. 5)
[5] r1 H = r1M
[ 6] r2H = -r2T
[7] r1 X = -r1 M
[8] r2X = -r2T
Increasing 9H decreases 'opt and
- CH
.. Ci\I
ex
0.008
0. 004
00000.000
0.086
O.I72
0.258
tau
0.344
0.430
S xrr··
P6-2 (g) Example 6-8
Using equation from example 6-8:
Polymath code:
See Polymath program P6·2··g.poL
rOLYMA TUResults
NLES Solution
Variable
CH
CM
CX
tau
K1
K2
CHo
CMo
Value
4.783E-05
0.0134353
0.0023222
0.5
55.,2
30.2
0,,016
0.016
f (x)
-4,,889E-ll
-1. 047E--ll
-9.771E-12
Ini Guess
1 . 0E-04
0 . 013
0.OQ2
NLES Report (safenewt)
Nonlinear equations
[1]
[.'2]
[3]
f(CH) = CH-CHo+K1*(CM*CHA.5+K2*CX*CHA.5)*tau = 0
f(CM) = CM-CMo-tK1*CM*CHI\,,5*tau = 0
f(CX) = (K1*CM*CHI\.5-K2*CX*CHI\0,,5)*tau-CX = 0
0.020,-·-----
Explicit equations
tau = 0,,5
K1 = 55 . 2
K2 = 30 . 2
[4] CHo=0016
[5] CMo=0 . 016
[1]
[2]
[3]
0.016
'
'- -- -.,-,
........ ..........'".........
......
.. -'""...~.--........... " ....
--.,...'"'"--........ ,~...
0.012
Df]
CH
0.008
A plot using different values of T is given .
For T =0.5, the exit concentration are
CH = 4 . 8 xW5 lbmoVfe CM =00134IbmoVfe
Cx =0.00232 Ibmollft3
" Cl\,,1
- ex
0.004
0.000 lL:"'-_ _~-=:::==:::====:d
0.00
0. 04
0.08 tau 012
0.16
0.20
6-5
The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for T =0..5:
YMX
Fx
= --.:..:...-F MO -FM
Cx
CMO
= - -0.00232
---0.016-0.0134
-eM
0.89mole· xylene· produced
mole· mesitylene . reacted
The overall selectivity of xylene relative to toluene is:
=.Fx = 8.3mole· xylene· produced
S
FI
x IT
mole· toluene· produced
This Question
4.8 x 10-5
Ex 6-8
0..0.0.89
0..0.0.29
0..0.0.33
0..5
0..41
0..7
CH
CM
Cx
T
Y MX
SXff
0..0.134
0..0.0.232
0..5
0..89
8.3
---
-_c--.
P6-2 (h) Example 6-9
(1)
SDIU
Original Problem-
P6-2 h
1.0.1
0..20.8
SDIU
Membrane Reactor
2.58
PFR
0..666
..
Doubling the mcommg flow rate of specIes B lowers the selectIvIty.
(2) The selectivity becomes 6.52 when the first reaction is changed to A+2B
-~
D
P6-2 (i) Example 6-10
Original Case -. Example 6-10
15
P6-2 i
---_..
---
15
- FA
12
.. FA
9
..
..
..
6
-
FC
FD
FE
..
IfIf
- Fe
- FD
FB
9
6
3
o0
FB
12
3
2
4
V
6
8
10
0
0
1
2
V
3
5
The reaction does not go as far to completion when the changes are made. The exiting concentration of D,
E, and F are lower, and A, B, and C are higher.
See Polymath program P62i.pol.
P6-2 (j)
6-6
O.Oe+O
140
O
280 t
420
560
700
At the beginning, the reactants that are used to create TF-VIla and TF-VIIaX are in high concentration. As
the two components are created, the reactant concentration drops and equilibrium forces the production to
slow. At the same time the reactions that consume the two components begin to accelerate and the
concentration of TF-VIla and TF-VIlaX decrease . As those reactions reach equilibrium, the reactions that
are still producing the two components are still going and the concentration rises again . Finally the reactions
that consume the two components lower the concentration as the products of those reactions are used up in
other reactions.
P6-2 (k)
Equal-molar feed
Base case
10
-
8
10
..
Fx
..
Ii'!
-
8
6
4
4
2
2
40
80
FlU
Fh
Fx
FlUe
6
o0
-
FlU
Fh
V
120
160
200
o
o
40
80
V
120
160
200
Increasing YMO will increase the production of m-xylene and methane, but will result in a large amount of
un-reacted mesitylene .
P6-2 (I) Individualized solution
P6-3 Solution is in the decoding algorithm given with the modules ( ICM problem)
P6-4 (a)
Assume that all the bites will deliver the standard volume of venom. This means that the initial
concentration increases by 5e-9 M for every bite .
6-7
After 11 bites. no amount of anttvenom can keep the number of free SIteS above 66 7% of total Sites, ThIs
means that the imttal concentration of venom would be 5 5e-8 M The best result ocelli'S when a dose of
anttvenom sucb that the Iruttal concentration of antivenom In the body IS 5.7e-8 M, wIll result in a mmimum
of 66 48% free sites, wincb IS below the allowable nurumurn.
See Polymath program P6-4-a.pol.
P6·4 (b)
The VICtim was bitten by a hatmless snake and anttvenom was Injected. ThiS means that the tmttal
concentration of venom 18 O. From the program below. we see that if an amount of antivenom such that the
tmttal concentration in the blood is 7e-9 M, the pattent Will dIe.
See Polymath program P6-4-h.pol.
POLYMATn Results
Calculated valges of the DEQ variables
Variable
t
initial value
o
o
.
fsv
fs
Cv
o
Ca
7 OE-09
1
fsa
Cp
kv
ksv
ka
kia
eso
ksa
kop
9
h
m
o
o
5
final value
0.5
o
0.6655661
0.6655661
1
o
o
o
4.503E-·09
0.3344339
4 503E-09
7.0E-09
0.3344339
o
o
2 OE+OS
6.0E+OS
2.0E+08
5.0E·-09
6.0E+08
1.2E+09
2.0E+08
6.0E+08
2 OE+OS
1
S.OE-09
6.0E+08
1.2E+09
2.0E+08
6.0E+OS
2.0E+08
1
5 OE-09
6 OE+OS
1 2E+09
2.0E+08
6.0E+08
2.0E+08
1
5.0E-·09
6.0E+08
1. 2E+09
0.3
o 3
0.3
0.3
3
0,,3
0.3
o 3
o
o
o
o
o
o
j
maximal value
o
o
o
kov
koa
o
o
o
o
1
kp
minimal value
-2.1E-09
o
o
o
o
o
o
o
o
-2.1E··09
-1 351E-- 09
o
o
-1. 3S1E-09
ODE Report (STIFf)
Differential equations as entered by the user
[ 11 d(fsv)/d(t) kv * fs * Cv • ksv * fsv * Ca
[2 J d(fs)/d(t) -kv*fs*Cv - ka * fs • Ca + kia * fsa + 9
[3] d(Cv)/d(t) Cso • (-kv * fs • Cv - ksa * fsa * Cv) + h
[4 J d(Ca)/d(t) Cso*(-ka * fs • Ca + kia * fsa) + j
[ 5 ] d(fsa)/d(t) ka * fs * Ca • kia • fsa - ksa * fsa • Cv
[6] d(Cp)/d(t) Cso • (ksv· fsv· Ca ,+ ksa * fsa * Cv) + m
=
=
=
=
=
=
100..-------ExpliCit equations as entered by the user
[1] kv=2e8
[2J ksv=6e8
[3J ka=2e8
[4J kia=1
rSJ Cso 5e-9
[6] ksa=6e8
[7J kp=12e9
0.92 ---"-- -- - .. -.------.-.-----.-._----- .----- ... --.-.-0.84 '-'--'--'--
=
Q
--.----.-------..-----.--.. --
0.76 ------ .--------.068 ---------------------.-.-..---""'-....::-......
0.60L----------0.0
0.1
02 t 03
6··8
0.4
0.5
[8] kov= 0
[9] koa=O . 3
[10] kop=O.3
[11] 9 = ksa * fsa * Cv + ksv * fsv * Ca
[12] h = -kp * Cv * Ca - kov * Cv
[ 1 3 J m = kp * Cv * Ca - kop * Cp
[14 J j = -Cso * ksv • fsv * Ca - kp * Cv * Ca - koa * Ca
P6-4 (C)
The latest time after being bitten that anti venom can successfully be administerd is 27.49 minutes. See the
cobra web module on the CDROMlwebsite for a more detailed solution to this problem
I
09 'rI
:2.-::
I::
<::>
'J:
"'" 0.8
.;::
.:::...
'"g::
....
l-<
Amivcnom Injected at t
2749 min
OJ
o ()
tO
02
04
()6
t· ·4'!
OJl
1.4
16
Time (houri
P6-4 (d)
Individualized Solution
P6-5 (a)
Plot of C A , CD and C u as a function oftime (t):
See Polymath program P6·Sa.poJ.
POL YMA Tn Results
Calculated values of the DEQ variables
Variable
t
Ca
Cd
Cu
k1
initial value
o
1
minimal value
maximal value
final value
15
1
0.7995475
0 . 5302179
1
15
0.0801802
0 .. 7995475
0.1202723
1
o
0 . 0801802
o
o
o
o
1
1
6-9
.1 8
2
k2
K1a
K2a
Cao
100
10
1.5
100
10
1.5
100
10
1.5
100
10
1.5
1
1
1
1
0.9198198
0.9198198
o
o
X
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -(k1 *(Ca-Cd/K1 a)+k2*(Ca-Cu/K2a))
[2) d(Cd)/d(t) = k1 *(Ca-Cd/K1 a)
[3] d(Cu)/d(t) k2*(Ca-Cu/K2a)
=
Explicit equations as entered by the user
[1] k1
1.0
[2] k2 = 100
[3] K1a = 10
[4] K2a = 1.5
[5] Cao = 1
[6) X = 1-CaiCao
=
1.0
1.0
0.8
C~l
- Cd
0.8
0. 6
0. 6
0.4
0.4
0.2
0.2
0.0
J--_~
o
__
3
~_.
____
6
9
0.0
~.
12
15
l~
o
----'
---~-~-,-.--'-.
3
6
9
12
15
To maximize CD stop the reaction after a long time . The concentration of D only increases with time
P6-5 (b)
Conc. Of U is maximum at t = 0.31 min.(C A= 0.53)
P6-5 (C)
Equilibrium concentrations:
3
CAe = 0.08 mol/dm
3
CDe = 0 . 8 mol/dm
CUe = 0.12 mol/dm3
P6-5 (d)
See Polymath program P6-S-d.poL
POLYMATH Results
NLES Solution
Variable
Ca
Value
0.0862762
f(x)
-3.844E-14
lni Guess
1
6-10
0 . 7843289
0 . 1293949
1
1
100
10
1.5
100
Cd
Cu
CaO
k1
k2
K1a
K2a
t
-2 . 631E-14
6 . 478E-14
0
0
NLES Report (safenewt)
Nonlinear equations
[1] f(Ca) = CaO-t*(k1 *(Ca-Cd/K1 a)+k2*(Ca-Cu/K2a))-Ca = 0
[2] f(Cd) = t*k1 *(Ca-Cd/K1 a)-Cd = 0
(3] f(Cu) = t*(k2*(Ca-Cu/K2a))-Cu = 0
T
Explicit equations
[1]
[2]
[3]
[4]
[5J
[6]
[7]
10 min
100min
0.295
0.133
0..0862
CDexit
0.2684
0.666
0.784
CUexit
0.436
0.199
0.129
X
0 ..705
0.867
0.914
C Aexit
=
1----
CaO 1
k1 = 1
k2 = 100
K1a=10
K2a = 1 . 5
t=100
X = 1-Ca/CaO
1 min
P6-6 (a)
kl
= 0.004(
;;3 J1I2
I
k2 =0.3min-1
A·~B
dm 3
k3 =0.25----I .
mO.mm
Sketch SBX, SBY and
SBIXY
as a function of CA
See Polymath program P6. ·6-a.poL
1)
6-11
min
40~----------------------~
32
24
16
8
o
0.0001 0.0326
0.065\.~.p.0976
0.1301 0.1626
2)
120o.1-----------------------,
960
720
480
240
o
0.0001 0.0326
0.065\~.P0976
0..1301 lU626
0.065\~J).0976
0.1301 01626
3)
.-1
2
o
0.0001 0.0326
P6-6 (b)
Volume of first reactor can be found as follows
6-12
We have to maximize SBIXY
3
From the graph above, maximum value of SBXY = 10 occurs at CA' = 0 . 040 mol/dm
So, a CSTR should be used with exit concentration CA*
3
Also, C AO = P AIRT = 0.162 mol/dm
1/2
+rB+ry = (k 1C A
And-rA=YX
=
=
>V
2
+k2 C A +k 3C A )
*
*
vO(CAO -CA ) =
vO(CAO -CA )
= 92.4dm 3
*
112
*
*
2
-rA
(k 1(C A ) +k 2C A +k3 (C A ) )
P6-6 (C)
Effluent concentrations:
.
CB
We know, 't = 9..24 mm => 1: = -
k2 CA
rB
Similarly: C X
*
mol
= 0.007 -"3
dm
mol
= -C B- ~ CB * = 0.11-3
dm
*
mol
Cy = 0.0037 --3
and
dm
P6-6 (d)
Conversion of A in the first reactor:
CAO -C A = CAOX
~
X =0.74
P6-6 (e)
A CSTR followed by a PFR should be used.
Required conversion = 0.99
dV
FAO
=> For PFR, Mole balance: - - = - dX
-YA
~V
= lOxO.162x
dX
0.99
f
2
1/2
074 (k 1C A
+k2 C A +k3 C A )
= 92.8dm
3
P6-6 (f)
If we notice that E2 is the smallest of the activation energies, we get a higher selectivity at lower
temperatures. However, the tradeoff is that the reaction rate of species B, and therefore production of B,
decrease as temperature drops . So we have to compromise between high selectivity and production . To do
this we need expressions for k[, k2, and k3 in terms of temperature. From the given data we know:
ki =
A exp(I:~~~ )
Since we have the constants given at T = 300 K, we can solve for Ai'
.004
" = - - - - - - - = 1.4ge12
--20000
exp ( 1.98 (300)
~ = ___ . 3
··-10000
exp ( 1.98 (300)
J
=5.7ge6
J
6-13
(.25
)~1.798e21
-30000
A, ~
exp 1.98(300)
Now we use a mole balance on species A
V= FAO -FA
-rA
V
v( CAO -CA)
=---'-----=-::..::.-.-:.:.:...
A mole balance on the other species gives us:
F; =vC =ljV
j
Ci =7lj
Using these equations we can make a Polymath program and by varying the temperature, we can find a
maximum value for CB at T = 306 K. At this temperature the selectivity is only 5"9,, This may result in too
much of X and Y, but we know that the optimal temperature is not above 306 K. The optimal temperature
will depend on the price ofB and the cost of removing X and Y, but without actual data, we can only state
for certain that the optimal temperature will be equal to or less than 306 K.
See Polymath program P6-6-f.pol.
POLYMA TIl Results
NLE Solution
Variable
Ca
T
R
k1
k2
Cao
Cb
k3
tau
Cx
Cy
Sbxy
Value
0.01.70239
306
1. 987
0.0077215
0.4168076
0.1
0.070957
0,,6707505
10
0,,0100747
0,,0019439
5,,9039386
f(x)
3.663E-10
Ini Guess
0.05
NLE Report (safenewt)
Nonlinear equations
[1] f(Ca} = (Cao-Ca}/(k1 *CaA ,,5+k2*Ca+k3*CaA 2}-1 0 = 0
Explicit equations
T=306
[2] R= 1.987
[3] k1 = 1.4ge12*exp(-20000/Rff)
[ 4] k2 = 5790000*exp( -1 OOOO/Rff}
[5] Cao =.1
[1]
6-14
6] Cb = 10*k2*Ca
71 k3 = 1.798e21*exp(-30000IRfT)
8] tau = 10
9] Cx = tau*k1 *Ca/\5
lO} Cy = tau*k3*Ca"2
11] Sbxy = Cb/(Cx+Cy)
P6-6 (g)
Concenttation is proportional to pressure in a gas-phase system Therefore:
S B / XY
~
~
p
Vp+p2
which would suggest that a low pressure would be ideal. But as before the ttadeoff is
lower production of B. A moderate pressure would pIObably be best
P6-7
US legal limit: 0 8 gil
Sweden legal limit: 0.5 gil
A~B
k2 >C
Where A is alcohol in the gastrointestinal tract and B is alcohol in the blood stteam
dCA =-kC
dt
) A
dCB =kC -k
dt
) A
2
k)
= lOhr- 1
g
2.0
k2 =0.J92-L hr
Two tall martinis = 80 g of ethanol
Body fluid = 40 L
C
AD
16
1.2
= 80g =2 g
40L
----------.-------
L
Now we can put the equations into Polymath
0.8
See Polymath program P67 .poL
04
PO'LY~fAT!LReslilts
Calculated values of the DEO variables
0 . 0 0'---2----,--4~·
Vaziable initial value minimal
value maximal value final
value
----
t O O
2
7 .131E-44
Ca
Cb
o
o
kl
10
k2
0 . 192
10
0.192
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ··k1 *Ca
6-15
6
10
10
2
1 . 8901533
10
0.192
0.08
10
0 . 192
7 131E-44
8
10
[2]
d(Cb)/d(t)
=-k2+k1 *Ca
Explicit equations as entered by the user
[1] k1 = 10
[2] k2 =0 . 192
P6-7 (a)
In the US the legal limit it 0.8 gIL.
This occurs at t = 6.3 hours ..
P6-7 (b)
In Sweden C B
= 0.5 gil , t = 7.8 hrs.
P6-7 (C) In Russia CB = 0 . 0 gil, t = 10.5 hrs
P6-7 (d)
For this situation we will use the original Polymath code and change the initial concentration of A to 1 gIL .
Then run the Program for 0.5 hours. This will give us the concentration of A and B at the time the second
martini is ingested . This means that 1 gil will be added to the final concentration of A after a half an hour.
At a half an hour CA = 0.00674 gIL and CB = 0.897 gIL . The Polymath code for after the second drink is
shown below.
See Polymath program P6·7··d.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
k1
k2
initial
- - - -value
--0.5
1.0067379
0.8972621
10
0 . 192
minimal value
0.5
5 . 394E-42
0.08
10
0.192
maximal value
10
1.0067379
1. 8069769
10
0.192
final value
10
5.394E-42
0 . 08
10
0.192
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) =-k1 *Ca
[2] d(Cb)/d(t) = ··k2+k1*Ca
Explicit equations as entered by the user
[1] k1 = 10
[2] k2 = 0.192
for the US t = 62 hours
Sweden: t = 7 . 8 hours
Russia: t =10.3 hours .
P6-7 (e)
The mole balance on A changes if the drinks are consumed at a continuous rate for the first hour.. 80 g of
ethanol are consumed in an hour so the mass flow rate in is 80 glhr·. Since volume is not changing the rate of
change in concentration due to the incoming ethanol is 2 gIL/hr.
For the first hour the differential equation for CA becomes:
6-16
dC A = -kl CA + 2t after that it reverts back to the original equations .
dt
See Polymath program P6-7 -e . pol.
POLYMATH Results
Calculated values of the DEQ variables
initial value
Variable
t
Ca
°
°10°0.192
Cb
k1
k2
minimal value
°
°
-1.1120027
10
0 . 192
maximal value
11
0.1785514
0 . 7458176
10
0.192
final value
11
6.217E-45
-1.1120027
10
0 . 192
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t)
if(t<1 )then(-k1 *Ca+2*t)else( -k1 *Ca)
[2] d(Cb )/d(t) = -k2+k1 *Ca
=
Explicit equations as entered by the user
[1) k1 = 10
[2) k2=O.192
US: C B never rises above 0 . 8 gIL so the is no time that it would be illegaL
Sweden: t = 26 hours
Russia: t = 5.2 hours
P6-7 (0
60 g of ethanol immediately -7 C A = 1.5 gIL
C B = 08 gIL at 0 . 0785 hours or 4 . 71 minutes.
So the person has about 4 minutes and 40 seconds to get to their destination.
P6-7 (g)
A heavy person will have more body fluid and so the initial concentration of C A would be lower. This
means a heavier person will reach the legal limit quicker. The opposite is true for a slimmer person. They
will take longer to reach the legal limit, as their initial concentration will be higher.
P6-8 (a)
Let A be the tarzlon in the stolllach and B be the tanion in the blood.
iVlole Balances:
dC~
_ .. _ _t_
..
dt
=r
A
deB
-·-=r
dt
B
6-17
Rate Laws:
·.. rA
ra
;;;;;;
k/-~A
-+- kZC A
= kiC A -
k, _., k~Cs
All k values .are given in the proble m statement. It must be noted, however, that
for CB < 0, k, must be equal to O.
These equations when entered in POLTh!ATH generate the following results:
See Polymath program P6-8·,·a.po1.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
k1
k2
k4
k3
initial value
o
6,,25
o
0 . 15
0.6
0.2
0.1
minimal value
o
0.3111692
o
0.15
0.6
0.2
0.1
maximal value
4
6.25
0.5977495
0.15
0.6
0.2
0.1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -k1 *Ca-k2*Ca
[2] d(Cb)/d(t) = k1 *Ca-k3-k4*Cb
Explicit equations as entered by the user
[1] k1 =0,,15
[2] k2 = 0.6
[3] k4 = 0,,2
[4] k3 = if(Cb<O)then (k1*Ca-k4*Cb) else (0,,1)
P6-8 (b)
6-18
final value
4
0.3111692
0.4057018
0.15
0.6
0.2
0.1
From the following graph generated using the above program in POLY1v1ATH, We
can see the proper doses of the drug:
.
1.
First take two doses of the crug.
2,
Six hours later take one dose.
),
Take one dose every four hours from then on.
Tr\
6·8b
ii \
~:1 \(~YVf~V~~~V\
!
I
. . . . . ;- __ .. "!.
~~
£~.Gcri1
~'""_'o. _ _ _ _ .~_....
2D.CCG
.
+.
'3lJ.,CGG
P6-8 (c)
If one takes initially two doses of Tarzlon, it is not recommended to take another dose within the first six
hours Doing so will result in build up of the drug in the bloodsueam that can cause harmful effects .
P6-8 (d)
If the drug is taken on a full stomach most of it will not reach the wall at all. The processed food can also
drag the drug to the intestines and may limit its effectiveness . This effect can be seen in the adsorption
constant kl and elimination constant k2 values If kl decreases this means that the adsorption process is slow
6-19
and if k2 increases means that the rate of elimination of Tarzlon increases. The next graph shows the
concentration profiles for kl = 0.10 h- I and k2 = 0.8 h- I " Note that the maximum amount of the drug in the
bloodstream is reduced by two.
O:lncentration Profiles
-+- Ca (1l§'dm3)
8~--------------------------~Cb(Il§'dm3)
0.6
'IiIm(h)
Concentration profile for Tarzlon in the stomach (A) and bloodstream
(B).. The maximum amount of Tarzlon in the bloodstream is 03 mg/dm 3•
P6-9 (a)
Reactor selection
A+B~D
=1Oexp(-8000K / T)C A CB
112 31Z
r ZA = lOOexp(--lOOOK IT)C A C B
'iA
A+B-~U
rD
SDU= - =
ru
lIZ
exp(-8000K IT)C A
lOexp(--8000K IT)CAC B
.
lOOexp(--lOOOK IT)CAI/2C/IZ
lOexp(-lOOOK IT)C B 112
AtT =300K
kl = 2.62
X
10- 11 &
k2 = 3.57
At T = 1000K
kl = 335
X
10-3 &
k2 =36.78
9.2xlO-S
SD/U =
CB
C/
IZ
IIZ-
Hence In order to maximize SDU, use higher concentrations of A and lower concentrations of B . This can be
achieved using:
1) A semibatch reactor in which B is fed slowly into a large amount of A
2) A tubular reactor with side streams of B continually fed into the reactor
3) A series of small CSTR's with A fed only to the first reactor and small amounts of B fed to each reactor..
8 '--_._.... _..._.._. ···-T........... __.....A·······_·__···_
B~b/'J
-j
·"l·-·
cLJLll
'-
~~;' ___ Pure A
intitially
Semi batch
f
T
Tubular reactor
with side streams
-
l
T
l:"'-
·_··--·-l
i
"d--'::..::0
lll I
Series 01 small: CSfRs
6-20
T
Also, since ED> Eu, so the specific reaction rate for D increases much more rapidly with temperature.
Consequently, the reaction system should be operated at highest possible temperature to maximize SDU
Note that the selectivity is extremely low, and the only way to increase it is to keep
6
c )Yz
( C: < 10- and
add B drop by drop.
P6-9 (b)
A+B~D
and
A +B ~U
and
r
D
SDU= --=
ru
rIA
=100exp(-1000KIT)CA C B
r2A = 10 6 exp( -8000K IT)C A CB
100exp(-1000K IT)CAC B
6
10 exp(-8000KIT)C A CB
=
exp(-1000K IT)
10 4 exp(-8000KIT)
AtT= 300K
6
kl = 3.57 &
k2 = 2.623
SDU = 1.14 x10
AtT = 1000K
kl = 36.78 &
k2 = 3354,.6
SDU = 0.103
Hence we should keep the temperature low to maximize SDU but not so low that the desired reaction doesn't
proceed to a significant extent
P6-9 (C)
A+B-·-~D and
B+ D
S
-7 U
and
r2A = 109 exp(-lO,OOOK IT)CBCD
= 'iA = lOexp(-8000K I1)CA\~
DU r2A 109 expC-10000K IT)CBCD
=
S
DU
exp(-8000IT)CA
10 exp(-10000IT) CD
8
Therefore the reaction should be run at a low temperature to maximize SDU, but not too low to limit the
production of desired product. The reaction should also take place in high concentration of A and the
concentration of D should be limited by removing through a membrane or reactive distillation
P6-9 (d)
A~D
and
'iA = 4280exp(-12000K IT)C A
D~Ul
and
r2D = 1O,100exp(--15000K IT)C D
A~U2 and
s
DIUIU2
r 3A
= 26 exp( --·10800
KIT)C A
= _~_ = 4280exp(--:12000K IT)C A -1O,100exp(-150~OK IT)C D
rVI + rV2
1O,100exp(-1.5000K IT)CD + 26exp(-10800K IT)C A
AtT= 300K
14
kl = 1.18 X 10- &
k2 = 194 X 10- 18 & k3 = 6.03
X
10- 15
If we keep C A > 1000CD
S
=~~XlO-I4CA -L94xlO- CD ""~~~=1.96
D/UIU2 1.94 X 10-18 CD + 6.03x 10-15 CA .603
18
6-21
At T= 1000K
k j = 0.026 &
k2 = 3.1 X 10-3 & k3 = 5.3
X
10-4
If we keep C A > 1000CD
S
=
DIUlUZ
3
0.026C A -3.1xlO- C D
3.1xlO- 3 CD +5.3xlO-4 C A
.026 =49
""
.00053
Here, in order to lower U j use low temperature and high concentration of A
But low temperature and high concentration of A favours U z
So, well have to optimize the temperature and concentration of A.
Membrane reactor in which D is diffusing out can be used.
P6-9 (e)
A+B ~ D
and
D ~ A + Band
A +B ~U
rD
=ru
SDIU
and
'iA
= 10 9 exp(-10000K IT)CAC B
rzv = 20 exp( -2000 K IT)C D
r3 A
= 10 3 exp( -- 3000 KIT) cAe B
9
10 exp(-10000K IT)CACB -20exp(-2000K IT)C D
10 3 exp(-3000K IT)CACB
AtT =300K
k j = 3.34 X 10-6 &
k2 = 0.025
& k3 = 0.045
The desired reaction lies very far to the left and CD is probably present at very low concentrations so that:
SDIU
",,0
AtT= 1000K
k j = 4.5399 ..9 & k2 = 2.7 & k3 = 49.7
If we assume that CACB > Q..OOICD then,
- 45399.9CAC B -2.7CD
S DIU
-
49.7CACB
45399 -913
_
- -
49.7
Here we need a high temperature for a lower reverse reaction of D and lower formation of U
Also we need to remove D as soon as it is formed so as to avoid the decomposition.
P6-9 (I)
A + B --7 D
S
and
-'iA = 10exp(-8000K IT)CACB
A ---7 U
and
-'2A = 26 exp(-IO, 800K IT)CA
U ---7 A
and
-r3U =1000exp(-1.5,000K IT)Cu
- rD
DIU -
-_____
lOexp(-8000K IT)CACB
ru - 26exp(-'}0,800K IT)CA-1000exp( -15,000K IT)Cu
6-22
We want high concentrations ofB and U in the reactor. Also low temperatures will help keep the
selectivity high.
If we use pseudo equilibrium and set -rA = 0 .
r-
-rA =lOexp ( -SOOO) CACB + 26exp (-lOS00)
T
CA-1000exp (-15000)
T
Cu =0
C, _ lOex p ( -8~OO
)C + 26exp ( -1~OO)
8
lOooex p ( -1~OO)
Cu -
CA =_1_ exp (7000)C
Cu 100
T
B
+~exp(4200)
1000
P6-9 (g)
A+B--7D
and
T
-·liA =SOOexp(-SOOOK IT)C~5CB
A+B---tU1
and
--r2B = 10exp(-300K IT)CACB
D+B--7U 2
and
-13D
= 106 exp(-SOOOK IT)CDCB
(1)
SOOexp( -SOOOIT) C~5CB . SO exp ( -SOOOIT)
SDlU = lOexp(-300IT)CACB - = exp( -300IT)C~5
j
AtT = 300
S
DIU
2.09S *10- 10
j
=----0.36SC~5
At T = 1000
= ____ 29.43
S
DlU
0.740SC~5
j
To keep this selectivity high, low concentrations of A, and high temperatures should be used.
S
DlU2
=
SOO exp ( -SOOO IT) C~5 CB SOOC~5
=106 exp(-SOOOIT) CDCB
106 CD
To keep this selectivity high, high concentrations of A and low concentrations of D should be used. Try to
remove D with a membrane reactor or reactive distillation . The selectivity is not dependant on temperature .
To keep optimize the reaction, run it at a low temperature to maximized
allows only D to diffuse out.
(2)
SOOexp(-SOOOIT)C~5CB
_
SDIUIU2 -
(
)
6
(._--)
lOexp -3001T CACB + 10 exp -SOOOIT CDCB
SOO exp ( -SOOO IT) C~5
SDlU1U2
= lOexp-300lT
(
)C
6 ---'--'--(...:..:........_-)-
A +10 exp -BOOOIT CD
At T = 300
6-23
SDiUl
in a membrane reactor that
S
DIV1U2
2.09 *10-9 COS
A::::O
3.67 CA+ 2.62 *10-6 CD
=
At T = 1000 and very low concentrations of D
=
S
DIU1U2
0.268C~s
7.408CA+335CD
.03617
C~5
If temperatme is the only parameter that can be varied, then the highest temperatme possible will result in
the highest selectivity. Also removing D will help keep selectivity high.
P6-9 (h) No solution will be given
P6-9 (i)
exp(-7000K IT)CA1I2
-r 1OC/ 2
rD
v
dF
- -A= r
dV
A
dFv
--=r.
dV
v
dF
dV
B
dF
dV
D
+R
B
_.-=r,
B
- -D= r
R = FAO
B
v:
T
A+B~D
liA = --1Oexp(-8000K IT)CACB
A+B~U
r2A = -1OOexp(-1OOOK IT)CA1I2C/12
rA = rB = liA + r2A
FA
C A =Cro -FI
mol
CIO =0.4--3
dm
FAo = Crovo
s
These equations are entered into Polymath and the plots below are for the membrane reactor . The code can
be modified to compare with the PFR results.
See Polymath program P6··9i.poJ.
POL YMA TH Results
Calculated values. of the DEQ variables
Variable
V
Fa
Fb
Fd
Fu
Cto
initial value
o
4
minimal value
o
0.5141833
o
o
o
o
o
o
0.4
600
4
Cb
o
0.4
600
4
Ca
0.4
0.0241566
T
Ft
maximal value
10
4
4.5141833
3.034E-06
.3.4858137
0 .. 4
600
8.5141833
0.2120783
0.4
o
6-24
final value
10
0.5141833
4.51418.33
3 . 034E-06
3.4858137
0.4
600
8.5141833
0.2120783
0 . 0241566
0
4,S7SE-07
0
0
0
0 ,4461944
S
4
0.8
S '. 423E+OS
-4.S7SE-07
0
-0.4461944
-0.4461948
-0.4461948
0
S
4
0.8
2.894E-07
0
0
0
0
0
0
S
4
rIa
rd
r2a
ra
rb
ru
Vt
Fao
Rb
Sdu
o '. 8
S . 423E+OS
-8 . 297E-08
8.297E-08
-0.2867066
-0.2867066
-0.2867066
0,.2867066
S
4
0.8
2.894E-07
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(V) = ra
[2] d(Fb )/d(V) = rb+Rb
[3] d(Fd)/d(V) = rd
[4) d(Fu)/d(V) = ru
Explicit equations as entered by the user
[1] Cto =..4
[2) T=600
[3) Ft = Fa+Fb+Fd+Fu
[4) Cb = Cto*Fb/Ft
[5) Ca = Cto*FalFt
[ 6] r1 a = -1 O*exp( -8000IT)*Ca*Cb
[7) rd= -r1a
[8) r2a = -1 OO*exp( -1 OOOIT)*CaA 5*CbI\1 ,,5
[ 9) ra = r1 a+r2a
[10] rb = ra
[ 11] ru = -r2a
[12] Vt=5
[1.3] Fao = 4
[14] Rb = FaolVt
[IS] Sdu = exp(-7000IT)*CaA ,5/(10*Cbl\,,5+,,00000000001)
5
6.0e+5
4
48e+5
3
3,6e+5
2
2 -Ie+5
1
1.2e+5
2
.:I
V
6
8
10
O,Oe+O
P6-9 (j) No solution will be given
P6-9 (k) No solution will be given
P6-9 (I) No solution will be given
6-25
0
.-.......--..
2
---.--.-~-
.:I
V 6
. ---.-.
s
10
P6-10 (a)
Spect<::5 A:
de
--"-:r
dt
-\
-ro. = ktC,-\
Species B:
Species C:
,
de c _ ..
--·c
dt
,
. . fC::::::
klC B
Pluggmg mto POLYMATH: gets the following,
dice) Id{:::) =rc
cb
k,1-=.4
cc
!E!b,8:~1:.Y~!~ MaximUlll ..~ Minimum.'y~~ !'i~!. . ~
a
100
1.00
0
6' .. 789Be .. 18
5.7898e-18
1.6
1.6
()
n.6037
1.4556
0
() .9963
0 .. 9963
0
0
k2"'.{)1
kl
0.4
0.4
04
0.4
ra", . kl"ca
rc"k2"cb
k2
0.01
ra
·0 64
0
0.54
0.01
·2.'71592et8
0.014556
0.01
-0.54
0 .. 01
-2.71.S92e-18
0
0.00603'1
O.6~
-O.OU2417
ESI::!=a.ti~:§!:
var~!!:
d(calldP::) "'rat
ci{co) /0.( t) =z·o
t
. '"
ca
rb"'k:;:'·ca· .. k2 "cb
·0
0,
t"...
'"
I'C
1.00
rb
key
~Ca
. Cb
•.. ·Cc
P6-10 (b)
For CSTR, T = O.Sh
First calculate kJ and k2 :
6-26
0.006037
k =O.Olex (20'OOO(_1__ ~)J
2
P
R
373 T
See Polymath program P6-1O-b.pol.
POLYMA)'H Results
NLES Solution
Variable
Ca
Cb
Cc
tau
T
k1
ra
k2
rb
rc
Value
1 . 5268515
0.0319385
0 . 04121
0.5
760
0 . 0958161
-0.146297
2 . 580579
0 . 0638771
0.0824199
lni Guess
1.5
f (x)
-6.446E-13
7.28E-14
0
o
o
NLES Report (safenewt)
Nonlinear equations
[1] f(Ca) = tau*(··ra)-(1 . 6-Ca) = 0
[2] f(Cb)
[ 3] f(Cc)
=tau*(rb)·Cb =0
=tau*(rc)-Cc =0
Explicit equations
[1] tau = . 5
[2] T = 760
[3] k1 = 72*exp(-10000/(1 . 987*T))
ra = -k1 *Ca
k2 = 1457152*exp(-20000/(t.987*T))
[6] rb k1 *Ca-k2*Cb
[7] rc = k2*Cb
[4]
[5]
=
Cb vs" temperature
350E-02
3 . 00E-02 + - - - - . - - - - - -
I
2.50E-02
~ 2.00E02
II
~
tl 1.50E·02
---f------\--~
I
100E-02
.0...._1
5.00E-03 +-_________-+--_________
o.OOE+OO--·-·-----.-·----··
o
200
600
400
------.-----------,-----J
800
temperature (K)
Therefore, CB is maximum at T=760K.
6-27
1000
1200
P6-10 (c)
(e) ?art is similar to part b except for two rate laws:
fA -k
_II *('·s
*('
- kI
'A
rB =k 1 *C-k
-,-\
-·1 *C B -k,;. *C a
Using those rate laws inPOLYThtlATH produce the following:
___
...
.......-..,.,
Eauations:
l!ti t ~a 1 ..~1::~
d{ca) /d(t) ;::::r:-a
1.6
d(cb)Jd(tl=rb
d(cc) Id(tl ::::e'C
a
o
klr·=8.33e·S
2.JC::
-_.
k1£"'.OOOl
k2=2.78e-5
\ ..ti::::
-.
!.,2IJC
1\
rc:::k2"cb
ra=klr""cb··klf"ca
!\
o.oeo
c"cco
var:iable
0.808
J.
-_-.-. _
_
----
t
value
In,itial value Maximu.m value Minim1.4'11
....... ...........value
_........ ............... -:inal
.,. .......... _-, ......
.......
-------.-'"'--350000
0
350000
0
ca
1.6
1.6
0.436316
0.4.16316
cb
0
0.833237
0
0.5159
cc
(J
0.647'784
0
0.64'1784
k.lr
8 3.3e·"05
0.0001
2.78e06
S.33e··05
13. Be-OS
8 . 33e-OS
0.0001
0.0001
0.0001
2.78e·06
2.78e-06
2.78e-·06
rc
ra
0
2 _3164e-- 06
a
1.4342e·-06
0.00016
··0.00016
--1.20632e··06
-6. 57168e-- 07
:r:b
klf
k2
~--
-0.00016
6 _S7168e--07
0.00016
P6-10 (d)
Th isis similar to
part d except for one rate law:
rc -k'
--"1 *C.... 8: -_ok . -2 *C"'c
Using that in POLTh1ATH produces the following:
6-28
. 0' .7703 4e-... 07
EqlJ.§!ci~~.:
2
d{ca)/d(t)=ra
d(cb)/d(t)=:::b
ce:
~ 5.:lC
T
i
.!..
d(cc)/d(t);r;c
k1:::=8 _33E!-'S
k1.f= _, 0001
k2:=2,78e--6
k2r=1.. 3 %-6
ra"k1:r:-·cb-· kIf'cd
rc'"'k2f*cb-·k2:::*cc
r'b=kl f"ca '" klr 'cb-- k2 f "cb+k2.::'cc
to '"
(),
t:l!
.L
Variable
--t
'"
350000
~~..s~,~! __ ::-al':l:~ 1:!~im1.l!:LvallJ.e Minim~ya1ue Final._y~!ue
350000
0
0
150000
Cd
1. 6
1 5
a,490306
0 490306
cb
0
0 833769
0
0.583662
cc
0
0.525032
0
0,526032
8,33e'OS
8.33e05
k1r
8,,33e,05
8 33e-05
kif
0.0001
()
kl:
2,788 06
k2:
1. 3%06
oootS
00001
0.0001
2 730 06
2. ii3e' 06
2.,78e-·06
1. 3ge06
1 39806
L '3%,,1)6
0001
··4 115738·-0'1
'·0 ,00016
--,4 .11573e .. 0 7
l:C
0
2 25569806
0
8.91396,0·-07
r;b
0, 00Gl6
0 GOO16
0
::::d
-1..124.93806
"4,, 79824eO)
P6-10 (e)
(e)
Vlhen kl>l00 and k2<O.1 the concentration of B immediately shoots up to 1,6 and
then slO\vly comes back down. while CA drops off irnmediate1y and falls to zero. This is
because the first reaction is so fast and the second reaction is slower with no reverse
reactions.
When k2 ;:;;; 1 then the concentratlon of B spikes again and remains high, while very
little of C is formed.. This is because after R is fonned it will not got to C because the
reverse reaction is faster.
Vlhell k-2 ::::: 0.25, B shoots up, but does not stay as high because the second reverse
reaction is a slightly slower than seen before, but still faster than the forward reaction..
P6-11 (a)
InteImediates (primary K-phthalates) are fOlmed flOm the dissociation of K-benzoate with a CdCh catalyst
reacted with K-terephthalate in an autocatalytic reaction step:
C
AD
:::::~ =
RT
A-~R~S
Series
R + S -~ 2S
Autocatalytic
1l0kPa
=0 02moll dm'
(8.314 kPa.dm')( 683K)
.
mol.K
6-29
Maximum in R occurs at t = 880 sec. See Polymath program P6-11-a. pol,
POLYMATH Results
Variable
t
A
initial value
0
0,02
minimal value
o
0.003958
S
0
o
o
k1
k2
k3
0.00108
0.00119
0.00159
0,,00108
0.00119
0.00159
R
0
maximal value
1500
0.02
0,0069892
0,0100382
0.00108
0.00119
0.00159
final value
1500
0.003958
0.005868
0.0100382
0.00108
0.00119
0.00159
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(A)/d(t) = -k1 *A
[2] d(R)/d(t) = (k1*A)-(k2*R)-(k3*R*S)
[3) d(S)/d(t)
(k2*R)-(k3*R*S)
Explicit equations as entered by the user
[1] k1 = 108e-3
[2] k2 = 1,1ge-3
(3) k3 = 1 ,,5ge-3
=
P6-11 (b)
1) T= 703 K
3
CAO = 0,,019 mol/dm
Similarly,
k~ =3.3xlO- s- and k~ =3.1xlO-3 dm 3 Imol.s
3
1
Maxima in R occurs at around t =320 sec" See Polymath program P6·11·b i pol
POLYMATH Results
Calculated values of the DEQ variables
Variable
initial value
minimal value
t O O
A
0,019
3,,622E-04
R
S
0
0
o
kl
k2
k3
0 00264
0.0033
0.0031
0,,00264
0.0033
0.0031
o
maximal value
1500
0,,019
0,,0062169
0,0174625
0.00264
0,,0033
0,0031
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(A)/d(t) = -k1 *A
[2] d(R)/d(t) = (k1 *A)-(k2*R)-(k3*R*S)
[3] d(S)/d(t) = (k2*R)-(k3*R*S)
Explicit equations as entered by the user
[1] k1 = 2,,64e-3
[2) k2 = 3,,3e-3
6-30
final value
1500
3,622E-04
8,856E-04
0,0174625
0.00264
0.0033
0.0031
[ 3 J k3 = 3,1 e-3
2) T = 663 K
3
CAO =0,19 moIJdm
(_I_-_I_)J
k; = (1.08xlO-3 s-l)ex p ( (42600cal / mol)
(1.987callmol.K) 683K
k~
3
=0.4xlO- s-
663K
1
k~ = 0.78xlO- dm
3
= 0.42 X10-3 S-1
3
/
mol.s
See Polymath program P6-11-b2.pol.
POLYMA Tn Results
Calculated values of the DEQ variables
Variable
t
A
initial value
0
0.019
R
0
S
k1
k2
k3
0
4.2E-04
4.0E-04
7.8E-04
minimal value
o
2.849E-04
o
o
4.2E-04
4.0E-04
7.8E-04
maximal value
10000
0.019
0.0071414
0,016889
4.2E-04
4.0E-04
7.8E-04
final value
10000
2.849E-04
0.0012573
0.016889
4.2E-04
4,OE-04
7.8E-04
ODE Report (RKF45)
Differential equations as entered by the user
l l ] d(A)/d(t) = -k1 *A
l2] d(R)/d(t) = (k1 *A)-(k2*R)-(k3*R*S)
[3] d(S)/d(t) = (k2*R)-(k3*R*S)
Explicit equations as entered by the user
[1] k1 = OA2e-3
[2] k2 =OAe-3
[3] k3 =0,,78e-3
Independent variable
variable name: t
initial value: 0
final value: 10000
0.020.------,-·------·-·-------,
0,004
2000
4000 t
6000
8000
10000
Maxima in R occurs around t = 2500 sec,
P6-11 (C)
Use the Polymath program from part Ca) and change the limits of integration to 0 to 1200, We get:
3
CAexit = 0.0055 moIJdm
3
CRexit = 0.0066 mol/dm
6-31
CSexit = 0,0078 mol/dm'
P6-12 (a)
P6-12 (b)
. mol
== -_(.7)(.1)
.....__ ... _:;;:: 0.023-""--,,,
3.3
dm 3 _$
-riA
fIB :;;:: ....-
r"ll
= 0 * ro_D =: 0-dm'
m<:t._
...
_s
1:,B ::: 0 * fj€
P6-12 (c)
_ -riA _
""
0
m~l_..
dm' .s
P6-12 (d)
(,7)(.1)
f lC - "-""'- - -,,--- :::
3
j
_-2*r
,
mol
0,023 "·-1"
<inr*s
_,2(3)(50 1 (.I)
2D
f:c - - ...... -'
-_..._--,. __.,_.-
3
_
mol
== -"Q,O(h2·"t--
3
dm' *5
lic =rJ £ ;:·-(.2)("049)( .51) = ··O.005-!~I_.,
dm'
*5
P6-12 (e)
Ii" ::::: 0 '" liA
.
mol
=O..dm
·~-3......*5
mol
'iE ::: O~· liD::: 0,:: -".
dm'·s
r,E ::: (.2)(.049)(.51):::;: O.OOS ....!!!.ol
3
dm .s
P6-12 (I)
fA :::
P6-12 (g)
F,F
V=, . M-, . ~ .
--0.07 - 0.0026 : : : -0.0726 mol
.-~"
::::::~!{~·~~_: . ~:.~2
Ie:::
dm-
fD
·"·fA
0.023 . . ,0.0052· 0.005 = O.0128 .... ~~~.
_ 100(3 ,0.1) _
3
- ..,_. . _""' ...."........ ::::: 4000dm
0,0726
.5
= 0.0078 ······0.033 == 0 OOll"""il:..?L
..
1
dm _s
P6-12 (h)
Mole balance:
Rate law,
CAD
·-
= (_. r A)r
Cc = (rc}r
CD = (rD)r
CA
rA = -[k"CA
+~kWCAC~ ]
6-32
Solving in polymath:
SBID
=I'B/rD = 247
See Polymath program P6-12··h. pol.
POL YMATH Results
NLES Solution
var'iable
Ca
Cb
Cc
Cd
Ce
kd
ka
rb
ra
ke
rc
rd
re
tau
Cao
Value
0.0068715
0.9620058
0.5097027
0.0038925
0,2380808
3
7
0.0160334
-0.0498855
2
0,008495
6,,488E-05
0,003968
60
3
fIx)
-2.904E-10
-1 . 332E-15
--1,67E-08
-2.391E-08
1.728E-08
lni Guess
3
o
o
o
o
NLES Report (safenewt)
Nonlinear equations
[1] f(Ca) = Cao-Ca+ra*tau = 0
[2] f(Cb) = Cb - rb*tau = 0
[3] f(Cc) = Cc-rc*tau =0
[4]
[5]
=
=
f(Cd) Cd-rd*tau 0
f(Ce) = Ce - re*tau = 0
Explicit equations
[1] kd = 3
[~2]
ka=7
rb = ka*Ca/3
ra = -(ka*Ca+kd/3*Ca*CcJ\2)
ke =2
rc = ka*Ca/3 - 2/3*kd*Ca*CcJ\2 ,. ke*Cd*Cc
rd = kd*Ca*CcJ\2 - 4/3*ke*Cd*Cc
re = ke*Cd*Cc
19] tau = 60
[10] Cao = 3
[3]
l4]
[5]
[6]
[7]
[8]
P6-12 (i)
For PFR and gas phase:
Mole balance:
dF
dV
- - A-
= rA
dF
dV
- -B= r
B
dF
dV
-.i:..·=r
6-33
C
dF
dV
_D_=r
D
dFE =r
__
dV
E
Rate law:
Stoichiometry:
FT =FA +FB +Fc +FD +FE
dy
dV
= -a
FT
2y FTO
Plot of CB and Cc are overlapping.
See Polymath program P6··12·i pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
initial value
minimal value
V
o
Fa
Fb
Fe
Fd
Fe
20
9 . 147E-04
o
o
o
o
o
o
o
o
o
y
1
Ft
Cta
Ce
20
0.2
0.9964621
13.330407
0.2
o
o
ka
kd
ke
7
3
7
3
2
Ca
rb
ra
Cd
Fta
re
rd
re
alfa
X
2
0.2
04666667
-1.4
o
20
0.4666667
o
o
1.0E-04
o
maximal value
100
20
6 . 6638171
6 . 6442656
0 . 02012S8
0.0043322
1
20
0.2
0 . 0993605
7
3
2
1.367E-05
3.191E-05
0.2
o 4666667
-9.S86E-05
3 . 0E-04
20
0.4666667
8 . 653E-04
S . 908E-OS
1.0E-04
0 . 9999543
-1..4
o
20
-1.923E··05
-7 . 012E- 05
o
1.0E-04
o
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(V) = ra
[2] d(Fb)/d(V) = rb
[3] d(Fc)/d(V) = rc
f 4] d(Fd)/d(V) = cd
[5] d(Fe)/d(V) = re
6-34
final value
100
9.147E-04
6.6638171
6 . 6442167
0 . 0171261
0 . 0043322
0 . 9964621
13 .330407
0.2
0.0993325
7
3
2
1 . 367E-OS
3 . 191E-OS
-9.S86E-05
2.S6E-04
20
-1 . 923E-05
-6.742E-OS
5 . 087E-·05
1.OE-04
0.9999S43
d(y)/d(V):::: -alfa*Ft/(2*y*Fto)
[6]
Explicit equations as entered by the user
[1] Ft:::: Fa+Fb+Fc+Fd+Fe
[2] Cto:::: 0.2
[3] Cc:::: Cto*Fc/Ft*y
[4] ka:::: 7
[5] kd:::: 3
[6] ke:::: 2
[7] Ca:::: Cto*Fa/Ft*y
[8] rb:::: ka*Ca/3
[9] ra:::: -(ka*Ca+kd/3*Ca*CcA 2)
[1. 0 1 Cd:::: Cto*Fd/Ft*y
[Ill Fto::::0 . 2*100
[12J rc:::: ka*Ca/3 - 2/3*kd*Ca*CcA 2·' ke*Cd*Cc
[13] rd:::: kd*Ca*CcA 2 - 4/3*ke*Cd*Cc
[lA] re:::: ke*Cd*Cc
[15J alfa:::: 0.0001
[16] X::::1-Fa/20
·40.
. V 60
..
12
!l
>4
\I
0
(I.. !l$(f
#.,1124
(I,out
iJAltZ
1).• IlM
n.IiOO
Il
41l
V
H)O
6(1
P6-12
G) Changes in equation from PaIt (i):
dFc
dV
rc -Rc
=
Rc
=
kdiffilSeCC
k diffilSe
= 2 nun
. -1
See Polymath program P6···12-j.po1.
6-35
20r----------------------------,
0,020.--------------------------,
16
0.016
F~
lBJ
12
8
4
40
V
60
80
100
40
V
60
P6-13 (a)
m-xylene --> benzene + methane
A
-->B
+M
m-xylene --> p-xylene
A
--> P
See Polymath program P613 . a.pol.
POLYMA TH Results
Calculated values of the DEQ variables
Variable
'-V
Fa
Fb
Fp
Fm
Fi
Ft
k1
k2
Cta
Ca
rl
Cb
r2
Cp
Spb
tau
y
X
initial value
o
75
o
o
o
minimal value
o
6.1072984
o
o
o
25
100
0.22
0.71
0.05
0.0375
-·0.00825
25
100
0.22
0 . 71
0.05
0.0026257
-·0.00825
-0.026625
-0 . 026625
o
o
o
o
o
o
o
o
o
o
o
o
maximal value
6000
75
16.297198
52.595503
16.297198
25
116.2972
0.22
0.71
0.05
0.0375
-5 . 777E-04
0.0070067
-0.0018643
0.0226125
3.2272267
3
0,7634409
0.9185694
ODE Report (RKF45)
Differential equations as entered by the user
[ 1) d(Fa)/d(V) = r1 +r2
[2) d(Fb)/d(V)=-r1
[3) d(Fp)/d(V) = -r2
Explicit equations as entered by the user
[1) Fm = Fb
[2) Fi =25
[3) Ft = Fa+Fb+Fp+Fm+Fi
[4J k1 = ,,22
6-36
final value
6000
6.1072984
16 . 297198
52.595503
16 . 297198
25
116.2972
0.22
0.71
0.05
0.0026257
-5.777E-04
0.0070067
-0.0018643
0.0226125
3.2272267
3
0 . 7634409
0.9185694
80
100
[5] k2 = . 71
[6] Cto = . 05
[7] Ca = Cto*Fa/Ft
[8] r1 = -k1*Ca
[9] Cb = Cto*Fb/Ft
[ 10 ] r2 =-k2*Ca
[11] Cp = Cto*Fp/Ft
[12] Spb = Cp/(Cb+ . 0000001)
[13] tau = V/2000
[14] Y = Fp/(75 . 00000001-Fa)
[15] X = (75-Fa)/75
a l' =2.8 is necessary to achieve 90% conversion
10r--------------------------,
4.0 .-----------------------------,
0.8
3.2
0. 6
2t
. - - - . - - - -.....- - - . - - - -•.•.---.•- - - - -
I~ ~Pj
L6
r........ .
0. 8
2tOO V
4800
3600
6000
0. 0
0
1200
2400 V
3600
4800
P6-13 (b)
CSIB
\
FM =1"MV
Mole Balances:
:c
~.? == rPV
Rate La\vs.·
.,,"
"E .r -."1 - -.I ". . . ')
k :;:; k. exp.· ,,_.d
l
,0
R IT
946·
\.
./
Using these equations and Polymath we find the optimal temperature is 1194 K. The maximum
concentration of p-xylene is 0.013 mol/dm3
See Polymath program P6·13b.pol.
pOLYMA THResul!§
NLES Solution
Variable
Ca
Cb
Cm
Cp
Cao
tau
k10
E1
R
k20
E2
Value
0 . 012197
0 . 0122301
0 . 0122301
0 . 0130729
0.0375
0.5
0 . 22
2 . 0E+04
1.. 987
0.71
10000
f(x)
7
5
5
-3
. 15E-ll
. 284E-12
. 284E-12
. 069E-ll
Ini Guess
0.0375
o
o
o
6-37
6000
Yp
0.5166557
1194
2.0054175
0.0244601
0.0244601
2.143628
0.0261459
-0.050606
T
k1
rb
rm
k2
rp
ra
NLES Report (safenewt)
Nonlinear equations
[1]
[2]
[3]
[4]
f(Ca) = Ca-Cao-ra*tau = 0
f(Cb) = Cb-rb*tau = 0
f(Cm) = Cm-rm*tau = 0
f(Cp) = Cp-rp*tau = 0
Explicit equations
[1] Cao = .0375
[2] tau = ,,5
[3] k1o=,,22
[4] E1 = 20000
[5] R = 1.987
[6] k20 = .71
[ 7] E2 = 10000
[8] Yp = Cp/(.03750000001-Ca)
[9] T= 1194
[10] k1 = k1o*exp((E1/R)*(1/946-1/T»
rb = k1*Ca
rm = k1 *Ca
k2 = k2o*exp((E2/R)*(1/946-1/T»
rp = k2*Ca
[ 15] ra = -k1 *Ca·k2*Ca
[11]
[12]
[13]
[14]
-_._-----------
---------------------
P6-14(a)
50dm 3 PFn
Mole balance:
dCA
f';\
<IV
Vo
dCc
rc
dll
Vo
11(:£
./ B
=
dV
Vo
[
deB
dl T
dev
dV
.:::::::
f'13
,,~
Uo
tlOr
Fp
dV'
=~~;
6-38
Rate laws:
r . ' 1 ' 1 'U 1-- ~li
Fa
·2rVI
£:;2
l'F:"
l'c
rlJ 1
I' [::2 .
"21'1' 3
tn
TO]
2TB2
I TF:)
'IE
TE2
Til(} equation for I.he conversion
ur A is :
C-"I
)( = .G. AU ----.
C AO
See Polymath program P6-14··a. pol.
POLYMATH Results
No Title 08-01-2005, RevS 1.233
Calculated values of the DEQ variables
Variable
t
Fa
Fb
Fe
Fd
Fe
Ff
va
Caa
Cba
Ft
Cta
kd1
ke2
kf3
Ce
Cd
Cb
Ca
rd1
re2
rf3
re
rf
rd
ra
rb
re
initial value
0
15
20
0
0
0
0
10
1..5
2
35
3.5
0 . 25
0.1
5
0
0
2
1.5
1.5
0
0
0
0
1..5
-.1. 5
-3
1..5
minimal value
----0
0 . 2090606
1. 3440833
0
0
0
0
10
1.5
2
15 . 582463
3.5
0 . 25
0.1
5
0
0
0 . 3018965
0 . 0469574
0 . 0010699
0
0
0
0
-0.042376
-1.5
-3
-0 . 0962952
maximal value
50
15
20
1 . 9655663
7 . 2554436
2.5920934
4.6265981
10
1.5
2
35
3.5
0 . 25
0. 1
5
0 . 228291
1.. 4503322
2
1..5
1.5
0.1004639
0.3632767
0.1004639
0.3632767
1..5
-0 . 0215011
-0 . 0116593
1..5
ODE Report (RKF45)
Differential equations as entered by the user
6-39
final value
50
0.2090606
1.. 3440833
0.3535564
6 . 4570707
2 . 5920934
4.6265981
10
1.5
2
15 . 582463
3.5
0.25
0. 1
5
0.0794128
1 . 4503322
0 . 3018965
0.0469574
0.0010699
0.0068104
0 . 0095194
0 . 0068104
0 . 0095194
-0 . 0030314
-0 . 0215011
-0 . 0116593
-0.0111585
[1]
[2]
[3]
[4]
[5]
[6]
d(Fa)/d(t) = ra
d(Fb)/d(t) = rb
d(Fc)/d(t) = rc
d(Fd)/d(t) = rd
d(Fe)/d(t) = re
d(Ff)/d(t) = rf
Explicit equations as entered by the user
[1] vo= 10
[2] Cao = 1.5
[3] Cbo = 2
[4] Ft = Fa+Fb+Fc+Fd+Fe+Ff
[ 5] Cto = Cao+Cbo
[6] kd1 = 0.25
[7] ke2 =.1
[8] kf3=5
[ 9] Cc = Cto*Fc/Ft
[ 10] Cd = Cto*Fd/Ft
[ 11] Cb = Cto*Fb/Ft
[12] Ca = Cto*Fa/Ft
[13] rd1 = kd1 *Ca*CbA 2
[14] re2 = ke2*Ca*Cd
[15] rf3 = kf3*Cb*CcA2
[16] re = re2
[17] rf = rf3
[18] rd = rd1-2*re2+rf3
[19] ra = -rd1-3*re2
[20] rb=-2*rd1-rf3
[21] rc = rd1 +re2-2*rf3
20
-
16
12
Fa
Fh
Fe
Fd
Fe
8
-4
()
o
10
20
V 30
40
50
P6-14 (b)
(b) Det.ermine the effiuellt concentration and eOllversion from a. 50dm:3 CSfI'll.
IVlo]e Balanee:
}~iO·-- f:4
-TA V
F HO ·PH
Fc'
Fl.)
~~
THV
TC V
TDV
6-40
F1:;
J'E V
FI"
1'FT-l
== £.
combining'J rate law and
110'
I
-
fleA)
CA
.t(C H )
CB
f(Cd
f"'\ C... /),)
f(C FJ )
f(CF)
--,Cc:
C.fto
--
lllUlc
halancc,.
rAi
C'BO-- rBl
+ reT
--Co I rv T
,
+ t ET
--(/p + 'fFT
--C'B
Polyumth code,
(ca)=ca-caO""xa*tau
f(cb)-cb-cbO-rb*tau
f (cc) =r c*tau-cc
f(cd)=rd*tau-cd
f
f(ce)=tau*re-ce
f(cf)=tau*rfcf
tau""V/vo
V=50
\'0=10
caO=1,5
cbO=2.0
rd1=kdl*ca*cb"2
re2"'ke2*ca*cd
rf3=kf3*cb*cc'2
kdl"'0,25
ke2=O.1
kf3=5
ra=rdl·3*re2
rb=2*rdl·rf3
rC"'l'd1.+re2·2*rf3
rd=rdl 2*re2+rf3
re=re2
rf=rf3
ca(O).,;;l
cb(O)=!
cc(O);l
cd(O)=l
ce(O)=l
cf(O)=1
(Ans) C/I = 0,6 t,
eu =
0 79, C c
() 11, CD
= 045, CEO:; 0 14, C p
= 025
P6-14 (C)
(c) V(J = ilOdlll:; S(,wj·,Batch tendor, (l) A is fed to B, (2) B h" fed to A
(Catle 1) A itl fod to 13,
6-41
dCA
elt
CAO
-
rA
+ tiO
···a£"· -
t'jj
--voV'
Ce,
.... vo·_V
deu
dCC)
·~lt
=
1'C
deD
dt
.
l'J)
dCs
dt
=
rp;
--
l"f'
-~
deF
elt
.t'{
CA
en
..
."yo.
CD
vo··v·
CE
tJo··V
OF
.... vo--·
V
(C80
(:8)
_._...._----CBO
alld mudifying eorrespondiug ]>OlYlllHt.h code.
d(ca)!d(t)=ra·vo!V*ca
d(cb)!d(t)"'.rb+vo!V*(cbO·cb)
ca(O)=1.5
cb(O)"O
x"'(caO·ca)!caO
Diffol cntes
(1) Becallse Cs o is highel than CAO (i.e ;10% higher molar flow rate), case (2) fenches
(2) \Vitll the snme reason, C(ISC (2) produces D and F mote
(3) With the same reason,
eB
(in
tilSC
X~:C
2) increases more dwstkally(exeessivcly) than GA
1 ill e!ulier time.
(ill
ease 1)
(<.:) case I concentration vs., lime
G1dl)h litle
"'"
''''
1.<10
\
,\
\
110
.
.,
.
",
""
.
••!••~.~~~~~~~.s~.~!m~~!;~~~~!flI'(~Q
'''' .~..~~;::"oo~'~"~~"'~.c~.~~t~;¢~. oo~~~~""~"'~~~".~o",~~~t~t~
,
6-42
conversion
Case
VS,
time
HIO
1
041}
case
ttl
016
,__.",.,_"-____..._ _ _-..<......_ _ _....._ _ _-'-_ _ _ _' -_ _ _..1
woo
0.00
2000
3j) flO
41} 00
5000
t
Case 2 concentration vs. time
lW
111
!"S2'
lH
IH
\
\
\I
\
:P
~-j
\
\
;I
d
"
6-43
&0 DO
70.00
eliDa
90
on
100 llil
Case 2 conversion vs. time
--_.._._--
1.00
090
OSO
Q 70
01>0
050
case (2)
040
0.30
02f1
D.li!
0.00
' - - -.........- - - ' - - - . - - ' - - - - - ' ! ' - -..-.-... --......---~.-........- - - ' - - - -.......- - - - '.....- -....
0.00
1000
2DOO
3000
40.00
50.00
60.00
701)0
80.00
gaoa
10000
P6-14 (d)
As aB increases the outlet concentration of species D and F increase, while the outlet concentrations of
species A, C, and E decrease . When aB is large, reactions 1 and 3 are favored and when it is small the rate
of reaction 2 will increase .
P6-14 (e)
When the appropriate changes to the Polymath code from part (a) are made we get the following.
See Polymath program P6-14·e.poL
POLYMATH Results
Calculated values of the DEQ variables
Variable
V
Fa
Fb
Fe
Fd
Fe
Ff
vo
Ft
Cto
kd1
ke2
kf3
Ce
Cd
Cb
Ca
rd1
re2
rf3
re
rf
initial value
0
20
20
0
0
0
0
100
40
0.4
0.25
0.1
5
0
0
0. 2
0. 2
0 . 002
0
0
0
0
minimal value
0
18.946536
18 . 145647
0
0
0
0
100
38.931546
0.4
0.25
0.1
5
0
0
0.1864364
0.1946651
0.0016916
0
0
0
0
maximal value
500
20
20
0.9342961
0.8454829
0.0445942
0.0149897
100
40
0. 4
0.25
0. 1
5
0.0095994
0.0086869
0.2
0.2
0.002
1. 691E-04
8.59E-05
1 . 691E-04
8 . 59E-05
6-44
final value
500
18.946536
18.145647
0.9342961
0.8454829
0.0445942
0.0149897
100
38.931546
0.4
0 . 25
0.1
5
0.0095994
0 . 0086869
0.1864364
0 . 1946651
0.0016916
1.691E-04
8.59E--05
1.691E·-04
8.59E-05
rd
ra
rb
re
Sed
Sef
0.002
-0 . 002
-0.003469
0.002
1.1734311
83 . 266916
0 . 0014393
-0.0021989
-0 . 004
0 . 0016889
1
0
0 . 002
-0.002
-0 . 004
0 . 002
1
0
0 . 0014393
-0.0021989
-0.003469
0 . 0016889
1 . 1734311
1.9686327
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(V) = ra
[2] d(Fb )/d(V) = rb
[3] d(Fc)/d(V) = rc
[ 4] d(Fd)/d(V) = rd
[5] d(Fe)/d(V) = re
[6] d(Ff)/d(V) = rf
Explicit equations as entered by the user
[1] vo = 100
[2] Ft = Fa+Fb+Fc+Fd+Fe+Ff
[3] Cto = A
[4] kd1 = 0..25
[5] ke2 = . 1
[6J kf3=5
[ 7 J Cc = Cto*Fc/Ft
[8] Cd = Cto*Fd/Ft
[ 9 J Cb = Cto*Fb/Ft
[10] Ca = Cto*Fa/Ft
[11] rd1 = kd1 *Ca*CbJ\2
[12] re2 = ke2*Ca*Cd
[ 13] rf3 = kf3*Cb*CcJ\2
[14] re = re2
[15] rf=rf3
[16] rd=rd1-2*re2+rf3
[ 17] ra = ··rd1-3*re2
[18] rb = -2*rd1-rf3
[19] rc = rd1+re2-2*rf3
[20] Scd = rc/(rd+..0000000001)
[21] Sef = re/(rf+ . 00000000001)
1.20
~~~ .... ~...-".~...."...~
.---------
90 , - - - - - - - - -
_.__
0.96
72
0.72
54
048
36
C-SI"
~
0.2-1
18
(100
o
o
100
400
200 V 300
500
P6-14 (1)
dF
--.---~
The only change from part (e) is: d~-
= rD -
kcDCD
6-45
.............................~".- ......
0
100
21)0 V 300
400
500
See Polymath program P6-14·-f.pol.
1.15
-----------" -.•
..•.-~--•..
-
125,.--------------,
100
~~
1-
0.69
Scid
J
75
046
50
0,23
25
0.OOO'-----1~0-0--20-0-V~3-00--4~0-0---'500
o0
100
200 V 300
400
500
P6-14 (g)
dF
dV
The only change from part (e) is: _---1L
=r
F
VT
--1rQ..
B
3
where V r = 500 dm and F BO = 20 moVmin
See Polymath program P6--14-g,pol.
1.2 , . - - - - - - - - - - - - - .
700.------
1.0
560
07
420
0. 5
280
0.2
140
flO ' - - - - - - .
o
100
200 V
300
400
500
o
P6-1S (a)
6-46
I...---=~--.---
I)
100
......-..•.-.-..".-""......"--..-., . . -.. . .
200 V
300
400
500
(a) Enter the given program into POLY?vlATH Equations for the concentrations
must be added.
The following maximums can be seen in the graph given (More exact values can
be found in (he corresponding table in POLY1VIATH )
CCm.u.
== 0 . 0434 and CDtrcu ;:;: 0.0033
~gu<,t:: i o~s .:
.!~~. S.~ al_.,!~.J:!:::~
d{fc)!d(v)=kl.{fa/ft}*(fb/ft).·ll!2)~k3.(fc/ft}+k4·{fw1ft
0
,*(fd/ft)
d(fal/d(v)=-kl O (fa/fcl*lfb/ft)"(1!21-k2*(fa/ftl*W2
9.8.3
d{fbl/dlv'=-kl/2*'fa/ft)-,fb/fcj··11/2)
4.91
o
o
o
o
d(fw}!d(v)=k3*lfc/ftl-k4*(fw/fc)*(fd/it)
d(fd)!dlv)=k2IZ*lfa/fc'··Z-k4· (fd/ft)* (fwiftl
d(fe) /d{v} ::ok)'" (re/ft)
d(fg) Id{v} ",k4* (fw/ft.)
* (Ed/ft)
kl=O . 04
f t=fa ·,·fb+ fc+fd+ fetfw+fg
~ ..
k2=O.007
k3=O 01.4
k4=O 45
vo=lOO
cto",O . 147
ca"'cto" (fal it)
C.2ca
cb"'cto*(fb/it)
cc,,"cto*(fc/ftl
c.aca
cci=cto*(fd/ft)
ce=cto*(fe/ft)
cw=cto' (fw/ft)
cg",ct:o~(fg/ft)
VI)
""
0,
1000
."'., ...•......•.... ,.....
;
a.oco
o
CDO
P6-15 (b)
6-47
C:.2:::'C
:::. :soc
(b) Overall yield of HCOOH:
Selectivity of HCHO to CO:
- - F
S
.. ~~..
,\E ~ F,
E
Selectivity of HCOOCH 3 to CH.;OH:
Selectivity of BCaOH to HCOOCH J
8 02
:
ell
=C-·
A
Add these equations to the previous program and use it to generate the desired plots .
o. acc
"S.tioo
1
OSCJ
o. "CO
O'10C
f.
t
i
r
I
l
?.·ne
T
-:--¥
C ... S~
1.
1
LL __ . ___ .
/".~.;
~,sa;
)tp2
P6-15 (c)
6-48
Modify the original POLY1-IATH program by adding y to each of the
concentration terms Also add (he followil1g equation:
(c)
1~~~~(~~)
a
::= 0002
Fro"" 15
The graphs of concentration down the reactor are very similar to those generated
in Palt (a),. The only major difIerence is that with the change in pressure, the
maximum reactor volume is significantly smaller.
E~~£ion~-.:
d{ fe)/d (vl =kl* (fa/fe) .. (fb/fe.) "* (1/2 )*y k3" (fe/ft) "'y+·k4" (f
:!.9::\:~i~~, ,. Y~lue
0
w/ft)"(fd/ft)"y
d(fa) Id(v) '=-kl* (fa/ft) * (fb/ft) ** O/2)·y·k2* (fa/ft) 'H2*y
10
d(fb} Id(v) ;;:·kL/2* (fa/ft) * (fb/n.) ** (1/2) *y
5
d (fw) Id (v) =k3'" (fel ft) *y .. k4" Cfw/ ft) .. (fd! ft) *y
d (fd) Id (vl ;k2/2" (fa/ft.) '"*2 *y··k4" (f'.... /ft)
* (fd/ft)
0
*y
a
d(fe)/d(v)=k3*Cfc/ft)*y
0
d(fgl/d(vl=k4*(fw/ft)"(fd/ft)*y
0
1
d(y)/d(v)=-O.002/2/y*(ft/fto)
k2=O,,007
kl",Q,04
COlICentralion profile with pre •• Ul'C .ha:nge
ft=fa+fb+tctfd+fe+fw+fg
k3=O.014
k4;;;O.45
vo=1011
ca",fa/vQ
cb=fblvo
cc=fc/vo
cd::;ofd/vo
-J.fJCO
ce=fe./vo
c.coo
cg=fg/vo
Vo '"
COlICentIalion profile with pre"Ul'C .hange
0,
P6-15 (d)
6-49
·
r fJ ---"-""'
1
1 \'\
. l.)
k"., "'" k,,n eXt11.
~I
\ R\ T
1',.)
Subsdtute this equation in for all of the k values. Vary 'r and find Dut wbat
temperature maximizes the yield of C,
The 'best temperature at \vhich to ron the reactor is 523 K or 250"(;.
P6-16 (a)
(a)
Mole Balances:
dFc
_·,"-'-=r
dV
C
dFp
dV
dFA
·-···-·=f
··,-·=f
p
dV
A
dF
o
--,-=t
dV
0
Ie ::;:: --kl C c -- k 2 C e
Rate Laws:
fA
;;;;:;k1C C +kJC p ,--k,;C" "kJC A
c o Cro (.FF{.>, )
., (FI')
C" =L!'o
F~'
Stoichiometry:
=:
'I'
l'
FI =F:+F
C P+F
A +F0
Use these equations in POLYMATH
to
generate aplol of the flow rates vs.
't ..
;:r:;.,-:::;t; ..
\
15.:::<: . \
'.
f,e
\
ip
fa
fo
..
.. -"
'
P6-16 (b)
(b)
For a CSTR
Mole Balances:
Fc
=Fco + Ie V
Fo =ro V
!'p = ktC c -- k3Cp+ k 4 C A
Rate Laws:
fo
6-50
== kSC A
Swichiometry :
f(C c) =: 0 = CcC e" -;-
Combine:
fCC?) = 0 = C p
f(C A ) =O=C A
(k 1 + kzTc'"
(kiC c- K)C p + k ,eA)r:
-
.....
(ktC c ... kJC p --k4CA-ksC~:,Jr
f(C o )= 0= Co ...... kSCAr
Use these equations in POLYMATH to generate values for the flow lates at different
values of 1: Use these values to generate the desired etHve.
E;~~.Sior~~.:.
f (eel "'CC "000+ (kltk2)
I;niS1:a1. vah:e
2
~cc*t:au
o
£ (cp) "'cp" (kl *co-kJ"cp+k4"'cal ""tau
£ (ca} =ca·· (kl""cc-·k3 ·cp·k4··ca·-·kS "'cal' tau
o
f (co) =co··-kS"ca"tau
o
cco=2
kl",O.12
Sol wt 1 0<1
k2=O .046
k3",O. {)2
ll. 039:370 1
0 .. 211711
O. Df.J633865
O. Cl?S0638
cc
k~",O.O:J4
c:p
vocolO
Cil
co
cco
kS=() . 04
V",3DOO
fc;:,:vo·cc
fp=vo*cp
fa-:vo*ca
£o;vo"co
2
D.t2
(] 046
C. D2
CU134
10
k 1
k .?
k .,
.J
.~
k4
\/0
k5
tat~=V/vo
U
3COU
fc
ip
fa
C. '393701
2 .. ,171 t
0.(;63:3865
fa
D.76()638
tau
300
(b) Flow rates vs. 1;
200
!80
Wl
14()
. Fe
120
. . . FP,
100
80
60
-FA'
Fe) .
:j()
20
()
0
50
leo
150
1;
200
350
250
(nJill)
P6-16 (C) Individualized solution
6-51
··8.54':5e"·'7
.- 1. ()78e--16
·.. 2 .. :30ge··15
'8,518e
18
P6-17 Individualized solution
P6-18 (a)
Blood coagulation living example
See Polymath program P6-18.po[
OLYMA TH Results
Example CD Solved Problems - Blood Coagulation 08-25-2005, Rev51233
Calculated values of the DEQ variables
Variable initial value
t
0
TF
2. SE-ll
VII
1. OE-08
TFVII
0
VIla
1. OE--10
TFVIIa
0
Xa
0
IIa
0
X
1.6E--07
TFVIIaX
0
TFVllaXa
0
IX
9.0E-08
TFVllaIX
0
IXa
0
II
1.. 4E-06
VIII
7 . 0E-I0
VIlla
0
IXaVllla
0
IXaVlllaX
0
VIIlalL
0
VIIla2
0
V
2 . 0E-08
Va
0
XaVa
0
XaVaII
0
mIla
0
TFPI
2.SE-09
XaTFPI
0
TFVIIaXaT
0
ATIII
3 . 4E-06
XaATIII
0
mllaATIII
0
IXaATIII
0
TFVlIIaAT
0
IIaATIII
0
kl
3.2E+06
k2
0.0031
k3
2 . 3E+07
k4
0.0031
kS
4 . 4E+OS
k6
1. 3E+07
k7
2 . 3E+04
k8
2.SE+07
k9
1. OS
k10
6
kll
2 . 2E+07
k12
19
k13
1.0E+07
k14
2.4
minimal value
0
8.24E-14
3.S13E-10
0
1. OE-10
0
0
0
1. 426E-07
0
0
8.994E-08
0
0
-3.41E-24
-2.024E-28
0
0
0
0
0
-1.SSE-S2
0
0
-3.938E-26
-8 . 77E-2S
2.094E-09
0
0
2 . 001E-06
0
0
0
0
0
3.2E+06
0.0031
2.,3E+07
0.0031
4 . 4E+OS
1. 3E+07
2.3E+04
2.SE+07
1. OS
6
2.2E+07
19
1.OE+07
2.4
maximal value
700
2 . SE-ll
1. OE--08
2.027E-ll
9.724E--09
3.361E-13
1.481E-09
2.487E-07
1. 6E-07
1.869E-13
S.673E-14
9.0E-08
7.2E-14
3 . S79E-ll
1. 4E-06
7.0E-IO
S.3S2E-I0
2.988E-12
S.372E-12
6.S8SE-10
6.S8SE-10
2.0E-08
1.943E-08
1 . 492E-08
2.281E--10
3.788E-07
2.SE-09
3.867E-I0
1 . 881E-ll
3.4E--06
6.073E-IO
8 . 247E-07
1.301E-ll
8.,3S4E-14
S.,734E-07
.3 . 2E+06
0.0031
2.3E+07
0.0031
4.4E+OS
1 . 3E+07
2 . 3E+04
2.SE+07
LOS
6
2.2E+07
19
1.0E+07
2.4
6-52
final value
700
8.24E-14
3.S13E-10
S.71E-12
9.724E-09
1.. 66SE-13
1.481E-09
1. 846E-09
1.426E-07
8.423E-14
2.608E-14
8.994E-08
3.S68E-14
3. S79E-ll
-1. OSE-2S
-1.026E-38
3.366E-ll
2.873E-12
4.99SE-12
6.S8SE-10
6.S8SE-10
2.793E-90
S.077E--09
1.492E-08
-6.977E-27
1. 663E-2S
2.094E-09
3.867E-IO
1. 881E-ll
2.001E-06
6.073E-10
8.247E-07
1.301E-ll
8.,3S4E-14
S.,734E-07
3 . 2E+06
0.0031
2.3E+07
0 . 0031
4.4E+OS
1. 3E+07
2 . 3E+04
2.SE+07
1. OS
6
2.2E+07
19
1.0E+07
2.4
k15
k16
k17
k18
k19
k20
k21
k22
k23
k24
k25
k26
k27
k28
k29
k30
k31
k32
k33
k34
k35
k36
k37
k38
k39
k40
k41
k42
r1
r2
r3
r4
r5
r6
r7
r8
r9
r10
r11
r12
r13
r14
r15
r16
r17
r'18
r19
r20
r21
r'22
r23
r24
r25
r26
r27
r28
r29
r30
r31
r32
r33
r34
r35
r36
r37
r38
r39
1.8
7500
2.0E+07
1.0E+07
0.,005
1 . 0E+08
0.001
8.2
0.006
2.2E+04
0 . 001
2.0E+07
4.0E+08
0,,2
1.0E+08
103
63 . 5
1.5E+07
9.0E+05
3.6E-04
3.2E+08
1 . lE-04
5 . 0E+07
1500
7100
490
7100
230
8,OE--13
0
5 . 75E-14
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1..8
7500
2 "OE+07
1 . 0E+07
0.,005
1. OE+08
0.001
8.2
0.006
2.2E+04
0,,001
2 . 0E+07
4.0E+08
0.2
1.0E+08
103
63.5
1 . 5E+07
9 . 0E+05
3.6E-04
3 . 2E+08
1.1E-04
5 . 0E+07
1500
7100
490
7100
230
8 . 0E-13
6.283E-14
5.75E-14
1 . 042E-15
9 . 923E-16
1_444E-ll
4 _073E-ll
1_318E-12
1.962E-13
1 . 121E-12
5 . 381E-15
1.078E-12
3.024E-13
1,728E-13
1 . 296E-13
2.883E-13
1.159E-ll
4 . 674E-14
1 . 494E-14
4.406E-ll
5,372E-15
4.405E-ll
3 . 211E-12
9 . 527E-15
5 . 372E-15
3 . 312E-10
2,999E-09
2,,974E-09
3 . 764E-08
2,35E-08
1,449E-08
7.132E-09
2.762E-12
1 . 372E-13
4 . 492E-14
2.065E-15
3 . 19E-15
4,387E-12
7 . 708E-09
1..8
7500
2"OE+07
1 . 0E+07
0 . 005
1.0E+08
0.001
8.,2
0 . 006
2 . 2E+04
0 . 001
2 . 0E+07
4.0E+08
0.,2
1. OE+08
103
63.5
1 . 5E+07
9 . 0E+05
3.6E--04
3.2E+08
1.1E-04
5 . 0E+07
1500
7100
490
7100
230
9 . 728E-17
0
L 735E-14
0
0
0
0
0
0
0
0
0
0
0
0
-1. 6E-29
-9 . 831E-28
0
0
0
0
0
0
0
0
-2 . 108E-52
0
0
-3.763E-24
-4 . 056E-24
-2.5E-24
-1 . 733E--25
0
0
0
0
0
0
-1,248E-26
6-53
1..8
7500
2 . 0E+07
1 . 0E+07
0 . 005
L OE+08
0.001
8.2
0.006
2 . 2E+04
0 . 001
2.0E+07
4.0E+08
0.2
1 . 0E+08
103
63.5
1.5E+07
9.0E+05
3.6E-04
3.2E+08
1.lE-04
5.0E+07
1500
7100
490
7100
230
9.728E-17
1. 782E-14
1. 855E-14
5.189E-16
2.696E-17
6 . 953E-12
1_602E-14
5 _971E-13
8.896E-14
5.083E-13
5 . 381E-15
4 . 982E-13
1_ 506E-13
8.609E-14
6.456E-14
4 . 511E-30
-4 . 231E-40
1.217E-14
1. 439E-14
4.106E-ll
5.006E-15
4 _105E-ll
2 . 044E-13
9 . 527E-15
5.006E-15
1 . 152E-91
2 . 999E-09
2.974E-09
6 . 12E-25
1,,105E-24
6,81E-25
7.062E-26
2,762E-12
1.372E-13
1.762E--14
2.065E-15
3.19E-15
4.387E-12
4 . 499E-27
r40
r4I
r42
Total
o
o
o
o
o
o
o
o
3 . S02E-14
4.224E-09
1. 78E-16
S.749E-07
3.S02E-14
2.704E-ll
7 . 70SE-17
1.903E-09
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(TF)/d(t) = r2-r1-r3+r4
[2] d(VII)/d(t) = r2-r1-r6-r7-r5
[ 3] d(TFVII)/d(t) = r1-r2
[4] d(Vlla)/d(t) = -r3+r4+r5+r6+r7
[5] d(TFVlla)/d(t) = r3-r4+r9-r8-r11 +r12-r13+r14-r42-r37 +r15
[ 6] d(Xa)/d(t) = r11 +r12+r22-r27+r28-r33+r34-r38
[7] d(lIa)/d(t) = r16+r32-r41
[8] d(X)/d(t) = -r8+r9-r20+r21 +r25
3.40E-13
[ 9] d(TFVllaX)/d(t) = r8-r9-r10
[ 10] d(TFVllaXa)/d(t) = r10+r11-r12-r35+r36
[11] d(IX)/d(t) = r14-r13
2.72E-13
[12] d(TFVllaIX)/d(t) = r13-r14-r15
[ 13] d(lXa)/d(t) = r15-r18+r19+r25-r40
2 . 04E-13
[14] d(II)/d(t) = r30-r29-r16
[15] d(VIII)/d(t) = -r17
[16] d(Vllla)/d(t) = r17 -r18+r19-r23+r24
136K·13
[17] d(lXaVllla)/d(t) = -r20+r21 +r22+r18-r19
[18] d(IXaVlllaX)/d(t) = r20-r21-r22-r25
[19] d(Vllla 1L)/d(t) = r23-r24+r25
6.80E-14
[20] d(Vllla2)/d(t) = r23+r25-r24
[21] d(V)/d(t) = -r26
O.OOEO 0. 00
[22] d(Va)/d(t) = r26-r2TH28
[23] d(XaVa)/d(t) = r27-r28-r29+r30+r31
[24] d(XaVall)/d(t) = r29-r30-r31
[25] d(mlla)/d(t) = r31-r32-r39
[26] d(TFPI)/d(t) = r34-r33-r35+r36
[27] d(XaTFPI)/d(t) = r33··r34-r37
[28] d(TFVllaXaTFPI)/d(t) = r35·r36+r37
[ 29] d(ATIII)/d(t) = -r38-r39-r40-r41-r42
[30] d(XaATIII)/d(t) = r38
r31] d(mllaATIII)/d(t) = r39
[32] d(IXaATIII)/d(t) = r40
[33] d(TFVlllaATIII)/d(t) = r42
[34] d(lIaATIII)/d(t) = r41
r;:::====:;--r,----,
Explicit equations as entered by the user
[1] k1 = 3 . 2e6
[2] k2 = 3.1 e-3
[3] k3=2.3e7
[4] k4 = 3 . 1e-3
[ 5] k5 = 4..4e5
[6] k6 = 1 . 3e7
[7] k7 = 2.3e4
[8] k8=2.5e7
[9] k9 = 1.05
[10] k10=6
[11] k11 = 2 . 2e7
[12] k12 = 19
[13] k13=1.0e7
[14] k14 = 2 . 4
[15] k15 = 1.8
[16] k16=7 . 5e3
6-54
138.93
277.8~
416.79
555.72
694.64
[17]
[18]
[19]
[20]
[21]
[22]
[23]
[24]
[25]
[26]
[27]
[28]
[29]
[ 30]
[31]
[32J
[33]
[34J
[35]
[36]
[37]
[38]
[39J
[40]
[ 41J
[42J
[43J
[44]
[45J
[46]
[47]
[ 48J
[49]
[50J
[51J
[5'2J
r 5 3]
[54]
[55]
[56]
[57]
[58]
[59]
[60]
[61]
[62]
[63 J
[64]
[65J
[ 66]
[67]
[68]
[69]
[70J
[71J
[72]
[73 J
[74]
[75]
[76]
k17=2e7
k18 = 1..0e7
k19 = 5e-3
k20 = 1e8
k21 = 1e-3
k22 = 8,.2
k23 = 6e-3
k24 = 2 . 2e4
k25 = 1e-3
k26 = 2e7
k27 = 4e8
k28 = 0,.2
k29 = 1e8
k30 = 103
k31 = 63 . 5
k32 = 1 . 5e7
k33 = ge5
k34 = 3,.6e-4
k35 = 3.2e8
k36 = 11e-4
k37 = 5e7
k38 = 1 . 5e3
k39 = 7..1e3
k40 = 4 . ge2
k41 = 7.1e3
k42 = 2 . 3e2
r1 = k1 *TF*VII
r2 = k2*TFVII
r3 = k3*TF*Vlla
r4 = k4*TFVlla
r5 = k5*TFVlla*VII
r6 == k6*Xa*VII
r7 = k7*lIa*VII
r8 = k8*TFVlla*X
r9 = k9*TFVllaX
r10 = k10*TFVllaX
r11 = k11 *TFVlla*Xa
r12 = k12*TFVllaXa
r13 = k13*TFVlla*IX
r14 = k14*TFVllaIX
r15 = k15*TFVllaIX
r16 = k16*Xa*1I
r17 = k17*lIa*VIIi
r18 = k18*IXa*Vllla
r19 = k19*IXaVllla
r20 = k20*IXaVllla*X
r21 = k21*IXaVlliaX
r22 = k22*IXaVlliaX
r23 = k23*Vllla
r24 = k24*Vllla1L*Vllla2
r25 = k25*IXaVlllaX
r26 = k26*lIa*V
r27 == k27*Xa*Va
r28 k28*XaVa
r29 = k29*XaVa*1I
r30 = k30*XaVall
r31 = k31*XaVall
r32 = k32*mlla*XaVa
r33 k33*Xa*TFPI
r34 = k34*XaTFPI
=
=
6-55
[77]
[78]
[79]
[80]
[81]
[82]
[83]
[84]
[85]
=
r35 k35*TFVllaXa*TFPI
r36 =k36*TFVllaXaTFPI
r37 =k37*TFVlla*XaTFPI
r38 =k38*Xa*ATIII
r39 = k39*mlla*ATIII
r40 = k40*IXa*ATIII
r41=k41*lIa*ATIII
r42 = k42*TFVlla*ATIII
Total lIa+1.2*mlla
=
P6-18 (b) No solution will be given
P6-19
C2H4 + 11202
E
+ 1120
(1)
FlO
~
~
C2H40
D
C2H4 + 302
E + 30
(2)
~
~
2C02 + 2H20
2U I + 2U2
= 0.82Fm = 0.007626
P6-19 (a)
Selectivity of D over CO 2
s= FD
FUl
See Polymath program P6·19···a.po1.
POLYMATH Results
Variable
W
Fe
Fo
Fd
Ful
Fu2
Finert
Ft
Kl
K2
Pto
Pe
Po
kl
k2
X
S
rle
r2e
initial value
-0----5 . 58E-04
0.001116
0
1.. OE-07
0
0.007626
0.0093001
6.5
4.33
2
0_1.199987
0.2399974
0.15
0.088
0
0
-0.0024829
-0.0029803
minimal value
0
1. 752E--I0
4.066E--·05
0
1 . OE--07
0
0 . 007626
0.0091804
6.5
4.33
2
3.817E-08
0.008858
0.15
0.088
0
0
-0.0024829
-0.0029803
maximal value
2
5.58E-04
0.001116
2.395E-04
6.372E-04
6.371E-04
0.007626
0.0093001
6.5
4.33
2
0.1199987
0 . 2399974
0.15
0.088
0_9999997
0.,4101512
-3.692E-I0
-8.136E-IO
Differential equations as entered by the user
[1] d(Fe)/d(W) = r1 e+r2e
[ 2] d(Fo)/d(W) = 1/2*r1 e + 3*r2e
[3] d(Fd)/d(W) = -r1 e
[4] d(Fu1 )/d(W) = -2*r2e
[5] d(Fu2)/d(W) = -2*r2e
Explicit equations as entered by the user
[1 J Finert = 0.007626
[2] Ft = Fe+Fo+Fd+Fu1 +Fu2+Finert
[3J K1 = 6 . 5
6-56
final value
2
1. 752E-10
4,,066E-05
2 . 395E-04
6.372E-04
6.371E-04
0 . 007626
0.0091804
6.5
4.33
2
3.817E-08
0.008858
0.15
0.088
0.9999997
0_3758225
-·3.692E-I0
·-8 _136E-·I0
[4] K2 == 4 . 33
[5] Pto == 2
[ 6] Pe == Pto*Fe/Ft
[7] Po == (Pto*Fo/Ft)
[8] k1 == 0.15
[9] k2 == 0.088
[10] X == 1 - Fe/0 . 000558
[11] S == Fd/Fu1
[12] r1e == -k1*Pe*Po"O . 58/{1+K1*Pe)"2
[13] r2e == -k2*Pe*Po"O.3/{1 +K2*Pe)/\2
x == 0..999 and S == 0.376(mol of ethylene oxide)/(mole of carbon dioxide)
P6-19 (b)
Changes in equation fr'Om part (a):
a
dF1 +3r +R
=-r
and Fa ()
a =0
a
IE
2E
dW
2
R = 0.12xO.0093 = 0.001116 mol
a
W
2
kg.s
From Polymath program: X == 0..71
S = 0..0.4 (mol of ethylene oxide)/(mole of carbon dioxide)
See Polymath program P6- t9-b . pol
P6-19 (C)
Changes in equation from pmt (a):
d~
- = rlE
dW
+r 2E +R E
and
FE ()
a =0
R = .0.06 x 0.0~9~ = 0.000558 mol
E
W
2
kg.s
X = 0. . 96
S == o.Al(mol of ethylene oxide)/(mole of carbon dioxide)
See Polymath program P6-19·c .pol
From Polymath program:
P6-19 (d) No solution will be given .
P6-20
Solved on web
Go to http://www.wits.ac.zalfac/engineering/procmatiARHomepage/frame.htm
P6-21 (a)
Isothermal gas phase reaction in a membrane reactor packed with catalyst.
A
~-7
B +C
lic,
1
c
= lsc [ CA - -CBC
_.
KIC
A-7D
2C+D -7 2E
6-57
CAD
=~ =
RT
24.6atm
(O.082dm atmlmol.K) (500K)
3
= O.6mol I dm 3
See Polymath program P6··21-a.po]
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value
WOO
Fa
Fb
Fe
Fd
Fe
10
0
0.349438
0
0
0
0
0
0
0
y
1
0.4
0.2
10
0.6
0" 3404952
0.4
0.2
9.7581913
0.6
k2d
K1e
Ft
Cta
Cb
Ca
Cd
Ce
kb
k1e
r2d
k3e
rIc
ra
r3e
rd
rb
re
re
Ce
alfa
Fto
0
0
0.6
0
0
0.0073158
0
0
1
2
0.24
5
1.2
-1.44
0
0.24
1..2
1.2
maximal value
100
10
3,,2375418
4.9873025
2.7304877
1.3722476
1
0.4
0,,2
13.220737
0.6
0.1403618
0.6
0.1019635
0.2117037
1
1
2
0,,0029263
5
0,,0051635
-1,,44
0
0.0014457
0,,0051635
-0.0042625
0
0
0
0
0.008
10
0.008
10
2
final value
100
0.349438
0,,4443151
4.8617029
2,,7304877
1.3722476
0.3404952
0.4
0.2
9.7581913
0.6
0.0093022
0.0073158
0.057l654
0.1017844
1
2
0,,24
0.0029263
5
5
1.2
-0.0080898
0,,0216828
0.24
1.2
1.2
0,,0216828
0.0398819
0.008
10
0.0051635
··0.0080898
0.0029612
0.0014457
0.0051635
0.0022023
0,,0029612
0.0287293
0.008
10
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(W) = ra
[2] d(Fb)/d(W) = rb·(kb*Cb)
(3) d(Fc)/d(W) = rc
[ 4] d(Fd)/d(W) = rd
(5) d(Fe)/d(W) = re
[6) d(y)/d(W) = -alfa*Ft/(2*Fto*y)
10
Explicit equations as entered by the user
[1J k2d = 0..4
[2J K1c=0 . 2
[3] Ft = Fa+Fb+Fc+Fd+Fe [4] Cto = 0 . 6
[5 J Cb = Cto*(Fb/Ft)*y
[6] Ca = Cto*(Fa/Ft)*y
[7] Cd = Cto*(Fd/Ft)*y
[8] Cc = Cto*(Fc/Ft)*y
[9] kb=1
[10] k1c=2
[11] r2d = k2d*Ca
[12) k3e 5
[13) r1 c = k1 c*(Ca-(Cb*Ce/K1 e»
[14] ra = -r1 e-r2d
- Fa
8
•. li'b
.. Fe
- Fd
6
4
2
-"', .....".........
=
_
...........
o "."..•... -====--~-.----~----I
80
100
o
20
40 \V 60
6-58
[15]
[16]
[17]
[18]
[19]
[20]
[21]
[22]
r3e = k3e*(CcA 2)*Cd
rd r2d-(r3e/2)
rb = r1c
rc=r1c-r3e
re = r3e
Ce = Cto*(Fe/Ft)*y
alfa = 0 . 008
Fto = 10
=
P6-21 (b)
The interesting concentrations here are species C and D, both of which go through a maximum Species C
goes through a maximum for two reasons: (1) it is an intermediate product which is formed and then
consumed, and (2) there is pressure drop along the length of the reactor and as pressure drops, so does
concentration.
The concentration of species D goes through a maximum because of reason (2) above . Species D is formed
but then the pressure drops, which causes the concentration to falL
P6-21 (e) Individualized Solution
P6-22 (a) What factors influence the amplitude and frequency of the oscillation reaction?
Ans: k and the initial conditions
P6-22 (b) Oscillations eventually cease because the CA is decreasing and becomes the limiting factor.
P6-22 (ae) Observation 1: 'tl and 't2 decreased
Observation 2: 'tpI increased
Observation 3: 'tp2 decreased
Now, Tl =
~£ In( JLlJL~ J
and T 2 =
~£ In(PO*
J
JL2
2n
Tpl
=(JLl*2- + KU )112
_ ko
£--,
2n
and
Tpz
= (JL2*2-; K
-)l/Z-
u
* _[(1-2Ku )±f!=8Ku ]112
JL12 -
k
'
2
2
From observation 2 and 3, we get
Decreasing III * and increasing 112* => Ku = (ku/k2) will increase
Also, from observation 1, E increased => koIk2 should be increased
Now,
P -7 A
k = ko ....... (1)
A-7B
k=ku.
(2)
A+2B -7 3B
k = kl ....... (3)
B-7C
k=k2 ............ (4)
Hence the reaction (1) and (2) are more temperature sensitive than reaction (4)
P6-22 (d) Individualized solution
6-59
P6-23 Individualized solution
CDP6-24
t).. ::
-u() Ca + ra W
•
·kl CA - k3 CA
ra = k1 CA
=0
..
k2 CB
rc:: k2 CB
-Uo Cc + rc W ;:: 0
"Uo CD +rD W =0
113' • .9'.9'
KEY:
1 -
ca
2 - cb
:] .. cc
..... cd
CDP6-25 (a)
PFR:
6-60
r-,.'lo1e balance:
dC},t = r
--_._.
dr
M
Rate laws
e KC ::::: 0 . 021
C MO ;;;;; 00105
T ==.5
Plugging those into POLYMATH gels the following:
6-61
CSTR:
Mole balance:
FHO '" Ffj == (.- rtB -. f;H····
FMO-FM
r,H) V
=-"I'I:-I V
Fx =(rtx+f1X)V
FT =(r21+ f;r) V
F;-'1< == (£'1 M~ + fz ),10 t·
r.l.\k )
V
Fa =: f'll
Rate Law:
Stoichiometry:
FH ::::: 'lOCH
~"1::::
VOCM,
Fx == voCx.
FT == voC r
F:\1e :::: VOC Mc == vo(C l10
...•
ell)
ell
FEI :: \'oC 3= vo[(C)"m '" eM)" ex·'
Combining all of these:
·'
CH::;;;;.·l
(k COH~(''M + k-:: (~l)5C
)
C
. -HO ·,tt 'K t'. k ; c,Q,$c"
HI r
CM
¢
::::
eHG -C H
ell = (C;"iO - eM ) . . C x .- CT
The following i:s the POLTh1.-'\TH program and
tIle concemrar,ions.
E:'i'!a;.i 0E!::!!:'
me summary table showing all of
f (en] :eh-eno+ (k1""ch ~ . 5 ·c:n-·k2 ~cn" . 5 'cx+kl "ct: 'ch~ . 5) "cau
£ (c:n) =cm· 'crne i kt ~c~ - ~ 5 .... c:tr'" 'C.a.!l
Uartaolt?
~(cCI;lk2·ch-.5·cxk3·c:·chA.5)·~aa-ct
c:'0=.021
ch
em
ex
k1.=5S.2
k2",3a.2
Solutl.O["'f
--
Value
........... .
"
.......,.,~
..... .......
0.00776519
0.00301658
0.00:317467
0.00286611
0.021
ld=l1.2
1;8.u"".51.
erne"',
.......
kl
55.2
k2
"30.2
11.2
1<:3
alaS
0.5l
erne
cmt?
6..62
0 .. 0105
0.01323-;0
0.00114254
~"".--
..
~"
-
.. fO
..
.....
--~
"
2.993.-15
I .463",
14-
2 .. 79t? -·14
1.495e- i"
The con\.crston of Hydrogen and i'v[esitylene are (hen:
X. = S'.:HQ ..
0 . 11 -~g.9gLS. == 0.63
H
C
O
O '. )- 1
HO
_Sll : ;: .2:
X "'1
_ .....e --......-C- - . =--.0010)··,0.0030
. . . . . _._ . . . . . . . . -.. . ----- -_ .~.f 1
-
MO "
SI
O.OW5
'C!'-IO
C H = 0,0078 Ib mol/frl
C r = 0.0029 ib mollft'
0.0030 lb molJfr\
Crr.::;::;;; 0.Q13 Ib mollft'
eM:;:;
C x = 0.00.32 Ib mol/fr'
C~ ::::: 0,,0014 Ib mollfr'
CDP6-25 (b)
When eM is reduced to 1.5, it now takes a L of 0.24 h to achieve a maximum of xylene, Increasing eM to 10
now requires a L of only 0 . 08 h
CDP6-25 (C)
To find out the reactor schemes needed, use the attainable region to get these graphs
Using fi PFR wouldmaxirnize C,. If ---'Ie used at rl1ri{) of mesil:ylt!ne to hydrogen of
10, tben we l.vQuld only have to have a t of .08 hours. So our voulme of the reactor
would bt; only 38,08 ft\ So our entering cOl1centraiol'ls would be .0105 Ib mollir'
of Hydrogen and, 105 111 moUff of Mt;s[~yler;e.
CDP6-25 (d)
Ftfst fmd the protJortionality ';;OEst;;;mts of the rate consr.ants usin2 (he
Anheniu$ equation.' •
's
A, :::: 12674
Once that is done, by t.rial fu'1d erw[ corne up w;th the temperature where SXT ""STB
and th~~ r.msw~r is then 862K or 155 L6°R.
6-63
~qu~9.!.::!.:
dIem) Id (t);:Dn
0.0105
0.021
d(chl/ci(cl =!h
o
o
o
d(cx)/d (t) =rx
d(eme) Id(t) ;;rme
d(ccl/d(c)=rt:
dlcb}/d(t)=r'b
Scale: 10
rm=-55 . 2 "cm"ch". 5
rb::::l1 . 2 "ct"chA. S
o
2 2.:::
K&'£.;.
,,""
em.
!x:;:~rm-30.2*cx*ch~.S
ch
rt;.=30 ,,2"'c:X:"'ch". 5-rb
rh:::rm-30.2"c:x:"ch".5-rb
e'X
..... -"
cme
1.:lCC
.
rme;::·-.t1l\+30 . 2"cx"'ch" .S+rb
,,
'
..
~.
ct
cb
O.JCO
t
!E:bE!~l . Y~l~ Max!~ Ye.!:!~ ~:h..t!l:!!!:!!!!.Y!:1~s:
() 51
(}
0
em
{}.O!.OS
0.01.05
(}
en
o. 021.
0.021
0 . 00381.84"1
YeE~~e.!.~
. 0001599836
~i::~! .. ~!:l~!:
0 .51
a
000699836
0 00381847
ex
0
()
0
0.00372332
erne
0
0.0111.815
0
0,,017181.5
c::.
Q
0.OO4772.U
0
0.00477233
cb
a
Q . 00130452
0.00130452
00506614.
::m
--0 .. 083992
.. Q.OO:n871S
0
-Q .083992
::b
0
000.3521.27
0
0,00330298
::x
0.083992
(}.08.1992
.• 1.1 _. 00523484
-0.0045612
-~
0
0 01.46868
°
0 00364547
.. 0 . 0126384
0.083992
0.0126384
0.0126384
!-h
0.083992
l::me
0.08,3992
-0.083992
The maximum concentration of xylene occurs at I: ;:;;: 0 . 19 h .
6-64
···0.00238715
-0 .0126384
CDP6-26 (a)
Starr wicb ihe mok balances:
dF,
_dEl
.....,,"- ;;;;;: C
dV
,"',","'-::: r
It
dF,
'
If.!::::
9
dF
dV
H
H
---,-,:;;;:
r
-''''''''.- ::: r,
dV
dV
(" «""'} +- k" ('"}j'("~3 + k 5 C'"ti
" C-,
k'~ C-ii«".~H + k ~ (~'c'
"f~i() -+ k-1'fl
?
Finally ,the stoichiometry:
Putting all of (hose together and put it into POLYMAnr and get the following
program and answers
d(cll)/d(V) = rl/vo
# cll(O)=0.l37
d(c9)/d(V) = (-r2+r3)/vo
# c9(O)=0
d(clO)/d(V) = (-rl+r2)/vo
# c10(O)=0
d(c8)/d(V)
(--r3+r4) /vo
# c8(O)=O
d(c7)/d(V)
(-r4+rS) /vo
# c7(O)=O
d(c6)/d(V)
-rS/vo
# c6(O)=0
d(ch) Id(V)
(rl+r2+r3+r4+rS)/vo
# ch(O)=0.389
890
c9/(cl0+c8+c7+c6+0.0000001)
887
c8/(c7+0.0000001) #
889
c8/(c9+0.0000001) #
8910= c9/(cl0+0.000000l)#
kS*10
kl
kS*17.6
# k2
kS*2.7
kS*4.4
# k4
k3
2.1
vo
1
# kS
rl
-kl*ch A O.S*cl1 #
r2
-k2*ch A O.S*cl0 #
r3
-k3*ch A O.S*c9 #
r4
-k4*ch A O.S*c8 #
rS
-kS*ch A O.S*c7 #
X = l-cll/cllo
# cll0 = 0.1
V(O)=O
# V(f)=0.8
#
6-65
006
0.05
0.05
0.04
0.03
0.03
0.02
0.02
0 . 01
0.01
0.00
0 . 00
0 . 03
0 . 16
0 . 24
0 . 32
0.40
0.43
0.56
0.64
072
0
V
•
The ratio of hydrogen to pentamethylbenzene is 2.83 and the volume is 0.8 m
polymath solution
3
•
CDP6-26 (b)
The polymath program is the same as the first, we see that the value of c11(o)=0.092 and ch(o)=0.434 and
the ratio now becomes 4 . 8 to 1 and the volume increase to 6.8 m3 to maximize SS9' To maximize SS7 it
follows that the volume would be smaller because earlier the reaction ends the less C 7 is formed.
67.0
60.3
53 . 6
46 . 9
40 . 2
33 . 5
26.8
201
13.4
6 ..7
0.0
0.00
0.67
1.35
2 . 02
2.70
3 . 37
V
CDP6-26 (C) Individualized solution
CDP6-27
6-66
4.05
4.72
5 . 40
6.07
674
I;:;.~'::' tftJ
1<=-09
o. S
~(fca)==ca·V-fca
V·"" 10
no
fho=lOOOO
fb=4320
fbo:..;:7200
k::",,2 . 7
Kl=.0264
k2=:.07
T=40J
Poo""1400
R;:::8.]09
PO;::: I?bo" , 6.
vo=2 .. 41e7/Poo
ph=fh-S.309*Q03,vo
Y"" fc / ( fbc.)~ fb}
pc_fc*8.309-T/vQ
~h=(:b)-k2·K2*Pc·Ph/(1.K2-Pc)
rc::-irb) ~k2"'K2"'Pc*Ph! (1+K2".l;>C)
Val UE?
f C:,8
5:3.5618
82:32.08
€l52.1S:
U
umu
: ~1 0
1 Ul)ClO
ib
4:320
,ba
72DO
2,,7
k 1
Kl
k2
K2
Pbo
Pb
va
Pt,
Pho
5~826@
1]
L8",7\'? 12
28-12"" 14
rc
O. ':1264
U .. 07
0.04
4U3
:400
8 .. 105
84(]
'.7214.3
2.3541
2.85836
16Ul.31
The highest yield occurs at pressures: P flO:::: 1400 kPa
PHa
::::
19452 kPa
6-67
19";5,2
,9. OS-j 26
0 .. 852181
-9.93644
8.232U8
CDP6-28
(a)
Mole Balances:
dF
+ r2A
dV
D
..dF
_=r.
dV
ZA
--~ ::: riA
Rate Laws:
kl(c",Ci -- CcIK t )
-I'y. := k1 (c A CD - C1i;C SIK2 )
-riA::
-f3C ;;;
k)Cc:
Stoichiometry :
C,
=(:,.(~J;)
Fy == FAt- Fat- Fe + Fo + FE + Fo
p
e'G =R'i'a
Use these equations in POLThIATII- Vary Po and To to fmd the optimal
conditions. We determine these to be:
T" =315.8 K
Po ::;: 160 atm
<:;:<1;(:to" (fa/:;;;) • (';,'o/'rl
cb$Cto . . {fblf':)·(Tol:')
cc=Cto"IEc/!c) "tTo/T)
al!.]
Id{v).~r2&
6
d (:0) lcl(v) "'4*rla· :::-2e· ::-3c
14
d( fa) Id(v) "r1a .. ;:2a
:5
dl!clJd(v).~:la·:Jc
Q
c.(:gJid(v):.;;-rJc
0
co.::;C-;,;o'(fd/f-;,;) "(To/T)
ce~C~o~(fejf~)V(To!T)
:::3c"" .. kJ"cc
rla=-k:*(ca~cb~2~CC/~1)
r'2a=-k2'" (ca."cd-ce/:~2)
4
1'0:=31.5
a
T"''I:o
k1.:=. 9::n"e;<;;i (2.5' (3140011. 5a7~ C./J30 "liT)))
K:,.13J.567" (0. DOl 98 "1:)"" 2 "'e;-cp LHl520.i L 987'" (1/1>1.,1 298) i
k2=. 6J6"t?:xpi 1300011.98'7" (U :'SOD-liT) l
K2 =103943 *el':,;;J 198J4/!. . 987'" n. iTo, 1/298) )
~:J::;
.. 244 *@.xp (1.5 w 28956 /! . 98 '{,. (1./ 32S-1./T) }
6-68
v
G
200
o
6
3.99S't2
5
J.A
1:.3038
1.4.
o
"!. .. 13754
£d
it
1;'Q
k1
o
o
O.68201H
1.13754
[} . 00:'0827
G.00441.766
30
27 A837
27,';:933
315.8
].15.8
'H58
:t60
l50
1.00
315.3
")
0.00428571
0.00,128571
O .. 0042857'\.
0.0042857:.
279::'12
279:112
2791.,12
2.SH122
2 881.22
2,831.22
4076: . S
40761.5
D.034 ..l899
0 .. 0.14]399
(} .0.HJ899
6.17366
6 ]:,865
o
O.I:l4:S214
15 g
') L5 .. S
40761..5
kJ
ca
l) ..
}lS 3
2.831.22
Ct.o
lB4792
0.:84792
3D
K1.
X2
9 . 99558
6 178,,5
1. . B'573
1 23$73
2,88337
041.5214
2.88337
O.20249.B
L 48::'36
n.1S32S6
cd
O~82Ja2l
0.323821
0.000236652
o
cO:
1.23573
2.24735
1.23S73
2.24593
·c
.. 0.00596389
·"·,0.0:)22,5647
-{j _ :~440299
ria
-1.9JJ0'5
08099:251.5
.(1. [HJ527:J54
-0 . 1)02:564']
4 GOG·He·OS
MeChanQl Syncbesig
v ................. _._._
Ij
()
20
40
O.6357();an
60
0.Si:334782
0.4<1820445
ao
O . 8'l()427)
100
O.90B2S74j
O.sn76642
o .8554.1262
o . 30470055
(} . 74650071
20a
(b)
o . 5a:S".l95
Use the same POLYMATH program as aoove and vary the ratio of entering
reactants. The optimal ratio would be :
fs
hydrogen gas,
TI
carbon monoxide, and ~ carbon dioxide
These results are similar to those in paxt (a) in tllat the optimal volume is still
100 dm\ and the cQucentrarion profile is very similar in shape. 'nle primary
difference is that the Fe values are more thar) doubled.
6-69
Mechanol synthesis
equations:
:tnitl.al value
d{Fa)/d(V)=~la+r2a
8
d{Fb)/d(V)=2·rla-~2a-r3c
16
~(Fc)/d(V)=-rla+r3c
o
d(Fe) IdCV) =-::2a
6
d(Fg}/d(V)=-r3c
o
o
;1(Fd) Id (V)
~::2a
1:0=315.8
Po=160
TeTo
J.CCC
r
,
I
2.~
t
t
t1
o.~o
1_.__._. .___
+__.___ _
. - , - - - + _ _ , -........;._.__
1().:C~
ea a:c
12~.a:{;
I ~.oc;:
2:)0::
u
kl=.933-exp«(2.S-(31400/l.987-(1/330-1/T}))
Kl=131667-(O.00198·T)~2-exp(30620/1.9a7·(1/T-1/298})
k2=O.636-~(18000/1.9a7-(1/300-1/T})
3:2::.10] 943"ex;,') (9834/1. 987· (1/T-1129B) I
~J=O.244·exp(1.S·289S6/1.9a7~(1/325-1/T»
Cto=?0/(.082-To)
Ca=C~o·(Fa/F~)·(TO/T)
C~=Cto·(!,b/Ftl·(To/T)
.Kee.h4l>ol. ~i.:I
.,
~val ...
!!:i.u-
0
200
0
lOa
.f-
a
a
1 94105
r.>
U
li
7 IIS7.a
1 U~Q5
7_aanS
Ce=Cto*(Fe/ft}·(To/TI
Fe
0
2 flU
Q
Y.
5
41
5 .. 9''''1
::3c=-kJ ·Cc
f'1i
I)
4 00756
\)
N
Q
n.
J,
o OGOStoln
lO
0
11
To
llS I
Po
15.:1
l'
lLS i
CI 0041ISU
2191 12
CC=Cto*C?c/ft}·(To/T)
cd=Cto·(Fd/:~I·(To/T)
rla=-kl·(Ca·Cb~2-Cc/~)
r2a=-k2-(Ca*Cd-Ce*Cb/Kl)
~;',:;i&l
vsi.able
kl
~~
value
1.5
l60
a
5.""1
4 D07S4I
II 0005'JOala
lasS
US.
:1:...as
3l.S ••
0.004215'11
US.S
115 .. 8
O.. OO41IS71
2791.12
2 !la~22
40161 5
Q OJ4J."
nn
o 00421511.
2791..1l
2 IBU2
Ct.Q
, 178i&
, nan
C'""
<:t>
!. o4'15~
1.29528
Cc:
0
C<!
0:.
0
~.Zl'7)
•. U7i4
J lf528
o 7243"
o Oool.Snn
~ Sil5l
.:lc:
.:l..t.'
-0
-0
"0.Q7S07a
-".Ol.lUQ8
6-70
lOU"
ll.Sa
160
IeJ
El
F~l_l...
laO
l '41.l2
4~'Gl S
o O14lU'
lcl
v.he
12
2 .. 111.22
401&1..5
o 0143899
, %.7866
0 . 5(7917
2
0
22U~
,0751.S
o OlUU9
, ::raSi
Q
547'H7
:z 11,;a
0
o 57nJ3
o OOOUi7a.
I. :U571
1 S93'S!.
-0 Oa.uU
-0 .. 076578
-a.Ol".,.
.. 0 . Qll.l .. oe
Met:hanol Synthesis
v
-.-.--... ....... ..... .. ..
~
~
~"
"
Fc
o
20
1.2H3568
40
:: .. 0129151
60
2.4406221
30
2 6365715
100
2.680932
120
2.6292203
1'1.0
2 .. 5193659
160
2 376518
180
2.2171.669
2DO
2.0519865
CDP6-29 No solution will be given
CDP6-A
a) A + B-1> C + D
C-+B~E+D
First, find 'r. To do this use the original desigl1 equation for a CSTR:
V == .~AO~
-··rA
Then since F . . a ==
CAOVO and 1;;:::
Vivo then the design equation becomes:
. _ CAQX
1--'
·-rA
Using the rate law and stoichiometry we find:
m·-rA = k i CAO .(1 .... X)
Combining aU these and solving for't when X = .3, CAQ::::: .1 and kl ::::: .412
1: :;:::
~M!'F:.. . . . :;: : -.. --.~-~. ~)~-. : : : 1.04b
k\CAQ(l-X) .412*.1*.7
Once that has been calculated, redo all the mole balances:
CAl) -., C A
::;;; --rA 1:
C BO · · · · C B ::;;; -rB1:'
Cc
==rc1:'
CD == Io1:'
6-71
Then do the rate laws:
-fA
= klC A
,,·rB :::: klC A + k 2CC
fC ;:;;;;;
fD
ktC A
-
k 2C C
== klC A -1- kzC c
fg =: k 2 C C
Combining and rearranging into a function:
fCC A );:;:: 0 == C A - CAO + k,cA-r
f( Ca):::: 0 == Cn ..... Coo + (k\C ... + k 2Cc)-r
f(cd==O=(k,C ... -"kzCc)-r-Cc
feeD) = 0:;;: (k1C ... + klCC)-r~· CD
f(C g )::::. a
=k C
2
C "f .....
CE
Plug those into POLyp.,IATII:
6-F
ESFlations;
f(ca)=ca-cao+kl·ca"cau
f (cb) =cb·'-cbo";. (kl"'ca..-k2 "cc) .... tau
f(ccJ;(kl*ca-k2·cc)*tau-cc
Solution
Uanab!e
ca
cb
cc
f(cd}·(kl·ca~k2·cc)·tau-cd
cd
f{c.J=k2~cc·tau-ce
Cf;'
cao=.l
kl=.412
tau=1.04
C.ao
kl
__
U<l!IJE'.,"""'.....,,_. fO....._-
0.0700015
0.10B382
0.0283726
Il03i6ia"
O.OO16229i
III
0.112
l.tM
cbo==.14
tau
r.be
O,ii
k.2=.055
t2
o.ffi5
Cc ;::: .028 need Fco == 10
v
o
= Fe
V=
C·c
'tVa
;;::;: -~~0"8
.....
=357 gall h
= 1.04 *' 357 = 371.3gal
b)No solution will be given.
CDP6-B
kt
~/I D
A---76/'
~C
BatCh r~:.or: YAO == 1
6-72
3.J6le-15
-t.02Je-! I
9. 112e--13
"' 9. iOSe-· i 3
LSi'S!?! !
kl
(a)
;;
s.:C- 1
O.Ot
T
t
== 1..5 ruin::: 90 s
A-7B
NAO!2X = -fA V
dt
CAO~= kl
CA
::::
ki CAo(i.X}
in Loo.1__);;:;; ktt
\l-X t
.--1- =: ct,t
I-Xl
CA = C.-...n{l-Xd = CAO e-k\t
.fA. = cook,t ::;; c-{O.Ol}(90) ::: 0.41
CM
B--7 D
rSl=-r",=kIC",
XI
fC
=:
k:z CB
X2
fC>
: :; ; k3 Cg
X3
-ra ""'-rsl 'r fC + I'D::::: - k t CA + leI CB +- k::3 Cs
. ~~ ::: -ki CA.O
e-lc,t ....
(k2 + k3) CB
~oodCB . ::::: -{0.01)(O.2) e·~·Olt + (0. 003
d!
::d~B. == -0.002 (1
t·
0.0(2) CB
- c·.o Oll} + 0.005 Cs
Using a Runge.,Kut:u Gill numerical so1.unon. we find that. for t::::: 2 min;: 120 s. CB:::
O. I 36 gmolJdm3..
(c)
!!;:.
= 'c = x, C. ;
£~. = k, C.
; C"
="
L
c. dl
From the solunon in patt (b). we have values of C:s at intervals as small as 5 sec, so we can
use Simpson's rule to obcain
t ""
(:'c:
1 min =: 60 sec.
Cc "" 0.003 Pf)[O + 4(0.0449) + 2(0.0782) + 4{O.102j + 0.1183]
Cc =0.0129 gmolJdm3
6-73
t
= 2 min == 120 sec
Cc:= 0.003 (3f)(0 + 4(0.0782)'1- 2{0.1l83) + 4(0.134) + 0.1355]
Cc == 0.0366 gmol/dm3
T
0
20
(d)
I
40
60
KO
C",
CB
0.2
0.164
0.134
0
a.lOK
0.01:199
0.0736
0.0602
100
120
140
0.0572
0.0950
0.118
0.131
0.136
0.135
0.131
24(
0.0493
0.0404
0.0331
0.0271
0.0222
0.0181
260
280
0.0149
0·9122
JUU
--0:00.)(:\
0.00996
0.0082
0.0566
0.0067
0.0499
160
lsu
2()(
I
22!
:320
340
I
0.125
O.lib
0.107
0.0982
0.0890
---o.()~Ol
0.0717
Cc
0
Co
(J.(J()1~
0~12
0.()()65
O.OLlY
O,t.J(J43
0.0081)
0.0205
0.0285
0.0366
0.0447
0.0523
O.Oij{)
0.O19()
u.u~95
U.0392
0.0442
0.041:12
0.0520
0
0.0244
0.029g
u.~~
O~Q§g
0.0724
0,r}}7,!
0.Q~~9
U~5~
O·tl,:s~
0.U1:I74
0.U914
O.()bU':/
0.0949
0.0632
0.0979 .-1~_. 0.0635'-·
020 •
o.ts
012
FAo -, FA +fAV =0
1,)
,..
CAO - 1.) CA - kl CA V ::::: 0
CAO -
CA ~
CAO -
CA (l -+- kl
CA
CA.O
:::::
kl
CA 1: ::; 0
---1__, __
(1 +
ki
t)
1:) ;;:: 0
FBO -
FB
+ fa V ::::: 0
·u CB + (kl CA " kl CH - k3 C:a) V::::: 0
,CHi-' kl C A t .. k2 Cg
1: -
k3 CB 1: == 0
Cg (1 + kz t + k3 "t) ;;;; kl t C A
CA =~a <.~+ k2.,'C+ k3 t)
kl t
6-74
~; __1_ z:< CB (1
+ \::2 1: + \::3 1:)
CM \::1 1:
CAo
1 + \::1
~=
\::11:
(l + kl "C)(1 + kl1: + kJ 1:)
CAQ
1:
t(s)
10
Ca/C,Ml
0.08
50
0.26
100
0.33
150
0.34
200
0.33
CDP6-C
First~ find the values for k.
kl :::: 298ge("·IO,~7/(OO19S7"100»
k1, ::;;: 9466e(-l5ll1(OOI9&"7-100}}
In this problem
= L2
::::
k J = 11127eC'1500(OOI9117°100l)
Isoburylene :::: I
Merhacrolein :::: ,M
.18
:::;; .22
CO,::::D
4
C0 ::::C
Oxygen:::: 0
The mole balances for these species are as follows:
F~o ... FI :::: ('I'll ,.- fu - IJr)V
F.'>i = riM V
Fo =r:m V
Fe = flO V
Oxygen is in excess so we will assume that F" ::: F 00'
The rate laws for these reactions are as follows:
. "IiI :::::--110 = riM = kLC1C
OO
-lil::;;:
25
(" ('""'00
::;;:41 rm:;;;: k 2"1
6-75
300
0.3
-··r = 4rJO = . ~.r. == k C C
-17 4 3C
3 1. 00
Combine all of these and come up with the following:
Cxo -C 1 = (kIC[C oo +. kZC1C oo + k 3C rC oo )'r
J!
Coo - Co = (k1C1C OO +~t-'k.tC!COO
+~}k3CIc()(,)'r
c,1.f == kIC/Cort'
CD =4 * kzC\Coo't
C c == 4'" k3CrCoo't
Before plugging into POL YMATH, evaluate the parameters and derive equations
for conversion.
C . == ...P. . : ;: : __........3__.__ : ;: : 0.034
1
RT .082 * 700
C/O == YAoCr .1 *' .034::;::: 0.0034
Ccq ::::; 0.031
=
X [ ::::: S;lQ:;:5':1
C[O
X
u
= .c
. . :.\':!
('
'[0
X ;;; _~::!L
21
4('
'\.o
X){::;::: XI ······X 11 ······X2!
Then plug into POL Y~'1ATH to get the following:
!n!ti~!-....y!!!.~~
Equa.t!~.:..
f (ci.) ""c:i·"ci.o+ (kl. -c:i. "c:oo+k2 ~ci. ·.:;:oo+1<:.J "ci "coo) ·cau
0 . OU1
f (cmJ =kl *ci. ~coo"c:au· . cm
'0,.001
f (cd) ='1 "k2 "'ci "coo'"::;au.··cd
f(cc)=4,"ci"coo"tau-cc
f {col =co···coo+- (lel. ·ci."c:oo+25/ 4 "k2*c:i"coo+17/2 "'k:3 "ci "'cool "cau
O.Q04.
0.004
0.01
c:i.o=.0034
kl.=l. . :2
coo=.03l.
}<2;; .. 1.8
k3== .22
t;.au=lO
xl.=cm/cio
x2"".2S"cd/ci.o
x=={cio'ci)/ci
x.3=x ··:x1 .... x2
6-76
valuE'
Cl
0.00227273
-1_775e-12
ern
0.Q0084-5"155
Cd
0.000507273
1" 0 11 e--12
-! 38:&-12
J.B3::'€' i:3
C.Cl028:818
D.CC3"'
1.2
O~C31
Coo
k2
0, L8
C.22
10
D.2"8663
Cl.03?29S5
C . 495
0.21 DC]?
x2
x
xJ
Xl;;; 0.496
X2! ::;::
0.037
X~l
::;:: 0.21
CDP6-D
A-··tB
C--3<[)+E
Fr :::: 10 Ibmollsec
dF
.,••• ~, :::::: fA
dV
;;;·k j C A
F.~::= CAV o
"" -.. -g:::?---.--.
C
10
:. Vo
0.73*900
= 13140
::: 7,.61 * l{r{i
V == l005ft'
See POLYMATH solution below
d{ fal Jd (VI z;ra
5
d (fe) /d(V) =rc
5
d (fb) /d {V) "'ra
d{fd) /d(V)",z:-c
o
o
d(fe)/d(V)::="rc
Q
kl"'lQ
k2"" .. OJ
6·77
ft=fa~·fc t·fb+fd+ fe
fao=5
cto=S/(900*.73)
fto==1.0
rc::;-k2
X'" (fao·~ f.a)
J fao
avo=fto/cto
ca=cto'" (fal ft)
ra=-kl*ca
y~~
:nJ:Sial"
V
0
fa.
fc
MinimU\~,,_~~b.~
F i~!Y£!h\±~
lOS
0
5
5
:2.49621
lOS
2.49621
5
5
1. 65
l.8S
fb
0
2.50379
0
2.50379
Ed
0
3.15
0
fa
0
3.15
0
3.15
3.15
kl
10
0.03
10
10
10
10
0 . 03
0.03
0.03
13 15
10
13.15
1<2
.......
""~
:ao
y~~ ~~i~~,,,.~~l lie
5
0 . 00761(}35
5
::;
5
cto
ft.o
0 . 00761035
O. 00761035
10
10
0.00761035
10
rc
··0 03
·0.03
-0.03
-0.03
x
avo
ca
0
0.500758
0
0.500758
1314
{) .. 00.3 8051.8
1314
131.4
1314
0.00330518
0.00144464
0OOl444€4
CDP6-E
a) Using the equation for the equilibdum constants:
We can come up with the equations for C A , CD, and ex.
CA =. ~~·f.q D .
KelCH
CD :::: .!!!~ACB.
Cc
ex = !;!.1.Cc~fL
Cy
The test can be found with stoichiometry.
Cy =Cx
CD:;:: RO _·CD,·Cy
e
Cc =cAO·······cA -ex
6-78
10
POLYlv1ATH
6--5
6--5
.-:1.
~ffildt.~.
Sotl".)':lon
nc 'cb)
f (ca) "'ca-··cc 'cdl
Var!abl~
_ _ ·_···N.' ____
E (cdl :cd-'K: ·ca.·cb/cc
t ~cx) -=cx"'K2""cc "cb/cy
.~_._
.•
_~"_.
ca
K2=1
c.y=cx
__
Ualue.•" ••.
.,,_,~_._
iO
_~
cx
Cl.CJ306182
0.109652
0.714386
Kl
4
cd
Kl=4
3.~J4ge-19
!.08 4 e-,t8
L 1 05e-' 16
K2
cbo=LS
c'iO~l
d
.. :;
cc=cao"'ca--c:x
Cl.J
0.714'386
cbo
1.5
1.5
};;"(ca.o-"c3.1 Jcao
CoO
cb"'coo-cd-,cy
cc
0,754995
0.979588
0.675961
x
cb
C,4, =0_0306
ex
=0,71
C e =0,68
C y =0.'11
We also fmd that X :::: 0.98
b) With the new equatjon we must fmd the new equilibrium equations.
ex =Cy _···Cz
C,A
= CAli ······CD·CZ
E:g;uat:i£s~ ;,
flc4)=c~-k3·ca·cx
10
f (c:yl :=cy--k2 ·Cc: .. c:b/c.x
cz
flcd)=cd-kL·ca-cbicc
eLJ
k;3",5
cd
cx
cx:;.r;:;:y~-cz
CdC
1<2:..1,,51.
ct:>Q
T
sdz
cao=l
S
cbo=,L 5
sdz;-;tcdicz
0 . 990725
t .07:384
CL 2681 0 1
0.0831 135
! .5
1.5
:300
U.27C6 i 1
1.08389
k3
c"
k2
49.4255
0241174
D . 480271
cb
0. ; 5806
ca=cao'-cd--cz.
k 1
cb:;:cbo-c:d-cy
x
cc
8.26888
0.839217
: . 17571
1
1.
<~
-r '.
6-79
......
~
1"" ......
'''':;J:;:'
! 09212--13
Lt"''1e-12
2 .s07c'-15
Wefmd that
Cz =0.65
C y =0.91
C a =0.22
C/o. = 0.48
ex =0.27
CD =0.37
C c =0.76
X =0.68
Sex =2.8
SDZ
Syz =1.41
=057
c) When the temperature is raised from 300 K to 500 K, Sex goes down,
SDZ
goes
up and SyZ goes up.
d) First find the proportionality constants fwm the A. rrhenius equation.
0.002 :::::
A~
A1e-n()Q(X)/1 911YHX})
== 38603
O.. 06 -
A
'''"\l
e
-(1(,(,'00/1.987' ;(0)
A2 == 2..24 x IOu
0.3 ::::: AJe,Pt)()OO/1 987".1(0)
AJ
::::
2.16 X 10 21
Once those are known, come up with equations for the equations in terms of
space-time 'to
.w . -. . . . .C
A . .,..._
'r: :::: ..............,C
. . ._
.. _
..
kj + k 2C,;, +- k3C~
rkl + rkJCA t· ti:~C~ - C1{)
+ C,4 :::: 0
r'·· . . ' '· . ''''''', . . ;...
.
.... "." ............,.....................
··. ··(;:/{,+1}+
I(rk, +1)-·4(rkJ )(rk·C
)
. ,·. ··. . ·\J
'.
1
AO
. __···. . . _-..·.
(.'~,\ = _ . . . ··_.. _
;;·;..i··. ·. . ·. . . _. ·. . . ·. _. -.. . . . . . . . . . . . . . . --.-.
- \.. 3
··c
I"= .,..... ZL
·. --k 1
-.. en
r == ...k1CA
'i
=--. --c
. <~;
··k>C;
ex ;:; kl r
CiJ :::: k:.1.CA
Cy = k,'t't:1
Use EXCEL solver in order to find the temperature that maximizes Ca
T
Ca
312.5336841
0.016513937
k1
0,Q03919233
Gx
0.039192325
k2
Cb
0.230891512
k3
2.260443602
0.038129279
Cy
0 . . 006164458
CDP6-F (a)
Mole balances:
6-80
dFA/dV = -rD-IU
dFn/dV =rD
dFu/dV =IU
Rate laws:
= klC~
ru = k 2 CA
kl = 15 fellbmol.s
rD
k2 = 0.015
S·I
Stoichiometry: C A = CTo(FA/FT),
F1 = FA + FD + Fu
Cost = 60FB - 15Fc -lOFAo
Using these equations in polymath to find the necessary volume to maximize Cost:
v = 1425 fe
POLYMATH Results
Calculated values of. the DEQ variables
Variable initial value minimal value
V
0
0
fu
fa
0
0.06705
0
0.0018652
fd
0
kl
k2
faa
ft
cao
ca
Cost
rd
ru
0
15
0.015
0.06705
0.06705
0.00447
0.00447
-0.6705
2.997E-04
6.705E-05
15
0.015
0.06705
0.06705
0.00447
1.243E-04
-0.6705
2.319E-07
1.865E-06
maximal value
2000
0.0237312
0.06705
0.0414537
15
0.,015
0.06705
0.06705
0.00447
0.00447
1. 46684
2.997E-04
6.705E--05
final value
2000
0.0237312
0.0018652
0.0414537
15
0.01.5
0.06705
0.06705
0.00447
1..243E-04
1. 4607519
i.319E-07
1.865E-06
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(fu)/d(V) = ru
[ 2] d(fa)/d(V) = -ru-rd
[ 3] d(fd)/d(V) = rd
Explicit equations as entered by the user
[1] k1 = 15
[2] k2=0 . 015
[ 3] fao = 0 . 06705
[4] ft = fu+fa+fd
[5] cao = 0 . 00447
[6] ca = cao*(fa/ft)
[7] Cost = (60*fd-15*fu-1 O*fao)
[8] rd = k1 *(caI\2)
[ 9] ru = k2*ca
Independent variable
variable name: V
initial value: 0
final value: 2000
CDP6-F (b)
Mole balances:
FA =FAo + IAV
FD =rDV
Rate laws: fA = -ID - IU
ki
Fu = IUV
rD = klC~
= kio exp ( ~ )(~-
ru = k 2 CA
91;.67)
F; = VOCi
Cost = v o(60FB -- 15Fc ·,lOFAo )
Use these equations in Polymath and vary T from 860 0 R to 1160 0 R and find maximum value of cost.
6--81
Cost steadily rises with temperature and reaches a maximum at 970 0 R(51O.33 0 F).
pOLYMATH Results
NLES Solution
Variable
ca
cd
cu
cao
E1
R
T
k20
E2
k10
vo
V
tau
cost
k1
k2
rd
ru
ra
Value
0.0023213
0.0016225
5.262E-04
0.00447
10000
1.987
970
0.015
2.OE+04
15
15
400
26.666667
0.6713603
11.292187
0.0085009
6.085E-05
1.973E-05
-8.058E-05
f(x)
5.354E-08
-5.354E-08
-2.76E-15
Ini Guess
0,,00447
o
o
NLES Report (safenewt)
Nonlinear equations
[1] f(ca) = ca-cao-ra*tau = 0
[2] f(cd) = cd-rd*tau = 0
[3] f(cu) = cu··ru*tau = 0
Explicit equations
[1] cao = 0.00447
[2] E1 = 10000
[3] R = 1..987
[4] T= 970
[5] k20 = 0,,015
[6] E2 = 20000
[7 J k10 = 15
[8J vo=15
[9J V = 400
[ 10 J
[11]
[12]
[13]
tau = Vivo
cost = vo*(60*cd-15*cu-1 O*cao)
k1 = k1o*exp«E1/R)*(11T-1/919.67))
k2 = k2o*exp«E2/R)*(11T-1/919,,67))
[14] rd = k1*(caA 2)
[15] ru = k2*ca
[16] ra = -k1 *(caA 2)-k2*ca
-----_._---------------------CDP6-G3
6-82
Mole Balances:
dC,
at
:= f\
de
....: ..,.::;: I'f
dt
dt
Rate Laws:
C
=.wk,C.
.". ,. . C~
..:.,.-~.:.."
It·KAC~
Stoichieme try :
-4 k,meJ)'rlurn''
C~'Eo == (';'Y
41
:=)...
0 mo1Id
. m3
from Henry's Law:
C ~,' ';;:;; 5.9 MPa(O.058 kmol/m J . MPa)= 0,3422 mol/dm!
Use these equations in the following POLYMATH program to generate graphs of CA' C,,_
and C F as a fUIlction of time
eL'
d(ea) /d(e) "":co.
o
d(cb) .ld(e) "'1:0
540
o
o
o
di,ec) Id(:.:.) "':::c
d(ce)/d(t)=.:;e
d(c:::) /d(e)=;::
kl",O . 000468
Ka=:22 . 76
k2",O . 000227
k4=O.00147
k3=;O .00282
rl:w*kl*ca-cb/(l.Xa-ca)
r2=w*k2*ca*cc/ll+Ka*ca)
:4=w~k4·ca·ce!(1+Ka·cal
ro"'· r1
20
6-83
3422
20
ll.3422
o
ca
o
o 3422
9 l.B51'7e-20
20
9 1851-7",--20
c~
540
540
539 658
SJ9 <iSS
cc
o
0,311641
o
0.205615
ce
o
II 136398
o
cf
o
6 .. 37928e-05
o
0116398
6 ,.3 7926e-·05
10
10
0 .. 000468
10
10
kl
0,000468
o
Ita
22,76
k2
22.76
0.000227
0.000468
22.76
0.000227
0 . 00022'1
0.000227
k4
0 .. 001.4.7
O.00H'7
0.00147
0.00147
k3
0.00282
0.00282
0.00282
0.00282
o
o
0.00580002
2.319S1e,,19
2,31981",-l.9
o
o
1 .. 84167e--22
o
r3
r1
o
000468
00873821
0.0984025
0984025
22,,76
1::'2
o
1. 621e-·05
::'4
o
4.96363e .. 06
rb
··0 .. 0984025
"2 .. 31981e-1.9
.. 0 09811025
-231981.", .. 19
re
o
a
000878695
0.00560002
.. 2.32208e .. 19
n
o
··n .091H025
0 .. 09811025
·'0.,00834598
-0.00580002
rf
2 .. 07778e··05
0,0984025
:::a
0.0984025
~.
-;:
~.~.
:.n:
ca
ec
eEl
Ca, Ce. Ce vs
t
;:, ;,,::
:l.,lea
cf vs
! /""
t /
'rj
t
j
/
/
1.iL/..
• •• ~ .... _.~._ ••• ~ ••• B ...... _ _ _ _ ............ .
~.
ago::
2,27051e··22
·2,32208e·l'}
Cb vs
t
::Lt-:
cf
4.2S838e·.;n
B ::lC::
CDP6-H
6-84
t
The equilibrium constant for the reaction
+::.
trOL
cs--OL
can be csrimaccd as a function of tcmpcratuIe from (he mole fraccion data below t(X)0c.
_ Yes
K --_.
P
Ya:
T
323
375
Y:t
.8
. 75
Yc'!;
.2
. 25
Kp
.,-)
,
33
In Kp
1
lCooff
3.09
-, -
39 ··1 11
2.63
}
__ ~39J::J:.!.n.x 1000 -:= -683 "" ~~it
3_09 - 268
R
6.H '" . . I 3 :51 _.S£l
rool
CDP6-I 2,9-17
CDP6-J
CDP6-K 2, 9-13
CDP6-L
Given: liquid feed
to CSTR
L1tttid Peed.
CAll ... 0.4 tm.clJI
Cl'O ...
0 ..4
[ttlOlil
V"", 1201
6-85
With the following re:lCcon s.equence
)::1
A --~ C
.. rl
k2
:;;:
k! C A
)::1 :::
0 . 0 t min- 1
=- 0.02 min- t
A·--,1- B
k3
B -..,. C
-£'z ;: )::2 C A
)::2
·r3 ::: );:3 CB
k3 = 0.07 min-I
F -..,. B + D i
-r4 ::: k... C~
k.. ::: 0 ..50 Vgmol - min
k."
1<2 k3
(a) Since C is an ~ product. formed through intermediate B by either A -...,r B -7 Cor
k.".
1::]
~l
F -...,r B + 0 i -.;. c. or din:ctJ.y from A ~ C. the m.aximum concenmuion of C occurs
when all A and F have b:.:en convened to C:
(b)
Cm.u;: CAO + CFOr (with E == 0).
Both A and F an: only decomposed by the above scheme:
Bahnee on A: ~ CAO· 'U() CA",:(.rA) V;: (-rl • fZ) V ={kl + k~d c" V
CAO:.C~
CA
;: (kl + kl)·Y.. "" {kt + kt} 1:
Uo
C~ __ ==
or C" ;:
1 ... (k! :
)::2) 't
0.4 gmol!!
= __0.4 gmol/l
1 + (0.01 + 0.02) min~l (21J~') t -+- (0.03) (60)
C,,;: 0.143 gmat /1
Balance on F: \10 Cr.J· Uo C;; ;: (-fF) Y ;: ~ C~ V
.
V.:.r. .
wnen::
1: :::--::; \N" rrun
Uo
Solving we get: CF = O.lgmoVI
Balance forB:
O-voCB
= (-rB)V = (k3CB -k 2 C A +k4 C;)V
(k 2 C A + k 4 C;)T (0.01min"lxO.143gmo1/l + 0.51.mor l min-'l 0.lgmo1 /l2)60min
CB =
1 + k3T
-=
1 + 0.07min-l 60 min
C B = 0.0907 gmo1 / 1
= (--r~.)V = (-k3CB -klCA)V
= (k3C B + klCA)T = (O.07min- l xO.0907gmoll1 +O.01min-1 xO.143gmo11l)60min
Balance for C:
Cc
O-voCe
C c = 0.467 gmol/ I
=
Mole fraction of C =
CA+CB+CC+C F
0.467
0.143 + 0.09 + 0.467 + 0.1
6-86
= O.~~z. = 0.583
0.8
CDP6-M
4A + 5B -:7 4C + 6D
-ljA
= k1ACAC;
2A +1.5B -:7 E + 3D
-rZA
= kZACACB
2C + B -:7 2F
-r3B
=k3BC~CB
4A + 6C -:7 5E + 6D
-Y4C
= k 4C CCC1 13
Rate laws:
=fIA + f2A + (2/3)f4C
-fC = -fIA + 2f3B + f4C
-fB
-fA
-fE
= -O.Sf2A -
=
1.2SfIA + O.7Sf2A +f3B
= -l.SfIA -fp = -2f3B
-fD
(S/6)f4C
Using these equations inpolymath to find the exiting molar flow rates.
POLYMATH Results
NLES &.lutioll
NLES Re,Qort {safenewtl
Value
Variable
Nonlinear equations
0.998927
ca
[ 1 ) f(ca) ::::; vo*ca-fao-ra*W ::::; 0
[ 2)
[3)
[4)
[5)
[6)
f(cb) ::::; vo*cb-fbo-rb*W ::::; 0
f(cc) ::::; vo*cc-rc*W ::::; 0
f(cd) ::::; vo*cd-rd*W ::::; 0
f(ce) ::::; vo*ce-re*W ::::; 0
f(cf) ::::; vo*cf-rf*W ::::; 0
Explicit equations
[1) vo::::; 10
fao::::; 10
[ 3) W::::;3
[4] fbo::::; 10
[5) rho::::; 00012
[6] k1 ::::; 5
[7] k2 ::::;2
[8) k3::::; 10
[9] k4::::;5
[10J fa::::; vo*ca
[ 11] fb::::; vo*cb
[12] fc::::; vo*cc
[1.3 J fd::::; vo*cd
[ 14] fe::::; vo*ce
[ 15] ff::::; vo*cf
[16] 1'1 ::::; rho*k1 *ca*(cb"2)
[17] r2 ::::; rho*k2*ca*cb
[18J r3::::; rho*k3*cb*(cc"2)
[19J r4::::; rho*k4*cc*(ca"(2/3»
[20] rf::::; 2*r3
[21] re ::::; O. 5*r2+(5/6)*r4
[22J ra ::::; -r1 +r2-(2/3)*r4
[23J rb ::::; -1 .25*r1-0. 75*r2-r3
[24] rc ::::; r1-2*r3-r4
[25] rd ::::; 1 . 5*r1 + 1 . 5*r2+r4
[2)
cb
cc
cd
ce
cf
vo
fao
0.997227
0.0017849
0.0037612
3.613E-04
1. 285E-12
10
10
W
3
fbo
rho
k1
k2
k3
k4
fa
fb
fc
fd
fe
ff
rl
r2
r3
r4
rf
re
ra
rb
rc
rd
10
0.0012
5
2
10
5
9.98927
9.9722697
0.0178487
0.0376119
0.0036129
1. 285E-ll
0.0059604
0.0023908
3.812E·-08
1.07E-05
7.625E-08
0.0012043
-0.0035767
-0.0092436
0.0059496
0.0125374
6-87
l.Sf2A - f4C
f(x)
-1. 05 6E-15
1. 08 2E-15
-1.422E-16
-1.041E-16
-1. 73 5E-18
1. 254E-13
Ini Guess
1
1
0
0
0
0
3
(b)
Y AE
FE
3.6x100336
=
=
= .
FAa -- FA 10-9.989
Y BF
=
7
FF
F Ba - FB
YAC
=
_
2.29xl010-9.973
= 8.25xl0-6
Fc
_ 0.0178 = 1.663
FAa - FA 10-9.989
CDP6-N 3, 6-21
CDP6-0 3, 6-25
6-88
Solutions for Chapter 7 - Reaction Mechanisms,
Pathways, Bioreactions and Bioreactors
P7-1 (a) Example 7-1
The graph of loll will remain same if CS 2 concentration changes. If concentration of M increases the slope
of line will decrease .
P7 -1 (b) Example 7-2
For t = 0 to t = 035 sec, PSSH is not valid as steady state not reached .
And at low temperature PSSH results show greatest disparity.
See Polymath program P7-1-h.pol.
POL YMA TH Results
Calculated values of the DEQ variables
Variable
t
C1
C2
C6
C4
C7
C3
C5
C8
CP5
CP1
k5
T
k1
k2
k4
k3
initial value
minimal value
0
0.1
0
0
0
0
0
0
0
0
o. 1
3.98E+09
1000
0 . 0014964
2.283E+06
9 . 53E+08
5.71E+04
0
2 . 109E-04
0
0
0
0
0
0
0
0
2.166E-04
3 . 98E+09
1000
0 . 0014964
2 . 283E+06
9.53E+08
5 . 71E+04
maximal value
final value
12
12
2 . 109E-04
L311E-09
3.602E-09
1.276E·-08
0 . 0979179
0.0012475
0 . 0979179
6.237E-04
0 . 0979123
2 . 166E-04
3 . 98E+09
1000
0 . 0014964
2 . 283E+06
9.53E+08
5 . 71E+04
o. 1
L311E-09
3.602K-09
2 . 665E-07
0.0979179
0 . 0012475
0.0979179
6 . 237E-04
0 . 0979123
0.1
3 . 98E+09
1000
0.0014964
2 . 283E+06
9.53E+08
5.71E+04
ODE Report (STIFF)
.t.Oe-9 , . . - - - - - - - - - - . - - . - - - ,
Differential equations as entered by the user
[1 J d(C1 )/d(t) = -k1 *C1-k2*C1 *C2-k4*C1 *C6
[2] d(C2)/d(t) = 2*k1 *C1-k2*C1 *C2
[3] d(C6)/d(t) = k3*C4-k4*C6*C1
[4] d(C4)/d(t)
k2*C1 *C2-k3*C4+k4*C6*C1··
k5*C41\2
[5] d(C7)/d(t) k4*C1 *C6
[6] d(C3)/d(t) = k2*C1 *C2
[7] d(C5)/d(t) = k3*C4
[8] d(C8)/d(t) Q..5*k5*C41\2
[9] d(CP5)/d(t) = k3*(2*k1/k5)A05*CP1I\O.5
[ 101 d(CP1 )/d(t) = -k1 *CP1-2*k1 *CP1(k3*(2*k1 /k5)1\Q . 5)*(CP 11\Q . 5)
32e-9
=
~
2Ae-9
=
16e-9
•
(''l
..
('6
=
S.lle-HI
OOe+O
P7-1 (C) Example 7-3
The inhibitor sho'NS competitive inhibition .
See Polymath program PT··l·e.poL
7-1
"'---~------
0.0
24
·tS t
72
9.6
120
0.40..--------------..
0.32
- I'_inhibitor
0. 08
r
-
_.-
--'
O.OOL------~--~--------l
S.Oe-3
7.0e-~~llr~ge-3
6"Oe-3
9.0e-3
LOe-2
P7-1 (d) Example 7-4
l)Now Cure. = 0.001mol/dm3 and t = 10 min = 600 sec.
t
= '~ln(_I_)+.CureaX
VMAX
I-X
_
600 sec _
VMAX
3
3
0.0266mol / dm
1n ( -1-) + --L
(O.OOlmol
/ dm
)X
_ ____
_..L-_
3
3
0.000266mol / s.dm
1- X
0.000266mol / s.dm
Solving, we get X =0,,9974.
2) For CSTR,
r = t = 461.7 sec
CureaX
T=----_... r
.- rurea
-
urea
= CureaX
r~rea
VMAXCurea
=K
C
M + urea
T
0.000266mol / s.dm 3 xO.lmol
/ dm 3 (1- X) -_ -(O.lmol
/ dm )X
.
-,---._-3
~
O.0266mol/ dm 3 + O.lmol/ dm 3 (1- X)
Solving, we get X = 0.,675
See Polymath program PT, I·e.pol.
POLYMA TH Results
NLE Solution
Variable
X
Value
0.6751896
fIx)
-8.062E-10
Ini Guess
0.5
NLE Report (safenewt)
Nonlinear equations
[1]
f(X)
=O,,0000266*(1-X)/(O.0266+0.1 *(1-X))-O" 1*Xl461. 7 = 0
3) ForPFR,
T=
CO
dC
Cum
-
f
urea
r
and
Curea
7·2
= CureaO (1- X)
461.7 sec
=>
Curea X
= -k M- In ( -1-) + --"--""--
r
VMAX
VMAX
1- X
x=o ..8
Same as batch reactor, but t replaced by r
P7-1 (e) Example 7-5
Y
= - t,.cs = - (238.7 - 245) = 2 18 /
SIP
5.03-2.14
J'j.C p
. g g
1
1
YplS =--=--=0.46g/g
YSIP 2.18
1
YSIC +P = y - C+PIS
1
0.075+0.46
= 1.87 g / g
Yes there is disparity as substrate is also used in maintenance .
P7-1 (0 Example 7-6
1) if we go for 24 hIS, fermentation will stop at 13..2 hIS as Cp
= Cp '
.
See Polymath program PT-l-f-1.pol.
POLY'!yIATH Results
Calculated values of the DEQ variables
Variable
.
--t
Cc
Cs
Cp
r'd
Ysc
Ypc
Ks
m
urnax
rBm
kobs
rg
initial value
minimal value
maximal value
final value
o
o
13 . 2
13 . 2
1
1
250
39 . 292786
0 . 01
12.5
5.6
1.7
0.03
0.33
0.03
0.33
0.3277712
0.01
12.5
5.6
1.7
0 . 03
0.33
0.03
0 . 0039386
0 . 0624825
16 . 558613
250
92 . 981376
0.16559
12.5
5.6
1.7
0 . 03
0.33
0 . 4967701
0 . 33
2.1455962
16.550651
39 . 292786
92.981376
0 . 1655065
12.5
5.6
1.7
0.03
0 . 33
0.4965195
0.0039386
0.0624825
o
o
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-rd
[2] d(Cs)/d(t) = Ysc*(-rg)-rsm
[3] d(Cp)/d(t) = rg*Ypc
Explicit equations as entered by the user
[1] rd =Cc*0 . 01
[2] Ysc = 1/0.08
[3] Ypc = 5 . 6
[4] Ks = 1 . 7
[5] m =0 .. 03
[6] umax = 0 . 33
[7] rsm = m*Cc
[8] kobs = (umax*(1-Cp/93)"O.52)
[9] rg =kobs*Cc*Cs/(Ks+Cs)
7-3
2)Semi-Batch reactor:
See Polymath program P7-1··f-2.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Cc
Cs
Cp
rd
Ysc
Ypc
Ks
m
umax
rsm
kobs
rg
Cso
vo
Vo
V
initial value
0
1.0E-04
1.. OE-04
0
1. OE-06
12.5
5.6
1.7
0.03
0.33
3.0E-06
0 . 33
1.941E-09
5
0.5
1
1
minimal value
0
1.0E-04
1. OE-04
0
1. OE-06
12 . 5
5. 6
1.7
0.03
0.33
3.0E-06
0.3294925
1.941E-09
5
0.5
1
1
maximal value
24
0.0474697
12.206266
0.2748298
4.747E-04
12.5
5.6
1.7
0.03
0.33
0.0014241
0.33
0 . 0137289
5
0.5
1
13
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-rd
[2] d(Cs)/d(t) = vo*CsoN + Ysc*(-rg)-rsm
[3] d(Cp)/d(t) = rg*Ypc
Explicit equations as entered by the user
[1] rd = Cc*0 . 01
[2] Ysc = 1/0 . 08
[3] Ypc = 5 . 6
[4] Ks = 1.,7
[5] m 0 . 03
[6] umax 0.33
[7] rsm =m*Cc
[8] kobs = (umax*(1-Cp/93)1\().52)
(9) rg =kobs*Cc*Cs/(Ks+Cs)
r10] Gso = 5
[11] vo = 0 . 5
[12] Vo = 1
[13] V = Vo+vo*t
=
=
7-4
final value
24
0.0474697
12.206266
0.2748298
4.747E-04
12.5
5.6
1.7
0.03
0.33
0.0014241
0.3294925
0.0137289
5
0.5
1
13
0,30
15
0,24
12
018
9
~
0.12
6
- C)
0.06
(100
0.0
3
4.8
96
t
14.4
19,2
0
140
0,,0
4.8
9.6 t
144
19.2
24.0
3)Changes from part(2)
See Polymath program PT-l·f·3.po1.
POL YMA TH Results
Calculated values of the DEQ variables
Variable
-t
Cc
Cs
Cp
rd
Ysc
Ypc
Ks
m
umax
rsm
kobs
Ki
Cso
vo
Vo
V
rg
initial value
minimal value
maximal value
final value
0
1 . 0E-04
1.0E-04
0
1.0E-06
12 . 5
5. 6
1.7
0.03
0.33
3 . 0E-06
0.33
0. 7
5
0.5
1
1
1,941E-09
0
1. OE-04
1.. OE-04
0
1.0E-06
12,5
5.6
1.7
0.03
0.33
3,OE-06
0,,3299991
0. 7
5
0.5
1
1
1,941E-09
24
1.. 514E-04
12 . 823709
4.669E-04
1.. 514E-06
12.5
5 "6
1..7
0.03
0,,33
4.541E-06
0 . 33
0.7
5
0.5
1
13
8 . 215E-06
24
1.514E-04
12.823709
4.669E-04
1 .. 514E-06
12.5
5.6
1..7
0.03
0.33
4.541E-·06
0.3299991
0.7
5
0. 5
1
13
2 . 568E-06
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-rd
[2] d(Cs)/d(t) = vo*CsoN + Ysc*(-rg)-rsm
[3] d(Cp )/d(t) = rg*Ypc
15
Explicit equations as entered by the user
[1] rd = Cc*Q.,o.1
(2] Ysc=1/o.,o.8
[ 3] Ypc = 5,,6
[4] Ks = 1 . 7
[5] m=o."o.3
[6] umax = 0.,,33
[ '7 1 rsm = m*Cc
9
12
6
3
0
7-5
~---~----~----~----~----~
0,0
48
9,,6 t
14 4
19,2
240
[8] kobs:;:;: (umax*(1-Cp/93)J\().52)
[9] Ki 0.7
[10) CSO 5
[11) vo = 0.5
[12] Vo=1
=
=
[13] V = VO+VO*t
[14] rg kobs*Cc*Cs/(Ks+Cs+CSA2/Ki)
=
1.00e-4
O.Oe+O
9.6 t
0.0
14.4
19.2
IL-_~_~
0.0
24.0
4. 8
4)After 9.67 rus, C s = O.
300,--------·
240
180
c~c
-
c.s
120
C)
60
oL-=:::::;;;;;;;~====::::::::::=U
o
P7 -1
P7-1
P7-1
P7 -1
P7-1
P7-1
P7-1
P7 -1
P7-2
2
4
t
6
8
10
(g) Individualized solution
(h) Individualized solution
(i) Individualized solution
(j) Individualized solution
(k) Individualized solution
(I) Individualized solution
(m) Individualized solution
(n) Individualized solution
Solution is in the decoding algorithm given with the modules.
7-6
_ _ _ _~
9.6 t
14.4
19.2
24 . 0
P7-3
Burning:
GO' '~OH'
y~
-+i,)H·
then
-reI.
= 0 =ks(HCl)(H) -- k6(CI)(H)
7-7
Cl _ k} (Hell
• k6
.+kj (o!};. 3k4 (H·) (O:d' kz (B- }(02l' 2ks (HCl) (H-)
fH =:
.:_ .
r'-------.. . . . . .,,----·,---··,'''--.. ,,·,,·----lr,I_!,:_:, 4k ~~=~ ),.~._~,~,~ ~.~.) _~.~_{~.1~: l.~ .( f:.:.U
ConstamVoiurne
e'.btdCH··
-' (""H *b*"hle -ae··hl
dt
Using the integraring f:lctor
d(e-!:l!C )
---"........-..H,-,-:::: a e-,be
dt
C£+
t ::
0
CH
=: ..
:::;
'~' +. KI eot
therefore K 1 == ~
0
,--'_.,,-:' ..
....
' ' -]
tc~~ :. t ~~~.~~l~
'
'''
> ks {HCI)
) then "b" is positive
> kd02}
, then"b" is negative
If
k4 (C'}z)
If
kj (HCt)
,
..".
,,,e"
1/
t .........". "" . ,."."_..".....,,'-.. _ .. , ,............, '"''''''
7-8
=·a··p(C\-{.)·
p(CH )= 3k,! (02 '[:f{
= CCJo
3k4 (OJ
a::::: 4ls (0 2 )
p:::::
t "'"
····..
0
Ceo
·····~··-·-··························-··········~··························-·-······
r,..
~~
i~·..CO :::; Cc.O-o -
.1
at·· p _aLt·
..; c" .... pi'1
._.~_~.__ . . . . . ._._J
L. . __.. .
Reaction Pathway with HGL
Reaction Pathway 'lifo HCL
[• . J. . . . . . . . . . . . . . . . . •
~)~
[~;~~
1-_~_ . 1
J:1
[1_ . ]
I2:,
-lH-
-
'
..
[ OR
--T
.I
[c~J [J~J
····I ]
l CO
P7 -3 (d) Individualized solution
P7-3 (e) Individualized solution
P7-4
The re::!crion sequence is
I"l
= k!
CH3CBO-.~···>CHJ --teBO .
!2
= kz CAt: COh-
CHI·+CH,CHO-~CH3· +-CO+CH4
I)
= kl
\4
= k4 ('.J;.,
· ..... 1J"
C'HO·· +(JIi~'HO---·~·-~ CIl3 -+2CO + H2
7CH
..···_ .. k ,·.·--4C
H
~'3
' "2 '6
7-9
CAe
eCHO- CAe
,2:
=CAC (k !... k'2 (~.. G~;· + ,KJ( ,-CHO.i
Active imc:n:ned.iates:
(1)
Cfh·, CHO·
·-rc:h;:;:; -r! .... r2 .. r2" r3 +l..r ::: 0
2 4 1
"rCHo- :::: -rl
:=·k1C,4C + k2CACC(11)
-
== ..,kle"AC +- k2 CAC('"CH,
"•. "
+ 1'3
:::: .. k!
k3C CHO C,\C .,' k;.CACCCHl
k ('"C'flO ('~'-'{C+ 1 k4 (,,2
'eH,'
CAe + k3 eCHO. CAe;::: 0
2
+2' k C
'4
2
CH}
(2)
(3)
Substituting (3) in (2) gives:
C2Hs
P7-4 (d) Individualized solution
P7-4 (e) Individualized solution
P7-5 (a)
7-10
Gas phase reaction
with third order kinetics and an apparent negative activation energy . Consider the
following mechanism, in which NO} is an active intermediate.
fj "'" kll NOJ[02J -k 1[N03 ]
==kzINOlfNOJ]
T2
"" k![NO][ O2 ] k 1!N03 ]· ·ls[N0J[NO,]
[N03 ] =5I~<.?le~1
kl ·+~[NO]
"T
NO
== . . k iNO][O,] + (k
1
k
~l..2
.
[NOJ)
~~P,'V.~(?l(?:J
l k [NO]
k
1
1
== ... ~!5I~~)tL?2 ]::~5~j\.foy 19.21
"_I
+.s[NO]
_ 2k1k2 [NOY[Oz]
-"--k~;·-t-iSfi~iorkl »k2 [NO)
. ····r
No
2k1k,
= _.- . . ·k·
··-"-[NO]41]
.()z
·1
For the overall activation energy to be negative,
···(E1 + EJt
=::!;>
E~l == EDt, < 0
E -<l(E+E)
iZ
A.s long as all energies are positive values.
P7-5 (b)
k3
Cl+CO~COCl
ks
COCl + Cl z -7 COCl z + Cl
7-11
= 0 = k3 (Cl)( CO)-k4 (COCI)-ks (COCI)( CIJ
r COCt
(COC!) = k3 (Cl)( CO)
k4 + k5 (Cl2 )
rcoCt2
r ct
= ks ( COCI ) ( Cl2) =
klk3 ( CO) ( Cl) ( Cl 2)
()
k4 + ks Cl2
=0 = kl (C1 2)-k2 (CI)2 -k3 (Cl)( CO)+ k4 (COCI) + ks (COCI)( C12)
add I'COCI to rCI
r Ct
+ r eoct = 0 + 0 = kl ( C12) - k2 ( Cl)2
( Cl) = ~ ( Cl2 )
k2
2
~(CIJ
(Cl)=
k2
:1 (C12t (CI2)
5
klk3 (CO)
rCOCt2
=
k4 + ~s ( Cl )
2
kl k3
:12 (CO)( CI2)%
.- --k-=4'--+-k-(·-C-1-)s 2
k4 » k5 (Cl 2 )
2k
_ k1 3 ( CO )( Cl )~2
2
k2k4
P7-5 (C) Individualized solution
rCOCt2 - - -
P7 -5 (d) Individualized solution
P7-6 (a)
K[
N02 + hv
-~
NO + 0
K2
O2 +O+M -7 0 3 +M
K3
0 3 +NO
-~
N02 + O2
UsingPSSH,
P7-6 (b)
7--12
0"..... M and O} aonear
in [he denoI11inacot 2c,;ve
~r~·e"";~s
..
...
....
~
,,~
~Jt"
~.....
Sl,,,,,,,,.r(M1
(') '
".;:,.;:::,'-';' ..... .....;"
...:>
~
fl..pplying rule '3 of t2.b!e 71 to 0::. and 0,:
0]
-i-
(h';-
0··.,. 202
0;.0 3
0 3 and M appear in the numerator. Applying rule 1 of Table 7·1 to 0;: 0,.-7 0;: + 0
If the second and third equations are combined, and M is added to each side of the
equation:
A mechanism is proposed which satisfies all the rules of thumb:
ro,::::' r l t r2
frJ
[01
'i '''2'
:;:o_
..
kz [()JO 1(M] k; [0 ][03 ]::::' 0
0 ·][M j'-k;,lo '10, ]= 0
rJ ::::. k, [lrf][Ol]···· k2 [0 2 1
rJ
::::. .•
kl [lvll[C?11
j.
k,[MJ[~~L._
kz[OJM]+'k;,[OJ
P7 -6 (d) Individualized solution
P7 -6 (e) Individualized solution
P7-7(a)
7-13
LOW TEMPERATURES • NO ANTIOXIDAm'
~] = ri
c(~
==
c(R~-]
= 2ko(I2] - ki[I-J[RH1
(A)
-~(I-][RH] - kpl[ROz-J[RH]
(B)
= lep, [R-] [Oil - IcI'1[ROze][RH] - k, (ROz-P
(C)
c:{R-]
Cit = ki(RHJ(I-] - kp, [R-] [021 + kpl[ROz-][RH]
dRADlCALS]
PSSA. ...
(A)
=0
(D)
= 0
dt
2ko[h]
rj
= 0 [I-] = ki [RH]
c(!.J = 0 = ki [RH][I-] - kpdR-J[021 + kpl(R<h-][RH]
i
c(~~-] = 0 = kp: [R-][Ch] - kp%[R02-][RH] - kt [R02·j~
Substitute for [R·] :
2ko [Iti - k t [R~·F =
0
[ROrJ = _!lko [Ill
'V
kt
Now substitute the expressions for [he radicals into d[:~1 the expression for the degradation of
the oil.
c{RHJ = _k"[2ko[hJl(RH] _k [2ko[hn llZ[RHl
cIt
I ki [RH]J
P:z
kl J
.
J.
[1-]
= 2ko (I:zJ- [2ki~koJr[I:zJ1n [RH]
7-14
ODE to be solved- low temperarures~ no antioxidant:
_c{RH] =
cit
.
.,1,._ "I?1 + [2~1 [koJ
e.AQ
L -..
kt
llnf'lI ]112 [RHl
J
2
..
P7-7(b)
Low temperatures with anti-oxidant
<{~e]
= ki[RH](J.]-kp1 lR.][Oll+kp:(R02.][RH] (same)
dfAe]
dt
=
kAl [AH][ R02-] - kA 2[A·][ ROr]
c(RHJ
= _ k.l [1-1.[RH1kp.. [Ro,·] rRH]
dt
....
_~
ApplyPSSH:
d[I-]
-=0
cit
fIe] = 2ko[h]
..
kj [RH]
d(R·l = 0
----0:.
dt
d(
2
R
2
.]
== 0
2ko[h] - kt[ROr]2 .. 2(l<Al [AH][R~.])
:. - kt [R0:2·p- 2kAl [AH] (RO:z·] + 2ko [12] = {}
k,[Reh·P+ (2kAl [AH]) [ROr] - 2ko[h]
=0
[ROr] = • 2kAl [AH] ± ../(2kAl [AH])2 + 8kt ko [h]
2k t
Now (FINALLY!)~ let'S substitute into d(:~] :
7-15
Quadradtic in [Rehe ]
[R~.] MUST be positive
t{RH] = _ 2ko [12] _kp [RH] - 2kAl [AM] + ../{2k A1 [AHJ)2 + 8kt
dt
?k
t
1
leo (Ill
P7-7(c)
If the radicals are formed at a constant rate, then the differential equation for the concentration of the
radicals becomes:
d ~ e]
and
=ko _ ki [Ie] [RH] = 0
[Ie]=-~-o-
ki [RH]
The substitution in the differential equation for R" also changes. Now the equation is:
d
~~e] = ki [Ie][ RH] -kpi [Re][ 02]+ kP2 [R02e][ RH] = 0
.
..,
and solvmg and substltutmg gIves:
[Re] =
ko+kP2[R02e][RH]
[ ]
kpi 02
Now we have to look at the balance for R02
d[R0 e ] .
2
_. _ -2- = kpi [Re][02]--k p [R02e][RH]-kt [R02e] = 0
dt
and if we substitute in om expression for [R] we get
0= ko - kt [R02 e
[R02 e] =
t
which we can solve for [R02·]·
I~!L
Vkt
Now we are ready to look at the equation for the motor oiL
d [~~e] = -ki [Ie][RH]-k P2 [R02e][ RH]
and making the necessary substitutions, the rate law for the degradation of the motor oil is:
d[RHe]
~~
di= r = --ko - k
[RH]
p2VT
RH
P7-7(d)
With antioxidants
Without antioxidants
7-16
-----------------------------------------------
-- -- --------------------------
----- --------------
- ----- ---- - - -
-'I }
\1(".
it
-+R H
)I-
I
Ii·
H1.
P7 -7 (e) Individualized solution
P7 -7 (f) Individualized solution
P7 -7 (g) Individualized solution
P7-8 (a)
Given: Illness mechanism
H
-7
r
k:2
r + E-, 21
{Healthy person g<:rs
(healthy per;;on cormac:s
from ill oerson i
(Hl person getS v. cH)
I ···t D
(Ill person dies)
7-17
rD:::
k.t [lJ
Applying pseudo steady state hypothesis
l:O 1:
Ii = fl + f: ~
I:"} .•
ft=;:C 0
P7-8 (b)
Jt
b+k.&
(H] = - - the demh i.H.e becomes infinite .
k2
P7-8 (c)
It is enlightening
fH "'" -rl - f2
to
calculate rtf:
+ f3 =:·k 1 [H) ~ k2 [H] [IJ + kJ [fj ::: ·k! [H] . {kz [H] .. kJHI]
(!!4--:-- ;;: _
rH;;: --- -k t k.t
(1<3 + k.t ~ k2 tH)}
fl)
<0
This expression states rhat for every person who is ill ilnd dies. a healthy person becomes ill For
the population as a whole, "illness is but a 'way station' on [he road to 'death'." Note fur..her that
IR < 0, and therefore the population will die off evenmally_ Initially, the death rate will be slow.
until [H].,,. {k3 + ~} I k2. I11e model neglects the possibility of birth. In practice, it would appear
to be useful in describing epidemic-like diseases, which occur over a short time so that the birth
:rate can be neglected.
P7-8 (d)
See Polymath program P7-8-d.pol.
1.0\"+9
1. 00E9
fi"======:: ....
,
8. 0e+8
8.00E8
6,OH8
6.00E8
_.---...._--_.
CJ
- I
". n
4.0e+8
400E8
2,Oe+8 .
2.00E8
O"Oe+O
O.OOEO l_"""",===O. OOEO 2.00E9 4.00Et9 6.00E9 800E9 1.00EIO
--- -
-~~-.~.-~~.....
0.01'+0
2.0H9
4.01'+f
6.01'+9
8.0e+9
LOe+10
.....
Everyone becomes ill rather quickly, and the rate at which an ill person recovers to a healthy person is much
slower than the rate at which a healthy person becomes ilL Eventually everyone is ill and people stru:t dying.
7-18
P7 -8 (e) Individualized solution
P7 -8 (0 Individualized solution
P7-9 (a)
Starting with the design equation for a batch reactor
de
--.. -.£.
dt
== r
P
Tp=kC
c= ((~o -(~{l"""~i~J
de
.
r
C 1
·····d. . ·r·!:· =kfC so . . C ] 1--·£
•
1.:
r __ .~
1 4.
I,'
integrating
Finding k, l:~40
y '" 0 C02x + 1. ],697
1
.......
o
~
.. -
20
--~.-:-
"
60
40
tlme
------r'- ...... , .. .
Using the data at 40"F and 45"F the following graphs ll!~l!1:ade:
Finding k
o
o
'5
y :O.CG2x t· l.l697
o
20
40
60
80
time
L:::~.:.:.:::::::::::: :::::::::::::::T:.:::.:.::::: .......... _." . . '".........!.."... ~~::~~.~:::.-~.~.:.~:_'"_ .. ~.. ,~""~ ... ".. ,..
., ... _.. ,_~....... "" ........ ,.. _:...: .. ...::::=~~.:"_"
7-19
. . . . ..
k<t):;'::· 2.8*10-~
C.\():;.:: 1.1697
k~5:;':: 8.26*10'" C45 = 1.1883
The activation energy then is:
from these:
1n
(
C -1.5) ::;::7.14kt + C
.933..;..a.·~. ···--
C p ··-1.4
E = _Rln(k.2lkll :::: 108120 Bt~
lfT1, --'l/1~
mol
P7-9 (b)
k3&
:=
( -E
k4s1 0'" ~ 2.3R
(1"4-9;7 -'- 5041)) =.18'
104;
'I'
Using interpolation, C:::: 1.18086
Using the same equation used on the graphs we can solve for t when C p :;':: 1.1
t = 49 days.
P7-9 (c)
kn = k4s lO'"
(2'~fR (5·16' ,_. ':5'64)) = .0023
C::::: 1.211
t::::: 2 days
P7-9 (d)
The data appears that it may fit the Monod equation for substrate consumption at the stationary phase.
P7-9 (e) Individualized solution
P7-9 (I) Individualized solution
P7-10 (a)
E+S
kJ
)E-S
E-S
E·S-E..~P+S
rE • S =
P+S'-~E-S
kJ(E)( s) - k2 (E-S) -k3 (E-S) + k4 (p)(s)
since S is not consumed:
rE • S
k2 )E+S
Sy
= S + E-S
or
S = Sy - E-S
= kJ(E)[( ST) -( E-S)]-k2 (E-S) -k3 (E-S) + k4 (P)[( Sy ) -( E-S)]
(E-S) = kJ(E) (SrJ+ k4 (P)(ST)
kJ( E) + k2 + k3 + k4 ( p)
-rE =kJ(E)(S)-k 2(E-S)
_
( )()[
S 1-
-r -k E
E
J
7
kJ(E)+k4(P)] - kJ k2(E)(Sy)+k2k4(P)(ST)
kJ(E)+k2+k3+ k4(P)
kJ(E)+k2+k3+ k4(P)
7-20
P7-10 (b)
E+S~E-S
E-S
E-S
E-P
k3 )E-P
k2 )E+S
k4 )E-S
E-P~P+E
rEop = 0 = k3 (E-S)- k4 (E-P) - k5 (E-P)
(E-P)
= k3 (E-S)
k4 -+-ks
rEoS = ~ (E)(S)-k2 (E-S)-k3(E-S)+k4 (E-P)
since E is not consumed:
ET
= E + E-S + E-P or E = ET -
k3)
E = ET - E-S - - - E-S
( ~+~
E-S - E-P
. (1+-""----k3)
E = E[ - E-S
~+~
Insert this into the equation for rE s and solve for the concentration of the intermediate:
(E-S)
r
=
=k
kl (S)(Ey) _ __
k
(S)+k2 +k3 - k4 3
k4 + ks
k4 + ks
[ l+~-]kl
(E-P) = k3 k5(E-S)
k 4 +ks
s
P
klk3k5 (S)( E[ )
=
r
P
--.--=-.::......:::..~~::.-!.---
[k3 +k4 +k5]kl (S)+k2k4 +k3k5
P7-10 (C) No solution will be given
P7-10 (d)
k3
E-S1 -+- S2
p
k4
E-S1S 2
k5
E-S1S2 --7 P+ E
(1)
(2)
= 0 = kl (E)( SI) - k2 (E-S) - k3 (E-S)( S2) + k4 (E-S 1S 2)
rEoS,sz = 0 = k3 (E-S) (S2 )_. k4 (E-S 1S2 ) - k5 (E-S 1 S2)
rEoS,
If we add these two rates we get:
(3)
rEoS, + 'Eos,s2
= 0 = kl (E)( S) -
k2 (E-S) - k5 (E-S 1S2)
Plug this into equation 3 and we get:
7-21
P7-10 (e)
kl
E+Sp(E.S)
~
j
k3
(E.S)j p( E.S)2 + ~
k4
ks
(E.S)2 -7Pz +E
(1) ~E.S)1 =O=k j(E)(S)-k 2(E.S)j-k 3(E.S)j +k4(E.S)2(1n
(2) ~E.S)2 = 0 = k3 (E.S)j - k4 (E.S)2 (p)j - k5 (E.S)2
k3 (E.S)j
(EoS)2 4 (R)+k5
="
j
Add (1) and (2) to get ~E.S)1
+ ~E.S)2
= 0 = kj
(E)(S)-k2 (E.S)j -k5 (E.S)2
kj(E)(S)
7-22
k
2
kSk3
+ kS +k4(In
=
(EoS)
k/S(E)(S)
2
k2k4 (Pr) + k2 kS + kSk3
r =k (EoS) =
kr k3kS(E)(S)
2
P
S
2
k2k4 (Pr) + k2kS + kSk3
(Ey) =(E)+(EoS)r +(EoS)2
P7-10 (0
ko
k,
E+S~EoS-~P
k2
k4
E+P~EoP
k5
k6
EoS
+P~EoSoP
k,
ks
EoP+S~EoSoP
""
r=k(EoS)
P
3
(1)
~EoS)
(2)
~EoP) =O=k4(E)(P)--k s (EoP)-k s (Eop)(S)+k9(EoSoP)
(3)
~EoSoP) = 0 = k6 (EoS)( p) - k7 (EoSoP) + kg (Eop)( S) - k9 (EoSoP)
=0 =kr (E)(S) -
(2) + (3): '(EoP)
+ ~EoSoP)
k2 (EoS) - k3 (EoS) + k6 (EoS)( p) + k7 (EoSoP)
= 0 = k4 (E)( p) - ks (EoP) + k6 (EoS)( p) - k7 (EoSoP)
now add (1) to this:
7-23
~E.S) + ~E'P) + ~E.S.P) = 0 = kl (E)( S) - k2 (Ee S )- k3 (Ee S )+ k4 (E)( p) - k5 (Ee p )
(Ee ) = kl (E)(S)+k4 (E)(P)-k5(Ee p )
S
k2 +k3
(Ee P) = k4 (E)(P)+k9 (EaSa P)
k5 + kg (S)
(EeSe p ) = k6 (EeS)(p)+kg (Eep)(S)
k7 +k9
k4 (E)(P)+k9 (k6 (EeS)(p)+k g(Eep)(S)]
(Ee P) =
k7 + k9 ._--"-k5 + kg (S)
(Ee ) = k7k4 (E)(P)+k9k4 (E)(P)+k9k6 (EeS)(P)+k g(EeP)(S)
P
k5 + kg (S)
(Ee P ) = k7 k4(E1( p) +k9 k4(E)( p) +k9k6 (Ee S )( p).
k5
kl (E)( S) + k4 (E)( p) - k5 (k7 k4(E)( p) + k9 k4(E)(P) + k9 k6(EeS)( P)]
(EeS)=
_____
k5
k2 +k3
(EeS) = kl (E)(S)+k4 (E)(P)·-k 7k4(E)(~)+k9k4 (E)(P)
k2 +k3 -k6 k9(p)
rp = k3 (Ee S ) = k3 (E) kJ (S) + k4 (p) -. k7 k4(p) + k9~4 (p)
k2 + k3 - k6 k9(p)
(Er )=(E)+(EeS)+(EeP)+(EeSeP)
(E] ) = (E)+ kJ (E)(S)+k4 (E)( p)- k7 k4(E)( P)+k9 k4(E)(P)
k2 +k3 ·-k6k9(p)
k7 k4(E)( p) + k9 k4(E)( p) + k9 k6(kJ (E)(S) + k4 (E)( p) -_. k7 k4(E)( p) + k9 k4(E)( P)J( p)
+__
.
_ k2 +k3 --k6~9 (p)
.-'----~
+
k (kJ (E)( S) + k4 (E)(P) - k7 k4(E)( p) + k9k4 (E)( P)]
6
k2 +k3 -k6 k9(P) (p)
-
7-24
-
kJ +k9
all of the terms in the numerator have (E) in it and so the (E) can be factored out and an expression for (E)
in terms of (El)' (P), and (S) can be made and plugged back into the equation for rp.
P7-10 (g)
kl
k3
E+SpEeS~p
k2
k4
E+ppEep
k5
~E.S)
=O=kl(E)(S)-k2(Ee S )-k3(Ee S )
(Ee S) =5lS)(E)
k2 +k3
~E.P) =O=k4(E)(P)-k5(Ee P)
(Ee p )= k4(E)(P)
k5
rp = k3 (EeS) - k4 (E)( p) + k5 (Eep)
rp
= klk~ (~~( ~1 __ k4 (E)(P)+k4 (E)(P)
2
3
Isk3 (S)(E)
k2 +k3
(ET ) = (E) + (EeS) + (EeP)
rp
=--'--:"""":
(ET )=(E)+ kl(S)(E)+ k4(E)(P)
k2 + k3
k5
(ET )=(E)[1+_ k1 (S) + k4(P)]
k2 + k3
k5
Er )
r - - - - klk3 (S)(
-'--'--c-"-'--_ _
P - (k2 +k )[1+ kl. (S) + k4(P)]
3
k2 + k3
k5
_
k3 (S)( ET )
r
p
(k2 +k3 +(S)+ k4 (k z +k3 )(P))
kl
klk5
--------'-'--'--'--:"-'---~
7-25
P7 -10 (h) No solution will be given
P7-10 (i) No solution will be given
P7-10 (j) No solution will be given
P7-10 (k) No solution will be given
P7-11 (a)
The enzyme catalyzed reaction of the decomposition of hydrogen peroxide . For a batch reactor:
~ ~~s
;;
~.~~ = fS
""
·:Ky-m~x~~
·r LS
at t
til
=:
0, Cs "" CS o
Rearranging and integrating:
Cs .,..
•Km In -....
eS o
or J In
t
S"t)
'"
\-5
Cs· CSo
"" • V . t
max
.~Sil:~l+- ':' m1!~..
Km t
Km
A plot of L I CSt) . Cso - Co;;
.'
t n C
vs _. -i........c- should be lwear with s:ope ..•J ...
s
Km
CS!L:CS
t
10
.01775
1.1268
.01193
20
.0158
.01179
50
.0106
.0050
12654
1.8867
4.0000
.01386
100
OI27()
.00225
.0042
.0094
.000225
.00021
. 000188
.0150
.00015
00140
0.0135
IIn~Jl
t
Cs
0.0130
0.0125
00120
0.0115
15
16
1 7 18
19
20
21
22 23
(C{so) - C(S»)lt X 1 E5
7·26
24
25
·
013· 012
(
min .j
"~
F rom t h e gnarl, slope "" ...-~--. -'--.--.--, -,,.-------"1
-
(17.5 ,. 206) x 10 5 ,g mol I min; J
,
:::: ,.0310 g mot!l
Km
C
Cs
C
C
1 75 x 10-4
I·min
mol
Art In ,~. '" ..Q~l,. ~.Q, . ,,,.2 ::;: """''''''
t
rom
mu = (01"
V
5 64.
'K;
, .> +- ,,'
Vm.z:s.
_·
x
10"»
'I
mlD
=, O'8
l ' 0) mm
r
-,.\
- ,,1){031·)gmOl\
-. ( --'l""'-j
:::;:; (..018 6J X Hun
=:;
-78 X '0
),
I
4
gmol
---':-"'I,nnn
P7-11 (b)
V m= a. (EJ. If the enzyme concentration is incrc:;lscd by a factor of th:ee. then
t/
-'3(-"78*10'4 gmol \-1'"
gmol
0-4 'L-;-;~ri;~'
.) "l*~n' ) - 7,.>4 * 1
v rna=< - .
r
I
_
, _ * 20 min
L' mm
km In·~·L +- Cs '" Cso :::;:; '·-1734 * 10 4,,~!pq.
C so
=:
,.-34.7 * 10
-ntis equation shouid be sOlved by ni;),l and error, Rearr;:mging,
in .Cs., :::: CSO,:::, Cs_.::...:.Q?:~"
Cso
Km
:. Cs
=:
C so exorCSQ.~:~,,~,,~.J)3~JJ '" ,,02 cxo f~:Q!47,~~d
.L
Km
1
"l,
.0310 1
Assume a Cs • caicut:lte a new (lme from the RHS of the :l.t~ove el1!luation.
Assume Cs
New Cs
% difference
O.0l5·-"1r.r;OO't\'7"'7r---;-;~:4"""l'"8;----
O.®077
0.0097
0.0091
0.®®93
Q,.Cl®92
O.®®97
O.®®91
O.®®93
20.~
~.,
2.2
0.0092
L1
0.0092
0
P7 -11 (c) Individualized solution
P7 -11 (d) Individualized solution
P7-12 (a)
7-27
4F.".,!::)1
L
Given me reacrion sequence:
E+S =£·5
The plat of ·r$ vs
~
is s.'lown below.
the rnec:banism for the aI:xwe :reaction is:
-r,. ::::; k Cs lit
1 + Kl Cs + K,
low values ofCs.
r.;
; For ('-S « I: -rs'" k C Et in qualitative agreement with the graph at
fur Cs »
1: -rs ""
;2~ .
This is also in agreement with the gr:aph.
=0
I
2I'l'fs ) -= _..._. -_...... ·.·.·~_)k._El~S
!<-.1..... _. ___
or C s2 == -K-·widi:
~- - - ld~(~~
(
C'
-,
1 +- Kl s~' Kl ('\2
~:~I
2kE t
r- Kl - 3{K~' +-:-i K11 < 0,,',
~.
.~~_?_dL:_~1.~ni~L.:.~K2~~1.
(1 • K j Cs + Kz c~f
(-r,) goes through:l maximum.
2•
This observation also agrees with (ile above graph
P7-12 (b)
7-28
For a CSTR operatbg with V.::; 1000 I; Uo ;:; 3 1. IJmin
v
Do
f""
with CAD"" 50 m mo\es/l .
= 50~~i~':::: this (linear) equacion is plotted on the accompanying
,)
.
.)
graph, the equari?n imersects the (.!,) v s C s curve from .he rare of renccion at
1~2 mID.Qks.
...• ('-~ "" --.
? 1 ll.rr.ill.e.s.
-(5
··r,; -- 0 .;)
.
I
..
Il1ID
·1
Stability of points: J.Ssume u'1ar a pe:,urbllrion
as
==
c so . Cs occurs where the overb:;r designarcs
the steady-state condition. ~1ated:ll balance for any time:
V ~~2. == rsV + Cso \)0' Cs
dt
or
1:
·c
~-~.
cit
:: IS:
+
eso . . Cs
at Steady stare
r5 ,.
rs
{ d rs \
\:-l
\d Cs IC-·.;:,
(Cs"
Uo
.'\
}
Combining
-
csl ""
"1
11 6s + 6s[)
:1:. t ""
0.
i
,1
For this solution of the 3.0ove eqtw.rion
!.~,r~~. I ,
I.
\
~ C' ~\.d
-s Ie:;
10
< I
as
-= 0
be stable:
or
t"L
~-,~:~
('"
~
d,S
r > ·1
les
At C. = 2. 1 m moies/!"fs "" O. tS~ 111 moles! I·min. rhe ~doce of rhe re:lction rare curve is posithe
'Therefore, rhis oper-:J.ring poim is stable. For the other (WO poi:1ts,. the s(;),bi!icy may be exarnined
by estimating the derivative gr;J.phically:
L\CS
7-29
1
'" • ~L5..Q
2. min·
'
:. t.'1e point Cs :::; 9.4.
''is::=
(~.L.::~!.\
At Cs"'" 2.1 m molesll.
( ~i::~lL
aCs les
At
,
f(
Cs
1: =: ( ..
0132 is unsrable ..
aCs
fE.s
=2.1
.=
~L::.~1. ;:;:
D"Cs
Q,-Qfi.12.. :.D_J218
= - Q~~.
24 - 44
min-I
20
min'1) (3 n min) ::: - 0.775> .,! ; s.abk
2"
u,.D!!25.
=5.1 m moleSJ1.
S
el(or
o -_
.....- )).IS necessary
(1 C s '
-rs :::; 0.154 rrLrTI.Q~~~. appears to be sc:!ole. but more accurate calculation
~o
I-mm
bl"
..
l'
d ,... ,
esra Isn tillS conC,USIon ennmvelV.
.
•
X = Cso -Cs = 50-2.1 =.958
Cso
50
P7-12 (C)
-rs =--z
1+ K1C, + KzCs
!fEr is reduced by 33%, -rs will also decrease by 33%. From the original plot, we see that if the curve -rs is
decreased by 33%, the straight line from the CSTR calculation will cross the curve only once at
approximately Cs = 40 mmol/L
X=O..2
P7 -12 (d) Individualized solution
P7 -12 (e) Individualized solution
P7-13 (a)
Data on Ba.ker's Yeast at 13.4 °c
no
su Ha:1i b.midc.
23.5
33.,0
"'!,-7
.5
42..0
43.0
·:+3.0
~\~
2C mg
._L
$u[[';mllarnidc!rrH
adccc to medium
17.::t
Q~
no
$ulfan!l.;mlidc
2..0
30.8
36439.6
.0425
.0:303
.02.666
02.38
.02.33
. ""~)
;lOJ)
(n,3)
.200
25.6
7-30
1.0
.66
AO
'1"'~
.0575
.0391
.03246
.02747
.0'2.53
.0250
...
.1.
::;
fp
_K!!!_ + ___L __
Vma.x S
YmJ.x
;::;;
K IIL \(1'
___
____ "_)
+ ______.1! - ._____
Po,
V ma;x
V m.;u
::
_1_
QC'z
I versus -p1---- wt'11 h
Km .., anu-an
!'
lave a sI
ope '
or ---;.;'...
Imercept ot--Un
P l~ot ot~ n:,-~
V ron
O;l
'-<\.h
0.06
.... WItt!
ODS
O/tlln
(C!5ClI C{:s»
0.04
OM
om
0.01
0.00
0.0
1..0
t5
1/1){O) (rnm HQ)
From the graph, slope'" JllQJ_:_J1J1. =0165 IlHcrCcpt;;;:: 0.019
1
... V m.:u = 5263, ml
nr rng cells
Km "" 0.0165 Vm:1X ::; (0_,0165)(52.63)
~
0
mrnHg
P7-13 (b)
Now, with competitive lnhibidon:
E +S
C;::l
E•S
Km( 1 +
Rare law becomes:
f
K It
In this C:lse, (he slope is :
(1)
+
"S
\mu
111
~
+_.L}
K"
----"---""--. '.vhile (he imer::epr is the S;;lme as in c;).se fa)
V ma ;>;
7-31
For the case of uncomperitive inhibition:
E+ S
= E· S
E·$+1=1-E-5
E'SC;)P+E
Rate law becomes:
1
In this case, the slope is the same. bur the inrercept is
+-'
·v.JSL
maA
And for the case of non-competitive inhibition:
E+I=E-1
E"I+S~I·E·S
E+S~E·S
I·E·S¢;:)E·S+I
E"S¢;:)P+E
In this case both the slope and intercept change. Plotting the data of
t- in mmlig versus r-!:~
'<VI;
with sul1anilamide on the same plot as was ploned the data for the case with no sulfanilamide. it is
seen that the slopes are different. but the intercept is the same. 11lerefore the inhibition is
cornparari ve.
P7 -13 (c) Individualized solution
P7 -13 (d) Individualized solution
P7-14
For No Inhibition, using regression,
Equation model:
---.!= aO + a1(.l..)
-r
S
s
aO = 0 . 008
al = 0.0266
For Maltose,
Equation model: _1_
-rs
= aO + a1(.l..)
aO= 0.0098
S
al = 0.33
For a-dextran,
7-32
Equation model: _1-
-rs
=
aO + al(!)
S
aO = 0008
al = 0.0377
=> Maltose show non-competitive inhibition as slope and intercept, both changing compared to no
inhibition case.
=> a-dextran show competitive inhibition as intercept same but slope increases compared to no inhibition
case.
P7-15
=k(EHS)
(EHS) = KM (EH)(S)
rp = kKM (EH) (S)
(EH;) = K2 (H+ )(EH)
(EH) = Kl ( H+ ) ( E- )
rp
(E-)=~EH)
Kl (H+)
(E] )=(E-)+(EH)+(EH;)
(EH)
+
(Er)=-K1(H+) +(EH)+K2(H )(EH)
(EH) = _
1+ K,
(E] )
_
(~') + K2 ( H' )
Now plug the value of (EH) into rp
kKM (E] ) Kl ( H+ ) ( S )
rp = kKM (EH) (S) =· · - - - - - - - - - - 2
1+ Kl (H+) + KIK2 (H+)
At very low concentrations of H+ (high pH) rp approaches 0 and at very high concentrations of H+ (low pH)
rp also approaches 0 . Only at moderate concentrations ofH+ (and therefore pH) is the rate much greater than
zero.. This explains the shape of the figure .
P7 -15 (a) Individualized solution
P7 -15 (b) Individualized solution
P7-16 (a)
For batch reaction,
7-33
dCs
dt
rs --
- - = rs
&
See Polymath program P7-16-a.poI.
POLYMA TH Results
Calculated values of the DEQ variables
initial value
0
20
20
0.25
0.5
0.1
0.1
1
-0.0987654
0.0493827
Variable
t
Cs
eso
KIn
Yes
Ceo
Ce
umax
rs
re
minimal value
0
6.301E-ll
20
0.25
0.5
0.1
0.1
1
-8,,0781496
1.273E-09
maximal value
10
20
20
0.25
0.5
0.1
10.1
1
-2.546E-09
4.0390748
final value
10
6.301E-ll
20
0.25
0.5
0.1
10.1
1
-2.546E-09
1.273E-09
ODE Report (RKF4S)
Differential equations as entered by the user
[1) d(Cs)/d(t) = rs
Explicit equations as entered by the user
[1) Cso = 20
[2] Km:::: 0 . 25
[3] Ycs = 0 . 5
[4] Cco=O.1
[5] Cc = Cco+Ycs*(Cso-Cs)
[6] umax:::: 1
[7] rs = -umax*Cs*Cc/(Km+Cs)
[8] rc -Ycs*rs
=
0.0
20
··1.8
16
c:J
-3.6
-5.4
12
8
-7.2
-9.0
.:I
0
2
.:I
6
8
0
10
P7-16 (b)
For logistic growth law:
dCc
dt
---=r
g
7-34
0
2
.:I
6
8
See Polymath program P7-16--b.pol.
POLYMATH Results
Calculated values of the DEQ variables
initial value
0
0. 1
1
Variable
t
Cc
umax
Coo
rg
1
0 . 09
minimal value
0
0.1
maximal value
7
0 . 9918599
1
1
1
0 . 0080739
1
0_2499857
final value
7
0.9918599
1
1
0.0080739
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) == rg
Explicit equations as entered by the user
[1] umax == 1
[2] Coo == 1
[3] rg == umax*(1-Cc/Coo)*Cc
030.---------------
1.00
0.82
0.64
0.46
0.28
0.10
0. 0
1.4
28
4_2
t
5.6
O. OOL---~-------0.0
1.1
2_8 t
4.2
5.6
7. 0
7.0
P7-16 (C)
ForCSTR,
rg
=_Y
r
CIS s
= YClSf.imaxCsCC
- rate at Wh-IC h was hout occurs =
D I-I utlOn
K
M
+C
s
YClSf.imaxCSO
KM +Cso
1
3
O.5xlhr- x20g / dm
O.25g / dm 3 + 20g / dm 3
= 0.494hr-1
P7-16 (d)
~ -J = O.5xlhr- X[l- ~O.25g/dm
r O.25~ +20g/dm
/ dm
~~o
3
Dmaxprod = YClsf.imax[lDmaxprod
1
= 0.44hr- 1
7-35
]
3
P7-16 (e)
Ifrd= k.tC c
rh
Divide by CcV,
= Cc v0 = (rg
- rd )v
(YC/s,umaxCs~_kd
D= (rg -rd ) =
Cc
KM +Cs
C -_.JD+kd)~_~
Yc/sf.Jmax -- D
~s
~
- rs = Ys/crg
Now
Cc = YC,sD(Cso - Cs )
D+kd
For dilution rate at which wash out occur,
~
Cso=Cs
~
DMAX.
Cc = 0
- (D+kd )K M_.
C so-
YC/sf.Jmax - D
1
3
O.02hr- xO.25g / dm
20g/dm 3 +O.25g/dm 3
= YC/S,umaxCSO -- kdKM = ~:5xlhr-l X 20g / dm 3 C so +KM
= 0.493hr-1
There is not much change in Dilution rate while consideling cell death to one where cell death is neglected.,
2
DC =YC/SD (c:so- Cs)
C
D + kd
d(DC c )
For D max .prod' .
=0
dD
D max prod = 0,446 hI-I
Now
C =_(D+kd)K M
s
P7-16 (0
Now -rm=mCc
DCc=rg
rh
&
D(Cso-Cs)=-rs-rm
= Ccvo = (rg)v = f.JCcV
7-36
YC/S,umax - D
D
Divide by CcV,
Solving
= Ji = (YC/sJimaxCs)
KM +C s
DKM
YC/sJi max .- D
- r"s = Ys1Crg
C -
~
S -
Now
~
= [D(C so -
Cc
Cs )]
YclsD+m
For dilution rate at which wash out occur,
Cc = 0
Cs
~
C
~
so
=
Now
For
YCISJimax C so
C so +KM
D max
D max
DK--=M__
YC/sJimax - D
DKM
YC/sJimax- D
DMAX
~
=
prod'
= 0.494hr-1
d(DC c )
=0
dD
---.-
= 0.4763 In·
prod
3
1
O.5xlhr- X 20g / dm
20g/dm 3 +O.25g/dm 3
l
P7 -16 (g) Individualized solution
P7 -16 (h) Individualized solution
P7-17
Tessier Equation,
. - Jimax (1 - e-Cs I k )c
rg
-
C
(a) For batch leaction,
dC_
_
s =r
dt
cc
. - Jimax (1 - e -Cslk)c
rg
s'
= Ceo
C
-
+ Yc I 5 ( C so - C 5 )
See Polymath program P7-17·a.poL
POLYMATH Results
Calculated values ofthe DEQ variables
Variable
t
Cs
Ceo
Yes
Cso
Ce
k
umax
Yse
rg
rs
RateS
initial value
o
20
0. 1
0. 5
20
0.1
8
1
2
0 . 0917915
-0.183583
0 . 183583
minimal value
-0-----
maximal value
final value
7
7
0 . 0852675
20
0. 5
0.1
o. 5
20
10.057366
0 . 0852675
0.1
0.5
20
10.057366
8
1
8
1
o. 1
20
0.1
8
1
2
3.8563479
-0.183583
7.7126957
2
0.0917915
-7 . 7126957
0 . 183583
7-37
2
0.1066265
-0 . 213253
0.213253
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cs)/d(t) = rs
Explicit equations as entered by the user
[1] Cco =0.1
[2] Ycs = 0.5
[3] Cso = 20
[4] Cc = Cco+Ycs*(Cso-Cs)
[5] k= 8
[6] umax= 1
[7] Ysc=2
[8] rg =umax*(1-exp(-Cs/k))*Cc
[9] rs = -Ysc*rg
[ 10 1 RateS = -rs
8.0
16
12
6.4
F-~
W
4.8
8
3.2
4
1.6
0.0
o 0. 0
(c)
2.8
1.4
DCc
4.2
5.6
b::::2::::::'--'_~
0. 0
7.0
14
_ _~ __~.
2.8
t
5.6
4.2
D(Cso--C s )= rs
= rg
m= Ccvo = {rg)v = ,uCcV
Divide by C cV,
D = ,u = ,urnax (1-. e -C s I k )
Cs = -kln(l-
,u:-)
Now
For dilution rate at which wash out occur,
(d)
=>
Cso = Cs =
=>
DMAJ(
--·k
Cc = 0
In(I--~)
=,urnax (1-
,urnax
e-Csolk)=
Ihr- I (I- e-20gldm3 ISgldm#
Cs=-kln(l-
DCc=DYm(Cso-C s )
DCc = DYos ( C so
+k+- ,u:-)J
7-38
)= O.918hr-
.u:-)
1
7.0
Now
For
d(DC c )
Dmaxprod '
Dmax
prod =
dD
:::
0
0,628 hr-!
P7-17 (a) Individualized solution
P7 -17 (b) Individualized solution
P7-18 (a)
rg = JlC c
Jl::: Jlmax CS
KM +C s
ForCSTR,
DCc:::rg
D(Cso-Cs):::-rs
-rs:::YsICrg
s
C :::Cso (1-X):::lOg/dm 3(1-0.9)=lg/dm 3
Cc ::: YCIS (c so - C s )::: 0.5(10 -l)g / dm 3
:::
D=
v;{
4.5g / dm 3
DC ::: r ::: JlmaxCsCc
c
g
K M +C s
Vo
=>
= 0.8hr- 1 xlg / dm 3 x4.5g / dm 3
4.5g / dm3
(4+1)g/dm 3
V
V ::: 6250dm 1
P7-18 (b)
Flow of cells out = Flow of cells in
Fc ::: VoCc
Cell Balance:
dC
Fc + r~ V - Fc ::: V· __cdt
dCc=r
__
dt
g
dC
Substrate Balance: _2_::: voCso
dt
r
g
- voCs - Ys/ r~
/lC
= JlmaxCsCC
~=----"---"'-
KM +Cs
This would result in the Cell concentration growing exponentially., This is not realistic as at some point
there will be too many cells to fit into a finite sized reactor., Either a cell death rate must be included or the
cells cannot be recycled"
P7-18 (C)
TwoCSTR's
For 1sl CSTR,
V = 5000dm3 ,
DCc
= rg
D(C-C
so
s )=-rs
7-39
POLYMA TH Results
NLES Solution
Variable
Ce
Cs
urnax
KIn
Csoo
Cso
Yse
rg
r's
V
vo
D
X
Ceo
fIx)
9 . 878E-12
1.976E-ll
Value
4.3333333
1. 3333333
0.8
4
10
10
2
0.8666667
-1.7333333
5000
1000
0.2
0 . 8666667
4 . 33
Ini Guess
4
5
NLES Report (safenewt)
Nonlinear equations
f(Cc) = D*(Cc)-rg = 0
f(Cs) = D*(Cso-Cs)+rs = 0
[1)
[2)
Explicit equations
[1) umax = 0.8
[2) Km=4
[3) Csoo = 10
Cso = 10
Ysc = 2
(6) rg = umax*Cs*Cc/(Km+Cs)
[7) rs=-Ysc*rg
[8) V = 5000
(9) vo = 1000
[10) D=voN
[ 11) X = 1-Cs/Csoo
[12] Ceo = 4 . 33
(4)
(5)
x = 0 . 867
Ccl = 4.33 g/dm3
3
CSt = 1.33 g/dm
CPt
= YP/CCCt =0 . 866 g/dm3
nd
D(CC2 -CCI)= rg
See Polymath program P7-18·. c-2cstr..pol.
For2 CSTR,
POL YMATH Results
NLES Solution
Variable
Ce
Cs
urnax
KIn
Csoo
Cs1
Value
4.9334151
0 . 1261699
O. B
4
10
fIx)
3 . 004E-I0
6.00BE-10
Ini Guess
4
5
1.333
7-40
2
0.120683
-0.241366
5000
1000
0. 2
0.987383
4.33
Yse
rg
rs
v
vo
D
X
eel
NLES Report (safenewt)
Nonlinear equations
[1]
[2]
f(Cc) = O*(Cc-Cc1 )-rg = 0
f(Cs) = O*(Cs1-Cs)+rs = 0
Explicit equations
[ 1]
[2]
[3]
[4]
[51
[6]
[7]
[8]
[9]
umax = 0.8
Km=4
Csoo = 10
Cs1=1.333
Ysc=2
rg = umax*Cs*Cc/(Km+Cs)
rs = -Ysc*rg
V = 5000
vo = 1000
[10] 0
[ 11]
=voN
X = 1-Cs/Csoo
[121 Cc1 = 4 . 33
CC2 = 4.933 g/dm
3
C S2 = 126 g/dm
3
X= 0.987
C P1 = YP/CCCl =0 . 9866 g/dm3
P7-18 (d)
For washout dilution rate, Cc = 0
=
D
max
DMAXPROD
= Jimax
[
1-
JimaxCso
KM
+ C so
~--]
~
-KM +C so
Production rate = Ccvo(24hr)
3
1
J!.:_8hr- x 109 / dm
4g/dm 3 +lOg/dm 3
= O. 8hr
=4 . 85
= O.57hr-1
-I[1-- ~·-·4g/dm3
._-]
4g/dm +lOg/dm
3
3
---·---3
3
g / dm xlOOOdm Jlnx24hr = 116472.56g/day
P7-18 (e)
For batch reactor,
dC c_=r
__
dt
g
= 0.37hr
dCs
---=r
dt
s
Ceo = 0.5 g/dm3 C so = 109/dm}
ro = .J!:max CS C
o
See Polymath program P7-18e.pol.
POL Y1\1ATH Results
7-41
K
M
+C S
C
-I
Calculated values of the DED variables
Variable
t
Cc
Cs
Km
Ysc
umax
rg
rs
initial value
o
0.5
10
4
2
0.8
0.2857143
-·0.5714286
minimal value
o
0 .. 5
0.1417155
4
2
0.8
0.1486135
-2.8064061
maximal value
6
5.4291422
10
4
2
0.8
1. 403203
-0.2972271
final value
6
5.4291422
0.14171.55
4
2
0.8
0.1486135
-0.2972271
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg
[2] d(Cs)/d(t) = rs
Explicit equations as entered by the user
[1] Km=4
[2] Ysc=2
[ 3 1 umax = 0.8
[4] rg = umax*Cs*Cc/(Km+Cs)
[5] rs = -Ysc*rg
For t =6hrs,
C c =5.43g1dm3•
So we will have 3 cycle of (6+2)hrs each in 2 batch reactors of V =500dm3 .
Product rate =Cc x no. of cycle x no. of reactors x V =5.43 gldm3 x 3 x 2 x 500dm3
= 16290glday.
P7 -18 (g) Individualized solution
P7 -18 (f) Individualized solution
7-42
Fogler 7 -19 Solution
Problem Statement:
Lactic acid is produced by a Lactobacillus species cultured in a CSTR. To increase the
cell concentration and production rate, most of the cells in the reactor outlet are recycled
to the CSTR, such that the cell concentration in the product stream is 10 % of cell
concentration in the reactor. Find the optimum dilution rate that will maximize the rate
of lactic acid production in the reactor. How does this optimum dilution rate change if
the exit cell concentration fraction is changed? [rp = (a /-l + ~)Cc J
/-lmax = 0.5 h-1,
~ = O.lg/g..h,
Ks =2.0 giL,
Y x/s = 0.2 gig,
a
=0.2 gig,
Yp/ s =
Cs,o= 50 gIL
0.3 gig
Solution:
CS,o = 50 IL
0.1 Cc
CS,Cc
A steady state material balance on the Bioreactor (including the recycle device) gives:
Accumulation =
outlet +
inlet
Cells:
o
=
Substrate:
o
= D *Cs o
Product:
o
=
o
- D*O.l * Cc +
-D
* Cs
o
/-l*Cc
(7 ..19.1)
-/-l*Cc/Yx/s -- r p /Y p/s(7.19.2)
+
7- 43
generation
(7..19.3)
From equation (7.19.1) :
C
s
= 0.1 * D* Ks
/I
rmax
.
Fromeq uatlOn(7.19.2):Cc
,
D*(Cso-Cs )
=(- - +
J=(
J1
aJ1 + fl
YX1S
YPIS
I
(7.19.4)
-0.1 *D
D*(Cso-Cs )
0.1 * D a * 0.1 * D + flJ'
+ --------"'YXIS
Y P1S
,
(7.19.5)
Rate of production
rp = (a Jl + B)Cc =
(a*O.l*D+jJ)*D*(C s,o- O.l*D*:s)
J1max -0.1 D
(7.19.6)
(9.1Y * D + a * 0.1Yp/s* D + flJ
XIS
Differentiating equation (7.19.6) w.r.t. the dilution rate D, one can determine the optimum
dilution rate that will maximize the rate of production.
For the given parameter values in the problem statement, the substrate and cell
concentrations and the rate of lactic acid production can be calculated from the above
equations and plotted versus the dilution rate. The optimum dilution rate = 3.76.5 In,-I.
-_._-------------------------
Recycle CSTR - Fogler 7-19
----------_._--_._--_._----------
80
70
-----------
60
~'7_"""--'----
50
40
30
20
10
-
=EUbstrate cone
Cell Cone
------
J
-"'- Rate of Production
--------1-
------
_~_._._
o ~---~=--='-:-::..-:-~========~~======~====::=
- - , - · - - - - - , - 1 ..--.---'L,
o
1
2
3
4
5
Dilution Rate, 1/h
----------
7-43a
P7-20 (a)
XI + S -.-> More XI + PI
X 2 + XI -> More X 2 + P2
ForCSTR,
dCs
- -_ D (C50
dt
-
.
C S ) - Ys I x rgX
I
I
dC x? =D-'C
(
) +r
._-dt
Xl
gX 2
r gX,
= Jil C XI
See Polymath program P7-20-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
initial value
minimal value
maximal value
t O O
1
7-44
final value
1
Cs
Cx1
Cx2
Km1
Km2
u1max
u2max
rgx1
rgx2
Yx1s
Ysx1
Yx2x1
Yx1x2
Cso
D
10
25
7
10
10
0.5
0 . 11
6.25
0.55
0.14
7.1428571
0.5
2
250
0.04
1. 2366496
25
7
10
10
0.5
0.11
1 . 4008218
0.55
0.14
7.1428571
0. 5
2
250
0.04
10
25.791753
7.2791882
10
10
0.5
0.11
6.25
0.5748833
0.14
7.1428571
0. 5
2
250
0.04
1.2366496
25.456756
7.2791882
10
1.0
0. 5
0.11
1.4008218
0.5748833
0.14
7.1428571
0.5
2
250
0.04
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cs)/d(t) = D*(Cso-Cs)-Ysx1 *rgx1
[2] d(Cx1 )/d(t) = D*(-·Cx1 )+rgx1-Yx1 x2*rgx2
[3] d(Cx2)/d(t) = D*(-Cx2)+rgx2
Explicit equations as entered by the user
[1] Km1=10
[2] Km2 = 10
[3] u1max = 0.5
[ 4] u2max = 0.11
[5] rgx1 = u1 max*Cs*Cx1/(Km1+Cs)
[ 6] rgx2 = u2max*Cx1 *Cx2/(Km2+Cx1)
[7] Yx1s = 0.14
[8] Ysx1=1lYx1s
[9] Yx2x1 = 0 . 5
[101 Yx1x2 = 11Yx2x1
[11] Cso = 250
[12] D = 0 . 04
8.2
6.4
4. 6
2. 8
10
0.0
0.2
04
t
0.6
0.8
10
7.30,---
7.24
~J
718
712
7.06
25.00'---~-.
0.0
0.2
_ _ _ _ _ _ _ ._ _ _...J
04 t 0.6
0.8
10
7.00
IL--_~. _
0.0
7-45
Q2
_~_
Q4 t
Q6
0.8
10
0..580 r - - - - - - - - - - - - - - - - - - ,
7.0
0.574
0.568 .
0562
0.556
0.550 t<--_ _ _ _
0.0
02
~
1.0 00
0. 2
04 t
0.6
P7 -20 (b) When we increase D, C s
0.8
1.0
~
__
~
0.4 t
__
0. 6
~
_ ___'
0.8
increases, C X ! decreases, and C X2 has very little decrease.
P7-20 (C)
When C so decreased, Cs and C X ! both decreases, C X2 has no noticeable change .
When C Si increased, C X ! increases, C X2 has no noticeable change for large t
P7 -20 (d) Individualized solution
P7 -20 (e) Individualized solution
P7-21 (a and b)
Run
#1
#2
No Yeast Extract
Yeast Extract
TIle percem volume of the gro'W1h product H 2S collected above [he broth was reponed as a
function of time:
cell t· nutrient." more cells + plOducr
1 ......... _......... ,........ ,; "j~.;..'""
....._
...
.-
o
... . •
..
(a and b)
ex ;::: Cxo&(t .1<....)
or
ex ""
CXQ ¢ill
e
j.l. 1;..
curve fit exponential CHIve;
7-46
o~
...... -._.,.
1.0
where A ::::: CXQ e iJ. ~
Run 1 (between 15, 20, 30 hrs)
Run 2 (poims 10, 15,20 ill'S)
= 0.2125 hrl
!lmu
!lOla;; ::
A ::: 7492.6
tlog
== In
tlag
::=
(~{4§2~r.)
A
0.3124 hr· l
= 557L7
: : 0.21252
6.0 hr
:;: 5.1 hr
il ag
P7-21 (C)
Stationary
tstationary between
TIme
45 to 55
length of time
25 to 15
20hr
10 ill
P7-21 (d) Production struts at the end of the exponential (for both runs)
P7-21 (e)
dCc :::: D {.(.~-co'· C'
-_.
-{;J"
dt
jJ.Cc•
= 0
D{C'..co· Cd + !lee;:;:: 0
-D Cc + jJ.f=<: "" 0
Cc :::
0 or D :::::
jJ.
wash om occurs when D :>
gm;u
P7 -21 (0 Individualized solution
P7-21 (g) Individualized solution
P7-22 (a)
dC..- ::: 0 (Cr . . . C..r) + fg
dt
....... C ' . . . . . . _
--~
=
}lma;;,
1.5 hrl
Cs~ == CeQ ::: 0.5 g/dm3
r __
d_'-l.
'-cit
"" ()
~o
£'.; = )
p'
f'
:;
g.,em
Ks == 1 gJdm3
D == 0.75
eso
=:
.30
YCiS
== v.08
7-47
o=
!-lrnax Cs Cc
\
.. DC'-c + ---.-·....--.----·--l
(
\ KS .,.
CS) JI
(' (l . 1<:-1
,,$
..,.
[----·----------------·---1
'
~l:r.ax Cs
'I'
1,D = / ~~:~~$ (~-.-~-~~-\).
I
\
'- .......... w_" ............ _ ••• _
••
D
,Ku,
_.-L._w. _ _ ...... w~ _ _ ._._._•• ,
C:;'Q..
12_5
Cc
pick Cs , caiculate D and
eso
=::
::0
YclS
(Cso' ('51
(~:
30.0 g.!dm->
D= (;:~~5( ;~;;jl
Cc "" (300··, C$)(0.08)
For D = 0 . 876hr-! (C s = 2..5g/dm\ production rate is maximum
P7-22 (C)
Cc
"'-'-- Cs
For D = 0.27hr-! , Ce = 0 if Ceo = 05 g1dm3 .
For D = 0.J14hr-! , Ce = 0 if Ceo = 0 g1dm3 ,
And for maximum production rate, D = 0,,876hr- l
P7-22 (d)
For batch reactor,
7-48
dC
dt
- -c= r
g
See Polymath program P7-22-d.pol.
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
Ce
Cs
Ki
umax
Ks
rg
Yes
Yse
rateS
initial value
o
minimal value
o
final value
2
maximal value
2
2. 9
2. 9
30
50
1.5
2.01E-07
50
1.5
1
1
1
8.744E-07
0.08
12 . 5
1.093E-05
2 . 9019168
0.08
12.5
36.27396
8.744E-07
0.08
12.5
1.093E--05
0.5
30
50
1.5
0.5
2.01E-07
50
1.5
1
0.4591837
0.08
12.5
5.7397959
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg
[2] d(Cs)/d(t) = -Ysc*rg
Explicit equations as entered by the user
[1] Ki=50
[2] umax= 1.5
(3] Ks= 1
[4] rg = umax*Cs*Cc/(Ks+Cs*(1+Cs/Ki»
(5] Ycs=O.08
[6] Ysc=1lYcs
[7] rateS = Ysc*rg
30 , . . . ; : : - - - - - - . - - - -
40
c:J
CC
24
..
18
. s.
('
24
12
16
6
8
o
~==========:=::!:=l
2. 0
c..0
04
0.8 t 1.2
1.6
0
P7-22 (e)
For semi-batch reactor,
dCc
dt
=
r
g
=r
g
~
~.~
32
_ VaCc_
V
JimaxCsCc
Ks+Cs(I+C s / K/)
--.---=~~...:=..~-
See Polymath program P7--22-e.pol.
7-49
..
0.0
04
0.8
-,
t
.. -.~~
1.2
1.6
2.0
POL YMA TH Results
Calculated values of the DEQ variables
variable
t
Ce
Cs
Ki
va
Va
V
Csin
umax
Ks
rg
Yes
Yse
rateS
VCe
initial value
0
o. 5
2
50
50
10
10
30
1.5
1
0.487013
0 . 08
12.5
6.0876623
5
maximal value
4. 5
2 . 2593341
24 . 016878
50
50
10
235
30
1.5
1
2 . 2464622
0.08
12.5
28.080778
530 . 94351
minimal value
0
0.2329971
0 . 8327919
50
50
10
10
30
1.5
1
0.2329022
0 . 08
12.5
2.9112771
5
final value
4 .. 5
2 . 2593341
0.8327919
50
50
10
235
30
1.5
1
1.5283423
0 . 08
12.5
19 . 104279
530.94351
ODE Report (RKF45)
Differential equations as entered by the user
[1) d(Cc)/d(t) = rg-vo*Cc/V
[2) d(Cs)/d(t) = -Ysc*rg+vo*(Csin-Cs)/V
30
Explicit equations as entered by the user
[ 1 ) Ki =50
[2] vo= 50
[3] Vo = 10
[4] V = Vo+vo*t
[5] Csin = 30
[6] umax = 1.5
[ 7] Ks = 1
[8] rg = umax*Cs*Cc/(Ks+Cs*(1 +Cs/Ki»
[9] Ycs = 0 . 08
[10 J Ysc = 1IYcs
[11 J rateS = Ysc*rg
[12J VCc = V*Cc
12
6
0
600
24
480
18
360
12
240
6
120
C'?" _,
0.0
~J
18
30
0
~.c
-
24
119
0.0
18
t
2.7
-------------------~
[vcc]
I)
09
1.8 t
2.7
36
-1.5
0. 0
7-50
0.9
1.8 t
2.7
3.6
4.5
P7-22 (I) Individualized solution
P7-22 (g) Individualized solution
P7-23 (a)
Fit the data to the equation:
D ; ___ 5:!it:
Ks+C s
Using POLYMATH, find the values for fi.,rw: and
Ks.
TIley are L98 and 0.97 respectively.
$·29
.--~,
CS
l i .OOC
......."., ._.-._;---~'~"'''''':'''~'----'''-''1
7.::lC:;J
9 G::JC
P7-23 (b)
Using this equation, solve for Y C,'5:
With the given inf<.umation Yc;s"'" 0.099, therefore, YS1C is equal to the inverse
of that, lOJ)'lS.
P7-24 No solution wiIl be given .
.--~---------------
P7-25 No solution will be given
P7-26 See Professional Reference Shelf 7..5 on the website for a sample solution.
CDP7-A
No solution wiIl be given.
----------------------------------------------------------CDP7-B
7-51
Given the following re:rc:lon scr:e::le:
i + 00 . _}
<
or··
+- C! .
with the following f:!te: law:
Active intermediates assumed to be HOCl and HOI
From table I, the: first rule of thumb suggests:
OH'
'"<
kl
HOCI .,.., oCt
< ;.
H20. -r\:.::: kl COH CWXI- k.\ Coo' CH:O
kl
kz
OH" + HOI <:..., 01· +- H20 ,<T'Z "" kz COH eHOI· k2 Cor CrhO
k .2
The third nlle of thumb suggests that the reverse reactions occur. A chain propagation step.
involving the conversion of one intermediate into the other might tie in both the reactions above.
k3
I +- HOC! ., HOI + Cl<
,r3 ""
k1 C!· CHOC1
This step makes the overa.ll reaction sequence 1-.;- OCl·· ....~ OI·· + C1 ' possible:
k
... TO! ;;;;:: r2 :.::::
[HOI ::=
--r2
k)C{ C HOCI
2COI
C[f~O
..
k;;,C[fO CHor
+ ') "" k:,.(-'OfJ C iJOI
...
k. 2C or ('H
k.2 Cor CfJp
:.:::: k2CO[fCHOJ
1
0 .~
k 3C
1
Cd
,
CHOCl :;;;; k.! C{Xl CHiJ ;
+ k t C{ . CHCX:J == 0
.
k -1 Ccx::l CH.,Q
i.e.,,
(HOC1 ::: ;-------<..- ...
Kl CoB· ·t k3 CI
.1 X3 ('1 CO::X Cn,o .
1'ben:<fm :::: k.-....!_
.. _ - ........<-.. .,-.-~ IS close; with Xl »
kl COH .,. \q Ci
r
Of
;;;;;; KC.
......,.-_COC'
. ......!-.
C. Oll
0
;:;;;;"()f
·lllClC..1 ::::: -rl - 1.3 :;;: kl COH· CH(X.l< k ·1 C(Xl CrhO
or (kl ('..OW .;- k3
CHOC! =:
k3
k_rkl
K :::;----.
k1
WheI~
An alternative approach assumes that reaction 1 quickly attains equilibrium, then:
•
fhen
<.
rO! ::;;;::
k
'3'-- r
~
(-t
'HOCI
','k 3 k .. C., . o
"'
;;;;'l -k
...r.,
1
"j Ct· C,C!
_(~_ _ .
1../ '''',
O-..
..
,
.
OR·
TIlcse two approaches are basically equivalent
CDP7·C
7-52
(a) Assumptions:
• Transfer rate from bubble buLle to f1 uid interface is not rate limiting, i.e., Ci is
tneequilibrium oxygen concemration.
• System is:ar pseudo steady-state with regard to the particle size, i.e., panicle
growth is slow compar'ed to oxygen transfer.
• Rate of oxygen consumption is direcl1y proponional to the cell growth rote:
ro,•
-=
_yl
rc where •vo,•
0,
=•vield of cells on oxv£en (.-.mmoe
g ce1lls )
o1
.~
This implies t.~ac any oxygen utilization required to maintain the cells is negligible,
and there is no significant metabolic product being synthesized.
Oxygen balance:
where:
kl ilb :: ove.all mass transfer lesiSlance from the bubble to the bulk.
ac
= surface area per gra."11 of cells
k
:: mass tronsier of cells on 02
yo,
::: yield of cells on 02
Rearrange {'}z balances:
Ro,
C·'i - (''u
--.-"::;
(1)
k;:~c.c ""
(2)
kl ab
,lo,.Ro.. =
Tlk C'.c
Cb - C"
c
.-$
(3)
Add equations 1, 2 and 3
~~. ::; k/ab + J-c (k(;I;c+ ~~)
(4)
0) When oxygen consumption by the cells is slow, the process is reaction rate limited.
C =: - ..........
I +. ..:..-.
Yo,
Thus Tl··~ 1 and: -1.
Ro.
k! ab <=<: k
{2} When oxygen consumption is much faster than In3.SS tr.lnsier. the mass transfer
becomes [he limiting factor.
..,9,,, "" .. J.... + --........ L._..
Ro, kl ab kc ac Cc
7-53
(b) To increase the growth rate, you could:
• Ifl(,,-rease kl ab by increasing fem1cntor agitarion,
• Increase the concemration of cells (since this is an autoca::alyuc reaction)
• Increase clump surface area and the effectiveness factor by decreasing the particle
size (also by increasing fermentor agitation).,
(c)
Re "" YOl R(h
From equation (4), have:
~c:
:=1
Ci
Yo., [ ____L
"'lkl
.;. ..L {:__ L,
<.tb
CC
+
Yfh.'ll
r;kt
KC 3.(:
yO;-~C; [k~;~ ,~.;: ~tl dec " Ldl
[
+
{1 _yo,) In Ceo
Cc
"" Y0 Cit
Cc .,. CeQ
~kl-ab -+ ~~;;; -t-- ~1k
~
2
(d) Assumptions:
• There is a constant number of particles
• Each pellet is roughly spherical and has const::!.nt density.
We do not know which resistances are controlling, so we know theIe are no reaction
limitations. but. may be either internal or external diffusion limitations.
~:
:=
Yo.
C[j;}a~ + de' (~~l~; + ~~)j
1
As particle growth increases, kc. ac and II will change as functions of the particle diameter.
Thus need to fmd particle diameter as a function of Cc
Cc "" npe V c
Vc "" TI;, d 3
,
where n "" number concennation of particles (Ill)
6 P
Pc ;;::: the density of the parTicles (gil)
Ve :::: particle volume (1)
Intemal diffusional resistance can be modded as:
;,
T)z(
:;::; (Xi
do.
External diffusional resis::aflce with or 'Withom shear is;
.. _L.
kc
3{:
where cr:! :;;:: at
iCn Pc
Dropping the primes and simplifying:
7-54
Cc,,~Cco
k[ <lb
+ 3<x,
,
(r.....c113 . CcoI(3) + 2£1
(C" b!3. Ccobn) ""
b'c
Y01
C
i t
(e) From part (c), we have:
For a vigorously stirred fen-nemor, assume [hac tluid shear is sufficiently high,
that transpOrt to the edge of the floc is negligible(l):
_. . 1.. . . .
--~
0
kcac
The mass transfer resistance from air bubble to bulk liquid depends on me fermentor
design, air flow rate, agitation rate and a number of other factors. For a 10 1 laoonuOlY
scale fermentor. kl ab was f()und to be~· 150 mMz. during the growth phase of the
I hI arm
fermentation . (2)
Dividing through by Hemy's low constant:
k1 ab ;;
(ISO mMQ!~)(0.88;Ur.n.L . .}(;:;JJl!....)
1 hr atm
mMole ..,600 s
"'" 3.67 x 10.2 S ·1
Effectiveness factor:
Microbial growth on multiple substrates (here oxygen and glucose)
is typically modelled using Moncxi type kinetics:
Rc ::
),l.(':c
By representing the reaction as first-order with respect to oxygen, we are essentially
assuming a low oxygen concentration, relative to the intrinsic rate parameter. Ko:
7-55
__,.1_
--t
0
kcac
The mass transfer resistance from air bubble to bulk liquid depends on the fermentor
design, air flow rate, agitation rate ::md a number of other factors. For a 10 1 laboratOIY
scale fennemoI', kl ab was found to be -- 150 mJ,1Qk during the growm phase of the
1hr atm
fermentation", (2)
Dividing through by Henry's low constant:
kl ab
:=
~~;;~HO.88 '~;~~i~']{3A~LJ
{1.50
= 3. 67 x 10.2 S·l
Effectiveness factor.
Microbial growth on multiple substrates (here oxygen and glucose)
is typically moocHed using Monoo type kinetics:
Rc ;;: ).lee
By representing the tt:action as first"order with respect to oxygen, we are essentially
assuming a low oxygen concentration, relative to lbe intrinsic rate parameter, Ku:
For a first· order reaction. the effectiveness factor is;
",~. ($ cosh ¢ .. 1)
¢(
n
11 ::: OA5
Reaction rate constant:
k ""
83 x 10
tI
5 S ,1
,.-.max "" ..... ,,,_.,, ................, ..,,,,,,._ .•
-K a
()
3.2 x 10.4 ~,-.~
:: 0.26·,··..1·, . ,.
g0 2 S
1
Finally, aSSume an initial cell concentration of 0.25 g cell/l, L-'le cell concem:."";;!.tlon equation
now becomes:
r
(Cc
367 x 10.2 s'
I
-
!
1.5
~~!~.)! 7:l x 10 ;) ~9.:d::)
; In 4Cc ;:;: (1. 5 2.g02
\
1
i.{(45)/O.26 0 "J,I
'.
\
g. 2
'" I
..
t
$,
2.7,25 (C-=C; ··0.2.5)
f
1281n4Cc;:;: 0.012t
Clearly, mass trilllsfer from the gas to lhe liquid phase ond internal diffusion play UnpOItilllt
roles in determining the cell growth rote.
7·56
Cell mas vs time, Stan ::n 0.25 gil
t
{hr)
Cc (g/l)
Cc (g./l)
1 thr)
-0--" 0.25 "'---"'8-113
2.37
3.74
S.17
6.64
1
2
3
4
5
'm
1 LIS
12.67
14.21
15.74
17.28
0.62
1..72
9
10
11
12
05
6
8. 13
L5
1
9.63
From the graph it can be seen that growth startS our exponentially and becomes linear as the
fennentation becomes limited by gas· liquid mass tmnsfer.
Sensitivity analysis:
TIle gas~liquid mass transfer coefficient is related to the agitation rate to the 0.95
powerO):
k!
au u
NO.95
What is the effect of increasing the agitation by 50%?
kl
ti2 ""
kJ
al
(1.5°·95)
::: (3.67 x 10 .2}s -I{ LSO.95}
;; {S.39 x !D,lls·1
(see graph)
Since cell growth has an exponential ponion. another way to increase the growth mte
would be to increase the innocuous size. C<:o. What happens when Ceo is quadrupled?
C'-cu
=:
1,0 gil
(see graph)
From the results shown in the graph. a relatively small increase in the agitation nne leads to
a significant increase in the cell growth rate. while an inCIe:lSe in innoculum size means that
the fennentation leaches a gas·liquid ttansfer,limited state more quickly, but the growth rate
remains the same.
Cell Mass 'IS Time
for a STR P("" ........ fermentation
18 '"
I
I
16 ..
I
14 .... •
cell
mass
12 ..
(gIl)
10·..
I
/
I
lq a ~ 3. 67 x 10-1 5 1
Ceo = O. 25 g/i
kl a = 5 39 x 10.1 S .;
/
I
Ceo " 0 25 g.'1
kl a '" 3 6.7 x 10
('':;C '"
1.0 gil
"-j . -T--r--"'r·T'-r·-"'·T'
2
4
fi
8 10 12
14
t (ttr)
7-57
Z $:
References:
(I) James E. Bailey and David F. Ollis, 5iQfh~rIlicalElJ..giDeerjng, FllndamemaIs,
(NY: McGraw-Hilf. 1977), Chapters "1 and 8
(2) (iF. Payne, PhD Thesis, University of Michigan, (1983)..
(3) D.Le Wang, et at, F~r:rm;.!l!illiQQ_l!nQ."Gnn'lIIt.TechD.Q102"y, (NY: John Wiley &
Sons, 1979), Chapter 9
CDP7-D No solution will be given.
CDP7-E
Since the deninification f6liow$ Michaelis Memon kinetics, first detemrine V mall;
and Km from Lineweaver BUlk plot.
Initial
. .:]
r-N-;:'ol' __ "-J'
t
.:2
Time for 50.6°
Rate of reaction
,
I .
1f!
_ N 01_fl?E!!~2..._. ___.._._.-:duc~!.~:~.: . (min )_r:...{E£!l1!E.~_~_._. . __.___._
[
25
50
7.5
100
100 (given)
.04
.02
.013
35
38
44
.010
.005
50
66 (given)
lIr
2.0··
{~~¥;;)
1 ..5 ..·..
852
1.173
1000
l.515
1.000
0660
rime for 50% reduction
'.j
2.5 .
2.800
1.520
...._J:::~.~al [N0;;.·l!.~_ ._...
where the n.He of reaction is fbl."lnd by the r:mo:
3.0
.357
.658
f
10
0. 5
0 .. 00
7-58
LineweaveI· BUlk:
From the plot, V ma>; :: _ _L_.... "" 2.73 EE~.
intercept
min
:. Km = 165 ppm
Next, need conversion as a function of rime.
NSO~·
m
Design equation:
;:;:
··rs V
_ V max S
Rate Law:
·IS -
R:+s
s
Stoichiometry:
== Sop··X)
Finally, VInax t =- So X • Km In (1 .. X)
Know V max • Km, t =- 68 hrs :; 4080 min
So "" 0..2.5
•
Iteration
[0
(!DOI.) (46.JL.) (.~Qg~rr:~) (!.PP!.::)
i
mol
,.
J
mgil
=:
11,500 ppm
find the conversion obtained after 68 hrs gives:
X:;: 0.930
The [N( 2 ) level is: (1·· 0.930) (11.500 ppm) == 805 ppm
Since t, Km. and So are fixed by the system. the change must be made in Vnu:x'
Desired conversion:
... Desired V r;1:U;
Since Vmax Ct [El ]
Vmu;
,
x=
(.9565)1651n
(l ... 9565)
_. _. __ ..._........
_. __ •40"80
. '-'.....................
_. __. =
= 11,500
2.823 E~~
mm
increasing the conceno·ation of whole cells in the emulsion will increase
(2...8.ll) 50 ... ,r:.:~.~en.~....
273
1 ... 200 _. :::: . 9565
11.500
ml emulsion
mg c:!~ . . _.
= Sl.7-
ml emulsion
Therefore, increasing the cell loading to 52 •. .::::£.~~n.s-- would results in a level of [NO£J <
ml emulslOn
500 ppm after 68 hours .
CDP7-F No solution will be given.
7-.59
CDP7-G
>folt bala.nces in a CSTR
DC so ..- LJ(:$ -;- fs
··DC p
C
~
?
·' II'
=:::;
0
C
r
,r.
D
the Lite la\-v as given is:
r~
;;;;; "Is
r
o=.Y~.~~ I(1· Ee.
1
(,,)
r
11'
• ("
1.\.5 '~"S \
'p )
Plugging tho$e into POLYMATH and using different values of D and C t , come up
with the following;
Using a volume of 200 dm' and a cell concentration of 50, get the best production
of the Lmalic acid.
f~cs]=cs ·cso~rp:D
c50=2
Sol ut:
Q"
f( )
cs
CJ ,35: 657
L 64834
cp
esc
LJ
cc
vrndx
76
krn
C. :.><;8
cpst ar
CDP7-H
Michaelis·l\·fentoll Kinetics:
E=40rng
7-60
3. 222E''" i 5
<3.2772·': 5
Mole Balance:
__ dC.s. == VWil.CS%
dt
K,.. +Cs~
CS~
== CSo%(1--· X)
=>
1.4 --0.2
14
](==-··. -==0.86
Where C S $ is the percent of fish oil.
For most oils:
density 0 . 9 3 xl 0.-3 mo1Jm1 :;:: 3 x 103 p.mo11m 1
C., :;::._---;;;:_._:;::
1
MW
300
Cs : ;: ; CrCs%
f . .K. . .
+C
{)2
=>
m(..
~-.~"(iC s%::::::
14"S%
(1.4)
Jtv(';'" dt == f 5.6xlO3
dt
0''1
0
5.6x 10 3 /3
CDP7-J
No solution will be given.
(b)
1
X ;;;;: . _-_.
n
1.- p
m
-
t
t:::::: 6.21n(7)+~:~ min == 7106 min == 118.4 hrs
CDP7-1 No solution will be given.
(a)
.
5.6xlO··3
=> ~ln 0.2 +- (1.4- 0.2);;;: -3--·
=>
3
t
For Xn :;:; 5:
p :;:: 0 . 80
X,=dO:
p:;::0..90
X" == 20 :
p ::;;: 0 95
7-61
Use these equations to generate the desired graphs:
Yl
5
o2
YS.
j for X. = 5, 10, and 20
.I
r-····--·--·'··'-·--'·'··-·~='~~~~~·~.. "~::~-=-- j
020I l '- -Xn;;;!O
;:-XneeZO
{I
---.-.-..-----.--... ---.-.--.
o 10 ~ "
;:: 040
!
030
1\\
S, 10, and 20
0,60
I
!
: t
=:
070
:'
o 15 i'
;>-~
j for X"
W J YS.
080
0.50
-,
I;
005~
0,00
t,._"~~,...;-:..=..::"~~:::;=_ _ _""..""'' .. ' ' ' ' ' _.
o
Pj
20
;::::
40
j
80
60
100
M,,(l . p)Zpjl =M o (1' p)2p9
Use the above equation to generate a graph of P IO V$, p:
°
D2
0.1.
P
0,6
0,8
{c)
Use these equations to generate graphs of Yj and
Wj VS,
W j
VS.
j.
j for p :::;; 0.80, 0.90, 0.95
0,,07 -r--·~"···"···"·-·-·--·'·'·---·'-·'·"'---"--·
0,,06
OSo.
;------1'=o,90
..-'-'" ."""'' ----,
+=---p:::;
l i\"
0.05 i
~
i
004
I
i
O.oJ
"
\"
,."
p =0.95
.. , ----,
',",."'-'"
\
t! /"'\,:" "
!
:: t~s::~~: ~-'~: ;~J
OO'J-i.
o
7-62
\"
20
40
60
80
100
We must find the value of p.
ill
~ =(~~:~,"nex{=~~) -I]; -0.067
wf = 2.803
I2 :::;:
12o cxp( -kat):::;: (O.OOl)exp[( -L4 X 10,3)(14,400)1 == 2.794x 10 . 9
k~M
P == fJ =: - - - - - - - -....- ..- ...- - - - - . . . .
kpM + kmM + kee + ksS + ~2ktkof(IJ
.
= 0.99991
This can then be used to calculate the desired values:
_.104
__ . 6
NI ::;:: .-_._--._ . . _. . . . . . . . :: l.b) x 10
Jl
1.- 0.99991
(e) Mole fraction of polystyrene of chain length 10 (y 10)'
Use the above equations t.o plot Yo VS.t::
7-63
y 10
VS.
t
o° ~: r------'--------------------'.--,.'----·------------------l
O
0.03
I
O.O!)
-
I
;'002
\
0.,015
~
I
O.Ol
0005
.
..:::::::~
2
________ .. _,J1
t (hr)
5
3
CDP7-K
Reaction
R.} + I . .", . &i", PJ
(a) and (b)
...;;-' r.:;:;
L}
r r +- R' (k
l"l+ k S S + k rt.[')
, m
.. [
_
l
l
X +~ +~. tk;h+ k,,~r ~ (1~rr0 +
N
(c)
k,
+
From the above derivation we know that
rp
Neglecting the solvent term and rearranging yields:
1
:::::
i
.\,·t
-1M
Substituting in for
-'1 M
and Ii and simplifying:
7-64
{:}1+ kk:fi
knI
z)!?vf(-r
=1. . . _- .2k,J(I
. . . . _ . . _-_
. . . . _. . . .M_,) + .krn- + _.,.-
XN'
k!(2kof(IzJlkl)
kp
kpM
---
. k m ..,.,.., --,_
krjI.....
1 _ .....kJ-r;.t)
-:;'."" "'--...,....
XN
k;(lVl) kp kpM
To determine rate law parameters experimentally from a CSTR both the [mal XI'!
value and the final concentrations of wI and I must be recorded. l11ese data can be
~-
(d)
..;,.,-
..
used in the above equation to frnd values for the parameters.
An increase in temperature would cause an increase in all three primary steps of
(e)
free-radical polymerization (initiation. propagation, and termination). By looking at
the overall rate law:
l2k:(I,W
r . ;;;:;; k M (._-_._.::_.
M
?~.
k~
it can be seem that the greatest effect of temperature would be on propagation.
Overall, there would be an increase in monomer disappearance and an increase in
pol ymerization.
CDP7-L
PFR:
a)
_dlvl
.._-:::::: r
dr
»',1
)"-~.--.-
12koL,f
riM : : : -kp~M ~~;;. Plug those into POLYMATH to get this graph.
§~ati2E:!':'
dIm) Id(tau);rm
!£!i t. i~!
d(i) /d(cau}=:·:.
n .01
kp=lG
ko""Je- :)
t.au
~~S!.!1..V~!.~
0
~~~. :;:!!s~
=; .
t:.au
n
kp"m·sqrt(2*ko~i"f/ktl
:; {)t
~b~~ ... :!~:·u~
~'in4-l
VAll.].f~
7'99$
,999~
'.1'H~
l
r:i;;-.. ko*i.
Y~±.:±~
:3
2 'S'!J.2"J.
0.0:
.J 01
S. %! ::. .. ·",8
S .06~:';,,,, .. Ja
:~
t·t-
10
0 OUl
14
0.001
0.5
{)
s,,·n7
$e+0 1
Se·,;].";
··t.e··05
•. :5
~:.@,
,,1
7-65
j4J)~~~ ..
i)$
~
-- -
C S
5
Ofjl.'1l~···41.
.. < 99lSa,,·23
-.
<)$
::'.1,i>4e·-,oS
0 .. .')01-
O.S
5,,·01
S.·06~ne·4~
·2 9g:5Bi:
n
:: :;c;;
1
~EJ~
-
.-,;
: i
J ZOO
0 .. ':00
CSTR:
I 20 -·· I 2:::: ····r1: *"...
M .- M :::: ..... r I *r
o
)<..
POI YM'ATH the followino- !rraphs are
The rate laws are the same so agam usrng
.
to to
J
generated.
Initiator Concentratlon ..-" Space Tlme CSTR
Monomer vs space time CS TR
·········. 1
•
'50000
) OOooc-
1 SO<lQC'
...... ~---.-""----+
2(101)(10
S:)O;)O
Tau 1J'::t
lQOOCO
l$C·OOO
20000-0
Z5~oao
Tau, H:,.
2.5
a~oc1.
:mo=]
t:.du=5el1
io.:=.Ol
kp=:10
ko"'le.·· 3
1':=.5
k:::·=5e"J
b)
Fm two CSTRs, the design equations change just a bit
'1"
M··M
= .. ! lvi, *.::-2
o
1
M .... ,. 1\1, :::: ,. L
j
•
.,,:
* 1:2
1'1
."" I,.., ::::
-.
*' "T2'. .
The rate laws are the same with the exception that instead of just 12 or M, 12 [> Iw
MI' or M;) are used depending on which reactor they came from and the following
graphs ale generated
7-66
..
..'.
.l •... ".,.'" ...., .. ~ ... " ...
., ..
I
1:_'1!1
I
I
.;2:
a
•
.. . .
a ..
•••
000.
r
.
o ,:)os L
000. [
{) 00" ~•
o{)'c;i: ~
r
~~
!loa t
Q,jQl :.
*!: ••
2. [
>.••
Q
Ii ..
0>01
roC
o ._1. ...................,._...•,........... _
........... .
'S-IKIQq
Q
lOOOQO
T--., ,,.
.-
c)
Making kg bigger causes [2 to decrease rapidly and M does not get formed
as much. Increasing kt> causes M to decrease sl.ightly but not by t.hat much.
Increasing ~, causes M nOt to decrease by very much staying very close to 3.
1
CDP7-M No solution will be given
CDP7-N
I +iH ..5_~ R1
R.J t·i.V!_.!:J:. ····7R.~J
--'I ::::; k/vlI
a)
, ..... ..,
!
~ OQ~ ~
Balance on I
L, .. ' I
10 .. I
. (1
k;iYll
'f;;;;;; .-...-----:;;;; -..... ..
10
=> 1;;;;;;",-·
l11k,M
~
'-I~1
;;;:;:
kiMI + kp!v(2':Rj
ASLR
J=l
j
-==
10 -- I
j~!
'-r~l :::: kMI
+ k PM(I 0 .- I)
1
7.. 67
........ ..
Balance on M:
r
_ l'tlfo-;lV!
A-fo - iYf
.,,','-----"" == ,----"------,, .-.---.---_.._.
~
·r;,!
kiA-1f + k/d( If)-
=> rk/vll +rkplYf( 10
-
!)
I) "'" Mo - M
rk"ly[Jo
-rk/oM( rk,lvt)
=> ----------'-,-- + .'"'_~,._.,_.,___ c, = 11,11) .- fH
I + 'rk,lv/
1 + rk,114
i
.
=> rkJiJAl +-r2klJoj'yf2 ::;;: (J"10
=>
-
M)(l+ rk,M)
,k, (1 + rk,?~) )M" + (1 + 'lk'!a -, rk,i'vio )1'.1 ,,',. A10 :::: 0
b)
--If<: :::::: --'kiJ\U + kl,;'vlRI
Balance on R1
Similm.!y,
,<
R,:::;
R21
Rj
:::;
1 -~'rk!
lv/X;'-
1
I
I + k l v1-'(·
If
ri<
\/j
1 + rk p
lrf)
\
k
,
p \
p
)
k"
Initiation Rate constant ~ < < propagation rate constant
Hence, nearly no change in the concentration of Initiator (I).
c)
Mo
10
1
t
xi
0015
kp
1QC()
7-68
......... w._ ••••• ~ ••••••••
0.5
+-. __.......
1.5
1
tau
d)
7-69
w ........ ~~ ........... w . . . . . . . ...
2
'~2
x(1+x)
j
= ·(-l:=~~-)T
As 2., J.X
r=l
CDP7-0
a)With the reaction self catalyzed the mole balance and rate law becomes:
dlCOOHl
,
_.. _J.:_:_.:..:.J
at
=k[COOHY
We can then get [COOH 1as a function of time. The following graph shows both
the given values of p and the calculated value as 11 function of time where
p ::= t CO,Ql!Jn:1CC!..l?!!J
.... __ ...,l~:~~~!J1. ,. ,.
P
1
•+
0,8
••
'IS
",.",,'"''
time
..... 'T...... ".""""""' .. "''',...--, ...... ""..""."","" .... - ••.. ,..• " ..•.•.".-._,,,.,,., ... ,"". """ •. " ....,""" ...."" .. """" .... "
41\
41\
06
0.4
02
o
41\ .."'',.""----•• '
o
t , , , , , , , , " " , · ·.. ··- , ....•.
t""""
15(:0
500
time
It appears to fonow this above 500 min"
7-70
2000
b)
The new mole balance and late law is:
-,!!l£.~f!!J == k[COOH][OHJ[lI+]
[OH]=[COOH]
COOH:~CO(r +H+
=Icq,~~~:J
K
[COOH]
eq
[COO"
]= [Bt ]
[Rtf == K~q[COOH]
.. .:1~q,g{:!J:= kICOOH]12
dt
'
Solving for [COOH] as a function of time gives the following graph:
p vs time
·~--;-'-t"--'-·t·-····"'I'···'-.··-··-'····
0.8 • •
•
0.6
0.
0,4 '
0,2
o
500
1500
1000
time
2000
It follows the data above 200 min.
e)
This mechanism can be made to fit either rate law, depending on whether UA
dissociates before or after the fust reaction.
CDP7-P
CDP7-Q
7-71
I + M--·~--7 RI
R.) +M ·--~---'tR-1+1
_. _~i!_ = kOMI =
(k M }~2!
P
k
dt
"
dI
k
."'-_._- == .._S!. I
k/r1.dt
k,v'
d8:::::k p Mdt
.. ~
dI
k
dB
kp
--- =.__ . '2. I => I "'" I e x,
0
dR
--_..l=-·-k MI +k MR
dt
0
p
I
dR
•. _.J.. ==
de
);,
~.
.-;-'~
1 e[ ',) ._- R
k
. . . '2..
kp
0
I
_~.{<~!) == .~Q.I e(l ~'J
dB
kp
dR
de
0
-- J. :;;;;
R.. ..- R
'
:!
CDP7-R
7-72
1 -+ ivf'~'--7 R1
R.)
-+ lvf "-'~"-7 R./+1
r:::: .I.11.._-... __[
ko/vfI
8' ::::. kpi'vFr:
8' :::: ~f!!L,::::
. !.l
koI
k/
I == -............
_"..o_.......-.
k o8' +kp
l"tl ..,·lvl
k/,;flo
l' :::: ....JL ...... ,... --
A'fo --A1 ;:;; -----..
kl,lor + 1
9';:;; _..~e!~!?~. "
k/o'f+ 1
R,;..
r :::: -----.--...-.....
kolvlI --- k,)rfR,
Ie'
R ;::: ...k_Q,....
!
k1',(1+ fJ')
1.'
&_ . . ,. ,_.-..
== -.. _.-.---.-..
k/,fR{- k/V{R2
R 8'
R2 :::
k k 1 8'2
k I 8 1z
i:~,{i; == k)i~~;1-~~e';':k~); (j . ~-e;Y~~e;·~-~k:)
7-73
Fogler 7-24 Solution
Problem Statement:
In biotechnology industry, E. coli is grown aerobically to highest possible concentrations in
batch or fed-batch reactors to maximize production of an intracellular protein product. To
avoid substrate inhibition, glucose concentration in the initial culture medium is restricted to
100 grams/liter in the initial charge of 80 liter culture medium in a 100 liter capacity
bioreactor. After much of this glucose is consumed, a concentrated glucose feed (500 gil)
will be fed into the reactor at a constant volumetric feed rate of 1.0 liter/hour. When the
dissolved oxygen concentration in the culture medium falls below a critical value of 0.5
mg/liter, acetic acid is produced in a growth-associated mode with an a of 0.1 g acetlg
cellmass. The by-product acetic acid inhibits cell growth linearly, with the toxic
concentration (no cell growth) at Cp * oflO g/liter. Find the optimum volumetric flow rate
that will maximize the overall rate of cell mass production when the bioreactor is filled up
and if the feed is turned on after glucose falls below 10 gil. Inoculum concentration is 1 g
cellslliter.
Additional parameters:
f.lmax
= 1.2 hr,-I, KG = 1.0 gil, Ko = 1 mg/l, Y xls = 0.5 gig, Yp/s = 0.3 gig,
qo/x
= 1000 mglg
Oxygen mass transfer rate kLa = 500 hr- I, Saturation oxygen concentration C02*= 7.5 mg/liter
Increase the value of mass transfer rate (up to 1000) or the saturation oxygen
concentration (up to 40 mg/liter) to see if higher cell densities can be obtained in the fed··
batch reacator"
(a) list ways you can work this problem inconectiy"
(b) How could you make this problem more difficult?
(Contibuted by Prof. D. S. Kompala, University of Colorado)
Solution:
This problem is solved numerically in three parts, using the following equations on
Berkeley Madonna package:
7-74
The first time period covers the simple batch culture, when glucose and dissolved oxygen
are being consumed for cell growth.
METHOD Stiff
STARTTIME = 0
STOPTIME = 3.4
DT=0.02
INITG =100
INIT X = 1.0
INITO =7.5
INITP=O
mumax = 1.2
KG = 1.0
KO = 1.0
Yxs = 0,,5
Yps = 0.3
q = 1000
kLa = 1000
alpha = 0
SATG = G/(KG+G)
SATO = O/(KO+O)
SATP = 1.0 - (PIlO)
mux = mumax * SATG*SATO*SATP*X
d/dt(X) = mux
d/dt(G) = - muxlYxs
d/dt(O) = kLa*(7.5 - 0) - q*mux
d/dt(P) = alpha*mux
The numerical simulation results shown below identifies the time at which the dissolved
oxygen concentration falls below the critical value of 0.5 mg/liter, triggering the
formation of the by-product acetic acid.
From the simulation results, we find that the dissolved oxygen concentration falls below
the critical values of 0.5 mg/l at the batch culture time of 3.64 hours. At that time, the
glucose concentration has fallen to 67.1 gil and cell mass concentration has growth to
17.45 gil. The by-product acetic acid concentration remains zero through the early batch
culture, as the dissolved oxygen concentration is above the critical level throughout this
time. In the program above, the parameter alpha is set to zero to ensure that no acetic acid
is produced.
7-75
Run 1: 43 steps in 0 seconds
18
100
16
80
14
70
12
60
10
0
..
~
.... ...- ...
50
,
1><
~
8
40
6
30
4
20
2
10
o
o
0.5
1
1. 5
2
2.5
3
3.5
4
TIME
In the second part of the batch culture, acetic acid is getting produced and glucose is still
above its set point of 10 gil, when the concentrated glucose feed is added to the
bioreactor.
7-76
The program equations are given above slightly modified to change the alpha value to the
given value of 0.1 g acetic acidlg g cell mass and integrated from the end of first part of
batch culture.
RlJn 1~ 2t; steps
in 0 s~90nds
~~~~~~~~~~~~~~~~-r~~~~,r45
.
c.o
20
3.5
4
4. 5
5
5 ..5
6
6.5
7
7.5
8
TIME
From the simulation results, we see that glucose concentration reaches the predetermined
value of 10 gil (for turning on the feed) at 7.65 hours of batch culture. At that time, the
cell mass concentration has reached 41.02 gil and the by-product acetic acid
concentration has reached 2.85 gil. Using these values as the initial conditions for the
third part of culture, a glucose feed is added and the balance equations are therefore
modified to include the dilution of all bioreactor contents with the fresh nutrient medium.
The modified program equations are shown below:
7-77
METHOD Stiff
STARTTIME = 7.65
STOPTIME = 209.5
DT=0.02
INIT G =10.0364
INIT X = 41.8957
INIT 0 = 0.1978
INIT P = 2.85834
INIT V = 80
mumax= 1.2
KG = 1.0
KO = 1.0
Yxs = 0 . 5
Yps = 0.3
q = 1000
kLa= 1000
alpha = 0.1
vin=O.1
SATG = G/(KG+G)
SATO = O/(KO+O)
SATP = 1.0 - (PIlO)
mux = mumax
* SATG*SATO*SATP*X
d/dt(X) = mux - X*vin/V
d/dt(G) = - muxlYxs +(vinlV)*(500 - G)
d/dt(O) = kLa*(7.5 - 0) - q*mux
d/dt(P) = alpha*mux - P*viniV
d/dt(V) = Yin
The constant value for yin, the volumetric feed rate can be systematically varied to find
the highest cell mass concentration, when the reactor volume gets filled, i.e. becomes 100
liters. Simulation results for different Yin values ar'e tabulated below:
yin
time, hrs
Volume
X cell con cent
0.,,0.5
40.7.8
10.0.
87,53
0.1
10.0.
8758
10.0.
87.,6
0.,3
20.8
10.7.,7
75,,3
0.4
574
0.,2
10.0.
87.,5
10.0.
87.,3
7-78
0.5
47.,7
0,6
41
0.7
36,2
100
100
100
87,5
87.,6
87,,5
It is clear from these simulation results that volumetric feed rate does not make a strong
difference in the final cell mass concentration. The time for filling up the reactor volume
to 100 liter is of course strong affected by the volumetric feed rate
It is expected if the kLa is smaller, then the acetic acid production will be higher. In that
case, the volumetric feed rate will have a significant effect on the maximum cell mass
concentration achieved in the fed batch reactor.
These simulations nevertheless provide a useful introduction to the concepts of fed-batch
culture.
7-79
Solutions for Chapter 8 - Steady-State Nonisothermal Reactor
Design
P8-1 Individualized solution
P8-2 (a) Example 8-1
ForCSTR
v = FAOX = __X__
Dok(l-X)
--fA
'tk
'tAe -E/RT
X=---=--,
l+'tk 1+ Ae- E/ RT
One equation, two unknowns
Adiabatic energy balance
T=To_MIRXX
CpA
In two equations and two unknowns
In Polymath form the solution
'tAe- E/ RT
f(X)= X --E/RT
l+Ae
f(T) = To -
~HRxX_
CPA
Enter X, A, E, R, C p A , To and i1HRx to find '( and from that you can find V.
P8-2 (b) Example 8-2
Helium would have no effect on calculation
%Error =
-~Cp (T - TR )
-[MI~x +~Cp(T-TR)]
1270
=----x100 = 5.47%
23,210
P8-2 (c) Example 8-3
(a)
V =0.8 m
3
See Polymath program pg-2-·c.pol.
8-1
POLYMA TH Results
Calculated values of the DEQ variables
Variable
initial value
o
V
o
x
Cao
Fao
T
Kc
k
Xe
ra
rate
minimal value
o
o
9.3
146 . 7
340
2.4595708
8.5452686
0.7109468
-110.4184
79.470998
9.3
146.7
340
2.8783812
8.5452686
0.7421605
-79.470998
79.470998
maximal value
0.8
0.5403882
9.3
146.7
363.39881
2.8783812
38.191248
0.7421605
-79.470998
110.4184
final value
0. 8
0.5403882
9.3
146.7
363 . 39881
2.4595708
38.191248
0.7109468
-85.208593
85.208593
ODE Report (RKF45)
Differential equations as entered by the user
[ 1) d(X)/d(V) = -ralFao
Explicit equations as entered by the user
[1] Cao=9.3
[2] Fao = . 9*163
[3] T = 340+43.3*X
[4] Kc = 3 . 03*exp(-830 . 3*«T-333)/(T*333)))
[5] k = 31..1*exp(7906*(T-360)/(T*360»
[6] Xe = Kc/(1+Kc)
[7] ra=-k*Cao*(1-(1+1/Kc)*X)
[ 8] rate = ora
PFR
T
330
X
0.26
340
054
350
0.68
370
0..66
390
0.65
420
0 . 62
450
0.59
500
0.55
600
0.48
----------_._-
0.7
0.6
0.5
[~J
0.4
0.3
02
330
384
-438 T
492
546
600
CSTR has the same trend.
P8-2 (d) Example 8-4
Counter-Current: Guess Ta at V =0 to be 330 and it will give an entering coolant temperature of 310
K.
See Polymath program P8-2-d.pol.
8-2
POL YMATH Results
No Title 08-17-2005, Rev51 233
Calculated values of the DEQ variables
initial value
Variable
V
X
T
Ta
Cao
Fao
Kc
0
0
310
330,,7
9,,3
14.67
3,6518653
0.9004084
0.7850325
-8,,3737978
-6900
5000
159
8.3737978
50
75
k
Xe
ra
dHrx
Ua
Cpo
rate
m
Cpc
minimal value
0
0
310
310,16835
9.3
14,,67
2.7812058
0,,9004084
0,7355341
-27,114595
-6900
5000
159
0.0460999
50
75
maximal value
5
0,,7797801
344,71423
335,79958
9,,3
14.67
3,6518653
11.. 763976
0.7850325
-0.0460999
-6900
5000
159
27.114595
50
75
final value
5
0.7797801
310.83085
310.16835
9.3
14.67
3,,6255777
0,,9639302
0,7838108
-0,0460999
-6900
5000
159
0.0460999
50
75
ODE Report (RKF45)
Differential equations as entered by the user
[11 d(X)/d(V) = -ralFao
[2] d(T)/d(V) = «ra*dHrx)-Ua*(T-Ta))/Cpo/Fao
[3] d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc
Explicit equations as entered by the user
[1 J Cao = 9.3
[2J Fao = ,,9*163*,,1
[3 J Kc = 3,,03*exp(-830,,3*«T-333)/(T*330)))
[4] k = 31,,1 *exp(7906*(T-360)/(T*360))
[5J Xe = Kc/(1+Kc)
[6] ra = -k*Cao*(1-(1+1/Kc)*X)
[ 7 J dHrx = -6900
[8] Ua = 5000
[9J Cpo = 159
[10] rate ,·ra
[11] m = 50
[12] Cpc = 75
=
P8-2 (e) Example 8-5
At V = 0, Ta = 995.15 and gives a counter current entering temperature of 1250 K.
See Polymath program P8-2·e.poL
POL YMATI{Results
Calculated values of the DEQ variables
variable
V
X
T
Ta
Fao
Cpa
delCp
Cao
To
initial value
0
0
1035
995,,15
0,,0376
163
-9
18.8
1035
minimal value
0
0
972.39417
986.00676
0,,0376
163
-9
18.8
1035
maximal value
0,,00;1.
0,,3512403
1035
1249,,999
0.0376
163
-9
18,,8
1035
8-3
final value
-----cf:001
0,,3512403
1034,,4748
1249,,999
0.0376
163
-9
18.8
1035
7 . 414E+04
-67 .. 304
1 . 65E+04
0.111
34 . 5
7 . 414E+04
-67.304
1 . 65E+04
0.111
34.5
dHrx
ra
Ua
me
Cpe
7 . 47E+04
-6.3363798
1.65E+04
0.111
34.5
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) =-ra/Fao
[2 j d(T)/d(V) = (Ua*(Ta-T)+ra*dHrx)/(Fao*(Cpa+X*deICp»
[3 j d(Ta)/d(V) = -Ua*(T-Ta)/mc/Cpc
Explicit equations as entered by the user
[1] Fao = .0376
[2] Cpa = 163
[ 3] delCp = -9
[4] Cao = 18 . 8
[5] To = 1035
[6] dHrx =80770+deICp*(T-298)
[7] ra = -Cao*3.58*exp(34222*(1!T0-1/T»*(1-X)*(TolT)/(1 +X)
[ 8] Ua = 16500
[9] mc = . 111
[ 10 1 Cpc = 34.5
(b)
P8-2 (f) Example 8-6
Energy balance will remain the same
X EB = 2 xlO-3 (T - 300)
for 2A-"72B
.... ,
\
\
1st Order
2nd Order
\
\
\
(c)
, ....
.... ........
P8-2 (g) Example 8-7
8-4
7.414E+04
-31.792345
1.65E+04
0.111
34.5
Both Xe and XEB will change. The slope of energy balance will decrease by a factor of 3.
x
T
Also Xe will be more temperature sensitive
K =K
e
e
exp~HRX
(~--~J
R
T T
1
The dotted line in the plot below shows an increase in -AHRx
\
\
\
\
x
\
\
\
,
" '" ....
T
(d)
(1)
(2)
P8-2 (h) Example 8-8
CAO will decrease but this will have no effect
't will decrease
't'=
401.1 ft 3
466.1 ft3
(3)
Is
In the energy balance the slope of the energy balance of X vs . T will be greater
8-.5
l:8 i C PC = 35 +(18.65X18)+4 x (1.67X19.5)= 35 + 335.7 + 130.2
=501 BTU
kmoloR
XM
,/
-- _.
XM,
T
I
I
I
I
I
I
I
I
/
I
I
Basecase
I
I
I
I
I
_
_ _
Change QM
I
I
I
I
I
l
8-6
Less Conversions
P8-2 (i) Example 8-9
Change C p =29 and -~H = 38700
POLYMA TH Results
NLES Solution
va:r::·iable
Value
x
fix)
2 . 444E-11
1. 2E-09
0.7109354
593 . 6885
0 . 1229
1.696E+13
3.24E+04
1. 987
20.01167
T
tau
A
E
R
k
Ini Guess
0.367
564
NLES Report (safenewt)
Nonlinear equations
[1 J f(X) = X-(397.3*(T-535)+92 . 9*(T-545))/(38700+7*(T-528)) = 0
[2J f(T) = X-tau*k/(1+tau*k)
=0
Explicit equations
[1] tau = 0 . 1229
[2J
[3]
[4]
[5 J
A=16.96*10A12
E = 32400
R = 1.987
k = A*exp(-E/(R*T))
Vary the heat exchanger area to find the effect on conversion.
P8-2 (j)
=
3
a 1.05 dm
See Polymath program P8·2··j.pol.
POL YMA TH Results
Calculated values of the DEQ variables
8-7
Variable
V
Fa
Fb
Fe
T
Y
kla
k2a
Cto
Ft
To
Ca
Cb
Cc
rla
r2a
Fto
alpha
initial value
0
100
0
0
423
1
482.8247
553.05566
0.1
100
423
0. 1
0
0
-48.28247
-5.5305566
100
1. 05
minimal value
0
2.738E-06
0
0
423
0.3120454
482.8247
553 . 05566
0. 1
77.521631
423
2.069E-09
0
0
--373.39077
-848 . 11153
100
1. 05
maximal value
1
100
55.04326
22.478369
812.19122
1
4.484E+04
1.48E+07
0.1
100
423
0.1
0.0415941
0 . 016986
-5.019E-05
-1.591E-ll
100
1. 05
final value
1
2.738E-06
55.04326
22.478369
722.08816
0 . 3120454
2.426E+04
3.716E+06
0.1
77.521631
423
2.069E-09
0.0415941
0.016986
-5.019E-05
-1. 591E-1l
100
1. 05
ODE Rel!ort (RKF45}
Differential equations as entered by the user
[1] d(Fa)/d(V)
r1 a+r2a
[2] d(Fb)/d(V) = -r1 a
[3] d(Fc)/d(V) =-r2a/2
[4] d(T)/d(V) (4000*(373-T)+(-r1 a)*20000+(-r2a)*60000)/(90*Fa+90*Fb+ 180*Fc)
[5] d(y)/d(V) =-alpha/21y*(FVFto)
=
=
Explicit equations as entered by the user
[1] k1a 10*exp(4000*(1/300-1/T»
[2 J k2a = 0 . 09*exp(9000*(1/300-1/T»
[3 J Cta =0.1
[41 Ft = Fa+Fb+Fc
[5J To 423
[6] Ca Cto*(Fa/Ft)*(To/T)
['7 J Cb =Cto*(Fb/Ft)*(To/T)
[8] Cc =Cto*(Fc/Ft)*(TolT)
[9] r1 a = -k1 a*Ca
[ 10] r2a = -k2a*Ca"2
[11] Fto = 100
[12] alpha = 1.05
=
=
=
(e)
P8-2 (k) Example 8-11
VaryUA
J/m
3
VA= 70,000
-s-K
only the lower steady state exists at T =318 K SBC = 0.05
J/m 3
VA=60,000
-s-K
only three steady states exist T =318, 380 (about) and 408 (about) depending how you read the
intersection on the graph.
3
VA= 700
-s-K
only three steady states T =300 (about), T =350 (about) and one are a very high temperature off the
scale of the R (T) and G(T) plot.
In all cases SBC remains low at 0.05, meaning that the reaction has neared completion to form
species C therefore reactor is too large.
J/m
8-8
To
= 275, very little effect.
Vary l'
't = 0.001 only the lower steady state at T = 316 about and other off scale SBC = 0.05
z =0.0001 only are steady state at T =316 and others off scale SBC =0.05
't = 0.00001, SBC =5
However, the upper steady state is off the graph and needs to be studied
(f)
P8-2 (1) Example PRS P8-4 ..l
1
(J.----
dpPo
U=U
2
j
:::::
=[~:2)(~)J=1
No effect for turbulent flow if both dp and P changed at the same time.
P8-2 (m) Example TS-3
fie = 200 g/s
See Polymath program P8·2·rn.pol.
POLYMATH Results
Calculated values of the DEQ variables
variable
W
Ta
Y
T
X
alpha
To
Uarho
Me
Cpme
Hr
Fao
theta I
CpI
CpA
the taB
CpB
Cto
Ea
Ke
ka
yao
xe
Cao
sumep
Ca
Cb
Ce
ra
initial value
0
320
1
330
0
2 . 0E-04
350
0.5
200
18
-2.0E+04
5
1
40
20
1
20
0.3
2.5E+04
66.01.082
0.046809
0.3333333
0 . 8024634
0. 1
80
0.1060606
0 . 1060606
0
-·5 . 265E--04
minimal value
0
320
0 . 3044056
330
0
2.0E-04
350
0.5
200
18
-2 . 0E+04
5
1
40
20
1
20
0.3
2.5E+04
0 . 8247864
0 . 046809
0.3333333
0 . 3122841
0. 1
80
0 . 0137198
0 . 0137l98
0
-0 . 0143957
maximal value
4500
334.77131
1
385 . 31436
0.5645069
2 . 0E-04
350
0. 5
200
18
-2.0E+04
5
1
40
20
1
20
0.3
2 . 5E+04
66.01082
11 . 205249
0 . 3333333
0 . 8024634
0.1
80
0 . 1060606
0.1060606
0.0724316
-1 . 745E-05
8-9
final value
4500
334 . 77l31
0.3044056
338.18498
0.5645069
2 . 0E-04
350
0.5
200
18
-2 . 0E+04
5
1
40
20
1
20
0.3
2.5E+04
31.. 551036
0.1177827
0 . 3333333
0 . 7374305
0. 1
80
0.0137l98
0 . 0137l98
0.0355685
-1 . 745E-05
ODE Report (RKF45)
Differential equations as entered by the user
[Ii d(Ta)/d(W) = Uarho*(T-Ta)/(Mc*Cpmc)
[2] d(y)/d(W) = -alpha/2*(T/To)/y
[3] d(T)/d(W) = (Uarho*(Ta-T)+(-ra)*(-Hr»/(Fao*sumcp)
[4 J d(X)/d(W) = -ra/Fao
Explicit equations as entered by the user
[ 1] alpha =.0002
(2) To=350
[3) Uarho = 0.5
[4] Mc = 200
[5] Cpmc = 18
[ 6 J Hr = -20000
[7) Fao=5
[ 8] thetal = 1
[9] Cpl = 40
[10J CpA =20
[l1J thetaB = 1
[12] CpB=20
[13J Cto = 0.3
[14] Ea = 25000
[15 J Kc = 1000*(exp(Hr/1.987*(1/303··1/T)))
[16J ka = .004*exp(Ea/1.987*(1/310-1/T»
[17 J yao = 1/(1 +thetaB+thetal)
[18] xe = Kc"O.5/(2+Kc"O.5)
[19] Cao = yao*Cto
[20 J sumcp = (thetal*Cpl+CpA+thetaB*CpB)
[21 J Ca = Cao*(1-X)*y*To/T
[22] Cb = Cao*(1-X)*y*To/T
[2.3] Cc = Cao*2*X*y*Torr
[ 24] ra = -ka *(Ca *Cb-CcJ\2/Kc)
P8-2 (n)
(1) The concentration of A near the wall is lower than in the center because the velocity profile is
parabolic. This means near the walls the velocity is much lower and therefore the time space near
the wall is much larger than in the center. This means the reaction has longer to take place and
conversion will be higher near the wall. Thus the concentration is lower.
Below is the FEMLAB solution.
1. Parameters in simulation on the tubular reactor from Example 8-12 (First Order reaction):
Reaction: A + B -7 C
A- propylene oxide; B- water; C- propylene glycol
(1) operating parameters
Reactants
•
Feed rate of A FAO = 0.1 molls
•
Inlet flow rate of A
VAO
=
FAOM A
PA
0.lx58.1xlO-3
-6
3
=
=7x10 mls
830
8-10
•
•
•
•
•
•
•
Inlet flow rate of B v BO
Inlet total flowrate
Vo
= 2.5 x 2 x V AO
6
= 35 X 10- rrhs
= 2v AO +v BO = 14xlO-6 +35xlO-6 = 49xlO-6
Inlet concentration of A C AO
FAO
= -Vo
0..1
49xlO-
.
F
v P
Inlet concentIatIOn of B C BO = ---.!!.Q.. = ~- =
Vo
Inlet temperature of the reactant To
6
MBVO
= 20.40. .8
3
mol/m
35xlO-6 x1000
3
= 39682.5 mol/rn
6
3
18xlO- x49xlO-
= 312K
Coolant flowrate, m] = 0.,,01 kg/s
Inlet temperature of the coolant, Tao = 298 K
•
(2) properties of reactants
• Heat of reaction, I1HRx , dHrx = -525676+286098+ 154911.6=-84666.4 J/mol
• Activation energy, E = 75362 J/mol
• Pre-exponential factor, A = 16.96x1012 13600 lis
• Specific reaction rate ko = 1.28/3600 lis @300K
• Reaction rate k = ko exp[ E
R
•
•
•
•
•
•
m 3/s
(-.~
- ~)] or k = A exp[- ~]
~
T
RT
Gas constant, R = 8.314 J/mol·K
Rate law - r A = kC A
Thelmal conductivity of the reaction mixture, ke = 0..559 W/mK
Average density of the reaction mixture, p, rho = WOo. kg/m3
Heat capacity of the reaction mixture, Cp = 4180 J/kg"K
Diffusivity of all species, Diff = 10-9 m%
(3) properties of coolant
2
• Overall heat transfer coefficient, Uk = 1300 J/m "s"K
• Heat capacity of the coolant, Cp] = 4180 J/kg-K
2. Size of the Tubular Reactor
• Reactor radius, Ra = 0.1 m
• Reactor length, L = 1.0 m
2. Femlab screen shots
(1) Domain
(2) Constants and scalar expressions
- Constants
8-11
~
tw·· . ··-·-··-·-··-···--····--···
f--··----····-·-----···--
-+----.---{"
.314
12
- - - - - - - - - - -..-------------.-
~---.-+------------~.Q................ -.-- . . . . . - ...... .
fBD __ ._ . __. _. __ . +_____._____._.__....__._..._......__..._._.....__...__.__..
_.._{----_._..._.... _....__...•._-_._...._----_........ -...__.._.
----·F=-=-··---------·------·-----+-=
r=-=---_.-.--.-.-.. -f'.---.--..-----.----------.-------..--.---fc:-::..----.--..-
- Scalar expressions
(3) Subdomain Settings
- Physics
(mass balance)
Equation
V.(-DVcA+cAu) =R, cA =concentration
Subdomain selection
Library material: .[ - - ; ..•.-
.~ .C~:.~j ,
,='
~"'-,-,-
T
Valueil'xpressionDestriptiOI1
.
[ ... ····---~~~:l~ime..~~~fing coefficient
0'5
G) D isotropic
~!!"
.•'..
-. IDiffusion coeffic~nt
~ ~;-i -I'
'
o D anisotropic._ .. --:::l Pilfus!9n (;oefficient
R
u
10=~===-:JJ'{ilaCiion rat!'
8-12
(Energy balance)
V*C-kVT + ~hiND,j) =Q - PCpu.VT, T=temperfrture
i PhYSics i Init ; Element '
Thermal properties and heat sourcesfsinks
~~
Library material
Quantity
ok
1\5
Valueibpression
Description
tC~~~=~-==~.~-~~--~]
Time-scaling coefficient
Thermal conductivity
(isotropic)
(':' k (anisotropic)
Thermal conductivity
Heat capacity
C
p
Q
hiND.1
l~p'~~~~~~i!~~~~?~ !~_~~~~
! [~iCial~fUSio~~:_J
(Cooling Jacket)
Equation
v.r = F
c Subdomain
selection
i Coefficients
In~
Element Weak
PDE coefficients
COefficient VilluelExpression
o
Qescription
..l§~..~'=] Flux vector
F;~g~:iT~!~j1~e:;~;;;~~l
Source term
da
Mass coefficient
[ ] Select by group
~ Active in this domain
(Source Term)
F = Taz-2*pi*Ra*Uk*(T-Ta)/(CpJ*m1)
- Initial values
(Mass balance) cA(tO) = cAO
(Energy Balance) T(tO) = TO
(Cooling Jacket) Ta(tO) = TaO
- Boundary Conditions
@ r = 0, Axial symmetry
@ inlet, cA = cAO (for mass balance)
T = TO (for energy balance)
@ outlet, Convective flux
@ wall, Insulation/Symmetry (for mass balance)
qO = -Uk*(T-TaO) (for energy balance)
8-13
(4) Results
(Concentration, cA)
.." ?,&'
1~3·aaa
1782.os5
660222
O.6~
0.6
0,4
0.2
'-0.2
- -..- - - - . - - . - - - - - - - - - - -..
..Ql5
..01
-0.05
-------~-=-
0.05
0.6
0.4
""
0.2
---
-0.2
·015
~====--,---.----
..0.1"
-0.05
0.05
8-14
01
0,,15
0.2
..--".--..025"
Second order reaction
[1] Domain
o Axis equal
[2] Constants & Scalar expressions
(1) Constants
, ~l'l.t'rlE;! .•..
• !;;::;Pf~ssiQn
uz
2·uO·C1-CrlRa).A2)
.t-....
_.-._-...." ..--_ ..._ ...
FkO;;""",.._~r°1e~~~~~2)
0--
txA
cB
'cC
1-- ~." V"·,.,_,
rf!:
J<eq
15--"
...__ .
(cAO-cA)/cAO
"-"-TBO~~~.Q~L~-·-·······
_".,_"._'"+2'cAO'xA
___
..
.~._~" ~._o.
v""''',.'''" __ •• _._ .. ~ .. ,,~'"_, .. " .•,, __
·_,,",v~·_,_·
__
..rf!:'~~!1~~'!r0.9.~~·(c~_"c_B-E~~~9J...... .
1303··1
-~
. . . . . -.. . .-.. .-.. . . .-.-.. . . . .
8-15
[3] Subdomain Settings
(1) Physics
(Mass balance)
Equation-' --,,V'(-D'lcA+cAu) =R, cA =concentration
i,Subdomain selection--
Ifilii:! Iii ffl 1111 imll1lllll ,±;
[1
library material:
i 1
Quantity
II
lh
II
~
!
ii I1
o D isotropic
Diffusion cQefficienl
o D anisotropic ii,ij,Q,:'.".=, _::.c._.J'DiffuSlon coefficient
II
!
ValuelExpressiofi Description;
[ ____ ..~~-~=] Time-s"aling coefficieoi
I
IL
R
~=--~-=-=='=:l Reacti':'Q;/'ate
v
~~, __7_-Jz~vel~qty
lo--·-·-------J,r,veloc~y
u
EJ Select by group
~rlifiCial Di!fu,~ion_,
G:'.I Active illIbis domain
]
(Energy balance)
I Equalion
,
I v.(:kVT ",:" Ilhi~D,I) = Q., pCp.....VT. T=t~mperatL!re
I
Subdomain selection ~'.
r'Physj~ in~ !Element 1
Tl:lermal properties and heat sourcesfsinkS
~ibr8ry material;
".~'.'..',~~J L.LOad ....
J
Value&press;OO De.cription
Quantity
h' ---- mm:"':::::':':.i.5;t:';-T Tjme-SC~ling coefficient
0kQsolroplC)
Therrr:taLcon~ivily
Ok (.niSoiropic)
Tberma.l.concWclivilY:-
~:c:;::::::c::::::::::::::::;
Dens~¥­
~::,::,==or;===::::,~ Heal Cop'CKY
Cp
Q
'''''''';;'''''''=c,;''''':::!
Healsoi1rce
r..velo~iiy
z-vel<ic~y
o Select by group
~~!!~~.~~
hiND ,I
[tJ Acti\ie in t~is '~6main
[ ~rtJ!ici~i~t~s~;]
(Cooling Jacket)
;Equation
lv·r =
F
L
EI~ment I W~~k:j
PDE coefficients-
II
Coefficient
I
r
ValueiEHpre~siQfJ
Des"ription
lQ_~_=-fo'cc,; _~-J Flux~ecto;
rl~Uk;(T~T~)~ceE~J~, ~o(.lr~e term
~-- -_,=:'~':~~:~~ :5' ~,ass coef!iCi~nt
F
_
I da
E]Selecl by group
~ Aclivein this dd!'lj~i!'l
(Source Term)
F = Taz-2*pi*Ra*Uk*(T-Ta)/(CpJ*m1)
8-16
(2) Initial values
(Mass balance) cA(tO) = cAO
(Energy Balance) T(tO) = TO
(Cooling Jacket) Ta(tO) = TaO
(3) Boundary Values
@ I' = 0, Axial symmetry
@ inlet, cA = cAO (for mass balance)
T = TO (for energy balance)
@ outlet, Convective flux
@ wall, Insulation/Symmetry (for mass balance)
qO = -Uk*(T-TaO) (for energy balance)
[4] Results
:~
_.
2
·.Q15
-01
··0 os
0.05
--0-1- - 0 - 1 5 ---0-'-2-·~-0.25-- 03
Min: 81735
xA is undefined at (--0 .. 1,0,,6)
8·-17
0.6
0:4
02
.0,05
·0.15
,.......
0.05
015
0.1
. 02
0.25 .
,;d''i~ ~d~~i~~d ~t t-9,,~,O.6j
P8-2 (0) Individualized solution
P8-3 Solution is in the decoding algorithm available in the beginning of this manual.
P8-4
Find the reactor temperature at steady state (prior to shutdown)
Let M = mass of the NH4N0 3 in reactor.
F AO =lbs/hr of NH4N0 3 fed to the reactor.
Mass balance: FAD - FAD (1- X) = -rAV = kM
X
= k(T)M
FAD
Energy Balance:
-FAD LBJHw - Hi (T)] + FADXMl Rxn(T) + FAD (1- X)MlVA
=0
Where Hj(T) is the enthalpy of i at the temperature of the reaction, Ml VA is the heat of vaporization
of A, and Ml Rxn (T) is the heat of reaction at the outlet temperature.
8-18
NH 4NO(l) ~ 2H2 0(g) + N 2 0(g)
A~2B+C
The last term FAO (1- X)t:Jl VA accounts for the unreacted N&N03, which exits as vapor rather than
the liquid. Now, we can make some substitutions
BB
= .17 ,BA =l,Bc =0
.83
Heat capacity of A is given, and the enthalpy change for water 200 0 F (I) ~ 500 0 F (g) is also given.
So, after dividing by F AO, we obtain
CpA (T -200)+ BB[H B(500° F) + C P1 (T - 500)]+ t:Jl RxnX + (1- X)t:Jl VA = 0
The previous equation assumes that the heat capacities are constant over a reasonable temperature
range.
the phase change NH4NO(aq) ----7 NH 4NO(1) is isenthalpic.
In addition, we must account for the effect of the temperature dependence of Ml Rxn .
44
Ml Rxn(T) = Ml Rxn(TR) + (T-- TR)[80 C P1 -CpA] = Ml Rxn(Tr) + (T -Tr)~Cp
Let Ml Rxn = HB (500) -- HB (200), we have
CpA(T -200)+OB(MlW +Cps(T -500)+ X [MlRxIl +i1Cp(T -500)]+(1- X)Ml VA =0
or,T[CpA -t0BCp> ]+[-CpA (200) +OB(Mlw --500Cp,)] + X[Ml Rxll + i1Cp(T -500)]+ (1-- X)MlVA
Numerical Substitution with
K(560) =:; 5.03 and k(51O) = .53
In(k2)
kj
-::::;.E
R
=-
(i_ -~J
E
R T2
~
=-In(k2)(~_~J-j =44499
kj
T2~
E
Ao = kj exp(-)
= 4.51424xlO19
R~
Btu
CpA =.38--;Cp1 =.466
Ih.R
(P = 1 atm over 450 - 500Dp Himmelblau)
Btu
C pc = .2521_
(Himmelblau, App E, over 230-265°C)
Ih.R
8-19
=0
8 =.:!2. = .205
.83
B
Ml = 1034 Btu .03
w
lb
F
= 257.3 lb NH 4 N03
hr
AO
ACp = 44.02 (.2521)+ 36.03 (.466)-.38=-.0316 Btu
80.05
80.05
Ib.R
MlRxn = C pA (41O-T)
Substituting all these into the mass and energy balances:
)M
4 ..51x1019 ex ( -44499
P T+459.67
Mass balance: X = ---------''----........:..257.3
Energy balance:
0= .48T +88.21 + X(-320.20-.03T) + (1- X)(1.55.80"-.38T)
Assume X = .96 and M=500. Then, from mass balance,
T = 5100P
P8-5
A+B~2C
_.-
A
B
C
F,. ( lb -h~Ole )
10
10
0.0
Tio(F)
80
80
C~Cb::e"F )
51
44
MW'(~--)
lbmol
128
p;,(; )
63
-,
'
47 ..5
94
222
67.2
65
, - - - + - ------- --
-----------'--
R
-----
-- f - - - - - -
-"""-----,--_t__
Ml = 20 000
--
-
Btu
lb mol A'
Energy balance with work term included is:
8-20
F
10
BA =1' B
=~=-=1 X
B
F
10 ' AF =1
AO
Q=UA(Ts-T)
Substituting into energy balance,
UA(Ts - T) - Ws - FAOM-J RX AF = FAO [ CpA + CpB ] [T - To]
~UA(Ts -T)--Ws -FAOM-JR ={FAO [ CpA +CpB]+UA}[T-Yo]
T
= T, + UA(Ts -T)- Ws o
-W
FAOM-J R
FAO [ CpA +CpB]+UA
=63525 Btu
hr
S
:. T
= 199°F
P8-6
A+B~C
Since the feed is equimolar, C AO = CBO =.1 mols/dm 3
CA = CAOO-X)
C B =CBo (I-X)
Adiabatic:
T =To + {[-_M!R(To)l
rACp; + X8Cp
8Cp =CpC-CpB-CpA =30-15--15=0
All R(T) = Hc - H B - H A = --41000-(-15000-(-20000) = -6000 cal/mol A
"L./i;C-
i
cal
= CpA +BBCpB =15+15=30--
mol K
T = 300 + ~~~~ X = 300 + 200 X
30
-rA = k C~o(l- X)2 = .0Ik(1- X)2
P8-6 (a)
8-21
For the PFR, F AO = CAovo = (.1)(2) =.2 mols/dm
3
See Polymath program P8-6-a.pol.
Calculated values of DEQ variables.., ..·-·--··t "-""""." "--''''---'1 ....
! IVariable nitial value Minimal value Maximal value i Final value
,---r-'---"
~---~~~----
--.~,-,--."""
I
I
-·-··--r·-·.·-.------·-·-----..·..
""-J:;'--'-""'--'
I
··T", •• ·-.., ·.. ··_ _·_-·····'·T--- .. -----.-. --,
illX
.
l
0
i V
0
.,', .•
10
10.85
_"."_~.,,._._'"
...1--".,.. , ,-. .. ,,,.,,.,, '-""-'
0
0.85
'" ". .......___.. _~ __ ...
1308.2917
1
_.JI
08.2917
1
'E-~~~f~=~]6~~===~-JF===:J~t=-=]I
151T
1300 .
_.~v~~.~'~",¥m~"'w""'~"___
16ik
10.01
f-""'T"'"' ,. ,-.~,~ "~.,, ""~·'I~~~'"-'~'~'''''
j71ra
1-0.0001
•. .,,, ..•. ,,
..• _
"
rw'-""~~
300 .
1
h
""'~.w '~~~N~'~'O
0
___
~~~~~~C
1470.1470..
~"
_,~'w,_=~ ~,_","
ww,~_,,
14.150375
14.150375
""''01" , ,.
10.01
w·,t·" '~'W~
'W~ ""."~,-" "'~l ~
1-0.0018941
1-0.0001
..
.
...
___ " ' ¥ ' ¥ ¥ _ N __
"'''"
"'W
',f'
"
,=,
YW
___
,~
.~~y,.,_~w~,~· ~ .•.." -----, ,~"W' ~., '~--"--'-'
••
1-0.0009338
, . . . '"
.......
I
;
I
1
Differential equations
:g,: d(V)/d(X) = -FaO / ra
Explicit equations
Cao = .1
FaO
= .2
T = 300 + 200 * X
k = .01 *exp((10000 / 2) * (1 / 300 - 1 / T»
$ ra = -k * (CaO /\ 2) * ((1 - X) /\ 2)
v = 308.2917dm3
For the CSTR,
X = .85, T = 300+(200)(85) = 470 K.
k =4.31 (Using T =470K in the formula).
-rA = .000971 mol/dm3/s
V=_l!AO X = .lx2x.85 =175 dm3
-rA
9.71xlO-4
The reason for this difference is that the temperature and hence the rate of reaction remains constant
throughout the entire CSTR (equal to the outlet conditions), while for a PFR, the rate increases
gradually with temperature from the inlet to the outlet, so the rate of increases with length.
P8-6 (b)
X[-Mi ]
T = T + - · -Ro
~)1;Cp;
For boiling temp of 550 k,
5.50 =To + 200
To = 3.50K
8-22
P8-6 (c)
0.90
500
0.72
460
GJ
0.54·
0,.36·
380
0.18
340
O.(lU
62
(I
[!]
:120
123 V 185
246
3U8
300
0
P8-6 (d)
FAoX
-rA
VCSTR = - -
=> X
= VCS7R (-rA )
FAO
=
3
For V 500 dm , F Ao=.2
-rA = k C~o(1- X)2 = .Olk(l- X)2
T =300+ 200 X
Now use Polymath to solve the non-·linear equations.
See Polymath program P8·6. d . ·1. pol.
Calculated values of NlE variables
I'
lvariablejvalue
if(x)
;Initial Guess'
liT
1484.4136 :0
21X
!0.9220681j-2.041E-09i o.9
1480.
iVariable t!Value
>
lr~
2 :ra
n
16.072856
I,(). 0003688
Nonlinear equations
1;ci f(T) = 300 + 200 * X - T
f(X)
= 500 - .2 * X /
=0
ra = 0
Explicit equations
1 k = .01 * exp(10000 /
1.98
* (1/300 - 1/ T))
8-23
62
123 V 185
246
308
= 0.01 * k * (1 - X)
ra
Hence, X = .922 and T
A
2
=484.41 K
For the conversion in two CSTR's of 250 dm3 each,
3
For the first CSTR, using the earlier program and V =250 dm ,
I rV~"~iable Ivalue
11 ]k""
15.105278
]~F~
I6.0007~59
Nonlinear equations
f(T) = 300 + 200 * X - T
i~ f(X)
= 250 "" .2 * X /
ra
=0
=0
Explicit equations
k = .01 * exp(lOOOO / 1.98
ra
= 0.01 * k * (1 - X)
A
* (1/300 - 1 / T))
2
T = 476.48 ad X = .8824
Hence, in the second reactor,
V
= FAO(X - Xl).
CSTR
X
~
-rA
V
=_CSTR
F
AO
(-r )+ X
A
I
= Tou/,CSTRl + 200 (X _. Xl)
T
See Polymath program P8-6"d·2 . pol.
Calculated values ,.-of"'rNlE variables
: IVariable;Value
if(x)
IInitial Guess
;1
1493.8738 fO" ...
1480. ..
.....
f"-~Y'"
~~.-~".-,
"~>-"-'f
~,-.~~-.~~.
~
."",~,~"
~~'t'·~~~-
IT "
121~
JO.9693688 t-1.359E-09 j0.8824
; lVariablelval~e "
~.
t,
'~"N '~ __ ·.'''''N·.·_"" T---"
jllk
r2r~a
~
."
,7.415252
'16.958E-05
.(,
8-24
31X1
;0.8824
Nonlinear equations
J~ f(T) = 476.48 + 200 * (X - Xl) - T
i
f(X) = 250 - .2
* (X - Xl) /
Explicit equations
ij k = .01 * exp(lOOOO / 1.98
= 0.01 * k * (1 - X)
Xl = .8824
::~; ra
A
=0
ra = 0
* (1 / 300 .. 1/ T))
2
Hence, final X = .9694
P8-6 (e) Individualized solution
P8-6 (0 Individualized solution
P8-7 (a)
For reversible reaction, the rate law becomes
-orA =k(CA
C- KCcJ
B
c
x =.~
l+Ke
e
T =300+200X
k = k(300)ex p
(ER (_l300___ .!.)J
T
(-.l___ .!.)]
K = K (450)exp [Ml Rxn
C
C
R
450
T
Stoichiometry:
Cc = CAOX
C A = CAo(l- X)
CB
= CAo(l- X)
See Polymath program P87-a . pol.
Calculated values of DEQ variables
"'lv~;i~bi'~lInitial val~~TMini;;':~1 value Maximal value Final valuel
,
l
i
'"
,.,
",...
".
i
" , .
'l1V
iOO
110.
10.
l?Ix.
\0
10.0050806
0.0050806
10
8-25
-;
!
Differential equations
d(X)/ dey) = ora / FaO
Explicit equations
T = 300 + 200 * X
= .2
i~i Kc = 10 * exp(6000 / 2 * (1 / 450 - 1 / T))
k = .01 * exp((10000 / 2) * (1/ 300 - 1/ T))
FaO
CaO =.1
i
= -k * (cao 2) * ((1 - X)
Xe = Kc / (1 + Kc)
ra
A
A
2 .. X / Kc)
302.0
0.30 , - - - - - - - - - - - - - - ,
30L6
0.24
--..-,----
...,.--......., . -,."., ...~"-... ,......
301.2
0. 18
300. 8
0..12
300.4
0.06
300.0
0
2
4
V 6
8
10
{tOO
IT]
- x
.,. ~e
(I
2
...
V
P8-7 (b)
When heat exchanger is added, the energy balance can be written as
dT _ Ua(Ta ·-T) + (-rA ) [-MIRxn (T)]
dV -
FAO
(I BiCpi +~Cp)
So with ~Cp= 0, I~CPi =30, MIRxn= -6000 cal/mol
dT
Ua(Ta -T) + (-rA ) [6000]
=-~---~~-~
dV
FAO (30)
Where Va = 20 cal/m3/s/K, T a = 450 K
8-26
6
8
10
See Polymath program P8-7-b.pol.
Calculated values of DEQ variables
""W'"r
IVariable !Initial value l Minimal value: Maximal value I Final value
~"~~""r-~AA~~_vn_'~_''''~~"l
~~C~~l~''''''''''-~-~vw-''~'
,-~.-4-~~"'''''.'~'~~''''''-''''~!''
"~,,
"'f'
1 Iv:O
:2
lxJo·
10
... fo
3Jr
i450.
i4 ICaO
's·IFao
!0.1
iO.2
10.1
/0.2
'110.
Tio.
i ..
16TKc
.
I?~Ik
..1
,,-~
... ·1~50.
12.586706. i~·58~?06
f-0.0258671 ;-0.027946
19 lxe
"]0.909090910.9090909
j10.
. 1450.
~
1
'463.99741452.3087
.1~:1
10.1
····]0.2
10.2
···-T1.2.22765
.-'
450.
1
flO.
10.9244008
TO.9118643
. lio .
[450.
i 10 ...
1
110 ..
,i
Differential equations
J, d(X)/d(V) = -ra / FaO
d(T)/d(V)
!10.34614
13.616753i2.737647
i-0.0035089
T-0.0035089
!1 0 . !10.
1
10 .
,. +~"."
".~!" ~-~~,
;10.
110.
···_····:0.5717112··10.5717112
18 ira
10 fUfo.
,"~'~
= (UA * (Ta - T) + (--ra) * 6000) /30/ FaO
Explicit equations
it CaO = .1
= .2
Kc = 10 * exp(6000 /2 * (1 / 450 - 1/ T))
k = .01 * exp((10000 / 2) * (1/ 300 - 1/ T))
~l ra = -k * (CaO /\ 2) * ((1 - X) /\ 2 - X / Kc)
Xe = Kc / (1 + Kc)
UA = 10
FaO
I~JTa
= 450
;al A = 10
8-27
470
1.0
466
0.8
462
0. 6
EiJ
458
0.4
454
0.2
450
2
0
4
V
6
2
10
8
4
V
6
8
P8-7 (C)
For a co-current heat exchanger,
Q=mcCpc(Tal-T)[l-.exp( . . UA
mcCpc
CpC = lcal/glK, Ta1 =450 K,
J]
m=50 L
sec
See Polymath program P87·c.pol.
Calculated values of DEQ variables
r~.~!variab'~li~~itia' v~i~~] Minimal value rMa~imal ~~lu~l~il1~i val~~]
11
12
13
:r"O"
. . 10
·····10
IV
Ix
IT
1450.
1"
T"
14 ICaO
1'15 I FaO
10.1
o
]0
[0.575571
]450.
,,,,
!0.2
11.
·1~·586706
1'.,•: .8•.. ·.•.. k. ·
ra
.9 '1-0.025867~
110jXe
11.1 jUA
112lT~
[131A
10.9090909
1141mc "150.
[lslq
I~
!0.1
'1'
10 .2
"ce,,,
10.1
!
;0.2
.\12.46524
11.
11.
[2.586706.
1-0.0281568
10.9090909
13.734633
1-0.0034309
16.9257347
Tlo.
1450./450.
110.
lio.
150.
•... ~O.
.0·····
!0.1
f
10 .2
.<..
hO.37812
150.
. ···..
I
I
·1
1
/1.1
I2.751.765 J
]-.0.0034309 J
!0.912112
····[10.
1
1
1450.
I
:10.
.. 150. .
1-139.420110
I,·
J.
I
J-22.833091
Differential equations
= -ra j FaO
12; d(T)jd(V) = (Q + (-ra) * 6000) j
I
j
..... 10.575571
/465.3828' 1452.51.92" .]
lio·rio.
1450.
110. ....
110 .
..... ·········r
··.Jl0.JlO.
17 kpc
rll.,· . .
110 .
~
l' . ".,. r
1611<c
jo
d(X)jd(V)
30 j FaO
8-28
Explicit equations
= .1
FaO = .2
3. Kc = 10 * exp(6000 /2 * (1/450 - 1/ T))
,fl· Cpc = 1
k = .01 * exp((10000 / 2) * (1/ 300 - 1/ T))
6" ra = -k * (CaO /\ 2) * ((1 - X) /\ 2 - X / Kc)
Xe = Kc / (1 + Kc)
8. UA = 10
1, CaO
9. Ta
= 450
= 10
.1l mc = 50
12 Q = mc * Cpc * (Ta - T) * (1 - exp(-UA /
10 A
mc / Cpc))
470
1.0
466
tl8
462
0. 6
458
0.4
454
lL2
450
0
2
"
V
8
6
-------------,
~
l~
2
10
4
V
6
8
Next increase the coolant flow rate and run the same program to compare results.
P8-7 (d)
For counter-current flow,
Q=
meC,e (I;, -T)[l +exp ( - m~~)]
See Polymath program P8,,:7 ·d.po!.
Calculated values of DEQ variables
!Variable'Initial value:Minimal valueiMaximal
value ' Final value
,
r~-l"--'"
1 iV
!
2
3
Ix
::r
<
'r~<'
~~.
~."
'~/'i
.'0'"
10
:450.
'
;~
;
, '~l
10
{
,~~"
~
[10.
io
1"450.
1
ilO.
'0.5395082
1451.6471
.. ,
8-29
10.5395082
... ~
.450.2745
10
f4'" r~o'''''-lo~i
is !FaO
10.2
" .......·'1-· ..·'.. · .. , ........ " Y
l
--'''''''
, .. '
.. ,
'"''''''''''''''''''''~'''
, , - , .. ,
.""TO:1-" . . . """.:_:~j,:~~"(,:.,., . ._,~~. ~~:'" ~.'",I~~~~. ,~.~ .
10.2..........." ......" . "110.2...." ......... ..
,t10.2, . -, ". . ."."""('.,
; 10.
110.24606
110.04072
, .............., ,
."...," "', 'T'......· . . . , ... , . , ....... ""'''1'''''''' . .,,--,. .
............. "
,. . . . . . .
.
"-, ''''-
I....Kc,. ._. . . . . . . .".-1110.
,....6,. . +
.......,'-, . ,
17 ICpc 11.
11.
11.
:1.
18'Tk-''-'''T2:s86706' "T2:586'706"'''''''T2~6936S7''-'''''' 1"2.604286
19' "F~'" . " ···'T~0.02"58671 r~0':0260956 ,." "T~o~"o04123i""" r~o:o04i231
lioTXe ""'10.9090909 10.9090909"'10:91108" 10.9094262
(' A 'Eo.'" "liD."" ···rio.'Tio.
1"12 !Ta ....t450.145o.
1450.
l4S0.
ii3IA"Tio.
,..,',. Fo: .
;10.
flO.
1i:!~~ro-o, --- Ji~.75n-~f~o,.
W~.96003
Differential equations
d(X)/d(V) = -ra / FaO
d(T)/d(V)
= (Q + (-ra) * 6000) / 30/ FaO
Explicit equations
CaO = .1
;2,'
= .2
Kc = 10 * exp( 6000 / 2* (1 / 450 - 1 / T»
Cpc = 1
is,~;; k = .01 * exp((lOOOO / 2) * (1 / 300 -1/ T»
ra = -k * (CaO /\ 2) * ((1 - X) /\ 2 - X / Kc)
Xe = Kc / (1 + Kc)
FaO
UA
= 10
Ta
= 450
10
50
l~1I A =
j,J, mc =
Q
= mc * Cpc * (Ta ,. T) * (1 + exp(-UA /
mc / Cpc»
8..30
452.0,.---------------,
10
0.8
~
4512
0.6
450.8
O·.j
450.4
02
2
4
V 6
00
10
8
U;J
0
2
.:j
V 6
8
P8-7 (e)
We see that it is better to use a counter-current coolant flow as in this case we achieve the maximum
equilibrium conversion using a lesser volume of the PFR.
(0
P8-7
If the reaction is irreversible but endothermic, we have
-rA =k C~o(1- X)2 =.Olk(1- X)2 as obtained in the earlier problem.
MI Rxn = 6000cal / mol
For co·-current flow of coolant,
Q~mcC,c (7;, -T)[l-exp ( - m~~J
See Polymath program P8-7·f·co.pol.
we use 8- 7f cocurrent.pol
Calculated values of DEQ variables
i"'fVariablehniti~1
v~i~'~l, Mini~al value I, Maximal value 1I Final value
,
,
, r i '.,
1 IV
10
~--;
i2 IX
I
13 ,IT
!4 CaO
15 FaO
16 Cpc
7 i!k
8 IIra
,
T
:l~lmc
.,
f
,.
:
.
10
,
1
10.10.
;
fO.5227896
;0.5227896
;0
10
"l
1
1450.
1439.9908
10.1
10.1
10.1
;0.1
10.2
10.2
"io. 2
11.
11.
·2.586706
:2.352492
1-0.0053573
!110
1-0.0053573
llO.
1450.
:450.
SO.
50.
i
:0.2
1.
11. .,
12.586706
[2.008972
1-0.0258671
1-0.0258671
:446.1887
1
'
~~
110.
,
9 !UA
10ha
Y
'
1450.
,
150.
A50.
'50.
I
8-31
c~····
T'·· ·······,····,······'··T·'· ...
10
i 12 iQ
1 ' ....,!., ..
~.,
.........,.. ',
~~, .. ,
~.
.'.......---..' ...
,. ,).".¥~ .." .. ~ ...." " ..
'···~r
.
.. 'c-'-
~
¥ ' •• ' ¥ ••••••
··-···-··-·'····-····~·--'~··'-T····'
...... ........ - .. ' .. --......,
~
J~O'~.?~779J~4.?4357
10
'¥.~.".¥ . .,-. •• ~~,: ..•. ~~.
Differential equations
d(X)jd(V) = -ra j FaO
d(T)jd(V)
= (Q + (-ra) * (-6000)) j
30 j FaO
Explicit equations
Cao = .1
FaO =.2
~a~ Cpc = 1
.41 k = .01 * exp«10000 j 2)
= -k * (CaO
UA = 10
ra
~Ta
, ,<
A
2)
* (1 j
* «1 - X)
A
300 - 1 j T))
2)
= 450
¥~: me
= 50
fii Q = me * Cpc * (Ta - T) * (1 .. exp(-UA j
450
me j Cpc))
-.------
0.60
OAS
0.36
·US
0.2-4
[J]
434
0. 12
430 01...--~2---4--\-!-6~-·-··8
~---'--~--~--~--~
10
2
4
V 6
For counter-current flow,
Q=rhcCpc(Tal--T)[l+exp(-
.UA
mcCpC
J]
See Polymath program P8--7 -·f-counteLp0l.
POLYMATH program 8-7f countercurrent.pol
Calculated values of-. DEQ
variables
-, '-..
., "'1
.... .......
IVariablelInitial value!Minimal value IMaximal valuelFinal value
---r'~~ ~~.~
r'
1 IV
~,~~.
... r
'~"~--"-
10
'·10
}150.
2lx
IT' . . .
3
'1V~""
~~.
,-~~ ~.
··r·
10
fa
f448.4634
.~
.
r'
:10.
10.5594826
..... 1450.
8-32
..
~
~«'~,
......
r~
,
0
~:.
...
110.
fo:s594826
l449.669
8
10
lcao
iO.1
iO.1
'5~1~~~
:0.2
0.2
:6 ICpc
1.
1.
'4
'ill<,
iO.1
11.
i
!2.586706
12.586706
2.4901
8T~~
1-0.0258671
-0.0258671
[ -0.0049788
9juA
110.
1450.
J
10.
110.
450.
A50.
i50.
:50.
50.
10
10
139.7022
101Ta
o
"
lli mc
12jQ
,
"
" [30.09601
Differential equations
~.•
d(X)/d(V)
~,d(T)/d(V)
= -ra /
FaO
= (Q + (-ra) * (-6000)) /30/ FaO
Explicit equations
cao = .1
= .2
Cpc = 1
k = .01 * exp((lOOOO / 2) * (1/ 300 _. 1/ T))
:5 ra = -k * (CaO /\ 2) * ((1 - X) /\ 2)
FaO
q. UA =
10
= 450
") Ta
8 me = 50
~. Q
= me * Cpc * (Ta - T) * (1 + exp(-UA /
450.0.--------
0.60
..-'~~~"
449.6
0,48
0.36
449.2
Q
1),.24
0,12
448,4
448.0
me / Cpc))
-.-~----~ --~
o
2
4
Y
6
8
10
000
()
P8-7 (g)
For a runaway reaction, the following must be true:
RT2
T, -TC >-'E
8-33
2
4
V
6
8
10
and T =
Yo + KTa
= 300+3*450 = 412.5
I+K'
1+3
So if we plug this value into the original equation we get:
c
1.987 T2 -T +450> 0
10000
Tr= 499 K
T
1
P8-8 (a)
A+B~C
Species Balance:
I
dX
rA
dW
FAO
--=--
= 20dm 3 / s
Po = lOatm
Vo
Stoichiometry:
CA =CAO ( I-X )I--,where £=1
1+£X To
~ C, =
CAO ( : :
~-):o
Rate Law is:
(-~- !)]
T
--r = kC with k = 0.133 exp [ E
A
A'
R 450
E=31400
!VIRxn = -20,000 J / mol
Energy Balance:
T
X[--·M!R(To)]_
o + '" ~B;Cp; + XI1Cp
_T
-.I
=15 + 24--40 =0
T = 450+ 20,0~0 X =450+ 500X
I1Cp
40
See Polymath program P8-8·-a.pol.
Calculated values of DEQ variables
r~lv~~i~bi~I~~iti~i;~i~~Tfv,i~i;;i:~~ lue fMaxim~lvalue ['~in~1 V~.llI~J
i1 1X
12.j~
10
. . I~.....
10
10
10.8
10.8
143.13711'"j43.1371.1
8--34
I
1
r-~
'3:T
1450.
'450.
,:20.
:20.
~ ..
~~
'~t~~
- '.~
1450 .
,5iTO
'850.
..
20.
,..
1
450.
1450.
'0.133
'6.904332
A'·l
'0.133
6ik
~
Differential equations
~ d(W)/d(X) = vO * (1 + X)
*T /
k / (1 - X) / TO
Explicit equations
:lin = 450 + 500 * X
= 20
3TO = 450
:4 k = .133 * exp(31400 / 8.314 * (1/ TO - 1/ T))
2. vO
0.8
0.6
0. 5
0. 3
0. 2
0.0
9
0
18
W 27
36
P8-8 (b)
Species Balance for CSTR:
FAOX
WeHR
=--,'-rA
T
= 450 + 500 X = 450 + 500(.8) = 850 K
k
= .133exp [}1400(_1__._1_)] = 6.9
8.314 450
WCSTR
850
= 39.42 kg
P8-8 (C) Individualized solution
P8-8 (d)
For pressure drop, an extra equation is added
8-35
dP
dW
CA
=_ a(~J-~(I+8X)
2 1'0 (PlPo)
= CAO
(I.-X)
+
I
T P
X To Po
See Polymath program P8-8-··d.pol.
Using POLYMATH program CRE_8_8d.pol
For a=.019
Calculated values of DEQ variables
i var;bi~tn;itii;l;';I~~ rMi~imal,;;'I~~lMaximal valuejFinal val~~1
?'W'l'~"
10
J~.8t~·8J
.
,
G*I~.o13E+06
ji.002E+06 ·r850.·····
K~~::~~ .. j~::::~~
r41i··1450.
..
1450.
1850.
......
151vo
"'120:120:"
161To
1450.
'1450.
',·"10.133'10.133
l!Jk,
,8IpOI1.013E+06
r
T .
,9lalpha
10.019
'fio:·120.·'1
1
450 . J
1450.
16.904332f6.904332
11.013E+06h.013E+06
r
10.019
10.019
Differential equations
l d(X)jd(W) = k j vO * (1- X) j (1 + X)
d(P)jd(W)
= -·alpha j
* TO j T * P j PO
2 * (T j TO) * PO A 2 j P * (1 + X)
Explicit equations
T = 450 + 500 * W
vO
~:TO
= 20
= 450
iJ k =
.133 * exp(31400 / 8.314 * (1 / TO - 1/ T))
PO = 1013250
alpha = .019
8-36
I
11.013E+06
. ···T···
i
10.019
.J
0 . 060
900
0.0-18
800
Q
0.036
I}, 02-1
600
0 . 012
500
0.000
000
016
032 W 0.48
G
700
400
0.80
0.6-1
0.00
0.16
0.32 W 0....8
0.64
0.80
P8-9 (a)
We use the same equations as problem P8-8, except that the energy balance changes as:
dT
I
-Ua ( Ta -T ) +(-rA)(-Mi
Rxn )
=p
---------
dW
FAOC pA
Where -Mi Rxn = 20,000 J/mol, T a=323 K, C pA =40 l/mollK
See Polymath program P8"·9"-a.pol.
Calculated values of DEQ variables
lV~riablellniti~1 ~~I~e~l Minimal~alue Maximal value 1Final value
i
Iw
,
;1
3
,
l
'
1010
.. ".
:50.
150."
'2jX
10
10
iO.1376181
;0.1376181
'3 IT
i450.
138 1.1888
1450.
)381.1888
1450.
i45O .
1450.
i450.
14
ITO
15 ivO
16
r
i20.
jk
iO.133
'{
1:
120 .
[0.0292331
i7 IUarho
10.08
0.08
is" ITa
1293.
1293.
,
i
19
ipo
1101cAO
1.013E+06
1270.8283
f""
1111CA
1121rA
I
20.
!20.
,
,/0.0292331
:0.133
r
l
08
1°.
i
10.08
" 1293.
j293.
r
11.013E+06
i1.013E+06
!1.013E+06
1270.8283
270.8283
12.42.3648
t
270.8283
!270.8283
i
1242 .3648
1·-36.02017
'-7.085084
..
1270.8283
1
"
1-36.02017
I
-7.085084
Differential equations
d(X)/d(W) = k * (1 - X) / (1 + X)
* TO / T / vO
d(T)/d(W) = (Uarho * (Ta - T) + rA * 20000) / vO /
CAO / 40
8-37
Explicit equations
TO = 450
= 20
ljl k = 0.133 * exp(31400 /8.314 * (1/ TO - 1/ T))
vO
~J Uarho
= 0.08
= 293
PO = 1013250
CAO = PO / 8.314 / TO
1m CA = CAO * (1 - X) / (1 + X) * TO / T
rA = -k * CA
Ta
450.----- - - - - - - - - - - - ,
0.20
436
0.16
422
0.12
40S
O.OS
394
0.1)4·
10
20
\V 30
40
50
0.00
0
10
20
W 30
40
50
If !!.-!2. was increased by a factor of 3000, we use the same program with the new value. The profiles
Ph
are in the graphs below.
----
450
0.20
0.16
434
EiJ
US
402
0. 08
386
0,,04
370
Q
0.12
_.
0
10
20
W 30
40
50
10
P8-9 (b)
8-38
20
W 30
40
50
For non-constant jacket temperature, the equation for incorporating the flow needs to be introduced.
co-current:
Tao = 50°C
~50r-----------------------~
0.20,----------------------------,
436
016
~22
012
408
0,,08
394
0,U4
380 L.-_ _
~
o
_ _ _ _~_ _ _ _ _ ,_~_,-=:S
10
20
W 30
40
10
50
20
W 30
40
50
countercurrent:
dTa
dW
=
mCCpC
Taf= 50°C
guess and check Tao until T a = 323 K at W = 50
Tao = 438.8 K
0.2U
016
422
0,12
.:J08
0,U8
394
n 114
380 L--_ _ _----~-----o
10
20 W 30
40
5U
ouo
10
0
P8-9 (C)
For a fluidized CSTR with W = 80 kg, VA = 500 J/s/K,
8-39
20
\V 30
40
50
W
-CAO
Species balance: X MB = ~,r = ..:..P-",b_ _
l+rk
FAO
UA (T-Ta) +C (T-T)
F
Energy balance: X EB
0
pA
=...:.P.-!:b:.----"A~O- - - - - - -MIRxn
X EB =XMB
Solving, X = .95, T
=323 K
P8-9 (d)
For a reversible reaction, we have all the previous equations, but the rate law is modified as:
--rA = kfCA -k,CBCC
C =C =C
B
C
AO
~1'0
l+X T
Plugging the equation for kl> and solving using POLYMATH program, we get the plots.
Only the co··current program and plots are shown.
See Polymath program P8-9-d.pol.
Calculated values of DEQ variables
!~~]ly~~i~~I~'lll~~i~ial v~'~~l Minimal valuE!1 Ma~,im~1 va'u~IFi~~' "-~IU,!!JI'
j1 _.,,,."
IW:O
f"
f-''''~'
!~-l~
13 IT
j..•... \...
J4
JTa
10
180.
1 .. ,
J~
. 1°. .
\450.
1420.7523
. t'< . " ' " .,
,P?~:
!~2~:
!?~!~" .. J450·l4?0.
i6 iVO
120.
l~f~~:hot::'3
191po
. li.013E+06
110'fCAo·'127o.82'83
42131
li·013E+06
[270.8283
flila!270.8283" "f 258.1071
fi21k'r
102"10.076962
(like"10"
r141'c8 ",'.'][0.
j15FA
'{
jO.057593
1450.
j
,1 4?6.1627
]450.
,20.
l20.
. ]it:.
180.
1
'"
.
1~·057~93
'"
1420.7523
... .
,
'v
,1
J
!
J
,I
j42.0.7565. ,
450 .
1
120.
I
1
··1~~~31~:'42131i
h:013E+06
'11.613E+06 1
1270.8283'12708283'
'j
b70.8283h58~io711
10.2
'10.076962
"'1
10t15.77362
h5.77362'j
[0'115.77362
115.77362"'1
1-36.02017' 1-36.02017
1-0.0062421
Differential equations
d(X)jd(W) = -rA j vO j CAO
8-40
. . . . ]-0:0062421]
= (Uarho * (Ta - T) + rA * 20000) / vO /
3 d(Ta)/d(W) = Uarho * (T - Ta) / .2/ 5000
2 d(T)/d(W)
+ sign = cocurrent, -ve sign
CAO / 40
=countercurrent in RHS of eqn
Explicit equations
1
TO
= 450
= 20
k = 0.133 * exp(31400 /8.314 * (1 / TO - 1/ T))
vO
~ Uarho == .08
4-
* 3000
S •PO = 1013250
6
CAO = PO / 8.314 / TO
= CAO * (1 - X) / (1 + X) * TO / T
aE'ili kr = 0.2 * exp(51400 /8.314 * (1/ TO - 1/ T))
CC = CAO * X / (1 + X) * TO / T
~m: CB = CAO * X / (1 + X) * TO / T
1;:~i rA = -(k * CA - kr * CB * CC)
"l...
CA
450..--------------.,
OU60
.:14.:1
I)
.:138
0.036
432
OJ124
.:126
U.II12
16
32
vV.:I8
64
8U
U.:l8
o.ouo
0
16
32
W 48
6.:1
80
P8-9 (e) Individualized solution
------------------P8-10 (a)
A-7B+C
C
A
=c_FFA
I
T
e[ = FF[
A
Cr
= CA +C[
Fr =FA +F[
8-41
C
=
CAO +CIO
BI + 1
AOI
P8-10 (b)
Mole balance:
dX = -rA
dV
FAO
-rA = kCA
Rate law:
. h·
C - C 1-- X 1'0
StOlC
lOmetry:
A AOI 1+£ X T
£ =
YAO
c5
8=1+1-1=1
Y = FAO .=
Fro
,AO
FAO
__
1_
FAO + F;o
1+ B;
1
£=--.
1+8.I
T=
-XMIRX
+ (CPA +B;CPi )1'o
CpA +B;CPi
Enter these equations into Polymath
See Polymath program P8--lO-b.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
V
X
Cao
Cio
theta
Fao
Cao1
e
To
dRrx
Cpa
Cpi
T
k
ra
initial value
0
0
0.0221729
0.0221729
100
10
4.391E-04
0.009901
1100
8 . 0E+04
170
200
1100
25.256686
···0 . Oll0894
minimal value
0
0
0 . 0221729
0.0221729
100
1.0
4.391E-04
0.009901
1100
8.0E+04
170
200
1098.3458
24.100568
-0.0110894
maximal value
500
0.417064
0.0221729
0.0221729
100
10
4.391E-04
0.009901
1100
8.0E+04
170
200
1100
25.256686
-0 . 0061524
ODE Report (RKF45)
Differential equations as entered by the user
[l] d(X)/d(V) = -ra/Fao
8-42
final value
500
0 . 417064
0.0221729
0.0221729
100
10
4.391E-04
0.009901
1100
8.0E+04
170
200
1098.3458
24.1.00568
-0,0061524
Explicit equations as entered by the user
[ 1] Cao =2/.082/1100
[2] Cio = Cao
[ 3] theta = 100
[4) Fao = 10
[5 J Cao1 = (Cao+Cio)/(theta+ 1)
[6] e 1/(1 +theta)
[7] To = 1100
[8] dHrx 80000
[9j Cpa = 170
[10] Cpi = 200
[11] T = (X*(-dHrx)+(Cpa+theta*Cpi)*To)/(Cpa+theta*Cpi)
[12] k exp(34 . 34-34222/T)
[13] ra =-k*Cao1 *(1-X)*To/(1 +e*X)/T
=
=
=
Conversion
1.0
--
1060
0.8
-
0.6
No inert
i\fodelate
Larae
980
0. 2
940
0
LOU
100
200 V 300
400
-
1020
004
0.0
Tell1per~lture
1100
500
900
--_..
100
(}
No inert
Moderate
Larae
-~---200 V 300
400
500
--_.
E~
040
0.20
0..0°0
3
6
Ihe~a
12
15
P8-10 (C)
There is a maximum at e =8. This is because when e is small, adding inerts keeps the temperature
low to favor the endothermic reaction. As e increases beyond 8, there is so much more inert than
reactants that the rate law becomes the limiting factor.
P8-10 (d)
The only change to the Polymath code from part (b) is that the heat of reaction changes sign. The
new code is not shown, but the plots are below.
See Polymath program P8··1O-d.pol.
8-43
1.0
Conversion
rr------=.......- - - - - - - ,
Tempel"nture
1600..-------'---------,
1500
0.8
- No inerts
- No inerts
l\Iodcrate
- Larue
0.6
0.4
1300
0.2
1200
0,0
w:::;..._~
o
__
100
~
_ _ _ _ _. _ _
200 V 300
- Moderate
- Laroe
1400
400
500
1100 ...----:::=.==~-.,-.-.,~-=,~=~,.,:~
0
100
200 V 300
400
.-.---'
500
1.00.-----~--------_,
0.92
0,,84
0.76
0.68
,~--~-~-,---'-.-.
10
20
TheiN
40
so
The maximum conversion occurs at low values of theta (8 < 8) because the reaction is now
exothermic. This means heat is generated during the reaction and there is no advantage to adding
inerts as there was in the endothermic case.
P8-10 (e)
We need to alter the equations from part (c) such that -rA
= kC~
and CAo =1
A plot of conversion versus theta shows a maximum at about 8 =5.
See Polymath program P81O··e.pol.
8-44
0.90.----------------,
-
0500' - - - - 2 - - - 4
rheIN
10
8
P8-10 (0
We need to alter the equations from part (c) such that
We already know that CA
-'A
~~c ]
= k [ CA _ C
I-X T
= CAO ----......Q.. Now weed need expressions for CB and Ce. From
I+EX T
stoichiometry we can see that CB =Ce. In terms of CAO we find that:
C =C =C
B e · AO
X To
I+EX T
We also need an equation for Ke: Kc = KCl exp
[MiR
RX
!J]
(~T.. T
= 2exp
When we enter these inot Polymath we find that the maximumn conversion is achieved at
approximately e = 8.
See Polymath program P8-1 o Lpol.
100
0.30 ' - - - - : 3 - - 6 - - - - 9 - - - 1 2
0
Iheta
15
P8-10 (g)
See Polymath program P8-1 Og.pol.
8--45
!)]
[8000~
(_1_ 8.314 1100 T
Fb
10
-~
8
- Part(d)
.. Part (e)
6
4
2
o0
100
200
V
300
400
500
P8-11 (a)
Start with the complete energy balance:
I
I
dE " s -"LEiF; in-"LEiF; out
-=Q_·W
dt
The following simplifications can be made:
It is steady state.
In part a, there is no heat taken away or added
There is no shaft work
That leaves us with
0= -"LEiF; lin -"LEiF; lout
Evaluating energy terms:
In: H AOFAO + H BOFBO + H eoFeo
Out: HA(FA + RAV) + HB(FB+RBV)+He(Fe + ReV)
Simplifying,
HA(FA+RAV)+HB(FB+RBV)+He(Fe + ReV)-HAoFAO -HBOFBO -HeoFeo = 0
If only C is diffusing out of the reactor we get:
HAFA +HBFB +He (FB +RBV)-HAOFAO -HBOFBO -HeoFeo =0
Now we evaluate Fi
FA = FAO .- FAOX
= FBO + FAOX
Fe = Feo + FAOX -
FB
ReV
Inserting these into our equation gives:
HAFAO -HAFAOX +HBFBO +HBFAOX + HcFeo + HeFAoX -HAOFAO -HBOFBO -HeoFeo =0
and note that FBo =Feo =0
H AFAO - H AFAOX + H BFAOX + H eFAoX - H AOFAO = 0 and combining and substituting terms gives:
FAO ( H A _. H AO ) + FAo X MJ.RX
=0
8-46
Differentiating with respect to V with i1Cp
=0
dT
dX
FAoC -+FAo-(LViRx(T))=O
P dV
dV
dT
(rA)[ LVi Rx (T) ]
dV
'i.~Cp
Combine that with the mole balance and rate law:
dFA =r dFB =-r dFe =-r -k C
dV
A dV
A dv
Ace
-rA =kCA
C =C
A
TO
FA To
F T
=C
C
B
FB
TO
T
Yo
C
FTC
=C
Fe
TO
T
Yo
F T
I
Forkc =10
See Polymath program P8-11a.pol.
Calculated
of DEQ
variables
-- .,-.-.---..values
,.
- _._ ..,;
lVariable :Initial value iMinimal value Maximal value Final value
..
1 IV
& "'
i2
Fe
\3
Fb
:0
:0
50.
50.
0
10
;0
0.0012968
0.0005261
:0
0.1978837
10.1978837
5.42
15.222116
548.9418
5.421297
:5.420526
;
I
i5 ;T
,450.
16 1Ft
,5.42
15.222116
,
1450 .
15 .42
i7
dHrx
'-2.0E+04
:-2.0E+04
-2.0E+04
;-2.0E+04
i8
lk
iO.133
iO.133
'10.
0.6036997
10.6036997
10.
i1O.
2.710027
12.710027
2.710027
12.610831
10.0006905
10.0002635
10
;0.0006482
10.000263
i4 lFa
{
9 ike
i5.42
~,
,
(
il0.
1
"r
'loiCto
12.710027
2.710027
:l1iCa
12.710027
;2.610831
f121Ke
,
10.0006905
,
10.0002635
1
;548.9418
+
~l
i13 ICe
,0
114icb
12.710027
12 .61 0831
12.710027
12.610831
l15lra
;
1-0.3604336
1-·0.3604336
1-0.0026249
1-0.0026249
116iCpa
:
~40.
140.
140.
140.
i
1
1
,,~
~
Differential equations
d(Fe)jd(V) = -ra - ke * Ce
d(Fb)jd(V)
= -ra
8-47
!
~J d(Fa)/d(V) = ra
~J d(T)/d(V) = ra * dHrx / Cpa
Explicit equations
Ft = Fa + Fb + Fc
dHrx = -20000
k = .133 * exp«31400 / 8.314) * (1 / 450 - 1/ T»
kc
= 10
Cto = 100 / .082 / 450
Ca = Cto * Fa / Ft
Kc = .01 * exp«dHrx / 8.314) * (1 / 300 - 1/ T»
Cc=Cto*Fc/Ft
Cb = Cto * Fa / Ft
;~Il ra = -k * (Ca - Cb * Cc / Kc)
Cpa = 40
6.0
4.,8
~ F~
[j
3,,6
Fa
2.-4
1.2
10
20
V 30
40
50
vary kc to see how the concentration profiles change.
P8-11 (b)
Now, the hear balance equation needs to be modified.
dT
dV
= Ua(Ta-T) + FAO(rA)[ LVIRx (T)]
FAO'X-B;Cp
See Polymath program P8·,1l-b.pol.
8-48
6.0.----------------,
4.8
[ill
FC
~ Fh
3. 6
Fa
2.4
1.2
10
20
40
V 30
50
P8-12(a)
To find the necessary heat removal, we start with the isothermal case of the energy balance
.lL_
Ws
F
F
AO
AO
--X[Ml RX +!J.CP (T-T,R )]="ac
.(T-T)
L,.IPI
0
Because there is no shaft work W s = O.
!J.Cp =60--25-35=0
And for isothermal operation T =To
If we simplify the energy balance using this information we get:
Q
--·--XMl RX =0
FAO
or Q = FAOXMl RX
= CAOvXMl RX
We now know everything except the heat of reaction to solve for the heat removed term. To find the
heat of reaction consider the adiabatic case:
!J.Cp = 60- 25-35 = 0
Q =0 and Ws = 0
--XMlRX = IB;CPi(T-To)
Because feed is equal molar in A and B, 88
- X Ml RX = (CpA + CPB )( T- To )
Ml RX
(25+35)(350-300)
kJ
-0.4
mol
=1
= ----.--.------ = -7500-
Now go back to the isothermal case:
.
Q=CAOvXMlRX
(mal)
mJ(0.2) (-7500~ol
kJ )
1000 m (.5 min
3
=
3
8-49
·
kJ
Q = -750000-.
mm
P8-12(b)
We start with the energy balance for the second CSTR (already simplified):
VA
-(Ta -T)-( X 2 -
XI)MI RX
FAO
= (CPA + CpB )(T --To)
This equation has two unknowns (T and X 2) and so we need another equation.
Now wee need the mole balance for the second reactor
FAO ( X 2- X I) FAO ( X 2- X I)
v: -
- ---'-'--'---
--rA2
2 -
-
kC~o (1- X 2 )2
This equation also brings in another unknown: k. We know that the specific reaction rate is
dependant on temperature and if we have the activation energy, we can make an implicit equation
for k as a function of T. To calculate the activation energy we will use the isothermal and adiabatic
information for reactor 1 and the mole balance for reactor 1.
=
V.
I
FAoX I
2 (
kCAO I-XI )2
k=
FAoX
VC~o (1- X)2
vX
-------VCAO (1- X)2
Solving for kat 300 and 350 K gives:
k(300 K) 0.00015625
k(350 K) = 0.0005555
=
If we plug these numbers into the Arrhenius equation we get
(kJ E(I IJ we get EIR = 2664.
In - 2 = -- -- -- kl
R 'Fr T2
Which means k(T) = k, ex p [
iU -~ J]
If we use a nonlinear equation solver to solve the energy balance and mole balance for reactor 2 we
find that the exit concentration is 0.423.
See Polymath program P8-12-h.pol.
POLYMATH Results
NLES Solution
Variable
T
X2
UA
Value
f (xl
Ini Guess
--~3~2~7~,,~6~87~1~2~~-2~.~.2~7~4E~--:1~3~~3~4~0~==
0.4214731
-6.666E-12
0.4
4
8-50
350
0. 5
0. 2
-7500
300
1
3 . 309E-04
1000
500
-110.73657
Ta
vo
Xl
dHrx
To
V
k
Cao
Fao
ra
NLES Report (safenewt)
Nonlinear equations
f(T) = (UA)*(Ta-T)/Fao-(X2-X1 )*dHrx·60*(T-To) = 0
[2] f(X2) = V-Fao*(X2-X1 )/(-ra) = 0
[1]
Explicit equations
[1] UA= 4
[2J Ta == 350
[3] vo=D..5
[4J X1
0.2
[5] dHrx = -7500
=
[6J To = 300
V =1
[8] k = . 00015625*exp(2663 ..8*(1/300-1/T»
[ 9] Gao = 1000
[101 Fao = Gao*vo
[11 J ra = -k*GaoJ\2*(1-X2)"2
[7J
P8-12(c)
Now we need the differential form of the energy balance
dT Ua(Ta -T)-rAMlRx Ua(Ta -T)-rAMlRx
dV-
FAOL~CPi
-- FAO(CPA+CPB)
we also need the mole balance for a PFR. For this case it simplifies to:
dC _ dCB _
- -A- - - - - r
dt
dt
A
with -rA = kCACB and we can use the same equation for k as in part (b).
When we put these equations into Polymath we get an outlet conversion of X
See Polymath program P8--12-c.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
--V
T
X
Cao
Ua
Ta
dHrx
Cb
initial value
0
300
0.2
1000
10
300
-7500
800
minimal value
0
283 . 98681
0.2
1000
10
300
-7500
671.87016
maximal value
1
300
0 . 3281298
1000
10
300
-7500
800
8-51
final value
1
283 . 98681
0 . 3281298
1000
10
300
-7500
671 . 87016
= 0.33
0.5
500
25
35
1.563E-04
800
-100
v
Fao
Cpa
Cpb
k
Ca
ra
0.5
500
25
35
1 . 563E-04
800
-42 . 749596
0.5
500
25
35
9.47E-05
671. 87016
-100
0.5
500
25
35
9.47E-05
671.87016
-42.749596
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb))
[2] d(X)/d(V) = -ralFao
Explicit equations as entered by the user
[1] Cao = 1000
[2] Ua = 10
[3] Ta=300
[4] dHrx = -7500
[5] Cb = Cao*(1-X)
[6] v =0 . 5
[7] Fao = Cao*v
[8] Cpa = 25
[9] Cpb 35
[10] k 0.00015625*exp(2664*(1/300-1/T))
[11] Ca Cao*(1-X)
[12] ra = ··k*Ca*Cb
=
=
=
P8-12(d)
In this case we need to replace the rate law we used in part (c)
-rA
=k
[c
A
CB _ K
Cc ]
C
We also need an equation to calculate Kc at different temperatures.
K
C
= K CI exp[~RX.(.!
__~J]
R
T.T
1
be careful of the units when entering Kcl into Polymath. Also note that the initial temperature is
different than in part (c)
We get an outlet conversion of X
=0.48
See Polymath program P8-12-d.pol.
POLYMA III Results
Calculated values of the DEQ variables
Variable
V
T
X
R
Ua
Ta
dHrx
Cao
initial value
0
350
0.2
0.0083144
10
300
-7500
1000
minimal value
0
314.93211
0.2
0.0083144
10
300
-7500
1000
maximal value
1
350
0.4804694
0.0083144
10
300
-7500
1000
8-52
final value
1
314.93211
0.4804694
0.0083144
10
300
-7500
1000
0. 5
500
25
35
5.556E-04
800
800
200
0.002
-300 . 01803
¥
Faa
Cpa
Cpb
k
Ca
Cb
Cc
Kc
ra
0. 5
500
25
35
5.556E-04
800
800
480.46942
8.63E+121
-64 . 253241
0. 5
500
25
35
2.381E-04
519 . 53058
519.53058
200
0 . 002
-335.38132
0. 5
500
25
35
2 . 381E-04
519.53058
519.53058
480.46942
8.63E+121
-64 . 253241
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb»
[.2] d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] R = 8 . 3144/1000
[2] Ua = 10
[3J Ta = 300
[4] dHrx=-7500
[ 5 j Cao = 1000
[6J v 0 . 5
[7] Fao = Cao*v
[8J Cpa = 25
[9] Cpb = 35
[10] k = 0.00015625*exp(2664*(1/300-1/T»
[11] Ca = Cao*(1-X)
[12] Cb=Cao*(1-X)
[13] Cc=Cao*X
[14] Kc = .002*exp«dHrxlR)*(1/350-1/T»
[15] ra = -k*(Ca*Cb-Cc/Kc)
=
P8-12( e) Individualized solution
P8-12(f)
For the gas phase the only the stoichiometry changes.
C
A
and
=C (~_)(To)
l+EX T
AO
E=
YAoJ=0.5(1-1-1) =·-0.5
From Polymath we see the exiting conversion is X = 0.365
See Polymath program P8-12fpol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
V
T
X
To
Ua
Ta
initial value
0
300
0. 2
300
10
300
minimal value
0
279 .3717
0.2
300
10
300
maximal value
final value
1
1
300
0.3650575
300
10
300
279 . 3717
0.3650575
300
10
300
8-53
-7500
1000
0.5
500
25
35
1.563E-04
-0.5
888.88889
888.88889
-123.45679
dHrx
Cao
vo
Fao
Cpa
Cpb
k
e
Ca
Cb
ra
-7500
1000
0.5
500
25
35
1.563E-04
-0.5
888.88889
888.88889
-56 . 423752
-7500
1000
0. 5
500
25
35
8.ll1E-05
-0.5
834.06666
834 . 06666
-123.45679
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(V) = (Ua*(Ta-T)-ra*dHrx)/(Fao*(Cpa+Cpb»
[2] d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] To = 300
[2] Ua = 10
[3] Ta= 300
[4] dHrx = -7500
[5] Cao = 1000
[6] vo 0 . 5
[7] Fao = Cao*vo
[8] Cpa = 25
[9] Cpb = 35
[10) k = 0 . 00015625*exp(2664*(1/300-1/T)
=
e =-.5
[11)
[12) Ca = Cao*(1-X)*To/(1+e*X)/T
[1.3] Cb = Ca
[14] ra = -k*Ca*Cb
P8-13
X;
= CCCD =
K
CACB
C
~X
.jK;
=
e
(1-. Xe)2
+.JK
1
T =To -
c
MIRX
CPA
+ CP
B
=300- (--30000). X = 300 + 600X
( 25 + 25)
See Polymath program pg··13.pol.
Calculated values of NLE variables
Tv~;i~b-I~lv~I"~;f(x) . ---- -~-r i~-itial Guess
-I
r~.999764~13.si8E-lilo:s .~ 6 <-x~·<
llxe
,,-
1. )
T-·
i IVariable IValue
!ll!
]300.
r'~'''"f A"_'''''~~'N~'
,,'
~","
.. " 1""0_",,,,.,,,.
,¥,,·,~.w_,
8-54
-7500
1000
0.5
500
25
35
8.1llE-05
-0.5
834.06666
834.06666
-56.423752
2 Kc
li8E+07'
.J
Nonlinear equations
f(Xe) = Xe - (1 - Xe) * Kc
A
0.5
=0
Explicit equations
= 300
Kc = 500000 * exp(-30000 /1.987 * (1/323 - 1/ T))
1. T
2
T
X
300
1
320
0 ..999
340
0 . 995
360
0.984
380
0 ..935
400
0.887
420
0.76
440
0 ..585
460
004
480
0.26
500
0 ..1529
520
0.,091
540
0.035
560
0.035
8-55
Xe
1
,-------~~-------.----------.----------.---------~
0.9
C 0.8
o
'en
... 0.7
~ 0.6
c
0.5
FxeJ
8
§. 0.4
!:!:!.. 0.3
Q)
><
0.2
0.1
o
·------·--···--···-·--·T--···--····--··-·----·-,--··--·-----·---,··-·-·-------·--,------·--··---····---1---·
300
350
400
450
500
550
Temp (K)
P8-14
For first reactor,
X
Kc
K = e l . or X c I-X el
el
I+KC
For second reactor
B X
K -B
K =2~orX = C
BZ
C
I-X eZ
eZ
I+KC
For 3rd reactor
K =.BB3+ X e3 orX _ Kc -BB3
C
I-X e3
e3
I+KC
1st reactor: in first reaction Xe =0.3
So, FB = F A01 (.3)
2nd reactor: Moles of A entering the 2nd reactor: F A02 = 2FA01 - F A01 (.3) = 1.7FA01
BBZ
= .2FA01 =.12
1.7 FAO
-FAOZ'LCp;~ (T -To) + FAOZX (-/ili R) = 0
X=
(CPA +BBCpB )(T -T 0)
-/ili R
Slope is now negative
3rd reactor:
8-56
X e2
=0.3
(.2FAOl ) + .3FA02 = FAOl (.2 + .3x1.8)
FA03 = FAOl + FA02 (1- XeJ = FAOl + 1.8FA01 (1- X e2) = 1 + 1.8(1- .3)FAOl
FA03 = 2.26FAOl
Feed to the reactor 3:
(2FA01 ) + .3FA02 = FAOl (.2 + .3x1.8) = 0.7FAo1
(}B =
Say
(} =
.74
2.26
Feed Temperature to the reactor 2 is (520+450)12 =485 K
Feed Temperature to reactor 3 is 480 K
Xfinal = .4
Moles ofB = .2FA01 + .3FAo 2 + .4FA03 =F Aol (.2 + .54 + (.4)(2.26))
X = FB/3FAOI = .54
B3
= 1.64 F Aol
Xe.
,
-----~~
............
__ -
1-/13
/"
.'f
.~
P8-14 (b)
The same setup and equations can be used as in part (a). The entering temperature for reactor 1 is
now 450 K and the outlet is 520 K. When the two streams are joined prior to entering reactor 2 the
temperature is (520+450)12 =485 K
Say that the outlet temperature for reactor 2 is 510 K. Then the entering temperature for reactor 3
would be (510+510+450)/3 = 490 K
For any reactorj,
-FAOl,CFj~ (T-To)+ FAOjX (--MI R )
X
=0
= (CpA + (}B C pB ) (T - To)
-MIR
and 9B for reactor 1 =O. For reactor 2, 8B > O. This means that the slope of the conversion line from
the energy balance is larger for reactor 2 than reactor 1. And similarly 8B for reactor 3> 8B for
reactor 2. So the line for conversion in reactor 3 will be steeper than that of reactor 2. The mass
8-57
balance equations are the same as in part (b) and so the plot of equilibrium conversion will decrease
from reactor 1 to reactor 2, and, likewise, from reactor 2 to reactor 3.
x
P8-1S (a)
Substrate -7 More cells + Product
S-7C+P
G(T) =
X*-~HRX
To solve for G(T) we need X as a function of temperature, which we get by solving the mass balance equation.
_ FAOX _ FsoX
V - - - - - - - and since -rs
-~
rg
= jiCc
-~
and ji
= -rg
~/!
then,
= ji(T) __C. _ 5 _
Ks +Cs
O.0038*Tex p( 21.6- 6~O)
where peT) = J4max
------
1+exP (153- 48~OO)
if we combine these equations we get:
v=
~/sFsoX
ji(T)CcCs
Ks +Cs
8-58
V=~/sFsoX(Ks+Cs)
p(T)CcCs
V= ~/sFsoX(Ks+Cso(1-X))
p(T)CSO~/l XCso (1- X)
Canceling and combining gives:
Now solve this expression for X:
Now that we have X as a function ofT, we can plot G(T).
To get R(T) we must calculate the heat removed which is the sum of the heat absorbed by reactants to get to the reaction
temperature and the heat removed from any heat exchangers .
The heat gained by the reactants =
Cpo (T - To )
The heat removed by the heat exchanger = UA(T-Ta)/Fso
VA
R(T) = Cps (T --Yo)+-F-(T -Ta)
so
Now enter the equations into polymath and specify all other constants_ The adiabatic case is shown below_ The nonadiabatic case would be with explicit equation [12] as A = 11.
See Polymath program P8-15--a.pol.
Differential equations as entered by the user
[ 1] d(T)/d(t) = 1
Adiabatic Case
Explicit equations as entered by the user
[lJ mumax=_5
[2J Ycs = . 8
[3] vo=1
[4J Ta = 290
[5] mu = mumax*( . 0038*T*exp(21 . 6-6700/T))/(1 +exp(15348000/T))
[6] Ks=5
[7] V = 6
[8] Cso = 100
[9 J Fso = vo*Cso
[10] Cps = 74
20000 . - - - - - - - - - - - - - - - ,
10000
o
-10000
-20000
-30000
L-_~
294
8-59
_ _~_~_ _~_---I
299
303 T 308
312
317
[ 11] dH = -20000
UA = 0*300
kappa = UA/(Cps*Fso)
[14] To = 280
[15] X = 1-«Fso*Ks)/«mu*V*(Cso"2))-Fso*Cso))
[16] Gt = X*(-dH)*Ycs
[17] Rt = Cps*(T-To)+UA*(T-Ta)/Fso
[12]
[13]
Non-Adiabatic Case
20000 . . . . - - - - - - - - - - - - - - - - ,
1-4000
Independent variable
variable name: t
initial value : 0
final value: 30
8000
2000
-4000
-10000
P8-1S (b)
_ _~_--,-_ _~__--l
299
303 r 308
312
317
Il-_~
2 4
To maximize the exiting cell concentration, we want to maximize
the conversion of substrate . If we look at G(T) from part A, we see that it is at a maximum at about 310 K. This
corresponds to the highest conversion that can be achieved. By changing the values of UA and Il1c we can change the
slope of the R(T). What we are looking to do is get R(T) to intersect with G(T) at 310 K.
Since we now have a limited coolant flow rate we will use a different value for Q.
and so,
-u~J) + Cps (T -To)
R(T) = mcCpc (T -Ta {1-exp(
\
mcCpc
Now we set R(T) equal to the maximum value of G(T) which is 15600 Jlh
f
u~.J) + FsoC
G(T) = 15600Fso = mcC pc (T - Ta 1- ex p(- \
mcCpc
PS
(T ._- To)
And now plug in the known values . Assume the maximum coolant flow rate and that will give the minimum heat
exchange area.
15600{(100 L )
g
hr
=
8··60
1560000
~.
= (88800000 ~)
hr
hr
-UA
1-- exp
[
4440000
_J_)
+ (162800
~)
hr
hr K
J
VA=70415-hrK
P8-1S (C)
There are two steady states for this reaction" There is an unstable steady state at about 294.5 K and a stable steady state
at 316 K.
P8-1S (d)
Increasing To enough will eliminate the lower temperature steady state point. It will also lower the outlet concentration
of cells. Decreasing To will increase the outlet concentration of cells.,
Increasing me increases the slope of R(T) and will increase the exit concentration of cells,
Increasing T a will lower the exit concentration of cells.
P8-16 (a)
G(T) = illi R X
X=
Tk
l+Tk'
-3
T=6.6xlO exp
[E[
1- T1)]
Ii 350
= cpo (1 + K)(T - Tc)
cpo = I. BP p; =BACpA +B1CpI =20+30=50
R(T)
K=
VA =_8000_=2
CpoFAO (50)(80)
= ToFAOCpO +UA~ = KT,,_+To
=350
VA+CpoFAO
l+K
To find the steady state T, we must set G(T) =R(T). This can be done either graphically or solving
the equations. We find that for To =450 K, steady state temperature 358 K.
Tc
P8-16 (b)
First, we must plot G(T) and R(T) for many different To's on the same plot. From this we must
generate data that we use to plot Ts vs To.
8-61
I
G('l) \S. R(1)
I
1;,'5. T.
10000 ,-------;-~,.........,,_r"7"__.
425
;----~t-)'#~(;C.:;;;;;I
400
I; ~+-------~"~~~~~--~
i-
8000
375
CD
E~r-----~~~~~~-----~
~ ~r---~~~~~~~----~
O~"
350
325
__~~U--L~~____~~
Z15
325
375
TOO
300
425
200
250
300
350
-----"
P8-16 (C)
400
T. (X)
450
500
550
For high conversion, the feed stream must be pre-heated to at least 390 K. At this temperature, X =
.98 and T =380 K in the CSTR. Any feed temperature above this point will provide for higher
converSIOns.
P8-16 (d)
For a temperature of 375 K, the conversion is .968
P8-16 (e)
The inlet extinction temperature is 375 K.
P8-17
The energy balance for a CSTR:
-FAO XllH Rx (T)
UA
= FAO ["'£.e.c pi.(T -.r,0 )+--(T
-.r,)]
F
0
I
AO
FAOX =-'rAV
Cps isindependentofT.
Cpo (T -- To ) = C p'
G(T)=(-llHRx)(-~VJ
F
AO
R(T) =
"f'l'
FAOCpS
+UATr -UAT
".
dR(T)
dT
dG(T) _ -llHRx d(-rA)
dT
FAO
dT
D 1 lerentlatmg: ---- = UA
where
d( -r )
A
E
= - - 2 ( -rA )
dt
RT
Setting these two equations equal to each other and manipulating,
8-62
In 1.421
1.127
=[E(323-313)]
R 323 *313
E=19474
FAOCpS + (T _ T ) > RT,2 ,andit will be runaway reaction
UA
T
a
E
2
At A < 2 ..5 m it will become a runaway.
P8-18 (a)
Mol Balance:
~ _ FAOX _
-
-rA
voCAO X
-kTC
A
-CB / Ke1
(:'}[(l-X)-X I K,]
X [1+rk(l+ 11 Ke)]
X
=
= rk
rk
l+rk(I+11 KJ
G(T) = -M/RxX = 80000X
k=lmin-- 1
T=lOmin
Ke =100
X =.
10
=.901
1+ 10(1.01)
G(400) = 72080 cal/mol
P8-18 (b)
UA
R(T)
= 3600 =9
10*40
= CpA (1 + K)(T -·Tc ) = 400(T -
Tc)
Tc =ToI+K
+KT .=31O
a
R(T)
= 400(T -
310)
The following plot gives us the steady state temperatures of 310,377.5 and 418.5 K
See Polymath program P8-18--b.pol.
8-63
-r-----
-
1_
OF------~~-_+_
-
_ _,_-__o_--+___...:=._
390
P8-18 (C)
3 10K and 418.5 K are locally stable steady-state points
P8-18 (d)
R{418.5) = 400(418.5-310)=43400
G(418.5} = 43400 = 80000 * x
43400
x = 80000 = .54
P8-18 (e)
-
The plot below shows Ta varied.
..
'
.....
......
._,.".".~." .."'''''-''''./''.''
t ....
..
Ta~
___._____.c=..__
• 100:--_ _-_~=----_, _~
m
_
_
~
m
m
_
GO
The next plot shows how to find the ignition and extinction temperatures. The ignition temperature
is 358 K and the extinction temperature is 208 K.
8-64
...
~5S
...
...
.----'---~---.-.---.---
1501.----150
-
P8-18 (f)
This plot shows what happens if the heat exchanger quits. The upper steady-state
temperature now becomes 431.5 K.
-,
70000
_50000
15
.l!.
i..
4DOOO
E
" 30000
20000
1DOOO
351
0
,.0
330
3IiO
TIKI
P8-18 (g)
At the maximum conversion G(t) will also be at its maximal value. This occurs at approximately T
=404 K. 0(404 K) =73520 cal. For there top be a steady state at this temperature, R(T) =OCT).
See Polymath program P8-18-g. pol.
8-65
VA
K=--FAOC pA
R(T) =
CpA
where T
(1 + K)(T - TC) = 73520
= T0 +KTa
I+K
C
If we plug in the values and solve for VA, we get:
VA 7421 cal/miniK
=
P8-18 (h) Individualized solution
P8-18 (i)
The adiabatic blowout flow rate occurs at T = 0.0041s
V
T = 0.0041s =-Vo
Vo
= TV = 0.0041 *10
Vo
dm 3
=0.041..
mIll
See Polymath program P8-18·-i.pol.
6000
4600
321.11.1
1801.1
-11.10tll-------~-
o
30
60
t
90
120
151.1
P8-18 (j)
Lowing T 1.1 or Ta or increasing VA will help keep the reaction running at the lower steady state.
P8-19
Given the first order, irreversible, liquid phase reaction:
A·-7B
VA = l.Ocal! minl°C
8-66
C 0, = CPo == 2eal/ g - mol/DC
Pure A Feed = o...5g mol / mm
I1CP
M1 R
= I1CpA -- ilCpB = 0
= -200
cal
gmofA
= canst
_ FA oX _
Vo
Designqn·
V -----.---E
-rA
k(l---X)
RateLaw..
-rA
StolchlOmetry
= kCA
C A = CAO(l- X)
EnergyBafance -UA(T -T,1)-- FA oX MR
Simplifying,
-rk
= FAOL~Cpi(T --1'0)
1
X=--=-l+-rk
1+_1
-rk
o
~g
=F X(--M ) == F40 (--M-!.R)_
AO
R
}
1+rk
ThIS IS the curve plotted m the prohlem statement From the equatIOn for heat generatIOn curve, we
get
X == __ Jl~_-_ =-~ = Qg
F 4o (--MI R )
5*200 100
The equatIOn for heat removal curve IS·
QR =FAoCp4.(T-To)+UA(T-T4)==·5*2*(T--I'u)+1 O*(T-T j )
QR = 2T--l;-·} 00
Plot thl:' along with the heat generatlon curve for vanous To This I:' shown m the figure
r=:::~:.---
Qg------1
I
1- -
I
.•- T0=15O I
I, . . . . . 1·0= 160 Ii
1-' -
·To=170
I
i-- ...- TO=180 \
l-
r 0=1901
. - -T0=200
I.____- ._.-___
To=210
L
._--"I
120
140
160
180
200
220
T fe)
The intersectlOn between Qg and QR can be used to
plepme the igmtion - extinctlOn CUlve :.hown m figure
P8-22-:2 The value~ fOl T~ a~ a flJnction of mlet
temperature, To, are tabulated below.
8-67
To (degree C)
150
157
160
170
190
202
210
Ts (degree C)
132
135, 172
137, 168,176
142.5, 167, 181
154.5, 165, 193.5
162, 199
204.5
P8-19 (a)
To obtain high conversion, the reactor must operate at or beyond point 14 in figure. So the
minimum inlet temperature for high conversion is To 2 202°C.
P8-19 (b)
The temperature of the fluid in the reactor corresponding to temperature in part 1 is Ts > 199°C
(point 14 in figure).
P8-19 (c)
The inlet temperature of the fluid is To = 202 + 5 = 207°C. This will be somewhere between points
14 and 15 in figure. Once the fluid is cooled from this temperature, it will follow the line formed by
points 15, 14, 12,9, etc .. Now To =207 - 10 = 197°C. From figure, Ts = 195.5 °C. From equation
(5),
Qs = 2*196.5-19-100 =95 call mol
QR
=Q =96 cal/mol
So X
,
g
= 100
Q =0.96
g
P8-19 (d)
Extinction temperature is the temperature corresponding to point 3 in the figure = 157 °c
P8-19 (e) Individualized solution
P8-20 (a)
The following are the explanations for the unexpected conversion and temperature profiles
Case 1: Broken preheater or ineffective catalyst
Case 2: The eqUilibrium conversion was reached due to a problem with the heat exchanger
Case 3: Broken preheater or ineffective catalyst
Case 4: The equilibrium conversion was reached due to a problem with the heat exchanger
Case 5: Ineffective catalyst
Case 6: The equilibrium conversion was reached due to a problem with the heat exchanger
P8-20 (b)
8-68
The following are the explanations for the unexpected conversion and temperature profiles
Case 1: Broken preheater or ineffective heat exchanger
Case 2: Ineffective catalyst
Case 3: Broken pre heater or ineffective heat exchanger
Case 4: Ineffective catalyst
P8-21
Below is the FEMLAB solution.
1. Parameters in simulation on the tubular reactor in Problem 8-6:
Reaction: A + B ~ C
(1) operating parameters
Reactants
•
Inlet concentration of A C AO
= 100
3
mol/m
•
Inlet concentration of B C BO = 100 moilln'
•
Inlet total flow rate
•
Inlet temperature of the reactant To = 300K
Vo
= 2xlO-3
m3/s
(2) properties of reactants
• Heat of reaction, t:.HRx , dHrx = -41+20+15=-6 kcallmol=-25100 J/mol
• Activation energy, E = 41840 J/mol
• Specific reaction rate ko = 0.0IxlO-3 m 3/mol.s @300K
• Reaction rate k
= ko exp[ E (~-!-)]
R To
T
• Gas constant, R = 8..314 J/mol·K
• Rate law - fA = kC A CB
Assumption:
• Thermal conductivity of the reaction mixture, ke
= 0,,68 W/mK
(needed in the mass halance and the energy balance)
• Average density of the reaction mixture, p, rho = 1000 kg/m3
(needed in the enelgy balance)
• Heat capacity of the reaction mixture, Cp = 4200 J/kgX
(needed in the energy baJance)
9
• Diffusivity of all species, Diff= 10- m%
2. Size of the Tubular Reactor
(1) Volume of reactor sized by a PFR = 0.317 m'
(2) From FEMLAB
• Reactor radius, Ra = 0 . 1 m
• Reactor length, L = 10.0 m
8-69
3. Femlab Screen Shots
(1) Domain
[ ] Axis equal
:\
[
OK ,,'
!I [
cane;]
,
[A;i¥___J
(2) Constants and Scalar Expressions
- Constants
- Scalar Expressions
[~'"~-:J
r'~-~
OK
Cancel
j
'"
(3) Subdomain Settings
- Physics
8-70
j
~z~·····,··,
.
-,
['---'''-J
Apply
_'_'
(Mass balance)
rEquation
I Vo(-DVcA+cAu) =R, cA =concentration
I
rSubdomain selection"'~"
'_-13]:1
Species
Library material:
ValueJExpression
Quantity
Time-scaling ...,.".ff;,~;",-,t
°ts
Diffusion coefficient
D isotropic
Diffusion coefficient
D anisotropic
R
[J Select by grou!"
__ "'" ___._N.____ ._"_____
~===;;;;;=~
Reaction rate
u
'=::;;;;;:;;:;;;;;;;;;;;;:;;==":: r-velocity
v
z-velocity
------.---J
._--_._-_._--
[ .__Artificial Diffusion ...
~ Active in this domain
(Energy balance)
Description
¥ _ _ _ _ _ · _ · _ _ _ , , _ _ _ _ • _ _ _ ' , · _ _ _ _ _· _ - - - - - . - - - - - - - - - - . - - -
;-Equation
Ii V.(··kVT +
~;h;HDJ) = Q - PCpu.VT, T=temperature
;
rsubdomain selection
In~.: Element
Physics
Thermal properiies and heat sources/sinks
I
library material: i.
ValueiEKpression
Quantity
[1
(V k (isotropic)
~k~
ok
Dllscription
1 Time-scaling coefficient
01S
•. , Thermal conductiv!y
Thermal conductiv~y
(anisotropic)
Dens!y
Heat capac!y
Heat source
u
.........................
o
hj"O,i
Select by group
~ Active in this domain
[
r-veloc~y
z.veloc!y
v
l~Eecie: ~~.!~~i"~~~~~~~ ¥~
:J
Artifici;oiff';-sio~ ..
- Initial Values
(Mass balance) cA(tO) = cAO
(Energy Balance) T(tO) = TO
- Boundary Conditions
r = 0, Axial symmetry
@ inlet, cA = cAO (for mass balance)
T = TO (for energy balance)
@ outlet, Convective flux
@ wall, InsulationiSymmetIy (for mass balance)
Thermal Insulation (for energy balance)
(4) Results
@
(Concentration, cA)
8-71
File Ed~ Options Draw Physics Mesh SOlve Poslp"ocessing Mulliphyslcs Help
_ti~ ~_ .iL~_~~I~I~~~J~_~~~_li[~~~_~L~~~~~~_~~l !. ____ _
____________R_________
[!]
___________________._____._
Max: 97.38
Contour: Concentration, cA
10
&~,
--------------- -------;===:;==:::;::~~;:;=;::;:':;
>--,
~
467
/
_~-1==:;=========~
01
02
-------0-.3- - - - - ' - 0 - _ 4 - - - - - -
2--~--<l1-----
.05"
'~
[SNAP i
(0,4.~)
10~'--'"----·----..····-'"--·-----·-··------378
333.361
331 344.
;.' 329 321
•• 327 31
4
03
8-72
P8-22 (a)
Liquid Phase: A + B ---7 C
= -rA
dX
dV
FAO
-rA =kCACB
k=.01*exp [- RE(~.T __
l )]
300
CA = CAo (l-- X) = CB
The energy Balance:
dT _ Ua(Ta -T)+(-rA)(-LlliR )
FAo [CPA
dV' -
+ CpB ]
4
D
a=-
AssumeD =4
U
=-~x2
m Ks
2
leal x 1m = .0120 cal
4.1841 lOdm 2
dm 2 Ks
~ee Polymath program pg·22·a.poJ.
Calculated values of DEQ variables
'-lv~~i;bleli~itial-~~I~e-)Mini~al valu~!Maximal valuelFinal value
,,1
i
i1 IV
f
,2 IT
'3
Y"~
"
.fo
IX
;300.
SR
11.988
,
16'E
; I!
r
17 leao
. "~~,,
~
0
1300.
:4 "Ta
;8
,.
10
'0
~
i1000.
11000.
1434.7779
1348.2031
10.9620102
10.9620102
r
-"t'"
300.
~
;300.
i300.
i
110000.
10 . 1
110000.
,
1.988
!10000.
10 . 1
iO.1
10.1
:0.1
1'6.003799
;0.003799
10.1
101
l •
10.003799
'0.01
.
'-0.0001
,0.01
11.808628
iO.1018746
-0.0010427
,-1.47E-06
15.
i1S.
,,i-1.47E-06
15.
15.
0.2
10.2
1
:1.988
11.988
,
110000.
1
t .
l'"
10.1
,
iO.003799
", r-
'[
lea
19 'f~b
.
I
iO. 2
~
r~6000.
.1
,
-6000.
..
...
i
i
115.
,
15 .
1
10.2
r-~
1-6000.
~
-
.1=6000:
8-73
Differential equations
d(T)jd(V) = (U * a * (Ta - T) + (-ra)
* (-Dhrl)) j
(faa
* (cpa + cpb))
2; d(X)jd(V) = -ra j faa
Explicit equations
Ta = 300
R = 1.988
E = 10000
caa
= .1
ca = caa
* (1 - X)
cb = caa
* (1 - X)
* exp(-E j R * (1 j T - 1 j 300))
ra = -k * ca * cb
k = .01
cpb = 15
~~r~l cpa
j~~;M
= 15
faa = .2
'~l' Dhrl = -6000
J3~a=1
1l;~.; U = .0120
500
1.0
460
0.8
EiJ
420
0.6
380
0.4
340
0. 2
300
0
200
400 V 600
800
1000
200
P8-22 (b)
8-74
400 V 600
800
1000
Gas Phase:A f=Z B + C
dX
= -rA
dW
FAD
= kjCA -
-fA
krCBCC
X Yo
C =C =C - - B e A D 1+ X T
k = .133*exp [- E(J:___
1 )]
R T
450
k. =.2exp [E, (_I__ !)],Er
,
R
450
T
= 51.4kJ Imol
The energy Balance:
dT Ua(Ta -T)+(--rA)(-Mi R )
dW
=
Mi R
= -40 -
50 + 70=-20 kllmol
U=5
See Polymath program P8-22-b.pol.
Calculated
values of ,-,
DEQ
variables
.. ", .....,.
.T
""T'
,-- i '
:Variable! Initial value IMinimal value, Maximal value! Final value
,1 jW
. :0 " " ' 1 0
150.'
. " "/50.
12 ,X
10
10
10.0560855' \0.0560855
,3 T
[400.
/371.902
1400.
,
1
IJl1.902
1
>
'4 iTO
!400.
1400.
1400.[400.
is ,!k
iO.133
10.0651696
,0.133
6 IvO
17
20.
'0.2
1
[kr
!
:0.0651696
,~
1"
:5.
18 IUarho
i20.
10.0622149
,5.
,;20.10.0622149
,
20.
t
iO.2
1
,~5.
!323.
15.
]"323.
11.013E+06
I~.,013E+06
1304.6819
"
1304.6819
,1304.6819
1304.6819
/304.6819
'292.8948
!304.6819
1292.8948
10
0
117.40323
!17.40323
14IC6
10
0
;17.40323
117.40323
15frA
'.
'-40.52269
!
-40.52269
1-0.2446624
1-0.2446624
19
IT~
b23.
··11.013E+06
10jPO
llicAO
12ICA
13 Icc
T
!
,
;
~
~
:323.
~
Differential equations
i d(X)jd(W) = -rA j vO j CAO
8-75
J1.013E+06
~j d(T)/d(W)
= (Uarho * (Ta - T) + rA * 20000) / vO / CAO /
40
Explicit equations
= 400
k = 0.133 * exp(31400 / 8.314 * (1 / TO - 1/ T))
vO = 20
kr = 0.2 * exp(51400 / 8.314 * (1/ TO - 1/ T))
Uarho = 5
;§;:Ta = 323
TO
PO = 1013250
= PO /8.314/ TO
CA = CAO * (1 - X) / (1 + X) * TO / T
$~l CC = CAO * X / (1 + X) * TO / T
CB = CAO * X / (1 + X) * TO / T
rA = -(k * CA - kr * CB * CC)
CAO
400
0.060
394
Q
388
382
376
0012
---~-- ...
0. 000
....J.-~---.I-_
0
10
20
W 30
:;0
40
370
0
10
20
W 30
---.--,
40
50
P8-23
First note that ~Cp = 0 for both reactions. This means that MIRx(T) = ~HRx 0 for both reactions.
Now strut with the differential energy balance for a PFR;
dT _ Ua(Ta - T)+
dV -
,L lij (-MiRxij) _ Ua(Ta -T)+ 'i.c (-MiRxIC) + r2D (_.Mi Rx2D)
,LFjCpj
-
,LFjCpj
If we evaluate this differential equation at its maximum we get
dT
--- = 0 and therefore, Ua(~ - T) -- 'i.c Mi RxIC -- r2D Mi Rx2D = 0
dV
We can then solve for rIC from this information.
8-76
MI RxIC
'ic
=
10(325 - 500) - 0.4 (0.2) (0.5) (5000)
-50000
=0.039
'ic = 0.039 = klCCA CB = klC (0.1)( 0.2)
klC =1.95
~C(500) = ~C(
~=
R
E
<ro) exp [
In(kl c(500)
)]
J
=7628
kIC (400)
(4~0 5~0)
=15,158
! (~ -5~O
cal
mol K
P8-24 (a)
See the additional homework problems in chapter 8 at bttp://www.engin.umich.edu/~cre for the full
solution.
,-----------------,
Iso1hermal: T = 321 K
RMAt\. = I, Q = 40 m~lIL
1
1
~
:
II
~.
S4-
~-~
!:~--~
.---"" ..,---~
.___'..-
i
-
o -t=----,-.--,.- --,----
o
-.2MIB
02
04
._-_....-,,2M2B
x
06
-Me0H
I
i
-l
0.8
--"'TAME
P8-24 (b)
See the additional homework problems in chapter 8 at http://w\yw.engin.umich.edu/~cre for the
equations to enter into Polymath. Then vary to see its effect.
8-77
P8-24 (c)
See the additional homework problems in chapter 8 at http://www.engin.umich.edu/-cre for full
solution
Non-Isothermal: To = 353 K, Tw = 298 K
RMIIA = 1.0, Q= 200 Llntin
7
-----------------
6
-
5
$ 4
o
.,5,3
U
2
0. -1"'"'----,.-----.- ---.------,-----1
o
0.2
08
0..4 X
0.6
P8-24 (d) No solution will be given
P8-25
Mole balance:
dF
- -A= r
dW
A
Rate Laws:
rA = --rZB + 'iA
dF
dW
- -B= r
B
dFc
--=r.
dW
C
+ ljA
rc = -r3A
-'iA = klCA
-rZB = kBCB
--r3A = k3 CC
Stoichiometry:
C
A
=C
1
_FA'Fo
FT
1
C =C FBTO
B
T FT
T
Energy balance:
_dT = Ua(Ta -- T) + (---'iA)( -LV!RIA) + (-rRZB ) + (-ljA)(-LV! R3A)
dW
FAC pA + FBC pB + FcCpc
8-78
·dT = 16(500 - T) + (-ljA)1800 + (-rR2B )1800 + (-'3A)1100
100(FA + FB + Fe)
dW
kJ =0.5exp[2(l-320IT)]
kJ
k =--
K
2
C
k3 = 0.005 exp [4.6(1-460IT)]
Kc =lOexp[4.8(430IT-1.5)]
See Polymath program P8"25.pol.
~alculated
"alues of D~Q variables
"
r
IInitial
value; Minimal value Maximal value i Final value
, Variable ,
<
l
""
1
iW
.~
'2 ,fb
o
11.
I
13 Ifa
r '
4 :fc
,5 IT
i1.
:0
330.
;100.
0
1
1.368476
10.9261241
jO.9261241
I
iO.6296429
iO
i1.
~
1
j
10.8694625
10.2044134
10.2044134
330.
1416.3069
;416.3069
:16.""
:1
,6 jua
16.
,16.
]16.
7 ITa
500.
500.
]500.
,
}
500.
8 /Dhrla
i- 1800.
·-1800.
1-"1800.
)-1800.
9 iDhr3a
i-llOO.
i-llOO.
:-1100.
'100.
100.
1~1l00.
,i100.
,100.
1100 .
<1
I 100
'100.
100.
1100.
.
1100.
,
;100.
1
1
'100.
1
:0.5312401
0.5312401
:0.7941566
:0.7941566
iO.0008165
'0.0008165
;0.0030853
:0.0030853
12.
2.
'2.
i2.
2.
Ir'2 .""""
:2.
'"2.
1
1330 .
;330.
17!To
1330.
181Kc
I1~·885029
r
11.062332
3.885029
19:k2
,0.1367403
10.1367403
0.74756
i21 cb
1.
~0.5682599
1.
i1.
)0.7341242
1.253213
Io.?341242
'22[rla
1-0.5312401
-0.5748799
-0.362406
" ;-0.5473402
-0.002138
]-0.0007594
i" .~,
f··
".~.
!201ca
J
f
,
f
!-0.0008165
231r3a
T'
241rc
;0.0008165
i25 jr2b
1-0.1367403
I
10.3944998
j
!261rb
~
i
127lra
,
,;0.002138
-0.5770243
-0.0522707
~<-~
i-0.3953164
!-0.1367403
,
10.3944998
,
-0.3953164
)~~~
--
""p30.
:1.062332
10.74756
"" 10.6892094
!-0.0021264
··T'
10.0021264
1-0.5488019
T
" --
1-0.0014617
.~"
10.0510521
8"79
1-0.0006647
Differential equations
d(fb)jd(w) = rb
= ra
d(fc)jd(w) = rc
d(T)jd(w) = (Ua * (Ta - T) + (-r1a) * (-Dhr1a) + (-r2b) * (Dhr1a) * (-r3a) * (-Dhr3a)) j
d(fa)jd(w)
cpb
(fa
* cpa + fb *
+ fc * cpc)
Explicit equations
Ua = 16
Ta
= 500
Dhr1a
= -1800
Dhr3a
= -1100
= 100
cpb = 100
cpc = 100
k1 = .5 * exp(2 * (1 - 320 j T))
k3 = .005 * exp(4.6 * (1 - (460 j T)))
l,~,~ ct = 2
!U:ift = 2
~Z~To = 330
m~ Kc = 10 * exp(4.8 * (430 j T - 1.5))
r~~; k2 = k1 j Kc
!t~~ ca = ct * fa j ft * To j T
cb = ct * fb j ft * To j T
r1a = -k1 * ca
[,1,;11 r3a = -k3 * ca
l~~: rc = -r3a
cpa
~~ r2b
= -k2 * cb
rb = -r1a + r2b
a~ ra = -r2b + r1a + r3a
P8-25 (a) As seen in the above table, the lowest concentration of o-xylene (A) = .568 mol/dm3
P8-25 (b) The maximum concentration of m-xylene (B) = 1253 molldm3
P8-25 (c) The maximum concentration of o-xylene = 1 molldm3
P8-25 (d) The same equations are used except that F BO =O.
=
The lowest concentration of o-xylene 0.638 mol/dm3. The highest concentration of m-xylene =
1.09 mol/dm3. The maximum concentration of o-xylene = 2 mol/dm3.
8-80
P8-25 (e)
Decreasing the heat of reaction of reaction 1 slightly decreases the amount of E formed. Decreasing
the heat of reaction of reaction 3 causes more of C to be formed. Increasing the feed temperature
causes less of A to react and increases formation of C. Increasing the ambient temperature causes a
lot of C to be formed.
P8-25 (0 Individualized solution
P8-26 (a)
AH B+C
A-7D+E
A+C-7F+G
We want the exiting flow rates B, D and F
Start with the mole balance in PFR:
dF
dFc
dFA
--=ra
- ·B= r
--=r.
B
dV
dV
c
dV
dF
dFE
dFG
_F·=r
-=
r
--=r.
dV
E
dV
F
dV
G
dF
dV
- -D= r
Rate Laws:
8-81
D
Fe
FT
Pe =--Pro
~=~+~+~+~+~+~+~+~
F[ = steamratio x .0034
Energy Balance:
dT
- ( 'is /lll RIA + r2B /lll R2A + r3T /lll R3A )
dV FA *299+ FB *283+ Fe *30+ FD *201 + FE *90+ FF *249+ Fo *68+ F[ *40
Kpi =exp(bi +
i
+b3 In(T)+[(b4T+b5 )T+b6 ]T)
See Polymath program P8-26.pol.
For TO =800K
Calculated values of DEQ variables
111~ariabl~~ti~1 val;;elrini;;;;;'~alu~lrjximal valu;l:~a! value!
"...
If
]2 a
13
14
15 Ifd
...,
.,
10.00344
Ifb '10
IreTa
1
16 rf~
17'
Iff
Is'l1=g
1
-['
;9;1
16
!
\0
"'[0
nip "
1,41p,~i
lsll<l
~~!:
118rr~,
j19 1Pa
.~
jO
10
.,.
h.078E-05
i3.588E-05
h.588E-05
"""1800'."
r1.18E~051i.18E+05
J
j
I
j
1
I
!1.078E-OSj
j3.588E-05j
]3.588E-05
',1765.237
I
[-s.39E+041-5.39E+041
12137.
J
hBl.
a.4
,jc):41
10:0459123 ... 10.0196554
10.0459123
1~:!988· ·1~:!988 ·16:!988
., "'1!~·05332
,10.05332
10.154838710.1104652
]i:991E-06
"... ....
jl.18E+051
k052E+05T1.052E+C)sji.oS2E+05!
1~.4
ri~I;;b j~.991E-06
Ed
10.00344
iO.0 02,496
10.0008974
1o.Q()08974
' jO.0008615
10.000861S'
i
11.078E-05';l 1.078E-05
f-5.39E+04' '!-5.39E+04
12137. .,".
12137.
i20lpbI~
]23
......
10.002496
/0
10
10
1010
":,i800.
T1765.237
10TH1al1.18E+05
11IH2a
. Ti:052E+05
121 H3~
, .
~
[0.0542282
10.1548387
lO.4
J
10.0196554 1
1~:!988· j
fO:c)542282 , j
10.1104652 J
fo
10.039715510.0397155 I
I~.16E-07
li.991E-06
tLE-07
!~:~!~~:1~:~!:~:;71
8-82
]5.16E-07
i24ire
f
2.991E-06
!2.991E-06Is.16E-07
2SF3t--Io
10
10-10
i26T~·To
10
271rg
J4.196E-06
,!4.196E-06
14. 196E-06
:?8.r~~-
ig·_0002138
-i2.481E-OS
!0.0002138
)2~! rb
10.0002138
12.481E-OS
:0.0002138
:.!3301._.•l..i rr.ac: _
10.0002138
12.066E-OS
~.
10.0002138
'0;
1-2.948E-OS
1-0.0002167]-0.0002167
... L
i2.481E-OS
L
~. t_.
12.066E-OS
'+ ...
j-2.948E-OS
Differential equations
= ra
7. d(fb)jd(v) = rb
,J d(fc)jd(v) = rc
d(fd)jd(v) = rd
~! d(fe)jd(v) = re
;p; d(ff)jd(v) = rf
d(fg)jd(v) = rg
J
d(fa)jd(v)
l
d(T)jd(v) = -(rls
•.. . * 68 + fi * 40)
* Hla + r2b * H2a + r3t * H3a) j
(fa
* 299 + fb * 273 + fc * 30 + fd * 201 + fe * 90 + ff
Explicit equations
Hla
= 118000
H2a
= 105200
H3a
= -53900
P = 2137
l5
phi
= .4
J§t. KI = exp(-17.34 fi
* In(T) + «-2..314e-1O * T + 1.302e-6) * T + -0.004931) * T)
= 14.5
7~: sr
8i~
1.302e4 j T + 5.051
= sr *
.00344
ft=fu+fu+k+~+fu+ff+~+fi
f~ Pa
= fa j
ft
* 2.4
= fb / ft * 2.4
re
··l pc-l'cjft*24
r2b = p * (1 - phi) * exp(13.2392 ... 25000 j
i~~] Pb
'<:.':~
II
•
T)
* Pa
rd = r2b
it$! re = r2b
!~qi r3t = p * (1 - phi) * exp(.2961 .- 11000/ T) * Pa * Pc
ll' rf = r3t
~~ rg
= r3t
rls = p * (1 - phi) * exp( -0.08539 - 10925 j T) * (Pa - Pb * Pc j
8-83
KI)
= rls
.~1.j rc = rls - r3t
;~Qi rb
= -rls - r2b - r3t
.~! ra
= 0.0008974
Fbenzene = 1.078B-05
Ftoluene =3.588B-05
Fstyrene
SS/BT
= 19.2
P8-26 (b)
To= 930K
=0.0019349
Fbenzene =0.0002164
Fstyrene
Ftoluene
SS/BT
= 0.0002034
=4.6
P8-26 (c)
To= 1100K
= 0.0016543
Fbenzene =0.0016067
Ftoluene =0.0001275
Fstyrene
SS/BT
=0.95
P8-26 (d)
Plotting the production of styrene as a function of To gives the following graph. The temperature
that is ideal is 995K
1. i=b (S~) VI. Tol
0.0023
r------:=--------.
0.0022,
C.0021
t
G.~02 f
0.001'
C.0:U8 i'
C.OO17
i
0.0018 T
0.001& ,.
.,/~
f/
.\
0.0014 -'-----...,...--~-__J
lOa
100
tooo
1100
1200
8-84
P8-26 (e)
Plotting the production of styrene as a function of the steam gives the following graph and the ratio
that is the ideal is 25:1
Q.OO2
--..:::=====::-___
0 . 0019 +0001' •
!
00017-
o OGIa0.0015 ....- - - -_ _ _ _ _ _ _ _,.j
to
20
40
P8-26 (0
When we add a heat exchanger to the reactor, the energy balance becomes:
dT
=
Ua(Ta -!)-('isLViRIA +r2BLViR2A +r3TLViR3A)
dV FA *299+ FB *283+ Fe *30+ FD *201 + FE *90+ FF *249+ FG *68+ F, *40
With Ta = 1000 K
Va = 100 kJ/min/K = 1.67 kJ/s/K
The recommended entering temperature would be To = 440 K. This gives the highest outlet flow
rate of styrene.
40e-3
300
3.2e-3
24£1
24e-3
1.6e-·3
80e-4
- fh
- fe
- fd
- fe
180
120
tt
60
2
4
v
6
8
10
4
P8-26 (g) Individualized solution
P8-26 (h) Individualized solution
P8-27 (a)
8-85
v
6
8
10
Adiabatic exothermic, adiabatic endothermic, exothermic with cooling, endothermic with heating.
All the profiles show the rate of reaction dropping toward the end of the reactor.
PS-27 (b)
The non-adiabatic profiles show an increase and a decrease in temperature profile, and the adiabatic
profiles do go from increasing temperature to decreasing temperature (or decreasing to increasing).
PS-27 (c)
Figure E8-5.3 shows a decrease in temperature while the reaction rate is large because the reaction
is endothermic. Once the reaction rate drops, the heat exchanger increases the temperature profile
because the reaction is no longer removing much heat.
Figure E8-3.1 shows a steady increase in temperature until the reaction rate drops to near zero. The
reaction rate increases at the beginning of the reactor because of the increase in temperature
affecting the specific reaction rate. At too high a temperature the equilibrium constant gets vbery
small and the reverse reaction becomes more prominent and thus the rate decreases as the
temperature rises above 350 K.
PS-27 (d)
In Figure E8-1O.1, the temperature increases quickly until the reactants are used up. Then there is no
more heat generated from the reaction, and the heat exchanger lowers the temperature. Figure E810.2 shows that the flow rate of A drops to about zero at about the same time the temperature
reaches a maximum. Once there is no reactant, the reaction ceases and the flow rates of the products
remains constant.
PS-2S (a)
2A-,'m"-'~ B
a) Design equation:
Rate law:
Stoichiometry:
Energy balance:
.~, == ,f!q,[!1.-:: T) ~L~!1X:,t:!!M..2
dV
FAOCpA
,4:£ = 5(7OQ.:: T )~ ~~?,~~' :gJ3{~~:~?~)X::~E1}
dV
5*122
Plugging those into POLYMATH gets the following program and the following
graphs. The conversion achieved is 0.36.
8-86
See Polymath program P8-28-a.pol.
POLYMATH Results
Calculated values of the DEQ variables
variable
V
T
X
Faa
Ua
Ta
dHr
Cpa
k
Caa
Ca
ra
initial value
0
675
0
5
5
700
-235.524
0.1222
0.0734336
1
1
-0 . 0734336
minimal value
0
675
0
5
5
700
-236.01067
0.1222
0 . 0734336
1
0 . 6419118
-0 . 3335975
maximal value
10
715 . 55597
0.3580882
5
5
700
-235 . 524
0 . 1222
0.3658299
final value
10
704 . 76882
0.3580882
5
5
700
-235 . 88123
0.1222
0 . 243008
1
0.6419118
-0 . 1001316
1
1
-0.0734336
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(V) = (Ua*(Ta-T)+(·ra*(-dHr)))/(Fao*Cpa)
[2] d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] Fao=5
[2] Ua = 5
[3] Ta=700
[4] dHr=-231-0 . 012*(T-298)
[5] Cpa = . 1222
[6] k = 1A8e11*exp(-19124fT)
[7] Cao = 1
[8] Ca=Cao*(1-X)
[ 9 ] ra = -k*CaA 2
0.32
700
0.24
690
016
680
_ ._ " _ ._ _ _ _ _ _ _
670 '----~-------------.o
2
4
V 6
8
10
2
P8-28 (b)
8-87
4
V
6
~_--.l
8
10
Using the same POLYMATH program we were able to change the entering
temperature and come up with this graph.
r
Conversion vs temperature
I
1.2.---------...-;..-----...,
)(
1·
0 .8
___J
5
eO.6
~0.4
8 0 .2
Or---~--~----~----~-~
o
200
400
600
Temperature, K
800
1000
P8-28 (c)
Again using the same POLYMATH program, we can vary the ambient
temperature until the reaction runs away. As the following sununary table will
show the maximum temperature is 708 K.
POLYMA TH Results
Calculated values of the DEQ variables
Variable
-V
T
X
Fao
Ua
Ta
dHr
Cpa
k
Cao
Ca
ra
initial value
-0-----675
0
5
5
708.2
-235.524
0.1222
0 . 0734336
1
1
·-0.0734336
minimal value
maximal value
0
675
0
5
5
708.2
-·236 . 5196
0.1222
0.0734336
1
0.5137156
-1 . 078707
10
758 . 02032
0 . 4862844
5
5
708.2
-235.524
0.1222
1.6320169
1
1
-0 . 0734336
final value
1()---712.2666
0.4862844
5
5
708.2
-235 . 9712
0.1222
0 .. 3233503
1
0.5137156
-0.0853333
P8-28 (d)
When the reaction becomes adiabatic the energy balance will then
become:
X-All
)
T = 1: + _(-r
..._A_,.,
__ ..,.L,
o
CpA
However, the heat of reaction is a function of temperature. This is a circular
reference, so we need to find T as a function of just X,
1~CpA -231X -3.576X
T =._-"'''--_.".,,'''''--_.._,.,'-' ' ',. .-
CpA··O.012X
Plugging that into POLYMATH gets the following program and graphs.
8-88
See Polymath program P8-28-d.pol.
POLYMA TH Results
Calculated values of the DEQ variables
initial value
0
0
5
5
708.2
675
0.1222
675
1
1
--235 . 524
0 . 0734336
-0.0734336
Variable
V
X
Fao
Ua
Ta
To
Cpa
T
Cao
Ca
dHr
,<
ra
minimal value
0
0
5
5
708 . 2
675
0.1222
618 . 33416
1
0.9695163
-235 . 524
0.0054738
-0 . 0734336
final value
10
0 . 0304837
5
5
708.2
675
0.1222
618.33416
1
0.9695163
-234.84401
0.0054738
-0.0051452
maximal value
10
0 . 0304837
5
5
708 . 2
675
0 . 1222
675
1
1
-234 . 84401
0 . 0734336
-0.0051452
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra/Fao
'
Explicit equations as entered by the user
[1] Fao = 5
[2] Ua = 5
[31 Ta=708.2
[4] To = 675
[ 5] Cpa = .1222
[6] T = (To*Cpa-231*X·3 . 576*X)/(Cpa-0.012*X)
[7] Cao = 1
[8] Ca = Cao*(1'X)
[9] dHr = -231-0012*(T-298)
[10] k= 1.48e11*exp(-19124/T)
[11] ra = -k*CaI\2
----
0.040
680
0.032
666
0.1)24
652
0 . 016
I
638
Q
0.008
[J
624
"-
0.000
0
2
4
V
6
8
10
610
0
2
4
V
6
S
P8-28 (e)
When it becomes reversible with inerts the two equations that change are the
rate law and the energy balance.
-r.=\c!-; )
8-89
I
10
We can use those equations in POLYMATH and the following graph is made:
1
~ 0.8
! 0.6
~ 0.4
02
8
Conversion vs. temperature,
reversible
~------~~~~----------,
_ _J
O+-----------~~----------~
o
500
1000
Temperature
There is no maximum. because the reaction is a runaway at a certain temperature
and the conversion goes to close to one at that point.
P8-28 (I) No solution will be given
P8-29
A+B
The elementary, reversible, gas phase teaCIion
~
2C
Feed: FAO =Fso =20 moVs =1200 mol/min.
Po
=580.5 kIno! =5.74 atm.
To = noc =3SOOK
YAO=YBO=0.5
CAO
=YAO Po =
RTo
Rate
(05) (5.74 atm.)
= 0.1 gmo1J1
10.082 1attn ) (3SOOK)
\
gtmlOK
conswlt: k = kl exp[f(i1 - ill
=0.035 exp [7~3~ (2j3 -i)]
:: 0.035 exp [8419.5
(2i3 -il]
(1)
8-90
Equilibrium constant:
Mia (Ta) =2Hc - HA - Hs =2(45.000) + 40,000 + 30.000 =-20.000 J/mole
= 2C?c - CpA - c;,. =2(20) - 25 -IS =0
ac;,
SO .MiR
en = constant = -20,000 l/mole
(2)
Kc=Kcl exp[~R{*_+)1
ICc =25,000 exp [-i~3~ (2~8 -
Kc = 25,000
k and
i)]
exp [-2405.6 (~- i)]
(3)
Kc can be calc:ulatcd by equation (1) and (3) ifT is given.
Kc=~
CACB
CA=CA«)(l-X)f
wheref=]:;}.
T
~=CAQ(l-X)f
Cc=2CAQXf
Substitute CA. Ca. and Cc into Kc
Kc =
4cio X2 f2
cio (1 • X)2 f2
=
4Xl
(1 - X)2
(4)
(5)
-JKc
Calculate
as a function of temperature from equation (3)~ substitute in equation (5)
to get Xe.g. as a function ofT_ Energy balance for adiabatic condition(6)
9A
=9B =
1 • 9c = 0 • To = 3SOK
Substitute: + 20,OOOX = (25 + 15) (T - 350)
or X
=
T = 500X + 350
=0.002 (T - 350)
(7)
Equations (3) = (4):
4X2 = 25000
(1 - X)2
exp [-2405•6 (....L
_ 1 )~
298 350 + SOOX ~
8-91
Xeq = 0.8667
0.85Xeq =0.7366
=0.7
(a) Plug flow reactor design equation:
Rate law:
orA =
(8)
Ic[CACa 4~1
-rA = Ie clAO f2 [(1 -X-y. -
(~~
(9)
(10)
To evaluate the integral, we need to evaluate fOO as a function of X. This is done in the
table below:
I!K
Kc
k
f.£Xl
(eqn.7)
(eqn. 3)
(eqn. 1)
(cqn.9)
350
7534.8
3191.1
30.96
626.• 1
1635.9
958.5
6491.5
4.216 x 1()4
0.0323
2.58 x 10-3
3.98 x 10-4
618.9
429.9
315.8
192.8
1.949 x lOS
A
0
0.1
400
450
500
550
0.2
0.3
0.4
600
0.5
0.6
0.7
650
700
6.978 x lOS
2.054 x 1()6
1.155 x 107
9.89 x 10-5
3.53 x
1.70 x
1.08 x
4.98 x
10-5
10-5
10-5
1(}-6
The integral can be evaluated using Simpson's rule for the fust six segmentS and
trapeziodal rule for the last segment.
1=[ f(X) dX '"
or-
+ 2(3.53 x 10..5) +
[0.0323 + 4(2.58 x
4( 1.70 x
10-') + 2(3.98 x 10-4) + 4(9.89 x 10.5)
10..5) + 1.08 x 1O.S] +
1- L5x 10-3
VPFR
= (1.2
x
HP)(1.5 x 10-3)= 180L
8-92
DzL[ 1.08 x 1O's + 4.98
X
10-6]
,Exothermic. adiabatic
T
~-------X
distance down the reactor
(e) CSTR. design equation:
v=
V _ FAO XA
-rA
FAO X
kcio f1 [(l-Xf -ifC]
X
_ =
(mol]
1 - 001"5
1200::
- . -
1500L X (0.1)2
kf2 [(l-Xf-~J
{l-xf· ~ = 80...x...
Kc
f2k
1-2X + X2 • 4X2 = .8.QX.
Kc
f 2k
(1. ic)X2-{2+ f~Qk)X+ 1 =0
Let bi
~
= 1 - -.!...and i>2 =2 +...&L
2
Kc
X =
f k
(11)
b2±.J~ - 4b}
2b2
(12)
Nonadiabaric energy balance:
-lO--YL. x 2
m2 K
x
1200
mm.
lkW
Ihr
3.6 x 1()6 J
lOOOW x 60min x lKW.hr (T - 290) + 20000X
=40(T - 350)
min
-T + 290 + 20.000X =40T - 14,000
T = 348.5 + 487.8 X
(13)
8-93
So the procedure to calcularc X is as follows:
1. Choose increments in T and calculate X as a function of T from equation (13). This is
the value given by energy balance.
2. Choose increments in T and calculate bl and b2 from equation (11). <Kc and k can be
calculated from equation (3) and cl) rcspcaively.)
3. Calculate X. discarding X > 1 or X < O. This is the value of X given by material
balance.
4. Plot X vs. T given by equation (13) and (11) on the same graph. The intcrSeC1ion'lives
the conversion in the reactor. A typical graph looks like the fonowing:
"
x
Operating paint
The aaual calculation gives: X - 0 which is the conversion in CSTR.
d) The same equations can be used except that ~ = 20000 and To
following graph shows the equilibrium. conversion for this case.
=SSOK. The
Conversion vs temperature
1 ·r-------------~----------_=======I
go.s·
. :!O.6
!
•
~O.4
gO.2
o ~--------------~~----~------~------~
o
50
100
150
Temperature
200
250 '
The following POLYMATH program gives the PFR. volume necessary to get a
conversion of .65.
See Polymath program P8··29.pol.
POL YMA TH Results
Calculated values of the DEQ variables
Variable
V
X
Fao
T
initial value
0
0
1200
550
minimal value
0
0
1200
225.22683
maximal value
1. OE+08
0.6495463
1200
550
8-94
final value
1.0E+08
0.6495463
1200
225.22683
0 . 064
550
1 . 949E+05
1 . 01E+06
1
0 . 064
0.064
0
-798.17344
Cao
To
k
Kc
f
Ca
Cb
Cc
ra
0 . 064
550
5.047E-05
1841 . 4832
1
0 . 0547713
0.0547713
0
-798 . 17344
0 . 064
550
1 . 949E+05
1 . 01E+06
2 . 4419826
0 . 064
0 . 064
0.2030311
-1 . 503E-07
0 . 064
550
5 . 047E-05
1841 . 4832
2.4419826
0.0547713
0.0547713
0 . 2030311
-1 . 503E-07
ODE Report (RKF45)
Differential equations as entered by the user
[ 1 J d(X)/d(V) = -ra/Fao
Explicit equations as entered by the user
[1] Fao = 1200
[2] T = -500*X+550
[3] Cao= . 064
[4] To = 550
[5] k = . 035*exp(8419 . 5*(1/273-1fT»
[6] Kc = 25000*exp(2405.6*(1/298-1fT»
[7] f = TofT
[ 8 j Ca = Cao*(1-X)*f
[9] Cb=Cao*(1-X)*f
[ 10 1 Cc = 2*Cao*X*f
[11] ra = -k*(Ca*Cb-Ccl\21Kc)
The CSTR conversion can be found similarly to the equilibrium conversion. The
following graph was made to find the conversion.
Conversion vs temperature
=:j
. 8f
~0
0 . 6 r-------------_
~
:0.4'
::==::::_
8o.~.
230
280
330
380
Temperature
The graph shows a conversion of .39 at a temperature of 351.8K
e) If the reaction is left in a large enough volume it will "tuIlaway·. If any of the
quantities get bigger. then it will run away even faster.
f) The ambient temperature around the CSTR has little effect on the conversion in
theCSTR.
P8-30 (a)
P8-6 is adiabatic so the radial reactor has no effect on it.
P8-30 (b)
8-95
dX
-r
dW
~o
_=--:1..
-rA =kCA
CA
1-- XL
= C".40 ·---·.Q
+X T.'V
1
-a
dy
dW
---=""--
dT
_
-
2y
u(r)~(TA ........
T)+ (-rA)(-MR )
._-
,-~--.-
._.,
dW
Since ~ and p are unknown, we will assume that they are both equal to one.
We wIll also assume that r varies asW 5 varies.
so:
dT _ U(r)(~~ - T) + (-,:~)(-L\HR)
-------~-.:..-~~-~
dW
F"oCiM
- -
w..s )'S( ToT)S
U = U{lQ);'
(
See Polymath program pg·JO·b.pol.
POLYMA Til Results
Calculated values of the DEQ variables
Variable
W
Y
X
T
Fao
alpha
Ta
dHr1
WO
To
E
R
Cao
Xl
y1
U
k
initial value
0
1
0
450
5
0,.007
300
-·2.0E+04
0,,01
450
3.14E+04
8,,314
0.2498645
0
1000
5"OE+04
1
minimal value
0
0.8062258
0
450
5
0.007
300
-2.0E+04
0.01
450
3,,14E+04
8.314
0.2498645
0
806,,22577
0.9244844
1
maximal value
50
1
0.7845538
756.08452
5
0.007
300
-2.0E+04
0,,01
450
3.14E+04
8.314
0.2498645
784.55376
1000
5"OE+04
29.893648
8-96
final value
50
0.8062258
0.7845538
756.08452
5
0.007
300
-2.0E+04
0.01
450
3.14E+04
8.314
0.2498645
784.55376
806,,22577
0.9244844
29.893648
Ca
Cc
Cb
Kc
ra
0 . 2498645
0
0 . 2498645
1635 . 8981
-0 . 0624323
0.2498645
0.2333455
0 . 2498645
1635 . 8981
-0 . 0220235
0 . 0320395
0
0 . 0320395
187.88715
-0 . 1234947
0.0320395
0.2333455
0 . 0320395
187.88715
-0 . 0220235
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(y)/d(W) = -alphal(2*y)
[2] d(X)/d(W) = -ralFao
[ 3] d(T)/d(W) = (U*(Ta-T)+(-ra)*(-dHr1 »/(Fao*40)
Explicit equations as entered by the user
[1] Fao=5
[2] alpha = .007
[3J Ta= 300
[ 4 ] dHr1 = -20000
[5] Wo = . 01
[6] To=450
[7 J E = 31400
[8] R=8.314
[9] Cao = 9 . 22/(..082*450)
[10] X1 = 1000*X
[11] y1 = 1000*y
[12] U = 5*(Wol\.5/(WI\.5+ . 000000001 »1\.5*(TlTo)I\ . 5*yl\ . 5*(1 +X)I\.5
[13] k = exp((E/R)*(1/450-1ff)
[14] Ca = Cao*(1-X)*TolT
[15] Cc = 2*Cao*X*TolT
[16] Cb = Ca
[17] Kc = 25000*exp((dHr1/R)*(1/298-11T»
[18] ra = -k*(Ca*Cb-CcI\2/Kc)
0.80
800
0.64
720
0.48
64U
Q
0.32
20
W 30
·to
[2J
560
400
'---~-.--~----------I
o
P8-30 (C)
8-97
10
20
W 30
40
50
= p v = p7tr2h
W
=
dW
= 2n:prhdr
Material balance:
Taking the limit as tJ.r ~ 0
~
FA
=
FAO (l-XA)
=
:: (rA) 21tprh
ar =
A
-FAO
~"
. dXA _ -2milp (rA)
FAO
(1)
so. dr
or: dXA :: _...!A..
. dW
(2)
FAO
Assuming ptessure drop is negligible. The rate equation is:
-rA = k(~)l/2
[P(h _{ KpPS(h
Pso, )2]
Pso,
The stoichiometry :
A+.LB=C
2
o=
-0.5, YAO
PsQz = PsOu
Then:.r
= 0.11 :. £ = YAOO = -0.055 ; ec = 0 ;
1- XA
; Pso,
1 +£XA
~k(t)(l-XA)lJ2[p
A
XA
SOu
= PsOu
XA
1 +£XA
; Paz
Sa
= ¥t = 0.91
A)
= Pso,.o (O.91-0.5X
~--..;..;.;..;;..;.:.::.
1 +£XA
(O.91-0.SX A )_
xl. 1
1-O.05SXA (l.XJ K~l
(3)
Equation (3) is aue for X A < 0.05
for X A ~ 0.05
- fA
- fA
= k(t) (4.3S>lr o.22 ( 0.91-9· 025 }_ 0,0025 1
1-0.05:>(0.05) (1-0.05fK~J
= k{t) [0.848 -
Q.Q~~O~l
(4)
8-98
Energy balance:
FAO (l:9i c;,i + X .6.C p) Tlr - FAo (Lei Cpi + .6.Cp X) Tlr+Ar
+ rA ~W(AHR) - (2) (AA) U(r) (T-T A) = 0
or: FAO (re j Cp; + x ACp) Tlr - FAO (rej CPi + ACp
x) Tlr+Ar
. + rA 2raIhp (.1HR) 1M - (2) (2xr tu) U(r} (T-T.J =0
Taking the limit as tu -+ 0:
FAO (rei ~ + x ~Cp) ~ = -U(r) 4m (T-TA) + (-.rA) (-MIR) 2xrhp
Rearranging:
dI.
=-U{r) 4m- (T-TA) + (-rA) (-AHR) 27trhp
dr
(5)
FAO(:r9i CPi +X~Cp)
-U(r) 1h.(T- TA ) + (-rAl (-~HR)
or: .dI..
dW
=------p~------'!""--
(6)
FAO{:rei CPi + X acp)
Assume that: U(r) = U(ro) (.It.)1f2 ;
~
•
Therefore: U(r) = U(ro}
l) = l)1-ro (1 +£X)J..
~o
(!f-)l'\f.yn {I - O.055XA)1f2
From example 8-10. we have:
K.p =exp (42R~ 1 _ 11.24] (Kp in ann- 1f2 • T in OR)
k = exp [
(1)
T - 110.1 In T + 912.84]
-176008
(8)
MiRm = -42,471 - (1.563)(T - 1260) + (1.36 x
1O-3)(T2 - i26(2)
- (2 459 x lO-7){T3 - 126(3) where AliR in Btu
.
3
l~le
8-99
(9)
l:8i Cy. = 57.23 + 0.014T - 1.788 x 1()-6 T2
(10)
Since equations (1) to (10) must be solved together as two pairs of coupled differential
equations. they must be solved on a computer. employing numerical methods such as
Runge-Kuaa. The results follow:
i
T
I
_ - - -~
1.,0".
,;'-
....---.·.-/·. .- - - - - -.. .E:
8
12
16
zo
24
n
'8
y"",.
T
---~~~--------------i~
--'
.-'
.'
.4
~
________________
•
I:
"
.0.24
12
%I
.,
,- /----------
0."
------------~O
16
~
...
U
48
"
60
(I~'I
i
I AGO
T
t
X
1"0_
t
12110
lOGO
.,;' --~
eoo
0
(J..:
I
a
:go
---
.
r.i.
.'
0 ..
'
I
~t.
I
:
"
4
a
12
16
:0
44
:t
n
16
.a
oM
41
~
." l....j
l71
P8-30 (d)
8-100
U
60
;
1I
.~~ = ::2m-~£t~.~4).
}~o
dr
~rA ={qB~f~ J
c = c =C
A
B
.'10
(1'"- X'l_!Q.
'T
TO
Cc =2CAO XT
dT _ U(r)47rr(TA -T)+(--rA)(-LlliRJ
dr FAO ( CpA + CPB )
u=u(ro)(;
)"(;,r
Now put these equations into Polymath to generate the plots.
See Polymath program P8 . 30·d.pol.
POL YMA TH Results
Calculated values of the DEQ variables
Variable
-r
X
T
Ta
dHrx
Fao
Cpa
Cpb
h
ro
To
E
R
Cao
U
k
Ca
ra
initial value
0. 5
0
350
373
-2.0E+04
1200
25
15
0.5
0.5
350
7.0E+04
8 . 314
0. 1
33.3
30 . 955933
0. 1
-3 . 0955933
minimal value
0.5
0
350
373
-2.0E+04
1200
25
15
0. 5
0. 5
350
7 . 0E+04
8 . 314
0.1
0 . 769425
30.955933
-2 . 608E-10
-3.0955933
maximal
- - -value
-1000
1
493.0235
373
-2 . 0E+04
1200
25
15
0.5
0. 5
350
7 . 0E+04
8 . 314
1
33.3
3 . 398E+04
0. 1
2.09E-06
°.
ODE Report (RKF45)
Differential equations as entered by the user
[ 1] d(X)/d(r) = -2*3 . 1416*h*ra/Fao
[2] d(T)/d(r) = (U*4*3 . 1415*r*(Ta-T)+(-ra)*(-dHrx))/(Fao*(Cpa+Cpb))
Explicit equations as entered by the user
[1] Ta = 373
[2] dHrx = ··20000
[ 3 ] Fao = 1200
[4] Cpa = 25
[5] Cpb = 15
8-101
final value
1000
1
373
373
-·2 . OE+04
1200
25
15
0. 5
0. 5
350
7 . 0E+04
8 . 314
0.1
0.769425
136.44189
-3 . 125E-17
4 . 264E-15
(6) h =.5
(7) ro =.5
To=350
(9) E = 70000
(10) R=8.314
(11) Cao =.1
(12) U = 33.3*(ro/r)I\.5*(TfTo)I\.5
[13] k
035*exp«E/R)*(1/273-1fT))
[14] Ca = Cao*(1-X)/(1+X)*(TofT)
[15] ra = -k*Ca
(8)
=.
P8-31 (a)
Mole balances:
v = F.4.0 - I:..
-rA
V=Fs=Fc
rB
re
rate laws:
= ","CA
rB =k.CA -~CB
rc = k"CB
-rA
Stoichiometry:
c.=F;I..
I
Va
To
Fa =10Fe
.5 = .05-~
0.5
F;. = .025
11Fe
=.025
Fe =.023
FB =.023
From this we can use two of the mole balances to solve for T
8"·102
~o-~
1G~
1
A1e- E1 KI
F;
=-
~Fs
0.1
-
~e-EI KI
T=269°F
P8-31 (b)
Knowing the temperature we can then solve for the Volume:
P8-31 (c)
We then need the energy balance:
UA(~ - T)- ~oCJHl(T - Yo)+ v[(MfR1)('iA) + MfR2 ('iB)] = 0
Solve for A and we get:
A = 399 ft2
P8-31 (d)
In order to get multiple steady-states, the kappa, tau and feed temperature had to
be changed. l( == 0.1, 't == 0.0005 and To would be changed around.
Tlus first graph is G(T), R(T) vs T at To == 2000 of
20000 " ,--,-,,----"""'-,,---.--..,------.
15000
10000
5000 "
-5000
1000
15 0
-,10000 .....- - - , . - - , - " -- - - - - - -.......
As can be seen there are three steady-states.
8-103
Ts vs To
1200 .
1000
800
~600
••
•
•
•• •
• •
•• • •
400 4
200 ....
0
10000
5000
0
To
P8-32 (a)
Energy balance
.!L= Ual PII(r. -T)+(-rAX-AHR(~)]
~o(L8iCpi+XACp)
dW
tIT _
Ual P.(7;, - T)+(-rAX-MIR(TR))
dW
I:..oC,..
dX
Mole balance: dW
-r
=-4.
~o
Pressure drop: dy =~1i(l+X)
dW 2y T
Rate law:
Stoichiometry: CA =
CAo(lXXI..}
l+X To
Evaluating the parameters:
(-1-_1.)1
=ex]3776.7J'2... _1.)1
T J
,.
\. 450 T J
k = ex] E
~R 450
Plugging these equations into POLYMAlH we get the following plots.
See Polymath program P8·32··a.pol.
OL YMATH Results
Calculated values of the DEQ variables
8-104
initial value
0
1
450
0
5
0.007
450
300
-2 . 0E+04
5
40
1
0.25
0 . 25
-0 . 25
},:ariable
W
Y
T
X
U
a
To
Ta
dHrx
Fao
Cpa
k
Cao
Ca
ra
minimal value
0
0 . 6823861
310 . 69106
0
5
0.007
450
300
-2 . 0E+04
5
40
0 . 0232094
0.25
0 . 0825702
-0 . 25
maximal value
50
1
450
0 . 175758
5
0.007
450
300
-2.0E+04
5
40
1
0 . 25
0 . 25
-0.0019164
final value
50
0 . 6823861
310 . 69106
0 . 175758
5
0.007
450
300
-2.0E+04
5
40
0.0232094
0 . 25
0.0825702
-0.0019164
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(y)/d(W) = -a/(2*y)*(TofT)*(1 +X)
[2 J d(T)/d(W) = (U*(Ta-T)-ra*dHrx)/Fao/Gpa
[3] d(X)/d(W) = -ra/Fao
Explicit equations as entered by the user
[IJ U = 5
[2] a = . 007
[3J To=450
[4] Ta = 300
[ 5 J dHrx = -20000
[6J Fao = 5
[7J Cpa = 40
[S J k = exp(3776 . 76*(1/450-1fT))
[9J Cao = . 25
[101 Ca = Cao*((1-X)/(1-+X))*(T/450)*y
[111 ra = -k*Ca
1.0
...-.=-------- -----..,
50U r - - - - - - - - - - - -
460
0. 8
IT]
- v
0.6
420
- X
380
0.4
0.2
.
---"-~
............. " ..."'" ... " ......................... ~ .... '~ .................. ~ ... ~ ..... """
340
~--.-..,.--10
20
W 30
40
50
300
(I
10
20
W 30
40
50
P8-32 (b)
From the Polymath summaty table, it is appatent that the maximum value for -rA occurs at the
beginning of the reactor.
P8-32 (c)
8-105
The maximum value for the temperature also occurs at the beginning of the reactor.
P8-32 (d)
Doubling the heat-transfer coefficient causes a decrease in the temperature, the conversion, and the
pressure drop. Halving the heat transfer coefficient casuses all three to increase.
P8-33
Mole balances
V = vo(C AO -CAl
-rA
Rate laws:
-rA
=k1CA +k2 CS
-rs =k1CA +k2C S
Energy balance:
ru =k2CS
-FAO[c;,A(T -TAO) +CpB(T - TBO)] + V(rAIX.MiR1(TR}+ 6Cp1 (T - TR») + V(rA2XllliR2(TR)+6Cp2(T - TR})
Evaluating the parameters:
T=400K
kl =1000exp(- 2~)=6.73
t..C p1 =50-20-30=0
k2 =2000exp( -
3~OO)=1.11
t..C p2 =40-30-20=-10
FAo 60
dm 3
vo =-=-=6000Coo =C AO
C AO .01
min
Simplifying:
C A = 't "rA +CAO
C s = 't "rs +C so
Co = 't" ro
C u = 't It ru
171000
V=-------20190· CA."!" 6660· C~
We can plug those into POLYMATH and find the exit concentrations of U
and D and find the volume of the CSTR.
See Polymath program P833.pol.
POLYMA TH Results
NLES Solution
8-106
variable
Ca
Cb
Cd
Cu
V
Cao
Cbo
vo
k1
k2
tau
rd
ru
ra
rb
Value
0.0016782
0.0016782
0 . 0071436
0.0011782
3794.94
0 . 01
0.01
6000
6.73
1.11
0.63249
0.0112944
0.0018628
-0 . 0131572
-0 . 0131572
f{x)
-1. 472E-16
-1.472E-16
-1.154E-16
-2 . 017E-17
1. 018E-09
Ini Guess
0 . 0017
0 . 0017
0.0072
0 . 0012
3794
NLES Report (safenewt)
Nonlinear equations
[ 1] f(Ca) = tau*ra+Cao-Ca = 0
f(Cb) = tau*rb+Cbo-Cb = 0
f(Cd) = Cd-tau*rd = 0
f(Cu) = Cu··tau*ru = 0
[ 5] f(V) = 171 000/(20190*Ca+6660*Cb)- V = 0
[2]
[3]
[4)
Explicit equations
[1] Cao = . 01
[2] Cbo = . 01
[3] vo = 6000
[4J k1 = 6..73
[5] k2=1..11
[6) tau = Vivo
[7J rd=k1*Ca
[ 8] ru = k2*Cb
[9] ra = -k1 *Ca-k2*Cb
[10] rb=ra
P8-33 (a)
Cu = .0012
Co = .0072
P8-33 (b)
V=3794dm3
P8-33 (c) Individualized solution
8-107
Solutions for Chapter 9 - Unsteady State Nonisothermal
Reactor Design
P9-1 Individualized solution
P9-2 (a) Example 9·1
The new TO of 20 of (497 OR) gives a new AHRn and T. With T=497+89 . 8X the polymath program of example 9-1 gives
t= 8920 s for 90 % conversion.
P9-2 (b)
Example 9·2
To show that no explosion occurred without cooling failure .
Isothermal operation throughout (T 17S0C)
=
~:laximum
cooling rate:
Qr == UAI448-· 298] == 142 * 150
::: 21300 BTU/min
Maximum Qg at t == 0 (maximum concentration and reaction rate)
Qg := k _liclQ1!!.!.Q.
V2 * V(··MI R~ )
== O.OOO1l67[.?.~'~~~J2..34 *' 10 6
. 5J 19 .
::: 15914.2 BTU/min
For all t:
Qg < Or
No explosion
9-1
To show Ll}at no explosion occurs with cooling shllt down for 10 mi after 12 P..IS
Isothermal operation for 12 hrs
(at T:::: 175°C)
t ==
[~~-Ie~~::2}n~(i:::~~)
') x__ == 1 276
__e,,·
iL-=::
8 s (l·-x)
3 64--2x == 1276*3.64(1····· x)
x:::: 0..38
Qg at
t :::;
12 hrs .
Qg "" k.~~JL
'-" 0..000
. X.llVzo{~l!.:_~.:l(._AH
)
V
&
11671··~:O~~{1.:~:~~X~..!:::_~*:~?112..34 '* 106
"
= 77844
5..119
.
BTU/min..
Adiabatic operation for 10 min. UA:::: 0.
After 10. minutes
x == 0.385, T == 184"C
Qg :::: 10000 BTU/min.
When we restart the cooling flow rate
Q,I"",. == 21,30.0 BTlJlrnin.
Temperature will dIop to 175"C
NQ~~lQ:iiQ11
P9-2 (c) Example 9-3
Decreasing the electric heating rate (Tedot in polymath program from example 9-3) by a factor of 10 gives a conversion
of 9.72 % at the onset temperature . For a decrease by a factor 10 the conversion is 2,49 %. The higher conversion of a
lower heating rate is logical since the time it takes to reach the onset temperature is longer and the reactants have a
longer time to react.
P9-2 (d) Example 9-4
Decreasing the coolant rate to 10 kg/s gives a weak cooling effect and the maximum temperature in the reactor becomes
315 K. An increase of the coolant rate to 1000 kg/s gives aT max of 312 K. A big change to the coolant rate has, in this
case, only a small effect on the temperature, and because the temperature does not change significantly the conversion
will be kept about the same.
P9-2 (e) Example 9-5
Using the same code as seen in Example 9",4, we were able to change the various
parameters. The two graphs we have show To;::: 70 and 120°C, T i :;;;; 160 and
40°C, and Caj
=:: .. 1 and ,,2. Each set of parameters had a Temperature time
trajectory and a teIuperature-<:oncentration phase plane. These 31'e the four graphs.
9-2
~:,lCOT
19'5_OOOi
KEY:
-T
'~-t
;::I5.oo0.-!-
'~-t
~.
~--- - - ..
~-----+I----------'i~·-------+.--~--~t
75.000
c..:~
~•
0.1iI'tlO
0.000
.soo
v.¢Q
:l..zou
'1 . oot).
T\
T
..!..
.:), ~2'C
T
KEY:
-Ca
0.::190
!
Q.co;a
1
;
t
o...~JO
Q.ooa
t
---~'-'---+--'--.--";os-coo
l~QOO
l::J~.QOQ
tiS;!i.COO
..,..~
2:2tl. to"Uu~·
Startup of a CSTR
!
1\
1'--!
'--i
~~.ac:
Tl =40)'
To;120~
.._.- -.. . . .--.----.. -.. . . . . . . . .
~-----.---
-
..
15':>.000
~---"--
....
f
1.0.0CO+ };
KEY:
-T
t
H,0.::I0011
60.000
-¥
r
zo..,oco
-1. .--------+,-=
. __
. . -.. ..".,.,.,..-+--....-~. ~
. .....,..-------t---.,...~"--Q.ooo
O. !lOa
:. q:a::I
:2. "\00
loa
".000
::I.
9.. 3
a CSTR To=120 Tl;40 Cal=.2
(L;;60
t·
1
t/~'y. q.):;!O
_"'_1:._::._
-Ca
T
P9-2 (0 Example 9-6
Using the code flom Example 9-5, we could produce the following graphs either by changing To and finding the steady
state conversion or changing the coolant flow rate and finding the steady state conversion and temperatme. These are the
graphs of those:
Conversion 'IS To
0,,9
------,-".,"',"---"--"'-,-
. ...--........................._ _ .............
0.8
0.'7
c
o 0.6
f
(1)
0.5
:> 0,4'
c:::
g
0,.3
--
0.2
..
0.1
0""------"'"---,---··.,.-----·,.,...·,,--,·---,.-··..-,,-·,,,····,,,. · · · · · . . ·. . ·"....""""'t----·. . . T···"·" ..
·~··
66
68
70
72
'74
To
9-4
76
78
80
82
_
..... ...........
_-_ _-_._-_ _
..
............
................
_._..........__....__--_...................................................._._-------..
x vs coolant flow
1 ...-...._ _................-._...............- .....--.
------.----~
0.9
0.8 .
c 0:7
.E 0.6
....coQ,) 0.5
c> 0.4
o
o
--_._-___ __ _--
0.3 .
•.
..
0,2 .
0.1
. . . . .-r------r,..-.. . . . . -----1
o -!---.----,------......,...,--,
o
2000
..
..
6000
4000
8000
10000
Coolant flow, mollh
180
...-_ ..•..-
T vs. coolant flow
---"'
...................._............................... __.,-------------
160 .
u.. 140
....c) 120 .
.aa:s
100··
~
80·
--_._-_ __ _---_._-....
M
....
E
60
Q,)
•
i-
40
20·
o ..-.. . . . . . . . . . . .--.,.....--_. _. . . . . · · ·. . . . · · . r······_·_··. ······. •·....·---r-·_-'--.,.....--- . . . ._ . . o
2000
------_
...
__._---"
6000
4000
8000
10000
Coolant flow mol/h
........................
P9-2 (g) Example 9-7
The temperature trajectory changes significantly., Instead of a maximum in temperature, there is now a minimum.
The concentration profiles also change. CB no longer goes through a maximum and Cc does not rise above 0 . It also
appears that there is no maximum for C A ..
It appears that because the reactor is kept much cooler with the increase heat transfer and lower coolant temperature the
second reaction does not occur .
P9-2 (h) Example RE9-1
Using the code from Example RE9·1 • we can determine the value of k. for which
the reacwi""willtall to the lower steady··state and when it becomes unstable. The
following two graphs show those points when ~;; ..2 and 24 respectively. The
third graph shows what happens when To;; 65. It becomes unstable ata much
lower temperature.
9-5
t
1"" •
KEY:
--T
,
.:J-co-'-
1
i
1 ~c",~cO;-·
t
T 0=65,
IntE'gr al
1<:c:=1
(i.90c
t
P9-2 (i) Example RE9-2
Using the code from Example RE9-2 ,we can change the values of kc and '1:[ and
find values that produce the lowest oscillations and the quickest return to steady~
state and getting kc =:; 150 and L J ;;;: • L The following graph shows the result.
.....3.:1:0(1
-+--..
t
P9-2 (j)
No solution will be given
P9-3
9-6
Find time to explosion
For the unsteady~su[e. we assume the following:
(i) The inlet is closed. but the ourlet is not closed
(ii) Operation is adiabatic, and PV terms are negligible so that H "'" U
Equarion 8-61 reduces to:
(-611R) (-fA V}= NAC'-PA
c;J
(-6HR) (kNA)": NACpA ~
ill. -; AHlhl'1
d!
Cp".
k(T)
k ::: .53 exp (44499 (9jO -
+)} / 60 min.
See Polymath program P9-3.pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
initial value
-0----
T
dH
Cpa
k
970
-336
0 . 38
0.53
minimal value
0
970
-336
0.38
0 . 53
maximal value
2.9
1.119E+23
-336
0.38
4.438E+19
ODE Report (RKF45)
Differential equations as entered by the user
[lJ d(T)/d(t) = -dH*k/Cpa
Explicit equations as entered by the user
[1 J dH =-336
[2] Cpa = .38
[3] k = .53*exp(44498*(1/970-1/T»
P9-4
Mole balance:
9-7
final
value
---2.9
1.119E+23
-336
0.38
4.438E+19
There is no A leaving the reactor so F" :::; O. 111ere is no Bore entering the reactor
so Fao and FQ) ::: 0.. The amount of B and C leaving is equal to the reaction rate so
Fa::: VIa and Fe::: VIc' Simplifying:
4:~!l.. == 0
dt
rf!!c
:=
0
dt
Rate law:
• Stoichiometry:
Simplifying:
Energy balance:
Evaluate the parameteIS:
InCji )=~(8(~j' .. ~£~)
E::: 3467
k = .19 exp
[i1i: (4}m . . -})1
tVi(400) ::8kjlmol = LVf(8(X)) = till (1 200) = MI(T)
Assume a heat capacity for A of 20 J/molK and assume there is no A already in the
reactor. We can plug these equations into POLYMATH and get this answer
See Polymath program P94 .pol
POLYMATH Results
No Title 08-11-2005, RevS I 233
Calculated values of the DEQ variables
Variable
-t
Na
T
Fao
dHr
Cpa
initial value
-0-----
1.0E-09
400
100
-8000
20
minimal value
o
1.. OE-09
400
100
-8000
20
maximal value
100
326,,10947
801.. 376
100
-8000
20
9-8
final value
100
312 . 51175
800
100
-8000
20
To
Vra
400
0.3202745
-1.9E-10
400
0.19
-101.76832
400
0.19
-1. 9E-10
k
400
0.319988
-100
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(Na)/d(t) = Fao+Vra
[2] d(T)/d(t) = «Vra*dHr)-Fao*Cpa*(T-To»/(Na*Cpa)
Explicit equations as entered by the user
[1] Fao = 100
[2] dHr = ··8000
[3] Cpa = 20
[4] To= 400
[5] k = . 19*exp(3467/8 . 314*(1/400-1/T»
[ 6] Vra = -k*Na
NA
=326.125
When the flow is turned off, F AO :::: 0 and we get this graph of T and N....
2000
1.0e··9
1600
Q
1200
6.0e-10
800
·t.Oe-10
400
2.0e.. l0
0
0
2
4
t
6
8
O.Oe+O
10
(I
2
4
t
8
6
10
P9-5 (a)
Mole balance :
dNa
--_. = ra.V
dt
dNb
- - - := fa.
dt
dNe
. . . . .---- =
Vr Fha
Rate law:
-ra:: k(T).Ca.Cb
Stoichiometry :
Ca::: .t:!a
dt
Cb
V
Nb
:=--....
V
IF (t<50) nffiN (V:;;;; Yo + u.t) ELSE (V=Vo + 50)
dm31miu
9-9
-Ta. V
Ne_.
Cc:: .-...
V
where Yo:;;:; 50 dm3 , '\) = 1
Combining:
Na Nb
-- ra ;:;: k (T) . ------
V V
k(T);:;:
ko.exp[-~(.L -"'~)\l
. R To
where E::: 10000 cal/mol
T_.
R == 1.987 callmoLK,
ko ;:;: 0.01 drn3/mol.min ,
To= 300 K
Energy balance:
dT _ Q. WS
d~-
'LFio.Lpi.(T-Tio)+(--IlHTX(T)).(--ra.V)
- ----····-·----··--·----··!Ni ..(;pT--.. -..- - - - · -
where Q=Ws=O
AHn: (f) == MIrx(fref) + ACp(T-Tref)
ACp = !:." Cp -"~'" C'pb·· Cpa
a
a
= 30 ·······15 . ·····15"'" 0
Mlrx(273);;:;; [(-41)-(-20}{15)]. lO J =6000 cal/mol
ilHrx(f) :::: -6000 cal/mol
2:Fio.Cpi.(T-Tio) =: fl)o.(15).(T-323)
LNLCpi:::::: NaCpat· f'.c'b.Cpb -+- Nc.Cpc
= 15Na + 15Nh + 30Nc
See Polymath program P9-5-a.pol
£OLYM~:rI!.Resul!§
08-11-2005, RevS. 1 233
Calculated values of the DEQ variables
Va:r:iable
-t
Na
Nb
Nc
T
X
Fbo
Nao
Cbo
k
vo
V
ra
initial value
0
500
0
0
298
0
10
500
10
0,,0089352
1
50
0
minimal value
0
0.1396707
0
0
298
0
0
500
10
0 . 0089352
0
50
-0.3365379
maximal value
120
500
42.43357
499.86033
510.44411
0.9997207
10
500
10
10 . 085112
1
100
0
ODE Report (RKF45)
9-·10
final value
120
0.1396707
0.1397745
499,,86033
510.44411
0,,9997207
0
500
10
10" 085112
0
100
-1.969E--05
Differential equations as entered by the user
[1] d(Na)/d(t) = ra*V
[2] d(Nb)/d(t) = ra*V+Fbo
[3] d(Nc)/d(t) = -ra*V
[41 d(T)/d(t) = «6000*(-ra*V»-(Fbo*1S*(T-323)))/(1S*Na+ 1S*Nb+30*Nc)
[5] d(X)/d(t) = -ra*V/Nao
Explicit equations as entered by the user
[1 J Fbo = if(kSO)then(10)else(0)
[2J Nao = SOO
[3] Cbo = 10
[4 J k = .01 *exp«10000/1 . 987)*(1/300-1/T»
[ 5 J vo = Fbo/Cbo
[6 J V = if(kSO)then(SO+(vo*t»else(100)
[7 J ra = -k*Na*Nb/(VI\2)
600.------
1.0
0.8
Q
0.6
04
280
200
0.2
o
24
48
t
96
72
24
120
48
t
72
96
120
P9-5 (b)
This is the same as part (a) except the energy balance .
Energy balance :
dT _ Q_.. Ws -
L Fio. Cpi.(T ... Tio)--::-...+ (--Mirx(T)),
(-ra. V)
-.-.-.....-..-""-..--..
-;j;" ---." ..-.--.-... ...............
,,·~~-~-C·
.£..; lVZ.
where Q :::: UA(Ta . T)
VA;;:: 100, 'fa::::: 323
See Polymath program 1'95 b .pol
9-11
pI.
-~-
1.0
r - - - - . -____---~.~.........- .......".............- .....--."-....... ..
500 r - - - - - - - - - - - - - - ,
0. 8
-t-t0'
0,6
380
0.0'1
320
0,2
260·
200
36
72
t
108
I-t4
L...-_~
o
180
_ _~_ _ _ _ _ ' _ _ _ - - l
36
72
t
108
144
180
P9-5 (C)
This is the same as part (b) except the reaction is now reversible
Rate law:
-ra:::: kI(1').Ca.Cb·· k2(T).Cc
Combining:
'·'ra::::
kl(T){~'.'~'}' k2 C
D.( i-)
k2(T)
= koexp[ ~{};, -
¥)]
where E:::: 16(X)(), R::;:; 1.987 ,
ko :::: 10 , To ::;: 300
See Polymath pIOgram P9··5-cIJol
0.9
..
500
--,...,.,...---.,,~.---------
-260
72
t
108
H-t
180
200
o
36
72
t
108
P9-6 (a)
Mol balance:
Rate law:
dNa
dNa
···········'·;:;;;;raV
dr
. . ra:::: k .. Ca
But Ca.V:::: Na
-_._-;:;;;; -kNa
dt
9-12
14-t
180
Na
In--;;;;-,k.t
Nao
Na:::: Nao.e·kt
Energy balance :
where Nao:::: 0,5 x 50 == 25
dT, ::;:; Q- Ws- LFio.Cpi(T
- Tio)
+ (---Mlrx)(-ra.v)- :::: 0
"._"""--",-,-.,-,-,,,,
..,,.,,,,-,,,,,-.,-,--..
dt
LNi.Cpi
constant T
Q=Ws=O
LFi.Cpi(T - Tio):::: (-AHrx)(-ra.V)
FCQ,Cpc.(T- Tio)::;:; (-Mfrx)(k.Na)
:::: ("L~Hrx)(k.Nao.ekt )
(-Mlrx).(kNao.e b )
Feo =.-.'-...-'----~-"''''''-Cpc.(T - Tio)
::;:; .~SOO'9,~ (WOO ~ 2 x ~?,,~::!<p( ::~:(~.~3_ x 2 x ???<?~
0.5. (l 00 - 80)
:::: 3.16 Ibis
P9-6 (b)
Av ;;; l000Btu Ilb
Nao::;:; 25 Ib/mol
Vo = 300 ft3 of which 250 if is solvent
dT
·-=0
dt
Fs.
Av = ::;:; (-Lllh'x) (-fa.V)
(-··AHrx). (kNao.e- b
F s ;;; _.,_.".__.......
)
,_,0""""""" .. ", •• _",-,,,
25000 x 0.00012 x 25 x exp( -.().OOO12 x 2 x 36{)O)
::;:; _
.. ,·..
_--"_··,,--'_·""""'''''·'''''''''1000--'··'--,·_..·. . ._-_·."""""""""'-.-".-.'.'.
/I."
Fs ::;:; 0.0316 Ibis
P9-7
9-13
Batch problem
dX
Nao~---
Mol balance:
' dX
C00--':::: '-Ta
dt
= ··ra V
dt
Rate law:'
Ca ~ Cao(lX) :::: 0 . 1(l-X)
Stoichiometry :
.t1(D =
Cb
:=
Cao( l25-X) :::: O.1(1.2S-X)
Cc
=:
Cao(O+X) :::: 0 IX
ko.expl'~(:/~' -" '~J]
E =: 100000 llmoi., R:::: 8.314 J/moLK
where
ko ;: ; : 0.002/$ > To;::;;: 373 K
E::= 150000 J/mol., R == 8.314 J/mol.K,
ko :::: O.O<XX)3 Is, To ::::: 373 K
Energy balance:
where INi.Cpi
L":.C"p
Y'
7
=:
=:
Nao(I8iCpi + L\CpX)
.,~." Cpc -.,.~" Cpb -, Cpa :;: 40 a
a
25 "". 25 = "'·10
Cbo,
b Ceo
Cpt ="Cao
.. '-, ('_pa +-.---.
Cpt--··
__·, Cpc
Cao
Cao
Cao
Q:"""
01..
~
= 25 +- 0 125/01 25 + 0 =56..25
L\Hrx(f) = .6Hrx(Tref) + L\Cp(T-Tlef)
::: '." 40000", lO,.(1'298)
Nao
V=="""""",·
Cao
dT
subs
NaO')
[ 40000 + 10" (T .,., 298)] ,,'
( ra, ,.-:;--..
""'' ' ' ' ' :::: '''' ,."",,",,. ,,,._._... "., ..._..."" .... __ ..._. __......______"....,..Cao.:....
dr
Nao(56.25 """ lOX)
9-14
>
(-ora)
r40000 + 10. (T -·298))...··
l
dT
.". . , " . . = .
dt
cancelling
._.__."".""".."",,". . . . . ,,_.. 0.1 __..
(56.25··· lOX)
See Polymath program P9-7.pol
POLYMA Tn Results
Calculated values of the DEQ variables
minimal value
initial value
Variable
t
o
o
°°
x
373
0.002
0.0749517
0 . 0999517
3.0E-05
373
0.002
0. 1
0.125
3.0E-05
T
k1
Ca
Cb
k2
o
o
Cc
ra
-·0.1344598
-2.236E-04
final value
maximal value
10
0.2504829
562.91803
106.13627
0.1
0.125
366.75159
0.0250483
8.644E-06
~--
0 . 2504829
562.91802
106.13612
0.0749517
0.0999517
366.75083
0.0250483
1.3E-07
ODE Report (RKF45)
Differential equations as entered by the user
[ 1 J d(X)/d(t) = -ral . 1
[2 J d(T)/d(t) = «40000+(1 0*(T-298)))*(-ra)*(1/.1 »/(56 . 25-(10*X»
Explicit equations as entered by the user
[1] k1 = .002*exp«100000/8.314)*(1/373-1/T»
[2] Ca = .1*(1-X)
[3] Cb = .1 *(1 . 25-X)
[4 J k2 = . 00003*exp«150000/8 . 314)*(1/373-1/T»
[5] Cc = .1*X
[ 6] ra = -«k1 *(Cal\.5)*(Cbl\.5»-(k2*Cc»
----
600
0.15,..-----·---· - - -
[J
480
0.09
300
DD
- eb
0.06
420
360
r-----_'. . -. .-
0 . 12 - - . -.. -..- - - - -..- - -
540
-""'- ....-....
0
-
c, ,
...-"'"
2
Check answer:
0.03
4
8
6
Cc
.~ Ca JCi;
-
10
0.00
2
0
kl
-'ki
0.02504
'JrE'0"?07""'4:=::9;;:::;;S';;..JO.09995
_",_',',
,~
tOO
=3.6~j == 0.289
9-15
4
t
~=~
..<¥=-~¥M
6
8
_J
10
P9-8 (a)
Use Polymath to solve the differential equations developed from the unsteady state heat and mass balances.
See Polymath program P9-8·a.pol
POLYMATH Results
No Title
08·11-2005. Rev5 1233
Calculated values of the DEQ variables
Variable
t
Ce
Cs
T
Iprime
KIn
mu1max
Yes
mu
Q
Cps
Hrxn
V
rho
rg
Cpe
rs
initial value
0
0,,1
300
278
0.0866522
5
0.5
0,,8
0.0426159
0
74
-2"OE+04
25
1000
0,,0042616
74
-0.005327
minimal value
0
0,,1
43.080524
278
6.499E-04
5
0.5
0,,8
2" 912E·-04
0
74
-2"OE+04
25
1000
0.0042616
74
-30 . 908876
maximal value
300
205.63558
300
333.55016
0,,3593146
5
0.5
0. 8
0" 1725264
0
74
-2"OE+04
25
1000
24.727101
74
-0.005327
ODE Report (STIFF)
Differential equations as entered by the user
[ 1 J d(Cc)/d(t) = rg
[2 J d(Cs)/d(t) = -rg!Ycs
[ 3] d(T)/d(t) = (O+(-Hrxn)*(rg))/(rho*Cps)
Explicit equations as entered by the user
[1] Iprime = (0,,0038*T*exp(21.6-6700/T))/(1 +exp(153-48000fT))
[2] Km =5.0
[3] mu1max= 0,,5
[4] Ycs=0 . 8
[5 J mu = mu1 max*lprime*(Cs/(Km+Cs))
[6] 0=0
[7J Cps = 74
[8] Hrxn = -20000
[9J V = 25
[10] rho = 1000
[111 rg = mu*Cc
[12] Cpc = 74
[13] rs=-rg!Ycs
9-16
final value
300
205.63558
43.080524
333,,55016
6.499E-04
5
0.5
0.8
2,,912E-04
0
74
-2.0E+04
25
1000
0,,0598648
74
-0 . 074831
340.--------------------------,
300 ...
326
240
312
180
298
120
284
60
60
120 t
180
240
300
o
--_=-----------,
~
W
---'-'o
2-40
60
300
0.40..---------------------------,
40 .--------------------------,
0 ..32
32
0.24
~
~l
24
0.16·
16
0.08\,..·----
8
0.000 -·---6~0----1-2(~)::'t-I 80
240
300
LJ
oo
----~~--
60
120 t
..-.--.- .•......
180
240
300
P9-8 (b)
When we change the initial temperature we find that the outlet concentration of species C has a maximum at To = 300
80
---------.--
64
48
32
16
o 280
300 To 310
320
330
P9-8 (C)
Cc can be maximized with respect to To (inlet temp), Ta (coolantiheating temperature), and heat exchanger area.
Therefore, if we are to find the optimal heat exchanger area the inlet temperature and coolantiheating temperature needs
to be specified. If we take To = 310 and T a = 290 we find that the optimal heat exchanger area is an infinite amount of
area . As A increases Cc increase without a maximum in 24 hows. Cc = 118..
9-17
P9-9
First order liquid phase, CSTR
First solve the steady state problem fin the heat exchange area A for notmal operation
T= 358 K.
Co
.- ......... J:\
Mol balance:
1:
V :::::-.dCs
__ ... =: 0
dt
In(}1·)
== ..-·-·-··l····--..·. ·-'~·-···-r . -·-··--......•- ..........................._..... _---
R.T2
8.314(323)
R.n
8.314(313)
94852 ::.: JI Inol K
k
= Llexp[ ?~~??(_l
8314
.......
313
l)t"l07A
Imin
J
T
Steady state solution:
CAo =2 M::;: 2*90 "" 180 g/dm'
C A :;;;; 180 107.4 (0.4) CA
180
H (0.4* 101.4)
9-18
for T;;;:.; 358 K
CB :: -
Energy balance:
( -
440) 04 = 175 . 9 gldm3
Q.- Ws- LFio.Cpi(T····Tio)+(-AHrx)(-··ra.V)
.dT
. . .-=----.
. . . . . . .- . . . . .-.. .-.-.-.--.. . . . . . . . . . . . . ._. . . . -.. . . . . . . . . . . . __.... == 0
LNi.Cpi
dt
Q == UA(fa - T) =(120*60)A(273-1)
FAOCpA.(f - TAO)
=90000*2*(T - 313)
J/min
J/min
A :::: .-180000.(358··313)
. . . . . _. . . . . _. . . . . . ._.•.--_. . -.. . .-..- +
-(250)(
- - .44Ox200)
- . - - = 22.7m2
7200(273 - 358)
Use the unsteady state equations to detennine what heat exchange area A will give a
l1maway reaction.
where
I
Ni. Cpi = Cpsol Psol V =2 x 900 x 200 =360000 11K
See Polymath program P9-9 . pol
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
T
tau
A
k
V
ra
Na
Nb
initial value
0
180
0
313
0.4
22.696
1..1
200
-198
0. 9
0
minimal value
0
4.1007565
0
313
0.4
22.696
1..1
200
-439 . 75367
0.0205038
0
maximal value
10
180
175 . 89924
357.97719
0.4
22.696
107.23721
200
-198
0.9
0.8794962
final value
10
4.1007565
175.89924
357.97719
0.4
22.696
107 . 23721
200
-439.75367
0.0205038
0.8794962
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = «180-Ca)/tau)+ra
[2] d(Cb)/d(t) = -ra-(Cb/tau)
[ 3] d(T)/d(t) = «7200*A*(273·T»-(90000*2*(T-313) )+( (-250)*ra*V) )/(2*900*200)
Explicit equations as entered by the user
[1 j tau =.4
[2] A = 22 . 696
[3] k=1.1*exp(11409*(1/313-1/T»
[4J V = 200
[ 5 1 ra = -k*Ca
[6] Na = Ca/V
[7] Nb = CbN
9-19
360,-----=======1
310 OI---2~--4--t-6---·8~--l10
A sensitivity analysis for different values of A show that a new steady state is reached
at every new A but at an increasingly higher T. The required A depends on the
definition of reactor runaway.
P9-10 (a)
X
=L 8;CPi [(T - To )/(-Illi Rx)]
(1)
X=1 7 T=T f gives
P9-10 (b)
CA
= CAO (1- X)
(2)
Insert (1) in (2)
(3)
For x= 1, T f =T expression (l) becomes
(4)
Insert the inverse of (4) in (3)
9-20
(5)
Remember that C BO is not constant here. Similar to C A, use equations (1) and (4)
IeicPi
CB
= CBO -- CAO • (_ Ml Rx) . (T - To) = CBO -
C
[eB ·T! -T + (To -eB 'To)]
(T-To) _
CAO • (Tj _ To) - CBO
•
e
B •
(T-To)
(T-To )_
(Tj _ To) - CAO • (Tj - To) -
T -T,
AO
f
0
For developing -rA use
(6)
8
C =C
B
AO
[
B
. T f- O
T + (T, B
- 8 . T, )]
0_
T -T,
f
(7)
0
(8)
Insert (7) and (8) in (9)
P9-10 (C)
The unsteady energy balance for a batch reactor is showed in equation 9-11.
dT
dt'=
Q+Ws + (-Ml Rx )· (-rA .V)
INtCPi
Adiabatic operation (
.
Q=0), neglecting W, and assuming ~Cp=O gives the following using the expression for -rA:
9·-21
P9-10 (d)
P= 1, E>B = 3 in the equation from part (c)
Put a = 1,
P9-10 (e)
Use the reaction constant expression from the great Swedish chemist Arrhenius and develop expression from 9-lOd.
dT [(--MI
---=
- - " -)]
Rx
dt
IepPi
.
k·CAO
(Tj
-T,)
2 [(
Tj -T ) . ( 3·Tf -T+ ( Yo -3·Yo ) ) ] ~
0
P9-10 (0
First use a plot ofT vs . t to get To (329 K) and Tf (439 K).. Checking the concentration of species B gives E>B=3. Make a
table showing t, T, dTldt, left hand side of equation P9-10 ..7 and liT.. Plot left hand side of equation P9-10.7 vs. liT in
Polymath and use linear regression to get E from the slope and kJ from the intercept.
Regression equation as shown in Polymath:
y = aO+ al· T _inverse =-8944.2· T _inverse + 17.534
Activation energy and Arrhenius constant from slope and intercept:
slope =
-% => E = -8944..2·8.314 = -74. 36
kJ/mol
9-22
~ ·C J+~'
= In(kl .6.7) => kl = 12100 ·exp
T-T)
R·Yo
12100
6.7
AO
aD =17.534 = In(
(
f
2
0
17534
= 1.298.1011
L-.,..--'
neglect
P9-10 (g)
Follow the procedme from example 9-3 gives heat of reaction.
X=l and T=T f gives
-MIRx
= I8 CPi .(T-To)=(CPA +8 B .CPBXTj-To)
j
=(189.7+3.75.4. J )(439-329K)=-45.75kJlmO[
mo[·K
P9-11 (a)
Mol balance:
.
dNa
'..' ...•. ::;: ra.V
dt
dX
Nao--·-·· ='raV
dt
dNb
.__.-.. :::;: 2. Ta. V + Fha
dt
Fba::;;: CbO.1)o=
dNc
. . --:::: -raY
dt
Rate law:
Stoichiometry:
.,
- ra == k.Ca.Cb
4.1)0
V==Vo +1)o .t
2
CQ="~c:..
Ne
C.C = ". . . . _.
O
V
V
Energy balance:
\\ hc:re Q:= UA.(Ta - T)
'vis::::: 0
= 250.(390 - T)
t~Hrx(T)
::::: -55000
2:Fio.Cpi ;;;; Fbo.Cpb :::: Cbo. Do.Cpb = 4*1)0*20:::: SODa
9-23
dT
250.(290 ----::: ---_
..
T) -80uu.{T -- 32S)~· (-55000). (-ra. V)
.. _.....,..",-_ ........"""""'".
- .- .
~,.,..,...,.......-.-----
35. Na + 20. Nb +- 75 ,vc
See Polymath program P9-11-a.pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
Na
Nb
Nc
X
T
vb
k
V
Fbo
Ca
Cb
Cc
ra
initial value
0
50
0
0
0
300
1.5
5.0E-04
10
6
5
0
0
0
minimal value
0
0 . 3324469
0
0
0
283 . 60209
1.5
2.3E-04
10
6
2.202E-04
0
0
-0.0028611
maximal value
1000
50
5900 . 6649
49.667553
0 . 9933511
300
1.5
5.0E-04
1510
6
5
3.9077251
0 . 0718765
0
final value
1000
0 . 3324469
5900.6649
49.667553
0.9933511
298.76383
1.5
4.73E-04
1510
6
2.202E-04
3 . 9077251
0 . 0328924
-1. 59E-06
ODE Report (RKF45)
Differential equations as entered by the user
[ 1 j d(Na)/d(t):::; ra*V
[2 J d(Nb)/d(t):::; 2*ra*V+Fbo
[3 J d(Nc)/d(t):::; -ra*V
[ 4 J d(X)/d(t):::; -ra*V/50
[5 J d(T)/d(t):::; ((250*(290-T)) .. (SO*vb*(T·325))+(-55000*(-ra*V)))/(35*Na+2O*Nb+75*Nc)
Explicit equations as entered by the user
[1J vb:::; 1.5
[2 J k:::; . 0005*exp((SOOO/1 . 9S7)*(1/300-1/T))
[3] V:::; 1O+(vb*t)
[4] Fbo:::; 4*vb
[5] Ca:::; Na/V
[6] Cb:::; NbN
[7] Cc:::; Nc/V
[8 J ra:::; -k*Ca*CbI\2
-l
[]
5
3
2
-..- - - -
296
292
288
28-t
1
o0
300
200
-ton t
600
800
1000
281.1
'----~---~---~--~---'
o
9-24
200
400 t
600
800
1000
1.0,------------==""'"
0,8
0.6
0.4
0.2
200
400 t
600
800
1000
T"ble ! Iterating \\ It Uo for X:::O.8 and T<403K and daily Nc " 110 mole
--'''·''·''-'-1.
~()
T
dm'/min ;
f"-·"'-·--_····_·,·"·,·",,·,,··,,_·,"-_··_'-""_·,,,,·,
.••-"-",
.._.",,Nc (X.::::Q 8)
Daily Nc
(X=O.8)
t
(X:O 8)
K
'min
mol
.. Nc(24*601t+30)
A flowfate of 1.5 mol B/min would produce 120 mol C/ day, with X ;;;;: 80 % and
T< 130°C at all times
P9-11 (b)
If the max. coolant rate falls to 200 molimin, then it may not be prudent to assume that
the coolant leaves at the entering arnbient temperature, Ta. It should be assumed that
the coolant temperature varies spatially along the heat exchanger pipes and the
required tenn for the heat exchange would be :
Q =: l'11;:; Cp.:ool (Ta 1 - Tal)
Ta2 = l' -
(1" -
where Tal:= ambient T coolant entering
Tal = ambient T coolant leaving
Tal).ex p(-_U:4 )
rTkCpcoJ/
The reduced flowrate and hence heat exchange. may increase the reactor temperature
to approaching 130o.C, the upper limit, at conversions approaching 80 %, and so more
caution is required. The incorporation of temperature control would be prudent.
P9-12
CSTR startup
Need steady-state values at To;::::: 75~
9-25
dNa
Mol balance: -;;;-;::: F ao . . . Fa + ra.v
dCa
(Cao - Ca).vo
dt
V
--=
-
deb
(Cbo - Cb)v.,
-=:
dt
V
+ra
+ra
dCc (Ceo·- Cc)u,
--.dt
V
dCM
---
...-.".
ra
__
V
(CMO -. CM)U,
....._. -
dt
Fio
C:ao == Faq. = 0.18157
Cia=:'Va
Ib moll ft3
1)"
Cbo = .!bo = 2.2696 Ib moll ff
u..
Cco=o
CMO
=
.!:.~?. ;: : 0.22696
v.,
Ib moll ftl
1
V ;::: ....--...;. X 500
7.484
Fat)
pao
Fba F.tfO
80
1noo 100.
3
.+ ._._....... ;::: ---'-- + - - + ---..- :::: 440 6 ft / hr
pbo PMO 0.932 3.45 L54
.
Vo ;::: .-..... _- + .,_....-
Rate law:
- fa;:::
k.ea
k = 16.96eI2.exp
(Co in excess)
(-i9s?1:ir460))
Energy balance:
dT
Q- Ws·- LFio.Cpi(T.- Tio) + (-·-t:.Hrx)(-ra_V)
°d;- = .---................ ··_········· ..--·I)ifi. C;i--·-·----·-·.·. . . --·
9-26
Ws=o
Q = IDc.Cpcool.(Tal - Ta2) == 1000 x 18 x (60 - Ta2)
l:Fio.CpL(T ~ Tio) == [Fao.Cpa + Fbo.Cpb + FMO.C,PM ].(f - 75)
AHrx ==
~36000
Btullb mol
I,..Ni.Cpi == Na.Cpa + Nb.Cpb + Nc.Cpc + NM'CPM
== 53Ca. V + 18Cb.V + 46Cc.V + 19.5CM.V
Ta2::: T- (T-- Tal).exJ _ V A
y~ l1kCpcool
== T -- (T - 60) exp(
.
.
J
l~_)
lOOOx18
Ta2 == T - 0.41111(T - 60)
tIT
-dt
18000(60- (T - 0.41111(T- 60)- 2275O(T -75) + (36000)(-ra.V)
..
.--...........
...-.35Ca.V + 18Cb.V + 46Cc.V + 195Cv.V
Initial conditions: To::.:; 75, T == 138.5 o:F. Ca:::: 0.03780 • Cb :;::: 3.3062 •
Cc;;: 0.0144,
If
eM::::
O.22691b moll ft
To drops from 75 to 70 OP
P9-12 (a)
P-control only:
manipulated variable:;: lIlc
controlled variable = T
where Illco = lOOOlb mollh
Tsp == 138.5 'F
~::: IIlco + kc.(T - Tsp)
kc= 10
See Polymath program 1'9· 12a..pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
initial value
minimal value
t O O
maximal value
4
9-27
final value
4
3
Ca
Cb
Ce
Cm
T
I
FaO
TO
V
Tsp
UA
Tal
ke
k
FbO
FmO
meO
ra
NCp
ThetaCp
vO
CaO
CbO
CmO
tau
X
me
Ta2
Q
0.03789
2 . 12
0 . 143
0.2265
138.53
0
80
70
66.809193
138 . 5
1.6E+04
60
10
24.990212
1000
100
1000
-0.9468791
3372 . 5882
284.375
441 . 46403
0 . 1812152
2 . 2651902
0.226519
0 . 1513355
0.7909116
1000 . 3
106 . 23674
8.325E+05
0.03789
2 . 12
0 . 1227539
0.2265
125 . 70694
-39 . 636625
80
70
66 . 809193
138 . 5
1 . 6E+04
60
10
13.770535
1000
100
1000
-0.9468791
3372.5882
284.375
441.46403
0.1812152
2 . 2651902
0.226519
0 . 1513355
0 . 6775218
872 . 17351
102 . 00022
6.594E+05
0 . 0584442
2 . 1423214
0.143
0.226519
138 . 53
0
80
70
66 . 809193
138.5
1.6E+04
60
10
24 . 990212
1000
100
1000
-0.7913918
3385 . 3399
284.375
441 . 46403
0 . 1812152
2 . 2651902
0.226519
0 . 1513355
0.7909116
1000.3
106.23674
8.325E+05
ODE Re~ort {RKF4S}
Differential equations as entered by the user
[1] d(Ca)/d(t) = 1/tau*(CaO-Ca)+ra
[2J d(Cb)/d(t) = 1/tau*(CbO-Cb)+ra
[ 3 J d(Cc)/d(t) = 1/tau*(0-Cc)-ra
[4J d(Cm)/d(t) = 1/tau*(CmO-Cm)
[5] d(T)/d(t) = (·Q-FaO*ThetaCp*(T-TO)+(·36000)*ra*V)/NCp
[6j d(I)/d(t) = T-Tsp
Explicit equations as entered by the user
[11 FaO = 80
[2] TO = 70
[3] V = (1/7484)*500
[4] Tsp = 138.5
[ 5 ] UA = 16000
[6j Ta1 =60
[7] kc= 10
[8J k = 16. 96e12*exp(-32400/1 . 987/(T+460))
[ 9 1 FbO = 1000
[10] FmO= 100
[11] mcO = 1000
[ 12] ra =·k*Ca
[13] NCp = Ca*V*35+Cb*V*18+Cc*V*46+Cm*V*19 . 5
[14] ThetaCp = 35+FbO/FaO*18+FmO/FaO*19 . 5
[15] vO = FaO/0.923+FbO/345+FmO/1 . 54
[16] CaO = FaOlvO
[ 17] CbO = FbOlvO
[18] CmO = FmO/vO
[19] tau = V/vO
[20] X = (CaO-Ca)/CaO
[21] mc=mcO+kc*(T-Tsp)
[22] Ta2 = T-(TTa1)*exp(-UA/(18*mc))
9-28
0 . 053671
2.1376461
0.1275442
0 . 226519
128 . 49299
-39.636625
80
70
66.809193
138.5
1.6E+04
60
10
15.702753
1000
100
1000
-0.8427832
3383.2355
284.375
441.46403
0 . 1812152
2 . 2651902
0.226519
0.1513355
0.703827
899.92991
102.98479
6 . 963E+05
[23] Q
= mc*18*(Ta2-Ta1)
140
1100
136
1040
E!J
132
980
-
128
920
860
124
120
Q
0.0
0.8
1.6
t
2,4
3.2
4.0
800
0.0
0.8
1.6 t
3.2
2.4
4.0
P9-12 (b)
I-control only :
manipulated variable ::: .~
controlled variabLe::: T
kc
mc:::mc... + - ]
where IDeo::: 1000 Ib mollh
1.1
kc ::: 10. 'tI::: 1
dI
- : (T-Tsp)
Tsp= 138_5 ~
tit
See Polymath program P9··12·b.pol
---
10001,.,...,-----------·----,
940
[mJ
880
820
760
110 '---~---~0.0
0.8
1.6 t 2.4
P9-12 (C)
3. 2
4. 0
700
0. 0
0. 8
L6 t
2.4
3.2
4.0
manipulated variable == me
PI-control only:
controlled variable = T
rrk
==
kc
meo
+
kc(T - Tsp) + ..- I
where nlco:;;:;: 1000 Ib molfh
'tl
kc ;;;; 10, 1:1 :;;:;:1
_.dl,'... = (T dt
Tsp ;;;; 138. 5 "F
Tsp)
See Polymath program P9-12·c.pol
1-'10,.------- ' - - - - ,
1100
1020
132
9-'1U
128
860
12-'1
780
120
'------~----------'
0.0
0,,8
1.6 t
24
3. 2
-'10
700
0,0
0,8
1.6 t
2.4
32
---,------,----
P9-13
dCa
-,-
Mol balance:
(Cao . . ,. Ca)
=~.
dt
,,_._. - - + ra
Cao ::: 0 ..1 kmollm 3
!
Cao :::::: 0.1 mol/dm3
deb (Cbo-- Cb)
--""= '--'_.'."'' ' ' ' .-'' ' ' ' -' -+ ra
dt
Cbo= Cao
1"
dec _ (0- Cc)
"
- ._ .. - -,-----,...... ra
dt
Rate law:
1:;;;;
t
50 s
,',' ra ;;;; k.Ca.Cb
k(T) ;;:;; 0.01 expr",,~_oooo.(_.
1 __ ..~)]
L1.987
9··30
300
T
-'1.0
Energy balance:
dT
Q-
w.~
. . . . L Fio, epi(T -
'dr" = .------.--......
Tio) + C,-t:JirxX-ra. V)
L~vi:-
Cpj---,··---·-----------
V == VI) • 't :: 2 x 50 == 100 dm3
=0.1 x 2 =.:: 0.2 molls:: Fbo (equimolar feed)
LFio.Cpi.(r - Tio) =2 x 0.2 x (15)(r - 300) =6(T- 300)
Fao:: Cao . 1)0
tJfrx(f)
=Lllirx(To) + ACp.(T - To)
but ACp;;;: 0
= - 41000 - (- 20000) - (-. 15(00)
= .' 6000 cal/mol
f!T =~(T.,::- 300) + (6000).(-ra. V)
dt
15.Ca.v+ 15.Cb. V +30.Cc. V
No control: See Polymath program P9·13 .. a.po!
POLYMA Til Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
Cc
T
V
k
ra
initial value
0
0,,001
0,,001
0
300
100
0.01
-·1" OE-08
minimal value
0
1.027E-04
1.027E-04
0
300
100
0.01
-0.0088037
maximal value
400
0.0732572
0.0732572
0.099864
2.574E+05
100
1.893E+05
-1.0E-08
final value
400---1.027E-04
1.027E-04
0,,099864
2.574E+05
100
1.893E+05
-0.001998
ODE Report (RKF45)
Differential equations as entered by the user
[1 J d(Ca)/d(t) = «.1-Ca)/SO)+ra
[2] d(Cb)/d(t) = «.1-Cb)/SO)+ra
[3 J d(Cc)/d(t) = (-Cc/SO)-ra
[4 J d(T)/d(t) = «6*(T'300»+(6000*(-ra*V)))/(1S*Ca*V+ 15*Cb*V+30*Cc*V)
Explicit equations as entered by the user
[1] V = 100
[2] k = .01*exp«10000/1 . 98'7)*(1/300-1rr»
[3] ra = -k*Ca*Cb
9-31
300000
rn
2-40000
.. ell
180000
- Cc
120000 .
60000
0.00
0
80
160
t
240
320
400
0
0
80
160 t
240
320
Tbe results without control, indicate a runaway reaction, as T continues to increase
after the concentrations have approached their steady-state values
Control aspects:
assume that the operating T should not exceed 550K, the
boiling point of the liquid.
Without given data for beat exchange:
Try
manipulated variable::::: Tio (iulet feed T)
conuolled variable::::; T
P-control
Tio::;: IF (T<550)'I1IEN (300) ELSE (Tioo +- kc.(f - Tsp))
where Tsp:::: 550 K, Tioo:::: 300 K, kc =-10
~
Tio rnanipuJation not feasible fc)[ T control for
any kc (requires Tio .. ···2000 K)
See Polymath program P9··13-b.pol
8001'----·------------,
700
600
500
400
300 L-_""""'==-__
o
Try
80
160 t
240
320
400
manipulated variable::::: Va (inlet volumetric flowrate)
controlled variable::::: T
9-32
400
P-control
U o ::::
IF (T<550) THEN (2) ELSE ('0 00 + kc.(f '- Tsp»
where Tsp :::: 550 K,
U OO
::::
2 dm 3, kc :::: 10
See Polymath program P9-13-c.pol
600~----------.....,
540
480
420
360
300
L...--~"---
o
_ _ ~_~_ _
80
160 t
240
320
400
P9-14 (a)
CSTR startup
Initial conditions: To:::: 75, T = 138.5 "f. Ca::: 0.03780 , Cb:::; 2.12 •
Cc::: 0.0144, eM:::: O.22691b mol f ft3
If
T drops froIn 138.5 to 133.5 OP
As in P9-12B except:
dCa
dt
Fao- Ca. Va
V
..---- = - - - - - . + ra
Fao
1000
3.45
100
1.54
Va ::::: ......... ,..................f. --"'. ._- .+ ... _-
0.932
dT
.--- :::;:;::
dE
_.
18000(60·(T ····0.41111(1'- 60) -(35Fao + 19950).(1'-75)
+ (36000)(·-ra. V)
............................ -...-.."-.--. "'.".~.- ....----.-.......................-~ ..,-_., ._.__ ...,,_._...... --_..._..
__............. ~- .. ,,- -. .-.~ .. - .............. _"._--.
".,.. ",,--~-
'""
35Ca. V+ 18Cb. V
+ 46Cc..1' + 19..5C~. V
manipulated variable = Faa
I-·control only:
controlled variable = T
Icc
Fao = Faoo+-I
11
where Faoo::::; 80 Ib roollh.
say kc = -0.2 ) 'tt == D~I
9--33
-----
dI
:::: (T - Tsp)
dt
-
Tsp = 138.5 OP
See Polymath program P9-14-a pol
POLYMATH Results
Calculated values of the DEO variables
Variable
t
Ca
Cb
Cc
em
T
I
FaOo
TO
V
Tsp
UA
Tal
ke
k
FbO
FInO
meO
ra
NCp
FaD
ThetaCp
vO
CbO
CmO
tau
CaO
me
Ta2
Q
X
initial value
o
0.03789
2.12
0.143
0.2265
138.53
o
80
70
66.809193
138.5
1.6E+04
60
0.2
24.990212
1000
100
1000
·-0.9468791
3372.5882
80
284 375
441 . 46403
2.2651902
0.226519
0.1513355
0.1812152
1000
106.24535
8.324E+'05
0.7909116
minimal value
o
,,·27 . 922973
2.12
'.. 1. 7469378
0.2265
17.999228
-307.34656
80
70
66.809193
1385
1.6E+04
60
0.2
0.0260082
1000
100
1000
·-0.9468791
3372.5882
·-534.4229
-3993.8255
·,,224.21625
-183,_ 76896
-18 376896
-12 277456
--118 36922
938.55771
34 . 305432
--4.341E+05
-65.857622
maximal value
4
0.0898852
95.168593
0.143
9.3421656
138.53
o
80
70
66.809193
138 5
1.6E+04
60
0.2
24.990212
1000
100
1000
0.723285
5.574E+04
80
9.686E+04
441 46403
364.28299
36.428299
24.337453
61.10206
1000
106.24535
8.324E+05
12.667586
final value
4
-27 922973
95.168593
-1.7469378
9.3421656
17.999228
-307.34656
80
70
66.809193
138.5
1 6E+04
60
0.2
0.0260082
1000
100
1000
0 . 723285
5.574E+04
-534.4229
-2.3299873
-224 21625
-4.45998
-0.445998
-0.2979677
2 3835154
938.55771
34.305432
-4.341E+05
12.667586
ODE Report (STIFF)
Differential equations as entered by the user
(1) d(Ca)/d(t)
1/tau*(Cao-Ca)+ra
(2) d(Cb)/d(t) = 1/tau*(Cbo-Cb)+ra
[3 J d(Cc)/d(t) = 1/tau*(Q-Cc)-ra
[4] d(Cm)/d(t) = 1/tau*(Cmo-Gm)
[5] d(T)/d(t) =(-Q-FaO*ThetaCp*(T-TO)+{-36000)*ra*V)/NCp
[6 J d(I)/d(t) =T-Tsp
150
-----.----,-----.--
=
144
138
132
126
120
9·14
0.0
08
1..6 t
2.4
3.2
4.0
FmO = 100
[ 11] meO = 1000
[12] ra = -k*Ca
[13] NCp = Ca*V*35+Cb*V*18+Cc*V*46+Cm*V*19,5
[14] FaO = FaOo+(kc/,,1)*J
[15] ThetaCp = 35+FbO/FaO*18+FmO/FaO*19.5
[16] vO = FaO/0.923+FbO/3.45+FmO/1 ,,54
[17] CbO = FbO/vO
[18] CmO = FmO/vO
[19] tau = VivO
[20] CaO = FaOlvO
[21] mc = mcO+kc*J
[22] Ta2 = T-(T-Ta1)*exp(-UA/(18*me))
[23] Q=mc*18*(Ta2-Ta1)
[24] X = (CaO-Ca)/CaO
[10]
90
88
86
...
82
80
P9-14 (b)
Tal
""""---.-- -----
84 .
0.8
0.0
1.6
t
3.2
2.4
4.0
= 55°P
I-control onl y :
nUl11ipulated variable ;;::: To
controlled variable::::: T
To;;::: Tooo
+-~~1
where Too
=75 OP ,
say kc,:;:; ·,0.2. 't'I
dl
=0.1
Tsp;:: 138.5 OP
dt ::::: (T - Tsp)
See Polymath program P9-14-b.pol
150
85
144
82
138
79
132
76
126
73
120 '---~--~
0.0
0.,8
1.6 t
2.4
3.,2
4.,0
70
01)
P9-14 (C) No solution will be given
P9-15
9-35
08
L6 t
2.4
3.2
4.0
ammonium nitr'ate
A
Feed, rno
nitrous oxide
stearn
C
B
= mao +- meo ; ; ;: 0,,83 mo + 0.17 mo
dfllfa
Mol balance:
(rnco
=::
0 "I- l'a. V
.._-:::;; mao"
dt
17% liquid water in feed)
Ma == Ib A, mao = lblb A
dMb
. _.._. :::: 0- mh···, rQ" V
dt
leaving:
m};;;::: mao. Gb
(
"\
t,;;b X ):::::
mao; Sb:::: 0, b/a::;;:: 1
(X;;;;; 1, i.e. all of A entering reacts, but there is some initially inside)
dlvlc
."'-..-~, :;:;;: mea- me -- 2. ra. V
dt
leaving: me:::: mao. ( 6t + ~-
x) :; ;:
mea + 2mao ;
e.
.;;t;(},
cia;;;;: 2
assuming all water entering with the ammonium nitrate leaves as steam.
Rate law:
-ra. V :;:;:; k.Ca. V ::::: k.Ma
Ln':))
0 -''')
( 5.03.
R,T2
1
1.987(560+460)
R.n
=::
1
1.987(510·+460)
88500 Btu Ilb mol
Energy balance :
aT Q- Ws~· ')' mio . Cpi(T·, Tio) ." .., meo. (Hi~ Hio)-+- (--Mlrx)( -"'la. V)
d;' = --,..--"-..~,,,-..where
....... ,-..,.---
I. l~fi:'L:;;i-''''-'''''''''''''''''''''''-'---'--'~ . .". . . . . . .". _.
Ws=o
Q =UA.(Ta - T)
9-36
Imio.Cpi.(T ~ Tio) == mao_CpaCT - Tio)
::::: 0.83 x 310 x 0.38 x (T - 200)
mco.(Hi - Hio) =:: mco.[Hg(T)* HI (Tio)]
=:: 0.17 x 310 x [(1202 + 0.47(f - 500)) - 168J
assumes all liquid water in feed leaves as steam.
Hg(T);;:; Hg(Tref) + Cpc.(T - Tref):::: 1202 + OA7.(T ~ 500)
HI (200);;;:: 168 Btu fIb
2:MLCpi ;::; 0.38 Ma + Cpb.Mb + 0.47 Me
mphysical properties systems:
1
Cpb (N,O. 516"F) = 1065 J/kg.K. (10 55 Btu ). ( 0.454
(~.)
From ChemCAD
:J
::: 044 Btullb N 20, K
P9-15 (a)
att=:O Ca=:5001b, T=516OP
__ ___
dT
10000.(515- T) ···,97..77(1' -, 200)
...... 52.7[(1202 + 0.47(T.,- 500»)·· ··1681 +- 336.(-ra
V)
... .. :;;:;::: .... .... .................. .......... ......... ..... ............-,-...
......... ...... .............. .,..................... ... ....... .. ......__ ....... _... ................... _..__.._..... ... ....... . .....
....................
OJ8Ma ·-j·OA4.Mb + OA7Mc
dt
"'" .. .,
~
~
,.,
,
.,
"
~-
"
-
-~-,-
"'
,
,
,
,
,
See Polymath program P9, IS-a. po]
POLYMATH Results
Calculated values of the DEQ variables
Variable
-t
Ma
Mb
Mc
T
mco
mao
k
r'aV
rob
mc
initial value
o
500
o
o
516
52.7
257.3
0 . 7028594
-351..4297
257.3
567.3
minimal value
o
---
399.06704
o
o
514.15817
52.7
257.3
0 . 6447579
-366.56963
257.3
567.3
maximal value
10
500
100.93296
201.86591
517.83877
52.7
257.3
0 . 7677849
-257.30162
257.3
567.3
final value
10
399 . 06704
100.93296
201.. 86591
514.15817
52.7
257.3
0.6447579
-257 . 30162
257 . 3
567.3
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ma)/d(t) =31 0*. 83+raV
[2] d(Mb)/d(t) = -mb-ra V
[3] d(Mc)/d(t) = mco-mc-2*raV
[4] d(T)/d(t) = ((1 0000*(515-T))-(99 . 77*(T·200))-(52.,7*(1202+(.47*(T-500)))-168)+(336*(-raV)))/(.,38*Ma+.44·'
Mb+.47*Mc)
Explicit equations as entered by the user
[1] mco = 310*.,1]
[2 J mao = 310*.83
[31 k = .,53*exp(44540*(1/970-1/(T+460)))
9-37
[4
[5
[6
raV = -k*Ma
mb = mao
me = meo+2*mao
soo ,------------------------,
S18~--~-------------------.
::I~
S17
S16
------------------~
S16
380
515
340
514 '------.- 2
4
0
t
6
8
10
300 '---~----------------------'
o
2
4
t 6
S
10
P9-15 (b)
lJA = 10000 Btu lh.ft2
P-{:ontrol only:
manipulated variable;:: Ta
controlled variable = T
Ta:;;;;; Tao + kc.(T - Tsp)
where Tsp == 516
of,
kc = -.5, Tao == 97.5 ~ == 515 ~
See Polymath program P9J5-b pol
SOO
460
420
380
--..._- _._
. _., .........
SIS.60
0
2
4
t
6
S
340
.................... .
10
300
0
P9-15 (C)
9-38
2
4
t
6
8
10
UA :::: 10000 Btu lh.ft2
PI-control:
manipulated variable :::: Ta
controlled variable:::: T
7'0.::::
Tao+kc.(T,-Tsp)+ kr:...I
1J
dI
dt
- ' ::::(T-Tsp)
where Tsp;;;: 516 OPt kc;;;: -5, 'tI
=:-
1. Tao =975 ~
=515 OP
See Polymath program P9·15·c,.po]
516.20.-----------------,
500
516.12
460
516.04
420
515.96
380
51588
340
515.80
2
0
4
6
t
8
10
300
0
---~---'---~
2
4
t
6
8
P9-15 (d)
UA;;;: 10000 Btu lh.ft2
PI-control1oop 1 :
manipulated variable = Ta
controlled variable:::: T
To.
= Tao + kcL(T -
kcl
Tsp) +--.11
1:11
dll
dl
-=(T-Tsp)
where Tsp
PI-cOIlu'olloop 2 :
=516 <>p, kcl =-5, crd
;:;;; 1. Tao;:;: 975 oR;:;;; 515 ~
manipulated variable;:;;;; mao
eontmlled variable == M (= Ma + Mb + Me)
9-39
10
kc2 I". . .
mao ;;;;; maoo + kc2 . (1\.1 - Jyfsp) +~.'
t/2
dI2
dt
~-
= (Ivf -
NIsp)
where Msp::::: 500 Ib kc2::;: 25, 't12:::: I, maoa :::: 310 Iblh
1
See Polymath program P9-15-d.pol
516.20...--------------,
500
516.12
460
51604
420
380
340
515.80 L -_ _ _
o
~
2
_ _~_~_----I
4
t
6
8
10
300
0
2
4
t
6
P9-16 (a)
Plot R(T) vs G(T):
R(T) = U.4(T ,,-···1:) + fJVoCp(T . . 1~)
G(T) =
-'L'\H~VkCA
Evaluat.e the parameters in those equations:
k
= 2 * 7.08;;: lOti exp( -30{}OOI 1.9871 T)
MI RX
;;;;;
-30000 BTU Ilbmol
p ::::: 50lb I .ftJ
[IA:;:: 150'* 250;;;;; 37500
See Polymath program P9-16-a pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
T
dHr
V
Caa
va
VA
initial value
0
520
-3 . 0E+04
48
0. 5
400
3.75E+04
minimal value
0
520
-3 . 0E+04
48
0.5
400
3 . 75E+04
maximal value
140
660
-3.0E+04
48
0. 5
400
3 . 75E+04
9-40
final value
140
660
-3 . 0E+04
48
0.5
400
3.75E+04
8
10
530
50
530
3.0E+04
0.3478362
0.12
0.0241085
0.75
2.404E+05
-5.25E+05
530
50
530
3.0E+04
0.3478362
0.12
0.479966
0.75
2.404E+05
-5.25E+05
Ta
rho
To
E
k
tau
Ca
Cp
Gt
Rt
530
50
530
3.0E+04
164.49672
0 . 12
0.479966
0.75
5.711E+06
6.825E+06
530
50
530
3.0E+04
164.49672
0.12
0.0241085
0.75
5.711E+06
6.825E+06
ODE Report (RKF45)
7.00E6 , . - - - - - - - - - - - - - - - . . . ,
Differential equations as entered by the user
[ 1] d(T)/d(t) = 1
5.40E6
Explicit equations as entered by the user
[ 1] dHr = -30000
[2] V=48
[3] Cao = ,,5
[4] vo = 400
[5] UA = 250*150
[6] Ta=530
[7] rho = 50
[8] To=530
[9] E = 30000
[10] k=2*7.08e11*exp(-E/1.987/T)
[11] tau = Vivo
[12] Ca = Cao/(1 +tau*k)
[13] Cp = .75
[ 14] Gt = -dHr*V*k*Ca
[15] Rt = UA*(T·Ta)+rho*vo*Cp*(T-To)
1-
G;T)
- Rn
1
3.80E6
2.20E6
6.00E5
-1.00E6
L-_~
520
_ _~_~_ _ _~_....J
548
576 T 604
P9-16 (b)
From part (a), we find the concentration and temperatures at the points where G(T) and R(T) intersect.
- .-
=547.t<'R
., - 04·'15
(;/\
C•A ::: 0.319
.
T
CA,= 0.068
T::: 628.6°R
T::: 57L3°R
The extinction temperature is To = 506°R. At this point R(T) is tangent to G(T) at the upper steady state.
P9-16 (C)
Using the unsteady-state equations for a CSTR we get the following
Mole balance:
d~;'_ = .~o{ C AO ....•. <:A} _ kC
dt
V
A
Energy balance:
See Polymath program P9-16-·c ..pol
9-41
632
660
,?Ol:-YI\llp'TH Report
Calculated values of DEQ
variables
"""1'''' """''''
; ' i
, ,Variablellnitial valuelMinimal value IMaximal value/Final value
"'~-"r"~'"
"" " " ' ' ' ' ' r ' '
'ifi"-"lo ,,","',"
[21Ca
'1'6"'16.
,~l
i3~J:('
)~~8.
A ICaO
'5 {FaO
'0.5
'200.
6 ITO
i530.
J
;J
J638.1791
0.5
,200.
'530.
11.416E+12
IR
11.987
j
1150 .
;150.
!1.416E+12
1.987
!1.987
1400.
530.
j48.
14iE
j3.0E+04
15/k
151.27669
16hau
;0.12
:17;rA
i
1150.
,
I
1400.
1250.
1400
i
.
1530 .
'530.
148.
i3.0E+04
'48.
TO~4253139
,j547.0711
10.5
'[200.
l530.
1.416E+12
i250.
1250.
"
<
10.4290427
JAr
"lu
"" '16.
1
11.416E+12
'11.987
t
1150 .
I
"
1250 .
:400.
:530.
148.
,
13.0E+04
r
!75.24179.
!3.0E+04
11.463353
:0.12
:0.12
10.12
!-3.491943
-3.971954
-0.4433316
[-0.6223844
i18 JI delH
j-3.0E+04
-3.0E+04
'-3.0E+04
i-3.0E+04
i19lNao
124.
24.
24.
24.
175.
'75.
~
1.165967
~
"f' ,
1
20 lCpa
;75.
Differential equations
d(Ca)jd(t) = rA + CaO j tau - Ca j tau
121 d(T)jd(t)
= (U * A * (Ta - T) - FaO * Cpa * (T -, TO) + (rA * V) * delH) j
Explicit equations
CaO = 0.5
2;1 FaO = 200
3/F TO
= 530
= 1.416 * 10
R = 1.987
U = 150
"7 A = 250
vO = 400
Ar
A
12
Ta = 530
'fijj V = 48
9-42
(NaO
* Cpa)
~tJ E = 30000
1'4) k = Ar * exp( -E / R / T)
i£~l tau = V / vO
= -k * ca
't~ delH = -30000
,161NaO = CaO * V
Cpa = 37.5 * 2
rA
""""""d
0.5
0.'1
0.3
0.2
0.1
IV
0.0
O.
0
1.2
204 t
3.6
48
When the upper steady state is used as the initial conditions, the unsteady-state mole balance shows that this steady state
is actually unstable. The concentration increases and the temperature drops to the lower stable steady state.
P9-16 (d)
When To or Ta is increases slightly, the upper steady state becomes stable . At these elevated inlet and coolant
temperatures, the lower steady state is no longer stable.
P9-16 (e)
Starting at the lower steady state, if To is increased to 550 OR, the lower steady state is no longer stable, but the upper
steady state is
0.50..---·-----------------,
800
0040
740
0.30
680
0.20
620
0.10
560
--6.0
0. 00 '---.-.-'----~-----....- .:1.8
0.0
1.2
2.4 t 3.6
500
""""""...........---------_._.-
0.0
1.2
2.4
t
3.6
48
6. 0
If we plot the concentration-temperature phase-plane trajectory, we see that increasing Ta will shift the trajectory to the
left. However, the final steady state is shifted to the right. This means that from the initial conditions, at any temperature
CA is lower for the larger Ta until the minimum in CA is reached.
9-43
0.5
0.5
0..-1
04
GJ
03
n.3
02
0. 2
0.1
tlJ
0.0
UO
'--------~-=-----'
500
560
P9-16 (f)
620 T 680
740
800
'--_~_~_._...._C:::...._
500
560
620
r
680
Individualized solution
P9-16 (g)
The following Polymath program gives the linear analysis of the plOblem
A = 1.175, B = 6.03, J = 1600, 1" =12
See Polymath program P9-16g.pol
~,~~¥ 1VIJ\~,f:!R,~portt,
Calculated values of
lVariable:lnitial value Minimal value Maximal value:Final value
''''
'm~'""l''~'~'''
i
1 't
;
1
0
0
6
12.
0.0176788
A:159719
iO:0176788
10.2
-0.0225558
0.2
!-0.0001258
:4 jea
!~.0681
0.04847
;5 IT
1628.
;542.6046
0.4290427,0.4253139
1
:638.1791
:547.0711
"6 )cpa
175.
175.
175.
~ [y
t,
ix
3
17k~0
6
l
)75.
18 ;Fao
fo.5
.1200.
9 lTO
1530.
1530.
/530.
i1.416E+12
h.416E+12
:1.416E+12h.416E+12
i 1.987
11.987
150 .
1.987
, f
1150.
:250.
10.50.5
. '.. 1200.
;200.
1
1400.
1250 .
1400.
(530.
1530.
:48.
17lE
i3.0E+04
181k
f
19ltau
151:27669
10.12
,
15!Ta
;~61v
.
;
L
'
:0.5
j200.
1530.
11.987
! l 5 0 · l i 5 0..
1250.
: 250.
:400.
1400.
/530.
t530.
J48.
t
+
J,
j
13.0E+04
i3.0E+043.0E+04
i
i
148.
148.
1.165967
175.24179
1.463353
0.12
!0.12
0.12
9-44
740
800
Differential equations
d(y)jd(t) = (-J * (1 - A)
* x + (B - C) * y)
= (-A * x - B * Yj J) j tau
d(Ca)jd(t) = rA + CaO j tau - Ca j tau
:ft d(T)jd(t) = (U * Area * (Ta - T) - FaO * Cpa * (T - TO) + (rA * V) * delH) j
d(x)jd(t)
Explicit equations
Cpa = 37.5 * 2
cao = 0.5
= 200
TO = 530
Ar = 1.416 * 10 /\ 12
1§~;;~i R = 1.987
FaO
U = 150
Area
= 250
= 400
:,~ITa = 530
V = 48
E = 30000
t~~l k = Ar * exp( -E j
vO
~4; tau
=V j
R j T)
vO
rA =-k * Ca
= -30000
NaO = CaO * V
!~~iJ J = -delHjCpa j
;~'~! delH
CaO
f~!C=6
~1:f;l)
t~ B = 6.03
A = 1.175
P9-16 (f)
Only the lower steady state plot of xl and yl will be shown.
9-45
(NaO
* Cpa)
0.50 , - - - - - - - - - - - - - - - - ,
0.50
038
0.38
r--------------,
~
l&J
0.26
0.26
OJ-I
(1.1-1
0.02 . +--~- .........::;;:=""""'.......--~-_1
0. 02 .
-0.10'---~----~--~-----'
0.0
12
24 t
3. 6
.:I S
60
·0.10
O.Oe+O 1.6e-3
3.2e-~1
.:I.Se-3
6Ae-3
ROe-3
P9-17
Batch reactor· series reaction
dea
-.._.":::: ral
dt
Mol balance on A :
···.ral ::: kl.Ca
Ca ::::: Cao.exp( ·k! .t)
deb
.... ;;;:; rbi + rb2
dt
Mol balance on B :
rbl :::;; . . ral
= kLCa
and
l'b2;:::; k2.Cb
deb
""ii·· = kLCao.exp( k1.t)· k2Cb
dCc
Mol balance on C;
dt
k2 ==
Energy balance :
·····,-b2
4.58.exp(-~::iif{·~) . . }))
Q+ (AHrxal)(-·raIY)+ (i.VJrxb2)(·rb2.v)
'dT
dt'" == '-'"..
..... ......- ..... 2..Ni:cpT· . ··_· . ··-_·_··. · , _· -, · . ·· . ·_·
9-46
Q = UA.(ra ... T)
LNi.Cpi = Ca V.Cpa + Cb.V.Cpb +Cc.V.Cpc
dT
dt
UA.{330- T)+55000(-ral.V)+ 71500(-rb2.v)
200.v,(Ca+ Cb+ Ge)
P9-17 (a)
dT
UA·:;:;O
•......•.... :::::
dt
V[55000{-"'raL)
_._-_._-_
__.- +__71500{-rb2.)]
_-.....
..•...
...........
200.v.(Ca+ Cb + Ge)
See Polymath program P9-17-a . pol
POLYMATH Results
Calculated values of the DEQ variables
Va:riable
t
Ca
Cc
Cb
T
UA
v
k1
k2
ra1
initial value
minimal value
o
o
0. 3
1. 34E-65
283
o
o
o
10
1.1172964
4.081E-09
-0.3351889
:rb2
o
rb1
0.3351889
maximal value
--
final value
-2.02E-44
283
0.2
0.3
0.3
0.2895784
915.5
0.2 .
1. 34E-65
0.3
-1.864E-65
915.5
10
1..1172964
4.081E-09
-35.016552
-27.963241
-6 . 345E-60
10
2.141E+05
1.04lE+06
6 . 345E-60
2.103E-38
35 . 016552
10
2.141E+05
1.041E+06
6.345E-60
-3.974E-59
-6.345E-60
o
o
o
o
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra1
[2 J d(Cc)/d(t) = -rb2
[3J d(Cb)/d(t) = rb1+rb2
[4] d(T)/d(t) = ((UA*(330-T))+(55000*(-ra1 *V))+(71500*(-rb2*V)))/(200*V*(Ca+Cb+Cc))
Explicit equations as entered by the user
[ll UA= 0
[2] V=10
[3] k1 = 3 . 03*exp((9900/1 . 987)*(1/300-1/T))
(4) k2 = 4.58*exp((27000/1.,987)*(1/500-1/T))
[5] ra1 = -k1*Ca
[ 6 ] rb2 = -k2*Cb
[7] rb1 = -ra1
9··47
1000,-------------,
8-10
0.22
- Ca
0.14
680
- Cc
Cb
0.06
-0.02
-0.10
520
1--==---.-"'----------.--
0.00
0. 0-1
0. 08 t
0.12
360
016
020
200
'---~--~----~----'
0. 00
0.04
0.08 t
0.12
0.16
0.20
P9-17 (b)
UA: 10000 llmin.K
0.30 '""""'---,--".....-.--.."....'--.'--.." ......".........---~...."' ....
1000,----------·---,
0..22
840
t
0.1-1
ll
680
("c
-.1D:.
-_
Cb
0.06
520
··0.02
-O . lH
360
0.00
0.04
0.08 t
fU6
0.12
0.20
200
000
00-1
008 t
0 . 12
0.30
900
0.22
760
Q
ell
-
('c
620
CI>
0.06
·480
--... .....-... -"--. ... ...."-. --". ...... .......
...."."" ... ...
~"
-002
-0.10
0.00
0. 04
"' .. ,,"",'
....
,
0 . 08 t
" '"'
0.12
UA: 100000 l/min.K
020
V::::::. 10 drn 3
UA :=: 40000 JJmiu.K
0.1-1
016
"
0 ..16
-
,
0.20
3·40
200
000
V;;;;;. lOdm3
9-48
0.04
008 t
.
012
016
020
030
I,
500
440
0.24
Em
- Cc
0.18
Q
380
- Cb
0.12
0.06 )
0.00 0.00
320
260
,
0.04
0.12
0. 08 t
0.16
0.20
200 o.
. 00
0.04
0.08 t
0.12
0.16
0.,20
At UA = 100000 J/min.K the 2nd reaction for C is totally suppressed
P9-17 (c)
UA = 40000 Jhnin.K
To =320 K
900 r--;--.-------------,
0040
0,,30
\
780
.
0.20
Ca
660
Cc
. ell
540
0.10
0.,00
lJ
-0.10 0.00
420
0.04
0.08 t
0.12
0.16
0.20
300 0.00
0.04
0.08 t
0.12
P9-18
Semibatch with parallel reactions
Mol balance;
dNa
-'"d;'; Fao+(rl+r2).V
dNd
dt
~,= Foo+ (rl +r2). V
.
dNu
-dt
=-r2. V
-·-=-rl.V
Rate laws:
-rl:;:;:kLCa
- r2:;:;: k2.Cb
9-49
0.16
0,,20
Na
Stoichiomeu), :
•
=_Nb
. --
Cb
Ca::::: ..- V
V
Nu
Nd
Cd:;::;-,V
Cu=V
v:::: Vo + OO.!+ vb.!
assume Vo=O
vcz:::;; . .Faa
....... -
w == .--.Fho
. . . . _Cbo
Cao
Ca6 : : ;: 5 molJdm3
Cbo == 4 mol/dm?
where
V =uO.t
But Fao == Fba (equimolar feed)
Fao Fba
+ .......
Cao Cbo
1)0;;;;;:; -.-.--
no
=GA5.Fao
v == 0.45Fao.t
dT
Energy balance:
Q" WS "LFio.Cpi.(T - Tio) + (-AHr:x(D).(-ra. V)
'dt' == -.--.-'----.- ..."., ....... ':2: Ni:C;;i---'·_ . . ···,·. ·. -'--_. ··_·,-'
where Q==Ws==O
MIrx(T) == .6.Hrx(Tref) + 1.\Cp(T-Tref)
1.\Cp1 == :!.. Cpd'" ,?.,Cpb- Cpa.:::: 50-20--30 = 0
a
a
1.\Hrx 1(T)
:=; '.,
3000 cal/mol A
1.\Gp2;:::: -Z:.Cpu ..~.. Cpb·· Cpa == 40- 20 -30:::: --10
a
a
&Irx2(T) ::: . , 5000, IOcr ... 3(0)
2.:Fio"Cpi.(r·Tio):::: 20 FaO.(T-Tao) + 30.Fbo.(T-Tbo)
:L.Ni,Cpi == NaCpa + Nb,.Cpb + Nd,Cpd + Nu.Cpu
.
dT. :::= _._20
. Fao(T
., Tbo ),f:,?5~~::t!'L Y.l~J?~.::}.gQ~.:: ~.25J1K~~~:YJ
.. ... -,-........ ...... . Tao) -+ 30. Fba.
dt
20. Na. + 30Nbt 50.1Vd + 40" Nu
".
".~
~
-~
Selectivity
rl
s= _.-.,
72
P9-18 (a)
9-50
The selectivity is proportional to CAIC B and therefore a small concentration of B in the reactor is preferable. To maintain
a low concentration of B it would be beneficial to run a semibatch reactor where B is slowly added to A.
P9-18 (b)
Let the rate laws be :
- r1 == kl.c'''b
-r2 ==k2.Ca
Now it would be best to slowly add A to B in a semibatch reactor .
Let the rate laws be :
-d == kLCaCb
····rl == kLCa
- r1 ==k1.Cb
- 1'2 ;;;;; k2.CaCb
- 1'2;;;;; k2.Ca
-r2= k2.Cb
For these rate law combinations a semibatch reactor will not improve selectivity.
Let the rate laws be :
. r1 :::: kl.Ca.Cb
.. 1'1 =kl.Cb
-1'2 =k2.Ca
• r2 :::: k:2.CaCb
Now it would be best to slowly add A to B in a semibatch reactor to maintain high concentrations ofB and low
concentrations of A.
Let the rat.e laws be :
- 1'1 == kl.Ca
- rl ; : : lel .Ca.Cb
- 1'2 :::: k:2. CaL"b
- r2=k2.0>
Now it would be best to slowly add B to A in a semibatch reactor to maintain high concentrations of A and low
concentrations of B .
P9-18 (C)
AHrxl
callmol A
aHlX2(300K) ca1lmol A
Temp. K
3000
If AIh-xl is constant, then if .1.Hrx 1 increases, T increases and S decreases.
If AIm! is constant, then if L1Hrx2(300K) increases, T increases and S decreases.
S is dependent on r1 and 1'2. whicb depend on kl and k2, which depend on the
temperature. The greater the temperature in the reactor the smaller the ratio between
kl and k2, hence reduced S.
P9-19
Plotting the data oft vs. T gives the initial and final temperature of the reaction, To = 52.5 °C and T t =1668
9-51
0c.
200~------------------------~
80
-10
oo
s
16
t
24
32
40
Recalling equation (8-29) with LlCp = 0, X = 1 and T == T f gives:
cal (
cal
!ill R = -18166.8 - 52.5 )K = -2057.4x
mol.K
mol
Make a table oft (min), T CC), dT/dt, liT (K\ In(dT/dt) and plot In(dT/dt) vs 11K (K) in Polymath. According to
equation (E9 . 3-14) the slope of the curve is E over R Linear regression gives:
See Polymath program P9-19. pol
~~.~'f\1!~,..tll~~p'otl·'
Model: InTdot
= aO + a1 *T_inverse_kelvin
VariableiValue
;95% confidence
aO
18.88162 . 2.425175
al
-6224.846 880.596
Statistics
RA2.
0.9755663;
:RA 2adj
0.9720758
•Rmsd
0.0597988
Variance 0.0413783
Remember that T s is the self· heating 1 ate, which occur s after the onset temperature is reached.
.
--6224
InT, =---+18.88
T
E
6224=R
9--52
E=6224K ·1.987
cal
cal
=-12367mol·K
mol
P9-20 No solution will be given
CDP9-A
CDP9-B
CDP9-C
Adiabatic. batch. reversible.
I~sign equation
:
Rate law:
Stoichiometry :
Combining :
kl
Ln"
Parameter evaluation: E =-1' b 1
............ ".
,~.,".
R.T2
_R.TI
0217
I.n--·--·::::: -'--T'-~. . __
03_2_4··-·T···-· = 8498J I mol
...............,......
,
...., ..... -,."' ..........
"
8314(340)
......., ...............,.... ... ...
"
8.314(300)
(I 1)
Ke(T)
In' .................
-....... ::::: _AHr:x
................ _. . ................. -....
Ke(300)
R
300
T
LlCp == 2 Cpa" CPA::::: 20 .. 12 ::::: 8 ]lmol.K
AHrx ::::: LlHIX (300) + LlCp(T· 300) == ., 75000 + 8 (1' " 3(0)
9-53
* 900 '" 254 = 272 127 mol A
N AO == 0 . 6667
Nro = 0..3333 * 900 *454 = 136064 moll
N AO I V
CAO :::=
:::=
T
Energy balance:
~
(1)(12)
=272127 I (50 *28.12) :::= 192 moll dm3
+ 0 + (136064/272027)*(15):::= 195 J/mol
• [.- Lllirx(3OO).X]
(.- 75(00). X
= 10+-···-·-·--··-----·== 300+·······································__•·········
CPA + tJ.Cp.X
195+8.X
POLYMATH
?~a t.i ons_,;.,
d {Xl / d ( c)
Ini::ial
""kl~
val>:i~
o
( ( 1-x) .. {4' cao· (x.... 2} I l:(e:} l
TQ",300
cao=192
TzTQ+11500Q·x/{19,5+{8~x}))
kl=. 217 "eX? (11322"
Dhrx=-75000 ...
(a
w
(l/HO~
llT) )
(T-3UO) )
Ke:z7QOaO·exp(Dh:::::-x/S.31.4."" /1J300-1,/T))
c:,.,
~
0,
"
9-8
variable
To
;::<:
:ni;:;ia,l._'~~
Final value
()
2
-----~~.-
o
o .O~':;9S73
JOQ
300
192
1.92
JOO
191
300
469."731
Joe
'io69.1S1
0_145345
0.49785
a . 145345
0 . 49735
·-7SQiJO
-7364.1 .8
--'75000
-73641.8
70000
loooa
1. 62'532
1..62532
o
O.04.:j,9S7]
300
.~.:;::::::
;,)I. . : : . : :
9-54
9-·3
"I:~"
,,,,,'r
"".---i'--"""'--'
O.~OO
"',.,"',.+".,-"·,,-'.",,..,,-,..,,+"'''··,....'''-'''. ,,-,''·. ''i''
o,.;;)QO
:,,200
1.600
:2 COC
Equilibrium conversion. Xe :: 4.5 %
90 % of Xe ::: 0.9
* 4.5 ;:: 4.05 % cOIlversion
Time for this conversion"" 0.11 min:: 6,,6 s
Check:
at equilibrium:
(4.CAo2).Xe 2
using the solution to a quadratic:
+ (Ke.CAo). Xe- Ke.(7Ao:: 0
_. Ke,. Cw ± ..JK~2~C~~i':.i6.K~:(:;'~3·
Xe = .--_._.
"""""""'''''''''''?'''-----''''.''''.. '''''''-''-
where at T ;:: Te ::: 470 K,
8.CAO-
Ke:: 1.625 moll drIl3
From POLYM.A.."TII, Xe:: 0.045
CDP9-D
9-55
CSTR startup
CSTR startup, gas phase, isobatic - no pressure drop
Part (a)
dFa
Mol balance:
._-.- =:
dt
(Fao-· Fa) + ra ..v
'dF'b
--" = (0- Fb) .. 2.m.V
dE
Fb
Cb= .. ···
Fa
Ca =, ...
VI
iJr
Fro ::::: Cro.vlQ =:
1)]" :;;;;:
P
- ' - : " . VIa
R.l0
Fr T
Fw To
111'0 .-.-. -...•..
Fr;:::; Fa + Fb
Fro::=: Fao + Tho
V=: lrr.'t
1:
=500 s
Energy balatlce :
UA
--- == 51 I kgcat.s. K
peat
UA
=5 x 50
=:
but kg catalyst =: 50 kg
2.."50 J I s.K
ACp=O
dt
40. Ca. V+ 20 Cb. V
Ta= 300 K, Tio:::: 450 K.; Fao = 5 moUs
No control:
!~itial
E9.::t,at.ions:
difa)!d(~l~{:ao-fal·(ra·VI
Ie-OS
d(::b)/d(c)"'(~:a)-(2"ra"V)
0
c',T) Id{t.};( {250" (Ta-T}} -t40"fao'" (T-Tio)" (20000" (-::::a"V})) / ((
450
40wCa·V)+{20~Cb·Vll
':'a=3Qa
7~.o=4.50
1:0.:;450
e,a.u=500
1:au:l;SQQ
k.,.;exp( (31400/8.3U}
,,< !1I450) -(lIt1]}
cao=O.2S
vo<.;fao/cao
Ca""fa/v
Cb;:fb/v
V=v"t:au
ra~--k"Ca
9-57
value
Mininr..:.tn value final value
o
30
a
30
Ie-OS
0_0106221
le-OS
0.00969749
o
299.Q98
a
299.098
450
451.548
445.634
451.548
s
s
5
'1'a
300
300
5
300
'1':.0
450
450
450
450
le-OS
299.107
le·-08
29:9.107
450
450
450
4sa
fa
300
taU
soo
500
sao
50{)
Tau
500
500
sao
500
1
1-0292
0 . .921054-
1.0292
0.25
0.25
0.25
Q.25
5
5
5
5
20
20
20
20
4e-08
1200.55
4.e~08
1200.55
va
Ca
0,25
0.251187
S .07756e··06
f1. Q77S6e-06
Cb
o
0,250078
a
-0.249135
V
la-OS
-0.25
600273
-8.31]]ge-06
2e-05
600273
-0.25
-8 ..HJ3ge·06
ra
.,
.. ;: \ . 200.i.
I
_....-...-~-
~
-"..,...=-...--~~.. .-"'" -_.,..,.."--- ..-....-----.--
~-""'-------"""'-'" ... --~"" .
.....e.scc.l(-"'
.:..;
""e,:;oo·I
4.·:i..,,-t-CO·"·~
j
~
•t
-4H
ao~ ~oa-"-'-"-'-"""+-""'-"---"-"'-+-"~""-""'-'+'---.-.~-- .....t--~-'----'1
........
6.000
::<.::00
19.000
2 ..... 000
30.00e
Part (b)
Mol balance:
dCa == (eao····
..................
...................,.Ca)
_. . . ._. + ra
dt
1:
Cao == 0.1 kmollm3
.~-,.
9·58
:::::
0.1 111olldm
3
P9-10 cont'd
Rate law:
=:;;
Cbo=Cao
dCc
(O··Cc)
_
. = -_
..._ ........ Ta
dt
1.
t=50s
"ra:: k.Ca.Cb
EneIgy balance:
v
deb :;: ---._
(Cbo''''
-_.
.......-Cb) + ra
dtz
t)o • 1" =
or
d~
Q_.• Ws·"LFio.Cpi(T··· Tio) + (-Mlrx)(·ra. V)
:;: -_...........__. . . . .-.._-LNi.Cpi··---·········_··········
m
•• -
2 X 50 ::: 100 dm3
Fao =:;; Cao " 'Uo = 0.1 x: 2 ::: 0.2 molls::: Fbo (equimolar feed)
IFio.Cpi.(T - Tio) =2 x 0.2 x (15)(1' - 300) = 6(1' - 3(0)
AIlrxCD;::: .6.H.Ix(To) + .6.Cp.(T - To)
but ACp ::: 0
::: - 41000 - (- 20000) - (- 15000)
::: - 6000 cal/mol
dT
6(T - 300)
(6000)
. ( -ra.
V) ..
- = ------_
. . ._ +
.. _
_.. _
-_
dt
15.Ca.v+15.Cb.V·t30.Cc.V
E;g;::ations:
Initial value
d(ca)/d(cl-«O.1-ca)/50)+ra
0.001
d(cb}/dICJ={(O.l-cbI/SD)+~a
>1),001
d(cc}fd(C)={(O-cc)/50}-ra
a
d(T) Id{c) =( (6" (T-JOO)) +{6000" (-ra*V) 1 } It (IS"ea"''\!)'" (lS"'cb*
)00
V) ... DO"cc"V) )
'\!",100
1<:=0.01 *exp { {1000011. 987) .. ( (lIJ 00 I,,· (liT) ) }
raz- k"ca",cb
to '" O.
t
f == 40.0
V<U;'iAble
~al v"alue
~im\!ft\
t:-
O
0.001
400
0.0732615
0.00010274
O. aOOlO::n4
O. OQ01021'4
0.00010274
ca
cb
value Mini.mum
v~lye
0
!ind
400
v~lu~
0.001
0.0132615
CC
I)
Q.0:99$64
{)
0.099864-
T
300
25.7:380
V
251380
100
300
100
100
100
k
0.01
lenSO
0.01
la9280
-Ie-OS
-ie-OS
--O.Oll(HHl
-0.00199796
:t'a
9·59
$c.lf':
10-- 52 ,;>00
~£!:.:...
.~.
T
-rT
2.\00
"i-
TI
t .. 5CO
T
o_.seo
"I'
0.300
I-
T
i
...... +--_...... _.............. ". ..;.-"" ._ . . -.------+ . ,,-,. __. _-_ . ---\
16ti,cca
2~O.:lOO
• .:fL!.:o.
9-60
320.0CC
400 or;
Solutions for Chapter 10 - Catalysis and Catalytic Reactors
PIO-I Individualized solution
PIO-2 (a) Example 10-1
(1) Pentane isomerization nP -pt--7 iP
Assume that Pt is the catalyst used.
Maximum f = 5 molecules/site/sec
Minimum f =3e-3 molecukes/site/sec
Maximum rate:
--rp
= fD(_l __ ) %Pt
100
MWpt
-rp = 5(0.5)(_1_)_~ = 1.28*10-4 mo~
195 100
s geat
Minimum:
--rp = 3*10-3 (0.5)(_1_)_~_ = 7.69*10- 7 mol__
195 100
s geat
1
(2) SO2 +--0
2 2
--7
SO3
No turnover frequency is given so this rate law cannot be determined by this method
(3) H2 + C2 H 4 -~ C2H 6
Assume Cobalt is the catalyst.
=
--r
H2
fD(_l _) %C~
MWCo
100
Maximum:
f =100 molecules/site/sec
1
mol
--rH =100 ( 0.5 )( - 1 ) ---=0.00849--2
58.9 100
s geat
Minumum:
f= 0.01
1 ) 1
-7
mol
( 58.9 -----=8.49*10
100
s geat
-rH =0.01 ( 0.5 ) 2
PIO-2 (b) Example 10-2
10-1
(1) Cv
=
1
Cr 1+ KrPr + KBPB
Ks = 1.39
Kr = 1.038
Pro = YroProtal = 0.3*40 = 12
For 60% conversion
Pr = Pro (1- X) = 12*0.4 = 4.8atm
PB =PTOX =12*0.6=7.2atm
Cv =
1
=_1_=0.063
Cr 1+(1.038)(4.8)+(1.39)(7.2) 15.99
6.3% of the sites are vacant
(2) X
=0.8
Cros = .CvKrPr _
KrPr
CT
Cr
1+ KrPr + KBPB
eros =
(1.038)(1)(1-0.8)
= 0.2076 =0 09
CT
1+(1.038)(1)(1-0.8)+(1.39)(1)(0.8) 2.3196
.
9% of the sites are covered by toluene
(3) Linearize the rate law to:
PH?Pr 1 KBPB KrPr
----=-+--+---r7
k
k
k
PIO-2 (c) Example 10-3
Increasing the pressure will increase the rate law.
p?
-r---T--R
A
7
l+KrPr
If the flow rate is decreased the conversion will increase for two reasons: (1) smaller pressure drop
and (2) reactants spend more time in the reactor.
From figure ElO-3.1 we see that when X =0.6, W
=5800 kg.
PIO-2 (d) Example 10-4
With the new data, model (a) best fits the data
(a)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1 +Kea*Peat·Ke*Pe)
10-2
Ini guess
3
0.1
2
variable
k
Kea
Ke
Value
3.5798145
0.1176376
2.3630934
95% confidence
0.0026691
0.0014744
0.0024526
Precision
RA2
= 0,,9969101
R A2adj
= 0 . 9960273
0 . 0259656
Rmsd
variance = 0.0096316
=
(b)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1+Ke*Pe)
Ini guess
3
2
variable
k
Ke
Value
2.9497646
1.9118572
95% confidence
0.0058793
0.0054165
Precision
RA2
0 . 9735965
RA2adj
= 0.9702961
Rmsd
= 0.0759032
Variance = 0.0720163
=
(c)
POLYMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/((1 +Ke*Pel"'2)
Ini guess
3
2
Variable
k
Ke
Precision
RA2
RA2adj
Rmsd
vari.ance
Value
1.9496445
0.3508639
95% confidence
0.319098
0 . 0756992
= 0.9620735
= 0 . 9573327
= 0 . 0909706
=
0 . 1034455
(d)
POL YMATH Results
Nonlinear regression (L-M)
Model: ReactionRate = k*Pel\a*Phl\b
Variable
Ini guess
a
3
1
1
-k---b
Precision
RA2
R A 2adj
Rmsd
Variance
Value
0.7574196
0 . 2874239
1.1747643
95% confidence
0 . 2495415
0.0955031
0.2404971
= 0 . 965477
= 0.9556133
= 0 . 0867928
= 0.107614
10-3
Model (e) at first appears to work well but not as well as model (a). However, the 95% confidence
interval is larger than the actual value, which leads to a possible negative value for Ka. This is not
possible and the model should be discarded. Model (f) is the worst model of all. In fact it should be
thrown out as a possible model due to the negative RII2 values.
(e)
POLYMA TH Results
Nonlinear regression (L·M)
Model: ReactionRate = k*Pe*Ph/((1 +Ka*Pea+Ke*Pe)ll2)
Variable
k
Ini guess
-3---
Ka
Ke
1
1
Value
2.113121
0,,0245
0 . 3713644
95% confidence
0 . 2375775
0.030918
0 . 0489399
Precision
RA2
RA2adj
Rmsd
Variance
= 0 . 9787138
= 0.9726321
0 . 0681519
= 0 . 0663527
=
(f)
POLYMATH Results
Nonlinear regression (L·M)
Model: ReactionRate
= k*Pe*Ph/(1+Ka'Pea)
Variable
k
Ini guess
3
Ka
1
Precision
RA2
RA2adj
Rmsd
Variance
Value
44.117481
101. 99791
95% confidence
7 . 1763989
16 . 763192
=·0 . 343853
= -0 . 5118346
= 0.5415086
= 3 . 6653942
PIO-2 (e) Example 10-5
(l)X =1-
1
klk
(l+kd t )
d
As t approaches infinity, X approaches 1.
10-4
large klta
x
t
(2) Second order reaction with first order decay.
dX
,W
- = - rA
-
dt
NAD
2
-rA '=ak'CA
a = exp[-kdt]
dX = Wk' C2 (1- X)2 exp[-·k t]
dt
N
AD
d
AD
dX =k(I-X)2 exp[-ki]
dt
X
k
1-- X = k(1-exp [-kd t ])
d
~. t...
X =kas t -y
m Imty -.-l--X kd
X=J{d
1+ k/
/kd
(3) First order reaction with first order decay
dX Wk'
di- =
CAD (1- X)exp[-kd t ]
""N
AD
dX
-
dt
= k(l- X)exp[-kdt]
InC~x)= ~ (I-exp[-kdtj)
t -7 infinity X = l-exp [ -
~]
PIO-2 (0 Example 10-6
10--5
Increasing the space time makes the minimum disappear. Decreasing the space time moves the
minimum to the left and the concentration is higher.
Increasing the temperature so that the rate constants are higher will cause the catalyst lifetime to be
shorter.
If tau =0.005 the minimum C A=0.607
If tau = O.Olthe minimum CA = 0.5088
PIO-2 (g) Example 10-7
(1) If the solids and reactants are fed from opposite ends,
da
kda
dW
Us
atW=We ,a=l
--=-
C
1
= kd~e
U
S
This gives the same expression for conversion as in the example.
(2) Second order decay
1
a=----kd W
1+--Us
_~_= kC~oUs In(I+'5dW J
I-X
kdFAO
Us
1.24= (0.6)(0.075)2 US_In
(0.72)(30)
(1 + (0.7~)2~~?OJ
Us
Solve for Us by trial and error or a non-linear equation solver.
Us 0.902
=
(3) If G =2
dX
dW
(l-x)2
AO (1+c X
2
r
F --- = kC ------a
AO
10-6
2
(
2£ (1 + £ ) In (1 - X ) + £2 X + 1+£)2 X = kC AOUS ( 1- exp [-kd W]J
e
I-X
Us
kdFAO
9X
12In(I-X)+4X +--=1.24
I-X
X =0.372
PIO-2 (h) Example 10-8
Uo = 0.25
1 . 0 -,- - - - - - - - - - - - - - ,
Uo = 0.025
0.8
~.'.""
L~J
0.6
0.6
0.4
0.4
0.2
0.2
"-",
4
1.0
z
8
6
10
Uo =2.5
0.0
1.0
0
2
4
z
6
8
10
Uo=25
.~
0.8
0.6
0.4
0.2
~-
4
z
6
8
10
0.0
~--'"".--=~
I)
2
4
z
6
PIO-2 (i)
For EA = 10 and Ed = 35, for first order decay we rearrange Eq 10-120 to:
In(l- kdotEd
EA
J
= Ed (!-'~J
R
T
To
=Ed
- ( ._--1 1
R I'o T
J
lO-7
8
10
1
R
-In
Ed
l-
T=
kdOtEd
1 1
=--1'a T
EA
1'a
400
=--------~--------~
1+0.07948In(
1
)
1-0.00286t
PIO-2 (j) Individualized solution
PIO-3 Solution is in the decoding algorithm given with the modules
PIO-4
W+S,·.,.w·s
TBA.S,(.,.TBA + S
PIO-4 (a)
Surface Rm Limited
r
".Wl.
=0
kA
(~
~7EA·S
C1,,.,,C,,.
'k,
V
{"
t
==-'j('-"
== £\.1BAI..'nl.·1 L y,
¥
.
1
K
l '
SInce
D
K"
.
';;;;;
D
10-8
'TBA
PIO-4 (b)
Adsorption of isobutene limited
rADl ;kAl
[Cr - ::~] Cv •
Bey Ridea1 Kinetics
PIO-4 (d)
10-9
W .. S + 19S1HTBA+S2+S1
2
r
-= k
S
S
C _.~1.BAKSC:~~'l=Y1l
[' C
'W·S2 1.$1
en == C V1 -+- CloSl
CT2 =c'V2'W-S2
+C
PIO-4 (e) Individualized solution
PIO-4 (0 Individualized solution
PIO-5 (a)
H..,+
C..,H
..
~ 4
H +E-4A
Bey Rideal.
E·S~4>A+S
E + S ii"~ E .. S • CE1OS :::: KE. PE. Cv
"
TS "" ks[ CE .. S PH]
CT""'CV+CE"S
f~ =C~~{-l:~~;l
PIO-5 (b) Individualized solution
PIO-5 (c)
02 +2S
p
Az +2S p
20·S
+ O·S ---) C3H 5 0H.S
C3 H 60H.S P C3 H 5 0H + S
C3H6
B + A·S ---) C·S
C·S---)C+S
-riJ = r's = k 3 PB C A • S
rAD =kA [ PAzCV2
-
2A·S
C~'S]
KA
10-10
rAD =0
kA
CAoS = CV~KAPA
-rB = rS
= k3PBCV~KAPA
re.s = kD [ Ce•s -
P~~v] = kD [Ce.s - KePeCv 1
rcos =0
kD
CCoS
= KcPcCv
Cr = Cv + CAoS + CCoS
-rB = rs
k 3 Cr PB
= Cv [1 +~KAPA + KcPc ]
.JK:P:
= l+~KAPA + KcPc
PIO-6 (a)
A (butanol) ;: B (butene) + C (warer)
Possible meChanism:
7 .. ::::
,~..
K (PC . . . .K~·AS_:)
AA
,4
S
AA.
rs "" ks (CA.s Cs • CaoS CcoS Ks)
rDB
C·S
t:! c+s
=:
knB
rDC;: klX
Assume stuface reaction controlling:
( ··a-s
'
- PB
Cs. -- PB K AS ('··S
-K
DB
' == -KPc Cs '" PC KAC (~-5
( ,<:'5
rx:
10-11
(esoS . PB Cs/Kos)
{Ccs - Pc
Cs/Kr:d
'-
-
(
.,!
..
pgpc C;._.) -_ kS KAi·s
-'Zl'pA
·r,J - rs - k s! P,tK,Li·s -- KKK
\
S
site bala:lce:
:. ···rA
;:: _.......
DB
DC
_.
}!ePc
h . K - KKK K
K Jl were
eq S AA DC DE
eq
('T == Cs +- CAS +- CBS + Ccs :;: Cs (l + KAA PA+ KAB PB + KAc Pc)
ks. KAA cf (P.A ~.........
F1l_ ......P~)
_ .. _ _ ._.
{I + P A K.A.A + PB K.A.B + Pc K A c)2
If PBO::: 0 and Pco == 0 , then
.;
:=
AO
~2.!MS1!:M.;::
{I -+ PAn KAA )2
Is1 PM
1 -+ k2
_.
Pio +- k3 PAO
This is consistent with the observation.
PIO-6 (b)
From r..iJ.c figure.
2
3
4
:;
6
0
0.275
0.5
0.77
0.77
0.5
0
4.5
27
54
112
229
405
6 . 45
8..14
L2.06
21.4
Point numb::::r
~I:
(_lmr~~)
AO hr'!b cat
PAC (arm)
.J.p-; . .
·"fAO
10-12
At large PAl):
-r~O =~.Ll'.~Q :;; ¥1...
k2
At small P AO:
.. rAO
. ;AO:= kl
=..... _...
1T
P~o
K2
(_l_) ; using point 6: k,
PAO
=(229) (0.5) ;; 114
k2"
P AO ; using point 2: kJ :: 0061
=
k2:::: 5.34xlO-4
0. 061. ~ AO
............... ; so
5.34x 10-4 PAo + k, P AO
Using point 3:
k3 :: 7.05xIO· 2
Using poim 4;
k3
Using rJOim 5;
k3 "" L05x 10.2
:=
3. 19x 10,,2
=
}
k3 = 3.19xlO-2
(The reason for the different
values of k, is from reading
the graph)
25.0·~ - - y "" 4.3103 + O.073937x R= 099583
20.0 -
lntercepb1/ko. 5
k::: 0.054
Slope
o.0
J
'~'j"'.:"""'~
o
KA
:::
:=
KA/kO'!
:::
4.31
= 0.074
0.32
····-~~i· .. ·~~ .. ·"':"·····':"· .. "':"."""';.... "1"..""":"...... ~ .... ~-.;" ..~-:"'--o:"' •• ~~\.' "';'-'''':''$''~''''"'~''''-i
50
100
P
ISO
200
250
a
PIO-6 (C)
Find the percent of vacant sites
Pb and Pc "" 0 so that red llces to:
%vacant ::::: ._-_ 1..... - .. _.:;::: .........._....1__ ...__. =: 0.41
l+K AAPAO 1+0.01596*90
Find the percent of sites occupied by A and B. No B will have occupied any
sites at X : : : O. So;
KAAP
~90 -- 0-9
A
i''''1:." -_ .-..........
_ . . _-._.
-_ ...... 0.01595
,. . . . . ,....._, ........,.....................
. ,J
1 + K,<\,APA "1 -+- 0.01595" 90
0/
PIO-6 (d) Individualized solution
PIO-6 (e) Individualized solution
---------------------------------------------------------------------PIO-7
10·13
ME
--'+
D}(fE + rhO
. -- ... _-----
~
100
300
The r.ue o~ formation ofDME is gream:r inirially. 11ris is a result of IJl(){e vacant siteS being
inirially available for reaction because water is not adsorbed on the sites. As time goes on
the equilibrium concenll:u:ion of Water sites is reached. WaI.C:J: is strongly adsorbed on this
catalyst.
Probable Mech;mism
ME+5
:;= ~fE·S
2ME· S""""1> W· S + DME
t
S
Assume Surface Reaction Controls
CME-S "" K/.u:. PME Cv
Cw.s == Kw Pv". Cv
PIO-8
Given: Kinetic r.:ue expression for the reduction of NO over :l solid catalyst:
kPNPC
t" ;: """"""'""""---"--------
{I +- Kl PN+- K z Per
~
== partial pressue NO
Pe =: partial pressure CO
Assume that overall reaction is of the form
NO +CO
"....,.t·N2 +COz
PIO-8 (a)
10-14
It is seen that neither Nz or C~ appear in the denominator. This infeIs that neither is
adsortxd on the catalyst, On the other hand. it can be infered that both NO and CO are
adsorbed on the surface. The squared denominaror suggests a dual site surface reaction of
the adsorbates of NO and CO. Therefore the following mechanism is proposed.
P,.. ;;;: P,vo
kA."i
NO(g) + S
<=:
NO· S
Pc::::: Pea
keN
CO(g) + S
<=:
CO· S
'-1"0/ ;;;:
k:cN [Pc Cs - Cco.s/KcN]
ks
NO • S + CO • S -1- ~ N2 + Cth + 25
With the surface reaction controlling
CNQ-S ;;;: KAN' PN Cs
4:o-S :: keN Pc Cs
Tuen CT :: Cs ,. CNQ.S + Cco.s :: Cs (1 +- KAN PN + Kc."'N Pc]
and therefore reaction is -rs:: ks CNo-S Cco-s:;; ks c..~ PN Pc KAN Ik"N
or.r$ ""!ss..!C~,~,,f~?s:.,gL
[1 +- K.~"I fIN + N.:N PcP
with
k\ :: ks
ct Km KeN
Kl :;;KAN
K1=KcN
kJ Pc PN'
"'1"s :::: ---,.",,,-,,,--'- ' "--'-.--~,,-..-.
[l + K t PN + K2 PcJ2
PIO-8 (b)
Assume chat Pc
»flN. Then Pc chattges very little during the course of the reaction
and remai.'1s consu.m,. A maximum in (-rs) rhen occurs. for a fixed value of PN at:
10-15
The rate of n::.action will. increase ..vim an inC're:l.Se in Pc until me: above value is reached.
• after which it '..vil.! decrease. It appears tbat there is an excess pressure which will mini:miz.e
reactOr YoiU!l'le .. Operating at excess pressure greater than t..'"tis value will. decrease (-rs), and
hence increase V. This analysis is exact if the catalytic reaCtor is a CSTR. If me reactor is
t:re:ued as PFR. the Cthical value of Pc is only approximarc. but the general observation is
qualitatively the same.
This analysis further assumes that the excess CO can be elimitmcd easily and economically
downstream from the NO converta'.
PIO-8 (C)
The conditions for which the rate law and mechanism are consistent are the
following.. DIe CO S surface reaction must be the rate limiting. P cdP NO must be
small. The mechanism must be a dual site mechanism (which it is).
PIO-9
Methyl ethyl ketone (MEK) is an important industrial solvent that can be plOduced {roll1lhc dehydrogenation
of outall·2-ol (Btl) over a zinc oxide catalysl
Bu·+MEK+ Hz
/i) 11 t C
The following data giving the reaction rate for MEK were obtained in a difterentialreactO! at 490°C,
_., - - - - , • - .,.... _,-
l~\Iu:
(atm)
PH, (atm)
_~;[J!:~ tIY1f:l!h x
-¥~~"'"
0.5
PBIt (atm)
cat)
5
o
o
o
2
1
0.044
O(W)
0069
0060
0
0
10
0.043
0.059
PIO-9 (a)
Suggest a rate law that is consistent with the experimental data
From data sets 2 and 5
PBIi (aim)
l~HBK (atm)
()
PII , (atm)
0
(molf}! x g cat)
. __.:c:= ...._. __.__.
0 . 040
0..043
-... _ .._ ....._..._......__ ._._ .........---.. - .... _ ... .
we can say that an increase in BlI partial pressure slightly increases the reaction rate.
10-16
From data sets 1 and 5
1
5
l~"le-K (aIm)
5
Pit, (aim)
0
o
o
Data Set
Pllu (aIm)
0.043
we can see that the MEK partial pressure has little if no effect 011 the rate law.
From data sets 4 and 6
...
DataSet..
-.-,.--,PSlI (arm)
,,~ ~~
P.WEK
..
.,-.-~,.-~~.~"
.. ..",.. ....
,~.~
"
'"~
..
6
1
4
1
(atm)
~/1 (aim)
r;~~!i .(~rl(){/ft~~JL ca~L
...
o
1
1
10
0.060
0.059
Is seems that the pattial pressure of Hz has no effect on the reaction rate.
If MEK and Hz are weakly adsorbed (or not adsorbed at all) we can proposed initially the following
.- r : : : r ::::: . . . . . kiP,
. ---.:..........
J
,
A
B
ItkP
2 A
But, from the complete data sel
"'
Data Set
2
2
0.1
0.044
0.040
'" ..•..•... __._ _ _ _ •.•
cat)
_~
.•• _•• _...................... _-...-_._.'_.'
_~
.......... __ ............. .
4
3
05
___ ••... __ ........................... _.• .·.M .. .. _____ ·_ ••·..."
~
2
0.069
0..060
We can see that the reaction rate goes through a maximum
008
0. 07
006
«i
~ 0.05
S0
,5,
~
- lii
...
004
003
002
001
0
0
15
0.5
,
~
2
PUu (lltm)
10-17
0.043
.........
-.-~-
....... 6 . -.- .... .
-
1
0.059
PIO-9 (b)
Suggest a reaction mechanism and rate limiting step consistent with the late law.
One possible mechanism is the following one
<-> A·S
(2) A S + S ,) B . S + C . S
(1) A r-S
(3) B S' "7 Bt, S
(4)C·S<)C+S
If the limiting step is #2 (irreversible surface reaction) and the others me at Pseudo Steady State (# L, 3and 4)
,
kC A.) ("/V
.... fA ;;:: '2
A site's balance wHl yield
TherefOlC,
Solving tm Cv
C"
(1
=
1 + P.K +
"
I
K3
+ f~.
K
.!
Substituting the expressions for Cv and CAS into the equation for -r' A
". r~
=.:
k 2 C;\ sC"
= k, K IPr\C..~ =,_,._~f..f<iZ.<l0.x,_,_. . . . . -;
(
1+ p. K + J>!L
il
I
K,
+!~.lK)
4
which for the case of weak adsorplion of MEK and Hz reduces to
PIO-9 (c) Individualized solution
10-18
PIO-9 (d)
First we need to calculate the rate constants involved in the equation for -r' A in
part (a). We can rearrange the equation to give the following
fE;*+ jtp,
YJk:' and intercept equal to Y.Jk:·' Shown below is the linear
which is a linear equation with slope equal to
regression we did using the problem data
g
"'-1..
7
6
V1
0
<;
~
I
~
4
e::.'"
3
~
I
------i
5
I
I
2
•••
(PaJr'~"Y{}5
I
'.H
= 2 7298l>flu +
I 3362,
,,': 0 .9991
1
I
I
I
0
0
05
f
15
2
2..5
Thus from the slope and intercept data
mol
kl ::::: 0.56·
.--._.
h· gcat·cum
and
k2
:::::
I
2.04 ---_.
atm
Thus,
The design equation for t.he PBR is
F ..cJ?i_. :::::--,'
AO
dW
II
From stoichiometry (gas phase)
C CA({i~-~~-\J¥ ~t1 :::::
From the reaction E:= 1+ L·· 1 _Assuming isobaric and isothermal operation and using ideal gas law
p,\
=PA{Li-)
Using equations 10 ..8-13,10-8-14 and 10-8·-16 together with Polymath we can solve for W at X::::: 90%..
10-19
See Polymath program P 109-d..pol.
POL YMA TH Results
Calculated values of the DEQ variables
initial value
Variable
o
o
W
X
maximal value
o
o
10
10
560
2.04
600
-12.228142
12.228142
final value
23
23
0.9991499
0.9991499
1
1
1
e
Paa
Pa
k1
k2
Faa
ra
rate
minimal value
1
10
10
560
2.04
600
-2 . 3403948
68 . 5622
10
0.0042521
560
2 . 04
600
-68 . 5622
2.3403948
10
0.0042521
560
2.04
600
--2 . 3403948
2 . 3403948
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = -ralFao
Explicit equations as entered by the user
[1] e=1
[2] Pao = 10
[3] Pa = Pao*(1-X)/(1 +e*X)
[4] k1 = 560
[5] k2 = 2 . 04
[6J Fao = 600
[7] ra = -k1 *PaI((1 +(k2*Pa))A2)
[8] rate = -ra
~
. mol
mol
it should be mentioned that I, . :::: to
. . ;;;: 600--tlU
min
hr
---
1.0
70
56
0.8
Q
06
-12
0.4
28
0.2
1-1
'---_ _ _ _ _ _ _ _ "_ _ _ _ _ _
"~
.0.0
(I
0.0
-1.6
9.2
\"
13.8
184
23.0
0.0
PIO-9 (e) Individualized solution
PIO-9 (0
10-20
4.6
9.2
W 13.8
_ _ .J
18.4
23.0
Now consider the change in pressure:
Stoichiometry:
P =C RT=C
Pressure:
. "".dy_. . . =.·'n
".". - (1+ X )
AA
dW
A"
(1-X }P
._--, -·..,RT
l+iX Po
2y
Use these new equations in the Polymath program from part (d).
See Polymath program P 1O-9-f.pol.
POLYMATH Results
Calculated values of the DEQ variables
Variable
W
initial value
o
X
o
Y
1
1
e
Pao
Pa
kl
k2
Fao
ra
rate
alpha
10
10
560
2.04
600
-12.228142
12.228142
0.03
minimal value
o
o
0.0746953
1
10
7.771E-05
560
2.04
600
-68.584462
0.0435044
0.03
maximal value
23
0.9997919
1
1
10
10
560
2.04
600
-0.0435044
68.584462
0.03
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(W) = -ra/Fao
[2] d(y)/d(W) = -alpha*(1 +X)/2/y
Explicit equations as entered by the user
[1 J e = 1
[2J Pao=10
[ 3 J Pa = y*Pao*(1-X)/(1 +e*X)
[4] k1 = 560
[5J k2 = 204
[6J Fao = 600
[7 J ra = -k1 *Pa/( (1 +(k2*Pa) )1\2)
[ 8 J rate = ora
[ 9] alpha = ,,03
10-21
final value
23
0.9997919
0.0746953
1
10
7.771E-05
560
2.04
600
-0.0435044
0.0435044
0.03
1.0
70
08·
56
0. 6
42
04
/,
[;J
./!
r:f.f{-
28
/\
I
/'
I
\
<",,"'
0.2
14
0.0
(I
9.2
0.0
W
13.8
1804
23,0
,,>P'
~ '"
...
'}
/
~#'V,-/t7
\
" ",
\
0.0
4.6
9.2
\V
13.8
18.4
"
"
23.0
PIO·IO (a)
Iso,ocrene +. Hydro gcn --? isCH)Ctanc
A
B
-t.
Discrirninarion of mcdds:
Assume
'fA ;:
k
C1 q
For runs 2 and J, 0 <
q:
a: < 1 ; 2 and 4, 0 < f)
< 1 ; and 2 and 5 , ·1
< r < O. From
Perry's handbook, 5th cd". p. 4-8; the reaction is probably surface reaction rate con'Q'"olling.
Mechanism I (H. Alvord):
A+S+!A.S
B2+2S ~ 2B·S
A·S,2B·S;:! C·S+2S
C·S ;:! C+S
Hence.
k fp;..
- PcfKuJ .... _-,._..
-rA, "'" _._., ..•,..._.
'--,•..PB
,_.,-""-,
[1 +'KAP;..+KfPft+ KcPcP
Mechanism II (S. L Mullick):
A+S +! A.S
B+S ~ B.S
A·S+B·S f:!: C·S+S
C·S ~ C+,S
Hence,
, _
k[P Ps - Pc/l<.,qJ
...... _ ••... , . , ....•.
"1;.. - ...." ._..._, ...• "...• _...A
,••.•.- ." •.._
[1 ,. K ... FA t· KB Ps -+- Kc PcP
10·22
From runs 2. 9
11. 12. P A:::: PB
=Pc "" P. a plot of ~rA vs. P shows a parabollic
behavior. therefore we Mil drop the second term in the denominator for easy linearization.
Tne readers can calculate ~ value by Gibbs free energy change 4'1 this equation (up to 650
K. the reverse reaction is negligible)
Tne linearized regression model is:
Using given 12 data pointS to solve for these four unknowns:
y :::: 3.0 -+ 1.42 P A
.;.
0.97 Pa
-t.
1.42 Pc
The tinal results are:
P A Fg
.... ". 0..1113
_._-,,--,,-,,_.
- _ ............... _ ...._"
(1 + 0.475 PAt 0322 PB .;. 0 . 414 Pcr
-fA:: ...-
The comparison of the pen:emage error between the model and the experimental data are:
r
(exp.)
(~A.!:lit .5
r
(calc.)
0
0..0362
1
1
1
0..0.239
0..0390
:'5.26
647
0.0345
0.0227
8.77
9.25
0..0410
3
O.D114
0.0534
Run
PA
Pa
Pc
2
3
1
1
1
3
1
1
4
5
6
7
8
9
10.
11
12
1
:3
1
1
10
1
0
0
10.
2-
2
2
0.2
0.1
5
0.2
0.1
5
0.2
0.1
5
I
10
1
1
o...Q351
0..0310
0.0033
0..0380
0.0032
0..0008
0.0566
expo
9.37
13,,69
17.96
lo.Al
10.26
3.54
3.54
21.02
% elmr
-4.8
-4.7
+5.0
A.a
0.0334
+5..7
0.0120
-5.4
0 . 0.505
-2.6
0.0302
-4.5
0.00315
0.0380
0
0.00288
·9.8
+10.7
0.00089
0.0599
....:.1:2 _
Isuml =59.2
avg.
=4.9
PIO-IO (b)
Discussion: The readers may check the validity of mechanism L To reduce the
accumulation of error in calculations, the readers should have used ail data points and
solved all unblowns simultaneously. To get the maximum informacion of complex kinetics
of a reaction from the least runs, it is advantageous to do planned experiments such as
factOrial design .. (W. G . Humer. and A. C. Atkinson. Olem1cai Engin~~rin!!, p . 159. June
6. 1966).
A paper discussing chemical reaction rare equations from experimental data is in:
C. n. \Vare 11., Summer Compmer Simulation Conference. Proceedings. 1975, Pa.."t 1. p .
368.
10-23
PIO-IO (c)
0.2223 C~o (RTr {l-Xr / (l-O.SXY
··rA =.-....- - -..- - . - - . - - . - . - -................
--.-~.-.--'J-
r1 + CAOO ~T {(0.475 + 0..322 (I-X)} + OA14x}11-
l
.)x
.
0 ..1 113 (RT CAot
-r!\ = ----.
(l·.xF
I {l-O.5Xr
..--.--.------..
~------
.
FAO:::
r 1 + c.:..o RT fO 79 7
.
2.5
1-0.SX \.
=:::
,
•
0383
.
x'\j2
'!
i50 I~l
CSTR:
w;;;;;; 21.380 g ::::: 21.4 kg
x
w::; 1501
o
2:1l13J1.5J:i!·X(LL0.5 x)Z
~
1
1 - 0 ..5 X
..,..
11+' _. . . . . .,..2. . . . . . {O.797- 0..383 X)I~
l.
'j
10-24
1
x
w= . .- . ..J~Q
G (X}dX
0.1113 x 2.25
0
. 1..5 (0..797' - 0..383 Xl]2
(1 - 0.5 xjZ
[...1-.. + -.-----------.
G(X):=
L.:Q~.X ___ . __.._._. . . _. . . . . . _. . .
where
(1-
xf
x
G{X)
0.
0 ..1
4.82
~l
5.38
~t
0.2
6.13
0..3
7.16
8.66
OA
05
0.6
0.7
0. 8
COO
11.00
:1
0
IS.03
0':'
a.l
I)
23.15
Using Simpson s rule.
area. under the curve ~ i 0
w:;: ..._____.J2!1_____. __ x.
10:;: 6000 g "" 6 kg
0.1113 x. 2.25
PIO-IO (d)
Consider the differential section between L and L
+ dL
In - Out + Generation"" Accumulation
4
,L)Ar Pc-H······ t/J)dL ==
-.dl{"1· Arv{L)CJL)]
But: A1 v(L)C4 (L):= N)L).:::: f~ll(I ... X)
:. FAD
f·:: TA pdl ..
$) rA {Pd
Relating the expression fA to X and L :
;; lFo (l·X)
2
1
:;: }Fo (l-X)
R, =-(2······
X)
2
X_•.
P __F
...:1.. _I...........
Fc=FAOX
l-Pr -2· . ·X
== tFox
PA == PA ... 1:.:X. Pr
2·X
0.4
"
44 . 62
r)C
c..J
Pc=-X-Pr
2-X
LetP=Pr,
10-25
as cu.
0::
lUI
From th.e Ergun equation. we: have:
All the things on R11S arc: constant. except far p
p=p,((L!x)
. :i = ~2X(~:)I~)gc%;;(1;1)l~50 ~-$)l: + [75 GJ
df. == 2.:X
dL
2P
~o where 6
.
0
=.!?Q.Q.. . . ~~:~?r.~. ~?,..0.,:d» 1: +Poge D? Q> L
Dp
Let L" ",., L..... : p. =..P. . then: dE"", l-X.
L-des
Po
dL 2p·
1.75 G
\3:
where
$=04
G :;:: t~ An fiA + Fso fiB) ==
AT
'"'''At: 9
G == J.,~
T
P :; (3 atm)
o
T.......
. ,.u.a.ll.•
frZ hr
(14.1 ?E~,·) (1441 frZm:.) =63487 .lb.
ft2
attn
p ;; C r m == (5ZJll.\ L
0-.'·
( I'sO gmol.) (2 + 112}~.-,{---.JL"-.)
hr
gmol 4)".6.g,
T K. {fr..25:I\2 f~
4
12}
lhmal J RT
(57 _Jh.,_.1 (3 ann)
:= ._ _ _ _ •_ _ _
(0,7301
\
.Jbmcl.!........"." . . . ,....:.....__ .. _ _...
ft~..rurrL\1 [(lOO + 273.15)( 1.8 jJ OR
IhmoloR,
10-26
IiJ ;:: 0.4
Dp ;: (_1 in) U . . fL} :: _1 ft
16
\12in
192
&<: = 32.174 lli.!IL.JL
Ibr sect
-JL-)
(150 grnol) (2 + 112) .....1..-.. {..
G;:: (FAO rnA +- Fae mB) =.
hr
gmol ~5.).6 g
AT
T
1L{n..2llj
f~
4 12
G=~ J.bm..
f~hr
T
Po =(3 atm)(l1:r:iHl~f~~-) = 6348.7!i'
__
(57
,.J.b-:-) (3 atm)
={l~:n~lJh= (0,,7301 ..fy3 9trU~IL(20~:;3.15H~ .all oR
Po:; Cr m
IbmoloR,
Po:: 0.2750 !wt
ft)
Ii (3 atID. 200°C);: 9.4719 x 10,3 cp
(t Cg,
~. He) }
-~.
:: 8.6211
=
X 10.3
PPROP
cp (all Ca)
(94719 X 1O·3j- 8.6211 x 10.3 )
-4
Ibm
"
!J. ;; ................- -..- -.........- - - - - cp 6.7197 x 10 ----2
ft sec cp
x JQQQ..z..;;.
hI"
Il "" 002188~n.
ft ru"
10·27
where
• rAT p~ (1 . 1j» K P;
Let: a. "" - . -.. . ------FAo
Then: dX:=
ell
(1.- _ ••.....
K~
=(KA + Ks) Po
J!.:::..~_
~K', K' xl1
lP:-+-l'1'
2 J
f2-X
For a given T. we
can
solve the
twO
ODEs to get X(L *)., We need to guess T until X :::; 0,8.
For the rate cOnStaIUS. we use the solution of 6.10 (a)
K :::; 0 . 1118 - .... ~:::~! . .-.. .gear'" hr • ~~.
;
KA"" 0.475 amr 1 ; KB
Kc "" 0.414 atm l
- (0 ..~ TAO
K ·2" ..... ,4"7.)
a:
:=
() .~~")}
- • 149
....
,.,.- ~
.)-"'1..
85.958 T
10-28
= 0.322 atnl 1
~. =(5...5.ll x 1(J4 f2..hr.) (378.09 + .ll2.Ql) J1'm..
o
T
ll:xn .
T
f~
hr
~: == _Q 2191 _ 8,~;5:i
A FORTRAN program is written to solve the equations. The results show that any
numbers 6f tubes !oufiicient to allow the given flow rates with a positive pressure provides
more than enough catalyst for the desired conversion The problem as stated, therefore
has no solution. However, we can choose a ditIerent L, and it only changes the
dimensionless parameters.. With L = 20, the problem is still unsolvable
For L ::;: 10 ft
, T:::: 2.16 rubes
=
1-::::
0.46
In
Note: Using the modified program with
Dp ::;i~ in • II "" 9 . 05 x 10"3 cp
L""' 10 ft
A:; 2.15 il'l2
=> X", 0.80 • p. == 0.7531
Using 1 1/2 in schedule 80 pipe (I.D. := 1..5 in) • A = 1.76715 in2 and me length to get
X "" 0 . 8 is impossible (p( < 0).
With 2 in schedule 80 pipe U.D. "" 1.939 in). A =- 2.952877 in1
This gives L "" 6.6Jft , X == 080 and p:::: 0.9113
1 1/2 in schedule 40 => A "" 2.03580 in2
L '"
1O.9~857
, X = 0 . 8 and P "" 0.6903
PI0-ll (a)
10·29
Assume a rate.,ljmiting step; start with surface reaction
A·S+S.....-··BS+H S
H·S ..... ·'H+S
Assume a rate·lit,wting step; start with surface reaction
~-~~~
.~4D.
:::
0
.:fl!2.. ::;;
0
kAD
knv
Find the expression for Cy
Combine all of those to get the following rale la\v suggestion.
CheckinO' to see if it fit>;, we see that for high P A' increases in P A cause decreases in
the rate. \Ve see that if PB or PH increase the rate will go down, which is consistent
with the rate law .
PI0-ll (b)
Now nsing POLYMATH's nonlinear equation regression we can find the values
for the parameters. We find that
k=O ..o0137
K·a = 4'16
•
I
0 ")-9
K B --._':>
Kc::= 0.424
In the problem it is given that K.. . is 1 or 2 orders of magnitude greater than KB and
~: which is tme so this is a good auswc{.
10·30
"aXl:~OO
3.200
T
o
~Ii'c;;r "li;;;~
on
:;lAta
J. .
[J Cdl;::~i
<i), ~d
vat de
2 .... 00
t.600
0.8CC
0.000
Model:
~
,. azi<.""Pa..... ' 1 ~Kil ... P.a·Kb.Pb··l(h)(Pt')-2
"
0.00137022
Ka .. '+.7608'11
? pOsatlv'@ r€'su::l,~ .. ls.
ICb .. 0.259382
1<1'\ '" O. "t2359
:3 "eqatlv~ .... 5lC:luals.
Sum o! squa r -E1'S '" 5.27033~-15
PIO-ll (C)
The estimates of the rate law parameters were given to simplify the search techniques to make sure
that it converged on a false minimum. In real life, one should make a number of guesses of the rate
lw parameters and they should include a large range of possibilities
PIO-12 (a)
Assume that the second reaction is the rate-limiting step.
Using PSSH, we know that
rS"'·H
",= 0
I 4'"
= k1CS·u
.
lD4·
sC·'V
10-31
Perform a site balance:
Combining all of these we find:
This rate law is consistent with the data. As the concentration gets larger, the
rate change gets srnaller which is consistent with the rate la\v as given.
PIO-12 (b)
No answer is right or wrong, but the points will probably be higher than
the ones given to see that the change in rate becomes even smaller.
PIO-13
Assume the rate law is of the form rD
ep
= kP::rl~O
1+ KPVI1PO
At high temperatures K J,. as T i and therefore KP::rIPO «1
'[Yep
= kPy~ [PO
'[Yep
--=k
Py~IPO
Run 1
0.028
----2
=11.2
( 0.0.5)
Run 2 0.45 = 11.28
(0.2)2
7.2
Run .5 - - 2 = 11.25
(0.8)
At low temperature and low pressure
'[Yep
= kP~TlPO
r
--.!l!L = k
Py~IPO
Run 1
.2. 004 = 0.4
(0.1)2
10-32
Run 2 0.015
(0.2)2
= 0.375
These fit the low pressure data
At high pressure KPv~1PO »1
rDep
_ kP~1PO _ k
KR2
- K
-
VT1PO
This fits the high pressure data
At PVTIPO = 1.5, r =0.095 and at P VTIPO =2, r =0.1
Now find the activation energy
At low pressure and high temperature k = 11.2
At low pressure and low temperature k =0.4
In(k2)
= E(~._._L) = E(!2 -7;)
kl
R 7; T2
R 7;T2
E(~3-~~)
In(I1.2) _
0.4 - R (473)(393)
E =7738
R
E
cal
= 15375----
mol
PIO-14
r1"iD,. :::::
ks.~.s
1;;:; Iv + ft.s;::: iv(l+
~1;
:
Kp
K,fj)
• Ii "" Kp Pfnp/PPI
_ ks Pr Kr
1 + PI Kr
I'iiOt----···
10-33
h-s:::; ~K[j~
f~.~, == 1:1:~'
ks
Pr Kl
I'Ti(~ ;:; 'r::;:: P~ Kr
kPirIP I pp!
rno, =i~r~~;?-I-'i~)K
Low
High
kP;'TP
=1>;1'" ~:P:;;;K
Pnn>:
Rxn is second orrier
Since
1»
K PfnpfPPl
PIT!P:
Reac:tion is ze:ro order
High Tempermure Kr very smail such that
PIO-15 (a)
Using Polymath non-linear regression few can find the parameters for all models:
See Polymath program PI 0 15. po I.
(1)
POLYMA TH Results
Nonlinear regression (L-M)
Model: rT = k*PMl\a*PH2I\b
Variable
k
a
1
0,,1
b
0. 1
PreciSion
RA2
RA2adj
Ini guess
Value
1 . 1481487
0 . 1843053
-0,,0308691
95% confidence
0.1078106
0.0873668
0,1311507
= 0,,7852809
= 0,,7375655
= 0,,0372861
Rmsd
Variance = 0,,0222441
a = 0.184 ~ = -0.031
k = 1.148
(2)
POLYMATH Results
Nonlinear regression (L-M)
10-34
Model: rT
= k*PMI(1+KM*PM)
Variable
Ini guess
k
1
2
KM
Value
12.256274
9,,0251862
95% confidence
2.1574162
1.8060287
Precision
R"2
= 0,,9800096
R"2adj
= 0.9780106
= 0,,0113769
Rmsd
variance = 0,,0018638
=12.26
k
KM = 9.025
(3)
OLYMATH Results
Nonlinear regression (L-M)
Model: rT = k*PM*PH2/((1+KM*PM)A2)
Variable
Ini guess
k
1
KM
2
Value
8.4090333
2.8306038
95% confidence
18.516752
4.2577098
Precision
R"2
=-4.3638352
R"2adj
= -4.9002187
Rmsd
= 0,,1863588
Vari.ance = 0.5001061
k
=8.409
KM=2.83
(4)
POL YMATH Results
Nonlinear regression (L-M)
Model: rT = k*PM*PH2/(1+KM*PM+KH2*PH2)
Variable
Ini guess
k
KM
1
2
2
KH2
Value
101,,99929
83.608282
67.213622
95% confidence
4.614109
7.1561591
5.9343217
Nonlinear regression settings
Max # iterations = 300
Precision
R"2
= -3,,2021716
R"2adj
=-4 . 1359875
Rmsd
= 0,,1649487
variance = 0.4353294
k =102
KM =83.6
KHZ =67.21
PIO-15 (b)
10-35
We can see from the precision results from the Polymath regressions that rate law (2) best describes
the data.
PIO-16
Using Polymath non-linear regression few can find the parameters for all models:
See Polymath program P1 0-·16.pol.
(1)
POL YMATH Results
Nonlinear regression (L·M)
Model: r = k*KNO*PNO*PH2/(1 +KNO*PNO+KH2*PH2)
variable
Ini guess
k--
-1---
KNO
KH2
1
1
Precision
R"2
=
R"2adj
=
Rmsd
=
Variance =
k = 0.0031
Value
0 . 0030965
57_237884
10l. 9967
95% confidence
3 . 702E-05
1.0353031
2 . 2870513
0 . 9709596
0 . 9645062
5 . 265E·07
4 . 436E-12
KNo = 57.23
KH2 = 102
(2)
•
POL YMATH Results
Nonlinear regression (L·M)
Model: r = k*KNO*PNO*KH2/(1 +KNO*PNO+KH2*PH2)
Variable
k
KNO
KH2
Ini guess
0:1-10
1
Value
----4.713E-06
-108.42354
1 . 046E-05
95% confidence
l.297E-05
4.9334604
2 . 878E-05
Nonlinear regression settings
Max # iterations = 300
Precision
R"2
= -9 . 6842898
R"2adj
= -12 . 058576
Rmsd
= 1. 01E-05
Variance = 1.632E-09
k = -0.00000471
KNO = -108.4
KH2 = 0..00001046
(3)
POLYMATH Results
Nonlinear regression (L·M)
10-36
Model: r = k*KNO*PNO*KH2*PH2/((1+KNO*PNO+KH2*PH2)"2)
KNO
Ini guess
0.1
10
KH2
1
Variable
k
Value
5.194E-04
13 .187119
18.487727
95% confidence
2 . 242E-04
7.659298
7.7652667
Nonlinear regression settings
Max # iterations = 300
Precision
= 0,,9809761
R"2
R"2adj
Rmsd
Variance
= 0,,9767486
= 4 . 262E-07
= 2 . 906E-12
k = 0.000519
KNO
= 13.19
KH2
=18.49
The third rate law best describes the data.
PIO-17 (a)
Mole balance:
-dX. . . . . =,_(-.,~)a
. .__ . . . . "
~
dW
FAC
Rate law:
Decay laVor:
.4.':. =._ kdc:.~:A.
dW
Us
Stoichiometry:
Evaluate the parameters:
8=.8
POLYMA'I11
m tial ..y!!!!!
~at.4:2E!.l.
o
d(x) la{w} :::a* (··ra) Ifao
d lal/d(w) ",··kd*a "ca!Us
1
kd:::9
fao=4000COO
10·37
Us=250000
(d;
Nov 1:::.g ,. Sed Reac cor
k::::90
cao-= .. 8
Ca
eps'" 8
ca=cao* (1. "xl f (l+eps*x)
-
x
50000
PIO-17 (b)
Mole balance:
{!~:,',4 == _(~dL
dlZ
Everything else is the same and we !Jt:cd to know"
r
V
= ,.,
= 0.004
Ii;)
5 CS'ffi' s in Series
!~~!:~51
d(a)/d(c)~-kd·a·ca
d (ca)
fa (t) ;ca4!t.au ( ( (1*ya4) I
1
(l +'ca/eto)
) t tau'a *k;) 'ca/t.au
d(ca41/d (t) "ca.,3/tau'" ( ( (1+ya3) I (1+ca4/eto) ) • tau· a*k) *ca4/t
O. 8
Q 8
d(call/d(tl"cao/tau'" « (l'-yao) I {l+callcto» ttau*a"k}*caJ./t
0.8
d(ca2) lalt) "'cal.ita\'!- « (l+yal) I (lica2/cto»
0,8
o. S
+tau*a"k) 'ca2/t
d (ea}) hilt) "ea2/tau- ( ( (1"'ya2ll (1+ca3/cto) ) +tatl'a"'k) 'ca3/t
ko"'9
10-38
value
tau=O 004
cto;:l
k"'45
.... q
Ca
ca,0=0.8
--- x .
ya4=ca4/cto
yal;cal/cto
ya,2;::ca2/cco
ya3:;ca3/cto
x=Ccao .. ca)/cao
yao=cao/cto
to = 0,
t
0.5
f
PIO-17 (C)
The only change from part (a) is the decay law:
'~='''k
aCA
dt
d
r == . W··W
_. _.llll!L.•••.•_ .•
Us
cit =: ;~~~.
Us
da
dW
Integrating:
w· _ _ _ •
=kd(~a. _
_""'--~"_._
Us
In a = .k.CW
. .!L_.:i............+k
Us
W=:WM.AX@a=l
k
=:
_._kdCA!~~AX.
Us
a = exp(~l;~{~~~~~~J)
.Ini:::!!::!. y~
Equa!::~£Ef!:l
o
d(x) Id(w) =a* ( . :raj Ifao
k=.90
fao=40000DO
kd;;;;S
Us",250000
==50000
cao"" .. 8
eps'='.8
ca=cao*(1.-x)/(l+eps"x)
tc)
LOOO
~
a
Ca
0"""
.. -- x
Q.6CO
Q,100
'Moving<Sed Reactor
TI
"T"",
t
l
.....
:ra",-·k"ca
a"'e:)Cp (kd*ca/Us" (w~=) )
PIO-17 (d)
10-39
(Count:er:::Ul:re.nt)
To find the Time-Temperature Trajectory we need to use the following equation
for first·,order decay,
r [ E' 1'1 I 'IT]
E
;;:k~~C~E:t 1 exp;_~RlT- 300JJ.
·r
t
"'d
Since no initial temperamre \vas given, we assumed one of 300K This is the graph
of thar equation,
Temperature- Time
trajectory
1000
-800
::,;:
~ ..... 600
E f: 400
~ ;:) 200
iii...
o
o
0,1
0.2
Time
0,3
(h)
PIO-I7 (e)
The two energy of activations are switched and this is {he new graph made,
Temperalure- Time
Trajectory
. " ., ......... "'- J
1000
-..800
¢i ~oo
:::J
iii
E w400
...
0.
ill
I-
-
200"
- ------------
-..
()
The graph looks the same just the time is much smaller.
PIO-I8 (a)
10-40
For all of the parts, the mole balances and rate laws are the same, 'They are:
de",
---.
=r
v
dvV
'" 0
'A::::: --·kaC",
Find the equation needed f()f a.
da
··--::::::kdaCA
dt
a:::::: exp(-.-kdC",t)
a:::::l when t=O
Assuming values for Vo. k, and ~ come up with the following graphs according to
the cases described,
y:!~ t.ia~.. ,.Y~!~
E:...quat.i.23~!.
o
d(cb)/d(w)~rb*vo
d{ca) Id{w) =ra*vo
vo=lO
(a)
a=l.
ko=l
..... Ca
t=20
Cb
_. a
k=l
0 a-oC
!Ll!\CO
ra=,,·k*a*ca
50
3 .. 200'
T
Q.~OO t:~0000
".. ., l
"*':"
KJ;;,X.
.......
C,~
,
Cll
-
Cas. I @ t-1000
1.000
ill
10-41
~
o~c
-
'-"-'.,
-..... ••..........~
--.--Q."'OO
~cc
-~.'"'~"'t-
Q
...... - ...
-.-.~
o.aoo
..... -.-"'"'!
PIO-IS (b)
Find the new equation for a:
,_!la...;;;:; k
dt
a:;;;::"···'·······
a2
d
e
i\
1
l+k d C A t
Using same values come up with these graphs:
g~.t:::i~l,Y~l~~
~~.i,.?E.E_-'
o
d(cbl/d{w)~xb'vo
1
dlcal/dlw)=ra*vo
vo",lO
(!:Ii
kd"'.OOl
t",lOOO
k"'l
a=11
e1;:;
-.~".
(l+kd~ca*t)
a
1
(h)
Cas,"" II @ t::;:80
Ca
Cb
a
10-42
Cc.:so I
@ t~
100e
{b)
Case IE
@
t=lG-
lSU
Ca
Cb
- a
PIO-18 (C)
Find the new equation for a;
__ .da = k aC
ilt
d
B
a;;;;; expC kd Cst)
The following graphs are made:
-_.__
Eouat.ions:.-
Ini~~~±,. y~:!~~
d(ca) Id(w) =:r:'a*vo
1.
d(cb) Id(w) '"l:'b*Vo
o
vo=lO
kd=.OOl
Ie)
t;;;;1000
k:::l
a=;exp( ··kd*cb*t)
z:b;;;;:k"a"ca
IOO::
..... Ca
. cb
-a
J:·a",··k*a*ca
10-43
Case I @ :-1000
l.Oc:;
K~
ea
en
a
~
-----
.. ooc
2.0ae
(e)
1.0':::
.......
-~-.,-'''.-
".000
.......
,.
'.'---.~.~;-,.-~-.---;
6.~CC
Case lXI @
B,.ooe
~-lO
KE'(
Ca
cb
:}"iJ!:n'!
Q.
0... 600
c,.lIce
-:d.2{;!:!
+
J ;)GG
PIO-18 (d)
~'iaking
a inlO a differential equation we come up with this:
, ,...4!!.. == kdCAG
dt
W
t :="
Us
. . . .,~a..", == .~:!,~~~..
dW
U,
10-44
JO .. OQC
Equati.~".
lnitia!
d{ca)/d{wj=ra~vo
1
d(cb)/d(w)=rb*vo
Q
d(a)!dtw)=·kd~a·ca/Us
1
va1.u~
vo",10
kd=.OOl
Us;1.0
(d)
Case I
~
k=l
.th::::::k*a*ca
Ca
Cn
.....- a
ra.t=:'""k*a 1rr ca
ld}
CaSe: II
KEY
Ca
C.b
a
-+-~.~o~ $P~
1 .00(/
(d)
-~.- .•..
K~X
Ca
Cb
T
r
t
a
(j.. QOO
1.
0 .. 000
-
Case III
....
--.-.--.- ~-.- -....
..
..
-
'............
~- ,
~~----
,,,,m~-"'r'''''' ....
::.000
.., ..... ""~
.... ... -,
.....
...
::..:.+=:-...;. ··· ..-..;·~r..-·.-----i-';--'.oeo
..
~
~.,
~ .-~ ~
a.ooo
000
PIO-18 (e)
10-45
,.",.,,~
=i
lo,ooe
Everything from part (d) is the same except for the decay law..
.~~ =-k.aC
dt
4'A
_ l~ttAX _. HI
t--_········,···,·
U,
dW
dt= - .....
Us
da ==: kJC, a
...
dW
Us
.,..".".~
'-'-~'-
Integrating:
In a =: ~:L~:'A t~ + k
Us
\'11 =- \V iv1A.'.: @ a == 1
~2::~t:iS'!?§~
d{ca)fd(w)"'ra~vo
o
d(cb)!d(w)"'rb~vo
vo"d,O
kd"l
k=,001
'..mlOlx;;10
a"'exp(kd.*ca/UsX (W'wmax) )
:t:bmlk'-,a,*ca
"'0 '" 0,
10
......... --- ... ".. _- +-- . .. .. .. --1
1> coo
G: ace
lC,;;;:~<';:
~.
...
10-46
~ac
~.
Case II
(€:)
EEX
_. Ca
", Cb
-
a
~-~~""il
a.ooo
'c_ooC
f;EY
Ca
... Cb
_. a
t,'T
O~2cn
I
-'
i
o~oo.c
~~:...:.:.::,;.~.:..:.r:~.:.::.:~::.:::.:
0,000
~ .. boo
2 .. CO(;
., .
,.,+
'¥ ....
~" ........ ' .. ~" .. -.'. ~~ ... ~, .......
... ~ _ . . . ,...". . ~ ~. . ~_ ....-.__ ._.......,
a. 00;]
tc,.o~~
:i" 10-:;
PIO-19(a)
da =-k
dt
W-Ut
- s
D
~__4..f!.. = =kD
dW dt
a
dW=Usdt
=1-- kD~
Us
Us
kDW t hen W =---=-=2.5
U s ·5
kg
If 0=1--Us
kD
.2
!X = -TA =~(-rA (0)) = akC:~o (l_X)2
dW
dX
dW
f
FAO
=
FAO
(1- .~D J (1-
dX
(I_X)2
W
Us
= kCAO
Vo
kCAO
FAO
X)2
Vo
fl- kDW dW
Us
Activity is zero for W > 2.5 kg, so the catalyst weight only goes to the effective weight.
10-47
2
_kCDWAO [
e ]_1(0.2)[
-x- W -k2.5- 0.2(2.5)2]_
-0.25
1- X
Vo
e
2U s
1
2 *0.5
X =0.2
PIO-19(b)
1
o ________ ______
~
~
2.5
o
5
PIO-19(c)
For infinite catalyst loading a = 1.
dX
= kCAO (1- X)2
dW
Vo
~=kCAoW=1
I-X
Vo
X=0.5
PIO-19(d)
~= kC
AO
I-X
Vo
[W _ kDW2]
_
2U,
-~=0.2[5-- 0 . 2*25]
1-0.4
2U,
_
kg
Us =1.5s
PIO-19(e)
kDW
Us
a=I----
0=1- kDW_
Us
10-48
Us =kDW=0.2*5=I.
kg
s
PIO-19(f)
a
=0 means there is no reaction is taking place. Activity can never be less than o.
PIO-19(g)
U=-U s
. da = kD
dW Us
kW
a=-D-+C
Us
_kDW; C
I ---+
Us
when W =W;, a =1
a = kDWe +1- kDW;
Us
Us
Now find We.
0= _kDWe.+ I·_.kDW;
Us
Us
We
= Us [kD"i. __ I]=.:~[:3*5
kD
.2.5
Us
-IJ
a
We =2.5
\Ve
dX.=(I_ kDW; +wJkCAo(I-xt
dW
Us
Vo
~=~~AO
I-X
Vo
\\'
1 - kDW; +W I-lW
Jl 1 Us r
~ =~~AO [(W; -W.)(I- kDW; J+.W;2 --We
2
I-X
Vo
_
Us
]
2.
~ = 0.2 [( 5 - 2.5) (1- 0.2 *~.) + 25 - 6.25]
I-X
X
--=1.875
I-X
X =0.65
0.5
2
PIO-19(h)
$ = I60FAo X -lOU s
10-49
\Vt
kDW
a=l---·
Us
dX =ka=k(l-!'DW]
dW
Us
2
X =kW _ kk DW
2U s
To maximize profit, a maximum in profit is reached and so we set the differential of profit equal to
O.
. d$ =O=160F dX_ lO
dU s
AO dU s
dX
kk D W 2
dUs - 2U~
2
160F kk DW = 10
AO 2U2
S
2
8FAO kk DW
Us
=U~
=~8FAOkkDW2 =~8(2)(.2)(.2)(25)
Us =4
k~
mm
PIO-19(i) No solution will be given
PIO-20 (a)
Stan with the mole balance for a balch reactor:
Rare lavv;
Decay law:
da
..-._-:z:
(i t
k J a"'
Stoichiomeny:
Plugging those into POLYrvlATH gets the fc)llov,.ing program and the following
graph
10-.50
2~ti2.!~:L'
dlx)/dlc)=-rt"w!cto
o
d{a}/d{ti~'kd~a~2
1
w=5
k;:=20
"'~.-~ ....
kd"'.l...6
r
v=l
:).&00
-1- /
T=735
p1:=pto·
(1 ··x)
rt"'·-kc·pt"a
II
Tl
lltO=pt.o"v/ (R"T)
=
//
T/
tI
Tf
2.;.082
::0
-
/,/
-!---..... -- -".--.-----,.~-~ ....." ...............- ...................... f - - - - - - . - - j
n.ctio
0,
a. ::u;;;::
.,...~
..
PIO-20 (b)
For the moving··bed reactor the mole balance now becomes:
dX
dW
.. /;
FM)
....-.-,:::;:;;-,,-~
The decay la\v nOw becomes:
da
k"a"
dW
Us
."""'----:""....:::::... ........ .
Everything stays the same.. Plug into POLThIl\TIl.
The conversion achieved is X:::: 0.266
Equas~~2.
d(x) Id(w)"'-!:t:./fao
o
d(a) td (w)
1.
", ..
kd"*aA2/0s
kt",,20
£ao,,600
l<d"1.. 6
~!lE.;E.!,£
Us=:2
'"
50
o
pto:2
pt"'pto* (l"x)
'NO'" O.
We
20
0 02439·02
2<';
500
600
teD
;;
1 6
5
20
x·c,,····kc"pc·a
fao
1<;d
50
266;6
2
,
2
2
4tr'~68
·-.:10
--0 .7:5i4:
,~c
PIO-20 (C)
c) Increasing Us \vill get us a higher conversion" Looking at this summary table,
U. = 10 kglh and X :: 0.6
10-51
50
a loS:.5
C .024.:130:
20
600
1 Ii
;;:
~labl~
~SciaJ_val~~ Ml1XimillU._"~5!f.~ Mi:!i~'!!::1~
()
50
0
f_~,.y~l'le
::.
1
o. : l l l l l
0
0
O,SS9-i:iSS
()
CL 599686
f~o
600
600
600
600
kd
LoG
]',6
1.6
Us
1.6
10
:),0
10
10
kt
20
20
20
20
pta
:2
:2
:2
:;:
pC
:2
:2
0.e00625
0,800625
t't
-40
-l. 77917
-4.0
--1.77917
\01
.,.
x
50
111111
PIO-20 (d)
For second-order decay:
For
s:::: 25 keal/rno! and E
4
:::::
10 kcalimol:
Use tlris equation in Excel to generate the following graph:
Temper ature-Time Trajectory
Temperarur·c.. Time Trajectory
-
~..
soo
r------- --.------- -, - ----------1
~ %500
i~
'"
'wo
--
------
~
£,-=10 lInd E;.;2S
1600 , ...... ,.--.-.... --.,----...
~ -
t-- -------------- -----------~
"]00
" , .... "' ....... ,." .......,.......
Q
100
I
.. .. ,.
400
bOD
o
tim" (h)
'lime (h)
Use this equation in Excel to generate the above graph.
PIO-20 (e)
10-52
In part e, the only thing that changes from (b) is the decay law and the decay
constant:
2
da
krJP/a
""-'"-=-..::......:....-
dW
Us
ka = 0.2
Plugging into POLYMATH we get the following summary tables for Us:;:: 2 and 10
kglh . X:;:: 0.50 and 0.88 respectively. X will again increase as Us increases.
Eq'-J,a~ions.;.
!ni tia1~1~~
d{x)/d(w)=-ra/fao
o
d(al/d(wl=~kd"'pt~2"aA2/Us
1
kd;;;.2
u5=2
fac=600
kt=20
a
~ ..~~ ~~.,.,~±~ fina:'_~
so
0
So
0 502431
0
0.502431
1.
1.
0 . 104635
0,,2
02
O.lC411J5
0.2
2
!.!'!±.ili!....yalue
w
pt.o.:Z
pt"'pto" (l-x)
Us
tao
:2
&00
600
sao
600
20
20
20
20
:2
:2
:;!
:2
0.9951.38
O.99Sl38
-40
··2.08253
2
,,40
f~g.!~l_Y~: \.I~
\)
o
2.08253
!,!,=immn.:2.!~~ ~2imt.:.r::...;;:alu~ fb~l.Y.§!;b~
50
0
SO
08']5709
(I
(I
37$'109
1
0553922
o
553922
0.2
kd
0 .. 2
0 2
02
uS
1()
10
10
€ao
600
600
600
600
kt
20
20
20
20
,/.
2
:2
Z
2;
2
0.248581
o
-40
-2 _ 75389
.. 40
2.75389
0. 2
2
PI0-21 (a)
A -7 B Elementary reaction with 1st order decay.
da
-=-k a
dt
D
a = exp( -kDt)
PI0-21 (b)
10-53
243581
t
The activity is never zero for first order decay. When a = exp (-kDt) there is no t such that a =O.
PI0-21 (c)
Mole balance:
dX= --r' A W = _
-r'_
-r' W
_
AW
_=_A._
dt
N Ao
CAOVO
Vo
CAO
Rate Law:
rA ' = -a [ --fA' (t
-rA '(t =
=0) J
0) = klCA
Decay:
a =exp( -kDt)
Stoichiometry:
C A = C AO (1- X )
Combine:
dX
W
dt- = exp( -kDt )kl (1-- X)~~-
In ( -1-)
kW [ l-exp(-kD t)
k Dv0
= _ 1-
1-·X
J
X=1-exp (_IsW
[1-exP(--k t)JJ
kDVO
D
PI0-21 (d)
klW
-=
kDVO
(0.2)(1)
=2
(0.1)(1)
X = l-exp( -2[ l-exp( -1)J) = 0.97
10-54
PIO-21 (e)
Decay rate increases more rapidly with temperature than does the specific reaction rate. Therefore,
conversion decreases with increasing temperature.
PIO-21 (1)
kjW (T)
kDVO
k
D=
= 2exP[-1500(_1___1_)] =0.57
310 400
0.lexp[2000(_~
__1_)] =0.53
300 400
X =l-exp( -0.57[1-exp( -5.3)J) =0.43
PIO-22 (a)
In order to get a high conversion the en.tering pressure should be as high as
possible since the rate is a sewnd order function of the pressure. U should be
kept low since the conversion is an indirect function of the flow rate.
PIO-22 (b)
11\e problem with such a low flow rate is that the activity will remain low
PIO-22 (c)
We (an use the same eCluations that are given in example 10,7 with a few
exceptions. For example the rate law, we use the one given in the problem:
I11e activity will be different because the equation given is different:
,,~:~, :;::; k naCcoke
da
dz
..... , ....... ,.....
kDaCcoke
.... ..
::;:,....... '""- ......•.... -. .---
U
To find the concentration of coke we use stoichiometry:
c'cot(.e
, == .~coke.
RT
We fmd that the value that gives the best conversion (X;:::: 0.337) and nses the
whole reactor is U :;::; 7. See the following POLYMATH program.
10-55
:sq~aS~
Inicial ... :alu€
.:
d(a} Id(z) =·~-kd"'a""'ccoke/U
1.
d(x)idlz)=-ra/U/Cdo
.)
cao~
22-
kc.;;;;1.OO
o
;;;
o
15
1
kp=i;ne;;;:5~···5
Q
LS
.a002S~34.g
00C2549';8
a
o
G 316731
cao
(} 22
0 .. 22
0.22
:) 22
eps~1.
;Cd
:00
100
leo
100
R=.OS2
k."1J::;-Z:;ne
5e: 05
5e-··CS
5,,-05
7
7
ao.:;:?
7
tlo
1
eps
pao=12
33673:
R
o
:,;-no=80
T
67)
6"1:!
673
U=:Uo*(l"e;:>s·xl
pao
12
12
pa=pao'
rho
80
Be
12
80
BO
)
9.35712
P"
L2
12
5.95425
s. 9S425
p-coke
o
}
0
) 02"288
.x:ap;:-i:ne
···0 0{)12
.~
(lox) I
(l+eps'x)
pcoke"'pao~xl (I Teps
032
082
35711
~
• xl
rapr ime=a' ( kpr ime 'p-,a ~2)
ccoke=pcoke/R/T
ra:::rho""r:aprirne
e·.:-ok~
.
0.2288
51934~···O7
0.082
-·0.0072
-4 S1.934e'··O'l
o .OS~7'162
0 0547762
·3.61547e·05
12
··0 51G
PIO-22 (d)
To fmd this the only change necessary was the values for the k' s because
they change with temperature.
_ pr" ED (1
1 \'1'
kD :;:: lOOeX
! --R6ij-i)
.....
....I
The POLYlvlATH program below shows the results. The temperature is 485K
and the conversion is 0 . 637 _
10-56
Equations:
Initia.l
d (x) !d(z} ;--ra!U/cao
a
d(a}/diz}~-kd*a~ccoke/U
1.
cao;.22
Uo:;;2.5
R=.082
T"'48S
pao"'12
!'ho::30
kd=lOO-e>..'P (15000/1. 987· (1/673-1/T) l
kpri.me ::oSe-5*exp(30001l • .987* (l/673-·l/Tl)
U:;;Uo*{1. ... eps*x}
pa=pao" (I--x) I
(l~eps"xl
pcoke"'pao*x/ (.l+·eps*x)
ccoke=pcoke/R/T
J:·apr.irne"'a* (-·kpri.me"'pa "2)
ra=;rho'raprime
'0 "" O.
:nit:ial_'ya1:!:!.! MaximlmLy!!ue Mi~i!!!..~l.,,~ UEill value
o
15
(}
o
0 6374aS
0
0 637285
() 605597
(}
C.22
0 22
2, S
1-
o
cao
()
Uo
;J .5
2-5
2.5
eps
1
1
1
.22
o
22
15
.605697
C.082
0 082
Q
485
485
485
4S5
12
80
12
12
12
80
80
80
1.29n
1.2932
1.2932
kprirr.e
1.29 ..31
2 .. 09558.-·05
.2 09558. 05
2.09558e-05
pa
2.5
12
I 09121
L2
2.5
2.65841
2.09558.-05
4.09321
2.65841
pcoke
ecok.
o
o
4.670&
(}
4,6708
0.lL7445
0
0.117145
-(}. 00J01 763
-897021e-05
·0.00301763
···0.24141
-0 00711616
)cd
082
0.24141
PIO-22 (e)
10-57
082
a 9703le-O!
-0 0071161&
val~
To find the temperature-·time trajectory, use equation (10-119) and add
'iNhere necessary:
C,o\<e
The following curve is generated from that equation.
o
() 015
01
·rOIme, t
0 . 005
PIO-23 (a)
~
... , , - '--'"" .... _ _ ..... ,,_..., , - - _................'T
I
·-·------·l
I
~-------.~--.-
..•.,.-""..............
t------------7
,.-......",
Design equation:
Assume W=lg
Cumene
(A)
FAD (lX)
r=FAoX
....
'"
Propene
(R)
FAoX
+
Benzene
(S)
FAoX
The amount of cume:le hydroperoxide does not com:ribme SignL.+J.caIldy co the roml number
ofmo!es.
or X =:_:!2.
1 ··Ys
10-58
X
a
.0204
1
t
0
.0165
.809
50
-
,0133
.652
100
,0107
.5145
150
,00851
.410
200
.OO563l .00311
.182
.276\
.00241
400
500
3001
1n a vs. t gives the best fit a "" c·-o.t
'Iberefore. decay is first order with decay constant of 4.27 x 10 3 (sec-I)
PIO-23 (b)
C
A
:; FA"",
~AO (I-X)
U
u
F AU
(1+X)
Assume no AP.. ...1L "" _ll. := ......
-.~.- Vo
no
FAO
=
V .::::: ( 1+X)
vo
C A ::: £::,,!dl-~2 "" CAO l!:X)
, '1,)0(1 tX)
(l+X)
PA
:=
CA RT =:
CAO
L~.~Xl RT
(l+X)
Mass balance: F AO dX ::;: rdW
10-59
.118
F AO
= 2.0 rnol/Il"Jn :: 1/3 mol/sec
CAD
'" 0,,06
kmoVm 3 == 0.06 mol/l
R =O . 082~!L
K'moi
T "" 273 + 420 :;; 693 K
k ::: 3.8xlO,·3 mol/g sec ann
Ct.
;:
4.27x 10--3 sec,·l
W
=:
100 kg
:=
10 kg/min:;; 1/6 kg/sec
Us
Equation (l) becomes:
2ln (l·X) -+ X =: ,CAn;:~: . !!~[e· (~t~l11
(O.06XO~082X69~~~.~.8~.~~~~!..P~J r
. ..
2m \ lX) -r X ::::: -
L~,3~~J.Q..3:.:.1.9.0.9 \. 1
Iexp \
L
J 11
-·. .···{I~) {4~'27}xl0.3
3
.. x ==
PIO-24 (a)
..
6
LOO
s ..:;. W
+. CO2
:
first order, irreversible..
. . da:;:;;
k<:J. an C',!1 == k.i an
ill
.">
··.fA
=KaC A
10-60
- -
--- -
- - --------------- ---------
-
---- -
-
---
-- -
-
-- - - - -
-
- - - - - -- -
-- - - - -
-- -
-
--------------------
For T :::::500K
o
20
X%
l!1ntL\
, iI-X!
99,5
60
80
120
0.,7
0.56
0,4.5
0,38
0..29
~
178.1
2..2L7
262.6
3443
5
to
1.5
20
30
40
,.-
0.89
0,,69
0.51
0,42
0.33
111.8
1444
174.9
237.6
302.5
142
For T::::: 550 K
o
1 .,
X%
,
I
-
\
11ln 1·..... ·,1
q .. X!
8 7 .....~
.!9,,5
4W,.
I
3QQ
In(lJ(l-X)
I
~
II
+T=500K
OT=550K
=1
aJ..I--'-II.
G
t
They axe Straight lines ."" n=2
For TI "" 500 K : slope:: 2.04 {t'~'h
}
z:::>
[l~t]=O.Dl =('tKh
For T 2 ;: 550 K ; siope;::
6.325
,
{-c K)z ;; 0.02
}
=:>
~I
""
0.02
~=0.126S
&::..Jl.QL.: exn {-.!fLf.....L ~ _l_l}
K.st 0.1265
• 8.314 \550 5001
~1 = ~1exp {- ~ (i l )} == K.i
K.it
t
exp {-
88~f~4 ('s60)} = 0.02
= 1.296 x 10 3
We want to maimain Ka = constant
10-61
K
Ko exp
h"~1:a t) Ko
:=
{f (f~" t)} ~ Ko (1
+
Ki t)
fE \f~"
,. l \JL 1 := v - t
. IR
To T I
,~
c;cp
rE( 1 IJ\']
exptR't~-- j ,- 1
t :;;;;; ,---""""-""- ,---------"""""""--""""':,'" --,
Kd
I
""",..1
480
485
490
495
500
l"""",,,
PIO-24 (b)
Since the equation for the acivation is:
a='""'
1 + lcd'
we cannot find a time for which a:::: 0, because it is mathematically impossible. We can, however,
find a time at wbich the activity is small enough that it can be considered to be zero, The following
graphs show the activation for'the two temperatures given
Activity
of
catalyst
~ 0 B
>
06
' .....·a (SaO)
::,";!Ii:",~i55~!
()
500
time
1000
1500
(days)
10-62
The graphs show that for sao K, the lifetime is about 1100 davs and for 550 K
the lifetime is about 450 days.
"
PIO-25
First we need to tlnd CAO '
. l!2_:::: 10
ETB
Y£TfJ
C
AO
=0.1
"" !l§.tu.. = ~_~_~_03(~.1.J =0.065
RT
(8.309)553
Start by guessing that the decay is fiIst order:
We were told that the reaction is zero order when the conversion is less than 0.75.
This is true at any time after 2 hours. We also need to fmd the denominator
as a function of conversion.
X =~:1!L~~.(~;t
CAO
CAO •.... C:<t ;: (~40X
So we graph this:
This is the graph that we get:
tn (1/cao*x» vs
time
i-~[~~·
o
5
10
Time (h)
15
As can be seen this graph is linear which means that the decay is first OldeL We
also know that the slope will be the decay law constant so kG :: 0.2024.
10-63
PIO-26 (a)
Mass balance:
Rate law:
Decay law:
.9a,. =':,~Q~
dW
Us
Energy balance:
dT
..Fa
:..-. (TA"'" T ) + {"fA '(
)·AH, Rx \)
-_., ....... _...".=
Stoichiometry
Evaluate the parameters:
k"'" 0.33 eXP[3777(.!-- "~,,,) 1
450 T J
['
k,. :::: o.,olexP!7000( 1
.)
L
450
1 \1
"T""')!J
,
Plug that into POL ThiA.TH and get the following program and answers Ug to get
maximum conversion will be 17 kg/g,
10-64
Equat;h2!'~.':'
a(al/d[wy=-kd"a/:J
Q
d(x) la(w) ,,··ralfao
dlT)
fa (w)" (Ua r (Ta"T) ... ! -::al
~
(··Dhrl) }! (U"cps+fao*C!;>d)
fao=5 .. 42
U;1 7
1'a=323
:;).. 27
Dr.rl=···80000
Y~E~2l.,~
<;:p5=100
~~~~&... =:~.;.:i~
~'!.:;:\'1t;..'!"._va:~
~:!~ Y~1.~~
,)
so
C
v
a
cpa=40
.So
~4~
0
5.U
11
5.42
1J
f."
Er=) 777
cao".27
SC
2-::-1596
o .>S~:2:
"l'
Ed=7QOO
I2~; ...Y£t±~
;; 211536
0 887222
6{S 6:3
He
5.42
S <12
1'1
17
321
···3£]oao
123
32}
··S()OOC
.. 8;10000
.. 61,3
1'a
n3
eps=l
Dhrl
.. seooo
U"-"'.8
c:ps
leo
100
.:CO
kd", . Ol-exp (Ed' (1/450-1/1') )
cpa.
100
40
40
40
40
Ed.
fOCO
loon
7000
7000
E:::;
J r17
37Ti
cao
0.21
a
21
3777
027
0 .. 21
Q
B
loa
O.S
0 O!
"ca
0.,33
<I
Q,;n
0.27
:a
,·0 069:
·0 C:Q;9:2S
k:: .. 33 'exp (Er" (1/450···1/T} )
Cd"cao" {t··x)! U .. eps·x)
l.
eps
,;;""
1.:,)4e·;.
H;:a
OS
oa
o.n
1.:"1401
4>" :~'j'lS
a
B
C Ol61349
.. 0 l.4S41
PIO-26 (b)
Using the same program we can see that the maximum conversion is 0.887
PIO-26 (c)
Everything is the same except the energy balance:
dT
........ ., •..
-~
dW
Uaw ('fa . . . T) + hap(Ts _.- T) + (r.~ )(llH Rx )
.:.::: .......
- . -.....-...-......-
......... .......
~
3';77
-.--.- ..
...
----
-
............... ,., .... "." ..... ............
FAOC pA
1<Ve also need an energy balance on the catalyst particles .
Choose values ofh and a!.. We fmd that Ts needs to be slightly Ingher
than l' in order to get a large conversion. The maximum conversion will
be different for each T5 that is \1se(i
CDPIO-A
10-65
0,0161349
·".OISJ92S
Given: The catalytic oxidation of ethanol
CH3 CH 2 OH
+}' O2 -1- C:FI3 eHO + H:cO
Denote: A[=]CH)CHzOH. B[=]0130IO, W[=] H20 , A'S[=]ethCfhO'S
Mecha111sm is belived to be the following:
,~
t-,
A-S +B-S
fA =leA [pA C§ "' CA-s CH.s!K.J
+ 23"
f:!
20-5'
-ro:= ko(P<4 (:-;, '" C~S'!Ko]
+o-S'
""'T
~
B +OH·S' +$
'TAO:::; kAo[CA'$ Co.s'· PE ({)ii-$" Cs/KAO]
~
W+S + S'
rw = kw rLCoH-$" Cli-S - Pw Cs Cs,!KwlJ
A
+ 25
(h
A·S
OH·S' +H·S
It is given also that CH,s =CA,s
'''fa
i..o.;:O
A
:ro,=O
k.:,-
cbs' ::: Ko Po: c~,
"
(-{)H-S'
or Cos,:: 'f~;p(;; c~.
Pw Cs es, ..
LH-S "" -'K.;-" wltn
'r?'-""--
'0
CH-S;::; Y!'.A. .t' A
Cs
PwCS'
COrV' c;;;;; -.-- ... - - .) Kw 'iKA,P A
CT;:; Cs + eH's + CA'S:::::; Cs [1 + HK;;"P;]
"""
('I ""
" I . . . . r . ,..--.-".--~ . -.,.
e' ' ' g• + C,)-S' .;. (iJH-S':;:::
(S\l+ YKo P Ch -+;:
Pw
~
Kw yKA,PA
L
,
--lAO"""
r",
1
PE c,:'.m.·s' Cs
."1' r----..,.------,,··, ,
Ps Pw Cs CS'
k.i\O i CA.S COS' ""--;::::--"1 = kAO 'i KA. PA Ko Po:: Cg (S· - - - -..;:~
!.
I\..AO
.:
L
Kw KAO 'i KA PA
r• . 'K"K"'C C _ ?-.AO
'i . A {) ·S·S·;
.,----·"tAO - "" .........
p A, ! Po"
=--_._--.
FA
l.
'I
K""1.::= K;.. Kw K..,o
~
kAO
.I
P-a Pw
.. -_ ...,,_._-_-._"'-",._-;
". K", Kw KA,cdKoi
{K;:;
iKAK:(:; Cr C r [ PA -{pc}.- . !:~w J
···r AO _ ...,...._.........................................................:"''''.....''' ...,. , ...................... - ...............
{P;'l'l
+.
2'1J{";A. PA.. . ] 1 + ·rKo-....· Pc-:-
,
Initially PB
=:;
.
-'7.
~
........,............·t.
Pw
I
Kwf"KA P",J
t ----.
Pw ::::; 0
10-66
(b)
With reaction 3 irreversible, '·IAO::;;; kAO CA-S (:Os', Since A·S. H·S, OH-S' are created
(and desrroyed) only at the reaction swface. A stoidriomerric relationship existS between
(a)
The same equilibrium expressions for reactions (1), (2) and (4) exist as before:
(b)
CoH-S' C~i-S
PwCsC s'
::;;; -,
Expressions for Cr and
•
,y>
(c)
Kw
C;. become
PwCsCs
wIth K AP.4 (, S » ··--K··"·-""".,-",
IV
CT == Cs (1 + 2fKAPAJ
C~ ~ c~1+ rK~;o,;:- +fx'x]
10-67
Rate expression becomes:
···T
A
=- ...... -.. ..
CDPIO-B
We can first try to come up with a rate law for this data. We can see that as P E
increases the rate law also increases but slowly the amount by which it
increases becomes smaller this tells us this:
We can then see a similar thing happens as PH increases so:
I
Fc!
""!A
"" ...,..,.....,.....
_ .•... _ .......
1 +KUPH
Sin,ce both reactants are adsorbed the me(hanism must be a dual site, This
makes the rate law:
.~££"~JL_ ........
1 +KEPE +KBPH
o::.,.".,.__
We can then plug this into POLYMATH. and we then get the follOWing values
for k, KEf and KH •
k= 14.6
Kl) =2 . 63
KH =1.76
10-68
O~90C
o.
lO"'8
--,.
t
~aa
~
i
~II
0.320
tt
f
!
c. ace
2
nOd~t;
~
'"
;(0/
1
rat"i"?;:;It;"O:e.Oh/~
t .... ·K€.')lfPE*+K~H(Ph)
~
3
~
............J?-._ ...... _...fF. . .
e
'~2
-rE'-SLdr;.;.atsl"
6
neq.,at!;\I(o 1"@$lOU.a!S ..
Sum: of
squaf9S::::
CDPIO·C
(1)
NzO + S ~ N20· S
N:zO • S -+ SiO • S ·4 Si0:2 -t. 25 + N2 (Rapid)
rSiOt ;:; ~ Pues PNf?
1 + KPN;;()
(3)
dF
-~:::::r a
dV
.
6
:<" '" 1.75797
1 "I .. 53-....
'" 2.63'17!
PQ5t':l\li:"
I!
J . ____ L.i . _. ._.
A . . __ . . _--..J............ ~."..... ""t~-.-......... !~.t...
... !.atec
vql~,,!i'"
I!
!
C .. 1SC
cc.a!~
A
'Plug Flow Reactor)
10-69
t,,"5236e~.06
k P'AO (l·X) (SB-2X)/O + EX)2
'fA
=---------:···-·--··{e- -2X}-'--1 +- KB PA~:!L-·
(1 t· EX)
k P).o(1 . X}{SB .. 2X)
~. K~P:~-{~ -2X) (l'~ eX)
:; (1':; eX?
.~ 0·7
as. ; : ; 1..1.:.2::
3 . 68 -' .
E:;
('--'}~"-)(2
+- 2· 1- 2):; 0.25
11.;) + ).68
1~~------------------------.r--'
Slope =: K. == 0.7465
k'
1:300
Inr.erceut
::;:; 1
~
k = 703
1200
!!l.rorr~"ITlin
k::;:; L49xlO- 3 NmTotr'-rr.in
11 00
K = Llx10 3 mTorr 1
A
~+---~~~--~--~---r--~--~
300
.00
500
600
"lOa
sao
900
1000
~)(m TorT)
KB::;:; 1.1 x 10. 3 In Tou l KB
PAl
=(1.1 x 10.3)(157) -= .173
k ~o:: 1.49 x 10..3 (1.57)2::;:; 36.68 Nmin
(3·9.?: ~lli:~L_ ...._._. . _._
-r* ; : ; 36.7 Nmin
(1 +- .. 25X}2 +-.173 (3.07. 2X)(1 +ZSX)
A
FAO :::: 3.68 x 10 J gmok/min
-r* ;;::; {Nrrim}:: 4.4
A
X
lO'-O r- (gmolel
A
a=:; 250
(m2min)
m2/rr.in
IX ~ Ar-.....a under curve of X
..Ya.. =:;
dX
F
·
AD
r
I)
A
YS.
L
r
4
A
10-70
x
.r~(Nmin)
.l/.r~(~)
l/.~ (m2/gmolel
...YlI...
0
0.025
0.5
0.1
0.15
020
0.25
0.30
0 . 35
73.6
0.0136
0. 0142
00149
0.0165
0.0182
0.020
00226
0.0255
0. 0288
00328
0.0378
0. 0439
00517
0.0617
0.0751
0.0936
001203
0.01617
3091
3235
3392
3739
4139
4545
5145
5784
6547
7465
8589
9983
il746
79 . 1
82.8
1783
196.9
217.1
242.3
273.2
3083
350.3
40L4
464.3
OAO
OA5
0.50
-0.55
060
0.65
0.70
0.75
0.80
70.3
67.0
60.8
549
494
44.2
39.3
34.7
304
26.5
22.8
19.3
16.2
133
10.7
83
6..2
V{m3 )
FAO
0
00012
0.0012
00026
0.0029
0.0032
0 . 0036
0,,0040
0. 0045
00052
00059
0.0068
0.0080
0.0095
0.0114
0.0141
5431
14{)29
6444
7774
958 . 2
1115
17067
21262
27338
36760
0.0179
0.0236
1602
Reacror Volume per wafer:;;; 9.8 x 104 m3/wafer
:. Number of wafers, n::::: V!9.&
X
10 4
A plot of conversion X. vs. number. of wafers can be obtained
80
200 --..•. - - - - - -
60
c
ri
;;;
;
"2"
100
40
-0
ci
:§
....~
';"
%
2Q
o 4"'"'"T-~--__,-...__1-..,.......,..........-+ 0
0.0
0 .. 2
0.4
O.S
O.S
1.0
10-71
Tne thickness on mest! wafers
C:l.l1
be obtained. from tb,e same plot,
Wafer#l ; X"" 0
or'A ;:;:: T3 .6 N'!nin
:. After 30 min: 12Q8~.
Wafer #50: X '" 0.52
.r~ "" 20 Nmin
:. After 30 tnl.'1: @ . b.
Wafer #110: X:: 0,78
-r~
A
:=;
6 Nmin
., .Aite:- 30 min:
J 80 ,J.,
CDPIO-D (3/e CDPIO-E)
CDPIO-E
.
(a) For simDlicitv,
. ' letters were substituted for the sDecies in the
Droblem sta tement.
"
The late equations for each of the three steps in the m.ecnanism
is given below.
Equation.s 1 and 2. are substituted into equation. 3 because the
third step (the deposition step) is assumed to be the rate
l.imiting step, we want to find the rate law of the deposition, and
we have data for the deposition rate.
10-72
.cuso, remember that we have the relation below:
.
f v=·
1
.
1+ KcP c
Substitute everything into the deposition rate law:
We now have to check if the above rate expression agrees with
experimental observations.
..
The rate of deposition is independent of AI' and 1h. .-YES
.
tJressures of TiCI. and NFL,
• At low partial
,
, the deposition
appears to be first order in TiCl" and second order in
~n)--YES
• A.t high partial pressures of l'·Hi], th.!'! rate varies inversely
with TiCl,.·YES
(b) To determine the reaction rate parameters we must
:earrange the rate expression to a linear £onn .
f
A plot of the experimental data is shown below.
:::::i',. ,'_. .". . ".__. . ". . _._. . . ._. _ ._.
'TItanium Nitride fUm Deposition
Model Veriiication
1200000 ..1
200000·
... l.2167%ii.... s:
R,o\l OIl 9 91t98UE.l
f(x):;;.- L~l6l£"'l"-l:
Pal'b"2
10-73
1.
- = Y .... mtcrc:;pt :: 2.2168 x 10 l
.Jk.!'
.
k. ::::: 2.035x 1O-1l_._
....!~~! . __'"
.
em-· min· mT'
k.
•
.
p~
,J
.
== slope = 14 42
1\..
k~ :::; :'i . 505 X 10-5
(c)
rnT3
The experimenrai data, when plotted with the rate law
deri,:ed
in this
problem, form a straiO'ht
line .,..........
Thor'e''''''''''
.
,
....
.::;
.. """
ttV2 ;)roposed. Iri.eC.:1arusm may O€ used to describe ti'..2
depOSition or titarl:l;,rr~ ru::ride fi.lms.
_
.
...-...;
CDPIO·F
Given: Ille dehydrogenation of ethyi benzt;';ne w st:;tTene:
E
St +- H2
Quanur.lUve d::ua suggests re.lc:ion rare is of rhe fonn
-r;; a
T11t
rJC~
Pj:;
-'~",'" ,'\-_-p-~_ :-.";::--p"'1 '!'" "L E"" ."">3 St
....
. .,-
where At, lu, A3 are consta..'1ts
that {orE) is independent of Hl suggesLS that
me reaction is i.::reversible, and that Ii2
is not adsorbed on the c;:;,ralysr Sth'iace. Also, the aoove expression for (-TE) suggests that
both E and
S~
are adsorbed on the su!face.". Try the following reaction scheme.
'E+S~·'}E.S
E "" S ·H· St .. S t 112
Sf • S ...;-)- SI t· S
-fA:::
kAlPECS· CE.sf'.r.<.A]
rs == ks fCE-s '. CSt-S PHJKsJ
I'D '" kD [CSt-S - Ps,CslKDl
10-74
To insure that PH2 does not appear-in the rate expression assume that the surface reaction
controls:
I~== 0
CE-S::: KA
: == 0
Cst-s
Cs PE
=C~~~f.
Then: CT Cs + CE-S + Cst-s :: Cs [ 1 + KA PE +
And ···rs =: ks
[e
E'S -
~~}
Cst-s PHll ::; ks KA Cs [FE _ PSt PH! J
Ks J
KAKS Ko]
~ith Ks »> 1, reaction can be considered ineversible, and its dependence on Pm drops.
kK",PE
··rs :::;; ---..-------.. . where k;: ks Cr
1;. K. P~ + EsJ..
out:
.-.
e
KD
Evaluation for k_ KA and Ko
Rearranging the nue expression: P e . "'" _1-. -+ fs.
-rs
kKA
k
+_..f~....
KokKA
with PS l ---1> 0, a oiot of ~. vs . PE should be linear with slope
.
"'5
A
only two points are given with Ps l ;::: 0, it is easy
1
I
C-'·~21.~·
~\
( ··rsJl
==
k KA
{1.}
and interceot (..•.L. . ).
k · .kK
1.00
O:Q.!=~"-·-·
-
[0
solve analytically:
L415 x 10 3 ._+-_7.:;:..06;;;..;...·! ..
0.214 x 10 3
46.73
.J
_.J
+ l (P-)l and f!:E..\ == ....L_ +.1 (PE'"
k e
\rsh k K." k I..
from aoove: 1..;:
k
'P>:~
'P~)
{~._.jrs~L:;;:
±6~.z'
. 7iliJ."Z : :; 666 7 ~~~~~~!~~
(PEh" {PEIl
0.01 1.0
.
gmole
k :;;: 1.50 x 10..3. _.~?l:_
gm:;at-mm
10-75
Since
1 _.. == (PlZ
- - - - "" 40 _a..Wl-g:mcat-min
ana,_.-;-..
- I\ - 1. (P)
E 1 "" '706 '7 .. bOO.1
..- - : - - _ .
-rs h
k KA
k
gmo!e
KA ::; 16.64 atm· l
Now, taking data points 1 and 3, in which PE is constant. the value ofKo k-KA a.ndhence
KD can be dete:n:ni.ned:
(~)1 =k'k"'+~(Pdl
+J~~t~-A
A
Forpoim L:
'-'-I)
Forpoim3: (PE) = .. L ... +1.(PEL
"IS 3 k KA
k
JJ
+j~~tn ....
Ko k KA
Substracting, and soiving for K~'h(~' noring mat (PE); "" (PEn
:~~l . Us),]
lo.166J~·i03 "i'4;51~'lo~31 cat ....
(PEB ((
1.0
gIn
lin
k K ... == ··-·(Ps-;;;~·lp·~;r~· =··················i~o·::·(rO··----··--~;jf-
KD
=:
0.0075.34
.'. Rate expression is
(a)
(~vf\.V)E
0-
= 106..1 ---£.......
gmole
..
1 grnole H..,O
Inen 8) = ._.: ---~.,--- . = 0..2
:;, grnole E
YEo=···_··L.
=0.833
; P""Po =OA15
1 + 8,
.'. <:: = VE {} (5 = 0. 833
. .
~ R ....
PE "" C·E
1
= F:=RT
---"" ....__. =
1.)
F, RT
PSt"" CstRT "" -~=u-'--
.
,
FSl( =
; XF=O . 60
; 15=2·1=1
~222.~~{.~ kg m::!)' "" 19.2
day
10.4 . 1 kg
kg ~~!. =
day
FE.COX)RT
CE.{}RT(1-X}
YE...oP,,{l-X)
...... _ .._ ......-.... -.--...• "" ........
-.~ ............--............. ::::; •.. __..._,. .•............
1.>0{ I +EX}
1 +EX
1+€X
"" ·1)Ji:;€.-x)"·- '1+£X
FE-o X RT ., YE.oPo X
10-76
; T=903"K
F~ eXt'
t.
Rate expression for a.."lY X is:
0.025
[YE.~~X}]
~I"E ='''''--6"64" l~YE.oPJ 1-X'~1l';""=1""'32-'.7:""-'YE.-G-p--o""'X
1 + 1.
- 1 ... eX j + -"," 'r'+ EX
Design expression for a C'.5TR (fluidized bed) is W
=F~~F
w """ ~.5:Q_x..l..r,,,,,,,.!,,..:eXF,__ + 16.64 + p2.7 XF 1
0,,025 lYE.a Po (I,,,XF)
l-XF
j
:: _FS tt ~E,j",,,,,!,,.:tE:XE__ + 16,64 + 132.7]<F.1
0.025 XF lYE.O Po (I-XF)
l-XF j
_ 19,2x103 gmoi
- -,- daY-"·_'"
W
1 day i 1 +- 0.833 (OA5)
,
A
132.7 (0.45)1
1 day
x 0.025 iO~833 (0.415)(1~-:-45)''' 16,,64 + ---i:~45'--j x 24x~
= 7,,06xl<r g
Cost"" 70.6 kg x Sk~' "" $T77
::>
(b)
Plug flow reac.O!. expression for YE.O. FSr;> PEt PS~ and ·rE are the same as for the
CSTR, Tbe design equation is:
w =- __~St/._,.. !-,...._.L_., (-EX - [l+E]lnO-X]) + 10.64X +
00.5 X;: \YE.O Po
W=-O""O·~:t'x__ "I:l_p_(.€XF'[l+eJll1[l"XF])'
,-)
F \Y::..O
132.7 (-X·, in (l"X)Jq;
f
16.64XF + 132.7[XF+ III (l-XF)]}
0
w =1?:2.:.~~: ~~?,~. x.."" L"""".",,,,,, ,gm:~min
0,025 x 0.45
day
gInol
1-.833(.45) +- 1.833 In (1-A5) _ _
r"
_ r
_ I
."l~.",..
""--"~08::;::;-YO-~""'-"""'''''''' 11.:l.64(O .... )}· b2"J l.o,45 + in (10.4))]1 x 24 *60 .-'
l ':UA .4'),
,
mm
1
W
=2.71 x
1()4
Cost:: 27.1 x Sk:11 "", .5298
g
CDPIO-G
Given: reduction of CO with hydrogen over Ni catalysr:
10-77
Kineric given by:
0.0183
ClL.)
r ::;; - - -PH~
- "Peo
---- (gmOle
-------,-..
-,-,
1 + 1,5 PH l
gIn car-min
From the above expression:
1) The appearance orF2 in the denominator suggest that H2 is adsorbed on the
surface.
2) T.ile fact ll:at CO does not appear in the der.ominator suggests that it. reacr.s as a gas
phase species"
3) 1be square f(x)t. dependence in the numerator suggests that Hz splits.
Therefore, suggested mechanism is:
kl
III
+
S
t-
S
-'J<
f"··
Ih'S
Kl
• H2'S
Ii·S
_..:;
2H·S
(0-
+
(X)
CHQ..S + H·S
k,
;!.
14
... --..
" ~
CHO-S
C·S + H20 + S
ks
2Hz
C·S
".',"
,
eXit -+ S
f',
f5 '" K5 l c.c-s
~-
p>
i-h -,- Pen, CsiK sl
Remarks made in Chapter 5 of the text suggest thar reactlon 3 is rare cCEU'oiling:
CT::; C 5 -t, C:H,S+ Ce,g + CCHOS
-, -_ Cs
-, 'I',1 ' Kl
• p}tz -,.. vKr
---7~,
Lr
K: Pi'£, -t
l
"1"-
.
PCE."
1
iKl K2 J
POL PlhO
"'''''-''-:;'''''>' . , . , - - - - - - - - , - - - - - ,
Ks Pfh
pit ~ Ks
10-78
Kinetic expression becomes
Reaction is irreversible .'. let Ks _.+
QQ
Then
with K1 PH~» "K 1 Kz Pit, we get
..,r3 "".,k;-
fK11G" CT
Peo·
' "Impbes that. at ml.l'l..\t;l<l.te pressures, most or~the
__Pre
.m__
. This expt-esslon
_...l_
~_,
1 + Kl PHl
active sites are occupied by the Hz mole...."Ules.
(a)
Design for a plug flow reactOr. Denote A[=] CO, B[=J Hz; then YAD. =0,25. YBO ;::
,
~
"
F·
.
0.75. as := pE:a!2.;::; 'tyJ~Q. =:: 3, XF = 0.80
AU
AD
F '0' Xc
,'>.
F AO
:=
=;
F("ll
."<4
=:
2000 J1t". x ~ "" 125 lbrnol.
day
16 Ib
day
ill.l.bmQl.;:: 156.25l.1mml.
day
0.8 day
DeSiEIl
couation
_.
F,o
dX :; -L'~ dW or ~
}l{". =:
n
f"U" dx.
~
o
___ 0.0183 FA plf
1+L5PB
-'A - - - - - . - ' . -..-.
P A :;; CART == .~ AU ~!.:.o ·X)
;:;:)1 Ai)
1+t:.X
PH. (1- X)
1+E.X
Pa "" CB RT = CAO R!J.~!!_:..,~~'!';::; fAO!!J~....:~~.~J;::;
l+s:X
I+eX
0.0183 [YM
.'. -rA;:::: .-••• " .. "." .....
f
\
~.lA{).!,O (I"X)
l+eX
P.:~!:X.l1f3 YAO P'JJ~,~}r2 0.018313·ri'AO,~2J~~~Jrl2
_._l,_tX.".,,_.·
t
of
fy~o Po (l;?C)1
1 + 1.5
fd:'Ao.~2.i!-Xl1
~
I+eX
' __ ;:: _"",,,,,.,,,,,,,,,__, ,__,,,,-_.J...!:.eX _"~.'",'
EX
1 + 1.5
I + LS[-Y~~~2.E~,~ll
. l+eX 1
.
)
____ ...."_. . ."l--1.±.~_JJ d.X
0.0183
1r[XAO Po (1- Xl 1312j
1 + E;;X
j
10-79
~Vf..... ==
FAo
0.0183 {'J(YAO pop!2
l
a~
1[L:LQ·25l?:J.312 +- 375 ri -0." 5X ]1I2 \ dX
\t
0
1·· X
L I··· X .
J
J
Tt:: integral can be solved using Si.'11pson's rule
Then
f.~ f{X) d.X ::; ~(f{O)
+ 6i(O.2) + 2f{OA) + 4f(O . 6) + f{O.8)]
3
o
X I loX T1-O.25X 11.:Q~25X 11.q~Z;X + 3.75 (l;:q:~x:ri2
o I 1 I
-.2:2 I 0.8 ,"
'lf4·To.·6 I
~-lj. .-~..0.8
(0)
0.2
1
I
1
••••••
0
1
0.95 I 1.1875
0.9
·500
0.85 T}·125 ,I ........
0.8Q.. _ 4.000
_....
m
4.150
1.000
4.750
4.974 ~.......
L090
5.420
5.250 ..._==~=·:::I':":.:2~~2'::'5·.-_-+--::6';:".4";;;3';:'0-+--=-"";'-';';;;;"';';":;'
5.875 -r-...._..!.~45=.~8;;;;.5,,:;,64.;,,+~-+;;.,;.;.;;;:;.;..!
7.950 ........1
2.000--1LS.5OO
Design of a CSTR
F eXt'
Desum cauanan
IS W:;:; .~.:....
..
-fA
••
' .
<f
~
.
G'"'l L-;J,'.
"un
\F
_ 1'1171 .l. h.
£.lQl
1;."
•• ,
rv -- .I, _.oJ ~
~,..
. X·
x
C.w.V
£!IIlOi
..
~
fOO
-454
p'mol
.;:,11 !
".--.~~.--
... X
,\ .
cav
...,---.....,.".;:. ....... __ ..,.
Ibmoi'.AX;::'O·
L ... v mm
w == 3900 g
CDPIO-H
10-80
(a)
.
-r~
-r;
I
/
I
0
'A
Runs 1. 3. 6, 7
. . -rts =
.
-r•
~.
Fa
Runs 4 . .5
Runs 1.2.4
t
I
. -. r. "" . . . . . . . -.-.-.---~
1+ KcP c +· ....
P", ..-...-.... ..
1t· KA PA+-· ... •
,.--.;'''"'"-"".~""~
~
~
(b) Numerator: P A and Pa
Denominator: P A and Pc
Power of Denominator: 1
(c) Proposed Rate Law:
(d) To find the rate law parameters., reauange the rate law so
that it can be plotted as a Hne with the rate law parameters
as the slope and intercept of the line.
First, hold Pc constant and plot
!'~~.!!. vs . Pi\
···r s
From the plot
K
Slope:: ." .....~"
k
=:
5.2,
l+K P
y.- Intercept =----.!:.-£. =: 3..59
k
10-81
Pa...~nett!r
Evaluation v.rith
Pc=const.ant=2 atJ:n
120
25
100
20
Pa..rameter Evaluation 'W'ith
P;;t:;const..ant=l atIn
.-.--.. --.--.-.--------------.-..- ..........--........-
gO
'"
::::
~
60
::~1
R..... 2
.=
J
'9 999'9'~E -I
-'i=--·--.,.-·----·-..'..·-·-T-......,..........·-,....---r-....,."""",-1.
0
4
6
12
$
14
o
.
"......--1"........... /" ...
o
16 18 20
1
Pa
Second, hold P", constant and plot
2-
·'!"T-,-.,-,.-··.....'f..,~~-~·~l"'f"'-r-r
:.> 4
5
"
7
8
9 10
Pc
_~,\PB. VS. Pc
-.{ s
from plot below,
1 -'~'J'"
. -'-.. =ntJ~.J.
51 ope = .K,-
.
k
v· ..· Imerc"'Pt :::: .1__....+K
__. . .~...P_.::. ~ )- '''i)k
- . -".
J
Using the four equations above
we get:
"'.
to
soive for k, K A, and Ke,
mol
k :;;:: 2.60 . --·-·-----·-
gear· sec" atrn-
(e) A and C are adsorbed on the surface of the catalyst
(f)
Proposed Mechanism:
c·s=c+s
The irrcver'sible reaction step was assumed to be the limitL'1g
step We check this mechanism and rate limiting step by
rearranging and combining the rate laws for each step. If the
mechanism is correct, we will obtain the rate law proposed in
pane
10-82
k5
is much
smaller than kA and
·
,
.
:::.r A. == .:r
~
k.
1/ -- therefore
~/k
A
=
0
C
··
(~"v:;: . _ - " " -C"r - - - .A f ter 511 b stltullon,
1 + K... P A + Kc Pc
Next, substitute the above equations into the reactions step to get
Because this is the same as the rate law in part c, the rate law and
limiting step assumptions have been verified.
(g) Ratio of sites of A to sites of B at 80% conversion:
Conversion at whid\ the Hnmber of sites of A equal the number
of sites oi C:
X=(J76
CDPIO-I
10-83
a) To detemline the mechanism and rate-limiting step we must come up with the
rate law. Looking at the rate dependence of A we See that between runs 1 and
2. P A increases fTom 1 to 1000 attn while the rate law only increases from 1 to
1.5. This tells us that as A gets larger it changes the rate law a good deal less .
This tells us that A is both in the numerator and the denominator.
.... f
' A
..~1 .
l+K A PA
Looking at the rate dependence of B we see that between runs 1 and 3, PB
incleases fmm 1 to 4.5. This teUs us that the rate law is directly related to B .
..... r~ .- PB
Looking at the rate dependence of C we see that between runs 7 and 9 Pc
increases from () to 4 arm and the rate increases from 4.5 to 4.8. Also in these
luns we see that P A increases from 1 to 4 atm. So one of two things is Due either
Pc is both in the numerator and denominator or just in the denominator. Since C
is a product it will not be in the numerator in an ineversible reactkm.
""
···r I ,................................
.
A
1+ K c P:c
So the rate law becomes
With that rate law the following mechanisrn exists:
Adsmption
A + S'-1- A S
Snrface reaction
A S+ B(g»
C· S
Dissociation
C ,S"'7 C+S
The surface reaction is the rate· limiting step.
10-84
b) In evaluating the parameters we can also see if our rate law is a good one.
Plugging into POLYMATH we can come up with the parameters.
~
r
.. aoo
r-ate llH0
4
"t.eoc
t
LoSee
O.soo
Model:
k 2
Ka
! !
r,
2.400
~
n
1
'r
3.200
n
I
I1
2
I
..,L . .
5
I
,I
II
t
!i
~
:3
~
I'
I I
n
I.
-ltJ· . . . ".u. . . . . .'ij,. . . . ·. ' . ·r
~
II
lL.
!
I
d.ata
[J C"I "'" at!>d
y.all.u?
I
~
iI I,
··..If·· ..
0;
~
o QeqrEl'SS~01""I
7
a
rate=k~Cd.Pb/(l+Ka~~a~Kc.P~)
O.. GOO·H'il'ilS7
K" = 0.:;
= 2.9SSaS
We find that the !ate law is a good one.
k:::: 0'(>0045
K A =3
Kc= ..5
c) The best places to add points would be where Pc is changed, but P A and PB are
not changed.
d) No solution will be given.
CDPIO-J
:2 C2 Hs OH? C2 H5 0 C2 Us + rhO
(A)
(E)
(W)
A+S
+:!
A·S
A-S + A-S ~ E·S + W·S
r2
0:::;
I"4
w+S
At steady srat.e r "" - ~~6.
= kl (pA Cs- CA-slKt1
k2 [G,.$ - CE'S Cw.slKzl
f3 :::: k3 [CE-S . . PE CsJK3J
E-S :;:: E +S
W·S ~!.
II
=
r == rt
=:
2r2
= k4. [Cw.s .. Pw CsJK4]
= 2r3 = 21'4
If swface reaction is controlling,
10-85
~:.== 0
=
CA-S:; Kl
:'" == 0
C;::.S :::: !'.ECs.
1<:2
== 0
I4
k4
P A Cs
~
K3
CWo:)
:::;:;!:w,. C;;i.
Kl.
c·'r == c··;.·
+ ("·'·w·s -.,s ' C.
,{os
.
r'
·'1
C's,,i 1T' Kr 1 } ,A + .---;:-.
Fe. . t· --.-..
Pw J
'
C:;
R]
K~.
=---___._c;.r.____.
1 +- Kl PA +!I{ + .1.1:.
K, K4
')k
..... r ~ 2r2, ::
I.... z p2
C~
111 +- K I PA
.
where k :::: 2k z Ct
Keq:;;
K1
PE
pwell
r,
P- Pwl
A·T klP.~ - ~-':
·····r""·"~-'-·-··"'--·--·······'·.. "···'·~· .. ·-::;;;,~ ........~~·"--:;-..::. ;::;: :::~~... . ., "' .........::..,., .....,........ ".__......._ ... "..:.~_ .. __.
- 2L l.\.l
K,-K;K:-i
.; JE
K3
+
~w 1~
K4J
I '1 .,- K··. P ,PE -+ Pw 12
."
ATK3 . K~'j
Kl
K2 K3 K4
Using points 9 and 13. PE "" 0 , Pw:: 0
... r == __...... ~!:1._
{l + Kl P,S:
°ll =:it i; +~£
r''''''''''
A
I ' fl
1
(,...
K
p at or "V r v'S . p-;..... produces a straight line "'1m sione 1 1,. and inteI"Ce')t __.L
A
• 1/ k
l:
k
slope
={f
4.945
... k "" O.04()9
interCept =:~. =: 85.59
•. K1 := 17 ..31
Using point
H),
=:
.•
P "" ()
~=0.0399
klleck
U sing point 11. PE "" 0
=
K4
=0.0368, dose enough
Using point 6, Pw "'" 0 = K3 "'" 0.659
Finally, using point 12
= K.,q "" 0.0975
__.J~i_..~g:~?§!'~_'!'~t
r=··. ....... 9._:.0_409
) + 17.31 PA t· L517 PE ;. 25.05 PwF
.........._.
Note: Keq may aiS{) be calculated using RT In K.,q "" ···ClCio.. Interested readers are
encomaged to check the goodness of fit of this rare law with the data .
---.-------------------------~--------------------------
10-86
--------.---------------
CDPIO-K
CDPIO-L
Rate law:
-rAe;::
kCcoC ACCNaOH
2
(l+KAcCAd
Proposed Mechanism:
Pd + COi--'Pd· CO
-,-.-.~
Pd· CO + NaOn--Pd· CO" NaOH
--'''''''-7
AC+Pd
-.~
AC·Pd
AC· Pd + Pd ,. CO· NaOH""--7C3H sCOOH + NaCl + 2Pd
Neither of the first two reactions can be limiting because they are reversible .
The rate step must be irreversible because there is no subtraction function in
the numerator.
We will first try the third equation as the ratelimit:ing step:
fA::::: kCYAC
Then
lACO _
""'-·-0
kACO
,~ACO:1':!aOU_ ;:: 0
k NaOH
Combine to find C"
.. " ...,_. C{ ...""."_.... "...... ,... ~_
1+CcoKco + KCOKNa()HCCOCNAOH
This is definitely not what is supposed to be on the bottom of the rate law so
reaction 3 can not be rate-limiting.
Cv
;::..__
Trying reaction 4:
r~ == k4CPdoACCPd<OoNaOH
CpeMe cannot be found so therefore this rate law is also not rate,·Umiting so
none of the mechanisms reactions agree with the rate law.
10-87
CDPIO-M
a) Start \!,lith a mole bahmce:
dX
--.::;;;;
Rare law comes next:
····r'A ::::
kr a
Then the decay law:
do
.. :;;; ---k,
dt
"
-YV'
t
::= .....
Us
da
dVV
Us
'We then come up with the equation
f()['
the profit:
Vvl1cre:
Then plug into POLThlATH and get the foilo\;villg program. The feed rate of
solids that gives a maximum profit is 4 kg/min .
•
~:!!S!-9.£§.'
~~....Y~1:::!::
d(a) Id(w) ,;;·}Cd/Us
1
d(x)/d(w);'ra/fao
0
fao",1
kd=2
Us=4
10 ··18 :).
;;.q~~l~~ . y.~i~'1£ ~~::nu..~" ..:5!:~~ !1i:1itt.UI~.-'y'alue ~_~~.L . Y~c;~~
l
fb=fao'x
P=1.601ffb~1.0~Us
a
x
0
';
75
c
;;
0 73
fao
l<d
-
2
2
1
:::-:1
··c 5
£1::
0 i5
?
4G
b) As seen above: X:::: 0.75 and a =.5
10-88
OiJ
-
"-'...
"
O. 75
"·40
SO
c) The only equation thai changes is the rate law:
.~~ --·k
dt -
d
W,······W
I =::_.M:1!.
__ .
U
··dW
dt =:: -_._-----
U
da
kddW
. . ".. == ._.-.
. . ". ,. . .dW
U
-~
~.
Integrating we get this:
W::::; W MAX @ a == 1
k ==
l-,~!l~
U
a == 1- .
~£!{~Y.u.-\X_.:~J
U
In~~~al.'.:'5.1u~
~t;.i2r:::!;,
o
d(x)/d{w)- ·ra/fao
kr;;:;5
kd=2
Us=.8
wmax"'l
fbaif{x<11 thenlfao·x) eLse(l)
0.",1.£ (kd/Us* (w-m<Lx'·w) <1) ::hen (lMkd/Us" ('HmilXW) ) else (0)
P"'150*iblO"Us
o
1)
G 8
1
'·3
ra
:42
-·0
·3
142
·s
\Ve find U:= 0.8 to ma.xirnize the profit X =:: I and a:: 0 exiting the reactor.
CDPIO·N
10-89
Design Equation:
~<\ssume at t
= 0,
\V
="~~ . \O~_
For n en order kinetics .
(-J:,\ )a(t)
a(O) "" 1 and X "" I.
Now
FB(t)::::: FAoX(t);::: RON(t)
X(t) == _~O~Q '" _~:2~.~!2.
106
RON{D)
Second order kinetics and second order decay rate tIt the data very welL
J
l+k .t:::: . . . . .kC~
. . . . L __
d
CAo-,C A
y == 0003x + 0.0855
Plot of -
- A____ vs t:
CA
C.A(J·-
-1.lnca; lExp
Data;
i
i
() t "..........". .......~"" ................... ,._"~ .. _- ....... ..,...". .... """.".. ,__....._., _ _".. ,_, ....-.J
o
100
200
300
~OO
500
t (h)
F.om the graph:
intercept
slope
(b)
= 1j k =: 0.0835
~
k ::::: 11.98
= kd J k "" 0003
Activation Energies both for rate constant and decay constant can be estimated from the
temperatu,re·time trajectory
CDPIO-O
10-90
Given
A
-+ R + S
Batch constant volume I"Cactor, P mCI""....ases with time
NAO dX =a fa Wdt
Assume
a = e..(lt
'0::;;
kPA
:. PA=CART
C A ;::: NAG 0··2.9.;::: .~AO (I
V
Va
lC
o
dX.:;:; W k
eM (i-X)
~AO RT J. ea.: dt:;::;; (W.) k RT J.' eat dt
•
NAo
I-X
· ~ Le~
=:
=We..(l' k CAO (1~X) RT dt
:. NAO <.LX
f.
·;.;.t
dt wl=e
Vo '
0
0
~ =(~-)k R1
.In(l.X) == ~ (le'<Xt),
a
Assume that there is very little deactivation in !he first. 10 sec
-In (i-X) =:
~.. (at) = ~t
Ct.
(for smail t)
~:;:; ~ln ( .. lJ :lln{--L-) '" 3. T7
t
l-XI
10
1-0. 037
X
10.3 sec
~=(~)kRT = k=dRT=i-~~%s-~~ioi
k = 6.63 x 10 5 sec 1
10-91
At t relatively large, e-<ll: = 0
~ -In (I-X)
-=
a
~
a:= ~ (l>o~~l ==
"
3.37
~'(1
~.
10.3
-Ts;J =
X
5.18 x 10'" sect
,1a(1-X) "" 0.128 (i"-e-S.1ax1 o-'1t)
" TIle assUt~rprion of a firs;: order :e2.Ction kinetics and a
justified
CDPIO-P
cyc1opem:ane
<
fl·
pemane
coke
Batch mole balam:."e :
AsStL."Ile q :::;;; 2
a = ..............L.". ._
10-92
order decay kinetics is
,
_I =_I_+5!t
Ifn=O
X
k-c
let
#
--1._ ;:;;: -L +.5t t
·In(l-X) lei kt'
Ifn::::l
X(%)
o
75
7 .7
20
1.33
1.414
~o. . . ,_. ;,,6 7.
_ ·
F
ou
112800
!
v
"'1
250
350
500
11 ..46~33
05
..,
---S--4.)8- '.~/· · ·"· ·
.__~/•. 80:,~j
_
'-
i
I
42
I
361
2.381
2.778
3.333
r-'
2i ',-"-'4.545
I
30
0.721!
0.333
0.815
I
OA15
0.902
II . ~;:~~--.J
1.076
O.653~1
._-_.
__
._.
I 1.245.__L_.Q;~E
1.498
I
L053
1.836
I . . 1.381
2.24T·--,
I
1.177
2.804
. . 2.333 --'
I
I '4~023"'-'3.545 - ,
···'1200··T-~. ! 6.3 _ ....L··6.TITI. ~. ·_"';'·); . .~;':;_.; .___?'6'; ;; 'O_·-_-_-+.;...-_-_··_5_·.1_3_5_:=1---,
1--';'800--
",,,l.
I
+lfX
<> -llln(l-X)
II1II (l-X)/X
f .
o
L.... ....~.~ ..... ~,~ .."'._.. "..... ~ ...._ ......... _ •.• ".. '"M_._ ......,
o
~
tOO
.' . -~....--.
~
ICQ
.--...
1=
From the above graph. a111ines are straight lines.. 'lberefore. q
= 2 is a good assumption.
We need to ex.amine the data to see which value of n having ~:::: constant n=:O and n "" 2
At
will have similar behavior of ~ because:
At
8.= 1.._ 1
X
X
n == 1 . Av.:;;.1 f.--L--l and.1t;;::"O min
l-In( l-X}J
'.
...
~ :;;; 0.094; 0.087; 0.087; 0.084;
At
0.081 ;
........
. 0.0797
lit
Therefore. ay is deo:easing gradually. It is not a constant.
At
•
10--93
For n
=:;
0, n.:::2: ~r
:::: 4 x
6,
10-3 = constant.
• If n::::{) :
_
_ QX, == !<::B_~~Q,
{lX)2
at
(l,Xr "ili.." ::: •... ,. L . . . . ;- --~Q.,.,.. t
.
dX
KR C AO
KR
c.-\o
n=2: Slope is negative. It is u.nr:::asonable
n:::{): ~. vs.
t
is a straight line.
10-94
<> dtldX
w =0.01 kg/m]
C W
't
(0.03
krlifl){0'<)1 kg)
k
= 22.-. '" ~_ ..................m... . . . :.......... .. _m 3 ... "" 20 _JL min
(b)
{m3 )2
1.5 x 10 5 kmol/min
FAO
The order of decay is q;;2
,
k
...Q. ""
kA
(c)
4x
1(},3 =:>
k: 3 x 10 3 mini
Moving bed reac~.or: F AO =2.~&!. ; X;;: 0.80
mm
*
lia. :;: k'
d1:
d
a2
In moving bed: t
.,
=if where
.
Subsnrure mea equanon (1): ..
u ==
=
~~
dW'
" = k.:!"
'1:1' a'"
. . ~.::;kd·dW
a2
u
10-95
SUDSritllIC in~o
equation (2), vie have:
W :::::455 kg
(d)
If
u
O - k~
= ) .---::".-
ttl!.(l
w ==48 . 6 kg
CDPIO-Q
10-96
a) Mole balance:
Rate law:
Decay law:
Stoichiometry:
CA = CAtI (1" X)
CD = (~O(e8 + X)
Evaluate the parameters:
d(x)/d(w)-.-'-ra)/fao
d (ia) Id (w) ",·J<d"cb/U
o
1.
kd,:6,
faOi,:20
:'
~
U;:8
k';350
1.. 000
f
cao=.l
ca"'cao"(1.·x)
, cb:cao * (thet.a +x)
r·a"·. ·k"ca ·CD
Wo '" o.
10-97
-, .. ,
b) The only change is in the stoichiomeuy:
POL'{;v1ATIl
Initial
d(xl Id(w) =a* (--ra) I£ao
o
d(a) /d(w) =··kd*co/U
1
v~~!:
kd=6
fao=20
!
~oo
U=S
i
T
K=3S0
cao::tl:"l,
C "'0..'1
1
J...
theta::: 1.
alp= . Q38
ca=cao' (1"",),
(l"alp~w)'·.
5
cb=cao* (theta+x)' (l--alp'w)
~O
.. 5
ra="'k*ca*cb
Wo = 0,
wf
= 24
CDPIO-R
Curoene
(A)
--,
+-
Propylene
(k)
Benz.ene
(S)
a:s .I. .
fo
(a)
We find a relationshipbetv.tcen a and C A
F X F
r '" -.~W = 2W . = F"-,
(W:= 19)
.R.ill:Ll.:
t
o
60
120
180
a
1
0,,594
0.491
CA
o
07.5
0.01
(LOIS
0.0243
Run 2:
PA
:::::
0.4
aL."'Il
t
0
100
200
300
400
a
1
0.833
0.733
0.583
CA
0
0.0057
0.0106
0.65
0.0148
10-98
0.0184
Plot of In a vs. C A gives a straight line passing thmugh the origin wirh slope a
(single ~te adsorbed. smface reaction controllingl
initiai
I
rate are used, PR == P s = 0 and 1 » KAPA (adsolption is small at high temper.rture), then
fo"" kt'KAPA·
Using data at time zao from runs 1 and 2 : k~ K.<\
=:
3.2x1O-3
Hence, overall apparent rare law is
r
= k~ KA e<t.c... P A
; k; KA "" 3.2x1O· 3
ex.::; 28.9
: : : - k:..:i amf{P A. PR, Ps) ==
-~ am P"A
Trym::::: -2
Since PAis almost constant. d~l.ring nms 1 and 2 (low conversion)
Run 1: ku PA. :;;; 5.767 x 10-3
Run 2: k<J P A = 1.769 x 10. 3
P A == i in nm 1 :::)
~:;::
PA ;;:; OA ill run 2 :::)
5.767 x 10.3
II ""
L29
.
Kl KA "'" 32xlO·'
~::
11 ::::;
5.767xlO:;
1.29
t L.'1 minutes
10-99
(c) Overall conversion = 0.60
(2.8 rr.oll:Jec cu...--nene j
\ 4.2 mollsec propene) L. "" 11.2 IT'.ol sec
1.4.2 IT'.o.llsec ber-.2e ne J
Cu:rnt!ne "'--t Propene -+ Benzene
(A)
(R)
(S)
COllIDosition at reactor oudet
.. YA: 2 &. =O?S
11.2'-A
?
VR = Vs ""..::!.;;.""-."
..
-
11.2
""
0 315
"
Composition at reactor inlet. Fa :::: 7 + 3 :.= 10 mol/sec
FAo ~ 1 , 3(0.25) "" 1.75
YA "'" 0.775
FRO::::: 3(03'15)
:=
Ll25
YR ;: 0.112.5
Fso
:=
LIZ5
Ys = 0 . 1115
=:;
3(0375)
Let Xf "" converS1on per pass
~rotal
flovi at :r~aC7or outlet before the recycle sc"ea.,s is:
At a.llY point along the reactor:
Assurne that rate law in (b) is srill good fOr the mO\o'ing tx:d operation (may not be rIue in
practice because of the high conversion)
I'::::;
• ~..
k, KA PA
a k, K.,., P A=:' .. .l..."•..•.•_-_..
,
..
Moving bed react()r:
l+k.:!?A,t
t """
l~;
F.,odX =rd W
lO~lOO
== ?_:15x5.7~~10-=~ (1,29+X.\O.Z9 W + _....1..7~. _._( 1.22+X.)
3.2xlO-'x2000
I-X J
3.2xlO-3
I-X
= 6.98 X 10.. 3 (1.2.2±Xj29 W + 2421 (122±X.)
I-X
With X::O
>
IX
W=o
If X:: 0.542. using digital computer W :: 3.1 kg
CDPIO-S
a) i'vlole balance:
dX
dW
a * -r'
:::::: .................• .::! .
rate law:
,
~r:4
::;: k'('',t
St.oichiometry:
Decay law:
da : : : ._kd. _. a
._............
dW Us
Evaluate the parameters:
?.;:;;; (1 . . o:wt5
Po
5~ "'" (1--1OOa)05
0:::::
0.0099
rOl da
kd dOO
'-'Jl -~- :::: 'i~!;' JodW
!:i..;:;;; 0.023
Us
10-101
POLYMATH
EguB!:ions-,.
d(x)/d(w)='~B'a/fao
o
d(a) Id(wj ;-·kd*a
I
kd" 023
Yi:~X~2!~
fao:=:4
w
k=.09
x
cao::o2
a
a2.p= 0099
kc
ca=cao*(I·x)*(1-alp·w)A.5
tao
"
0
!<:
:t'a;;;-·:t',llca
:.c.':'tial val.l:.f! ~~,..:£,elue Minirnu:r:. ·,ta,l>.!.e £l!~h-y~lue
l{W
100
{}
0 .75:;':51
C ,57161
0 :00251
C :)0259
D23
0 .023
0 023
C 023
~) 9
0 09
C 09
0 0099
0.,0099
C 0099
C 0099
0.04856"18
.. 0 004371.1
0 04.35678
-0 13
Cd
ra
{L1S
b) POLYMATH
J J2:)
:-
ceo
CDPIO-T
Design Equation:
FAn -1~
RaleLaw:
< ;;;:k'C
Decay Law:
(for sintering)
~ a(WX·· r~)
A
Stoichiometry:
Combine:
dX. = _... ___ L .... ~:(l' X)
dW
dX
.=.,k'
l+!'AWu
Us
0
l{Ilx)~ ~:~;ln(l+ tt w J
1
1
From the problem statement a - ::::; ................. :.:::
'e.u,
1+1:< W
4
Us
Plugging in 100 kg fIx W, we can solve
C 0'1
2
CdO
fort.:!
$
1
10-102
! ..
X
:=
3.43
dW
'<Ii 00437:1
CDPIO-U
The heat of activation is given in the problem as a function of the carbon
number so we can just graph that:
Heat of adsorption vs
number
carbon
c: 80
~ 70
1:
o
60
50 .
40
~
nI
'030
_ 20
<IS
~ 10
'r"
0
............
'H ...... _ _ """ . . . . . .
o
10
........ ,'
...........
, . , ' " ' ' . . . - - - - " . , . - - .- - _. . _ . _ . . . . . . . . . . .
40
30
20
Carbon number
To graph the activation energy we need to find its equation.
Arrhenius equation.
It is the
k:;;:Ae- E/RT
Solving for E we get:
We know that as the temperature increases when n S; 15, the rate increases so
k still gets larger with greater temperature so E is still positive. When the
temperature increases when n > 15, the rate decreases so k decreases making E
negative So we can come up with some equation with the above equation
that fits this criteria and we can corne up with the following graph
Activation energy vs carbon
number
6
>-
....• _ .•_--........ -._.-......- ... _-_..........................._ ......................... ..
5
0'1
:;;
c
.
4
3
o ---- ---.----~~
.
.. . . . . .
20-·-·-----·3"e-
10
4
1 . -------.. -----....-.------.-.... ---.. --......- -... --....- ..- -..- - -.....•-.----
Carbon number
The reason for this unusual temperature dependence is due to the fact that
the higher the carbon number the less it wants to add another carbon
10-103
CDPIO-V
(a)
a::: 0 at the end of the reac[Qr:
Us
:;;:?
(b)
= 1 kg/s
For Us::: 0.5 kg/s:
a
~ 1 il~ W ~ 1 {nOoN;;} ;1
(04kg')w
Catalyst Activity vs. Catalyst Weight
W (kg)
When a::: 0, the catalyst is inactive. In theory a can be negative, but in reality, once the
catalyst is inactive there can be no further decrease in activity_
(c)
For a catalyst feed rate of Us =0..5 kgls:
dX:;;:: a( . . r')
Mole Balance:
F
Rate Law ;
. r,
Decay Law ;
dW
Ao
A
..... ~~.. =.~:P. =g~~.~. . = 0.4 kg!
l.
dW
Stoichiometry:
A
k(·' (.,
:;;:: '"'A'S
Us
:;;:?
a == 1·· OAW
0.5 kgls
(Assume T = To, P z Po' and 1) "" 1) Q)
C\
=C s ==(~AoCl·X)
Combine:
FAG .~= (l.'.O.4W;{kC~,,(l . ." X)"]
(k6t}(i~)l+ t~w}w ~ (~Md~~~(W-2itw'l
10··104
(l~xr!'i: (w -~ w' J
From part (b) we know that the maximum catalyst weight (the point where a:::: 0) is 25 kg
We will find the conversion at this point:
(i?Xr QX~:i2'.{2.5 -$~:5) 2.5' )= 025
X=O.2
(d)
To achieve 4()% conversion:
(l~Xl =~~:{ w .. ·ittw ' )
f~~~=O{5-fJ;5') =>
lJ s
:=
O.667 = 1
2.5
Us
7.5 kg/s
(e)
ForU'==:~'?xj=!'¥:( W- 2(!jW')
(i·~X) = (0.2)w:::: (0.2X5)= 1
=>
X =0".50
=?
CDPIO-W
N~:=<aW
So dt
Design Equation:
,
k'C'z
-B
RateLaw:
- fll ;;;;
DecayLaw:
da
....• =k =0.05
dt
D
a
I
1
0
Jda =,0.05Jdt
~
a:: I,O.OSt
From tilL'> we can see that the maximum reaction time is 20 min, We will find the conversion at this point
Stoichiometry :
(Assume constant volume)
ell "" ClIo(l···X)= ,~§2(1" x)
V
Combine:
,~?c. =~/(l:~:~:?~)~~Q_::_~Z~ =: ,~~~.!t! (1 ""' 0.05t Xl····, X l
~
V-N 80
V2
_~_:;_ :: ~/~l!!!_ (1 . . O.05t}it ~--~ :: Y!~l!!! (t _O.025t
(1" X
Y
V I " "X
"'~, =~g:g!X!gX12 ('20 . . 0,025(20'f )= 1
I-X
(1)
X=050
10-105
V
2
)
CDPIO-X
a) Mole balances:
.dF
. . . . _. = r
A
dW
A
dF.l!.:; r
___
dFJ.:,. = r
____
dW
dW
B
C
rate 1a'N5:
stoichiometry:
(
,
F..
F4 (1- aW)Q5
V
Vo
Fa
Fs (1-'aW)QS
V
Vo
= ..._ .. "" ...'-., ....... --....._,......
'-'tt
,
CB
"'" ."--,, := ...,....---..,........... "-,-,,.
decay law:
da
dw
k a
U,
"" ,.12"_,,
Evaluate the parameteIs:
Plugging all of tllis into POLYMATH we can change values of Us. T, and
that will give us the most of product B.
Vo
CP... . o)
We find that at a tempcratllIe of 396K, a solids velodty of lOkg/s and <I gas volumetric velocity of
3
o 63'3 dm /s that con esponds to a p3ltial pressure of 5l 3 attn we can the maximum yield CO! B,
d (fa) fd (w) ""a
d(fb) Id (',") =,,10
Y!f,~bl(;!
d(Le) Idlw)=J::'c
;:::'~E.~~:L..y;e t~~:; ~eXit~1f!E_yalue £!~imum :!~;~~
100
d(al/d(w)=kd'a/Us
Us"lO
fa
f;;
T=396
Ec
Q
0
xci",
Os
633
oa'exp(~500011
937* (1/400·111')
J,O
11
:0
196
39G
0 .CO:iB
0 009'8
00093
0 631
0
C.066:':'49
Q
0 065::49
0.00775554
kl=, 02 'exp (lOOOOn 987* (1J400 .., 1 IT) )
kd
0 0661149
cb~fb~
(l·"alp*w)
A
,SIva
xa-;;:'-a*kl *ca
rc~a'*k2'Jrcb
:cb= ··af!'
(k2~ch-kl *ca)
0 '1l19
1 . 9Ji02
396
VQ
(J..···alp""t..t) . . . 5/vo
i:
0 . 4 n,,711.93102
390
k2= OJ..*e;<p{20000/1 ,987* (1/40Q··l/'.l'I)
ca,;!;.i.:a~
Q.0913942
0 ;4t>631
10
o. 009S
n 633
41p
Eb~.~.,?:~±~~
leO
3813942
.;) ,. 52'~43a
alp=9 <lEO",3
Vo=
0
633
0 6JJ
k2
OO,i55S4
o .cp:nSSS4
.00lil149
0 00715554
kl
Olj6Ln.
0,0116131
O,0176i3}
;)
Ol"l6l·.H.
1 579"; S
0.0131646
0
01Bla~S
c""
c".a
1 S'191S
0 631)42
.. 0 0278243
,c
d,
.. 0 0(10620401
.. ·0 .07.::6245
0 00641366
0 02 :82";8
10-106
0 02/'S246
0991~46
0
.;}. ':;OO6:!O403
Q C01,4$'~02
"-0.0021742:3
•. 1)
O;)O~nBi$2:
b) Using the same program we can find what it takes to get the most of C possible.
We find that at T:::; 396K, U,:;;;; 10, and vo:;;;; .03 we can get Imol Cis. P AO::::
1082.4 atm
fb
~~~~ ~~_~al:.!Z Minimu,tl1 v .. l" .. !:ir..al .. ~~;,!!!
0
100
0
lOO
102629 .... 23
1 . G252ge.:n
1.
l
Q
0 SZ4487
a
135042"",:0
~a£±~
'"
fa
fc
()
a
1
1
1 .9370:2
1
Us
10
10
1.0
lO
l'
396
396
J96
390'
0 .. 0096
1
1.93702
<>1.;;>
0 0098
0,03
0 .. 009S
0 0)
0.0098
vo
0 03
0,03
kd
0.05511.49
o .0651149
00;;61149
0.0661149
0.00775554
k2
0.00775554
0 .. 00715554
0 00175554
kl
ca
0.0116131.
33.33J3
0 0176131.
J3 3333
0 .. 01'76131
1. 83'798e .. 23
ch
ra
0
-0.581104
17 .. 26%
.. 1 65051" .. ·24
-0. 5871C4.
-1 6505"113 .. ·14
:c
0
rb
o .5,s'1104
0 .. 13615
05971.04
'·0. C7!$781
9 56332,,-::
·9 55332 .. ":'2
{)
0176131
4 .6379Se-23
6. 36S55e-:.n
c) To get the Time Temperature trajectory we can use the following equation to
create it:
This will give us the following graph:
Temperature~Time
Trajectory
1200
...
1U
.;!
.-........ -.......... - ....... -..-----.--... --..-,,-.-............ -....... .
1000
800'
lU -
d;Q. ::':::.600
,E
......
400
__ _ __ -_._ ....
..................... .......
...-
.......
1U
I-
200-
o ...............-.. . . . . ..
o
5
Time
{s}
to
d) For this we just add an energy balance We have to assume a heat capacity of the
catalyst since none is given Here it is assumed to be 100 J/kgcat
Plugging this into the POLYIvlATH gives the follow'jng program.
We find that the tempemture is 38SK, D, "" 10 kg/s and v 0
62 . 4 atIll.
;;::;
051 dmo/s. P AO
10-107
=
.Eq:..lationS,l,
d(fa)/d('w)=:r:a
1
d (fb) /d (w) ,,:tb
o
d(fc)/d(w)=rc
o
d(a)Jd(w)~kd'a!Us
d(1') Id(w) = (:::i;l~ (-16000) .. rh T (--32000»
kd= 08*exp(15000/1. S87*(lJ4QG-I/T»
1
/ (cp' (Us+'''cl+fbdc) 1
~ia::'l/;J
Us=10
t
c,,=lO()
k2=
388
~.i-;~;._·...alue ~ v.al~~ P~nimumwvaL~~ f~-,"l.~b.~£
100
Leo
G g 1:,155
Ol*exp(20000!1 987*(li400-1/T)}
o
21532&
o
o
61.1.155
t)
,215328
kl'" 02*exp(lOOOQ/l 987* (1}400·,,1/1')}
alp=9,8e" 3
Vo~
40~O!,
T
382 2.:.2
C 0446169
51
cp
cb"fb*(l"alp'w)A Siva
k:::!
ta.:::::-.a,'*kl"'ca
kl
to
:,:<)
lce
10;)
o
{) .C0459207
>10
004Sn{)1
':11J553
dlp
,;;,
0.0332906
10
10
ca;:fa* (l-,alp'*w) . . . S/vo
t;'c=a "<k2 '"'cb
358
,0·098
o
O~';553-
o
.,
-Q
OC29~-!19
O,(H085Z:
0093
0098
31
5:
0 S:
3i5D}S
96{)'8
-0 Q~a':lS~
7;6.3935
:1
o
:;c
:0
:3265'44
10-108
'''0,000'15307;5
0,001'73258
,·00255;.1\4
Q
:)03}'0577
0,0111477
() .. JOSS
C ,$1
0481:;.54
,':694'1'2
-C.00:)753076
o
00)139124
~,
18S:6~-
05
Solutions for Chapter 11 - External Diffusion Effects on
Heterogeneous Reactions
PI I-I Individualized solution
PII-2 (a)
A
2B
1,
Z 0,1'=0
WB =-2WA
cD AB dy A
(1 + YA)
WA = - - - - = -cD ABdin - - (1+ YA) dz
dz
Integrating with YA = 0 at z = ()
WA = .cD AB ·In(l + YA)
(S-z)
(1)
at z = 0 YA = YAO
cDAB (
)
WA =-S-l+YAO
(2)
Taking the ratio of Equation (I) to Equation (2) to eliminate W A and solving for YA
!~(1 + YA) = In(1 + YAO)
S-z
YA =1- (1+ YAO )
5
1-z/0
11-1
,,
,
"
'fA
/
EIVKD
"
", ,
,
Pll-2 (b)
(g)
kc2
kcl
Tl = 300K
T2 = 350K
=(PAB2J2/3(~J1/6(U2J1/2(dplJ1/2
DABl
'\)2
U1
(11-70)
dP2
As a first approximation assume
DAB2 =~
DABl 112
AtTl=300Klll ",,0.883cP
At T2 = 3.50K1l2 "" 0.380cP
Assume density doesn't change that much, '\) = Il
p
~=A=2.32
V2
112
1 d p1
U;-2' d P2
U2
_
_
1
-"2
rr
kl
= 4.61xlO-6 mj s
k,
~ 4.61x 10-<> m/s[2.32 J'" [~ [~
=4.65xlO- 6 mj s
WA = --r; = kc2C Ab =(4.61XlO-6 mj s )(103 mOllm 3 )
-.r; = 0.00465 mollm Is
2
11-2
Pll-2 (c)
(h)
A 50-50 mixture of hydrazine and helium would only affect the kinematic viscosity to a small extent.
Consequently the complete conversion would be achieved. Increase diameter by a factor of 5
k,2
=
kd(~:~r
=
2.9m/tr
=1.3mjs
X=
1.-ex1-1.3.1~~3 0.05]
= 1-exp(-4.6)
X=1-0.01=0.99
again virtually complete conversion.
Pll-2 (d)
Liquid phase: e.g.. water
See margin notes on page 786 and 787 for solution,
Pll-2 (e)
~:
=
-4.,;1-;'}-4.000( 7~3
e
-
8~3J
= e-0 .59 = 0.55
Assume
~:: - (~)- o.~5
kc2 _ ( DA2
-- kcl
DAl
6
)2/3 (-VI )1/
V2
_
-
( -flJ )4/6 ( -flJ )1/6 =( -flJ.)5/6
Ji2
Ji2
Ji2
k c2 . = (0.55)5/6 = 0.91
kcl
!!~~C2
U/s.
In
= (1.059)(0.91) = 0.96
1 =(0.96)(2)=1.92
1-X2
X 2 =1_e- 192 =0.85
11-3
Pll-2 (f)
CAD
Assume concentration in blood is negligible (C A2
=0 at ()2). Assume quasi steady state
d(VpCAp )
---=-WA
dt
A P
WA =.!!ABI
&.[CA --CAl ]
I
W
A
= DAB2
J [CAl -0]
2
adding
[~L+~_]=C
DABI
A
W
A DAB 2
_
CA
J
&.
_2_+ __1_
DAB2 DABI
CA = RCAP
WA
-
V dCAP
P
dt
=
ApRCAP
~
8
- - + -2-
DABI
DAB 2
R-(~+~J!AR
- DABI DAB
P
2
Flow into the blood
FAB
= CpA mol/time, R == [time]
RV
If CpA = constant = C pAO
- CpAO
FAB---
R
11-4
FB
t
If CAD varies
=.-1 CpA
dt
V R
C pAO
t
In--=··-_·
CA
VR
!lCAP
CA
= CAO e-t/ VR
FB
A
=--
C
81 and 82 are given in the side note.
R
FB
t
Pll-3
Mol balance on oxygen:
Fo~·
0 ...
to:! ::::
0
constant liquid composition .
where n is the reaction order.
. 'd
AssulIlmg
I eat gas Iaw app l'les: F02
P.l)"
=i?:y'
•
t
Assummg
hla.t
••..P
_ . = cons tan t. ('"
by correction of 1,)0 to some reference,
v
R.T
.
h
rt._
were
u,,:::: oxygen uptalce rate m1...ItU
Assuming Henry's law applies (low pressures)
where H is Henry's constant and Po:! is the oxygen pa.rtial pressure
11·5
Substituting into the mol balance:
Fo~
=- ['~
In u"
=:
nJn
P~
k.H"
+- in ._.". . .
C
A plot of in U o vs In
Po~
will give n as the gradient.
As the system pressure, P, given in the data is absolute:
if in the reactor. the xylene is at boiling point and dissolved oxygen and
oxidized xylene are at low levels, then P", =' 1 arm (open to the atmosphere)
In order to deduce the correct kinet.ics of the oxidation it is necessary to find the partial
oxygen pressures for rhe conditions where the rate is limited only by the reaction
kinetics and not by diffusional mass transfer.
Plot of stirrer speed, W, vs oxygen uptake rate, 1)", for each run will show the
conditions at which diffusion is negligible.
............. 1 "
.........-"'''.~.-
Ir. can be seen that at stirrer
speeds above 1200 rpm that
OUR is illsensitive to W and
hence the reactor is well mixed
ie.
no
liquid
diffusional
limitations.
....... ~." ....
atm-
..•..." 1.0 atm.
.. ··,..···.. 20 atm
:::~::]:.2 ~~.
4QO
900
1400
OUR: Oxygen Uptake Rate
W, rpm
Hence using the system pressure dam at 1600 rpm for the plot of in
give n uninfluenced by diffusioIl .
Po: ;:::: P, PjI{
=:
U
o
vs In Po;: will
P, I
At 1600 rpm
·:~Ef=B~fl*~~~~=
11-6
,----_._.,-'''..... ...
_-""' •... "
,
...".,..
'"'."-'"'''''""'"'--"",-"""'-""",","-"""-"'-"","_.,""'",".,,"""''''"
y =O.9992x + 4.,6422
-1
"l.S
o
.(),,5
0.5
LN{(02)
TIle grndient, n"" .999 ::: 1, so the rate law is:
Pll-4
Diffusion in adjacent skin layers
Skin
la~er
interface
P:-;~I
P;>cl
P:-I~:::::: 101 kPa
PH,,:::::: 0 kPn
Inner layer
Outer layer
Strotum
comeum
0, ::::::0.002 em
.,,'--,,-,_._---..
__ ."" .• _
P:-I1:::: 0 kPa
PH¢:::: 81 kPa
Epidermis
~::::::OJ)!
em
.•" _ . _ _ _ _ •••'''. _ _ •• " .•.••" ...... _''_m" '''''''''''''''''''''''''''''''''''' """_""""."""' ••••• _ _ _ _, _ .
Assnming dilute solution and constant total concentration in both layers gives:
d CA
""·"-::::::KI
dz
for each diffusing component in each layer..
N~
:
Outer layer:
boundary conditions
0 , C A :::::: CAO
Z :::: 01 , C,.\ : : : CAl
1.::::
c.v-C4.0
KI : : : -_.--_.-.-
01
11-7
c . == C,\O - ( C"O .- (AI).
Profile
Inner layer:
:1
boundary conditions z:::; Or ,C . . . ;: CAl
z=b:.,C,-\=O
CAl
K I == . _._._--
02 --Ot
Profiie
Total partial pressure profiles
Outer skin layer:
7
.",
;::; 1010 -. (10 1.0 '" t008)..······:::·······t·lO·-·~· .. 0.002
0.002
::::: WIO·-lOOOz + 5000.;::
=::
Inner skin layer:
p, + Pa ::::
1010
+ 4000.;:
O' . . ~.)
(b; -·o! + Pao'"
Pv-;::···~·· .. ...::~·
(O,-.~.)
(PSD ..... Pal) ._.' ....-=:..
\0 :"01
::::; 810 +- 126000(0.01··. · :::) . . . 100000(0.01 ... z)
=;::
Cht.:ck:
SlO+ 26000(0.01
i\t interface (2 == 0.002 em) :
::)
olllerlaver :::: inner/cryer
p., +- Pa::: p, .-t. Pa
1010 + 4000:; == 810 + 26000(0.0 1·····
1018:::: lOl8
Plot these two profiles across the skin from z :::; 0 to
l ::::
11-8
OO! ern
conect!
zJ
He :
Inner layer:
boundary conditons
z
=01 ,es =Cal
z = 0:: • Co ::: eso
K.I
C80 -
CST \
K z =: CBO - -( - - - - - ._.
o! --01
U51.
\ 51"01 )
0' . . .-7')
0: -"01
Profile
Outer layer:
C80'
C81
:::::: ... _-._- ... _ ..
CB :::::: CIJO'- (C80 -- CSI)( -:.-.--":::'..
boundary couditons
z:;::: 01 • C a ::: Cal
z:;:::
0, e a :::: 0
Profile
Total concentration profiles:
Outer skin layer:
.....
CA';'
"'t..,
e8 ::::: CAD -.- (CAO'-
.....
ell).
s;Z +
., Z
(8/ ~f;'
Inner skin layer:
Evaluation of C,\l and
N::: :
em :
Outer layer:
IhNl [..
W,U ::: --.---CAl .-. 0].
Inner layer:
02-01'
.
Assume that flux in inner layer == flux in outer layer i.e. W AI
w"I N=?[C'
.D
---AO- C"]
AI
'
.
-
:::
W A2
......... ....
lh-Ol
w<{
~;;; ~~,. =c.o
+
Conversion of kPa to kglCIll*s2
101kPa=101OOOPa==101OOOkg/ms2:::101Okg/cms2 ,-.lkPa::::::lOkg/cms2
11-9
w
::::o _ _ _ ••
E~-_ ... -.:::: .-.- . . . . . ~} 0 _.....____ :;: 50 * 10-2...~f_.
" [~: +~£~:1 h*~~~·~%~i':1
He:
DIN, [. _
Wa ::;;:-...... (af --
Outer I:lyer :.
'"'
O.
..
an's'
°
]
lrU1ef layer:
Pressure Profile of skin Jayers
1050
'!
E
1000
inner layer
.~
26 950
e::>
...e
900
0.
]j
.., 850
interface
I-
800
0004
0002
(J
0,006
0008
001
z(cm)
...,.....
6:·· 61 .•
Hence .
P~I
; :;:;
== lO08kg I em . s:
a,
0002
= lOkg I em.s:!
11·10
The maximum sum of partial pH~ssures occurs at ilie skin layer interface
z
== 0002 em
PrOIa! :::: 10 18 kg/ems!
Hence the maximum sum of the partial pressures is slightly greater than the saturation
partial pressure and so gas will fonn bubbles at rhe skin layer interface causing
blisters.
Pll-S (a)
Part (a & b)
Packed bed. mass tranfer limited, gas phase
Mol balance:
.-.. 1._4.!:!: + ra"
. a"
:=
0
where Fa: U.Ca.Ac
Ac dz
1 FaodX
.............. .•••...• _ ... _ ••• _ .••. =:;:;
Ac
ra.. . a,
dz
U . Cao.dX
..
U : const.ant superficial gas velocity
········::::ra . a.:
dz
Rate law:
Mass transfer limited boundary condition
but Cas:::: 0, rapid reaction
fa" :::: kc . (Ca ... Cas)
Stoichiometry •
Assume constant T. p. gas phase
• ...Cao.{l······X)
Ca;;:::
_...... _ .........._._ . . . .
(l 'fE.X)
Combining :.
(j
Cao.dX
where e:::: yaoo:::: 0.05 x 3:::: 0.15
..
......... ---: ra . a.;
dz
Cao.uo..~!:..;::;.; a.;.kc. Ca
dz
.dX
'. . --,,, =- a... kc.(l. . . . . _. . "........ ",X)
.
dz
(Jo.(1 '+' O.l5X)
-"~'-""'"
,,> - -
Use Thoenes & Kramers correlation for ke :
11-11
Parameter evaluation:
d
p
r1
=
[?v J [·~~~~lg::~·l =
1C
.
=
n'
P
~
O.238cm
..
Diffusion of cydohexane in hydrogen (assuming constant T, P)
I'll'
O.OOIT175L·~~j;;+iii;J [Jab == ---,-.,..-.............-....-,----.. -.,,",--""'p[(l Va yn "1 (I Vb yny
Fuller, Schettler, Giddings for binalY mixture, low pressure, non
polar (Peny's handbook chem.eng,)
:VIa, Me ::::: molecular masses::::: 84,2 respectively
Va, Vb :::: diffusion volumes;:;:;; 122, 1.07 cmJ/mol respect.ively
4:" 0.4
Y :::'
05: +05tOj)
..... _. -"
4. _._._. _....."".. __ .. ""._ , 0
::: l.lLi L
-'7')
O.. )/~
rr..dp!
Uo ",,"?'~,., = 60000
:::: 509cm / s
Ac
4
u :; ;: Uo.( I + EX)::.:: 50.9(1 + O.15X)
11-12
!l =: 0.00017 g/cm.s (Hz. 500 K , 2 atm)
Sc' =: --_Q~~:!'.. ,...0.000 19 xO.857
=:
1.044
kc
1 [, (
_ ')']112 [
'Ji/3
=OJ'8S'''
47.3 1+' OJ)X
1.044J
a.
=:
=:
17.98(1 + O.l5X)
III
~.Q.::~.~.;~~ == 6.29cm'l
0.572
!1 =: 0.00017 g/cm.s (H 2 • 500 K , 2 atm)
+ O.l5X)0.572xO . OO019
.
Re• =: 50.9(1
....... _._...... _............ _ _......__. _ - - ::= 47.3(1 + O.l5X)
0.00017(1·· 0.4 )1.146
Sc' =: _~~~.:l.._ =: 1.044
0.000 19 xO.857
kC
=: .... 1
_.., ....
r'47.3(1+ o.])X- )'l1/z[
1 1.
.
Q44'JiIl
J
0.388 .
6(1·0.4)
0572
Q, =: ._-_...._.. =:
dX
.."..... .. '=
=:
5X' liZ
l ..,... 98( 1+ OJ)
.1
6.29cm
(1 ... Xl
~
t1. + O.! .5X-)ili·
POLYMATH
Pll-5 (b)
11-13
§gU~bons.:.
~Edal ~
d{xl/dlzl=2_22·(1-x)/(1~.15·xIA_5
o
Zf '"
5
~~iable
:z;
o
5
o
0.99997
o
o
5
0.99997
(.000
~sr;..-' x
a.BOO
Q.600
C.4CO
O.2CD
o.oeo
-+ ceO
os. 000
z
!: ...........
.2 625
"',
(l
9S591419
:2 .5875
0,99541018
75
0.9';)081594
2 8125
;: S'IS
O.99722ilS
.2 .9375
0.99786072
~
Q
39756518
0.993:2039
.3 .. 1.25
3.1875
25
" 3125
j
3'75
.4375
O.99<l:l4ilS3
O. '33354898
0,9937251
o
99g87983
G 99901579
a
o
9991:3525
99924Q2
3
O.99'i33242
J. 5525
0.99941344
The results show that only :3 :3 c:c;: of [he mbe is reqllired for conversion of 99.9%,
much tess rhan the fun 10 £1:
Pll-S (C)
11-14
J);.bh="$h~
kc . dp
lj)
I
U ,dp· P -11/2
[_;'(1-$).
!l
']113
[p Dab
Assuming the porosity remains the same, factors in the correlation affected by the size
of the catalyst pellets are:
d!
2,1t.-···, +red.l
0,.25:
_
_
.- ....... +0.2.)(0..2))
y : : ; -_._._:1._-,_.. -. ::::;
0.286
rt.dp-
~!~
= 1.145
;; l~~?? ::~?.::?~( 1 + O~:?~~~~ ,.J!,.:_:'<l.._
50.9
dX
.
(i--X)
-" ... ::::; 6 '8-",-"·",,--,,,_·. ,-dz
-- (1 + OJ5X)II:
. (1 + OJ 5X)
POLThlATH
Problem Pll-18 parI a/b
l.OoO
.".
!
L,
.!Ss.'!'..;".
c,acc
I
·. ·. x
o . 6ca
0,,-100
z
As can be seen from the above graph. rhe affect of reducing the dimensions of the
pellets by half results in the conversion reaching 99.9 % U8 em from the entrance.
This seems reasonable because reducing the size of the catalyst particles is one of the
methods for increasing mass transfer and hence kc.
11-15
Pll-5 (d)
rf pure cyclohexane feed were used at the same volumetric flowrar.e 60 dm3/min , then
the initial bulk concentration would be greater and there would be a greater
concentration gradient across the sragnant film on the pellets However the mass
transfer coefficient will be reduced as the products must diffuse away fmm the surface
which will be harder in higher concentrations of cydohex<lne.
The equations used In Part (a) do not incorporate Cao, the initial cyclohexane
concentration (Cao is cancelled out), so the only affect this change would have is to
alter [he physical properties of the bulk flow:
(84:<2)
' _ em·.1
== 2647. 11m J I == (
).0026)8/
(t0821x773
,"
mm
• • • • •_
••••••••
)J.=? g!cm.s (cyclohexam:, 500 K, 2 atm)
Pll-5 (e)
This problem gives an indication as to how changes in parameters may affect a packed
bed.
Pll-6
Given,
L=O
•
•
•
Minimum respiration rate of chipmunk,
FAL = L5 /lmol 02/min
Breathing rate of Chipmunk,
Vo= 0.05 dm3 gas/min
Diameter of hole, D = 3cm
:-.,-...,.........................
"" , ,"',
,""',
:-..,"""""",..
"'............"".......,""",..
:-..,....'··s...........,"",.·.,:·
;......"'...,""',....,',..
:-......... ............... ....
..
:--.....
.........,",..
".
x .......,""",......
" . . ,"",. . . . ,"""',..........
","'!.,. . . . ,""',. . .
,...."""""""",...."".......""
."""··.s,,,,,,,,,,,,,,,,"'-:..
Assuming,
• A represents oxygen
•
B represents nitrogen
•
Chipmunk has a constant breathing rate of 0.05 dm3 of gas/min
.....""""",. . .,···s..." ........-.. ;
,-:-.,............""""""""",-..;.
,"""',.....:...,"',....,'"
:-,....."""",...",...
..""...,,
j
""'
.....,""""""""",.......
:..."""""""" ... ...,,,
..."""",. . . . . ."
.,,'\.,."",···s
........,',................,",....:......., ............,
........,........................,............, ........, ...., ........-..;,
........, ...., ...................., ........" ........, ........,"-:..
",
,
,
,"',
....
....,"',....,",....,'"
....,"',....,"',............,""....;.
""""""",.
:-......" ...,:.......,",........" ..
"""""",..
:-,.....
"""""""',..
:-.,....."""""",..
..:......."""""" ..
........,""""" ...., ..
:--
. . . . . . . . . . . . . . . . . . . . . . . . ." . . .
:-..................................':--......................
:.............................................................
:............... ............ .... .... ..
, , , ,',
"""",....,"',..
: ,.............................,........"..
,. . " . . , . . . . . . . ." . . , . . . .,',. . . . . . . . ;.
. . ' L = L ". . . . . . . . .
~~~~:~AL~ PT ~~~~~~_..::.._"" ~~. . . ., . ., . . . . . . ~~~~~:
"
"
Minimum flow rate of oxygen to the bottom,
:-..,,·..·s...:..x ........"...:...,',..
FAL
= Minimum respiration rate of chipmunk
= 15 /lmo} of 02/min
11-16
Flow rate of A down the hole
= Flux of Ax Cross - sectional Area
C AO -CAL
= ~B
:. FAL
L
CAO _. CAL
or L=~B-
X
Ac
----(1)
xAc
FAL
~B
= 0.18 cm 2 / s = 0.18x1O-4m 2 / s
Ac= 1'lf)2 = JZ"x(0.03m)2 = 7.069x1O-4m2
4
4
6
FAL = 1.50x1O- mol /min = 2.5x1O-8 moll s
Pll-6 (a) At Pasadena, California
C AO
PI
= -RT
X Y
.A
(Ideal gas law)
where,
Py = 1.013 x 10 5 N / m 2 at Pasadena, California ( situated at sea level)
R = 8.314 J / mol.K
T
= 25°C = 298K
(Mole fraction of oxygen in air)
YA =0.21
C AO
=
5
2
f'.! / m xO.21
( 8.314-~-X298K)
1.013x10
mol.K
:.CAO =8.59mol/m 3
Now,
Flow rate of A = (Concentration of A at the bottom) x (Volumetric intake of gas)
= CAL xVo
3
Vo = O.OS dm /
FAL
. C
..
min
gas/min
8
-
FAL _
AL - - Vo
= 0. OS X 10-3 m3
2.S X 10- moll s
3.
(0.OSX1O-3
~x Imm
min
60s
_ 0 03
--.
J
1/ 3
mo m
Solving for the length from (1),
L=~B
CAO -CAL
xAc
FAL
11-17
3
L = 0.lSx1O-4 m2 I SX((S.59 -0.03 0) moll m ]X7.069X1O-4 m2
2.5x1O-8 moll s
:. L=4.36m
Pll-6 (b) At Boulder, Colorado
Boulder, Colorado is 5430. feet above sea··level The corresponding atmospheric pressure is 0.,829 x 10-5 N/m2
=0.S29x10 5N I m 2 at Boulder, Colorado
PT
Py
C AO =-xy
RT ,A
5
2
0.S29x10 N 1m
=
C
xO.21
AO
J X29SK)
( S.314 mol.K
:. C AO
= 7.03 moll m 3
Solving for the length from (1),
L
= ~B ,~AO -, CAL xAc
FAL
--O.O~)
3
L = 0.lSx1O-4 m 2 I sx((7·03
mol 1m ]X7.069X1O- 4 m 2
2.5x1O- moll S
:. L =3.56m
Pll-6 (C)
During winter at Ann Arbor, Michigan
T
= 00 F = -17.7SoC = 255.37 K
Py
C AO
C
AO
='RT
xYA
= __1.013X1~5 N Im~_-XO.21
( S.314 -J-x255.37
mol.K
K)
.. CAO = 10.02 moll m 3
Solving for the length from (1),
L
=~B CAO -
CAl X
Ac
FAL
11-18
J
L=0.18xlO-4m21 sx 255 175 x ( (10.02 -0.03) mollm3 x7.069xlO-4 m2
( 298 )
2.5xlO-8 mol / s
:. L=3.87m
During winter at Boulder, Colorado
T
= 0° F = -17.78°C = 255.37 K
CAO = RT
Pr xy. A
CAO =
:. CAO
5
0.829x10 N I m
2
( 8.314 -~-x255.37
mol.K
xO.21
K)
=8.20 moll m 3
Solving for the length from (1),
V0r
CAO - CAl
L -- 4B
.11
x''C
FAL
J
L = 0.18xlO-4m 2 1 sX 255 L75 x ( (8.20 -0.03) moll m 3 x7.069xlO-4 m2
(
298 )
2.5xlO-8 moll s
:.L=3.17m
PII-6 (d) Individualized solution
PII-7 (a)
Given: P y = 510 nun Hg
@
35°C (from plot of In P y vs lIT)
B
InPv = A - (T + C)
DAB =
DAB
0 . 120 cm2/sec
Antoine Equation
(from equation of Fuller, et aL)
0.00 IT' 75 ~~ + _1_
MA MB
=---P(Vl': +VV;;)2
Fuller Equation
where V A and VB are the Fuller molecular diffusion volumes which are calculated by summing the atomic contributions.
This also lists some special diffusion volumes for simple molecules.
Fuller diffusion volumes
Atomic and structural diffusion volume increments
C
15..9
F
14.7
H
2..31
CI
2LO
o
6..11
Br
21.9
N
454
I
29.8
11-19
Ring
-18.3
S
22,9
Diffusion volumes of simple molecules
He
267
CO
18.0
Ne
5,.98
C02
26.,9
Ar
16.2
N20
35,.9
Kr
245
NH3
20,,7
Xe
32.7
H20
13.1
H2
6.12
SF6
71.3
D2
6.84
Cl2
382
N2
18..5
Br2
69,,0
02
16.3
S02
4L8
Air
19,,7
PII-7 (b)
By CS z molar flow rate balance
dNA =0
- - Z=20cm
dz
And
NA =K1
Fick's First Law (For NAIR = 0)
t
dXA
NA =-CDAB--+XA(N
A +0)
dz
- - Z=Ocm
-CDAB dXA
1-XA
CS 2
U
dz
R=0,5 cm
Equating the results of Fick's First Law and molar flow balance, then rearranging and writing in the integral fOIm
Kl1
°
J
dz = CDAB X dln(l- X A )
Xo
=.C;!?AB 1{1 ,- X2]
Kl
1- Xo
Z2
Evaluating for Z2 = 20, X2 = 0
C
= ..!..- = _,____ ....:(l....:)_at_m_ __
RT
(82.6) atm.em' (308)oK
gmol.oK
= 3.95 *10-5 gmol / em 3
Xo
=
Pv -. = 510
PTotal
760
= 0.671
N = K = i~·95 * 1O~5)(Q..1221{ 1- 0.0 ]
A
1
(20)
1,- 0.671
= 2.64
* 10 ··7 gmol
/ em 2.sec
PII-7 (c)
11-20
Fo, "';f:e;:) between z=0and z=20
z: = I-X, l{__
l ]
l- X o
(I-X )=(I-X) [_I_I/Z2 =(1-X) (1-ZIZ 2)
A
OAI_X
OA
X B =(1- X)A
o A
=(X Bjl-Z/20)
~=~C
M
~=~C
= XAMA + XBM B
CB =XBC
VA =NAICA
V* = NA +NB.=7.17*1O-3
VB =NBICB =0
C
n +n
V= A B =nAlp
p
fA =CA(VA - V*) =NA - XANA = XBNA
fB = CB(VB .- V*) =-XBNA =-fA
(*10 3)
(*105 )
Z
XB
XA
CB
CA
PB
PA
0
5
10
15
18
20
0.329
0.434
0.573
0.757
0.895
1
0.671
0.566
0.427
0.243
0.105
0
1.3
1.71
2.26
2.99
3.54
3.95
2.65
2.24
1.69
0.96
0.415
0
0.376
0.495
0.655
0.865
1.02
1.14
2.02
1.71
1.29
0.73
0.316
0
(*103 )
Z
VB
VA
.
V
0
5
10
15
18
20
0
0
0
0
0
0
9. 9
11.8
15.6
27.5
63.5
1E+06
7.17
7.17
7.17
7.17
7.17
7.17
(*10 6)
V
8.32
9.05
10 . 25
12.5
14.9
17.5
P
2.4
2.21
1.95
1.6
1.34
1.14
.J:1.1O
jA=-jB
JA=-J B
3.16
4.52
6.77
10.9
14.9
20
8.7
11.45
15.15
20
23.6
26.4
PII-7 (d)
11-21
ffis
ooA
00
0.158
0.226
0.338
0.544
0.764
1
0.842
0.774
0.661
0.456
0.236
0
60.7
56
49.4
40.5
34
28.97
Mole fraction
0. . 9
0. . 8
0..,7
0..,6
0.,,5
6
...~
g
o.A
Gl
0.,3
0.,,2
0.,,1
0.
......... ·.... ·····T····
0.
10.
5
15
20.
Z (cm)
Co ncentraiion
0
E
~ 3,5
~
3
0
~ 2,5
c
2
0
~
C
2!
1.5
1
c 0..5
0
u
Ill.
1
,--,--,-,~
0.
5
0.
10.
15
20.
Z (crn)
Diffusion Flux
30
CiO
<
0
,...
•
25
cO
0
0
,...
cJ,
<
Q)
III
20
•
15
IiIE
.2~
III
c ra 10
. ..
11.
0
'iii
...
::J
E
5
:!::
'D
0
.................. ··T· .. N . . . . . . . . .
0
5
10
15
20
Z (em)
PII-7 (e)
Evaporation Rate of CS 2 (Pliq
@
20°C = 126 gm/cc)
11-22
PII-7 (f) No solution will be given
PII-7 (g)
Molecular diffusion of air is taking place
NB = J B + XB(NB + N A)
O=JB +XB(NA)
J B =-XB(NA) = (l-XA)NA
=-JA
PII-8 (a)
Quasi-steady-state - no accumulation in the capillary tube or in either chamber.
Volume 1:
Accumulation = in - out.
dNA d(CA~)
dt=-dt-=O--WAAc
WA =-DABCt
+ YA (WA +WB)
:
EMCD:WA=-WB
- DAB-dCA ~ WAZoIL -- DAB CACICA2
WA -dt
Ai
WA =DAB
- - ( CAl -CA2 )
L
Volume 2:
d(CA~)
-.- dt
=WAAc- 0
dCAI =. AcDAB
dt
V,L
(c -c )
Al
I
dCA2 = AcDAB
dt
VL
A2
(c .- C )
2
Al
A2
Subtraction gives us:
d( CAl -CA2 ) = _ AcDAB (~+~J(c
dt
L
V, V
Al
I
2
-c )
A2
Now integrate:
11-23
AcDAB( - 1 _. - 1 Jt + Constant
V; V2
L
Pll-8 (b)
(CAl -CA2 )
= kMC
IJ t+Constant
AcDAB(1
----
Ink+lnMC=
V; V2
L
J
AcDAB ( -1 ' - 1- t+C 2
InMC=--L
InMC =
V; V2
-DAB (O.01025)t+C 2
if we plot In(~IC) as a function of time and find the slope, that would give us -DAB
See Polymath program PIl··8··b.poL
POL YMA TH Results
Linear Regression Report
Model: InDIC = aO + a1 *t
Variable
aO
a1
Value
3 . 7024708
-0.0107008
95% confidence
0.0078035
9,,83E-05
General
Regression including free parameter
Number of observations = 9
Statistics
R"2 =
R"2adj=
Rmsd=
Variance =
0 . 9998944
0,,9998793
0.0015013
2.608E-05
DAB = 0.0107
Pll-9
Dissolution of monodisperse solid particles in exceSs solvent.
Di "'" D +._..
Dil ..... [)~) =
2D*
!. .--{
Ct..!
Define conversion in tenus of volume dissolved:
gives
D:: Di (t·· X) II:;
11-24
Substituting for D :
. .Dl
. (l - X) lil. +._........
1 (D':!
D/t
2D*
.. ,. [D'1 (I -X
) 113]2) -== fl.. t
.
terIll 1
Surface reaction controls:
term 2
D* is large. term 2 is small cf teEm 1
Di - Di (1· X)II} ::: fI..t
1- (l .. X)
Mass transfer controls:
IIJ
fI.,(
-== --_.
Di
D* is small. term 1 is small cf term 2
1_. ( Dl·2 - [D1
. (l - X)lJ)'f).•.•.
- fl.. t
2D*
.
Di... ( 1·- (1-· X) 2/3 )
_. __
2D*
Borh regimes appare?t :
= .fI._ .... tDi
Tenn 1 and term 2 apply
J
Di . . . (1- (1. X)-"13 ).' ::::"l"')"'~'
fl..!
[ 1 ..· (t .. X). 10] + ['--.
.2D*
•
Pll-lO (a)
Mass transfer limited
A in excess, C,"" :::: C AO
T he reaction nue is equal to the mass transfer rate
For small particles and negligible shear stress:
kc ""
3:pe
D
11-25
;'101 balance on solids:
For t mole A dissolving l mole B then . . c-'./' =.
(3," ;;;:;
kcC AO as e A, is undefined
Boundary conditions, t = 0 , D ::: Di
J
(D)dD
=:
-.~:P::~~?J de
p
0,
0
. . . ". = . . . 4.De.Cw.t
. . . . ". "'"
2
p
~,
~
~
Time for dissolution, te , at. D =: 0, and assuming particle density, p:::: 2000 kg/m3
tc::::
Di 2
2000 x (lOe - 5)2
='""'""",_..
8. De. CAD 8 x (We, to) x (2000)
_',""0"0::::;
, ~
0.1255
ie. vinuully instama..'1eous dissolution
PII-IO (b)
Surface reaction limited
Mass transfer effects are not imporlant when the surface reaction rate is limiring.
~1ot
balance ()n solids:
For I mole A dissolving L mole B then
LA;'
= '" la,"
Boundary conditions, t. :::: 0 , D =: Di
C
JdD ::::;.:::. .....k:.~.-:~.?
Jdt
D
Dr
)
I
P
0
11-26
D-Di = _ 2.kr.CAo:.!...
p
D::: Dj _!::!-r_~".:.?t
p
Time for dissolurion,
(C ,
at D:: 0, and assuming particle density, p:: 2000 kglm3
tc == . p.pi ==
2000~..QE.::::_?'!
== 5xlOel2s
2.kT. C~O 2 x (toe -10) x (2000)
a very long time.
....
PII-IO (C)
At
t;;::
0, total moles A in tan.lc:: 0.1 x 100 == 10 mol A
Le. there is just enough mols of A to completely dissolve all of B in a well mixed tank.
Acid not in excess and dissolution is mass transfer limited (CA:I:;O: 0. and CA ;: CAo)
assume zero order in B
and
,.
-rAJ
kr
..
="'-0' C"
1+ -.~..
D*;::: .~:..I?e
where
kr
D*
Mol balance on solids:
For 1 mole A dissolving 1 mole B then
"r.~.,";::;:
.~e -u.r·.· · · .· · ·~.·.- ]
dt
D
l+ .-_.
=:
"
. . lss"
where a.
2.kr.Ct
= -_.j)
D*
11-27
kc
2.De
= --_.
D
Boundary conditions.
t
= 0 , D == Di
Di······Dt
DP . . . [)J.)
:::: eLl
2D'"
As a function of radius:
2ri2
'">r2
:'...._..- 2r + at····· 2ri .....-.. _. =0
D
D'
,.
r-
[··D*.a
t
..
.,]
+ D* r -+ 1.'."'-'2'-'.' "-' D'" n- rC
""
0
·-b ± Ji}':"4·7.t.·~
· ·. · . . · .iD· .-;. -.. . . .---.-..-.. -.
f
.1
r = ...................... _. . . . . . . . . . . . . . . . . . . . . . . -
Using the quadratic solver:
2.a
~
at D 'n. . . . . n ·2 '\I
..,. [) • ± (Of
I,D ...............
r :::; ......................."'........... ................... . . .... ...... ~.................._............................
2
Time for complete dissolution, te, at r:= 0 ;
tc =
assume p == 2000 kg/m.'.. and
2kr. C, = 2x(lOe····
18)x(OJe3)
......... _...........
_.... "......_.
p
2000
0:;;:;: -. __ ...........
.
D'"
gives
~{Vi ~ ·~t;;;]
2. De
2x(iOe·1O)
kr
10e ······18
::::=.._. ;;;;;............ -.::::
I
r"
tc :::: lo·~::~i9~ (lOe
-
)) +
=lOe ..... 19
_
2xlOe8
(IOe ···5)::'1
:2. ;(:i·;io~-~::"8)J:::: 10e14s
again a very long time .
PII-I0 (d)
tc ::::: ._._.P . . .
2.kr.'C,
i2
[Di +e
'1
2D*
To reduce te, increase C, and lor decrease Di.
To increase tc, decrease C", and I Of increase Di .
PII-II (a)
Irreversible, gas-phase, adiabatic. no pressure drop. packed bed.
Isorhermal
Mol balance :
moUgcat
$
11-28
where F Ao
Rate law:
= C ..ou" = lOe-3 x lOe4::: 10 mollem'
- to\,':;; k' .C ...,
Bur C,,-, is unknown .
Assume reaction rate is mass transfer limited.
"-·f:.~,\
cm' mol
"'-1
s.gecu em
k' .kC.CA
::::: ", ........ _-"." ... _ •.•
kc+k'
where k'
=0,01 cm~/s.gcat at 300 K (constant . . isothermal)
converting:
70.7 x acal;;;:: 70.7 x 60 =4242 cm3/s . gcat
0..0 lx4242xCt
4242 +0.01
······/"A. :::; •..................•.•. - ... - ...-. . - .
Stoichiometry:
gas .phase, constant pressure and temperature
C", = CAO (1- X)
and CAO = I mol/dm'
:=!
lOe·J mol/em"
POLYMATH
~~ ...Y.!1:!;!~
~S:!:2!E~.
o
d!x) fd (w) =··::a/fao
k=O.Ol
fao=10
kc.. 4242
cao=O.OO1.
ca"'cao"U···xl
ra=· .. tk;*k;c"ca) j
Wo
= 0,
We
<k; ..kc)
= 1e+05
n·29
t-*.axi.mtJ.m ,.~~
Minburn. value ~.~~~
0
1.e+06
0
le+06
0.63212
0
0 63212
k
0 C1
0.01
0.01
0.01
tao
kc
cao
1.0
10
10
4.242
10
4242
4242
4242
0.001
0.001
{LOOl
Q
Cd,
0 .. 001
0.000.36783
0.Qa0367S8
ra
··9.99998e-06
0.001
··3.6787ge-06
·'9.99993e···06
·-3.6787ge··06
yariabl~
Initial
'"x
0
valu~
001
a.aco
...•. :x
0 .. :320
0 .. 15<:]
+ ..
C.GCO
0 .. 600
..... t······
a.sob:
w
Fur a c{)(l\c:r::;ion of X :::;: 60 %. \\ :~;
:=
9:::5 kg
Scale:
!$£;Y.; ...
·······ca
PII-II (b)
\ In! bal~l!KC ;lr:..i rare: b'.:I.
:.lS
in Part (a)
\vber~
11·30
E;:::: 0 ~nd To :::: 300 K
and C->,o::: lO~·-3 mollem'
dF Q - W - "\.'
F.tQ'P'
C . (T . . . . T. )+ F X '-rT)~
L..J_
I.: -AH
. .._
xx\;
.._....... -_ ........_ ........ _s. . . . . _
....__ . ._ . . . . . . .-w.... _._ . . . . .,\0
__ ..............
_.-'} ::; 0
Energy bal;mcc: :
LN/:pi
dt
' ) Fio Cpi.(T -- Tio) :: [to x ::!.5 IT - 300)] + [t 0 x 75 (T. 7.50)]
.........
'.
:: 1000 (T·· 30(})
.. tOOO IT-. 300} + to.X.( tOOOO) :: 0
POLY:;"'!ATH
!!!.!!::!~...~~
~!~:
o
d(xl Id{",) ".·-::.al fao
1<:",o . 01
fao .. 10
k.::",4242
cao.. 0 .. 0 0 1
To=JOO
T=o(lO·x"lOOOO/lOOOl+-To
= 300 K
ca~caO'{l-xl*(To/T)
ra",-(k*kc"ca) I (k+-kc)
Wo
~ 0,
wf = 1,2e+06
value ?~~
variabb!!.
rnj.t:.l,.:!!;!;.yal!±~
Max~ .. yal.u~ MinimUll\
w
0
1.2e~()6
0
1.2 .... Q6
x
a
0.654825
0
O.654S25
k
0.01.
0.01
0.01
o.n
fao
10
10
1.0
1Q
ko:;
4242
4242
424.2
eao
~l.OOl.
0.001
0.00l.
4242
0 001
To
300
]00
.lao
3an
T
)00
365493
300
ell.
O.OOl.
O.Oell
\}
000283.331
365.483
o .000283331
··2 . S33.:le-·06
-9.39998e-06
,2. 333.3e···06
·9 .. 9939Se··05
:::a.
J.900
~b:r.! ..
0 ... <0
····x
0."80
0.320
0.160
0.000
T
1
T
t
I
I
t
t
O.COO
..... ~~._ ...
-.~·.t···.· .. ·~~-¥·~·· ... ·..·" .. "·· ..
Cl.300
+--...,.---.. .,.....~.t-.'.
0.600
O.SOO
W
11-31
,~ ....... ~... ~ ........ ,. .. t····
1.200
..... ~'"'!-"--- ····'·1
1.500-
For a convc:rsion of X
= 60 % , W,at :::: 10.:0 kg
376.000T
~
3bc.aoo ..~.
!5s.:r:~
-,. r
T
J~1.coa.-.L.
I.
i
,
349.000-ii
3\~..
cco·t
T
r
296. 000 .,,-.-...... _-..••...•.-t.. -, .----~. . ...
C.,CGO
~,JCO
+ .......,.. ~.-......+................... ----,. 1-........- •.--------j
C.. ~GO
O~3CO
l,2CC
L='JOO-
w
Scale!
~S;.t.!_.
.. ca
"'-''''
w
Pll-ll (C)
It is possible to generalize that the addition of'temperature variation in adiabatic
operation does not affect the conversion. cqncentration profIles in form, but the
numerical values are sli,ghlly different. Because the reaction is exothermic, isothennal
operation enhances the conversion profile along the packed bed, so that for a given bed
diameter less catalyst is required. It is clear that adiabatic operation inhibits conversion
as heat is not removed from the system so more catalyst is needed for the required
conversion, The removal of the heat generated in the reaction allows a reduction of 85
kg of catalyst in the bed.. More detailed economics will indicate whether isothermal
operation is worth it.
PI0-12 (a)
Dissolution of pills
Complete dissolution .
11-32
WdruJfttstUm.t.fCh
Col ;;:::: .... ----..".--.V£Ioma~h
des
dt
-_.. ==
~.
"c.,
c". -....V,la-.......".-
W.o,,,;,
_--
...-...... gtl"'!r
Relate C" to rime:
T11ree piUs.. each with different thickness OtHer layers. the inner cores dissolving at
differenr times (but at the same rate) to each other, will each contri.bute to C" in the
$romach.
Pill 1 D1 ::: 5 mm. Dl == 3 mm
1
Time for outer layer to dissolve tl
O.0354x(0.5 ··"OJ!)
.
=----....... -.----... - = 1.18 nun
8x(6:clOe- 4)xLO
Pill .2 :
.,..
"..
Ilme for outer layer to dlssolve
Pill 3 :
0.03:54x(OA"··0.3!)
8x(6:dOe·. ··4)xLO
Cl::: ._--. _ _......._._".......... _ ........ _.:::
0.. )2
- rrun
.
D1 ::: 3..5 mm, Dl ::: 3 mIll
Time for outer layer to dissolve
tl:::
g:??~.~::j~.:?..:~~.:~:g~~!.::: 024 min
8.t(6xlOe ..... 4)xl.O
mol balance on drug:
and
dC,
k.:.S;".TC.D!
2.D~8SIfl.it [)1
. . ...."".. :;;;:: ....... "."" .. _-,--_ ....... = ... . .,. . ....... . - ........... ....._..
~-
dt
and boundary conditions:
~
V
~,,"-
~
~
DV
where W ::: mass of drug in stOmach, g
V ::: volume stomach fluid, cm3
t ::: tl • C", :::: 0
t::: 12, C", ==
WJV = 0.5/1.200::: 4J7xl0e4 g/cmJ
c.",,=
11-33
---'-----~
((0)
D2 =4mm
DI:;:;;3mm
~--
".-... -.-=,~"..,...
Outer layer:
~ (~_1C_D3p{)
Mol balance on outer laver: 0 :,. 0 +- rA" ,.1C, D- ;;::----'------.--, ..dt
and fA'· rate of dissolution of outer layer;;:: rate of mass transfer from pill surface W A
2.,DA8
Sh ::::: 2 , k. ="""
D
.
Iayer;::: mner
"
1ayer ::;::-----.-_._
500xlOe'
6 = () . 0".,)4Kg
- I I em 3
1
Assume uensltv
0 f outer
... '.--.
~,: ()..3X_
"
6
Sour::;:: 1.0 kg I em)
. dD
. . ". . . = .--.2.(-'r,"")
..
. -,, : ; : : : -- "-2x2xDAiJ"Sout
_. . . -". . . . .
,.~
-~~~~"-.~
-"~,,
p
l),p
dt
Boundary conditions :
t== 0, D == I~
t= tl ,
D12 -
2
D::::: D!
._ ..-= 2x2DAB.S"",
--"------,,t,
DII
P
where tl ;::: time for outer layer to dissolve
BC)undnc'y comli::ions :
(;;.;:0,
t
= (:,
D,:
D=D,
D;;;;; 0
D '.01, s,,"
. . . . -.", == --..-.. -,--.-..-.. -". . .
2,
p
-~
-r!
11-34
".~-.-----"'
t:
= .. ~:g~.~:~9:.~~ __._ == L66 min
8x(6xlOe ... 4)xO.4
time for complete dissO:lmion , tr == tl + tl == 0 ..52 + 1.66 == 2 . 18 min
PIO-12 (b)
blood stream
stomach
......--adsoI"ption
e s conc. in blood
,tA conc . in stomach
:@
@@
dissolutjon
stomach wall
Let rare of adsorption into bloo<isrrearll
Relate concentration in stomach C A • to concentlation in blood Ca :
Mol balance on drug in bloodstream:
.. deB
--·rA ==
dt
Scale;
~g;t!.
·-Ca
11·3.5
2.,000
--'-"~
1.6CO
.;. ..
I-~-
1.i
~--~---~---
- --
...
1
Seal e;
1.:100
~...l.-
..!$s.'t-!
/
~
·. ··Cb
0.900
J.
j
o.. ~oo
T
i
0.::00
..
/
f
.1j. . --..
0,,;::00
-----t-.---.----.. -..
~C .. :iCO
----i-----------.-~. ----.~----+-.-...----.---""~
20.DCC
30.CCC
-+c~c~a
50,,00C
PIO-12 (C)
The graph of C s against time shows how the drug concentration in the bloodstream
initially varies \\oith rime and then becomes independent of t.ime as aU the drug initially
in the stomach has absorbed into the blood (consumption of the drug within the
bloodstream has not been modelled, unrealistic but no data). If a certain drug revel is
specified, and assuming a constant size of inner core (drug) and that the pills were to
be taken similtaneously, then the way to achieve this would be to use pills of differem
outer layer thicknesses - to maintain an even sromach concentration and hence
absorption rale over the whole period.
Relate D with t.ime :.
de -= .._._~.I:::_~i~ = . . 4.
de
and boundary conditions:
P
DAB
Sin
D.p
t::::: t[ ,D == Dl
t::::: tz, D:= 0
Using logic to obtain the corr~ct tinting for the drug concentration profiles inside the
stomach ror each pill, the toral profile is used in the relation with the concentration
profile in the bloodstream.
11-36
Pll-12C
, §'.gUations:
d{tll.) Id( eJ =if (e>O .24) then (i.e (01.>0. 00001.) ehen( --4 "Dab-Sinl (
'Ini tial_.y!:!~
0.:3
Ol.*rho»else{O)lels~(OJ
d (CO) Id( c) "'ka"'ea ·V/'Hbody
0
d(Cal.l/d(el=if{t>O.24)ehen(2 w Oab*Sio*3.14*DL/V}else(0}
0
d{Ca2)/d{t)=;if(t>0.52)ehen{2*Pab*SiIl"'3.l4*D2./V)else(O}
0
d(02)/d( tl =if (e>0.52) t:nen{i!: (;:)2>0. 00001) t:hen{ '4*Dab"S;in1 (
0.:3
n2·rno»else{O})else{0)
d(Ca3) Id(c} ""i.f (e>1.1.8) th.en,(2"Qaj;,"Sin"3.14"DJ/V) else(O)
0
den)} Id (el ,die (t;:>l .18) then (it' (0.3>0. 00001) then ( .... 4"'Dab"'Sinl (
0.]
D3"rhollelse(O»)else(O)
t).ab;O.0006
Sin.. 40Q
::'!:lo==)5 .4
ka·;O .156657
V=1200
Wbody=7S000
Ca,=Ca.1.+Ca2 ... Ca3 .... (CO·'Nbody/V)
Co '"
0,
t
e
..
co
4.5
~!:.~1~ ~i.'Uum ...Y:!!1~
45
a
0 ...3
0.3
2 .. 001640-05
0
Cal.
0
0.0004.3.7376
0
9. 9965e-06
2:. OO1.64e··OS
0.000417370
0
0.0004.1'1373
0
0.0,00411:373
0.3
0.3
'3.99T74e-Q6
ca:a
0
0.000411364
0
03
0,]
0.3
9 _9.9S41.e- 06
9. 99774e-01i
0.0004.1,7364
9. 9984.le· . 06
Dab
0.0006
0_0006
0 .. 0005
Sin
;;ho
400
0.0006
400
35.4
y~le
t::
D1
Ca2
D2
Mini.'11um v<l.lue Fi~b...yaJ.~
45
0
9.996Se,,·06
0
4.00
400
.35.4-
35 4.
lea
35.40.166667
V
1.200
1200
1200
1200
Wbody
75000
75000
75000
75000
Cae
0
0.000991.:1.8
0
1.0S0Ue-01S
Relate D with time:.
0.1.66667
0 . 166667
0.166661
:!!?,,;, -!:'-~~~~"-:= -.-~~.!:?~~~§~
dt
and boundary conditions:
p
t=tl,
t::: tz ,
D.p
D=D1
D::: 0
Using logic to obtain the correct timing for the drug concentration prof11es inside the
stomach for each pill, the total profile is used in the relation with the concentration
profile in the bloodstream.
PIO-12 (d)
11-37
To maintain constant drug level by maintaining a constant stomach concentration,
time needs to be allowed for the dissolution of the outer layer. for a given period of
say 3 hours, a size disuibution of outer layers is needed, with thin layers for initial
response and thicker layers for delayed response. This disrribution would be back
,calculated given the necessary stomach cOllcenu'ation for the required bloodstream
concentration accounting for bloodstream drug consumption.. Optimization of the
stomach concentration will indicate the times at which complete dissolution of the
outer layer of the <next' pill is requited to maintain this level. The range of pill sizes
der.->ends on the number of pills which can be reasonably consumed in one sitting, the
period for effect and the limits of practical pill size
PII-13
----.. . . . . . . . --.. . . . . . . . . . __. _-._._. _. . . -._. . . ._. . . . . . . . _·_. _·__·__· __· . ·.· .1
The plot of the data is shown below
Particle Diameter vs Time
10-..
9
.
;:
..
'-
4>
4>
8
7
E
6
i5
5
"0
4
3
tiS
a.
2
1
0
2D
30
50
'70
60
80
00
100
110
Tlme
Initially the rate of incine:ration of the droplet in terms of diameter, is non ..linear, but
apparently becomes linear after . ~ 50 time units. 'TIle linear form of the data indicates
that the diameter is directly proportional to time and the rate of decrease in diameter is
constant and hence not a function of diameter. TIus relationship should make it easier
to estimate the reauiredincineration time for complete destruction (zero diametei).
Assuming that rate of diameter decrease continues at the linear rate 1ll1til complete
destruction (at time !ct and hence complete decomposition of the POlle's, the
equation for the linear relation indicates 41 .~ 160 u111ts .
._--_._-----PII-14
11-38
Mol balance on layer of earth control volume:
but FA == A:..W",
dW~
,--,
= . ._.dC~
. . , _. . . .
Z -70
dz
dt
For dilute solution and constant total concentration:
Gives;
Let
gives
Boundary conditions:
t ::: t (present day)
0 (surface), 'V == 0
Z. :::
z::: 0 0 .
t
=0 (end
'V= I
of glacial) • z > 0 ,
Gives the error function:
let
It is defined that at Tl = 1.82 , C A = 0.01
CAO
and for 1'\ > 1.82 • C", is negligibly small.
This defmes the penetration thickness, 0 (as a function of time) :
Concentration Profile
1.2
..
~.-.--.'" ~
o 0.8
- _._"._,,--
«I
~ 0.0
0 0 .4
o
o
5
10
15
Depth (m)
11-39
1.82
= _-;:==0=
../4.DAlJ.t
But the graph gives at, C\ = 001 C\O --»
penetration thickness after time L
z::= 0,4 hence (5::::: 18 - 0,4 m , the
The time taken for 0 to reach this thickness, is the time since the difftlsion started (i.e
at the end of the last glacial)
CDPII-A
CDPII-B
CDPII-C
CDPII-D
CDPII-E
CDPII-F
CDPII-G
CDPII-H
CDPII-1
CDPII-J
CDPII-K
CDPII-L
CDPII-M
11-40
Solutions for Chapter 12 - Diffusion and Reaction in Porous
Catalysts
P12-1 Individualized solution
P12-2 (a)
(i)
t=5
1
D
ex
-D
e
l-~
175
( J
_T2
P Tl
(lines and angles not to scale)
P12-2 (b)
(1) First Order Reaction Kinetics
2
A. = 0
W= 1
D d CA
e dz 2
_
kC =0
A
'
1jf
=C A
C AO'
A=~
b
1jf = A cosh..JDaA + B sinh"DaA
-dW = 0 _._ _ _-----1 A. =1
dA.
symmetry:' = 0
B=--A
1jf=1
sinhrnacosh..JDa
A=O
@
.~ = A..Jij-;; sinh ..JDaA + B"Da cosh A
A = 1 ..
~
=-A Tanh-vDa
@
12-1
1 = A coshvlDa
A=
1
B = Tanh-v'llil.
cosh viDa '
cosh viDa
cosh ..JDa A Tanh..JDa . h ~D '\
sm -yua fI.
cosh ..JDa
cosh ..JDa
(2)
Monod Kinetics
\jf =
2
D d C A _ !.!max C ACc = 0 Use
e dz 2
Ks + C A
Quasi
Steady
State
dCc =!.!max C ACe = 0
Analysis
dz 2
Ks +C A
No further solution to Monod Kinetics will be given.
(3)
Variable Diffusion Coefficient
clFw = Dc Wo2 1_oAc
dt
zWA =--De dC A.=_ DeC AO d~1
dz
L
dA }.=O
?Fw = D~~AO d~1
dt
L
Mole balance
d~:
Dc Ac/ V
dA }.=o
]
{De
-------k=O
dz
for hindered diffusion
=
D
e
2
DAB
1+ a F;/(1-- Fw)
As a first approximation, assume no variation in De with A
2
d \jf
2
kL
- -2 - - - = 0
dA DeC AO
as before
2
kL
<1>0=--2D eC AO
~f\jf
--2<1>0=0
dA2
Solution the same as before Equation (E12-2.13)
12-2
w
°2
=W =(
A
DeCAoY_~=-J=kL
L A. DeC AO
The flux of 02 in does not depend upon De which is not uprising since this reaction is zero order.
For the build up of material that hinders diffusion
dFw_=u A kL=A Lk=kV
dt
C
C
C
Fw =kVt
From a quasi steady state approximation as time goes on
increases.
Fw
increases De decreases and <1>0
However, the point at which the oxygen concentration is equal to zero has to be found. We can
parallel the analysis used in P12-10 switching the coordinates of A = 0 and A = 1 (see solution to
P12-1O(c) in which the solution manual) we will find
1
A =--<1>0
C
~=O
5
Increasing
t
o
-------~-------~~------~~
A =1
We see that as t increases Ac decreases, that is the point at which the oxygen concentration is zero
moves toward the top of the gel.
P12-2 (c)
12-3
(1) For Rl 111 =0.182 18.2% Surface reaction limited and
81.8% Diffusion limited
For R2 11 = 0.856 85.6% Surface reaction limited and
14.6% diffusion limited
R2
,j-,2
(12-59)
(2) Cwp = -ri\(obs)pc
= 11'f'1
DeC As
= (0.95)(0.9)2 = 0.77
which is less than 1 so there are no significant diffusion limitations.
1
N
k SaPb
=
1
-+---=--=11
kcac
Q = 0.059 So 5.9% surface reaction resistance and 94/1 % external and internal diffusion limited
%R =
0.941
N
.!.. + ~aPb
11
-
0.941
0.941
6.0+ 10.96
16.96
= ------.--
kcac
% Internal diffusion reaction 0.941 x 100 x 6.0 -- = 33.3%
16.96
% External diffusion resistance = (0.941)(100) x 10.96 = 60.8%
16.96
Summary of Resistances
60.8%
External Diffusion
Internal Diffusion
33.3%
5.9%
Surface Reactor
100.0%
Increase temperature significantly. Surface reaction % resistance decreases. Increase gas velocity
external resistance decreases decrease pellet size both internal and external resistances decrease. For
99.99% Conversion
In
1
1
1
L2 = L 1 l n ---ln-- - / l n -- = 0.16 (1O,000)
1- X 2 1- X 2
1- Xl
In.500
=0.16x~~=0.24m
6.21
P12-2 (e)
12-4
(j)
From Mears's Criterion
-LlliRx (-rA)PbRE < 0.15
hT2R g
1)
The value from the question
Llli Rx = -25kcal/mol = -104.6k1 Imol
h = 100BTU Ih· ft2.o F = 0.567kJ Ism2 . K
E = 20kcal/mol = 83.682k1 Imol
Rg = 8.3144 *1O-3 k1 ImolK
From example 12.3
-rA =kSaC NO
k = 4.42 * 1O-lO m 3 1m2. sec
2
Sa = 530m Ig
Pb = (1- <I> )P c = (1- 0.5)(2.8 * 10 6 ) = 1.4 * 10 6 g/m 3
R = 3*1O-3 m
T=1173K
At the inlet of the reactor the fraction of NO =0.02
From ideal gas law
n
P
V RT
= ~01325 * 1O~ = 10.39mol/m3
8.3144 * 1173
C NO = 0.02 * 10.39 = 0.2078mol/m 3
Substituting all value in the first equation
(104.6)(( 4.42 *!.9-.~O)(.530)(0.2078»(1.46 * 10 )(3 * I_Q-3)(83.~_82) = 2.88 * 10-4
(0.567)(1173 2 )(8.3144 * 10-3 )
As the calculated result is lower than 0.15, there is not the temperature gradient. The bulk fluid
temperature will be virtually the same as the temperature at the external surface of the pellet.
6
P12-2 (0
(k)
Foq = 30 use Figure 12-7.
If you draw a vertical line up from <1>1 =0.4 it should be tangent (or very, very close) to the J3 =0.4
curve. Any slight increase in temperature will cause the reaction to go to upper steady state.
P12-2 (g)
(1)
For large <I>
12-5
S'
=
a
Sa
1+ kDt
.
Area 2
(1) Pore closure. ConsIder De As t -7 00 pore throat closes ~ =
-7 00 ,
Area 1
°
-r~ --7
(2) Loss of surface area Sa. As t -700 then S~
<1>1 -7 00
-7
°
then
<I>
-7
°
Ci c -70,
11 -71, but -r~
De -70, and
=.JS:: -7 °
P12-2 (h)
(m)
The activation energy will be larger than that for diffusion control and hence the reaction is more temperature
sensitive, If the apparent reaction order is greater than one half, then the rate of reaction will be less sensitive to
concentration. If it is less than one half, the true order will be negative and the rate will increase significantly at low
concentration .
P12-2 (i)
In example CDRI2-1, the reactor is 5 m in diameter and 22 m high, whereas the reactor in CDR12-2 is only 2 m3 in
volume. The charge is much different In CDR12-1 the charge is 100 kg/m3 and in CDRI2-2 it is only 3.9 kg/m3
P12-2 (j) No solution will be given at this time
P12-2 (k)
With the increase in temperature, the rate of reaction will increase. This will cause the slope of C/R j vs, 11m and,
therefore, the resistance to decrease"
P12-2 (I) No solution will be given at this time"
P12-3 (a) Yes
P12-3 (b)
All temperatures, F H ) = 10 moJJhr" The rate of reaction changes with flow rate and increases linearly with temperature
P12-3 (C) Yes
P12-3 (d)
T < 367 K, F H ) = 1000 moJJhr, 5000 moJJhr.
T < 362 K, Fro = 100 mol/hI.
P12-3 (e) Yes
P12-3 (0
T > 367 K, Fro = 1000 moJJhr, 5000 moJJhr"
12-6
T > 362 K, FTO = 100 moVhr
P12-3 (g)
actual rate of reaction
-rA (at362K, Fro = 10 moll hr)
- --=..:...;'-------=--=-------'--,-- ideal rate of reaction - -rA (at 362 K, Fro = 5000 mol / hr)
Q -
Q
= 0.26 = 0.37
0.70
P12-3 (h)
At FTO = 5000 moVhr, there is non external diffusion limitation, so the external effectiveness factor is I.
1]=
actual rate of reaction ( at 362 K, FTO = 5000 mol / hr)
.
extrapolated rate of reaction ( at 362 K, Flo = 5000 mol / hr )
1]=~~=0.86
1.4
P12-3 (i)
1] =
3[¢cosh¢-1]
--7= 0.86
by iterative solution
¢ =1.60
CA
1 sinh ( ¢A )
rp=--=
_.CAS
A sinh¢
P12-4 (a)
External mass transfer limited at 400 K and dp = 0.8 cm. Alos at all Fro < 2000 moVs
P12-4 (b)
Reaction rate limited at T
= 300 K and dp =0.3 cm. When T =400 K: dp = 0.8, 0..1, and 0.03 cm.
P12-4 (C)
Internal diffusion limited at T = 400 K and 0.1 < dp < 0.8
P12-4 (d)
1] = -rate with d p = 0.8 = 10 = 0.625
rate with d p = 0.03 16
--_._--P12-5
Cmve A is reaction-rate limited . This is so because of the way the curve bends, implying an exponential function which
is the equation form for the specific reaction rate with respect to temperature.
Cmve B is inner-diffusion limited . This is because it has a dependence on temperatUIe, but that dependence is small.
12-7
Cmve C is outer-diffusion limited., This cmve has a much larger dependace on temperatme than cmve B,
P12-6 (a)
_ (1 \ sin h <b.t _A
Th en <P-'--,
].) sin h 01
Effectiveness factor:
n
cP
conditions:
., :.;:; "
(A. :::: 1) = 1
= (A. :::: 0) :: finite
cP ;:;:
Boundary
3 (A
h'
1)
'4'1 cos IPl - , flrst order reaction.
q:.t
At •'" -- R
1 > C-A
2' ~~:::: 7)'
::::
0.1
CAS
where CAS = 1 x 10 -3 rnpl~,
I
7 X 10-4
A. = '-.-.---- = 0.7
R = 1 X 10,3 cm1
Dc= 0.1 cm2/sec
1 x 10 3
<P =
(f-}(Si~,.~.~l 1)
sm h 01
A.
r
At A = !
1. ' .<D = 0 '1". 0 1 -- 2 sin h ($1/2)'1
-
L sinh($l)
J
::::::::>
. ;: :; : 60
q>.
1
f
q> :.;:; __
1 sin h (6 x 0.7)1
0.7 L
C.~
1 x 10 -3
sin h 6
J
= -L r~4.2 - e 4:21
0.7 l eO ~ e ~ J
P12-6 (b)
" ::.: 1[4>1 cos h ~1 - 1]
- 0.80
~
<PI -
2.04
<1>1
<1>12
= ~~~ SA PB
De
At A.
=~ .
P12-6 (c)
-
cp :.;:; 0.1 : <PI :.;:; 6 (see part (a»)
Individualized solution
12-8
"
P12-7 (a)
Start with a mole balance:
WAI. _".WAL,..~ +. TAALlz = 0
Divide by A& and take the limit as z --> 0.,
dW
dz
·__ ··· ..···,·,··,T
=0
A
From the flux equation:
w = -_. D .~~:i.
e
d:..
Combining the two equations we get:
d[-De dCA/dz] + k =0
--_.,,-,_._dz
Dividing by -De we get:
d 2 C +---=0
k
dz
De
_._A
1
We need boundary conditions
de
__ ._.:1"
B.C. (1):
dz
= 0 @ z =L
B"C. (2):
We can then solve for the concentration profile:
dC" =~+Cl
dz
De
Using boundary condition (1):
"kL
O=-+C1
De
kL
C1 =--
De
dC"
dz
kz
kL
De
-=---
De
12-9
Integrating again:
dCA
=(kzDe _kLDer
1-1-.
k 2 kL
C =-Z
--Z+ C2
A
2De
De
From boundary condition 2
CAO = C2
k 2 --Z+
kL
C
C =-Z
AO
A
2D,
De
P12-7 (b)
11
=
rate of reaction with diffusion
rate of reaction without diffusion
Tl
=
2D{CA - CAo)
z{z - 2L)
_ 2D (CA - CAO)
-~---k
k z(z- 2L)
P12-7 (c)
Boundary conditions:
o=
k5 (~ -L)+CAO
CA = 0 atz
=L
CAO = k L2
2D
L =
L 1(2
~2DF''- = Vr~:~::~~~6::~?-::):::)
:;:::
0.0640
L :;::: 0.0041 em :;::: 41
~m
P12-7 (d)
The answer in pan (c) is equal to the average tail length.
11 = 1 in this problem. Ifl1 = 1, then it con1I'3dicts the assnmption of diffusion
being rate-limiting.
P12-8
12-10
=
z 0
CA = CAS
I
Z
z=L
CA
= 2L
= CAS
CA = _1 C ...
10
""->
First-order irreversible reaction:
A-+B
.r~ = k CA
WA = _D~CA
dz
~fole
L
balance:
.1ld
= lO-3c~
dl
2L
(d~A) + n: d r~ :: 0
,dz
= 2 x 10-3 em
CAS = 10-3 g moleJl
D
= 0.1 crrjl/sec
tl')
At z = ~ CA = ..L CAS = CAS ( COS h['(l - ~
10
cosh,
...L
10
=
1
cosh,
cp =2.9932
12-11
P12-8 (a)
Atz = 1/2 L:
= CAS (COS h[q,{l
9A
.
- ~)])
coshcp
C
= (0.001) (COS h [209932(0 05)])
cos h (2.9932)
A
C..!\. = 2.345 X 10 4 g mol/l
P12-8 (b)
ct>
= 2L VDkd ==
l10ld
=
tan h <bold
A
~old
TInew = 0.8 =
Q a (2L)
tan h (2.9932)
- . ,......99 ...~.,..
tan h Onew
==
.
Qnew
-
0.3324
= 0.8880
Onew
P12-8 (c)
At z = L. CA = _C
__AS-=-_
cos h 4>new
= cos gltfJ880)
= 7.038 x 10-4 g mo1ll
Thus minimum CA is now 70.38% of CAS- Therefore the suggestion is plating entire
surface of the inside of the pore.
P12-8 (d) Individualized solution
P12-9 (a)
12-12
Define W A > 0 in direction of increasing z.
Material balance:
WAAp~ - WAApIZ.4Z+-r~aApAz = 0
..'
=: kCA
-rA
;
WA
::;
dC
dC
-De dzA +CA V == -De dzA forsmallCA
Taking t:J.--; 0: -Ap d:A + r~ a Ap = 0
Apt-(Ded~~)- k CA aAp
2
For constant De; De ,d C A
-
dz2
Boundary conditions: z
z
=L
= 0 d<?A
az
2
d CA _ 0:2 CA = 0
dz2
0
k CA a ::; 0
CA = CAS
= 0 by symmetry
where (1.2 ;;; k a
De
Assume CA = e rz then r2
y
CA
=
- (X2 ::;
0
r
= Al e -a.z + A2 eaz
Atz = L: CAS ::; Al e-aL +A2&L
deAl
dz .,.0 ::; 0
::;.C!.
Al + a A2 ::; 0
...
12-13
= ±a.
Therefore, concentration profile can be written as:
c = C Jeas h(az)l
AS \ COS h (aL)
A
WA A;- = -De (d~A
~eL) Ap =
- Ap De CAS ex ( : :
l~) Iz.&.}
WA Ap =:= -ApDe a CAS tan h (aLl
By the sign convention:
= ApDe a4-.s tanh(cxL)
Tl = (~)tanh(aL)
='VfIT
(De}~tanh{' Ifi'L)
leaL
0; lea L
'V 0;
:. 11 kCASaApL
1\
= (-v'~ t}tanh(aL)(.Jft L)
Overall effectiveness factor:
or:
•W A Ap =
n k CAO a ApL = 11 k CAS a Ap L
.n. = CAS
11
CAO
-WAAp = kcAp(CAo-CA,S) = aApDeCAs tanh(aL)
CAO = CAS [I +
a:::.
tan h (aLl]
P12-9 (b)
A-7B
WAltrLI r - W AllrLI r+t.r + rAllr/).rL = 0
1d
--rWAr+rA =0
r dr
12-14
EMCD therefore,
dC
WA = -De __
A
dr
d
rdCA
.!. D dr = D
r
dr
e
2
(d C2A -.!. dCA
e
dr
r dr
dlf/
At A. = I, 'l' = 1 and at A. = I, -
dA
J
=0
Bessel Function Solution
(n) P12-10 (a)
(0)
EMCD
dC A
wA =-D--
dz
, fA =-k
In - Out + Gen = 0
WAAcl z - WAAcl z +Llz + rA~zAc = 0
dW
A
---+rA=O
dz
'
1\,= 0
12-15
'-_ _ _ _ el1{J = 0
ell\,
J
Integrating equation (1)
kL2).?
\jf =
+ C2
2D A C As
at 11,=1 \jf=1
kL2
C2 = 1 - - - 2D A C As
th. 2
_
'1'0 -
2
kL
2DaC As
P12-10 (b)
(p)
Now let's find what value of A that 'If = 0 for different
<1>0
For <1>5 = 1 : 0 = 1 + 1[11,2 -.1]= 1 + 11,2 ---1
11,2=0
Therefore the concentration is zero (i.e., \jf = 0) at
111,=01
For <1>5=16 : 0=1+ [11,2-- 1]=1+1611,2- 16
11,2 = 15 = 0.938
16
Seems okay, but let's look further and calculate the concentration ratio \jf at 11,= 0 for <1>0 = 4.
\jf = 1 + 16 ~0.2)2
-1]= 1 + 16[0.04 -1] = --14.9
Negative concentration.
(q)
P12-10 (c)
(1)
Let's try again with
\jf = 1 + (10/
I
<1>0 =
10
~.12 -1]= 1-10 2(0.99)
\jf = 1- 99 = -99
not possible
\jf will be negative for any v-a-Iu-e-o-f-·<I>-o-gr-e-at-er-t-h--an-o-ne-.-.===_~~-_.-_~-J
P12-10 (d)
12-16
We now need to resolve the problem with the fact that there is a critical value of A, Ac, for which both 'If = 0
(s)
and
d",
-=0
dA,
2
d ",
2
-dA,2 -2<1>0 =0
d", _ 2<1>5A+ C1
dA,2
'" = <1>5A2 + CIA + C2
At A = AC ' '" =0
0= <1>5At +C1AC +C 2
Subtracting
1 = <1>5 -<1>5At +C 1(I- Ac)
Solving for CI
_
1-<1>5~-At)
C1 --
l- Ac
Solving for C2
2
~-<1>5~-At»)
C2 =1-<1>0-------l- Ac
12-17
1.0
<1>0=2
<1>0=1
o
0.5
LO
Sketch of concentration profile for different values of <1>0
<1>0 = 1 then Ac =0
<1>0 = 2 then Ac = 0.5
That is for <1>0 =2, the concentration of A is zero half way (A =0 ..5) through the slab.
1
1- Ac = <1>0
\11= <I>~A2 + [<1>0'-
<I>~(l + Ac)]+ 1- <I>~ - [<1>0 -<I>~(1 + Ac)]
212 ~
2 2
n
2
2
• = <1>0/\
+ l<l>o - <1>0 - <1>0 + <1>0 r + 1- <1>0 - 2<1>0 + 2<1>0
\II = <I>~A2 + 2<1>0(1- <1>0)11, + 1- 2<1>0 + <I>~
ForA >Ac
P12-10 (e)
P12-10 (f)
12-18
In
1.0
1 - -......
T)
0.001
'-_.....ii....-_ _ _ _;;a".,_ _ _
In
1.0
P12-10 (g) No solution will be given
P12-10 (h) No solution will be given
P12-10 (i) Individualized solution
P12-11
Given: second-order decomposition reaction: A -P B + 2C
= SO m6/g sec mol;
k
P = 500 kPa
dp :: O.4cm; U :: 3 mls; T
= 4.936 atm;
X
= 0.80;
= 2S()OC = 523 K;
Dc :: 2.66 x 10..a frills; Eb = 0.4;
Pb = 2 X 106 g/m3 ; Sa = 400 m2/g.
CAo = L =
4,936 atm
= 0.115 g mol
RT
(0.082 g
S23 OK
I
!!'flK)
Rate law:
-r;'"
k
cl
Mole balance:
D
1
d CA _ U dCA + r' p :: 0
~ ctz2
dz
1
A B
DAB d CA - U dCA - Q k SA PB
dzl
dz
0
ci = 0
Neglecting axial diffusion with respect to forced axial convection. we have:
12-19
d;A = _(Qk~APBlcl
-1e.. dC~ = Ink~APBll~dz
CoM
CA
0
. l .. --L
CA
= IOkSA PB)Z
CAO
At z = 1.;
u
\
--L = ( 1 _ I) = (-2PB k SA) L
CAO
L =
In
I-X
j(
U
1 - 1)
Ps k SA CAO 1 .. X
" =
Internal effective factor:
~
(-Lyn
n+l
~
wheren =2
(50 gSm~Ol}(4001-}{2X
cp.z = 0.2 x 10-2 m
t.z
U
106~){0.1l5 ~)(~)
2.66 x 1O..g I1J2
= 2.63 X 107 very large
11 =
(-L}112
2+ 1
3
2.63 x
107
= 9.313 x 10..a
Intcmal-diffusion limited:
n
= 11
= 9.313 x
10-8
Reactor length:
L =
( 9.313 x 10 -8) 2 X
L
3 m/s (1 _10 .8 .. 1)
106 L(SO roO )
m3
gsmol
(400 m!)
(lIS g mol)
g
tal
= 2.80 x lO-Sm
-------------------------------P12-12 (a)
12-20
Start with the mole balance taken on a shell
Divide by -21tlM and take the limit as llr approaches zero to get:
d(lf.t,.r)
dr
'i.Pc.r = 0
Next fmd the equation for equimolar counterdiffusion and plug it into· the above
equation:
tt({-D_dC~))
dr
-r;PtT=(}
Next differentiate to get the following differential .equation:
We can then set the following:
.
C
rp:; .---:L
CAS
r
It=-
R
Solution:
({J
= Clo(<P.A.)+C2 K o(cf>il)
Boundary Conditions:
rp=l@l=l
drp =O@l=O
dA.
12-21
~~ =ct>[Cl (ct>1)-C K(ct>J)]
1
2
1
11(0)= 0
Kl{O) = 00
as
dqll ~O =>Cz =0
dl ~=o
9> = C/O(Cl>A)
t= C.I.{<I»=> C, =
()
10 ct>
10(<1>11)
ffJ
= l~ (<.Il)
P12-13 (a)
(t)
r;
!.(r2kt ~:J+
=
(flHRx)(-rA) 0
(1)
(2)
Dividing by flHRx and using Equation (2) to substitute for -IA
dT)+~~(r2De dCA)=o
(_1--J~(~r2kt
--flH Rx r2 dr
dr r2 dr dr
12-22
CA +
ktT
De(-~HRx)
=C A
kTs
DemRx
+-~-
s
T = De (-mRx) (CAs -C A)+ TS
kT
T=Tmax at CAs =0
P12-13 (b) No solution will be given
P12-15 (a)
CAS
Given:
A H B on the walls of a cylindrical catalyst pore.
zl = length of poisoned section
-r~ = kCA
In the poisoned section:
WAxriz - WA~lz.~ +r~Pdz
For 0 S z S
WA
WA
Z 1,
~ Iz
=0
r;" = 0, since this area is poisoned.
- WA
m2 IZ .62:
= 0 or
-<1:
= -CDAR ~A +XA(WA +WB)
12-23
A
= 0
But WA
WA
At z
=.WB , since. for each mole of A consumecL one mole of A is reacted.
= .CDAB~ and iz(CDAB~.t,) = 0
=0
At Z =
= XAS C = CAS
CA = XAI = CAl
CA
t
Zl
t
.
Integraang:
XAI
Atz=O,. XA =
atZ=Zl,XA
=! Kl
C~ :
Z
+ K2
K2 =
C~
C~l
= XAI :: C
= kl %1 + ~
C = kl
=
CAl-CAS
CZl
CAl" CAS Z + CAS
CZl
CA XA = C
-
C
CA = (CAl - CAS) (it) + CAS
The flux is WA = -C DAB dX.... = .DAB dCA (CAl - CAS)
dz
dz
P12-15 (b)
Before poisoning, 11
= tan h q,1
q,1
where 11
= q,1
= L(
r
~. )112
AB
After poisoning. the differential equation and boundaIy conditions are the same for the
unpoisoning region of the pore, Zl ~ Z S L if Z is replaced by L - z; and if we let
CA
= CAl at z' =0 and d;A =0 at z =L - Zl.
then. for the unpoisoned section of me
catalyst pore 11 appli~s if q, is ~p1aced by :
2k"
(L - z) (r DAB
)112 -_.(1 - LZ) tPl • l.e.,
. 11 -
tan h
[4>1 (1 i)l
CPt (1
t)
The effectiveness factor for the unpoisoned section of catalyst pore is defined as :
WA = ll1tr{L-z)CA l
1bis can be related to the overall effectiveness factor for the entire pore by
12-24
WA ::; 11' x r L CAS = ll1trL CAl
11' CAS = 11 CAl (1 - ~)
:.
But W A = - ~;S (CAl - CAS) from pan (a)
:.
CAl = CAS _ WA Zl
DAB
:.
l1'CAS =
11( 1 -
r}[CAS - ~::1]
WAxr2 = Tt'2ltrLCASk"
But
t
(1 ZI)(1 21'1' Ll CAS k" Z1 \
11 = 11 - L
- ~--:rD~AB~L--';;'1
-------_. ' - P12-16
.The reaction is A -+
t - t;
a=
c
1 = -
E
- CAoe!· X) where
A -
(1 - 0.5X)
t ~2
= YAO 0 = -0.5
CAO
= L
RT
=
8.2 atro
0.082 I atm (227 + 273 K)
gmolOK
12-25
= 0.2 gmol
1
= .l-+1.FAO
.l
2
YA
FAO = 2
FA
2 FA
(...L _1.)
YA
or FA
1 _ YA
F AO = .l.. _1 - 2 - YA
2
YA
= 1... FA
X
FAO
r~ = FA~X
In {-r;J
where W
.
M
2 - YA
= 4x40g =
160g
= In ko;
At =
ex; and N
= Ao+A1N
D
i
i
L Mi = nAO+Al L Ni
I, ~ti Ni
(2)
n
n
n
= Ao
L Ni + At L Ni
1
2
3
4
F-r-n
1
2
4
6
5
6
11
20
2
(3)
i
i
i
= In CA
(1)
n
i
= 0.16 kg
= In ko + ex 1n CA
Let M = In (-r;j; A(1
.o.
= 2 .. 2YA
= 1... YA
2 ... YA
'1A
X
C~
0.21
0.33
0.40
0.57
0.70
0.81
0.88
0.80
0.75
0.60
0.46
0.32
4.285 x 10 -2
6.666 x 10 -2
8 x 10 ·2
1.143 x 10 ·1
1.403 x 10 ·1
1.619 x 10 -1
-r~
M
5.5
1.705
10
2.303
18.75 2.931
22.5
3.114
31.625 3.454
40
3.689
-3.150
-2.708
-2.526.
-2.169
-1.964
-1.821
N2
9.923
7.333
6.381
4.704
3.857
3.315
-5.371
-6.236
-7.404
-0.754
-6.784
-6.717
-14.338
35.513
·39.266
N
n
2:=
Equations (2) and (3) become:
17.196 = 6Ao - 14.338 Al
-39.266 = -14.338 Ao -+ 35.513 Al
12-26
17.196
MN
:. Ao
= 6.36
= In leo
-?
ko = 578.25
I
At = 1.46 == I.S ::
(l
At Tl = 237 C = 510 K: kl
-r'
A
= FAGWX
.
where X = 2 .. 2(0.097) = 0 9490
2 - 0.097
.
(9)(0.9490)
0.16
=
-rA
C
0
A
RT
53,38
.
where:
°
8.2
- 196 g mol
0.082(SI0} - .
1
= 0019 g mol
1 - 0.5(0.9490)
A
kl =
= P AO =
= 53.38 = kl cl.5
= 0.196 (1 - 0.9490)
C
.
:: ..:!:6..
cks
.
1
= 2.035 x 1()4
,0.019)1.5
kl
= lco exp{f{1\--t)}
2.035 x 10' = 578.25 exp (E(llmole)
8.314
...
E = 7.55 x lOS
(-L_-L)}
500 510
J 1
gmo
P12-16 (b)
= 2naw .. 1 = 2{ 1.5) .. 1 = 2: second order
Euue = 2Eapp = 15.1 x loS~
ntrue
P12-16 (c)
~ = R ... / ~ Sa Ph C,o\O =
~V
(1
X
w·2f (2.035 X 1(4)49 (2.3 X
De
0.23
12-27
X
10'"
10') 0.196
= 1.40 x 1()6
$2
(--L)I12
(-l.:){--L}l12 (1.40 3x 1()6 ) -n+ 1
$n - 2 + 1
11 -
1.75 x 10-6
P12-16 (d)
To make the catalyst more effective, we should use a smaller diameter.
P12-16 (e)
CA = 0.01
R To T
=
•
-rA
=
527 C
leo exp {E.{..L -..l)l
k =
k
gmol
1 ;T
I
= 800 K
lOS (-L --L))
\ 8.314 500 800
= 578.25 exp 17.55 x
2.19 X 1()32
= k CA105 = 2.19 x
1()32 (0.01)1.5
= 2.19 x
Is
P12-17
(u)
2
d y
~2
n
-_. <l>nY = 0
Multiply by 2 Y ~.
2y dy _!(dY) = <l>2yn2dy_
dAdA dA
n
dA
Manipulating the L.H.S.
2
-.!(dy )2 = 2 dy d y
dA dA
dA dA,2
d(d )2 = <l>nY 2dy
y
dA dA
n
)2 =2<1>
yn+1
--+C 1
nn+l
C
d'JI
Y='JIA=~ , A=O y=O --·=0 thereforeCl =0
CAO
dA
Taking the deri vati ve of y and evaluating at A = 1
dy
.
( dA
•
~YI
dA A=1
=
[2~~yn+1
~- n + 1
=
)..,=1
1()29 g mole
ri<l>~
V~ + 1
12-28
The effectiveness factor is
nR2(DA dC A )
11 =
dr
r==k
kCn 4 nR3
As
3
In dimensionless form
_{~L
<I>~
11 -
A = \IfA , differentiating gives
dy = Ad\lf + \If
dA
dA
at A=1
= d~+1
dyl
dA 1..==1
dljl =
dA
dA
~24>~ -1
n+l
For larger <l>n
~=3~t
CDPI2-A
-------------, ----,
CDPI2-B (a) 3rd ed. 12-19 (a)
CDPI2-B (b) 3rd ed. 12-19 (b)
CDPI2-B (c) 3rd ed. 12-19 (c)
CDPI2-C (a) 3rd ed. 12-20 (a)
CDPI2-C (b) 3rd ed. 12-20 (b)
CDPI2-C (c) 3rd ed. 12-20 (c)
12-29
CDPI2-D (a) 3 rd ed. 12-21 (a)
CDPI2-D (a) 3 rd ed. 12-21 (b)
CDPI2-D (a) Individualized solution
CDPI2-E 2nd ed. 11-18
CDPI2-F 2nd ed. 11-19
CDPI2-G 2 nd ed.II-20
CDPI2-H 2nd ed. 11-21
CDPI2-1 2nd ed. 11-22
CDP12-J (a) 2nd ed. 12-7 (a)
CDPI2-J (b) 2 nd ed. 12-7 (b)
CDPI2-J (c) 2nd ed. 12-7 (c)
CDPI2-J (d) 2nd ed.12-7 (d)
CDPI2-J (e) 2nd ed. 12-7 (e)
CDPI2-K 2nd ed.12-9
CDPI2-L (a) 2nd ed. 12-8 (a)
CDPI2-L (b) 2nd ed. 12-8 (b)
CDPI2-L (c) 2nd ed. 12-8 (c)
CDP12-L (d) 2 nd ed. 12-8 (d)
CDPI2-M (a) 3 rd ed. CDPI2-L (a)
CDPI2-M (b) 3 rd ed. CDPI2-L (b)
CDPI2-M (c)
CDPI2-M (d)
12-30
CDP12-N 3rd ed. CDP12-M
CDP12-0
CDP12-P
CDP12-Q
CDP12-R (a) 3rd ed. CDP12-Q (a)
CDP12-R (b) 3rd ed. CDP12-Q (b)
CDP12-S
CDP12-T
CDP12-U
12··31
Solutions for Chapter 13 - Distributions of Residence Times
for Chemical Reactors
PI3-I No solution will be given,
P13-2 (a)
The area of a triangle (h=0,,044, b=5) can approximate the area of the tail :0.11
P13-2 (b)
(3)
(4)
CSTR
LG~[CSTRtCSTR
(6)
(5)
13-1
----...
I
J
l
AI
1
--.I
-I PFR
J
PFR
r-A~
PFR
el
PFR
AJ
J
P-
or
PFR
Recycle
(8)
~~~
(10)
(9)
~~
LFR
(11)
(12)
P13-2 (c)
For a PFRICSTR Series
13-2
.
IfkSTlU
Ifksrn
P13-2 (d)
X=O.75
For a PFR first order reaction.:
Da
= In(_I_) = In(4) = 1.39
l--X
where Da
= kr
For a CSTRfirst order reaction.:
Da
=
(_1_.) -1 =3
where Da
I-X
= kr
For a LFR first order reaction:
Solving iteratively Hilder approximate formula with an initial value
Dao (i.e . DapFR<Dao< DacSIR)
Daex p(
~a) + Da
0.75 = ---
.
4 + Da exp (
~c.!.. ) + Da
where Da
= kr
Da=258
The ratio of the Damkohler numbers is equal to the ratios of the sizes.
V PFR
Relative sizes: -----
V CSTR
LFR
= 0.46, -V- = 0.86
V CSTR
;
P13-2 (e)
For a PFR, r=5J5min, first order; liquid phase, irreversible reaction with k=O.lmin- 1
X
=1- e -k1· = 0.402
13-3
For a CSTR, r=.5.1.5min, first order; liquid phase, irreversible reaction with k=O.lmin-
X
l
.
=~~=0.402
l+kr
0.385
0.402
X CSTR
0.340
Page 851, only the RTD is necessary to calculate the conversion for a first-order reaction in any type of reactor. Not
good when the RTD has a long tail that is difficult to interpret or interpolate.
P13-2 (0
Decrease of 10 't' in temperature
See Polymath program P13-2-f.pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
Xbar
k
Cao
initial value
minimal value
0.0025446
0.75
0.0025446
0.75
o
o
o
o
X
o
o
tau
1000
500
5.0E+l0
1000
500
6.25E-08
tl
E2
E
o
o
maximal value
2.0E+04
0.6023837
0.0025446
0.75
0.9744692
1000
500
5.0E+lO
0.0023564
ODE Report (RKF45)
Differential equations as entered by the user
[ 1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[1 j k = .00493*exp(13300/1.9872*(1/323.15-1/313 . 15))
[2] Gao = .75
[3 J X = k*Gao*t/(1 +k*Gao*t)
[4J tau = 1000
[ 5 J t1 = tau/2
[ 6 J E2 = tauI\2/2/(tA3+.00001)
[7 J E = if (1<t1) then (0) else (E2)
13-·4
final value
2.0E+~
0.6023837
0.0025446
0.75
0.9744692
1000
500
6.25E-08
6.25E-08
0.70~--------------,
0.56
0.42
0.28
014
O.OOU--~-·-~----------'
o
-1000
8000 t
12000 16000 20000
The decrease of lOoe in temperature has the effect ofreducing the mean conversion by 14%.
Decrease in reaction orderfrom 2nd to pseudo ]'t
See Polymath program P 13-2-[:-2 .pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
Xbar
k
Cao
initial value
o
o
0 . 004
0 . 75
minimal value
o
o
0 . 004
0.75
X
o
o
tau
t1
E2
1000
500
5 . 0E+10
1000
500
6.25E-08
E
o
o
maximal value
2.0E+04
0 . 9391084
0 . 004
0.75
final value
2.0E+04
0.9391084
0 . 004
0.75
1
1
1000
500
5.0E+10
0 . 0024915
1000
500
6.25E-08
6 . 25E-08
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Xbar)/d(t) = X*E
Explicit equations as entered by the user
[1] k = 0 . 004
[2] Cao = . 75
[3] X = 1-exp(·k*t)
[4J tau = 1000
[5] t1 = tau/2
[ 6] E2 = tauA 2/21(tA 3+ . 00001)
[7] E = if (t<t1) then (0) else (E2)
13-5
1.0 , - - - - - - - - - - - - - - - - ,
0.8
0.6
0.4
0.2
0.0
U--_~
o
____
4000
8000
~
t
_ _ _ __ _ '
12000
16000
20000
The decrease in reaction order from 2 nd to pseudo 1st has the effect of increasing the exit conversion by 20%. The
smaller the dependency of the rate on C A means that when C A is below 1 moVdm3 then the rate of consumption of A is
larger and hence resulting in a larger conversion.
Exothermic reaction in adiabatic reactor:
See Polymath program P 13-2-f-3 . pol
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Xbar
X
T
Cao
tau
t1
E2
initial value
o
o
minimal value
o
o
o
o
323.15
0.75
1000
500
5.0E+10
323.15
0.75
1000
500
6.299E-·08
E
o
o
k
0.01
0.01
maximal value
2.0E+04
0.999375
final value
2.0E+04
0.999375
1
1
823.15
0.75
1000
500
5 . 0E+10
0.0022142
19 . 337202
823 . 15
0.75
1000
500
6.299E-08
6.299E-08
19.337202
ODE Report (STIFF)
Differential equations as entered by the user
[ 1 J d(Xbar)/d(t) = X*E
[2] d(X)/d(t) = k*(1-X)
Explicit equations as entered by the user
[ 1] T = 323.15+500*X
[2J Cao = .75
[ 3] tau = 1000
[ 4 ] t1 = tau/2
[5] E2 = tauJ\2/2/(tA3+.00001)
[6] E = if (t<t1) then (0) else (E2)
[7 J k = 0 . 01 *exp(8000/1.9872*(1/323.15-11T))
13-6
1.0
r--====-----------,
08
0.6
04
0.2
0. 0
U-_~
o
_ _ _ _ _ _ _ _ _ _____'
.tOOO
8000
12000
t
1600020000
The mean conversion Xbar, the integral, is estimated to be 99.9%. The reaction is adiabatic and exothermic as the
temperature increases to a maximum of 1373.15 K once the batch conversion within the globules has reached 100%
which occurs after only - 4 seconds . Hence, the adiabatic increase in temperature considerably increases the rate at
which conversions increases with time and hence also the final value.
P13-2 (g)
For a PFR, '!=40min, second order, liquid phase, irreversible reaction with k=O.OI dm3 /mol·min- I
X
= kTC~=0.76
1+ KTC Ao
For a CSTR, '!=40min, second order, liquid phase, irreversible reaction with k=O.OI dm3 /mol·min- 1 .
X
-(--)2 = kTC Ao
1·-X
--)
X = 0.58
Maximum Mixedness Model and Segregation model are given in E13-7
Maximum Mixedness :Model
See Polymath program P13-2-g . pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
z
x
cao
k
lam
ca
E1
E2
F1
F2
ra
E
F
EF
initial value
0
0
8
0.01
200
8
0 . 1635984
2 . 25E-04
5 . 6333387
0.9970002
-0.64
2 . 25E-04
0.9970002
0.075005
minimal value
-0-------0
8
0 . 01
0
3 . 2490705
0.0028734
2_25E-04
0
0.381769
--0 . 64
2 . 25E-04
0
0 . 0220689
maximal value
200
0 . 5938635
8
0_01
200
8
0 . 1635984
0 . 015011
5.6333387
0 . 9970002
-0.1055646
0.028004
0.9970002
0.075005
ODE Report (RKF45)
Differential equations as entered by the user
13-7
final value
200
0.5632738
8
0.01
0
3.4938093
0.028004
0 . 015011
0
0 . 381769
-0 . 122067
0 . 028004
0
0.028004
[1]
d(x)/d(z) = -(ra/cao+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[2] k = .01
[ 3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-1 0*lam"4-1.1802e-7*lam"3+ 1.35358e-5*lam"2-.000865652*lam+.028004
[6] E2 = -2.64e-9*lam"3+ 1,,3618e-6*lam"2-.00024069*lam+.015011
[7] F1 = 4..44658e-1 0/5*lam"5-1 ,,1802e-7/4*lam"4+ 1. 35358e-5/3*lam"3-.000865652/2*lam"2+,,028004 *Iam
[8] F2 = -( -9.3076ge-8*lam"3+5,,02846e-5*lam"2-.00941 *lam+.618231-1)
[ 9] ra = -k*ca"2
[ 1 0 ] E = if (Iam<=70) then (E1) else (E2)
[11] F = if (Iam<=70) then (F1) else (F2)
[12] EF = E/(1-F)
0.60
0.48
0. 36
0.24
0.12
0.00
0
40
80
z
120
160
200
56%
P13-2 (h)
Liquid phase, first order, Maximum Mixedness model
1
Rate Law: ' - fA =k1C A where kl = CAok =O.08minC A = CAo(l-X)
fA _
= -k1(1-X)
CAo
_dX =
dA
rA
+ E(A).X
CAO
I--F(A)
dX
E(A)
d)'- = --k(l- X)+ 1- F(A) X
dX = k(1- X)--
dz
E(A) X
1·_· F(A)
See Polymath program P13-2-h-l.pol
13-8
POLYMATH Results
Calculated values of the DEQ variables
Variable
z
x
cao
k
lam
ca
E1
E2
F1
F2
ra
E
F
EF
initial value
0
0
8
0 . 08
200
8
0 . 1635984
2.25E-04
5 . 6333387
0.9970002
-0.64
2.25E-04
0 . 9970002
0.075005
minimal value
0
0
8
0 . 08
0
1 . 7365447
0 . 0028731
2.25E-04
0
0.381769
-0.64
2.25E-04
0
0 . 0220691
maximal value
200
0 . 7829342
8
0 . 08
200
8
0.1635984
0.015011
5.6333387
0.9970002
-0.1389236
0.028004
0.9970002
0.075005
final value
200
0 . 7463946
8
0 . 08
0
2 . 0288435
0.028004
0.015011
0
0.381769
-0.1623075
0.028004
0
0.028004
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+EI(1-F)*x)
Explicit equations as entered by the user
[1 J cao = 8
[2J k = 0 . 08
[3J lam=200-z
[4 J ca = cao*(1-x)
[ 5 J E1 = 4..44658e-1 O*lam"4-1.1802e-7*lamI\3+ 1.35358e·5*lamI\2-.. 000865652*lam+ . 028004
[6] E2 = -2.64e-9*lamI\3+ 1.3618e-6*lamI\2-.00024069*lam+ . 015011
['7] Fl = 4.44658e-1 O/5*lamI\5-1 . 1802e-7/4*lam"4+1.35358e-5/3*lamI\3-.. 000865652/2*lamI\2+ . 028004*lam
[ SJ F2 = -(-9 . 3076ge-8*lamI\3+5 . 02846e-5*lamI\2-.00941 *lam+ . 618231-1)
[ 9 J ra = -k*ca
[ 10] E = if (lam<=70) then (E1) else (E2)
[11] F = if (lam<=70) then (F1) else (F2)
[12] EF = E/(1-F)
0.80 1
-.--:::::::::::======::::::===
0. 64
CJ
tU8
0.32
fJ 16
O.OOl-r_ _
o
~40 _ _~---_-~
___ J
80 z 120
160
200
At z = 200, i.e. A = a (exit), conversion X = 75 %.
The decrease in reaction order from 2nd to 151 has the effect of increasing the exit conversion by 19%. Once the
concentration of A drops below 1 mol/dm3 then the rate of consumption of A does not fall as rapidly (as the 2nd order
reaction) and hence resulting in a larger conversion.
Liquid phase, third order, Maximum Mixedness model
Rate Law: -
fA
= kC A
3
13-9
C A = CAo (l- x)
~A =-k'C Ao 2 (1-X?
Where
k'C A / =k=O.08min- 1
Ao
dX =~+ E(A) X
dA
CAo
1-F(A)
2(1_ X)3 _
dX = k'C
dz
Ao
E(A) X
1- F(A)
See Polymath program P L3-2-h-2,pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
z
x
cao
k
lam
ca
El
E2
F1
F2
E
F
EF
initial value
0
0
8
0.08
200
8
0.1635984
2.25E-04
5,,6333387
0,,9970002
2.25E-04
0.9970002
0,,075005
minimal value
0
0
8
0.08
0
4.1061501
0.0028733
2.25E-04
0
0.381769
2.25E··,04
0
0.0220689
maximal value
200
0.4867311
8
0.08
200
8
0.1635984
0.015011
5.6333387
0.9970002
0.028004
0.9970002
0.075005
final value
200
0.4614308
8
0.08
0
4.3085534
0.028004
0.015011
0
0.381769
0.028004
0
0.028004
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = ·'(-k*(1-x)1\3+E/(1-F)*x)
Explicit equations as entered by the user
[lJ cao = 8
[2] k = 0.08
[ 3] lam = 200-z
[4] ca = cao*(1-x)
[ 5] E1 = 4..44658e-1 0*lam"4-1.1802e-7*lamI\3+ 1,,35358e-5*lamI\2-.000865652*lam+.028004
[ 6 J E2 = -2,,64e-9*lamI\3+ 1,,3618e-6*lamI\2-,,00024069*lam+,,015011
[ 7 ] F1 = 4..44658e-1 O/5*lamI\5-1.1802e-7/4 *lam"4+ 1,,35358e-5/3*lamI\3-.,000865652/2*lamI\2+.028004*Iam
[8 J F2 = -(-9,,3076ge-8*lamI\3+5,,02846e-5*lamI\2-.00941 *lam+,,618231-1)
[9 J E = if (lam<=70) then (E1) else (E2)
[10] F = if (lam<=70) then (F1) else (F2)
[11] EF = E/(1-F)
At z = 200, ie., A = 0 (exit), conversion X = 46..1 %"
The increase in reaction order from 2nd to 3rd has the effect of decreasing the exit conversion by 10%. Once the
concentration of A drops below 1 moVdm3 then the rate falls rapidly and CA is not consumed so quickly, resulting in a
smaller conversion.
Liquid phase, half order, Maximum Mixedness model
Rate Law:
-fA
= k 'C A 112
13-10
CA
= C Ao (1- x)
~
C
= -k'C
Ao
-112(1_ X)1/2
Where k
008 nnn
. -1
= k 'C Ao -1/2 =.
Ao
=~+ E(/L) X
d/L CAo 1- F(/L)
dX
dX
dz
=kC
-1I2(1_X)1I2 _
Ao
E(/L) ,X
1- F(/L)
See Polymath program P13·2··h,J.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable
z
x
cao
k
lam
ca
E1
E2
F1
F2
E
F
EF
initial value
°°
8
0.08
200
8
0.1635984
2,,25E-04
5.6333387
0.9970002
2,25E-04
0.9970002
0,075005
minimal value
°°
°
8
0,,08
0,532148
0.002873
2.25E-04
°
°
0,,381769
2.25E-04
0,0220677
maximal value
200
0,,9334778
8
0,,08
200
8
0,,1635984
0.015011
5.6333387
0,,9970002
0.028004
0.9970002
0,075005
final value
200--0.9038179
8
0,,08
°
0.7694568
0,,028004
0.015011
°
°
0.381769
0.028004
0.028004
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)A(.5)+EI(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[2] k = 0.,08
[31 lam = 200-z
[4J ca=cao*(1-x)
[ 5] E1 = 4,44658e-1 0*lam"4-1, 1802e-7*lamA3+ 1.,35358e-5*lamA2-,,000865652*lam+028004
[ 6 J E2 = -2.64e-9*lam A3+ 1,,3618e·,6*lamA2-,,00024069*lam+.015011
[ 'I ] F1 = 4,44658e-1 0/5*lamA5-1, 1802e-7/4 *lam"4+ 1,35358e-5/3*lamA3-000865652/2*lamA2+,,028004*Iam
[8] F2 = -(-9.,3076ge,'8*lamA3+5.,02846e,·5*lamA2··.00941 *lam+ . 618231-1)
[9] E = if (lam<=70) then (E1) else (E2)
[10] F = if (lam<=70) then (F1) else (F2)
[11] EF=E/(1-F)
13-11
1.0
~---------------,
0.8
0,6
0.4
0.2
0.0 4)'---~4-0---80---1~2-0--1-6f-)---l200
z
At z = 200, i.e. A = 0 (exit), conversion X =90 %.
The decrease in reaction order from 2nd to Y2 has the effect of increasing the exit conversion by 34%. The smaller the
dependency of the rate on C A means that when C A falls below 1 moVdm3 then the rate of consumption of A does not fall
as rapidly ( as the 2nd order reaction) and hence resulting in a larger conversion.
P13-2 (i)
Assymetric RTD:
See Polymath program P IJ·2-i-l"pol
rn
POLYMA
Results
Calculated values of the DEO variables
variable
t
ca
cb
cc
cabar
cbba:r
ccbar
cd
ce
cdbar
cebar
T
k1
k2
E1
E2
rc
k3
ra
re
E
rb
Sed
Sde
rd
initial value
0
1
1
0
0
0
0
0
0
0
0
350
1
1
-0.004
-27.402
1
1
-2
0
-0 . 004
-1
0
0
1
minimal value
0
0.0228578
0.2840909
0
0
0
0
0
0
0
0
350
1
1
--27.414373
-27.402
0.0064937
1
-2
0
-0.0272502
-1
0
0
-0.0522659
maximal value
2.52
1
1
0.3992785
0.1513598
0.4543234
0.3570959
0 . 3178411
0.3166306
0.3029636
0.1782569
350
1
1
0.958793
0 . 9557439
1
1
-0.0293515
0.1762951
0.958793
-0 . 0807076
1.5284379
42.398031
1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(ca)/d(t) = ra
[2 J d(cb)/d(t) = rb
13-12
final value
2.52
0 . 0228578
0.2840909
0.3992785
0.1513306
0 . 4539723
0.3566073
0 . 2612331
0.3166306
0.3026417
0.1778722
350
1
1
-27.414373
-0.0272502
0.0064937
1
-0.0293515
0.0742139
-0.0272502
-0.0807076
1. 5284379
0.8250406
"·0.0513561
[3] d(cc)/d(t) = rc
[4] d(cabar)/d(t) = ca*E
[5] d(cbbar)/d(t) cb*E
[6] d(ccbar)/d(t) = cc*E
[7] d(cd)/d(t) = rd
=
[8] d(ce)/d(t) = re
[9 j d(cdbar)/d(t) = cd*E
[10] d(cebar)/d(t) = ce*E
Explicit equations as entered by the user
[1] T= 350
[2 J k1 = exp((5000/1 . 987)*(1/350-1/T))
[3 J k2 = exp((1 000/1.987)*(1/350-1/T))
[4] E1 = -2.104*t"4+4 . 167*tI\3-1 . 596*tI\2+0 . 353*t-0.004
[5] E2 = -2 . 104*t"4+17 . 037*tI\3-50.247*tI\2+62.964*t-27.402
[ 6 J rc = k1 *ca*cb
[7] k3 = exp((9000/1 . 987)*(1/350-1/T))
[8] ra = -k1 *ca*cb-k2*ca
[9] re = k3*cb*cd
[ 10] E = if(k=1 . 26)then(E1 )else(E2)
[ 11] rb = -k1 *ca*cb-k3*cb*cd
[12] Scd = cc/(cd+.000000001)
[13] Sde = cd/(ce+ . 000000000001)
[14] rd = k2*ca-k3*cb*cd
If the temperature is raised, the conversion of A increases. The selectivity Sc/d increases with temperature and Sd/e
decreases with increasing temperature
Bimodal RTD
See Polymath program P 11-2-i-·2.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable
'z
ca
cb
cc
F
cd
ce
cbo
cao
ceo
cdo
ceo
lam
T
k2
ki
rc
k3
E1
E2
E3
re
ra
rb
E
EF
initial value
°
°0.99
0
1
1
0
1
1
0
0
0
6
350
1
1
1
1
346.34561
6737 . 4446
0 . 00911
0
-2
-1
0.00911
0 . 911
minimal value
°0.5350642
0.2660482
-0°. 0033987
0
°11
0
0
0
0
350
1
1
0 . 1424065
1
0.1019474
0.0742397
0.0061156
0
-2
--1
0 . 0061156
0 . 2083818
maximal value
6
1
1
0.2872257
0 . 99
0.2692177
0.1929233
1
1
0
0
0
6
350
1
1
1
1
346.34561
6737.4446
1.84445
0.1694414
-0 . 4084547
-0.28635
0.6288984
1.8694436
13-13
final value
--6--.--
0.2660482
0.5352659
0.2745726
-0.0033987
0 . 2692177
0.1901615
1
1
0
0
0
0
350
1
1
0 . 1424065
1
0 . 20909
925.46463
1 . 84445
0 . 144103
-0.4084547
-0.2865096
0 . 20909
0 . 2083818
rd
Sed
Sde
1
0.1219452
1
o
1 . 0856272
17.698981
o
o
o
0.1219452
1.0198908
1. 4157321
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(ca)/d(z) = -(-ra+(ca-cao)*EF)
[2] d(cb)/d(z) = -(-rb+(cb-cbo)*EF)
[3] d(cc)/d(z) = -(-rc+(cc-cco)*EF)
[ 4] d(F)/d(z) = -E
[5] d(cd)/d(z) = -(-rd+(cd-cdo)*EF)
[6] d(ce)/d(z) = -(-re+·(ce-ceo)*EF)
Explicit equations as entered by the user
[1] cbo= 1
[2) cao = 1
[3) cco=O
[4) cdo=O
[5) ceo=O
[6) lam=6-z
[7) T=350
[8) k2 = exp«1000/1 . 987)*(1/350-1/T))
[9] k1 = exp«5000/1 . 987)*(1/350-1/T))
[1.0) rc=k1*ca*cb
[11) k3 = exp«9000/1 . 987)*(1/350-1/T))
[1.2) E1 = 0.47219*lam"4-1.30733*lamA3+0.31723*lamA2+0.85688*lam+0.20909
[13) E2 = 3.83999*lamA6-58.16185*lamA5+366.2097*lam"4-1224 . 66963*lamA3+2289.84857*lamA22265.62125*lam+925.46463
[14] E3 = 0.0041 0*lam"4-0.07593*lamA3+0 . 52276*lamA2-1.59457*lam+ 1.84445
[ 15) re = k3*cb*cd
[16] ra = -k1 *ca*cb-k2*ca
[17) rb = ·k1 *ca*cb-k3*cb*cd
[18] E = if(lam<=1 . 82)then(E1 )else(if(lam<=2 . 8)then(E2)else(E3))
[19) EF=EI(1-F)
[20] rd = k2*ca-k3*cb*cd
[21] Scd = cc/(cd+.0000000001)
[22) Sde = cd/(ce+.0000000001)
If the temperature is raised, the conversion of A increases. The selectivity Sc/d increases with temperature and Sdle
decreases with increasing temperature
P13-2 G)
Exothermic Reaction: E=45Kj/mol
See Polymath program P 13-2+ I . pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
z
x
cao
T
lam
ca
E1
E2
F1
F2
initial value
o
o
8
320
200
8
0 . 1635984
2.25E-04
5.6333387
0 . 9970002
minimal value
o
o
8
320
o
0 . 2971785
0.0028744
2.25E-04
o
0.381769
maximal value
200
0.9628524
8
464.4279
200
8
0.1635984
0.015011
5.6333387
0.9970002
13-14
final value
200
0.9579239
8
463.68858
o
0.3366089
0.028004
0.015011
o
0.381769
0.01
2.25E-04
0
0.0220681
-0,8227063
0 . 01
2.25E-04
0 . 9970002
0.075005
-0.64
k
E
F
EF
ra
1,924805
0.028004
0 . 9970002
0.075005
-0.1699892
1. 8893687
0,028004
0
0,028004
-0.2140759
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[ 2 1 T = 320+ 150*x
(3] lam = 200-z
[ 4] ca = cao*(1-x)
(5 J E1 = 4.44658e-1 0*lamA4-1.1802e-7*lamA3+ 1.35358e-5*lamA2-,000865652*lam+,028004
( 6] E2 = -2,,64e-9*lam A3+ 1.3618e-6*lamA2-,,00024069*lam+.015011
(7] F1 = 4..44658e-1 0/5*lamA5-1 ,,1802e-7/4*lamA4+ 1,35358e-5/3*lamA3-,000865652/2*lamA2+.028004*lam
[ 8] F2 = -(-9,,3076ge-8*lamA3+5,,02846e-5*lamA2- . 00941 *lam+,,618231-1)
(9] k= .01*exp(45000/8.314*(1/320-1/T»
(10] E = if (lam<=70) then (E1) else (E2)
(11] F = if (lam<=70) then (F1) else (F2)
(12J EF = E/(1-F)
( 1 3] ra = -k*caA2
--_._--_ _---
LO , - - - . - - - . - - - - - - - - - - - - - ,
.
0.8
0. 6
0.4
0,,2
80
40
160
120
200
Endothermic Reaction: E=45Kj/mol
See Polymath program P 13·2+ 1"po)
POLYMATH Results
Calculated values of the DEQ variables
Variable
z
initial value
minimal value
x
o
o
cao
8
8
T
320
200
289.82835
8
5.5862683
0.0028739
2.25E-04
lam
ca
El
E2
F1
0,.1635984
2 . 25E-04
5.6333387
o
o
o
o
maximal value
200
0,,3017l58
8
320
200
8
0 . 1635984
0.015011
5,6333387
13-15
final value
200
0 . 2860515
8
291. 39485
o
5.7l15879
0.028004
0,015011
o
F2
k
E
F
EF
0.9970002
0.01
2.25E-·04
0.9970002
0.075005
-0.64
ra
0.381769
0.0017191
2 . 25E-04
0
0 . 0220679
-0.64
0.9970002
0.01
0.028004
0.9970002
0.075005
-0.0536459
0.381769
0.0019006
0.028004
0
0.028004
-0.0620023
ODE Report (RKF45)
Differential equations as entered by the user
[1 J d(x)/d(z) = -(ra/cao+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[ 2] T = 320-1 OO*x
[3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-1 0*lam"4-1.1802e-7*lamA3+ 1. 35358e-5*lamA2-.000865652*lam+.028004
[ 6] E2 = -2.64e-9*lamA3+ 1.3618e-6*lamA2-.. 00024069*lam+ ..Q15011
[7] F1 = 4.44658e-1 0/5*lamA5-1 . 1802e-7/4*lam"4+1.35358e-5/3*lamA3-.000865652/2*lamA2+.028004*lam
[ 8] F2 = -( -9.3076ge-8*lamA3+5 . 02846e-5*lamA2-.00941 *lam+.618231-1)
[9] k = .01*exp(45000/8 . 314*(1/320-1rr»
[10J E = if (lam<=70) then (E1) else (E2)
[11] F = if (Iam<=70) then (F1) else (F2)
[12] EF = E/(1-F)
[ 13 1 ra = -k*caA2
0.32
GJ
0..16
0.08
0.000
40
80
z
120
160
200
P13-2 (k)
Base case:
ODE Report (RKF45)
Differential equations as entered by the user
[ 1 ] d(Ca)/d(t) = ra/vo
[2 J d(Cb)/d(t) = rb/vo
[ 3 J d(Cc)/d(t) = rclvo
Explicit equations as entered by the user
[1] vo= 10
[2] k1 = 1
[3] k2=1
[ 4] tau = 1.26
[5] ra = -k1 *Ca
[ 6 ] rc = k2*Cb
13-16
[7] Cao = 1
[ 8] x = 1-Ca/Cao
[9] rb = k1 *Ca-k2*Cb
PFR
CA
CB
Cc
X
K 1/K2=2
0.080
0.406
0.513
0.919
K 1/K2=1
0.284
0.357
0.359
0.716
K 1/K2=O.5
0.284
0.203
0.513
0.716
See Polymath program P13·2-k·-2. pol
POLYMATH Results
NLES Solution
Variable
ca
cb
cc
cao
tau
cbo
ceo
k1
k2
ra
rc
rb
Value
0 . 4424779
0.2466912
0.3108309
1
1..26
0
0
1
1
--0.4424779
0 . 2466912
0 . 1957867
lni Guess
fIx)
4.704E-10
-3 . 531E-10
o
1
o
o
NLES Report (safenewt)
Nonlinear equations
[1] f(ca) = caOHa*tau-ca = 0
[2] f(cb) '" CboHb*tau-cb = 0
[3 J f(cc) = CCOHc*tau-cc = 0
Explicit equations
[1]
[2 1
[3]
[4J
[51
[6]
[7]
[ 8]
[9]
CSTR
CA
CB
Cc
~-
cao = 1
tau = 1. 26
cbo = 0
cco = 0
k1 = 1
k2=1
ra=-k1*ca
IC = k2*cb
rb = k1 *ca-k2*cb
K 1/K2=1
-
0.443
0.247
0.311
0.557
-
KI/K2=2
---
0.284
0.317
0.399
0.716
K 1/K2=O.5
0.443
0.158
0.399
0.557
See Polymath program P i3-2-k-3 pol
POLYMA TH Results
Calculated values of the DEQ variables
13-17
--
variable
t
ca
cb
cc
cabar
cbbar
ccbar
k1
k2
cao
E1
E2
ra
rc
x
rb
E
initial value
0
1
0
0
0
0
0
1
1
1
-0.004
-27 . 402
-1
0
0
1
-0.004
minimal value
0
0.0804596
0
0
0
0
0
1
1
1
-·27.462382
-27.402
-1
0
0
-0.1353314
-0 . 0272502
maximal value
2 . 52
1
0.3678466
0.7167822
0.3050655
0 . 3350218
0.3476989
1
1
1
0.9523809
0 . 9568359
-0.0804596
0.3678269
0 . 9195404
1
0.9568359
final value
2.52
0.0804596
0.2027582
0.7167822
0.304964
0.3347693
0.3468313
1
1
1
-27 . 462382
-0.0272502
-0 . 0804596
0.2027582
0.9195404
-0.1222986
-0.0272502
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(ca)/d(t) = ra
[2] d(cb)/d(t) = rb
[3] d(cc)/d(t) = rc
[4] d(cabar)/d(t) = ca*E
[5] d(cbbar)/d(t) = cb*E
(6) d(ccbar)/d(t) = cc*E
Explicit equations as entered by the user
[1] k1 = 1
[2J k2=1
[3) cao = 1
[ 4] E1 = -2 . 104*tll4+4.164 *tA3-1 . 596*tA2+0.353*t·0.004
[5 J E2 = -2.1 04*tll4+ 17.037*tA3-50 . 247*tA2+62.964*t-27.402
[6) ra = -k1*ca
[ "7 ] rc = k2*cb
[ 8] x = (cao..ca)/cao
[9] rb = k1 *ca-k2*cb
[10 J E = if(k=1 . 26)then(E1 )else(E2)
Asymmetric RTD
~~gation Model
CA
CD
Cc
--
._-
X
KJ1K2=2
0.110
0.390
0.486
0.89
KJ1K2=1
0.306
0.335
0.347
0.694
--_.
---,_._-K}1K2=O.5
0.306
0.195
0.486
0.694
See Polymath program P13-2-k-4 . pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
ca
cb
cc
cabar'
initial value
o
minimal value
o
1
9.1l9E-04
o
o
o
o
o
o
maximal value
7
1
0 . 3678325
0.9927049
0.3879174
13-18
final value
7
9 . 119E-04
0 . 0063832
0 . 9927049
0.3879174
--
cbbar·
ccbar
0
0
1
1
1
0.20909
925.46463
-1
0
0
1
0
1 .. 84445
0.20909
k1
k2
cao
E1
E2
°r·a
rc
x
rb
E4
E3
E
0
0
1
1
1
0.1017293
0.074462
-1
0
0
-0 . 1353022
0
0 . 0061369
0
0.2782572
0.3271713
1
1
1
707 . 06552
3.072E+04
-9.1l9E-04
0 . 3675057
0.9990881
1
0
1.84445
0 . 628067
0 . 2782572
0 . 3271713
1
1
1
707 . 06552
3 . 072E+04
-9.1l9E-04
0.0063832
0.9990881
-0 . 0054713
0
0.09781
0
ODE Rel10rt {RKF45}
Differential equations as entered by the user
[ 1 J d(ca)/d(t) = ra
[2 J d(cb)/d(t) = rb
[3] d(cc)/d(t) = rc
[4 J d(cabar)/d(t) = ca*E
[5 J d(cbbar)/d(t) = cb*E
[6 J d(ccbar)/d(t) = cc*E
Explicit equations as entered by the user
[lJ k1 = 1
[2] k2=1
[3] cao = 1
[4] E1 = 0.47219*t"4-1 . 30733*tA3+0 . 31723*tA2+0 . 85688*t+0.20909
[ 5] E2 = 3.83999*tA6-58 . 16185*tA5+366 . 20970*t"4-1224.66963*tA3+2289.84857*tA2-2265.62125*t+925.46463
[ 6 J ra = -k1 *ca
[7 J rc = k2*cb
[ 8 J x = (cao-ca)/cao
[9 J rb = k1 *ca-k2*cb
[10] E4 = 0
[ 11 J E3 = 0 . 0041 0*t"4-0 . 07593*tA3+0 . 52276*tA2-1 . 594S7*t+ 1. 84445
r12] E = if(k=1 . 82)then(E1 )else(if(k=2 ..8)then(E2)else(if(k6)then(E3)else(E4)))
BimodalRTD
.-~--
St~gregation
Model
K 1/K2=1
0.388
0.278
0.327
0.612
Kl/K2=2
0.213
0.350
0.430
0.787
Maximum
Mixedness
K 1/K2=1
KI/K2=2
K 1/K2=O.5
CA
CB
Cc
0.306
0.335
0.347
0.694
0.110
0.390
0.486
0.89
0.306
0.195
0.486
0.694
CA
CB
Cc
X
._----_.
-
K 1/K2=O.5
0.388
0.175
0.430
0.612
._---
Asymmetric RTD
,.------------
X
'-'--.
-.
P13-2 (l-r) No solution will be given at this time.
13-19
..
PI3-3
Equivalency Maximum Mixedness and Segregation model for first order reaction:
dC A
dA
Rearranging:
-[k + 1-F(A)
E(A) ]c
A
Using the integration factor:
by definition
gives:
e
=
CAoE(A)
1-F(A)
k E(A) ]dA]
d CAe -I[ +l-F(A)
_ [k+ E(A) JdA
[ CAo
E(A)e I I-F(A)
dA
=1-F(A)
E(A )d(A) = dF(A)
-I[ I-F(A)
E(A) ]dA
I dF
= e - I-F(A) = ed[Ln(l-F(A))) = (1- F(A))
changing the variables from')... to t in the RHS integral:
kt
~e-kA(1-. F(A)) = f[E(t )e- (1- F(t ))]dt
C Ao
1- F(t)
_
Ak
CA = CAo e ( ) fE(t)e-ktdt
1-F A
(1)
r
Exit concentration is when ')...=0, F(O)=O hence eqn (1) becomes:
CA = CAo
E(t)e-ktdt
This is the same expIession as for the exit concentration for the Segregation model.
13-20
P13-4 (a)
Mean Residence Time
By definition
I(
E t)dt
= 1.
_
_ llr
The area of the semicircle representing the E(t) is given by A
2
o
T
~ J!r
Fm const,nt volumetrk flow t m
~ T ~ J!r ~ 0.8 m; n .
P13-4 (b)
Variance
00
0'2
= I(t o
~
r? E(t)dt = It 2 E(t )dt - r2
0
21"
5
0
It 2 E(t )dt = It 2 ~r2 - (t -- r)2 dt = _r 4 J[cos 2(x)+ 2cos(x)+ llsin 2(x)dx = ~r4
o
0
1l
8
0'2
= 51l r 4 _r2 =_1_=0.159
8
21l
Using Polymath:
See Polymath program P13-4-b pol
POL,YMA TH Results
Calculated values of the DEQ variables
Variable
t
sigma
tau
t1
E2
E
initial value
o
o
0.7980869
1.5961738
o
o
minimal value
o
o
0 . 7980869
1..5961738
o
o
maximal value
1..596
0.15931.61
0 . 7980869
1.5961738
0 . 7980614
0.7980614
ODE Report (RKF45)
Differential equations as entered by the user
[ 1) d(sigma)/d(t) = (t-tau)A2*E
Explicit equations as entered by the user
[1] tau = (2/3 . 14)"0 . 5
[ 2 ] t1 = 2*tau
[ 3 J E2 = (t*(2*tau-t»A( 112)
[4] E = if (1<t1) then (E2) else (0)
13-21
final value
1. 596
0.1593161
0 . 7980869
1.5961738
0.0166534
0.0166534
2
--1
and
0.20
016
1-
0.12
sigma
0.08
0.64 t
1.28
0.96
1.60
P13-4 (C)
Conversion predicted by the Segregation model
x
=
J (t)E(t )dt
X
o
X(t) =l_··e-kt
21"
X=l-
Je-kt(r2-(t-lljl2dt
o
See Polymath program P 13-4·c . po]
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
Xbar
tau
t1
initial value
-0-----
o
0.7980869
1. 5961738
minimal value
o
o
0.7980869
1. 5961738
E
o
o
o
o
k
0.8
0.8
E2
X
o
o
maximal value
1. 596
0,,4447565
0.7980869
1. 5961738
0,,7980671
0.7980671
0.8
0,7210716
ODE Report (RKF45)
Differential equations as entered by the user
[ 1] d(Xbar)/d(t) =X*E
Explicit equations as entered by the user
[ 1 J tau = (2I3.14)1I().,5
[2 J t1 = 2*tau
[3 J E2 = (t*(2*tau-t))A(1/2)
[4 J E = if (kt1) then (E2) else (0)
[5J k=.8
[6 J X::: 1-exp(-k*t)
13-22
final value
1. 596
0.4447565
0.7980869
1. 5961738
0.0166534
0.0166534
0.8
0,7210716
0.5 , - - - - - - - - - - - - - - - - - ,
0.4
OJ
0.2
OJ
0.0
L..-.......,~
0.000
_ _ _ _ _ _ _ _ _____'
0319
o 638t
0958
1277
1.596
X =44.5%
P13-4 (d)
Conversion predicted by the Maximum Mixedness model
dX =.3L+ E(A) X
dA
rA
CAo
1-·F(A)
= -·kCA = -kC Ao(l- X)
dX
E(A)
-dA = -"k(l- X)+ 1- F(A) X
dX
E(A)
d~- = k(l- X)-l-F(A) X
See Polymath program PI J-4-d pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
z
x
F
k
lam
tau
E1
E
initial value
0
0
1
0. 8
1. 596
0.7980869
0 . 0166534
0 . 0166534
minimal value
0
0
-5 . 053E-04
0.8
0
O. 7980869
0
0
maximal value
1..596
0 . 4445289
1
0. 8
1..596
0 . 7980869
0.7980666
0.7980666
ODE Report (RKF45)
Differential equations as entered by the user
[1 J d(x)/d(z) = -(-k*(1-x)+E/(1-F)*x)
[2 J d(F)/d(z) = -E
•Explicit equations as entered by the user
k= . 8
[2] lam = 1 . 596-z
[3] tau = (2/3.14)A0 . 5
[4] E1 = (tau A2-(lam-tau)A2)A0.5
[ 5] E = if (lam<=2*tau) then (E1) else (0)
[1]
13-23
final value
1":5%"-0.4445289
--5.053E-04
0.8
0
0 . 7980869
0
0
0.5 , - - - - - - - - - - - - - - - ,
0.4
D
0.3
0. 2
OJ
0.0
~-~--~-~--~-----'
0.000
0.319
x = 44.5%
0.638
z
0.958
1.277
1.596
as for the Segregation Model, but we knew this because for first order reactions
XMM
P13-5 (a)
The cumulative distribution function F(t) is given:
I
F(t)
-
0.5 -
I
20
40
The real reactor can be modelled as two parallel PFRs:
Therelative
I
E(t) = { --J(t -
1"1
4
3
)+-J(t 4
1"2)
Mean Residence Time
1
tm = JtdF = (10 min * 1) + (20min*O.75) = 25 min
o
~)""·························T--·-·····
t
F
13-24
Xseg =
or
P13-5 (b)
Variance
=
(j2
Jrt -T)2 E(t)dt
= S{t -
!
!
tm )2[ ~o{t - Tj)+ o{t - T,}Jdt = ~ {Tj - tm )2 + {T2 - tm)2 =
o
= 75min 2
P13-5 (C)
For a PFR, second order, liquid phase, irreversible reaction with k = 0.,1 dm3 /mol min- l , t
mol/dm3
X
=
k'lC Ao
1 + k'lC Ao
= 25 min and C Ao = 1..25
= 0.758
FOI a CSTR, second order, liquid phase, irreversible reaction with k = 0 ..1 dm3 /mol"min- l , t
mol/dm3
X
2
(l-X )
= k'lC Ao
-t
= 25 min and C Ao = 1.25
= 0.572
X
For two parallel PFRs, tl = 10 min and tz = 30 min, Faol = 1/4Fao and Fao2 = 3/4Fao , second order, liquid phase,
ineversible reaction with k = 0..1 dm3 /mol,min- l and C Ao = 1.25 mol/dm3
CAl
= C Ao -
C A2
= C Ao -
X
=
kT1C Ao
1+ kTIC Ao
kT2 C Ao
CAo = 0.556mol / dm
--C Ao
1+ kT 2 C Ao
vC Ao -!VCAI -iVC A2
4
4
vC Ao
3
= 0.263mol dm 3
= 0.731
P13-5 (d)
I-Conversion predicted by the Segregation Model
13-25
x=
1
X (t)E(t )dt =
o
1
0
[.!
kC Aot
8(t 1+ kC Aot 4
1"1 )
+ ~. 8(t - 1"z )]dt =
4
1 kCAo 1"1 +3 kCAoTz =0.731
4 1 + kC Ao 1"1 4 1 + kC Ao 1"Z
2-Conversion predicted by the Maximum Mixedness model
dX =~+ E(A) X
dA CAo I-F(A)
Z
r A =-kC/ =-kC Ao (I-X?
dX = -kC (1- X)Z + E(A) X
dA
Ao
I-F(A)
We need to change the variable such the integration proceeds forward:
.dX =kCAo(l-X? __E(T-~X
dz
I-F(T-z)
See Polymath program P 13-5-d.pol
POLYMATH Results
Calculated values of the DEO variables
Variable
--z
x
F
cao
k
lam
ca
t1
t2
E3
ra
E2
El
E
EF
initial value
0
0
0.9999
1. 25
0.1
40
1..25
10
30
0
-0.15625
1..25
1.25
0
0
minimal value
-0---0
-1. 081E-04
1. 25
0.1
0
0 . 3614311
10
30
0
-0.15625
1. 25
1.25
0
0
maximal value
40
0.7125177
0.9999
1.25
0.1
40
1.25
10
30
0
·-0 . 0130632
1.25
1.25
1.25
2.8604931
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) =-(ra/cao+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[ 1 J cao = 1 . 25
[2J k =.1
[ 3 J lam = 40-z
[4 J ca = cao*(1·x)
[5 J t1 = 10
[6] t2=30
[7] E3 = 0
[ 8] ra = -k*caA 2
[9 J E2 = 0 . 75/(t2*2*(1-0.99))
[10] E1 =0.25/(t1*2*(1-0.99))
13-26
final value
40
0,,7061611
-1. 081E-04
1.. 25
0.1
o
0.3672986
10
30
o
-0.0134908
1.25
1.25
o
o
[11]
[12]
E = if ((Iam>=0 . 99*t1 )and(lam<1.01 *t1» then (E1) else( if ((lam>=0.99*t2)and(lam<1.01 *t2» then (E2) else (E3))
EF = E/(1-F)
0.731
0.706
0.731
0.758
X CSTR
0.572
P13-5 (e)
Adiabatic Reaction E=10000cal/mol and T
= 325 -
500X
= 325 - 500X and the constitutive equation for
k
.
(45000/8314*(l/325-1IT)). th MM
d I
k - 325 e
III e
mo e .
Introducing the enthalpy balance: T
The conversion is drastically reduced.
P13-5 (0
Conversion Predicted by an ideal laminar flow reactor
For a LFR, second order, liquid phase, ineversible reaction with k = 0..1 dm3 imolmin- 1,
mol/dm3
T
We apply the Segregation model, using Polymath:
X
=
ktC A
0
1 + ktC Ao
and E(t)
{Ofor t < 12.5min
=
625/(2t 3 )min-1 for t ~ 12.5min
See Polymath program P ].3-5-e .po]
POLYMA ~H Results
Calculated values of the DEQ variables
Variable
t
xbar
cao
k
tau
El
t1
E
x
initial value
0
0
1. 25
0.1
25
3_125E+06
12,,5
0
0
minimal value
0
0
1.. 25
0.1
25
1.157E-05
12.5
0
0
maximal value
300
0.7077852
1.. 25
0.1
25
3.125E+06
12 . 5
0.0991813
0.974026
ODE Report (RKF45)
Differential equations as entered by the user
[ 1] d(xbar)/d(t) = x*E
Explicit equations as entered by the user
[ 1 ] cao = 1 . 25
[2l k = . 1
[3] tau = 25
[4 J E1 = tauJ\212/(tJ\3+0 . 0001)
[ 5 ] t1 = tau/2
[ 61 E = if (1<t1) then (0) else (E1)
[7 1 x = k*cao*tJ(1 +k*cao*t)
13-27
final value
300
0" 7077852
1.. 25
0.1
25
1,,157E-05
12 . 5
1.157E-05
0.974026
= 25 min and C Ao = 1..25
We can compare with the exact analytical formula due to Denbigh .
X~ D+ -(D;}n(l + 21 Da)] ~ 0.709
with
na= kC Ao'
P13-6 (a)
E(t)
Mean Residence Time
00
By definition
JE(t)dt
= 1.
The area of the triangle representing the E(t) is given by A = 2t1 0.2 = 1
2
o
tl =5min.
----7
t if t < tl
Z'l
tl
E(t)
2t1 )if tl 5: t 5: 2tl
tl
o otherwise
= - -;. (t -
P13-6 (b)
Variance
00
a
2
00
= J(t-t m ?E(t)dt = Jt 2 E(t)dt-t;
o
0
and
a
2
7
6
=-t
2
m
-t
2
m
t;
25
. 2
=-=--=4.167mm
6
6
See Polymath program P 13-6-b vol
POLYMA TH Results
13-28
Calculated values of the DEQ variables
Variable
initial value
sigma
tau
tl
El
E2
minimal value
o
o
o
o
5
10
5
o
o
o
t
10
0.4
o
o
E
maximal value
final value
10
10
4.1666667
5
10
0.4
0.4
0,,1989341
4,,1666667
5
10
0.4
o
o
ODE Report (RKF45)
Differential equations as entered by the user
[1 J d(sigma)/d(t) = (t-tau)1\2*E
Explicit equations as entered by the user
[1] tau = 5
[2] t1 = 2*tau
[3] E1 = tltaul\2
[ 4] E2 = -(t-2*tau)/tauI\2
[5] E = if (ktau) then (E1) else (if(k=t1 )then(E2)else(O))
P13-6 (C)
For a PFR, second order, liquid phase, irreversible reaction with kCAo =02 min- 1,"C=5 min
X
=
ktC Ao
= 0.5
1+ktCAo
For a CSTR, second order, liquid phase, ineversible leaction with kC Ao =0.2 min- 1,"C=5 min
.
X
(l-X
)2 = ktC Ao --"* X = 0.382
P13-6 (d)
I-Segregation model
See Polymath program Pl3-6cl·l ..pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
Xbar
kl
X
tau
t1
El
E2
E
initial value
0
0
0.2
0
5
10
0
0.4
0
minimal value
0
0
0. 2
0
5
10
0
-0 . 2
0
maximal value
15
0,,4767547
0,,2
0 . 75
5
10
0.6
0. 4
0 . 1978698
ODE Report (RKF45)
Differential equations as entered by the user
[ 1. ] d(Xbar)/d(t) = X*E
13-29
final value
15
0,,4767547
0. 2
0,,75
5
10
0.6
-0 . 2
0
Explicit equations as entered by the user
[1] k1 =.2
[2] X=k1*t/(1+k1*t)
[3] tau = 5
[4] t1 = 2*tau
[5] E1 = t/tauA 2
[6] E2 = -(t-t1 )/tau A 2
[7] E = if (t<tau) then (E1) else(if(t<=t1 )then(E2)else(O))
0.,5
r---------------,
0.4
0.3
0.2
0.1
12
15
X =47.7%
2-Maximum Mixedness Model
(-"..!L.
dX __
+ E(T - Z)
dz C Ao I-F(T-z)
xJ
rA =-kCA 2 = -kCAo 2 (1-- X )2
dX =k'(I-X?-. E(T-_~X
dz
1,- F(T _. z)
See Polymath program P13-6·d-2,pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
z
x
F
k
lam
tau
El
t1
E2
E
initial value
0
0
1
0.2
20
5
0.8
10
-0.4
0
minimal value
0
0
-7.513E-06
0.2
0
5
0
10
-0.4
0
maximal value
20
0.6642538
1
0.2
20
5
0.8
10
0.4
0 . 1994823
ODE Report (RKF45)
13-30
final value
20
0.4669205
-7.513E-06
0. 2
0
5
0
10
0.4
0
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)1I.2+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] k = 0.2
[2] lam = 20-z
[3 J tau = 5
[4) E1 = lam/taull.2
[5] t1 = 2*tau
[ 6] E2 = -(lam-t1 )/tauIl.2
['7] E = if (Iam<tau) then (E1) else(if (lam<=t1) then(E2)else (0))
X =46.7%
P13-6 (e)
Laminar Flow Reactor
For a LFR, 2nd order, liq., phase, irreversible reaction kC Ao =0.2 min- l ,T=5 min"
We apply the segregation model, using Polymath:
ktC A
{O for t < 2.5 min
X =.
0
and E(t) =
1 + ktC Ao
25 /(2t 3 ) min -} for t ~ 2.5 min
See Polymath program P 13-6-e ..pol
POLYMA TH Results
Calculated values of the DEO variables
Variable
t
xbar'
kcao
tau
El
t1
E
x
initial value
0
0
0.2
5
1.. 25E+05
2.5
0
0
minimal value
0
0
0. 2
5
4 . 63E-07
2. 5
0
0
maximal value
300
0 . 4506243
0. 2
5
1 . 25E+05
2.5
0.0549822
0.9836066
ODE Report (RKF45)
Differential equations as entered by the user
[ 1 J d(xbar)/d(t) = x*E
Explicit equations as entered by the user
[1] kcao = 0.2
[2] tau = 5
[3] E1 = taull.2/2/(tIl.3+0 . 0001)
[ 4] t1 = tau/2
[5] E = if (1<t1) then (0) else (E1)
[6] x = kcao*tJ(1 +kcao*t)
We can compare with the exact analytical formula due to Denbigh"
13-31
final value
._-----
300
0 . 4506243
0.2
5
4 . 63E-07
2. 5
4.63E-·07
0.9836066
XMM
0.451
XcSIR
0.467
0.5
0.477
0.382
P13-7
Irreversible Liquid phase, half order; Segregation model.
Mean conversion
X=
f X (t )E(t )dt = 0.1
(1)
Assume a Gaussian distribution for E(t):
E(t) =
ex p[- (t -
1
a.j271
rt]
2.a
1
=.
3.j2.71
ex p[- (t-
112
dX
CAo
.
-= k- . - ()112
1- X
dt
CAo
51~]
2·3
3
and CAo=l mol/dm
The only unknown kl is estimated solving with a trial and error method Eq(1)..
Using POLYMATH: kJ =0.0205 moI 1l2/dm312 . s
P13-8 (a)
The E(t) is a square pulse
I
EXt)
-
OJ -
o
I
o
05
2
2j
3
t
60
Third order liquid··phase reaction: fA = kCA3 with CAo = 2mol/dm3 and k = 0.3 dm6/moe/min ( Isothermal Operation)
I-Conversions
Segregation Model
1
X (t) = 1- -;:;====:::::;;:
~(2kCAo 2t + 1)
X· = jX(t)E(t)dt
o
=~1I 1
Jdt=
IV2kC~ot + 1
.
2
~2kC~ot +1
t -=-----=-2- - = 0.53
kC Ao
1
See Polymath program P13·-8-a-l.pol
13-32
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
Xbar
k
Cao
t1
E2
E
X
initial value
0
0
0. 3
2
1
1
0
0
minimal value
0
0
0.3
2
1
1
0
0
final value
2
0.5296583
0.3
2
1
1
1
0 . 5847726
maximal value
2
0.5296583
0. 3
2
1
1
1
0 . 5847726
0. 60
ODE Reuort {RKF45)
Differential equations as entered by the user
[ 1] d(Xbar)/d(t) = X*E
0.48
Explicit equations as entered by the user
[lJ k=.3
[2] Cao = 2
[3] t1 = 1
[4J E2=1
[5] E = if (t>=t1) then (E2) else (0)
[ 6] X = 1-1/( 1+2*k*CaoA 2*t)A( 1/2)
036
0..2-t
0.12
o.oo'---~-----,--L-~------l
2-Maximum Mixedness Model
_dX __ (~+_ E(T-z)
dz - C Ao I-F(T-z)
dX
dz
0.0
OA
xJ
=kC~o(I-X)3 _ E(T-z)
X
1- F(T -z)
See Polymath program P13-8-a-2.pol
POL)''MA 1]1 Res!!lts
Calculated values of the DEQ variables
Variable
---.
z
x
F
cao
k
lam
ca
E1
ra
t2
t1
E
EF
initial value
0
0
0.9999
2
0.3
2
2
1
-2.4
2
1
1
1 . 0E+04
minimal value
0
0
-9 . 999E-05
2
0.3
0
0 . 9569223
1
-2 . 4
2
1
0
0
maximal value
2
0 . 5215389
0 . 9999
2
0. 3
2
2
1
-0 . 2628762
2
1
1
1.0E+04
13-33
final value
2
0 . 5215389
-9.999E-05
2
0.3
0
0.9569223
1
-0 . 2628762
2
1
0
0
0.8
t
12
1.6
2.0
2.0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(ra/cao+E/(1-F)*x)
[2] d(F)/d(z) = -E
1.6
1. 2
Explicit equations as entered by the user
[1] cao = 2
[2] k= . 3
[3] lam = 2-z
[4] ca = cao*(1-x)
[5] E1 = 1
[ 6] ra = -k*ca"3
[7] t2=2
[8] t1 = 1
[9] E = if ((lam>=t1 )and(lam<=t2)) then (E1) else(O)
[10] EF=EI(1-F)
0. 8
0.4
0.0
0.0
0. 8
0.4
z
1.2
d2X
The conversion shows an inflection point in cOIrespondence of z = 1, where start the pulse --2- = 0 .
dA
P13-8 (b)
Introducing in the Segregated Model and in the MM Model :
k -- k 300' e (20000*(l I 300-11 T)
See Polymath program P I3-S-b- Lpol and P13-8·b-2.pol
--:::-----
X Se2
X MM
300K
0.530
0.521
310K
0.82
0.806
--
320K
0.933
0.924
330K
0.974
0.97
340K
0.989
0.987
--,------
350K
0.995-0.994
The discrepancy is greatest at 300K
P13-8 (C)
Adiabatic Reaction
Introducing the enthalpy balance:
T = To + (- Ml RX ) X = 305 + 40000 X
LBiCPi
50
See Polymath program P 13-8-\.: pol
POLYMA TIl Results
Calculated values of the DEQ variables
Va:dable
t
Xbar
initial value
minimal value
ko
0.8948702
0.8948702
t1
E2
t2
Cao
1
1
1
1
2
2
E
k
To
o
o
o
o
maximal value
5
0.8949502
0.8948702
o
o
1
1
2
2
1
7.6864651
305
7 . 6864651
305
7.6864651
305
2
2
13-34
final value
5
0.8949502
0.8948702
1
1
2
2
7.6864651
305
o
1.6
2.0
X
o
o
T
305
305
0 . 9430621
1059.4497
0 . 9430621
1059.4497
110U
ODE Report (RKF45)
Differential equations as entered by the user
[ 1] d(Xbar)/d(t) = X*E
940
Explicit equations as entered by the user
[1 J ko = 0.3*exp(20000*(1/300-1/30S))
[2 J t1 = 1
[3] E2=1
[4] t2=2
[5] Cao = 2
[61 E = if((t>=t1) and (1<=t2)) then (E2) else (0)
[7] k = ko*exp(20000*(1 1300-1 131 0))
[8] To = 30S
[ 9] X = 1-1 I( 1+2*k*Cao"2*t)".S
[101 T = To+800*X
780
Q
620
460
300
0
P13-9 (a)
(LFR, PFR, CSTR with r=lOOs)
PFR
Design equation
CA
dX
2
C Ao = k CA CB
dV
= CAo (1- X)
CB = C Bo (lx
V0
dX
X)
J-(-)3 =kCBo
o 1-·X
2
V
-
V
1
2V
- - - = 2kC Bo -·+1
(1- X)2
V
Using the quadratic solution
X
1
=1---;====== = 1·- 0.168 = 0.832
1+2kC Bo
2
V
V
The conversion for a PFR X=83..2%
13-35
1
2
t
3
5
CSTR
Design equation
V0
(C Ao -, C A) = -rAV
2
-rA =kCAC B
CA = CAo (1- X)
C B = C Bo (1- X)
CAo-CA =X
C Ao
X =kCBo
2
(I-XY V
V
The conversion for a CSTR X=66.2 %
LFR (completely segregated)
t<1:12
=1:2 l2e
1:=V/v=1000/l0=100s=1.67min
E(t)=O
X
for
for
1>= 1:12
= fX(t )E(t )dt
o
d~=kC2 (I-X)3
dt
Ao
WhcreX(t)=l-
~
1· ",
1+ 2kCBo t
See Polymath program PI3-9-a,.pol
POLYMA TH Results
Calculated values of the DEQ variables
variable
t
xbar
cbo
k
tau
E1
x
E
initial value
1. OE-05
0
0.0313
175
1. 67
1.394E+15
1.714E-06
0
minimal value
1. OE-05
0
0" 0313
175
1. 67
1.403E-,06
1. 714E-06
0
maximal value
100
0,,1827616
0,,0313
175
1. 67
1.394E+15
0.8315008
0,,1480215
ODE Report (RKF45)
Differential equations as entered by the user
[ 1] d(xbar)/d(t) = x*E
Explicit equations as entered by the user
[1] cbo = 0.0313
[2] k = 175
[3] tau = 1..67
13-36
final value
100
0.1827616
0,,0313
175
1. 67
1.403E-06
0.8315008
1. 403E-06
[5
E1 = tauJ\2/(2*tJ\3)
x = 1-(1/(1 +2*k*cb0"'2*t))I\() . 5
[6
E = if(t>=tau/2) then (E1) else (0)
[4
The integral X
J
= X (t )E(t )dt
gives mean conversion=18%
o
P13-9 (b)
(Segregation Model and Maximum Mixedness Model applying RTD of Example 13-1)
Segregation model
dNA
dt
---=-r V
A
Batch reactor
CA
= C Ao (1-- X)
CB
= C Bo (1 -
X)
rX dX
2
.b (1- i)3 = kC Bo t
Similarly
X(t)=l--
vl1 -=-]
1
~kCB02t
=
=
f
X (t )E(t )dt and E(t) from the given data, fitted using Polymath
o
See Polymath program P13·9·b-regressioll..poi
X
POLYMATH Results
Polynomial Regression Report
Model: C02
Variable
a1
a2
a3
a4
= a1 *C01
+ a2*C01J\2 + a3*C01J\3 + a4*C01J\4
Value
0.0889237
-0.0157181
7.926E-04
-8.63E-06
95% confidence
0 . 0424295
0 . 0163712
0.0019617
7 . 288E-05
General
Order of polynomial = 4
Regression not including free parameter
Number of observations = 13
13-37
Statistics
R/\2 =
R/\2adj =
Rmsd=
Variance =
0.8653673
0 . 8204897
0.0065707
8.107E-04
See Polymath program P13-9-b-l.pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
-t
initial value
o
xbar
F
cbo
k
x
o
o
o
o
o
0.0313
175
0 . 0313
175
o
o
o
o
E
minimal value
maximal value
14
0.4106313
1.1137842
0 . 0313
175
0.5847898
0.1566631
ODE Report (RKF45)
Differential equations as entered by the user
[ 1] d(xbar)/d(t) = E*x
[2] d(F)/d(t) = E
Explicit equations as entered by the user
[lJ cbo = 0.0313
[2] k = 175
[3) x = 1-(1/(1+2*k*cboA2*t»AQ.5
[4] E = 0 . 0899237*t-0.0157181 *t/\2+0.000792*t/\3-0.00000863*tA4
0.5
r---------.------,
0.4
0..3
0.2
0.1
0.0
__
2.8
5.6 t
'--..::;...~
0. 0
_ _ _ _J
8.4
11..2
14.0
~_._~
~
X =41%.
Maximum Mixedness
~X =.!L+ E(A)
dA
CAo
X
I-F(A)
Rate Law: - r A = kC A CB
2
13-38
final value
14
0.4106313
1.1137842
0.0313
175
0.5847898
0.0199021
lC = kC 80 2 (1- X)3
Ao
-dF = -E(z)
dz
where z=14-A
See Polymath program P13-9-b-2.pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
z
X
F
Cbo
k
lam
Cao
Ca
E
EF
Cb
ra
initial value
0
0
0,9999
0.0313
175
14
0" 0313
0" 0313
0.0199021
199.0212
0.0313
-0,0053663
minimal value
0
0
-0.1138842
0.0313
175
0
0,,0313
0.0202322
0
0
0.0202322
-0.0053663
maximal value
14
0.3536026
0.9999
0" 0313
175
14
0,0313
0,,0313
0.156664
199.0212
0,0313
-0.0014493
final value
14
0.3536026
-0.1138842
0" 0313
175
0
0.0313
0.0202322
0
0
0.0202322
--0.0014493
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(z) = -(ra/Cao+E/(1-F)*X)
[2] d(F)/d(z) = -E
.Explicit equations as entered by the user
[1] Cbo=,0313
[21 k= 175
[3] lam = 14-z
[41 Cao=.0313
[5] Ca=Cao*(1-X)
[61 E = O.0899237*lam-O,,0157181 *lamA 2+0,,000792*lamA 3·,000000863*lam"4
[7] EF = E/(1-F)
[8] Cb=Cbo*(1-X)
[9] ra = -k*Ca*CbA 2
X=35.4%
P13-9 (c)
Exit time(t), internal age(a) and life expectancy;'
rE(t)dt = F(t)~ dF(t)
= E(t)
dt
.t
where E(t) is obtained from the polynomial fit in Part (b),
13-39
J(a) = ~[1- F(a)]
V
A(A) = E(A)
1-F(A)
Intensity Function
P13-9 (d)
Adiabatic reaction
Segregation model
dX
dt
= kC 2(1- X )3
80
Where
k(T) = koexp[E (~.--~)] = 175exp[3000~(_1_
R To
8.314 320
T
--!)]
T
T(K)=TO+(..:::::.MlrxJx =320+150X
CPA
and
=
J
X = X (t )E(t )dt and E(t) from the given data.
o
See Polymath program P 13-9··d-l ,pol
POLYMA Til Results
Calculated values of the DEQ variables
Variable
t
xbar
F
x
cbo
initial value
o
minimal value
o
o
o
o
o
o
o
0_0313
320
0.0313
320
T
E
o
o
k
175
175
maximal value
14
0.7585435
1.1137842
0.8973303
0 . 031.3
454.59954
0.1565966
4931. 8727
ODE Report (RKF45)
Differential equations as entered by the user
[ 1 ] d(xbar)/d(t) = E*x
[2] d(F)/d(t) = E
[3] d(x)/d(t) = k*cboA2*(1-x)A3
Explicit equations as entered by the user
[1) cbo = 0,,0313
[2] T = 320+150*x
[3] E = 0,,0899237*t-0,,0157181 *tA2+0,,000792*tA3-0,,00000863*tA4
[4] k = 175*exp(30000/8.314*(1/320-1/T»
X =76%
Maximum Mixedness Model
13-40
final value
-.:r:a-----0.7585435
1.1137842
0 . 8973303
0.0313
454.59954
0.0199021
4931. 8727
dX =~+ E(A) X
dA CAo I-F(A)
~ = -kC 2(I_X)3
C
80
Ao
where
and
[E(
1 1)] =175exp[30000(-1- - 1)]
k(T)=koexp - ._-R To T
8.314 320 T
T(K) = To + (-
Mlrx)x
=
320 + 1.50X
CPA
See Polymath program P 13-9-d-2 . pol
POLYMATH Results
Calculated values of the DEQ variables
Variable
z
X
F
Cbo
T
lam
Cao
Ca
E
EF
Cb
k
ra
initial value
0
0
0 . 9999
0.0313
320
14
0.0313
0.0313
0 . 0199021
199.0212
0.0313
175
-0.0053663
minimal value
0
0
-0.1138842
0.0313
320
0
0 . 0313
0.0087977
0
0
0 . 0087977
175
-0 . 007411
maximal value
14
0.7189248
0 . 9999
0.0313
427.83873
14
0 . 0313
0 . 0313
0.1566233
199 . 0212
0 . 0313
3001 . 8809
-0 . 0020441
final value
14
0.7189248
-0.1138842
0 . 0313
427.83873
0
0.0313
0.0087977
0
0
0.0087977
3001 . 8809
-0 . 0020441
ODE Report (RKF45)
Differential equations as entered by the user
[1 J d(X)/d(z) = -(ra/Cao+E/(1-F)*X)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] Cbo= . 0313
[2] T = 320+ 150*X
[3] lam = 14-z
[4] Cao = . 0313
[ 5] Ca = Cao*(1-X)
[6] E = 0 . 0899237*lam··O . 0157181 *lamI\2+0 . 000792*lamI\3-0.00000863*lam"4
[7] EF = E/(1-F)
[8] Cb=Cbo*(1-X)
[9] k = 1l5*exp(30000/8 . 314*(1/320-1/T»
[101 ra = -k*Ca*CbI\2
13-41
0.80.------------------.,
0.64
0.-18
5.6
z
11.2
8.4
14.0
gives X=72%
If the reaction is carried out adiabatically the conversions are more than doubled .
P13-10
Irreversible, first order, long tubular reactor; constant volume, isothermal
dF
A
ForaPFR - =-r
dV
A
x = l_e- kt
For X = 0.865
=> kr = 2.0
For laminar flow with negligible diffusion (LFR), the mean conversion is given by:
=
JX {t )E{t )dt = JX (t )E{t )dt
~
X
=
o
X{t) = 1- e-
,,/2
kt
"('2
E(t) for laminar flow = - 3 where
t
2t
J(l-e
=
Therefore
X
=
,,/2
r
'?2
2
J
2 =
-b
"('
e
-,
dt=I---dt
3
3
2t
2 ,,/2 t
-kt) "('
We can apply the approximated solution due to Hilder:
13-42
x = (4+ Da)eO SDa +Da-4 =0.782
(4+ Da)eO SDa +Da
where Da=kr=2
x = 0.782 < X
PFR
= 0.85
P13-11 (a)
First Moment about the mean: by definition is always equal to zero.
~
~
~
= f(t - T)E(t )dt = ftE(t )dt - T fE(t )dt = T- T= 0
ml
°=
mlCSTR
° =0
= m lLFR
m lPFR
°
P13-11 (b)
Second-order liquid-phase reaction Da= rkCAo =I.O,r=2min and kCAo =05min-1•
CSTR
FAo -- FA = -'A V
FAo - FA = FAoX
V
=
FAoX
(- r A ) exit
Liquid-phase T
V
CAo -C
= - = ( .) A
Vo
Second-order - r A
-'A
A
= kC A2 and T =-CAo -C
2
kC A
X
-----
kC Ao (l- X)2
Solved
X
=
(1 + 2Da)-.Ji+ 4Da =0382
2Da
PFR
-dFA
dV
dX
F Ao - - '- - rA
dV
X dX
V =FAo
0-'11.
f--
Second-order
13-43
V = F Ao
fdX
X
--2
where
O kC A
(I-X)
C A = C Ao -'---..t(1+EX)
Liquid-phase E = 0 and integrating
1 [
r= kC
X ]
I-X·
or
X
Ao
Da
= I+Da
=0.5
LFR
In the ring globule of radius r
dC A
- = rA
dt
ktC Ao
X =
= kC A2
(2order batch)
I+ktC Ao
E(t) =
=r
Where - rA
ofor t < 1min
{4 1(2t 3 ) min -1 for t ~ 1min (E(t) LFR)
2
kC Ao
2
[._! + In(I +
kC Ao
ktC AO.J]
t
Evaluate for Da=l,
t
00
= Da[I- Da
r /2
2
In(I +DaDaI2)]
12
X = 0.451
See Polymath program P 13-11-b.pol
~
_. --f..::.p.
~1~=~~5.::..;1------._-_~
.::..;FR::..:...----.
~O.5
P13-12
The criteria
a r
ac
2
--
___
A
A
>0
Xseg>XMM
2
13-44
The following figure shows the reaction rate as function of the concentration .
•
0.01 ,---,..,..-----,.-----.,.------y-------,
0.005
0"-----·
3
The second derivati~ is initially fitgative (X tf.6('''vl), then ptSkitive (Xseg>~M)' The flex point is for C A=8molJdm
(Xseg=XMM).
CA
In the limit of low concentration
- rA = kC A + o(C A)2
(Fir st order) and Xseg=XMM
in the limit of high concentration
- fA
=
KA
P13-13
~C . + o(~J
(Reaction order=-l) and Xseg>XMM
C
A
A
No solution will be given
P13-14 (a)
Liquid phase, Segregation Model, second order; non-ideal CSTR, adiabatic.·
00
X=
fX(t,T)E(t )dt = 0.67
o
E(t) = C(t)
-00
fC(t )dt
. and
f C(t )dt = Img mini dm 3
o
o
E(t)=IF (t<=I) THEN (t) ELSE ( IF (1)=2) THEN (0) ELSE (2-t))
For a batch globule: C Ao
dX
= -rA
dt
-
13-45
-rA = kC A
2
dX = kC (1- X)2
dt
Ao
Where C Ao=2 mol/dm3
k(T) = 0.5exp[~(-1
8.314 300
T =To+
Where
Cps
-!)]
T
-M/ X
rx
Cps +1'1CpX
=I
tJ;C pi
1'1Cp = 1I2C pb - Cpa
= Cpa + 0 = 50J 1mol· K
= 10012 - 50 = 0
T=300+1S0X
Iterate with Ea, using the ODE solver for values of X(t,T) and substitute these into the polynomial regression to evaluate
~
the integral X
= fX(t,T)E(t)dt =0.67
o
An activation energy Ea of 10000 J/mol gives approximately the correct mean conversion, X =0.67. Inaccuracy lies in
the polynomial fit and hence the integral area.
See Polymath program P IJ·14a.pol
POLY~ATH Results
Calculated values of the DEQ variables
variable
t
xbar
x
cao
T
El
E2
E
Ea
k
initial value
0
0
0
2
300
0
2
0
1.25E+04
0.5
minimal value
0
0
0
2
300
0
-1
0
1.25E+04
0.5
maximal value
3
0.6673438
0.9198721
2
437 . 98082
3
2
0.9933851
1.. 25E+04
2.4247011
ODE Report (RKF45)
Differential equations as entered by the user
[ 1 J d(xbar)/d(t) = E*x
[2 j d(x)/d(t) = k*cao*«1-x)A2)
Explicit equations as entered by the user
[1J cao =2
[2 J T =300+150*x
13-46
final value
3
0 . 6673438
0.9198721
2
437.98082
3
-1
0
1.25E+04
2 . 4247011
(3
(4
(5
[6
[7
E1 = t
E2 = 2-t
E = if (t<=1) then (E1) else (if (t>=2) then (0) else (E2))
Ea = 12500
k = 0,5*exp(Ea/8,,314*((1/300)-(1fT)))
P13-14 (b)
Parallel reactions, isothermal, segregation model
Batch globules
dC A = rA = rAl + rA2 = -k1A C A2 -k 2C C AC B
dt
dCc
- = rC = -rA2 = k 2C C ACB
dt
Exit concentrations
dC Aba!
dt
=C
Bba
dt
dC- -! dC Cba!
dt
E{t)
A
C B E(t )
= CCE{t)
E(t)=IF (t<=l) THEN (t) ELSE ( IF (t>=2) THEN (0) ELSE (2-t))
Selectivity
s = CBba = 2.38
!,
CCbar
Iteration with k2C until S = 2..38 gives k2C = 0..3755 dm 3 /mol min
See Polymath program P13··14·b.,pol
,----, ,----_._---,--------
P13·-15
Reactor: fluidised CSTR (V=lm3 ; F=lOdm 3/s, Cco=2 Kmollm3)
The system of complex reactions for the Kentucky coal n,.9 is given by
13-47
P13-15 (a)
Segregation Model
See Polymath program P l3-IS-a,pol
POLYMA Til Results
Calculated values of the DEQ variables
Variable
t
ca
cp
cc
cabar
cpbar
ccbar
co
cobar
k1
k2
k3
k4
k5
rc
ra
rp
tau
ro
E
Spo
initial value
0
0
0
2000
0
0
2000
0
0
0.012
0.046
0,,02
0.034
0.04
0
92
24
1. 667
0
0.59988
2.4E+08
minimal value
0
0
0
1383.5708
0
0
2000
0
0
0.012
0.046
0.02
0_034
0.04
-49.695094
10.804749
24
1. 667
0
3,,695E-06
0.9280349
maximal value
20
900,,42918
689.61265
2000
137.10239
46.011527
3988.5578
466.78696
9.1391124
0.012
0.046
0.02
0.034
0,04
0
92
37.542445
1. 667
36.017167
0.59988
2.4E+08
ODE Report (RKF4S)
Differential equations as entered by the user
[1] d(ca)/d(t) = ra
[2] d(cp)/d(t) = rp
[3] d(cc)/d(t) = rc
[ 4] d(cabar)/d(t) = ca*E
[5] d(cpbar)/d(t) = cp*E
[6] d(ccbar)/d(t) = cc*E
[7 j d(co)/d(t) = ro
[8] d(cobar)/d(t) = co*E
Explicit equations as entered by the user
[lJ k1 = 0.012
[2 J k2 = 0 . 046
[3 J k3 = 0.020
[4J k4 = 0034
[5] k5 = 0.04
[ 6 J rc = -k1 *cp-k2*ca
[7 J ra = k2*cc+k3*cp-k4*ca-k5*ca
[ 8] rp = k1 *cc+k4 *ca-k3*cp
[ 9] tau = 1.667
[ 1 0 1 ro = k5*ca
[11] E=exp(-tltau)/tau
13-48
final value
20
900.42918
689.61265
1383,,5708
137.10239
46.011527
3988.5578
466.78696
9" 1391124
0.012
0.046
0.02
0.034
0.04
-49.695094
10.804749
33.425188
1. 667
36.017167
3 . 695E-06
0.9280349
[12 J Spo = rp/(ro+0,0000001)
The exiting selectivity is 0.,928
P13-15 (b)
Maximum Mixedness model
See Polymath program P 13-15-b,pol
POLYMA Til Results
Calculated values of the DEQ variables
Variable
z
Ca
Cp
Ce
Co
F
k1
k2
k3
k4
k5
re
ra
rp
tau
ro
E
Spo
EF
sigma
Cao
Cpo
Ceo
Coo
initial value
0
0
0
2000
0
1
0,,012
0.046
0,,02
0.034
0.04
0
92
24
1. 667
0
0,59988
2,,4E+08
5,,999E+07
3
0
0
2000
0
minimal value
0
0
0
1463,,3111
0
6.149E-06
0,,012
0.046
0,,02
0.034
0.04
-47,,836902
15,,190162
24
1.. 667
0
3.695E-06
0.9906544
3,,695E-06
3
0
0
2000
0
maximal value
20
875,,11273
631,,80972
2000
408.43732
1
0.012
0.046
0.02
0.034
0.04
0
92
37,,327161
1. 667
35.004509
0,,59988
2,,4E+08
5.999E+07
3
0
0
2000
0
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(z) = -(-ra+(Ca-Cao)*EF)
[21 d(Cp)/d(z) = -(-rp+(Cp·Cpo)*EF)
[3] d(Cc)/d(z) = -(-rc+(Cc-Cco)*EF)
[4] d(Co)/d(z) = -(-ro+(Co·,Coo)*EF)
[ 5] d(F)/d(z) = -E
Explicit equations as entered by the user
[1 J k1 = 0.012
[2J k2 = 0.046
[ 3 J k3 = 0,,020
[4J k4 = 0,,034
[5J k5 = 0,04
[ 6 J rc =·k1 *Cp-k2*Ca
[7] ra = k2*Cc+k3*Cp-k4*Ca-k5*Ca
O[ 3 J rp = k1 *Cc+k4*Ca-k3*Cp
[ 9 J tau = 1,,667
[10] ro = k5*Ca
[11] E = exp(-Z/tau)/tau
[12] Spo = rp/(ro+0.0000001)
[13] EF=E/(1-F)
[14] sigma = 3
13-49
final value
20
875,11273
631. 80972
1463.3111
408,,43732
6.149E-06
0.012
0.046
0.02
0.034
0.04
-47,,836902
15.190162
34,677371
1.. 667
35.004509
3.695E-06
0.9906544
3.695E-06
3
0
0
2000
0
[15) Cao = 0
[16) CpO = 0
[17) Ceo = 2000
[18) Coo = 0
P13-15 (C)
The selectivities are reported in the following table:
I
LSifMpFR
I
XMMCSTR
I
XSffiPFR
XSffiCSTR
~. _ _ _ _ _-l.. ...;:0~.9:..::.9_ _ _ _ _-1._4;.:.;.1~7....:.4_ _ _ _ _-----l.---,0:...;.:.9...c:2,--_ _ _ _--,
P13-15 (d)
Normal Distribution with T = 5min and (J = 3min
E(t) =
1
a.J21t
ex p[-
r]
(t - T
2.a
1
=
3.J2.J[
ex p[-
r]
(t - 5
2·3
Segregation Model
See Polymath program P13·15-d-l.pol
POLYMA TH Results
Calculated values of the DEQ variables
variable
t
ea
ep
ee
eabar
epbar
eebar
co
eobar
k1
k2
k3
k4
k5
re
ra
rp
sigma
ro
tau
Spo
E1
E
initial ---.value
0
0
0
2000
0
0
0
0
0
0.012
0.046
0.02
0.034
0.04
0
92
24
3
0
5
2.4E+08
10
0.0331675
minimal value
0
0
0
1991.2989
0
0
0
0
0
0.012
0.046
0.02
0.034
0.04
-8.5350648
79.947816
24
3
0
5
4.1743032
10
0.0331675
maximal value
2
171.75027
52 . 879376
2000
11.116671
3.2689487
221. 39065
7.0306771
0.3200147
0.012
0.046
0.02
0.034
0.04
0
92
28 . 677508
3
6.8700107
5
2.4E+08
1.0
0.0806774
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(ea)/d(t) = ra
[2] d(ep)/d(t) = rp
[3) d(ee)/d(t) = re
[4] d(eabar)/d(t) =ea*E
[ 5] d(epbar)/d(t) = ep*E
[6) d(eebar)/d(t) = ee*E
[7 J d(eo)/d(t) = ro
[8] d(eobar)/d(t) = co*E
13-50
final value
2
171. 75027
52.879376
1991. 2989
11.116671
3 . 2689487
221 . 39065
7.0306771
0 . 3200147
0.012
0.046
0.02
0.034
0.04
·-8.5350648
79.947816
28.677508
3
6.8700107
5
4.1743032
10
0.0806774
Explicit equations as entered by the user
[1] k1 = 0 . 012
[2 j k2 = 0 . 046
[3] k3=0 . 020
[4] k4 = 0 . 034
[5} k5 = 0 . 04
[6] rc = -k1 *cp-k2*ca
[7 J ra = k2*cc+k3*cp-k4*ca··k5*ca
[8] rp = k1 *cc+k4*ca-k3*cp
[9] sigma = 3
[ 10] ro = k5*ca
[11] tau = 5
[12] Spo = rp/(ro+0 . 0000001)
[13] E1 = 1/(tau*2*(1-0.99»
[14] E = exp(-(t-tau)A2/(2*sigmaA2»/(sigma*(2*3 . 14)AO.5)
Maximum Mixedness Model
See Polymath program P 13- J 5-d-2.pol
~OLYl\1A TH
Results
POLYMA TH Report
0825-2005, RevS.! 233
Calculated values of the DEQ variables
Variable
z
Ca
Cp
Ce
Co
F
kl
k2
k3
k4
k5
re
ra
rp
lam
r·o
tau
Spo
sigma
E
Cao
Cpo
Ceo
Coo
EF
initial value
0
0
0
2000
0
1
0 . 012
0.046
0.02
0 . 034
0.04
0
92
24
2
0
5
2.4E+08
3
0 . 0296795
0
0
2000
0
2.968E+06
minimal value
0
0
0
1995.2809
0
0.9770814
0.012
0 . 046
0 . 02
0 . 034
0.04
--5 . 986657
83.563477
24
0
0
5
5.6574287
3
0.0020622
0
0
2000
0
0 . 089981
maximal value
2
120 . 77791
35.906087
2000
3.8182818
1
0 . 012
0.046
0.02
0 . 034
0.04
0
92
27.331698
2
4 . 8311165
5
2.4E+08
3
0 . 0296795
0
0
2000
0
2.968E+06
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(z) = -(-ra+(Ca-Cao)*EF)
[2 j d(Cp)/d(z) = -(-rp+(Cp-Cpo)*EF)
[3] d(Cc)/d(z) = +rc+(Cc-Cco)*EF)
[4] d(Co)/d(z) = +ro+(Co-Coo)*EF)
[5] d(F)/d(z) = -E
13-51
final value
2
120.77791
35.906087
1995.2809
3.8182818
0.9770814
0.012
0.046
0.02
0 . 034
0.04
-5.986657
83.563477
27.331698
0
4 . 8311165
5
5 . 6574287
3
0 . 0020622
0
0
2000
0
0.089981
Explicit equations as entered by the user
[1] k1 = 0 . 012
(2) k2=0.046
(3) k3 = 0.020
(4) k4 = 0.034
(5) k5=0.04
[ 6 ) rc = -k1 *Cp-k2*Ca
[7) ra = k2*Cc+k3*Cp-k4*Ca-k5*Ca
(8) rp = k1*Cc+k4*Ca-k3*Cp
(9) lam = 2-z
[10] ro = k5*Ca
[l1J tau = 5
[12) Spo = rp/(ro+0.0000001)
[ 13] sigma = 3
[14] E = exp(-(lam-tau)/\2/(2*sigma))/(sigma*(2*3.14)"O.5)
[15] Cao = 0
[16] Cpo = 0
[ 17] Cco = 2000
[18] Coo = 0
[19] EF = E/(1-F)
P13-16
Multiple parallel reactions, isothermal
E(t)=0.0279693 - 0.0008527t + 1.2778e-5 t2 - 1.0661e-7
e+ 4..5747e·-1O t -7.73108e-13 t
P13-16 (a)
Segregation model
dC A =--k C C 2 -·k C C
dt
lAB
2AB
·_3.3 k 4CCCA 2/3
13-52
4
5
dC D = CDE(t)
dt
See Polymath program P13-16-a.pol
Selectivities:
Exit concentrations:
CA
= 0.026mol / dm 3
SeD = 0.00196
CB
= 0.008mol / dm 3
SDE
Cc
= 6.955e -
CE
= 0.012mol/ dm 3
-
= 4.156e --7mol/ dm 3
CF
5mol / dm 3
= 3.009
SEF = 28433
P13-16 (b)
Maximum Mixedness model
E(t)=O . 0279693-0 . 0008527A+12778e-5A2-L0661e.. 7 A?+4.5747e-lO A4-7..73108e-13
where z=t, tp200min (extent ofE(t».
_dC A
-
dz -
r -((C -- C ) E(A)
A
A
Ao
I-F(A)
J
and
and so on for other species.
dF{A) = - E(A )
._dz
gives F(A)
See Polymath program PI 3-J6-b.. pol
13-53
1.,5
Exit concentrations:
Selectivities:
CA
= 0.028mol I dm 3
SeD
= 0.0009
CB
= O.OlOmol I dm 3
SDE
= 3.004
S EF
= 1343110
Cc = 2.927e -, 5moll dm 3
CD
= 0.033mol I dm 3
CE
= O.Ollmol I dm 3
C F =7.985e-9molldm 3
P13-16 (C)
Ideal CSTR
-tIT
tm=t and
E(t) = _e_
T
Mol balances:
~F:1_ = F - (- r
dt
Ao
FAo
A
)
= CAoVo = 0.05·10 =0.5mol I min = FBo
and so on for the other species"
CA
CTo
FA
FT
= C Ao + CBo
=CTo--~etc ....
= 0.05+0.05 = 0.1
FT = FA +FB +Fe +FD +FE +FF
See Polymath program P13-16-c"po!
Exit concentrations:
Selectivities:
CA
= 0.050mol I dm 3
SeD = 0.068
CB
= 0.049mol I dm 3
SDE
= 3.342
Ce =5.63ge-5molldm 3
SEF
=815844
CD = 0.0008mol I dm 3
CE
= 0.0002mol I dm 3
13-54
CF
= 3.03e -lOmol/ dm 3
Ideal PFR
tm=T and RTD function
E{t) = 8{t - 't')
Selectivities:
Exit concentrations:
CA
= 3.98e -
CB
9mol / dm 3
SCD
= 0.004
= 0.0065mol / dm 3
S DE
= 3.368
Cc
= O.000277mol / dm 3
SEF
= 4 . 286
CD
= 0.068mol / dm 3
CE
= 0.0202mol / dm 3
CF
= 0.0047mol/ dm 3
Segregation
Maximum Mixedness
SCD = 0.00196
SCD
S DE = 3.009
S DE = 3.004
SEF
=28433
SEF
= 0.0009
=1343110
CSTR
PFR
SCD
= 0.068
SDE
=3.342
S EF = 815844
SCD
= 0.004
SDE
=3.368
S EF = 4.286
S EF For the PFR is very much smaller than for the others, because CF is not so small at the exit of the PFR in turn due
to exit Cc is not so small either., The conversion of CA in the PFR is virtually complete at the exit of the PFR, hence
greater Cc
P13-16 (d)
See Polymath program P13-16-d.pol
E(t)=IF (t<=lO) THEN (0,,01) ELSE (IF(t>=20) THEN (0) ELSE (02-0,01 t))
13-55
Segregation
Maximum Mixedness
PFR
CSTR
= 0.004
SCD
=0.004
SCD
= 0.0035
SCD
= 0.068
SDE
= 3.109
SDE
=3.106
SDE
=3.342
SDE
=3.368
S EF
= 24078
SEF
= 41503
SEF
= 815844
SEF
= 4.286
For the Segregation and Maximum Mixedness models,
S EF
SCD
is much lower than
for the CSTR but still far greater than for the PFR. The CSTR and PFR values are unchanged as they do not depend on
E(t).
P13-17
Multiple parallel reactions, isothermal
Asymmetric RTD: E=IF (t <=1.26) THEN (El) ELSE (E2)
4
E1=-2.104t +4" 167t3-1.596t2-0.353t.. O,004
4
E2=-2,,104t + 17 .037t3-50.247t2+62,,964t-27.402
P13-17 (a)
Segregation model
dC A
--= -kDlCAC B 2 -3k E2 C AC D
dt
!:C B = C E{t)
dt
B
dCC
()
--=CcEt
dt
13-56
dC F
(t)
--=CFE
dt
See Polymath program P13-17-a.pol
Exit concentrations:
Selectivities:
CA
= 0.819mol / dm 3
SCD
=0.272
CB
= 0.767moll dm 3
SDE
= 11.330
Cc
-
= 0.163moll dm 3
SEF
= 0.267
CD
= 0.600mol / dm 3
C~
= 0.053mol / dm 3
CF = 0.199moll dm 3
P13-17 (b)
Maximum Mixedness model
As the RTD is asymmetric we can use the same equations for E(A) as we did for E(t), with: E(A)=IF (A<=1.26) THEN
(EI) ELSE (E2)
dC A
__ " "
dA -
r
~ iA
+ (C _ C ) E(A)
A
Ao I-F(A)
The same applies to the equations for the other species as in Part (a)..
See Polymath program P t 3- t 7 -b .pol
Exit concentrations:
CA
= 0.847mol / dm 3
Selectivities:
SCD
= 0.281
13-57
C B = 0.824mol I dm 3
SDE
=10.7
Ce =0.162molldm 3
SEP
= 0.280
CD
= 0.576mol I dm 3
CE
= 0.054mol I dm 3
Cp
= 0.192moll dm 3
P13-17 (c)
Ideal CSTR
-tIT
tm=L and
E(t) = _e'['
dC
(
'--=v
CAo -C A) +rA
A
Where r A =
dt
(-kD1CAC~ -2kE2 C AC D )
vo =v
and so on for the other species.
See Polymath program P 13·17c-1.pol
Exit concentrations:
Selectivities:
CA
= 1.386mol I dm 3
SeD
= 0.832
CB
= 1.774moll dm 3
SDE
= 71.100
Ce
= 0.095moi I dm 3
SEP
= 0.198
CD = 0.114moll dm 3
C E = 0.002mol / dm 3
C P = 0.D08mol I dm 3
Ideal PFR
=.
tm and RTD function
E(t) = g(t - '[')
and so on for the other species .
13-58
See Polymath program P 13-17-c-:2.pol
Selectivities:
Exit concentrations:
CA
= 0.31Omol / dm 3
SCD
= 0.162
CB
= 0.338mol / dm 3
SDE
= 3.618
Cc
= 0.106mol / dm 3
SEF
= 0.497
CD
= 0.652mol / dm 3
CE
= 0.180mol / dm 3
CF
= 0.362mol / dm 3
Segregation
Maximum Mixedness
CSTR
PFR
= 0.832
= 0.162
SCD
= 0.272
SCD
= 0.281
SCD
SDE
= 11.330
SDE
= 10.7
S DE = 71.100 S DE = 3.618
SEF
= 0.237
SEF
= 0.280
SEF
= 0.198
SCD
SEF
= 0.497
S CD is significantly greater than for the others, because exit CD is lower.
Similarly S DE is much greater in the CSTR than for the others, because exit CE is so low.. This is because the achievable
conversion in a CSTR is not so high .
P13-18 (a)
P13-18 (b)
P13-18 (c)
P13-18 (d)
P13-19 (a)
External age distribution E(t)
By plotting C 105 as a function of time, the curve shown is obtained
13-59
C{t)
----------------.----.
9oor----800
700
600
500
u
400
300
200
100
o --.....-.......... -------,--........--..-.--........-....--r-----.. --.-.-.......
a
10
20
30
- r '..•_..
=.-.=--:=-.:::
...... ::::::;::=""'=""""""''''l''"---~------.'''----"'--::::=...
40
50
60
70
tlmin]
To obtain the E(t) curve from the C(t) curve, we just divide C(t) by the integral
f C(t)dt = f C(t)dt + rC(t)dt + CC(t)dt + CC(t)dt
f C(t)dt
r
C(t)dt
= %(1)[1(0) + 3(622) + 3(812) + 2(831) + 3(785) + 3(720) + (650))0-5 = 4173.4 .10-5
= j(2)[(650) + 4(523) + (418))0-5 = 2106.7 .10-5
+2(136)+4(77)+ 2(44)+4(25) + 2(14)+4(8)+5)0-5 =3671.7 .10-5
JofOC(t)dt =!(5)[418+4(238)
3
roC(t)dt = !(1O
)[1(5) + 1(1))0-2
5
!o
f C(t)dt = 9981.7 .10-
5
= 30 .10-5
== 0.1
We now calculate:
E(t)
= ~QL = C(t)
f C(t)dt
0.1
13-60
E(I)
009
008
0.07
006
coos
~
m:004
003
002
001
o
2
4
6
8
10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64
I[minj
Using Excel we fit E(t) to a polynomial:
for 0 5: t 5: 3 E1 (t)
= -1.1675 .10-3 t 4 + 1.1355 .10-2 t 3 -
4.7492 .10-2 t 2
+ 9.9505 .10-2 t
for 3 5: t 5: 20 E2 (t)
E(t) = -1.7979.10-4 t
= -1· 8950 .10-6 t 4 + 8.7202 .10-5 t 3 -1.1739 .1O-3 t 2
+ 0.092343
for 20 5: t 5: 60 E3 (t)
= 1.2618 .1O-8 t 4
_
2.4995 .1O-6 t 3 + 1..8715 .1O-4 t 2
6.3512.10- 3 t + 0.083717
for t > 60
0
P13-19 (b)
External age cumulative distribution F(t)
F(t)
=
!E(t)dt
Integrating the E(t), we obtain the F(t):
l3-61
-
- - - ------------------------1
12 - , - - - - - - - - - - - - -
e
1
08
~ 06
04
": __
o
~- - - - - ~- - - - - - - - -~- - - - - - - - - - ~- - - -J
20
10
40
30
50
60
\[min]
P13-19 (C)
Mean residence time and variance
r
tm =
E(t)tdt
The area under the curve of a plot tE(t) as a function of t will yield tm •
IE(I)
045
04
035
03
025
::
'I:l
02
015
01
005
0
10
0
20
30
40
50
Hm
0
(t-t~2E(t)
-10
60
t[lIIin)
E(t)
C(t)
t
0
0
tE(t)
0
13-62
D
,--,
.....
~
,--,
'-'
Jl
.....
t m = '-'
I
tm
0.4
1
2
3
4
5
6
8
10
15
20
25
30
35
40
45
50
60
)tdt =
329
622
812
831
785
720
650
523
418
238
136
77
44
25
14
8
5
1
0 . 0329
0.0622
0.0812
0.0831
0.0785
0.072
0.065
0 . 0523
0.0418
0.0238
0.0136
0.0077
0.0044
0.0025
0 . 0014
0.0008
0.0005
0.0001
-9.6
-9
-8
-7
-6
-5
-4
-2
0
5
10
15
20
25
30
35
40
50
0.01316
0.0622
0.1624
0.2493
0.314
0.36
0.39
0.4184
0.418
0.357
0.272
0.1925
0.132
0.0875
0 . 056
0.036
0.025
0
f E(t)tdt + rE(t)tdt + LO E(t)tdt + CE(t)tdt
= T = 9.88min == 10 min
We can calculate the variance by calculating the area under the Clive of a plot of:
o
10
20
30
= [(t -
tJ2 E(t)tdt
=
f (t - tJ2 E(t)dt + r(t - tJ2 E(t)dt +
C(t--tJ2 E(t)dt+ C(t-tmY E(t)dt
13-63
(t-tniE(t)
40
llmin)
(72
3.032064
5.0382
5.1968
4.0719
2 . 826
1.8
1.04
0.2092
0
0.595
1.36
1.7325
1.76
1.5625
1.26
0.98
0.8
0.25
50
60
P13-19 (d)
Fraction of the material that spends between 2 and 4min in the reactor
E(t)
009
008
0.07
006
::s 005
J~
;::.
~ 004
003
002
001
0
o
2
4
6
8
10 12
14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46
48 50 52 54 56 58 60
r
E(t)dt = shaded area = j[I(0.0812) + 4(0.0831) + 1(0.0785)] = 0.16
P13-19 (e)
Fraction of the material that spends longer than 6min
E(t)
o
2
4
6
8
10 12 14 16
18 20 22
24 26 28 30 32 34 36 36 40
tlmin)
13-64
42 44 46 48 50 52 54 56 58 60
f E(t)dt
r
E(t)dt
= shaded area =
r
E(t)dt +
CE(t)dt + fo°E(t)dt
= ~(1(0.065) + 4(0.0523) + 1(0.0418») = 0.210
fO E(t)dt = 1(1(0.0418) + 4(0.0238) + 2(0.0136) + 4(0.0077) + 2(0.0044) + 4(0.0025)
10
3
+ 2(0.0014) + 4(0.0008) + 1(0.0005») = 0.367
{'°E(t)dt
= tail = 10 (0.0005 + 0.0001) = 0.003
f E(t)dt
= shaded area = 0.581
!o
2
P13-19 (f)
Fraction of the material that spends less than 3min
E(t)
009
008
007
006
:§" 005
'-~
.=-
~ 004
003
002
001
o
:::;:
o
2
4
6
8
10 12
14 16 18 20 22
24 26
28 30
32 34 36
38 40
42 44 46
48 50
52 54
t[min)
! E(t)dt = shaded area =%[1(0) + 3(0.0622) + 3(0.0812) + 1(0.0831)] =0.192
P13-19 (g)
Normalized distributions
13-65
56
58 60
Normalized RTD
e=':' E{e)=rE{t)
T
09
08
07
E(8)
6
05
04
03
02
0.1
0
0
2
4
3
8
Normalized cumulative RTD
F{e)=
f E{e)de= fE{t)dt
13-66
6
7
F(e)
/'
08
----.--'-'-'
_.--.-----'
./'
06
04
02
o .'---"'-"'-"'---"'-"'-~"'----"--'"
.......... ___....___...__ .___ .__ ..._____ .________.... _. . _...._.... _______________" ________."."_____..1
"'
"'
"'
.
e
P13-19 (h)
Reactor Volume
F= 10 dm3/min
V
= F 'T=100dm 3
P13-19 (i)
Internal age distribution
I(t)=2.[l-F(t)]
T
01
008
g006
004
002
o -..--.."'----..------..
o
-~-" -"'- "-"'- -~--- ---"'-~-=---==. ~==!?=~_o__o____~--------J
..----..
10
..
20
40
30
I[min]
13-67
50
60
70
P13-19
U)
Mean internal age
am =
r
I (t)tdt
=
r
I (t)tdt +
r
I (t)tdt + rI(t)tdt + CI(t)tdt =1min
P13-19 (k)
Intensity function
E(t)
A(t)=--1- F(t)
P13-19 (1)
Mean catalyst activity
Integrating the decaying rate law:
f -~~ = ! k Ddt
a
1
a=--l+kDt
Applying the Segregation Model with the previous RTD data:
a mean
=
r
aE(t)dt
See Polymath program P 13--1 9 "l.pol
POLYMA Tn Results
Calculated values of the DEQ variables
Variable
t
amean
initial value
o
o
minimal value
o
o
maximal value
70
0.5778625
0.1
0.0838847
0,,092343
final value
70
0.5778625
0.1
-2.436E+04
-21.261016
0.092343
0,,1
-2.436E+04
-21.261016
o
o
o
0.083717
1.724E-05
0.0017977
o
0.125
kd
0.1
E1
E2
E4
E3
o
o
E
o
o
0.083717
0.0838847
a
1
0.125
1
ODE Report (RKF4S)
Differential equations as entered by the user
[1] d(amean)/d(t) = a*E
Explicit equations as entered by the user
[1] kd = 0 . 1
[2] E1
-0.0011675*t"4+0,,011355*tA3-0 . 047492*tA2+0,,0995005*t
[3 J E2 = -1,,8950*1 OA(-6)*t"4+8.7202*1 QA(-5)*tA3-1 . 1739*10A(-3)*tA2-1.7979*1 QA(-4)*t+0 . 092343
[4] E4 = 0
[5 J E3 = 1.2618*1 QA(-8)*t"4-2 . 4995*1 QA(·6)*tA3+1 ,,8715*1 OA(-4)*tA2-6 . 3512*1 OA(-3)*t+0.083717
[6 J E = if(t<=3)then(E1 )else(if(t<=20)then(E2)else(if(t<60)then(E3)else(E4)))
[7] a = 1/(1 +kd*t)
=
P13-19 (m)
Ideal PFR
13-68
2nd order, liquid phase, kC Ao =0. 1min' I, CAo=lmol/dm3
.=10 min (from P 13..1 9 (c))
X
=
k'lC Ao
1+ k'lC Ao
= 0.5
P13-19 (n)
LFR
Laminar Flow Reactor
2nd order, liq. phase, ineversible reaction kC Ao =0.1 min· I,.=1O min.
We apply the segregation model, using Polymath:
X
=
ktC A
0
and E(t)
I+ktC Ao
={O for t < 5.00min
3
1
(l0)2/(2t )min- fort~5min
See Polymath program P13-19-n.pol
POLYMAT.!I Results
Calculated values of the DEQ variables
Variable
t
xbar
cao
k
tau
E1
t1
E
x
initial value
0
0
1
0. 1
10
5.0E+05
5
0
0
minimal value
0
0
1
0.1
10
1 . 852E-06
5
0
0
maximal value
300
0 . 4504221
1
0. 1
10
5 . 0E+05
5
0.2284987
0.9677419
ODE Report (RKF45)
Differential equations as entered by the user
[ 1 ] d(xbar)/d(t) = x*E
Explicit equations as entered by the user
[1] cao = 1 . 0
[2] k =.1
[3 ] tau = 10
[4] E1 = tauI\2/2/(tA3+00001)
[5 ] t1 = tau/2
[6] E = if (1<t1) then (0) else (E1)
[7] x = k*cao*t/(1 +k*cao*t)
We can compare with the exact analytical formula due to Denbigh .
X
~ Da[l- (~a }n(1 + 21 Da)] ~ 0.451 w;th Da=kC"",=l
P13-19 (0)
Ideal CSTR
2nd order, liquid phase, kCAo =0.1 min-I, C Ao=lmol/dm3
X
-(--)2 = k'lCAo ~ X = 0.382
I-X
13-69
final value
300
0.4504221
1
0. 1
10
1 . 852E-·06
5
1.852E-06
0 . 9677419
P13-19 (p)
Segregation Model
2nd order; liquid phase, kCAo =0.1 min-I, CAo =1 molldm3
x=
J (t )E(t )dt
X
o
kCAot
1+ kCAot
Where X (t ) = ---"=--
See Polymath program P 13-19-p. pol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
Xbar
kCao
E1
E2
E4
E3
initial value
-0------
E
X
o
0.1
minimal value
o
o
0.092343
0.1
-2.436E+04
-21 . 261016
0.083717
1.949E-05
o
o
o
o
o
maximal value
70
0.4224876
0.1
0.0836855
0.092343
o
o
o
0.083717
0.0836855
0.875
final value
70
0.4224876
0.1
-2.436E+04
-21.261016
o
0.0017977
o
0.875
ODE Report (RKF45)
Differential equations as entered by the user
[ 1) d(Xbar)/d(t) = E*X
Explicit equations as entered by the user
[IJ kCao=O.1
[2 J E1 = -O.0011675*t"4+0 . 011355*tA3-0.047 492*tA2+0.0995005*t
[3] E2 = -1 . 8950*1QA(-6)*t"4+8 . l202*1 QA(-5)*tA3 . 1. 1739*1 QA(-3)*tA2··1 . 7979*10A(-4)*t+O.092343
[4] E4 = 0
[5 J E3 = 1.2618*1 QA(··8)*t"4-2.4995*1 OA(-6)*tA3+ 1.8715*1QA(-4)*tA2-6 . 3512*1 QA(-3)*t+O.083717
[6] E = if(t<=3)then(E1 )else(if(k=20)then(E2)else(if(k60)then(E3)else(E4)))
[7] X = kCao*t/(1+kCao*t)
P13-19 (q)
Maximum Mixedness Model
2nd order; liquid phase, kCAo =0.1 min-I, CAo =1moZldm3
Rate Law :- r A =
CA
kC A
2
= CAo (1- X)
13-70
= kC Ao 2 (1- X Ywhere k=O.l dm3 Imol min
rA
~=kC
C
Ao
(1-X)2
Ao
'dF
-
dz
= -
E( z)
where Z=60-A
See Polymath program P13-19-q.pol
POLYMATH Results
Calculated values of the DEQ variables
Variable
z
X
F
cao
lam
E1
E2
E3
E4
Ca
k
ra
E
initial value
0
0
0 . 99
1
60
-1.284E+04
-9.8680524
2.228E-05
0
1
0.1
-0.1
0
minimal value
0
0
-0 . 010344
1
0
-1..284E+04
-9.8680524
1.806E-05
0
0.52273
0.1
-0 . 1
0
maximal value
60
0.4773052
0.99
1
60
0.0832647
0.092343
0.083717
0
1
0.1
-0.0273247
0.0832647
final value
60
0.4047103
-0.010344
1
0
0
0.092343
0.083717
0
0 . 5952897
0.1
-0.035437
0
ODE Report (RKF45)
DifferentiaJ equations as entered by the user
[1] d(X)/d(z) = -(ra/cao+E/(1-F)*X)
[ 2] d(F)/d(z) = -E
Explicit equations as entered by the user
[lJ cao = 1
[2 J Jam = 60-z
[ 3] E1 = -0.0011675*Jam"4+0.011355*Jam"3-0047 492*Jam"2+0 . 0995005*Jam
[4] E2 = -1.8950*1 0"(-6)*Jam"4+8 . 7202*1 0"(-5)*Jam"3-1 ..1"739*1 0"(-3)*Jam"2-1..7979*1 0"(-4)*Jam+0 . 092343
[5] E3 = 1 . 2618*1 0"(-8)*Jam"4-2A995*1 0"(-6)*Jam"3+ 1. 8715*1 0"(-4)*Jam"2-6 . 3512*1 0"(-3)*Jam+0.083717
[6J E4 = 0
[ 7] Ca = cao*(1-X)
[8] k = 0.1
[ 9] ra = -k*Ca"2
[101 E = if(lam<=3)then(E1 )eJse(if(Jam<=20)then(E2)eJse(if(lam<60)then(E3)eJse(E4)))
~
1X
1 Xse
I' X
~ _ _--,-.:..O•.4.;.O.::..::5_ _ _ _ _.-,-..:.O.:..:.4.::.22::" _ _ _ _ .;.:O.::..::.5_ __
MM
PFR
u
..L
.
13-71
Solutions for Chapter 14 - Models for Non-ideal Reactors
P14-1
Individualized solution
P14-2 (a)
Approximatedformula for Segregation Model (pt reaction)
2 2
k t
e = 1- kt + - - + Error. The enOl is then o((kt)\
2
kt 3
Approximating the Error = - - .
-k/
3!
kT
ffit
~
Error
rO~110·_I--,~~~~=-------------------:=~~~:O~I~~~~~7~16~6~7=_________________
_-166.7
20~--.----.----~---.----r---~
o
Ln~.... el1'orl)
I.,r
-20
10
0.1
100
k'T
P14-2 (b)
Parameters Dispersion Model
Closed-Closed dispersion model
X = 1 - - . - - - - 4qexp(Pe,/2)
(1 + qYexp(Pe,q 12) - (1-- qYexp(- Pe,q 12)
Where
q = ~1+4Da/ Pe,
Da
= kr = k --.I Damkohler number
U
14-1
(14-26)
_
UZ
Pe , =D-
Peclet number
a
Pe,q
= ~ Pe2 + 4DaPe,
2
Where
kZ2 Rate oj consumption oj Aby reaction
DaPe. = - =
, Da
Rate oj transport by diffusion
Numerical example
T
= ~ = const = 5.I5min
U
= r *U = 5.I5min*0.Icm/ s = 0.309m
Sc =L =1000 (Liquids region in Fig.. 14 . 11)
L
DAB
= kr = 1.288
Da
dt
lcm
Re
10
1 dm
100
--
WOO
1m
Lldt
30.9
D/(V*d t )
0 ..18 From
Fig14.10
3.09
40 From Fig..
105
14.11
0..309
3000 From
106
_ _ _. _ _ _ _ ....!ig.14.11
ReSc
104
----------
--
D (cm2/s)
0.018
40
30000
,---._---
Per
172
0.077
1.0310-4
Q
1.015
8.226
223.609
X ._--- _
0.721
0.567
0.563
..
---
Is there a diameter that would maximize or minimize conversion in this range?
According Fig . 14.10 there will be a radius that maximizes the conversion
P14-2 (C)
(1) Vary the Damkohler number for a second-order reaction (Example 14-3(b»
For a second order reaction
Da
L
= TkC AO = - kC AO
Uo
R=0.05m
L=6.36m
k = 0.5dm 3 /moZ.min
U o = I.24m/min
C AO = 0.5mol / dm 3
2
D aris = 1.05m / min
Damkohler numberl Da
0.1603
0.3205
-
Conversion
0.l38
0.239
14-2
Parameter
8*VO
4*VO
0.641
1.282
2.564
5.129
10.258
2*DO
DO
DO/2
DO/4
DO/8
0.377
0.523
0.644
0.732
0.795
(2) Vary the Peclet and Damkohler numbers for a second-order reaction in laminar flow (Example 14-3(c»
Da = 'ikC AD
For a second order reaction
L
=-
UD
kC AD; P e = U 0. L / DAB
R=O.05m
L = 6.36m
k = O.5dm 3 / mol. min
U D = I.24m/min
= O.5mol / dm 3
5
2
DAB = 7.6xlO- m /min
-Da
CAD
-
0.1603
0.3205
0.641
1.282
2.564
5.129
f-----------10.258
Pe
8.32xIO'
4.16x105
.------;3"""---- 1-2.08xlO
----i04xl0r
--
0.52xlO'
0.26xlO'
5
0.13x10
-
Conversion
0.132
0.229
0.369
0.536
0.699
0.825
0.906
..-
-
Parameter
8*DO
4*DO
2*DO
DO
DO/2
DO/4
DO/8
1.,0.
0.,9
0..6
c
o
"[!? 0.,5
~
§
0.4
U
0.,3
0.,2
0.,1
0.,0.
DO.
5
2,DxlD
-,-- i i i
5
4 DxlD
6,DxlD 5
8 DxlD5
Pe
When Peclet Number decreases less than 2 x105, the conversion is influenced significantly.
Below is a FEMLAB anaylsis of the problem.
(1) Vary the Damkohler number for a second-order reaction (Example 14-3(b»
14-3
-----
--
.
Da = TkC AO
For a second order reaction
L
Uo
= -kCAO
R =O.05m
L=6.36m
k = O.5dm 3 I mol. min
U o =1. 24m I min
= O.5mal I dm 3
CAO
DaTi'
=1.05m
2
I min
Damkohler number/ Da
0.1603
.0.3205
0.641
1.282
2.564
5.129
10.258
Conversion
0.138
0.239
0.377
0.523
0.644
0.732
0.795
- Femlab Screen Shots
[1] Domain
Ir?~i~f1 Grid I
[ ] Axis equal
[2] Constants and scalar expressions
- Constants
2""~=~~.=
..""--.J;xpr&~sipn
. m• • • • • • • • • • •
.".. -
]1:9?
0.5
"~05"
...
. . . . -.. .-.. --.. --j:~
.............
"'.... -""..
~~,,-.~''''~
~-
. . ,. .. r..--.. ,,, ...
·t
24
" .... ,+
-
.·····f····
t ..
- Scalar expressions
14-4
Parameter
8*UO
4*UO
2*UO
UO
UO/2
UO/4
UO/8
COi<"'"J [c;;~J I
Apply
[3] Subdomain Settings
- Physics
(Mass Balance)
'Subdomain selection" '"
r Species""" ,
"
I
Library material: ['
i QllantltY
I
I®
0
1
'[J Sele()t by group
I
=-:,::='_~C~ li""'.-L-"o~a.d-~~.]ValueJExpre~sion Description
5""'"'''''' --",' ..,,'
0ts
0 isotropic
°:"v
!,[~rtifi~;i~i!i~~i;~::~]
- Initial Values
(Mass balance) cA(tO) = cAO
- Boundary Conditions
@ r = 0, Axial symmetry
@ inlet,
rBoundary
conlS!ition~
, Boundary Glonditi9n:
Ii=i~; .
I
V~lueIEHpre$sion
Descl\iplion
!2;i\d'
Concentration
I
Qu~ntity
! cAo
I
@
@
No
""
.,--"
~2;£~Q'~~"':---']lhward flux
outlet, Convective flux
wall, Insulation/Symmetry
14-5
Time-scaling coefficient
"]' Diffusiol) coefficient
DiffusiOI) coefficient
OptiOns Draw Physics Mesh Solve PostproceSSR1g Multiphysics Help
!ill ~rl.~
. . . . i[J~.#=.~L~~. ~'gI~l? ~.~_':!I.~ • ~. .~ n .=.
!..: .. c,.c.................................... .
Max: 0 523
Sur1ace: xA
0,2
0.15
-----_._-----_._---01
0.05
··005
Min: 0 110
· ,5)
(2) Vary the Peelet and Damkohler numbers for a second-order reaction in laminar flow (Example 14-3(c))
For a second orderreaction
Da = ikC AO
L
= --. kC AO;
Uo
Pe = U oLI DAB
R =O.05m
L= 6.36m
k = O.5dm 3 I mol. min
U o = 1. 24m I min
= O. 5mol I dm 3
DAB = 7.6xlO-5 m 2 I min
C AO
f---.
'---.
Da
0.1603
0.3205
0.641
1.282
2564
5.129
10.258
--.
Pe
8.32xl05
4. 16x105
2.08x105
L04x105
052x105
0.26x105
0.13x10 5
-
-_...- f---.
Conversion
0.132
0.229
0.369
0.536
- - - f-0.699
.. 0.825
0.906
--'--.
14-6
-
Parameter
8*DO
4*DO
- 2*DO
DO
DO/2
DO/4
DO/8
---
10
0.9
08
0.7
c:
0.6
0
§
(])
>
c:
0
0.5
0.4
•
()
03
0.2
0.1
0.0
00
2.0x10'
4.0x10'
6.0x10'
8.0x10'
Pe
When Peclet Number decreases less than 2 x10 5, the conversion is influenced significantly.
- Femlab Screen Shots
[1] Domain
[2] Constants and scalar expressions
- Constants
.................................... .1
- Scalar expressions
14-7
[3] Subdomain Settings
- Physics
(Mass Balance)
: Vo(-DVcA+cAu)
=R, cA =concentration
rCA
Subdomain selection
Intt
Elemerd
I
Species
[Lo~d~~-'-l
Library material:
ValuelElIpression Description
Cits
] Time-scaling coefficient
D isotropic
DAB
D anisotropic
R
u
v
[J Select by group
~ Active in this domain
J
("---'-'--"--'-"'-'---.
-.
Artificial Diffusion ..
- Initial Values
(Mass balance) cA(tO) = cAO
- Boundary Conditions
@ r = 0, Axial symmetry
@ inlet,
Boundary conditions
r
Boundary condition: rl~)(
ValuelExpression
Description
J Concentration
.j
] Inward flux
@
outlet, Convective flux
14-8
'I Diffusion coefficient
1 Diffusion coefficient
j
@
wall, illsulation/Symmetry
[4] Results
('-'u '.''''-','HI.<'Uun,
s
;}
- - .- --C.05
--------------0-,-·-·------- . 0,05
P14-2 (d)
Two parameters model
T=~=lOmin
va
I=2 and S=4min- 1"
fJ = .Vb_ = (~ -_!2 = 0.5
va
I
a= V,.=Q-fJ)_=O.013
V
'is
m3
v, = (1- fJ)v a = 0.05-.
nnn
V, = (aT)v a = O.013m 3
T,
=y,
=0.25min
v,
14-9
-~-----'-"
-0-;---..-.-----'
CA =
s
Jl+4r,kC Ao -1
kmal
= 1.779-32rk
m
s
X = 1- CAS = 0.111
CAO
Ideal CSTR
CSTR with Dead Space and
Bypass
(1=2.0 and S=4 min-I)
X=O.lll
CSTR with Dead Space and
Bypass
(1=1.25 and S=0.115 min-I)
X=0.51
X=0.66
P14-2 (e)
Two CSTR with interchange ( 1st order reaction)
L ____________
X
.~
CAl Vo
= (fJ + ark)[p + (1- a)tk]- fJ2
(1- fJ)+ ark
See Polymath program P14--2-e.pol
POL YMA TII_Re§!!lts
Calculated values of the DEQ variables
Variable
t
CTI
CT2
beta
alpha
tau
CTel
CTe2
tl
CTe
k
X
initial value
0
2000
921
0,,15
0.75
40
2000
921
-80
2000
0.03
0.5134788
minimal value
0
31_814045
164 . 15831
0.15
0.75
40
-1.275E+04
13
-80
13
0.03
0_5134788
maximal value
200
2000
1048.4628
0.15
0.75
40
2000
921
120
2000
0.03
0,,5134788
ODE Report (RKF45)
14-10
final value
200
31.814045
164.15831
0.15
0,,75
40
-1.275E+04
13
120
13
0,,03
0,,5134788
Differential equations as entered by the user
[1] d(CT1 )/d(t) = (beta*CT2-(1 +beta)*CT1 )/alphaltau
[2] d(CT2)/d(t) = (beta*CT1-beta*CT2)/(1-alpha)/tau
Explicit equations as entered by the user
[1] beta = 0.15
[2] alpha =0.75
[3] tau =40
[4] CTe1 = 2000-596*t+0 . 64*tA2-0.00146*tA3-1.047*1 QA(-5)*t"4
[5] CTe2 = 921-17.3*t+0.129*tA2-0.000438*tA3+5 . 6*1QA(-7)*t"4
[6] t1 = t-80
[7] CTe = if(k80)then(CTe1 )else(CTe2)
[8] k = 0 . 03
[9] X = «beta+alpha*tau*k)*(beta+(1-alpha)*tau*k)-betaA2)/«1 +beta+alpha*tau*k)*(beta+(1-alpha)*tau*k-betaA2»
800
~~
~J
400
o
o
40
80
t
120
160
200
Comparison experimental and predicted (13=0.15 a=0.75) concetration .
For small deviations from the original value of the parameters concentration and conversion are not significantly
affected.The following table show the conversion for different combinations of 13 and a..
a.
0.75
0.8
ro.s.
~.
0.5
0.5
r-o.-I0.9
_--
.
a
0.15
0.1
0.5
0.8
0.1
0.9
0.5
0.5
x ---0.51
..
0.51
0.54
0.46
0.41
0.83
0.34
0.76
14-11
'J~
'Ji.i
0;=0.5
Ot
X
O:>::!O
""'
IJ~
~i
-1
iU
\12
'"
Il"-
Q"
116
X
IJ5
0..
EJ
113
o.:z
0.1
------r-
0
0.1
o.:z
03
OJ5
06
0,9
Given the interchange flowrate, the conversion is increased with the increasing of the volume of the highly agitated
reactor" Given the volume of the highly agitated reactor, the conversion is increased with the increasing of the
interchange.
P14-2 (f)
Tubular Reactor Design
The correlations between Re and Da show what flow conditions (characterised by Re) give the greatest or smallest Da
and hence dispersion"
14-12
pud
To minimise dispersion a Re number of -10-20 gives the lowest value for Da. Because Re = - - , the design of the
J.l
vessel could be altered (i.e. diameter) for a given fluid and flowrate,.
To maximise dispersion, either a very low Re «0 . 1) or Re -2300 will give the maximum Da values .
For a packed bed, the dispersion also depends on the Schmidt number as well as Re number.
P14-2 (g)
Linearizing non-first-order reactions
May be a good approximation if CA does not change very much with time, Le . A is in excess, in which case C Ao should
not be divided by anything.
Linearizing the non 1st order reactions may give significantly inaccurate results using Equation 14-27, which can be
tested experimentally by recording tracer concentrations with time and using the tanks in series model for a conversion
comparison.
P14-2 (h)
Figures 14.3 and 14. ]b
The curves in Fig. 143 represent the residence time distributions for the Tanks in Series model as function of the number
of reactors.
Given a CSTR of volume 1 (V=V 1), we divide the CSTR in two CSTRs (V2=1I2).. The mean residence time is
unchanged (VIF) but the molecules going out of the second reactor will be delayed by the time (distribution) that occur
to pass the first reactor (n=2, shift of the maximum) . In the limit of infinite division (Vn=O) , so in the single CSTRs the
residence time goes to zero for all the molecules (zero variance), but their summation is the mean residence time (n=oo,
PFR behaviour).
The model in Fig. 14.1, assumes a "faster" (channelling) reactor and a "slower" uniform reactor.. There is an exit age
distribution for the faster reactor which occurs as a distinct pulse clearly before the exit age distribution of the second
reactor . The fraction of effluent which has been in the real reactor for less than time t shows a step up frum zero when
flow leaves from the "faster" reactor,. This fraction is the fraction of flow in the "slower" reactor. When the flow leaves
the "slower" reactor this fraction becomes one .
The model in Fig. 14.1 is PFR and CSTR in paralleL The exit age distribution for the CSTR is a negative gradient curve,
interrupted by the distinct exit age distribution pulse of the PFR. The CSTR will always provide a fraction of effluent
which has been inside the reactor for less than time t, which increases with time. But when the effluent exits the PFR at a
specified time after zero, this increased effluent is superimposed upon F(t) of the CSTR, giving a combined F(t)..
Conversion
I-a
82 =--=1.5
1- fJ
v:
v
a=--L fJ=-1.
V
v
dX
PFR mol balance: - -
dV
rA
= --FAa
Rate law: - r A = kC~
14-13
Stoichiometry: liquid phase
CA
= CAo (1- X)
V = _1_[2£(1
+ £)In(l- X)+ £2 X +-->-(I_+_£2-<-)X]
2
2nd order PFR:
1- X
kC Ao
V
c
a
1
2
b
1
2
8 = - + - -1 = - + - -1 = 0
where
a
£=0
a
Determining
and
fJ:
a == VI ~=0 ..5
fJ
V
VI
1- ~ =Ls(I- ~ J
V =1OvI
_ 0 25
- VI
/2.5 - 5a - 0 5
a-_ 2.5 _Vi . and fJ - - - - - - •.
Gives
10
1"1 + 1"2
1" =
Now
2
Hence
VI IVI
But
VI
=
V/5
VI
VI I VI + V2I V2
2
= 2..5vI
1 2
kC Ao
V
5 .
= lll1n
+ V2 IV2 =10
-
Gives 1"1 = 2.5 min
Substituting into
~=
2•.5
V2
= 7..5v 2
1"2 = 7.5min
2
[2c(l+c)ln(I-X)+c2X + (l+c )X]
1- X
Reactor 1:
2.5
=0.1
~ 2' [1 ~~,-]
1=[-~]
1-- XI
CAl
-XI =0 ..5
= C AJl- XI) = 2(1-0.5) = Imall dm
3
Reactor 2:
75= __1_[
X2 ]
2
..
0.1*2 l-X2
1 [X2]
3='0.1*22
i-x2
-X2
=0.75
C A2 = C Ao (1- X 2) = 2(1- 0.7.5) = 0.5mal I dm
C . =
AexlI
CAl +.EA2
2
3
= 1+0.5=075 lid 3
2
.ma m
14-14
x = CAO -CA = 2-0.75 =0.625molldm 3
2
CAo
P14-3 (a)
Money for buying reactors
Using the tank in series model:
X -_ 2Da + 1- .J4i)a + 1 wheI'e Da--k""Cao
.
Second order reaction
Assume that 7
=7 t
.L
2Da
and that in reactors medelled as mOle than one tank, that 7
7t
=-.
Number of tanks
n
7
n =-
a2
rounded to the nearest integer.
Reactor
Maze & blue
Green & white
Scarlet & grey
Orange & blue
Pmple & white
Silver & black
Crimson &
white
_.
~(min)
2
4
3.05
2.31
5.17
2.5
2.5
:r(min)
2
4
4
4
44
2
-
n
1
1
2
3 ..
1
3
1
X
0.50
0.61
0.69
0.72
0.61
0.72
0.5
---
.--
Where
Scarlet & grey: Xl = 0 ..5, C AI =0.5 -7 X 2= 0.38, CA2=0.31--t X = 0.69
Orange & blue: Xl = 0.43, C AI =0.57 ---t X2 = 0.34, C A2=0.38-7 X3 = 0.27,
CA3 =0.28-7X=0.72
Using the combination of maze & blue followed by crimson & white reactor (same overall conversion either way)
Xl = 0.5, CAI =O.5 -7 X 2 = 0.17, CA2=0.41-7 X=0.59
The orange & blue or silver & black reactors which both approximate to 3 tanks in series give the greatest conversion.
P14-3 (b)
More money for buying reactors
Try:
Green & white and Maze & blue: Xl = 0 . 61, C AI =0.39 -7 X2= 0.34, C A2 =0.26-7 X= 0 . 74
Scarlet & grey and Maze & blue: Xl = 0.69, CAI =0.31-7 X 2 = 0.42, C A2=0.18-7 X= 0.82
Orange & blue and Maze & blue: Xl = 0 . 72, CAI =0.28 -7 X 2= OAO, C A2=0.17-7 X = 0 . 83
The highest conversion is now obtained from the Orange & blue reactor combined with the Maze & blue reactor..
P14-3 (C)
Ann Arbor, MI
East Lansing, MI
Columbus, OR
Urbana, IL
14-15
2
Evanston, IL
West Lafayette, IN
Madison, WI
P14-4
Packed bed reactor with dispersion
1st order, k1=O.0167/s, 1::=0.5, dp=O.l cm,
V
= ,u =0.Olcm2 / s
L=lO cm, U= 1 cmls
P
Re =
pUd p
,u
V
= 10 and Sc = - - no data concerning
DAB
DAB
From packed bed conelation for D a ' and liquid phase region of graph,
Da c
Gives - - - =
Ud p
p -_ 2*1*0.1 __ 0.4cm2/s
2 approx -7 Da ___2_U,_d_
c
0.5
UL 1 *10
Pe., =-=--=25
Da
0.4
4qexp(Pe,/ 2)
X = 1-
.---------
[(I+q)2 exp(Pe,q/ 2)]-[(I- q)2
q=
f
exp(-pe,%)]
4Da
+--
Pe,
L
Da='tK and 1: = -
U
10
=-.
1
= lOs
-7
Da=O.l67 and q=L013
X = 0.15
Conversion X=15%.
P14-5 (a)
Number of tanks in series
Bo
+1
2
Assuming the Peclet-Bodenstein relation: n = Where
UL
Bo = . -
Da
To estimate Bo ,
diU
Re = V
5*2
V
0.01
=- = 1000 and Sc = ._=- =2
0.01
From gas phase dispersion correlation chart,
Gives
Da
DAB
Da
ud t
=8
= 80cm 2 / s
14-16
0.005
Eo= 2*200 =5
80
n=~+1=3.5
2
P14-5 (b)
Conversion
Using individual reactor material balances:
Reactor 1:
X
Mol balance:
rAV
= -.F Ao
Rate law: - r A
= kC A
Stoichiometry:
CA
2
= CAo (1- X I)
0=0 and £=0 hence no volume
change
XI = kC Ao
2
(I-XJ2
vC Ao
=_25*0.01(I_XJ2
0.039
XI =6.366(I-X I )2 ~0.674
3
CAl = C Ao (1- XI) = 0.00326moi / dm
Reactor 2:
XJ2
kCAI(lX 2 = -------- ~ X 2 = 0.507
v
CA2 = CAl (1- X 2) = 0.00326(1-· 0.507) = 0.001607moi / dm 3
Reactor 3:
X3 = kC A2 (1- X 3)2 ~ X 3 = 0.387
v
C A3 = C A2 (1- X 3) = 0.001607(1-- 0.387) = 0.000985moi / dm 3
Reactor 4:
C A4
kC
(1 ~ X)2
4
~ X 4 = 0.305
v
= CA3 (1- X 4) = 0.000985(1-··· 0.305) =0.000685moi / dm 3
X4 =
A3
Bounds on conversion:
3 tanks
X
= C Ao -. CA3
= 0.9015
CAo
4 tanks
X
= C Ao -
C A3. = 0.9315
CAo
P14-5 (C)
Change of the fluid velocity
Let U=O.lcmls
Re=50 and Sc=2
14-17
D
From gas phase dispersion correlation chart,
._a
udt
= 0.5
= 0.5ud t = 0.5 * 0.1 * 5 = 0.25cm 2 / s
Bo = UL = 0.1 * 200 = 80
Gives
Da
Da
0.25
Bo
n=-+1=41
2
The conversion is close to the one PFR 2nd order reaction:
k=25dm3/(mol's)
't=l!U=2/0.00 1=2000s
Da=k'tC Ao=500
X
=
Da =0.998
l+Da
Let U=100cmls
Re=50000 and Sc=2
From gas phase dispersion conelation chart,
D
_a
udt
= 0.21
=0.21udt = 0.21 *100* 5 = 0.25cm 2 / s
Bo = UL = 100 * 200. = 190.5
gives
Da
Da
105
_ Bo 1 - 190.5
n--+
- + 1- 96
2
2
nd
The conversion is close to the one PFR 2 order reaction:
k=25dm3/(mol's)
't=l!U=2/1=2s
Da=k'tC Ao=05
Da
X = - - - = 0.333
l+Da
P14-5 (d)
Change of the superficial velocity
Re
= dtu = 0.2*4 = 80
V
0.01
From packed bed dispersion con elation chart,
DE
ud p
_ a-
= 0 ..55
0.55ud p 0.55 * 4 * 0.2
D a=---=
=1.1
E
0.4
Bo = UL = 4 * 200 = 727
Da
1.1
Bo
727
n = .- + 1 = - + 1 = 364.5
2
2
14.. 18
The conversion is close to the one PFR 2 nd order reaction:
check
k=25dm3/(mol's)
.=1!U=2/1=2s
Da=k'ICAo=O.5
Da
X = - - - = 0.333
I+Da
P14-5 (e)
Liquid instead of gas
Part (a)
Re = dtu = 5 * 4 * 0.001 = 0.2
V
0.1
Sc=~-=~=2*107
DAB
5e-6
This is off the scale of the graph for liquid phase dispersion, hence Da cannot be evaluated .
P14-6 (a)
Peelet numbers
From Example 13.2 cr2=6.J9min2 and tm =5.15min
Closed:
=_~
__
2_(I_e-pe')---:;Pe
P
2
r
(j2
2
tm
er
Per
=7414
•
Open:
(J"2
2
8
Per
Per
--2 =--+---2 -~Per =11.68
tm
P14-6 (b)
Space-time and dead volume
T
tm
=. -----.= 4.40
1+2/ Per
= T*V o = 263.8dm 3
VD =V - Vs = 156.2dm 3
Vs
156.2
%deadvolume = --- = 37.2%
420
P14-6 (C)
Conversions for ]'t order isomerization
Dispersion model
Da=k't=O. 927
14-19
FW
Da
q = 1+-- = 1221
Per
= _
X
4qexp(Pe r 12)
= 0 570
1 (1 + q exp(Perq 12)- (1- q)2 exp(- Perq 12)
.
Y
Tanks-in-series
'['2
n = - 2 =4.35
(J
P14-6 (d)
Conversions PFR and CSTR
PFR: X = 1- e -k1 X=0,604
CSTR: X
= 1--1-
X=OA81
1+1k
I~~~~ __________~I_~_.~_~_-~_
-?
+X~CS1R
XPFR
I 0.604
.--~.--------------
0.481:=J
.----------~
P14-7
Tubular Reactor
1st order, irreversible, pulse tracer test --7
(J'2
= 65
S2
and t m = 10 s
For a 1 order reaction, PFR: X =1- e-k~ =0.98
We need 't' and k. There being no data for diffusivity (Schmidt number) Daand hence Per cannot be obtained using
tubular flow correlations.
st
Calculate
V= 3 m * 25 dm2 * (01 m/dm)2 = 0.,75 m3
2
3
't' = 0.75 m / (3*10 m3/s) = 25 s
This is greater than tm so channeling is occuning
Calculate, k
1
In-k=
1- X = 391125 s = 0.,156 S-I
'['
Therefore assume closed vessel dispersion model:
tm ='t' = 10 s space-time
Calculate, Pe
14-20
2
(Y2
-2
= ---
Per
tm
2
--2
(l_e-per ) = 65/102 = 065
Per
Iterating -7 Per = 1.5
Calculate, X using the measured V and
4q
X = 1-
Vo
giving 't = 25 s
exp( P;,)
------=
[(I+q)' exfe~*q )]+-q), ex{ -p~ *q)]
Da = 't k = 25 * 0.156 = 3.9
q=
FW
Da
Pet
1+ -- =
Pf.1*3.9
1+
= 3.376
1.5
4 *3.376eXpC;5)
X = 1--
[(I + 3.376)'
ex{1.5 *~.376)] -[(1- 3.376)' ex{ -1.5 :3.376)]
= 0 . 88
Conversion for the real reactor assuming the closed dispersion (X=0.88) model is less than for the ideal PFR (X=0_.98)
V
'tideal
= - =25 s
Vo
V=VD+Vs
Vs= a *V
Note,
'tRTD = 10 s = a * 't , a=10125 =0.4
./DisperSion Volume
~---------------------------7~---~
Dead Volume = (1- a) V= 0_6 V, Vs = OAV
Use tm = lOs, Da = tm * k = 10*0.156 = 1.56
4*1.56
1.5
q= 1+----=2.27
14-21
XD = 1- (19.22/58.38) = 0.67
P14-8 (a)
E(t):
06~~--·~------~
0.4
E(t)
0.:2
:2
4
t
Conversion T-J-S and Maximum Mixedness Model
X T_I _S=05
1
r
For a fir st order reaction Xmm=X I -I-S= 1-- - - - -
(1+: k
=
= JE(t)tdt=
1:=tm
2
4
2
Jo.25.t dt+ J(1-O.25.t)tdt=2min
0 0 2
(Y2
=
=
o
0
2
2
= JE(t)(t-t m)2dt = JE(t)t dt-1: 2 =-min 2
1:
3
2
n=-=6
(Y2
From the conversion it is possible to determine k at 300K:
14-22
(
k=
-1)
1
~
. = 0.367 min-I
T
n
The conversion at T=31O K is given by:
k310
.
= k300 exp(
R = 1.986
E
_
E
)
R300K R· 310K
= 1.422 min -I
ca"-
molK
P14-8 (b)
Complex Reactions
dC A
dT-=-klC A -k3 C A
dCB
- = klC A -k 2 C B
dT
!-CC
=k C
dr
2 B
dCD.=k C
dT
3 A
Where k l = k2= k3=Ol min· l
dC A
dC A
"r
L.
__
d/l, -
+ (C _ C ) E(/I, )
iA
-;iT = -(-kICA
A
Ao
--k3 C J+(C A
dCB _
----(kICA -k 2 C B )+(CB
d/l,
I-F(/I,)
E(/I,)
-C J1="F(/I,)
A
-cBJ
E(/I,)
()
1-·-F A
14-23
Changing variable: A=z-4
See Polymath program P 14-8-bpol
POLYMA TH Results
Calculated values of the DEQ variables
Variable
z
ca
cb
cc
cd
F
k1
k2
k3
rc
ra
rb
t2
rd
cao
cbo
ceo
cdc
t1
lam
E1
E2
E
EF
initial value
0
1
0
0
0
0 . 9999999
0.1
0.1
0.1
0
-0 . 2
0.1
4
0.1
1
0
0
0
2
4
1
0
0
0
minimal value
0
0.6793055
0
0
0
2.854E-06
o. 1
0.1
0.1
0
-0 . 2
0.0537147
4
0.0679305
1
0
0
0
2
0
0
0
0
0
maximal value
4
1
0 . 142158
0 . 0181892
0.1603473
0.9999999
0.1
o. 1
0. 1
0 . 0142158
-0.1358611
0.1
4
0. 1
1
0
0
0
2
4
1
1
0 . 4965476
25 . 277605
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(ca)/d(z) = -(-ra+(ca-cao)*EF)
[2] d(cb)/d(z) = -(-rb+(cb .. cbo)*EF)
[3] d(cc)/d(z) = +rc+(cc-cco)*EF)
[4] d(cd)/d(z) = -(-rd+(cd-cdo)*EF)
[5 J d(F)/d(z) = -E
Explicit equations as entered by the user
[1] k1 = 0.1
[2J k2=0 . 1
[3J k3=O . 1
[4J rc=k2*cb
[5] ra = -k1 *ca-k3*ca
[6] rb = k1 *ca··k2*cb
[7] t2=4
[8] rd=k3*ca
[9 J cao = 1
[10] cbo=O
[11] cco=O
[12) cdo = 0
[13) t1 =2
[14J lam=4-z
[15 J E1 = D..25*lam
[16 J E2 = 1-0 . 25*lam
[17) E = if (lam<t1) then (E1) else ( if (lam<=t2) then (E2) else (0))
[18) EF=E/(1-F)
14-24
final value
4
0.6793055
0.142158
0 . 0181892
0.1603473
2.854E-06
0. 1
0.1
0.1
0.0142158
-0.1358611
0.0537147
4
0.0679305
1
0
0
0
2
0
0
1
0
0
0.20r-------------------,
0.16
0.12
0.08
0,04
O.OO"--~--~----~----'
0,,0
0.8
1.6 z
2.4
3.2
4.0
P14-8 (C)
Use FEMLAB for thefull solution
Complex reactions and Dispersion Model.
dC A
dT
= -klCA -- k 3 C A
dCB
- - = klC A -k2 C B
dT
dCc=k C
dT
2 B
dC:!2. = k C
dT
3 A
P14-9 (a)
FromP13-4:
tm
=T = {2' min,
V;
0"2
= -~ = O.159min2 and k= 0 . 8 min- 1
21l
Tanks-in-series
For a flrst order reaction the conversion is given by:
From Example 14-1
X T-/-S
= X seg = 0.447
P14-9 (b)
Closed-closed vessel dispersion model
For a flrst order reaction the conversion is given:
4qexp(Pe,I_2)
____
(1 + q)2 exp(Perq 12) -- (1-- q)2 exp(- Perq 12)
X =1-.---
14-25
2
2 (
-Pe )
0.25=----1-e
'
Pe, Pe, 2
Per=683
Da = rk
= 0.638
1l=J1+ 4Da = 1+ 4*0.638 =1.885
Per
6.83
X=0.41
I
XDispersion
0.41 .
Referling to P14.2 the approximate formula for the T-I-S is not good approximation.
P14-10 (a)
Combination of ideal reactors
The cumulative distribution function F(t) is given:
F(t)
05
0
0
I
20
~J
40
t[min]
The real reactor can be modelled as two parallel PFRs:
The relative E(t)
= {l_&(t -- 1'1)+ ~ &(t - 1'2)
4
4
P14-10 (b)
Model parameters
For two parallel PFRs, the parameters are L1=10 min and LF30 min,
P14-10 (C)
14--26
Conversion
Fao! =1I4Fao and Fao2=3/4Fao ,second order, liquid phase, in eversible reaction with k=O ..l dm3 /molmin-! and C Ao=1.25
mol/dm3
CAl
= CAo -
krICAo
CAo
1+ krIC Ao
= 0.556mol I dm
3
kr2 C Ao
3
CAo = 0.263mol dm
1+kZ"2 C Ao
1
3
vC Ao --VC A1 --VC A2
X=
4
4
= 0.731
vC Ao
C A2 = CAo -
P14-11 (a)
FromP13-6
tl =
5 min
E(t)
t lif t < tl
·-2
tl
E(t) = - - \ (t - 2t1 )if tl 5, t 5, 2tl
tl
tm
= tland
a2
= t~ = ~5 = 4.16'
6
0.1
6
o otherwise
Tanks--in-series
10
Z"2
n = -= 6 (As in
2
a
PI4.8)
Second order reaction, liquid phase, kC AO=0.2min-!
C Ai -C Aout
kC Aout
2
=Z"
Solving for CAou1
C
Aout =
-1 + ~1 + 4k-xC Ain
--
2kr
Six-Reactor System: T6=T/6
The Damkohler number is given by Dao= kC Ao't6=0.167
Da.
I
=
--I +
F+4Da"
n
V-.l}
2
with i=1..6
In the following table the exit concentrations fOI each reactor are reported:
Da!
Da2
Da3
Da4
Das
Da6
--
-
0.145
0.129
0.116
0.105
0.095
0.088
14-27
--
--
x = CAO -CA6 = Da o -Da6 =0.473
CAO
Dao
P14-11 (b)
Peelet number
The Peclet number
Ul
Pe, = Da
For Closed-Closed System (no dispersion at the entrance and at the exit of the reactor P13 . 6)
~2 =~=~.--~(l-e-pe,)
r2
p.2
er
Pe ,
6
Pe..=lO..875
P14-11 (C)
Conversion
Linearizing the 2st order according 14..2 (g):
2 ~ k C AO C - k'C
, -- kC A=
-rA
- - AA
2
it is possible to obtain an approximated solution;
k'= k C AO = 0.lmin-1
2
Da =rk'=0.5
Pe..=10.875
(flom Part(b»
q=~l+ 4Da
Per
-1-
X-
=
~*0.5·=I.088
Vl-rlO.875
4qexp(Pe, 12)
(1 + q)2 exp(Pe,q 12) - (1- q)2 exp(- Pe,q 12)
X=0.382
Full solution can be obtained with FEMLAB
P14-12 (a)
Remembering the physical meaning of the Peclet number:
Pe =.
r
rateo! transport by convection
rateo! transport by diffusion or dispersion
lim X Disp
Pe,---7=
Ul
Da
= X PFR
For a first order reaction:
14-28
4qexp(Pe,/2)
X=l-·----------~--~~--~-----------
(1)
(1 +qY exp(Pe,q 12) --(l-qY exp(- Pe,q 12)
q~~1+-4D-a.
Pe,
Da = 'lk
For Per»1 (or for small vessel dispersion number)
~
4Da
q = 1+·--Pe,
( J3
= 1+2----2(J
--- 2+0---.
Da
Per
Da
Per
Da
Pe,
Introducing (2) into (1) and neglecting terms
2kr (
4 .( 1 +--- ) e
Pe
=1-e
0(~-J2
Pe,
4{1+~:)Jn
X = 1---
X
(2)
[
-kH (tt)'
Pe,
pe,(!+ 2D!!.. 2Da'_))
2
Pe
"
Pe
J
P14-12 (b)
_k1:+{k1:j2
X pFR =1_e-k 1: and
X DisoPe»! = 1- e
Per
In order to achieve the same conversion:
P14-12 (C)
Defining:
Pe, = UI PFR = 0.1
Da
J 2J
4q exp(per ( ___
I 1
0.99(1- e-
k
1:)
IpFR
= 1-
(1 + qYexP(pe,(_1-J q I2) - (l-qYexp(- pe,(_I_Jq I2J
IpFR
IpFR
14-29
The following figure shows the solution as function of ktpFR fOl different values of lJlpFR ·
P14-12 (d)
Starting from 14 . 12.1 and subtracting the conversion for a PFR:
InS.L= kr
CA p1ug Pe
For small deviations from the plug flow:
P14-12 (e)
According P12.3 dispersion doesn't affect zero order reaction..
P14-13
From P13.J9:
2
2
T=lO min and a =74min
for 0 ~ t ~ 3 El (t)
f+ 9.9505.10-
2
= ---LI675 .1O-3 t 4 + L1355 .10-2 t 3 -
4.7492 .1O-2 t 2
t
for 3 ~ t ~ 20 E2 (t)
= -1· 8950 .10-6 t 4 + 8.7202 .10-5 t 3 -- L1739 .10-3 t 2
E(t) = - L7979 .1O-4 t + 0.092343
for 20 ~ t S 60 E3 (t)
= L2618 .10-8 t 4 -
2.4995 .1O-6 t 3 + L8715 .1O-4 t 2 -
6.3512.10- 3 t + 0.083717
for t > 60 0
P14-13 (a)
2nd order, kCAO=O 1min-!, C Ao=lmolJdm3, Segregation Model
Segregation Model
00
X
= fX(t)E(t)dt
o
kC Aot
1+ kCAot
Where X (t ) = --.:.=--
See Polymath program P 14-13-apoJ
POLYMA TH Results
Calculated values of the DEQ variables
Variable
t
Xbar
initial value
o
o
minimal value
o
o
maximal value
70
0.4224876
14-30
final value
70
0.4224876
0.1
0
0.092343
0.083717
0
0
0
kCao
E1
E2
E3
E4
E
X
0.1
-2.436E+04
-21.261016
L949E-05
0
0
0
0. 1
0.0836855
0 . 092343
0.083717
0
0.0836855
0.875
0.1
-2.436E+04
-21.261016
0.0017977
0
0
0.875
ODE Report (RKF45)
Differential equations as entered by the user
[ 1] d(Xbar)/d(t) = E*X
Explicit equations as entered by the user
[1] kCao = 0.1
[2] E1
[3] E2
[4] E3
=-0.0011675*t"4+0.011355*tA3-0.047492*tA2+0 . 0995005*t
=-1.8950*1 QA(-6)*t"4+8.7202*1 OA(-5)*tA3-1 . 1739*1 QA(-3)*tA2-1 . 7979*1 OA(-4)*t+0.092343
= 1.2618*1 OA(-8)*t"4-2.4995*1 OA(-6)*tA3+ 1.8715*1 QA(_4)*tA2-6 . 3512*1 QA(-3)*t+0 . 083717
[5] E4 =0
[6] E = if(t<=3)then(E1 )else(if(t<=20)then(E2)else(if(t<60)then(E3)else(E4)))
[7] X = kCao*t/(1 +kCao*t)
P14-13 (b)
2nd order; kCAo=O.lmin·1, CAo=lmolldm3, Maximum Mixedness Model
Rate Law: - r A
= kC A 2
C A =C Ao (1-X)
= kC Ao 2 (1- X Ywhere k=O.l dm3 /molmin
rA
~=kC (1_X)2
C
Ao
Ao
dF
.--= --E(Z)
where z=6Q.··).
dz
See Polymath program P 14··13-b.pol
POL YMA TH Results
Calculated values of the DEQ variables
Variable
--z
X
F
Cao
lam
Ca
k
ra
E4
initial value
0
0
0.99
1
60
1
0. 1
-0.1
0
minimal value
0
0
-0.010344
1
0
0 . 52273
0.1
-0.1
0
maximal value
60
0.4773052
0 . 99
1
60
1
0.1
-0.0273247
0
14-31
final value
6()"""--0 . 4047103
-0.010344
1
0
0.5952897
0.1
-0 . 035437
0
-1.284E+04
-9.8680524
2.228E-05
E1
E2
E3
o
E
-1 . 284E+04
-9.8680524
1.806E-05
o
0.0832647
0.092343
0.0837l7
0 . 0832647
o
0 . 092343
0.0837l7
o
ODE Report (RKF45)
DifferentiaJ equations as entered by the user
[1) d(X)/d(z) = -(raiCao+EI(1-F)*X)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] Cao = 1
[2] Jam = 60-z
[3] Ca=Cao*(1-X)
=.
[4] k
1
[ 5 j ra = -k*CaA2
[6) E4 = 0
[7 J E1 = -0.0011675*Jam"4+0 ..Q11355*Jam A3-0.047 492*JamA2+0.0995005*Jam
[8) E2 = -1.B950*1 OA(-6)*Jam"4+B.7202*1 QJ\(-5)*JamA3-1 . 1739*1 OA(-3)*JamA2-1 . 7979*1 OA(-4)*Jam+0 . 092343
[9 j E3 = 1 . 261B*1 OA(-B)*Jam"4-2A995*1 OA(-6)*Jam A3+ 1.B715*1 OA(-4)*JamA2-6.3512*10A(-3)*Jam+0 . OB3717
[1 ()) E = if(Jam<=3)then(E1 )eJse(if(Jam<=20)then(E2)eJse(if(Jam<60)then(E3)eJse(E4)}}
(Check F(t))
P14-13 (C)
Tanks in series and 1st order reaction with k=O.lmin-1
For 1st order reaction is acceptable a non integer value of n for calculating the conversion
1
r
X =1----=o.527
(1+ : k
P14-13 (d)
Dispersion models and 1sf order reaction with k=O.lmin-1
Peclet number open system
(J"2
-
t m2
2
8
= --.. + - ~ Pe = 4.906
Pe Pe 2
FEMLAB application
Peclet number closed system
2
(J"2
2 (
-=·-----I-e
t'/'n Pe Pe 2
X
= 1-
q=
-pe) ~Pe=0.98
4q exp(Pe /2)
r
-.--"----"-'----'-----
(1 + qy exp(Pe,q / 2) -
(l-qy exp(- Pe,q /2)
~+ 4Da
V~
Per
14-32
Da
= 'lk,
X=4.59
P14-13 (e)
Dispersion models and 2nd order reaction with k=O.ldm3Imol·s and CAo=lmolldm3
Conversion
Linearizing the 2 s1 order reaction rate according 14.2 (g):
2
CAO
- k- r A -- kC A =
- C A -- k'C A
2
it is possible to obtain an approximated solution;
k'= k
CAO
= 0.05s-1
2
Da
= 'lk,'= 30
Pe..=O.98
q=
(fromPart(b))
~+ 4Da =~1+ 4*30 =11.111
V Per
0.98
J
x = 1.-
4qexp(Pe,/2:-:-)_-:-___
(1 + qYexp(Pe,q 12) - (1- qYexp(-- Pe,q 12)
X=O . 998
Full solution can be obtained with FEMLAB.
P14-13 (I)
Two parameters model (backflow)
For tracer pulse input
dCI
~--=VICI2 -VICI l
dt
dC 2
V2 ----= vlen dt
V OC T2
Defining:
a=
~
,fJ=}2. and
V
r=~
Vo
Vo
We arrive at two coupled differential equations:
For tracer pulse input
ar dCI = fJ(C T2
dt
(1- a) dC 2
dt
-
ell)
= fJ(C n
- CT2 )
See Polymath program P 14-1 3-f.pol
14-33
P14-13 (g)
3
Two parameters model models and 2nd order reaction with k=O,ldm'lmol 's and CAo=lmolldm
= -ka'ZC~1
= -kC~2 (1- a}r
C AO + j3C A2 - j3CAl
j3C Al
-
j3C A2 - CA2
P14-13 (h)
Table of the conversions
~X~l~I~S________I~X~M~M~______-4~X~s~eoL-________~X~_D~i~Sp~ers~io~n ____~~X~t~wo~M=~et~~s~--~
~0.527
0.405
0.422
0.5
?
P14-13 (i)
Adiabatic reaction with Segregation Model
2nd order, kC Ao=O.lmin- l , CAo=lmoVdm3, Segregation Model
Segregation Model
ax
= kC (I-xl
at
AO
T
= 320+ 150X
k =0.1 *exp(45000/8.314 * (1/320-lIT))
x = fX(t)E(t)dt
o
See Polymath program P141J-j ..pol
P14-14 (a)
Product distribution for the CSTR and PFR in series
T
k2/kl
'tkICAO
5.0
0.2
TJ
-~.
2.0--2
0.5
20
T3
0.1
200
T4
_.
Considering Anhenius equation and applying the following notation:
k'1
E,
In--=--=-E
A,
RT
r1
We can write this linear system of 5 equations for the 5 unknowns (Ell' E 12 , E 13 , E 14 , AI):
k
-
-
_.
-
In.-..!L = EI2 - Ell =
kl2
k
In·-1.!.. = E13 - Ell =
k13
14-34
P14-14 (b-d) No solution will be given at this time .
P14-15 (a)
Two parameters model
Omin::;; t < 10 min C(t) = 10· (1- exp(-O.lt))
10 min ::;; t C(t) =.5 + 10· (l-exp(-O.lt))
Omin ::;; t < 10 min
10 min ::;; t
Fz (t)
FI (t)
1
3
= 2 . (1--- exp( -O.lt))
3
2
3
= - +-. (l-exp(-O.lt))
F(t)
t(min)
The F (t) can be representative of the ideal p~ and ideal CSTR in parallel model:
14-35
node1
C{= Fractional volume
13= Fraction of flow
f3 = !3' a = ~-3'
7
= 10 min
•
7 PPR
= -Vi- = -a7 =10mIn
V pPR
f3
_ (I-a) _
_
(
)7 -lOmm
1- f3
7CSIR -
P14-15 (b)
Conversion
2 nd order, vo=l dm 3/min, k=O ..ldm3/mol·min, CAo=L25molldm 3
Balance around node 1
fJv oC PFR + (1- f3)vOCCSTR
= VoC A
For the PFR:
Second-order
= C AO .- C PFR =
X PFR
Where Da PFR
C PFR
Da PFR -- = 0556
1+ Da pFR
C AO
a
= k f3 1C Ao =1.25
= 0.556mol / dm 3
For the CSTR:
X CSTR
=
C AO
- CCSTR
C AO
=
I-a
(1 + 2Da cSIR ) - ~1 + 4Da cSTR2Da cSTR
Where Da CSTR = k --1C Ao =
1- f3
CCSTR
Where
1.25
= 0.725mol / dm 3
CA = fJC PFR + (1- f3)CCSTR = 0.668mol / dm 3
X = CAO -- CA = 0.465
C AO
14-36
= 0.42
P14-15 (c)
Two parameters model
omin ~ t < 10 min
10 min ~ t
F2 (t)
Fl (t)
=0
2
r
n
="31 +"3'
{1- exp-- LO.2(t -lO)Jf
F(t)
The F (t) can be representative of the ideal PFR and ideal CSTR in series with a bypass between them:
Vb=~Vo
.--.----~
node 1
a= Fractional volume
(3= Fraction of flow
1
3
P = 3' a = 4'
40.
T
= -3 rrun
TpFR -- -~ -- aT -10'
rrun
VO
.
TCSTR -- (1( a))T -- 5 rrun
I-P
Conversion
2nd order, vo=1 dm3/min, k=O.1dm3/molmin, C Ao=1.25moVdm3
Balance around node 1
PVoCb + (1- P)vOCCSTR
= PVoCPFR + (1- P)vOCCSTR
= VOC A
For the PFR:
Second-order
14-37
pFR - Da pFR - 0 556
X PFR -- CAD - C
- .
CAD
1+ DapFR
Where Da pFR =karC Ao =1.25
C PFR = 0.556mol / dm
3
For the CSTR:
X CSTR
=
C PFR ·-CCSTR
CPFR
=
(1+2DacsTR)-~1+4DacsTR
2Da cSTR
= 0.185
I-a
= k--rC
PFR = 0.278
1- /3
Where
Da CSTR
CCSTR
= 0.453mol / dm 3
C -C
X=~_A =0.61
CAD
Where
C A = j3C PFR + (1- /3)CCSTR
=0.487mol / dm 3
P14-16
PFRj--+--+1 :::r.
l~~J
VI,
l[
PFR
;L
(4)
(3)
CSTR
v.
CS1R
t·,,.")
~
PFR
t-EJ
(6)
14-38
...
...-,
.... .1
A1
I
_-: PFR
J
,
PFR
Al
--+
I
I
PFR
I
I
PFR
-
I As
J
.~
or
(7)
PFR
Recycle
II
...
(8)
--1_-....J~+--~--.....
z=o
Dead Zonas
z=l
(10)
(9)
LFR
CSTR
LFR
(12)
(11)
P14-17
The presence of a minimum (mn2) imply the presence of a radial temperature profile that effects the reaction rate and
determines the deviation from the concentration profile given by Eq . (14.51) .
The enthalpy balance allows the identification of the dimensionless thermal parameters:
Defining
T
iJ=---~sot
C
If/=._A
CAO
U
U =Uo
14-39
qJ=O
where
013
OqJ
hR
_ 013
-=0
Bi = -
Rt
= Bi(l- 13)
oqJ
ratio convection-radial conduction
Ak
ratio conduction-enthalpy flux
p
~Hreac
--...:.="'--
dimensionless adiabatic heat
CpTcoolant
Da(tJ) = k(tJ)C~o'l"
A pruametric study of the numerical solutions ofEq . (1451) and the enthalpy balance would give the "exact" conditions
for a minimum in the concentration.
P14-17 (a)
A decrease in the thermal conductivity of the reaction mixtures determines an increase of the bulk fluid temperature and
can determine a minimum in concentration .
P14-17 (b)
Overall heat transfer coefficient increases
For a given flux this determines less deviation from the isothermal case: no minimum
P14-17 (C)
Overall heat transfer coefficient decreases ."
For a given flux this determines more deviation from the isothermal case
and the possibility of a minimum.
P14-17 (d)
The coolant flow rate increases;·
The external heat transfer coefficient increases, this implies a lower wall temperatme and a resulting lower temperatme
in the reacting mixtme and less deviation hom the isothermal case: no minimum
P14-17 (e)
The coolant flow rate decreases:
The external heat transfer coefficient increases, this implies an increase in the wall temperatme and a resulting higher
temperatme in the reacting mixtme that can determine a minimum in the concentration .
Below shows how to use FEMLAB for this problem.
Femlab Screenshots for the baseline case
(1) Domain
14-40
o
Axis equal
>-r-z limits··----·· ..· ---'- ... '
phi
~Auto
[[--OK ..... ,II
(2) Constants and scalar expressions
- Constants
- Scalar expressions
14-41
canceCJ
I
Apply
!
N~me
~~
~--~x;;r~~;i90- -~- -~~-~~-~~- ~-~ .-~~-~--~-
----~~---
~~-~~-------------------l
--+ti\
~?:i;~~::aY2)---~-----~-------~
------.----. - -·-------·-·---·---·-·-1
kA-------[~A-O.:-~A)/CAO~-·---·-··-·-·-·-·-----····---.
_____ .____ y____--_--"._ ..-.-.. ---.------.-".-..".--m-'.----'-------..-------..
. . ---.'-----"1
pB . ___ .... _~~~O·X~--_- .. ______ .. _ ...____________________
~
CC
i2'cAO'xA
i
i-----·--i
---.-~-.".---~----~---------~~---------.-,-.---~----~----4
~
A' exp( -ElRfTl!h..~at·(cA·cB-c~JKeq) _____.___ ~ ____________ _l
Keq
KeqO'exp(dHrxlR'(1 1303-1 fT))
I'
;-..,..;
--------<
p
l:-rA)'(-dHrx)
I
;~
--.~.-.---~--.--
n-~'
____ r
r---·-----~·-·I-~--·---·
~_._.
i
--.-.-.~-
. ----- -.-----.
..--..--..-.-..
~----.-~- ~----~"1
__.
_____
.•___""'_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
I'
--"'u;;;.;;;;;;
b_,,_~
r
OK
l t cancel] [
(3) Subdomain settings
Physics
(Mass balance)
V'(-DVcA+cAu) = R, cA = concentration
Sub domain selection
r--L~-~d:~J
Library material: i
Valuell'xilression Description
[1
Cits
(~1 D isotropiC
~!.!
() 0 anisotr opic
R
u
[ ] Select by group
v
r;;'] Active in this domain
(Energy balance)
14-42
Time-scaling coefficient
.1Ditfusion coetficient
Ditfusion coeUicient
Apply
,'Subdomain selection~-
':1
r;;;;;;i~~llnit [,,,,,,s,',t
i ,-Thermal properties and heal souI'ceslsink:s-'
. I
!I
i!
II
i
II
[LMCLJ
Library malerial:
I Quantity
,
15
I
1
ValueJl;xpression
ts
Thermal conductivity
10k (isotropic)
I
i!
I
i!
-II
() k (anisotropic) c=::..::.'"-.=-..______.J Thermal conductivity
I
II
I ,
! I
I!
i
ii i , '
p
;:====;
U
'1,1
: i
i 'D'~~I:ct:y group
l: b
',
I
1
, J,
L!'JAoIive in thiSd!,~ain,
II
hiN o .;
r.,_A,rt,ificial,_Dlt,fu,~i;__.~..J"
~.~..:_ _ _ _•__~ ... '..
-- Initial Values
(Mass balance) cA(tO) = cAO
(Energy balance) T(tO) =TO
-- Boundary Conditions
@ I' = 0, Axial symmetry
@ inlet, cA = cAO (for mass balance)
(for energy balance)
Boundary cpnditions"
Boundary condition: [Heat flux
-
!, ,,---
,,~-~-
ValuelExpression Description
k~;~E~!!h:l~-
jlnward heat flux
Temperature
@
@
outlet, Convective flux
wall, Insulation/Symmetry (for mass balance)
(for energy balance)
I'-Boundary
,"" conditions
'"
'
IBoundary condition:
~~~f~l!..~·:""
Qi:t~ntitY
,ValuelExpression Description
qo
[~~~f~f~r
To
Description
Time"scaling c,oefficienl,'
Inward heat flux
1 Temperature
.. ,J
14·43
Heal source
P14-18
Vary the Peclet number and the reaction order in laminar flow (Example 14-3(c))
n- 1
....l·CAO
Da --UI-
-
n- 1
-'£kCAO
Uo
Pe = UoLI DAB
L = 6.36m
R=O.05m
3
k = O.25dm I mol. min
U o = I.24mlmin
C AO =O.5molldm 3
DAB
= 7.6xlO-5 m 2 I min
n
-t-0.1
Attempt to evaluate noninte gral power of negative
number!
0..5
Attempt to evaluate nOllinte gral power of negative
number!
08
1--
-
1
15
1--
-r----
2
2.5
--- 1----
Conversion
Pe
1.04xl03
I
1.04x104
I
1.04xlOs
I
--1.04xld'
I
1.04x107
I
1.04x103
I
1.04x104
I
1.04xlOs
I
1.04xlO°
I
--1.04x107
I
j
1.04xlO
0.824
4 1.04x10
0.818
1.04xlOs
0.783
--- 1.04xlO°
I
1.04x107
I
1.04x103
- 0.721
4
1.04x10
0.717
0.687
1.04x105
-1.04x106
0.653
1.04x107
0.641
--1.04xlOJ
0.526
1.04x104
0.522
1.04xlOs
0.503 1.04x106
0.481
1.04x107
0.472
j
1.04xlO
0.390
4
1.04x10
0.388
1.04xlOs
0.374
1.04x106
- 0.359
7
1.04x10
0.352
1.04x103
0.292
1.04x104
0.290
1.04xlOs
0.281
14-44
Parameter
100* DAB
---10* DAB
DAB
O.I*D AB
O.OI*D AB
100* DAB
-10* DAB
DAB ------O.I*DAB
-------O.OI*DAB
-100* DAB
10* DAB
DAB
O.l*DAB
-O.OI*D AB
100* DAB
10* DAB
--DAB
0.1 *DAB
0.01 *DAB
100* DAB
10* DAB
- -DAB
0.1 *DAB
0.01 *DAB
100* DAB
10* DAB
DAB
0.1 *DAB
O.OI*D AB
100* DAB
10* DAB
DAB
-.~-----
-
3
-
1.04xl00
1.04x107
1.04x103
1.04x104
1.04x105
0.270
0.265
0.219
0.218
0.211
0.203
0.199
0.164
0.163
0.159
0.153
0.150
0.123
0.122
0.119
0.115
0.113
1.04xl06
3.5
1.04x107
1.04x103
1.04x104
1.04xlO~
1.04xlO°
1.04x107
4
-
1.04xl03
1.04x104
1.04xlO'
1.04xlO°
1.04x107
-
---
1.0
n
0 ..8
--*-·1 . 0
1..5
---y--- 2 . 0
-.--11-
09
-_.!!!-__•
08
0.7
c
o
.i!1
..• ·111 ................._..• __ III
06
05
~
§
O.I*DAB
O.OI*DAB
100* DAB
10* DAB
DAB
O.I*DAB
O.OI*DAB
100* DAB
10* DAB
DAB
O.I*DAB
O.Ol*DAB
100* DAB
10* DAB
DAB
0.1 *DAB
O.OI*D AB
0.4
2. 5
•. ·3..0
3..5
•
4. 0
·--·-----····T------·T·-·---T_···_··__ T
<.)
..
.....
03
...........
02
0.1
00·
1if
1·-
_........
..
, '''''''I
1if
...--. -.-...
, .. ""'I
1if
1if
III
, .. ""'I
1d
, .. ""'I
1if
Pe
(1)
(2)
With the increase of the reaction order n, the conversion will decrease. The conversion will also decrease when
the Peclet number increases.
5
When n is less than 1.5 and Pe is 10 , the diffusion in the system becomes more important.
Below show FEMLAB screenshots useful for this problem.
Femlab Screenshots
(1) Domain
14-45
DAxis equal
phi limits
~Auto
rmin:
rmax:
zmin:
zmax:
(4) Constants and scalar expressions
- Constants
Li-J;me .
. .Expr~55iQn
~
,--=f:~~~
~·-·····,-··-t~~5
C·:---"--. :.~::.·=:·.~~:~,36 _. _. __..
k
,_~,~~
... ___N'_'_. __
.....
,05
__
~t
~
_____ ,_._____ "_
36
b,25
~~==~-=f~
-il:~~:::::~:··~ .-.
.~_",.
__
A_
................... _ ..•_
L~~<~~~"~W"~H'_ ~:l~.
.25
.24
~.~_
_
.8 .,,,._.
.L__".~.__ .
[~---J L_ can~:~J [_AJ:~L.J
- Scalar expressions
..;~
.-.-~-
1
[_.
r
+. -,_ . "
-1
. -t-- .
L.... .....____.. __ .~t~
..._,,__.
(5) Subdomain settings
Physics
14-46
s;;bd~n S~ttin~ - Convection flOd Diffusi~
V.(-DVcA+cAu)
=R, cA =concentration
rSpecies'~--""'------ C"
I
Library material:
I Quantity
-,'.'
[. Load...
[~=::::.~====J
IQAB~::::~~::-.-
D :nisotroPic
I
. [ ] Select by group
~
l_C.!~!i5i;i Diff~SiO~~J
1.1.
- Initial Values
(Mass balance) cA(tO) = cAO
- Boundary Conditions
@ r = 0, Axial symmetry
@ inlet,
r
,
I
Boundary conditiohs
i Boundary condition:
I
I Quantity
I
II
@
@
c;:
No
I
~!t:::
I
Valuel1:xpre$$ion pescfiption
~~~~~~'::~·_-:_~-~·:.=.]concerrtration
~~~Aq'
_. '] In~~r~flux
outlet, Convective flux
wall, Insulation/Symmetry
P14-19 (a)
First order reaction
Input Parameters:
n =1
R=0.05m
L=6.36m
k = 0.251 min
U o = 1. 24m I min
CAO
I
ValuelExpression Description
°ts
@ D isotropic
I0
~ Active in this domaih
•• - •.•••••' - . - - . - - - - - - - -•••• --- •••-- •• - ••.• -- ..
= 0.5mall dm 3
14-47
Time-scaling
co~fficient
Diffusion coefficient
Diffusion coefficient
Conversion:
xA= 0..687 @ Open-vessel Boundary: -Nin =2*UG*(I-(r/Ra)"2)*CAG
xA= 0. . 726 @ Close-vessel Boundary: -Ni·n =UG*CAG
Conversion:
xA= 0..755 @ Open-vessel Boundary: -Nin =2*UG*(I-(r/Ra),,2)*CAG
xA= 0..778 @ Close-vessel Boundary: -Ni·n =UG*CAG
I Variation of Da number
Damkohler numberl Da
1--.
-.
_.
_.
0..449
0..898
1.795
3.59
7.18
14.36
28.72
II. Variation of Pe numbel
Peclet numberlPe
f--.
I.G4e3
1.G4e4
1.G4e5
l.G4e6
1.G4e7
Parameter
Conversion
Closed-vessel
0..40.4
0..527
0..658
0..778
0..870.
_.- 0..930.
0..964
-
--
Open··vessel
0..287
0..439
0..60.6
0..755
0..862
0..928
0..964
8*UG
4*UG
2*UG
UG
DG/2
UG/4
DG/8
_.Conversion
Closed-vessel
0..781
0..781
0..778 -0..772
0..770.
-
OJ)en-vessel
0..781
0..776
..- i---------0..755
---- ' - - - - - -0..740.
0..737
-----
IlL Femlab Screen Shots
(1) Domain
,;;;;";:;;;;,'1.
[lAxls!i Grid
t:."."...... ,.;,
[ ] Axis equal
phi
lim~s
~l Auto
(2) Constants and scalal eXPlessions
- Constants
14-48
Parameter
11~~*g::-DAB ----0..1 *DAB
0..0.1 *DAB
--
. . ._--_.........._-_ ..-+.._ .._.....__ ...........__...._--_ .............. __. . . . _-_. -.--..- . --....--.-
[
OK
'1 [c~ncel
1C!fPiY~
- Scalar expressions
(3) Subdomain settings
- Physics
I Vo( -ElVcA+cAii) = R. cA = concentration
;
Subdomain
f;Al;~itll Element I
rSpecies
["l~;d~::-J
Library material:
Quantity
ValuelExp(ession
DSelect~y group
rtJ Active in this domain
Description
Time-scaling CQefficient
Qts
El isotropic
Eliffu.sion coefficient
El anisotropic
Eliffll~ion coefficjent
R
Re.action rate
u
r ..velocity
v
z.:velocily
[-.-.--------:J
.. ArtificiaIEiffusion .. : ....
14-49
File
Edit·
Options Draw. Physics Mesh Solve Postprocessirig. I';IulfiphysJcs Help
Dri~j~I!_~~lfII~:i.~J~:~_~_gTi !~E~.ifL~~~. .~. ~~L!____ :-_.-
-.f!l...,
Max: 4.319
Sl61'ace: Concentration cA
\~,~:
._ _ _ _ .. _ _ _ )oc'
.---.-.-.------
@
~
;;.::
g
*'
35
~:!
iii
/
05
004
002
0.06
0.Q8
0,1
rSNAP f
(0.12,6)
P14-19 (b)
Third order reaction with k*C 2AO= 0,,7 min-1
n - 1 = _!::-..kC n - 1
Da = rkC AO
U
AO
o
Pe
= UoLI DAB
Input Parameters:
n=3
R=O.05m
L =6.36m
k = 2.8dm 3 I mol. min
U o = 1.24m I min
CAO
= O.5mal I dm 3
DAB
= 7.6xlO- m I min
5
2
I Variation of Da numbeI
Damk6hler numberl Da
-1-.
1----
0.255
0.51
1.02
2.03
4.06
-_.
Conversion
-
-
Closed-vessel
0.381
-----0.466
0.560
0.655
0.741
---
14-51
Parameter
Open-vessel
0.253
0.375
0.506
0.628
0.730
-- f---
8*UO
4*UO
2*UO
UO
UO/2
II Variation of Pe number
Peelet numberJPe
•
1.04e3
1.04e4
1.04e5
1.04e6
1.04e7
Parameter
Conversion
Open-vessel
0.649
0.645
0.628
0.617
0.615
Closed-vessel
0.650
0.654
0.655
0.652
0.650
-
100* DAB
10* DAB
DAB
O.I*DAB
O.OI*DAB
(c) Half order reaction with k= 0,495 (mol/dm3)1I2rnin-l
n--1
D a = rkC AO
-
~kcn-l
U
AO
o
Pe = UoL/ DAB
Input Parameters:
n=1I2
R=O.05m
L
=6.36m
= 0.495(mal / dm 3 )1/2 min-1
U o = I.24m/min
k
CAO = O.5mal / dm
DAB
3
=7.6xlO-5 m 2 /min
I Variation of Da number
Damkohler numberl Da
1-------
1----------"-.--.-- 0.69
1.38
2.76
5.52
11.04
22.08
44.16
II Variation of Pe number
Peelet numberlPe
1-L.....-'-.
Open-vessel
0.378
0.641
0.911
1
I
I
I
8*DO
4*DO
2*DO
DO
DO/2
DO/4
DO/8
Conversion
_.
1.04e3
1.04e4
1.04e5
1.04e6
1.04e7
Parameter
Conversion
Closed-vessel
0.432
0.713
0.959
1
I
I
I
Closed-vessel
1
1
1
I
I
Parameter
Open-vessel
1
1
1
0.9995
0.9994
14-52
-
100* DAB
10* DAB
DAB
0.1 *DAB
O.Ol*D AB
--
--
._-
(d) The radial conversion profiles for various order of reaction
(1) First order
n =1
L=6.36m
R=O.05m
k = O.25dm 3 / mol. min
U o = I.24m/min
= O.5mol / dm 3
2
5
DAB = 7.6xlO- m I min
CAO
L Open-vessel:
a--
- " ! - .
[
,!
~-L---L
o
0005
001
0015
002
0025
003
Note: 1: Z= LllO; 2: z=3L11O; 3: z=Ll2; 4: z=7L110; 5: z=9L11O
II. Closed-vessel:
14-53
0035
0.04
0.045
005
~
3
."+ 4
-lr"-·5
-1--
1
2
-+-3
,:. 4
-'if-
08
-.-9---
--JC-
5
06
04
02
-0.2
o
0005
001
0015
002
0025
003
0035
004
0.045
005
Note: 1: Z= LllO; 2: z=3L11O; 3: z=Ll2; 4: z=7L11O; 5: z=9L11O
Something is not correct with z= LIlO because of the FEMLAB!
(2) Second Older
n=2
R=O.05m
L= 6.36m
k = l.4dm 1 / mol. min
Vo = 1.24m/min
= O.5mol / dm 3
5
2
DAB = 7.6xlO- m /min
C AO
1. Open-vessel:
14-.54
08
0.7
06
05
';l;
04
03
02
L...---L-.............L _ _- ' - - _ - ' - - '
---L.
o
00050010015002002500300350.040.045005
Note: 1: Z= UIO; 2: z=3UIO; 3: z=U2; 4: z=7UlO; 5: z=9UlO
II. Closed-vessel:
-->;f-
1
--8- 2
08
--+- 3
4
--*-, 5
07
06
05
04
Note: 1: Z= LIlO; 2: z=3UlO; 3: z=U2; 4: z=7UlO; 5: z=9UlO
(3) Third order
n=3
R=O.05m
14-55
L = 6.36m
k = 2.8dm 3 f mol. min
U 0 = 1. 24m f min
CAO = O.5moll dm
DAB
3
= 7.6xlO-5 m 2 fmin
L Open vessel:
-'if--
1
-_.13-... 2
-1-3
-4
-"*- 5
06
05
~04
03
Note: 1: z= LllO; 2: z=3L11O; 3: z=Ll2; 4: z=7L11O; 5: z=9L110
II. Closed- vessel
14-56
--"r-
0.7
1
-a--2
-1-3
-c, 4
-It- 5
065
0.6
0.55
0.5
~
045
0.4
035
03
0.25
o
I
I
0005
001
0.015
002
0025
003
0035
004
0045
005
Note: 1: Z= UlO; 2: z=3UlO; 3: z=U2; 4: z=7UI0; 5: z=9UlO
P14-20 (a)
Higher Peclet number
Curve 1 ("closer" to the one of an ideal PFR) has a higher Peclet number, because the cumulative distribution is more
concentrated around the mean residence time.
P14-20 (b)
Higher dispersion coefficient
Curve 2 has a higher dispersion coefficient: a higher Peclet number corresponds to a lower dispersion coefficient
P14-20 (b)
Large number ofT-loS
Curve 1: Increasing the number ofT-I-S corresponds to increase the Pec1et Number (i.e . Bo=2(n-l))"
P14-21 (a)
14-57
node 1
F(t) and E(t) curvesfor
a = -\1]
= O.4and
fJ = --Vb
= 0.2
\1
Vo
The analytical expression for E (t) is given by:
fJc5(t) t < 7]
E(t)
=
-(t-r1)t-'2
{ (l-fJ)e
t'27]
72
14-58
1
E(t)
0.5
O~~~-----L~~~====~==~
o
T./2
2
4
6
8
10
Integrating we can obtain the analytical expression for F (t):
T
P t<F(t) =
2
1- (1- p)e +-~)f:)
T
t~-
2
1-(1- (3)/e
{3=O.2
o 'T.I_
iJ
2 514'7
4
6
14-59
8
10
P14-21 (b)
Conversion
2nd order; kCAO =0.5min- 1, r=2min
Balance around node 1
(1- fJ)voCCSTR + fJvOC b = voC A
For the PFR:
Second-order
_ CAO - C pFR -_ DapFR -- 0333
.
C AO
1+ Da pFR
X pFR Where
DapFR = kC Ao (
a
1- fJ
) T = 0.5
kC PFR = 0.333 min-1
For the CSTR:
X CSTR =
C PFR -- CCSlR
.-'-'---=::':'-
(1 + 2Da cSTR ) - ~1 + 4Da cSTR-
- - - = 0.268
=
C PFR
I-a
Where DaCSTR = k - - rC PFR
1- fJ
2DacSTR
= 0.5
kCCSTR =0.366 min-1
kC A = (1- fJ)kC cslR + jJkCb = (1- fJ)kC CSTR + jJkC Ao = 0..393 min -1
x = kCAO-kC A =0.214
kC AO
In absence of bypass (P=O) the conversion would be X=0245
P14-22
Two parameters model
A possible two-parameter model is the PFR with Bypass (Vb) and Dead Volume (VD)
node 1
~-----_._14-60
To evaluate the conversion we write a balance around node Ion species A:
(I-P)voC PFR +PVOCb =VOCA
PFR, 2nd order, liquid phase
r A =-kCAC B
Where k=L5 m3/(krnoJ.rnin) and C Ao
X PFR --
= C Bo = 2mol / dm 3
Da pFR -0857
.
1+ Da pFR
a
(
) l' = 6
I-P
CAO -C PFR CAO
Where Da pFR = kC Ao
C PFR = C Ao
-
X PFRC Ao = 0.286mol / dm
3
CA = (1- p)CPFR + PCb = (1- p)C PFR + Pc Ao = 1. 143moll dm 3
X = _CAO - C A = 0.429
CAO
P14-23
0.1
E(t)
o
o
10
20
t (nun)
00
By definition: JE(t)dt
=1
o
OOJE(t )dt = 0.1 * tl = 1 ~ tl = 20 mm
.
o
2
2
For a triangular E(t): t1 = - E(O)
The analytical expression for E(t) is given by:
14-61
For 0 5, t 5, t} E(t) = -0.1
- t + 01
.
t}
For t > t} E(t)
=0
P14-23 (a)
Mean residence time
tm = =fE ()
t tdt=
o
IR --t+0.1
0.1
Jtdt= [-0.lt
0
(For a triangular E(t): t m
t1
3
3t1
2
2
2
0.1 2 +--=-6-=
0.lt1
0.lt1 667
'
+0.lt
- -]11 =--t
. mIn
1
2 0
3
2
= -t} )
3
Variance
=
0-
2
= JE(t)(t-tJ2 dt =
o
( 0.1 0 1J 2d
Jr ---t+ . t t-t
I.
o
2 _
m -
[-.0.lt
4
0.lt
3
=_~Jt3 + 0.ltI _tm2 0.1· 20
4 13
]tI=20ffiill
+-3
4tI
tl
3
2
-tm
0
3
_
6.67 2
=
22.2 min
12
Assuming closed-closed system: '[' = t m
P14-23 (b)
Peclet number
(j2
I'I m2- - - ' 2 - (1 ·-·e
-=1=
'['2
Pe-tO
Pe r
Pe r.2
-pe)
r
And Per = 0
P14-23 (C)
Tanks in series
The triangular E (t) can be interpreted as an approximation for t«1' of a CSTR where the continuity impose E(O)*T=2/3
instead of 1.
P14-24
The F(t) con be representative of a CSTR in parallel with two PFRs:
14-62
no del
First order reaction
r::;;;;-.--Id-ea-/-c-s-r.-'R_-_-_
--.--_l-de-a-/-/am"-in-a-r--,c_
se-g-re-g-a-tio-n--
t====
__
flow reactor
Maximum
Mixedness
14-63
Disper,sion
Tanks in series
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