Solutions Manual for
Elements of Chemical
Reaction Engineering
Sixth Edition
H. Scott Fogler
Ame and Catherine Vennema, The Arthur F. Thurnau, Ann Arbor,
Michigan
Table of Contents
Screen Shot of Web Home Page ....................................................................................... i
Screen Shot of Interactive Computer Games (ICGs) ........................................................ ii
Screen Shot of Polymath Living Example Problems (LEPs) ........................................... iv
Algorithm to Decode ICGs .............................................................................................. v
Sample Course Syllabus ................................................................................................. ix
Solutions to Chapter 1 – Mole Balances
Questions Q1-1 through Q1-12
Problems P1-1 through P1-8
Solutions to Chapter 2 – Conversion and Reactor Sizing
Questions Q2-1 through Q2-6
Problems P2-1 through P2-11
Solutions to Chapter 3 – Rate Laws
Questions Q3-1 through Q3-7
Problems P3-1 through P3-16
Solutions to Chapter 4 – Stoichiometry
Questions Q4-1 through Q4-9
Problems P4-1 through P4-13
Solutions to Chapter 5 – Isothermal Reactor Design: Conversion
Questions Q5-1 through Q5-13
Problems P5-1 through P5-26
Solutions to Chapter 6 – Isothermal Reactor Design: Moles and Molar Flow Rates
Questions Q6-1 through Q6-7
Problems P6-1 through P6-13
Solutions to Chapter 7 – Collection and Analysis of Rate Data
Questions Q7-1 through Q7-7
Problems P7-1 through P7-12
Solutions to Chapter 8 – Multiple Reactions
Questions Q8-1 through Q8-5
Problems P8-1 through P8-18
Solutions to Chapter 9 – Reaction Mechanisms, Pathways, Bioreactions and Bioreactors
Questions Q9-1 through Q9-4
Problems P9-1 through P9-23
Solutions to Chapter 10 – Catalysis and Catalytic Reactors
Questions Q10-1 through Q10-9
Problems P10-1 through P10-23
Solutions to Chapter 11 – Nonisothermal Reactor Design: The Steady-State Energy Balance
and Adiabatic PFR Applications
Questions Q11-1 through Q11-14
Problems P11-1 through P11-10
Solutions to Chapter 12 – Steady-State Nonisothermal Reactor Design: Flow Reactors with
Heat Exchange
Questions Q12-1 through Q12-6
Problems P12-1 through P12-27
Solutions to Chapter 13 – Unsteady State Nonisothermal Reactor Design
Questions Q13-1 through Q13-5
Problems P13-1 through P13-11
Solutions to Chapter 14 – Mass Transfer Limitations in Reacting Systems
Questions Q14-1 through Q14-8
Problems P14-1 through P14-17
Solutions to Chapter 15 – Diffusion and Reaction
Questions Q15-1 through Q15-8
Problems P15-1 through P15-17
Solutions to Chapter 16 – Residence Time Distributions of Chemical Reactors
Questions Q16-1 through Q16-5
Problems P16-1 through P16-13
Solutions to Chapter 17 – Predicting Conversion Directly from the Residence Time
Distribution
Questions Q17-1 through Q17-4
Problems P17-1 through P17-18
Solutions to Chapter 18 – Models for Nonideal Reactors
Questions Q18-1 through Q18-6
Problems P18-1 through P18-20
Synopsis for Chapter 1 – Mole Balances
Mole balances are the first building block of the chemical reaction engineering algorithm.
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 5.
Key to Nomenclature
l = Always assigned
AA = Always assign one from the group of
alternates
O = Often assigned
I = Infrequently assigned
S = Seldom assigned
G = Graduate level
N = Never assigned
E.g., means problem l P1-3B will be assigned every time I teach the course, problem AA P1-8 means that
this problem or one of the other problems with the prefix AA is always assigned for this chapter, Problem
l P1-2 will be infrequently assigned, Problem O P1-6B will often be assigned, and Problem S P3-16B is
seldom assigned.
Alternates: In problems that have a dot in conjunction with AA means that one of the problems, either
the problem with a dot or any one of the alternates are always assigned.
Time: Approximate time in minutes it would take a B student to solve the problem.
l Q1-1A (9 seconds) Questions Before Reading (QBR).
(a) John Falconer at the University of Colorado gives workshops on Teaching in which he points out
that students have a better comprehension if they ask themselves a question before reading
the text. The first question of each chapter, Q1, is just such a question.
(b) The students are asked, at a minimum read through the Questions to help put the chapter and
their studies in perspective.
(c) I encourage using the i>Clicker questions.
l Q1-2A (8-10 min) i>Clicker
l Q1-5A (5-75 min) through Q1-12A. To get a “feel” of the resources available, the students should
spend a total of about 50-75 minutes on these questions.
Computer Simulations and Experiments (5-15 minutes per simulation)
These problems are interactive and are a minor paradigm shift in the way we use homework problems.
Here the students are asked to explore the reaction and the reactor in which they occur to get an
intuitive feel and understanding of the reactor system. This procedure is called Inquiry Based Learning
(IBL).
l P1-1A (10-15 min) Good introduction to the use of Wolfram and Python.
Problems
I P1-2B (60 min) Problem reinforces wide range of applications of CRE and problem is given in the
web module which can be accessed from the Web Home Page (www.umich.edu/~elements). Many
students like this straight forward problem because they see how CRE principles can be applied to
S1-1
an everyday example. It is often assigned as an in-class problem where parts (a) through (f) are
printed out from the web and given to the students in class. Part (g) is usually omitted.
l P1-3B (45 min) I always assign this problem so that the students will learn how to use
Polymath/MATLAB, Wolfram and Python before needing it for chemical reaction engineering
problems. Most problems will use either Polymath or MATLAB to solve the end of chapter problems.
l P1-4A (30 min) The Interactive Computer Games (ICGs) have been found to be a great motivation
for this material. This ICG will help student AIChE chapters prepare for the Jeopardy Competition at
the Annual AIChE Meeting.
l P1-5A (10 min) Old Exam Question (OEQ) to reinforce the convention and stoichiometry in mole
balances.
O P1-6B (30 min) A hint of things to come on sizing reactors. Fairly straight forward problem to make
a calculation. 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.
I P1-7A (30 min) Helps develop critical thinking and analysis.
AA P1-8A (20 min) Puzzle problem to identify errors in the solution. Many students especially those
who enjoy working Sudoku or crossword puzzles enjoy working these types of problems.
S1-2
Solutions for Chapter 1 – Mole Balances
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-1
http://umich.edu/~elements/5e/01chap/obj.html#/
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q1-1 Individualized solution.
Q1-2 Individualized solution.
Q1-3
For CSTR:
V=
10 * 0.9
v0 C A 0 X A
v0 X A
FA0 X A FA0 X A
=
=
=
=
kC A
kC A0 (1 - X A ) k (1 - X A ) 0.23 * (0.1)
- rA,exit
=391.3 dm3
Q1-4 Individualized solution.
Q1-5 Individualized solution
Q1-6 Individualized solution
Q1-7 (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
Q1-7 (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
Q1-7 (c)
The assumptions made in deriving the design equation of PFR are:
- Steady state
- No radial variation in properties of the system
1-1
Q1-7 (d)
The assumptions made in deriving the design equation of PBR are:
- Steady state
- No radial variation in properties of the system
Q1-7 (e)
For a reaction
•
•
•
•
Aà B
-rA is the number of moles of A reacting (disappearing) per unit time per unit volume [=]
moles/ (dm3.s).
-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.
Q1-8 Individualized solution.
Q1-9 Individualized solution.
Q1-10 Individualized solution.
Q1-11 Individualized solution.
Q1-12 Individualized solution.
P1-1 (a) Example 1-3
1-2
The above graph represents intial Ca and Cb profiles for k=0.23 and v0 = 10.
(i)
With an increase in k (lets take k =0.35) for same volume and v0 , Ca decreases and Cb increases
Now lets make k=0.23(initial value) and make an increase in v0 (change from 10 to 15) for same volume.
We notice that now Ca increases and Cb decreases.
All of these graphs of concentration profiles are taken from Wolfram player by shifting the sliders.
1-3
We can observe that varying rate constant has more effect on concentration profiles as compared to
varying volumetric flow rate.
(ii) CA decreases and CB increases with an increase in k and Ke, and a decrease in v0 for the same
volume.
(iii) Individualized solution
(iv) See the following polymath code:
Polymath Code:
d(Ca)/d(V) = ra / v0
d(Cb)/d(V) = rb / v0
k = 0.23
Ke=3
ra = -k * (Ca-Cb/Ke)
rb = -ra
v0 = 10
V(0)=0
V(f)=100
Ca(0)=10
Cb(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
10.
2.849321
10.
2.849321
2 Cb
0
0
7.150679
7.150679
3 k
0.23
0.23
0.23
0.23
4 Ke
3.
3.
3.
3.
5 ra
-2.3
-2.3
-0.1071251
-0.1071251
6 rb
2.3
0.1071251
2.3
0.1071251
7 V
0
0
100.
100.
8 v0
10.
10.
10.
10.
Differential equations
1 d(Ca)/d(V) = ra / v0
2 d(Cb)/d(V) = rb / v0
1-4
Explicit equations
1 k = 0.23
2 Ke = 3
3 ra = -k * (Ca-Cb/Ke)
4 rb = -ra
5 v0 = 10
P1-2
Given
A = 2*1010 ft 2
TSTP = 491.69R
H = 2000 ft
V = 4*1013 ft 3
T = 534.7 ° R
PO = 1atm
R = 0.7302
atm ft 3
lbmol R
yA = 0.02
FS = CO in Santa Ana winds
C S = 2.04*10−10
FA = CO emission from autos
lbmol
ft
v A = 3000
P1-2 (a)
Total number of lb moles gas in the system:
PV
N= 0
RT
N=
1atm×(4 ×1013 ft 3 )
= 1.025 x 1011 lb mol
"
3%
$ 0.73 atm. ft ' ×534.69R
$
lbmol.R '&
#
P1-2 (b)
Molar flowrate of CO into L.A. Basin by cars.
FA = y AFT = y A ⋅v A CT
3
•%no.%of%cars
STP
3000 ft 1lbmol
×
×400000 cars
hr car 359 ft 3
FA = 6.685 x 104 lb mol/hr
FT =
(See appendix B)
P1-2 (c)
Wind speed through corridor is U = 15mph
W = 20 miles
The volumetric flowrate in the corridor is
vO = U.W.H = (15x5280)(20x5280)(2000) ft3/hr = 1.673 x 1013 ft3/hr
P1-2 (d)
Molar flowrate of CO into basin from Santa Ana wind.
FS := v0 ⋅C S
= 1.673 x 1013 ft3/hr ×2.04 ×10−10 lbmol/ft3
= 3.412 x 103lbmol/hr
1-5
3
C = 4*105 cars
ft 3
per car at STP
hr
P1-2 (e)
Rate of emission of CO by cars + Rate of CO in Wind - Rate of removal of CO =
dC
FA + FS − voCco = V co
dt
dNCO
dt
(V=constant, Nco = CcoV )
P1-2 (f)
t = 0 , Cco = CcoO
t
Cco
0
CcoO A
dC
co
∫ dt = V ∫ F + F −v
C
S
o co
V "$ FA + FS −voCcoO %'
t =
ln
vo $# FA + FS −voCco '&
P1-2 (g)
Time for concentration to reach 8 ppm.
lbmol
2.04
lbmol
CCO0 = 2.04 ×10−8
×10−8
, CCO =
3
4
ft
ft 3
From (f),
%
V " F + F −v .C
t = ln$$ A S O CO0 ''
vo # FA + FS −vO .CCO &
"
%
3
$ 6.7×104 lbmol + 3.4 ×103 lbmol −1.673×1013 ft ×2.04 ×10−8 lbmol '
$
hr
hr
hr
4 ft 3
ft 3 '
=
ln$
'
3
3
$
13 ft
4 lbmol
3 lbmol
13 ft
−8 lbmol '
1.673×10
6.7×10
+ 3.4 ×10
−1.673×10
× 0.51×10
'
hr $#
hr
hr
hr
ft 3 &
t = 6.92 hr
P1-2 (h)
(1)
to = 0
tf = 72 hrs
C co = 2.00E-10 lbmol/ft3
a = 3.50E+04 lbmol/hr
vo = 1.67E+12 ft3 /hr
b = 3.00E+04 lbmol/hr
Fs = 341.23 lbmol/hr
V = 4.0E+13 ft3
! t$
dC
a + bsin# π & + Fs − voCco = V co
dt
" 6%
Now solving this equation using POLYMATH we get plot between Cco vs. t
See the following polymath code:
1-6
Polymath Code:
v0 = 1.67*10^12
A= 35000
B = 30000
F = 341.23
V = 4*10^13
d(C)/d(t) = (A+B*sin(3.14*t/6)+F-v0*C)/V
C(0)=2.0e-10
t(0)=0
t(f)=72
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 A
3.5E+04
3.5E+04
3.5E+04
3.5E+04
2 B
3.0E+04
3.0E+04
3.0E+04
3.0E+04
3 C
2.0E-10
2.0E-10
2.134E-08
1.877E-08
4 F
341.23
341.23
341.23
341.23
5 t
0
0
72.
72.
6 V
4.0E+13
4.0E+13
4.0E+13
4.0E+13
7 v0
1.67E+12
1.67E+12
1.67E+12
1.67E+12
Differential equations
1 d(C)/d(t) = (A+B*sin(3.14*t/6)+F-v0*C)/V
Explicit equations
1 v0 = 1.67*10^12
2 A = 35000
3 B = 30000
4 F = 341.23
5 V = 4*10^13
! t$
dC
a + bsin# π & − voCco = V co
dt
" 6%
Now solving this equation using POLYMATH we get plot between Cco vs t
(2) tf = 48 hrs
Fs = 0
1-7
Polymath Code:
v0 = 1.67*10^12
A= 35000
B = 30000
F = 341.23
V = 4*10^13
d(C)/d(t) = (A+B*sin(3.14*t/6)+F-v0*C)/V
C(0)=2.0e-10
t(0)=0
t(f)=48
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 A
3.5E+04
3.5E+04
3.5E+04
3.5E+04
2 B
3.0E+04
3.0E+04
3.0E+04
3.0E+04
3 C
2.0E-10
2.0E-10
1.921E-08
1.71E-08
4 F
341.23
341.23
341.23
341.23
5 t
0
0
48.
48.
6 V
4.0E+13
4.0E+13
4.0E+13
4.0E+13
7 v0
1.67E+12
1.67E+12
1.67E+12
1.67E+12
Differential equations
1 d(C)/d(t) = (A+B*sin(3.14*t/6)+F-v0*C)/V
Explicit equations
1 v0 = 1.67*10^12
2 A = 35000
3 B = 30000
4 F = 341.23
5 V = 4*10^13
P1-2 (i)
Changing a è Increasing ‘a’ reduces the amplitude of ripples in graph. It reduces the effect of the
sine function by adding to the baseline.
1-8
Changing b è The amplitude of ripples is directly proportional to ‘b’. As b decreases amplitude
decreases and graph becomes smooth.
Changing v0 è As the value of v0 is increased the graph changes to a “shifted sin-curve”. And as v0 is
decreased graph changes to a smooth increasing curve.
P1-3 (a)
Initial number of rabbits, x(0) = 500
Initial number of foxes, y(0) = 200
Number of days = 500
dx
= k x − k2 xy …………………………….(1)
dt 1
dy
= k3 xy − k4 y …………………………….(2)
dt
Given,
k1 = 0.02day −1
k2 = 0.00004 / (day × foxes)
k3 = 0.0004 / (day × rabbits)
k4 = 0.04day −1
See the following polymath code:
Polymath Code:
d(x)/d(t) = (k1*x)-(k2*x*y)
d(y)/d(t) = (k3*x*y)-(k4*y)
k1 = 0.02
k2 = 0.00004
k3 = 0.0004
k4 = 0.04
t(0)=0
t(f)=500
x(0)=500
y(0)=200
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 k1
0.02
0.02
0.02
0.02
2 k2
4.0E-05
4.0E-05
4.0E-05
4.0E-05
3 k3
0.0004
0.0004
0.0004
0.0004
4 k4
0.04
0.04
0.04
0.04
5 t
0
0
500.
500.
6 x
500.
2.962693
519.4002
4.219969
7 y
200.
1.128572
4099.517
117.6293
Differential equations
1 d(x)/d(t) = (k1*x)-(k2*x*y)
2 d(y)/d(t) = (k3*x*y)-(k4*y)
1-9
Explicit equations
1 k1 = 0.02
2 k2 = 0.00004
3 k3 = 0.0004
4 k4 = 0.04
When, tfinal = 800 and k3 = 0.00004 /(day ´ rabbits )
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 k1
0.02
0.02
0.02
0.02
2 k2
4.0E-05
4.0E-05
4.0E-05
4.0E-05
3 k3
4.0E-05
4.0E-05
4.0E-05
4.0E-05
4 k4
0.04
0.04
0.04
0.04
5 t
0
0
800.
800.
6 x
500.
377.9769
2086.088
1467.831
7 y
200.
114.6959
1341.876
143.6569
Differential equations
1 d(x)/d(t) = (k1*x)-(k2*x*y)
2 d(y)/d(t) = (k3*x*y)-(k4*y)
1-10
Explicit equations
1 k1 = 0.02
2 k2 = 0.00004
3 k3 = 0.00004
4 k4 = 0.04
Plotting rabbits vs. foxes
P1-3 (b)
By increasing k4 and decreasing k2, foxes verses rabbits plot tends to become circular
P1-3 (c)
Below are the graphs when death rate is taken in to account
1-11
P1-3 (d)
To solve the system of equation, we can use Polymath Nonlinear Equation Solver
Polymath Code:
f(x) = (x^3)*y-4*y^2+3*x-1
x(0) = 2
f(y) = 6*y^2-9*x*y-5
y(0) = 2
Polymath Output:
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 x
2.385039 2.53E-11 2.
2 y
3.797028 1.72E-12 2.
Nonlinear equations
1 f(x) = (x^3)*y-4*y^2+3*x-1 = 0
2 f(y) = 6*y^2-9*x*y-5 = 0
P1-4 Individualized solution
P1-5
P1-6 (a)
– rA = k with k = 0.05 mol/h dm3
CSTR: The general equation is
F −F
V = A0 A
−rA
Here CA = 0.01CA0 , v0 = 10 dm3/min, FA = 5.0 mol/hr
Also, we know that FA = CAv0 and FA0 = CA0v0, CA0 = FA0/ v0 = 0.5 mol/dm3
Substituting the values in the above equation we get,
C v −C v
(0.5)10 − 0.01(0.5)10
V = A0 0 A 0 =
k
0.05
à V = 99 dm3
1-12
PFR: The general equation is
dFA
= rA = k , Now FA = CAv0 and FA0 = CA0v0 =>
dV
Integrating the above equation, we get
v0
CA
∫
k
C A0
V
dC A =
dC Av0
dV
= −k
v
∫ dV => V = k0 (C A0 −C A )
0
3
Hence V = 99 dm
Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of
concentration.
P1-6 (b)
–rA = kCA with k = 0.0001 s-1
CSTR:
We have already derived that
C v −C v
v C (1− 0.01)
V = A0 0 A 0 = 0 A0
−rA
kC A
k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1
àV=
(10dm3 / hr)(0.5mol / dm3 )(0.99)
3
−1
(0.36hr )(0.01*0.5mol / dm )
=> V = 2750 dm3
PFR:
From above we already know that for a PFR
dC Av0
dV
Integrating
v0
k
CA
= rA = −kC A
= − dV
A
0
∫ C
C A0
V
dC A
∫
v0
C
ln A0 =V
k CA
Again k = 0.0001s-1 = 0.0001 x 3600 hr-1= 0.36 hr-1
Substituting the values in above equation we get V = 127.9 dm3
P1-6 (c)
–rA = kCA2 with k = 300 dm3/mol.hr
CSTR:
C v −C v
v C (1− 0.01)
V = A0 0 A 0 = 0 A0
−rA
kC 2
A
1-13
Substituting all the values we get
V=
(10dm3 / hr)(0.5mol / dm3 )(0.99)
3
32
(300dm / mol.hr)(0.01*0.5mol / dm )
=> V = 660 dm3
PFR:
dC Av0
dV
Integrating
CA
v0
∫
k
C A0
=> V =
= rA = −kC 2A
V
v 1
1
) =V
= − dV => 0 ( −
2
k C A C A0
C
dC A
A
∫
0
10dm3 / hr
1
1
−
) = 6.6 dm3
3
0.01C
C
300dm / mol.hr
A0
A0
(
P1-6 (d)
CA = 0.001CA0
t=
NA0
dN
A
A
∫ N −r V
Constant Volume V=V0
t=
C A0 dC A
∫C
A
−rA
Zero order:
.999C Ao
1
t = "#C A0 − 0.001C A0 $% =
= 9.99h
k
0.05
First order:
! 1 $
1 !C $
1
t = ln## A0 && =
ln#
& = 69078s = 19.19h
k " C A % 0.0001 " .001 %
Second order:
1" 1
1 % 1 "
1
1 %
'=
t= $ −
−
$
' = 6.66h
k $# C A C A0 '& 300 # 0.5⋅0.001 0.5 &
P1-7 Enrico Fermi Problem
P1-7 (a) Population of Chicago = 4,000,000
Size of Households = 4
Number of Households = 1,000,000
Fraction of Households that own a piano = 1/5
Number of Pianos = 200,000
Number of Tunes/year per Piano = 1
Number of Tunes Needed Per Year = 200,000
Tunes per day = 2
1-14
Tunes per year per tuner =
250!days 2
×
= 500/yr/tuner
yr
day
200,000!tunes
1
×
= 400 Tuners
yr
500!tunes / yr / tuner
P1-7(b) Assume that each student eats 2 slices of pizza per week.
Also, assume that it is a 14” pizza, with 8 pieces.
Hence, the area of 1 slice of pizza = 19.242 inch2 = 0.012414 m2
Thus, a population of 20000, over a span of 4 months, eats
20000 * 2 slices * 4 months * 4 weeks/month = 640000 slices of pizza, with a total area of
640000 * 0.012414 m2 = 7945 m2 of pizza in the fall semester.
P1-7(c) Assume you drink 1L/day
Assume you live 75 years*365days/year = 27375 days
1L/day*27375 days = 27375 L drank in life
Bathtub dimensions: 1m*0.7m*0.5m = 0.35m3 = 350L/tub
Bathtubs drunk = 27375L*1tub/350L = 78 tubs
P1-7(d) Jean Valjean, Les Misérables.
P1-8
Mole Balance:
F − "F
V"=" A0 A
−rA
Rate Law :
−rA = kC 2A
Combine:
F − "F
V"=" A0 A
kC 2A
FA0 = v0C A = 3
FA = v0C A = 3
dm3 2molA 6molA
.
=
s
s
dm3
dm3 0.1molA 0.3molA
.
=
s
s
dm3
mol
3
s
V"="
= 19000
1900dm
dm3
3
dm
mol 2
(0.03
)(0.1
)
mol.s
dm3
The incorrect part is in step 6, where the initial concentration has been used instead of the exit
concentration.
(6 − 0.3)
1-15
Page intentionally blank
1-16
Synopsis for Chapter 2 – Conversion and Reactor Sizing
General: The overall goal of these problems is to help the student realize that if they have –rA = f(X) they
can “design” or size a large number of reaction systems. It sets the stage for the algorithm developed in
Chapter 5.
See Chapter 1 Synopsis for nomenclature guide to problem assignments.
Questions
l Q2-1A (4 seconds) Questions Before Reading (QBR).
O Q2-2A (20 min) Secondly, I also encourage going through the i>Clicker questions AFTER the students
have completed all the reading and homework associated with this chapter.
l Q2-4A (20-25 min) Firstly, I encourage students to take the Solomon/Felder Inventory of Learning
Styles test (https://www.engr.ncsu.edu/stem-resources/legacy-site/learning-styles/) and then use
Appendix I.2 to see how they can best use the text and interactive web materials.
O Q2-5A (15 min) If a student did not visit the University of Colorado’s LearnChemE site in Chapter 1,
I recommend they view one or two screencasts now.
I Q2-6A (7 min) NFPA. Now is also the time to visit the tutorials of the Safety Website
(http://umich.edu/~safeche/) to become acquainted with the wealth of safety resources available
on the safety website.
Interactive Computer Games (ICG)
l P2-1A (20-25 min) Because this interactive game has so many choices of reactions to maximize the
conversion, the time to play the game is a little longer than other ICGs.
Problems
O P2-2A (45 min) Helps the student explore the example problems in this chapter. Parts (d) and (e)
take a little longer than the other parts.
O P2-3B (35 min) Reinforces use of the Levenspiel plots.
AA P2-4B (40 min) Requires the student to construct a Levenspiel plot. Alternative to problems P2-5B,
P2-7B, and P2-10C.
AA P2-5B (30 min) This problem is a reasonably challenging trial and error problem that reinforces
Levenspiel’s plots and reactor staging.
O P2-6B (45 min) Novel application of Levenspiel plots from an article in CEE by Professor Alice Gast
formerly at Massachusetts Institute of Technology, now President of Imperial College, London.
AA P2-7B (30 min) Straight forward problem alternative to problems P2-4B and P2-10C. The answer gives
ridiculously large reactor volume. The point is to encourage the student to question their numerical
answers. Alternative to P2-4B, P2-5B and P2-10C.
I P2-8A (30 min) Helps the students get a feel of real reactor sizes.
l P2-9D (2 min) Great motivating problem. Students from all universities around the world remember
this problem long after the course is over.
S2-1
AA P2-10C (45 min) Alternative problem to P2-4B, P2-5B, and P2-7B.
l P2-11B (45 min) This problem is a departure from the other problems in this chapter because it is a
batch reactor.
S2-2
Solutions for Chapter 2 – Conversion and Reactor Sizing
Q2-1 Individualized solution.
Q2-2 Individualized solution.
Q2-3 Individualized solution.
Q2-4 Individualized solution.
Q2-5 Individualized solution.
Q2-6 Individualized solution.
P2-1 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer
Games (ICGs) is given at the front of this Solutions Manual.
P2-2 (a) For a batch reactor,
𝑑𝑋
𝑡 = 𝑁%& '
& −𝑟% 𝑉
400 dm3= 400/1000 m3 =0.4 m3
For 10% conversion,
5&
𝑑𝑋
𝑡 = (100/0.4) '
& −𝑟%
The area under the curve of 1/-rA vs X needs to be found out till X=0.1
Using trapezoidal rule,
Area= 0.5*(2.22+2.7)*0.1= 0.246
Thus, time= 250*0.246 = 61.5 s
To find BR times for X=0.5 and X=0.8, apply Simpson’s rule and find time in similar way as described
above
P2-2 (b) Example 2-1 through 2-3
For Example, 2-1
If flow rate FAO is cut in half.
v1 = v/2 , F1= FAO/2 and CAO will remain same.
Therefore, volume of CSTR in example 2-1,
2-1
If the flow rate is doubled,
F2 = 2FAO and CAO will remain same,
Volume of CSTR in example 2-1,
V2 = F2X/-rA = 12.8 m3
For Example, 2-2,
If flow rate is cut to half,
V= 0.5*2.165 m3 =1.083 m3
If flow rate is doubled,
V=2*2.165= 4.33 m3
For Example, 2-3,
If flow rate is cut to half,
VCSTR= 0.5*6.4 m3 =3.2 m3
VPFR= 0.5*2.165 m3 =1.08 m3
If flow rate is doubled,
VCSTR= 2*6.4 m3 =12.8 m3
VPFR=2*2.165= 4.33 m3
In a 4.5 m3 PFR,
We need to find the value of X at which area under the curve of (Fao/-rA) and X is 4.5
From Figure E2-2.1 it can be seen that 80% conversion is achieved in 2.165 m3. Since data is only available
till X=0.8, it is expected that with 4.5 m3, conversion achieved would be near to 100 %
In a 4.5 m3 CSTR,
𝐹%& ∗ 𝑋
4.5 =
−𝑟%
So, we need to find a value of (FA0/-rA) and X which when multiplied gives 4.5
We can see that it is achieved at approximately 74 % conversion at which (FA0/-rA) is 6.1
P2-2 (c)
𝐹%& ∗ 𝑋
𝑉=
−𝑟%
So, we need to find a value of (FA0/-rA) and X which when multiplied gives 2.5
We can see that it is achieved at approximately 63.5 % conversion at which (FA0/-rA) is 3.9
P2-2 (d)
Conversion achieved in a 2.4 m3 CSTR is 62.5 %. Now we add a 1 m3 PFR volume to it.
We need to find X at which area under the curve of (Fao/-rA) and X is 1. The starting point is X=0.625.
So, by trial and error, using trapezoidal rule, we find at 80 % conversion, below equation is satisfied
1= 0.5 (8+3.7)*(0.8-0.625)
So, Conversion achieved is 80 %.
2-2
P2-2 (e) Example 2-4
x
Now, FAO = 0.4/2 = 0.2 mol/s,
Table: Divide each term
in Table 2-3 by 2.
X
[FAO/-rA](m3)
0.1
0.545
0
0.445
Reactor 1
V1 = 0.82m3
V = (FAO/-rA)X
0.2
0.665
0.4
1.025
0.6
1.77
0.7
2.53
0.8
4
Reactor 2
V2 = 3.2 m3
!F $
0.82 = ## A0 && X1
" −rA %X
( )
1
"F %
3.2 = $$ A0 '' X2
# −rA &X
( )
2
By trial and error, we get:
X1 = 0.546
and
X2 = 0.8
Overall conversion XOverall = (1/2)X1 + (1/2)X2 = (0.546+0.8)/2 = 0.673
P2-2 (f) Example 2-5
(1) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the smaller
CSTR.
Conversion
X1 = 0.20
X2 = 0.60
X3 = 0.65
Original Reactor Volumes
V1 = 0.188 (CSTR)
V2 = 0.38 (PFR)
V3 = 0.10 (CSTR)
For PFR,
X1 = 0.2
Using trapezoidal rule,
XO = 0, X1 = 0.2
2-3
Worst Arrangement
V1 = 0.23 (PFR)
V2 = 0.53 (CSTR)
V3 = 0.10 (CSTR)
1 (𝑋5 − 𝑋& )(𝑓(𝑋& ) + 𝑓(𝑋5 ))
2
−𝑟%
= 0.2(1.28+0.94)/2
= 0.222 m3
𝑉5 =
For CSTR,
F
F
V2 = A0 X2 − X1 = 1.32 0.6 − 0.2 = 0.53-m3
−rA
3
At X2 = 0.6 A0 = 1.32 m ,
(
−rA
)
(
)
For 2nd CSTR,
F
F
V3 = A0 X3 − X2 = 2)m3 0.65− 0.6 = 0.1)m3
−rA
At X3 = 0.65, A0 = 2 m3 ,
(
−rA
)
(2) For first CSTR,
At X=0.2,
FA0
= 0.94 m3 ; V1 = 0.94&m3 (0.2) = 0.188&m3
- rA
(
)
From previous example; V1 (volume of first CSTR) = 0.188 m3
Also, the next reactor is PFR, Its volume is calculated as follows
For next CSTR,
X3 = 0.65,
, V3 =
(3) Now the sequence of the reactors remain unchanged.
But all reactors have same volume.
First CSTR remains unchanged
Vcstr = .1 = (FA0/-rA )*X1
ð X1 = .088
Now
For PFR:
By estimation using the Levenspiel plot
X2 = .183
2-4
(
)
For CSTR,
VCSTR2 =
=> X3 = .316
P2-2 (g)
Residence time, 𝜏 is given as 2 sec
=
Since, 𝜏 = >
?
3
𝑚C
𝑣& = = 1.5
= 75 𝑚 C /𝑚𝑖𝑛
2
𝑠
P2-3
X
FAO/-rA (m3)
0
0.89
0.1
1.08
0.2
1.33
0.4
2.05
0.6
3.54
0.7
5.06
0.8
8.0
V = 1 m3
P2-3 (a) Two CSTRs in series
For first CSTR,
V = (FAo/-rAX1) X
So, we have to find a value of (FAo/-rAX1) and X in such a way that on multiplication it gives 1
So at (FAo/-rAX1) = 2.3, X= 0.435
=> X1 = 0.435
For second CSTR,
V = (FAo/-rAX2) (X2 – X1)
Again, by trial and error, we have to find a condition which satisfy
1=(FAo/-rAX2)*( X2-0.435)
=> X2 = 0.66
P2-3 (b)
Two PFRs in series
2-5
For X1, we need to find area under the curve such that area =1. We can use trapezoidal rule to estimate
the conversion taking two points at a time to get
X1 = 0.565
X2 = 0.775
P2-3 (c)
Two CSTRs in parallel with the feed, FAO, divided equally between two reactors. FANEW/-rAX1 = 0.5FAO/-rAX1
V = (0.5FAO/-rAX1) X1
2=(FAo/-rAX1) X
Solving in similar manner as part a, we get, Xout = 0.585
P2-3 (d)
Two PFRs in parallel with the feed equally divided between the two reactors.
FANEW/-rAX1 = 0.5FAO/-rAX1
So, similar to part b, the area under the curve must be 2 in this case i.e. V=2
Xout = 0.775
P2-3 (e)
A PFR followed by a CSTR,
XPFR = 0.565 (using part(b))
V = (FAo/-rA-XCSTR) (XCSTR – XPFR)
Now, 1=(FAo/-rA-XCSTR) (XCSTR – 0.565)
By looking at graph we get, XCSTR = 0.732
P2-3 (f)
A CSTR followed by a PFR,
XCSTR = 0.435 (using part(a))
1= Area under the curve (with starting point as 0.435)
By calculating area, we get XPFR = 0.71
P2-3 (g)
A 1 m3 PFR followed by two 0.5 m3 CSTRs,
For PFR,
XPFR = 0.565 (using part(b))
CSTR1: V = (FAo/-rA-XCSTR) (XCSTR – XPFR) = 0.5 m3
XCSTR = 0.673
CSTR2: V = (FAo/-rA-XCSTR2) (XCSTR2 – XCSTR1) = 0.5 m3
XCSTR2 = 0.75
Two PFR in parallel or series was a better option as exit conversion was higher
P2-3 (h)
A CSTR and a PFR are in parallel with flow equally divided
2-6
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 = (0.585 + 0.775) / 2 = 0.68
P2-4
Exothermic reaction: A → B + C
X
0
0.20
0.40
0.45
0.50
0.60
0.80
0.90
r(mol/dm3.min)
1
1.67
5
5
5
5
1.25
0.91
1/-r(dm3.min/mol)
1
0.6
0.2
0.2
0.2
0.2
0.8
1.1
P2-4 (a) For a batch reactor,
To solve this problem, first plot 1/–rA vs. X from the chart above.
- I-
t=NA0∫&
NA0=400 moles
V=400 dm3
t=(400 /400)*(area under the curve)
For X=0.4, area under curve is
=0.4*0.2+0.5*0.4*0.8=0.24 dm3.min/mol, thus
t= 0.24 mins
JKL =
Similarly,
For X=0.8, area under curve is
=0.4*0.2+0.5*0.4*0.8+0.2*0.2+0.2*0.2+0.5*0.2*0.6= 0.38 dm3.min/mol, thus
t= 0.38 mins
P2-4 (b)
Use mole balance as given below.
CSTR:
Mole balance:
PFR:
=>
=>VCSTR = 24 dm3
Mole balance:
= 300(area under the curve)
VPFR = 72 dm3
P2-4 (c)
For a feed stream that enters the reaction with a previous conversion of 0.40 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.
2-7
P2-4 (d)
VCSTR = 105 dm3
Mole balance:
Use trial and error to find maximum conversion.
At X = 0.70,
1/–rA = 0.5, and X/–rA = 0.35 dm3.min/mol
Maximum conversion = 0.70
P2-4 (e)
From part (a) we know that X1 = 0.40.
Use trial and error to find X2.
Mole balance:
Rearranging, we get
At X2 = 0.64,
Conversion = 0.64
P2-4 (f)
From part (a), we know that X1 = 0.40. Use trial and error
to find X2.
Mole balance:
At X2 = 0.908, V = 300 x (area under the curve)
=> V = 300(0.24) = 72dm3
Conversion = 0.908.
2-8
P2-4 (g)
P2-5
We must first find a CSTR up to X = 0.2 to minimize the volume, and hence the cost of reactor.
We can either use a CSTR or a PFR for X = 0.2 to X = 0.6 since the rate is independent of X in that range.
A PFR for X > 0.6 is preferable since it will reduce the volume required and the cost.
Let’s calculate the volumes for the different cases.
i)
X = 0 to X = 0.2
Volume for CSTR = 20*0.2 = 4 dm3
Volume for PFR = 0.5*(50+20)*0.2 = 7 dm3
ii)
X = 0.2 to X = 0.6
Volume for CSTR = Volume for PFR = 0.4*20 = 8 dm3
iii) X > 0.6
Volume for PFR < Volume for CSTR; for the same conversion.
Hence a PFR would minimize the cost.
First, we use the 4 dm3 CSTR worth $2000. Conversion at this stage = 0.2
After this, we have $8000, and we see that the PFR minimizes the cost (So, we can have more
conversion with the same money)
Hence, we can use a 12 dm3 PFR followed by a 4 dm3 PFR, totally worth $8000, which will give the
highest conversion.
The first 8 dm3 of PFR would give conversion up to 0.6, and the rest (16 – 8) dm3 = 8 dm3 would give the
conversion of:
Therefore, X = 0.8
Final conversion = 0.8
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) = ½ Fao (mother). The Levenspiel Plot is shown:
2-9
Autocatalytic Reaction
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.
P2-6 (b)
2) If Vmax and mao are both one half of the mother’s then
and since
then
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 0.40
P2-6 (c)
Vstomach = 0.2 m3
From the web module we see that if a polynomial is fit to the autocatalytic reaction we get:
2-10
= 127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499
And since Vstomach =
X,
solve V= 127X5 - 172.36X4 + 100.18X3 - 28.354X2 + 4.499X = 0.2 m3
Xstomach = .067
For the intestine, the Levenspiel plot for the intestine is shown below. The outlet conversion is 0.178.
Since the hippo needs 30% conversion to survive but only achieves 17.8%, the hippo cannot survive.
P2-6 (d)
PFR→ CSTR
PFR:
Outlet conversion of PFR = 0.111
CSTR:
We must solve
V = 0.46 = (X-0.111)(127X4 - 172.36X3 + 100.18X2 - 28.354X + 4.499)
X=0.42
Since the hippo gets a conversion over 30% it will survive.
2-11
P2-7
2A + B → 2C
P2-7 (a)
PFR volume necessary to achieve 50% conversion
Mole Balance
Volume = Geometric area under the curve of
(FA0/-rA) vs X)
V = 150000 m3
P2-7 (b)
CSTR Volume to achieve 50% conversion
Mole Balance
V = 50000m3
P2-7 (c)
Volume of second CSTR added in series to achieve 80% conversion
V2 = 150000m3
P2-7 (d)
Volume of PFR added in series to first CSTR to achieve
80% conversion
VPFR = 90000m3
P2-7 (e)
For CSTR,
V = 60000 m3 (CSTR)
Mole Balance
X = 0.463
For PFR,
V = 60000 m3 (PFR)
2-12
Mole balance
X = 0.134
P2-7 (f)
Real rates would not give that shape. The reactor volumes are absurdly large.
P2-8
Problem 2-8 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:
CSTR
h = 56ft
d = 9 ft
V = πr2h = π(4.5 ft)2(56 ft) = 3562 ft3 = 100,865 L
PFR
Length of one segment = 23 ft
Length of entire reactor = (23 ft)(12)(11) = 3036 ft
D = 1 ft
V = πr2h = π(0.5 ft)2(3036 ft) = 2384 ft3 = 67,507 L
Answers will vary slightly for each individual.
P2-9 No solution necessary.
P2-10 (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 following figure.
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.
2-13
P2-10 (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.
The area of the rectangle is approximately 23.2 kg of catalyst.
P2-10 (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.
The area of the rectangle is approximately 7.6 kg of catalyst.
P2-10 (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.
The necessary catalyst weight is approximately 22 kg.
2-14
P2-10 (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.
The necessary catalyst weight is approximately 13 kg.
P2-10 (f)
P2-11
X
0
0.20
0.40
0.60
0.80
-r(mol/dm3.min)
4
3
2.2
1.5
1
1/-r(dm3.min/mol)
0.25
0.33
0.454
0.666
1
To solve this problem, first plot 1/–rA vs. X from the chart above
2-15
For calculating the area from the above graph, approximate the above curve as polynomial of degree 2
and its equation can be obtained from trendline option in excel. Then do the definite integral from limits
you wish.
We know that for a batch reactor, from mole balance we get the following equation:
The volume term can be taken out from integral and then Nao/V can be substituted as Cao which is
equal to 2.
P2-11 (a)
The time for 10% conversion can be found out by evaluating the integral from 0 to 0.1
t = 0.051min for conversion of 10%
t = 0.34min for conversion of 50%
t = 0.783min for conversion of 80%
P2-11 (b)
Plot the rates for the above time which you have got and get –r vs t plot
2-16
Synopsis for Chapter 3 – Rate Laws
General: The second building block is the rate law. The purpose of these problems is to help students
understand how to write a rate law and how activation energy affects the rate of reaction.
See Chapter 1 Synopsis for nomenclature guide to problem assignments.
Questions
l Q3-1A (20 seconds) Questions Before Reading (QBR).
l Q3-2A (15-18 min) The i>Clicker questions are an excellent review for the student chapter Jeopardy
competition held at the Annual AIChE Meeting in the Fall. It is recommended that the student read
through the questions and then spend 1 minute (2 minutes at most) thinking about the answer to
the question.
O Q3-3C – Q3-6A (10-15 min each)
I Q3-7B (5 min) Not fearing for my own safety from attack from the crickets, I collected this data from
my yard and put it on the audio in the Summary Notes. The students should compare it with Sheldon
Cooper’s analysis in the Big Bang Theory and on government websites.
Computer Simulations and Experiments
l P3-1B (a) (10 min maximum) Wolfram and Python explore change in the energy distribution of the
collisions with temperature.
O P3-1B (b) (8 min) This stochastic simulation was developed by Kaushik Nagaraj of IIT Bombay.
Students should change the initial concentrations and run at least one simulation.
Problems
O P3-2B (a) – (b) (10 min) Lead the student to explore the example problems.
I P3-2B (c) (75 min) Requires additional reading on collision theory on PFR on the web and is
considered advanced material or graduate material.
AA P3-3B (15 min) Uses Figure 3-4 and LEP 3-1B to study collision frequencies at different temperatures.
O P3-4B (15 min) Helps the student understand reaction coordinate plots.
AA P3-5B (30 min) Expands on P3-3B and P3-4B.
AA P3-6A (30 min) Further discussion of the activation energy.
I P3-7A (30 min) Picture of cricket and the table may encourage the student to work this problem
without it being assigned. When making the Arrhenius plot for each of the insects, one finds they
all have virtually the same activation energy.
AA P3-8B (15 min) Real industrial example of competing effects between concentration and
temperature.
l P3-9B (20 min) Coupled with YouTube video “Black Widow,” this is a great motivating problem for
the students (https://www.youtube.com/watch?v=CE7b9teZwqs).
AA P3-10B (35 min) This problem is not too straight forward to derive this rule of thumb analytically.
S1-1
AA P3-11C (15 min) Very short, if the student has read the text before starting the problem. Straight
forward calculation to find the activation energy.
AA P3-12B (6-10 min) Straight forward problem. 5-10 minutes if the student has read the text before
starting to write the rate law.
l P3-13A (10 min) Straight forward problem. 5-10 minutes if the student has read the text before
starting to write the rate law.
AA P3-14A (20 min) Somewhat tricky problem to find the rate law.
AA P3-15B (10 min) Find the rate law for reversible reactions.
S P3-16B (10 min) Open-ended problem to guess how to incorporate light into rate law.
S1-2
Solutions for Chapter 3 – Rate Laws
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Chapter 3:
http://www.umich.edu/~elements/5e/03chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram codes):
http://www.umich.edu/~elements/5e/tutorials/Wolfram_LEP_tutorial.pdf
Q3-1 Individualized solution.
Q3-2 Individualized solution.
Q3-3 (a) Individualized solution.
(b) Individualized solution.
(c) kB = 50 (dm3/mol)2/s and kC = 50 (dm3/mol)2/s
(d) Individualized solution.
(e) Individualized solution.
Q3-4 For ethylene oxide, the following GHS symbols are applicable:
1. Health hazard
2. Gas cylinder
3. Exclamation mark
4. Flame
5. Skull and crossbones
Q3-5 Individualized solution.
Q3-6 NFPA diamond colors and GHS symbols that can be categorized into that color:
1. Blue: Health hazard, exclamation mark, corrosion, skull and cross bones
2. Red: Flame, flame over circle
3. Yellow: Explosive bomb
4. White: Environment, exclamation mark, gas cylinder
Q3-7 (a) From P3-7,
52150
ln(𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐ℎ𝑖𝑟𝑝𝑠/𝑚𝑖𝑛𝑢𝑡𝑒) = 179 −
𝑇
In the recording, approximately 20 chirps were heard in 10 seconds i.e, 120 chirps/min
This gives T = 300 K
3-1
(b) The correlation given in the website is,
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐ℎ𝑖𝑟𝑝𝑠
+ 37 = 𝑐𝑇
15
Where T is in Fahrenheit and c is a proportionality constant
Arrhenius equation: 𝑘 = 𝐴 𝑒 −𝐸𝑇
𝑘 = 𝐴(1 − 𝐸𝑇) (approximation of the exponent term)
P3-1 (a)
(i)
For a low value of E, irrespective of value of T the fraction of molecules is almost equal to 1 (can
be seen by below graphs)
For activation energy as high as 10 Kcal/mol, the fraction of molecules with energy greater than
activation energy increases as temperature increases but even at high values of T=5000K , still the
fraction approximates to be 0.5-0.6
, Fraction of molecules with
energy greater than
Activation Energy is, 0.5667
3-2
(ii)
2550 K (can be seen by varying Wolfram parameters)
(iii)
Individualized solution
P3-1 (b)
(i)
The equilibrium concentration changes, but the equilibrium conversion of A remains the same in
all the three cases (50%). The time required to attain equilibrium decreases with increasing
particle number.
1. We can conclude that conversion of A does not depend on no of A or B molecules present
initially provided number of molecules of A is less than or equal to number of molecules of B.
2. Equilibrium concentration and equilibrium time will for sure vary depending on the no of A
and B molecules present initially
(ii) The trajectories remain similar, but the time taken to attain equilibrium changes, as the rate
constants are lower.
(iii) The forward/reverse reaction goes to completion.
(iv) This is the nature of a stochastic simulation. The number of molecules in the simulation are very
less, as compared to the large number in the deterministic model (number of molecules are of the
order of the Avogadro number). The fluctuations reduce as you increase the number of molecules,
and the stochastic model is identical to a deterministic model when there are infinite molecules.
(v) The fluctuations in the trajectories reduce.
(vi) No, equilibrium only means that the forward reaction rate is equal to the reverse reaction rate. It
does not mean that the reactions stop occurring. This can be seen from the fact that there is
always a small fluctuation about the equilibrium concentration.
(vii) 1. By this whole experiment one can conclude that equilibrium conversion of A is independent of
no of reacting molecules present initially provided number of molecules of A is less than or
equal to number of molecules of B.
2. We can conclude that by increasing the no of molecules, one moves to a more realistic model.
3. Both forward and backward rate constants affect the time to attain equilibrium.
3-3
P3-2 (a)
𝑙𝑛𝑘 = 𝑙𝑛𝐴 −
𝐸 1
( )
𝑅 𝑇
From the graph of ln k vs 1/T above, we get:
𝑙𝑛𝐴 = 28.817,
Thus,
𝐴 = 3.27 ∗ 1012 𝑠 −1 ,
𝐸
= −11299 𝐾
𝑅
𝐸 = 93.94 𝑘𝐽/𝑚𝑜𝑙
𝑘 = 3.27 ∗ 1012 𝑒𝑥𝑝 (−
P3-2 (b) Example 3-1
For, E = 60kJ/mol
11299 −1
)𝑠
𝑇
For, E = 240kJ/mol
T (K)
k (1/sec)
1/T
ln(k)
T (K)
k (1/sec)
1/T
ln(k)
310
1023100
0.003226
13.83918
310
4.78E-25
0.003226
-56.0003
315
1480488
0.003175
14.2087
315
2.1E-24
0.003175
-54.5222
320
2117757
0.003125
14.56667
320
8.77E-24
0.003125
-53.0903
325
2996152
0.003077
14.91363
325
3.51E-23
0.003077
-51.7025
330
4194548
0.00303
15.25008
330
1.35E-22
0.00303
-50.3567
335
5813595
0.002985
15.57648
335
4.98E-22
0.002985
-49.0511
3-4
(1) For lower activation energy, we observe that the increase in rate constant is gradual. The slope of
the plot: ln(k) vs 1/T is steeper for higher activation energy.
(2) From the plots we can see that higher the activation energy smaller the rate constant at a given
temperature. Therefore, higher activation energy will lead to slower reaction rates. The molecules
will have to collide with greater energies for the reaction to occur. Activation energy could be a
measure of the affinity molecules have to react with each other.
P3-2 (c) Solution will not be given
P3-3 (a)
Refer to Fig 3-4
The fraction of molecular collisions having energies less than or equal to 2.5 Kcal is given by the area
under the curve, f(E,T)dE from EA = 0 to 2.5 Kcal. From wolfram code, answer is
At 300 K, Area= 1-0.0388 =96.1 %
At 500 K, Area =1-0.17 =83 %
P3-3(b)
The fraction of molecular collisions having energies between 3 and 4 Kcal is given by the area under the
curve f(E,T) from EA = 3 to 4 Kcal.
At 300 K, Area = 0.018-0.00383 = 1.4 %
At 500 K, Area = 0.11-0.045 = 6.5 %
P3-3 (c)
The fraction of molecular collisions having energies greater than the activation energy E A= 25 Kcal is
given by the area under the curve f(E,T) from E A =25 to 50 Kcal.
At 300 K, Area = 0 %
At 500 K, Area = 6.97E-9 %
3-5
P3-4 (a)
Considering the molecules to be linear harmonic oscillator:
P3-4 (b,c) Use Figure 3-1(b) to know the position of RAB = 2.8 A° in Figure 3-1(a)
P3-5 (a) Use the following equation to find out the fraction of molecules with energy greater than
EA = 25 kcal,
1
𝐹(𝐸 > 𝐸𝐴 , 𝑇) =
2
𝜋
𝐸𝐴 2
𝐸
) 𝑒𝑥𝑝 𝑒𝑥𝑝 (− 𝑅𝑇𝐴 )
1(
𝑅𝑇
2
3-6
(1)
Substituting 𝐸𝐴 = 25 kcal/mol and T = 300 K, 500 K, 800 K and 1200 K
𝑘𝑐𝑎𝑙
𝐹 (25 𝑚𝑜𝑙 , 300) = 5.844× 10-18
𝑘𝑐𝑎𝑙
𝐹 (25 𝑚𝑜𝑙 , 500) = 7.835× 10-11
𝑘𝑐𝑎𝑙
𝐹 (25 𝑚𝑜𝑙 , 800) = 7.303× 10-7
𝑘𝑐𝑎𝑙
𝐹 (25 𝑚𝑜𝑙 , 1200) = 1.09× 10-4
P3-5 (b) We need to find the ratio of the quantity, 𝐹(25 𝑘𝑐𝑎𝑙/𝑚𝑜𝑙, 𝑇) − 𝐹(15 𝑘𝑐𝑎𝑙/𝑚𝑜𝑙, 𝑇) at 300 K
and 1200 K
The ratio of the faction of energies between 15 and 25 kcal at 300 K to that for the same energy range
(15–25 kcal) at 1200 K =
𝑘𝑐𝑎𝑙
𝑘𝑐𝑎𝑙
𝑘𝑐𝑎𝑙
𝑘𝑐𝑎𝑙
𝐹(25 𝑚𝑜𝑙 ,300𝐾)−𝐹(15 𝑚𝑜𝑙 ,300𝐾)
𝐹(25 𝑚𝑜𝑙 ,1200𝐾)−𝐹(15 𝑚𝑜𝑙 ,1200𝐾)
= 1.467× 10-8
P3-5 (c) The fraction of collisions with energy greater than E A is given by the equation,
𝐹(𝐸 > 𝐸𝐴 , 𝑇) =
This equation is plotted as a function of
2
1
2
𝐸
𝐸
𝐴
) 𝑒𝑥𝑝 𝑒𝑥𝑝 (− 𝑅𝑇𝐴 )
1(
𝑅𝑇
𝜋2
𝐸𝐴
𝑅𝑇
𝐸
for 𝐴 > 3 using MATLAB
𝑅𝑇
At 700 K, the fraction of collisions with energy greater than 25 kcal/mol is given by:
𝐸𝐴
25000
= 2×700 = 17.86
𝑅𝑇
Substituting in (1) gives,
𝑘𝑐𝑎𝑙
𝐹 (𝐸 > 25 𝑚𝑜𝑙 , 700𝐾) = 8.354 × 10-8
3-7
(1)
P3-5 (d) The fraction of collisions with energy greater than E A is given by the equation,
1
𝐹(𝐸 > 𝐸𝐴 , 𝑇) =
2
𝜋
𝐸𝐴 2
𝐸
) 𝑒𝑥𝑝 𝑒𝑥𝑝 (− 𝑅𝑇𝐴 )
1(
𝑅𝑇
2
(1)
At 500 K, the fraction of collisions with energy greater than 15 kcal/mol is given by:
𝐸𝐴
15000
= 2×500 = 15
𝑅𝑇
Substituting in (1) gives,
𝑘𝑐𝑎𝑙
𝐹 (𝐸 > 15 𝑚𝑜𝑙 , 500𝐾) = 9.713 × 10-5
P3-5 (e) The plots were generated using MATLAB
As can be seen from the plots, as the activation energy increases the fraction of molecules with energy
greater than the activation energy at a given temperature decreases.
P3-6 (a)
The fraction of collisions having energy between E = 3 and E = 5 is the area under the graph between
those two boundaries
For Figure P3-6 (a), Area =2*0.5(sum of parallel sides)*height
= 2*0.5*(0.1875 + 0.25)*1 = 0.4375
For Figure P3-6 (b),
Area=0.5*(0.2+0.15)*2=0.35
3-8
P3-6 (b)
For Figure P3-6 (a),
Fraction of collision having energy grater than 5 kcal/mol= 0.5*3*0.1875=0.281
For Figure P3-6 (b)
Fraction of collision having energy greater than 5 kcal/mol= 0.5*3*0.15=0.225
P3-6 (c) For both the figures
8
∫ 𝑓(𝐸, 𝑇)𝑑𝐸 = 1
0
P3-6 (d)
Since f(E,T) = 0 for E > 8 kcal, the fraction with energies greater than 8 kcal = 0
P3-6 (e)
The fraction with energies greater than EA ,8 kcal is zero
P3-6 (f)
At 400K, activation energy will be higher than the value at 300 K. So, X axis will have higher nos. Since
total area under curve has to remain constant to 1, so y axis value has to be lower. The graph may look
like this
P3-7 (a)
Note: This problem can have many solutions as data fitting can be done in many ways.
Using Arrhenius Equation
For Fire flies:
T(in K)
1/T
294
0.003401
298
0.003356
303
0.003300
Flashes/min
9
12.16
16.2
ln(flashes/min)
2.197
2.498
2.785
Plotting ln (flashes/min) vs. 1/T,
We get a straight line.
For Crickets:
T(in K)
1/T x103
chirps/min
ln(chirps/min)
287.2
293.3
300
80
126
200
4.382
4.836
5.298
3.482
3.409
3.333
Plotting ln (chirps/min) Vs 1/T, we get a straight line.
➔Both, Fireflies and Crickets data follow the Arrhenius Model.
ln y = A + B/T , and have the similar activation energy.
3-9
P3-7 (b)
For Honeybee:
T(in K)
1/T x103
298
3.356
303
3.300
308
3.247
V(cm/s)
0.7
1.8
3
ln(V)
-0.357
0.588
1.098
Plotting ln (V) vs. 1/T, almost straight line.
ln (V) = 44.6 – 1.33E4/T
At T = 40oC (313K)
V = 6.4cm/s
At T = -5oC (268K)
V = 0.005cm/s (But bee would
not be alive at this temperature)
P3-7 (c)
For Ants:
T(in K)
283
1/T x103
3.53
V(cm/s)
0.5
ln(V)
-0.69
293
303
3.41
3.30
2
3.4
0.69
1.22
311
3.21
6.5
1.87
Plotting ln (V) vs. 1/T,
We get almost a straight line.
So activity of bees, ants, crickets and fireflies follow Arrhenius model. So activity increases with an increase
in temperature. Activation energies for fireflies and crickets are almost the same.
Insect
Cricket
Firefly
Ant
Honeybee
Activation Energy
52150
54800
95570
141800
P3-7 (d)
There is a limit to temperature for which data for any one of the insects can be extrapolate. Data which
would be helpful is the maximum and the minimum temperature that these insects can endure before
death. Therefore, even if extrapolation gives us a value that looks reasonable, at certain temperature it
could be useless.
P3-7 (e)
1) The rate at which the beetle can push a ball of dung is directly proportional to its rate constant,
therefore
–rA =c*k, where c is a constant related to the mass of the beetle and the dung and k is the rate constant
−𝐸𝑎
𝑘 = 𝐴𝑒𝑥𝑝 ( 𝑅𝑇 )
From the data given
-rA
6.5
13
18
T(K)
300
310
313
1/T
0.003333
0.003226
0.003195
ln k
1.871802
2.564949
2.890372
3-10
Refer to P3-8 (similar procedure)
Therefore, A = 1.299X1011
E = 59195.68 J/mol
k = 1.299X1011 exp(-7120/T)
Now at T = 41.5 C = 314.5 K
k = 19.12 cm/s
Therefore, beetle can push dung at 19.12 cm/s at 41.5 C
P3-7 (e)
2) Individualized solution
P3-8 (a)
There are two competing effects that bring about the maximum in the corrosion rate: Temperature and
HCN-H2SO4 concentration. The corrosion rate increases with increasing temperature and increasing
concentration of HCN-H2SO4 complex. The temperature increases as we go from top to bottom of the
column and consequently the rate of corrosion should increase. However, the HCN concentrations (and
the HCN-H2SO4 complex) decrease as we go from top to bottom of the column. There is virtually no
HCN in the bottom of the column. These two opposing factors results in the maximum of the corrosion
rate somewhere around the middle of the column.
P3-8 (b)
The component ‘Corrosion’ can be applied to this problem
P3-9 Antidote did not dissolve from glass at low temperatures.
P3-10 (a)
If a reaction rate doubles for an increase in 10°C, at T = T1 let k = k1 and at T = T2 = T1+10, let k = k2 = 2k1.
Then with k = Ae-E/RT in general,
and
3-11
, or
or
Therefore:
which can be approximated by
. Consequently, for this doubling rate rule of thumb to be
valid, the temperature at which the doubling will take place must be related to the activation energy by
this relationship.
P3-10 (b) Individualized solution
P3-11
From the given data
-rA(dm3/mol.s)
T(K)
1/T (𝐾 −1 )
0.003333
ln (k)
-8.00637
0.002
300
k= -rA/ (4*1.5)
0.00033333
0.046
320
0.00766667
0.003125
-4.87087
0.72
340
0.12
0.002941
-2.12026
8.33
360
1.38833333
0.002778
0.328104
Plotting ln(k) vs (1/T), we have a straight line:
𝐸𝑎
Since,𝑙𝑛𝑘 = 𝑙𝑛𝐴 − 𝑅𝑇 therefore ln A = 41.99 and E/R = 14999
P3-11 (a)
Activation energy (E),
 E = 14999*8.314 =124700 J/mol = 124.7 kJ/mol
3-12
P3-11 (b)
Frequency Factor (A), ln A = 41.99
𝑑𝑚6 1
⇒ 𝐴 = 1.72 ∗ 1018 𝑚𝑜𝑙2 . 𝑠
P3-11 (c)
𝑘 = 1.72 ∗ 1018 𝑒𝑥𝑝 (
−14999
𝑇
)
(1)
Given T0 = 300K
𝑑𝑚6 1
Therefore, putting T = 300K in (1), we get 𝑘(𝑇0 ) = 3.33 ∗ 10−4 𝑚𝑜𝑙2 . 𝑠
Hence,
𝐸 1
1
𝑑𝑚6 1
𝑘 = 𝑘(𝑇0 )𝑒𝑥𝑝 ( (
− ))
.
𝑅 300 𝑇
𝑚𝑜𝑙 2 𝑠
P3-12 (a) –ra=kCACB
P3-12 (b) –ra=kPAPB
P3-12 (c) –ra=kCA
P3-12 (d) –ra=k
P3-13 (a) C2H6 → C2H4 + H2
Rate law: -rA =
P3-13 (b) C2H4 + 1/2O2 → C2H4O
Rate law: -rA =
, k = [ s-1 ]
, k = [ dm1.5/s.mol0.5 ]
P3-13 (c) (CH3)3COOC(CH3)3 ↔ C2H6 + 2CH3COCH3
A
↔ B
+
2C
Rate law: -rA = k[CA – CBCC2/KC] , k = [ s-1 ]
P3-13 (d) n-C4H10 ↔ I- C4H10
Rate law: -rA = k[
–
/Kc], k = [ s-1 ]
P3-13 (e) CH3COOC2H5 + C4H9OH ↔ CH3COOC4H9 + C2H5OH
A
+
B
↔
C
+
D
Rate law: -rA = k [CACB – CCCD/KC], k = [ dm3/mol.s ]
P3-13 (f) 2CH3NH2 ↔ (CH3)2NH + NH3
Rate law: -r’A = kP2CH3NH2, k = [ mol/kg cat.s. Pa2]
P3-13 (g) (CH3CO)2O + H2O → 2CH3COOH
A+
B →
2C
Rate law: -rA = kCACB, k = [ dm3/mol.s ]
3-13
P3-14 (a)
(1) –rA = kACACB2
(2) –rA= kACB
(3) –rA= kA
(4) –rA= kACA/CB
P3-14 (b)
(1) H2 + Br2 → 2HBr
(2) H2 + I2 → 2HI
P3-15 (a)
Rate law: -rHBr =
Rate law: -rA =
A→B A→B
–rA = kCA
We need to assume a form of the rate law for the reverse reaction that satisfies the equilibrium
condition. If we assume the rate law for the reverse reaction (B → A) is rArev = k–1CB = then:
From Appendix C we know that for a reaction at equilibrium: KC
At equilibrium, rnet 0, so:
Solving for KC gives:
P3-15 (b)
If we assume the rate law for the reverse reaction 2D → A + 2B is r–A = k–1 CD then:
From Appendix C we know that for a reaction at equilibrium: KC
At equilibrium, rnet 0, so:
3-14
Solving for KC gives:
P3-15 (c)
If we assume the rate law for the reverse reaction 𝐶 + 𝐷 → 𝐴 + 𝐵is
then:
From Appendix C we know that for a reaction at equilibrium: KC
At equilibrium, rnet 0, so:
Solving for KC gives:
P3-16
Assuming the reactions to be elementary:
2 Anthracene →Dimer
Similarly, for the second reaction:
Norbornadiene → Quadricyclane
where β is greater than 1.0 and is related to the intensity of the light.
3-15
Page intentionally blank
3-16
Synopsis for Chapter 4 – Stoichiometry
General: This chapter adds our third building block, stoichiometry. The goal is to express concentrations
in terms of either molar flow rates or conversion and then to write the rate laws in terms of these variables
for both liquid phase and gas phase reactions.
Questions
Most often (O) I require the student to think through O Q4-3A, O Q4-4A and O Q4-5A.
l Q4-1A (20 seconds) Questions Before Reading (QBR).
Q4-2A (10-12 min) I usually assign i>Clicker questions.
l Q4-6A (2 min) Queen is the answer.
Q4-8A (15 min) As usual, if a student is having difficulty with stoichiometry the LearnChemE
screencasts are encouraged.
Q4-9A (12 min) Ties in safety to CRE.
Computer Simulations and Experiments
l P4-1A (5-10 min) Use Wolfram for part (a) and part (b). Part (vi) is always assigned for Example 4-5
that compared the equilibrium conversion for batch and flow systems.
Interactive Computer Games (ICG)
l P4-2A (15 min) This ICG is excellent preparation for the Student Chapter Jeopardy Competition at
the Annual AIChE meeting.
Problems
O P4-3A (25 min) Straight forward problem on chemical equilibrium.
AA P4-4B (20 min) Very easy problem on stoichiometry, but parts (c) and (d) may surprise you.
l P4-5A (20-45 min) I always assign two or three parts (a) through (d). Gives practice at writing the
rate law as a function of conversion and help reinforce the CRE algorithm.
O P4-6A (30 min) this problem will be continued in an Example Problem in Chapter 13.
AA P4-7B (30 min) Old Exam Question (OEQ). Similar to Problem P4-4B.
l P4-8B (30 min) Old Exam Question (OEQ). Good problem! Students need to choose their basis of
calculation carefully.
AA P4-9B (45 min) Old Exam Question (OEQ). This problem is a good problem as it relates to an industrial
process and an industrial reaction.
AA P4-10B (30 min) Old Exam Question (OEQ). Good problem on equilibrium calculation. I usually assign
one or two problem P4-10B to P4-12B on equilibrium conversion.
AA P4-11B (15 min) Another good equilibrium calculation.
AA P4-12B (10 min) Equilibrium conversion calculation.
S4-1
AA P4-13C (50 min) Fairly challenging problem. Requires thinking, not just formula plugging.
S4-2
Solutions for Chapter 4 – Stoichiometry
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-4
http://umich.edu/~elements/5e/04chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q4-1 Individualized solution
Q4-2 Individualized solution
Q4-3 Individualized solution
Q4-4 For part (a):
CD=CA0*(X/3)=3*(0.2/3)=0.2 mol/dm3
CB=3*0.133=0.4 mol/dm3
CD and CB decreases when initial concentration is lower, and A becomes the limiting reactant.
Q4-5 Individualized solution
Q4-6 Individualized solution
Q4-7 Individualized solution
Q4-8 Individualized solution
Q4-9 Individualized solution
P4-1 (a) Example 4-4
(i)
By increasing thetaB, the rate profile is affected only for the initial levels of conversion. Increase in
thetaB makes the rate to fall down. This is because increase in thetaB causes the partial pressure of
O2 to rise, hence decreasing the rate of SO2 oxidation.
4-1
(ii)
Kp has the least effect, and KSO3 has the greatest effect (can be seen by varying Wolfram
parameters)
Increased the value of Kp to 30 times (from 64 to 1967) but still the shape of graph remains almost
same
Increased the value of KSO3 to just 10 times and the see how the graph completely changed
(iii)
1. By varying the sliders, one can say that Kp and epsilon have almost a negligible effect on the
rate vs conversion profile. The greatest effect the rate profile is of K SO3.
2. By increasing thetaB, the rate profile is affected only for the initial levels of conversion.
3. Increasing PSO2 and k increases rate for a given conversion while opposite is the case for thetaB, KO2
and KSO3.
4-2
(iv) Polymath:
Polymath code:
d(X)/d(t) = 1
Fao = 3
PSO20 = 4.1
epsilon = -0.14
PSO2 = PSO20* (1-X)/(1+epsilon*X)
PSO3 = PSO20* X/(1+epsilon* X)
thetaB = 0.54
PO2 = PSO20* (thetaB-X/2)/(1+epsilon* X)
k = 9.7
KO2 = 38.5
KSO3 = 42.5
KP = 930
ra = -k *(PSO2* Sqrt(PO2)-PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2
yaxis = Fao/(-ra)
X(0)=0
t(0) = 0
t(f) = 0.4
Polymath Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
epsilon
-0.14
-0.14
-0.14
-0.14
2
Fao
3.
3.
3.
3.
3
k
9.7
9.7
9.7
9.7
4
KO2
38.5
38.5
38.5
38.5
5
KP
930.
930.
930.
930.
6
KSO3
42.5
42.5
42.5
42.5
7
PO2
2.214
1.476695
2.214
1.476695
8
PSO2
4.1
2.605932
4.1
2.605932
9
PSO20
4.1
4.1
4.1
4.1
10 PSO3
0
0
1.737288
1.737288
11 ra
-0.5651727
-0.5651727
-0.0045241
-0.0045241
12 t
0
0
0.4
0.4
13 thetaB
0.54
0.54
0.54
0.54
14 X
0
0
0.4
0.4
15 yaxis
5.308112
5.308112
663.1111
663.1111
Differential equations
1 d(X)/d(t) = 1
Explicit equations
1
Fao = 3
2
PSO20 = 4.1
3
epsilon = -0.14
4
PSO2 = PSO20* (1-X)/(1+epsilon*X)
5
PSO3 = PSO20* X/(1+epsilon* X)
6
thetaB = 0.54
7
PO2 = PSO20* (thetaB-X/2)/(1+epsilon* X)
8
k = 9.7
4-3
9
KO2 = 38.5
10 KSO3 = 42.5
11 KP = 930
12 ra = -k *(PSO2* Sqrt(PO2)-PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2
13 yaxis = Fao/(-ra)
At X = 0.4
FA0/–rA’ = 663.11
W = (FA0/–rA’)X at X = 0.4
Hence,
W = 265.24 g
(v) This program is to find Xe:
Polymath code:
f(X) = -k *(PSO2* Sqrt(PO2)-PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2
PSO20 = 4.1
epsilon = -0.14
PSO2 = PSO20* (1-X)/(1+epsilon*X)
thetaB = 0.54
PSO3 = PSO20* X/(1+epsilon* X)
k = 9.7
KO2 = 38.5
KSO3 = 42.5
KP = 930
PO2 = PSO20* (thetaB-X/2)/(1+epsilon* X)
X(min) = 0.5
X(max) = 1
Polymath Output:
POLYMATH Report
Nonlinear Equation
Calculated values of NLE variables
Variable Value
1 X
f(x)
Initial Guess
0.997577 -1.251E-09 0.5 ( 0 < X < 1. )
Variable Value
1
epsilon
-0.14
2
k
9.7
3
KO2
38.5
4
KP
930.
5
KSO3
42.5
6
PO2
0.1963961
7
PSO2
0.0115472
8
PSO20
4.1
9
PSO3
4.754015
10 thetaB
0.54
Nonlinear equations
1 f(X) = -k *(PSO2* Sqrt(PO2)-PSO3/KP)/(1+Sqrt(PO2* KO2)+PSO3 *KSO3)^2 = 0
4-4
Explicit equations
1
PSO20 = 4.1
2
epsilon = -0.14
3
PSO2 = PSO20* (1-X)/(1+epsilon*X)
4
thetaB = 0.54
5
PSO3 = PSO20* X/(1+epsilon* X)
6
k = 9.7
7
KO2 = 38.5
8
KSO3 = 42.5
9
KP = 930
10 PO2 = PSO20* (thetaB-X/2)/(1+epsilon* X)
Hence,
Xe = 0.9976
Now we will use Xe in the following Polymath Program
Polymath code:
d(X1)/d(t) = 1
d(X2)/d(t) = 0.99*Xe/0.3
Fao = 1000
PSO20 = 4.1
epsilon = -0.14
PSO21 = PSO20* (1-X1)/(1+epsilon*X1)
PSO31 = PSO20* X1/(1+epsilon* X1)
thetaB = 0.54
PO21 = PSO20* (thetaB-X1/2)/(1+epsilon* X1)
k = 9.7
KO2 = 38.5
KSO3 = 42.5
KP = 930
ra1 = -k *(PSO21* Sqrt(PO21)-PSO31/KP)/(1+Sqrt(PO21* KO2)+PSO31 *KSO3)^2
yaxis1 = Fao/(-ra1)
Xe = 0.9976
PSO22 = PSO20* (1-X2)/(1+epsilon*X2)
PSO32 = PSO20* X2/(1+epsilon* X2)
PO22 = PSO20* (thetaB-X2/2)/(1+epsilon* X2)
ra2 = -k *(PSO22* Sqrt(PO22)-PSO32/KP)/(1+Sqrt(PO22* KO2)+PSO32 *KSO3)^2
yaxis2 = Fao/(-ra2)
X1(0)=0
X2(0)=0
t(0) = 0
t(f) = 0.3
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
epsilon
-0.14
-0.14
-0.14
-0.14
2
Fao
1000.
1000.
1000.
1000.
3
k
9.7
9.7
9.7
9.7
4
KO2
38.5
38.5
38.5
38.5
5
KP
930.
930.
930.
930.
6
KSO3
42.5
42.5
42.5
42.5
7
PO21
2.214
1.669102
2.214
1.669102
8
PO22
2.214
0.2197559
2.214
0.2197559
4-5
9
PSO20
4.1
4.1
4.1
4.1
10 PSO21
4.1
2.995825
4.1
2.995825
11 PSO22
4.1
0.0588832
4.1
0.0588832
12 PSO31
0
0
1.283925
1.283925
13 PSO32
0
0
4.698973
4.698973
14 ra1
-0.5651727
-0.5651727
-0.0092831
-0.0092831
15 ra2
-0.5651727
-0.5651727
-5.276E-06
-5.276E-06
16 t
0
0
0.3
0.3
17 thetaB
0.54
0.54
0.54
0.54
18 X1
0
0
0.3
0.3
19 X2
0
0
0.987624
0.987624
20 Xe
0.9976
0.9976
0.9976
0.9976
21 yaxis1
1769.371
1769.371
1.077E+05
1.077E+05
22 yaxis2
1769.371
1769.371
1.895E+08
1.895E+08
Differential equations
1 d(X1)/d(t) = 1
2 d(X2)/d(t) = 0.99*Xe/0.3
Explicit equations
1
Fao = 1000
2
PSO20 = 4.1
3
epsilon = -0.14
4
PSO21 = PSO20* (1-X1)/(1+epsilon*X1)
5
PSO31 = PSO20* X1/(1+epsilon* X1)
6
thetaB = 0.54
7
PO21 = PSO20* (thetaB-X1/2)/(1+epsilon* X1)
8
k = 9.7
9
KO2 = 38.5
10 KSO3 = 42.5
11 KP = 930
12 ra1 = -k *(PSO21* Sqrt(PO21)-PSO31/KP)/(1+Sqrt(PO21* KO2)+PSO31 *KSO3)^2
13 yaxis1 = Fao/(-ra1)
14 Xe = 0.9976
15 PSO22 = PSO20* (1-X2)/(1+epsilon*X2)
16 PSO32 = PSO20* X2/(1+epsilon* X2)
17 PO22 = PSO20* (thetaB-X2/2)/(1+epsilon* X2)
18 ra2 = -k *(PSO22* Sqrt(PO22)-PSO32/KP)/(1+Sqrt(PO22* KO2)+PSO32 *KSO3)^2
19 yaxis2 = Fao/(-ra2)
At X = 0.3
FA0/–rA’ = 107,700
W = (FA0/–rA’)X at X = 0.3
Hence,
W = 32,310 g = 32.31 kg
4-6
At X = 0.99Xe
FA0/–rA’ = 1.895 x 108
W = (FA0/–rA’)X at X = 0.99Xe
Hence,
W = 1.872 x 108 g = 1.872 x 105 kg
P4-1 (b) Example 4-5
(i)
Area under the figure E4-5.1 multiplied by the mole flow rate will be the volume of PFR required.
(ii) Area under the figure E4-5.1 multiplied by the initial concentration will be the time required for
the reaction to occur in a batch reactor.
(iii) Kc = 0.4 and CA0 = 0.09 (Red and blue curves are farthest apart)
(iv)
When Kc is minimum and CA0 is maximum (can be seen from below graph). Both red and blue curves
merge together.
4-7
(v)
(v-vi)
1) We can see that when Kc is set to minimum and CT0 at maximum, the line becomes parallel to X-axis
taking value equal to 1 for all values of mole fraction. Under these conditions both of the
conversions obtained take their minimum values.
2) Another case when Kc is at its maximum and CT0 at minimum, the ratio of conversion is still equal to
1 but this time both of the conversions are individually equal to 1 for almost all values of mole
fraction.
3) The above results can easily be justified as when you have a high initial concentration but a low rate
constant, no matter which reactor we choose, conversion would be very low but in the other
extreme case the conversion is very high irrespective of the reactor we choose.
P4-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer
Games (ICGs) is given at the front of this Solutions Manual.
P4-3
2A  B
CA0 = 4 mol/dm3
Xe = 0.6
P4-3 (a)
Kc = CB/ CA2
ε = –1/2
CA = CA0(1–X)/(1+ εX) = 4*0.4/0.7 = 2.286 mol/dm3
CB = 4*0.3/0.7 = 1.714 mol/dm3
Therefore Kc = 0.328 dm3/mol
P4-3 (b)
For a liquid phase reaction, CA = 1.6 mol/dm3
CB = 1.2 mol/dm3
Thus, Kc = 0.469 dm3/mol
4-8
P4-3 (c)
CA = CA0(1–X)/(1+ εX)
CB = CA0(0.5*X)/(1+ εX)
–rA = k(CA^2– CB/Kc)
–rA = k((CA0(1–X)/(1+εX)) ^2 – (CA0(0.5*X)/(1+ εX) )/Kc)
–rA = 2((2.0(1–X)/(1– 0.5X)) ^2 – (2.0(0.5X)/(1–0.5X) )/0.5)
P4-3 (d)
For a constant volume reaction, ε = 0
Therefore,
–rA = k((CA0(1–X)) ^2 – CA0(0.5*X)/Kc)
P4-4
1
1
B→ C
2
2
1
 1  1 1 
 = y A0 =    − − 1 = −
2
 2  2 2 
A+
P4-4 (a)
C B0 = C A0 = 0.1
P4-4 (b)
C A = C A0

mol
dm 3
(1− X) = 0.1 (1− 0.25) = (0.1)(0.75) = 0.086 mol
( ) 1 
0.875
dm 3
(1+ X)
1− X 

2 
X
 0.25 
 


( 0.1)( 0.125) = 0.0143 mol
2
2 
CC = C A0   = ( 0.1) 
=
0.875
dm3
 1 
(1 +  X )
1
−
X


 2 

P4-4 (c)
 1 
 1 
1 − X  C A 0 1 − X 
2 
 2  = C = 0.1 mol
C B = C A0 = 
=
A0
(1 + X )
dm 3
 1 
1 − X 
 2 
P4-4 (d)
C B = 0.1

mol
dm 3
P4-4 (e)
rA rC
= , rA = −2rC = −4 mol dm 3 min
−4 2
4-9

P4-4 (f)
2A + B  C
ε = δyA0 = –0.5
CA0 = 2 mol/dm3
Irreversible rate of reaction
–rA = kCA2CB
3
−𝑟𝐴 = 𝑘𝐶𝐴0
(
1−𝑋 2
)
1 − 0.5𝑋
P4-4 (g)
–rA = 7.11 mol/dm3s
P4-5 (a)
Liquid phase reaction,
O
CH2–OH
CH2 – CH2 + H2O → CH2–OH
A
+ B → C
CA0 = 16.13mol/dm3
Stoichiometric Table:
Species
Ethylene oxide
Water
Glycol
CB0 = 55.5 mol/dm3
Symbol
A
B
C
Initial
CA0=16.13 mol/dm3
Change
–CA0X
CB0= 55.5 mol/dm3,  B = 3.441
–CA0X
Remaining
CA= CA0(1–X)
= (1–X) mol/dm3
CB = CA0(  B –X)
CA0X
=(3.441–X) mol/dm3
CC = CA0X mol/dm3
0
Rate law:
–rA = kCACB
Therefore,
2
–rA = k C AO
(1–X) (  B –X) = k (16.13)2(1–X) (3.441–X)
At 300K
 CSTR =
E = 12500 cal/mol,
k = 0.1dm3/mol.s
X = 0.9,
(16.13)( 0.9 )
C AO X
= 2.196sec
=
2
−rA
( 0.1)(16.13) (1 − 0.9 )( 3.441 − 0.9 )
and, V = τ *vo = 2.196 sec X 200 liters/sec = 439.2 liters
At 350K,
k2 = k exp((E/R)(1/T–1/T2))= 0.1exp((12500/1.987)(1/300–1/350))
= 1.99 dm3/mol.s
Therefore,
 CSTR =
(16.13)( 0.9 )
C AO X
= 0.110sec ,
=
2
−rA
(19.99 )(16.13) (1 − 0.9 )( 3.441 − 0.9 )
and, V = τ* vo = 0.110 X 200 liters = 22 liters
4-10
P4-5 (b)
Isothermal, isobaric gas-phase pyrolysis,
C2H6 C2H4 + H2
A → B + C
Stoichiometric table:
Species symbol Entering
C2H6
A
FA0
C2H4
B
0
H2
C
0
FTO=FA0
Change
–FA0X
+FA0X
+FA0X
Leaving
FA=FA0(1–X)
FB=FA0X
FC=FA0X
FT=FA0(1+X)
 = yao  = 1(1+1–1) = 1
v = vo(1+  X)
=> v = vo(1+X)
P
CA0 = yA0 CTO = yA0
RT
=
(1)(6atm )

m3 atm 
0.082

 (1100 K )
K .kmol 

= 0.067 kmol/m3 = 0.067 mol/dm3
CA =
(1 − X ) mol/dm3
FA FAO (1 − X )
=
= C AO
v
vO (1 + X )
(1 + X )
CB =
F (X )
FB
X
= AO
= C AO
mol/dm3
v vO (1 + X )
1
+
X
(
)
CC =
FC
F (X )
X
= AO
= C AO
mol/dm3
v vO (1 + X )
(1 + X )
Rate law:
–rA = kCA= kCA0
(1 − X ) =0.067 k (1 − X )
(1 + X )
(1 + X )
If the reaction is carried out in a constant volume batch reactor, =>(  = 0)
CA = CA0(1–X) mol/dm3 CB = CA0 X mol/dm3
CC = CA0 X mol/dm3
P4-5 (c)
Isothermal, isobaric, catalytic gas phase oxidation,
1
O2 → C2H4O
2
1
+
B →
C
2
C2H4 +
A
–rA’ = kApApB0.5
pA = yA0P = yA0CRT = CART
pB = yB0P = yB0CRT = CBRT
Therefore, rA’ = kACACB0.5(RT)1.5
4-11
(1) For a fluidized batch reactor, V is constant
CA = CA0(1–X)
CB = CA0(θB – b/a X)
θB = 0.5 (stoichiometric feed)
b/a = 0.5
Therefore, CB = CA0(1–X)/2
Therefore, –rA’ = kACA01.5(1–X)1.5(RT)1.5/20.5
(2) Stoichiometric table:
Species Symbol Entering Change Leaving
C2H4
A
FA0
–FA0X
FA=FA0(1–X)
O2
B
FB0
1
FB=FA0(  B –X/2)
– FA0X
2
C2H4O
1
FBO 2 FAO 1
B =
=
=
FAO
FAO
2
C
y AO =
0
+ FA0X
FC=FA0X
FAO
FAO
2
=
=
FTO FAO + FBO 3
(6atm )
P 2
mol
=
= 0.092 3
3
RT 3 
dm
atm.dm 
 0.082
 (533K )
mol.K 

F (1 − X ) C AO (1 − X ) 0.092 (1 − X )
F
C A = A = AO
=
=
v vO (1 +  X ) (1 − 0.33 X ) (1 − 0.33 X )
C AO = y AO CTO = y AO
X

FAO   B − 
F
2  0.046 (1 − X )

CB = B =
=
v
vO (1 +  X )
(1 − 0.33 X )
CC =
0.092 ( X )
FC
FAO X
=
=
v
vO (1 +  X ) (1 − 0.33 X )
If the reaction follow elementary rate law
Rate law: –rA’ = kACACB0.5(RT)1.5, T = 533 K
 0.092 (1 − X )  0.046 (1 − X )
 −rA = k 


 (1 − 0.33 X )   (1 − 0.33 X ) 
P4-5 (d)
Isothermal, isobaric, catalytic gas-phase reaction in a PBR
C6H6 + 2H2 → C6H10
A + 2B → C
Given:
 0 = 50 dm 3 / min
4-12
0.5
(RT)1.5
Stoichiometric Table:
Species
Symbol
C6H6
A
H2
B
C6H10
C
Entering
FA0
FB0=2FA0
0
FB 0 2 FA0
=
=2
FA 0
FA 0
FA 0
FA 0
1
=
=
FT 0 FA0 + FB 0 3
B =
C A 0 = CT 0 y A 0 =
CA =
FA
CB =
FB
CC =
FC



y A0 =
Change
–FA0X
–2FA0X
FA0X
Leaving
FA=FA0(1–X)
FB=FA0(θB–2X)
Fc=FA0X
1
2
 = y A0 = (1 − 2 − 1) = −
3
3
P 1
6atm
mol
1
=
0
.
055
 =


3
RT  3  0.0821 dmmolatm
3
dm 3
 K * ( 443K )  
=
FA0 (1 − X )
(1 − X )
= C A0
 0 (1 + X )
(1 − 23 X )
=
FA0 ( B − 2 X )
(2 − 2 X )
(1 − X )
= C A0
=
2
*
C
A
0
 0 (1 + X )
(1 − 23 X )
(1 − 23 X )
=
FA 0 X
X
= C A0
 0 (1 + X )
(1 − 23 X )
Rate Law:
NOTE: For gas-phase reactions, rate laws are sometimes written in terms of partial pressures instead of
concentrations. The units of the rate constant, k, will differ depending on whether partial pressure or
concentration units are used. See below for an example.
− rA ' = kPA PB
2
mol
mol
=
* atm * atm 2
3
kgcat  min
kgcat  min atm
Notice that if you use concentrations in this rate law, the units will not work out.
PA = y A0 P = ( y A0 C ) * RT = C A RT
− rA = kPA PB = kC A C B ( RT ) 3 = 4kC A0
2
2
3
(1 − X ) 3
( RT ) 3
3
2
(1 − 3 X )
Design Equation for a fluidized CSTR:
W=
W=
W=
FA0 X
− rA '
FA0 X (1 − 23 X ) 3
4kC A0 (1 − X ) 3 ( RT ) 3
3
 0 X (1 − 23 X ) 3
4kC A0 (1 − X ) 3 ( RT ) 3
2
Evaluating the constants:
k = 53
mol
at 300 K
kgcat  min atm 3
4-13
At 170°C (443K),
J


80000

 EA  1


1
mol
mol  1 − 1  = 1663000
 = 53 exp 
k 443 = k 300 exp  
−

J
kgcat  min atm
 300 K 443K 
 R  T300 T443 
 8.314
mol  K


Plugging in all the constants into the design equation:
X = 0.8
3
2
50 dm
min  0.8  (1 − 3 0.8)
W=
= 5.25  10 −7 kgcat
3
3
mol
mol 2
dm3 atm
4  1663000 kgatmin
(
0
.
055
)
(
1
−
0
.
8
)
(
0
.
0821

443
K
)
mol  K
atm3
dm3
3
At 270°C (543K),
J


80000

 EA  1


1
mol
mol  1 − 1  = 90790000
 = 53 exp 
k 543 = k 300 exp  
−

J
kgcat  min atm
 300 K 543K 
 R  T300 T543 
 8.314
mol  K


Plugging in all the constants into the design equation:
X = 0.8
3
2
50 dm
min  0.8  (1 − 3 0.8)
W=
= 5.22 10 −9 kgcat
3
3
mol
mol 2
dm 3 atm
4  9079000 kgatmin atm3 (0.055 dm3 ) (1 − 0.8) (0.0821 molK  543K )
3
P4-6 (a)
Let A = ONCB
B = NH3
C = Nibroanaline
D = Ammonium Chloride
A + 2B ⎯⎯
→ C+D
-rA = kC ACB
P4-6 (b)
Species
A
B
C
D
P4-6 (c)
For batch system,
CA=NA/V
Entering
FA0
FB0 = ΘBFA0
=6.6/1.8 FA0
0
0
–rA = kNANB/V2
4-14
Change
–FA0X
–2 FA0X
FA0X
FA0X
Leaving
FA0(1–X)
FB=
FA0(ΘB – 2X)
FC=FA0X
FD=FA0X
P4-6 (d)
-rA = kC ACB
FA =
N A N A N A0
F
F
=
=
(1 − X ) = C A0 (1 − X ), C A = A = A = C A0 (1 − X )
V
V0
V0
v
v0
FB =
N B N B N A0
F
=
=
( B − 2 X ) = C A0 ( B − 2 X ), CB = B = C A0 ( B − 2 X )
V
V0
V0
v0
−rA = kC A2 0 (1 − X )( B − 2 X )
B =
CB 0 6.6
=
= 3.67
C A0 1.8
C A0 = 1.8
kmol
m3
−rA = k (1.8 ) (1 − X )(3.67 − 2 X )
2
P4-6 (e)
1) At X = 0 and T = 188°C = 461 K
2
− rA0 = kC A2 0  B = 0.0017
−rA0 = 0.0202
m3
kmol
 kmol 
1.8 3  3.67 = 0.0202 3
kmol min 
m 
m min
kmol
m3 min
2) At X = 0 and T = 250C = 298K
 E  1 1 
k = k O exp 
−  
 R  TO T  
cal


11273


m
mol  1 − 1  
k = 0.0017
exp


cal  461 298  
kmol. min

1
.
987


mol.K


m3
= 2.03  10 −6
kmol. min
3
2
m3
kmol 
 kmol  
−rA = 2.03 10
1.8 3   3.67 3 
kmol min 
m  
m 
−6
–rA0 = kCA0aCB0 = 2.41 X 10–5 kmol/m3min
3)
 E  1 1 
k = k0 exp   −  
 R  T0 T  
4-15


cal
11273




m
1
1
mol
k = 0.0017
exp 
−



kmol min
1.987 cal  461 K 561 K  


mol K
3
k = 0.0152
m3
kmol min
−rA0 = kC A0 2CB 0
2
m3
kmol 
 kmol  
−rA = 0.0152
1.8 3   3.67 3 
kmol min 
m  
m 
−rA = 0.1807
kmol
m3 min
P4-6 (f)
rA = kCA02(1–X)(θB–2X)
At X = 0.90 and T = 188C = 461K
1) at T = 188 C = 461 K

 kmol 
m3
1.8 3  (1 − 0.9 )(3.67 − 2(0.9 ))
− rA =  0.0017
kmol. min 
m 

kmol
= 0.00103 3
m min
2
2) at X = 0.90 and T = 25C = 298K

 kmol 
m3
−6
1.8 3  (1 − 0.9 )(3.67 − 2(0.9 ))
− rA =  2.03  10
kmol. min 
m 

kmol
= 1.23  10 −6 3
m min
2
3) at X = 0.90 and T = 288C = 561K

 kmol 
m3
1.8 3  (1 − 0.9 )(3.67 − 2(0.9 ))
− rA =  0.0152
kmol. min 
m 

kmol
= 0.0092 3
m min
2
P4-6 (g)
FA0 = 2 mol/min
1) For CSTR at 25oC –rA = 1.23  10
V =
=
FAO (1 − X )
− rA X =0.9
−6
kmol
m 3 min
2mol / min 0.1
= 162.60m 3
mol
1.23  10 −3 3
m min
4-16
2) At 288oC, –rA = 0.0092
V =
=
FAO (1 − X )
− rA X =0.9
kmol
m 3 min
2mol / min 0.1
= 21.739m 3
mol
0.0092 3
m min
P4-7 (a) It is given that the feed is equimolar, so CB0 = 2 mol/dm3
P4-7 (b) For every mole of B, two moles of A are required. Therefore, A is the limiting reagent.
P4-7 (c) From the stoichiometry of the reaction, we can write
(NA0 – NA)/2 = (NB0 – NB)
As NA0 = NB0 and conversion of A = 0.25 we get,
NA = (3/4) NA0
NB = (7/8) NA0
NC = (1/8) NA0
Let the initial volume be V0 , CA0 = NA0 /V0
As the reaction is in gas phase, volume of the reaction will change which will affect the concentrations.
NT= (7/4) NA0
V= (7/4) V0
Therefore, CB= NB /V = (½) CA0 = 1 mol/dm3
P4-7 (d) For the given reaction,
1
1
𝛿 = 2 − 2 − 1 = −1
𝑦𝐴0 = 0.5
𝜀 = 𝑦𝐴0 ∗ 𝛿 = −0.5
Therefore, concentrations of each species is given by,
CA = CA0 (1–X)/(1+ εX)
CB = CA0 (1–X/2)/(1+ εX)
CC = CA0 X/(2(1+ εX))
Rate expression is given by,
–rA = kf CA 2 CB – kb CC
–rA = kf (CA 2 CB – CC /Kc)
Substituting the equations derived for concentrations,
–rA = kf CA0 ((CA0 (1–X)/(1+ εX))2 (1–X/2)/(1+ εX) – X/(2(1+ εX))/Kc) – (1)
Substitute the given values to obtain the final expression for rate.
P4-7 (e) equilibrium conversion is given by, rA = 0
(CA0 (1–X)/(1+ εX))2 (1–X/2)/(1+ εX) – X/(2(1+ εX))/Kc = 0
2 Kc CA02 (1 –X3/2+2X2 –5X/2) = X (1+ 2εX+ε2X2)
CA0 = 2.0 (units)
ε = –1/2
Kc= 0.5
4-17
Substituting gives a cubic equation,
–2.25 X3 +9 X2 –11 X +4 =0
Whose roots are given by,
X = 2, 0.67 and 1.33
Clearly, the only practical root is Xe = 0.67 which is the equilibrium conversion.
P4-7 (f)
(1) Substitute X = 0 in (1)
–rA=2*2(2*2*(1)*(1)–0)= 16 mol/dm3.s
(2) Substitute X= 0.5 in (1) to get –rA =4.45 mol/dm3.s
(3) Substitute X=0.99*0.67 =0.663 in (1) to get –rA =0.099 mol/dm3.s
P4-8 (a)
Isothermal gas phase reaction.
1
3
N 2 + H 2 → NH 3
2
2
Making H2 as the basis of calculation:
1
2
H 2 + N 2 → NH 3
3
3
1
2
A+ B → C
3
3
Stoichiometric table:
Species Symbol Initial
change
H2
A
FA0
–FA0X
N2
B
FB0=  B FA0 –FA0X/3
NH3
C
0
+2FA0X/3 FC=(2/3)FA0X
P4-8 (b)
2 1



Leaving
FA=FA0(1–X)
FB=FA0(  B –X/3)
2
 =  − − 1 = −
3 3
3
1
 2


16.4
( atm )
C AO = 0.5
= 0.2 mol/dm3

atm.dm3 
 0.082
 (500 K )
mol.K 

 = y AO = 0.5   −  = −
3
3
C AO (1 − X ) 0.2 (1 − 0.6 )
=
= 0.1mol / dm3
0.6


(1 +  X )
1 −

3


2 C ( X ) 2 0.2 ( 0.6 )
CNH3 = CC =  AO
= 
= 0.1mol / dm3
3 (1 +  X ) 3  0.6 
1 −

3 

CH 2 = C A =
4-18
P4-8 (c)
kN2 = 40 dm3/mol.s
(1) For Flow system:
3
1
−rN2 = k N2 CN2  2 CH 2  2
1
3
 X   2 
 2
1
−


2
3    (1 − X ) 
 

= 40 (C AO )  
 1 − X    1 − X  



 
3    
3  
3

2
 (1− X ) 

−rN 2 = 1.6
1− X  
 3  
(2) For batch system, constant volume.

P4-9 (a)
2A + 9B
2C + 4D + 4E
9
A+ B
C + 2D + 2E
2
We are going to rework the class problem for the case of 3.5% naphthalene and 96.7% air and for a
pressure of 10 atm and a temperature of 500 K. A feed under these conditions’ naphthalene is the
limiting reactant.
4-19
Stoichiometric Table
Species
Symbol
In
Change
Out
Naphthalene
A
FA0
–FA0X
FA = FA0 (1 – X)
Oxygen
B
FB0 = B FA0
–
9
2
FA0X
FB = FA0 (B –
9
2
Phthalic Anhydride
C
FC = 0
+FA0X
FC = FA0 X
Carbon Dioxide
D
FD = 0
+2 FA0X
FD = 2FA0 X
Water
E
FE = 0
+2 FA0X
FE = 2FA0 X
Nitrogen
I
FI = I FA0
–––
FI = I FA0
FT = FT0 +  FA0X
IT0
 = [1 + 2 + 2 –
P4-9 (b)
Species
9
2
– 1] = –
Symbol
In
Change
Out
A
NA0
–NA0X
NA = NA0 (1 – X)
B
NB0 = B NA0
Naphthalene
Oxygen
–
9
2
NA0X
NB = NA0 (B –
9
2
Phthalic Anhydride
C
NC = 0
+NA0X
NC = NA0 X
Carbon Dioxide
D
ND = 0
+2 NA0X
ND = 2NA0 X
Water
E
NE = 0
+2 NA0X
NE = 2NA0 X
Nitrogen
I
NI = I NA0
–––
NI = I NA0
IT0
For isothermal constant volume batch
𝑁
𝑁
0
0
∗𝐹
(0.965)∗(0.21)
𝑉 = 𝑉0, 𝐶𝐴 = 𝑉𝐴, 𝐶𝐵 = 𝑉𝐵 , Isothermal T = T0
Now,
𝐹
𝑦
𝐴0
𝐴0
𝐹
𝑦 ∗𝐹
𝐴0
𝐴0
𝑇0
𝜃𝐼 = 𝐹 𝐼0 = 𝑦 𝐼0 ∗𝐹𝑇0 =
0.035
(0.965)∗(0.79)
𝑇0
0.035
=5.79
=21.78
For constant Volume,
(1) Concentration
9
2
𝐶𝐵 = 𝐶𝐴0 (𝜃𝐵 − 𝑋)
𝐶𝐷 = 𝐶𝐴0 (0 + 2𝑋)
(2) Total pressure
𝑃𝑇 = 𝑃𝐴 + 𝑃𝐵 + 𝑃𝐶 + 𝑃𝐷 + 𝑃𝐸 + 𝑃𝐼
4-20
1
2
X)
NT = NT0 +  NA0X
 = [1 + 2 + 2 –
𝜃𝐵 = 𝐹𝐵0 = 𝑦𝐵0 ∗𝐹𝑇0 =
X)
9
2
– 1] = –
1
2
Where
𝑃𝐴 = 𝐶𝐴0 (1 − 𝑋) ∗ 𝑅𝑇
9
2
𝑃𝐵 = 𝐶𝐴0 (𝜃𝐵 − 𝑋) ∗ 𝑅𝑇
𝑃𝐶 = 𝐶𝐴0 (0 + 𝑋) ∗ 𝑅𝑇
𝑃𝐷 = 𝐶𝐴0 (0 + 2𝑋) ∗ 𝑅𝑇
𝑃𝐸 = 𝐶𝐴0 (0 + 2𝑋) ∗ 𝑅𝑇
𝑃𝐼 = 𝐶𝐴0 (𝜃𝐼 ) ∗ 𝑅𝑇
Where T=500 K and R =0.082
𝑃𝑜
(0.035)(10 𝑎𝑡𝑚)
𝐶𝐴0 = 𝑦𝐴0
=
= 0.0085 𝑚𝑜𝑙/𝑑𝑚3
𝑑𝑚3 𝐾
𝑅𝑇
(0.082
) (500 𝐾)
𝑎𝑡𝑚. 𝑚𝑜𝑙
𝑃𝑇 = 𝐶𝐴0 (1 + 𝜃𝐵 + 𝜃𝐼 − 0.5 ∗ 𝑋)𝑅𝑇=0.0085*(1+5.79+21.78–0.5X)*0.082*500
𝑃𝑇 =0.36(28.57–0.5*X)
(3) Rate law
−𝑟𝐴 = 𝑘𝐴 𝐶𝐴2 𝐶𝐵
−𝑟𝐴 = 𝑘𝐴 ∗ (𝐶𝐴0 ∗ (1 − 𝑋))2 ∗ 𝐶𝐴0 ∗ (𝜃𝐵 − 9⁄2 𝑋)
9
3 (1
−𝑟𝐴 = 𝑘𝐴 𝐶𝐴0
− 𝑋)2 (5.79 − 𝑋)
2
(4) Time to achieve 90 % Conversion
0.9 𝑑𝑋
t=(NA0/V)*∫0
−𝑟𝐴
0.9
1
𝑑𝑋
57.5
t=𝐶 2 ∗𝑘 ∫0
9 ==
2 ∗0.01= 921 days
2
(0.0085)
(1−𝑋)
∗(5.79−
𝑋)
𝐴0
𝐴
2
P4-9 (c)
Rate Law:
−𝑟𝐴 = 𝑘𝐴 𝐶𝐴2 𝐶𝐵
Gas Phase Flow System:
𝑣 = 𝑣0 (1 + 𝜀𝑋)
𝑃0 𝑇
𝑃 𝑇0
𝐹𝐴 𝐹𝐴0 (1 − 𝑋) 𝑃0 𝑇
=
𝑣
𝑣0 (1 + 𝜀𝑋) 𝑃 𝑇0
(1 − 𝑋) 𝑃0 𝑇
𝐶𝐴 = 𝐶𝐴0
(1 + 𝜀𝑋) 𝑃 𝑇0
𝑃𝑜
(0.035)(10 𝑎𝑡𝑚)
𝐶𝐴0 = 𝑦𝐴0
=
= 0.0085 𝑚𝑜𝑙/𝑑𝑚3
𝑑𝑚3 𝐾
𝑅𝑇
(0.082
) (500 𝐾)
𝑎𝑡𝑚. 𝑚𝑜𝑙
𝐶𝐴 =
9
2 𝑋) 𝑃0 𝑇
𝐶𝐵 = 𝐶𝐴0
(1 + 𝜀𝑋) 𝑃 𝑇0
(𝜃𝐵 −
𝜖 = 𝑦𝐴0 ∗ 𝛿= (0.035)*(–0.5)=–0.0175
9
(5.79− 𝑋)
2
𝐶𝐵 = 0.0085 (1−0.0175
𝑋)
For isothermal flow system with no pressure drop
9
3 (1
𝑘𝐴 𝐶𝐴0
− 𝑋)2 (5.79 − 2 𝑋)
−𝑟𝐴 =
(1 − 0.0175𝑋)3
4-21
P4-9 (d)
For CSTR,
𝐹
∗𝑋
𝐴0
V= −𝑟
𝐴
At X=0.98, –rA=3.6e–12
V=10*0.98/(3.6e–12)=2.72*1012 dm3
For PFR,
0.98 𝑑𝑋
V= (FA0)*∫0
−𝑟𝐴
𝐹𝐴0
0.98 (1−0.0175∗𝑋)3 𝑑𝑋
10∗29.76
V=
*∫
=
9
0.01∗(0.0085)^3
𝑘∗CA0 3 0
(1−𝑋)2 ∗(5.79− 𝑋)
2
V=4.85*1010 dm3
P4-9 (e) Individualized solution
P4-10
P4-10 (a)
P0 = PA0 = CA0RT
PA = PA0(1–Xe)/(1 + ɛXe)
PB = PA0Xe/2/(1 + ɛXe)
CA = PA / RT
CB = PB / RT
Kc = CB/CA2
CA0 = 4 mol/dm3
Xe = 0.6
1
ɛ =𝑦𝐴0 ∗ 𝛿 = 1 ∗ (2 − 1) = −1/2
Substituting the values of CA0, Xe and ɛ, we get
CA = 4*(1–0.6)/(1–0.5*0.6) =2.28 mol/dm3
CB = 4*(0.6/2)/(1–0.5*0.6)=1.71 mol/dm3
Kc = CB/CA2 = 0.329 dm3/mol
P4-10 (b)
Kc = CB/CA2
CA = CA0(1–Xe)
CB = CA0Xe
CA0 = 4 mol/dm3
Xe = 0.6
Substituting the values of CA0 and Xe, we get
CA = 1.6 mol/dm3
CB = 1.2 mol/dm3
Kc = CB/CA2 = 0.46875 dm3/mol
P4-11
Rate law:
–rA=kA*(CA–Cc^3/Kc)
4-22
P4-11 (a)
Constant Volume:
At equilibrium,
–rA=0, X=Xe, therefore,
𝐶𝑐3
𝐾𝑐 =
𝐶𝐴
For constant volume,
𝐶𝑐 = 3 ∗ 𝐶𝐴0 ∗ 𝑋
𝐶𝐴 = 𝐶𝐴0 ∗ (1 − 𝑋)
𝐶𝐴0 = 𝑃𝐴0 /(𝑅𝑇) =10/(0.082*400)=0.3 mol/dm3
Now, Kc is
2
27 ∗ 𝐶𝐴0
∗ 𝑋3
𝐾𝑐 =
(1 − 𝑋)
Using the value of Kc=0.25, CA0=0.3,
X= Xe= 0.4
P4-11 (b)
Constant Pressure:
𝐶𝑐 = 3 ∗ 𝐶𝐴0 ∗ 𝑋/(1+𝜖𝑋)
𝐶𝐴 = 𝐶𝐴0 ∗ (1 − 𝑋)/(1+𝜖𝑋)
𝜖 = 𝑦𝐴0 ∗ 𝛿
𝑦𝐴0 = 1,
𝛿 = 3−1 = 2
Thus,
𝜖=2
Now, Kc is
2
27 ∗ 𝐶𝐴0
∗ 𝑋3
𝐾𝑐 =
(1 − 𝑋) ∗ (1 + 2𝑋)2
Solving for X,
X=Xe= 0.585
P4-12
Rate law:
–rA=kA*(CA^3–Cc/Kc)
P4-12 (a)
Constant Volume:
At equilibrium,
–rA=0, X=Xe, therefore,
𝐶𝑐
𝐾𝑐 = 3
𝐶𝐴
4-23
For constant volume,
1
𝐶𝑐 = ( ) ∗ 𝐶𝐴0 ∗ 𝑋
3
𝐶𝐴 = 𝐶𝐴0 ∗ (1 − 𝑋)
𝐶𝐴0 = 𝑃𝐴0 /(𝑅𝑇) =10/(0.082*400)=0.3 mol/dm3
Now, Kc is
1
(3) ∗ 0.3 ∗ 𝑋
𝐾𝑐 =
0.33 (1 − 𝑋)3
Using the value of Kc=2.5,
X= Xe= 0.27
So, equilibrium conversion is lower in P4-12B compared to P4-11B
Constant Pressure:
𝐶𝑐 = (1/3) ∗ 𝐶𝐴0 ∗ 𝑋/(1+𝜖𝑋)
𝐶𝐴 = 𝐶𝐴0 ∗ (1 − 𝑋)/(1+𝜖𝑋)
𝜖 = 𝑦𝐴0 ∗ 𝛿
𝑦𝐴0 = 1,
1
𝛿 = − 1 = −2/3
3
Thus,
𝜖 = −2/3
Now, Kc is
1
(3) ∗ 0.3 ∗ 𝑋/(1 − 2/3𝑋)
𝐾𝑐 =
0.33 (1 − 𝑋)3 /(1 − 2/3𝑋)3
Solving for X,
X=Xe= 0.136
P4-13
Given: Gas phase reaction A + B → 8C in a batch reactor fitted with a piston such that V = 0.1P0
( ft )
3
k = 1.0
2
lb mol 2 sec
−rA = kC A2 CB
NA0 = NB0 at t = 0
V0 = 0.15 ft3
T = 140°C = 600°R = Constant
4-24
P4-13 (a)
y A0 =
N A0
= 0.5
N A0 + N B 0
 = 8 −1−1 = 6
 = y A0 = 3
V0 P0
(1 +  X ) and T = 1 , P0 = 10V0 , and P = 10V
T 
T0
P 
 T0 
Now V =
Therefore
V=
10V02
(1 +  X )
10V
V 2 = V02 (1 +  X )
or
N B = N A0  B − X 
N A = N A0 1 − X 
kN A2 N B kN A0 1 − X 
− rA = kC CB =
=
3
V3
V03 (1 +  X )2
Therefore
3
N B0
=1
N A0
y P 
N A0 =  A0 0  V0
 RT 
3
2
A
 y P  1 − X 
−rA = k  A0 0 
3
 RT  (1 +  X )2
3
1 − X  lb mol
−rA = 5.03*10
3
3
(1 + 3 X )2 ft sec
3
−9
P4-13 (b)
V 2 = V02 (1 +  X )
0.22 = 0.152 (1 +  X )
X = 0.259
− rA = 8.63*10−10
B =
lb mol
ft 3 sec
4-25
Page intentionally blank
4-26
Synopsis for Chapter 5 – Isothermal Reactor Design: Conversion
General: This chapter combines all the building blocks to solve isothermal reaction
engineering problems.
Questions
l Q5-1A (19 seconds) Questions Before Reading (QBR).
I Q5-3A (10-60 min) Gives Guidelines and Ground Rules for formulating an original problem on
Chemical Reaction Engineering.
O Q5-4A (15-30 min) Takes 30 minutes if you cook the spaghetti and is an excellent motivational
problem in which a one or more students have actually brought cooked spaghetti to their instructor.
I Q5-9A (5 min) Based on the disparity in the numbers in that one would stop after making ethylene
and sell that instead?
O Q5-10A (5 min) This video is a fun and inventive video on the letters, “CSTR”. In the song the letters
“YMCA” are replaced by “CSTR”.
Computer Simulations and Experiments
l P5-1B (a) – (f) (10-15 min per simulation) Uses Wolfram and ask the students to spend between 10
and 15 minutes max on the simulation and writing conclusions. Also, go to the Chapter Summary
Notes in the left hand column in the Chapter 5 page of the Website there is a number of self-tests
that will help the student get a feel on how they are doing.
Interactive Computer Games (ICG)
l P5-2B (25 min) World Famous Game. This game is based on a real murder case. However, in the
game, the murderer and the victim are chosen randomly each time a student signs on. One of the
most popular problems in the book.
Problems
l P5-3A (8 min) Multiple choice, short and sweet. Always assigned.
l P5-4B (8 min) Multiple choice, short and sweet. Most always assigned.
AA P5-5A (30 min) Excellent problem similar to a California Exam Problem in that it requires an
intermediate calculation to find k.
I P5-6C (40 min) Straight forward calculation and plot that can be obtained using Polymath.
l P5-7B (25 min) Old Exam Question (OEQ). Excellent problem requiring an intermediate calculation
similar to California Exam Questions.
l P5-8B (25 min) Old Exam Question (OEQ). California Professional Engineers Registration Exam
Problem. The problems are necessary to show the student the level of understanding they are
expected to achieve if they want to become a registered professional engineer.
S5-1
AA P5-9A (40 min) Old Exam Question (OEQ). Straight forward calculations using the CRE algorithm.
Shows the effect of temperature on the size of the reactors. For the exam, parts (f) and (g) were
deleted and Part (d) CSTR and PFR reactors in parallel was deleted.
lAA P5-10B (60 min) Old Exam Question (OEQ). Excellent problem. I usually assign all three parts.
Somewhat tricky questions based on real (case study) troubleshooting problems.
lAA P5-11B (25 min) Old Exam Question (OEQ). Straight forward PBR with pressure drop. I always assign
a problem with pressure drop from this group of problems (P5-11B, P5-13B (k), P5-20B,
P5-21B, P5-23B or P5-24B).
AA P5-12B (30 min) Good problem. Leads the student step by step through the CRE algorithm.
O P5-13B (40 min) Both reactor calculation to solve a murder.
For the California Exam Problems below, P5-14B through P5-19B) although 30 minutes for each of the six
problems is allotted, students should strive to solve them in 25 minutes and apply the remaining 5 minutes
to the thermo problems which are usually move difficult.
AA P5-14B (45 min) Straight forward, but uses partial pressures rather than concentration. Many of the
rate laws in Chapter 10 on catalysis will be formulated in terms of partial pressures.
AA P5-15B (25 min) Old Exam Question (OEQ). California Registration Example Problem. I usually assign
2 or 3 California problems from this group of problems (P5-15B through P5-19B).
AA P5-16B (25 min) Old Exam Question (OEQ). California Registration Example Problem. I usually assign
2 or 3 California problems from this group of problems (P5-15B through P5-19B).
AA P5-17B (25 min) Old Exam Question (OEQ). California Registration Example Problem. I usually assign
2 or 3 California problems from this group of problems (P5-15B through P5-19B).
AA P5-18B (25 min) Old Exam Question (OEQ). California Registration Example Problem. I usually assign
2 or 3 California problems from this group of problems (P5-15B through P5-19B).
AA P5-19B (25 min) Old Exam Question (OEQ). California Registration Example Problem. I usually assign
2 or 3 California problems from this group of problems (P5-15B through P5-19B).
AA P5-20B (40 min) Intermediate calculation requires to find reaction rate law parameters and then to
use them in a PBR with pressure drop.
AA P5-21B (60 min) Micro reactor with real reaction data and with pressure drop. Very comprehensive
problem. I always assign a problem with pressure drop from this group of problems (P5-11B, P5-13B
(k), P5-20B, P5-21B, P5-23B or P5-24B).
l P5-22A (5 min) Fun problem, always assigned.
AA P5-23B. (20-25 min) This problem is straight out of a popular TV series “Breaking Bad” an interesting
scenario for students. Leads the student step by step through the algorithm. Uses partial pressure
rather than conversion. Motivational problem on using basic CRE principles to solve crimes.
AA P5-24B (35 min) Old Exam Question (OEQ). PBR micro reactor with DP alternative to P5-21B. I always
assign a problem with pressure drop from this group of problems (P5-11B, P5-14B (k), P5-20B,
P5-21B, P5-23B or P5-24B).
AA P5-25B (30 min) Good problem of PBR with pressure drop. I always assign a problem with pressure
drop from this group of problems (P5-11B, P5-13B (k), P5-20B, P5-21B, P5-23B or P5-23B).
S P5-26B (20 min) Interactive Reactor Laboratory problem of Professor Herz, Department of Chemical
Engineering, University of California, San Diego.
S5-2
Solutions for Chapter 5 – Isothermal Reactor Design: Conversion
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-5
http://umich.edu/~elements/5e/05chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q5-1 Individualized solution.
Q5-2 Individualized solution.
Q5-3 Individualized solution.
Q5-4
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
Boiling Point (°C)
99.2
94.6
88.3
Assume reaction is zero order with respect to spaghetti conversion:
so that
For complete conversion (i.e.: well cooked) CA = 0 at time t.
Therefore
5-1
Time (min)
11
14
?
Now, plot the natural log of the cooking time versus 1/Tb and get a linear relationship. Extrapolation to
Tb = 88.3°C = 361.45 K yields t = 20.4 minutes.
Q5-5 There would be no error as the initial liquid phase concentration remains the same.
Q5-6 For three CSTRs in series,
X = 1 – 1/(1 + 2.167)3 = 0.96
For three CSTRs in parallel with feed equally divided, the conversion will remain same.
This is true for n CSTRs in parallel with feed equally divided.
Therefore, adding a CSTR in series increases the conversion. But, having more reactors may require
more maintenance and may add to operational cost.
Q5-7
X = 0.5, k = 0.311 min-1 = 0.0052 s-1
FA0 = Fc/X = 46.4/0.5 = 92.8 mol/s
V = v0X/(k(1-X))
vA0 = FA0/CA01 = 92.8/16.1 = 5.76 dm3/s
vB0 = vA0
v0 = vA0 + vB0 = 11.52 dm3/s
So, V = 11.52*.5/(0.0052*.5) = 2215.38 dm3 = 585 gal
Q5-8
DP2 = 3DP1/4
Laminar Flow:
𝛼 2 = 16𝛼 1/9
𝛽02 = 16 𝛽01/9 where 𝛽01 = 𝛽0
p = (1-16𝛼 1W/9)1/2
Turbulent Flow:
𝛼 2 = 4𝛼 1/3
𝛽02 = 4 𝛽01/3 where 𝛽01 = 𝛽0
p = (1-4𝛼 1W/3)1/2
5-2
Q5-9 Individualized solution.
Q5-10 Individualized solution.
Q5-11 Individualized solution.
Q5-12 Individualized solution.
Q5-13 Individualized solution.
P5-1 (a) Example 5-3
(i)
By varying the sliders it can be seen that increasing rate constant, temperature and activation
energy increases the conversion for a given volume. T, k1, E influence the rate of reaction which
affects the conversion.
(ii) The rate constant k and temperature have the greatest effects on the conversion of the plug flow
reactor.
P5-1 (b) Example 5-4
(i)
After varying each of the parameters in the Ergun equation, it is found Ac(area of catalyst) has the
most effect on pressure drop. By varying Ac, one can vary the G term that appears in the equation
for beta which further has an influence in the pressure vs z equation.
(ii) 1. We can conclude that for decreasing the pressure drop, one needs to take utmost care of the
area of catalyst that one chooses. Changing viscosity has the least effect on pressure drop and
hence can be get rid of.
2. The other parameters that significantly affect the pressure drop are the mass flow rate and
particle diameter. So that’s need to be taken care of as well while concluding about the
pressure drop.
P5-1 (c) Example 5-5
(i)
It can be seen that conversion profile changes by varying parameters k, Cao and Vo whereas
pressure drop depends upon the parameter alpha.
1. Increasing k and Cao increases the conversion for a given weight and vice-versa for the case of
volumetric flow rate.
2. Increasing alpha also increases the pressure drop. This is evident by seeing the differential
equations and the explicit equations on the next page
5-3
(ii)
(iii)
We observe that all 3 of concentration curves start to decrease as we approach the catalyst
weight 25kg. This is because as we approach the end of reactor, conversion becomes almost
constant and pressure drops rapidly. Cb and Cc both being directly dependent on conversion and
pressure also act in the same way and starts to fall.
The entering volumetric flow rate has the greatest effect on conversion whereas pressure drop is
just a function of alpha.
P5-1 (d) Example 5-6
(i)
1. We can observe that the terms affecting pressure drop mostly are void fraction, diameter of
particle, initial pressure and area of catalyst
2. The terms mostly affecting the conversion profile are rate constant, volumetric flow rate and
initial concentration
(ii) Left hand graph shows base case. Increased the pressure to 5 times (5005 kpa) and decreased
diameter of particle by 5 times (0.0012) results in the graph shown on right side.
The required catalyst weight for new case to achieve 70% conversion is only 15 kg.
(iii)
One can conclude from the above example that depending on whether the flow is turbulent or
laminar or no pressure drop, there is a different conversion. The first term in the Betao expression
corresponds to the laminar flow while the next term is for turbulent one.
P5-1 (e) Example 5-7
(i)
X increases and f decreases with an increase in k’ for the same W, while p remains unchanged, and
vice versa. This is an expected observation, because as k’ increases, the rate of the reaction
increases, and naturally, X increases.
(ii) 1. As α and FA0 increase, X and p decrease for the same W, while f increases.
2. Since the conversion down the reactor increases, more product is formed, lesser amount of
reactants are present. Since ɛ < 0, the volumetric flow rate decreases with an increase in
conversion.
(iii) Individualized solution
(iv) Individualized solution
P5-1 (f) Example 5-8
1. The major effect on the pressure drop curve is of parameter alpha. It is obvious as the
differential equation involving the pressure drop is directly related to alpha.
2. Conversion rises with an increase of rate constant k and Kc while falls with an increase of Cao,
Yao and Vo. Volumetric flow rate(Vo) has the greatest effect on the conversion profile. Vo
5-4
appears in the denominator term of the differential equation for conversion. As we can see it
is inversely related to conversion. Fao = Cao*Vo
3. Rate profile rises with an increase of k, Cao, Kc, Vo but falls with an increase of Yao.
As we can rate is directly proportional to k, the expressions of Ca, Cb, Cc itself contains the terms of Cao
and Vo. Also increasing Kc reduces the negative term, hence increasing rate.
P5-1 (g) Example 5-3, Using ASPEN, we get
(1) At 1000K, for the same PFR volume we get only 6.2% conversion. While at 1200K, we get a
conversion of nearly 100%. This is because the value of reaction constant ‘k’ varies rapidly
with reaction temperature.
(2) Earlier for an activation energy of 82 kcal/mol we got approx. 81% conversion. For activation
energy of 74 kcal/mol keeping the PFR volume the same we get a conversion of 71.1%. While
for an activation energy of 90 kcal/mol we get a conversion of 89.93%.
(3) On doubling both flow rate and pressure we find that the conversion remains the same.
P5-1 (h) Individualized solution.
P5-1 (i) Individualized solution
P5-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer
Games (ICGs) is given at the front of this Solutions Manual.
P5-3 (a) (3) 50% (Liquid phase reactions do not depend on pressure)
P5-3 (b) (2) <50% (Low concentration if we lower the pressure)
,
low rate
P5-3 (c) (1) >0.234
2
−rA = kC 2A0 1− X p
(
)
−rA =∝ kp2
p2 < 1
ktrue p2 = k = 0.234
0.234
ktrue =
> 0.234
p2
5-5
P5-4 (a)
Ans. (1) X > 0.5
1
α~
D
Increase D, decrease 𝛼, increase X.
P5-4 (b)
Ans. (3) Xe = 0.75
P5-4 (c)
Ans. (3) remain the same
Kc =
CA
C Ae
=
(
)
C A0 1− Xe p 1− X
=
C A0 Xe p
Xe
Xe is not a function of reactor diameter
P5-4 (d)
Ans. (3) remain the same
Xe is not a function of particle size
P5-4 (e)
Ans. (4) insufficient information to tell
P5-5
r = - k C A2
Thus, k = 5
For a CSTR,
P5-6
To = 300K
KCO (300K)= 3.0 V = 1000gal = 3785.4 dm3
5-6
Mole balance:
Rate law:
Stoichiometry:
and
𝑉=
𝐹'(
,
𝑘* 𝐶'*
.(1
𝑋
𝑋,
− 𝑋), − 𝐾 5
Z = 2902.2 dm3, Let z =
4
( "
+
k
E 1 1%
= exp* $$ − ''*) R # T0 T &-,
ko
Now using:
where
and
Solving using polymath to get a table of values of X Vs T. See the following Polymath program
Polymath Code
f(X) = (Z/z)*X/((1-X)^2 - X^2/Kc) -V
To = 300
T = 305.5
Z = 2902.2
V = 3785.4
E = 15000
R=2
z = exp(E/R*(1/To-1/T))
Kco = 3
Hrx = -25000
Kc = Kco*exp(Hrx/R*(1/To-1/T))
X(0) = 0.5
X(min) = 0
X(max) = 1
Polymath Output
Calculated values of NLE variables
Variable Value
1 X
f(x)
Initial Guess
0.4229453 3.638E-12 0.5 ( 0 < X < 1. )
Variable Value
1
E
1.5E+04
2
Hrx
-2.5E+04
3
Kc
1.416906
5-7
4
Kco
3.
5
R
2.
6
T
305.5
7
To
300.
8
V
3785.4
9
z
1.568441
10 Z
2902.2
Nonlinear equations
1 f(X) = (Z/z)*X/((1-X)^2 - X^2/Kc) -V = 0
Explicit equations
1 To = 300
2 T = 305.5
3 Z = 2902.2
4 V = 3785.4
5 E = 15000
6 R=2
7 z = exp(E/R*(1/To-1/T))
8 Kco = 3
9 Hrx = -25000
10 Kc = Kco*exp(Hrx/R*(1/To-1/T))
T(in K)
X
300
301
303
304
305
305.5
305.9
307
310
315
0.40
0.4075
0.4182
0.4213
0.4228
0.4229
0.4227
0.421
0.4072
0.3635
We get maximum X = 0.4229 at T = 305.5 K.
P5-7
PFR
5-8
CSTR
The exact value of activation energy comes out to be 8002 cal/mol.
P5-8
Base Case PFR above. Find k
67
9:
Mole Balance 68 = < ;
;=
Rate Law
Stoichiometry
Combine
CSTR added upstream of PFR
5-9
X1 = 0.5
Now find X2
P5-9 (a)
Using the Arrhenius equation at the CSTR temperature of 300 K yields the new specific reaction rate.
The blades makes two equal volumes zones of 500gal each rather than one ‘big’ mixing zone of 1000gal.
So, we get 0.57 as conversion instead of 0.5.
5-10
X=0.85
So, considering the above results, we will choose a CSTR.
P5-9 (b)
P5-9 (c)
P5-9 (d)
1) CSTR and PFR are connected in series:
5-11
Solving the quadratic equation, XCSTR = 0.44
For PFR,
𝑑𝑋 =
𝑑𝑚 D
, (
?0.07 𝑚𝑜𝑙 𝑚𝑖𝑛I 𝐶'(
1 − 𝑋),
10 𝑚𝑜𝑙/𝑚𝑖𝑛
𝑑𝑉
X = 0.865
2) when CSTR and PFR are connected in parallel,
XCSTR = 0.56
For PFR,
XPFR = 0.92
Hence, final conversion X =
= 0.74
P5-9 (e)
To process the same amount of species A, the batch reactor must handle
If the reactants are in the same concentrations as in the flow reactors, then
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.
, and since
5-12
Assume that it takes three hours to fill, empty, and heat to the reaction temperature.
tf = 3 hours
ttotal = tR + tf
ttotal = 2.14hours + 3 hours = 5.14 hours.
Therefore, we can run 4 batches in a day and the necessary reactor volume is
Referring to Table 1-1 and noting that 3600 dm3 is about 1000 gallons, we see that the price would be
approximately $85,000 for the reactor.
P5-9 (f)
The points of the problem are:
1) To note the significant differences in processing times at different temperature (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.
P5-9 (g)
Individualized solution
P5-10 (a)
The blades makes two equal volumes zones of 500gal each rather than one ‘big’ mixing zone of 1000gal.
So, we get 0.57 as conversion instead of 0.5.
5-13
P5-10 (b)
CAO = 2 mol/dm3
A®B
Assuming 1st order reaction,
For CSTR,
–rA = kCA = kCAO(1-X)
=>
For PFR,
,
=>
= 1 – exp(-0.67) = 0.486
Now assuming 2nd order reaction,
For CSTR,
Now, assuming 2nd order reaction,
For CSTR,
=
=>
For PFR,
=>
So, while calculating PFR conversion they considered reaction to be 1st order. But actually it is a second
order reaction.
P5-10 (c)
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 C, which have same conversion as particle size at A.
But when we decrease the particle size by 20%, we reach at point D, so a decrease in conversion is
5-14
noticed. Also, when we increase the particle size from position A, we reach at point B, again there is a
decrease in the conversion.
P5-11
Reaction:
A+B®C+D
Mole Balance:
dX
= −rA"
dW
Rate Law:
FA0
−rA" = KCACB
Stoichiometry:
𝐹' 𝐹'( (1 − 𝑋)
𝐶' =
=
= 𝐶'( (1 − 𝑋)𝑝
𝑣
𝑣
𝑃
𝑝=
= (1 − 𝛼𝑊 )O/,
𝑃(
𝐶' = 𝐶'( (1 − 𝑋)(1 − 𝛼𝑊 )O/,
( )(
Equimolar flow rate: CB = CA = CA0 1− X 1− αW
12
)
Combine:
−rA" = kC2A0 1− X
(
2
) (1− αW)
2
2
dX kCA0 (1− X) (1− αW)
=
dW
FA0
Evaluate:
"
%
" αW %
X $ FA0 '
= W $1−
'
1− X $ kC2 '
2 &
#
# A0 &
5-15
Solution:
First, we must solve for k*W:
Replacing alpha*w = 1-(P/Po)^2 = 0.99
And putting values of X=0.8, Fao=10 mol/min and Cao = 0.4mol/dm^3
We get k*W = [(0.8/1-0.8)*(10/0.4^2)] / (1-0.99/2) = 495.04
Now we can calculate the effect of increasing particle size: d2= 2*d1 and since alpha is inversely
proportional to diameter, we can say that alpha2=0.5 * alpha1
Solving for X with k*W = 495.04 and alpha2*W = 0.495
"
%
" αW %
X $ FA0 '
= W $1−
'
1− X $ kC2 '
2 &
#
# A0 &
X = [(0.4^2/10)*495.04*(1-0.495/2)] / {1+ [(0.4^2/10)*495.04*(1-0.495/2)] } = 0.856
Hence conversion is 85.6%
P5-12
P5-12 (a)
5-16
P5-12 (b)
P5-12 (c)
Note : The value substituted for 𝐹'( should be 200 but it is currently written as 500. This affects final
answer, which should be 110.94 dm3
P5-12 (d)
The conversion will remain unchanged since the reactor is well-mixed ideal CSTR.
P5-13 (a)
V = 5 dm3
Mass of one capsule = 18 g
M.wt. of Iocane = 56.25 g/mol
OP
Moles of Iocane in one capsule (NA0) = QR.,Q = 0.32 mol
CA0 = NA0 / V = 0.064 mol/dm3
5-17
P5-13 (b)
For a batch reactor,
Solving the differential equation, we get:
𝑑𝐶'
= 𝑟'
𝑑𝑡
𝑑𝐶'
= −𝑘' 𝐶'
𝑑𝑡
𝐶'
= 𝑘' 𝑡
𝐶'(
For concentration of Optoid to be 0.01 mol/dm3, CA = 0.064 – 0.01 = 0.054 mol/dm3
Substituting the values of kA, CA and CA0, we get t = 18 hr
ln
So, Ambercromby should arrest Ms. Patel.
P5-13 (c)
𝑘'
𝐸 1 1
= exp Z− ? − I^
𝑘'(
𝑅 𝑇 𝑇(
Substituting values of kA0 = 0.00944 hr-1, E = 20500 cal/mol, R = 1.986 cal/mol K, T = 311.7 K and T0 = 310
K, we get:
kA = 0.01131 hr-1
𝐶'
= 𝑘' 𝑡
𝐶'(
For concentration of Optoid to be 0.01 mol/dm3, CA = 0.064 – 0.01 = 0.054 mol/dm3
Substituting the values of kA, CA and CA0, we get t = 15 hr
ln
If Shoemaker had a fever, then Ambercromby arrested the wrong suspect. He should arrest Mr. Jenkins.
P5-14 (a)
Polymath Code
d(x)/d(w) = -rBu / FBuO
FBuO = 50
PBuO = 10
PBu = PBuO * (1 - x) / (1 + x)
k = 0.054
K = 0.32
rBu = -k * PBu / (1 + K * PBu) ^ 2
w(0) = 0
w(f) = 2000
x(0)=0
Polymath Output
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1 w
0
0
2000.
2000.
2 x
0
0
0.9978192
0.9978192
3 FBuO
50.
50.
50.
50.
4 PBuO
10.
10.
10.
10.
5 PBu
10.
0.0109159
10.
0.0109159
6 k
0.054
0.054
0.054
0.054
7 K
0.32
0.32
0.32
0.32
8 rBu
-0.0306122
-0.0421872
-0.0005854
-0.0005854
5-18
Differential equations
1 d(x)/d(w) = -rBu / FBuO
Kgcat-1
Explicit equations
1 FBuO = 50
Kmol.hr-1
2 PBuO = 10
atm
3 PBu = PBuO * (1 - x) / (1 + x)
atm
4 k = 0.054
kmol.kgcat-1.hr-1.atm-1
5 K = 0.32
atm-1
6 rBu = -k * PBu / (1 + K * PBu) ^ 2
Kmol.kgcat-1.hr-1
Hence, weight of catalyst required for 80% conversion is 1054.1 kg.
5-19
P5-14 (b)
Differential Mole Balance:
FA0 =
PBu = PBuO
Here,
Rate equation:
dX
"
= −rBu
dW
–rBuʹ = K*PBu
For Fluidized CSTR:
Therefore,
w = FBuo*X
Given,
X = 0.8
FBuO = 50 Kmol.hr-1
KBu = 0.32 atm-1
PBuO = 10 atm
Therefore, putting the values in the above equation
w = 50*0.8
= 1225 Kg
Therefore, 1225 Kg of fluidized CSTR catalyst weight is required to achieve 80% conversion.
P5-14 (c) With pressure drop parameter alpha = 0.0006 kg-1, we will add pressure drop equation in
terms of p (p=P/P0)
Polymath Code:
d(x)/d(w) = -rBu / FBuO
d(p)/d(w) = -alpha * (1 + x) / 2 / p
FBuO = 50
f = (1 + x) / p
PBuO = 10
PBu = PBuO * (1 - x) / (1 + x) * p
k = 0.054
K = 0.32
rBu = -k * PBu / (1 + K * PBu) ^ 2
alpha = 0.0006
w(0) = 0
w(f) = 1100
x(0)=0
p(0)=1
Polymath Output
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.0006
0.0006
0.0006
0.0006
2
f
1.
1.
6.600157
6.600157
3
FBuO
50.
50.
50.
50.
4
K
0.32
0.32
0.32
0.32
5
k
0.054
0.054
0.054
0.054
5-20
6
p
1.
0.2689385
1.
0.2689385
7
PBu
10.
0.3408456
10.
0.3408456
8
PBuO
10.
10.
10.
10.
9
rBu
-0.0306122
-0.0421871
-0.0149635
-0.0149635
10 w
0
0
1100.
1100.
11 x
0
0
0.7750365
0.7750365
Differential equations
1 d(x)/d(w) = -rBu / FBuO
2 d(p)/d(w) = -alpha * (1 + x) / 2 / p
Explicit equations
1 FBuO = 50
2 f = (1 + x) / p
3 PBuO = 10
4 PBu = PBuO * (1 - x) / (1 + x) * p
5 k = 0.054
6 K = 0.32
7 rBu = -k * PBu / (1 + K * PBu) ^ 2
8 alpha = 0.0006
5-21
So, we see the maximum rate in case with pressure drop is at catalyst weight equal to around 600 Kg.
To achieve 70% conversion, catalyst weight required is 932.3 kg .
In case of (a), 915.5 kg of catalyst is required to achieve 70% conversion.
P5-14 (d) Individualized solution
P5-14 (e) Individualized solution
P5-15
Gaseous reactant in a tubular reactor: A → B
5-22
For a plug flow reactor:
At T2 = 260°F = 720°R, with k1 = 0.0015 min-1 at T1 = 80°F = 540°R,
Therefore 14 pipes are necessary.
P5-16 (a)
A → B/2
5-23
Combining
(for the integration, refer to Appendix A)
from the Ideal Gas assumption.
Substituting Eqn. (5), X = 0.8 and ε = –1/4 to Eqn. (4) yields,
……….(6)
P5-16 (b) Individualized solution.
P5-17 (a)
Given: The metal catalyzed isomerization
liquid phase reaction
with Keq = 5.8
For a plug flow reactor with yA = 1.0, X1 = 0.55
Case 1: an identical plug flow reactor connected in series with the original reactor.
Since yA = 1.0, ÈB = 0. For a liquid phase reaction
5-24
and
For the first reactor,
or
Take advantage of the fact that two PFR’s in series is the same as one PFR with the volume of the two
combined.
VF = V1 + V2 = 2V1 and at VF, X = X2
X2 = 0.74
P5-17 (b)
Case 2: Products from 1st reactor are separated and pure A is fed to the second reactor,
5-25
The analysis for the first reactor is the same as for case 1.
By performing a material balance on the separator, FA0,2 = FA0(1-X1)
Since pure A enters both the first and second reactor CA0,2 = CA0, CB0,2 = 0, ÈB = 0
for the second reactor.
and since V1 = V2
or
Overall conversion for this scheme:
P5-17 (c) Individualized Solution
P5-18
Given: Meta- to ortho- and para- isomerization of xylene.
(neglect)
5-26
Pressure = 300 psig
T = 750°F
V = 1000 ft3 cat.
Assume that the reactions are irreversible and first order.
Then:
Check to see what type of reactor is being used.
Case 1:
Case 2:
Assume plug flow reactor conditions:
or
CM0, k, and V should be the same for Case 1 and Case 2.
Therefore,
The reactor appears to be plug flow since (kV)Case 1 = (kV)Case 2
As a check, assume the reactor is a CSTR.
or
Again kV should be the same for both Case 1 and Case 2.
5-27
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.
For the new plant, with v0 = 5500 gal / hr, XF = 0.46, the required catalyst volume is:
This assumes that the same hydrodynamic conditions are present in the new reactor as in the old.
P5-19
A→ B in a tubular reactor
Tube dimensions: L = 40 ft, D = 0.75 in.
nt = 50
with
or
5-28
Assume Arrhenius equation applies to the rate constant.
At T1 = 600°R, k1 = 0.00152
At T2 = 760°R, k2 = 0.0740
so
From above we have
so
Dividing both sides by T gives:
Evaluating and simplifying gives:
Solving for T gives:
T = 738°R = 278°F
5-29
P5-20
5-30
P5-20 (d)
P5-20 (e)
P5-21
Production of phosgene in a micro reactor
CO + Cl2
COCl2 (Gas phase reaction)
A + B
C
The equations that need to be solved are as follows:
d(X)/d(W) = -rA/FA0
d(p)/d(W) = -α*(1+Ɛ*X)/(2*p)
FB0 = FA0;
Fb = FB0-FA0*X;
Fc = FA0*X
p= P/Po
See the following Polymath program.
5-31
Polymath code:
d(X) / d(W) = -rA/FA0
d(p) / d(W) = -a*(1+e*X)/(2*p)
e = -0.5
FA0 = 2E-05
FB0 = FA0
Fa = FA0*(1-X)
Fb = FB0*(1-X)
v0 = 2.83E-07
v = v0*(1+e*X)/p
Fc = FA0*X
Ca = Fa/v
Cb = Fb/v
a = 3.55E05
k = 0.004
rA = -k*Ca*Cb
Cc = Fc/v
X(0) = 0
p(0) = 1
W(0) = 0
W(f) = 3.5E-06
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
a
3.55E+05
3.55E+05
3.55E+05
3.55E+05
2
Ca
70.67138
9.163871
70.67138
9.163871
3
Cb
70.67138
9.163871
70.67138
9.163871
4
Cc
0
0
53.53098
33.25957
5
e
-0.5
-0.5
-0.5
-0.5
6
Fa
2.0E-05
4.32E-06
2.0E-05
4.32E-06
7
FA0
2.0E-05
2.0E-05
2.0E-05
2.0E-05
8
Fb
2.0E-05
4.32E-06
2.0E-05
4.32E-06
9
FB0
2.0E-05
2.0E-05
2.0E-05
2.0E-05
10 Fc
0
0
1.568E-05
1.568E-05
11 k
0.004
0.004
0.004
0.004
12 p
1.
0.3649802
1.
0.3649802
13 rA
-19.97777
-19.97777
-0.3359061
-0.3359061
14 v
2.83E-07
2.444E-07
4.714E-07
4.714E-07
15 v0
2.83E-07
2.83E-07
2.83E-07
2.83E-07
16 W
0
0
3.5E-06
3.5E-06
17 X
0
0
0.7839904
0.7839904
Differential equations
1 d(X)/d(W) = -rA/FA0
2 d(p)/d(W) = -a*(1+e*X)/(2*p)
Explicit equations
1 e = -0.5
2 FA0 = 2E-05
3 FB0 = FA0
4 Fa = FA0*(1-X)
5 Fb = FB0*(1-X)
6 v0 = 2.83E-07
7 v = v0*(1+e*X)/p
8 Fc = FA0*X
9 Ca = Fa/v
10 Cb = Fb/v
11 a = 3.55E05
5-32
12 k = 0.004
13 rA = -k*Ca*Cb
14 Cc = Fc/v
P5-21 (a)
P5-21 (b)
The outlet conversion of the reactor is 0.784
The yield is then MW*FA*X = 99 g/mol * 2 e-5 mol/s * 0.784 = .00155 g/s = 48.95 g/ year.
Therefore 10,000 kg/year / 48.95 kg/ year = 204 reactors are needed.
P5-21 (c)
Assuming laminar flow, α ~ Dp-2, therefore
P5-21 (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.
P5-21 (e) Individualized solution
P5-21 (f) Individualized solution
P5-21 (g) Individualized solution
5-33
P5-22 Individualized solution
P5-23 (a)
P5-23 (b)
Elementary with respect to partial pressure, so the rate law is:
P5-23 (c)
Feed is pure A, so
Therefore
There is no change in P or T, and
= 0, so
Stoichiometric Table:
Species
Methylamine
Dimethylamine
Ammonia
Symbol
A
B
C
Initial
FA0
0
0
P5-23 (d)
By ideal gas law & Stoichiometric table:
P5-23 (e)
Substituting for the partial pressures:
5-34
Change
- FA0X
0.5 FA0X
0.5 FA0X
Remaining
FA0(1-X)
0.5 FA0X
0.5 FA0X
Finally,
P5-23 (f)
At equilibrium,
, so
Therefore:
P5-23 (g)
Mole Balance for PBR:
Rate Law:
Stoichiometry:
Feed is pure A, so
& A is the limiting reactant;
Therefore
There is no change in P or T, so
;
Therefore:
Combine:
5-35
Separate and integrate:
Let’s look at the left hand side first:
Notice that the equation is now in quadratic form, where
The solution to the integral is, from Appendix A (equation A-12):
Where p and q are the roots of the quadratic equation:
p, q =
So
Now we move the constants from the right hand side to the left:
So
(*)
P5-23 (h)
We want the catalyst weight at X = 0.9 XeXe = 0.684
Using equation (*), we have:
P5-23 (i)
For a fluidized bed reactor (CSTR), when X = 0. 9 XeXe = 0.684:
5-36
We can put smaller CSTRs in series to reduce the total catalyst weight.
P5-23 (j)
𝜀=0
`
𝑝 = ` = (1 − 𝛼𝑊)(.Q
=
It can be easily seen that
Pressure fall below 1 atm:
P5-23 (k)
Pressure fall below 1 atm:
5-37
P5-24
1. Mole Balance – Use Differential Form
dX −rA
=
dW FA0
2. Rate Law – Pseudo Zero Order in B
−rA = kCACB0 = kCA
" T %" P %
3. Stoichiometry – Gas: v = v0 1+εX $$ ''$$ 0 '' Isothermal, therefore T = T0
# T0 &# P &
(1 − 𝑋)
𝐹' 𝐹'( (1 − 𝑋) 𝑃
𝑃
𝐶' =
=
= 𝐶'(
𝑝, 𝑝 =
(1 + 𝜀𝑋)
𝑣
𝑣( (1 + 𝜀𝑋) 𝑃(
𝑃(
(
)
1
𝜀 = 𝑦'( 𝛿 = ? I (1 + 1 − 1 − 1) = 0
2
𝐶' = 𝐶'( (1 − 𝑋)𝑝
6f
6g
=−
h
,f
where W = 0 then p = 1 and X = 0
Integrating
(
p = 1− αW
12
)
P5-24 (a)
Now since P/P0 >0.1 .
So, p= .1 for minimum value of W
Now (1 – αW)1/2 = .1 => W = 99 kg
P5-24 (b)
4. Combine
( )
( )(
12
dX kCA0
=
1− X) (1− αW)
(
dW F
−rA = kCA = kCA0 1− X p = kCA0 1− X 1− αW
A0
5-38
12
)
Separate Variables
when W = 0, X = 0
2𝑘
2 ∗ 1.3
=
= 3.47
3𝛼𝑣( 3 ∗ 0.01 ∗ 25
1
𝑙𝑛
= 3.47m1 − (1 − 𝛼𝑊 )D/, n
1−𝑋
𝑝 = (1 − 𝛼𝑊)O/,
𝑃(
𝑣 = 𝑣(
𝑃
𝑣
= 1/𝑝
𝑣(
W
0
1
2.5
5
15
25
50
75
90
99
p
1
1
0.99
0.975
0.92
0.87
0.71
0.5
0.32
0.1
1/p
1
1
1.01
1.02
1.08
1.16
1.41
2.0
3.16
10
X = 0.9
2.3
= 1 − (1 − 𝛼𝑊 )D/,
3.47
W = 51.5 kg
P5-24 (c) Polymath
𝑑𝑋 −𝑟𝐴𝑝𝑟𝑖𝑚𝑒
=
𝑑𝑊
𝐹'(
5-39
X
0
0.47
0.11
0.21
0.5
0.67
0.87
0.94
0.955
0.96
P
10 atm
1 atm
𝑑𝑝
−𝛼
=
𝑑𝑊 2𝑝
𝑝 = 𝑃/𝑃(
𝐶' = 𝐶'( (1 − 𝑋)𝑝
𝑑𝑚D
𝑘 = 1.3
𝑘𝑔. 𝑐𝑎𝑡. 𝑚𝑖𝑛
See the following Polymath program:
Polymath code:
d(X) / d(W) = -raprime/Fao
d(p) / d(W) = -alpha/2/p
k = 1.3
Cao = 10/2/(118+273)/0.082
Ca = Cao*(1-X)*p
alpha = 0.01
raprime = -k*Ca
f = 1/p
vo = 25
Fao = vo*Cao
P = 10*p
X(0) = 0
p(0) = 1
W(0) = 0
W(f) = 99
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.01
0.01
0.01
0.01
2
Ca
0.1559479
0.0004886
0.1559479
0.0004886
3
Cao
0.1559479
0.1559479
0.1559479
0.1559479
4
f
1.
1.
9.999994
9.999994
5
Fao
3.898696
3.898696
3.898696
3.898696
6
k
1.3
1.3
1.3
1.3
7
p
1.
0.1000001
1.
0.1000001
8
P
10.
1.000001
10.
1.000001
9
raprime
-0.2027322
-0.2027322
-0.0006351
-0.0006351
10 vo
25.
25.
25.
25.
11 W
0
0
99.
99.
12 X
0
0
0.9686707
0.9686707
Differential equations
1 d(X)/d(W) = -raprime/Fao
2 d(p)/d(W) = -alpha/2/p
Explicit equations
1 k = 1.3
2 Cao = 10/2/(118+273)/0.082
3 Ca = Cao*(1-X)*p
4 alpha = 0.01
5 raprime = -k*Ca
6 f = 1/p
7 vo = 25
8 Fao = vo*Cao
9 P = 10*p
5-40
p
P5-24 (d)
for first 5% conversion, W = 1.0 kg
for 85 % conversion, W= 41.2 kg
for last 5% (85 to 90%) W = 51.5 – 41.2 = 10.3 kg
Ratio = 10.3/1 = 10.3
5-41
P5-25
δ = 0 ","ε = 0
(
∴p = 1− αW
12
)
Mole Balance/Design Equation
Rate Law
Stoichiometry
(
)
CA = CA0 1− X p
Combining
2
dX
FA0
= kC2A0 1− X p2
dW
( )
P5-26 Individualized solution
5-42
Synopsis for Chapter 6 – Isothermal Reactor Design:
Moles and Molar Flow rates
General: Uses molar flow rates and concentation rather than converison to solve CRE problem. Polymath
is necessary for all problems except P6-2, P6-3, P6-10 and P6-13.
Questions
l Q6-1A (20 seconds) Questions Before Reading (QBR).
O Q6-2A (15 min) Again, I always ask the students to go through a number of i>Clicker question in
preparation for their exam and also for the Student Chapter Jeopardy Competition at the Annual
AIChE meeting.
O Q6-4A (5 min) This question requires a little time, but it requires a lot of thought.
I Q6-5B (5 min) This question ties safety to CRE.
O Q6-7A (8 min) Screencasts to reinforce concepts of this chapter.
Computer Simulations and Experiments
l P6-1B Parts (a) through (f) (10-15 min per simulation) I always assign (a) – (f) for the students to
explore each simulation using Wolfram and then write a set of conclusions. Part (b) the membrane
reactor both Wolfram and Python should be allotted 15-20 minutes.
O P6-1B Part (g) (25 min) This module problem is interactive, but is a little more or less user-friendly
than the other Computer Simulations and Experiments problems because the student has to copy
and paste the code into Polymath and then vary the parameter k using Polymath.
AA P6-1B Part (h) (45 min) This module is graduate level and the model was at the forefront when the
manufacture of nanoparticles was a hot research area.
Interactive Computer Games (ICG)
l P6-2B (15-18 min) Tic-Tac-Toe Answer question correctly and you get an “X”. However, if an
incorrect answer is given a “O” goes in the square.
Problems
O P6-3C (30 min) Old Exam Question (OEQ). Conceptual problem. Many of the students initially have
difficulty with this problem, but once they see it, it takes a very, very short time to calculate.
AA P6-4B (30 min) Fairly straight forward problem to calculate molar flow rates in a PFR and a CSTR.
O P6-5B (25 min) Continuation of P6-4B, but this time in a membrane reactor.
l P6-6C (40 min) Old Exam Question (OEQ). Membrane reactor.
S P6-7B (50 min) Real data for membrane reactor fuel cell.
O P6-8C (50 min) Old Exam Question (OEQ). Rather difficult semibatch reactor problem because CO2
leaves the reactor and needs to be accounted for in the mole balance.
AA P6-9C (50 min) Straight forward semibatch reactor problem.
S6-1
S P6-10B (10-30 min) Interactive Reactor Laboratory problem of Professor Herz, Department of
Chemical Engineering, University of California, San Diego.
l P6-11B (30 min) Old Exam Question (OEQ). Straight forward semibatch reactor problem.
I P6-12C (30min) Safety problem related to and a precursor to Problem P11-3.
S P6-13C (30 min) Old Exam Question (OEQ). Difficult problem, but all of Professor Shankar’s (IIT
Bombay) problems are difficult.
S6-2
Solutions for Chapter 6 – Isothermal Reactor Design:
Moles and Molar Flow Rates
Useful Links:
Click on the link given below to download Wolfram/python codes for Chapter 6:
http://www.umich.edu/~elements/5e/06chap/live.html
Click on the link given below to view Wolfram tutorial (for running Wolfram codes):
http://www.umich.edu/~elements/5e/tutorials/Wolfram_LEP_tutorial.pdf
Click on the link given below to view Polymath tutorial (for running Polymath codes):
http://www.umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q6-1 Individualized solution.
Q6-2 Individualized solution.
Q6-3 Individualized solution.
Q6-4 The mole balances and rates remain the same. Only the stoichiometry changes.
(a) For a constant volume reactor, the pressure changes as the number of moles of gases changes
which is the same as what is shown in Table 6.2
(b) For a variable volume reactor, the pressure can remain constant. Therefore, p (P/Po) = 1.
Q6-5 Some of the similarities and differences between the BowTie Diagram, the Safety Analysis of the
Incident and the Swiss Cheese model are listed below:
Similarities: Help in identifying the hazard, mitigating actions etc., All the three show us which layers of
protection need to fail for an accident to occur
Differences: Swiss Cheese model does not indicate the initiating event. Safety Analysis lacks the visual
representation (in terms of pictures) that the other two methods provide.
Q6-6 Individualized solution.
Q6-7 (a) Individualized solution.
(b) Removing hydrogen from the membrane reactor helps in producing more products as
equilibrium is never reached as we keep removing part of the products.
(c) Individualized solution.
6-1
P6-1 (a) Example 6-1
(i) The conversion increases with an increase in both E and T. This is because the rate constant is higher
for higher E and T.
1. Setting E at its max and T at its min value makes the conversion almost negligible for all value of
Volume whereas in the opposite case conversion reaches up to 0.5.
2. Setting both E and T at their max values boosts the rate constant so much that conversion
reaches 1 rapidly and setting at their min values puts conversion to 0 for all volume.
(ii) See the following Polymath program
Polymath code:
d(Fa)/d(V) = ra #
d(Fb)/d(V) = rb #
d(Fc)/d(V) = rc #
Cto = 1641/8.314/T #
k = 0.29*exp(E/1.987*(1/500-1/T))
E = 24000 #
ra = -k*(Ca^2- Cb^2*Cc/ Kc) #
rb = -ra #
rc = -ra/2 #
vo = Fao/Cto #
Fao = 0.0000226 #
Tau = V/vo #
rateA = -ra #
X = 1-Fa/Fao #
T = 698 #
Kc = 0.02 #
Ca = Cto*Fa/Ft #
Cb = Cto*Fb/Ft #
Cc = Cto*Fc/Ft #
Ft = Fa+Fb+Fc #
V(0)=0
Fa(0)=0.0000226
Fb(0)=0
Fc(0)=0
V(f)=0.00001
#
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.2827764
0.1426944
0.2827764
0.1426944
2
Cb
0
0
0.093388
0.093388
3
Cc
0
0
0.046694
0.046694
4
Cto
0.2827764
0.2827764
0.2827764
0.2827764
5
E
2.4E+04
2.4E+04
2.4E+04
2.4E+04
6
Fa
2.26E-05
1.366E-05
2.26E-05
1.366E-05
7
Fao
2.26E-05
2.26E-05
2.26E-05
2.26E-05
8
Fb
0
0
8.94E-06
8.94E-06
9
Fc
0
0
4.47E-06
4.47E-06
10 Ft
2.26E-05
2.26E-05
2.707E-05
2.707E-05
11 k
274.4284
274.4284
274.4284
274.4284
12 Kc
0.02
0.02
0.02
0.02
13 ra
-21.94397
-21.94397
-1.497E-10
-1.497E-10
14 rateA
21.94397
1.497E-10
21.94397
1.497E-10
15 rb
21.94397
1.497E-10
21.94397
1.497E-10
16 rc
10.97199
7.487E-11
10.97199
7.487E-11
6-2
17 T
698.
698.
698.
698.
18 Tau
0
0
0.1251223
0.1251223
19 V
0
0
1.0E-05
1.0E-05
20 vo
7.992E-05
7.992E-05
7.992E-05
7.992E-05
21 X
0
0
0.3955739
0.3955739
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = rb
3 d(Fc)/d(V) = rc
Explicit equations
1
T = 698
2
Cto = 1641/8.314/T
3
E = 24000
4
Kc = 0.02
5
Ft = Fa+Fb+Fc
6
Ca = Cto*Fa/Ft
7
Fao = 0.0000226
8
vo = Fao/Cto
9
Tau = V/vo
10 Cb = Cto*Fb/Ft
11 X = 1-Fa/Fao
12 k = 0.29*exp(E/1.987*(1/500-1/T))
13 Cc = Cto*Fc/Ft
14 ra = -k*(Ca^2 - Cb^2*Cc/Kc)
15 rb = -ra
16 rc = -ra/2
17 rateA = -ra
General
Total number of equations
20
Number of differential equations 3
Number of explicit equations
17
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
6-3
(iii)
-
.
𝑃
𝑝=
= %1 − ∝* 𝑉,
𝑃$
𝐶0 =
𝐶1$ 𝐹0 𝑝𝑇$
𝐹1 𝑇
For isothermal operation,
-
(1 − 99000𝑉 ). 𝐶1$ 𝐹0
𝐶0 =
𝐹1
See the following Polymath program
Polymath code:
d(Fa)/d(V) = ra #
d(Fb)/d(V) = rb #
d(Fc)/d(V) = rc #
Cto = 1641/8.314/T #
k = 0.29*exp(E/1.987*(1/500-1/T))
E = 24000 #
p = (1-99000*V)^0.5 #
ra = -k*Ca^2 #
rb = -ra #
rc = -ra/2 #
vo = Fao/Cto #
Fao = 0.0000226 #
Tau = V/vo #
rateA = -ra #
X = 1-Fa/Fao #
T = 698
Ca = p*Cto*Fa/Ft #
Ft = Fa+Fb+Fc #
V(0)=0
Fa(0)=0.0000226
Fb(0)=0
Fc(0)=0
V(f)=0.00001
#
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.2827764
0.005108
0.2827764
0.005108
2
Cto
0.2827764
0.2827764
0.2827764
0.2827764
3
E
2.4E+04
2.4E+04
2.4E+04
2.4E+04
4
Fa
2.26E-05
5.616E-06
2.26E-05
5.616E-06
5
Fao
2.26E-05
2.26E-05
2.26E-05
2.26E-05
6
Fb
0
0
1.698E-05
1.698E-05
7
Fc
0
0
8.492E-06
8.492E-06
8
Ft
2.26E-05
2.26E-05
3.109E-05
3.109E-05
9
k
274.4284
274.4284
274.4284
274.4284
10 p
1.
0.1
1.
0.1
11 ra
-21.94397
-21.94397
-0.0071603
-0.0071603
12 rateA
21.94397
0.0071603
21.94397
0.0071603
13 rb
21.94397
0.0071603
21.94397
0.0071603
14 rc
10.97199
0.0035801
10.97199
0.0035801
15 T
698.
698.
698.
698.
6-4
16 Tau
0
0
0.1251223
0.1251223
17 V
0
0
1.0E-05
1.0E-05
18 vo
7.992E-05
7.992E-05
7.992E-05
7.992E-05
19 X
0
0
0.7514895
0.7514895
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = rb
3 d(Fc)/d(V) = rc
Explicit equations
1
T = 698
2
Cto = 1641/8.314/T
3
E = 24000
4
p = (1 - 99000*V)^0.5
5
k = 0.29*exp(E/1.987*(1/500-1/T))
6
Ft = Fa+Fb+Fc
7
Fao = 0.0000226
8
vo = Fao/Cto
9
Tau = V/vo
10 Ca = p*Cto*Fa/Ft
11 X = 1-Fa/Fao
12 ra = -k*Ca^2
13 rb = -ra
14 rc = -ra/2
15 rateA = -ra
General
Total number of equations
18
Number of differential equations 3
Number of explicit equations
15
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
6-5
P6-1 (b) Example 6-2
(i)
On increasing KC, FA decreases and Fc increases. The peak value of maxima in B increases and
maxima shifts to right.
On increasing k, there is a sharp decrease in FA and sharp increase in Fc. The maxima in B shifts to
left.
On increasing CT0 , FA decreases and Fc increases. The maxima in B shifts to left.
On increasing kc, FA decreases and Fc increases. The peak value of maxima in B decreases.
In each of the cases Fa decreases as, the rate of consumption of A increases.
(ii) If we set Kc at its maximum value and then vary k and kc, conversion increases as we increase
both of these parameters.
(iii) 1. RB is maximum when KC, k, CT0, and kc are at their maximum values
2. Kc has the greatest effect on conversion profile and rate constant k has the least effect.
3. With an increase in Kc, molar flow rate of A decreases whereas molar flow rate of B and C
increases. This behavior can also be visualized by the rate and differential equations for Fa,
Fb,Fc
(iv) Individualized solution
(v) See the following Polymath program
Polymath code:
d(Fa)/d(V) = ra #
d(Fb)/d(V) = -ra-kc*Cto*(Fb/Ft)*p #
d(Fc)/d(V) = -ra #
ra = -k*Cto*((Fa/Ft)*p-Cto/Kc*(Fb/Ft)*(Fc/Ft)*p*p)
Ft = Fa+Fb+Fc #
p = (1 - 0.002*V)^0.5
k = 0.7 #
Cto = 0.2 #
Kc = 0.05 #
kc = 0.2 #
V(0)=0
Fa(0)=10
Fb(0)=0
Fc(0)=0
V(f)=500
Rate=-ra
Rb=kc*Cto*(Fb/Ft)
Fao=10
X=(Fao-Fa)/Fao
#
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Cto
0.2
0.2
0.2
0.2
2
Fa
10.
3.338718
10.
3.338718
3
Fao
10.
10.
10.
10.
4
Fb
0
0
3.687651
3.421463
5
Fc
0
0
6.661282
6.661282
6
Ft
10.
10.
13.68765
13.42146
7
k
0.7
0.7
0.7
0.7
8
kc
0.2
0.2
0.2
0.2
9
Kc
0.05
0.05
0.05
0.05
10 p
1.
0
1.
0
11 ra
-0.14
-0.14
0
0
12 Rate
0.14
0
0.14
0
13 Rb
0
0
0.0107766
0.010197
6-6
14 V
0
0
500.
500.
15 X
0
0
0.6661282
0.6661282
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = -ra-kc*Cto*(Fb/Ft)*p
3 d(Fc)/d(V) = -ra
Explicit equations
1
p = (1 - 0.002*V)^0.5
2
Ft = Fa+Fb+Fc
3
k = 0.7
4
Cto = 0.2
5
Kc = 0.05
6
kc = 0.2
7
ra = -k*Cto*((Fa/Ft)*p - Cto/Kc*(Fb/Ft)*(Fc/Ft)*p*p)
8
Rate = -ra
9
Rb = kc*Cto*(Fb/Ft)
10 Fao = 10
11 X = (Fao-Fa)/Fao
General
Total number of equations
14
Number of differential equations 3
Number of explicit equations
11
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
The conversion goes up from 60% to 66.6%
(vi) Individualized solution
(vii) Individualized solution
P6-1 (c) Example 6-2
See the following Polymath program
Polymath code: (modified code)
d(Fa)/d(V) = ra #
d(Fb)/d(V) = -3*ra-kc*Cto*(Fb/Ft) #
d(Fc)/d(V) = -ra #
ra = -k*Cto*((Fa/Ft)-Cto^4/Kc*((Fb/Ft)^3)*(Fc/Ft))
Ft = Fa+Fb+Fc #
k = 0.7 #
Cto = 0.2 #
Kc = 0.001 #
kc = 0.2 #
V(0)=0
Fa(0)=10
Fb(0)=0
Fc(0)=0
V(f)=500
Rate=-ra
Rb=kc*Cto*(Fb/Ft)
Fao=10
X=(Fao-Fa)/Fao
#
6-7
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Cto
0.2
0.2
0.2
0.2
2
Fa
10.
2.376562
10.
2.376562
3
Fao
10.
10.
10.
10.
4
Fb
0
0
14.88615
11.80458
5
Fc
0
0
7.623438
7.623438
6
Ft
10.
10.
24.88615
21.80458
7
k
0.7
0.7
0.7
0.7
8
kc
0.2
0.2
0.2
0.2
9
Kc
0.001
0.001
0.001
0.001
10 ra
-0.14
-0.14
-0.0028323
-0.0028323
11 Rate
0.14
0.0028323
0.14
0.0028323
12 Rb
0
0
0.0239268
0.0216552
13 V
0
0
500.
500.
14 X
0
0
0.7623438
0.7623438
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = -3*ra-kc*Cto*(Fb/Ft)
3 d(Fc)/d(V) = -ra
Explicit equations
1 Kc = 0.001
2 Ft = Fa+Fb+Fc
3 k = 0.7
4 Cto = 0.2
5 ra = -k*Cto*((Fa/Ft)-Cto^4/Kc*((Fb/Ft)^3)*(Fc/Ft))
6 kc = 0.2
7 Rate = -ra
8 Rb = kc*Cto*(Fb/Ft)
9 Fao = 10
10 X = (Fao-Fa)/Fao
P6-1 (d) Example 6-3
(i)
As we increase vo, k, Cbo conversion increases as a function of time while an increase in Vo makes
the conversion to dip. This can be understood as increasing vo rises the rate at which C and D are
produced, thus increasing conversion. Rise in Cbo makes the rate of reaction to increase, hence
increasing conversion.
(ii) 1. An increase in Vo also increases V but as Volume is a function of time so Volume rises faster
compared to Vo. The conversion expression contains both V and Vo. But V has a higher
impact. Hence conversion decreases wrt time as Vo increases.
2. An increase in Cbo or vo raises the maxima value but an increase in k makes maxima to shift
towards left. Also, an increase in Vo causes the maxima to dip.
(iii) See the following Polymath program
Polymath code: (Modified for reversible reaction)
d(Ca)/d(t) = ra- vo*Ca/V #
d(Cb)/d(t) = ra+ (Cbo-Cb)*vo/V #
d(Cc)/d(t) = -ra-vo*Cc/V #
d(Cd)/d(t) = -ra-vo*Cd/V #
V = Vo+vo*t #
Vo = 5 #
vo = 0.05 #
k = 2.2 #
6-8
Cbo = 0.025 #
Kc = 0.1 #
ra = -k*(Ca*Cb-Cc*Cd/Kc) #
X = (Cao*Vo-Ca*V)/(Cao*Vo) #
rate = -ra #
Cao = 0.05 #
t(0)=0
Ca(0)=0.05
Cb(0)=0
Cc(0)=0
Cd(0)=0
t(f)=500
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.05
0.0052864
0.05
0.0052864
2
Cao
0.05
0.05
0.05
0.05
3
Cb
0
0
0.0177864
0.0177864
4
Cbo
0.025
0.025
0.025
0.025
5
Cc
0
0
0.0041087
0.0030469
6
Cd
0
0
0.0041087
0.0030469
7
k
2.2
2.2
2.2
2.2
8
Kc
0.1
0.1
0.1
0.1
9
ra
0
-0.0001295
0
-2.62E-06
10 rate
0
0
0.0001295
2.62E-06
11 t
0
0
500.
500.
12 V
5.
5.
30.
30.
13 Vo
5.
5.
5.
5.
14 vo
0.05
0.05
0.05
0.05
15 X
0
0
0.3656277
0.3656277
Differential equations
1 d(Ca)/d(t) = ra- vo*Ca/V
2 d(Cb)/d(t) = ra+ (Cbo-Cb)*vo/V
3 d(Cc)/d(t) = -ra-vo*Cc/V
4 d(Cd)/d(t) = -ra-vo*Cd/V
Explicit equations
1
vo = 0.05
2
Vo = 5
3
V = Vo+vo*t
4
k = 2.2
5
Cbo = 0.025
6
Kc = 0.1
7
ra = -k*(Ca*Cb - Cc*Cd/Kc)
8
Cao = 0.05
9
rate = -ra
10 X = (Cao*Vo-Ca*V)/(Cao*Vo)
General
Total number of equations
14
Number of differential equations 4
Number of explicit equations
10
6-9
Elapsed time
0.000 sec
Solution method
RKF_45
Step size guess. h
0.000001
Truncation error tolerance. eps 0.000001
P6-1 (e) Example 6-4
(i)
Cao has the highest effect on conversion
(ii) CA0 = 0.043 for exit pressure drop to go to atmospheric pressure
(iii) Individualized solution.
1. We can observe that curves of molar flow rates, rate and concentration for A and B were
overlapping. It can be justified as both of them were equally identical in terms of initial conc
and the rate expressions. The major factor is that both of them have a stoichiometric
coefficient equal to 1.
2. The alpha parameter plays a major role only in the pressure drop curve. It is because alpha
appears in the differential equation for pressure drop.
3. Even if we increase the initial conc for any of A or B, still their conc curves and the molar flow
rates remain the same as they are being consumed in equal quantity.
P6-1 (f) Example 6-5
(i)
When all variables are set to their minimum vales, the rate constant is very low. This means that
reaction will proceed very slowly and hence conversion will be almost negligible for initial few
minutes.
Also X= 1-Ca*V/(Cao*Vo). Decreasing Vo makes the –ve term higher resulting in less conversion.
(ii) Decrease in Kc decreases conversion whereas increase in rate constant k enhances conversion.
Overall, conversion has a mixed effect and it increases or decreases depending upon how much k
or Kc is varied. There is no effect on maximum concentration of B as Cbo is independent of k or Kc.
(iii) Rate constant k has the greatest effect on conversion
(iv) 1. Rate of formation of C is twice the rate of consumption of A and B. This can be seen from the
rate curves.
2. Increase in rate constant makes reaction more spontaneous and thus reaction completing at
an early time. Rate is mostly affected by k and a little bit by the volumetric flow rate vo.
3. Increase In k shifts the maxima and minima in the conc curves to the left. It is because
increasing k fastens up the reaction and hence production or consumption becomes early or
fast.
4. Increasing volumetric flow rate have deep impacts on conc curves. This is because as vo rises,
conc dips to keep molar flow rates same.
P6-1 (g) Individualized Solution
P6-1 (h) Individualized Solution
P6-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer
Games (ICGs) is given at the front of this Solutions Manual.
P6-3
Polymath
d(Cb) / d(t) = rb
Cb(0) = 2
d(Ca) / d(t) = ra
Ca(0) = 2
d(Cc) / d(t) = rc
6-10
Cc(0) = 0
t(0) = 0
t(f) = 100
k = 0.0445
Ca0 = 2
ra = -k*Ca*Cb
rb = ra
rc = -ra
X = (Ca0-Ca)/Ca0
P6-3 (a)
The conversion after 10, 50, 100 minutes are:
At 10 mins: X = 0.47
At 50 mins: X = 0.82
At 100 mins: X = 0.9
For reactor 1:
Where
89
8:
= −𝑣$ , then
For reactor 2:
Where, 𝑉 = 𝑉$ + 𝑣$ 𝑡 , thus
89
8:
= 𝑣$, then
Therefore, the design equation and concentration profiles are the same as in part a) for the two
reactors. As they have the same concentration profile at any time, the conversions are the same for the
two reactors as in part a).
6-11
P6-3 (b)
At 10 mins: X = 0.47, Ca = Cb = 1.06 mol/dm3, Cc = 0.94 mol/dm3.
At 50 mins: X = 0.82, Ca = Cb = 0.37 mol/dm3, Cc = 1.63 mol/dm3.
At 80 mins: X = 0.9, Ca = Cb = 0.25 mol/dm3, Cc = 1.75 mol/dm3.
P6-3 (c)
At 10 mins: X = 0.47, Ca = Cb = 1.06 mol/dm3, Cc = 0.94 mol/dm3.
At 50 mins: X = 0.82, Ca = Cb = 0.37 mol/dm3, Cc = 1.63 mol/dm3.
P6-3 (d)
The conversion in the two reactors are the same.
At 50 mins: X = 0.82
P6-3 (e) Individualized Solution
P6-4 (a)
The algorithm for this problem can be developed taking reference from LEP Example 6-1
Polymath Code
d(Fa) / d(V) = ra
d(Fb) / d(V) = rb
d(Fc) / d(V) = rc
T=400
E=85000
k=(10^-4)*exp(E/8.314*(1/323-1/T))
ra=-k*Ca
rb=-ra
rc=-2*ra
Ft=Fa+Fb+Fc
Cto=(10*101.325)/(8.314*T)
Ca=Cto*Fa/Ft
X=1-Fa/Fao
Fao=2.5
Fa(0) = 2.5
Fb(0) = 0
Fc(0) = 0
V(0) = 0
V(f) = 950
vo=Fao/Cto
tau=V/vo
Polymath Output
6-12
From the above graph, 950 dm3 of volume is required for 90 % conversion. The required space time is
116 min.
P6-4 (b)
P6-5 (a)
6-13
Using Polymath to solve the differential equation gives a volume of 290 dm3
See the following Polymath program
d(V)/d(X)=Fao/(-ra)
Kc = 0.025
Fao = 2.5
Cao = 0.3
k = 0.044
e=2
ra = -(k*Cao/(1+e*X))*((1-X)-(4*Cao^2*X^3)/((1+e*X)^2*Kc))
X(0)=0
X(f) = 0.47
V(0) = 0
P6-5 (b) See the following polymath program
Polymath Code
d(Fa)/d(V) = ra #
d(Fb)/d(V) = -ra-Rb #
d(Fc)/d(V) = -2*ra #
Ca=Cto*Fa/Ft
Cb=Cto*Fb/Ft
Cc=Cto*Fc/Ft
Ft=Fa+Fb+Fc
ra = -k*(Ca-(Cb*Cc^2)/Kc) #
T=300
E=85000
k=(10^-4)*exp(E/8.314*(1/323-1/T))
Cto = 0.3 #
Kc = 0.025 #
kc = 0.08*60 #
#
6-14
Rate=-ra
Rb=kc*Cto*(Fb/Ft)
Fao=2.5
X=(Fao-Fa)/Fao
V(0)=0
Fa(0)=2.5
Fb(0)=0
Fc(0)=0
V(f)=2.3e6
B has a maxima in flow rate. However, in the above graph, the flow rate of B is almost zero and thus
maxima can’t be seen.
P6-5 (c) Individualized solution
P6-6 (a)
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
Next, we express the rate law:
First-order reversible reaction
Rename
Transport out the sides of the reactor:
6-15
RA = k A C A =
RB = k B C B =
Stoichiometry:
–rA = rB =1/2 rC
Combine and solve in Polymath code:
See the following Polymath program
Polymath code:
d(Fa)/d(v) = ra - Ra
d(Fb)/d(v) = -ra -Rb
d(Fc)/d(v) = -2*ra
Kc = 0.01
Ft = Fa+Fb+Fc
Co = 1
K = 10
Kb = 40
ra = -(K*Co/Ft)*(Fa-Co^2*Fb*Fc^2/(Kc*Ft^2))
Ka = 1
Ra = Ka*Co*Fa/Ft
Rb = Kb*Co*Fb/Ft
v(0) = 0
v(f) = 20
Fa(0) = 100
Fb(0) = 0
Fc(0) = 0
POLYMATH Results
Calculated values of the DEQ variables
Variable
v
Fa
Fb
Fc
Kc
Ft
Co
K
Kb
ra
Ka
Ra
Rb
Fao
X
initial value
0
100
0
0
0.01
100
1
10
40
-10
1
1
0
100
0
minimal value
0
57.210025
0
0
0.01
100
1
10
40
-10
1
0.472568
0
100
0
maximal value
20
100
9.0599877
61.916043
0.01
122.2435
1
10
40
-0.542836
1
1
2.9904791
100
0.4278998
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Fa)/d(v) = ra - Ra
[2] d(Fb)/d(v) = -ra - Rb
[3] d(Fc)/d(v) = -2*ra
6-16
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
Explicit equations as entered by the user
[1] Kc = 0.01
[2] Ft = Fa+ Fb+ Fc
[3] Co = 1
[4] K = 10
[5] Kb = 40
[6] ra = - (K*Co/Ft)*(Fa- Co^2*Fb*Fc^2/(Kc*Ft^2))
[7] Ka = 1
[8] Ra = Ka*Co*Fa/Ft
[9] Rb = Kb*Co*Fb/Ft
P6-6 (b)
The setup is the same as in part (a) except there is no transport out the sides of the reactor.
See the following Polymath program
Polymath code:
d(Fa)/d(v) = ra
d(Fb)/d(v) = -ra
d(Fc)/d(v) = -2*ra
Kc = 0.01
Ft = Fa+Fb+Fc
Co = 1
K = 10
ra = -(K*Co/Ft)*(Fa-Co^2*Fb*Fc^2/(Kc*Ft^2))
v(0) = 0
v(f) = 20
Fa(0) = 100
Fb(0) = 0
Fc(0) = 0
POLYMATH Results
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
minimal value
0
84.652698
0
0
0.01
100
1
10
-10
100
0
maximal value
20
100
15.347302
30.694604
0.01
130.6946
1
10
-3.598E-09
100
0.153473
6-17
final value
20
84.652698
15.347302
30.694604
0.01
130.6946
1
10
-3.598E-09
100
0.153473
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- Co^2*Fb*Fc^2/(Kc*Ft^2))
P6-6 (c) Conversion would be greater if C were diffusing out.
P6-6 (d) Individualized solution
P6-7 (a)
Assuming catalyst distributed uniformly over the whole volume
Mole balance:
Rate law:
Stoichiometry:
6-18
Solving in polymath:
See the following Polymath program
Polymath code:
d(Fa)/d(W) = r
d(Fb)/d(W) = r
d(Fc)/d(W) = -r
d(Fd)/d(W) = -r-Rh
Keq = 1.44
Ft = Fa+Fb+Fc+Fd
Cto = 0.4
Ca = Cto*Fa/Ft
Cb = Cto*Fb/Ft
Kh = 0.1
Cc = Cto*Fc/Ft
Cd = Cto*Fd/Ft
Rh = Kh*Cd
k = 1.37
r = -k*(Ca*Cb-Cc*Cd/Keq)
W(0) = 0
W(f) = 100
Fa(0) = 0
Fb(0) = 0
Fc(0) = 0
Fd(0) = 0
POLYMATH Results
Calculated values of the DEQ variables
Variable
W
Fa
Fb
Fc
Fd
Keq
Ft
Cto
Ca
Cb
Kh
Cc
Cd
Rh
k
r
initial value
0
2
2
0
0
1.44
4
0.4
0.2
0.2
0.1
0
0
0
1.37
-0.0548
minimal 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
maximal value
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] d(Fa)/d(W) = r
[2] d(Fb)/d(W) = r
[3] d(Fc)/d(W) = -r
[4] 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] Ca = Cto*Fa/Ft
[5] Cb = Cto*Fb/Ft
[6] Kh = 0.1
6-19
final value
100
0.7750721
0.7750721
1.2249279
0.5536716
1.44
3.3287437
0.4
0.0931369
0.0931369
0.1
0.147194
0.0665322
0.0066532
1.37
-0.002567
[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
P6-7 (b)
In a PFR no hydrogen escapes and the equilibrium conversion is reached.
solve this for X,
X = .5454
This is the maximum conversion that can be achieved in a normal PFR.
P6-7 (c)
If feed rate is doubled, then the initial values of Fa and Fb are doubled. This results in a conversion of
0.459
P6-8 (a)
A+ B → C + D +CO2
Mole Balance
dNA
= rAV
dt
dNB
= FB0 + rBV
dt
dNC
= rCV
dt
ND = NC
Where A=NaHCO3, B= Ethylene chlorohydrin
Overall Mass Balance
Accumulation = In – Out
6-20
dm
 CO
= ρυ0 − m
2
dt
m = ρV !Assume!constant!density
 CO
m
dV
2
= υ0 −
dt
ρ
The rate of formation of CO2 is equal to the rate of formation of ethylene glycol (C).
 CO = FCO •MWCO2
m
2
2
FCO •MWCO2
dV
= υ0 − 2
dt
ρ
FCO •MWCO2
υCO = 2
2
ρ
= υ0 − υCO
2
Rate Law and Relative Rates
−rA = kC ACB
rB = rA
rC = −rA
Stoichiometry
N
CA = A
V
NB
CB =
V
8>?
8:
8>H
8:
=
8>E
8:
= 𝑟A −
BI ∗>HI
9
%BC DBEFG ,>?
9
KBCD𝑣𝐶𝑂 N>H
2
−
+ 𝑟A
9
= −𝑟A −
%BC DBEFG ,>E
9
𝐶O = 𝐶>
𝑑𝑁>RG
= 0 = −𝐹>RG − 𝑟A 𝑉
𝑑𝑡
𝐹>RG = −𝑟A 𝑉
Polymath code:
d(Ca)/d(t) = ra+((vo-vCO2)/V)*(-Ca)
d(Cb)/d(t) = ra+vo*Cbo/V + ((vo-vCO2)/V)*(-Cb)
d(Cc)/d(t) = -ra+((vo-vCO2)/V)*(-Cc)
d(V)/d(t) = vo-vCO2
Cd = Cc
Fbo = 6
Cao = 0.75
Vo = 1500
Na = Ca*V
k = 5.1
ra = -k*Ca*Cb
Nao = Cao*Vo
rate = -ra
rho = 1000
MWCO2 = 44
6-21
FCO2 = -ra*V
NC = Nao-Na
Cbo = 1.5
vo = Fbo/Cbo
vCO2=MWCO2*FCO2/rho
CC = NC/V
Nc = Cc*V
X = 1-Ca*V/(Cao*Vo)
Ca(0) = 0.75
Cb(0) = 1.5
Cc(0) = 0
t(0) = 0
t(f) = 8
V(0) = 1500
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.75
5.096E-15
0.75
5.096E-15
2
Cao
0.75
0.75
0.75
0.75
3
Cb
1.5
0.7794262
1.5
0.791231
4
Cbo
1.5
1.5
1.5
1.5
5
Cc
0
0
0.7714923
0.7588533
6
CC
0
0
0.7714923
0.7588533
7
Cd
0
0
0.7714923
0.7588533
8
Fbo
6.
6.
6.
6.
9
FCO2
8606.25
3.048E-11
8606.25
3.048E-11
10 k
5.1
5.1
5.1
5.1
11 MWCO2
44.
44.
44.
44.
12 Na
1125.
7.555E-12
1125.
7.555E-12
13 Nao
1125.
1125.
1125.
1125.
14 NC
0
0
1125.
1125.
15 Nc
0
0
1125.
1125.
16 ra
-5.7375
-5.7375
-2.056E-14
-2.056E-14
17 rate
5.7375
2.056E-14
5.7375
2.056E-14
18 rho
1000.
1000.
1000.
1000.
19 t
0
0
8.
8.
20 V
1500.
1454.788
1500.
1482.5
21 vCO2
378.675
1.341E-12
378.675
1.341E-12
22 vo
4.
4.
4.
4.
23 Vo
1500.
1500.
1500.
1500.
24 X
0
0
1.
1.
Differential equations
1 d(Ca)/d(t) = ra+((vo-vCO2)/V)*(-Ca)
2 d(Cb)/d(t) = ra+vo*Cbo/V + ((vo-vCO2)/V)*(-Cb)
3 d(Cc)/d(t) = -ra+((vo-vCO2)/V)*(-Cc)
4 d(V)/d(t) = vo-vCO2
Explicit equations
1 Cd = Cc
2 Fbo = 6
3 Cao = 0.75
4 Vo = 1500
5 Na = Ca*V
6-22
6 k = 5.1
7 ra = -k*Ca*Cb
8 Nao = Cao*Vo
9 rate = -ra
10 rho = 1000
11 MWCO2 = 44
12 FCO2 = -ra*V
13 NC = Nao-Na
14 Cbo = 1.5
15 vo = Fbo/Cbo
16 vCO2 = MWCO2*FCO2/rho
17 CC = NC/V
18 Nc = Cc*V
19 X = 1-Ca*V/(Cao*Vo)
P6-8 (b) Suppose the volumetric flow rate could be varied between 0.6 mol/hr and 12,000 mol/h and
the total time to fill, heat, empty and clean is 4.5 hours.
Using the polymath code to adjust the flow rate of B, we find that decreasing flow rate of B results in
more production of C.
P6-8 (c)
FB0 = 0.15 mol/min = 9 mol/hr
vo = FBo/cbo = (9 mol/hr) / (1.5 mol/dm3) = 6 dm3/hr
1000 dm3 is needed to fill the reactor. At 6 dm3/hr it will take 166.67 hours
Now solving using the code form part (a) with the changed equations:
6-23
See the following Polymath program
d(Ca)/d(t) = ra+((vo-vCO2)/V)*(-Ca)
d(Cb)/d(t) = ra+vo*Cbo/V + ((vo-vCO2)/V)*(-Cb)
d(Cc)/d(t) = -ra+((vo-vCO2)/V)*(-Cc)
d(V)/d(t) = vo-vCO2
Cd = Cc
Fbo = 9
Cao = 0.75
Vo = 1500
Na = Ca*V
k = 5.1
ra = -k*Ca*Cb
Nao = Cao*Vo
rate = -ra
rho = 1000
MWCO2 = 44
FCO2 = -ra*V
NC = Nao-Na
Cbo = 1.5
vo = Fbo/Cbo
vCO2=MWCO2*FCO2/rho
CC = NC/V
Nc = Cc*V
X = 1-Ca*V/(Cao*Vo)
Ca(0) = 0.75
Cb(0) = 1.5
Cc(0) = 0
t(0) = 0
t(f) = 167
V(0) = 1500
P6-8 (d) Individualized solution
P6-9
(
2
)(
)
π D L π 1.5m 2.5m
VT =
=
4
4
3
VT = 4.42m = 4,420)dm3
2
6-24
We don’t know V0 or v0. First try equal number of moles of A and B added to react.
NA = NB
C A0V0 = CB0 ΔV = CB0 #$VT −V0 %&
VT
4420
V0 =
=
= 2456
C A0
0.8 +1.0
+1
CB0
ΔV = 1964
υ0 =
ΔV
tR
k0 = 0.000052*dm3 mol s = 0.187*dm3 mol h
For 1 batch tR = 24 – 3 = 21
V −V
υ0 = T 0
21
(1) For one batch we see that only 198 moles of C are made so one batch will not work.
(2) For two batches, we have a down time of 2 × 3 = 6 h and therefore each batch has a reaction time of
18h/2 = 9 h. We see that 106 moles of C are made in one batch therefore 2 × 106 = 212 moles/day
are made in two batches.
See the following Polymath program
Polymath code:
d(Ca)/d(t) = ra+(vo/V)*(-Ca)
d(Cb)/d(t) = ra+vo*(Cbo-Cb)/V
d(Cc)/d(t) = -ra+(vo/V)*(-Cc)
d(Cd)/d(t) = -ra+(vo/V)*(-Cd)
Cbo = 0.25
Cao = 0.2
tr = 9
Vo = 4420/1.8
ko = 5.2e-5*3600
DeltaV = 4420-Vo
vo = DeltaV/tr
T = 37+273
k = ko*exp((42810/8.3144)*(1/293-1/T))
ra = -k*Ca*Cb
Fao = 0.04*3600
V = Vo+vo*t
Nbf = Cbo*DeltaV
Nao = Cao*Vo
Nc = Cc*V
Na = Nao-Nc
Ca(0)=0.2
Cb(0) = 0
Cc(0) = 0
Cd(0) = 0
t(0) = 0
t(f) = 9
6-25
P6-9 Two Batches
P6-10 Individualized solution
P6-11 (a)
C C
At equilibrium, r = 0 => C ACB = C D
KC
V = V0 + vot
CA =
(
NAO 1− X
V
) = C AO (1− X ) = C AO (1− X )
V
VO
" v %
$1+ o t '
$ V '
#
O &
6-26
v
CBO O t −C AO X
VO
CB =
" v %
$1+ o t '
$ V '
#
O &
CC = C D =
C AO X
! v $
#1+ o t &
# V &
"
O %
(
"
% C X
v
AO
C AO 1− X $$CBO O t −C AO X '' =
VO
KC
#
&
(
)
2
)
"
%
VOC AO $ X 2
t=
+ X'
'
vOCBO $ K 1− X
# C
&
(
)
See the following Polymath program
Solving in Polymath
f(x) = 2825.25*(x^2/1.08/(1-x)+x)
x(0) = 0.495
x(max) = 0.99
x(min) = 0
POLYMATH Report
Nonlinear Equation
Calculated values of NLE variables
Variable Value f(x) Initial Guess
1 x
0
0
0.495 ( 0 < x < 0.99 )
Nonlinear equations
1 f(x) = 2825.25*(x^2/1.08/(1-x)+x) = 0
If we solve in Excel
6-27
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
0.99
X
0
148.1466374
311.591358
493.0338235
695.8486111
924.3101852
1183.914286
1481.84765
1827.692593
2234.515909
2720.611111
3312.40216
4049.525
4994.264683
6250.42963
8004.875
10631.31111
15001.7287
23732.1
49902.28611
259188.435
t (sec)
P6-11 (b)
See the following Polymath program
Polymath code:
d(Ca)/d(t) = ra-Ca*vo/V
d(Cb)/d(t) = ra-(Cb-10.93)*vo/V
d(Cc)/d(t) = -ra-Cc*vo/V
d(Cd)/d(t) = -ra-Cd*vo/V
Kc = 1.08
k = 0.00009
ra = -k*(Ca*Cb-Cc*Cd/Kc)
vo = 0.05
Vo = 200
V = Vo+vo*t
X = 1-Ca/7.72
t(0)= 0
t(f) = 1.5e4
Ca(0) = 7.72
Cb(0) = 10.93
Cc(0) = 0
Cd(0) = 0
6-28
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
Cc
Cd
Kc
k
ra
vo
Vo
V
X
initial value
0
7.72
10.93
0
0
1.08
9.0E-05
-0.0075942
0.05
200
200
0
minimal value
0
0.2074331
7.6422086
0
0
1.08
9.0E-05
-0.0075942
0.05
200
200
0
maximal value
1.5E+04
7.72
10.93
3.2877914
3.2877914
1.08
9.0E-05
-1.006E-05
0.05
200
950
0.9731304
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra - Ca*vo/V
[2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93)
[3] d(Cc)/d(t) = -ra - vo*Cc/V
[4] d(Cd)/d(t) = -ra - vo*Cd/V
Explicit equations as entered by the user
[1] 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
Polymath solution
P6-11 (c)
As ethanol evaporates as fast as it forms:
Now using part (b) remaining equations,
Polymath code:
CD = 0
6-29
final value
1.5E+04
0.2074331
9.51217
1.41783
1.41783
1.08
9.0E-05
-1.006E-05
0.05
200
950
0.9731304
See the following Polymath program
Polymath code:
d(Ca)/d(t) = ra-Ca*vo/V
d(Cb)/d(t) = ra-(Cb-10.93)*vo/V
k = 0.00009
ra = -k*Ca*Cb
vo = 0.05
Vo = 200
V = Vo+vo*t
X = 1-Ca/7.72
t(0)= 0
t(f) = 6000
Ca(0) = 7.72
Cb(0) = 10.93
POLYMATH Results
Calculated values of the DEQ variables
Variable
t
Ca
Cb
k
ra
vo
Vo
V
X
initial value
0
7.72
10.93
9.0E-05
-0.0075942
0.05
200
200
0
minimal value
0
0.0519348
6.9932872
9.0E-05
-0.0075942
0.05
200
200
0
maximal value
6000
7.72
10.93
9.0E-05
-3.69E-05
0.05
200
500
0.9932727
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = ra - Ca*vo/V
[2] d(Cb)/d(t) = ra - vo/V*(Cb- 10.93)
Explicit equations as entered by the user
[1] k = 0.00009
[2] ra = -k*Ca*Cb
[3] vo = 0.05
[4] Vo = 200
[5] V = Vo + vo*t
[6] X = 1 - Ca/7.72
6-30
final value
6000
0.0519348
7.8939348
9.0E-05
-3.69E-05
0.05
200
500
0.9932727
P6-11 (d)
Change the value of vo, CAO, etc. in the Polymath program to see the changes.
P6-11 (e) Individualized solution
P6-11 (f) Individualized solution
P6-12 (a)
𝑉 = 𝐹A$ ∗ 𝑋/−𝑟A
X=0.8*Xe=0.8*0.994=0.795
Use the polymath program of P6-11 (a), to get the value of -rA at X=0.795
Vary time in code such that X final is 0.795 and then find the value of -rA as
-rA=0.0002651=2.65e-4
FA0=Ca0*vo=10.93*0.05
FA0=0.54 mol/s,
Substitute value of FA0,X, and -rA in CSTR volume equation
V=
$.VW∗$.XYV
$.$$$.ZV-
V=1619.5 dm3
P6-12 (b) 1% of rate in part (a)
New required rate=2.65e-6
For water addition, polymath program needs to be modified to include inlet water term and then tfinal
needs to be adjusted such that Rate= 2.65 e-6
Polymath code:
d(Ca)/d(t) = ra - 2*Ca*vo/V
d(Cb)/d(t) = ra - vo/V*(Cb- 10.93)-Cb*vo/V
d(Cc)/d(t) = -ra - 2*vo*Cc/V
d(Cd)/d(t) = -ra - 2*vo*Cd/V
Kc = 1.08
k = 0.00009
ra = -k*(Ca*Cb - Cc*Cd/Kc)
vo = 0.05
Vo = 200
V = Vo + 2*vo*t
X = 1 - Ca/7.72
Ca(0)=7.72
Cb(0)=10.93
Cc(0)=0
Cd(0)=0
t(0)=0
t(f)=2.1e4
Polymath Output
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
7.72
0.0685332
7.72
0.0685332
2
Cb
10.93
5.314301
10.93
5.337446
6-31
3
Cc
0
0
2.487708
0.6027712
4
Cd
0
0
2.487708
0.6027712
5
k
9.0E-05
9.0E-05
9.0E-05
9.0E-05
6
Kc
1.08
1.08
1.08
1.08
7
ra
-0.0075942
-0.0075942
-2.644E-06
-2.644E-06
8
t
0
0
2.1E+04
2.1E+04
9
V
200.
200.
2300.
2300.
10 Vo
200.
200.
200.
200.
11 vo
0.05
0.05
0.05
0.05
12 X
0
0
0.9911226
0.9911226
Differential equations
1 d(Ca)/d(t) = ra - 2*Ca*vo/V
2 d(Cb)/d(t) = ra - vo/V*(Cb- 10.93)-Cb*vo/V
3 d(Cc)/d(t) = -ra - 2*vo*Cc/V
4 d(Cd)/d(t) = -ra - 2*vo*Cd/V
Explicit equations
1 Kc = 1.08
2 k = 0.00009
3 ra = -k*(Ca*Cb - Cc*Cd/Kc)
4 vo = 0.05
5 Vo = 200
6 V = Vo + 2*vo*t
7 X = 1 - Ca/7.72
P6-13
Reaction:
𝐴 ⇄ 𝐵 with
𝑘_ = 𝑘 = 0.4
-
𝐾> = 4
a
Reactor - mass balance (1): 𝑚̇A,0f = 𝐶A$ ∙ 𝑣$ + 𝐶A$ ⋅ 𝑣i = 100 ⋅ (12 + 𝑣i )
Reactor - component A balance (2): input = output + reaction; 𝑚̇A,0f = 𝑚̇A,jk: + (−𝑟A 𝑉)
>
Rate law in the reaction term (3): −𝑟A 𝑉 = 𝑘 K𝐶A − lH N 𝑉
E
Stoichiometry (4): 𝐶m = 𝐶A$ − 𝐶A
6-32
Combine (3) and (4), substitute for numerical values of CA0, k and KC to get (5): −𝑟A 𝑉 = 30𝐶A − 600
Combine (5) and (1) with (2) and evaluate numerically to get: 𝑐A =
𝑚̇m,jk: (𝑣i ) = 𝐶m ⋅ (𝑣$ + 𝑣i ) =
-q$$r-$$Bs
W.rBs
2400(12 + 𝑣i )
42 + 𝑣i
𝑃𝑟𝑜𝑓𝑖𝑡(𝑣i ) = 𝐼𝑛𝑐𝑜𝑚𝑒(𝑣i ) − 𝑂𝑝𝑒𝑟𝑎𝑡𝑖𝑛𝑔 𝐶𝑜𝑠𝑡𝑠(𝑣} )
𝑃𝑟𝑜𝑓𝑖𝑡(𝑣i ) = [𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒 𝑜𝑓 𝐵] × [𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐵] − „
𝑐𝑜𝑠𝑡 𝑜𝑓 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛
† × [𝑣𝑜𝑙𝑢𝑚. 𝑓𝑙𝑜𝑤𝑟𝑎𝑡𝑒]
𝑚…
𝑃(𝑣i ) = 𝑚̇m,jk: (𝑣i ) × 2 − 50 × (𝑣$ + 𝑣i )
𝑃 (𝑣i ) =
2400(12 + 𝑣i )
× 2 − 50 × (12 + 𝑣i )
42 + 𝑣i
Optimization: extremum (maximum) in profit when the derivative 𝑑𝑃(𝑣i )/𝑑𝑣i is zero:
𝑑𝑃(𝑣i )
4800 ⋅ 30
=
− 50 ⇒ 𝑣i = 11.6656 𝑚 … /ℎ
(42 + 𝑣i ).
𝑑𝑣i
Into the definition of the recycle ratio, f:
𝑓=
𝑣i
= 0.89
𝑣$ + 𝑣i − 𝑣m
Where 𝑣m is calculated from the overall mass balance of A :
(𝑣$ + 𝑣‹ )𝑐A = (𝑣$ + 𝑣‹ − 𝑣m ) ⋅ 𝑐A$ ⇒ 𝑣m = 10.58 𝑚 … /ℎ
6-33
.W$$
; 𝑐m = 100 − 𝑐A = W.rB
s
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6-34
Solutions for Chapter 7 – Collection and Analysis of Rate Data
General: Different methods of data analysis are presented. I give at least one problem that uses
Polymath’s nonlinear least square analysis.
Questions
l Q7-1A (16 seconds) Questions Before Reading (QBR).
O Q7-2A (5-10 min) i>Clicker questions.
I Q7-3A (30 min) Three very short problems and an open-ended experiment to be designed by the
student.
O Q7-4B (6 min) Question about experimental error.
I Q7-5A (4 min) Once alpha is fixed, need to recalculate the parameters.
O Q7-6A (10 min total for all 4 videos) These 1-2 minute videos are really cleaver.
Interactive Computer Games (ICG)
l P7-2A (30 min) Interactive Computer Game (ICG) with an ecology scenario.
Problems
l P7-3A (25 min) A problem related to hand warmers for skiing.
I P7-4A (30 min) Straight forward analysis of data on blood oxygen in a plug flow liquid phase reactor.
O P7-5A (25 min) Very straight forward of the application of the differential method of analysis.
l P7-6B (25 min) Old Exam Question (OEQ). Very tricky problem as the students rush to use the
differential method of analysis which is incorrect because reaction is in a CSTR operated at steady
state.
AA P7-7A (35 min) Straight forward differential analysis of data. Good practice at analyzing data.
AA P7-8A (30 min) Old Exam Question (OEQ). Straight forward analysis once the student realizes they
don’t need the first data point. Good practice at analyzing data.
l P7-9A (30 min) California Professional Engineers Registration Exam Problem. Excellent problem, Part
(a) is straight forward differential analysis of data, but part (b) is difficult because the student must
carry out a mass or mole balance and realize the volume of the body of water does not change
“virtually” at all.
O P7-10C (60 min) Need to postulate a rate law then find the parameters. Somewhat difficult.
O P7-11A (10 min) Old Exam Question (OEQ). Once you see the problem it takes 10 minutes to
calculate the answers.
O P7-12A (40 min) Similar to P7-10C.
S7-1
Page intentionally blank
S7-2
Solutions for Chapter 7 – Collection and Analysis of Rate Data
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-7
http://umich.edu/~elements/5e/07chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q7-1 Individualized solution.
Q7-2 Individualized solution.
Q7-3 Individualized solution.
Q7-4
–rA = kCACB
CB = CB0 – CA0 + CA
dCA/dt = kCA2(CB0 – CA0 + CA)
$
"#$%
+
'( "$%( )$% )$%
∫$ % ($
%(
= kt
$
"#$%
+
%( ($'( "$%( )$% )$%
y = kt, where y = ∫$ %
CA (mol/dm3)
y (dm3/mol)2
t (min)
0.05
0
0
0.038
12.79981
50
0.0306
25.94781
100
0.0256
39.30235
150
0.0222
51.92862
200
0.0195
65.17656
250
0.0174
78.38982
300
Plot of y vs t ( y on “y axis” and t on “x axis”)
k = Slope of line = 0.26 (dm3/mol)2/min
7-1
Q7-5 To find the value of k¢ and k when α = 2.
Q7-6 Individualized solution.
Q7-7 Individualized solution.
P7-1 (a) Example 7-3
In the initial settings, s2 is minimized at alpha = 2.
(i)
As we increase beta keeping A fixed, the minimal value of s2 decreases and point at which minima
occurs also moves to left.
(ii) As we decrease A keeping beta fixed, the minimal value of s2 decreases and point at which minima
occurs also moves to left.
(iii) To suggest a rate law that minimizes s2, we need to obtain the model parameters.
The parameters at which minimum value of s2 occur can be found by varying the sliders as shown below
Value of alpha comes to be around 0.45
So the rate law that minimizes s2 is
−𝑟. = 1.5 exp(−𝐸/𝑅𝑇)𝐶.;.<= 𝐶>?
P7-1 (b) Example 7-4
Visit the following link for Polymath tutorial on nonlinear regression:
http://umich.edu/~elements/5e/07chap/Nonlinear_Regression_Tutorial.pdf
7-2
Polymath code:
R
5.2e-3
13.2e-3
30e-3
4.95e-3
7.42e-3
5.25e-3
PCO
1
1.8
4.08
1
1
1
PH2
1
1
1
0.1
0.5
4
Model:
r = k*(PCO^alfa)*(PH2^beta)
Guess Value:
k =0.006 alfa = 1.14 beta = 0.01
Polymath Output:
Model: r = k*(PCO^alfa)*(PH2^beta)
Variable Initial guess Value
95% confidence
k
0.006
0.0060979 0.0014096
alfa
1.14
1.138146
beta
0.01
0.0103839 0.2658648
0.1854698
Nonlinear regression settings
Max # iterations = 64
Precision
R^2
0.9869709
R^2adj 0.9782849
Rmsd
0.0004176
Variance 2.093E-06
P7-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer
Games (ICGs) is given at the front of this Solutions Manual.
P7-3
P7-3 (a)
Assuming the reaction to be first order w.r.t to O2 (A)
𝑑𝐶.
−
= 𝑘 ∗ 𝐶.
𝑑𝑡
Or,
𝑑𝐶.
= 𝑘 ∗ 𝑑𝑡
𝐶.
Integrating gives
𝐶. = 𝐶.; ∗ exp (−𝑘 ∗ 𝑡)
Assuming total volume to be constant and total initial air moles to be 100
𝑁. = 𝑁.; ∗ exp (−𝑘𝑡)
From the table, at t=0, NA0=21
So, 𝑁. = 21 exp (−𝑘𝑡)
−
7-3
P7-3 (b)
From the rate law equation (P7-3.1), at t=0, 𝑁. = 3
from the table, at t=0, 𝑁. = 21
So, the rate law is incorrect
P7-3 (c)
Using polymath non-linear regression to regress the data in table
We get A=20.6, k=0.097
So rate law is
%𝑂J = 20.6 exp (−0.097𝑡)
POLYMATH Report
Model: O2 = A*exp(k*t)
Variable
Initial guess
Value
95% confidence
A
21.
20.58151
1.031828
k
1.
-0.0968008
0.0102254
Precision
R^2
0.9960872
R^2adj 0.9953047
Rmsd
0.1514985
Variance 0.2249275
Source data points and calculated data points
t
O2
O2 calc
Delta O2
1
0
21
20.58151
0.4184858
2
3
15
15.3942
-0.3942018
3
5
12
12.68461
-0.6846088
4
8
10
9.487612
0.5123877
5
15
5
4.818109
0.1818914
6
20
3
2.969452
0.0305478
7
25
2
1.830105
0.1698946
P7-3 (d) with acetic acid, rate constant k will increase and thus O2 will decrease faster
P7-4 (a)
For the reaction, HbO2 → Hb + O2
Limiting Reactant,
A, or HbO2
Reactor Type
tubular
Dependent Variable
X
Design Equation
dFA/dV = rA
Rate Equation
Assume power law, -rA = kCAn
Stoichiometry
FA = voCA, dV = ACdz
Combine
dCA/dz = -(kAC /vo)CAn
Assume the reaction 1st order, n = 1.0. Integrate with initial conditions X(z=0) = 0, we get
Ln(1/(1-X)) = (kAC /vo)z
7-4
Electrode
Position
Position (cm)
Conversion(XA)
ln(1/(1-XA))
1
2
3
4
5
6
7
0
0
0
5
0.0193
0.0195
10
0.0382
0.0389
15
0.0568
0.0585
20
0.0748
0.0777
25
0.0925
0.0971
30
0.110
0.1165
Plot ln(1/(1-XA) vs. z, we find a perfect linear relationship
Slope = kAC /vo = 0.0039
AC = πd2 /4= 0.0196 cm2
k = 0.0039 vo/AC = (0.0039 cm-1)X 19.6 cm3/s/(0.0196cm2)
k = 3.9 s-1
Hence, the rate law is, –rA = 3.9CA mol/L/s
P7-4 (b)
First, we fit a polynomial to the data. Using Polymath, we use regression to find an expression for X(z)
(See Polymath tutorials on regression
(http://www.umich.edu/~elements/6e/tutorials/Polymath_tutorials.html))
Polymath Output:
Model: X = a0 + a1*z + a2*z^2 + a3*z^3 + a4*z^4 + a5*z^5 + a6*z^6
Variable
Value
95% confidence
a0
2.918E-14
0
a1
0.0040267
0
a2
-6.14E-05
0
a3
7.767E-06
0
a4
-5.0E-07
0
a5
1.467E-08
0
a6
-1.6E-10
0
Next, we differentiate our expression of X(z) to find dX/dz and knowing that
æ dX ö
ln ç
÷ = ln a + n ln (1 - X )
è dz ø
kC n A
where, a = Ao c
FAo
æ dX ö
÷ as a function of ln (1 - X ) gives us similar values of slope and intercept
è dz ø
Linear regression of ln ç
as in the finite differences. Using Polymath for linear regression, we get the following result
7-5
Polymath Output:
Model: ln(dxdz) = a0 + a1*ln(1-X)
Variable
a0
a1
Value
-5.531947
1.2824279
95% confidence
0.0241574
0.3446187
n = 1.28
ln a = -5.53, a = 0.00396
k=
FAO a
n
C AO
AC
=
45.7 ×10−6 moles / s
3.96 ×10−3 cm = 4.0s−1
−6
3
2
2.3×10 moles / cm × 0.0196cm
(
)
Hence rate law is,
−rA = 4.0C 1.28
A
mol
dm3 is
P7-5
Constant volume batch reactor:
Mole balance:
-
Aà B
dC A
= kC Aa
dt
Integrating with initial condition when t = 0 and CA = CAO for a ¹ 1.0
t=
(1-a )
- CA(1-a ) ……………substituting
1 CAO
1 (2)(1-a ) - CA(1-a ) for initial concentration CAO = 4 mol/dm3.
=
k
(1 - a )
k
(1 - a )
CA (mol/dm3)
4
2.25
1.45
1
0.65
0.25
0.06
0.008
t (min.)
0
5
8
10
12
15
17.5
20
(See Polymath tutorials on regression
(http://www.umich.edu/~elements/6e/tutorials/Polymath_tutorials.html))
Polymath Output:
Model: t = ((4^(1-alpha))-Ca^(1-alpha))/(k*(1-alpha))
Variable Initial guess Value
95% confidence
alpha
2.
0.532479
0.034828
k
1.
0.1963504 0.0031774
Nonlinear regression settings
Max # iterations = 64
7-6
Precision
R^2
0.9989802
R^2adj 0.9988102
Rmsd
0.0699673
Variance 0.0522179
Source data points and calculated data points
Ca
t
t calc
Delta t
1 4
0
0
0
2 2.25
5
4.912288 0.0877122
3 1.45
8
7.867621 0.1323793
4 1
10
9.934279 0.0657208
5 0.65
12
11.92141 0.0785877
6 0.25
15
15.13018 -0.1301777
7 0.06
17.5 17.90411 -0.4041085
8 0.008 20
19.688
0.3120008
Hence,
k= 0.2 (mol/dm3)0.5min-1 and α = 0.5
P7-6 (a)
Liquid phase irreversible reaction:
A à B + C ; CAO = 2 mole/dm3
C AO - C A
t
= kC aA
æ C - CA ö
ln ç AO
÷ = ln k + a ln C A
t
è
ø
Space time (t )min.
15
38
100
300
1200
C (mol/dm )
1.5
1.25
1.0
0.75
0.5
A
3
ln(CA)
0.40546511
0.22314355
0
-0.28768207
-0.69314718
ln((CAO-CA)/ t )
-3.4011974
-3.9252682
-4.6051702
-5.4806389
-6.6846117
By using linear regression in polymath:
(See Polymath tutorials on regression
(http://www.umich.edu/~elements/6e/tutorials/Polymath_tutorials.html))
Polymath Output:
Model: y = a0 + a1*lnCa
æ C - CA ö
ln ç AO
÷ = ln k + a ln C A
t
è
ø
Variable Value 95% confidence
a0
-4.6080579
0.0162119
a1
2.9998151
0.0411145
7-7
Statistics
R^2 =
R^2adj =
Rmsd =
Variance =
0.9999443
0.9999258
0.003883
1.256E-04
Hence,
a = slope º 3
ln(k) = intercept = -4.6
therefore, k = 0.01 mole-2min-1.
Rate law:
-
dC A
= 0.01C A3 mol / dm3 min
dt
P7-6 (b) Individualized solution
P7-6 (c) Individualized solution
P7-7 (a)
Constant volume batch reactor:
Mole balance:
-
Aà B +C
dC A
= kC Aa
dt
Integrating with initial condition when t = 0 and CA = CAO for a ¹ 1.0
t=
(1-a )
- CA(1-a ) 1 (2)(1-a ) - CA(1-a )
1 CAO
=
……substituting for initial concentration CAO = 2 mol/dm3.
k
(1 - a )
k
(1 - a )
t (min.)
0
5
9
15
22
30
40
60
CA (mol/dm3)
2
1.6
1.35
1.1
0.87
0.70
0.53
0.35
7-8
By using nonlinear regression in polymath:
(See Polymath tutorials on regression
(http://www.umich.edu/~elements/6e/tutorials/Polymath_tutorials.html))
Polymath Output:
Model: t = (1/k)*((2^(1-alfa))-(Ca^(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
R^2 = 0.9997773
R^2adj = 0.9997327
Rmsd = 0.1007934
Variance = 0.0995612
K= 0.03 (mol/dm3)-0.5min-1 and a = 1.5
P7-7 (b) Individualized solution
P7-7 (c) Individualized solution
P7-7 (d) Individualized solution
P7-8 (a)
At t = 0, there is only (CH3)2O. At t = ∞, there is no(CH3)2O. Since for every mole of (CH3)2O consumed
there are 3 moles of gas produced, the final pressure should be 3 times that of the initial pressure.
P(∞) = 3P0
931 = 3P0
P0 ≈ 310 mm Hg
P7-8 (b)
Constant volume reactor at T = 504°C = 777 K
Data for the decomposition of dimethylether in a gas phase:
Time
PT(mm Hg)
0
312
390
408
777
488
(CH 3 ) 2 O ® CH 4 + H 2 + CO
y A0 = 1
d = 3 -1 = 2
e = d y A0 = 2
æP ö
V = V0 ç 0 ÷ (1 + e X ) = V0 because the volume is constant.
èPø
P = P0 (1 + e X )
at t = ∞, X = XAF = 1
7-9
1195
562
3155
799
¥
931
N dX
1 dN A
= - A0
= rA
V dt
V0 dt
Assume -rA = kCA (i.e. 1st order)
CA = CA0 (1 - X ) (V is constant)
dX
= kC A0 (1 - X )
dt
P - P0
and X =
e P0
Then: C A0
Therefore:
dX
1 dP
=
dt e P0 dt
é P - P0 ù k
1 dP
= k ê1 ([1 + e ]P0 - P )
ú=
e P0 dt
e P0 û e P0
ë
or
dP
= k ([1 + e ]P0 - P )
dt
P
t
dP
òP [1 + e ]P0 - P = ò0 kdt
0
é
ù
é 2 P0 ù
e P0
é 624 ù
ú = ln ê
ú = ln ê
ú = kt
ë 936 - P û
ë 3P0 - P û
ëê (1 + e )P0 - P ûú
Integrating gives: ln ê
Therefore, if a plot of ln
624
versus 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:
-rA = 0.00048C A
1.6
y = 0.00048x - 0.02907
1.2
0.8
0.4
0
-0.4
0
1000
2000
3000
4000
P7-8 (c) Individualized solution
P7-8 (d) The rate constant would increase with an increase in temperature. This would result in the
pressure increasing faster and less time would be need to reach the end of the reaction. The opposite is
true for colder temperatures.
7-10
P7-9 (a)
Photochemical decay of bromine in bright sunlight:
t (min)
CA (ppm)
10
2.45
20
1.74
30
1.23
40
0.88
50
0.62
60
0.44
Mole balance: constant V
dC A
dt
= rA = −kC αA
" dC %
ln$$ − A '' = ln k + αln C A
# dt &
()
( )
Differentiation
T (min)
Δt (min)
C (ppm)
ΔC (ppm)
A
A
DC A æ ppm ö
ç
÷
Dt è min ø
10
20
10
2.45
30
10
1.74
40
10
1.23
50
10
0.88
60
10
0.62
-0.71
-0.51
-0.35
-0.26
-0.18
-0.071
-0.051
-0.035
-0.026
-0.018
7-11
10
0.44
After plotting and differentiating by equal area
-dC /dt
ln(-dC /dt)
ln C
0.082
-2.501
0.896
A
A
A
0.061
-2.797
0.554
0.042
-3.170
0.207
0.030
-3.507
-0.128
0.0215
-3.840
-0.478
0.014
-4.269
-0.821
Using linear regression: α = 1.0
ln k = -3.3864
k = 0.0344 min-1
P7-9 (b)
dNA
dt
= −VrA −FB = 0**,**FB = –rA V
rA = -0.0344
ppm
mg
at CA = 1 ppm
= -0.0344
min
l min
æ
mg ö æ min ö æ 1 g öæ 3.7851l öæ 1lbs ö
lbs
FB = (25000 gal )ç 0.0344
÷ ç 60
֍
֍
÷ = 0.426
֍
l min ø è
hr ø è 1000 mg øè 1 gal øè 453.6 g ø
hr
è
P7-9 (c) Individualized solution
P7-10
For the reaction,
Rate law:
Oz + wall à loss of O3
k1
Oz + alkene à products
k2
-rO z =
dCO z
dt
m n
= k 1CO z + k 2CO
C
z Bu
Using Polymath nonlinear regression, we can find the values of m, n, k1 and k2
(See Polymath tutorials on regression
(http://www.umich.edu/~elements/6e/tutorials/Polymath_tutorials.html))
7-12
Polymath Output:
7-13
P7-11
Given: Plot of percent decomposition of NO2 vs V/FA0
X=
%DecompositionofNO2
100
Assume that - rA = kC An
For a CSTR V =
or
FA0 X
-rA
V
X
X
=
=
FA0 -rA kC An
with n = 0, X = k
V
FA0
X has a linear relationship with
V
as
FA 0
shown in the figure.
Therefore, the reaction is zero order.
P7-12
(See Polymath tutorials on regression
(http://www.umich.edu/~elements/6e/tutorials/Polymath_tutorials.html))
Polymath Output:
7-14
7-15
Page intentionally blank
7-16
Synopsis for Chapter 8 – Multiple Reactions
General: Polymath is required for most all solutions except P8-2, P8-4, P8-5, P8-6, P8-7, P8-8 and P8-18.
Questions
l Q8-1A (17 seconds) Questions Before Reading (QBR).
O Q8-2A (10 min) i>Clicker questions.
I Q8-3A (15 min) Part (a) This question would go better in Chapter 10.
I Q8-4B (30 min) This simulation is not as complete as it should be because we could not find out if
the antivenom is adsorbed on the receptor sites.
O Q8-5B (8 min) Useful screencasts to reinforce the material in Chapter 8.
Computer Simulations and Experiments
I usually assign all Wolfram LEP Problems in each chapter.
l P8-1A (a) (10 minutes max) The point here is that when we increase the conversion the selectivity
decreases.
O P8-1A (b) (15 min) Not a Wolfram problem, but the table the students construct is a very useful table
comparing the various reactors.
l P8-1A (c) – (h) (10-12 min) The students should spend between 10 and 12 minutes on each Wolfram
simulation including writing conclusions.
l P8-1A (m) and (n) (Time Varies) Requires studying material from the web. For the (m) blood
coagulation and the (n) oscillating reactions, these are quite straight forward and give interesting
results.
Interactive Computer Games (ICG)
l P8-2A (25 min) This game is based on the TV show The Amazing Race.
Problems
O P8-3B (30 min) Interesting problem because of the two different time scales for the parallel
reactions. One very fast reaction and one very slow reaction.
I P8-4A (45 min) Problem is very very similar to the Example 8-1.
l P8-5B (40 min) Old Exam Question (OEQ). Excellent problem. Brings awareness to drinking and
driving. Almost always assigned
AA P8-6B (35 min) Old Exam Question (OEQ). A more complicated version of P8-5B. The students take a
long time to realize they need to take two doses immediately followed by one dose at regular
intervals. One should note the next time they get a prescription if it says take 2 pills at once followed
by 1 pill every 4 hours.
AA P8-7C (20 min) Suggest reactor schemes to minimize undesired product.
O P8-8B (30 min) Old Exam Question (OEQ). Straight forward problem to calculate selectivity in a CSTR.
S8-1
O P8-9B (30 min) Old Exam Question (OEQ). Straight forward batch reactor problem if one uses
Polymath.
AA P8-10B (50 min) Conceptually a little difficult because of the autocatalytic effect and reaction
stoichiometry.
AA P8-11A (45 min) Very important to ask the student to sketch the results before solving numerically.
Very good first problem with multiple reactions because a computer is not necessary. Problem leads
the students through the algorithm
AA P8-12B (75 min) Very thorough problem on multiple reactions. Both gas and liquid phases are
considered. I don’t always assign all parts (d) through (g).
AA P8-13B (45 min) Sketch results first. Similar to P8-12B, but for a membrane reactor with pressure
drop.
O P8-14B (55 min) Sketch results first. Real reactors and real data used in a PFR. One part requires
pressure drop analysis.
I P8-15C (50 min) Real data using partial pressures in the rate law.
l P8-16B (50 min) Old Exam Question (OEQ). Current application of solar energy to carryout analysis
of multiple reactions in a PFR. I usually assign this problem.
AA P8-17B (50 min) Similar to P8-16B.
S P8-18A (Time Varies) Interactive Reactor Laboratory problem of Professor Emeritus Herz,
Department of Chemical Engineering, University of California, San Diego.
S8-2
Solutions for Chapter 8 – Multiple Reactions
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-8
http://umich.edu/~elements/5e/08chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q8-1 Individualized solution
Q8-2 Individualized solution
Q8-3 (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.
After 11 bites, no amount of antivenom can keep the number of free sites above 66.7% of total sites. This
means that the initial concentration of venom would be 5.5e-8 M. The best result occurs when a dose of
antivenom such that the initial concentration of antivenom in the body is 5.7e-8 M, will result in a
minimum of 66.48% free sites, which is below the allowable minimum.
Q8-3 (b)
The victim was bitten by a harmless snake and antivenom was injected. This means that the initial
concentration of venom is 0. From the program below, we see that if an amount of antivenom such that
the initial concentration in the blood is 7e-9 M, the patient will die.
See the following Polymath program:
Polymath Code:
d(fsv)/d(t) = kv * fs * Cv - ksv * fsv * Ca
d(fs)/d(t) = -kv*fs*Cv - ka * fs * Ca + kia * fsa + g
d(Cv)/d(t) = Cso * (-kv * fs * Cv - ksa * fsa * Cv) + h
d(Ca)/d(t) = Cso*(-ka * fs * Ca + kia * fsa) + j
d(fsa)/d(t) = ka * fs * Ca - kia * fsa - ksa * fsa * Cv
d(Cp)/d(t) = Cso * (ksv * fsv * Ca + ksa * fsa * Cv) + m
kv = 2e8
ksv = 6e8
ka = 2e8
kia = 1
Cso = 5e-9
ksa = 6e8
kp = 1.2e9
kov = 0
koa = 0.3
kop = 0.3
g = ksa * fsa * Cv + ksv * fsv * Ca
h = -kp * Cv * Ca - kov * Cv
m = kp * Cv * Ca - kop * Cp
j = -Cso * ksv * fsv * Ca - kp * Cv * Ca - koa * Ca
8-1
t(0)=0
t(f)=0.5
fsv(0)=0
fs(0)=1
Cv(0)=0
Ca(0)=7e-9
fsa(0)=0
Cp(0)=0
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
t
0
0
0.5
0.5
fsv
0
0
0
0
fs
1
0.6655661
1
0.6655661
Cv
0
0
0
0
Ca
7.0E-09
4.503E-09
7.0E-09
4.503E-09
fsa
0
0
0.3344339
0.3344339
Cp
0
0
0
0
kv
2.0E+08
2.0E+082.0E+08
2.0E+08
ksv
6.0E+08
6.0E+086.0E+08
6.0E+08
ka
2.0E+08
2.0E+082.0E+08
2.0E+08
kia
1
1
1
1
Cso
5.0E-09
5.0E-095.0E-09
5.0E-09
ksa
6.0E+08
6.0E+086.0E+08
6.0E+08
kp
1.2E+09
1.2E+091.2E+09
1.2E+09
kov
0
0
0
0
koa
0.3
0.3
0.3
0.3
kop
0.3
0.3
0.3
0.3
g
0
0
0
0
h
0
0
0
0
m
0
0
0
0
j
-2.1E-09
-2.1E-09
-1.351E-09
-1.351E-09
ODE Report (STIFF)
Differential equations as entered by the user
[1] d(fsv)/d(t) = kv * fs * Cv - ksv * fsv * Ca
[2] d(fs)/d(t) = -kv*fs*Cv - ka * fs * Ca + kia * fsa + g
[3] d(Cv)/d(t) = Cso * (-kv * fs * Cv - ksa * fsa * Cv) + h
[4] 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
Explicit equations as entered by the user
[1] kv = 2e8
[2] ksv = 6e8
[3] ka = 2e8
[4] kia = 1
[5] Cso = 5e-9
[6] ksa = 6e8
[7] kp = 1.2e9
[8] kov = 0
[9] koa = 0.3
[10] kop = 0.3
[11] g = ksa * fsa * Cv + ksv * fsv * Ca
[12] h = -kp * Cv * Ca - kov * Cv
[13] m = kp * Cv * Ca - kop * Cp
[14] j = -Cso * ksv * fsv * Ca - kp * Cv * Ca - koa * Ca
8-2
Q8-3 (c)
The latest time after being bitten that antivenom can successfully be administered is 27.49 minutes. See
the cobra web module on the CDROM/website for a more detailed solution to this problem
Q8-3 (d) Individualized Solution
Q8-4 Individualized Solution
Q8-5 Individualized Solution
P8-1 (a) Example 8-1
(i)
Selectivity increases with an increase in k2 and decrease in k1 and k3 for both PFR and CSTR. It is
obvious after seeing the expression for selectivity below.
(ii)
(iii)
Conversion increases with increase in all of k1, k2 and k3.
1. Volumetric flow rate has no effect on selectivity.
2. k1 has the most effect on conversion as well as selectivity profiles. This is because rate
involving k1 is independent of any concentration term whereas k2 and k3 have Ca multiplied
with them and Ca being small reduces the overall term.
For PFR (gas phase with no pressure drop or liquid phase),
dC A
= - k1 - k 2 C A - k 3 C A2
dt
dC B
= k 2C A
dt
dC X
= k1
dt
dC Y
= k 3 C A2
dt
In PFR with V = 1566 dm3 we get τ = V/v0 = 783
X = 0.958 , SB/XY (instantaneous) = 0.244, and SB/XY (overall) = 0.624
also, at τ = 350, SB/XY (instantaneous) is at its maximum value of 0.84
8-3
Polymath code:
d(Ca) / d(tau) = -k1-k2*Ca-k3*Ca^2
d(Cx) / d(tau) = k1
d(Cb) / d(tau) = k2*Ca
d(Cy) / d(tau) = k3*Ca^2
Cao = 0.4
X = 1-Ca/Cao
k1 = 0.0001
k2 = 0.0015
k3 = 0.008
Sbxy_overall = Cb/(Cx+Cy)
Sbxy_instant = k2*Ca/(k1+k3*Ca^2)
Ca(0) = 0.4
Cx(0) = 1.0E-07
Cb(0) = 0
Cy(0) = 1.0E-07
tau(0) = 0
tau(f) = 783
Polymath output
Calculated values of DEQ variables
Variable
Initial value Minimal value Maximal value Final value
1 Ca
0.4
0.0166165
0.4
0.0166165
2 Cao
0.4
0.4
0.4
0.4
3 Cb
0
0
0.1472919
0.1472919
4 Cx
1.0E-07
1.0E-07
0.0783001
0.0783001
5 Cy
1.0E-07
1.0E-07
0.1577917
0.1577917
6 k1
0.0001
0.0001
0.0001
0.0001
7 k2
0.0015
0.0015
0.0015
0.0015
8 k3
0.008
0.008
0.008
0.008
9 Sbxy_instant 0.4347826
0.2438609
0.8385255
0.2438609
10 Sbxy_overall 0
0
0.6437043
0.6238755
11 tau
0
0
783.
783.
12 X
0
0
0.9584587
0.9584587
Differential equations
1 d(Ca)/d(tau) = -k1-k2*Ca-k3*Ca^2
2 d(Cx)/d(tau) = k1
3 d(Cb)/d(tau) = k2*Ca
4 d(Cy)/d(tau) = k3*Ca^2
8-4
Explicit equations
1 Cao = 0.4
2 X = 1-Ca/Cao
3 k1 = 0.0001
4 k2 = 0.0015
5 k3 = 0.008
6 Sbxy_overall = Cb/(Cx+Cy)
7 Sbxy_instant = k2*Ca/(k1+k3*Ca^2)
(iv) PFR - Pressure increased by a factor of 100.
(a) Liquid phase: No change, as pressure does not change the liquid volume appreciably.
(b) Gas Phase:
Now CA0 = P/RT = 100 (P0_initial)/RT = 0.4 × 100 = 40 mol/dm3
Running with CA0 = 40 mol/dm3,we get:
X = 0.998 , SB/XY (instantaneous) = 0.681, and SB/XY (overall) = 0.024
As a practice, students could also try similar problem with the CSTR.
P8-1 (b) Example 8-2
(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.
(f) and (g) Tubular with side streams: selectivity i.e. 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, for highly exothermic reactions
(j) CSTR with recycle: Low conversions are achieved to reuse reactants, temperature control, liquid
reactants, for highly exothermic reactions
(k) Membrane Reactor: yield i.e. series reactions that eliminate a desired product, used for
thermodynamically limited reactions where the equilibrium shifts on left side.
(l) Reactive Distillation: when one product is volatile and the other is not; for thermodynamically
limited gas phase reactions.
8-5
P8-1 (c) Example 8-3
(i)
After setting E1 and E2 equal to zero, both selectivity and yield decreases as k1 and k2 increases.
K2 has no effect on Ca while it increases Cc and decreases Cb. With an increase in k1, Ca decreases
whereas Cb and Cc increases for a given time.
(ii) With an increase in E1 and E2, both selectivity and yield decreases E2 creating a higher impact.
Only E1 affects conversion as it affects the rate constant k1 and conversion is defined on the basis
of conc of A.
(iii) 1. If temp increases, then conc of all A, B and C reaches their steady state value at an early point
indicating that reaction is progressing at a faster rate because of increase in rate constants.
This is the reason for an increase in conversion.
2. From the rate expressions, we can conclude that rate of formation of B and C is dependent on
both k1 and k2 while for A it is only dependent on k1.
P8-1 (d) Example 8-4
(i)
After setting E1 and E2 both equal to zero, increase in k2 decreases both selectivity and yield while
k1 has no effect on both. CA decreases with an increase in k1 and vice-versa for both Cb and Cc. k2
has no effect on Ca while Cc increases with an increase in k2 whereas Cb decreases.
(ii) 1. Concentration of B increases with k1 while decreases with k2. Thus, as temperature is
increased, both k1 and k2 increases but k2 having a greater effect makes the Cb to decrease.
2. Selectivity and yield are almost affected by the same set of parameters.
(iii) According to the plot of CB versus T the maximum CB is produced at temperature of 310.52°C.
CB =
t ¢k1C A0
(t ¢k 2 + 1)(t ¢k1 + 1)
Polymath code:
d(T) / d(t) = 1
Cao = 5
tau = 0.5
k1o = 0.4
k2o = 0.01
E1 = 10
E2 = 20
R = 0.001987
k1 = k1o*exp((-E1/R)*(1/T-1/300))
k2 = k1o*exp((-E2/R)*(1/T-1/300))
Cb = (tau*k1*Cao)/(tau*k2+1)/(tau*k1+1)
T(0) = 300
t(0) = 0
t(f) = 100
8-6
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Cao
5.
5.
5.
5.
2 Cb
0.6944444
0.0052852
0.8039654
0.0052852
3 E1
10.
10.
10.
10.
4 E2
20.
20.
20.
20.
5 k1
0.4
0.4
26.51303
26.51303
6 k1o
0.4
0.4
0.4
0.4
7 k2
0.4
0.4
1757.352
1757.352
8 k2o
0.01
0.01
0.01
0.01
9 R
0.001987
0.001987
0.001987
0.001987
10 T
300.
300.
400.
400.
11 t
0
0
100.
100.
12 tau
0.5
0.5
0.5
0.5
Differential equations
1 d(T)/d(t) = 1
Explicit equations
1 Cao = 5
2 tau = 0.5
3 k1o = 0.4
4 k2o = 0.01
5 E1 = 10
6 E2 = 20
7 R = 0.001987
8 k1 = k1o*exp((-E1/R)*(1/T-1/300))
9 k2 = k1o*exp((-E2/R)*(1/T-1/300))
10 Cb = (tau*k1*Cao)/(tau*k2+1)/(tau*k1+1)
P8-1 (e)
(i)
When Cto increases, selectivity decreases whereas yield Yc decreases and Yd increases upto a
value of 0.2
(ii) Increase in k has also the same effects on both selectivity and yield as Cto.
(iii) 1. Cto has the greatest effect on conversion profile, also with an increase in both rate constants
conversion tends to increase.
2. Alpha term has effect on only pressure drop term which is very significant too as it appears in
differential equation of pressure drop.
3. Increase in rate constants as well as initial total conc decreases the molar flow rates of A, B
and C while increasing Fd. This is due to an increase in conversion.
(iv) Modify the Polymath code of LEP-8-5
Only expression of r1a needs to be changed to include reversible reactions as shown below
Polymath Code
d(Fa)/d(W) = ra #
d(Fb)/d(W) = rb #
d(Fc)/d(W) = rc #
d(Fd)/d(W) = rd #
d(p) / d(W) = -alpha/2/p*(Ft/Fto)
Fto=20
alpha=0.0019
Ft = Fa+Fb+Fc+Fd #
r1a = -k1a*(Ca*Cb^2-Cc/Kc)
8-7
Kc=0.002 #
r2c = -k2c*Ca^2*Cc^3 #
r1b = 2*r1a #
r1c = -r1a #
r2a = 2/3*r2c #
r2d = -1/3*r2c #
ra = r1a+r2a #
rb = r1b #
rc = r1c+r2c #
rd = r2d #
Ca = Cto*(Fa/Ft)*p #
Cb = Cto*(Fb/Ft)*p #
Cc = Cto*(Fc/Ft)*p #
Cd = Cto*(Fd/Ft)*p #
k1a = 100 #
k2c = 1500 #
v = 100 #
Fbo = 10
Fao = 10
Xb = (Fbo-Fb)/Fbo
Xa = (Fao-Fa)/Fao
Cto = 0.2 #
Scd =if(W>0.0001)then(Fc/Fd)else(0)
W(0)=0
Fa(0)=10
Fb(0)=10
Fc(0)=0
Fd(0)=0
W(f) =1000
p(0) = 1
Run the code and then compare the results with Example 8-5
P8-1 (f) Cbo has the greatest effect on selectivity
P8-1 (g)
(i)
Increasing Fao increases conversion but decreases selectivity.
(ii) Molar flow rate Fao has the highest effect on selectivity.
(iii) 1. Volumetric flow rate mostly has effect on flow rates of A, B and C. Na and Nb increases with
an increase in vo while Nc decrease.
2. Fao has greatest effect on Na. It is pretty obvious after seeing the differential equations below
P8-1 (h) Example 8-8
(i)
With an increase in CT0, FD increases while other flow rates FA, FB and FU decreases. With an increase
in k1a, formation of desired product, FD increases while increase in k2a increases molar flow rate of
undesired product FU. Selectivity increases with an increase of k1a and CT0 but decreases with an
increase in k2a. Conversion increases with an increase in k1a, k2a and CTO
(ii) 1) Despite of increase in molar flow rate of undesired product by increase in K2a, conversion still
increases with K2a although slightly. This is because increasing K2a pushes reaction to forward
direction, hence conversion rises.
2) Total Volume Vt has the greatest effect on conversion profile while K1a has greatest effect on
selectivity profile.
8-8
(iii) FA0=2*4=8 mol/s
Change the value of FA0 in the polymath code of LEP-8-8 to 8 mol/s
Doubling the incoming flow rate of species A increases the selectivity to 5.36.
If instead of FA0, Fbo is doubled then,
Doubling the incoming flow rate of species B decreases the selectivity to 1.01.
Membrane Reactor
SD/U Original Problem
2.58
SD/U P8-2 h
5.36
Doubling the incoming flow rate of species A lowers the selectivity.
when the first reaction is changed to A+2B à D
The answers will not change since the rate law is asked to be kept same
P8-1 (i) Aspen Problem: No solution will be given.
P8-1 (j)
For equal molar feed in hydrogen and mesitylene.
CHO = yHOCTO = (0.5)(0.032)lbmol/ft3 =0.016 lbmol/ft3
CMO = 0.016 lbmol/ft3
Using equations from example, solving in Polymath, we get
t opt = 0.38 hr. At t = 0.5 hr all of the H2 is reacted and only the decomposition of X takes place.
XH
CH
CM
CX
t
S"#⁄$
CD-ROM
example
0.50
0.0105
0.0027
0.00507
0.2hr
0.596
See the following Polymath program:
Polymath code:
d(CH) / d(tau) = r1H + r2H
CH(0) = 0.016
d(CM) / d(tau) = r1M
CM(0) = 0.016
d(CX) / d(tau) = r1X + r2X
CX(0) = 0
k1 = 55.2
k2 = 30.2
r1M = -k1*CM*(CH^0.5)
r2T = k2*CX*(CH^0.5)
r1H = r1M
r2H = -r2T
r1X = -r1M
r2X = -r2T
tau(0) = 0
tau(f) = 0.43
8-9
This question
0.99
0.00016
0.0042
0.0077
0.38hr
1.865
Polymath Output
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
tau
0
0
0.43
0.43
CH
0.016
1.64E-06
0.016
1.64E-06
CM
0.016
0.0041405
0.016
0.0041405
CX
0
0
0.0077216
0.0077207
k1
55.2
55.2
55.2
55.2
k2
30.2
30.2
30.2
30.2
r1M
-0.1117169
-0.1117169
-2.927E-04
-2.927E-04
r2T
0
0
0.0159818
2.986E-04
r1H
-0.1117169
-0.1117169
-2.927E-04
-2.927E-04
r2H
0
-0.0159818
0
-2.986E-04
r1X
0.1117169
2.927E-04
0.1117169
2.927E-04
r2X
0
-0.0159818
0
-2.986E-04
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(CH)/d(tau) = r1H+r2H
[2] d(CM)/d(tau) = r1M
[3] d(CX)/d(tau) = r1X+r2X
Explicit equations as entered by the user
[1] k1 = 55.2
[2] k2 = 30.2
[3] r1M = -k1*CM*(CH^.5)
[4] r2T = k2*CX*(CH^.5)
[5] r1H = r1M
[6] r2H = -r2T
[7] r1X = -r1M
[8] r2X = -r2T
Increasing θH decreases τopt and S"#⁄$ .
P8-1 (k)
From web example
Polymath code:
f(CX) = (K1*CM*CH^.5-K2*CX*CH^0.5)*tau-CX
CX(0) = 0.002
f(CM) = CM-CMo+K1*CM*CH^.5*tau
CM(0) = 0.013
f(CH) = CH-CHo+K1*(CM*CH^.5+K2*CX*CH^.5)*tau
CH(0) = 1e-04
tau = 0.5
K1 = 55.2
K2 = 30.2
CHo = 0.016
CMo = 0.016
Polymath Output
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 CH
0.00004784 -4.8886E-11 0.0001
2 CM
0.01343528 -1.0467E-11 0.013
3 CX
0.00232219 -9.7709E-12 0.002
8-10
Variable Value
1 CHo
0.016
2 CMo
0.016
3 K1
55.2
4 K2
30.2
5 tau
0.5
Nonlinear equations
1 f(CX) = (K1*CM*CH^.5-K2*CX*CH^0.5)*tau-CX = 0
2 f(CM) = CM-CMo+K1*CM*CH^.5*tau = 0
3 f(CH) = CH-CHo+K1*(CM*CH^.5+K2*CX*CH^.5)*tau = 0
Explicit equations
1 tau = 0.5
2 K1 = 55.2
3 K2 = 30.2
4 CHo = 0.016
5 CMo = 0.016
A plot using different values of t is given.
For t =0.5, the exit concentration are
CH = 4.8 ×10-5lbmol/ft3 CM =0.0134 lbmol/ft3
CX =0.00232 lbmol/ft3
The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for t =0.5:
YMX =
FX
CX
0.00232
0.89mole × xylene × produced
=
=
=
FMO - FM C MO - C M 0.016 - 0.0134
mole × mesitylene × reacted
The overall selectivity of xylene relative to toluene is:
F
8.3mole × xylene × produced
~
S X /T = X =
FT
mole × toluene × produced
CH
CM
CX
t
YMX
SX/T
CD-ROM example
0.0089
0.0029
0.0033
0.5
0.41
0.7
This Question
4.8 x 10-5
0.0134
0.00232
0.5
0.89
8.3
P8-1 (l) Individualized solution
8-11
P8-1 (m)
At the beginning, the reactants that are used to create TF-VIIa and TF-VIIaX are in high concentration. As
the two components are created, the reactant concentration drops and equilibrium forces the production
to slow. At the same time the reactions that consume the two components begin to accelerate and the
concentration of TF-VIIa and TF-VIIaX decrease. As those reactions reach equilibrium, the reactions that
are still producing the two components are still going and the concentration rises again. Finally the
reactions that consume the two components lower the concentration as the products of those reactions
are used up in other reactions.
P8-1 (n) Individualized solution
P8-2The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer
Games (ICGs) is given at the front of this Solutions Manual.
P8-3 (a)
Plot of CA, CD and CU as a function of time (t):
Polymath code:
d(Ca) / d(t) = -(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))
d(Cd) / d(t) = k1*(Ca-Cd/K1a)
d(Cu) / d(t)= k2*(Ca-Cu/K2a)
Cu(0) = 0
Ca(0) = 1
Cd(0) = 0
t(0) = 0
t(f) = 15
k1 = 1.0
k2 = 100
K1a = 10
K2a = 1.5
Cao = 1
X = 1 - Ca/Cao
8-12
Polymath Output:
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
t
0
0
15
15
Ca
1
0.0801802
1
0.0801802
Cd
0
0
0.7995475
0.7995475
Cu
0
0
0.5302179
0.1202723
k1
1
1
1
1
k2
100
100
100
100
K1a
10
10
10
10
K2a
1.5
1.5
1.5
1.5
Cao
1
1
1
1
X
0
0
0.9198198
0.9198198
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Ca)/d(t) = -(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))
[2] d(Cd)/d(t) = k1*(Ca-Cd/K1a)
[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-Ca/Cao
To maximize CD stop the reaction after a long time. The concentration of D only increases with time
P8-3 (b)
Conc. Of U is maximum at t = 0.31 min.(CA = 0.53)
P8-3 (c)
Equilibrium concentrations:
CAe = 0.08 mol/dm3
CDe = 0.8 mol/dm3
CUe = 0.12 mol/dm3
P8-3 (d)
Polymath code (for t = 100 min):
8-13
f(Ca) = Ca0-t*(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))-Ca
Ca(0) = 1
f(Cd) = t*k1*(Ca-Cd/K1a)-Cd
Cd(0) = 0
f(Cu) = t*(k2*(Ca-Cu/K2a))-Cu
Cu(0) = 0
Ca0 = 1
k1 = 1
k2 = 100
K1a = 10
K2a = 1.5
t = 100
X = 1-Ca/Ca0
Polymath Output:
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 Ca
0.0862755 1.031E-05 1.
2 Cd
0.7843227 2.075E-08 0
3 Cu
0.1293938 -2.312E-06 0
Variable Value
1 Ca0
1.
2 k1
1.
3 K1a
10.
4 k2
100.
5 K2a
1.5
6 t
100.
7 X
0.9137245
Nonlinear equations
1 f(Ca) = Ca0-t*(k1*(Ca-Cd/K1a)+k2*(Ca-Cu/K2a))-Ca = 0
2 f(Cd) = t*k1*(Ca-Cd/K1a)-Cd = 0
3 f(Cu) = t*(k2*(Ca-Cu/K2a))-Cu = 0
Explicit equations
1 Ca0 = 1
2 k1 = 1
3 k2 = 100
4 K1a = 10
5 K2a = 1.5
6 t = 100
7 X = 1-Ca/Ca0
t
CAexit
CDexit
CUexit
X
1 min
0.295
0.2684
0.436
0.705
10 min
0.133
0.666
0.199
0.867
100min
0.0862
0.784
0.129
0.914
8-14
P8-4 (a)
1/2
1/2
A
! mol $
k1 = 0.004 # 3 & min
" dm %
A→ X
rX = k1C
A→ B
rB = k2C A
k2 = 0.3min −1
A→Y
rY = k3C A2
k3 = 0.25
dm3
mol.min
Sketch SBX, SBY and SB/XY as a function of CA
It shows the table of SBX, SBY and SB/XY in correspondence with values of CA.
(1)
r
k C
SB/X = B = 2 A
=
(2)
r
k C
SB/Y = B = 2 A =
k2
k 3C A
(3)
SB/XY =
rX
rY
k1C A1 / 2
k 3C A 2
k2
C A1 / 2
k1
rB
k 2C A
=
1
rX + rY k1C A / 2 + k 3C A 2
8-15
P8-4 (b)
Volume of first reactor can be found as follows
We have to maximize SB/XY
From the graph above, maximum value of SBXY = 10 occurs at CA* = 0.040 mol/dm3
So, a CSTR should be used with exit concentration CA*
Also, CA0 = PA/RT = 0.162 mol/dm3
1/ 2
And - rA = rX + rB + rY = (k1C A
+ k 2 C A + k 3C A 2 )
v0 (C A0 - C A* )
v0 (C A0 - C A* )
=> V =
=
= 92.4dm 3
* 1/ 2
*
* 2
- rA
(k1 (C A ) + k 2C A + k3 (C A ) )
P8-4 (c)
Effluent concentrations:
CB
C
mol
= B Þ C B * = 0.11 3
rB k 2C A
dm
mol
mol
*
*
C
=
0
.
0037
Similarly: C X = 0.007
and
Y
dm 3
dm 3
We know, τ = 9.24 min => t =
P8-4 (d)
Conversion of A in the first reactor:
C A0 - C A = C A0 X Þ X = 0.74
P8-4 (e)
A CSTR followed by a PFR should be used.
Required conversion = 0.99
=> For PFR, Mole balance:
0.99
Þ V = 10 ´ 0.162 ´ ò
dV FA0
=
dX - rA
dX
1/ 2
+ k 2 C A + k 3C A 2 )
0.74 ( k1C A
8-16
= 92.8dm 3
P8-4 (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 k1, k2, and k3 in terms of temperature. From the given data we know:
æ - Ei ö
ki = Ai exp ç
÷
è 1.98T ø
Since we have the constants given at T = 300 K, we can solve for Ai.
A1 =
.004
= 1.49e12
æ -20000 ö
exp çç
÷÷
è 1.98 (300 ) ø
.3
A2 =
= 5.79e6
æ -10000 ö
exp çç
÷÷
è 1.98 (300 ) ø
.25
A3 =
= 1.798e21
æ -30000 ö
exp çç
÷÷
è 1.98 (300 ) ø
Now we use a mole balance on species A
FA0 - FA
- rA
v (C A0 - C A )
V=
-rA
C - CA
C A0 - C A
t = A0
=
0.5
-rA
k1C A + k2C A + k3C A2
V=
A mole balance on the other species gives us:
Fi = vCi = rV
i
Ci = t ri
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
8-17
will depend on the price of B 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.
Polymath Code
f(Ca) = (Cao-Ca)/(k1*Ca^.5+k2*Ca+k3*Ca^2)-10
Ca(0) = 0.05
T = 306
R = 1.987
k1 = 1.49e12*exp(-20000/R/T)
k2 = 5790000*exp(-10000/R/T)
Cao = .1
Cb = 10*k2*Ca
k3 = 1.798e21*exp(-30000/R/T)
tau = 10
Cx = tau*k1*Ca^.5
Cy = tau*k3*Ca^2
Ca(min) = 0.005
Ca(max) = 0.1
Sbxy = Cb/(Cx+Cy)
Polymath Output
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 Ca
0.0170216 0.0016427 0.0525 ( 0.005< Ca < 0.1 )
Variable Value
1 Cao
0.1
2 Cb
0.0709474
3 Cx
0.010074
4 Cy
0.0019434
5 k1
0.0077215
6 k2
0.4168076
7 k3
0.6707505
8 R
1.987
9 Sbxy
5.903734
10 T
306.
11 tau
10.
Nonlinear equations
1 f(Ca) = (Cao-Ca)/(k1*Ca^.5+k2*Ca+k3*Ca^2)-10 = 0
Explicit equations
1 T = 306
2 R = 1.987
3 k1 = 1.49e12*exp(-20000/R/T)
4 k2 = 5790000*exp(-10000/R/T)
5 Cao = .1
6 Cb = 10*k2*Ca
7 k3 = 1.798e21*exp(-30000/R/T)
8 tau = 10
9 Cx = tau*k1*Ca^.5
10 Cy = tau*k3*Ca^2
11 Sbxy = Cb/(Cx+Cy)
8-18
P8-4 (g)
Concentration is proportional to pressure in a gas-phase system. Therefore:
S B/ XY ~
PA
PA + PA2
which would suggest that a low pressure would be ideal. But as before the tradeoff is
lower production of B. A moderate pressure would probably be best.
We know that:
SB/XY =
rB
k 2C A
=
1
rX + rY k1C A / 2 + k 3C A 2
Substituting PA = CA R T ;
in the above co – relation we get
SB/XY =
rB
k2 ( PA / RT )
=
rX + rY k1 ( PA / RT )1/2 + k3 ( PA / RT ) 2
Thus, we find that the maximum is obtained at P= 1 atm .
P8-5
US legal limit: 0.8 g/l
Sweden legal limit: 0.5 g/l
k1
k2
A ¾¾
® B ¾¾
®C
Where A is alcohol in the gastrointestinal tract and B is alcohol in the blood stream
dC A
= -k1C A
dt
dCB
= k1C A - k2
dt
k1 = 10 hr -1
k2 = 0.192
g
L hr
8-19
Two tall martinis = 80 g of ethanol
Body fluid = 40 L
80 g
g
=2
40 L
L
Now we can put the equations into Polymath.
Polymath code:
C A0 =
d(Ca) / d(t) = -k1*Ca
Ca(0) = 2
d(Cb) / d(t) = -k2+k1*Ca
Cb(0) = 0
k1 = 10
k2 = 0.192
t(0) = 0
t(f) = 10
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
2.
7.037E-44
2.
7.037E-44
2 Cb
0
0
1.88665
0.08
3 k1
10.
10.
10.
10.
4 k2
0.192
0.192
0.192
0.192
5 t
0
0
10.
10.
Differential equations
1 d(Ca)/d(t) = -k1*Ca
2 d(Cb)/d(t) = -k2+k1*Ca
Explicit equations
1 k1 = 10
2 k2 = 0.192
P8-5 (a)
In the US the legal limit it 0.8 g/L.
This occurs at t = 6.3 hours.
P8-5 (b)
In Sweden CB = 0.5 g/l , t = 7.8 hrs.
P8-5 (c) In Russia CB = 0.0 g/l, t = 10.5 hrs
P8-5 (d)
For this situation we will use the original Polymath code and change the initial concentration of A to 1 g/L.
Then run the Program for 0.5 hours. This will give us the concentration of A and B at the time the second
martini is ingested. This means that 1 g/l will be added to the final concentration of A after a half an hour.
At a half an hour CA = 0.00674 g/L and CB = 0.897 g/L. The Polymath code for after the second drink is
shown below.
Polymath code:
d(Ca) / d(t) = -k1*Ca
Ca(0) = 1.0067379
d(Cb) / d(t) = -k2+k1*Ca
Cb(0) = 0.8972621
k1 = 10
k2 = 0.192
t(0) = 0.5
t(f) = 10
8-20
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
1.006738
5.342E-42
1.006738
5.342E-42
2 Cb
0.8972621
0.08
1.808554
0.08
3 k1
10.
10.
10.
10.
4 k2
0.192
0.192
0.192
0.192
5 t
0.5
0.5
10.
10.
Differential equations
1 d(Ca)/d(t) = -k1*Ca
2 d(Cb)/d(t) = -k2+k1*Ca
Explicit equations
1 k1 = 10
2 k2 = 0.192
For the US t = 6.2 hours
Sweden: t = 7.8 hours
Russia: t =10.3 hours.
P8-5 (e)
The mole balance on A changes if the drinks are consumed at a continuous rate for the first hour. 80 g
of ethanol are consumed in an hour so the mass flow rate in is 80 g/hr. Since volume is not changing the
rate of change in concentration due to the incoming ethanol is 2 g/L/hr.
For the first hour the differential equation for CA becomes:
dC A
= −k1C A + 2 after that it reverts back to the original equations.
dt
Polymath code:
d(Ca) / d(t) = if(t<1)then(-k1*Ca+2*t)else(-k1*Ca)
Ca(0) = 0
d(Cb) / d(t) = -k2+k1*Ca
Cb(0) = 0
k1 = 10
k2 = 0.192
t(0) = 0
t(f) = 11
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
0
0
0.1736012
4.679E-45
2 Cb
0
-1.154784
0.707101
-1.154784
3 k1
10.
10.
10.
10.
4 k2
0.192
0.192
0.192
0.192
5 t
0
0
11.
11.
Differential equations
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
1 k1 = 10
2 k2 = 0.192
8-21
US: CB never rises above 0.8 g/L so there is no time that it would be illegal.
Sweden: t = 2.4 hours
Russia: t = 5.0 hours
P8-5 (f)
60 g of ethanol immediately à CA = 1.5 g/L
CB = 0.8 g/L at 0.0785 hours or 4.71 minutes.
So, the person has about 4 minutes and 40 seconds to get to their destination.
P8-5 (g - h)
A heavy person will have more body fluid and so the initial concentration of CA would be lower. This
means a heavier person will reach the legal limit slower. The opposite is true for a slimmer person. They
will take shorter time to reach the legal limit, as their initial concentration will be higher.
P8-6 (a)
Let A be the tarzlon in the stomach and B be the tarzlon in the blood.
Mole Balances:
Rate Laws:
All k values are given in the problem statement. It must be noted, however, that for CB < 0, k3 must equal 0.
These equations when entered in POLYMAT generate the following results:
Polymath Code:
d(Ca)/d(t) = -k1*Ca-k2*Ca
d(Cb)/d(t) = k1*Ca-k3-k4*Cb
k1 = 0.15
k2 = 0.6
k4 = 0.2
k3 = if(Cb<0)then(k1*Ca-k4*Cb)else(0.1)
Ca(0)=6.25
Cb(0)=0
t(0)=0
t(f)=4
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
6.25
0.3110786
6.25
0.3110786
2 Cb
0
0
0.597863
0.4057264
3 k1
0.15
0.15
0.15
0.15
4 k2
0.6
0.6
0.6
0.6
5 k3
0.1
0.1
0.1
0.1
6 k4
0.2
0.2
0.2
0.2
7 t
0
0
4.
4.
8-22
Differential equations
1 d(Ca)/d(t) = -k1*Ca-k2*Ca
2 d(Cb)/d(t) = k1*Ca-k3-k4*Cb
Explicit equations
1 k1 = 0.15
2 k2 = 0.6
3 k4 = 0.2
4 k3 = if(Cb<0)then (k1*Ca-k4*Cb) else (0.1)
P8-6 (b)
From the following graph generated using the above program in POLYMATH, we can see the proper
doses of the drug:
1. First take two doses of the drug.
2. Six hours later take one dose.
3. Take dose every four hours from then on.
8-23
P8-6 (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 buildup of the drug in the bloodstream that can cause harmful effects.
P8-6 (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 k1 and elimination constant k2 values. If k1 decreases this means that the adsorption process is
slow and if k2 increases means that the rate of elimination of Tarzlon increases. The next graph shows
the concentration profiles for k1 = 0.10 h-1 and k2 = 0.8 h-1. Note that the maximum amount of the drug
in the bloodstream is reduced by two.
Concentration Profiles
8
Ca
(mg/dm3)
Cb
(mg/dm3)
0.6
CB (mg/dm3)
CA (mg/dm3)
6
0.4
4
0.2
2
0
0
0
2
4
6
8
10
12
Time (h)
Concentration profile for Tarzlon in the stomach (A) and bloodstream (B). The maximum
amount of Tarzlon in the bloodstream is 0.3 mg/dm3.
P8-7 (a)
Reactor selection
A + B ® D rD = -r1A
A + B ® U rU = -r2 A
- r1 A = 10 exp(-8000 K / T )C AC B
- r2 A = 100 exp(-1000 K / T )C A
1/ 2
CB
3/ 2
1/ 2
r
10 exp(-8000 K / T )C AC B
exp(-8000 K / T )C A
SDU = D =
=
1/ 2
3/ 2
1/ 2
rU 100 exp(-1000 K / T )C A C B
10 exp(-1000 K / T )C B
At T = 300K
1/ 2
k1 = 2.62 x 10-11&
k2 = 3.57
SD/U =
k2 =36.78
SD/U =
7.35 ´10 -12 C A
1/ 2
CB
At T = 1000K
k1 = 3.35 x 10-3&
9.2 ´ 10 -5 C A
CB
1/ 2
1/ 2
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.
8-24
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.
1
æ CB ö 2
-6
Note that the selectivity is extremely low, and the only way to increase it is to keep ç C ÷ < 10 and add
è Aø
B drop by drop.
P8-7 (b)
A + B ® D and
– r1 A = 100 exp( -1000 K / T )C AC B
A + B ® U and
- r2 A = 10 exp(-8000K / T )C ACB
SDU =
6
rD 100 exp(-1000 K / T )C ACB
exp(-1000 K / T )
= 6
= 4
rU 10 exp(-8000 K / T )C ACB 10 exp(-8000 K / T )
At T = 300K
k1 = 3.57 &
k2 = 2.623
SDU = 1.14 x106
At T = 1000K
k1 = 36.78 & k2 = 3354.6
SDU = 0.103
Hence temperature should be kept as low as possible to maximize SDU, but at the same time care should
be taken to form significant reaction product.
P8-7 (c)
A + B ® D and -r1A =10 exp(−1000 𝐾/𝑇)𝐶3 𝐶4
B + D ®U
and
- r2 A = 109 exp(-10, 000 K / T )CB CD
𝑆6/7 =
𝑟:3
10 exp(−1000 𝐾/𝑇)𝐶3 𝐶4
= <
𝑟;3 10 exp(−10000 𝐾/𝑇)𝐶4 𝐶6
𝑆6/7 =
exp(−1000 /𝑇)𝐶3
=
10 exp(−10000 /𝑇) 𝐶6
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.
P8-7 (d)
A® D
and
- r1A = 4280 exp(-12000K / T )C A
D ® U1
and
- r2 D = 10,100 exp(-15000K / T )CD
A ®U2
and −𝑟>3 = 26 exp A−18800 D E 𝐶3
C
8-25
:;HHH
4280 exp A− D E 𝐶3 − 10100 exp (−15000/𝑇)𝐶6
𝑟6
𝑆6/7:7; =
=
:KHHH
C
𝑟7: + 𝑟7;
E 𝐶6 + 26 exp A−18800 E 𝐶3
10,100 exp A−
D
D
At T = 300K
k1 = 1.82 x 10-14 , k2 = 1.94 X 10-18 & k3 = 1.58 x 10-26
If we keep CA> >CD
1.82 ∗ 10N:O 𝐶3 − 1.94 ∗ 10N:= 𝐶6 1.82 ∗ 10:;
𝑆6/7:7; =
≈
= 1.15 ∗ 10:;
1.94 ∗ 10N:= 𝐶6 + 1.58 ∗ 10N;Q 𝐶3
1.58
At T = 1000K
k1 = 0.026 &
k2 = 3.1 X 10-3& k3 = 0.18 x 10-6
If we keep CA> >CD
𝑆6/7:7; =
0.026 ∗ 𝐶3 − 3.1 ∗ 10N> 𝐶6
0.026 ∗ 10Q
≈
= 0.14 ∗ 10Q
3.1 ∗ 10N> 𝐶6 + 0.18 ∗ 10NQ 𝐶3
0.18
Here, in order to lower U1 use low temperature and high concentration of A
But low temperature and high concentration of A favors U2
So, it’s a optimization problem with the temperature and concentration of A as the variables .
Membrane reactor in which D is diffusing out can be used.
P8-7 (e)
A + B ® D and
- r1 A = 10 9 exp(-10000 K / T )C A C B
D ® A + B and
- r2 D = 20 exp(-2000K / T )C D
A + B ® U and
- r3 A = 10 exp(-3000 K / T )C AC B
3
r
10 9 exp(-10000 K / T )C A C B - 20 exp(-2000 K / T )C D
S D /U = D =
rU
10 3 exp(-3000 K / T )C A C B
At T = 300K
k1 = 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:
S D /U =
3.34 ´ 10 -6 C A C B - 0.025C D 0.000334
»
= 0.000074
0.045C A C B
4.5
SD /U » 0
At T = 1000K
k1 = 45399.9 & k2 = 2.7 & k3 = 49.7
If we assume that CACB> 0.001CD then,
S D /U =
45399.9C AC B - 2.7C D 45399
»
= 913
49.7C AC B
49.7
8-26
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.
P8-7 (f)
A+ B®D
and
-r1 A = 800 exp(-8000 K / T )C A0.5CB
A + B ® U1
and
-r2 B = 10exp(-300K / T )CACB
D + B ® U2
and
-r3 D = 106 exp(-8000 K / T )CD CB
(1)
S D /U1 =
800exp (-8000 / T )C A0.5CB
10exp (-300 / T )C ACB
=
80exp (-8000 / T )
exp (-300 / T )CA0.5
At T = 300
2.098*10-10
S D /U1 =
0.368C A0.5
At T = 1000
S D / U1 =
29.43
0.7408C A0.5
To keep this selectivity high, low concentrations of A, and high temperatures should be used.
800exp (-8000 / T )C A0.5CB
800C A0.5
SD /U 2 = 6
= 6
10 exp (-8000 / T )CDCB
10 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 dependent on
temperature.
To keep optimize the reaction, run it at a low temperature to maximized SD/U1 in a membrane reactor
that allows only D to diffuse out.
(2)
S D / U 1U 2 =
S D / U 1U 2 =
800 exp (-8000 / T )C A0.5CB
10 exp (-300 / T )C ACB + 106 exp (-8000 / T )CD CB
800 exp (-8000 / T )C A0.5
10 exp (-300 / T )C A + 106 exp (-8000 / T )CD
At T = 300
S D / U 1U 2 =
2.09*10-9 C A0.5
»0
3.67C A + 2.62*10-6 CD
At T = 1000 and very low concentrations of D
S D / U 1U 2 =
0.268C A0.5
.03617
=
7.408C A + 335CD
C A0.5
If temperature is the only parameter that can be varied, then the highest temperature possible will
result in the highest selectivity. Also removing D will help keep selectivity high.
8-27
P8-8
Cao= 1 mol/dm3
V=1 dm3
FA0=1 mol/min
Mole balance on A
𝐹3H − 𝐹3 + 𝑟3 𝑉 = 0
𝑣H 𝐶3H − 𝑣H 𝐶3 + 𝑟3 𝑉 = 0
𝑟3 = −𝑘: 𝐶3
Solving for 𝐶3
𝐶3H
𝐶3 =
1 + 𝜏𝑘:
Using the calculations shown in Example 8-4, mole balance on D and U gives
𝜏𝑘: 𝐶3H
𝐶6 =
(1 + 𝑘: 𝜏)(1 + 𝑘; 𝜏)
𝜏 ; 𝑘: 𝑘; 𝐶3H
𝐶7 =
(1 + 𝑘: 𝜏)(1 + 𝑘; 𝜏)
P8-8 (a)
Conversion
Z
𝑋3 =1 − Z [
[\
𝜏𝑘:
𝑋3 =
1 + 𝜏𝑘:
For, 𝑘: = 1
Instantaneous Selectivity
𝑟6
𝑆6/7 =
𝑟7
𝑟6 = 𝑘: 𝐶3 − 𝑘; 𝐶6
𝑟7 = 𝑘; 𝐶6
𝑘: 𝐶3 − 𝑘; 𝐶6
𝑆6/7 =
𝑘; 𝐶6
For 𝑘: = 𝑘; = 1
Substituting the values of 𝐶3 , 𝐶6 in selectivity equation, we get
𝑆6/7 = 1/𝜏
8-28
Instantaneous Yield
𝑟6
𝑌6 =
−𝑟3
𝑘: 𝐶3 − 𝑘; 𝐶6
𝑌6 =
𝑘: 𝐶3
For 𝑘: = 𝑘; = 1
Substituting the values of 𝐶3 , 𝐶6 in selectivity equation, we get
𝑌6 = 1/(1 + 𝜏)
P8-8 (b) for 𝑋3 = 0.5, 𝜏 = 1
0.5 =
Solving above equation, we get 𝑘: = 1
For 𝑆6/7 = 0.5
^
0.5 =
Solving above equation for 𝑘;
𝑘:
1 + 𝑘:
_
0.5 − ;∗(:`^
)
^_
;∗(:`^_)
𝑘; = 2
P8-9 (a)
Species A :
dC A
= rA ;
dt
-rA = k1 * Ca
8-29
_
Species B:
dC B
= rb
dt
rB = k1 * Ca – k2 * Cb
Species C:
dCC
= rc
dt
rC = k2*Cb
Polymath Code
d(Ca) / d(t) = ra
Ca(0) = 1.6
d(Cb) / d(t) = rb
Cb(0) = 0
d(Cc) / d(t) = rc
Cc(0) = 0
k1 = 0.4
k2 = 0.01
ra = -k1*Ca
rb = k1*Ca - k2*Cb
rc = k2*Cb
t(0) = 0
t(f) = 100
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
1.6
6.796E-18
1.6
6.796E-18
2 Cb
0
0
1.455275
0.6036996
3 Cc
0
0
0.9963004
0.9963004
4 k1
0.4
0.4
0.4
0.4
5 k2
0.01
0.01
0.01
0.01
6 ra
-0.64
-0.64
-2.719E-18
-2.719E-18
7 rb
0.64
-0.0132395
0.64
-0.006037
8 rc
0
0
0.0145527
0.006037
9 t
0
0
100.
100.
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
Explicit equations
1 k1 = 0.4
2 k2 = 0.01
3 ra = -k1*Ca
4 rb = k1*Ca - k2*Cb
5 rc = k2*Cb
8-30
P8-9 (b)
Now the rate laws will change ra = k1-1Cb – k1*Ca
rb =k1*Ca – k1-1Cb – k2*Cb
rc = k2*Cb
Polymath Code:
d(Ca) / d(t) = ra
Ca(0) = 1.6
d(Cb) / d(t) = rb
Cb(0) = 0
d(Cc) / d(t) = rc
Cc(0) = 0
k1 = 0.4
k2 = 0.01
k1b = 0.3
ra = k1b*Cb - k1*Ca
rb = k1*Ca - k1b*Cb - k2*Cb
rc = k2*Cb
t(0) = 0
t(f) = 100
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
1.6
0.3949584
1.6
0.3949584
2
Cb
0
0
0.8736951
0.5191343
3
Cc
0
0
0.6859073
0.6859073
4
k1
0.4
0.4
0.4
0.4
5
k1b
0.3
0.3
0.3
0.3
6
k2
0.01
0.01
0.01
0.01
7
ra
-0.64
-0.64
-0.0022431
-0.0022431
8
rb
0.64
-0.0047669
0.64
-0.0029483
9
rc
0
0
0.008737
0.0051913
10 t
0
0
100.
100.
8-31
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
Explicit equations
1 k1 = 0.4
2 k2 = 0.01
3 k1b = 0.3
4 ra = k1b*Cb - k1*Ca
5 rb = k1*Ca - k1b*Cb - k2*Cb
6 rc = k2*Cb
P8-9 (c)
rC = k2*CB – k-2*Cc
Polymath Code:
d(Ca) / d(t) = ra
Ca(0) = 1.6
d(Cb) / d(t) = rb
Cb(0) = 0
d(Cc) / d(t) = rc
Cc(0) = 0
k1 = 0.4
k2 = 0.01
k1b = 0.3
k2c = 0.005
ra = k1b*Cb - k1*Ca
rb = k1*Ca - k1b*Cb - k2*Cb + k2c*Cc
rc = k2*Cb - k2c*Cc
t(0) = 0
t(f) = 100
8-32
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
1.6
0.4500092
1.6
0.4500092
2
Cb
0
0
0.8742432
0.5953978
3
Cc
0
0
0.554593
0.554593
4
k1
0.4
0.4
0.4
0.4
5
k1b
0.3
0.3
0.3
0.3
6
k2
0.01
0.01
0.01
0.01
7
k2c
0.005
0.005
0.005
0.005
8
ra
-0.64
-0.64
-0.0013843
-0.0013843
9
rb
0.64
-0.0044688
0.64
-0.0017967
10 rc
0
0
0.0085186
0.003181
11 t
0
0
100.
100.
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
Explicit equations
1 k1 = 0.4
2 k2 = 0.01
3 k1b = 0.3
4 k2c = 0.005
5 ra = k1b*Cb - k1*Ca
6 rb = k1*Ca - k1b*Cb - k2*Cb + k2c*Cc
7 rc = k2*Cb - k2c*Cc
P8-9 (d) Individualized Solution
8-33
P8-9 (e)
When k1 > 100 and k2 < 0.1 the concentration of B immediately shoots up to 1.5 and then slowly comes
back down , while CA drops off immediately and falls to zero . This is because the first reaction is very fast
and the second reaction is slower with no reverse reactions.
When k2 = 1 then the concentration of B spikes again and remains high, while very little of C is formed.
This is because after B is formed it will not further get converted to C because the reverse reaction is fast.
When k-2 = 0.25 , B shoots up , but does not stay as high because for the second reaction in the reverse
direction is a slightly slower than seen before , but still faster than the reaction in the forward direction.
P8-9 (f) Individualized Solution
P8-10 (a)
Intermediates (primary K-phthalates) are formed from the dissociation of K-benzoate with a CdCl2
catalyst reacted with K-terephthalate in an autocatalytic reaction step:
k1
k2
A ¾¾
® R ¾¾
® S Series
k3
R + S ¾¾® 2S Autocatalytic
C AO =
P
110kPa
=
= 0.02mol / dm3
3
RT æ
ö
kPa.dm
ç 8.314
÷ (683K )
mol.K ø
è
Maximum in R occurs at t = 880 sec.
Polymath Code:
d(A) / d(t) = -k1*A
A(0) = 0.02
d(R) / d(t) = (k1*A)-(k2*R)-(k3*R*S)
R(0) = 0
d(S) / d(t) = (k2*R)-(k3*R*S)
S(0) = 0
k1 = 1.08E-3
k2 = 1.19E-3
k3 = 1.59E-3
t(0) = 0
t(f) = 1500
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 A
0.02
0.003958
0.02
0.003958
2 k1
0.00108
0.00108
0.00108
0.00108
3 k2
0.00119
0.00119
0.00119
0.00119
4 k3
0.00159
0.00159
0.00159
0.00159
5 R
0
0
0.0069894
0.005868
6 S
0
0
0.0100382
0.0100382
7 t
0
0
1500.
1500.
8-34
Differential equations
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
1 k1 = 1.08E-3
2 k2 = 1.19E-3
3 k3 = 1.59E-3
P8-10 (b)
(1)T = 703 K
CAO = 0.019 mol/dm3
æ E æ 1 1 öö
k1' = k1 exp ç 1 ç - ' ÷ ÷
è R è T T øø
æ (42600cal / mol ) æ 1
1 öö
-3 -1
k1' = (1.08 ´10-3 s -1 ) exp ç
ç
÷ ÷ = 2.64 ´10 s
è (1.987cal / mol.K ) è 683K 703K ø ø
Similarly,
k2' = 3.3 ´ 10-3 s -1
And, k3' = 3.1´10-3 dm3 / mol.s
Maximum in R occurs at around t =320 sec.
Polymath Code:
d(A) / d(t) = -k1*A
A(0) = 0.019
d(R) / d(t) = (k1*A)-(k2*R)-(k3*R*S)
R(0) = 0
d(S) / d(t) = (k2*R)-(k3*R*S)
S(0) = 0
k1 = 2.64E-3
k2 = 3.3E-3
k3 = 3.1E-3
t(0) = 0
t(f) = 1500
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 A
0.019
0.0003622
0.019
0.0003622
2 k1
0.00264
0.00264
0.00264
0.00264
3 k2
0.0033
0.0033
0.0033
0.0033
4 k3
0.0031
0.0031
0.0031
0.0031
5 R
0
0
0.0062169
0.0008856
6 S
0
0
0.0174625
0.0174625
7 t
0
0
1500.
1500.
8-35
Differential equations
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
1 k1 = 2.64E-3
2 k2 = 3.3E-3
3 k3 = 3.1E-3
(2) T = 663 K
CAO = 0.19 mol/dm3
æ (42600cal / mol ) æ 1
1 öö
-3 -1
k1' = (1.08 ´10-3 s -1 ) exp ç
ç
÷ ÷ = 0.42 ´10 s
è (1.987cal / mol.K ) è 683K 663K ø ø
k2' = 0.4 ´ 10-3 s -1
k3' = 0.78 ´10-3 dm3 / mol.s
Polymath Code:
d(A) / d(t) = -k1*A
A(0) = 0.019
d(R) / d(t) = (k1*A)-(k2*R)-(k3*R*S)
R(0) = 0
d(S) / d(t) = (k2*R)-(k3*R*S)
S(0) = 0
k1 = 0.42E-3
k2 = 0.4E-3
k3 = 0.78E-3
t(0) = 0
t(f) = 10000
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 A
0.019
0.0002849
0.019
0.0002849
2 k1
0.00042
0.00042
0.00042
0.00042
3 k2
0.0004
0.0004
0.0004
0.0004
4 k3
0.00078
0.00078
0.00078
0.00078
5 R
0
0
0.0071413
0.0012573
6 S
0
0
0.016889
0.016889
7 t
0
0
10000.
10000.
Differential equations
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
1 k1 = 0.42E-3
2 k2 = 0.4E-3
3 k3 = 0.78E-3
Maxima in R occurs around t = 2500 sec.
8-36
P8-10 (c)
Use the Polymath program from part (a) and change the limits of integration to 0 to 1200. We get:
CAexit = 0.0055 mol/dm3
CRexit = 0.0066 mol/dm3
CSexit = 0.0078 mol/dm3
P8-11 (a)
P8-11 (b)
−𝑟:3 −7 ∗ 0.1
𝑚𝑜𝑙
𝑟:4 = −
=
= 0.23
3
3
𝑑𝑚 > . 𝑚𝑖𝑛
𝑚𝑜𝑙
𝑟;4 = 0 ∗ 𝑟;6 = 0.0
𝑑𝑚> . 𝑚𝑖𝑛
𝑚𝑜𝑙
𝑟>4 = 0 ∗ 𝑟>h = 0.0
𝑑𝑚 > . 𝑚𝑖𝑛
P8-11 (c)
−𝑟:3 −7 ∗ 0.1
𝑚𝑜𝑙
𝑟:Z = −
=
= 0.23
3
3
𝑑𝑚> . 𝑚𝑖𝑛
P8-11 (d)
𝑚𝑜𝑙
𝑟:3 = −𝑘:3 𝐶3 = (−7)(0.1) = −0.7
𝑑𝑚 > . 𝑚𝑖𝑛
(3) ∗ 0.51; ∗ (0.1)
𝑟;6
𝑟;3 = −
=−
3
3
𝑚𝑜𝑙
= −0.026
𝑑𝑚 > . 𝑚𝑖𝑛
𝑚𝑜𝑙
𝑟>3 = 0 ∗ 𝑟>h = 0.0
𝑑𝑚 > . 𝑚𝑖𝑛
𝑟;Z = −
𝑟:6 = 0 ∗ 𝑟:3 = 0.0
2𝑟;6 −2 ∗ (3) ∗ 0.51; ∗ (0.1)
=
3
3
𝑚𝑜𝑙
= 0.052
𝑑𝑚> . 𝑚𝑖𝑛
𝑟;6 = 3 ∗ 0.51; ∗ 0.1 = 0.078
𝑟>6 = −
𝑟>Z = −𝑟>h = −(2)(0.049) ∗ 0.51
= −0.05
𝑚𝑜𝑙
𝑑𝑚> . 𝑚𝑖𝑛
P8-11 (e)
𝑚𝑜𝑙
𝑑𝑚 > . 𝑚𝑖𝑛
𝑚𝑜𝑙
𝑟;h = 0 ∗ 𝑟;6 = 0.0
𝑑𝑚 > . 𝑚𝑖𝑛
𝑚𝑜𝑙
𝑑𝑚 > . 𝑚𝑖𝑛
4𝑟>h
4 ∗ (2) ∗ 0.049 ∗ (0.51)
=−
3
3
𝑚𝑜𝑙
= −0.067
𝑑𝑚> . 𝑚𝑖𝑛
P8-11 (f)
𝑟:h = 0 ∗ 𝑟:3 = 0.0
𝑟>h = 2 ∗ 0.049 ∗ 0.51 = 0.05
𝑚𝑜𝑙
𝑑𝑚 > . 𝑚𝑖𝑛
𝑟3 = −0.7 − 0.026 = −0.726
𝑟4 = 0.23
𝑚𝑜𝑙
𝑑𝑚 > . 𝑚𝑖𝑛
𝑚𝑜𝑙
𝑑𝑚 > . 𝑚𝑖𝑛
𝑚𝑜𝑙
𝑑𝑚 > . 𝑚𝑖𝑛
𝑚𝑜𝑙
𝑑𝑚> . 𝑚𝑖𝑛
𝑚𝑜𝑙
𝑟6 = 0.078 − 0.067 = 0.011
𝑑𝑚 > . 𝑚𝑖𝑛
𝑚𝑜𝑙
𝑟h = 0.05
𝑑𝑚 > . 𝑚𝑖𝑛
𝑟Z = 0.23 − 0.052 − 0.05 = 0.128
8-37
P8-11 (g)
𝐹3H − 𝐹3
𝑉=
−𝑟3
𝑣H (𝐶3H − 𝐶3 )
V=
−𝑟3
100(3 − 0.1)
𝑉=
0.726
𝑉 = 400 𝑑𝑚 >
P8-11 (h)
Mole balance: C AO - C A = (- rA )t
CC = (rC )t
CD = (rD )t
Rate law:
1
é
ù
rA = - êk1 A C A + k 2 D C A CC2 ú
3
ë
û
2
é1
ù
rC = ê k1 A C A - k 2 D C A CC2 - k 3 E CC C D ú
3
ë3
û
4
é
ù
rD = êk 2 D C A CC2 - k 3 E CC C D ú
3
ë
û
See the following Polymath program:
Polymath Code:
f(Ca) = Cao-Ca+ra*tau
f(Ce) = Ce - re*tau
Ce(0) = 0
f(Cd) = Cd-rd*tau
Cd(0) = 0
f(Cc) = Cc-rc*tau
Cc(0) = 0
f( Cb) = Cb - rb*tau
Cb(0) = 0
Ca(0) = 3
kd = 3
ka = 7
rb = ka*Ca/3
ra = -(ka*Ca+kd/3*Ca*Cc^2)
ke = 2
rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc
rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc
re = ke*Cd*Cc
V=400
vo=100
tau = V/vo
Cao = 3
Output:
8-38
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 Ca
0.0997816 -3.206E-12 3.
2 Cb
0.931295 2.22E-15
3 Cc
0.5161542 -4.127E-10 0
4 Cd
0.0490343 -5.683E-10 0
5 Ce
0.2024742 4.191E-10 0
0
Variable Value
1 Cao
3.
2 ka
7.
3 kd
3.
4 ke
2.
5 ra
-0.7250546
6 rb
0.2328238
7 rc
0.1290385
8 rd
0.0122586
9 re
0.0506186
10 tau
4.
11 V
400.
12 vo
100.
Nonlinear equations
1 f(Ca) = Cao-Ca+ra*tau = 0
2 f(Ce) = Ce - re*tau = 0
3 f(Cd) = Cd-rd*tau = 0
4 f(Cc) = Cc-rc*tau = 0
5 f(Cb) = Cb - rb*tau = 0
Explicit equations
1 kd = 3
2 ka = 7
3 rb = ka*Ca/3
4 ra = -(ka*Ca+kd/3*Ca*Cc^2)
5 ke = 2
6 rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc
7 rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc
8 re = ke*Cd*Cc
9 V = 400
10 vo = 100
11 tau = V/vo
12 Cao = 3
Exit molar flow rates are
𝐹3 = 𝐶3 ∗ 𝑣 = 9.97 𝑚𝑜𝑙/𝑚𝑖𝑛
𝐹4 = 𝐶4 ∗ 𝑣 = 93.1 𝑚𝑜𝑙/𝑚𝑖𝑛
𝐹Z = 𝐶Z ∗ 𝑣 = 51.6 𝑚𝑜𝑙/𝑚𝑖𝑛
𝐹6 = 𝐶6 ∗ 𝑣 = 4.9 𝑚𝑜𝑙/𝑚𝑖𝑛
𝐹h = 𝐶3 ∗ 𝑣 = 20.24 𝑚𝑜𝑙/𝑚𝑖𝑛
8-39
P8-11 (i)
For PFR and gas phase:
Mole balance:
dFA
= rA
dV
dFB
= rB
dV
Rate law:
⎡
⎤
1
rA = −⎢k1ACA + k2DCACC2 ⎥
⎣
⎦
3
dFC
= rC
dV
dFD
= rD
dV
⎡1
⎤
rB = ⎢ k1AC A ⎥
⎣3
⎦
⎡1
⎤
2
rC = ⎢ k1AC A − k2 DC ACC2 − K 3ECC C D ⎥
3
⎣3
⎦
⎡
⎤
4
rD = ⎢k2 DC ACC2 − k3ECC C D ⎥
3
⎣
⎦
rE = ⎡⎣k3ECC C D ⎤⎦
Stoichiometry: C A = CTO
FA
p
FT
CC = CTO
FC
p
FT
FT = FA + FB + FC + FD + FE
dp −α FT
=
dV 2 p FTO
Where p =P/P0
Plot of CB and CC are overlapping.
Polymath Code:
d(Fa) / d(V) = ra
Fa(0) = 20
d(Fb) / d(V) = rb
Fb(0) = 0
d(Fc) / d(V) = rc
Fc(0) = 0
d(Fd) / d(V) = rd
Fd(0) = 0
d(Fe) / d(V) = re
Fe(0) = 0
d(p) / d(V) = -alfa*Ft/(2*p*Fto)
p(0) = 1
Ft = Fa+Fb+Fc+Fd+Fe
Cto = 0.2
Cc = Cto*Fc/Ft*p
ka = 7
kd = 3
ke = 2
Ca = Cto*Fa/Ft*p
rb = ka*Ca/3
ra = -(ka*Ca+kd/3*Ca*Cc^2)
Cd = Cto*Fd/Ft*p
Fto = 0.2*100
rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc
rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc
re = ke*Cd*Cc
alfa = 0.001
X = 1-Fa/20
V(0) = 0
V(f) = 100
SDE=Fd/(Fe+0.000001)
SCD=Fc/(Fd+0.000001)
8-40
C D = CTO
FD
p
FT
dFE
= rE
dV
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
alfa
0.001
0.001
0.001
0.001
2 Ca
0.2
1.58E-05
0.2
1.58E-05
3 Cc
0
0
0.0970427
0.0961036
4 Cd
0
0
0.0002905
0.0002466
5 Cto
0.2
0.2
0.2
0.2
6 Fa
20.
0.0010922
20.
0.0010922
7 Fb
0
0
6.6638
6.6638
8 Fc
0
0
6.644696
6.644681
9 Fd
0
0
0.0198259
0.0170495
10 Fe
0
0
0.0041045
0.0041045
11 Ft
20.
13.33073
20.
13.33073
12 Fto
20.
20.
20.
20.
13 ka
7.
7.
7.
7.
14 kd
3.
3.
3.
3.
15 ke
2.
2.
2.
2.
16 p
1.
0.9640276
1.
0.9640276
17 ra
-1.4
-1.4
-0.0001107
-0.0001107
18 rb
0.4666667
3.686E-05
0.4666667
3.686E-05
19 rc
0.4666667
-1.083E-05
0.4666667
-1.083E-05
20 rd
0
-6.554E-05
0.0008497
-6.276E-05
21 re
0
0
5.614E-05
4.74E-05
22 V
0
0
100.
100.
23 X
0
0
0.9999454
0.9999454
24 SCD
0
0
2.441E+04
389.7056
25 SDE
0
0
502.0227
4.152833
Differential equations
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(p)/d(V) = -alfa*Ft/(2*p*Fto)
Explicit equations
1 Ft = Fa+Fb+Fc+Fd+Fe
2 Cto = 0.2
3 Cc = Cto*Fc/Ft*p
4 ka = 7
5 kd = 3
6 ke = 2
7 Ca = Cto*Fa/Ft*p
8 rb = ka*Ca/3
9 ra = -(ka*Ca+kd/3*Ca*Cc^2)
10 Cd = Cto*Fd/Ft*p
11 Fto = 0.2*100
12 rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc
8-41
13 rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc
14 re = ke*Cd*Cc
15 alfa = 0.001
16 X = 1-Fa/20
17 SDE = Fd/(Fe+0.000001)
18 SCD = Fc/(Fd+0.000001)
P8-11 (j) Changes in equation from part (i):
dFC
= rC - RC ,
dV
RC = k diffuse C C
Polymath Code:
d(Fa) / d(V) = ra
Fa(0) = 20
d(Fb) / d(V) = rb
Fb(0) = 0
d(Fc) / d(V) = rc - Rc
Fc(0) = 0
d(Fd) / d(V) = rd
Fd(0) = 0
d(Fe) / d(V) = re
Fe(0) = 0
d(p) / d(V) = -alfa*Ft/(2*p*Fto)
p(0) = 1
Ft = Fa+Fb+Fc+Fd+Fe
Cto = 0.2
Cc = Cto*Fc/Ft*p
ka = 7
8-42
kd = 3
ke = 2
kdiff = 10
Rc = kdiff*Cc
Ca = Cto*Fa/Ft*p
rb = ka*Ca/3
ra = -(ka*Ca+kd/3*Ca*Cc^2)
Cd = Cto*Fd/Ft*p
Fto = 0.2*100
rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc
rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc
re = ke*Cd*Cc
alfa = 0.001
X = 1-Fa/20
V(0) = 0
V(f) = 100
SDE=Fd/(Fe+0.000001)
SCD=Fc/(Fd+0.000001)
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alfa
0.001
0.001
0.001
0.001
2 Ca
0.2
8.681E-09
0.2
8.681E-09
3 Cc
0
0
0.0297883
6.748E-09
4 Cd
0
0
0.0001152
0.0001143
5 Cto
0.2
0.2
0.2
0.2
6 Fa
20.
2.958E-07
20.
2.958E-07
7 Fb
0
0
6.666223
6.666223
8 Fc
0
0
2.029507
2.3E-07
9 Fd
0
0
0.0039055
0.0038969
10 Fe
0
0
6.933E-05
6.933E-05
11 Ft
20.
6.67019
20.
6.67019
12 Fto
20.
20.
20.
20.
13 ka
7.
7.
7.
7.
14 kd
3.
3.
3.
3.
15 kdiff
10.
10.
10.
10.
16 ke
2.
2.
2.
2.
17 p
1.
0.9785808
1.
0.9785808
18 ra
-1.4
-1.4
-6.076E-08
-6.076E-08
19 rb
0.4666667
2.025E-08
0.4666667
2.025E-08
20 rc
0.4666667
2.025E-08
0.4666667
2.025E-08
21 Rc
0
0
0.2978826
6.748E-08
22 rd
0
-8.995E-07
0.0002765
-2.058E-12
23 re
0
0
3.676E-06
1.543E-12
24 SCD
0
0
2.661E+04
5.9E-05
25 SDE
0
0
438.8506
55.41238
26 V
0
0
100.
100.
27 X
0
0
1.
1.
Differential equations
1 d(Fa)/d(V) = ra
2 d(Fb)/d(V) = rb
3 d(Fc)/d(V) = rc – Rc
8-43
4 d(Fd)/d(V) = rd
5 d(Fe)/d(V) = re
6 d(p)/d(V) = -alfa*Ft/(2*p*Fto)
Explicit equations
1 Ft = Fa+Fb+Fc+Fd+Fe
2 Cto = 0.2
3 Cc = Cto*Fc/Ft*p
4 ka = 7
5 kd = 3
6 ke = 2
7 kdiff = 2
8 Rc = kdiff*Cc
9 Ca = Cto*Fa/Ft*p
10 rb = ka*Ca/3
11 ra = -(ka*Ca+kd/3*Ca*Cc^2)
12 Cd = Cto*Fd/Ft*p
13 Fto = 0.2*100
14 rc = ka*Ca/3 - 2/3*kd*Ca*Cc^2 - ke*Cd*Cc
15 rd = kd*Ca*Cc^2 - 4/3*ke*Cd*Cc
16 re = ke*Cd*Cc
17 alfa = 0.0001
18 X = 1-Fa/20
19 SDE = Fd/(Fe+0.000001)
20 SCD = Fc/(Fd+0.000001)
8-44
P8-12 (a)
Polymath Code:
d(Ca) / d(V) = ra/vo
Ca(0) = 1.5
d(Cb) / d(V) = rb/vo
Cb(0) = 2
d(Cc) / d(V) = rc/vo
Cc(0) = 0
d(Cd) / d(V) = rd/vo
Cd(0) = 0
d(Ce) / d(V) = re/vo
Ce(0) = 0
d(Cf) / d(V) = rf/vo
Cf(0) = 0
vo = 10
kf3 = 5
ke2 = 0.1
kd1 = 0.25
rf3 = kf3*Cb*Cc^2
rd1 = kd1*Ca*Cb^2
re2 = ke2*Ca*Cd
rf= rf3
re = re2
rd = rd1-2*re2+rf3
ra = -rd1-3*re2
rb = -2*rd1-rf3
rc = rd1+re2-2*rf3
V(0) = 0
V(f) = 50
8-45
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
1.5
0.3100062
1.5
0.3100062
2 Cb
2.
0.3384818
2.
0.3384818
3 Cc
0
0
0.1970424
0.105508
4 Cd
0
0
0.6751295
0.6516261
5 Ce
0
0
0.1801018
0.1801018
6 Cf
0
0
0.3621412
0.3621412
7 kd1
0.25
0.25
0.25
0.25
8 ke2
0.1
0.1
0.1
0.1
9 kf3
5.
5.
5.
5.
10 ra
-1.5
-1.5
-0.0694818
-0.0694818
11 rb
-3.
-3.
-0.0365985
-0.0365985
12 rc
1.5
-0.0490698
1.5
-0.0085994
13 rd
1.5
-0.0128607
1.5
-0.0126825
14 rd1
1.5
0.0088793
1.5
0.0088793
15 re
0
0
0.0523415
0.0202008
16 re2
0
0
0.0523415
0.0202008
17 rf
0
0
0.2550077
0.0188398
18 rf3
0
0
0.2550077
0.0188398
19 V
0
0
50.
50.
20 vo
10.
10.
10.
10.
Differential equations
1 d(Ca)/d(V) = ra/vo
2 d(Cb)/d(V) = rb/vo
3 d(Cc)/d(V) = rc/vo
4 d(Cd)/d(V) = rd/vo
5 d(Ce)/d(V) = re/vo
6 d(Cf)/d(V) = rf/vo
Explicit equations
1 vo = 10
2 kf3 = 5
3 ke2 = 0.1
4 kd1 = 0.25
5 rf3 = kf3*Cb*Cc^2
6 rd1 = kd1*Ca*Cb^2
7 re2 = ke2*Ca*Cd
8 rf = rf3
9 re = re2
10 rd = rd1-2*re2+rf3
11 ra = -rd1-3*re2
12 rb = -2*rd1-rf3
13 rc = rd1+re2-2*rf3
8-46
0.9
0.8
0.7
0.6
X
0.5
0.4
Conversion
0.3
0.2
0.1
0
0
20
40
60
Time
P8-12 (b)
Determine the effluent concentration and conversion from a 50 dm3 CSTR.
Mole Balance:
8-47
P8-12 (c)
V0 = 40 dm3 Semi-Batch reactor. (1) A is fed to B, (2) B is fed to A (Case 1) A is fed to B,
8-48
Case 1 concentration vs. time
Case 1 conversion vs. time
8-49
Case 2 concentration vs. time
Case 2 conversion vs. time
P8-12 (d)
As θB increases the outlet concentration of species D and F increase, while the outlet concentrations of
species A, C, and E decrease. When θB is large, reactions 1 and 3 are favored and when it is small the
rate of reaction 2 will increase.
P8-12 (e)
When the appropriate changes to the Polymath code from part (a) are made we get the following.
Polymath Code:
d(Fa) / d(V) = ra
Fa(0) = 20
d(Fb) / d(V) = rb
8-50
Fb(0) = 20
d(Fc) / d(V) = rc
Fc(0) = 0
d(Fd) / d(V) = rd
Fd(0) = 0
d(Fe) / d(V) = re
Fe(0) = 0
d(Ff) / d(V) = rf
Ff(0) = 0
vo = 100
Ft = Fa+Fb+Fc+Fd+Fe+Ff
Cto = .4
kd1 = 0.25
ke2 = .1
kf3 = 5
Cc = Cto*Fc/Ft
Cd = Cto*Fd/Ft
Cb = Cto*Fb/Ft
Ca = Cto*Fa/Ft
rd1 = kd1*Ca*Cb^2
re2 = ke2*Ca*Cd
rf3 = kf3*Cb*Cc^2
re = re2
rf = rf3
rd = rd1-2*re2+rf3
ra = -rd1-3*re2
rb = -2*rd1-rf3
rc = rd1+re2-2*rf3
Scd = rc/(rd+0.0000000001)
Sef = re/(rf+0.00000000001)
V(0) = 0
V(f) = 500
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
0.2
0.1946651
0.2
0.1946651
2 Cb
0.2
0.1864364
0.2
0.1864364
3 Cc
0
0
0.0095994
0.0095994
4 Cd
0
0
0.0086869
0.0086869
5 Cto
0.4
0.4
0.4
0.4
6 Fa
20.
18.94654
20.
18.94654
7 Fb
20.
18.14565
20.
18.14565
8 Fc
0
0
0.9342961
0.9342961
9 Fd
0
0
0.8454829
0.8454829
10 Fe
0
0
0.0445942
0.0445942
11 Ff
0
0
0.0149897
0.0149897
12 Ft
40.
38.93155
40.
38.93155
13 kd1
0.25
0.25
0.25
0.25
14 ke2
0.1
0.1
0.1
0.1
15 kf3
5.
5.
5.
5.
16 ra
-0.002
-0.0021989
-0.002
-0.0021989
17 rb
-0.004
-0.004
-0.003469
-0.003469
18 rc
0.002
0.0016889
0.002
0.0016889
8-51
19 rd
0.002
0.0014393
0.002
0.0014393
20 rd1
0.002
0.0016916
0.002
0.0016916
21 re
0
0
0.0001691
0.0001691
22 re2
0
0
0.0001691
0.0001691
23 rf
0
0
8.59E-05
8.59E-05
24 rf3
0
0
8.59E-05
8.59E-05
25 Scd
1.
1.
1.173431
1.173431
26 Sef
0
0
76.74074
1.968633
27 V
0
0
500.
500.
28 vo
100.
100.
100.
100.
Differential equations
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
1 vo = 100
2 Ft = Fa+Fb+Fc+Fd+Fe+Ff
3 Cto = .4
4 kd1 = 0.25
5 ke2 = .1
6 kf3 = 5
7 Cc = Cto*Fc/Ft
8 Cd = Cto*Fd/Ft
9 Cb = Cto*Fb/Ft
10 Ca = Cto*Fa/Ft
11 rd1 = kd1*Ca*Cb^2
12 re2 = ke2*Ca*Cd
13 rf3 = kf3*Cb*Cc^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+0.0000000001)
21 Sef = re/(rf+0.00000000001)
8-52
P8-12 (f)
The only change from part (e) is:
dFD
= rD - kcD CD
dV
P8-12 (g)
The change from part (e) is
𝑑𝐹4
𝐹4H
= 𝑟4 +
𝑤ℎ𝑒𝑟𝑒 𝑉D = 500 𝑑𝑚> 𝑎𝑛𝑑 𝐹4H = 20 𝑚𝑜𝑙/𝑚𝑖𝑛
𝑑𝑉
𝑉D
𝐹4 (0) = 0
Polymath code:
d(Fa) / d(V) = ra
Fa(0) = 20
d(Fb) / d(V) = rb+FB0/Vt
Fb(0) =0
d(Fc) / d(V) = rc
Fc(0) = 0
d(Fd) / d(V) = rd
Fd(0) = 0
d(Fe) / d(V) = re
Fe(0) = 0
d(Ff) / d(V) = rf
Ff(0) = 0
vo = 100
Ft = Fa+Fb+Fc+Fd+Fe+Ff
Cto = .4
kd1 = 0.25
8-53
ke2 = .1
kf3 = 5
FB0= 20
Vt=500
Cc = Cto*Fc/Ft
Cd = Cto*Fd/Ft
Cb = Cto*Fb/Ft
Ca = Cto*Fa/Ft
rd1 = kd1*Ca*Cb^2
re2 = ke2*Ca*Cd
rf3 = kf3*Cb*Cc^2
re = re2
rf = rf3
rd = rd1-2*re2+rf3
ra = -rd1-3*re2
rb = -2*rd1-rf3
rc = rd1+re2-2*rf3
Scd = rc/(rd+0.0000000001)
Sef = re/(rf+0.00000000001)
V(0) = 0
V(f) = 500
P8-13 (a)
Isothermal gas phase reaction in a membrane reactor packed with catalyst.
¾¾
® B + C , r1'C = k1C éêC A - CBCC ùú
A ¬¾
¾
K1C û
ë
8-54
AàD
r2' D = k2 D C A
2C + D à 2E
r3' E = k3 E CC2 CD
C AO =
P
24.6atm
=
= 0.6mol / dm3
3
RT (0.082dm atm / mol.K )(500 K )
Fa(0) = 10 mol/min
Fb(0) = Fc(0) = Fd(0) = Fe(0) = 0
Pressure ratio, p=P/P0
See the following Polymath program:
Polymath Code
d(Fa)/d(W) = ra
d(Fb)/d(W) = rb-(kc*Cb)
d(Fc)/d(W) = rc
d(Fd)/d(W) = rd
d(Fe)/d(W) = re
d(p)/d(W) = -alfa*Ft/(2*Fto*p)
k2d = 0.4
K1c = 0.2
Ft = Fa+Fb+Fc+Fd+Fe
Cto = 0.6
Cb = Cto*(Fb/Ft)*p
Ca = Cto*(Fa/Ft)*p
Cc = Cto*(Fc/Ft)*p
Cd = Cto*(Fd/Ft)*p
kc = 1
k1c = 2
r2d = k2d*Ca
k3e = 5
r1c = k1c*(Ca-(Cb*Cc/K1c))
ra = -r1c-r2d
r3e = k3e*(Cc^2)*Cd
rd = r2d-(r3e/2)
rb = r1c
rc = r1c-r3e
re = r3e
Ce = Cto*(Fe/Ft)*p
alfa = 0.008
Fto = 10
Fa(0)=10
Fb(0)=0
Fc(0)=0
Fd(0)=0
Fe(0)=0
p(0)=1
W(0) = 0
W(f) = 100
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alfa
0.008
0.008
0.008
0.008
2 Ca
0.6
0.0073158
0.6
0.0073158
3 Cb
0
0
0.140361
0.0093022
4 Cc
0
0
0.2117202
0.1017844
5 Cd
0
0
0.1019606
0.0571654
6 Ce
0
0
0.0398806
0.0287293
7 Cto
0.6
0.6
0.6
0.6
8-55
8 Fa
10.
0.349438
10.
0.349438
9 Fb
0
0
3.237542
0.4443151
10 Fc
0
0
4.987302
4.861703
11 Fd
0
0
2.730488
2.730488
12 Fe
0
0
1.372248
1.372248
13 Ft
10.
9.758191
13.2205
9.758191
14 Fto
10.
10.
10.
10.
15 k1c
2.
2.
2.
2.
16 K1c
0.2
0.2
0.2
0.2
17 k2d
0.4
0.4
0.4
0.4
18 k3e
5.
5.
5.
5.
19 kc
1.
1.
1.
1.
20 p
1.
0.3404952
1.
0.3404952
21 r1c
1.2
0.0051635
1.2
0.0051635
22 r2d
0.24
0.0029263
0.24
0.0029263
23 r3e
0
0
0.0216836
0.0029612
24 ra
-1.44
-1.44
-0.0080898
-0.0080898
25 rb
1.2
0.0051635
1.2
0.0051635
26 rc
1.2
-0.0042614
1.2
0.0022023
27 rd
0.24
0.0014457
0.24
0.0014457
28 re
0
0
0.0216836
0.0029612
29 W
0
0
100.
100.
Differential equations
1 d(Fa)/d(W) = ra
2 d(Fb)/d(W) = rb-(kc*Cb)
3 d(Fc)/d(W) = rc
4 d(Fd)/d(W) = rd
5 d(Fe)/d(W) = re
6 d(p)/d(W) = -alfa*Ft/(2*Fto*p)
Explicit equations
1
k2d = 0.4
2
K1c = 0.2
3
Ft = Fa+Fb+Fc+Fd+Fe
4
Cto = 0.6
5
Cb = Cto*(Fb/Ft)*p
6
Ca = Cto*(Fa/Ft)*p
7
Cd = Cto*(Fd/Ft)*p
8
Cc = Cto*(Fc/Ft)*p
9
kc = 1
10 k1c = 2
11 r2d = k2d*Ca
12 k3e = 5
13 r1c = k1c*(Ca-(Cb*Cc/K1c))
14 ra = -r1c-r2d
15 r3e = k3e*(Cc^2)*Cd
8-56
16 rd = r2d-(r3e/2)
17 rb = r1c
18 rc = r1c-r3e
19 re = r3e
20 Ce = Cto*(Fe/Ft)*p
21 alfa = 0.008
22 Fto = 10
P8-13 (b)
Species B, C, D, and E all go through a maximum. The concentration of species are affected by two factors:
reaction and pressure drop. We can look at species B for example:
Species B:
Production by reaction 1 => tends to increase concentration of B
Pressure drop => tends to decrease concentration of B
At W<10, concentration of B rises because the production of B by reaction outweighs the dilution of B by
pressure drop. Further down the reactor, the concentration of B drops because the effect of pressure
drop outweighs the production of B by reaction.
Species E:
The reasoning for species E is similar to species B.
Species C and D:
Species C and D are produced in reactions 1 and 2, respectively. However, species C and D are also
reactants in reaction 3; this contributes to their decline at later times (in addition to the pressure drop).
P8-13 (c)
Polymath Code:
d(Fa)/d(W) = ra
d(Fb)/d(W) = rb
d(Fc)/d(W) = rc-(kc*Cc)
d(Fd)/d(W) = rd
d(Fe)/d(W) = re
8-57
d(p)/d(W) = -alfa*Ft/(2*Fto*p)
k2d = 0.4
K1c = 0.2
Ft = Fa+Fb+Fc+Fd+Fe
Cto = 0.6
Cb = Cto*(Fb/Ft)*p
Ca = Cto*(Fa/Ft)*p
Cd = Cto*(Fd/Ft)*p
Cc = Cto*(Fc/Ft)*p
kc = 1
k1c=2
r2d = k2d*Ca
k3e = 5
r1c = k1c*(Ca-(Cb*Cc/K1c))
ra = -r1c-r2d
r3e = k3e*(Cc^2)*Cd
rd = r2d-(r3e/2)
rb = r1c
rc = r1c-r3e
re = r3e
Ce = Cto*(Fe/Ft)*p
alfa = 0.008
Fto = 10
Fa(0)=10
Fb(0)=0
Fc(0)=0
Fd(0)=0
Fe(0)=0
p(0)=1
W(0) = 0
W(f) = 100
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 alfa
0.008
0.008
0.008
0.008
2 Ca
0.6
0.0068286
0.6
0.0068286
3 Cb
0
0
0.2249202
0.1017619
4 Cc
0
0
0.1395468
0.0079127
5 Cd
0
0
0.1095929
0.0548084
6 Ce
0
0
0.0057877
0.0026986
7 Cto
0.6
0.6
0.6
0.6
8 Fa
10.
0.4144848
10.
0.4144848
9 Fb
0
0
6.176815
6.176815
10 Fc
0
0
3.217123
0.4802904
11 Fd
0
0
3.326801
3.326801
12 Fe
0
0
0.1637987
0.1637987
13 Ft
10.
10.
13.22836
10.56219
14 Fto
10.
10.
10.
10.
15 k1c
2.
2.
2.
2.
16 K1c
0.2
0.2
0.2
0.2
17 k2d
0.4
0.4
0.4
0.4
18 k3e
5.
5.
5.
5.
19 kc
1.
1.
1.
1.
20 p
1.
0.290017
1.
0.290017
21 r1c
1.2
0.005605
1.2
0.005605
8-58
22 r2d
0.24
0.0027314
0.24
0.0027314
23 r3e
0
0
0.0051612
1.716E-05
24 ra
-1.44
-1.44
-0.0083364
-0.0083364
25 rb
1.2
0.005605
1.2
0.005605
26 rc
1.2
0.0055879
1.2
0.0055879
27 rd
0.24
0.0027228
0.24
0.0027228
28 re
0
0
0.0051612
1.716E-05
29 W
0
0
100.
100.
Differential equations
1 d(Fa)/d(W) = ra
2 d(Fb)/d(W) = rb
3 d(Fc)/d(W) = rc-(kc*Cc)
4 d(Fd)/d(W) = rd
5 d(Fe)/d(W) = re
6 d(p)/d(W) = -alfa*Ft/(2*Fto*p)
Explicit equations
1
k2d = 0.4
2
K1c = 0.2
3
Ft = Fa+Fb+Fc+Fd+Fe
4
Cto = 0.6
5
Cb = Cto*(Fb/Ft)*p
6
Ca = Cto*(Fa/Ft)*p
7
Cd = Cto*(Fd/Ft)*p
8
Cc = Cto*(Fc/Ft)*p
9
kc = 1
10 k1c = 2
11 r2d = k2d*Ca
12 k3e = 5
13 r1c = k1c*(Ca-(Cb*Cc/K1c))
14 ra = -r1c-r2d
15 r3e = k3e*(Cc^2)*Cd
16 rd = r2d-(r3e/2)
17 rb = r1c
18 rc = r1c-r3e
19 re = r3e
20 Ce = Cto*(Fe/Ft)*p
21 alfa = 0.008
22 Fto = 10
8-59
Major Differences:
(1) When C diffuses out instead of B, the concentration of C is lower and the concentration of B is
higher
(2) In case of diffusion of C, the concentration of E is lower because third reaction is slow owing to
lower concentration of C.
P8-13 (d) Individualized Solution
P8-14 (a)
Equations:
Rate law:
-rA = k1CACB1/2 + k2CA2
rB = rA/2
rC= k1CACB1/2 - k3CC + k4CD
rD = k2CA2 – k4CD
rE = k3CC
rw = k3CC
rG = k4CD
8-60
Mole Balance:
See the following polymath problem:
Polymath Code
d(Fa)/d(V) = ra
d(Fb)/d(V) = rb
d(Fc)/d(V) = rc
d(Fd)/d(V) = rd
d(Fe)/d(V) = re
d(Fw)/d(V) = rw
d(Fg)/d(V) = rg
Cto = 0.147
k1 = 0.014
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
k2 = 0.007
Cb = Cto*(Fb/Ft)
k3 = 0.014
k4 = 0.45
Ca = Cto*(Fa/Ft)
Cc = Cto*(Fc/Ft)
ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
rb = ra/2
re = k3*Cc
Cd = Cto*(Fd/Ft)
Ce = Cto*(Fe/Ft)
Cg = Cto*(Fg/Ft)
Cw = Cto*(Fw/Ft)
rg = k4*Cd
rw = k3*Cc
rd = k2*(Ca)^2-k4*Cd
rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
Yc = Fc/Fa
Scd = if(V>0.0001)then(Fc/Fd)else(0)
Sae = if(V>0.0001)then(Fa/Fe)else(0)
Sdg = if(V>0.0001)then(Fd/Fg)else(0)
Fa(0)=10
Fb(0)=5
Fc(0)=0
Fd(0)=0
Fe(0)=0
Fw(0)=0
Fg(0)=0
V(0) = 0
V(f) = 2e5
8-61
Polymath Output:
Maximum values can be found out from this table:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.098
0.0064314
0.098
0.0064314
2
Cb
0.049
0.0032157
0.049
0.0032157
3
Cc
0
0
0.0138769
0.0004265
4
Cd
0
0
0.0001148
6.459E-07
5
Ce
0
0
0.0642751
0.0642751
6
Cg
0
0
0.0083756
0.0083756
7
Cto
0.147
0.147
0.147
0.147
8
Cw
0
0
0.0642751
0.0642751
9
Fa
10.
0.9041349
10.
0.9041349
10 Fb
5.
0.4520674
5.
0.4520674
11 Fc
0
0
1.463177
0.0599575
12 Fd
0
0
0.0116183
9.08E-05
13 Fe
0
0
9.035817
9.035817
14 Fg
0
0
1.177438
1.177438
15 Ft
15.
14.87618
20.66532
20.66532
16 Fw
0
0
9.035817
9.035817
17 k1
0.014
0.014
0.014
0.014
18 k2
0.007
0.007
0.007
0.007
19 k3
0.014
0.014
0.014
0.014
20 k4
0.45
0.45
0.45
0.45
21 ra
-0.0003709
-0.0003709
-5.395E-06
-5.395E-06
22 Rb
-0.0001855
-0.0001855
-2.698E-06
-2.698E-06
23 Rc
0.0003037
-3.328E-05
0.0003037
-5.744E-07
24 Rd
6.723E-05
-8.355E-07
6.723E-05
-1.1E-09
25 re
0
0
0.0001943
5.971E-06
26 Rg
0
0
5.166E-05
2.906E-07
27 Rw
0
0
0.0001943
5.971E-06
28 Sae
0
0
25.89133
0.1000612
29 Scd
0
0
660.3424
660.3424
30 Sdg
0
0
0.0501992
7.711E-05
31 V
0
0
2.0E+05
2.0E+05
32 Yc
0
0
0.2299208
0.0663148
Differential equations
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(Fw)/d(V) = rw
7 d(Fg)/d(V) = rg
8-62
Explicit equations
1
Cto = 0.147
2
k1 = 0.014
3
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
4
k2 = 0.007
5
Cb = Cto*(Fb/Ft)
6
k3 = 0.014
7
k4 = 0.45
8
Ca = Cto*(Fa/Ft)
9
Cc = Cto*(Fc/Ft)
10 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
11 rb = ra/2
12 re = k3*Cc
13 Cd = Cto*(Fd/Ft)
14 Ce = Cto*(Fe/Ft)
15 Cg = Cto*(Fg/Ft)
16 Cw = Cto*(Fw/Ft)
17 rg = k4*Cd
18 rw = k3*Cc
19 rd = k2*(Ca)^2-k4*Cd
20 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
21 Yc = Fc/Fa
22 Scd = if (V>0.0001)then (Fc/Fd) else (0)
23 Sae = if (V>0.0001)then (Fa/Fe) else (0)
24 Sdg = if (V>0.0001)then (Fd/Fg) else (0)
8-63
P8-14 (a) continued
8-64
P8-14 (b)
8-65
Polymath Code:
d(Fa)/d(V) = ra
d(Fb)/d(V) = rb
d(Fc)/d(V) = rc
d(Fd)/d(V) = rd
d(Fe)/d(V) = re
d(Fw)/d(V) = rw
d(Fg)/d(V) = rg
Cto = 0.147
k1 = 0.014
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
k2 = 0.007
Cb = Cto*(Fb/Ft)
k3 = 0.014
k4 = 0.45
Ca = Cto*(Fa/Ft)
Cc = Cto*(Fc/Ft)
ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
rb = ra/2
re = k3*Cc
Cd = Cto*(Fd/Ft)
Ce = Cto*(Fe/Ft)
Cg = Cto*(Fg/Ft)
Cw = Cto*(Fw/Ft)
rg = k4*Cd
rw = k3*Cc
rd = k2*(Ca)^2-k4*Cd
rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
Yc = Fc/Fa
Scd = if (V>0.0001)then (Fc/Fd) else (0)
Sae = if (V>0.0001)then (Fa/Fe) else (0)
Sdg = if (V>0.0001)then (Fd/Fg) else (0)
Sce = if (V>0.0001)then (Fc/Fe) else (0)
Fa(0)=10
Fb(0)=5
Fc(0)=0
Fd(0)=0
Fe(0)=0
Fw(0)=0
Fg(0)=0
V(0) = 0
V(f) = 2e5
8-66
P8-14 (b) continued
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.098
0.0951293
0.098
0.0951293
2
Cb
0.049
0.0475647
0.049
0.0475647
3
Cc
0
0
0.0032152
0.0032152
4
Cd
0
0
0.0001412
0.0001411
5
Ce
0
0
0.0002229
0.0002229
6
Cg
0
0
0.0005038
0.0005038
7
Cto
0.147
0.147
0.147
0.147
8
Cw
0
0
0.0002229
0.0002229
9
Fa
10.
9.63739
10.
9.63739
10 Fb
5.
4.818695
5.
4.818695
11 Fc
0
0
0.3257277
0.3257277
12 Fd
0
0
0.0143026
0.014299
13 Fe
0
0
0.0225834
0.0225834
14 Fg
0
0
0.0510394
0.0510394
15 Ft
15.
14.89232
15.
14.89232
16 Fw
0
0
0.0225834
0.0225834
17 k1
0.014
0.014
0.014
0.014
18 k2
0.007
0.007
0.007
0.007
19 k3
0.014
0.014
0.014
0.014
20 k4
0.45
0.45
0.45
0.45
21 ra
-0.0003709
-0.0003709
-0.0003538
-0.0003538
22 Rb
-0.0001855
-0.0001855
-0.0001769
-0.0001769
23 rc
0.0003037
0.0003037
0.000336
0.000309
24 rd
6.723E-05
-1.674E-07
6.723E-05
-1.674E-07
25 re
0
0
4.501E-05
4.501E-05
26 rg
0
0
6.352E-05
6.351E-05
27 rw
0
0
4.501E-05
4.501E-05
28 Sae
0
0
7.875E+05
426.7466
29 Scd
0
0
22.77975
22.77975
30 Sce
0
0
594.3536
14.42332
31 Sdg
0
0
18.10358
0.2801563
32 V
0
0
1000.
1000.
33 Yc
0
0
0.0337983
0.0337983
Differential equations
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(Fw)/d(V) = rw
7 d(Fg)/d(V) = rg
8-67
Explicit equations
1
Cto = 0.147
2
k1 = 0.014
3
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
4
k2 = 0.007
5
Cb = Cto*(Fb/Ft)
6
k3 = 0.014
7
k4 = 0.45
8
Ca = Cto*(Fa/Ft)
9
Cc = Cto*(Fc/Ft)
10 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
11 rb = ra/2
12 re = k3*Cc
13 Cd = Cto*(Fd/Ft)
14 Ce = Cto*(Fe/Ft)
15 Cg = Cto*(Fg/Ft)
16 Cw = Cto*(Fw/Ft)
17 rg = k4*Cd
18 rw = k3*Cc
19 rd = k2*(Ca)^2-k4*Cd
20 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
21 Yc = Fc/Fa
22 Scd = if (V>0.0001)then (Fc/Fd) else (0)
23 Sae = if (V>0.0001)then (Fa/Fe) else (0)
24 Sdg = if (V>0.0001)then (Fd/Fg) else (0)
25 Sce = if (V>0.0001)then (Fc/Fe) else (0)
Then taking different values of FA0 and FB0, we find the following data:
FA0
FB0Θ= FA0/FB0
YC
SC/E
SD/G
0.01
1
5
10
100
1000
5
5
5
5
5
5
0.002
0.2
1
2
20
200
SC/D
0.139038
4.424789
0.070678
3.48E+04
0.111343
0.057955
5.379092
9.342066
0.088876
0.168248
321.5455
52.61217
0.033798
0.001968
14.42332
107.1563
0.280156
2.87001
22.77975
1.931864
5.99E-05
1001.268
30.05269
0.413009
8-68
Yc
ΘO2
SC/E
ΘO2
SD/G
ΘO2
SD/G
ΘO2
8-69
P8-14 (d) Now we have a pressure drop parameter α = 0.002 with pressure ratio, p=P/P0, so we modify
our polymath program as follows:
Polymath Code
d(Fa)/d(V) = ra
d(Fb)/d(V) = rb
d(Fc)/d(V) = rc
d(Fd)/d(V) = rd
d(Fe)/d(V) = re
d(Fw)/d(V) = rw
d(Fg)/d(V) = rg
d(p)/d(V) = -alpha/2/p*(Ft/Ft0)
alpha = 0.002
Ft0 = 15
Cto = 0.147
k1 = 0.014
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
k2 = 0.007
Cb = Cto*(Fb/Ft)*p
k3 = 0.014
k4 = 0.45
Ca = Cto*(Fa/Ft)*p
Cc = Cto*(Fc/Ft)*p
ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
rb = ra/2
re = k3*Cc
Cd = Cto*(Fd/Ft)*p
Ce = Cto*(Fe/Ft)*p
Cg = Cto*(Fg/Ft)*p
Cw = Cto*(Fw/Ft)*p
rg = k4*Cd
rw = k3*Cc
rd = k2*(Ca)^2-k4*Cd
rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
Yc = Fc/Fa
Scd = if (V>0.0001)then (Fc/Fd) else (0)
Sae = if (V>0.0001)then (Fa/Fe) else (0)
Sdg = if (V>0.0001)then (Fd/Fg) else (0)
Sce = if (V>0.0001)then (Fc/Fe) else (0)
Fa(0)=0
Fb(0)=5
Fc(0)=0
Fd(0)=0
Fe(0)=0
Fw(0)=0
Fg(0)=0
p(0)=1
V(0) = 0
V(f) = 2e5
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.002
0.002
0.002
0.002
2
Ca
0.098
0.0041647
0.098
0.0041647
3
Cb
0.049
0.0020823
0.049
0.0020823
4
Cc
0
0
0.0004581
3.917E-05
5
Cd
0
0
5.56E-05
3.278E-06
8-70
6
Ce
0
0
8.655E-06
9.533E-07
7
Cg
0
0
3.75E-05
3.755E-06
8
Cto
0.147
0.147
0.147
0.147
9
Cw
0
0
8.655E-06
9.533E-07
10 Fa
10.
9.896873
10.
9.896873
11 Fb
5.
4.948437
5.
4.948437
12 Fc
0
0
0.0930717
0.0930717
13 Fd
0
0
0.0082669
0.0077896
14 Fe
0
0
0.0022655
0.0022655
15 Fg
0
0
0.0089238
0.0089238
16 Ft
15.
14.95962
15.
14.95963
17 Ft0
15.
15.
15.
15.
18 Fw
0
0
0.0022655
0.0022655
19 k1
0.014
0.014
0.014
0.014
20 k2
0.007
0.007
0.007
0.007
21 k3
0.014
0.014
0.014
0.014
22 k4
0.45
0.45
0.45
0.45
23 ra
-0.0003709
-0.0003709
-2.782E-06
-2.782E-06
24 Rb
-0.0001855
-0.0001855
-1.391E-06
-1.391E-06
25 Rc
0.0003037
3.587E-06
0.0003037
3.587E-06
26 Rd
6.723E-05
-4.638E-06
6.723E-05
-1.354E-06
27 re
0
0
6.413E-06
5.483E-07
28 Rg
0
0
2.502E-05
1.475E-06
29 Rw
0
0
6.413E-06
5.483E-07
30 Sae
0
0
3.344E+06
4368.521
31 Scd
0
0
11.94813
11.94813
32 Sce
0
0
1218.255
41.08224
33 Sdg
0
0
37.46977
0.8729093
34 V
0
0
500.
500.
35 p
1.
0.0428242
1.
0.0428242
36 Yc
0
0
0.0094042
0.0094042
Differential equations
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(Fw)/d(V) = rw
7 d(Fg)/d(V) = rg
8 d(p)/d(V) = -alpha/2/p*(Ft/Ft0)
8-71
P8-14 (d) continued
Explicit equations
1
alpha = 0.002
2
Ft0 = 15
3
Cto = 0.147
4
k1 = 0.014
5
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
6
k2 = 0.007
7
Cb = Cto*(Fb/Ft)*p
8
k3 = 0.014
9
k4 = 0.45
10 Ca = Cto*(Fa/Ft)*p
11 Cc = Cto*(Fc/Ft)*p
12 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
13 rb = ra/2
14 re = k3*Cc
15 Cd = Cto*(Fd/Ft)*p
16 Ce = Cto*(Fe/Ft)*p
17 Cg = Cto*(Fg/Ft)*p
18 Cw = Cto*(Fw/Ft)*p
19 rg = k4*Cd
20 rw = k3*Cc
21 rd = k2*(Ca)^2-k4*Cd
22 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
23 Yc = Fc/Fa
24 Scd = if (V>0.0001)then (Fc/Fd) else (0)
25 Sae = if (V>0.0001)then (Fa/Fe) else (0)
26 Sdg = if (V>0.0001)then (Fd/Fg) else (0)
27 Sce = if (V>0.0001)then (Fc/Fe) else (0)
8-72
P8-14 (d) continued
8-73
P8-14 (d) continued
8-74
P8-14 (d) continued
P8-14 (e)
Since C(Formic acid) is our desired product, so temperature corresponding to the maximum yield of C
should be recommended.
Polymath Code:
d(Fa)/d(V) = ra
d(Fb)/d(V) = rb
d(Fc)/d(V) = rc
d(Fd)/d(V) = rd
d(Fe)/d(V) = re
d(Fw)/d(V) = rw
d(Fg)/d(V) = rg
d(T)/d(V) = 0.25
k20 = 0.007
k10 = 0.014
k1 = k10*exp((10000/1.987)*(1/300-1/T))
k2 = k20*exp((30000/1.987)*(1/300-1/T))
k30 = 0.014
k3 = k30*exp((20000/1.987)*(1/300-1/T))
k40 = 0.45
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
Cto = 0.147
Ca = Cto*(Fa/Ft)
Cb = Cto*(Fb/Ft)
Cc = Cto*(Fc/Ft)
Cd = Cto*(Fd/Ft)
Ce = Cto*(Fe/Ft)
Cg = Cto*(Fg/Ft)
ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
rb = ra/2
re = k3*Cc
8-75
k4 = k40*exp((10000/1.987)*(1/300-1/T))
Cw = Cto*(Fw/Ft)
rg = k4*Cd
rw = k3*Cc
rd = k2*(Ca)^2-k4*Cd
rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
Yc = Fc/Fa
Scd = if(V>0.0001)then(Fc/Fd)else(0)
Sae = if(V>0.0001)then(Fa/Fe)else(0)
Sdg = if(V>0.0001)then(Fd/Fg)else(0)
Fa(0)=10
Fb(0)=5
Fc(0)=0
Fd(0)=0
Fe(0)=0
Fw(0)=0
Fg(0)=0
T(0)=300
V(0) = 0
V(f) = 1000
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.098
4.457E-08
0.098
4.457E-08
2
Cb
0.049
2.228E-08
0.049
2.228E-08
3
Cc
0
0
0.0150399
2.047E-12
4
Cd
0
0
0.008913
1.303E-10
5
Ce
0
0
0.0496473
0.0496473
6
Cg
0
0
0.0477053
0.0477053
7
Cto
0.147
0.147
0.147
0.147
8
Cw
0
0
0.0496473
0.0496473
9
Fa
10.
8.977E-06
10.
8.977E-06
10 Fb
5.
4.489E-06
5.
4.489E-06
11 Fc
0
0
1.847816
4.123E-10
12 Fd
0
0
1.022982
2.625E-08
13 Fe
0
0
9.999991
9.999991
14 Fg
0
0
9.608841
9.608841
15 Ft
15.
14.91792
29.60884
29.60884
16 Fw
0
0
9.999991
9.999991
17 k1
0.014
0.014
28.69217
28.69217
18 k10
0.014
0.014
0.014
0.014
19 k2
0.007
0.007
6.026E+07
6.026E+07
20 k20
0.007
0.007
0.007
0.007
21 k3
0.014
0.014
5.88E+04
5.88E+04
22 k30
0.014
0.014
0.014
0.014
23 k4
0.45
0.45
922.2483
922.2483
24 k40
0.45
0.45
0.45
0.45
25 ra
-0.0003709
-0.0618773
-1.199E-07
-1.199E-07
26 rb
-0.0001855
-0.0309386
-5.994E-08
-5.994E-08
27 rc
0.0003037
-0.0237815
0.0232582
-9.059E-12
28 rd
6.723E-05
-0.009135
0.0117881
-4.832E-10
8-76
29 re
0
0
0.0763914
1.204E-07
30 rg
0
0
0.0619064
1.202E-07
31 rw
0
0
0.0763914
1.204E-07
32 Sae
0
0
5.953E+05
8.977E-07
33 Scd
0
0
3.456581
0.0157098
34 Sdg
0
0
19.45999
2.732E-09
35 T
300.
300.
550.
550.
36 V
0
0
1000.
1000.
37 Yc
0
0
0.446161
4.593E-05
Differential equations
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(Fw)/d(V) = rw
7 d(Fg)/d(V) = rg
8 d(T)/d(V) = 0.25
Explicit equations
1
k20 = 0.007
2
k10 = 0.014
3
k1 = k10*exp((10000/1.987)*(1/300-1/T))
4
k2 = k20*exp((30000/1.987)*(1/300-1/T))
5
k30 = 0.014
6
k3 = k30*exp((20000/1.987)*(1/300-1/T))
7
k40 = 0.45
8
Ft = Fa+Fb+Fc+Fd+Fe+Fw+Fg
9
Cto = 0.147
10 Ca = Cto*(Fa/Ft)
11 Cb = Cto*(Fb/Ft)
12 Cc = Cto*(Fc/Ft)
13 Cd = Cto*(Fd/Ft)
14 Ce = Cto*(Fe/Ft)
15 Cg = Cto*(Fg/Ft)
16 ra = -(k1*Ca*(Cb)^0.5 + k2*(Ca)^2)
17 rb = ra/2
18 re = k3*Cc
19 k4 = k40*exp((10000/1.987)*(1/300-1/T))
20 Cw = Cto*(Fw/Ft)
21 rg = k4*Cd
22 rw = k3*Cc
23 rd = k2*(Ca)^2-k4*Cd
24 rc = k1*Ca*(Cb)^0.5 - k3*Cc+k4*Cd
25 Yc = Fc/Fa
8-77
26 Scd = if (V>0.0001)then (Fc/Fd) else (0)
27 Sae = if (V>0.0001)then (Fa/Fe) else (0)
28 Sdg = if (V>0.0001)then (Fd/Fg) else (0)
Temperature corresponding to maximum yield = 367.8 K
P8-15
(1)
C2H4 + 1/2O2 → C2H4O
E + 1/2O → D
(2)
E
FIO = 0.82 FTO = 0.007626
P8-15 (a)
Selectivity of D over CO2
S=
FD
FU 1
8-78
C2H4 + 3O2 → 2CO2 + 2H2O
+ 3O → 2U1 + 2U2
See the following Polymath program:
Polymath code
d(Fe)/d(W) = r1e+r2e
d(Fo)/d(W) = 1/2*r1e + 3*r2e
d(Fd)/d(W) = -r1e
d(Fu1)/d(W) = -2*r2e
d(Fu2)/d(W) = -2*r2e
Finert = 0.007626
Ft = Fe+Fo+Fd+Fu1+Fu2+Finert
K1 = 6.5
K2 = 4.33
Pto = 2
Pe = Pto*Fe/Ft
Po = (Pto*Fo/Ft)
k1 = 0.15
k2 = 0.088
X = 1 - Fe/0.000558
S = Fd/Fu1
r1e = -k1*Pe*Po^0.58/(1+K1*Pe)^2
r2e = -k2*Pe*Po^0.3/(1+K2*Pe)^2
Fe(0)=5.58E-04
Fo(0)=0.001116
Fd(0)=0
Fu1(0)=1e-7
Fu2(0)=0
W(0) = 0
W(f) = 2
Polymath Output
Variableinitial valueminimal valuemaximal valuefinal value
W
0
0
2
2
Fe
5.58E-04
1.752E-10
5.58E-04
1.752E-10
Fo
0.001116
4.066E-05
0.001116
4.066E-05
Fd
0
0
2.395E-04
2.395E-04
Fu1
1.0E-07
1.0E-07
6.372E-04
6.372E-04
Fu2
0
0
6.371E-04
6.371E-04
Finert
0.007626
0.007626
0.007626
0.007626
Ft
0.0093001
0.0091804
0.0093001
0.0091804
K1
6.5
6.5
6.5
6.5
K2
4.33
4.33
4.33
4.33
Pto
2
2
2
2
Pe
0.1199987
3.817E-08
0.1199987
3.817E-08
Po
0.2399974
0.008858
0.2399974
0.008858
k1
0.15
0.15
0.15
0.15
k2
0.088
0.088
0.088
0.088
X
0
0
0.9999997
0.9999997
S
0
0
0.4101512
0.3758225
r1e
-0.0024829
-0.0024829
-3.692E-10
-3.692E-10
r2e
-0.0029803
-0.0029803
-8.136E-10
-8.136E-10
Differential equations as entered by the user
[1] d(Fe)/d(W) = r1e+r2e
[2] d(Fo)/d(W) = 1/2*r1e + 3*r2e
[3] d(Fd)/d(W) = -r1e
[4] d(Fu1)/d(W) = -2*r2e
[5] d(Fu2)/d(W) = -2*r2e
Explicit equations as entered by the user
[1] Finert = 0.007626
[2] Ft = Fe+Fo+Fd+Fu1+Fu2+Finert
[3] K1 = 6.5
[4] K2 = 4.33
[5] Pto = 2
[6] Pe = Pto*Fe/Ft
[7] Po = (Pto*Fo/Ft)
8-79
[8] k1 = 0.15
[9] k2 = 0.088
[10] X = 1 - Fe/0.000558
[11] S = Fd/Fu1
[12] r1e = -k1*Pe*Po^0.58/(1+K1*Pe)^2
[13] r2e = -k2*Pe*Po^0.3/(1+K2*Pe)^2
X = 0.999 and S = 0.376(mol of ethylene oxide)/(mole of carbon dioxide)
P8-15 (b)
Changes in equation from part (a):
dFO 1
and FO (o ) = 0
= r1E + 3r 2 E + RO
dW 2
0.12 ´ 0.0093 0.001116 mol
RO =
=
W
2
kg .s
𝐹n (0) = 0.0093(1 − 0.12) ∗ 0.06 = 4.91𝐸 − 04 𝑚𝑜𝑙 𝑒𝑡ℎ𝑦𝑙𝑒𝑛𝑒
From Polymath program:
X = 0.83
S = 0.05 (mol of ethylene oxide)/(mole of carbon dioxide)
P8-15 (c)
Changes in equation from part (a):
dFE
= r1E + r 2 E + R E and FE (o) = 0
dW
0.06 ´ 0.0093 0.000558 mol
RE =
=
W
2
kg .s
𝐹r (0) = 0.0093 (1 − 0.06) ∗ 0.12 = 1.05𝐸 − 03 𝑚𝑜𝑙 𝑒𝑡ℎ𝑦𝑙𝑒𝑛𝑒
From Polymath program:
X = 0.95
S = 0.46(mol of ethylene oxide)/(mole of carbon dioxide)
P8-16
The reactions are
1. C + W à 6 H2 + 6CO
2. L +7W à 13 H2 + 10 CO
Rate laws are :
- r1C = k1C CC CW
-r2L = k2L CL Cw2
CT0= P0/RT = ( 1 atm )/(0.082)(1473) = 0.00828 mol/dm3
FC0 = 0.00411 mol/s
FL0 = 0.00185 mol/s
FW0= 0.02 mol/s
P8-16 (a)
Polymath Code
d(Fc)/d(V) = r1c
d(Fl)/d(V) = r2l
d(Fw)/d(V) = r1c+7*r2l
d(Fh)/d(V) = -6*r1c-13*r2l
d(Fco)/d(V) = -6*r1c-10*r2l
T = 1473
8-80
R = 0.082
P=1
k2l = 14000000
k1c = 30000
Ft = Fc+Fl+Fw+Fh+Fco
Ct0 = P/(R*T)
Fc0 = 0.00411523
Fl0 = 0.00185185
Mc = Fc*162
Ml = Fl*180
Mh = Fh*2
Mco = Fco*28
Mw = Fw*18
r1c = -k1c*(Ct0*(Fc)/(Ft))*(Ct0*(Fw)/(Ft))
r2l = -k2l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2)
ConvC = (Fc0-Fc)/Fc0
ConvL = (Fl0-Fl)/Fl0
Fc(0)=0.0041152
Fw(0)=0.02
Fh(0)=0
Fl(0)=0.0018519
Fco(0)=0
V(0) = 0
V(f) = 0.417
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
ConvC
0
0
0.8337198
0.8337198
2
ConvL
0
0
0.7379696
0.7379696
3
Ct0
0.0082791
0.0082791
0.0082791
0.0082791
4
Fc
0.0041152
0.0006843
0.0041152
0.0006843
5
Fc0
0.0041152
0.0041152
0.0041152
0.0041152
6
Fco
0
0
0.0342518
0.0342518
7
Fh
0
0
0.0383516
0.0383516
8
Fl
0.0018519
0.0004852
0.0018519
0.0004852
9
Fl0
0.0018519
0.0018519
0.0018519
0.0018519
10 Ft
0.0259671
0.0259671
0.0807757
0.0807757
11 Fw
0.02
0.0070028
0.02
0.0070028
12 k1c
3.0E+04
3.0E+04
3.0E+04
3.0E+04
13 k2l
1.4E+07
1.4E+07
1.4E+07
1.4E+07
14 Mc
0.6666673
0.1108536
0.6666673
0.1108536
15 Mco
0
0
0.9590499
0.9590499
16 Mh
0
0
0.0767032
0.0767032
17 Ml
0.333333
0.0873434
0.333333
0.0873434
18 Mw
0.36
0.1260502
0.36
0.1260502
19 P
1.
1.
1.
1.
20 R
0.082
0.082
0.082
0.082
21 r1c
-0.2509955
-0.2509955
-0.0015102
-0.0015102
22 r2l
-0.3361048
-0.3361048
-0.0003587
-0.0003587
23 T
1473.
1473.
1473.
1473.
24 V
0
0
0.417
0.417
8-81
Differential equations
1 d(Fc)/d(V) = r1c
2 d(Fl)/d(V) = r2l
3 d(Fw)/d(V) = r1c+7*r2l
4 d(Fh)/d(V) = -6*r1c-13*r2l
5 d(Fco)/d(V) = -6*r1c-10*r2l
Explicit equations
1 T = 1473
2 R = 0.082
3 P=1
4 k2l = 14000000
5 k1c = 30000
6 Ft = Fc+Fl+Fw+Fh+Fco
7 Ct0 = P/(R*T)
8 Fc0 = 0.00411523
9 Fl0 = 0.00185185
10 Mc = Fc*162
11 Ml = Fl*180
12 Mh = Fh*2
13 Mco = Fco*28
14 Mw = Fw*18
15 r1c = -k1c*(Ct0*(Fc)/(Ft))*(Ct0*(Fw)/(Ft))
16 r2l = -k2l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2)
17 ConvC = (Fc0-Fc)/Fc0
18 ConvL = (Fl0-Fl)/Fl0
Plot of Fc vs V
8-82
P8-16 (a) continued
Plot of Fl vs V
Plot of Fw vs V
Plot of Fh vs V
8-83
Plot of Fco vs V
P8-16 (b)
W is the key reactant
YC = (-dCC/-dCW) = -r1C/(-r1C -7*r2L)
The modification is made in the polymath code for Problem 8.16(a)
The plot of YC versus V is as follows;
For plotting YW versus volume;
since W is our key reactant then,
Yw = -dW/(-dW) = 1
Now YL = -dL/(-dW) = -r2L/(-r1C – 7*r2L)
The plot is as follows :
8-84
P8-16 (b) continued
Overall selectivity
SCO / H 2 = N CO = exit molar flow rate of CO/exit molar flow rate of H2 =
N H2
CCO |exit /CH2 | exit at any point in the reactor.
P8-16 (c) Individualized solution
P8-17 (a)
The reactions are
(i) L + 3 W à 3 H2 + 3 CO + char
(ii) Ch + 4 W à 10 H2 + 7CO
rate laws at 1200 0C are :
-r1L = k1LCLCw2
-r2CH = k2ChCchCw2
; k1L = 3721 (dm3/mol)2/sec
; k2CH = 1000 ( dm3/mol)2/sec
8-85
Polymath Code
d(Fl)/d(V) = r1l
d(Fch)/d(V) = r2ch-r1l
d(Fw)/d(V) = 3*r1l+4*r2ch
d(Fh)/d(V) = -3*r1l-10*r2ch
d(Fco)/d(V) = -3*r1l-7*r2ch
k2ch = 1000
k1l = 3721
Ft = Fch+Fl+Fw+Fh+Fco
Ct0 = 0.2
r1l = -k1l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2)
r2ch = -k2ch*(Ct0*(Fch)/(Ft))*((Ct0*(Fw)/(Ft))^2)
Fl0 = 0.0123
ConL = (Fl0-Fl)/(Fl0)
Fl(0)=0.0123
Fch(0)=0.0
Fw(0)=0.111
Fh(0)=0
Fco(0)=0
V(0) = 0
V(f) = 0.417
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
ConL
0
0
0.9899505
0.9899505
2
Ct0
0.2
0.2
0.2
0.2
3
Fch
0
0
0.0075887
0.0047166
4
Fco
0
0
0.0887476
0.0887476
5
Fh
0
0
0.1111269
0.1111269
6
Fl
0.0123
0.0001236
0.0123
0.0001236
7
Fl0
0.0123
0.0123
0.0123
0.0123
8
Ft
0.1233
0.1233
0.2493464
0.2493464
9
Fw
0.111
0.0446317
0.111
0.0446317
10 k1l
3721.
3721.
3721.
3721.
11 k2ch
1000.
1000.
1000.
1000.
12 r1l
-2.406642
-2.406642
-0.0004728
-0.0004728
13 r2ch
0
-0.0997755
0
-0.0048484
14 V
0
0
0.417
0.417
Differential equations
1 d(Fl)/d(V) = r1l
2 d(Fch)/d(V) = r2ch-r1l
3 d(Fw)/d(V) = 3*r1l+4*r2ch
4 d(Fh)/d(V) = -3*r1l-10*r2ch
5 d(Fco)/d(V) = -3*r1l-7*r2ch
Explicit equations
1 k2ch = 1000
2 k1l = 3721
3 Ft = Fch+Fl+Fw+Fh+Fco
4 Ct0 = 0.2
5 r1l = -k1l*(Ct0*(Fl)/(Ft))*((Ct0*(Fw)/(Ft))^2)
8-86
6 r2ch = -k2ch*(Ct0*(Fch)/(Ft))*((Ct0*(Fw)/(Ft))^2)
7 Fl0 = 0.0123
8 ConL = (Fl0-Fl)/(Fl0)
P8-17 (b)
P8-17 (c) SCO/Ch
Let S1 = S~CO/Ch = Fco/Fch
8-87
The plot is as follows
Considering W as the principle reacting species;
Yw = 1
and Yl = (rl)/(rw) = (r1l)/(3*r1l + 4*r2ch)
the plot is as follows:
P8-17 (d)
The molar flow rate for char is maximum at about V= 0.02085 dm3.
P8-18 (a) Individualized Solution
P8-18 (b) Individualized Solution
8-88
Synopsis for Chapter 9 – Reaction Mechanisms, Pathways,
Bioreactions and Bioreactors
General: Bioreaction problems on enzymes and micro-organism growth which require Polymath.
Questions
l Q9-1A (20 seconds) Questions Before Reading (QBR).
O Q9-3A (15 min) These screencasts are quite relevant to this chapter. View two screencasts at 7
minutes/case.
Computer Simulations and Experiments
l P9-1A (a) – (e) (10 min) Wolfram simulations and the student should spend 10-15 minutes on each
one including writing a set of conclusions. Always assigned.
O P9-1A (g) (35-45 min) Excellent problem on alcohol metabolism with actual data from the literature.
Interactive Computer Games (ICG)
l P9-2A (a) (30 min) Interactive Computer Game (ICG) Enzyme Man. Based loosely on Pac Man Video
Game.
Problems
I P9-3C (30 min) Apply PSSH to the oxidation of CO.
AA P9-4A (30 min) Straight forward application of PSSH to derive a rate law.
l P9-5B (30min) Old Exam Question (OEQ). (b) I always assign Part (b) and sometimes part (c).
Excellent and challenging problem on the application of the PSSH.
AA P9-6C (50 min) PSSH problem that serves as motivation as to why one should change their motor oil
by applying CRE to real everyday problems.
AA P9-7A (25 min) Excellent problem that is applied to COVID-19 in Chapter 13 because Chapter 9 had
been finalized by the time the virus pandemic struck. Apply PSSH (which in this case is NOT a good
assumption) to make predictions on the spread of disease.
O P9-8B (30 min) Write rate laws for a series of enzymatic reactions.
AA P9-9B (35 min) Old Exam Question (OEQ). Very straight forward problem to use the data to find the
Michaelis–Menten parameters.
O P9-10B (40 min) Old Exam Question (OEQ). Requires a bit of thinking to realize that one must plot
the mass balance on the same graph to find the intersections.
S P9-11B (35 min) Straight forward problem where the Lineweaver Burk plot is used to find the rate
law parameters.
AA P9-12B (30 min) Old Exam Question (OEQ). Fairly straight forward problem where the Lineweaver
Burk plot can be used to find the type of inhibition.
S9-1
S P9-13B (15 min) Not completely straight forward to use the reactions given to arrive at Figure
P9-13B.
l P9-14B. (20 min) Short conceptual problem to match type of inhibition with data.
AA P9-15B (25 min) A quick plot of the data of k2 versus CS will reveal the answers.
O P9-16B (25 min) Short problem to use equation for maximum production rate to find uo
corresponding volumetric flow rate.
AA P9-17B (45 min) Straight forward problem using Polymath to solve mass balance equation.
I P9-18B (50 min) More complex version of P9-17B.
AA P9-19B (45 min) Straight forward application of Monod kinetics.
AA P9-20A (25 min) Short conceptual problem to find umax and Ks.
AA P9-21C (50 min) Real life drug example. It is a rather advanced problem.
AA P9-22B (45 min) Current application of cell growth to the production of biofuels.
AA P9-23A (35 min) Shorter version of P9-22B.
S9-2
Solutions for Chapter 9 – Reaction Mechanisms, Pathways,
Bioreactions and Bioreactors
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Chapter 9:
http://www.umich.edu/~elements/5e/09chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram codes):
http://www.umich.edu/~elements/5e/tutorials/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath codes):
http://www.umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q9-1 Individualized solution
Q9-2 Individualized solution
Q9-3 (a) Individualized solution
(b) The concentration of substrate decreases with time, so we need to put a negative sign before the
rate of change in substrate concentration. 𝑟" indicates rate of generation of cells and hence there is no
negative sign before it.
Q9-4 Individualized solution
P9-1 (a) Example 9-1
(i)
The intensity increases with an increase in CS2 and M for both the cases: presence and absence of
Alcohol
(ii) Intensity increases with increase in concentration of M.
(iii) 1) Intensity in presence of alcohol depends on all parameters k1, k2, k3 and k4.
2) Intensity in absence of alcohol does not depend on parameter k3.
3) When ratio of both intensities is taken k1 gets cancelled. Hence slope of line Io/I does not
depend on k1.
P9-1 (b) Example 9-2
(i)
The rate of substrate consumption is almost a constant and independent of substrate
concentration at very low KM
(ii) There could be several combinations of KM and Vmax at which rate becomes zero. One such value
is Vmax = 0.2 kmol/m3s, and KM = 0.0035 kmol/m3
(iii) 1. Apart from Michaelis – Menton plot, all the other 3 curves are straight lines with slopes and
intercept depending on Vmax and Km.
2. Vmax and Km both have greatest effect in Lineweaver – Burk plot. This is because in that
curve reciprocals of both rate and concentration are plotted.
9-1
(iv)
(v)
To check type of inhibition, we need to plot the data of “without inhibition” and “with inhibition”
in the form of Lineweaver-Burk Plots.
The data for “without inhibition” is taken from Example 9-2
From the above plot, the intercept for the both the graphs appears to be merging with each other.
However, slope is different for both the graphs.
So, the inhibitor shows competitive inhibition.
'
In a Hanes- Woolf plot, we need to plot 𝑆 𝑣𝑠 ()
The equation for no inhibition is
'
()*
,
'
/01
/01
=. - +.
The equation for Competitive inhibition is
'
()*
=
,./01
31 +
5
,6
7+
'
./01
The equation for Uncompetitive inhibition is
'
()*
=
,./01
+
'
./01
31 +
5
,6
7
The equation for Noncompetitive inhibition is
'
()*
,
5
'
/01
6
/01
= . - 31 + , 7 + .
31 +
5
,6
7
The equation for Substrate inhibition at low substrate concentration is
'
()*
=(𝐾: )/𝑉>?@
Therefore, in a Hanes-Woolf plot for,
Competitive inhibition: slope remains same as no inhibition case but intercept changes
Uncompetitive inhibition: slope changes but intercept remains constant
Noncompetitive inhibition: intercept and slope both changes
Substrate inhibition: a straight line parallel to X axis
9-2
Hanes-Woolf plot
)B
In an Eadie- Hoftstee plot, we need to plot − 𝑣𝑠 − 𝑟𝑠
B
The equation for no inhibition is
()
−𝑟' = 𝑉>?@ − 𝐾: 3 * 7
'
The equation for Competitive inhibition is
5
()
−𝑟' = 𝑉>?@ − 𝐾: 31 + 7 3 * 7
,6
'
The equation for Uncompetitive inhibition is
.
,
()
−𝑟B = /016 − -6 3 * 7
CD
E6
CD
E6
'
The equation for Noncompetitive inhibition is
.
()
−𝑟B = /016 − 𝐾: 3 * 7
CD
E6
'
Therefore in a Eadie- Hoftstee plot for,
Competitive inhibition: slope changes but intercept remains constant
Uncompetitive inhibition: slope and intercept both changes
Noncompetitive inhibition: Slope remains constant but intercept changes
For plot, refer Figure P9-14B (Problem P9-14B)
P9-1 (c) Example 9-3
(i)
The curve asymptotically approaches X = 1. For the conversion to approach ‘1’ almost it takes
900 s.
(ii) As we increase Km, conversion decreases for a particular time t (or) The time required to reach a
particular conversion X increases.
(iii) Parameters that increase conversion for a given time: Vmax2, Et2. Hence Increasing Et2 and
Vmax1 increases conversion while conc of urea and Km which are in direct multiplication with
conversion terms, increasing them decreases conversion wrt time. This can also be seen by
differential equations below.
9-3
(iv)
Now Curea = 1 mol/dm3 and t = 15 min = 900 sec.
H.HJKK
(v)
C
R
900 = J.L∗CHNO ln 3C(R7 + J.L∗CHNO
Solving, we get X = 0.2136
For t = 30 min = 1800 sec
,C
H.CR
3 7
1800 = J.L∗CH
NO ln C(R + J.L∗CH NO
Putting X = 0.5
KM = 0.551 mol/dm3
Thus, KM must be greater than 0.551 mol/dm3 for conversion to be below 50% after half an hour.
P9-1 (d) Example 9-4
Mass of substrate consumed per mass of cells,
VW
JCY(JZH
𝑌* = − VWX = − K.J[(C = 13.97 g/g
U
U
U
U
Mass of product formed per mass of cells
VW
Y.[](H
𝑌\ = − VW\ = K.J[(C = 3.97 g/g
Total = -13.97 + 3.97 = -10.06 g/g
Yes, there is disparity as substrate is also used in maintenance of cells which we have not accounted.
P9-1 (e) Example 9-5
(i)
YP/C
This profile has been obtained by varying
Ypc. When Ypc is set at its minimum value, a
maxima can be seen at t=9hr.
(ii)
(iii)
(iv)
(v)
Keeping all the parameters same as given, the minimum value of KS such that CS is always higher
than CP for the given time range is 107
Keeping all the parameters same as given, the maximum value of µmax so that rg continuously
increases up to t = 12 hrs. is 0.27
Keeping all the parameters same as given, the value of YS/C such that virtually 100% conversion is
achieved at the end of the reaction is 17.0
1) Increase in Ysc, m and Umax increases conversion wrt time. Other parameters have a negative
effect on conversion profile.
2) The concentration of substrate and product increase with an increase of Ypc and decrease of
Ysc. This can be seen from the differential equations below.
9-4
3) Rates Rg, Rd and Rsm vary according to their individual formulas. For the case Rd follows the
same trends as for Cc. The curve of Rg clearly has a maxima at some point while profiles of Rd
is quite low in comparison to Rg. Rsm is highly sensitive to ‘m’ and at higher values of ‘m’ it
becomes comparable to Rg.
(vi) See the following Polymath program
Polymath code:
d(Cc)/d(t) = rg-rd- vo*Cc/V #
d(Cs)/d(t) = Ysc*(-rg)-rsm+(Cso-Cs)*vo/V #
d(Cp)/d(t) = rg*Ypc-vo*Cp/V #
rd = Cc*.01 #
Ysc = 1/.08 #
Ypc = 5.6 #
Ks = 1.7 #
m = .03 #
umax = .33 #
rsm = m*Cc #
kobs = (umax*(1-Cp/93)^.52) #
rg = kobs*Cc*Cs/(Ks+Cs) #
V = Vo+vo*t #
mp = Cp*V #
t(0)=0
Cc(0)=0.0002
Cs(0)=0.0005
Cp(0)=0
t(f)=24
vo = 0.5
Cso = 5
Vo = 1
Calculated values of the DEQ variables
Variable
t
Cc
Cs
Cp
rd
Ysc
Ypc
Ks
m
umax
rsm
kobs
rg
vo
Cso
Vo
V
mp
initial value
0
2.0E-04
5.0E-04
0
2.0E-06
12.5
5.6
1.7
0.03
0.33
6.0E-06
0.33
1.941E-08
0.5
5
1
1
0
minimal value
0
1.276E-04
5.0E-04
0
1.277E-06
12.5
5.6
1.7
0.03
0.33
3.83E-06
0.3299706
1.941E-08
0.5
5
1
1
0
maximal value
24
0.002742
4.5794959
0.0159354
2.742E-05
12.5
5.6
1.7
0.03
0.33
8.226E-05
0.33
6.598E-04
0.5
5
1
13
0.20716
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(Cc)/d(t) = rg-rd-vo*Cc/V
[2] d(Cs)/d(t) = Ysc*(-rg)-rsm+(Cso-Cs)*vo/V
[3] d(Cp)/d(t) = rg*Ypc-vo*Cp/V
9-5
final value
24
0.002742
4.5794959
0.0159354
2.742E-05
12.5
5.6
1.7
0.03
0.33
8.226E-05
0.3299706
6.598E-04
0.5
5
1
13
0.20716
Explicit equations as entered by the user
[1] rd = Cc*.01
[2] Ysc = 1/.08
[3] Ypc = 5.6
[4] Ks = 1.7
[5] m = .03
[6] umax = .33
[7] rsm = m*Cc
[8] kobs = (umax*(1-Cp/93)^.52)
[9] rg = kobs*Cc*Cs/(Ks+Cs)
[10] vo = 0.5
[11] Cso = 5
[12] Vo = 1
[13] V = Vo+vo*t
[14] mp = Cp*V
(vii) Change the cell growth rate law:
All other equations are the same as in part (v).
9-6
P9-1 (e) continued
Notice that uncompetitive inhibition by the substrate causes the concentration of cells to decrease with
time, whereas without uncompetitive inhibition, the concentration of cells increased with time.
Consequently, the concentration of product is very low compared to the case without uncompetitive
inhibition.
(viii) Change the observed reaction rate constant:
All other equations are the same as in part (v).
Since CP is so small, the factor
whether CP* = 93 g/dm3 or CP* = 10,000 g/dm3.
Thus, the plots for this part are approximately the same as the plots in part (v).
P9-1 (f)
(i)
By varying Vmax between its minimum and maximum, the rate of substrate consumption varies
between zero and infinity respectively.
When KM is minimum the lines become horizontal (slope = 0). When KM is maximum the rate of
substrate consumption tends to zero.
(ii) For high values of KI, all the three inhibitions behave in the same way as no inhibition case. For
low values of KI, the slopes of the lines increase resulting in lower rates at a given substrate
concentration. The rates are least for noncompetitive inhibition followed by uncompetitive
inhibition and then competitive inhibition.
(iii) 1. The higher the value of KI, the lower the effect of inhibition on the reaction
2. Noncompetitive inhibition has the highest effect on rate among the three types of inhibitions.
3. Increase in KM results in smaller value of rate of substrate consumption.
P9-1 (g)
(i)
Individualized answer.
(ii) 1. Initial concentration of alcohol Cso greatly effects the concentration of alcohol in blood. It also
effects the time duration for which high concentrations of alcohol and acetaldehyde remain in
the blood. Concentration of acetaldehyde is not very sensitive to the initial concentration of
alcohol.
2. Without ALDH, the metabolism of acetaldehyde into acetic acid will almost not happen (as the
enzyme ALDH is the catalyst for this metabolism). Metabolism of acetaldehyde follows
Michaelis-Menten kinetics and Vmax depends on the concentration of enzyme. In case of ALDH
deficiency Vmax,De will be very small.
As can be seen in the plots below, the concentration of alcohol in blood decreases very slowly and
the concentration of acetaldehyde shoots up and will take a very long time to reduce.
9-7
(iii)
Normal value of Vmax,De = 2.75 mmol/min
Set the slider between 0.275 and 1.375 mmol/min.
Time taken for the concentration of alcohol to decrease increases when Vmax,De decreases. The
concentration of acetaldehyde in blood increases drastically.
Base case:
For Vmax,De = 0.275 mmol/min:
9-8
P9-1 (g) continued
For Vmax,De = 1.375 mmol/min:
The maximum enzymatic rates (Vmax,Ac and Vmax,De ) effect the concentration profiles to a great extent. Also,
the initial concentration of alcohol in stomach also plays an important role in the time taken for
metabolism. Increase in blood flow rates in the liver or muscle help in faster metabolism but the effect is
small.
P9-2 (a) The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer
Games (ICGs) is given at the front of this Solutions Manual.
P9-2 (b) Individualized Solution
P9-2 (c)
Applying the pseudo-steady state hypothesis,
J
𝑟^_`∗ = 𝑘C 𝐶^_`
+ 𝑘L 𝐶^_` 𝐶: − 𝑘J 𝐶^_` 𝐶^_`∗ − 𝑘K 𝐶^_`∗ − 𝑘Z 𝐶^_`∗ 𝐶: = 0
𝐶^_`∗ =
J
𝑘C 𝐶^_`
+ 𝑘L 𝐶: 𝐶^_`
𝑘J 𝐶^_` + 𝑘K + 𝑘Z 𝐶:
𝑟Wc de = 𝑘K 𝐶^_`∗ = 𝑘K
P9-3 (a), (b)
Burning:
9-9
J
𝑘C 𝐶^_`
+ 𝑘L 𝐶: 𝐶^_`
𝑘J 𝐶^_` + 𝑘K + 𝑘Z 𝐶:
9-10
9-11
9-3 (c) To observe the trends, the initial concentrations of oxygen, water, carbon mono-oxide and HCl
are taken to be 1 (arbitrary). The initial concentrations of the rest of the compounds/intermediates is
taken to be zero. k2, k3 are taken to be 0.01
Polymath code:
d(H2)/d(t) = k5*Hdot*HCl
d(HCl)/d(t) = -k5*Hdot*HCl+k6*Hdot*Cldot
d(CO2)/d(t) = k3*CO*OHdot
d(CO)/d(t) = -k3*CO*OHdot
d(Hdot)/d(t) = -k4*Hdot*O2+k3*CO*OHdot-k5*Hdot*HCl-k6*Hdot*Cldot
d(H2O)/d(t) = -k2*H2O*Odot
d(O2)/d(t) = -k1*O2-k4*O2*Hdot
k2 = 0.01
k3 = 0.01
k6 = 0.005
k1 = 0.00001
k4 = 0.02
k5 = 0.05
Cldot = k5*HCl/k6
Odot = ((2*k1+k4*Hdot)/(k2*H2O))*O2
OHdot = ((4*k1+3*k3*Hdot)*O2)/(k3*CO)
t(0) = 0
t(f) = 100
CO(0) = 1
CO2(0) = 0
H2(0) = 0
H2O(0) = 1
HCl(0) = 1
O2(0) = 1
Hdot(0) = 0
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Cldot
10.
10.
10.
10.
2 CO
1.
0.9948203
1.
0.9948203
3 CO2
0
0
0.0051797
0.0051797
4 H2
0
0
0.0019737
0.0019737
5 H2O
1.
0.997213
1.
0.997213
6 HCl
1.
1.
1.
1.
7 Hdot
0
0
0.0004436
0.0004436
9-12
8 k1
1.0E-05
1.0E-05
1.0E-05
1.0E-05
9 k2
0.01
0.01
0.01
0.01
10 k3
0.01
0.01
0.01
0.01
11 k4
0.02
0.02
0.02
0.02
12 k5
0.05
0.05
0.05
0.05
13 k6
0.005
0.005
0.005
0.005
14 O2
1.
0.9982121
1.
0.9982121
15 Odot
0.002
0.002
0.0028901
0.0028901
16 OHdot
0.004
0.004
0.005349
0.005349
17 t
0
0
100.
100.
Differential equations
1 d(H2)/d(t) = k5*Hdot*HCl
2 d(HCl)/d(t) = -k5*Hdot*HCl+k6*Hdot*Cldot
3 d(CO2)/d(t) = k3*CO*OHdot
4 d(CO)/d(t) = -k3*CO*OHdot
5 d(Hdot)/d(t) = -k4*Hdot*O2+k3*CO*OHdot-k5*Hdot*HCl-k6*Hdot*Cldot
6 d(H2O)/d(t) = -k2*H2O*Odot
7 d(O2)/d(t) = -k1*O2-k4*O2*Hdot
Explicit equations
1 k2 = 0.01
2 k3 = 0.01
3 k6 = 0.005
4 k1 = 0.00001
5 k4 = 0.02
6 k5 = 0.05
7 Cldot = k5*HCl/k6
8 Odot = ((2*k1+k4*Hdot)/(k2*H2O))*O2
9 OHdot = ((4*k1+3*k3*Hdot)*O2)/(k3*CO)
P9-4 (a)
Active intermediates:
9-13
Plugging in expression for
:
Now, substitute expressions for
P9-4 (b)
When,
into equation for
W
7 ≪ 2 , −𝑟^W ~ 𝑘𝐶^W
𝑘J 𝑠𝑟𝑞𝑡 3i hU
i
j O
P9-4 (c)
9-14
(power law)
:
P9-5 (a) The PSSH is valid in all the three reactions given in (b), (c) and (d).
P9-5 (b)
P9-5 (c)
9-15
add rCOCl to rCl
P9-5 (d)
The proposed mechanism for this reaction is:
Through this mechanism, it may be deduced that the net rate of formation of HBr (product) is given by:
where the concentrations of the intermediates may be determined by invoking the steady state
approximation, i.e.:
9-16
and:
i.e. from (ii) + (i) we obtain:
i.e.:
which substituted back in (i) gives:
i.e.:
i.e.:
or: dividing 'top' and 'bottom' by k’b :
Thus, the net rate of formation of HBr may be written solely in terms of reactants, i.e.,:
9-17
which simplifies to:
as predicted by the empirical rate law:
where:
P9-6 (a)
9-18
.
P9-6 (b)
Low temperatures with anti-oxidant
9-19
P9-6(c)
If the radicals are formed at a constant rate, then the differential equation for the concentration of the
radicals becomes:
and
The substitution in the differential equation for R· also changes. Now the equation is:
and solving and substituting gives:
Now we have to look at the balance for RO2·.
and if we substitute in our expression for [R·] we get
which we can solve for [RO2·].
9-20
Now we are ready to look at the equation for the motor oil.
and making the necessary substitutions, the rate law for the degradation of the motor oil is:
P9-6 (d)
Without antioxidants
With antioxidants
P9-6 (e) Individualized solution
P9-7 (a)
9-21
P9-7 (b)
P9-7 (c)
P9-7 (d)
See the following Polymath program
d(H)/d(t) = -k1*H-k2*I*H+k3*H
d(I)/d(t) = k1*H+k2*H*I-k3*H-k4*I
d(D)/d(t) = k4*I
k1 = 1e-8
k2 = 1e-16
k3 = 5e-10
k4 = 1e-11
t(0) = 0
t(f) = 10
H(0) = 1e9
I(0) = 0
D(0) = 0
9-22
Everyone becomes ill rather quickly, and the rate at which an ill person recovers to a healthy person is
much slower than the rate at which a healthy person becomes ill. Eventually everyone is ill and people
start dying.
P9-7 (e) Individualized solution
P9-8 (a)
By applying PSSH for the complex [E.S], we have
9-23
Since E is not consumed, we have
where
is a constant
So,
where
P9-8 (b)
since E is not consumed:
or
Insert this into the equation for rE·S and solve for the concentration of the intermediate:
9-24
P9-8 (c)
(1)
(2)
If we add these two rates we get:
(3)
From equation (2) we get
Plug this into equation 3 and we get:
9-25
P9-8 (d)
P9-8 (e) Reaction (b) can be converted into a Lineweaver–Burk plot.
9-26
P9-9 (a)
The enzyme catalyzed reaction of the decomposition of hydrogen peroxide. For a batch reactor:
9-27
P9-9 (b)
P9-9 (c) Individualized solution
P9-9 (d) Individualized solution
P9-10 (a)
9-28
P9-10 (b)
9-29
9-30
P9-10 (c)
If ET 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
P9-10 (d) Individualized solution
P9-10 (e) Individualized solution
P9-11 (a)
9-31
P9-11 (b)
In this case both the slope and intercept change. Plotting the data of o
C
pc
C
in mmHg versus q
pc
with
sulfanilamide on the same plot as was plotted the data for the case with no sulfanilamide, it is seen that
the slopes are different, but the intercept is the same. Therefore, the inhibition is competitive.
P9-11 (c) Individualized solution
P9-11 (d) Individualized solution
9-32
P9-12
For No Inhibition, using regression,
Equation model:
a0 = 0.008
a1 = 0.0266
a0 = 0.0098
a1 = 0.33
For Maltose,
Equation model:
For α-dextran,
Equation model:
a0 = 0.008
a1 = 0.0377
Þ Maltose show non-competitive inhibition as slope and intercept, both changing compared to no
inhibition case.
Þ α-dextran show competitive inhibition as intercept same but slope increases compared to no
inhibition case.
P9-13 (a)
Now plug the value of (EH) into rP
9-33
At very low concentrations of H+ (high pH) rS approaches 0 and at very high concentrations of H+ (low
pH) rS also approaches 0. Only at moderate concentrations of H+ (and therefore pH) is the rate much
greater than zero. This explains the shape of the figure.
P9-13 (b) Individualized solution
P9-13 (c) Individualized solution
P9-14 (a) We know that in an Eadie - Hofstee Plot,
Competitive inhibition: slope changes, intercept remains same
Uncompetitive inhibition: slope and intercept change
Noncompetitive inhibition: slope remains same, intercept changes
Therefore, Line A = Competitive inhibition
Line B = Noncompetitive inhibition
Line C = Uncompetitive inhibition
P9-14 (b) In a Hanes- Woolf plot, we need to plot 𝑆 𝑣𝑠
'
()
The equation for no inhibition is
'
()*
=
,./01
+
'
./01
The equation for Competitive inhibition is
'
()*
,
5
'
/01
6
/01
= . - 31 + , 7 + .
The equation for Uncompetitive inhibition is
'
()*
,
'
/01
/01
=.- +.
31 +
5
,6
7
The equation for Noncompetitive inhibition is
'
()*
=
,./01
31 +
5
,6
7+
'
./01
31 +
5
,6
7
Therefore, in a Hanes-Woolf plot for,
Competitive inhibition : slope remains same as no inhibition case but intercept changes
Uncompetitive inhibition : slope changes but intercept remains constant
Noncompetitive inhibition : intercept and slope both changes
Hanes-Woolf plot
9-34
P9-15 Modifying the given equation in the form of line given by Eadie–Hofstee Plot,
C
)
i W
−𝑟 = , 3− '7 + c, rs
-
-
We know that only k2 is changing with inhibitor concentration, i.e., only intercept of the line is changing.
Therefore, it is Noncompetitive inhibition. From the text we know the general form of Noncompetitive
inhibition is given by,
)
.
/01
−𝑟 = 𝐾: 3− '7 + (CD5/,
)
6
P9-16 Neglecting death and maintenance for this question
We know,
This would imply,
u
W
X
µ = ,/01
DW
X
𝐶B = u
and µ = D (for CSTR)
X
,X v
/01 (v
𝑟" = 𝐷𝐶x = 𝐷𝑌W/' (𝐶BH −
,X v
u/01 (v
)
For max ‘rg’,
𝑑𝑟"
=0
𝑑𝐷
Solving this gives,
𝐾'
𝐷>?@z){| = 𝜇>?@ ~1 − •
€
𝐾' + 𝐶'H
Substituting values of KS, 𝜇>?@ and 𝐾' in above equation, gives D=0.2764
Volumetric flow rate = 3.317 g/s
P9-17
Given,
−𝑟" =
µ>?@ 𝐶B 𝐶W
𝐾' + 𝐶B
µ>?@ = 1hr-1
𝐾> = 0.25 gm/dm3
𝑌x/B = 0.5g/g
P9-17 (a)
𝐶xH = 0.1g/dm3
𝐶BH = 20 g/dm3
𝐶x = 𝐶xH + 𝑌• (𝐶BH − 𝐶B )
X
See the following Polymath problem
Polymath code:
d(Cs)/d(t) = rs
Cc0 = 0.1
Ycs = 0.5
Cs0 = 20
umax = 1
Cc = Cc0+Ycs*(Cs0-Cs)
9-35
Ks = 0.25
rs = -umax*Cs*Cc/(Ks+Cs)
rc = -rs*Ycs
t(0) =0
t(f) = 10
Cs(0) =20
Calculated values of DEQ variables
Variable Initial value
Minimal value
Maximal value
Final value
1
Cc
0.1
0.1
10.1
10.1
2
Cc0
0.1
0.1
0.1
0.1
3
Cs
20.
6.301E-11
20.
6.301E-11
4
Cs0
20.
20.
20.
20.
5
Ks
0.25
0.25
0.25
0.25
6
rc
0.0493827
1.273E-09
4.043487
1.273E-09
7
rs
-0.0987654
-8.086974
-2.546E-09
-2.546E-09
8
t
0
0
10.
10.
9
umax
1.
1.
1.
1.
0.5
0.5
0.5
0.5
10 Ycs
Differential equations
1 d(Cs)/d(t) = rs
Explicit equations
1 Cc0 = 0.1
2 Ycs = 0.5
3 Cs0 = 20
4 umax = 1
5 Cc = Cc0+Ycs*(Cs0-Cs)
6 Ks = 0.25
7 rs = -umax*Cs*Cc/(Ks+Cs)
8 rc = -rs*Ycs
Plot of Cc and Cs versus time
9-36
Plot of rs and rc with time
P9-17 (b)
Cs0 = 20g/dm3
Cc0 = 0
The dilution rate at which wash-out occurs will be by setting Cc=0 in equation;
Cc = Ycs*(Cs0 – (DKs)/(umax – D))
𝜇>?@ ∗ 𝐶'H
𝐷>?@ =
𝐾' + 𝐶'H
= 0.987 hr-1
⇨ Dmax =
Thus, dilution rate at which washout occurs is 0.987hr-1.
P9-17 (c)
,
,
,
u
W
X
µ = ,/01
DW
X
u
X
W
X
Divide by CCV, 𝐷 = µ = ,/01
DW
X
𝐾' ∗ 𝐷
𝐶' =
𝜇>?@ − 𝐷
X
Now
, ∗v
Þ DCc = D YC/S (CS0 – ƒ * (v)
/01
Now, for
,
𝐾'
€
𝐷>?@z){| = 𝜇>?@ ~1 − •
𝐾' + 𝐶'H
Substituting values of KS, 𝜇>?@ and 𝐾' in above equation,
Þ Dmax,prod = 0.88 hr-1
9-37
Using this value of D we can find the value of Cs
, ∗v
𝐶' = ƒ * (v = 1.83 g/dm3
/01
= 9.085g/dm3
𝐶ˆ = 𝑌ˆ/W (𝐶x − 𝐶x{ ) = 1.36 𝑔/𝑑𝑚3
rs = D(Cs0 – Cs) = 15.98 g/dm3/hr
P9-17 (d) Cell death cannot be neglected.
Kd =0.02 hr-1
DCc = rg –rd
And D(CS0 –Cs)= rS
For steady state operation to obtain mass flow rate of cells out of the system, Fc
FC =CCv0 = (rg-rd)V= (µ-kd) CCV
After dividing by CcV;
D=µ-kd
Now since maintenance is neglected.
Substituting for µ in terms of substrate concentration;
Cs =
The stoichiometry equation can be written as :
-rs = rg YS/C
CC = Yc/s
Now the dilution rate at can be found by substituting Cc =0;
= 0.96 hr-1
Thus Dmax =
Similarly, the expression for dilution rate for maximum production is given by solving the equation ;
(D+kd) Cc = (D+kd ) YC/S (CS0 -𝐾' (𝐷 + 𝑘| )/(𝜇>?@ − (𝐷 + 𝑘| )) )
Now for
,
Thus, we obtain Dmax,prod = 0.86 hr-1
P9-17 (e)
In this case the maintenance cannot be neglected.
m = 0.2 g/hr/dm3
The correlation for steady substrate concentration will remain the same.
Cs =
But the cell maintenance cannot be neglected. Thus, the stoichiometry equation will be changed. The
equation will be –
9-38
-rs = Ys/Crg + mCc
=> -rs = rg YS/C + mCc
=> (Cs0 – Cs) = Cc/YC/S +
Also, we know that by mass balance Using this relation, the stoichiometric equation for substrate consumption changes to -
⇨ Cc =
Now for finding the dilution rate at which wash out occurs,
Cc =0;
So Dwashout = 0.98 hr-1
Similarly, to calculate
,
Thus, we obtain Dmax,prod = 0.74 hr-1
P9-17 (f) Change the polymath code to include
rg = μmax*(1-Cc/C∞)*Cc
C∞= 1 g/dm3
Polymath code:
d(Cs)/d(t) = rs
Cc0 = 0.1
Ycs = 0.5
Cs0 = 20
Cinf = 1
umax = 1
Cc = Cc0+Ycs*(Cs0-Cs)
Ks = 0.25
rg = -umax*Cc*(1-Cc/Cinf)
rs = -rg*Ycs
rc = -rs*Ycs
t(0) =0
t(f) = 10
Cs(0) =20
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Cc
0.1
0.1
0.5751209
0.5751209
2
Cc0
0.1
0.1
0.1
0.1
3
Cinf
1.
1.
1.
1.
4
Cs
20.
19.04976
20.
19.04976
5
Cs0
20.
20.
20.
20.
6
Ks
0.25
0.25
0.25
0.25
7
rc
0.0225
0.0225
0.0624999
0.0610892
8
rg
0.09
0.09
0.2499995
0.2443569
9
rs
-0.045
-0.1249997
-0.045
-0.1221784
10 t
0
0
10.
10.
11 umax
1.
1.
1.
1.
12 Ycs
0.5
0.5
0.5
0.5
9-39
Differential equations
1 d(Cs)/d(t) = rs
Explicit equations
1 Cc0 = 0.1
2
Ycs = 0.5
3
Cs0 = 20
4
umax = 1
5
Cinf = 1
6
7
Cc = Cc0+Ycs*(Cs0-Cs)
Ks = 0.25
8
rg = umax * (1-Cc/Cinf) * Cc
9
rs = -Ycs*rg
10 rc = -rs*Ycs
Plot of Cc with time
P9-17 (g) Individualized solution
P9-17 (h) Individualized solution
P9-18
Tessier Equation,
Redoing P9-17 part (a)
For batch reaction,
,
,
See the following Polymath program
9-40
Polymath code:
d(Cs)/d(t) = rs
Cco = 0.1
Ycs = 0.5
Cso = 20
Cc = Cco+Ycs*(Cso-Cs)
k=8
umax = 1
Ysc = 2
rg = umax*(1-exp(-Cs/k))*Cc
rs = -Ysc*rg
RateS = -rs
t(0) =0
t(f) =7
Cs(0) = 20
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t
0
0
7
Cs
20
0.0852675
20
Cco
0.1
0.1
0.1
Ycs
0.5
0.5
0.5
Cso
20
20
20
Cc
0.1
0.1
10.057366
k
8
8
8
umax
1
1
1
Ysc
2
2
2
rg
0.0917915
0.0917915
3.8563479
rs
-0.183583
-7.7126957
-0.183583
Rates
0.183583
0.183583
7.7126957
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] RateS = -rs
9-41
7
0.0852675
0.1
0.5
20
10.057366
8
1
2
0.1066265
-0.213253
0.213253
Redoing P9-17 part (b)
m = CC v0 = rg V = µCCV
( )
Divide by CCV,
Now
For dilution rate at which wash out occur, CC = 0
CSO = CS
Redoing P9-17 part (c)
Now
For
,
= 0.628 hr-1
For redoing part (d), follow the procedure as in P9-17 (d)
P9-18 (a) Individualized solution
P9-18 (b) Individualized solution
P9-19 (a)
rg = µCC
u
W
X
µ = ,/01
DW
X
X
For CSTR,
𝐷𝐶W = 𝑟" =
9-42
µ𝑚𝑎𝑥 𝐶𝑠 𝐶𝐶
𝐾𝑠 + 𝐶 𝑠
P9-19 (b)
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.
P9-19 (c)
Two CSTR’s
For 1st CSTR,
V = 5000 dm3,
µ
𝐶𝑠 𝐶𝐶
𝑟" = 𝑚𝑎𝑥
𝐾 +𝐶
𝑠
𝑠
See the following Polymath program
Polymath code:
f(Cc) = D*Cc-rg
Cc(0) = 4
f(Cs) = D*(Cso-Cs)+rs
Cs(0) = 5
umax = 0.8
Ks = 4
Csoo = 10
Cso = 10
Ysc = 2
rg = umax*Cs*Cc/(Ks+Cs)
rs = -Ysc*rg
V = 5000
vo = 1000
D = vo/V
X = 1-Cs/Csoo
Cco = 4.33
POLYMATH Results
NLES Solution
Variable Value
f(x)
Ini Guess
Cc
4.3333333
9.878E-12
Cs
1.3333333
1.976E-11
umax
0.8
Ks
4
Csoo
10
Cso
10
Ysc
2
rg
0.8666667
rs
-1.7333333
V
5000
vo
1000
D
0.2
X
0.8666667
Cco
4.33
4
5
NLES Report (safenewt)
Nonlinear equations
[1] f(Cc) = D*(Cc)-rg = 0
[2] f(Cs) = D*(Cso-Cs)+rs = 0
9-43
Explicit equations
[1] umax = 0.8
[2] Ks = 4
[3] Csoo = 10
[4] Cso = 10
[5] Ysc = 2
[6] rg = umax*Cs*Cc/(Ks+Cs)
[7] rs = -Ysc*rg
[8] V = 5000
[9] vo = 1000
[10] D = vo/V
[11] X = 1-Cs/Csoo
[12] Cco = 4.33
CC1 = 4.33 g/dm3
CS1 = 1.33 g/dm3
For 2nd CSTR,
X = 0.867
CP1 = YP/CCC1 =0.866 g/dm3
See the following Polymath program
Polymath code:
f(Cc) = D*(Cc-Cc1)-rg
Cc(0) = 4
f(Cs) = D*(Cs1-Cs)+rs
Cs(0) = 5
umax = 0.8
Ks = 4
Csoo = 10
Cs1 = 1.333
Ysc = 2
rg = umax*Cs*Cc/(Ks+Cs)
rs = -Ysc*rg
V = 5000
vo = 1000
D = vo/V
X = 1-Cs/Csoo
Cc1 = 4.33
POLYMATH Results
NLES Solution
Variable
Cc
Cs
umax
Ks
Csoo
Cs1
Ysc
rg
rs
V
vo
D
X
Cc1
Value
f(x)
Ini Guess
4.9334151
3.004E-10
0.1261699
6.008E-10
0.8
4
10
1.333
2
0.120683
-0.241366
5000
1000
0.2
0.987383
4.33
4
5
NLES Report (safenewt)
Nonlinear equations
[1] f(Cc) = D*(Cc-Cc1)-rg = 0
[2] f(Cs) = D*(Cs1-Cs)+rs = 0
9-44
Explicit equations
[1] umax = 0.8
[2] Ks = 4
[3] Csoo = 10
[4] Cs1 = 1.333
[5] Ysc = 2
[6] rg = umax*Cs*Cc/(Ks+Cs)
[7] rs = -Ysc*rg
[8] V = 5000
[9] vo = 1000
[10] D = vo/V
[11] X = 1-Cs/Csoo
[12] Cc1 = 4.33
CC2 = 4.933 g/dm3
CS2 = 1.26 g/dm3
X = 0.987
CP1 = YP/CCC1 =0.9866 g/dm3
P9-19 (d)
For washout dilution rate,
CC = 0
𝐷>?@ =
𝜇>?@ 𝐶'H 0.8 ∗ 10
=
= 0.57 ℎ𝑟 (C
𝐾' + 𝐶'H
4 + 10
𝐷’“” z){| = 𝜇>?@ •1 − •
𝐾'
–
𝐾' + 𝐶'H
Production rate = CCvO (24hr) = 4.85 g/dm3 x1000dm3/hrx24hr = 116472.56g/day
P9-19 (e)
For batch reactor,
V = 500dm3,
CCO = 0.5 g/dm3 CSO = 10g/dm3
See the following Polymath program
Polymath code:
d(Cc)/d(t) = rg
d(Cs)/d(t) = rs
Ks = 4
Ysc = 2
umax = 0.8
rg = umax*Cs*Cc/(Ks+Cs)
rs = -Ysc*rg
t(0) = 0
t(f) = 6
Cc(0) = 0.5
Cs(0) = 10
9-45
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t
0
0
6
Cc
0.5
0.5
5.4291422
Cs
10
0.1417155
10
Ks
4
4
4
Ysc
2
2
2
umax
0.8
0.8
0.8
rg
0.2857143
0.1486135
1.403203
rs
-0.5714286
-2.8064061
-0.2972271
6
5.4291422
0.1417155
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] Ks = 4
[2] Ysc = 2
[3] umax = 0.8
[4] rg = umax*Cs*Cc/(Ks+Cs)
[5] rs = -Ysc*rg
For t = 6hrs,
CC = 5.43g/dm3.
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 g/dm3 x 3 x 2 x 500dm3
= 16290g/day.
CSRT production rate (considering the max rate case) = 116472.56 g/day
Therefore, we will need around 15 batch reactors.
P9-19 (f) Individualized solution
P9-19 (g) Individualized solution
P9-20 (a)
9-46
CS gm /dm2
D(day-1)
CS/D
Slope = 0.5 day
1
1
1
3
1.5
2
4
1.6
2.5
10
1.8
5.6
intercept = 0.5 gm/dm3
µmax = 2 KS = 0.5*2 = 1
P9-20 (b)
Inserting values from dataset 4
P9-20 (c) Individualized solution
P9-20 (d) Individualized solution
P9-21 (a)
Rate of substrate consumption without inhibition is given by,
−𝑟—˜ˆ =
𝑉>?@ [𝐺𝐿𝑃]
𝐾: + [𝐺𝐿𝑃]
In case of competitive inhibition,
−𝑟—˜ˆ =
𝑉>?@ [𝐺𝐿𝑃]
𝐽
𝐾: (1 + ) + [𝐺𝐿𝑃]
𝐾Ÿ
𝑟—˜ˆ (𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑖𝑛ℎ𝑖𝑏𝑖𝑡𝑖𝑜𝑛)
−
=
−𝑟—˜ˆ (𝑤𝑖𝑡ℎ 𝑖𝑛ℎ𝑖𝑏𝑖𝑡𝑖𝑜𝑛)
𝐽
[𝐺𝐿𝑃] + 𝐾: ¦1 + 𝐾 §
Ÿ
[𝐺𝐿𝑃] + 𝐾:
In case of uncompetitive inhibition,
−𝑟—˜ˆ =
𝑉>?@ [𝐺𝐿𝑃]
𝐽
[𝐺𝐿𝑃](1 + 𝐾 ) + 𝐾:
Ÿ
𝑟—˜ˆ (𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑖𝑛ℎ𝑖𝑏𝑖𝑡𝑖𝑜𝑛)
−
=
−𝑟—˜ˆ (𝑤𝑖𝑡ℎ 𝑖𝑛ℎ𝑖𝑏𝑖𝑡𝑖𝑜𝑛)
9-47
𝐽
𝐾: + [𝐺𝐿𝑃] ¦1 + 𝐾 §
Ÿ
[𝐺𝐿𝑃] + 𝐾:
Here,
J is concentration of Januvia (inhibitor)
[GLP] is concentration of GLP
𝐾: and 𝑉>?@ are Michaelis parameters
The values of [𝐺𝐿𝑃], 𝐾: and I can be varied across solutions. Depending on whether [𝐺𝐿𝑃] is greater or
less than 𝐾: , the plots vary. In the plot given below, 𝐾: is taken to be greater than [𝐺𝐿𝑃].
P9-21 (b)
For competitive inhibition,
𝑉>?@ 𝑆
𝐼
𝐾: (1 + 𝐾 ) + 𝑆
5
−𝑟5 = −𝑘L 𝐼𝐸 + 𝑘Z 𝐸. 𝐼
−𝑟ª = −𝑘C 𝐸𝑆 + 𝑘J 𝐸. 𝑆 + 𝑘K 𝐸. 𝑆 − 𝑘L 𝐼𝐸 + 𝑘Z 𝐸. 𝐼
𝑘L 𝐸𝐼
𝐸. 𝐼 =
𝑘Z
𝑘C 𝐸𝑆
𝐸. 𝑆 =
𝑘J + 𝑘K
−𝑟' =
These set of equations form the model for the dosage schedule for Januvia if the type of inhibition is
taken to be competitive.
9-48
P9-22 (a)
See the following Polymath program
Polymath code:
d(Cc)/d(t) = f*Cc*0.9/24
f = (If(((t>0) And (t < 12)) Or ((t > 24) And (t < 36))) Then (sin(3.14*t/12)) Else (0))
t(0) = 0
t(f) = 48
Cc(0) = 1
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Cc
1.
1.
1.774022
1.774022
2 f
0
0
0.99985
0
3 t
0
0
48.
48.
Differential equations
1 d(Cc)/d(t) = f* Cc * 0.9 / 24
Explicit equations
1 f = (If (((t > 0) And (t < 12)) Or ((t > 24) And (t < 36))) Then (sin(3.14 * t / 12)) Else (0))
P9-22 (b)
Given the initial concentration as Co = 0.5 mg/liter we have,
CC dC
t
# πt &
∫ C C C = ∫ 0 sin%$ 12 (µ+dt
'
o
C
#
# πt &&
C # 12 &
ln C = % (µ+dt %%1− cos % (((
Co $ π '
$ 12 ''
$
*,# 12 & #
# πt &&.,
CC = Co exp +% (µ %%1− cos % (((/
,-$ π ' $
$ 12 '',0
µ = 0.9 day -1 = 0.0375 hr–1
From 6am to 6pm, t= 12 hrs
9-49
Thus, the concentration progresses as = 1.332(conc. of previous day)
Therefore, the time taken to reach the concentration of 200 mg/dm3 using MS Excel is 22 days.
P9-22 (c) Now 𝜇 is a function of Cc.
"
C %
rg = µCC = µ0 $$1− C ''CC
# 200 &
rg decreases with increasing cell concentration.
It will take an infinitely long time to reach 200 mg/dm3 because the cell density blocks the sunlight,
inhibiting growth. If CC ever actually reached 200 mg/dm3, the growth rate would be zero. As we approach
200 mg/dm3, the growth rate slows so much that it is effectively zero, so we would never reach 200
mg/dm3 in a reasonable timescale.
P9-22 (d) From part (b)
Concentration at the end of 1st day = 1.33Co
Concentration at the start of 2nd day =1.33Co/2 + 100 (mg/lit)
Concentration at the end of 2rd day= 1.33(1.33Co/2 + 100)
And so forth.
Thus, the concentration progresses as = 1.332(conc. of previous day)+100
Solving for the progression on MS Excel we find the concentration to converge to 299.4011mg/lit
Thus, at the end of 365 days the conc. becomes = 299.4011mg/lit
P9-22 (e) Assuming that dilution and removal of algae is done at the end of the day after the growth
period is over. Then,
dCC
Rate of decrease of conc. =
= –k CC where, k = 1 (day–1)
dt
9-50
At t=0,
Co = 200mg/lit (i.e. maximum concentration allowed for maximum productivity at the end of the day)
So, removal per day:Cc= Co exp (-kt) = 200 exp (-1 day-1 x 1 day) = 200 exp (-1)
Þ Cc= 73.5 mg/dm3/day
Þ Volume of the pond = 5000 gallons or 18940.5 dm3
Þ Total mass flow rate of algae = 1392.1 g/day
P9-22 (f) Now, since CO2 will affect the rate of growth of algae too, then let the reaction be:
CO2 + Algae !!
→ more algae
→2C
Hence, A +C !!
Where,&&A :&&CO2
C :&&algae
Now, rate of growth of algae should be
rc = k sin (pt/12) CaCc
Earlier, Ca = 1.69gm/kg of water = 1.69 gm/dm3 of water (density of water = 1kg/lit)
We have rate law as
rc = µ sin (pt/12) Cc (CO2 concentration was assumed to be a constant = 1.69g /dm3 of water)
Hence, µ = k ´ 1.69 gm / dm3
k = (.9 per day) ÷ (1.69 gm / dm3)
k = 0.5325 dm3/g day or 0.022 dm3/g hr
A +C !!
→2C
At t=0 Ca0 Cc0 0
Let conversion with respect to C is X
Then, at t=t Ca0 – XCc0 Cc0 (1-X) 2Cc0X thus, total number of moles of C
Cc= Cc0 (1 + X)
Ca= Ca0 – XCc0 = Cc0 (M – X) where, M= Ca0/Cc0
dC
Hence, C = k sin (pt/12) CaCc
dt
Þ
( ( )) = k sin (pt/12) Cc0 (1+X) Cc0 (M-X)
d CC 0 1+ X
dt
dX
= Cc0 k sin (pt/12) (1+X) (M-X)
dt
X
t $ πt '
dX
Þ ∫
= k Cc0 ∫ sin& )dt
0 1+ X M− X
0
% 12 (
Þ
( )
( )( )
)+
M(1+ X)
# πt &-+# 12 &
= (M+1)kCc0 *1− cos % (.% ( - ln(M)
Þ ln
$ 12 '/+$ π '
,+
(M− X)
9-51
0
)
# πt &+
+# 12 &3
exp2k M+1 Cc0 *1− cos % (.% (5 −1
+
21
$ 12 '+
,
/$ π '54
Þ X=
0
)
# πt &+
+# 12 &3
1+ exp2k M+1 Cc0 *1− cos % (.% (5 M
+
21
$ 12 '+
,
/$ π '54
(
)
(
Now, X= 1 –
)
Cc
Cc0
0
)+
# πt &-+# 12 &3
exp2k M+1 Cc0 *1− cos % (.% (5 −1
21
$ 12 '/+$ π '54
Cc
,+
Hence,
= 1−
0
)+
Cc0
# πt &-+# 12 &3
1+ exp2k M+1 Cc0 *1− cos % (.% (5 M
+,
21
$ 12 '+/$ π '54
We have, Ca0 = Ks = 2g/dm3 & Cc0 = 1mg/ dm3 (assuming initial seeding value of algae from (c))
M= Ca0/Cc0 = 2000
(
)
(
)
Substituting the values, we have the concentration profile i.e.,
Cc
vs. t
Cc0
We know that only the plot above y=0 is practical. From the plot we can see that the maximum
conversion achievable is 0.4 for these conditions.
P9-22 (g) Let A: the species of unwanted algae
C: the species of desired algae
Then,
Rate of growth of A = 2 ´ rate of growth of C
dCa
dCc
=2´
dt
dt
Þ µ for A = 2 ´ µ for C
Þ
Now overall density of the medium at any time t
= Ca + Cc
Hence, from given conditions,
Ca=0.5(Ca + Cc)
9-52
So, at that time
Ca=Cc
Now,
dCc
= µ Cc
dt
Þ Cc = Cc0 exp (µt)
Þ
dCa
= 2 µ Ca
dt
Þ Ca = Ca0 exp (2 µt)
Þ
Assuming, Cco = initial seed concentration of desired algae= 1 mg/liter & Cao = 0.1 mg/liter (given).
Now, since the concentration is very less assuming there is no constraint of sunlight
Þ Ca = Cc
Þ Cc0 exp (µt) = Ca0 exp (2 µt)
Þ ln (Cco/Cao) = exp (µt)
where, µ = 0.9 per day
Putting the values, we get
t = 2.56 days or 61.4 hrs.
Since the number of days is coming less than 4.347 days, which was calculated in part (c), so the effect
of daylight can be neglected. Hence, the initial assumption is verified.
P9-23 (a)
Rate law is given by
𝑟" =
𝜇>?@ 𝐶W 𝐶'
𝐾' + 𝐶'
Since algae is formed in a pond, substrate concentration , 𝐶' >> 𝐾'
Thus, rate law equation reduces to
𝑟" = 𝜇>?@ 𝐶W
Cell mass balance,
𝑑𝐶W
= 𝑟" = 𝜇>?@ 𝐶W
𝑑𝑡
Or,
𝑑𝐶W
= 𝜇>?@ ∗ 𝑑𝑡
𝐶x
Integrating above equation
𝐶x = 𝐶WH ∗ exp(𝜇>?@ ∗ 𝑡)
𝑟" = 𝜇>?@ 𝐶WH ∗ exp(𝜇>?@ ∗ 𝑡)
Now,
𝐶WH = 0.5
𝐶W® = 200
ln (200/0.5)=umax*t
6=umax*t
Assuming umax=1/day
Time =6 hrs
9-53
P9-23 (b)
As the concentration of cell increases, specific growth rate (𝜇) decreases. Since growth rate is directly
proportional to 𝜇, so growth rate also decreases with increase in cell concentration.
Now,
𝐶W
𝑟" = 𝜇H ∗ ¦1 −
§ ∗ 𝐶W
200
𝑑𝐶
𝐶W
= 𝜇H ∗ ¦1 −
§ ∗ 𝐶W
𝑑𝑡
200
Or,
𝑑𝐶
= (𝜇{ /200)𝑑𝑡
(200 𝐶𝑐 − 𝐶WJ )
On integration from Cc=0.5 to 199.9 (at Cc=200 solution doesn’t converge) and t=0 to t
0.9t/200=0.06794
Or, t= 15 days
P9-23 (c)
For the invading algae species, rate is twice the rate of original species. The equation for invading
species can be derived similar to part (a)
𝑑𝐶W
= 2 ∗ 𝜇>?@ ∗ 𝑑𝑡
𝐶x
Let’s assume that total cell concentration reaches 50 mg/dm3 at time t, so invading species need to
reach 25 mg/dm3 to become over 50 %
ln (Cif/Cio)=2*umax*t
Or ln (25/0.01)=2*t
Or t= 3.91 days
For cultivating species, using equation from part a
ln(25/0.5)=t
Or, t=3.91 days
So, at 3.91 days, the invading species and cultivating species concentration are equal at 25 mg/dm3.
After this time, the concentration of invading species increases more than that of cultivating species
So, the required time is 3.91 days
9-54
Synopsis for Chapter 10 – Catalysis and Catalytic Reactors
General: Problems given to gain an understanding on how to develop catalytic rates, interpret data and
design PBR.
Questions
l Q10-1A (12 seconds) Questions Before Reading (QBR).
O Q10-4B (7-8 min) This question points out that if you have an adsorbing inert for adsorption limiting
reaction, the form of the partial pressure rate law will be very similar to that when surface reaction
limits.
Computer Simulations and Experiments
l P10-1B (a) – (b), (d) – (f) (10-12 min per simulation) All the Wolfram simulations are always assigned.
The student should spend 10-12 min per simulation including writing conclusions.
Interactive Computer Games (ICG)
AA P10-2A (60-70 min) A realistic on the job interactive problem to find the rate law. Very detailed and
thorough.
Problems
AA P10-3A (40 min) Very straight forward problem using the stops to develop a rate law.
l P10-4B (20 min) Old Exam Question (OEQ). Very short conceptual problem to determine rate law
from rate versus pressure plots. I usually like P10-3A and P10-4B, P10-7B and P10-8B for helping to
understand how to develop rate laws from catalytic data.
O P10-5A (35 min) Sort of the reverse of P10-3A, as here we are given the rate law and need to find
the mechanism.
O P10-6B (25 min) Similar to P10-3A.
AA P10-7B (35 min) Old Exam Question (OEQ). Using real data to find both the rate law and mechanism.
Reasonably straight forward problem.
O P10-8B (40 min) Old Exam Question (OEQ). Excellent Problem. Conceptual problem using data taken
in different form to postulate a rate law and mechanisms with rate limiting step. No calculations
necessary.
l P10-9B (50 min) Application of catalytic mechanism and data analysis to a real system.
AA P10-10B (40 min) Old Exam Question (OEQ). Challenging problem to develop a catalytic rate law and
parameters from reaction rate - pressure data.
I P10-11B (a) (60 min) More complicated version of P10-10B. Very short problem if the students are
only asked to find the rate law.
AA P10-12B (50 min) Old Exam Question (OEQ). Straight forward analysis of data to find rate law and
rate law parameters.
S10-1
AA P10-13B (45 min) Application of catalytic principles in this chapter to the hot topic of Water splitting,
more challenging than the other problems.
CVD
O P10-14A (45 min) Straight forward problem in CVD to determine mechanisms.
O P10-15A (30 min) Old Exam Question (OEQ). Very straight forward problem where student needs to
suggest a rate law for given CVD mechanism.
O P10-16B (50 min) Need to use Polymath non-linear least squares to find rate law and mechanisms
which best fit the data.
Catalyst Decay
AA P10-17A (25 min) Old Exam Question (OEQ). Catalyst decay problem where rate and decay laws are
used to make sketches of various profiles. No calculations involved.
AA P10-18B (60 min) Straight forward application of CRE algorithm with catalyst decay for a moving bed
reactor.
O P10-19B (40 min) Old Exam Question (OEQ). Similar in concepts and difficulty to P10-18B.
AA P10-20A (30 min) Use Polymath LEP to this straight forward extension of Example Problem P10-6B.
AA P10-21C (45 min) Use conversion time data to find decay law and then find conversion in a moving
bed reactor.
AA P10-22C (45 min) Wulfrene is the desired product. Use conversion time data to find decay law and
then develop a conversion equation.
O P10-23B (50 min) Rather open-ended problem as you are not told in detail what to do.
S10-2
Solutions for Chapter 10 – Catalysis and Catalytic Reactors
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-10
http://umich.edu/~elements/5e/10chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q10-1 Individualized solution
Q10-2 Individualized solution
Q10-3 Individualized solution
Q10-4 Initial rate will be independent of total pressure.
Q10-5 Individualized solution
Q10-6 No solution will be given
Q10-7 Individualized solution
Q10-8 Individualized solution
Q10-9 Individualized solution
P10-1 (a) Example 10-1
(i)
Critical value of KB is 1 and critical value of KT is 1.4
(ii) KB = 1.2. Keeping the value of Kt and Pto same, one can see that double differential of this equals
to zero at Kb =1.2. Hence shape changes from convex to concave
(iii) 1. One important observation is that except from all other curves, only the graph of ratio of
toluene to benzene occupied site is independent of inlet total pressure of toluene (Pto).
2. Ratio of toluene to vacant occupied site is the only curve which is a straight line for all values
of conversion. This can be easily understood by viewing the following equation.
10-1
P10-1 (b) Example 10-2
(i)
With an increase in 𝛼, the PH2, PB, PT decrease.
With an increase in KT, PH2, PT increase and PB decreases.
With an increase in k, PH2, PT decrease and PB increases.
(ii) If the flow rate is decreased the conversion will increase for two reasons:
(a) Smaller pressure drop
(b) Reactants spend more time in the reactor.
(iii) Individualized solution
(iv) 1. The term alpha has a significant effect on both pressure as well as conversion curves. This is a
unique observation as in this example the differential equation involving conversion has a
term which contains rt and this rt is dependent on alpha.
2. Another unique observation is that no matter whichever parameter you change, the
terminating pressure for all three hydrogen, toluene and benzene decreases to zero
simultaneously.
(v) From figure E10-2.1 we see that when X = 0.6, W = 5800 kg.
(vi) k has to be varied
P10-1 (c) Example 10-3
Visit the following link: http://www.umich.edu/~elements/5e/10chap/Polymath_Tutorial_LEP_10_3.pdf
(1)
With the new data, model (a) best fits the data
(a)
Polymath Output:
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1+Kea*Pea+Ke*Pe)
Variable
Ini guess
Value
95% confidence
k
3
3.5798145
0.0026691
Kea
0.1
0.1176376
0.0014744
Ke
2
2.3630934
0.0024526
Precision
R^2
R^2adj
Rmsd
Variance
= 0.9969101
= 0.9960273
= 0.0259656
= 0.0096316
(b)
Polymath Output:
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1+Ke*Pe)
Variable
Ini guess
Value
95% confidence
k
3
2.9497646
0.0058793
Ke
2
1.9118572
0.0054165
Precision
R^2
R^2adj
Rmsd
Variance
= 0.9735965
= 0.9702961
= 0.0759032
= 0.0720163
10-2
P10-1 (c) Example 10-3 continued
(c)
Polymath Output:
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/((1+Ke*Pe)^2)
Variable
Ini guess
Value
95% confidence
k
3
1.9496445
0.319098
Ke
2
0.3508639
0.0756992
Precision
R^2
R^2adj
Rmsd
Variance
= 0.9620735
= 0.9573327
= 0.0909706
= 0.1034455
(d)
Polymath Output:
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe^a*Ph^b
Variable
Ini guess
Value
95% confidence
k
3
0.7574196
0.2495415
a
1
0.2874239
0.0955031
b
1
1.1747643
0.2404971
Precision
R^2
R^2adj
Rmsd
Variance
= 0.965477
= 0.9556133
= 0.0867928
= 0.107614
P10-1 (c) Example 10-3
(2)
(e)
Polymath Output:
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/((1+Ka*Pea+Ke*Pe)^2)
Variable
Ini guess
Value
95% confidence
k
3
2.113121
0.2375775
Ka
1
0.0245
0.030918
Ke
1
0.3713644
0.0489399
Precision
R^2
R^2adj
Rmsd
Variance
= 0.9787138
= 0.9726321
= 0.0681519
= 0.0663527
(f)
Polymath Output:
Nonlinear regression (L-M)
Model: ReactionRate = k*Pe*Ph/(1+Ka*Pea)
Variable
Ini guess
Value
95% confidence
k
3
44.117481
7.1763989
Ka
1
101.99791
16.763192
Precision
R^2
R^2adj
Rmsd
Variance
= -0.343853
= -0.5118346
= 0.5415086
= 3.6653942
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
10-3
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 R^2 values.
P10-1 (d) Example 10-4
(i)
About 99%
Conversion
1.0
0.8
0.6
0.4
0.2
0.0
0
(iii)
(iv)
(v)
100
200
300
400
time
500
1. Increase in k1d decreases Xd while increasing k1 increases both conversion with and without
catalyst decay.
2. On increasing Ed conversion with catalyst decay increases. On increasing Ea, both the
conversion decreases
3. Ea has an effect on both conversions while Ed only has effect on conversion without catalyst
decay.
As the activation energy increases, it becomes more difficult for the reaction or decay process to
occur (from Arrhenius law, rate constant reduces). Hence, with increase in Ed, the conversion
increases and with an increase in E, conversion decreases.
1st order reaction, 2nd order decay
Xd =1-
1
(1 + kd t ) k / k
d
(vi) 2nd order reaction, 1st order decay
Rate Law:
N A0
dX d
= -rA'W
dt
10-4
For 2nd order reaction:
-r ' = k 'a(t )C 2
A
A
For 1st order decay
da
= -kd a , at t = 0, a = 1.
dt
So a = e d
Stoichiometry:
-k t
C A = C A0 (1 - X d ) =
N A0
(1 - X d )
V
Combine:
dX d
N W
= k '(1 - X d )2 a A02
dt
V
Let k = k ' N A0W / V . Substituting for we have
2
dX d
= k (1 - X d ) 2 e - k t
dt
X
t
1
-k t
dX
=
k
ò0 (1 - X )2 d ò0 e dt
d
d
d
X
k
= (1 - e - k t )
1 - X kd
d
X=
1 - e- k t
kd / k + 1 - e - k t
d
d
(vii) 1st order reaction, 1st order decay
Rate Law:
N A0
dX d
= -rA'W
dt
10-5
For 1st order reaction:
-r ' = k 'a(t )CA
A
For 1st order decay
da
= -kd a , at t = 0, a = 1.
dt
So a = e- kd t
Stoichiometry:
C A = C A0 (1 - X d ) =
N A0
(1 - X d )
V
Combine:
dX d
W
= k '(1 - X d )a
dt
V
Let k = k 'W / V . Substituting for we have
dX d
−k t
= k(1− X d )e d
dt
X
t
∫
1
−k t
dX = k e d dt
1− X d d
ln
1
k
−k t
=
(1− e d )
1− X kd
0
∫
0
X = 1− e
k −kd t
(e
−1)
kd
(viii) 2nd order reaction, 2nd order decay
Rate Law:
N A0
dX d
= -rA'W
dt
10-6
For 2nd order reaction:
-r ' = k 'a(t )C 2
A
A
For 2nd order decay
da
= -kd a 2 , at t = 0, a = 1.
dt
1
So a =
1 + kd t
Stoichiometry:
C A = C A0 (1 - X d ) =
N A0
(1 - X d )
V
Combine:
dX d
N W
= k '(1 - X d )2 a A02
dt
V
Let k = k ' N A0W / V . Substituting for we have
2
dXd
(1− Xd )2
=k
dt
1+ kd t
X
t
1
dXd = k
dt
2
1+ kd t
(1−
X
)
d
0
0
∫
1
∫
X
k
= ln(1+ kd t)
1− X kd
X=
ln(1+ kd t)
kd / k + ln(1+ kd t)
P10-1 (e)
(i)
When kd and Us are set at their maximum values, increasing k, Cao and decreasing Fao can
increase conversion. But among these, Cao has the greatest affect.
10-7
(ii)
1. One important observation, see the expression for rate which is equal to k*Ca^2. It increases
with increase in k but when Cao is increased, rate increases till a particular value of Cao and
then starts decreasing. This is because Ca= Cao*(1-X) and when Cao increases, X also
increases. Hence overall product determines the complete Ca.
2. Activity term(a) is dependent only on kd and Us. It increases with increase in Us and decrease
in kd.
(iii) Polymath Code:
d(X) / d(W) = a*(-raprime)/Fao
Us = 10
kd = 0.72
Fao = 30
Cao = 0.075
Ca = Cao*(1-X)
k = 600
raprime = -k*Ca^2
a=1
X(0) = 0
W(0) = 0
W(f) = 22
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
a
1.
1.
1.
1.
2
Ca
0.075
0.0215827
0.075
0.0215827
3
Cao
0.075
0.075
0.075
0.075
4
Fao
30.
30.
30.
30.
5
k
600.
600.
600.
600.
6
kd
0.72
0.72
0.72
0.72
7
raprime
-3.375
-3.375
-0.2794886
-0.2794886
8
Us
10.
10.
10.
10.
9
W
0
0
22.
22.
10 X
0
0
0.7122302
0.7122302
Differential equations
1 d(X)/d(W) = a*(-raprime)/Fao
Explicit equations
1 Us = 10
2 kd = 0.72
3 Fao = 30
4 Cao = 0.075
5 Ca = Cao*(1-X)
6 k = 600
7 raprime = -k*Ca^2
8 a=1
Hence,
X = 0.712
10-8
(iv)
If the solids and reactants are fed from opposite ends,
da kd a
=
at W = We, a = 1
dW U S
k
ln a = d W + C1
US
C1 =
kdWe
US
ék
ù
a = exp ê d (W - We ) ú
ëU S
û
dX
2
FA0
= kC A2 0 (1 - X ) a
dW
We
é -k W ù U
ék W ù
kC A2 0
X
=
exp ê d e ú S exp ê d ú
1- X
FA0
ë U S û kd
ë US û 0
é -k W ù ö
kC 2 U æ
X
= A0 S ç1 - exp ê d e ú ÷
÷
1- X
kd FA0 çè
ë US û ø
(v)
This gives the same expression for conversion as in the example.
Second order decay
a=
1
kW
1+ d
US
kC A2 0U S æ kdW ö
X
=
ln ç1 +
÷
1- X
kd FA0
US ø
è
( 0.6)( 0.075) U S æ ( 0.72) 22000 ö
1.24 =
ln ç1 +
÷
US
( 0.72)( 30)
è
ø
2
(vi)
Solve for US by trial and error or a non-linear equation solver.
US = 0.902
If ε = 2
(1 - X ) a
dX
FA0
= kC A2 0
2
dW
(1 + e X )
2
1 + e ) X kC A2 0U S æ
é -kdWe ù ö
(
2e (1 + e ) ln (1 - X ) + e X +
=
çç1 - exp ê
ú ÷÷
2
2
1- X
12ln (1 - X ) + 4 X +
X = 0.372
kd FA0 è
ë US û ø
9X
= 1.24
1- X
P10-1 (f)
(i)
Conversion increases with an increase in temp while temp has negligible effect on activity.
(ii) X decreases and a increses with an increase an U0.
(iii) 1. Conversion increases with an increase of k, temp and bulk density while decreases or remains
same with all other parameters. Rate constant k has the greatest affect on conversion.
2. Activity(a) is majorly affected by either epsilon or activity constant(A).
3. Initial flow rate(Uo) has the greatest affect on rate profile. Rate decreases with an increase of
all equlibrium constants Ka, Kb and Kc.
10-9
(iv-ix)
Polymath Code:
d(X) / d(z) = -ra/Uo/Cao #
Ka=0.05
Kb=0.15
Pao=12
eps=1
A=7.6
R=0.082
T=400+273
rho=80
kprime=0.0014
D=1.5
Uo=2.5
Kc=0.1
U=Uo*(1+eps*X)
Pa=Pao*(1-X)/(1+eps*X)
Pb=Pao*X/(1+eps*X)
vo=Uo*3.1416*D*D/4
Cao=Pao/R/T
Pc=Pb
a=1/(1+A*(z/U)^0.5)
raprime=a*(-kprime*Pa/(1+Ka*Pa+Kb*Pb+Kc*Pc))
ra=rho*raprime
X(0)=0
z(0) = 0
z(f) = 10
Use the Polymath program given in the example problem and vary all the sated parameters. k’ has the
largest effect on the conversion, followed by A, U0, and PA0. When the ratio of k to U0 and k to A is
increased, the conversion is increased. When these are decreased the conversion decreases. Use the
Polymath program to generate graphs of the conversion and activity profiles for U0 = 0.025 m/s, 0.25
m/s, 2.5 m/s, and 25 m/s:
It is apparent that as U0 increases, activity increases and converison descreases.
10-10
Use the same program to vary U0 and generate values for the exit conversion and activity. Use these values in Excel
to generate the following graph:
P10-2 The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer
Games (ICGs) is given at the front of this Solutions Manual.
P10-3
P10-3 (a)
10-11
P10-3 (b)
Adsorption of isobutene limited
P10-3 (c) Site balance: CT = CV ( 1 + KI CI + KTBACTBA)
P10-3 (d)
10-12
P10-3 (e) Individualized solution
P10-4 (a) Species A & C.
P10-4 (b) Figure B tells us that the reaction is irreversible because when PC0 = 1 atm, increasing the
product B does not change the rate. If the reaction were reversible with B and C present increasing B
would drive the reaction in the reverse direction.
P10-4 (c) Assume surface reaction is the rate limiting step
→ A •S
A + S←
A •S + A •S → B(g) + C •S
→C + S
C •S←
$
&
&
rA0
~0
&
€k A
&
&
C A•S = K APAC V
&
&
rDI
kPA2
& −rA =
≈0
&
k DI
[1+ K APA + KC PC ] 2
%
&
CC•S = KC PC C V
k = k SK 2AC 2t
&
&
−rA = rS = k SK 2APA2C 2V
&
&
C t = C V + K APAC V + KC PC C V &
&
Ct
&
CV =
&'
1+
K
P
+
K
P
[
A A
B C]
−rA = rS = k SC 2A•S
€
10-13
P10-4 (d)
PB 0 = PC 0 = 0
-rA0 =
kPA2
(1 + K APA ) 2
1
1 + K APA
=
-rA0
k PA
1
1 1 KA
=
+
-rA0
k PA
k
So one can plot
1
1
versus
to linearize the intial rate data in Figure A.
PA
-rA0
P10-4 (e)
C A•S = CV K A PA = CV K A PA 0
(1 - X )
(1 + e X )
CC •S = CV K C PC = CV K C PA 0
X
(1 + e X )
C A•S = CC •S
K A (1 - X ) = K C X
1 - X K C 0.25
=
=
= 0.5
X
K A 0.5
1 = 0.5 X + X = X (1 + 0.5 )
X=
1
= 0.66
1.5
P10-5 (a)
H = H2
E = Ethylene
A = Ethane
H2 + C2H4 !cat!
→ C2H6
H+E !→
!A
Because neither H2 or C2H6 are in the denominator of the rate law they are either not adsorbed or
weakly adsorbed. Assume H2 in the gas phase reacts with C2H6 adsorbed on the surface and ethane goes
directly into the gas phase. Then check to see if this mechanism agrees with the rate law
Eley Rideal
!→
!
E+ S ←!
! E•S
"
C %
rE = k AD $PEC V − E•S '
KE '&
$#
E•S +H !→
! A+S
rS = k SCE•SPH
2
10-14
Assume surface reaction is rate limiting:
CE•S = KEPEC V
rS = k S !"CE•SPH #$
C T = C V + CE•S
! PP #
−rA& = k SKEC T ' E H (
'" 1+KEPE ($
P10-5 (b) Individualized solution
P10-6
¾¾
® 2O • S
O2 + 2S ¬¾
¾
¾¾
®2A• S
A2 + 2S ¬¾
¾
C3 H 6 + O • S ® C3 H5OH • S
B + A• S ® C • S
¾¾
® C3 H 5OH + S
C3 H 6OH • S ¬¾
¾
C•S ®C +S
-rB = rS = k3 PBC A•S
é
C A2•S ù
2
rAD = k A ê PA2 CV ú
KA û
ë
Assume surface reaction is the rate-limiting step
rAD
=0
kA
C A! S = CV K A PA2
-rB = rS = k3 PBCV K A PA2
é
PC ù
rC •S = kD êCC •S - C V ú = kD [CC •S - KC PC CV ]
KD û
ë
rC • S
=0
kD
CC •S = KC PC CV
CT = CV + C A•S + CC •S = CV éë1 + K A PA 2 + K C PC ùû
-rB = rS =
k3CT PB K A PA2
1 + K A PA2 + KC PC
10-15
P10-7 (a)
P10-7 (b)
From the figure,
10-16
P10-7 (c)
P10-7 (d) Individualized solution
P10-7 (e) Individualized solution
P10-7 (f) Individualized solution
10-17
P10-8
2ME → DME + W
P10-8 (a)
From the plot we can conclude:
1. The partial pressure of DME is greater initially, while the generation rate of water and DME are the
same according to stoichiometry.
→ Water must be adsorbed on the surface.
→ DME is not adsorbed on the surface, otherwise it would take up sites and would not exit for a
while.
2. When we reach steady state after a period of time, the exit concentration of DME and water are the
same, which is consistent with the stoichiometry.
3. The partial pressure of DME decreases after ~ 80 minutes, this is because water occupies a number
of the sites so that there are fewer for ME to adsorb upon and react.
P10-8 (b) H2O & ME took longer than others to exit the reactor.
For ME, this is because it’s adsorbed on the surface and consumed in the reaction.
For H2O, this is because it’s adsorbed on the surface.
P10-8 (c) H2O & ME are adsorbed on surface.
P10-8 (d) Yes, DME is not adsorbed on the surface.
P10-8 (e) Ans: (4)
P10-8 (f) With the information above, a probable mechanism could be (assuming surface reaction is
irreversible):
ME + S ↔ ME•S
2ME•S → DME + W•S + S
W•S ↔ W + S
Assume surface reaction is the rate-limiting step:
,
𝑟% = 𝑘% 𝐶)*•%
𝐶)*•% = 𝐾)* 𝑃)* 𝐶/
𝐶0•% = 𝐾0 𝑃0 𝐶/
10-18
𝐶1 = 𝐶/ + 𝐶)*•% + 𝐶0•%
3
Therefore, 𝑟)*
= 𝑟% =
8
4567
(:;<= 5= ;<67 567 )8
,
, where 𝑘 = 𝑘? 𝐾)*
𝐶1,
P10-9
P10-9 (a)
P10-9 (b)
Assume that PC >> PN. Then PC changes very little during the course of the reaction and remains
constant. A maximum in (–rS) then occurs, for a fixed value of PN at:
10-19
The rate of reaction will increase with an increase in PC until the above value is reached, after which it
will decrease. It appears that there is an excess pressure which will minimize reactor volume. Operating
at excess pressure greater than this value will decrease (–rS), and hence increase V. This analysis is exact
if the catalytic reactor is a CSTR. If the reactor is treated as PFR, the critical value of PC is only
approximate, but the general observation is qualitatively the same.
This analysis further assumes that the excess CO can be eliminated easily and economically downstream
from the NO converter.
P10-9 (c)
In hybrid vehicle, the conventional internal combustion engine is combined with electric propulsion
system. So, the emission from vehicle will be greatly reduced and relevance of this problem will be less.
With driverless car by 2020,assuming it to be completely electrical, there would be zero CO, NO emissions
and this problem would be irrelevant.
P10-10
P10-10 (a)
10-20
P10-10 (b)
10-21
Substituting the expressions for CV and CA·S into the equation for –r’A
P10-10 (c) Individualized solution
P10-10 (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
10-22
Thus, from the slope and intercept data
Polymath Code:
d(X)/d(W) = -ra/Fao
e=1
Pao = 10
W(0) =0
W(f)=23
X(0) = 0
Pa = Pao*(1-X)/(1+e*X)
k1 = 560
k2 = 2.04
Fao = 600
ra =- k1*Pa/((1+(k2*Pa))^2)
rate = -ra
10-23
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
e
1.
1.
1.
1.
2
Fao
600.
600.
600.
600.
3
k1
560.
560.
560.
560.
4
k2
2.04
2.04
2.04
2.04
5
Pa
10.
0.0042521
10.
0.0042521
6
Pao
10.
10.
10.
10.
7
ra
-12.22814
-68.57357
-2.340395
-2.340395
8
rate
12.22814
2.340395
68.57357
2.340395
9
W
0
0
23.
23.
10 X
0
0
0.9991499
0.9991499
Differential equations
1 d(X)/d(W) = -ra/Fao
Explicit equations
1 e=1
2 Pao = 10
3 Pa = Pao*(1-X)/(1+e*X)
4 k1 = 560
5 k2 = 2.04
6 Fao = 600
7 ra = - k1*Pa/((1+(k2*Pa))^2)
8 rate = -ra
P10-10 (e) Individualized solution
P10-10 (f)
Now consider the change in pressure:
Pressure:
𝑑𝑝
𝑑𝑊
=
−𝛼
2𝑝
(1 + 𝑋), where 𝑝 = 55
H
10-24
Polymath Code:
d(X)/d(W) = -ra/Fao
d(p)/d(W) = -alpha*(1+X)/2/p
e=1
Pao = 10
Pa = p*Pao*(1-X)/(1+e*X)
k1 = 560
k2 = 2.04
Fao = 600
ra = -k1*Pa/((1+(k2*Pa))^2)
rate = -ra
alpha =0.03
W(0)=0
W(f)=23
X(0)=0
p(0)=1
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.03
0.03
0.03
0.03
2
e
1.
1.
1.
1.
3
Fao
600.
600.
600.
600.
4
k1
560.
560.
560.
560.
5
k2
2.04
2.04
2.04
2.04
6
p
1.
0.0746953
1.
0.0746953
7
Pa
10.
7.771E-05
10.
7.771E-05
8
Pao
10.
10.
10.
10.
9
ra
-12.22814
-68.61447
-0.0435044
-0.0435044
10 rate
12.22814
0.0435044
68.61447
0.0435044
11 W
0
0
23.
23.
12 X
0
0
0.9997919
0.9997919
Differential equations
1 d(X)/d(W) = -ra/Fao
2 d(p)/d(W) = -alpha*(1+X)/2/p
Explicit equations
1 e=1
2 Pao = 10
3 Pa = p*Pao*(1-X)/(1+e*X)
4 k1 = 560
5 k2 = 2.04
6 Fao = 600
7 ra = -k1*Pa/((1+(k2*Pa))^2)
8 rate = -ra
9 alpha = 0.03
10-25
P10-11 (a)
10-26
P10-11 (b)
Now using Polymath’s non-linear equation regression, we can find the values for the parameters. We
find that
k = 0.00137
KA = 4.76
KB = 0.259
KC = 0.424
In the problem it is given that KA is 1 or 2 orders of magnitude greater than KB and KC which is true so this
is a good answer.
P10-11 (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 law
parameters and they should include a large range of possibilities
P10-11 (d)
A ¾¾
®B+ H
ra =
-kPA
(1 + K A PA + K B PB + K H PH )
PA =
2
PA0 (1 - X )
P X
P X
here, e = y A0d = 1(2 - 1) = 1
, PB = A0 , PH = A0
(1 + X )
(1 + X )
(1 + X )
Where,
k = 0.00137
K A = 4.77
K B = 0.259
K H = 0.424
PA0 = 15atm
FA0 = 10mol / sec
Now, the design equation for the PFR,
10-27
dX -rA
=
dW FA0
Solving the above equations using Polymath
Polymath Code:
d(X)/d(W) = -ra/Fao
Pao = 15
k = 0.00137
Ka = 4.77
Kb = 0.262
Kh = 0.423
Fao = 10
Ph = Pao*X/(1+X)
Pa = Pao*(1-X)/(1+X)
Pb = Pao*X/(1+X)
ra = -k*Pa/((1+Ka*Pa+Kb*Pb+Kh*Ph)^2)
W(0)=0
W(f)=2.0E+06
X(0)=0
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Fao
10.
10.
10.
10.
2
k
0.00137
0.00137
0.00137
0.00137
3
Ka
4.77
4.77
4.77
4.77
4
Kb
0.262
0.262
0.262
0.262
5
Kh
0.423
0.423
0.423
0.423
6
Pa
15.
3.064E-09
15.
3.064E-09
7
Pao
15.
15.
15.
15.
8
Pb
0
0
7.5
7.5
9
Ph
0
0
7.5
7.5
10 ra
-3.904E-06
-1.26E-05
-1.114E-13
-1.114E-13
11 W
0
0
2.0E+06
2.0E+06
12 X
0
0
1.
1.
Differential equations
1 d(X)/d(W) = -ra/Fao
Explicit equations
1
Pao = 15
2
k = 0.00137
3
Ka = 4.77
4
Kb = 0.262
5
Kh = 0.423
6
Fao = 10
7
Ph = Pao*X/(1+X)
8
Pa = Pao*(1-X)/(1+X)
9
Pb = Pao*X/(1+X)
10 ra = -k*Pa/((1+Ka*Pa+Kb*Pb+Kh*Ph)^2)
10-28
From the above graph, we get the weight of catalyst required for 85% conversion is 1.2083x106 kg.
The required catalyst weight is so high because of very low value of reaction constant ’k’.
P10-12 (a)
Changing Variable
Run
Pa
1
1
3
10
6
20
7
0.1
Pa
Pb
1
1
1
1
Pc
2
2
2
2
reaction rate
0.114
0.18
0.186
0.0243
Changing Variable
Run
Pa
1
1
2
1
4
1
Pb
Pb
1
10
20
Pc
2
2
2
reaction rate
0.114
1.14
2.273
10-29
-rA' (mol/gcat s)
2.5
Pa = 1 atm
Pc = 2 atm
2
1.5
1
0.5
0
0
Changing Variable
Run
Pa
4
1
5
1
5
10
15
Pb (atm)
Pc
Pb
20
20
Pc
2
10
20
25
reaction rate
2.273
0.926
-rA' (mol/gcat s)
2.5
2
1.5
Pa = 1 atm
Pb = 20 atm
1
0.5
0
0
2
4
6
Pc (atm)
8
10
12
Suggested rate law according to the plots:
-rA ' =
kPA PB
1 + K APA + K C PC
P10-12 (b) Species A and C are on the surface
P10-12 (c)
𝐴 + 𝑆 → 𝐴. 𝑆
𝐴. 𝑆 + 𝐵 → 𝐶. 𝑆
𝐶. 𝑆 → 𝐶 + 𝑆
10-30
Finally,
-rA ' = rS = kS C A! S PB
-rA ' =
kS Ct K A PA PB
kPA PB
=
1 + K A PA + KC PC 1 + K A PA + KC PC
Where k = k S Ct K A
P10-12 (d)
C A•S = CV K A PA = CV K A PA 0
(1 - X )
(1 + e X )
CC •S = CV K C PC = CV K C PA 0
X
(1 + e X )
CA•S K A (1 - X ) (4)(1 - 0.8) 1
=
=
=
CC •S
KC X
(13)(0.8) 13
P10-12 (e)
C A•S = CV K A PA = CV K A PA 0
(1 - X )
(1 + e X )
CC •S = CV K C PC = CV K C PA0
X
(1 + e X )
C A•S = CC •S
K A (1 - X ) = K C X
1 - X K C 13
=
=
X
KA 4
X = 0.235
P10-12 (f)
10-31
17
26
1
X + ln(
)]0.9
0
5
5 1- X
= 8.9 gcat
= 2[-
P10-13 (a)
Proposed Single site Mechanism:
¾¾
® S '• H 2O
S '+ H 2O ¬¾
¾
1
S '• H 2O ¾¾
® S '• H 2 + O2
2
¾¾
® S '+ H 2
S '• H 2 ¬¾
¾
1
S '+ O2 ¾¾
®S
2
Rate of adsorption: rAD = k AD (Cv ' PH O 2
CS '• H2O
K H 2O
)
Rate of surface reaction: rs = k s CS '• H 2O ---------- (1)
Rate of desorption: rD = k D (CS '• H 2 - K H 2 PH 2 Cv ' )
Assuming, surface reaction to be rate limiting, we get
Þ
rAD rD
@
@0
k AD k D
\ CS '• H 2O = K H 2O Cv ' PH 2O and CS '• H 2 = K H 2 PH 2 Cv '
From total site balance, we get
Cv ' =
Ct
1 + K H 2O PH 2O + K H 2 PH 2
Putting all the values in equation (1), we get
rs =
ks K H 2OCt PH 2O
1 + K H2O PH 2O + K H 2 PH 2
Þ rs =
kPH 2O
1 + K H2O PH 2O + K H 2 PH 2
Where, k = k s K H 2O Ct
10-32
P10-13 (b)
Proposed Mechanism:
¾¾
® S '• H 2O
S '+ H 2O ¬¾
¾
¾¾
® S • H2
S '• H 2O ¬¾
¾
¾¾
® S + H2
S • H 2 ¬¾
¾
Rate of adsorption: rAD = k AD (Cv ' PH O -
CS '• H2O
2
K H 2O
)
Rate of surface reaction: rs = ks (CS '• H 2O - CS • H 2 / K S )
Rate of desorption: rD = k D (CS • H 2 - K H 2 PH 2 Cv )
Assuming, surface reaction to be rate limiting, we get
Þ
rAD rD
@
@0
k AD k D
\ CS '• H 2O = K H 2O Cv ' PH 2O and CS • H 2 = K H 2 PH 2 Cv
From total site balance, we get
Cv ' =
Ct '
1 + K H 2O PH 2O
Cv =
Ct
1 + K H 2 PH 2
Putting all the values in equation (1), we get
K H O PH OCt '
K H 2 PH 2 Ct
rs = ks ( 2 2
)
1 + K H2O PH2O K S (1 + K H 2 PH 2 )
P10-13 (c)
Proposed Mechanism:
Step (1)
¾¾
® S '• O2
S + hu ¬¾
¾
¾¾
® S '+ O2
S '• O2 ¬¾
¾
Step (2)
¾¾
® S '• H 2O
S '+ H 2O ¬¾
¾
¾¾
® S • H2
S '• H 2O ¬¾
¾
¾¾
® S + H2
S • H 2 ¬¾
¾
For Step (1)
Rate of adsorption of O2: rAD O 2 = k D O 2 (Cv -
CS '•O2
KS
)
10-33
---------- (1)
Rate of Reaction: rSO 2 = k SO 2 (CS '•O2 - K O2 PO2 Cv ' )
! Rate of surface reaction of water is rate controlling, we get
Þ
rAD O 2
kD O 2
rS O 2
@
kS O 2
@0
\ CS '•O = Cv K S ----------- (2)
2
CS '•O2 = K O2 PO2 Cv ' ------- (3)
Equating (2) and (3), we get
Cv =
K O2 PO2 Cv '
KS
-------- (4)
Now, here the site balance gets modified because oxygen is also getting adsorbed.
Ct = Cv + Cv ' + CS '•O2 + CS '• H 2O + CS • H 2
Putting, the values and replacing Cv using equation (4), we get
Cv ' =
Ct
KO PO KO K H PO PH
(1 + KO2 PO2 + K H 2O PH 2O + 2 2 + 2 2 2 2 )
KS
KS
And, Cv =
K O2 PO2 Cv '
KS
=
KO2 PO2 Ct
KO PO KO K H PO PH
K S (1 + KO2 PO2 + K H 2O PH 2O + 2 2 + 2 2 2 2 )
KS
KS
Putting all the values in equation (1), we get
rs = ks (
K H 2O PH 2OCt
K H 2 PH 2 KO2 PO2 Ct
)
KO2 PO2 KO2 K H 2 PO2 PH 2
KO2 PO2 KO2 K H 2 PO2 PH 2
(1 + KO2 PO2 + K H 2O PH 2O +
+
) K S (1 + KO2 PO2 + K H 2O PH 2O +
+
)
KS
KS
KS
KS
P10-14
Assume the rate law is of the form rDep =
2
kPVTIPO
2
1 + KPVTIPO
2
At high temperatures K ¯ as T ­ and therefore KP VTIPO
<< 1
2
rDep = kPVTIPO
rDep
2
PVTIPO
Run 1
Run 2
Run 5
=k
0.028
(0.05)
2
0.45
(0.2 )
2
7.2
(0.8)
2
= 11.2
= 11.28
= 11.25
10-34
At low temperature and low pressure
2
rDep = kPVTIPO
rDep
=k
2
PVTIPO
Run 1
Run 2
0.004
(0.1)
2
0.015
(0.2 )
2
= 0.4
= 0.375
These fit the low-pressure data
2
>> 1
At high pressure KP VTIPO
rDep =
2
kPVTIPO
k
=
2
KPVTIPO K
This fits the high pressure data
At PVTIPO = 1.5, r = 0.095 and at PVTIPO = 2, r = 0.1
Now find the activation energy
At low pressure and high temperature k = 11.2
At low pressure and low temperature k = 0.4
æ k ö E æ 1 1 ö E æ T -T ö
ln ç 2 ÷ = ç - ÷ = ç 2 1 ÷
è k1 ø R è T1 T2 ø R è T1T2 ø
æ 11.2 ö E æ 473 - 393 ö
ln ç
÷
÷= ç
è 0.4 ø R çè (473)(393) ÷ø
E
= 7743
R
E = 15330.65
cal
mol
P10-15
10-35
P10-16 (a)
Using Polymath non-linear regression, we can find the parameters for all models:
(See Polymath regression tutorials at
http://www.umich.edu/~elements/5e/tutorials/Polymath_tutorials.html)
(1)
Polymath Output:
Nonlinear regression (L-M)
Model: rT = k*PM^a*PH2^b
Variable
k
a
b
Precision
R^2
R^2adj
Rmsd
Variance
Ini guess
1
0.1
0.1
Value
95% confidence
1.1481487
0.1078106
0.1843053
0.0873668
-0.0308691
0.1311507
= 0.7852809
= 0.7375655
= 0.0372861
= 0.0222441
α = 0.184 β = -0.031 k = 1.148
(2)
Polymath Output:
Nonlinear regression (L-M)
Model: rT = k*PM/(1+KM*PM)
Variable
Ini guess
k
1
KM
2
Value
12.256274
9.0251862
95% confidence
2.1574162
1.8060287
Model: rT = k*PM*PH2/((1+KM*PM)^2)
Variable
Ini guess
Value
k
1
8.4090333
KM
2
2.8306038
95% confidence
18.516752
4.2577098
Precision
R^2
R^2adj
Rmsd
Variance
= 0.9800096
= 0.9780106
= 0.0113769
= 0.0018638
k = 12.26 KM = 9.025
(3)
Polymath Output:
Nonlinear regression (L-M)
Precision
R^2
R^2adj
Rmsd
Variance
= -4.3638352
= -4.9002187
= 0.1863588
= 0.5001061
k = 8.409 KM = 2.83
10-36
(4)
Polymath Output:
Nonlinear regression (L-M)
Model: rT = k*PM*PH2/(1+KM*PM+KH2*PH2
Variable
Ini guess
Value
k
1
101.99929
KM
2
83.608282
KH2
2
67.213622
95% confidence
4.614109
7.1561591
5.9343217
Nonlinear regression settings
Max # iterations = 300
Precision
R^2
R^2adj
Rmsd
Variance
= -3.2021716
= -4.1359875
= 0.1649487
= 0.4353294
k = 102 KM = 83.6
KH2 = 67.21
P10-16 (b)
We can see from the precision results from the Polymath regressions that rate law (2) best describes the
data.
P10-16 (c) Individualized solution
P10-16 (d)
We have chosen rate law (2)
Proposed Mechanism
¾¾
®M • S
M + S ¬¾
¾
M • S ® H 2( g ) + T( g )
Rate of adsorption: rAD = k AD (Cv PM -
CM • S
)
KM
Rate of surface reaction: rs = ks CM •S ---------- (1)
Assuming, surface reaction to be rate limiting:
rAD
@0
k AD
Þ CM •S = Cv PM K M
Þ
Putting, the value of CM •S in Equation (1)
rs = ksCv PM K M ----------------- (2)
Now, applying site balance
Ct = Cv + CM •S
Þ Ct = Cv + Cv PM K M
Þ Cv =
Ct
1 + PM K M
Putting in Equation (2), we get
10-37
Þ rs =
ks K M Ct PM
1 + K M PM
Þ rs =
kPM
Where, k = ks K M Ct
1 + K M PM
P10-17 (a)
For all of the parts, the mole balances and rate laws are the same. They are:
Find the equation needed for a.
––A better solution is to put this equation directly into polymath.
a = 1 when t = 0
Assuming values for v0, k and kd come up with the following graphs according to the cases described.
10-38
P10-17 (b)
Find the new equation for a:
assume CA changes very slowly w.r.t. “a” changing
Using the same values come up with these graphs:
P10-17 (c)
Find the new equation for a: again, assuming CB changes slowly w.r.t. “a”. A better solution is to put
da
= –kaC B into the Polymath program
dτ
10-39
The following graphs are made:
P10-17 (d)
Making a into a differential equation we come up with this:
10-40
P10-17 (e)
Everything from part (d) is the same except for the decay law.
Integrating:
10-41
Not a good solution. Just put
da kd C Aa
=
into Polymath program.
dW
Us
P10-18 (a)
da
= -kD
dt
W = USt
dW = U S dt
10-42
dt da -k D
=
dW dt U S
If 0 = 1 -
a = 1-
k DW
US
U
k DW
.5
then W = S = = 2.5 kg
US
k D .2
akC A2 0 (1 - X )
dX -rA
a
=
=
( -rA ( 0 ) ) =
dW FA0 FA0
FA0
dX æ kDW ö kC A0 (1 - X )
= ç1 ÷
dW è
US ø
v0
dX
kC A0
k DW
0
S
ò (1 - X ) = v ò 1 - U
2
2
2
dW
Activity is zero for W > 2.5 kg, so the catalyst weight only goes to the effective weight.
2
0.2 ( 2.5 ) ù
kC A0 é
k DWe2 ù 1( 0.2 ) é
X
=
ê 2.5 ú = 0.25
êWe ú=
1- X
v0 ë
2U s û
1 ê
2*0.5 ú
ë
û
X = 0.2
P10-18 (b)
P10-18 (c)
For infinite catalyst loading a = 1.
dX kC A0 (1 - X )
=
dW
v0
2
kC
X
= A0 W = 1
1- X
v0
X = 0.5
P10-18 (d)
kC é
k W2 ù
X
= A0 êW - D ú
1- X
v0 ë
2U s û
é 0.2*25 ù
0.4
= 0.2 ê5 ú
1 - 0.4
2U s û
ë
U S = 1.5
kg
s
10-43
P10-18 (e)
a = 1-
k DW
US
0 = 1-
k DW
US
U S = k DW = 0.2*5 = 1
kg
s
P10-18 (f)
a = 0 means there is no reaction is taking place. Activity can never be less than 0.
P10-18 (g)
U = -U S
da k D
=
dW U S
a=
k DW
+C
US
1=
k DWt
+C
US
a=
k DWe
k W
+1- D t
US
US
when W = Wt , a = 1
Now find We.
0=
k DWe
k W
+1- D t
US
US
We =
U S é kDWt ù .5 é .2*5 ù
- 1ú = ê
- 1ú
ê
kD ë U S
û
û .2 ë .5
We = 2.5
ö kC (1 - X )
dX æ kDWt
= ç1 + W ÷ A0
dW è
US
v0
ø
2
æ k W
ö
kC
X
= A0 ò ç1 - D t + W ÷dW
1- X
v0 è
US
ø
æ k W ö W 2 - We2 ù
kC é
X
= A0 ê(Wt - We ) ç1 - D t ÷ + t
ú
1- X
v0 êë
US ø
2
úû
è
é
X
æ 0.2*5 ö 25 - 6.25 ù
= 0.2 ê( 5 - 2.5 ) ç1 ÷+
ú
1- X
0.5
2
è
ø
ë
û
X
= 1.875
1- X
10-44
X = 0.65
P10-18 (h)
$ = 160 FA0 X - 10U S
a = 1-
k DW
US
æ k Wö
dX
= ka = k ç1 - D ÷
dW
US ø
è
kkDW 2
X = kW 2U S
To maximize profit, a maximum in profit is reached and so we set the differential of profit equal to 0.
d$
dX
= 0 = 160 FA0
- 10
dU S
dU S
kk DW 2
dX
=
dU S
2U S2
160 FA0
kkDW 2
= 10
2U S2
8FA0 kk DW 2 = U S2
U S = 8FA0 kk DW 2 = 8 ( 2 )(.2 )(.2 )( 25 )
US = 4
kg
min
P10-19 (a)
Start with the mole balance for a batch reactor:
10-45
P10-19 (b)
For the moving-bed reactor the mole balance now becomes:
Everything stays the same. Plug into Polymath.
The conversion achieved is X = 0.266
P10-19 (c)
Increasing US will get us a higher conversion. Looking at this summary table, US = 10 kg/h and X = 0.6
10-46
P10-19 (d)
P10-19 (e)
In part e, the only thing that changes from (b) is the decay law and the decay constant:
Plugging into Polymath we get the following summary tables for US = 2 and 10 kg/h. X = 0.50 and 0.88
respectively. X will again increase as US increases.
10-47
P10-20 (a)
In order to get a high conversion, the entering pressure should be as high as possible since the rate is a
second order function of the pressure. U should be kept low since the conversion is an indirect function
of the flow rate.
P10-20 (b)
The problem with such a low flow rate is that the activity will remain low.
P10-20 (c)
We can use the same equations that are given in example 10-7 with a few exceptions. For example, the
rate law, we use the one given in the problem:
The activity will be different because the equation given is different:
To find the concentration of coke we use stoichiometry:
We find that the value that gives the best conversion (X = 0.337) and uses the whole reactor is U = 7. See
the following Polymath program.
10-48
P10-20 (d)
To find this the only change necessary was the values for the k’s because they change with temperature.
The Polymath program below shows the results. The temperature is 485K and the conversion is 0.637.
10-49
P10-20 (e)
10-50
P10-21 (a)
P10-21 (b)
10-51
FA0= 200 mol/min = 10/3 mol/s
CA0 =0.06 kmol/m3= 0.06 mol/l
R= 0.082 atm.l/ (K.mol)
T=273+420= 693 K
k= 3.8*103 mol/kg cat.s.atm
α= 4.27*10-3 s-1
W= 100 kg
Us= 1 kg/min =1kg/60s
Equation (1) becomes:
𝐶OP 𝑅𝑇𝑘𝑈% WXY0
\
U𝑒 Z[ − 1]
2 ln(1 − 𝑋) + 𝑋 =
𝐹OP 𝛼
no
l.,m∗:P ∗:PP
⎡
⎤
q
:
0.06 ∗ 0.082 ∗ 693 ∗ (3.8 ∗ 10d ) ∗ (1/60) ⎢ Wk
( )
pP
(
)
2 ln 1 − 𝑋 + 𝑋 =
− 1⎥⎥
⎢𝑒
10
X \ ∗ (4.27 ∗ 10Wd )
⎢
⎥
3
⎣
⎦
2ln (1-X)+ X=-1.5e7
Therefore, X~1
P10-22 (a)
10-52
10-53
P10-22 (b)
Since the equation for the activation is:
we cannot find a time for which a = 0, because it is mathematically impossible. We can, however, find a
time wat which the activity is small enough that it can be considered to be zero. The following graphs
show the activation for the two temperatures given.
The graphs show that for 500 K, the lifetime is about 1100 days and for 550 K, the lifetime is about 450
days.
P10-23
First, we need to find 𝐶Ou
𝐻,
= 10
𝐸𝑇𝐵
1
𝑦*yz =
= 0.091
11
𝑃 ∗ 𝑦*yz 3 ∗ 10d ∗ 0.09
𝐶OP =
=
= 0.0594
𝑅𝑇
8.314 ∗ 553
Start by guessing that the decay is first order
𝐶O•
𝑘| 𝑡 = ln(𝑘~ ) + 𝑙𝑛
𝐶OP − 𝐶O
It is told that 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 find the denominator (𝐶OP − 𝐶O ) as a function of conversion.
10-54
As it can be seen that graph is linear, so decay is first order. We know that slope of this graph is decay
law constant, so 𝑘| = 0.2499
10-55
Page intentionally blank
10-56
Synopsis for Chapter 11 – Nonisothermal Reactor Design: The
Steady-State Energy Balance and Adiabatic PFR Applications
General: This chapter develops the user friendly form of the energy balance and then applies them to
adiabatic operation of reactors.
Questions
l Q11-1A (10 seconds) Questions Before Reading (QBR).
O Q11-2A (10-15 min) A good connection with A Word From Our Sponsor—Safety.
O Q11-3A (4 min) Good question to make the “thought link” between batch and steady state.
l Q11-4A (2 min) Always assigned this question as it is a favorite with the students.
I Q11-5A (4 min) Good question for discussion in the class room.
O Q11-8A (4 min) Excellent short question requiring thinking.
O Q11-10A (15-20 min) Excellent–Excellent video that leads to a good class discussion after viewing.
I Q11-12A (10 min) Oops, this question is really part of Q11-14A.
Computer Simulations and Experiments
l P11-1A (10-12 min) Parts (a), (b) and (f) are excellent Wolfram simulations, 10-12 minutes each
including writing conclusions. Part (e) makes a good connection with the heat transfer course.
Problems
O P11-2A (15 min) Very short problem to calculate the heat of reaction.
l P11-3B (15 min) Old Exam Question (OEQ). Conceptual problem. No calculations.
l P11-4A (40 min) Old Exam Question (OEQ). Straight forward problem applying the adiabatic energy
balance to a PFR, CSTR and a BR.
AA P11-5A (40 min) Similar calculations and level of difficulty to those in P11-4A. I sometimes don’t
assign the pressure drop part, part (d).
O P11-6B (40 min) Old Exam Question (OEQ). Rather difficult problem to learn effect of inerts on the
exit conversion of a PFR.
AA P11-7B (40 min) Old Exam Question (OEQ). Rather straight forward Polymath application of the
energy and mole balance algorithm to an adiabatic reaction in a PBR with pressure drop.
AA P11-8B (55 min) Similar calculations to those in P11-7B. It takes time to work through part (c).
AA P11-9A (55 min) Only chapter with home problem on reactor staging – fairly straight forward.
I P11-10A (45 min) Qualitative description of reactor staging with feed streams along the reactor
length.
S11-1
Page intentionally blank
S11-2
Solutions for Chapter 11 – Nonisothermal Reactor Design:
The Steady-State Energy Balance and Adiabatic
PFR Applications
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-11
http://umich.edu/~elements/5e/11chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q11-1 Individualized solution.
Q11-2 Individualized solution.
Q11-3 Step 1( mole balance ) will be different.
Q11-4 No solution necessary.
Q11-5 Individualized solution.
Q11-6 Mole balance(design equation) would change, other equations remain the same.
Q11-7 (1) Heat of reaction will remain unchanged.
(2) ΔCp(T-TR)=10.12*(423-298)=1265
% error= ΔCp(T-TR)/ ΔHrxn=(1265/23310)*100%=0.0542*100% = 5.4%
Q11-8 The graph of FA/-rA vs. XA is a decreasing curve until a sufficient XA. Hence, the volume for CSTR is
less than volume for PFR.
Q11-9 Individualized solution
Q11-10 Individualized solution
11-1
Q11-11 Individualized solution.
Q11-12 Individualized solution
Q11-13 Individualized solution
Q11-14 Individualized solution
P11-1 (a) Example Table 11-2
(i)
Xe reduces and X increases for the same distance down the reactor with an increase in T0 up to a
point when X and Xe becomes equal. Any further increase in To results in decrease in both Xe and
X. Since the reaction is exothermic, according to Le Chatelier’s principle, the equilibrium
conversion reduces. However, with an increase in inlet temperature, the rate of the reaction
increases. Hence, the conversion is expected to increase.
(ii) T0 has the greatest effect on the temperature profile. 𝐸" has greatest effect on conversion profile
(iii) The maximum of the rate increases with an increase in T0 and the distance down the reactor
where this maximum is achieved decreases.
(iv) 1. Equilibrium conversion depends mainly on To and heat of reaction. But other parameters have
a significant effect on conversion.
2. With an increase in Cao, the maximum value of rate starts to increase with rate declining to
zero at less volume (early stage of reactor).
3. Increasing the initial temperature (To) raises the temperature throughout the reactor. It is also
evident by seeing the equation following the temp change.
P11-1 (b) Example 11-3
(i)
Increasing To overlaps both of the conversion profiles. Xe comes down and X goes up. Increasing
Cao causes Xe to decrease slightly while it increases X significantly. Increasing Kc increases both X
and Xe
(ii) A minimum feed temperature of 320 K must be maintained.
(iii) On increasing heat of reaction, the reactor temperature increases.
(iv) T0 affects the rate the most. The maximum of the rate increases with an increase in T0 and the
distance down the reactor where it is achieved reduces.
(v) 1. With an increase in Cao, the maximum value of rate starts to increase with rate declining to
zero at less volume(early stage of reactor).
2. This example is pretty much similar to the last one. Temperature here also is mainly
dependent on To as the equation remains the same. Hence the conclusions from previous
example follows
(vi) Set Vfinal = 0.8 m3
See the following polymath code:
Polymath Code:
d(X)/d(V) = -ra/Fa0
Ca0 = 9.3
Fa0 = .9*163
T = 330+43.3*X
Kc = 3.03*exp(-830.3*((T-333)/(T*333)))
k = 31.1*exp(7906*(T-360)/(T*360))
11-2
Xe = Kc/(1+Kc)
ra = -k*Ca0*(1-(1+1/Kc)*X)
V(0)=0
V(f)=0.8
X(0)=0
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
V
0
0
0.8
0.8
X
0
0
0.2603491
0.2603491
Ca0
9.3
9.3
9.3
9.3
Fa0
146.7
146.7
146.7
146.7
T
330
330
341.27312
341.27312
Kc
3.099466
2.852278
3.099466
2.852278
k
4.2238337
4.2238337
9.3196276
9.3196276
Xe
0.7560658
0.7404133
0.7560658
0.7404133
ra
-39.281653
-56.196156
-39.281653
-56.196156
rate
39.281653
39.281653
56.196156
56.196156
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra/Fa0
Explicit equations as entered by the user
[1] Ca0 = 9.3
[2] Fa0 = .9*163
[3] T = 330+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*Ca0*(1-(1+1/Kc)*X)
[8] rate = -ra
(vii) Change the entering temperature T0 in the polymath code above (highlighted in yellow), and
record the exiting conversion:
T
X
330
0.26
340
0.54
350
0.68
370
0.66
390
0.65
420
0.62
450
0.59
500
0.55
600
0.48
X
We see that there is a maximum in the exit conversion corresponding to T0 = 350K. Therefore, the
inlet temperature we would recommend is 350K.
(viii) Optimum inlet temp is 350 K
11-3
(ix)
P11-1(c) Aspen Problem
No solution will be provided.
P 11-1(d)Example 11-4
(i)
As T0 increases, equilibrium conversion decreases.
As Hrxn increases(magnitude decreases), equilibrium conversion increases.
As Ke1 increases, eq. conversion increases
As ΘI increases, equilibrium conversion increases
(ii) As T0 increases, adiabatic equilibrium temperature increases. Hence according to the range of T0
possible on the sliders,
T0=300 K gives the lowest value of adiabatic equilibrium temperature of ~ 460 K
T0=650 K gives the highest value of adiabatic equilibrium temperature of ~ 670 K
(iii) 1) It is observed that at the same inlet temperature, the equilibrium conversion on the addition
of inert is greater. This also follows from the Le Chatelier’s principle.
2) As inlet temperature T0 increases, the equilibrium conversion and decreases and the adiabatic
equilibrium temperature increases.
3) As the heat of reaction increases, equilibrium conversion increases. This implies that a
reaction which is more exothermic( Hrxn negative) will have a lower value of equilibrium
conversion.
P11-1(e) Example 11-5
 ΔT
Q = mC
P
Q = 220kcal / s
=
⇒m
220 ×103
18(400 −270)
= 854.7mol / s
For the2nd case:
Hot Stream: 460K à 350K
Cold Stream: 270K à 400K
LMTD = 69.52 oC
11-4
⇒ A=
220 ×103
100 ×69.52
= 31.64m2
P11-1 (f) Example 11-6
(i)
Initially X and Xe both increase upto ΘI=1.3 . On increasing ΘI beyond this, Xe increases, but X
decreases.
(ii) At low values of -5000 cal/mol, optimum T0 is 650 K and conversion achieved is ~1. As Hrxn
inceases, optimum To decreases and conversions also decrease.
(iii) The optimum feed temperature is 480 K. This temp cannot be too high or low because at these
two extremes, low conversion at high temperatures due to equilibrium limitations and low
conversion at low temperatures because of a small rate of reaction.
(iv) The optimum entering temperature does not change. However, the conversion increases when
compared to the base case.
(v) When θI is set to 3, the new optimum feed temperature becomes 525 K. At this temperature we
can achieve maximum possible conversion.
(vi) Initially X and Xe both increase upto ΘI=1.3 . On increasing ΘI beyond this, Xe increases, but
decreases
(vii) 1. Finding optimum temperature was the most important task of this example. Optimum
temperatures are found by varying inlet temp To and then seeing the conversion profiles X
and Xe. The temperature which gives the maximum possible conversion at given set of
parameters is the optimum feeding temperature.
2. The start of the temperature curve is determined by the temp To. As the reactions are
exothermic, hence the temperatures are always greater at the end of reactor for all sets of
parameters.
P11-2
11-5
P11-3 (a)
True. Endothermic & adiabatic so T decreases, for endothermic reaction, Xe decreases as T
decreases. Once X reaches Xe, T doesn’t change and therefore Xe reaches a plateau.
P11-3 (b)
False. If the reaction is endothermic and heated along the length, T will decrease and then
increase along the reactor. Therefore, Xe will decrease, reach a minimum, and then increase
along the reactor for an endothermic reaction.
P11-3 (c)
False. If the reaction is exothermic and cooled along the length, T will increase and then
decrease along the reactor. Therefore, Xe will decrease, reach a minimum, and then
increase along the reactor for an exothermic reaction.
P11-3 (d)
True. Exothermic & adiabatic so T increases, for exothermic reaction, Xe decreases as T
increases. Once X reaches Xe, T doesn’t change and therefore Xe reaches a plateau.
P11-4
A+B → C
Since the feed is equimolar, CA0 = CB0 = .1 mols/dm3
CA = CA0(1-X)
CB = CB0(1-X)
Adiabatic:
T = T0 +
X[−ΔHR (T0 )]
θ C + XΔC
∑ i pi
P
ΔCP = C pC −C pB −C pA = 30 −15−15 = 0
(
)
ΔHR (T) = HC − HB − HA = − 41000 −(−15000 − −20000 = −6000 cal / mol A
cal
∑θiCpi = CpA +θBCpB = 15+15 = 30 mol
K
T = 300 +
6000 X
= 300 +200X
30
−rA = kC 2A0 (1− X)2 = .01k(1− X)2
VPFR = FA0
dX
∫ −r
A
P11-4 (a)
For the PFR, FA0 = CA0v0 = (.1)(2) = .2 mol/s
See the following Polymath code:
11-6
Polymath Code:
d(X)/d(V) = -ra/(Fa0 )
Ca0 = 0.1
Fa0 =0.2
T0 = 300
T = T0 + 200 * X
k =0.01*exp((10000 / 2) * (1 / 300 - 1 / T))
ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)
V(0)=0
V(f)=310
X(0)=0
Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca0
0.1
0.1
0.1
0.1
2 Fa0
0.2
0.2
0.2
0.2
3 k
0.01
0.01
4.29714
4.29714
4 ra
-0.0001
-0.0018941
-0.0001
-0.0008701
5 T
300.
300.
471.5403
471.5403
6 T0
300.
300.
300.
300.
7 V
0
0
310.
310.
8 X
0
0
0.8577017
0.8577017
Differential equations
1 d(X)/d(V) = -ra/Fa0
Explicit equations
1 Ca0 = .1
2 Fa0 = .2
3 T0 = 300
4 T = T0 + 200 * X
5 k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))
6 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)
The conversion and temperature both increase down the reactor volume because the reaction is
exothermic. They increase faster because the reaction rate increases as the temperature increases.
11-7
P11-4 (b)
T = T0 + 200X
T0 = T - 200X = 550 – 200(1) = 350 K
P11-4 (c)
T and therefore k is constant under isothermal condition
V = FA0
dX
∫ −r
A
2
−rA = kC A0 (1− X)2
V=
V=
FA0
2 ∫
kC A0
FA0
(
dX
(1− X)2
1
−1)
kC 2A0 1− X
1
X=
FA0
kVC 2A0
+1
Qr = FA0 X(−ΔHR (T0 )) =
−FA0 ΔHR (T0 )
FA0
kVC 2A0
=
+1
−(0.2)(−6000)
1200
=
(0.2)
200,000
+1
+1
2
V
(0.01)V(0.01)
P11-4 (d)
𝑟" = 𝑘 &𝐶" 𝐶) −
𝐶+
-=0
𝐾+
0
𝑟" = 𝑘 &𝐶"/
∗ (1 − 𝑋)0 −
𝐶"/ 𝑋
𝐾+
𝑟" = 0, then
:
𝑋7 =
2 + ; ∗+
<
=>
0
:
/.E
− ?@2 + ; ∗+ A − 4C
<
=>
2
11-8
Polymath Code:
d(X)/d(V) = -ra / Fa0
Ca0 = .1
Fa0 = .2
T0 = 300
T = T0 + 200 * X
k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))
dHrx = -6000
Kc = 10*exp((dHrx/8.314)*(1/450-1/T))
ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2-Ca0*X/Kc)
Xe = 0.5*(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)
V(0)=0
V(f)=10
X(0)=0
Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca0
0.1
0.1
0.1
0.1
2
dHrx
-6000.
-6000.
-6000.
-6000.
3
Fa0
0.2
0.2
0.2
0.2
4
k
0.01
0.01
0.010583
0.010583
5
Kc
22.29685
22.11524
22.29685
22.11524
6
ra
-0.0001
-0.0001047
-0.0001
-0.0001047
7
T
300.
300.
301.0234
301.0234
8
T0
300.
300.
300.
300.
9
V
0
0
10.
10.
10 X
0
0
0.005117
0.005117
11 Xe
0.518003
0.5166571
0.518003
0.5166571
Differential equations
1 d(X)/d(V) = -ra / Fa0
Explicit equations
1 Ca0 = .1
2 Fa0 = .2
3 T0 = 300
4 T = T0 + 200 * X
5 k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))
6 dHrx = -6000
7 Kc = 10*exp((dHrx/8.314)*(1/450-1/T))
8 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2-Ca0*X/Kc)
9 Xe = 0.5*(2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)
11-9
The conversion and temperature are both lower than in part (a), because the reaction rate is lower due
to reversible reaction.
P11-4 (e) CSTR volume
𝑉=
𝐹"/ 𝑋
−𝑟"
T=300+200*X
T=300+200*0.9=480 K
k =0.01*exp((10000 / 2) * (1 / 300 - 1 / T))
k=5.2 dm3/mol.s
-ra = k * (Ca0 ^ 2) * ((1 - X) ^ 2)
-ra=5.2*(0.01)*(0.1^2)
-ra=0.00052 mol/dm3.s
V=0.2*0.9/(0.00052)
V= 346 dm3
11-10
P11-4 (f) BR volume
𝑑𝑋 −𝑟" ∗ 𝑉
=
𝑑𝑡
𝑁"/
T = T0 + 200 * X
k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))
0
−𝑟" = 𝑘𝐶"/
∗ (1 − 𝑋)0
𝑁" = 𝑁"/ ∗ (1 − 𝑋)
Polymath Code:
T0=300
T=T0+200*X
k=0.01*exp((10000/2)*(1/300-1/T))
ra=-k*(Ca0)^2*((1-X)^2)
V=25
NA0=10
NA=NA0*(1-X)
Ca0=0.1
d(X) / d(t) = -ra*V/NA0
t(0) = 0
t(f) = 2500
X(0) = 0
Polymath Output
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca0
0.1
0.1
0.1
0.1
2 k
0.01
0.01
7.770572
7.770572
3 NA
10.
0.0278508
10.
0.0278508
4 NA0
10.
10.
10.
10.
5 ra
-0.0001
-0.0018926
-6.027E-07
-6.027E-07
6 T
300.
300.
499.443
499.443
7 t
0
0
2500.
2500.
8 T0
300.
300.
300.
300.
9 V
25.
25.
25.
25.
10 X
0
0
0.9972149
0.9972149
Differential equations
1 d(X)/d(t) = -ra*V/NA0
Explicit equations
1 T0 = 300
2 T = T0+200*X
3 k = 0.01*exp((10000/2)*(1/300-1/T))
4 Ca0 = 0.1
5 V = 25
6 NA0 = 10
7 NA = NA0*(1-X)
8 ra = -k*(Ca0)^2*((1-X)^2)
11-11
P11-5 (a)
For the reaction, A ® B + C
Limiting reactant
A
Reactor type
PFR
Dependent variable W
Design equation
dX/dW –rA'/FA0
Rate equation
–rA = kCA
k = 0.133 exp[E/R(1/450–1/T)]
E = 31400 J/mol
Stoichiometry
CA CA0(1 – X)/(1 + eX) T/T0, e = 1
CA CA0(1 – X)/(1 + X) T/T0
Combine
dX/dW = k(1 – X)/(1 + X) T/T0/u0
Energy Balance
T = T0 + X[–DHR(T0)]/ Sqicpi + XDCp)
DCp = 15 + 25 – 40 = 0
T = 450 + 20000X/40 = 450 + 500X
See the following Polymath code:
Polymath Code:
d(X)/d(W)=k*(1-X)*T0/T/(1+X)/v0
T=450+500*X
v0=20
T0=450
k=0.133*exp(31400/8.314*(1/T0-1/T))
W(0)=0
W(f)=43.13711
X(0)=0
11-12
Output:
P11-5 (b) In the following Polymath code, vary T0 to obtain T0 vs X data
d(X)/d(W)=k*(1-X)*T0/T/(1+X)/v0
T=T0+500*X
v0=20
T0=425
k=0.133*exp(31400/8.314*(1/450-1/T))
W(0)=0
W(f)=43.13711
X(0)=0
Results:
T0 (K)
300
400
425
450
500
Conversion, X
0.005
0.18
0.44
0.80
0.99
11-13
Thus, increasing inlet temp increases the conversion as the rate of reaction increases. Below 400 K,
there is virtually no conversion.
P11-5 (c)
V = FA0
dX
∫ −r
A
−rA = kC A0 (1− X) / (1+ X)
V=
FA0
kC A0
1+ X
∫ 1− X dX
v
1
V = 0 (2ln
− X)
k
1− X
Qr = FA0 X(−ΔHR (T0 )), so X=
v
V = 0 (2ln
k
1−
1
Qr
Qr
FA0 (−ΔHR (T0 ))
−
FA0 (−ΔHR (T0 ))
Pv
),where FA0 = 0 0
FA0 (−ΔHR (T0 ))
RT0
Qr
Qr vs V can be plotted according to the above equation.
Qr vs. V
30
25
Qr
20
15
10
5
0
0
10
20
30
40
50
V
P11-5 (d)& (e)
Polymath Code:
d(X)/d(W)=k*(1-X)*T0/T/(1+X)/v0*P/P0
d(P)/d(W)=-0.5*alpha*T/T0*P0^2/P*(1+X)
T=450+500*X
v0=20
T0=450
k=0.133*exp(31400/8.314*(1/T0-1/T))
alpha=0.019
P0=1013250
P(0)=1013250
W(0)=0
W(f)=39.1
X(0)=0
11-14
Output:
For pressure drop, an extra equation is added
Limiting reactant
A
Reactor type
PFR
Dependent variable W
Design equation
dX/dW –rA'/FA0
dP/dW = –a/2(T/T0)P0/(P/P0)(1 + eX)
Rate equation
–rA = kCA
k = 0.133 exp[E/R(1/450–1/T)]
E = 31400 J/mol
Stoichiometry
CA CA0(1 – X)/(1 + X) T/T0 P/P0
Combine
dX/dW = k(1 – X)/(1 + X) T/T0 P/P0/u0
Energy Balance
T = 450 + 500X
Using Polymath, a = 0.019, we have
Pressure drops to almost zero when weight of catalyst is 39.1 kg, X=35.7 %
α=0.0075:
Polymath Code:
d(X)/d(W)=k*(1-X)*T0/T/(1+X)/v0*P/P0
d(P)/d(W)=-0.5*alpha*T/T0*P0^2/P*(1+X)
T=450+500*X
v0=20
11-15
T0=450
k=0.133*exp(31400/8.314*(1/T0-1/T))
alpha=0.0075
P0=1013250
P(0)=1013250
W(0)=0
W(f)=50
X(0)=0
Output:
P=1atm=101325 Pa, α=0.019
Polymath Code:
d(X)/d(W)=k*(1-X)*T0/T/(1+X)/v0*P/P0
d(P)/d(W)=-0.5*alpha*T/T0*P0^2/P*(1+X)
T=450+500*X
v0=20
T0=450
k=0.133*exp(31400/8.314*(1/T0-1/T))
alpha=0.019
P0=101325
P(0)=101325
W(0)=0
W(f)=39
X(0)=0
11-16
Output:
P=1atm=101325 Pa, α=0.0075
Polymath Code:
d(X)/d(W)=k*(1-X)*T0/T/(1+X)/v0*P/P0
d(P)/d(W)=-0.5*alpha*T/T0*P0^2/P*(1+X)
T=450+500*X
v0=20
T0=450
k=0.133*exp(31400/8.314*(1/T0-1/T))
alpha=0.0075
P0=101325
P(0)=101325
W(0)=0
W(f)=39
X(0)=0
Output:
d(X)/d(W)=k*(1-X)*T0/T/(1+X)/v0*P/P0
d(P)/d(W)=-0.5*alpha*T/T0*P0^2/P*(1+X)
T=450+500*X
v0=20
T0=450
k=0.133*exp(31400/8.314*(1/T0-1/T))
alpha=0.0075
P0=101325
P(0)=101325
W(0)=0
W(f)=50
X(0)=0
11-17
P11-5 (f)Individualized Solution
P11-6 (a)
A → B +C
F
C A = CT A
FT
θI =
FI
FA
CT = C A + C I
FT = FA + FI
(
C A01 = C A0 + C I 0
C A01 =
FA0
)F +F
A0
I0
C A0 + C I 0
θ I +1
11-18
P11-6 (b)
Mole balance:
dX −rA
=
dV FA0
Rate law:
−rA = kC A
T
Stoichiometry: C A = C A01 1− X 0
1+ ε X T
ε = y A0δ
δ =1+1−1=1
y A0 =
ε=
FA0
FA0
1
=
=
FT 0 FA0 + Fi0 1+ θ i
1
1+ θi
T=
(
)
− X ΔH RX + C PA + θ iC Pi T0
C PA + θ iC Pi
See the following Polymath code for theta I=100
To plot graphs for different values of theta I, change thetaI value in the following code
Polymath Code:
d(X)/d(V) = -ra/Fao
Cao = 2/(.082*1100)
Cio = Cao
theta = 100
Fao = 10
Cao1 = (Cao+Cio)/(theta+1)
e = 1/(1+theta)
To = 1100
dHrx = 80000
Cpa = 170
Cpi = 200
T = (X*(-dHrx)+(Cpa+theta*Cpi)*To)/(Cpa+theta*Cpi)
k = exp(34.34-34222/T)
ra = -k*Cao1*(1-X)*To/(1+e*X)/T
V(0)=0
V(f)=500
X(0)=0
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
V
0
0
500
500
X
0
0
0.417064
0.417064
Cao
0.0221729
0.0221729
0.0221729
0.0221729
Cio
0.0221729
0.0221729
0.0221729
0.0221729
theta
100
100
100
100
Fao
10
10
10
10
Cao1
4.391E-04
4.391E-04
4.391E-04
4.391E-04
e
0.009901
0.009901
0.009901
0.009901
To
1100
1100
1100
1100
dHrx
8.0E+04
8.0E+04
8.0E+04
8.0E+04
Cpa
170
170
170
170
11-19
Cpi
T
k
ra
200
1100
25.256686
-0.0110894
200
1098.3458
24.100568
-0.0110894
200
1100
25.256686
-0.0061524
200
1098.3458
24.100568
-0.0061524
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] Cao = 2/(.082*1100)
[2] Cio = Cao
[3] theta = 100
[4] Fao = 10
[5] Cao1 = (Cao+Cio)/(theta+1)
[6] e = 1/(1+theta)
[7] To = 1100
[8] dHrx = 80000
[9] 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
P11-6 (c)
There is a maximum at θ = 8. This is because when θ is small, adding inerts keeps the temperature high
to favor the endothermic reaction. As θ increases beyond 8, there is so much more inert than reactants
that the rate law becomes the limiting factor.
11-20
P11-6 (d)
The change to the Polymath code from part (b) is that the heat of reaction changes sign. Also, since the
reaction is reversible, rate equation and equation for Cb and CC will change.
"
CC %
We need to alter the equations from part (c) such that −rA = k $C A − B C '
KC '&
$#
1− X T0
. Now we need expressions for CB and CC. From stoichiometry
1+ ε X T
X T0
we can see that CB = CC. In terms of CA0 we find that: CB = CC = C A0
1+ ε X T
We also need an equation for KC:
We already know that C A = C A0
∆𝐻PQ 1 1
−80000
1
1
𝐾+ = 𝐾+: ∗ 𝑒𝑥𝑝 ?
& − -C = 2 ∗ 𝑒𝑥𝑝 ?
&
− -C
𝑅
𝑇: 𝑇
8.314 1100 𝑇
Polymath code for theta I =0
d(X)/d(V) = -ra/Fao
Cao = 2/(.082*1100)
Cio = Cao
theta = 0
Fao = 10
Cao1 = (Cao+Cio)/(theta+1)
e = 1/(1+theta)
To = 1100
dHrx =-80000
Cpa = 170
Cpi = 200
T = (X*(-dHrx)+(Cpa+theta*Cpi)*To)/(Cpa+theta*Cpi)
k = exp(34.34-34222/T)
CA=Cao1*(1-X)*To/(1+e*X)/T
CB=Cao1*(X)*To/(1+e*X)/T
CC=Cao1*(X)*To/(1+e*X)/T
ra=-k*(CA-CB*CC/Kc)
Kc=2*exp(-9622.3*(1/1100-1/T))
V(0)=0
V(f)=500
X(0)=0
By changing the value of theta I in above code, following table is generated
Theta I
0
0.5
2
3
5
8
X
0.917
0.96
0.987
0.9915
0.995
0.997
From the above table, it can be concluded that conversion increases rapidly after adding inert up to theta
I =3. After that, adding inert has little advantage on conversion. This is because small quantity of inert
helps in heat removal and has less impact on reaction rate and thus helps reaction to go in forward
direction. Adding too much inert will hinder rate of reaction more and thus benefit of heat removal will
be partially compensated by decrease in reaction rate.
11-21
P11-6 (e)
Volumetric flow rate
vo= FT0/CA01
vo=(FA0+theta*FA0)/CA01
FB=CB*vo
P11-6 (f)
For pure feed (theta=0) and exothermic reaction, heat release is
Qg=rA*deltaHrx
So the heat that must be removed (to maintain isothermal condition) is equal to heat generated by
exothermic reactions.
Polymath code
d(X)/d(V) = -ra/Fao
d(T) / d(V) = 0
Cao = 2/(.082*1100)
Cio = Cao
theta = 0
Fao = 10
Cao1 = (Cao+Cio)/(theta+1)
e = 1/(1+theta)
To = 1100
dHrx = -80000
Cpa = 170
Cpi = 200
k = exp(34.34-34222/T)
ra = -k*Cao1*(1-X)*To/(1+e*X)/T
Qg=ra*dHrx
T(0) = 1100
Qr=-Qg
V(0)=0
V(f)=500
X(0)=0
11-22
Polymath Output
P11-7 (a)
Adiabatic
Mole Balance
dX
= −rAʹ FA0
dW
(1)
W = ρbV
rʹ ρ
r
dX
=− A B =− A
dV
FA0
FA0
Rate Law
Stoichiometry
(2)
⎡
C ⎤
rA = −k ⎢C A − B ⎥
KC ⎥⎦
⎢⎣
(3)
⎡ ⎛
⎤
E 1 1⎞
k = k1 exp⎢ ⎜⎜ − ⎟⎟⎥
⎢⎣ R ⎝ T1 T ⎠⎥⎦
(4)
⎡ ΔH ⎛
⎤
1 1⎞
KC = KC2 exp⎢ Rx ⎜⎜ − ⎟⎟⎥
⎢⎣ R ⎝ T2 T ⎠⎥⎦
(5)
(6)
𝐶" = 𝐶"/ (1 − 𝑋)𝑝 (𝑇/ /𝑇)
𝐶) = 𝐶"/ 𝑋 𝑝 (𝑇/ /𝑇)
𝑑𝑝
𝛼 𝐹Y 𝑇
𝛼 𝑇
=
& -=− & 𝑑𝑊 𝑝 𝐹Y/ 𝑇/
2𝑝 𝑇/
W=𝜌𝑉
[\
^_
Y
= − 0\` @Y A
[]
>
a
Where 𝑝 = a
>
Parameters
(7) – (15)
FA0 , k1 , E, R, T1 , KC2 , ΔHRx , T2 , C A0 , T0 , 𝛼𝜌b
11-23
Energy Balance
Adiabatic and DCP = 0
(16A)
T = T0 +
(−ΔHRx ) X
∑Θi CP
i
T0 , ∑Θi CP = CP + ΘICP
Additional Parameters (17A) & (17B)
Heat Exchange
i
( )(
A
)
I
(
dT −rA −ΔHRx −Ua T −Ta
=
dV
∑Fi CP
i
)
⎡
⎤
∑Fi CP = FA0 ⎢∑Θi CP + ΔCP X ⎥ , if ΔCP = 0 then
⎣
⎦
i
i
(16B)
( )(
)
(
dT −rA −ΔHRx −Ua T −Ta
=
dV
FA0 ∑Θi CP
i
Adiabatic
See the following Polymath code:
Polymath Code:
d(X)/d(V)=-ra/Fa0
d(T)/d(V)=(Ua*(Ta-T)+DH*ra)/(Fa0*(sumcp+DCp*X))
d(Ta)/d(V)=(Ua*(T-Ta))/m/Cpc
d(p)/d(V)=-alpharho/p/2*(T/T0)
T0=300
Ca0=2.
Cb=Ca0*(X)*p*(T0/T)
E=10000
Ca=Ca0*(1-X)*p*(T0/T)
T1=300
alpharho=0.02
k1=0.1
k=k1*exp((E/R)*(1/T1-1/T))
R=1.987
DH=-20000
Kc=1000*exp((DH/R)*(1/T1-1/T))
Fa0=5
ra=-k*(Ca-Cb/Kc)
ThetaI=2
CpI=18
Cpa=160
sumcp=Cpa+ThetaI*CpI
Xe=Kc/(1+Kc)
rate=-ra
Ua=0
Cpcool=20
DCp=0
m=100
Cpc=20
X(0)=0
T(0)=300
Ta(0)=0
p(0)=1
V(0)=0
V(f)=40
11-24
)
Output:
11-25
P11-7 (b) Varying “ratio of inert to A” in Polymath code gives below table
X
Xe
p
T
𝜽𝑰
0
0.58
0.58
0.15
373.0
2
0.67
0.67
0.20
368.2
4
0.74
0.74
0.23
363.6
6
0.80
0.80
0.26
359.4
8
0.84
0.84
0.28
355.4
10
0.88
0.88
0.30
351.7
Thus, increasing 𝜃f increases conversion. Also increasing 𝜃f increases pressure drop but decreases
reactor temperature.
Varying “entering temperature” in Polymath code gives below table
X
Xe
p
T
𝑻𝟎
300
0.67
0.67
0.2
368.1
350
0.36
0.36
0.34
386.3
400
0.105
0.105
0.42
410.7
From the above table, it can be seen that on increasing entering temp, conversion decreases, p
increases and T increases.
P11-7(c)
FA0
dX
= −rA '
dW
−rA ' = k(1− α W)0.5 C A0 [(1− X)−
∫ (1− α W)0.5 dW = FA0 ∫
X
]
KC
1
1
1−(1+ )X
KC
dX
F
2
1
[1−(1− α W)1.5 ] = − A0 ln[1−(1+ )X]
3α
1
KC
(1+ )
KC
2(1+
ln[1−(1+
1
)
KC
1
)X] = −
[1−(1− α W)1.5 ]
KC
3αFA0
2(1+
1−(1+
1
)X = exp(−
[1−(1− α W)1.5 ])
KC
3αFA0
2(1+
1−exp(−
X=
1
)
KC
1
)
KC
3αFA0
1+
[1−(1− α W)1.5 ])
1
KC
11-26
2(1+
Qr = FA0 X(−ΔHR (T0 )) =
FA0 (−ΔHR (T0 ))(1− exp(−
3αFA0
1+
Qr = FA0 X(−ΔHR (T0 )) =
[1−(1− α W)1.5 ]))
1
KC
2(1+
FA0 (−ΔHR (T0 ))(1− exp(−
1
)
KC
1
)
KC
3αFA0
1+
[1−(1− αρb V)1.5 ]))
1
KC
Qr vs V can be plotted according to the above equation.
Qr vs. V
120
Qr (kcal/min)
100
80
60
40
20
0
0
10
20
30
40
V (dm3)
P11-8 (a)
(1)
dX −rA"
=
dW FA0
(2)
"
CC2 %
$
−rA = k C ACB − '
KC '&
$#
(3)
( E " 1 1 %+
k = k1 exp* $$ − ''*) R # T1 T &-,
(4)
𝑘: = 0.004
(5)
T1 = 310
(6)
E = 25,000
(7)
) ΔH # 1 1 &,
KC = KC2 exp+ Rx %% − ((.
+* R $ T2 T '.-
(8)
ΔHRx = −20,000
(9)
𝑇0 = 303
(10)
KC 2 = 1,000
𝐾i = 1000𝑒𝑥𝑝 ?
11-27
−20,000 1
1
&
− -C
1.987 303 𝑇
(11)
δ = 2−1−1 = 0
(12)
ε = y A0δ = 0.5 2−1−1 = 0
(
)
T
C A = C A0 1− X p 0
T
(
(13)
)
ΘB = 2
(14)
(
T0
)T p
(15)
CB = C A0 2− X
(16)
T
CC = 2C A0 X 0 p
T
(17)
dp −α T
=
dW 2p T0
(18)
α = 0.0002
a
where 𝑝 = a
>
P11-8 (b)
2+
(
"
T0 % *
$2C Xp '
*
" T %
" T % $# A0 T '& 0
0
−rA = k *C A0 1− X $$ p ''C A0 2− X $$ p '' −
KC
*
# T&
# T&
*
*)
-,
(
)
(
)
2
"
4X 2 %( T0 +
'* p −rA = kC $ 1− X 2− X −
KC '&*) T -,
$#
2
A0
(
)(
)
P11-8 (c)
Equilibrium Conversion
KC =
2
4C A0
X e2
2
C A0
1− X e 2− X e
KC
4
(
)(
=
X e2
)
(1− X )(2− X )
e
e
2
e
K 3K
K X
2 C − C X e + C = X e2
4
4
4
"K
%
3K
K
$$ C −1'' X e2 − C X e +2 C = 0
4
4
#4
&
Equilibrium Conversion (19)
Xe =
3KC
−
4
2
" 3K %
"K
%
$$ C '' −2KC $$ C −1''
# 4 &
#4
&
"K
%
2$$ C −1''
#4
&
11-28
Xe
Xe
T
X
W
W
Case 1 Adiabatic
P11-8 (d)
Energy Balance
T = T0 +
(20)
(−ΔH ) X
Rx
∑Θi CP
i
Energy Balance
(21)
T0 = 330
(
ΔHRx = ΔHR + ΔCP T −TR
)
( )
ΔCP = 2CP −CP −CP = 2 20 −20 −20 = 0
C
A
B
ΔCP = 0
∑Θi CP = CP + ΘBCP + ΘC CP + ΘICP
i
A
B
C
I
=20+2*20+0*20+2*40= 140
Energy Balance
∑ 𝜃m 𝐶am = 140
(22)
P11-8 (e)
𝑇 = 𝑇/ +
ino
pqo
.𝑘
(−∆𝐻Pr )𝑋
20,000𝑋
= 330 +
= 330 + 142.9𝑋
∑ 𝜃m 𝐶am
140
P11-8 (f)
Adiabatic
See the following Polymath code:
Polymath Code:
d(X)/d(W)=-ra/Fa0
d(T)/d(W)=(Ua/rhob*(Ta-T)+DH*ra)/(Fa0*(sumcp))
d(Ta)/d(W)=(Ua/rhob*(T-Ta))/m/Cpcool*0
d(p)/d(W)=-alpha/2/p*(T/T0)
T0=330
yao= 0.2
Ca0=0.2*0.3
Cc=Ca0*2*X*p*(T0/T)
Cb=Ca0*(2-X)*p*(T0/T)
E=25000
Ca=Ca0*(1-X)*p*(T0/T)
alpha=0.0002
k1=0.004
k=k1*exp((E/1.987)*(1/310-1/T))
R=1.987
DH=-20000
Kc=1000*exp((DH/1.987)*(1/303-1/T))
Fa0=5
11-29
Everything that enters
ra=-k*(Ca*Cb-Cc^2/Kc)
sumcp=140
ThetaI=2
CpI=18
Xe=(3*Kc/4-((3*Kc/4)^2-2*Kc*(Kc/4-1))^0.5)/(2*(Kc/4-1))
rate=-ra
Ua=0
m=18
Cpcool=18
rhob=1400
W(0)=0
W(f)=1357
X(0)=0
T(0)=330
Ta(0)=300
p(0)=1
Output:
11-30
Gas Phase Adiabatic
P11-8 (g)
For isothermal solution, set dT/dW to be zero in the polymath code provided in P11-8(f) and add an
equation for heat that must be removed along the reactor as Q = -DH*X*Fa0 to get the below plot.
11-31
P11-9
KC =
CC CD
X e2
=
C ACB (1 - X e )2
Þ Xe =
T = T0 -
KC
1 + KC
(-30000) X = 300 + 600 X
DH R X
= 300 CPA + CPB
(25 + 25)
See the following Polymath code:
Polymath Code:
f(Xe) = Xe - (1 - Xe) * Kc ^ 0.5 == 0
T = 300
Kc = 500000 * exp(-30000 / 1.987 * (1 / 323 - 1 / T))
Xe(min)=0
Xe(max)=1
Output:
Calculated values of NLE variables
Variable Value
1 Xe
f(x)
Initial Guess
0.9997644 3.518E-11 0.5 ( 0 < Xe < 1. )
Variable Value
1 T
300.
2 Kc
1.8E+07
Nonlinear equations
1 f(Xe) = Xe - (1 - Xe) * Kc ^ 0.5 = 0
Explicit equations
1 T = 300
2 Kc = 500000 * exp(-30000 / 1.987 * (1 / 323 - 1 / T))
T
300
320
340
360
380
400
420
440
460
480
500
520
540
560
X
1
0.999
0.995
0.984
0.935
0.887
0.76
0.585
0.4
0.26
0.1529
0.091
0.057
0.035
11-32
Xe
1
Xe (Eqm Conversion)
0.9
0.8
0.7
0.6
Xe
0.5
0.4
0.3
0.2
0.1
0
300
350
400
450
500
550
Temp (K)
P11-10 (a)
For first reactor,
KC =
X e1
KC
or X e1 =
1 - X e1
1 + KC
For second reactor
KC =
q B 2+ X e 2
K -q
orX e 2 = C B 2
1 - X e2
1 + KC
For 3rd reactor
KC =
q B 3+ X e 3
K -q
orX e3 = C B 3
1 - X e3
1 + KC
1st reactor: in first reaction Xe = 0.3
So, FB = FA01(.3)
2nd reactor: Moles of A entering the 2nd reactor: FA02 = 2FA01 - FA01(.3) = 1.7FA01
qB2 =
.2 FA01
= .12
1.7 FA0
- FA02 SCPi qi (T - T0 ) + FA02 X (-DH R ) = 0
X=
(C + q C )(T - T )
pA
B
pB
0
-DH R
Slope is now negative
3rd reactor:
X e2 = 0.3
Say
q B = (.2 FA01 ) + .3FA02 = FA01 (.2 + .3 ´1.8)
FA03 = FA01 + FA02 (1 - X e 2 ) = FA01 + 1.8FA01 (1 - X e 2 ) = 1 + 1.8(1 - .3) FA01
FA03 = 2.26 FA01
Feed to the reactor 3:
(2 FA01 ) + .3FA02 = FA01 (.2 + .3 ´1.8) = 0.7FA01
q B3 =
.74
2.26
11-33
Feed Temperature to the reactor 2 is (520+450)/2 = 485 K
Feed Temperature to reactor 3 is 480 K
Xfinal = .4
Moles of B = .2FA01 + .3FAo2 + .4FA03 = FA01(.2 + .54 + (.4)(2.26)) = 1.64 FA01
X = FB/3FA01 = .54
P11-10 (b)
The same setup and equations can be used as in part (a). The entering temperature for reactor 1 is now
450 K and the outlet is 520 K. When the two streams are joined prior to entering reactor 2 the
temperature is (520+450)/2 = 485 K
Say that the outlet temperature for reactor 2 is 510 K. Then the entering temperature for reactor 3
would be (510+510+450)/3 = 490 K
For any reactor j,
- FA0 j SCPi qi (T - T0 ) + FA0 j X (-DH R ) = 0
X=
(C + q C )(T - T )
pA
B
pB
0
-DH R
and θB for reactor 1 = 0. For reactor 2, θB> 0. This means that the slope of the conversion line from the
energy balance is larger for reactor 2 than reactor 1. And similarly, θB for reactor 3 > θB for reactor 2. So,
the line for conversion in reactor 3 will be steeper than that of reactor 2. The mass 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.
11-34
Synopsis for Chapter 12 – Steady-State Nonisothermal Reactor
Design: Flow Reactors with Heat Exchange
General: The key point of the course and text where everything is brought together.
Questions
l Q12-1A (20 seconds) Questions Before Reading (QBR).
AA Q12-3A (3 min) Excellent discussion of an everyday example of striking a match to help the students
understand Figures 12-13 and the jump from the lower steady state to the upper steady state. Slow
movement causes only friction and no spark. Rapid movement causes a spark.
Computer Simulations and Experiments
The Wolfram and Python problems of this chapter require significant attention and time. Approximately
20-35 minutes per simulation. In the simulations the four different modes of heat operation are
considered: Adiabatic, Co-Current Heat Transfer, Counter Current Heat Transfer and Constant Heat
Exchanger Temperature, Ta.
l P12-1B (30-35 min per simulation) I always assign Parts (a), (b), (c), (e), (f), (g), (h), and (i). The time
spent on these simulations will help the student get an intuitive feel of the various modes of heat
transfer and how they affect both the reaction and reactors. In P12-1B (b) involves a real reaction of
industrial importance.
l P12-2B Similar to P12-1B, except the reaction is endothermic. (I suggest 8-10 min per simulation,
parts (i) through (vi)). The time spent on these simulations will greatly help the student get an
intuitive feel of the various modes of heat transfer and how they affect both the reaction and
reactors.
O P12-3B (40 min) This problem on Page 663 is a little out of the Problems section. However, it is an
excellent problem which I always assign. (60 min) Very important problem. Comprehensive problem
for co-current heat exchange. This problem solution is now included in the LEPs. Use Wolfram LEPs
to explore parameter variations.
Interactive Computer Games (ICG)
P12-4A (60 min) Interactive Computer Game (ICG). This game shows a novel derivation of the heat
transfer equations as the different colored symbols move around and talk to one another.
Problems
AA P12-5C (45 min) Old Exam Question (OEQ). Fairly straight forward problem if one follows part (b) to
find out what happens after the operation shuts off the feed stream which is also used to cool the
reactor. Still some unresolved issues as the G(T) and R(T) curves intersect at an unstable point.
O P12-6B (25 min) Old Exam Question (OEQ). California Registration Exam. Fairly straight forward
professional engineer’s registration exam problem.
l P12-7B (50 min) Old Exam Question (OEQ). Straight forward application of energy balance to the
four types of heat exchange operations. The instructor may choose to assign only some of the
modes of heat transfer, e.g., co-current and counter current.
S12-1
AA P12-8A (60 min) Similar to P12-7B in type of calculations and in difficulty.
AA P12-9A (60 min) Includes pressure drop, otherwise similar to problems P12-7B, P12-8A, P12-10B,
P12-11B, P12-12C and P12-18C.
AA P12-10B (60 min) Similar to P12-7B, P12-8A, P12-11B and P12-12C.
AA P12-11B (60 min) Old Exam Question (OEQ). Similar to P12-7B, P12-8A, P12-10B and P12-12C.
AA P12-12C (60 min) Similar to P12-7B, P12-8A, P12-10B and P12-11B.
l P12-13B (20 min) Old Exam Question (OEQ). Short conceptual problem to explain the figures given
in the problem. Students would have to complete the Wolfram simulations to answer these
questions.
AA P12-14A (30 min) Old Exam Question (OEQ). Short problem on multiple steady states in a CSTR.
O P12-15B (60 min) Similar in concepts and calculations to P12-11A, but here both the R(T) curve and
the G(T) curve must be developed.
AA P12-16B (75 min) Very similar to P12-15B except the reaction is reversible thereby changing the
shape of the G(T) curve.
l P12-17B (50 min) Old Exam Question (OEQ). Very good problem made easier than P12-16B because
the G(T) curve is given.
AA P12-18C (75 min) This problem is a c-level problem but it is “reasonably” straight forward similar to
problems P12-9A through 12-12C.
S P12-19C (75 min) C-level fairly difficult problem.
l P12-20B (20 min) Old Exam Question (OEQ). The students will have to work through the Wolfram
simulations to answer this question. Short conceptual problem to match figures for different type
of heat exchanger operations.
l P12-21B (45 min) Old Exam Question (OEQ). Excellent problem with few calculations to test
concepts of multiple reaction with heat effects.
AA (P12-22B through P12-24B) Similar straight forward problems of multiple reactions with heat effects.
AA P12-22B (30 min) Old Exam Question (OEQ). Short problem to teach concepts of heat effect with
multiple reaction without having to use Polymath.
→
AA P12-23B (75 min) Oops! There’s a typo in the problem statement, it should read 2A B. Straight
←
forward use of Polymath to solve multiple reactions with four types of heat exchange.
AA P12-24B (75 min) Same concepts and degree of difficulty as P12-23B.
AA P12-25C (120 min) Old Take Home Exam Question (OTHEQ). A major problem of the course and
meets one of the major goals of the course. multiple reactions with heat effects. Either P12-25C or
P12-26C are assigned every time I teach the course.
l P12-26C (120 min) Old Take Home Exam Question (OTHEQ). Major course problem along with P1225C. Either P12-25C or P12-26C assigned every time I teach the course. I have a slight preference for
P12-26 because some reactions are exothermic and some are endothermic.
O P12-27B (120 min) Old Take Home Exam Question (OTHEQ). Alternative to P12-25C and P12-26C.
Same method of analysis.
S12-2
Solutions for Chapter 12 – Steady-State Nonisothermal
Reactor Design: Flow Reactors with Heat Exchange
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Chapter 12:
http://www.umich.edu/~elements/5e/12chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram codes):
http://www.umich.edu/~elements/5e/tutorials/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath codes):
http://www.umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q12-1 Individualized solution.
Q12-2 Individualized solution.
Q12-3 By applying pressure and striking it rapidly the entering temperature of the pumice is increased
which gives it a greater chance to reach the ignition temperature point on the Fig 12-13. The temperature
then jumps to the point ‘11’ on Fig 12-13 and the match is ignited.
Q12-4 Individualized solution.
Q12-5 Individualized solution.
Q12-6 Individualized solution.
P12-1 (a)
(i)
Temperature plot: As the reaction is exothermic, heat released as the reaction proceeds. After a
point, the concentration of reactants will decrease and hence the reaction rate will be low. This
allows the inert gases and the coolant present in the system to reduce the temperature.
Equilibrium conversion: For an exothermic reaction, we know that as temperature increases the
equilibrium conversion decreases. The same is observed in the plot.
Conversion plot: We observe the rate of increase in conversion is highest near the maximum
temperature point, which is expected. Further, as the temperatures decrease the rate of increase
in conversion is decreasing. Due to low rates, equilibrium conversion is not achievable.
(ii) Increasing Fao decreases conversion. This is evident by seeing the differential equation for
conversion below.
12-1
Increasing thetaI from 0.5 to 4 increases the distance between X and Xe curves with X being
reduced to almost zero when thetaI approaches to 4. This is because increase in thetaT increases
the inerts in the reactor, hence affecting reaction a lot.
Increasing Hr(heat of reaction) from -30000 to -10000 increases both X and Xe till a particular Hr
but after that X decreases while Xe increases and both of these curves taking a constant value at
Hr around -10000. Increasing Hr makes Kc to rise which affects both X and Xe.
Increasing Ua/rhoB makes both of the conversion profiles to move apart. Xe increases while X
decreases as we increase Ua/rhoB.
(iii,iv) θI
(v) Individualized solution
(vi) As To is increased, final conversion remains the same but it is reached with less W. Therefore,
increasing ‘To’ can reduce the weight of catalyst required.
(vii) As Ta is increased from 300 to 335 K, the conversion increases as forward reaction rate constant k
dominates over Kc and thus increasing overall reaction rate. On increasing temp beyond 335 K,
conversion decreases as Kc will dominate over k and will control reaction rate.
For countercurrent flow, the only equation which changes is:
d(Ta)/d(W) = -Uarho*(T-Ta)/(mc*Cpcool)
Note that the right hand side of the equation has been multiplied by -1.
Now, we have to guess a value of Ta such that it matches Ta0 at W = 4500.
We get Ta = 323.1302 K, for which Ta = Ta0 = 320 K at W = 4500.
See the following Polymath program
Polymath code:
Xe = ((thetaB+1)*Kc- (((thetaB+1)*Kc)^2-4*(Kc-4)*(Kc*thetaB))^0.5)/(2*(Kc-4))
d(Ta)/d(W) = -Uarho*(T-Ta)/(mc*Cpcool) #
d(p)/d(W) = -alpha/2*(T/To)/p #
d(T)/d(W) = (Qg - Qr)/(Fao*sumcp) #
d(X) / d(W) = -ra/Fao
X(0) = 0
Qg = (-ra)*(-Hr)#
Qr = Uarho*(T-Ta)#
alpha = .0002 #
To = 330 #
Uarho = 0.5 #
mc = 1000 #
Cpcool = 18 #
Hr = -20000 #
Fao = 5 #
thetaI = 1 #
CpI = 40 #
CpA = 20 #
thetaB = 1 #
CpB = 20 #
Cto = 0.3 #
Ea = 25000 #
Kc = 1000*(exp(Hr/1.987*(1/303-1/T))) #
k = .004*exp(Ea/1.987*(1/310-1/T)) #
yao = 1/(1+thetaB+thetaI) #
Cao = yao*Cto #
sumcp = (thetaI*CpI+CpA+thetaB*CpB) #
Ca = Cao*(1-X)*p*To/T #
Cb = Cao*(thetaB-X)*p*To/T #
Cc = Cao*2*X*p*To/T #
ra = -k*(Ca*Cb-Cc^2/Kc) #
W(0)=0
Ta(0)=323.13
p(0)=1
T(0)=330
W(f) = 4500
12-2
P12-1 (a) Continued
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.0002
0.0002
0.0002
0.0002
2
Ca
0.1
0.0111851
0.1
0.0111851
3
Cao
0.1
0.1
0.1
0.1
4
Cb
0.1
0.0111851
0.1
0.0111851
5
Cc
0
0
0.0673836
0.0258241
6
CpA
20.
20.
20.
20.
7
CpB
20.
20.
20.
20.
8
Cpcool
18.
18.
18.
18.
9
CpI
40.
40.
40.
40.
10 Cto
0.3
0.3
0.3
0.3
11 Ea
2.5E+04
2.5E+04
2.5E+04
2.5E+04
12 Fao
5.
5.
5.
5.
13 Hr
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
14 k
0.046809
0.0207345
11.53755
0.0207345
15 Kc
66.01082
0.805727
126.6264
126.6264
16 mc
1000.
1000.
1000.
1000.
17 p
1.
0.2359323
1.
0.2359323
18 ra
-0.0004681
-0.012037
-2.485E-06
-2.485E-06
19 sumcp
80.
80.
80.
80.
20 T
330.
323.0995
385.7156
323.0995
21 Ta
323.1302
320.
323.1302
320.
22 thetaB
1.
1.
1.
1.
23 thetaI
1.
1.
1.
1.
24 To
330.
330.
330.
330.
25 Uarho
0.5
0.5
0.5
0.5
26 W
0
0
4500.
4500.
27 X
0
0
0.5358329
0.5358329
28 Xe
0.8024634
0.3097791
0.849089
0.849089
29 yao
0.3333333
0.3333333
0.3333333
0.3333333
Differential equations
1 d(Ta)/d(W) = -Uarho*(T-Ta)/(mc*Cpcool)
2 d(p)/d(W) = -alpha/2*(T/To)/p
3 d(T)/d(W) = (Uarho*(Ta-T)+(-ra)*(-Hr))/(Fao*sumcp)
4 d(X)/d(W) = -ra/Fao
Explicit equations
1
Hr = -20000
2
alpha = .0002
3
To = 330
4
Uarho = 0.5
12-3
5
mc = 1000
6
Cpcool = 18
7
Kc = 1000*(exp(Hr/1.987*(1/303-1/T)))
8
Fao = 5
9
thetaI = 1
10 CpI = 40
11 CpA = 20
12 thetaB = 1
13 CpB = 20
14 Cto = 0.3
15 Ea = 25000
16 Xe = ((thetaB+1)*Kc- (((thetaB+1)*Kc)^2-4*(Kc-4)*(Kc*thetaB))^0.5)/(2*(Kc-4))
17 k = .004*exp(Ea/1.987*(1/310-1/T))
18 yao = 1/(1+thetaB+thetaI)
19 Cao = yao*Cto
20 sumcp = (thetaI*CpI+CpA+thetaB*CpB)
21 Ca = Cao*(1-X)*p*To/T
22 Cb = Cao*(thetaB-X)*p*To/T
23 Cc = Cao*2*X*p*To/T
24 ra = -k*(Ca*Cb-Cc^2/Kc)
Plot of X, Xe and p versus W
12-4
P12-1 (a) Continued
Plot of T and Ta versus W
(viii) For constant Ta, we have to use the same Polymath program as in Text but multiply the right hand
side by 0 in the program;
i.e., d(Ta)/d(W) = Uarho*(T-Ta)/(mc*Cpcool)*0
See the following Polymath program
Polymath code:
Xe = ((thetaB+1)*Kc- (((thetaB+1)*Kc)^2-4*(Kc-4)*(Kc*thetaB))^0.5)/(2*(Kc-4))
d(Ta)/d(W) = Uarho*(T-Ta)/(mc*Cpcool)*0 #
d(p)/d(W) = -alpha/2*(T/To)/p #
d(T)/d(W) = (Qg - Qr)/(Fao*sumcp) #
d(X) / d(W) = -ra/Fao
X(0) = 0
Qg = (-ra)*(-Hr)#
Qr = Uarho*(T-Ta)#
alpha = .0002 #
To = 330 #
Uarho = 0.5 #
mc = 1000 #
Cpcool = 18 #
Hr = -20000 #
Fao = 5 #
thetaI = 1 #
CpI = 40 #
CpA = 20 #
thetaB = 1 #
CpB = 20 #
Cto = 0.3 #
Ea = 25000 #
Kc = 1000*(exp(Hr/1.987*(1/303-1/T))) #
k = .004*exp(Ea/1.987*(1/310-1/T)) #
yao = 1/(1+thetaB+thetaI) #
Cao = yao*Cto #
sumcp = (thetaI*CpI+CpA+thetaB*CpB) #
Ca = Cao*(1-X)*p*To/T #
Cb = Cao*(thetaB-X)*p*To/T #
Cc = Cao*2*X*p*To/T #
ra = -k*(Ca*Cb-Cc^2/Kc) #
W(0)=0
Ta(0)=320
p(0)=1
T(0)=330
W(f) = 4500
12-5
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.0002
0.0002
0.0002
0.0002
2
Ca
0.1
0.011675
0.1
0.011675
3
Cao
0.1
0.1
0.1
0.1
4
Cb
0.1
0.011675
0.1
0.011675
5
Cc
0
0
0.0659781
0.026488
6
CpA
20.
20.
20.
20.
7
CpB
20.
20.
20.
20.
8
Cpcool
18.
18.
18.
18.
9
CpI
40.
40.
40.
40.
10 Cto
0.3
0.3
0.3
0.3
11 Ea
2.5E+04
2.5E+04
2.5E+04
2.5E+04
12 Fao
5.
5.
5.
5.
13 Hr
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
14 k
0.046809
0.021188
8.254819
0.021188
15 Kc
66.01082
1.053204
124.4536
124.4536
16 mc
1000.
1000.
1000.
1000.
17 p
1.
0.2441145
1.
0.2441145
18 ra
-0.0004681
-0.0080133
-2.769E-06
-2.769E-06
19 sumcp
80.
80.
80.
80.
20 T
330.
323.2791
381.7968
323.2791
21 Ta
320.
320.
320.
320.
22 thetaB
1.
1.
1.
1.
23 thetaI
1.
1.
1.
1.
24 To
330.
330.
330.
330.
25 Uarho
0.5
0.5
0.5
0.5
26 W
0
0
4500.
4500.
27 X
0
0
0.5314832
0.5314832
28 Xe
0.8024634
0.3391177
0.8479767
0.8479767
29 yao
0.3333333
0.3333333
0.3333333
0.3333333
Differential equations
1 d(Ta)/d(W) = -Uarho*(T-Ta)/(mc*Cpcool) *0
2 d(p)/d(W) = -alpha/2*(T/To)/p
3 d(T)/d(W) = (Uarho*(Ta-T)+(-ra)*(-Hr))/(Fao*sumcp)
4 d(X)/d(W) = -ra/Fao
Explicit equations
1
Hr = -20000
2
alpha = .0002
3
To = 330
4
Uarho = 0.5
5
mc = 1000
6
Cpcool = 18
7
Kc = 1000*(exp(Hr/1.987*(1/303-1/T)))
8
Fao = 5
9
thetaI = 1
12-6
10 CpI = 40
11 CpA = 20
12 thetaB = 1
13 CpB = 20
14 Cto = 0.3
15 Ea = 25000
16 Xe = ((thetaB+1)*Kc- (((thetaB+1)*Kc)^2-4*(Kc-4)*(Kc*thetaB))^0.5)/(2*(Kc-4))
17 k = .004*exp(Ea/1.987*(1/310-1/T))
18 yao = 1/(1+thetaB+thetaI)
19 Cao = yao*Cto
20 sumcp = (thetaI*CpI+CpA+thetaB*CpB)
21 Ca = Cao*(1-X)*p*To/T
22 Cb = Cao*(thetaB-X)*p*To/T
23 Cc = Cao*2*X*p*To/T
24 ra = -k*(Ca*Cb-Cc^2/Kc)
Plot of X, Xe and p versus W
Plot of T and Ta versus W
Ta will remain constant with W
12-7
P12-1 (a) Continued
For adiabatic operation, using the polymath program of constant Ta and making the parameter UA=0,
we have
POLYMATH Report
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.0002
0.0002
0.0002
0.0002
2
Ca
0.1
0.0153303
0.1
0.0153303
3
Cao
0.1
0.1
0.1
0.1
4
Cb
0.1
0.0153303
0.1
0.0153303
5
Cc
0
0
0.0403759
0.0105141
6
CpA
20.
20.
20.
20.
7
CpB
20.
20.
20.
20.
8
Cpcool
18.
18.
18.
18.
9
CpI
40.
40.
40.
40.
10 Cto
0.3
0.3
0.3
0.3
11 Ea
2.5E+04
2.5E+04
2.5E+04
2.5E+04
12 Fao
5.
5.
5.
5.
13 Hr
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
14 k
0.046809
0.046809
22.60961
22.60961
15 Kc
66.01082
0.470375
66.01082
0.4703751
16 mc
1000.
1000.
1000.
1000.
17 p
1.
0.2456997
1.
0.2456997
18 ra
-0.0004681
-0.0120578
1.617E-08
-2.451E-18
19 sumcp
80.
80.
80.
80.
20 T
330.
330.
393.8384
393.8384
21 Ta
320.
320.
320.
320.
22 thetaB
1.
1.
1.
1.
23 thetaI
1.
1.
1.
1.
24 To
330.
330.
330.
330.
25 Uarho
0
0
0
0
26 W
0
0
4000.
4000.
27 X
0
0
0.2553537
0.2553537
28 Xe
0.8024634
0.2553537
0.8024634
0.2553537
29 yao
0.3333333
0.3333333
0.3333333
0.3333333
Differential equations
1 d(Ta)/d(W) = -Uarho*(T-Ta)/(mc*Cpcool) *0
2 d(p)/d(W) = -alpha/2*(T/To)/p
3 d(T)/d(W) = (Uarho*(Ta-T)+(-ra)*(-Hr))/(Fao*sumcp)
4 d(X)/d(W) = -ra/Fao
Explicit equations
1
Hr = -20000
2
alpha = .0002
3
To = 330
4
Uarho = 0
5
mc = 1000
6
Cpcool = 18
7
Kc = 1000*(exp(Hr/1.987*(1/303-1/T)))
12-8
8
Fao = 5
9
thetaI = 1
10 CpI = 40
11 CpA = 20
12 thetaB = 1
13 CpB = 20
14 Cto = 0.3
15 Ea = 25000
16 Xe = ((thetaB+1)*Kc- (((thetaB+1)*Kc)^2-4*(Kc-4)*(Kc*thetaB))^0.5)/(2*(Kc-4))
17 k = .004*exp(Ea/1.987*(1/310-1/T))
18 yao = 1/(1+thetaB+thetaI)
19 Cao = yao*Cto
20 sumcp = (thetaI*CpI+CpA+thetaB*CpB)
21 Ca = Cao*(1-X)*p*To/T
22 Cb = Cao*(thetaB-X)*p*To/T
23 Cc = Cao*2*X*p*To/T
24 ra = -k*(Ca*Cb-Cc^2/Kc)
Plot of X, Xe and p versus W;
Plot of T and Ta versus W
Ta remains a constant
12-9
P12-1 (b)
(i)
Of all the four types of heat exchanges we see that the temperature plot in the adiabatic case tends
to saturate. Whereas, in the other three cases the temperature profile peaks and then the due to
decreased reaction rates (due to low concentration) the temperature reduces as the coolant’s effect
becomes dominant. Without the coolant, the adiabatic reaction reaches completion (i.e,
equilibrium conversion) when the reaction rate is close to zero, which means no more heat release.
In counter current operation, the entry point of the reactants is the exit for the coolant. Clearly, at
the entry of coolant (V = 5m3) highest energy releases are seen (high rates) and hence the
temperature of the coolant is increasing in the backward direction. A similar argument can be given
for the co-current operation.
Counter current and constant Ta cases give the highest conversions and the adiabatic case gives the
least conversion.
(ii) In the Co-Current operation,
Temperature plot: As the reaction is exothermic, heat released as the reaction proceeds. After a
point, the concentration of reactants will decrease and hence the reaction rate will be low. This
allows the inert gases and the coolant present in the system to reduce the temperature. As the
coolant and the reactants flow in the same direction, at the starting the coolant is exposed to the
highest temperatures which result in increasing temperatures.
Equilibrium conversion: For an exothermic reaction, we know that as temperature increases the
equilibrium conversion decreases. The same is observed in the plot.
Conversion plot: We observe the rate of increase in conversion is highest near the maximum
temperature point, which is expected. Further, as the temperatures decrease the rate of increase
in conversion is decreasing. Due to low rates, equilibrium conversion is not achievable.
As ya0 is decreased (i.e., the initial concentration of reactants is decreased by adding inerts) the
reaction rate will decrease naturally, which results in low heat releases and hence the temperatures
will reduce. This is also reflected in the way conversion profiles behave.
(iii) Ua brings the temperature profiles close together.
(iv) FA0 separates X and Xe the most.
(v) Varying CA0 keeps X and Xe profiles closest together.
(vi) In a co-current operation involving an exothermic reaction, the overall temperature decreases as
we proceed along the flow.
Increase in heat transfer coefficient helps in keeping the temperature profile close to the coolant’s
temperature.
Increasing the flow rate, will result in smaller conversions. This can be used to our advantage
when runaway reactions are involved.
(vii) Ta = 306 K
(viii) Increasing ya0 can lead to a huge spike in temperature which the reactor may not be able to
handle. Decreasing heat transfer coefficient Ua can also lead to the same situation.
(ix) For isothermal operation, the same code in part (vii) is used, with one change:
d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cpo/Fa0*0
Note that the right hand side of the equation is multiplied by 0.
See the following Polymath program
Polymath code:
Cpc = 28
m = 500
d(Ta) / d(V) = Ua*(T-Ta)/m/Cpc
Ua = 5000
Ta(0) =315
d(X)/d(V) = -ra/Fa0 #
d(T)/d(V) = (Qg-Qr)/(Cpo*Fa0)*0 #
Qg = ra*deltaH #
Qr = Ua*(T-Ta) #
12-10
Ca0 = 1.86 #
Fa0 = 0.9*163*.1 #
Kc = 3.03*exp((deltaH/8.314)*((T-333)/(T*333))) #
k = 31.1*exp((7906)*(T-360)/(T*360)) #
Xe = Kc/(1+Kc) #
ra = -k*Ca0*(1-(1+1/Kc)*X) #
deltaH = -34500 #
Cpo = 159
rate = -ra #
V(0)=0
X(0)=0
T(0)= 305
V(f)= 5
POLYMATH Report
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca0
1.86
1.86
1.86
1.86
2
Cpc
28.
28.
28.
28.
3
Cpo
159.
159.
159.
159.
4
deltaH
-3.45E+04
-3.45E+04
-3.45E+04
-3.45E+04
5
Fa0
14.67
14.67
14.67
14.67
6
k
0.5927441
0.5927441
0.5927441
0.5927441
7
Kc
9.512006
9.512006
9.512006
9.512006
8
m
500.
500.
500.
500.
9
Qr
-5.0E+04
-5.0E+04
-8383.862
-8383.862
10 ra
-1.102504
-1.102504
-0.7278292
-0.7278292
11 rate
1.102504
0.7278292
1.102504
0.7278292
12 T
305.
305.
305.
305.
13 Ta
315.
306.6768
315.
306.6768
14 Ua
5000.
5000.
5000.
5000.
15 V
0
0
5.
5.
16 X
0
0
0.3075111
0.3075111
17 Xe
0.9048707
0.9048707
0.9048707
0.9048707
Differential equations
1 d(Ta)/d(V) = Ua*(T-Ta)/m/Cpc
2 d(X)/d(V) = -ra/Fa0
3 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cpo/Fa0*0
Explicit equations
1
Cpc = 28
2
m = 500
3
Ua = 5000
4
Ca0 = 1.86
5
Fa0 = 0.9*163*.1
6
deltaH = -34500
7
k = 31.1*exp((7906)*(T-360)/(T*360))
8
Kc = 3.03*exp((deltaH/8.314)*((T-333)/(T*333)))
9
Xe = Kc/(1+Kc)
10 ra = -k*Ca0*(1-(1+1/Kc)*X)
11 Cpo = 159
12 rate = -ra
13 Qr = Ua*(T-Ta)
12-11
P12-1 (b) Continued
Plot of Ta versus V to maintain isothermal operation
Plot of Qr versus V to maintain isothermal operation
(x)
In counter current operation, the entry point of the reactants is the exit for the coolant. Clearly, at
the entry of coolant (V = 5m3) highest energy releases are seen (high rates) and hence the
temperature of the coolant is increasing in the backward direction. Refer to (ii) of P12-1(b)
(xi) Increasing the initial temperature of the coolant can lead to a huge spike in reaction rates and
temperatures. Conversion approaches equilibrium conversion faster too.
The value of heat transfer coefficient determines the position of the rate/temperature spike. If
low heat transfer coefficient is used, the reactor temperature will be higher than the coolant
temperature throughout the reactor.
(xii) As ya0 is decreased (i.e., the initial concentration of reactants is decreased by adding inerts) the
reaction rate will decrease naturally, which results in low heat releases and hence the temperatures
will reduce. This is also reflected in the way conversion profiles behave.
(xiii) With increase in ya0, the initial concentration of reactants is high which results in higher rates and
hence more heat release. The coolant can not cool the system with high rates of energy release
and hence increase in Qr is observed.
Decrease in coolant flow rate will decrease Qr and this spirals to further increase in Qg. For the
intersection of Qg, Qr to occur more time will be required.
12-12
(xiv) 1. While increasing the initial temperature of the coolant can result in achieving equilibrium
conversion faster, it may lead to a runaway reaction because of the higher temperatures the
reaction touches due to the spike in reaction rates.
2. Coolant flow rate can be used to control the temperature rises in the system.
(xv) As the magnitude of heat of reaction increases, it is found that the point where X and Xe curves
come close together decreases. This can be explained from the Van’t Hoff equation. As the
enthalpy of the reaction becomes more exothermic, the equilibrium conversion is expected to
drop.
(xvi) As Fao is increased, both T and Ta increases
(xvii) Individualized solution
(xviii) Ua can be varied so as to have the desired conversion and keep the temperature under 370 K.
See the following Polymath program
Polymath code:
Cpc = 28
m = 500
d(Ta) / d(V) = -Ua*(T-Ta)/m/Cpc
Ua = 43000
Ta(0) = 340.3
d(X)/d(V) = -ra/Fa0 #
d(T)/d(V) = (Qg-Qr)/(Cpo*Fa0) #
Qg = ra*deltaH #
Qr = Ua*(T-Ta) #
Ca0 = 1.86 #
Fa0 = 0.9*163*.1 #
Kc = 3.03*exp((deltaH/8.314)*((T-333)/(T*333))) #
k = 31.1*exp((7906)*(T-360)/(T*360)) #
Xe = Kc/(1+Kc) #
ra = -k*Ca0*(1-(1+1/Kc)*X) #
deltaH = -34500 #
Cpo = 159
rate = -ra #
V(0)=0
X(0)=0
T(0)= 305
V(f)= 5
POLYMATH Report
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca0
1.86
1.86
1.86
1.86
2
Cpc
28.
28.
28.
28.
3
Cpo
159.
159.
159.
159.
4
deltaH
-3.45E+04
-3.45E+04
-3.45E+04
-3.45E+04
5
Fa0
14.67
14.67
14.67
14.67
6
k
0.5927441
0.5927441
54.76627
1.362998
7
Kc
9.512006
0.884224
9.512006
6.144184
8
m
500.
500.
500.
500.
9
ra
-1.102504
-43.8873
-0.319219
-0.319219
10 rate
1.102504
0.319219
43.8873
0.319219
11 T
305.
305.
369.5214
315.1228
12 Ta
340.3
314.8105
344.917
314.8105
13 Ua
4.3E+04
4.3E+04
4.3E+04
4.3E+04
14 V
0
0
5.
5.
15 X
0
0
0.751735
0.751735
16 Xe
0.9048707
0.4692775
0.9048707
0.860026
12-13
Differential equations
1 d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpc
2 d(X)/d(V) = -ra/Fa0
3 d(T)/d(V) = ((ra*deltaH)-Ua*(T-Ta))/Cpo/Fa0
Explicit equations
1
Cpc = 28
2
m = 500
3
Ua = 43000
4
Ca0 = 1.86
5
Fa0 = 0.9*163*.1
6
deltaH = -34500
7
k = 31.1*exp((7906)*(T-360)/(T*360))
8
Kc = 3.03*exp((deltaH/8.314)*((T-333)/(T*333)))
9
Xe = Kc/(1+Kc)
10 ra = -k*Ca0*(1-(1+1/Kc)*X)
11 Cpo = 159
12 rate = -ra
In the above report, it can be seen that Ua is the only changed parameter to get a conversion that is
greater than 75% and the temperature is maintained under 370 K.
(xix) As ya0 is decreased (i.e., the initial concentration of reactants is decreased by adding inerts) the
reaction rate will decrease naturally, which results in low heat releases and hence the
temperatures will reduce. This is also reflected in the way conversion profiles behave: as
equilibrium conversion depends on temperature and conversion depends on reaction rates and
temperature.
Increase in activation energy of the reaction shifts the peak of the rate/temperature spike.
(xx) Decreasing Ua will result in high temperatures throughout the reactor, this can be used in our
favor if high conversions are desired. Decreasing Ta will result in low temperatures in the reactor
and the spike in rate/temperature profile also vanishes as Qr is able to compensate Qg. However,
low temperatures lead to low rates and hence low conversion.
(xxi) As the heat of the reaction is decreased, the amount of heat released decreases resulting in lower
temperature rises. Lower temperature means low rates and low conversions.
(xxii) Parameters like ya0, Ua, Ta etc., can be used to our advantage to control the reaction. If there is a
possibility of runaway reaction or temperatures being too high for the reactor to handle these
external parameters can be controlled to either slow the reaction or to achieve high conversions.
(xxiii) As ya0 is decreased, the reaction rate is decreased due to low concentrations of reactants. This
means low saturation temperature in the reactor leading to high equilibrium conversion.
(xxiv) 1. The adiabatic operation gives the least conversion as compared to other heat exchanges
2. Equilibrium conversion is always achieved in the adiabatic operation.
3. High conversions can be achieved by lowering the initial concentration of reactants.
P12-1 (c)
(i)
In an endothermic reaction, heat is absorbed. As a result as the reaction proceeds the
temperatures and concentrations are lowered leading to lower reaction rates.
(ii) The reaction dies out near the reactor entrance when varying the heat capacity of A.
(iii) The heat capacity of A is the variable which most drastically changes the profiles.
(iv) It increases with an increase in heat capacity of A.
(v) Heat transfer coefficient could have changed.
(vi) In counter current operation, the entry of the coolant is the exit for the reactants. So, at the exit of
reactants the coolant is able to provide heat for the reaction thus increasing the rates and
12-14
temperatures. That is why, a dip in the profiles of reaction rate and temperature is seen. Also the
nature rate of increase in conversion also changes midway.
(vii) The const ant Ta type heat exchange achieves the highest conversion and the counter current
operation seems to achieve the lowest conversion.
(viii) Individualized solution
(ix) The terms Qg = ra*dHRx and Qr = Ua*(T-Ta) are added in the Polymath code for each of the four
cases.
See the following Polymath program for co-current case
Polymath code:
d(X)/d(V) = -ra/Fao #
d(T)/d(V) = (Qg-Qr)/(Fao*(Cpa+X*delCp)) #
d(Ta)/d(V) = Ua*(T-Ta)/(mc*Cpc) #
Qg = ra*deltaH #
Qr = Ua*(T-Ta) #
Fao = .0376 #
Cpa = 163 #
delCp = -9 #
Cao = 18.8 #
To = 1035 #
deltaH = 80770+delCp*(T-298) #
ra = -Cao*3.58*exp(34222*(1/To-1/T))*(1-X)*(To/T)/(1+X)
Ua = 16500 #
mc = .111 #
Cpc = 34.5 #
V(0)=0
X(0)=0
T(0)=1035
Ta(0)=1250
V(f)=1e-3
rate = -ra
Case1: adiabatic operation
Take Qr=0 in the code given above
12-15
#
P12-1 (c) Continued
Case 2: For constant heat exchange conditions
d(Ta)/d(V)=0
Case 3: Co-current heat exchange
Case 4 : counter current heat exchange :
We need to guess a value of Ta such that at exit Ta = 1250K
If we take Ta(0) = 995.15K then this can be done.
12-16
Add a negative sign to the expression of d(T)/d(V)
(x)
(xi)
Substitute, the volume V= 5 m3 in the Polymath code and get the results.
Plot of Qg for all 4 cases against volume
Plot of Qr versus volume for all cases
12-17
Plot of -ra (rate) versus V
It can be seen that the rate for constant heat exchange fluid temperature Ta, is higher than the rest of
cases because the difference between heat generated and heat removed in this case is highest.
The rate of reaction for all cases is decreasing because the temperature of the system is decreasing with
volume.
The rate of reaction for counter-current heat exchanger system is a U shaped curve plotted against
volume. At the front of the reactor, the reaction takes place very rapidly, drawing energy from the
sensible heat of the gas causing the gas temperature to drop because the heat exchanger cannot supply
energy at an equal or greater rate. This drop in temperature, coupled with the consumption of
reactants, slows the reaction rate as we move down the reactor.
(xii) Introduction of inert will introduce a change in energy balance equation and the value of Ѳ1 as well.
(xiii)
(1) ѲI = 0
All the plots will remain as for part (vi)
(2) ѲI = 1.5
Instead of the energy equation which was used previously
the equation will change.
Now we have to change the value of
Now the value will be ∑Cp = ѲI*CpI + CpA = 1.5×50 + 1×163 = 238
Value of Ѳ will change as well
Ѳ=
=
(3) Ѳ I = 3
∑Cp = ѲI*CpI + CpA = 3×50 + 1×163 = 313J/molA/K
Ѳ=
=
12-18
Incorporating these changes in the code and plotting X versus V for different cases.
The analysis is as follows:
Case I: adiabatic operation:
Case2: Constant heat exchange fluid temperature Ta
Case 3: Co-current heat exchange
The Polymath program for reference is for the case of co-current heat exchange with ѲI =3.
By changing values of ∑Cp and ε variables as shown we can change the program for various cases and
sub cases.
12-19
P12-1 (c) Continued
Case 4: Countercurrent heat exchange
Remarks: Thus, we can see that for all cases when inert gas concentration is more, then the reaction
proceeds faster but then the overall yield is less as well. In the case of adiabatic operation this
phenomenon is very significant. In case of constant heat exchange fluid temperature, the effect of inert
gas is negligible.
(xiv) Here we will change the Polymath program as entered in P12-2(c) part (vi).
The Ta value will be changed and the program will be tested for following values of Ta.
1000K, 1175K, 1350K
Case 1: adiabatic conditions
Case 2: Constant heat exchange fluid temperature Ta
12-20
Case 3: Co-current conditions
Case 4: Countercurrent conditions
We need to enter Ta (V =0) values such that at V=Vf, Taf = 1000K, 1175K and 1350K respectively
Ta (V=0) (K)
Ta(V=Vf) (K)
983.75
992
999
1000
1175
1350
P12-1 (d) Aspen problem
No solution will be provided
P12-1 (e)
(
(i)
There are at least two solutions for 16415 𝑅 < ) < 16500 𝑅
(ii) The conversion is 0.8 at T = 605 R
(iii) 525 K < T0 < 542 K
If the flow rate of methanol were increased by a factor of 4, the new operating range is:
527 K < T0 < 558 K
(iv) At V = 34.15 m3, tangency is observed at T = 560 K and intersection is observed at T = 610 K
(v) Individualized solution
12-21
P12-1 (f)
(i)
Increase in activation energy decreases the conversion obtained from mole balance, but has no
effect on the conversion obtained from energy balance. Increase in enthalpy of the reaction
decreases the conversion obtained from energy balance, but has no effect on the conversion
obtained from mass balance.
(ii) The feed temperature, reactor volume, water flow rate affect the intersection the most.
(iii) A set of values where 80% conversion is achieved while maintaining the temperature below 125F
is: E = 30,000 Btu/lb mol R; Enthalpy of reaction = -20,000 Btu/lb mol
(iv) Individual answer
(v) See the following Polymath program for tau=0.01. From the polymath report, it can be seen that
conversion is very small i.e. X=0.011. Hence tau=0.01 is not a realistic number
f(X)=X-(403.3*(T-535)+92.9*(T-545))/(36400+7*(T-528))
f(T)=X-tau*k/(1+tau*k)
tau=0.01
A=16.96*10^12
E=32400
R=1.987
k=A*exp(-E/(R*T))
X(0)=0.367
T(0)=564
POLYMATH Report
Nonlinear Equations
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 T
537.7053
2 X
0.0113353 1.369E-10 0.367
-8.515E-10 564.
Variable Value
1 A
1.696E+13
2 E
3.24E+04
3 k
1.146528
4 R
1.987
5 tau
0.01
Nonlinear equations
1 f(X) = X-(403.3*(T-535)+92.9*(T-545))/(36400+7*(T-528)) = 0
2 f(T) = X-tau*k/(1+tau*k) = 0
Explicit equations
1 tau = 0.01
2 A = 16.96*10^12
3 E = 32400
4 R = 1.987
5 k = A*exp(-E/(R*T))
If vo is to be decreased by a factor of 100, then tau will increase by a factor of 100.To decrease k by
factor of 100, let’s decrease A by 100. From the polymath output it can be seen that conversion remains
same as base case. This is because increasing tau and decreasing k cancels each other’s effect
12-22
Polymath code:
f(X)=X-(403.3*(T-535)+92.9*(T-545))/(36400+7*(T-528))
f(T)=X-tau*k/(1+tau*k)
tau=12.29
A=16.96*10^10
E=32400
R=1.987
k=A*exp(-E/(R*T))
X(0)=0.367
T(0)=564
POLYMATH Report
Nonlinear Equations
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 T
563.7289
-9.319E-10 500.
2 X
0.3636087 3.864E-11 0.5
Variable Value
1 A
1.696E+11
2 E
3.24E+04
3 k
0.0464898
4 R
1.987
5 tau
12.29
Nonlinear equations
1 f(X) = X-(403.3*(T-535)+92.9*(T-545))/(36400+7*(T-528)) = 0
2 f(T) = X-tau*k/(1+tau*k) = 0
Explicit equations
1 tau = 12.29
2 A = 16.96*10^10
3 E = 32400
4 R = 1.987
5 k = A*exp(-E/(R*T))
(vi) See the following Polymath program. Here we have fooled polymath to pot XEB and XMB as a
function of temperature
Polymath code
d(T) / d(t) = 1
T(0) = 400
XEB=(403.3*(T-535)+92.9*(T-545))/(36400+7*(T-528))
XMB=tau*k/(1+tau*k)
tau=0.1229
A=16.96*10^12
E=32400
R=1.987
k=A*exp(-E/(R*T))
t(0) = 0
t(f) = 300
12-23
P12-1 (f) Continued
Polymath Output
(vii) See the following Polymath program with change CP = 29 and ΔH = -38700.
Assume sumthetaICpI remains same
Polymath code:
f(X)=X-(403.3*(T-535)+92.9*(T-545))/(38700-29*(T-528))
f(T)=X-tau*k/(1+tau*k)
tau=0.1229
A=16.96*10^12
E=32400
R=1.987
k=A*exp(-E/(R*T))
X(0)=0.367
T(0)=564
POLYMATH Report
Nonlinear Equations
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 T
571.1234 -2.125E-11 564.
2 X
0.453823 -3.237E-11 0.367
Variable Value
1 A
1.696E+13
2 E
3.24E+04
3 k
6.760849
4 R
1.987
5 tau
0.1229
Nonlinear equations
1 f(X) = X-(403.3*(T-535)+92.9*(T-545))/(38700-29*(T-528)) = 0
2 f(T) = X-tau*k/(1+tau*k) = 0
Explicit equations
1 tau = 0.1229
2 A = 16.96*10^12
3 E = 32400
4 R = 1.987
5 k = A*exp(-E/(R*T))
12-24
(viii) Modify the polymath code to incorporate area as shown. Change the area in the above program
to get the values of conversion corresponding to the value of the area.
P12-1 (g)
(i)
Increasing k1a0, will make the first reaction more favorable and the molar flow rate of C will be
very low as k2 is not that high. Increasing both the rate constants will lead to exhaustion of A and
saturation is reached in the reactor very soon.
(ii) 0.178 mol/L
(iii) Increasing the heat transfer coefficient results in lowering of the temperatures at the end of the
reactor. This affects the highly exothermic reaction ‘2’ and its rate constant decreases more than
the what decrease is seen in the rate constant of first reaction. Therefore, the selectivity of C
decreases
(iv) Reducing Ua can lead to very high selectivity of C with conversion of A being 1.
(v) Individualized answer
(vi) At the start of reaction, Qg>Qr, so temp increases and then when reactant is virtually consumed,
Qr> Qg. So, there is a maximum in temperature.
See the following Polymath program with pressure drop included
Polymath code:
d(Fa)/d(V) = r1a+r2a #
d(Fb)/d(V) = -r1a #
d(Fc)/d(V) = -r2a/2 #
d(T)/d(V) = (Qg-Qr)/(90*Fa+90*Fb+180*Fc) #
d(p)/d(V) = -alpha/(2*p)*(Ft/Fto)*(T/To)
Qr = 4000*(T-373)#
Qg = (-r1a)*(-deltaH1)+(-r2a)*(-deltaH2) #
k1a = 10*exp(4000*(1/300-1/T)) #
k2a = 0.09*exp(9000*(1/300-1/T)) #
Cto = 0.1 #
deltaH1 = -20000 #
deltaH2 = -60000 #
Ft = Fa+Fb+Fc #
To = 423 #
alpha = 1.05
Ca = Cto*(Fa/Ft)*(To/T)*p #
Cb = Cto*(Fb/Ft)*(To/T)*p #
Cc = Cto*(Fc/Ft)*(To/T)*p #
r1a = -k1a*Ca #
r2a = -k2a*Ca^2 #
V(0)=0
Fa(0)=100
Fb(0)=0
12-25
Fc(0)=0
Fto = 100
T(0)=423
V(f)=1
p(0)=1
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
1.05
1.05
1.05
1.05
2
Ca
0.1
1.306E-07
0.1
1.306E-07
3
Cb
0
0
0.0208092
8.93E-07
4
Cc
0
0
0.0038445
1.739E-07
5
Cto
0.1
0.1
0.1
0.1
6
Fa
100.
9.521981
100.
9.521981
7
Fb
0
0
65.11707
65.11707
8
Fc
0
0
12.68047
12.68047
9
Ft
100.
87.31953
100.
87.31953
10 Fto
100.
100.
100.
100.
11 k1a
482.8247
482.8247
1.753E+04
1.706E+04
12 k2a
553.0557
553.0557
1.79E+06
1.683E+06
13 r1a
-48.28247
-136.5345
-0.0022276
-0.0022276
14 r2a
-5.530557
-90.93151
-2.87E-08
-2.87E-08
15 T
423.
423.
682.1122
678.9481
16 To
423.
423.
423.
423.
17 V
0
0
0.8
0.8
18 p
1.
1.922E-05
1.
1.922E-05
Differential equations
1 d(Fa)/d(V) = r1a+r2a
2 d(Fb)/d(V) = -r1a
3 d(Fc)/d(V) = -r2a/2
4 d(T)/d(V) = (4000*(373-T)+(-r1a)*20000+(-r2a)*60000)/(90*Fa+90*Fb+180*Fc)
5 d(p)/d(V) = -alpha/(2*p)*(Ft/Fto)*(T/To)
Explicit equations
1 k1a = 10*exp(4000*(1/300-1/T))
2 k2a = 0.09*exp(9000*(1/300-1/T))
3 Cto = 0.1
4 Ft = Fa+Fb+Fc
5 To = 423
6 Ca = Cto*(Fa/Ft)*(To/T)*p
7 Cb = Cto*(Fb/Ft)*(To/T)*p
8 Cc = Cto*(Fc/Ft)*(To/T)*p
9 r1a = -k1a*Ca
10 r2a = -k2a*Ca^2
11 Fto = 100
12 alpha = 1.05
12-26
P12-1 (g) Continued
(vii) In the Polymath program of example 12-5, rate equation will be changed as –
r1a = -k1a*(Ca-Cb/Kc)
with
Kc= 10*exp((-20000/8.314)*(1/450-1/T))
Polymath Code
d(Fa)/d(V) = r1a+r2a #
d(Fb)/d(V) = -r1a #
d(Fc)/d(V) = -r2a/2 #
d(T)/d(V) = (Qg-Qr)/(90*Fa+90*Fb+180*Fc) #
Qr = 4000*(T-373)#
Qg = (-r1a)*(-deltaH1)+(-r2a)*(-deltaH2) #
k1a = 10*exp(4000*(1/300-1/T)) #
k2a = 0.09*exp(9000*(1/300-1/T)) #
Cto = 0.1 #
deltaH1 = -20000 #
deltaH2 = -60000 #
Ft = Fa+Fb+Fc #
To = 423 #
Ca = Cto*(Fa/Ft)*(To/T) #
Cb = Cto*(Fb/Ft)*(To/T) #
Cc = Cto*(Fc/Ft)*(To/T) #
r1a = -k1a*(Ca-Cb/Kc) #
r2a = -k2a*Ca^2 #
12-27
Kc=10*exp((deltaH1/8.314)*(1/450-1/T))
V(0)=0
Fa(0)=100
Fb(0)=0
Fc(0)=0
T(0)=423
V(f)=1
(viii) The selectivity would increase with an increase in Ua, and vice versa.
P12-1 (h)
(i)
A little above 430 K the reactor temperature will jump to the upper steady state.
(ii) 250 K < To < 350 K
(iii) In the temperature range shown on the plot: As U is increased from 10000, the number of solution
change from zero steady states to two steady states and then to three steady states and then to
one steady state.
(iv) Individualized solution
P12-1 (i)
(i)
Increasing the coolant flow rate tends to keep the coolant temperature steady. The profile of
reactor temperature does not change much but the temperature at the end the reactor tends to
decrease.
12-28
(ii)
(iii)
Decreasing Ua increases the selectivity of C wrt D for a larger part of the reactor.
Increase in heat transfer coefficient will lead to lower temperature rise and hence lower rates. So,
the conversions will be low which means low flow rates of C (products).
(iv) CpD has the least affect on the profiles. As D is not produced much it’s heat capacity does not
affect the profiles much.
(v) CT0 has to be varied.
(vi) E1 and E2 affect SC/D the most
(vii) Individualized solution
(viii) See the following Polymath program
Polymath code:
d(Fa)/d(V) = ra #
d(Fb)/d(V) = rb #
d(Fc)/d(V) = rc #
d(Fd)/d(V) = rd #
d(T)/d(V) = (Qg-Qr)/sumFiCpi #
d(Ta) / d(V) = Ua*(T-Ta)/m/Cpco
ra = r1a+r2a #
rb = r1b #
rc = r1c+r2c #
rd = r2d #
r1a = -k1a*Ca*Cb^2 #
r2c = -k2c*Ca^2*Cc^3 #
k1a = 40*exp ((E1/R)*(1/300-1/T)) #
Ca = Cto*(Fa/Ft)*(To/T)*p #l
Cb = Cto*(Fb/Ft)*(To/T)*p #
Cc = Cto*(Fc/Ft)*(To/T)*p #
k2c = 2*exp((E2/R)*(1/300-1/T)) #
Cpco=10
m = 50
Ta(0) = 325
r1b = 2*r1a #
r1c = -r1a #
r2a = 2/3*r2c #
r2d = -1/3*r2c #
E1 = 8000 #
E2 = 12000 #
DH1b = -15000 #
DH2a = -10000 #
p=1#
sumFiCpi = Cpa*Fa+Cpb*Fb+Cpc*Fc+Cpd*Fd #
Cpa = 10 #
Cpb = 12 #
Cpc = 14 #
Cpd = 16 #
To = 300 #
Ua = 80 #
Cto = 0.2 #
Ft = Fa+Fb+Fc+Fd #
R = 1.987 #
Qg = r1b*DH1b+r2a*DH2a #
Qr = Ua*(T-Ta) #
V(0)=0
Fa(0)=5
Fb(0)=10
Fc(0)=0
Fd(0)=0
T(0)=300
V(f)=10
SCD=if(V>0.0001) then (Fc/Fd) else (0)
12-29
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.0666667
0.0077046
0.0666667
0.0077046
2
Cb
0.1333333
0.0156839
0.1333333
0.0156981
3
Cc
0
0
0.0909427
0.0909427
4
Cpa
10.
10.
10.
10.
5
Cpb
12.
12.
12.
12.
6
Cpc
14.
14.
14.
14.
7
Cpco
10.
10.
10.
10.
8
Cpd
16.
16.
16.
16.
9
Cto
0.2
0.2
0.2
0.2
10 DH1b
-1.5E+04
-1.5E+04
-1.5E+04
-1.5E+04
11 DH2a
-10000.
-10000.
-10000.
-10000.
12 E1
8000.
8000.
8000.
8000.
13 E2
1.2E+04
1.2E+04
1.2E+04
1.2E+04
14 Fa
5.
0.3890865
5.
0.3890865
15 Fb
10.
0.7927648
10.
0.7927648
16 Fc
0
0
4.592674
4.592674
17 Fd
0
0
0.003648
0.003648
18 Ft
15.
5.778173
15.
5.778173
19 k1a
40.
40.
2.861E+05
1.248E+04
20 k2c
2.
2.
1.21E+06
1.102E+04
21 m
50.
50.
50.
50.
22 p
1.
1.
1.
1.
23 Qg
1422.222
714.0015
9.589E+04
714.0015
24 Qr
-2000.
-2000.
3.863E+04
1450.125
25 R
1.987
1.987
1.987
1.987
26 r1a
-0.0474074
-3.196187
-0.0236907
-0.0236907
27 r1b
-0.0948148
-6.392374
-0.0473814
-0.0473814
28 r1c
0.0474074
0.0236907
3.196187
0.0236907
29 r2a
0
-0.0043154
0
-0.000328
30 r2c
0
-0.0064731
0
-0.000492
31 r2d
0
0
0.0021577
0.000164
32 ra
-0.0474074
-3.196833
-0.0240187
-0.0240187
33 rb
-0.0948148
-6.392374
-0.0473814
-0.0473814
34 rc
0.0474074
0.0231987
3.195219
0.0231987
35 rd
0
0
0.0021577
0.000164
36 SCD
0
0
1.277E+13
1258.968
37 sumFiCpi 170.
77.75984
170.
77.75984
38 T
300.
300.
885.7738
524.395
39 Ta
325.
322.6384
506.2685
506.2685
40 To
300.
300.
300.
300.
41 Ua
80.
80.
80.
80.
42 V
0
0
10.
10.
Differential equations
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(T)/d(V) = (Qg-Qr)/sumFiCpi
6 d(Ta)/d(V) = Ua*(T-Ta)/m/Cpco
12-30
Explicit equations
1 E2 = 12000
2 p=1
3 R = 1.987
4 Ft = Fa+Fb+Fc+Fd
5 To = 300
6 k2c = 2*exp((E2/R)*(1/300-1/T))
7 E1 = 8000
8 Cto = 0.2
9 Ca = Cto*(Fa/Ft)*(To/T)*p
10 Cc = Cto*(Fc/Ft)*(To/T)*p
11 r2c = -k2c*Ca^2*Cc^3
12 Cpco = 10
13 m = 50
14 Cb = Cto*(Fb/Ft)*(To/T)*p
15 k1a = 40*exp ((E1/R)*(1/300-1/T))
16 r1a = -k1a*Ca*Cb^2
17 r1b = 2*r1a
18 rb = r1b
19 r2a = 2/3*r2c
20 DH1b = -15000
21 DH2a = -10000
22 r1c = -r1a
23 Cpd = 16
24 Cpa = 10
25 Cpb = 12
26 Cpc = 14
27 sumFiCpi = Cpa*Fa+Cpb*Fb+Cpc*Fc+Cpd*Fd
28 rc = r1c+r2c
29 Ua = 80
30 r2d = -1/3*r2c
31 ra = r1a+r2a
32 rd = r2d
33 Qg = r1b*DH1b+r2a*DH2a
34 Qr = Ua*(T-Ta)
35 SCD = if(V>0.0001) then (Fc/Fd) else (0)
12-31
Addition of Inerts:
The adding of inerts will decrease the peak in the reactor temperature. By trial and error, an inert flow
rate of 5.25 mol/min is seen to be sufficient to keep the reactor temperature below 700K.
Polymath code:
d(Fa)/d(V) = ra #
d(Fb)/d(V) = rb #
d(Fc)/d(V) = rc #
d(Fd)/d(V) = rd #
d(T)/d(V) = (Qg-Qr)/sumFiCpi #
d(Ta) / d(V) = Ua*(T-Ta)/m/Cpco
E2 = 12000
p=1
R = 1.987
Ft = Fa + Fb+ Fc+Fd
To = 300
k2c = 2*exp((E2/R)*(1/300-1/T))
E1 = 8000
Cto = 0.2
Ca = Cto*(Fa/Ft)*(To/T)*p
Cb = Cto*(Fb/Ft)*(To/T)*p
Cc = Cto*(Fc/Ft)*(To/T)*p
r2c = -k2c*Ca^2*Cc^3
Cpco = 10
m = 50
k1a = 40*exp((E1/R)*(1/300-1/T))
r1a = -k1a*Ca*Cb^2
r1b = 2*r1a
rb = r1b
r2a = r2c
DH1b = -15000
DH2a = -10000
r1c = -r1a
Fi = 5.25
Cpd = 16
Cpa = 10
Cpb = 12
Cpc = 14
Cpi = 10
sumFiCpi = Cpa*Fa+Cpb*Fb+Cpc*Fc+Cpd*Fd+Cpi*Fi
rc = r1c+r2c
Ua = 80
r2d = -2*r2c
ra = r1a+r2a
rd = r2d
Qg = r1b*DH1b+r2a*DH2a
Qr = Ua*(T-Ta)
V(0) = 0
V(f) = 10
Fa(0) =5
Fb(0) =10
Fc(0) =0
Fd(0) =0
T(0) = 300
Ta(0) = 325
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.0666667
0.0091668
0.0666667
0.0091668
2
Cb
0.1333333
0.0185507
0.1333333
0.0185507
3
Cc
0
0
0.0729412
0.0729412
4
Cpa
10.
10.
10.
10.
5
Cpb
12.
12.
12.
12.
12-32
6
Cpc
14.
14.
14.
14.
7
Cpco
10.
10.
10.
10.
8
Cpd
16.
16.
16.
16.
9
Cpi
10.
10.
10.
10.
10 Cto
0.2
0.2
0.2
0.2
11 DH1b
-1.5E+04
-1.5E+04
-1.5E+04
-1.5E+04
12 DH2a
-10000.
-10000.
-10000.
-10000.
13 E1
8000.
8000.
8000.
8000.
14 E2
1.2E+04
1.2E+04
1.2E+04
1.2E+04
15 Fa
5.
0.5567431
5.
0.5567431
16 Fb
10.
1.126675
10.
1.126675
17 Fc
0
0
4.430068
4.430068
18 Fd
0
0
0.0131891
0.0131891
19 Fi
5.25
5.25
5.25
5.25
20 Ft
15.
6.126675
15.
6.126675
21 k1a
40.
40.
8.474E+04
3.096E+04
22 k2c
2.
2.
1.95E+05
4.306E+04
23 m
50.
50.
50.
50.
24 p
1.
1.
1.
1.
25 Qg
1422.222
1422.222
5.252E+04
2943.824
26 Qr
-2000.
-2000.
2.341E+04
9854.971
27 R
1.987
1.987
1.987
1.987
28 r1a
-0.0474074
-1.750429
-0.0474074
-0.0976594
29 r1b
-0.0948148
-3.500858
-0.0948148
-0.1953188
30 r1c
0.0474074
0.0474074
1.750429
0.0976594
31 r2a
0
-0.0034069
0
-0.0014043
32 r2c
0
-0.0034069
0
-0.0014043
33 r2d
0
0
0.0068138
0.0028086
34 ra
-0.0474074
-1.75115
-0.0474074
-0.0990637
35 rb
-0.0948148
-3.500858
-0.0948148
-0.1953188
36 rc
0.0474074
0.0474074
1.749708
0.0962551
37 rd
0
0
0.0068138
0.0028086
38 sumFiCpi 222.5
133.8195
222.5
133.8195
39 T
300.
300.
698.732
594.7904
40 Ta
325.
321.9934
471.6033
471.6033
41 To
300.
300.
300.
300.
42 Ua
80.
80.
80.
80.
43 V
0
0
10.
10.
Differential equations
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(T)/d(V) = (Qg-Qr)/sumFiCpi
6 d(Ta)/d(V) = Ua*(T-Ta)/m/Cpco
Explicit equations
1 E2 = 12000
2 p=1
3 R = 1.987
4 Ft = Fa + Fb+ Fc+Fd
12-33
5 To = 300
6 k2c = 2*exp((E2/R)*(1/300-1/T))
7 E1 = 8000
8 Cto = 0.2
9 Ca = Cto*(Fa/Ft)*(To/T)*p
10 Cb = Cto*(Fb/Ft)*(To/T)*p
11 Cc = Cto*(Fc/Ft)*(To/T)*p
12 r2c = -k2c*Ca^2*Cc^3
13 Cpco = 10
14 m = 50
15 k1a = 40*exp((E1/R)*(1/300-1/T))
16 r1a = -k1a*Ca*Cb^2
17 r1b = 2*r1a
18 rb = r1b
19 r2a = r2c
20 DH1b = -15000
21 DH2a = -10000
22 r1c = -r1a
23 Fi = 5.25
24 Cpd = 16
25 Cpa = 10
26 Cpb = 12
27 Cpc = 14
28 Cpi = 10
29 sumFiCpi = Cpa*Fa+Cpb*Fb+Cpc*Fc+Cpd*Fd+Cpi*Fi
30 rc = r1c+r2c
31 Ua = 80
32 r2d = -2*r2c
33 ra = r1a+r2a
34 rd = r2d
35 Qg = r1b*DH1b+r2a*DH2a
36 Qr = Ua*(T-Ta)
(ix)
(x)
The only way to make more of species D is to increase E2.
Use the polymath code of LEP 12-7 for different heat exchanger operation to carry out the analysis
12-34
(xi)
The molar flow rate of C does not go through a maximum because the activation energy of the
second reaction is higher than that of the first reaction, and the specific rate constant is higher for
the first reaction. As a result, the reactor volume is not large enough for the flow rate of C to go
through a maximum. For the case of co-current heat exchange, the molar flow rate of C goes
through a maximum when the inlet flow rate of A and the inlet temperature are doubled, and the
heat transfer coefficient is halved. Conditions for other modes of heat exchange can be found
similarly, by varying these parameters.
(xii) Include pressure drop in the polymath code as shown:
Refer to explicit equations 2 and 9, and vary alpha_rho in equation 2 to get the results.
Polymath code:
d(Fa)/d(V) = ra #
d(Fb)/d(V) = rb #
d(Fc)/d(V) = rc #
d(Fd)/d(V) = rd #
d(T)/d(V) = (Qg-Qr)/sumFiCpi #
d(Ta) / d(V) = Ua*(T-Ta)/m/Cpco
E2 = 12000
aplha_rho = 0.05
R = 1.987
k2c = 2*exp((E2/R)*(1/300-1/T))
Cto = 0.2
To = 300
E1 = 8000
Ft = Fa + Fb+ Fc+Fd
p = (1-aplha_rho*V)^0.5
Ca = Cto*(Fa/Ft)*(To/T)*p
k1a = 40*exp((E1/R)*(1/300-1/T))
Cpco = 10
m = 50
Cb = Cto*(Fb/Ft)*(To/T)*p
Cc = Cto*(Fc/Ft)*(To/T)*p
r1a = -k1a*Ca*Cb^2
r2c = -k2c*Ca^2*Cc^3
r1b = 2*r1a
rb = r1b
r2a =2/3* r2c
DH1b = -15000
DH2a = -10000
Cpd = 16
Cpa = 10
Cpb = 12
Cpc = 14
r1c = -r1a
sumFiCpi = Cpa*Fa+Cpb*Fb+Cpc*Fc+Cpd*Fd
Ua = 80
r2d = -1/3*r2c
ra = r1a+r2a
Qg = r1b*DH1b+r2a*DH2a
Qr = Ua*(T-Ta)
rc = r1c+r2c
rd = r2d
V(0) = 0
V(f) = 10
Fa(0) =5
Fb(0) =10
Fc(0) =0
Fd(0) =0
T(0) = 300
Ta(0) = 325
12-35
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
aplha_rho 0.05
0.05
0.05
0.05
2
Ca
0.0666667
0.0169454
0.0666667
0.0169454
3
Cb
0.1333333
0.0338918
0.1333333
0.0338918
4
Cc
0
0
0.0229633
0.0229633
5
Cpa
10.
10.
10.
10.
6
Cpb
12.
12.
12.
12.
7
Cpc
14.
14.
14.
14.
8
Cpco
10.
10.
10.
10.
9
Cpd
16.
16.
16.
16.
10 Cto
0.2
0.2
0.2
0.2
11 DH1b
-1.5E+04
-1.5E+04
-1.5E+04
-1.5E+04
12 DH2a
-10000.
-10000.
-10000.
-10000.
13 E1
8000.
8000.
8000.
8000.
14 E2
1.2E+04
1.2E+04
1.2E+04
1.2E+04
15 Fa
5.
2.122944
5.
2.122944
16 Fb
10.
4.246027
10.
4.246027
17 Fc
0
0
2.876882
2.876882
18 Fd
0
0
3.484E-05
3.484E-05
19 Ft
15.
9.245888
15.
9.245888
20 k1a
40.
40.
2.449E+04
2.449E+04
21 k2c
2.
2.
3.029E+04
3.029E+04
22 m
50.
50.
50.
50.
23 p
1.
0.7071068
1.
0.7071068
24 Qg
1422.222
1422.222
1.801E+04
1.43E+04
25 Qr
-2000.
-2000.
1.264E+04
1.255E+04
26 R
1.987
1.987
1.987
1.987
27 r1a
-0.0474074
-0.6003607
-0.0474074
-0.4766352
28 r1b
-0.0948148
-1.200721
-0.0948148
-0.9532704
29 r1c
0.0474074
0.0474074
0.6003607
0.4766352
30 r2a
0
-7.022E-05
0
-7.022E-05
31 r2c
0
-0.0001053
0
-0.0001053
32 r2d
0
0
3.511E-05
3.511E-05
33 ra
-0.0474074
-0.6003886
-0.0474074
-0.4767054
34 rb
-0.0948148
-1.200721
-0.0948148
-0.9532704
35 rc
0.0474074
0.0474074
0.6003189
0.4765299
36 rd
0
0
3.511E-05
3.511E-05
37 sumFiCpi 170.
112.4587
170.
112.4587
38 T
300.
300.
574.8775
574.8775
39 Ta
325.
322.5903
417.9689
417.9689
40 To
300.
300.
300.
300.
41 Ua
80.
80.
80.
80.
42 V
0
0
10.
10.
Differential equations
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(T)/d(V) = (Qg-Qr)/sumFiCpi
6 d(Ta)/d(V) = Ua*(T-Ta)/m/Cpco
12-36
Explicit equations
1
E2 = 12000
2
aplha_rho = 0.05
3
R = 1.987
4
k2c = 2*exp((E2/R)*(1/300-1/T))
5
Cto = 0.2
6
To = 300
7
E1 = 8000
8
Ft = Fa + Fb+ Fc+Fd
9
p = (1-aplha_rho*V)^0.5
10 Ca = Cto*(Fa/Ft)*(To/T)*p
11 k1a = 40*exp((E1/R)*(1/300-1/T))
12 Cpco = 10
13 m = 50
14 Cb = Cto*(Fb/Ft)*(To/T)*p
15 Cc = Cto*(Fc/Ft)*(To/T)*p
16 r1a = -k1a*Ca*Cb^2
17 r2c = -k2c*Ca^2*Cc^3
18 r1b = 2*r1a
19 rb = r1b
20 r2a = 2/3* r2c
21 DH1b = -15000
22 DH2a = -10000
23 Cpd = 16
24 Cpa = 10
25 Cpb = 12
26 Cpc = 14
27 r1c = -r1a
28 sumFiCpi = Cpa*Fa+Cpb*Fb+Cpc*Fc+Cpd*Fd
29 Ua = 80
30 r2d = -1/3*r2c
31 ra = r1a+r2a
32 Qg = r1b*DH1b+r2a*DH2a
33 Qr = Ua*(T-Ta)
34 rc = r1c+r2c
35 rd = r2d
P12-1 (j) We note that the derivation remains the same till Equation (12-2)
+,-
= 𝑟1 = 𝛴𝑣14 (−𝑟4 )
+.
+8+.
= 𝐶:1
+;
+.
From equation (12-2),
+;
𝑈𝑎(𝑇? − 𝑇) = 𝛴1 𝛴4 𝑣14 @−𝑟4 A𝐻1 + 𝛴1 𝐹1 𝐶:1 +.
(1)
(As, 𝛴𝑣1 𝐻1 = 𝛥𝐻)
𝛴1 𝛴4 𝑣14 @−𝑟4 A𝐻1 = 𝛴4 @−𝑟4 A𝛴1 𝑣14 𝐻1 = 𝛴(−𝑟)(𝛥𝐻);
Substituting (2) in (1) and rearranging (1) gives,
+;
+.
=
F(GH)(GI8)GJ?(;G;K )
F,- LM-
which is equation (P12-35)
12-37
(2)
P12-1 (k)
(1) Reducing the particle diameter, will result in increase of the parameters in the pressure and
temperature equations.
(2) When pressure is doubled, the conversion achieved is decreasing from 0.9 to 0.6 and the
temperatures achieved are also decreasing.
(3) Increasing the initial temperature makes the temperature plot steeper but does not seem to affect
the output temperature (i.e., temperature at z = 20 ft). Varying the temperature almost has no
effect on the conversion and pressure.
Increasing the pressure results in lower temperatures in the reactor, this can be used to our benefit
when the reactor body materials cannot go to high temperatures.
P12-1 (l) Individualized solution
P12-2 (a)
(i)
Increasing inert flow decreases conversion as inert absorbs the heat. Increasing inert flow causes
slight decrease in temp as rate of reaction decreases with addition of inert flow
(ii) The reaction rate at inlet of reactor is very high i.e. 70 mol/m3.s. The required rate at V=0.001 is
0.1*70=7 mol/m3. s. The required temp is >1250 K
(iii) Individualized Solution
(iv) Minimum required inlet temp of heating fluid is 1290 K to have exit conversion 100%
(v) Minimum value of inlet concentration is 40 mol/m3 for exit conversion to be virtually 100%
(vi) Individualized Solution
P12-3 (a) FA0
P12-3 (b) CT0
P12-3 (c) Ua/ρ
P12-3 (d) EA
P12-3 (e) θI
P12-4
The key for decoding the algorithm to arrive at a numerical score for the Interaction Computer Games
(ICGs) is given at the front of this Solutions Manual.
P12-5
NH 4 NO3 () → 2H 2O( g) + N 2O( g)
A() → 2W ( g) + B( g)
From Rate Data
# 2.912 & E
k
E #T − T &
(50)
ln 2 = % 2 1 ( = ln%
(=
$ 0.307 ' R (970)(1020)
€ k1 R $ T2T1 '
12-38
€
E
= 44518 R
R
)E # 1
1 &,
k = 0.307exp+ %
− (.
* R $ 970 T 'Mole Balance
€
V=
€
FA 0 X
−rA
M
−rAV k V ⋅ V kM
X MB =
=
=
FA 0
FA 0
FA0
€
Energy Balance
€
FA 0 H A0 + FW 0 HW 0 − FA H A ( g) − FW HW ( g) − FB H B ( g) = 0
FA 0 H A0 + FA 0ΘW HW 0 − FA 0 (1− X ) H A ( g) − ( FA0ΘW + 2FA0 X ) HW ( g) − FA0 XH B ( g) = 0
H A ( g,T ) = H A (,T ) + ΔHVap
ΔH Rx = 2HW ( g) + H B ( g) − H A ()
ΔH Rx
%
(

'
H A 0 − H A ( g) + ΘW ( HW 0 − HW ( g)) − 2HW ( g) + H B ( g) − H A () − ΔHVap *X = 0
'
*
&
)
C
(T −660)
A 
P
(H A (,T ) − H A 0 ) + (1− X )ΔHVap + ΘW H S (500°F ) − HW (200°F ) + CPS (T − 500) = −ΔH Rx X
[
XE =
[
]
CP (T − 660) + ΘW H S (500°F ) − HW (200°F ) + CPS (T − 500)
−ΔH Rx
ΘW =
€
]
FW (0.17) (18)
=
= 0.9103
FA (0.83) (80)
CPA = 0.38
BTU 80 lb
BTU
×
= 30.4
lb°R
mol
lbmol°R
CPS = 0.47
BTU12-39
18 lb
BTU
×
= 8.46
lb°R mol
lbmol°R
ΘW =
FW (0.17) (18)
=
= 0.9103
FA (0.83) (80)
CPA = 0.38
BTU 80 lb
BTU
×
= 30.4
lb°R
mol
lbmol°R
CPS = 0.47
BTU 18 lb
BTU
×
= 8.46
lb°R mol
lbmol°R
ΔH Rx = −336
BTU 80 lb
BTU
×
= −26,880
lb
mol
lbmol
H (200°F ) = 168
BTU
BTU
= 2,916
lb
lbmol
HW (500°F ) = 1,202
BTU
BTU
= 21,636
lb
lbmol
Polymath code:
d(T)/d(t) = 1
€
M= 500
k = 0.307*exp(44518*(1/970-1/T))
thetaw = (0.17/18)/(0.83/80)
Cp = 0.38*80
Fao = 310*0.83
Xm = k*M/Fao
DHrx = -336*80
Xe = (Cp*(T-660)+thetaw*(0.47*18*(T-960)+1034*18))/(-DHrx)
t(0) = 0
t(f) = 15
T(0) = 975
12-40
P12-5 (a)
The blast should have been caused by excessive heat release as the conversion is virtually 100 %
P12-5 (b)
From the above graph, it can be seen that reaction temp prior to shutdown wad approx. 981 R and a
conversion of 99.9 %
P12-5 (c) Individualized solution
P12-5 (d) Individualized solution
P12-5 (e) Individualized solution
P12-6
A+ B → 2C
A
B
C
10
10
0.0
80
80
-
!
"
Btu
CPio #
$
% lb mole° F &
51
44
47.5
! lb "
MW , #
$
% lb mol &
128
94
222
" lb #
ρi , $ 3 %
& ft '
63
67.2
65
ΔH R = 20, 000
Btu
,
lb mol A
" lb − mole #
%
& hr '
Fio $
Tio(F)
~
12-41
Energy balance with work term included is:
Q − WS
− X A ΔH R = ∑ θ i C Pi [T − To ]
FA0
θ A = 1,θ B =
FB 0 10
= = 1, X AF = 1
FA0 10
Q = UA(Ts − T )
Substituting into energy balance,
UA(TS − T ) − WS − FA0 ΔH R X AF = FA0 !%C pA + C pB "& [T − T0 ]
{
}
⇒ UA(TS − T ) − WS − FA0 ΔH R = FA0 !%C pA + C pB "& + UA [T − T0 ]
T = T0 +
UA(TS − T ) − WS − FA0 ΔH R
FA0 !%C pA + C pB "& + UA
−Ws = 63525
Btu
hr
∴T = 199° F
P12-7 (a)
Since the feed is equimolar, CA0 = CB0 = 0 .1 mol/dm3
CA = CA0(1-X)
CB = CB0(1-X)
Adiabatic:
T = T0 +
X [−ΔH R (T0 )]
∑φiC P + X ΔC P
i
ΔC P = C pC − C pB − C pA = 30 −15 −15 = 0
ΔHR (T) = HC −HB −HA = - 41000 - (-15000 ) - (-20000) = -6000 cal/mol A
cal
∑θ C = C + θ C = 15 + 15 = 30 mol K
i
i
pA
T = 300 +
B
pB
6000 X
= 300 + 200 X
30
−rA = k C A2 0 (1 − X ) 2 = .01 k (1 − X ) 2
VPFR = FA0 ∫
VCSTR =
dX
−rA
FA0 X
−rA
FA0 = CA0v0 = (.1)(2) = 0.2 mols/dm3
k = .01*exp((10000 / 2) * (1 / 300 - 1 / T))
12-42
See the following Polymath program
f(T) = T-300-200*X
T(0) = 700
f(X) = X+V*ra/Fa0
X(0) = 0.99
V = 500
k = 0.01*exp((10000/2)*(1/300-1/T))
Fa0 = 0.2
Ca0 = 0.1
ra = -k*(Ca0^2)*((1-X)^2)
Calculated values of NLE variables
Variable Value
f(x)
Initial Guess
1 T
483.8314 -1.421E-13 700.
2 X
0.919157 -1.516E-10 0.99
Variable Value
1 Ca0
0.1
2 Fa0
0.2
3 k
5.625546
4 ra
-0.0003677
5 V
500.
Nonlinear equations
1 f(T) = T- 300 - 200 * X = 0
2 f(X) = X + V*ra/Fa0 = 0
Explicit equations
1 V = 500
2 k = 0.01*exp((10000 / 2) * (1 / 300 - 1 / T))
3 Fa0 = 0.2
4 Ca0 = 0.1
5 ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2)
For 500 dm3 CSTR, X = 0.92
For two 250 dm3 CSTR in series, X = 0.967
P12-7 (b) Constant heat exchanger temperature Ta
When heat exchanger is added, the energy balance can be written as
So, with
=0,
,
= -6000 cal/mol
Where Ua = 20 cal/m3/s/K, Ta = 450 K
See the following Polymath program :
d(X)/d(V) = -ra/Fa0
d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
k = 0.01*exp((10000/1.987)*(1/300-1/T))
Kc = 10*exp((-6000/1.987)*(1/450-1/T))
Fa0 = 0.2
Ca0 = 0.1
12-43
ra = -k*(Ca0^2)*((1-X)^2-X/Ca0/Kc)
Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
DH = -6000
Ua = 20
Ta = 450
Fao = 0.2
sumcp = 30
V(0) = 0
V(f) =10
X(0) = 0
T(0) = 300
Calculated values of the DEQ variables
Variable initial value minimal value maximal value
V
0
0
10
X
0
0
0.3634806
T
300
300
455.47973
K
0.01
0.01
3.068312
Kc
286.49665
9.2252861
286.49665
Fa0
0.2
0.2
0.2
Ca0
0.1
0.1
0.1
Ra
-1.0E-04
-0.0221893
-1.0E-04
Xe
0.8298116
0.3682217
0.8298116
DH
-6000
-6000
-6000
Ua
20
20
20
Ta
450
450
450
Fao
0.2
0.2
0.2
sumcp
30
30
30
final value
10
0.3634806
450.35437
2.7061663
9.9473377
0.2
0.1
-0.0010758
0.3810642
-6000
20
450
0.2
30
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
[8] Ua = 20
[9] Ta = 450
[10] Fao = 0.2
[11] sumcp = 30
12-44
P12-7 (b) continued
P12-7 (c)
For a co-current heat exchanger,
 = 50
CpC = 1cal/g/K, Ta1=450 K, m
g
sec
See the following Polymath program
d(X)/d(V) = -ra/Fa0
d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
k = 0.01*exp((10000/1.987)*(1/300-1/T))
Kc = 10*exp((-6000/1.987)*(1/450-1/T))
Fa0 = 0.2
Ca0 = 0.1
ra = -k*(Ca0^2)*((1-X)^2-X/Ca0/Kc)
Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
DH = -6000
Ua = 20
Fao = 0.2
sumcp = 30
mc = 50
Cpc = 1
V(0) = 0
V(f) =10
X(0) = 0
T(0) = 300
Ta(0) = 450
Calculated values of the DEQ variables
Variable initial value minimal value maximal value
V
0
0
10
X
0
0
0.3611538
T
300
300
442.15965
Ta
450
434.90618
450
K
0.01
0.01
2.1999223
Kc
286.49665
11.263546
286.49665
Fa0
0.2
0.2
0.2
Ca0
0.1
0.1
0.1
Ra
-1.0E-04
-0.0160802
-1.0E-04
Xe
0.8298116
0.4023362
0.8298116
DH
-6000
-6000
-6000
Ua
20
20
20
Fao
0.2
0.2
0.2
Sumcp
30
30
30
Mc
50
50
50
Cpc
1
1
1
final value
10
0.3611538
442.15965
441.60853
2.1999223
11.263546
0.2
0.1
-0.0019246
0.4023362
-6000
20
0.2
30
50
1
12-45
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
[3] d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
[8] Ua = 20
[9] Fao = 0.2
[10] sumcp = 30
[11] mc = 50
[12] Cpc = 1
Next increase the coolant flow rate and run the same program to compare results.
P12-7 (d)
For counter-current flow, swap (T – Ta) with (Ta-T) in dTa/dV equation in the Polymath program given in
part (c)
Calculated values of the DEQ variables
Variable initial value minimal value maximal value
V
0
0
10
X
0
0
0.3647241
T
300
300
463.44558
Ta
440.71
440.71
457.98124
K
0.01
0.01
3.7132516
final value
10
0.3647241
450.37724
450.00189
2.7077022
12-46
Kc
Fa0
Ca0
Ra
Xe
DH
Ua
Fao
Sumcp
Mc
Cpc
286.49665
0.2
0.1
-1.0E-04
0.8298116
-6000
20
0.2
30
50
1
8.2274817
0.2
0.1
-0.0256436
0.3488462
-6000
20
0.2
30
50
1
286.49665
0.2
0.1
-1.0E-04
0.8298116
-6000
20
0.2
30
50
1
9.9439517
0.2
0.1
-9.963E-04
0.381006
-6000
20
0.2
30
50
1
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
[3] d(Ta)/d(V) = Ua*(Ta-T)/mc/Cpc
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 1.987) * (1 / 300 - 1 / T))
[2] Kc = 10 * exp(-6000 / 1.987 * (1 / 450 - 1 / T))
[3] Fa0 = 0.2
[4] Ca0 = 0.1
[5] ra = -k * (Ca0 ^ 2) * ((1 - X) ^ 2 - X /Ca0/ Kc)
[6] Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
[7] DH = -6000
[8] Ua = 20
[9] Fao = 0.2
[10] sumcp = 30
[11] mc = 50
[12] Cpc = 1
12-47
P12-7 (e) Adiabatic
In the polymath code given in part (c), put Ua = 0
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca0
0.1
0.1
0.1
0.1
2
Cpc
1.
1.
1.
1.
3
DH
-6000.
-6000.
-6000.
-6000.
4
Fa0
0.2
0.2
0.2
0.2
5
k
0.01
0.01
0.010587
0.010587
6
Kc
286.4967
276.8576
286.4967
276.8576
7
mc
50.
50.
50.
50.
8
Qg
0.6
0.6
0.6286162
0.6286162
9
Qr
0
0
0
0
10 ra
-0.0001
-0.0001048
-0.0001
-0.0001048
11 sumCp
30.
30.
30.
30.
12 T
300.
300.
301.0235
301.0235
13 Ta
450.
450.
450.
450.
14 Ua
0
0
0
0
15 V
0
0
10.
10.
16 X
0
0
0.0051176
0.0051176
17 Xe
0.8298116
0.827152
0.8298116
0.827152
Differential equations
1 d(X)/d(V) = -ra/Fa0
2 d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fa0*sumCp)
3 d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
Explicit equations
1
Kc = 10*exp(-6000/1.987*(1/450-1/T))
2
k = 0.01*exp((10000/1.987)*(1/300-1/T))
3
Fa0 = 0.2
4
Ca0 = 0.1
5
ra = -k*(Ca0^2)*((1-X)^2-X/Ca0/Kc)
6
Xe = (2+1/Kc/Ca0-((2+1/Kc/Ca0)^2-4)^0.5)/2
7
DH = -6000
8
Ua = 20*0
9
sumCp = 30
10 mc = 50
11 Cpc = 1
12 Qg = ra*DH
13 Qr = Ua*(T-Ta)
12-48
12-49
P12-7 (f)
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.
P12-7 (g)
12-50
P12-8
Refer to solution P11-7 . We need to add energy balance equations:
Heat Exchange
(16B)
A. Constant Ta (17B) Ta = 300
Additional Parameters (18B – (20B): Ta,
, Ua,
B. Variable Ta Co-Current
(17C)
C. Variable Ta Counter Current
(18C)
Guess Ta at V = 0 to match Ta = Tao at exit, i.e., V = Vf
P12-8 (a) Variable Ta Co-Current
Polymath Code
d(X)/d(V)=-ra/Fa0
d(T)/d(V)=(Ua*(Ta-T)+DH*ra)/(Fa0*(sumcp+DCp*X))
d(Ta)/d(V)=(Ua*(T-Ta))/m/Cpc
d(p)/d(V)=-alpharho/p/2*(T/T0)
T0=300
Ca0=2.
Cb=Ca0*(X)*p*(T0/T)
E=10000
Ca=Ca0*(1-X)*p*(T0/T)
T1=300
alpharho=0.02
k1=0.1
k=k1*exp((E/R)*(1/T1-1/T))
12-51
R=1.987
DH=-20000
Kc=1000*exp((DH/R)*(1/T1-1/T))
Fa0=5
ra=-k*(Ca-Cb/Kc)
ThetaI=2
CpI=18
Cpa=160
sumcp=Cpa+ThetaI*CpI
Xe=Kc/(1+Kc)
rate=-ra
Ua=150
Cpcool=20
rhob=1400
DCp=0
m=50
Cpc=20
X(0)=0
T(0)=300
Ta(0)=300
p(0)=1
V(0)=0
V(f)=40
P12-8 (b) Gas Phase Counter Current Heat Exchange Vf = 20 dm3
For counter-current, we have to adjust coolant inlet temp to match coolant outlet temp of 300 K at V= 20
dm3
Polymath Code
d(X)/d(V)=-ra/Fa0
d(T)/d(V)=(Ua*(Ta-T)+DH*ra)/(Fa0*(sumcp+DCp*X))
d(Ta)/d(V)=-(Ua*(T-Ta))/m/Cpc
d(p)/d(V)=-alpharho/p/2*(T/T0)
T0=300
Ca0=2.
Cb=Ca0*(X)*p*(T0/T)
E=10000
Ca=Ca0*(1-X)*p*(T0/T)
T1=300
alpharho=0.02
k1=0.1
k=k1*exp((E/R)*(1/T1-1/T))
R=1.987
DH=-20000
12-52
Kc=1000*exp((DH/R)*(1/T1-1/T))
Fa0=5
ra=-k*(Ca-Cb/Kc)
ThetaI=2
CpI=18
Cpa=160
sumcp=Cpa+ThetaI*CpI
Xe=Kc/(1+Kc)
rate=-ra
Ua=150
Cpcool=20
rhob=1400
DCp=0
m=50
Cpc=20
X(0)=0
T(0)=300
Ta(0)=345.767
p(0)=1
V(0)=0
V(f)=20
Polymath Output
12-53
P12-8 (c) Constant Ta
For constant Ta, only 1 change needs to be done in co-current code
d(Ta)/d(V)=0*(Ua*(T-Ta))/m/Cpc
P12-8 (d) Refer to problem P11-7
P12-8 (e) Individualized solution
P12-8 (f) Refer to problem P11-7
12-54
P12-9
Refer to solution P11-8
For Heat Exchanger
Energy Balance
Energy Balance
A
B
Constant Ta
Co-Current Exchange
Ta = 300 K
" Ua %
$ ' T−T
a
dTa $# ρb '&
=
 cCP
dW
m
(
)
Cool
C
Counter Current
, where mc = 18, CP
Cool
Sum cp= CpA+thetaB*CpB+thetaI*CpI= 20+2*20+2*18= 96
P12-9 (a) Co-Current Heat Exchange
Gas Phase Co-current Variable Ta
Polymath Code:
d(X)/d(W) = -ra/Fao
d(T)/d(W) = (Ua/rhob*(Ta-T)+DH*ra)/(Fao*(sumcp))
d(Ta)/d(W) = (Ua/rhob*(T-Ta))/m/Cpcool
d(p)/d(W) = -alpha/2/p*(T/To)
To = 325
Cao = 0.2
Cc = Cao*2*X*p*(To/T)
Cb = Cao*(2-X)*p*(To/T)
E = 25000
Ca = Cao*(1-X)*p*(To/T)
T1 = 300
alpha = 0.00015
k1 = 0.0002
k = k1*exp((E/1.987)*(1/300-1/T))
R = 1.987
DH = -20000
Kc = 1000*exp((DH/1.987)*(1/305-1/T))
Fao = 5
ra = -k*(Ca*Cb-Cc^2/Kc)
sumcp = 96
ThetaI = 2
CpI = 18
Xe = (3*Kc/4-((3*Kc/4)^2-2*Kc*(Kc/4-1))^0.5)/(2*(Kc/4-1))
rate = -ra
Ua = 320
m = 18
Cpcool = 18
rhob = 1400
W(0) = 0
12-55
= 18
W(f) = 5000
X(0) = 0
T(0) = 325
Ta(0) = 300
p(0) = 1
P12-9 (b) Counter Current Ta
Polymath code:
d(X)/d(W) = -ra/Fao
d(T)/d(W) = (Ua/rhob*(Ta-T)+DH*ra)/(Fao*(sumcp))
d(Ta)/d(W) = (Ua/rhob*(Ta-T))/m/Cpcool
d(p)/d(W) = -alpha/2/p*(T/To)
To = 325
Cao = 0.2
Cc = Cao*2*X*p*(To/T)
Cb = Cao*(2-X)*p*(To/T)
E = 25000
Ca = Cao*(1-X)*p*(To/T)
T1 = 300
alpha = 0.00015
k1 = 0.0002
k = k1*exp((E/1.987)*(1/300-1/T))
R = 1.987
DH = -20000
Kc = 1000*exp((DH/1.987)*(1/305-1/T))
Fao = 5
ra = -k*(Ca*Cb-Cc^2/Kc)
sumcp = 96
ThetaI = 2
CpI = 18
Xe = (3*Kc/4-((3*Kc/4)^2-2*Kc*(Kc/4-1))^0.5)/(2*(Kc/4-1))
rate = -ra
Ua = 320
m = 18
Cpcool = 18
rhob = 1400
W(0) = 0
W(f) = 5000
X(0) = 0
T(0) = 325
Ta(0) = 381.37
p(0) = 1
12-56
Gas Phase Counter Current Variable Ta
12-57
P12-9 (b) continued
Gas Phase Counter Current Variable Ta
Gas Phase Counter Current Variable Ta
P12-9 (c) Constant Ta
Polymath code:
d(X)/d(W) = -ra/Fao
d(T)/d(W) = (Ua/rhob*(Ta-T)+DH*ra)/(Fao*(sumcp))
d(Ta)/d(W) = (Ua/rhob*(Ta-T))/m/Cpcool*0
d(p)/d(W) = -alpha/2/p*(T/To)
To = 325
Cao = 0.2
Cc = Cao*2*X*p*(To/T)
Cb = Cao*(2-X)*p*(To/T)
E = 25000
Ca = Cao*(1-X)*p*(To/T)
T1 = 300
alpha = 0.00015
k1 = 0.0002
k = k1*exp((E/1.987)*(1/300-1/T))
R = 1.987
DH = -20000
12-58
Kc = 1000*exp((DH/1.987)*(1/305-1/T))
Fao = 5
ra = -k*(Ca*Cb-Cc^2/Kc)
sumcp = 96
ThetaI = 2
CpI = 18
Xe = (3*Kc/4-((3*Kc/4)^2-2*Kc*(Kc/4-1))^0.5)/(2*(Kc/4-1))
rate = -ra
Ua = 320
m = 18
Cpcool = 18
rhob = 1400
W(0) = 0
W(f) = 5000
X(0) = 0
T(0) = 325
Ta(0) = 300
p(0) = 1
Gas Phase Constant Ta
Gas Phase Constant Ta
P12-9 (d) Refer to solution P11-8 for comparison
12-59
P12-10 (a)
For reversible reaction, the rate law becomes
−𝑟N = 𝑘 P𝐶N 𝐶Q −
𝐶L 𝐶R
T
𝐾L
Stoichiometry:
𝐶R = 𝐶NU ∗ 𝑋
See the following Polymath program
d(X)/d(V) = -ra/Fa0
T = 300+200*X
k = 0.01*exp((10000/1.987)*(1/300-1/T))
Fa0 = 0.2
Ca0 = 0.1
Kc = 10*exp(-6000/1.987*(1/450-1/T))
ra = -k*(Ca0^2)*((1-X)^2-(X^2)/Kc)
Xe = 1/(1+1/sqrt(Kc))
V(0) = 0
V(f) = 10
X(0) = 0
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca0
0.1
0.1
0.1
0.1
2 Fa0
0.2
0.2
0.2
0.2
3 k
0.01
0.01
0.010587
0.010587
4 Kc
286.4967
276.8567
286.4967
276.8567
5 ra
-0.0001
-0.0001048
-0.0001
-0.0001048
6 T
300.
300.
301.0236
301.0236
7 V
0
0
10.
10.
8 X
0
0
0.0051181
0.0051181
9 Xe
0.9442158
0.9433075
0.9442158
0.9433075
Differential equations
1 d(X)/d(V) = -ra/Fa0
Explicit equations
1 T = 300+200*X
2 k = 0.01*exp((10000/1.987)*(1/300-1/T))
3 Fa0 = 0.2
4 Ca0 = 0.1
5 Kc = 10*exp(-6000/1.987*(1/450-1/T))
6 ra = -k*(Ca0^2)*((1-X)^2-(X^2)/Kc)
7 Xe = 1/(1+1/sqrt(Kc))
12-60
P12-10 (b)
When heat exchanger is added, the energy balance can be written as
So, with
= 0,
,
= -6000 cal/mol
Where Ua = 20 cal/m3/s/K, Ta = 450 K
See the following Polymath program
d(X)/d(V) = -ra/Fa0
d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
k = 0.01*exp((10000/1.987)*(1/300-1/T))
Fa0 = 0.2
Ca0 = 0.1
Kc = 10*exp(-6000/1.987*(1/450-1/T))
ra = -k*(Ca0^2)*((1-X)^2-(X^2)/Kc)
Xe = 1/(1+1/sqrt(Kc))
DH = -6000
Ua = 20
Ta = 450
Fao = 0.2
sumcp = 30
12-61
Qg=ra*DH
Qr=Ua*(T-Ta)
V(0) = 0
V(f) = 10
X(0) = 0
T(0) = 300
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca0
0.1
0.1
0.1
0.1
2
DH
-6000.
-6000.
-6000.
-6000.
3
Fa0
0.2
0.2
0.2
0.2
4
Fao
0.2
0.2
0.2
0.2
5
k
0.01
0.01
3.149565
2.775299
6
Kc
286.4967
9.081743
286.4967
9.797914
7
Qg
0.6
0.6
146.9165
26.28162
8
Qr
-3000.
-3000.
131.0667
27.46541
9
ra
-0.0001
-0.0244861
-0.0001
-0.0043803
10 sumcp
30.
30.
30.
30.
11 T
300.
300.
456.5533
451.3733
12 Ta
450.
450.
450.
450.
13 Ua
20.
20.
20.
20.
14 V
0
0
10.
10.
15 X
0
0
0.5637972
0.5637972
16 Xe
0.9442158
0.7508467
0.9442158
0.7578787
Differential equations
1 d(X)/d(V) = -ra/Fa0
2 d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
Explicit equations
1 k = 0.01*exp((10000/1.987)*(1/300-1/T))
2 Fa0 = 0.2
3 Ca0 = 0.1
4 Kc = 10*exp(-6000/1.987*(1/450-1/T))
5 ra = -k*(Ca0^2)*((1-X)^2-(X^2)/Kc)
6 Xe = 1/(1+1/sqrt(Kc))
7 DH = -6000
8 Ua = 20
9 Ta = 450
10 Fao = 0.2
11 sumcp = 30
12 Qg = ra*DH
13 Qr = Ua*(T-Ta)
12-62
P12-10 (c)
For a co-current heat exchanger,
 = 50
m
g
sec
CpC = 1cal/g/K, Ta1=450 K,
See the following Polymath program
d(X)/d(V) = -ra/Fa0
d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
k = 0.01*exp((10000/1.987)*(1/300-1/T))
Fa0 = 0.2
Ca0 = 0.1
Kc = 10*exp(-6000/1.987*(1/450-1/T))
ra = -k*(Ca0^2)*((1-X)^2-(X^2)/Kc)
Xe = 1/(1+1/sqrt(Kc))
DH = -6000
Ua = 20
Fao = 0.2
sumcp = 30
mc = 50
Cpc = 1
V(0) = 0
V(f) = 10
X(0) = 0
T(0) = 300
Ta(0) = 450
12-63
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca0
0.1
0.1
0.1
0.1
2
Cpc
1.
1.
1.
1.
3
DH
-6000.
-6000.
-6000.
-6000.
4
Fa0
0.2
0.2
0.2
0.2
5
Fao
0.2
0.2
0.2
0.2
6
k
0.01
0.01
2.434
2.434
7
Kc
286.4967
10.60052
286.4967
10.60052
8
mc
50.
50.
50.
50.
9
ra
-0.0001
-0.0171143
-0.0001
-0.0053027
10 sumcp
30.
30.
30.
30.
11 T
300.
300.
446.1228
446.1228
12 Ta
450.
434.915
450.
444.6542
13 Ua
20.
20.
20.
20.
14 V
0
0
10.
10.
15 X
0
0
0.5078713
0.5078713
16 Xe
0.9442158
0.765029
0.9442158
0.765029
Differential equations
1 d(X)/d(V) = -ra/Fa0
2 d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
3 d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
Explicit equations
1 k = 0.01*exp((10000/1.987)*(1/300-1/T))
2 Fa0 = 0.2
3 Ca0 = 0.1
4 Kc = 10*exp(-6000/1.987*(1/450-1/T))
5 ra = -k*(Ca0^2)*((1-X)^2-(X^2)/Kc)
6 Xe = 1/(1+1/sqrt(Kc))
7 DH = -6000
8 Ua = 20
9 Fao = 0.2
10 sumcp = 30
11 mc = 50
12 Cpc = 1
12-64
Next increase the coolant flow rate and run the same program to compare results.
For counter-current flow,
See the following Polymath program
d(X)/d(V) = -ra/Fa0
d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
d(Ta)/d(V) = -Ua*(T-Ta)/mc/Cpc
k = 0.01*exp((10000/1.987)*(1/300-1/T))
Fa0 = 0.2
Ca0 = 0.1
Kc = 10*exp(-6000/1.987*(1/450-1/T))
ra = -k*(Ca0^2)*((1-X)^2-(X^2)/Kc)
Xe = 1/(1+1/sqrt(Kc))
DH = -6000
Ua = 20
Fao = 0.2
Qg=ra*DH
Qr=Ua*(T-Ta)
sumcp = 30
mc = 50
Cpc = 1
V(0) = 0
V(f) = 10
X(0) = 0
T(0) = 300
Ta(0) = 446
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca0
0.1
0.1
0.1
0.1
2
Cpc
1.
1.
1.
1.
3
DH
-6000.
-6000.
-6000.
-6000.
4
Fa0
0.2
0.2
0.2
0.2
5
Fao
0.2
0.2
0.2
0.2
6
k
0.01
0.01
4.453505
2.764801
7
Kc
286.4967
7.377316
286.4967
9.82022
8
mc
50.
50.
50.
50.
9
Qg
0.6
0.6
197.7048
21.61714
10 Qr
-2920.
-2920.
187.3862
25.86831
11 ra
-0.0001
-0.0329508
-0.0001
-0.0036029
12 sumcp
30.
30.
30.
30.
13 T
300.
300.
471.3669
451.2199
14 Ta
446.
446.
463.7906
449.9265
12-65
15 Ua
20.
20.
20.
20.
16 V
0
0
10.
10.
17 X
0
0
0.5924961
0.5924961
18 Xe
0.9442158
0.7309022
0.9442158
0.7580873
Differential equations
1 d(X)/d(V) = -ra/Fa0
2 d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
3 d(Ta)/d(V) = -Ua*(T-Ta)/mc/Cpc
Explicit equations
1 k = 0.01*exp((10000/1.987)*(1/300-1/T))
2 Fa0 = 0.2
3 Ca0 = 0.1
4 Kc = 10*exp(-6000/1.987*(1/450-1/T))
5 ra = -k*(Ca0^2)*((1-X)^2-(X^2)/Kc)
6 Xe = 1/(1+1/sqrt(Kc))
7 DH = -6000
8 Ua = 20
9 Fao = 0.2
10 Qg = ra*DH
11 Qr = Ua*(T-Ta)
12 sumcp = 30
13 mc = 50
14 Cpc = 1
12-66
P12-10 (d) Use reference from P11-6 and P12-7 to answer this part
P12-10 (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.
P12-10 (f)
If the reaction is irreversible but endothermic, we have
as obtained in the earlier problem.
See the following Polymath program for co-current heat exchange
d(X)/d(V) = -ra/Fa0
d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
k = 0.01*exp((10000/1.987)*(1/300-1/T))
Fa0 = 0.2
Ca0 = 0.1
ra = -k*(Ca0^2)*(1-X)^2
DH = 6000
Ua = 20
Fao = 0.2
sumcp = 30
mc = 50
Cpc = 1
V(0) = 0
V(f) = 10
X(0) = 0
T(0) = 300
Ta(0) = 450
Calculated values of the DEQ variables
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca0
0.1
0.1
0.1
0.1
2
Cpc
1.
1.
1.
1.
3
DH
6000.
6000.
6000.
6000.
4
Fa0
0.2
0.2
0.2
0.2
5
Fao
0.2
0.2
0.2
0.2
6
k
0.01
0.01
1.539336
1.351674
7
mc
50.
50.
50.
50.
8
ra
-0.0001
-0.0136308
-0.0001
-0.0047286
9
sumcp
30.
30.
30.
30.
10 T
300.
300.
428.7105
424.0147
11 Ta
450.
425.3134
450.
425.3134
12 Ua
20.
20.
20.
20.
13 V
0
0
10.
10.
14 X
0
0
0.4085332
0.4085332
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(X)/d(V) = -ra / Fa0
[2] d(T)/d(V) = (Ua*(Ta-T)+(ra)*DH)/(Fao*sumcp)
[3] d(Ta)/d(V) = Ua*(T-Ta)/mc/Cpc
12-67
Explicit equations as entered by the user
[1] k = .01 * exp((10000 / 2) * (1 / 300 - 1 / T))
[2] Ca0 = 0.1
[3] Fa0 = 0.2
[4] ra = -k * (Ca0 ^ 2) *(1 - X) ^ 2
[5] DH = 6000
[6] Ua = 20
[7] Fao = 0.2
[8] sumcp = 30
[9] mc = 50
[10] Cpc = 1
For counter-current flow in irreversible reaction,
See the following Polymath program
d(X)/d(V) = -ra/Fa0
d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
d(Ta)/d(V) = -Ua*(T-Ta)/mc/Cpc
k = 0.01*exp((10000/1.987)*(1/300-1/T))
Fa0 = 0.2
Ca0 = 0.1
ra = -k*(Ca0^2)*(1-X)^2
DH = 6000
Ua = 20
Fao = 0.2
sumcp = 30
Qg=ra*DH
Qr=Ua*(T-Ta)
mc = 50
Cpc = 1
V(0) = 0
V(f) = 10
X(0) = 0
T(0) = 300
Ta(0) = 420.35
12-68
P12-10 (f) continued
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca0
0.1
0.1
0.1
0.1
2
Cpc
1.
1.
1.
1.
3
DH
6000.
6000.
6000.
6000.
4
Fa0
0.2
0.2
0.2
0.2
5
Fao
0.2
0.2
0.2
0.2
6
k
0.01
0.01
2.53913
2.53913
7
mc
50.
50.
50.
50.
8
Qg
-0.6
-86.91098
-0.6
-38.29409
9
Qr
-2407.
-2407.
-45.04464
-45.04464
10 ra
-0.0001
-0.0144852
-0.0001
-0.0063823
11 sumcp
30.
30.
30.
30.
12 T
300.
300.
447.8013
447.8013
13 Ta
420.35
420.35
450.0536
450.0536
14 Ua
20.
20.
20.
20.
15 V
0
0
10.
10.
16 X
0
0
0.4986421
0.4986421
Differential equations
1 d(X)/d(V) = -ra/Fa0
2 d(T)/d(V) = ((ra*DH)-Ua*(T-Ta))/(Fao*sumcp)
3 d(Ta)/d(V) = -Ua*(T-Ta)/mc/Cpc
Explicit equations
1 k = 0.01*exp((10000/1.987)*(1/300-1/T))
2 Fa0 = 0.2
3 Ca0 = 0.1
4 ra = -k*(Ca0^2)*(1-X)^2
5 DH = 6000
6 Ua = 20
7 Fao = 0.2
8 sumcp = 30
9 Qg = ra*DH
10 Qr = Ua*(T-Ta)
11 mc = 50
12 Cpc = 1
12-69
P12-11 (a)
(1) Reaction: 𝐴 + 𝐵 → 𝐶. Equal molar feed, irreversible elementary reaction.
Mole Balance:
+[
+\
= −
HK
,]^
( _
_
Rate Law: −𝑟N = 𝑘_ 𝐶N 𝐶Q , with 𝑘_ (𝑇) = 𝑘_ (𝑇_ )exp [) (; − ; )], 𝑘_ (𝑇_ ) = 0.01, E=10000 cal/mol,
_
R=1.98 cal/mol/K, 𝑇_ = 300𝐾
Stoichiometry (liquid):
𝐶N = 𝐶Q = 𝐶NU (1 − 𝑋)
Energy balance:
Reactor:
𝑇U
𝑇
𝑈𝑎
(
)
𝑑𝑇
𝜌 𝑇? − 𝑇 + (−𝑟N )(−Δ𝐻)[ )
=
𝑑𝑊
𝐹NU [𝐶kN + 𝐶kQ ]
Δ𝐻)[ = −6000 𝑐𝑎𝑙/𝑚𝑜𝑙
𝐶kN = 𝐶kQ = 15 𝑐𝑎𝑙/𝑚𝑜𝑙
𝐶NU = 0.1𝑚𝑜𝑙/𝐿
𝐹NU = 0.2𝑚𝑜𝑙/𝑠
𝑇? = 323𝐾
𝑈𝑎
𝐽
𝑐𝑎𝑙
= 0.08
= 0.0191
𝜌
𝑠 𝑘𝑔𝑐𝑎𝑡. 𝐾
𝑠 𝑘𝑔𝑐𝑎𝑡. 𝐾
Polymath code:
d(X)/d(W) = -ra/Fa0
d(T)/d(W) = (Ua*(Ta-T)+ra*DelH)/(2*Fa0*Cpa)
ra = -k1*Ca0^2*(1-X)^2*(To/T)^2
k1 = 0.01*exp((10000/1.987)*(1/300 - 1/T))
Qg = ra*DelH
Qr = Ua*(T-Ta)
Fa0 = 0.2
Cpa = 15
Ca0 = 0.1
DelH = -6000
Ua = 0.0191
Ta = 323
To = 300
W(0) = 0
W(f) = 50
X(0) = 0
T(0) = 300
12-70
J
(2) when yK increase by a factor of 3000:
𝑈𝑎
𝐽
𝑐𝑎𝑙
= 240
= 57.4
𝜌
𝑠 𝑘𝑔𝑐𝑎𝑡. 𝐾
𝑠 𝑘𝑔𝑐𝑎𝑡. 𝐾
In the code given in part (1) of P12-11(a) replace Ua = 0.0191 with Ua = 57.4
12-71
P12-11 (a) continued
12-72
(3) Taking the pressure drop into account:
𝑑𝑝
𝛼
𝑇
= − (1 + 𝜖𝑋)
𝑑𝑊
2𝑝
𝑇U
With: 𝜖 = 𝑦NU ∗ 𝛿 = 0.5 ∗ (1 − 2) = −0.5, 𝛼 = 0.019𝑘𝑔G_
In addition, the expression for concentration of A changed into:
𝑇U
𝐶N = 𝐶NU (1 − 𝑋) 𝑝
𝑇
Solving the system using the following Polymath code:
d(X)/d(W) = -ra/Fa0
d(T)/d(W) = (Ua*(Ta-T)+ra*DelH)/(2*Fa0*Cpa)
d(p)/d(W) = -alpha*(1-0.5*X)*(T/To)/(2*p)
alpha = 0.019
ra = -k1*Ca0^2*(1-X)^2*(To/T)^2*(p^2)
k1 = 0.01*exp((10000/1.987)*(1/300 - 1/T))
Qg = ra*DelH
Qr = Ua*(T-Ta)
Fa0 = 0.2
Cpa = 15
Ca0 = 0.1
DelH = -6000
Ua = 0.0191
Ta = 323
To = 300
W(0) = 0
W(f) = 50
X(0) = 0
T(0) = 300
p(0) = 1
12-73
P12-11 (b)
(1) Co-current Coolant:
Polymath code:
𝑈𝑎
𝑑𝑇?
𝜌 (𝑇 − 𝑇? )
=
𝑑𝑊
𝑚€ 𝐶k€
𝐽
𝑐𝑎𝑙
𝐶k€ = 5000
= 1195
𝑘𝑔 𝐾
𝑘𝑔 𝐾
d(X)/d(W) = -ra/Fa0
d(T)/d(W) = (Ua*(Ta-T)+ra*DelH)/(2*Fa0*Cpa)
d(Ta)/d(W) = Ua*(T-Ta)/mc/Cpc
mc = 0.2
Cpc = 1195
ra = -k1*Ca0^2*(1-X)^2*(To/T)^2
k1 = 0.01*exp((10000/1.987)*(1/300 - 1/T))
Qg = ra*DelH
Qr = Ua*(T-Ta)
Fa0 = 0.2
Cpa = 15
Ca0 = 0.1
DelH = -6000
Ua = 0.0191
Ta(0) = 323
To = 300
W(0) = 0
W(f) = 50
X(0) = 0
T(0) = 300
12-74
(2) Counter-current Coolant:
𝑈𝑎
𝑑𝑇?
𝜌 (𝑇? − 𝑇)
=
𝑑𝑊
𝑚€ 𝐶k€
12-75
P12-11 (b) continued
In code given in part (1), add a negative sign to the third equation i.e.,
d(Ta)/d(W) = -Ua*(T-Ta)/mc/Cpc
12-76
(3) For an adiabatic process, no heat transfer occurred, the reactor energy balance became:
𝑑𝑇
(−𝑟N )(−Δ𝐻)[ )
=
𝑑𝑊 𝐹NU [𝐶kN + 𝐶kQ ]
i.e., put Ua = 0
Calculated values of DEQ variables
Variable
Initial value
Minimal value
Maximal value
Final value
1
Ca0
0.1
0.1
0.1
0.1
2
Cpa
15.
15.
15.
15.
3
Cpc
1195.
1195.
1195.
1195.
4
DelH
-6000.
-6000.
-6000.
-6000.
5
Fa0
0.2
0.2
0.2
0.2
6
k1
0.01
0.01
0.0135654
0.0135654
7
mc
0.2
0.2
0.2
0.2
8
Qg
0.6
0.6
0.7416283
0.7416283
9
Qr
0
0
0
0
10 ra
-0.0001
-0.0001236
-0.0001
-0.0001236
11 T
300.
300.
305.5541
305.5541
12 Ta
323.
323.
323.
323.
13 To
300.
300.
300.
300.
14 Ua
0
0
0
0
15 W
0
0
50.
50.
16 X
0
0
0.0277704
0.0277704
P12-11 (c)
For a CSTR reactor:
Mole balance:
𝑊=
𝐹NU 𝑋
−𝑟N
Rate Law:
−𝑟N = 𝑘𝐶N•
𝑇U
𝐶N = 𝐶Q = 𝐶NU (1 − 𝑋)
𝑇
12-77
Combining the mole balance and the rate law:
𝑊=
Set:𝑘𝐶NU 𝑇U• 𝑊 = 𝑘 ⋅ 80 ⋅ 300• ⋅ 0.1 = 𝐾
𝐹NU 𝑋
𝑇
𝑘(𝐶NU (1 − 𝑋) 𝑇U )•
(2𝐾 + 𝐹NU 𝑇 • ) − „(2𝐾 + 𝐹NU 𝑇 • )• − 4𝐾 •
𝑋ƒQ =
2𝐾
( _
_
And: 𝑘 = 𝑘_ (𝑇_ )exp [) (; − ;)]
_
Energy balance gives:
𝑈𝐴
119
𝐹NU (𝑇 − 𝑇? )
𝑋(Q =
= 0.2 (𝑇 − 300)
−Δ𝐻𝑟𝑥𝑛(𝑇) ) 6000
Make plot of 𝑋(Q and 𝑋ƒQ on the same plot:
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
298
300
302
304
XEB
306
308
310
XMB
T=306K
X=0.66
P12-11 (d)
Now W=80kg
Rate Law: −𝑟N = 𝑘_ 𝐶N 𝐶Q − 𝑘H 𝐶L
( _
_
, with 𝑘_ (𝑇) = 𝑘_ (𝑇_ )exp [) (; − ; )], 𝑘_ (𝑇_ ) = 0.01, E=10000 cal/mol, R=1.98 cal/mol/K, 𝑇_ = 300𝐾
_
(‡
_
_
‰_.ˆ‹Œ
€?•
And 𝑘H = 0.2 exp [ ) (ˆ‰U − ;)], 𝐸H = •Ž• = 12285 •Ž•
Plots can be generated using polymath code.
P12-11 (e) Individualized solution
12-78
P12-12 (a)
Start with the complete energy balance:
dEˆ  
= Q − WS − ΣEi Fi in − ΣEi Fi out
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 = −ΣEi Fi in − ΣEi Fi out
Evaluating energy terms:
In: H A0 FA0 + H B 0 FB 0 + H C 0 FC 0
Out: H A ( FA + R AV ) + H B ( FB + RBV ) + H C ( FC + R CV )
Now we evaluate Fi
FA = FA0 − FA0 X
FB = FB 0 + FA0 X
FC = FC 0 + FA 0 X − RCV
FB0 = FC0 = 0
Inserting these into our equation gives:
Combining and substituting terms gives:
Differentiating with respect to V with ΔCP = 0
Combine that with the mole balance and rate law:
See the following Polymath program
Polymath code:
d(Fc)/d(V) = -ra-Rc
d(Fb)/d(V) = -ra
d(Fa)/d(V) = ra
d(T)/d(V) = (ra*dHrx-Rc*Hc)/(Fa0*Cpa)
k = 0.133*exp((31400/8.314)*(1/450-1/T))
12-79
To = 450
Ct0 = 10/0.082/450
Ft = Fa+Fb+Fc
Cc = Ct0*Fc/Ft*To/T
Ca = Ct0*Fa/Ft*To/T
Cb = Ct0*Fb/Ft*To/T
Fa0 = Ct0*20
dHrx = -20000
Kc = exp((dHrx/8.314)*(1/300-1/T))
Cpc = 15
ra = -k*(Ca-Cb*Cc/Kc)
kc = 1.5
Rc = kc*Cc
Xe = ((Kc*T/(Ct0*To))/(1+(Kc*T/(Ct0*To))))^0.5
Cpa = 40
Cpb = 25
Hc =40000+Cpc*(T-273)
V(0) = 0
V(f) = 100
Fa(0) = 5.42
Fb(0) =0
Fc(0) = 0
T(0) = 450
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.2710027
0.072777
0.2710027
0.072777
2
Cb
0
0
0.0498859
0.0498859
3
Cc
0
0
0.0273679
0.0060305
4
Cpa
40.
40.
40.
40.
5
Cpb
25.
25.
25.
25.
6
Cpc
15.
15.
15.
15.
7
Ct0
0.2710027
0.2710027
0.2710027
0.2710027
8
dHrx
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
9
Fa
5.42
3.215735
5.42
3.215735
10 Fa0
5.420054
5.420054
5.420054
5.420054
11 Fb
0
0
2.204265
2.204265
12 Fc
0
0
0.9446979
0.2664623
13 Ft
5.42
5.42
6.364698
5.686462
14 Hc
-3.735E+04
-3.735E+04
-2.988E+04
-2.988E+04
15 k
0.133
0.133
10.91169
10.91169
16 kc
1.5
1.5
1.5
1.5
17 Kc
0.0690539
0.0041692
0.0690539
0.0041692
18 ra
-0.0360434
-0.1033995
-0.0067748
-0.0067748
19 Rc
0
0
0.0410519
0.0090457
20 T
450.
450.
947.6106
947.6106
21 To
450.
450.
450.
450.
22 V
0
0
100.
100.
23 Xe
0.4506284
0.1771437
0.4506284
0.1771437
Differential equations
1 d(Fc)/d(V) = -ra-Rc
2 d(Fb)/d(V) = -ra
3 d(Fa)/d(V) = ra
4 d(T)/d(V) = (ra*dHrx - Rc*Hc)/(Fa0*Cpa)
12-80
Explicit equations
1
k = .133*exp((31400/8.314)*(1/450-1/T))
2
To = 450
3
Ct0 = 10/0.082/450
4
Ft = Fa + Fb + Fc
5
Cc = Ct0*Fc/Ft*To/T
6
Fa0 = Ct0*20
7
Ca = Ct0*Fa/Ft*To/T
8
dHrx = -20000
9
Cb = Ct0*Fb/Ft*To/T
10 Kc = 1*exp((dHrx/8.314)*(1/300-1/T))
11 Cpc = 15
12 ra = -k*(Ca-Cb*Cc/Kc)
13 kc = 1.5
14 Rc = kc*Cc
15 Xe = ((Kc*T/(Ct0*To))/(1+(Kc*T/(Ct0*To))))^0.5
16 Cpa = 40
17 Cpb = 25
18 Hc = -40000+Cpc*(T-273)
Concentration profile:
12-81
As the equilibrium constant Kc increases, more product will form and be removed through the
membrane.
P12-12 (b)
Now, the heat balance equation needs to be modified.
Polymath code:
d(Fc)/d(V) = -ra-Rc
d(Fb)/d(V) = -ra
d(Fa)/d(V) = ra
d(T)/d(V) = (Ua*(Ta-T)+ra*dHrx-Rc*Hc)/(Fa0*Cpa)
Ua = 30
Ta = 323
k = 0.133*exp((31400/8.314)*(1/450-1/T))
To = 450
Ct0 = 10/0.082/450
Ft = Fa+Fb+Fc
Cc = Ct0*Fc/Ft*To/T
Ca = Ct0*Fa/Ft*To/T
Cb = Ct0*Fb/Ft*To/T
Fa0 = Ct0*20
dHrx = -20000
Kc = exp((dHrx/8.314)*(1/300-1/T))
Cpc = 15
ra = -k*(Ca-Cb*Cc/Kc)
kc = 1.5
Rc = kc*Cc
Xe = ((Kc*T/(Ct0*To))/(1+(Kc*T/(Ct0*To))))^0.5
Cpa = 40
Cpb = 25
Hc =40000+Cpc*(T-273)
V(0) = 0
V(f) = 100
Fa(0) = 5.42
Fb(0) =0
Fc(0) = 0
T(0) = 450
12-82
The difference between three cases are not obvious, because the reaction rate is very low due to the
low temperature caused by cooling. The counter-intuitive raising trend of concentration of A is caused
by the re-movement of C in the reactor (The decreasing of Ft is faster than Fa).
P12-13 (a)
A) Increase conversion.
If it is an adiabatic system, then it has to be endothermic, because
the temperature decreases and the heat of reaction is positive.
Increasing inerts increases the temperature, increasing k,
increasing the rate and hence increasing conversion.
P12-13 (b)
Answer: C and D
A) The above reaction could be adiabatic. False.
Xe could not increase if adiabatic.
B) The above reaction could be exothermic with constant cooling temperature. False.
T would increase then decrease follow slope temperature curve.
C) The above reaction could be endothermic with constant heating temperature. True.
D) The above reaction could be second order. True.
12-83
P12-13 (c) Part A is false. For endothermic reaction in adiabatic system, temp will continuously decrease
and conversion will continuously increase. Once temp is very low, reaction will proceed at a very slow rate
and hence there will be minor increase in conversion only and graph should flatten out. So once the graph
has flattened, there should not be any second peak in conversion.
P12-14
Assume Adiabatic
P12-14 (a) Heat of reaction is obtained from G(T) curve. Since G(T) = -∆𝐻)‘ ∗ 𝑋. We can see that at high
temp G(T) saturates to 12000 cal/mol. At this temp, X would also approach 1. Hence,
∆𝐻)‘ = −12000 𝑐𝑎𝑙/𝑚𝑜𝑙
P12-14 (b)
For this problem, we need to draw parallel line for R(T) curve to determine temp at tangent lines to G(T)
curve
Ignition = 260°C (at this temp, reaction will ignite and operate only at upper steady state value)
Extinction = approx. 190°C (at this temp, reaction will operate only at lower steady state value)
P12-14 (c)
For this part, we need to find intersection of R(T) and G(T) curve at ignition and extinction temp
At Ignition (260 C), T = 300°C, 475°C
At Extinction (190 C), T = 215°C, 375°C
P12-14 (d)
For ignition,
If reactor operates at 300 C, G(T) =~ 2500, X= 2500/12000= 0.21
If reactor operates at 475 C, G(T) =12000, X= 12000/12000= 1
For extinction,
If reactor operates at 215 C, G(T) =~ 1000, X= 100/12000= 0.083
If reactor operates at 375 C, G(T) =~10000, X= 10000/12000= 0.83
P12-15 (a)
G(T) = ΔHR X
X=
) E # 1 1 &,
τk
,k = 6.6×10−3 exp+ %
− (.
1+ τ k
+* R $ 350 T '.-
(
)(
R(T ) = C p0 1 + κ T −Tc
)
C p0 = ∑Θi C pi = 50
κ=
UA
8000
=
=2
C p0FA0 50×80
Tc =
κTa +T0
1+ κ
= 350K
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 T0 = 450 K, steady state temperature 399.94 K.
12-84
Use the following Polymath code:
d(T)/d(t) = 0.1
RT = 150*(T-350)
EoR = 40000/1.987
k = 6.6*0.001*exp(EoR*(1/300-1/T))
tau = 100
X = tau*k/(1+tau*k)
GT = 7500*X
at = GT-RT
t(0) = 0
t(f) = 1000
T(0) = 350
Graphical method
Polymath Results 02-22-2006, Rev5.1.233
Variable initial value Minimal value
t
0
0
T
350
350
RT
0
0
EoR
2.013E+04
2.013E+04
k
0.0066
0.0066
tau
100
100
X
0.3975904
0.3975904
GT
2981.9277
2981.9277
at
2981.9277
-7500.032
maximal value
1000
450
1.5E+04
2.013E+04
2346.7972
100
0.9999957
7499.968
4336.6841
final value
1000
450
1.5E+04
2.013E+04
2346.7972
100
0.9999957
7499.968
-7500.032
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(T)/d(t) = 0.1
Explicit equations as entered by the user
[1] RT = 150*(T-350)
[2] EoR = 40000/1.987
[3] k = 6.6*0.001*exp(EoR*(1/350-1/T))
[4] tau = 100
[5] X = tau*k/(1+tau*k)
[6] GT = 7500*X
[7] at = GT-RT
- Equation
12-85
Polymath code:
f(T) = RT-GT
T(0) = 300
T(max) = 1000
T(min) = 0
RT = 150*(T-350)
EoR = 40000/1.987
k = 6.6*0.001*exp(EoR*(1/350-1/T))
tau = 100
X = tau*k/(1+tau*k)
GT = 7500*X
Polymath Results 02-22-2006, Rev5.1.233
NLE Solution
Variable Value
f(x)
Ini Guess
T
399.9425
3.181E-09 300
RT
7491.375
EoR
2.013E+04
K
8.6856154
tau
100
X
0.99885
GT
7491.375
NLE Report (safenewt)
Nonlinear equations
[1] f(T) = RT-GT = 0
Explicit equations
[1] RT = 150*(T-350)
[2] EoR = 40000/1.987
[3] k = 6.6*0.001*exp(EoR*(1/350-1/T))
[4] tau = 100
[5] X = tau*k/(1+tau*k)
[6] GT = 7500*X
P12-15 (b)
First, we must plot G(T) and R(T) for many different T0’s on the same plot. From this we must generate
data that we use to plot Ts vs To.
P12-15 (c)
For high conversion, the feed stream must be pre-heated to at least 404 K. At this temperature, X = .991
and T = 384.2 K in the CSTR. Any feed temperature above this point will provide for higher conversions.
12-86
P12-15 (d)
For a temperature of 369.2 K, the conversion is 0.935
P12-15 (e)
The extinction temperature is 360 K (87oC).
P12-16 (a)
Mole Balance:
P12-16 (b)
The following plot gives us the steady state temperatures of 310, 377.5 and 418.5 K
See the following Polymath program
12-87
P12-16 (c)
P12-16 (d)
P12-16 (e)
The following plot shows Ta varied.
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.
12-88
P12-16 (f)
P12-16 (g)
At the maximum conversion G(t) will also be at its maximal value. This occurs at approximately T = 404
K. G(404 K) = 73520 cal. At steady state at this temperature, R(T) = G(T).
12-89
If we plug in the values and solve for UA, we get:
UA = 7421 cal/min/K
P12-16 (h) Individualized solution
P12-16 (i)
The adiabatic blowout flow rate occurs at
P12-16 (j)
Lowing T0 or Ta or increasing UA will help keep the reaction running at the lower steady state.
P12-17
V = 12 dm3, v0 = 10 dm3/h implies τ = 1.2 h
𝐺 (𝑇) = (−𝛥𝐻)Ž )𝑋
Writing mass balance based on example 12-6
𝑉=
𝑣Ž ∗ (𝐶NU − 𝐶N )
@𝑘” ∗ 𝐶N − 𝑘• ∗ 𝐶Q A
Where,
𝐶N = 𝐶NU ∗ (1 − 𝑋)
𝐶Q = 𝐶NU ∗ 𝑋
Solving for X gives,
𝑋=
Where,
Let T = 350 K,
Calculated G(T=350K) is,
𝜏𝑘”
1
1 + 𝜏𝑘” —1 + 𝐾˜
30000 1
1
𝑘” = 0.001 exp (
P
− T)
1.987 300 𝑇
42000 1
1
𝐾 = 5000000 exp (−
P
− T)
1.987 300 𝑇
𝑘” = 1.326
𝐾 = 212.6
𝑋 = 0.6123
𝐺 (𝑇) = 42000 × 0.6123 = 25720 𝑐𝑎𝑙/𝑚𝑜𝑙
Which is close to the value (27000 cal/mol) shown on the plot.
12-90
P12-17 (b)
𝑅(𝑇) = 𝐶:Ž (1 + ĸ)(𝑇 − 𝑇€ )
Where,
_UU›_‰U
𝐶:Ž =
reactor)
•
ĸ=
𝑈𝐴
𝐶:Ž 𝐹NŽ
𝑇€ =
ĸ𝑇? + 𝑇Ž
1+ĸ
= 125𝑐𝑎𝑙/𝐾𝑚𝑜𝑙 (as initially equimolar concentrations of inert and A are sent into the
From the part a, G(T) is known. Using Polymath, we can find out that X = 0.708 and T = 377.5 K
Polymath code:
f(T) = delH*X-Cpo*(1+k)*(T-Tc)
tou = 1.2
X = tou*kf/(1+tou*kf*(1+1/K))
To = 330
Ta = 330
UA = 5000
Cpo = 125
Fao = 10
k = UA/Cpo/Fao
Tc = (k*Ta+To)/(1+k)
kf = 0.001*exp((30000/1.987)*(1/300 -1/T))
K = 5000000*exp((-42000/1.987)*(1/300-1/T))
delH = 42000
T(0) = 320
T(max) = 500
T(min) =100
Polymath report:
Calculated values of NLE variables
Variable Value
1 T
f(x)
Initial Guess
377.5563 -3.309E-08 300. ( 100. < T < 500. )
Variable Value
1
Cpo
125.
2
delH
4.2E+04
3
Fao
10.
4
k
4.
5
K
2.590141
6
kf
30.885
7
Ta
330.
8
Tc
330.
9
To
330.
10 tou
1.2
11 UA
5000.
12 X
0.7076834
Nonlinear equations
1 f(T) = delH*X-Cpo*(1+k)*(T-Tc) = 0
12-91
Explicit equations
1 tou = 1.2
2 K = 5000000*exp((-42000/1.987)*(1/300-1/T))
3 To = 330
4 Ta = 330
5 UA = 5000
6 Cpo = 125
7 Fao = 10
8 k = UA/Cpo/Fao
9 Tc = (k*Ta+To)/(1+k)
10 kf = 0.001*exp((30000/1.987)*(1/300 -1/T))
11 X = tou*kf/(1+tou*kf*(1+1/K))
12 delH = 42000
P12-17 (c) To find out the temperature at which maximum conversion is achieved, the following
polymath code can be used (alternatively, the equation given for conversion in part (a) can also be used
to solve analytically)
Polymath code:
d(T)/d(t) = 1
E = 30000
k = 0.001*exp((E/1.987)*(1/300-1/T))
DHrx = -42000
Kc = 5000000*exp((DHrx/1.987)*(1/300-1/T))
tau = 1.2
X = tau*k/(1+tau*k+tau*k/Kc)
GT = X*(-DHrx)
t(0) = 0
t(f) = 1000
T(0) = 300
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 DHrx
-4.2E+04
-4.2E+04
-4.2E+04
-4.2E+04
2 E
3.0E+04
3.0E+04
3.0E+04
3.0E+04
3 GT
50.33959
6.083E-13
3.599E+04
6.083E-13
4 k
0.001
0.001
6.5E+13
6.5E+13
5 Kc
5.0E+06
1.448E-17
5.0E+06
1.448E-17
6 T
300.
300.
1300.
1300.
7 t
0
0
1000.
1000.
8 tau
1.2
1.2
1.2
1.2
9 X
0.0011986
1.448E-17
0.8569502
1.448E-17
Differential equations
1 d(T)/d(t) = 1
Explicit equations
1 E = 30000
2 k = 0.001*exp((E/1.987)*(1/300-1/T))
3 DHrx = -42000
4 Kc = 5000000*exp((DHrx/1.987)*(1/300-1/T))
5 tau = 1.2
6 X = tau*k/(1+tau*k+tau*k/Kc)
7 GT = X*(-DHrx)
12-92
P12-17 (c) continued
From the plot, the maximum conversion achieved, Xmax = 0.86
At Xmax, T = 364.12 K and G(T) = 36000 cal/mol
Also,
R(T) = G(T) = 36000 cal/mol
𝑅(𝑇) = 𝐶:Ž (1 + ĸ)(𝑇 − 𝑇€ )
Where,
𝑈𝐴
5000
ĸ=
=
=4
𝐶:Ž 𝐹NŽ 125 × 10
ĸ𝑇? + 𝑇Ž 1320 + 𝑇Ž
𝑇€ =
=
1+ĸ
5
1320 + 𝑇Ž
125(4 + 1) P364.12 −
T = 36000
5
Solving gives,
𝑇Ž = 212.6 𝐾
P12-17 (d) Replace UA = 5000 with UA = 0 in the polymath code given in part (b)
Polymath Report:
Calculated values of NLE variables
Variable Value
1 T
f(x)
Initial Guess
394.3924 1.366E-08 300. ( 100. < T < 500. )
12-93
Variable Value
1
Cpo
125.
2
delH
4.2E+04
3
Fao
10.
4
k
0
5
K
0.2373543
6
kf
170.2643
7
Ta
330.
8
Tc
330.
9
To
330.
10 tou
1.2
11 UA
0
12 X
0.1916441
Nonlinear equations
1 f(T) = delH*X-Cpo*(1+k)*(T-Tc) = 0
Explicit equations
1 tou = 1.2
2 K = 5000000*exp((-42000/1.987)*(1/300-1/T))
3 To = 330
4 Ta = 330
5 UA = 0
6 Cpo = 125
7 Fao = 10
8 k = UA/Cpo/Fao
9 Tc = (k*Ta+To)/(1+k)
10 kf = 0.001*exp((30000/1.987)*(1/300 -1/T))
11 X = tou*kf/(1+tou*kf*(1+1/K))
12 delH = 42000
Final conversion X = 0.192 and exit temperature T = 394 K
We can see an increase in the exit temperature due to the absence of coolant and higher temperatures
for an exothermic reaction manifest in the form of lower conversions, which is as expected.
P12-17 (e) Using the following polymath code, plot of G(T) and R(T) can be obtained
Polymath code:
d(T)/d(t) = 1
E = 30000
kf = 0.001*exp((E/1.987)*(1/300-1/T))
DHrx = -42000
Kc = 5000000*exp((DHrx/1.987)*(1/300-1/T))
tau = 1.2
X = tau*kf/(1+tau*kf+tau*kf/Kc)
GT = X*(-DHrx)
RT = Cpo*(1+k)*(T-Tc)
Cpo = 125
k = UA/Cpo/Fao
Tc = (k*Ta+To)/(1+k)
To = 330
Ta = 330
UA = 5000
Fao = 10
t(0) = 0
t(f) = 1000
T(0) = 300
12-94
Clearly, only one steady state is possible for this system. Therefore, it is not possible to find ignition and
extinction temperatures.
P12-17 (f) Individualized solution
P12-18
Polymath code:
d(X)/d(W) = -ra/Fa0
d(p)/d(W) = -alpha/2/p*(1+epsilon*X)*T/T0
d(T)/d(W) = (ra*(DH+deltaCp*(T-298))-Ua*(T-Ta))/Fa0/Cp0
d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc
Cp0 = 40
DH = -20000
epsilon = -1
T0 = 450
Ca0 = 1.9
Fa0 = 5
alpha = 0.019
Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*p
Ua = 0.8*0
deltaCp = 1/2*20-40
Kc = 10000*exp(DH/8.314*(1/450-1/T))
Cc = Ca0*X/2/(1+epsilon*X)*T0/T*p
m = 0.05
Cpc = 4200
k = 0.01*exp(8000/8.314*(1/450-1/T))
ra = -k*(Ca^2-Cc/Kc)
W(0) = 0
W(f) = 50
X(0) = 0
p(0) = 1
T(0) = 450
Ta(0) = 500
Adiabatic
POLYMATH Report
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.019
0.019
0.019
0.019
2
Ca
1.9
0.7510709
1.902453
0.7510709
3
Ca0
1.9
1.9
1.9
1.9
12-95
4
Cc
0
0
0.1974831
0.1113793
5
Cp0
40.
40.
40.
40.
6
Cpc
4200.
4200.
4200.
4200.
7
deltaCp
-30.
-30.
-30.
-30.
8
DH
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
9
epsilon
-1.
-1.
-1.
-1.
10 Fa0
5.
5.
5.
5.
11 k
0.01
0.01
0.0261675
0.0261675
12 Kc
10000.
902.809
10000.
902.809
13 m
0.05
0.05
0.05
0.05
14 ra
-0.0361
-0.0627617
-0.014758
-0.014758
15 T
450.
450.
817.9727
817.9727
16 T0
450.
450.
450.
450.
17 Ta
500.
500.
500.
500.
18 Ua
0
0
0
0
19 W
0
0
50.
50.
20 X
0
0
0.4949381
0.4949381
21 p
1.
0.2174708
1.
0.2174708
Differential equations
1 d(X)/d(W) = -ra/Fa0
2 d(p)/d(W) = -alpha/2/p*(1+epsilon*X)*T/T0
3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0
4 d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc
Explicit equations
1
Cp0 = 40
2
DH = -20000
3
epsilon = -1
4
T0 = 450
5
Ca0 = 1.9
6
Fa0 = 5
7
alpha = 0.019
8
Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*p
9
Xe = Kc/(1+Kc)
10 deltaCp = 1/2*20-40
11 Kc = 10000*exp(DH/8.314*(1/450-1/T))
12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*p
13 m = 0.05
14 Cpc = 4200
15 k = 0.01*exp(8000/8.314*(1/450-1/T))
16 ra = -k*(Ca^2-Cc/Kc)
12-96
P12-18 continued
Now, constant temperature Ta = 300K
Polymath code:
d(X)/d(W) = -ra/Fa0
d(p)/d(W) = -alpha/2/p*(1+epsilon*X)*T/T0
d(T)/d(W) = (ra*(DH+deltaCp*(T-298))-Ua*(-Ta))/Fa0/Cp0
d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc*0
Cp0 = 40
DH = -20000
epsilon = -1
T0 = 450
Ca0 = 1.9
Fa0 = 5
alpha = 0.019
Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*p
Ua = 0.8
deltaCp = 1/2*20-40
Kc = 10000*exp(DH/8.314*(1/450-1/T))
Cc = Ca0*X/2/(1+epsilon*X)*T0/T*p
m = 0.05
Cpc = 4200
k = 0.01*exp(8000/8.314*(1/450-1/T))
ra = -k*(Ca^2-Cc/Kc)
W(0) = 0
W(f) = 50
X(0) = 0
p(0) = 1
T(0) = 450
Ta(0) = 500
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.019
0.019
0.019
0.019
2
Ca
1.9
0.7615715
1.906858
0.7615715
3
Ca0
1.9
1.9
1.9
1.9
4
Cc
0
0
0.1991675
0.1180575
5
Cp0
40.
40.
40.
40.
6
Cpc
4200.
4200.
4200.
4200.
12-97
7
deltaCp
-30.
-30.
-30.
-30.
8
DH
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
9
epsilon
-1.
-1.
-1.
-1.
10 Fa0
5.
5.
5.
5.
11 k
0.01
0.01
0.0248669
0.0248669
12 Kc
10000.
1025.524
10000.
1025.524
13 m
0.05
0.05
0.05
0.05
14 ra
-0.0361
-0.0610755
-0.0144197
-0.0144197
15 T
450.
450.
783.9972
783.9972
16 T0
450.
450.
450.
450.
17 Ta
500.
500.
500.
500.
18 Ua
0.8
0.8
0.8
0.8
19 W
0
0
50.
50.
20 X
0
0
0.4848172
0.4848172
21 p
1.
0.2300674
1.
0.2300674
Differential equations
1 d(X)/d(W) = -ra/Fa0
2 d(p)/d(W) = -alpha/2/p*(1+epsilon*X)*T/T0
3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0
4 d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc *0
Explicit equations
1
Cp0 = 40
2
DH = -20000
3
epsilon = -1
4
T0 = 450
5
Ca0 = 1.9
6
Fa0 = 5
7
alpha = 0.019
8
Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*p
9
Ua = 0.8
10 deltaCp = 1/2*20-40
11 Kc = 10000*exp(DH/8.314*(1/450-1/T))
12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*p
13 m = 0.05
14 Cpc = 4200
15 k = 0.01*exp(8000/8.314*(1/450-1/T))
16 ra = -k*(Ca^2-Cc/Kc)
12-98
P12-18 continued
Now, Co-current heat exchanger
Polymath code:
d(X)/d(W) = -ra/Fa0
d(p)/d(W) = -alpha/2/p*(1+epsilon*X)*T/T0
d(T)/d(W) = (ra*(DH+deltaCp*(T-298))-Ua*(T-Ta))/Fa0/Cp0
d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc
Cp0 = 40
DH = -20000
epsilon = -1
T0 = 450
Ca0 = 1.9
Fa0 = 5
alpha = 0.019
Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*p
Ua = 0.8
deltaCp = 1/2*20-40
Kc = 10000*exp(DH/8.314*(1/450-1/T))
Cc = Ca0*X/2/(1+epsilon*X)*T0/T*p
m = 0.05
Cpc = 4200
k = 0.01*exp(8000/8.314*(1/450-1/T))
ra = -k*(Ca^2-Cc/Kc)
W(0) = 0
W(f) = 50
X(0) = 0
p(0) = 1
T(0) = 450
Ta(0) = 500
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.019
0.019
0.019
0.019
2
Ca
1.9
0.7614153
1.906781
0.7614153
3
Ca0
1.9
1.9
1.9
1.9
4
Cc
0
0
0.1990505
0.1177523
5
Cp0
40.
40.
40.
40.
6
Cpc
4200.
4200.
4200.
4200.
12-99
7
deltaCp
-30.
-30.
-30.
-30.
8
DH
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
9
epsilon
-1.
-1.
-1.
-1.
10 Fa0
5.
5.
5.
5.
11 k
0.01
0.01
0.0249108
0.0249108
12 Kc
10000.
1021.01
10000.
1021.01
13 m
0.05
0.05
0.05
0.05
14 ra
-0.0361
-0.0610733
-0.0144393
-0.0144393
15 T
450.
450.
785.126
785.126
16 T0
450.
450.
450.
450.
17 Ta
500.
499.0741
521.1534
521.1534
18 Ua
0.8
0.8
0.8
0.8
19 W
0
0
50.
50.
20 X
0
0
0.4849408
0.4849408
21 p
1.
0.2296895
1.
0.2296895
Differential equations
1 d(X)/d(W) = -ra/Fa0
2 d(p)/d(W) = -alpha/2/p*(1+epsilon*X)*T/T0
3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0
4 d(Ta)/d(W) = Ua*(T-Ta)/m/Cpc
Explicit equations
1
Cp0 = 40
2
DH = -20000
3
epsilon = -1
4
T0 = 450
5
Ca0 = 1.9
6
Fa0 = 5
7
alpha = 0.019
8
Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*p
9
Ua = 0.8
10 deltaCp = 1/2*20-40
11 Kc = 10000*exp(DH/8.314*(1/450-1/T))
12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*p
13 m = 0.05
14 Cpc = 4200
15 k = 0.01*exp(8000/8.314*(1/450-1/T))
16 ra = -k*(Ca^2-Cc/Kc)
12-100
P12-18 continued
Now, counter – current heat exchanger
Polymath code:
d(X)/d(W) = -ra/Fa0
d(p)/d(W) = -alpha/2/p*(1+epsilon*X)*T/T0
d(T)/d(W) = (ra*(DH+deltaCp*(T-298))-Ua*(T-Ta))/Fa0/Cp0
d(Ta)/d(W) =- Ua*(T-Ta)/m/Cpc
Cp0 = 40
DH = -20000
epsilon = -1
T0 = 450
Ca0 = 1.9
Fa0 = 5
alpha = 0.019
Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*p
Ua = 0.8
deltaCp = 1/2*20-40
Kc = 10000*exp(DH/8.314*(1/450-1/T))
Cc = Ca0*X/2/(1+epsilon*X)*T0/T*p
m = 0.05
Cpc = 4200
12-101
k = 0.01*exp(8000/8.314*(1/450-1/T))
ra = -k*(Ca^2-Cc/Kc)
W(0) = 0
W(f) = 50
X(0) = 0
p(0) = 1
T(0) = 450
Ta(0) = 500
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
alpha
0.019
0.019
0.019
0.019
2
Ca
1.9
0.7671508
1.910238
0.7671508
3
Ca0
1.9
1.9
1.9
1.9
4
Cc
0
0
0.2023329
0.1203647
5
Cp0
40.
40.
40.
40.
6
Cpc
4200.
4200.
4200.
4200.
7
deltaCp
-30.
-30.
-30.
-30.
8
DH
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
9
epsilon
-1.
-1.
-1.
-1.
10 Fa0
5.
5.
5.
5.
11 k
0.01
0.01
0.025358
0.025358
12 Kc
10000.
976.5939
10000.
976.5939
13 m
0.05
0.05
0.05
0.05
14 ra
-0.0361
-0.062846
-0.0149206
-0.0149206
15 T
450.
450.
796.6909
796.6909
16 T0
450.
450.
450.
450.
17 Ta
520.
500.0379
521.8302
500.0379
18 Ua
0.8
0.8
0.8
0.8
19 W
0
0
50.
50.
20 X
0
0
0.4958573
0.4958573
21 p
1.
0.2280604
1.
0.2280604
Differential equations
1 d(X)/d(W) = -ra/Fa0
2 d(p)/d(W) = -alpha/2/p*(1+epsilon*X)*T/T0
3 d(T)/d(W) = (ra*(DH+deltaCp*(T-298)) - Ua*(T-Ta))/Fa0/Cp0
4 d(Ta)/d(W) = - Ua*(T-Ta)/m/Cpc
Explicit equations
1
Cp0 = 40
2
DH = -20000
3
epsilon = -1
4
T0 = 450
5
Ca0 = 1.9
6
Fa0 = 5
7
alpha = 0.019
8
Ca = Ca0*(1-X)/(1+epsilon*X)*T/T0*p
9
Ua = 0.8
10 deltaCp = 1/2*20-40
11 Kc = 10000*exp(DH/8.314*(1/450-1/T))
12-102
12 Cc = Ca0*X/2/(1+epsilon*X)*T0/T*p
13 m = 0.05
14 Cpc = 4200
15 k = 0.01*exp(8000/8.314*(1/450-1/T))
16 ra = -k*(Ca^2-Cc/Kc)
P12-19 No solution will be given.
P12-20 (a) Figure 1 matches Figure _C_
P12-20 (b) Figure 2 matches Figure _A_
P12-20 (c) Figure 3 matches Figure _D_
P12-20 (d) Figure 4 matches Figure _B_
12-103
P12-21
First note that ΔCP = 0 for both reactions. This means that ΔHRx(T) = ΔHRx° for both reactions.
Now start with the differential energy balance for a PFR:
dT Ua(Ta −T)+ ∑rij (ΔHRxij ) Ua(Ta −T)+ r1A (ΔH1A )+ r2B (ΔHRx2B )
=
=
dV
∑FjCPj
∑FjCPj
If we evaluate this differential equation at its maximum we get
dT
= 0 and therefore, Ua(Ta −T)+ r1A (ΔHRx1C )+ r2B (ΔHRx2B ) = 0
dV
We can then solve for r1A from this information.
r1A = −
r1A = −
Ua(Ta −T)+ r2B (ΔHRx2B )
(ΔHRx1A )
Ua(Ta −T)−2k2DCBCC (ΔHRx2B )
r1 A = −
(ΔHRx1A )
10(325 − 500) − 2(0.4)(0.2)(0.5)(5000)
= −0.043
− 50000
r1A = −0.043 = − 1 k1C C ACB = − 1 k1C (0.1)(0.2)
2
2
k1C = 4.3
(E " 1
1 %+
k1C(500) = k1C(400) exp* $
−
'*) R # 400 500 &-,
( E " 1
1 %+
4.3 = 0.043exp*
−
$
'*)1.987 # 400 500 &-,
E = 18300
cal
mol ⋅K
Alternate Solution:
12-104
P12-22
at V = 0, T = T0 = 400 K for
dT
=0
dV
At the reactor entrance,
CA=CA0= 3 mol/dm3
CB=CB0= 1 mol/dm3
CC=CC0=0 mol/dm3
So,
−𝑟_Q = 1 ∗ 3• ∗ 1 = 9
−𝑟•N = 1.333 ∗ 3 = 4
−𝑟œQ = 0
(−9) ∗ (−10,000) + (−4) ∗ (10,000)
Ÿ
𝑇? = 400 − •
100
𝑇? = 400 − 500 = −100 𝐾
Ta is coming less than 0 K which is not possible
P12-23 (a)
Polymath code for Adiabatic operation:
d(Ta)/d(V) = Ua*(T-Ta)/m/Cpcool
d(T)/d(V) = (Qg-Qr)/sumFiCpi
d(Fc)/d(V) = rc
d(Fb)/d(V) = rb
d(Fa)/d(V) = ra
Cpc = 100
Cpb = 80
Cpa = 20
k2c2 = 4000
k1a1 = 50
T3 = 315
T2 = 310
T1 = 305
E2 = 4000
12-105
E1 = 8000
R = 8.31
k2c = k2c2*exp(E2/R*(1/T2-1/T))
k1a = k1a1*exp(E1/R*(1/T1-1/T))
Ft = Fa+Fb+Fc
K1c1 = 10
Cto = 0.2
To = 300
Cc = Cto*Fc/Ft*To/T
DH2B = 30000
Cb = Cto*Fb/Ft*To/T
Ca = Cto*Fa/Ft*To/T
DH1A = -20000
Cpcool = 10
m = 50
Ua = 200*0
Qr = Ua*(T-Ta)
r2c = k2c*Ca*Cb^2
r2b = -2*r2c
K1c = K1c1*exp(DH1A/R*(1/T3-1/T))
r1a = -k1a*(Ca^2-Cb/K1c)
Qg = r1a*DH1A+r2b*DH2B
Qg2 = r2b*DH2B
Qg1 = r1a*DH1A
r2a = -r2c
r1b = -r1a/2
rb = r1b+r2b
rc = r2c
ra = r1a+r2a
sumFiCpi = Fa*Cpa+Fb*Cpb+Fc*Cpc
V(0) = 0
V(f) = 10
Ta(0) = 320
T(0) = 300
Fa(0) = 5
Fb(0) = 0
Fc(0) = 0
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.2
0.0911174
0.2
0.0911174
2
Cb
0
0
0.0115217
0.0068936
3
Cc
0
0
0.0323539
0.0323539
4
Cpa
20.
20.
20.
20.
5
Cpb
80.
80.
80.
80.
6
Cpc
100.
100.
100.
100.
7
Cpcool
10.
10.
10.
10.
8
Cto
0.2
0.2
0.2
0.2
9
DH1A
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
10 DH2B
3.0E+04
3.0E+04
3.0E+04
3.0E+04
11 E1
8000.
8000.
8000.
8000.
12 E2
4000.
4000.
4000.
4000.
13 Fa
5.
1.708403
5.
1.708403
14 Fb
0
0
0.382013
0.1292515
15 Fc
0
0
0.6066188
0.6066188
16 Ft
5.
2.444273
5.
2.444273
17 k1a
47.43767
47.43767
144.996
144.996
18 k1a1
50.
50.
50.
50.
19 K1c
14.65244
0.8970751
14.65244
0.8970751
20 K1c1
10.
10.
10.
10.
12-106
21 k2c
3798.235
3798.235
6640.459
6640.459
22 k2c2
4000.
4000.
4000.
4000.
23 m
50.
50.
50.
50.
24 Qg
3.795E+04
66.48175
3.795E+04
66.48175
25 Qg1
3.795E+04
1791.7
3.795E+04
1791.7
26 Qg2
0
-6202.384
0
-1725.218
27 Qr
0
0
0
0
28 R
8.31
8.31
8.31
8.31
29 r1a
-1.897507
-1.897507
-0.089585
-0.089585
30 r1b
0.9487534
0.0447925
0.9487534
0.0447925
31 r2a
0
-0.1033731
0
-0.0287536
32 r2b
0
-0.2067461
0
-0.0575073
33 r2c
0
0
0.1033731
0.0287536
34 ra
-1.897507
-1.897507
-0.1183386
-0.1183386
35 rb
0.9487534
-0.0462432
0.9487534
-0.0127148
36 rc
0
0
0.1033731
0.0287536
37 sumFiCpi 100.
100.
115.2805
105.1701
38 T
300.
300.
460.2464
460.2464
39 T1
305.
305.
305.
305.
40 T2
310.
310.
310.
310.
41 T3
315.
315.
315.
315.
42 Ta
320.
320.
320.
320.
43 To
300.
300.
300.
300.
44 Ua
0
0
0
0
45 V
0
0
10.
10.
Polymath code for constant Ta operation
d(Ta)/d(V) = 0*Ua*(T-Ta)/m/Cpcool
d(T)/d(V) = (Qg-Qr)/sumFiCpi
d(Fc)/d(V) = rc
d(Fb)/d(V) = rb
d(Fa)/d(V) = ra
Cpc = 100
Cpb = 80
Cpa = 20
k2c2 = 4000
k1a1 = 50
T3 = 315
T2 = 310
T1 = 305
E2 = 4000
E1 = 8000
12-107
R = 8.31
k2c = k2c2*exp(E2/R*(1/T2-1/T))
k1a = k1a1*exp(E1/R*(1/T1-1/T))
Ft = Fa+Fb+Fc
K1c1 = 10
Cto = 0.2
To = 300
Cc = Cto*Fc/Ft*To/T
DH2B = 30000
Cb = Cto*Fb/Ft*To/T
Ca = Cto*Fa/Ft*To/T
DH1A = -20000
Cpcool = 10
m = 50
Ua = 200
Qr = Ua*(T-Ta)
r2c = k2c*Ca*Cb^2
r2b = -2*r2c
K1c = K1c1*exp(DH1A/R*(1/T3-1/T))
r1a = -k1a*(Ca^2-Cb/K1c)
Qg = r1a*DH1A+r2b*DH2B
Qg2 = r2b*DH2B
Qg1 = r1a*DH1A
r2a = -r2c
r1b = -r1a/2
rb = r1b+r2b
rc = r2c
ra = r1a+r2a
sumFiCpi = Fa*Cpa+Fb*Cpb+Fc*Cpc
V(0) = 0
V(f) = 10
Ta(0) = 320
T(0) = 300
Fa(0) = 5
Fb(0) = 0
Fc(0) = 0
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.2
0.0295805
0.2
0.0295805
2
Cb
0
0
0.0227674
0.006257
3
Cc
0
0
0.1519534
0.1519534
4
Cpa
20.
20.
20.
20.
5
Cpb
80.
80.
80.
80.
6
Cpc
100.
100.
100.
100.
7
Cpcool
10.
10.
10.
10.
8
Cto
0.2
0.2
0.2
0.2
9
DH1A
-2.0E+04
-2.0E+04
-2.0E+04
-2.0E+04
10 DH2B
3.0E+04
3.0E+04
3.0E+04
3.0E+04
11 E1
8000.
8000.
8000.
8000.
12 E2
4000.
4000.
4000.
4000.
13 Fa
5.
0.1844489
5.
0.1844489
14 Fb
0
0
0.4721449
0.0390154
15 Fc
0
0
0.947504
0.947504
16 Ft
5.
1.170968
5.
1.170968
17 k1a
47.43767
47.43767
107.401
57.70313
18 k1a1
50.
50.
50.
50.
19 K1c
14.65244
1.899756
14.65244
8.978836
20 K1c1
10.
10.
10.
10.
12-108
21 k2c
3798.235
3798.235
5715.106
4189.092
22 k2c2
4000.
4000.
4000.
4000.
23 m
50.
50.
50.
50.
24 Qg
3.795E+04
-974.5289
3.795E+04
-85.48644
25 Qg1
3.795E+04
205.5906
3.835E+04
205.5906
26 Qg2
0
-1.878E+04
0
-291.0771
27 Qr
-4000.
-4000.
1.65E+04
-99.12652
28 R
8.31
8.31
8.31
8.31
29 r1a
-1.897507
-1.917491
-0.0102795
-0.0102795
30 r1b
0.9487534
0.0051398
0.9587454
0.0051398
31 r2a
0
-0.3130527
0
-0.0048513
32 r2b
0
-0.6261054
0
-0.0097026
33 r2c
0
0
0.3130527
0.0048513
34 ra
-1.897507
-1.953803
-0.0151308
-0.0151308
35 rb
0.9487534
-0.166039
0.9487534
-0.0045628
36 rc
0
0
0.3130527
0.0048513
37 sumFiCpi 100.
100.
118.8858
101.5606
38 T
300.
300.
402.4929
319.5044
39 T1
305.
305.
305.
305.
40 T2
310.
310.
310.
310.
41 T3
315.
315.
315.
315.
42 Ta
320.
320.
320.
320.
43 To
300.
300.
300.
300.
44 Ua
200.
200.
200.
200.
45 V
0
0
10.
10.
Polymath code for Co current operation:
d(Ta)/d(V) = Ua*(T-Ta)/m/Cpcool
d(T)/d(V) = (Qg-Qr)/sumFiCpi
d(Fc)/d(V) = rc
d(Fb)/d(V) = rb
d(Fa)/d(V) = ra
Cpc = 100
Cpb = 80
Cpa = 20
k2c2 = 4000
k1a1 = 50
T3 = 315
T2 = 310
T1 = 305
12-109
E2 = 4000
E1 = 8000
R = 8.31
k2c = k2c2*exp(E2/R*(1/T2-1/T))
k1a = k1a1*exp(E1/R*(1/T1-1/T))
Ft = Fa+Fb+Fc
K1c1 = 10
Cto = 0.2
To = 300
Cc = Cto*Fc/Ft*To/T
DH2B = 30000
Cb = Cto*Fb/Ft*To/T
Ca = Cto*Fa/Ft*To/T
DH1A = -20000
Cpcool = 10
m = 50
Ua = 200
Qr = Ua*(T-Ta)
r2c = k2c*Ca*Cb^2
r2b = -2*r2c
K1c = K1c1*exp(DH1A/R*(1/T3-1/T))
r1a = -k1a*(Ca^2-Cb/K1c)
Qg = r1a*DH1A+r2b*DH2B
Qg2 = r2b*DH2B
Qg1 = r1a*DH1A
r2a = -r2c
r1b = -r1a/2
rb = r1b+r2b
rc = r2c
ra = r1a+r2a
sumFiCpi = Fa*Cpa+Fb*Cpb+Fc*Cpc
V(0) = 0
V(f) = 10
Ta(0) = 320
T(0) = 300
Fa(0) = 5
Fb(0) = 0
Fc(0) = 0
Similarly, for Counter current operation, replace the first differential equation with minus sign and guess
Ta(0)= 359.35 K to get Ta (f)= 320 K
Polymath code
d(Ta)/d(V) = -Ua*(T-Ta)/m/Cpcool
d(T)/d(V) = (Qg-Qr)/sumFiCpi
d(Fc)/d(V) = rc
d(Fb)/d(V) = rb
d(Fa)/d(V) = ra
Cpc = 100
Cpb = 80
12-110
Cpa = 20
k2c2 = 4000
k1a1 = 50
T3 = 315
T2 = 310
T1 = 305
E2 = 4000
E1 = 8000
R = 8.31
k2c = k2c2*exp(E2/R*(1/T2-1/T))
k1a = k1a1*exp(E1/R*(1/T1-1/T))
Ft = Fa+Fb+Fc
K1c1 = 10
Cto = 0.2
To = 300
Cc = Cto*Fc/Ft*To/T
DH2B = 30000
Cb = Cto*Fb/Ft*To/T
Ca = Cto*Fa/Ft*To/T
DH1A = -20000
Cpcool = 10
m = 50
Ua = 200
Qr = Ua*(T-Ta)
r2c = k2c*Ca*Cb^2
r2b = -2*r2c
K1c = K1c1*exp(DH1A/R*(1/T3-1/T))
r1a = -k1a*(Ca^2-Cb/K1c)
Qg = r1a*DH1A+r2b*DH2B
Qg2 = r2b*DH2B
Qg1 = r1a*DH1A
r2a = -r2c
r1b = -r1a/2
rb = r1b+r2b
rc = r2c
ra = r1a+r2a
sumFiCpi = Fa*Cpa+Fb*Cpb+Fc*Cpc
V(0) = 0
V(f) = 10
Ta(0) = 359.35
T(0) = 300
Fa(0) = 5
Fb(0) = 0
Fc(0) = 0
P12-23 (b) The maximum temperature is 460.2 K and is reached at V = 10 m3.
P12-23 (c) The flow rate of B is maximum at V = 0.5 m3 and the value is 0.37 mol/min
12-111
P12-23 (d) The maximum temperature is 402.5 K and is reached at V = 0.6 m3.
P12-23 (e) The flow rate of B is maximum at V = 1 m3 and the value is 0.47 mol/min
P12-23 (f) Ta becomes greater than T at V = 2.8 m3. As the reactions proceed, reaction 2 becomes
prominent. Being a exothermic reaction, it heat ups the coolant.
P12-23 (g) Ta becomes greater than T at V = 3.5 m3.
P12-24
Polymath code:
d(Ca)/d(V) = (r1a+r2a)/vo
d(Cb)/d(V) = 2*r1a/vo
d(Cc)/d(V) = (-2*r1a+r2a)/vo
d(Cd)/d(V) = -2*r2a/vo
d(T)/d(V) = (r1a*dh1a+r2a*dh2a)/((Ca+Cb+3*Cc+4*Cd)*vo*Cpa)
vo = 10
dh1a = 20000
dh2a = -10000
Cpa = 20
k1a = 0.001*exp(-5000/1.987*(1/T-1/300))
k2a = 0.001*exp(-7500/1.987*(1/T-1/300))
r1a = -k1a*Ca*Cb^2
r2a = -k2a*Ca*Cc
V(0) = 0
V(f) = 10
Ca(0) = 2
Cb(0) = 4
Cc(0) =0
Cd(0) = 0
T(0) = 600
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
2.
1.053265
2.
1.053265
2
Cb
4.
2.6518
4.
2.6518
3
Cc
0
0
1.075564
1.075564
4
Cd
0
0
0.5452704
0.5452704
5
Cpa
20.
20.
20.
20.
6
dh1a
2.0E+04
2.0E+04
2.0E+04
2.0E+04
7
dh2a
-10000.
-10000.
-10000.
-10000.
8
k1a
0.0662826
0.036197
0.0662826
0.036197
9
k2a
0.5396338
0.2177756
0.5396338
0.2177756
10 r1a
-2.121043
-2.121043
-0.2680971
-0.2680971
11 r2a
0
-0.3140768
0
-0.2467081
12 T
600.
524.3635
600.
524.3635
13 V
0
0
10.
10.
14 vo
10.
10.
10.
10.
Differential equations
1 d(Ca)/d(V) = (r1a+r2a)/vo
2 d(Cb)/d(V) = 2*r1a/vo
3 d(Cc)/d(V) = (-2*r1a+r2a)/vo
4 d(Cd)/d(V) = -2*r2a/vo
5 d(T)/d(V) = (r1a*dh1a+r2a*dh2a)/((Ca+Cb+3*Cc+4*Cd)*vo*Cpa)
12-112
Explicit equations
1 vo = 10
2 dh1a = 20000
3 dh2a = -10000
4 Cpa = 20
5 k1a = 0.001*exp(-5000/1.987*(1/T-1/300))
6 k2a = 0.001*exp(-7500/1.987*(1/T-1/300))
7 r1a = -k1a*Ca*Cb^2
8 r2a = -k2a*Ca*Cc
By varying the inlet temp from 300 K to 675 K, we can generate the following table
T(K)
300
400
500
600
650
625
675
0.058
0.346
0.78
1.075
1.1025
1.099
1.088
Since the maximum allowable temp is 600 K, we would recommend to keep temp at 600 K to maximize
production of C
P12-25 (a)
Mole balance:
Rate Laws:
𝑟_N = −𝑘• 𝐶N
𝑟N = −𝑘• 𝐶N + 𝑘_ 𝐶Q
𝑟œQ = −𝑘œ 𝐶Q
𝑟L = −𝑟œQ
𝑟•Q = −𝑘_ 𝐶Q
𝑟Q = −𝑟_N + 𝑟•Q + 𝑟œQ
Stoichiometry:
𝐶L = 𝐶;
𝐹L 𝑇U
𝐹; 𝑇
12-113
Energy balance:
𝑑𝑇
𝑈𝑎(𝑇? − 𝑇) + (−𝑟N )(−∆𝐻)‘_N ) + (−𝑟œQ )(−∆𝐻)‘œN )
=
𝑑𝑊
𝐹N 𝐶kN + 𝐹Q 𝐶kQ + 𝐹L 𝐶kL
P12-25 (d)
Considering O-Xylene as A, M-xylene as B and P-xylene as C
Polymath Code:
d(fa)/d(w) = ra
d(fb)/d(w) = rb
d(fc)/d(w) = rc
d(T)/d(w) = (Ua*(Ta-T)+(r2b-r1a)*(-Dhr1a)+(-r3a)*(-Dhr3a))/(fa*cpa+fb*cpb+fc*cpc)
Ua = 16
Ta = 500
Dhr1a = -1800
Dhr3a = -1100
cpa = 100
cpb = 100
cpc = 100
k1 = 0.5*exp(2*(1-320/T))
k3 = 0.005*exp(4.6*(1-460/T))
ct = 2
ft = 2
To = 330
Kc = 10*exp(4.8*(430/T-1.5))
k2 = k1/Kc
ca = ct*fa/ft*To/T
cb = ct*fb/ft*To/T
cc=ct*fc/ft*To/T
r1a = -k1*ca
r3a = -k3*ca
rc = -r3a
r2b = -k2*cb
rb = -r1a+r2b
ra = -r2b+r1a+r3a
w(0) = 0
w(f) = 100
fa(0) = 1
fb(0) = 0
fc(0) = 0
T(0) = 330
12-114
The maximum concentration of M-xylene is 0.47 mol/dm3 at W= 2.5 kg
P12-25 (e) Individualized solution
P12-25 (f) Individualized solution
P12-26 (a)
We want the exiting flow rates B, D and F
Start with the mole balance in PFR:
Rate Laws:
12-115
Stoichiometry:
Energy Balance:
See the following Polymath program
For T0 = 800K
d(fa)/d(v) = ra
d(fb)/d(v) = rb
d(fc)/d(v) = rc
d(fd)/d(v) = rd
d(fe)/d(v) = re
d(ff)/d(v) = rf
d(fg)/d(v) = rg
d(T)/d(v) = -(rls*Hla+r2b*H2a+r3t*H3a)/(fa*299+fb*273+fc*30+fd*201+fe*90+ff*68+fi*40)
Hla = 118000
H2a = 105200
H3a = -53900
p = 2137
phi = 0.4
Kl = exp(-17.34-1.302e4/T+5.051*ln(T)+((-2.314e-10*T+1.302e-6)*T-0.004931)*T)
sr = 14.5
fi = sr*0.00344
ft = fa+fb+fc+fd+fe+ff+fg+fi
Pa = fa/ft*2.4
Pb = fb/ft*2.4
Pc = fc/ft*2.4
r2b = p*(1-phi)*exp(13.2392-25000/T)*Pa
rd = r2b
re = r2b
r3t = p*(1-phi)*exp(0.2961-11000/T)*Pa*Pc
rf = r3t
rg = r3t
rls = p*(1-phi)*exp(-0.08539-10925/T)*(Pa-Pb*Pc/Kl)
rb = rls
rc = rls-r3t
ra = -rls-r2b-r3t
v(0) = 0
v(f) = 10
fa(0) = 0.00344
fb(0) = 0
fc(0) = 0
fd(0) = 0
fe(0) = 0
ff(0) = 0
fg(0) = 0
T(0) = 800
12-116
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
v
0
0
10.
10.
2
fa
0.00344
0.002496
0.00344
0.002496
3
fb
0
0
0.0008974
0.0008974
4
fc
0
0
0.0008615
0.0008615
5
fd
0
0
1.078E-05
1.078E-05
6
fe
0
0
1.078E-05
1.078E-05
7
ff
0
0
3.588E-05
3.588E-05
8
fg
0
0
3.588E-05
3.588E-05
9
T
800.
765.237
800.
765.237
10 Hla
1.18E+05
1.18E+05
1.18E+05
1.18E+05
11 H2a
1.052E+05
1.052E+05
1.052E+05
1.052E+05
12 H3a
-5.39E+04
-5.39E+04
-5.39E+04
-5.39E+04
13 p
2137.
2137.
2137.
2137.
14 phi
0.4
0.4
0.4
0.4
15 Kl
0.0459123
0.0196554
0.0459123
0.0196554
16 sr
14.5
14.5
14.5
14.5
17 fi
0.04988
0.04988
0.04988
0.04988
18 ft
0.05332
0.05332
0.0542282
0.0542282
19 Pa
0.1548387
0.1104652
0.1548387
0.1104652
20 Pb
0
0
0.0397155
0.0397155
21 Pc
0
0
0.0381277
0.0381277
22 r2b
2.991E-06
5.16E-07
2.991E-06
5.16E-07
23 rd
2.991E-06
5.16E-07
2.991E-06
5.16E-07
24 re
2.991E-06
5.16E-07
2.991E-06
5.16E-07
25 r3t
0
0
4.196E-06
4.151E-06
26 rf
0
0
4.196E-06
4.151E-06
27 rg
0
0
4.196E-06
4.151E-06
28 rls
0.0002138
2.481E-05
0.0002138
2.481E-05
29 rb
0.0002138
2.481E-05
0.0002138
2.481E-05
30 rc
0.0002138
2.066E-05
0.0002138
2.066E-05
31 ra
-0.0002167
-0.0002167
-2.948E-05
-2.948E-05
Differential equations
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
7 d(fg)/d(v) = rg
8 d(T)/d(v) = -(rls * Hla + r2b * H2a + r3t * H3a) / (fa * 299 + fb * 273 + fc * 30 + fd * 201 + fe * 90 + ff * 68 + fi * 40)
Explicit equations
1 Hla = 118000
2 H2a = 105200
3 H3a = -53900
4 p = 2137
5 phi = .4
6 Kl = exp(-17.34 - 1.302e4 / T + 5.051 * ln(T) + ((-2.314e-10 * T + 1.302e-6) * T + -0.004931) * T)
7 sr = 14.5
12-117
8 fi = sr * .00344
9 ft = fa + fb + fc + fd + fe + ff + fg + fi
10 Pa = fa / ft * 2.4
11 Pb = fb / ft * 2.4
12 Pc = fc / ft * 2.4
13 r2b = p * (1 - phi) * exp(13.2392 - 25000 / T) * Pa
14 rd = r2b
15 re = r2b
16 r3t = p * (1 - phi) * exp(.2961 - 11000 / T) * Pa * Pc
17 rf = r3t
18 rg = r3t
19 rls = p * (1 - phi) * exp(-0.08539 - 10925 / T) * (Pa - Pb * Pc / Kl)
20 rb = rls
21 rc = rls - r3t
22 ra = -rls - r2b - r3t
P12-26 (a)
Fstyrene = 0.0008974
Fbenzene = 1.078E-05
Ftoluene = 3.588E-05
SS/BT = 19.2
P12-26 (b)
T0 = 930K
Fstyrene = 0.0019349
Fbenzene = 0.0002164
Ftoluene = 0.0002034
SS/BT = 4.6
P12-26 (c)
T0 = 1100 K
Fstyrene = 0.0016543
Fbenzene = 0.0016067
Ftoluene = 0.0001275
SS/BT = 0.95
P12-26 (d)
Plotting the production of styrene as a function of To gives the following graph. The temperature that is
ideal is 995K
12-118
P12-26 (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
P12-26 (f)
See the following Polymath program.
d(fa)/d(v) = ra
d(fb)/d(v) = rb
d(fc)/d(v) = rc
d(fd)/d(v) = rd
d(fe)/d(v) = re
d(ff)/d(v) = rf
d(fg)/d(v) = rg
d(T)/d(v) = (Ua*(Ta-T)-(rls*Hla+r2b*H2a+r3t*H3a))/(fa*299+fb*273+fc*30+fd*201+fe*90+ff*68+fi*40)
Hla = 118000
H2a = 105200
H3a = -53900
Ua = 100
Ta = 1000
p = 2137
phi = 0.4
Kl = exp(-17.34-1.302e4/T+5.051*ln(T)+((-2.314e-10*T+1.302e-6)*T-0.004931)*T)
sr = 14.5
fi = sr*0.00344
ft = fa+fb+fc+fd+fe+ff+fg+fi
Pa = fa/ft*2.4
Pb = fb/ft*2.4
Pc = fc/ft*2.4
r2b = p*(1-phi)*exp(13.2392-25000/T)*Pa
rd = r2b
re = r2b
r3t = p*(1-phi)*exp(0.2961-11000/T)*Pa*Pc
rf = r3t
rg = r3t
rls = p*(1-phi)*exp(-0.08539-10925/T)*(Pa-Pb*Pc/Kl)
rb = rls
rc = rls-r3t
ra = -rls-r2b-r3t
v(0) = 0
v(f) = 10
fa(0) = 0.00344
fb(0) = 0
fc(0) = 0
fd(0) = 0
fe(0) = 0
ff(0) = 0
fg(0) = 0
T(0) = 400
When we add a heat exchanger to the reactor, the energy balance becomes:
12-119
With Ta = 1000 K
Ua = 100 kJ/min/K = 1.67 kJ/s/K
The recommended entering temperature would be T0 = 440 K. This gives the highest outlet flow rate of
styrene.
P12-26 (g) Individualized solution
P12-26 (h) Individualized solution
P12-27
Let
a=A
b = A2
c = A4
Polymath code:
d(T)/d(V) = (Qg-Qr)/sumCp
d(Fc)/d(V) = rc
d(Fb)/d(V) = rb
d(Fa)/d(V) = ra
Cpc = 100
Cpb = 50
Cpa = 25
k2 = 0.35
Ta = 315
Ua = 1000
DH2b = -27500
DH1a = -32500
T1 = 300
E1 = 4000
k1 = 0.6
T2 = 320
E2 = 5000
k2b = k2*exp((E2/1.987)*(1/T2-1/T))
vo = 50
Cb = Fb/vo
Ca = Fa/vo
sumCp = (Fa*Cpa+Fb*Cpb+Fc*Cpc)
k1a = k1*exp((E1/1.987)*(1/T1-1/T))
r2b = -k2b*Cb^2
rc = -r2b/2
r1a = -k1a*Ca^2
12-120
r1b = -r1a/2
rb = r1b+r2b
ra = r1a
Qg = r1a*DH1a+r2b*DH2b
Qr = Ua*(T-Ta)
V(0) = 0
V(f) = 10
T(0) = 300
Fa(0) = 100
Fb(0) = 0
Fc(0) = 0
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
2.
0.155599
2.
0.155599
2
Cb
0
0
0.4963834
0.1995088
3
Cpa
25.
25.
25.
25.
4
Cpb
50.
50.
50.
50.
5
Cpc
100.
100.
100.
100.
6
DH1a
-3.25E+04
-3.25E+04
-3.25E+04
-3.25E+04
7
DH2b
-2.75E+04
-2.75E+04
-2.75E+04
-2.75E+04
8
E1
4000.
4000.
4000.
4000.
9
E2
5000.
5000.
5000.
5000.
10 Fa
100.
7.779951
100.
7.779951
11 Fb
0
0
24.81917
9.975439
12 Fc
0
0
18.06729
18.06729
13 k1
0.6
0.6
0.6
0.6
14 k1a
0.6
0.6
76.97387
6.133163
15 k2
0.35
0.35
0.35
0.35
16 k2b
0.2072016
0.2072016
89.46089
3.787125
17 Qg
7.8E+04
8971.335
1.569E+06
8971.335
18 Qr
-1.5E+04
-1.5E+04
7.696E+05
1.44E+05
19 r1a
-2.4
-42.01417
-0.1484903
-0.1484903
20 r1b
1.2
0.0742452
21.00708
0.0742452
21 r2b
0
-18.03538
0
-0.1507418
22 ra
-2.4
-42.01417
-0.1484903
-0.1484903
23 rb
1.2
-8.261933
17.21274
-0.0764966
24 rc
0
0
9.017691
0.0753709
25 sumCp
2500.
2500.
2500.
2500.
26 T
300.
300.
1084.641
459.0062
27 T1
300.
300.
300.
300.
28 T2
320.
320.
320.
320.
29 Ta
315.
315.
315.
315.
30 Ua
1000.
1000.
1000.
1000.
31 V
0
0
10.
10.
32 vo
50.
50.
50.
50.
Differential equations
1 d(T)/d(V) = (Qg-Qr)/sumCp
2 d(Fc)/d(V) = rc
3 d(Fb)/d(V) = rb
4 d(Fa)/d(V) = ra
12-121
Explicit equations
1 Cpc = 100
2 Cpb = 50
3 Cpa = 25
4 k2 = .35
5 Ta = 315
6 Ua = 1000
7 DH2b = -27500
8 DH1a = - 32500
9 T1 = 300
10 E1 = 4000
11 k1 = 0.6
12 T2 = 320
13 E2 = 5000
14 k2b = k2*exp((E2/1.987)*(1/T2-1/T))
15 vo = 50
16 Cb = Fb/vo
17 Ca = Fa/vo
18 sumCp = (Fa*Cpa+Fb*Cpb+Fc*Cpc)
19 k1a = k1*exp((E1/1.987)*(1/T1-1/T))
20 r2b = -k2b*Cb^2
21 rc = -r2b/2
22 r1a = -k1a*Ca^2
23 r1b = -r1a/2
24 rb = r1b + r2b
25 ra = r1a
26 Qg = r1a*DH1a+r2b*DH2b
27 Qr = Ua*(T-Ta)
12-122
P12-27 (b)
The required reactor volume = 3.5 dm3
P12-27 (c)
Polymath code:
d(T)/d(V) = (Qg-Qr)/sumCp
d(Fc)/d(V) = rc
d(Fb)/d(V) = rb
d(Fa)/d(V) = ra
Cpc = 100
Cpb = 50
Cpa = 25
k2 = 0.35
Ta = 315
Ua = 1000
DH2b = -27500
DH1a = -32500
T1 = 300
E1 = 4000
k1 = 0.6
T2 = 320
E2 = 5000
k2b = k2*exp((E2/1.987)*(1/T2-1/T))
vo = 50
Cb = Fb/vo
Ca = Fa/vo
sumCp = (Fa*Cpa+Fb*Cpb+Fc*Cpc)
k1a = k1*exp((E1/1.987)*(1/T1-1/T))
r2b = -k2b*Cb^2
rc = -r2b/2
r1a = -k1a*Ca^2
r1b = -r1a/2
rb = r1b+r2b
ra = r1a
Qg = r1a*DH1a+r2b*DH2b
Qr = Ua*(T-Ta)
V(0) = 0
V(f) = 10
T(0) = 450
Fa(0) = 100
12-123
Fb(0) = 0
Fc(0) = 0
12-124
P12-27 (c)
12-125
Page intentionally blank
12-126
Synopsis for Chapter 13 – Unsteady State Nonisothermal
Reactor Design
General: This chapter has a major focus on the safety through Examples 13-2 and 13-6.
Questions
l Q13-1A (20 seconds) Questions Before Reading (QBR).
O Q13-2A (10-15 min) i>Clicker – Good time to review questions.
AA Q13-3A (20-25 min) This list should be saved and perhaps memorized/learned for working with
potential runaway reactions in industry.
I Q13-5A (10 min) The Monsanto and Synthron explosions are excellent discussion problems and are
discussed on the safety website (http://umich.edu/~safeche/).
Computer Simulations and Experiments
All Wolfram simulations are always assigned.
l P13-1B (a) (10 min) Wolfram and Python. Good introduction to unsteady state. I would not spend
too much time trying get the cobra shape, i.e.,
mentioned in the problem statement.
l P13-1B (b) (30 min) Wolfram and Python. Outstanding simulations. Student should use Wolfram to
construct and explore the drawing in the text showing the regions of potential explosions.
l P13-1B (c) (10-12 min) Wolfram and Python. See if the student can find initial conditions when the
practical stability limit is exceeded. Students should also explore the COMSOL safety module.
l P13-1B (d) (10-12 min) Wolfram and Python. Students should also go to the Safety Website
(http://umich.edu/~safeche/) and carry out the Synthron Safety Module.
l P13-1B (e) (10-12 min) Wolfram and Python. Excellent multiple reaction example.
l P13-1B (f) (10-15 min) Wolfram and Python. Outstanding analysis of an actual tragic accident. The
students should first view the video on the Safety Website (http://umich.edu/~safeche/) and then
use the Wolfram sliders to explore the T2 Laboratories accident.
O P13-1B (g) – (i) (10-12 min) Wolfram should be used to explore these LEPs.
Problems
l P13-2B (60 min) Old Exam Question (OEQ) and Questions Before Reading (QBR). Actual Event. Safety
problem to calculate the time it took for the explosion to occur after the feed stream was abruptly
shut off, which it should not have been abruptly.
O P13-3B (60 min) Batch reactor with heat effects using data from problems in Chapters 11 and 12.
AA P13-4B (60 min) Old Exam Question (OEQ). This problem has a twist as it asks the flow rate of inert
as a function of time FI(t), necessary to keep the reactor isothermal. A computer is not necessary to
solve this problem.
O P13-5B (45 min) Batch reactor with heat effects. Rather straight forward problem using Polymath
S13-1
AA P13-6B (35 min) Somewhat open-ended problem of a semibatch reactor with heat effect to find
optimum volumetric flow rate.
AA P13-7B (40 min) Straight forward batch reactor with a single reaction. Somewhat open-ended as
students are asked to explore the problem using Polymath .
AA P13-8B (30 min) Start-up problem in which the students should be looking for initial conditions that
exceed the practical stability limit (i.e., temperature).
AA P13-9B (30 min) Old Exam Question (OEQ). Multiple reactions with heat effects which does not
require Polymath to find solutions.
O P13-10B (75 min) In order to work through this problem, you will need to go to the Safeche website
http://umich.edu/~safeche/ and read through the safety module.
O P13-11B (45 min) This problem belongs in Chapter 9, however that chapter had already gone to
press at the time Covid-19 had shut down most universities. Because the principles of Chapter 9 can
be applied to the spread of the virus, it is included here in Chapter 13.
S13-2
Solutions for Chapter 13 – Unsteady State Nonisothermal
Reactor Design
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-13
http://umich.edu/~elements/5e/13chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q13-1 Individualized solution.
Q13-2 Individualized solution.
Q13-3 Individualized solution.
Q13-4 Individualized solution.
Q13-5 Individualized solution.
P13-1 (a) Example 13-1
(i)
The reaction runs away at T0 = 282.2 K
(ii) (1) Runaway occurs at Ta1 = 292 K
(2) As Ta1 increases, the maxima of the Qr versus t, and Qg versus T curves occur at shorter times.
This is because the inlet temperature of the coolant is increased, due to which the driving
force for heat transfer becomes lesser. As NA0 increases, Qr,max and Qg,max increase. This is
because we are starting with a larger number of reactant molecules, which gives rise to higher
generated heat. As mc is increased, the heat generated decreases. This is because a larger
amount of heat is removed with an increase in flow rate of the coolant.
(iii) Coolant flow rate does not have a significant effect on the conversion. The temperature of the
reactor decreases with an increase in coolant flow rate. This is an intuitive observation.
(iv) The new T0 of 20 ˚F, T=276+80X (neglecting change in heat of reaction). The polymath program
gives t= 4908 s for 90 % conversion.
Polymath Code
T=276+80*X
d(X) / d(t) = k*(1-X)
X(0) = 0
t(0) = 0
t(f) = 4908
k = .000273*exp(9059*(1/297-1/T))
13-1
Polymath Output
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 k
2.681E-05
2.681E-05
0.0237739
0.0237739
2 T
276.
276.
347.9572
347.9572
3 t
0
0
4908.
4908.
4 X
0
0
0.8994652
0.8994652
Differential equations
1 d(X)/d(t) = k*(1-X)
Explicit equations
1 T = 276+80*X
2 k = .000273*exp(9059*(1/297-1/T))
(v) NA0= CA0*V= 1.2 lbmol= 544.8 gmol
UA=99.85 cal/mol.K
Cps=Sum NCP/NA0=403/1=403 Btu/lbmol/R= 403.27 cal/mol/K
Polymath Code
d(T)/d(t) = (Qg-Qr)/Cps/Nao
d(X)/d(t) = k*(1-X)
UA = 99.85
DeltaH = -20202
Ta1 = 498
Cpc = 4.18
Cps = 403.27
mc = 10
R = 1.987
E = 18000
k = .000273*exp((E/R)*(1/297-1/T))
Ta2 = T-(T-Ta1)*exp(-UA/mc/Cpc)
Qr = mc*Cpc*(T-Ta1)*(1-exp(-UA/mc/Cpc))
Qg = Nao* k*(1-X)*(-DeltaH)
DeltaQ = Qr-Qg
Nao = 544.8
T(0)=286
X(0)=0
t(0) = 0
t(f) = 4000
13-2
Polymath Report
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Cpc
4.18
4.18
4.18
4.18
2 Cps
403.27
403.27
403.27
403.27
3 DeltaH
-2.02E+04
-2.02E+04
-2.02E+04
-2.02E+04
4 DeltaQ
-8978.237
-6.729E+04
-2971.305
-2971.305
5 E
1.8E+04
1.8E+04
1.8E+04
1.8E+04
6 k
8.447E-05
8.447E-05
2.039612
2.039612
7 mc
10.
10.
10.
10.
8 Nao
544.8
544.8
544.8
544.8
9 Qg
929.6335
-0.5359172
6.168E+04
-0.1112958
10 Qr
-8048.604
-8048.604
-2971.417
-2971.417
11 R
1.987
1.987
1.987
1.987
12 T
286.
286.
419.733
419.733
13 t
0
0
4000.
4000.
14 Ta1
498.
498.
498.
498.
15 Ta2
305.4497
305.4497
426.9135
426.9135
16 UA
99.85
99.85
99.85
99.85
17 X
0
0
1.
1.
Differential equations
1 d(T)/d(t) = (Qg-Qr)/Cps/Nao
2 d(X)/d(t) = k*(1-X)
Explicit equations
1 UA = 99.85
2 DeltaH = -20202
3 Ta1 = 498
4 Cpc = 4.18
5 Cps = 403.27
6 mc = 10
7 R = 1.987
8 E = 18000
9 k = .000273*exp((E/R)*(1/297-1/T))
10 Ta2 = T-(T-Ta1)*exp(-UA/mc/Cpc)
11 Qr = mc*Cpc*(T-Ta1)*(1-exp(-UA/mc/Cpc))
12 Nao = 544.8
13 Qg = Nao* k*(1-X)*(-DeltaH)
14 DeltaQ = Qr-Qg
13-3
P13-1 (b) Example 13-2
(i)
As NA0 increases, the exponential rise in temperature to the ‘point of no return’ occurs at a shorter
time. This is because Qg becomes more than Qr, and this difference increases with an increase in
NA0.
(ii) Put t1 and t2 = 45 minutes and see that the desired temperature is maintained.
(iii) The graph of t1 vs (t2-t1) helps in identifying the region where explosion can occur. If t1 is known,
then from the following graph, one can determine the maximum allowable downtime at which no
explosion would occur.
13-4
(iv)
No explosion would occur for NA0 < 8.7 kmol. Similarly, no explosion would have occurred for NB0 <
32.2 kmol
(v) No explosion would occur for NCp > 2665 kCal/K
No explosion would occur for UA > 37.2 kCal/min C
(vi) Individualized solution.
(vii) 1. Investigate why heat exchanger failed in the past and implement suitable measures.
2. Before increasing charge, make sure you know how increased charge would affect the reaction
conditions.
3. Monitor increased temperature in the reactor and have an emergency shut down procedure.
(viii) To show that no explosion occurred without cooling failure, Put t1 = t2 = 45 in the polymath code
of LEP 13-2
Polymath Code
d(T)/d(t) = if (t<t1) then (0) else ((Qg-Qr)/NCp)
d(X)/d(t) = (-ra)*V/Nao
Nao=9.044 # kmol
Nbo=33 # kmol
Nw=103.7 # kmol
mao=Nao*157.55 #kg
mbo=Nbo*17.03 #kg
mw=Nw*18.03 # kg
NCp= Nao*40+Nw*18+Nbo*8.38 # kcal/K
Va = mao/1199 #m3
Vaqam= 3.9 #m3
V=Va+Vaqam #m3
DeltaHrx = -590000 # kcal/kmol
k = .00017*exp(11273/(1.987)*(1/461-1/T)) # m3/(kmol.min)
Qr = if(t>t1 and t<t2) then (0) else (UA*(T-298)) # kcal/min
UA= 35.85 #kcal/(min C)
t1=45 #min
t2=45 #min
Theata = Nbo/Nao
ra = -k*Nao^2*(1-X)*(Theata-2*X)/V^2 # kmol/(m3.min)
Qg = ra*V*DeltaHrx # kcal/min
t(0)=0 #min
t(f)=122 #min
T(0)=448 #K
X(0)=0
Polymath Output
As Qr > Qg, no explosion would have occurred
13-5
To show that no explosion occurred with cooling failure for 3.17-kmol ONCB charge,
Put Nao=3.17
As the temperature can be maintained at desired value, no explosion would have occurred.
(ix) Individualized solution.
(x) Mass flow rate 1/2-in disc is = 1/16 of mass flow rate 2-in disc ~ 51.88 kg/min. (Considering only
the effect of change in cross sectional area)
Qr = 𝑚̇ ∆𝐻%&' + UA(T-Ta) = 51.88x540 + 35.83(543-298) = 36794 kcal/min
Qr > Qg, (Qg ~ 27000 kcal/min). So, the explosion could have been still avoided.
P13-1 (c) Example 13-3
(i)
1650 lbmol/hr
(ii) No, it is not possible.
(iii) The temperature trajectory exceeds practical stability limit for T0 = 120 F, CAi = 0
(iv) See Polymath program
Polymath code:
d(Ca)/d(t) = 1/tau*(Ca0-Ca)+ra
d(Cb)/d(t) = 1/tau*(Cb0-Cb)+rb
d(Cc)/d(t) = 1/tau*(0-Cc)+rc
d(Cm)/d(t) = 1/tau*(Cm0-Cm)
d(T)/d(t) = (Qg-Qr)/NCp
Fa0 = 80
T0 = 75
V = (1/7.484)*500
UA = 16000
dh = -36000
Ta1 = 67
k = 16.96e12*exp(-32400/1.987/(T+460))
Fb0 = 1000
Fm0 = 100
mc = 1000
ra = -k*Ca
rb = -k*Ca
rc = k*Ca
Nm = Cm*V
Na = Ca*V
Nb = Cb*V
Nc = Cc*V
ThetaCp = 35+Fb0/Fa0*18+Fm0/Fa0*19.5
v0 = Fa0/0.923+Fb0/3.45+Fm0/1.54
Ta2 = T-(T-Ta1)*exp(-UA/(18*mc))
13-6
Ca0 = Fa0/v0
Cb0 = Fb0/v0
Cm0 = Fm0/v0
Qr2 = mc*18*(Ta2-Ta1)
tau = V/v0
NCp = Na*35+Nb*18+Nc*46+Nm*19.5
Qr1 = Fa0*ThetaCp*(T-T0)
Qr = Qr1+Qr2
Qg = ra*V*dh
Ca(0)=0.1
Cb(0)=3.45
Cc(0)=0
Cm(0)=0
T(0)=150
t(0) = 0
t(f) = 4
Polymath Output
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.1
0.0074672
0.1
0.0309593
2
Ca0
0.1812152
0.1812152
0.1812152
0.1812152
3
Cb
3.45
2.114934
3.45
2.114934
4
Cb0
2.26519
2.26519
2.26519
2.26519
5
Cc
0
0
0.1502559
0.1502559
6
Cm
0
0
0.226519
0.226519
7
Cm0
0.226519
0.226519
0.226519
0.226519
8
dh
-3.6E+04
-3.6E+04
-3.6E+04
-3.6E+04
9
Fa0
80.
80.
80.
80.
10 Fb0
1000.
1000.
1000.
1000.
11 Fm0
100.
100.
100.
100.
12 k
41.71035
31.59807
144.4831
32.06998
13 mc
1000.
1000.
1000.
1000.
14 Na
6.680919
0.4988795
6.680919
2.068369
15 Nb
230.4917
141.2971
230.4917
141.2971
16 Nc
0
0
10.03847
10.03847
17 NCp
4382.683
3372.614
4382.683
3372.614
18 Nm
0
0
15.13355
15.13355
19 Qg
1.003E+07
2.339E+06
1.003E+07
2.388E+06
20 Qr
2.586E+06
2.377E+06
3.578E+06
2.388E+06
21 Qr1
1.706E+06
1.564E+06
2.383E+06
1.571E+06
22 Qr2
8.798E+05
8.133E+05
1.195E+06
8.168E+05
23 ra
-4.171035
-4.171035
-0.9725584
-0.9928658
24 rb
-4.171035
-4.171035
-0.9725584
-0.9928658
25 rc
4.171035
0.9725584
4.171035
0.9928658
26 T
150.
143.7291
179.7336
144.0607
27 t
0
0
4.
4.
28 T0
75.
75.
75.
75.
29 Ta1
67.
67.
67.
67.
13-7
30 Ta2
115.8777
112.1848
133.3874
112.3801
31 tau
0.1513355
0.1513355
0.1513355
0.1513355
32 ThetaCp 284.375
284.375
284.375
284.375
33 UA
1.6E+04
1.6E+04
1.6E+04
1.6E+04
34 V
66.80919
66.80919
66.80919
66.80919
35 v0
441.464
441.464
441.464
441.464
Differential equations
1 d(Ca)/d(t) = 1/tau*(Ca0-Ca)+ra
2 d(Cb)/d(t) = 1/tau*(Cb0-Cb)+rb
3 d(Cc)/d(t) = 1/tau*(0-Cc)+rc
4 d(Cm)/d(t) = 1/tau*(Cm0-Cm)
5 d(T)/d(t) = (Qg-Qr)/NCp
Explicit equations
1
Fa0 = 80
2
T0 = 75
3
V = (1/7.484)*500
4
UA = 16000
5
dh = -36000
6
Ta1 = 67
7
k = 16.96e12*exp(-32400/1.987/(T+460))
8
Fb0 = 1000
9
Fm0 = 100
10 mc = 1000
11 ra = -k*Ca
12 rb = -k*Ca
13 rc = k*Ca
14 Nm = Cm*V
15 Na = Ca*V
16 Nb = Cb*V
17 Nc = Cc*V
18 ThetaCp = 35+Fb0/Fa0*18+Fm0/Fa0*19.5
19 v0 = Fa0/0.923+Fb0/3.45+Fm0/1.54
20 Ta2 = T-(T-Ta1)*exp(-UA/(18*mc))
21 Ca0 = Fa0/v0
22 Cb0 = Fb0/v0
23 Cm0 = Fm0/v0
24 Qr2 = mc*18*(Ta2-Ta1)
25 tau = V/v0
26 NCp = Na*35+Nb*18+Nc*46+Nm*19.5
27 Qr1 = Fa0*ThetaCp*(T-T0)
28 Qr = Qr1+Qr2
29 Qg = ra*V*dh
From the above program, we see that by varying Ta1, the maximum possible value such that practical
stability limit is not exceeded is 67 F.
13-8
(v) See the following polymath program
Polymath Code
d(Ca)/d(t) = 1/tau*(Ca0-Ca)+ra
d(Cb)/d(t) = 1/tau*(Cb0-Cb)+rb
d(Cc)/d(t) = 1/tau*(0-Cc)+rc
d(Cm)/d(t) = 1/tau*(Cm0-Cm)
d(T)/d(t) = (Qg-Qr)/NCp
Fa0 = 80
T0 = 70
V = (1/7.484)*500
UA = 16000
dh = -36000
Ta1 = 67
k = 16.96e12*exp(-32400/1.987/(T+460))
Fb0 = 1000
Fm0 = 100
mc = 1000
ra = -k*Ca
rb = -k*Ca
rc = k*Ca
Nm = Cm*V
Na = Ca*V
Nb = Cb*V
Nc = Cc*V
ThetaCp = 35+Fb0/Fa0*18+Fm0/Fa0*19.5
v0 = Fa0/0.923+Fb0/3.45+Fm0/1.54
Ta2 = T-(T-Ta1)*exp(-UA/(18*mc))
Ca0 = Fa0/v0
Cb0 = Fb0/v0
Cm0 = Fm0/v0
Qr2 = mc*18*(Ta2-Ta1)
tau = V/v0
NCp = Na*35+Nb*18+Nc*46+Nm*19.5
Qr1 = Fa0*ThetaCp*(T-T0)
Qr = Qr1+Qr2
Qg = ra*V*dh
Ca(0)=0.1
Cb(0)=3.45
Cc(0)=0
Cm(0)=0
T(0)=160
t(0) = 0
t(f) = 4
Polymath Output
13-9
(vi)
See the following graphs for different set of parameters. 4 sets of graphs are presented below
(vii) For the case with Cai=0, Ti= 297 K
(1) Practical Stability limit is exceeded if UA is decreased to 2800 kcal/h.k
(2) Practical Stability limit is exceeded if Fbo is decreased to 370 kmol/hr
(3) Practical stability limit is not exceeded if reactor volume, V, is increased or decreased by a
factor of 2
13-10
(viii) As the coolant flow rate is decreased compared to base case: (a) For C-t graph, Ca decreases to low
value in less time (b) For Temp-time graph, temp peak appears in less time and value of peak temp
also increases (c) For Conc-Temp graph, the reaction crosses practical stability limit once coolant
flow is below 150 kmol/hr
Conclusion: Since the reaction is exothermic, so decreasing coolant flow increases reaction temp
and thus rate of reaction is increased and concentration decreases in shorter period of time.
However, the reaction temp approaches practical stability limit as coolant flow is decreased.c rf c
(ix) Time to reach steady state increases with increase in tank volume. When the tank volume is at its
minimum, practical stability limit is not exceeded for any of the three sets of initial conditions.
However, as we increase the volume, it is observed that for Ti = 340 K and CAi = 1.4 M, the practical
stability is exceeded.
(x) The maximum feed temperature such that practical stability limit is not exceeded is 284 K.
(xi) Individualized solution
(xii) The number of oscillations increase with increasing EA initially, and then decrease later on.
(xiii) Individualized solution
P13-1 (d) Example 13-4
(i)
The maximum in CC increases with an increase in CB0.
(ii) mc and CPc have virtually no effect on the temperature trajectory. UA, CP, CPc and mc have virtually
no effect on the concentration trajectory.
(iii) Ta1
(iv) Minimum inlet temperature to achieve 100 % conversion is 240 K
(v) 1. mc and CPc have virtually no effect on the temperature and concentration trajectory.
2. The maximum in CC increases with an increase in CB0.
3. Varying Ta1 causes graph of T and Ta2 to overlap each other.
(vi) On running the following Polymath code, we find that NC attains its maximum at t = 270 s and
remains constant thereafter. CC is maximum at t = 250 s.
Polymath Code
d(Ca)/d(t) = ra-(v0*Ca)/V #
d(Cb)/d(t) = rb+(v0*(Cb0-Cb)/V) #
d(Cc)/d(t) = rc-(Cc*v0)/V #
d(T)/d(t) = (Qg-Qr)/NCp #
d(Nw)/d(t) = v0*Cw0 #
v0 = 0.004 #
Cb0 = 1 #
UA = 3000 #
cp = 75240 #
T0 = 300 #
dh = -7.9076e7 #
Cw0 = 55 #
k = 0.39175*exp(5472.7*((1/273)-(1/T))) #
Cd = Cc #
Vi = 0.2 #
Kc = 10^(3885.44/T) #
cpa = 170700 #
V = Vi+v0*t #
Fb0 = Cb0*v0 #
ra = -k*((Ca*Cb)-((Cc*Cd)/Kc)) #
Na = V*Ca #
Nb = V*Cb #
Fw = Cw0*v0
Nc = V*Cc #
rb = ra #
rc = -ra #
Qr1 = ((Fb0*cp) + (Fw*cp))*(T - T0)
Nd = V*Cd #
rate = -ra #
13-11
NCp = cp*(Nb+Nc+Nd+Nw)+cpa*Na #
Cpc = 18 #
Ta1 = 285 #
mc = 100 #
Qr2 = mc*Cpc*(T-Ta1)*(1-exp(-UA/mc/Cpc)) #
Ta2 = T-(T-Ta1)*exp(-UA/mc/Cpc) #
Qr = Qr1 + Qr2
Qg = ra*V*dh
t(0)=0
Ca(0)=5
Cb(0)=0
Cc(0)=0
T(0)=300
Nw(0)=6.14
t(f)=360
Polymath Output
(vii) 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 a Tmax of 312 K. A
big change to the coolant rate has, in this case, only a small effect on the temperature, and
because the temperature does not change significantly the conversion will be kept about the
same.
(viii) See the following Polymath program
Polymath Code
d(x)/d(t) = 1
Vo = 0.2
v = 0.004
V = Vo+v*t
Dr = 5
Pi = 3.14
Ar = Pi*(Dr^2)/4
h = V/Ar
Dpipe = 0.5
n = h/Dpipe
Acoil = Pi*Dpipe*Pi*Dr
Atotal = n*0.5*Acoil
x(0)=0
t(0) = 0
t(f) = 10000
13-12
Polymath Output
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Acoil
24.649
24.649
24.649
24.649
2
Ar
19.625
19.625
19.625
19.625
3
Atotal
0.2512
0.2512
50.4912
50.4912
4
Dpipe
0.5
0.5
0.5
0.5
5
Dr
5.
5.
5.
5.
6
h
0.0101911
0.0101911
2.048408
2.048408
7
n
0.0203822
0.0203822
4.096815
4.096815
8
Pi
3.14
3.14
3.14
3.14
9
t
0
0
10000.
10000.
10 v
0.004
0.004
0.004
0.004
11 V
0.2
0.2
40.2
40.2
12 Vo
0.2
0.2
0.2
0.2
13 x
0
0
10000.
10000.
Differential equations
1 d(x)/d(t) = 1
Explicit equations
1 Vo = 0.2
2
v = 0.004
3
V = Vo+v*t
4
Dr = 5
5
Pi = 3.14
6
Ar = Pi*(Dr^2)/4
7
h = V/Ar
8
Dpipe = 0.5
9
n = h/Dpipe
10 Acoil = Pi*Dpipe*Pi*Dr
11 Atotal = n*0.5*Acoil
13-13
This area doesn’t depend on jacket diameter as long as jacket diameter is sufficiently smaller than the
reactor diameter.
P13-1 (e) Example 13-5
(i)
One of the possible combinations is CA0 = 8 mol/dm3 and 𝑣) = 240 dm3/h.
(ii) Concentration of all the species A, B, and C increases with an increase in CA0 and 𝑣) .
(iii) The parameter is ∆𝐻*+,- . It occurs at a value of 32350 cal/mol B.
(iv) The conversion of A increases with an increase in the volumetric flow rate. We also find that the
final concentration of C increases with an increase in the flow rate. The temperature also
increases.
(v) See the following Polymath program
Polymath Code
d(Ca)/d(t) = ra+(Cao-Ca)*vo/V #
d(Cb)/d(t) = rb-Cb*vo/V #
d(Cc)/d(t) = rc-Cc*vo/V #
d(T)/d(t) = (Qg-Qr)/((Ca*30+Cb*60+Cc*20)*V+100*35) #
UA = 35000 #
dHrx1 = -6500 #
dHrx2 = 8000 #
Qg = (dHrx1*(-k1a*Ca)+dHrx2*(-k2b*Cb))*V #
Qr = UA*(T-298) + Cao*vo*30*(T-305) #
Cao = 4 #
vo = 240 #
k1a = 1.25*exp((9500/1.987)*(1/320-1/T)) #
k2b = 0.08*exp((7000/1.987)*(1/300-1/T)) #
ra = -k1a*Ca #
NA=Ca*V
NB=Cb*V
V = 100+vo*t #
rc = 3*k2b*Cb #
rb = k1a*Ca/2-k2b*Cb #
t(0)=0
Ca(0)=1
Cb(0)=0
Cc(0)=0
T(0)=290
t(f)=15
Polymath Output
We can see from the graph that NA and NB become constant at large times.
13-14
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
1.
0.0719104
2.017219
0.0719104
2 Cao
4.
4.
4.
4.
3 Cb
0
0
0.9839784
0.339083
4 Cc
0
0
4.753264
4.753264
5 dHrx1
-6500.
-6500.
-6500.
-6500.
6 dHrx2
8000.
8000.
8000.
8000.
7 k1a
0.2664781
0.2664781
33.26563
3.590753
8 k2b
0.0533618
0.0533618
1.870243
0.3626699
9 NA
100.
24.80002
413.0125
266.0685
10 NB
0
0
1254.607
1254.607
11 Qg
1.732E+05
1.732E+05
3.628E+07
2.57E+06
12 Qr
-7.12E+05
-7.12E+05
6.948E+06
2.753E+06
13 ra
-0.2664781
-24.8374
-0.2582125
-0.2582125
14 rb
0.133239
-0.3716666
10.74578
0.006131
15 rc
0
0
5.520837
0.3689256
16 t
0
0
15.
15.
17 T
290.
290.
410.059
344.3179
18 UA
3.5E+04
3.5E+04
3.5E+04
3.5E+04
19 V
100.
100.
3700.
3700.
20 vo
240.
240.
240.
240.
Differential equations
1 d(Ca)/d(t) = ra+(Cao-Ca)*vo/V
2 d(Cb)/d(t) = rb-Cb*vo/V
3 d(Cc)/d(t) = rc-Cc*vo/V
4 d(T)/d(t) = (Qg-Qr)/((Ca*30+Cb*60+Cc*20)*V+100*35)
Explicit equations
1 UA = 35000
2 dHrx1 = -6500
3 dHrx2 = 8000
4 vo = 240
5 V = 100+vo*t
6 Cao = 4
7 Qr = UA*(T-298) + Cao*vo*30*(T-305)
8 k1a = 1.25*exp((9500/1.987)*(1/320-1/T))
9 k2b = 0.08*exp((7000/1.987)*(1/300-1/T))
10 ra = -k1a*Ca
11 Qg = (dHrx1*(-k1a*Ca)+dHrx2*(-k2b*Cb))*V
12 NA = Ca*V
13 NB = Cb*V
14 rc = 3*k2b*Cb
15 rb = k1a*Ca/2-k2b*Cb
13-15
(vi)
We find that after running the code for a final time of 10 hours, we get NA = 266.07 mol and NB =
1254.61 mol.
𝐶/) 𝑣)
240
=4∗
= 267.3 𝑚𝑜𝑙
𝑘1/
3.590753
𝐶/) 𝑣)
240
=4∗
= 1323.5 𝑚𝑜𝑙
2𝑘,2 ∗ 0.3626699
A
%
BC C
(vii) Since NB = ,D
at long times, we can maximize NB by increasing the volumetric flow rate.
EF
P13-1 (f) Example 13-6
(i)
The pressure shoots up when the liquid volume is 3976 dm3.
(ii) UA has to be varied.
(iii) With the current sliders it is not possible that temperature limitation is reached before pressure
limitation.
(iv) -270,000 J/mol
(v) Individualized solution
(vi) Individualized solution
(vii) Individualized solution
(viii) The transition to runaway occurs over a very small range of UA. If UA value is changed such that
UA is taken from 2.1 ×104 to 1.9 ×104 J/hr/K the runaway occurs. Thus it occurred over a very
narrow range of UA values.
(ix) The modeling is done and the changes are incorporated in the polymath code for Example 13.6
A new switch type variable is introduced.
Sw2 such that Sw2 = 1
if 300<T<422
else 0
and a new term is added in the energy equation such that the equation becomes:
d(T)/d(t) = SW1*((V0*(r1A*DHRx1A+r2S*DHRx2S)-SW1*UA*(T – 373.15))/SumNCp) +
SW1*Sw2*4/60
Case 1: When Ua = 2.77 × 106 J/hr/K
But the cooling starts only for T>455K
Thus Ua = 2.77 × 106 J/hr/K, T> 455K
= 0 for T< 455K
Polymath Code
d(CA)/d(t) = SW1*r1A
d(CB)/d(t) = SW1*r1A
d(CS)/d(t) = SW1*r2S
d(P)/d(t) = SW1*((FD-Fvent)*0.082*T/VH)
d(T)/d(t) = SW1*((V0*(r1A*DHRx1A+r2S*DHRx2S)-SW1*UA*(T-373.15))/SumNCp) + SW1*Sw2*4/60
V0 = 4000
VH = 5000
DHRx1A = -45400
DHRx2S = -3.2E5
SumNCp = 1.26E7
A1A = 4E14
E1A = 128000
k1A = A1A*exp(-E1A/(8.31*T))
A2S = 1E84
E2S = 800000
k2S = A2S*exp(-E2S/(8.31*T))
SW1 = if (T>600 or P>45) then (0) else (1)
r1A = -k1A*CA*CB
r2S = -k2S*CS
FD = (-0.5*r1A-3*r2S)*V0
Cv2 = 53600
Cv1 = 3360
Fvent = if (FD<11400) then (FD) else(if (P<28.2) then ((P-1)*Cv1) else ( (P-1)*(Cv1 +Cv2)))
UA = if T> 455 then (2.77e6) else (0)
13-16
Sw2 = if ((T>300) and (T<422)) then (1) else (0)
CA(0)= 4.3
CB(0)=5.1
CS(0)=3
P(0)=4.4
T(0)=422
t(0) = 0
t(f) = 4
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
A1A
4.0E+14
4.0E+14
4.0E+14
4.0E+14
2
A2S
1.0E+84
1.0E+84
1.0E+84
1.0E+84
3
CA
4.3
0.0457129
4.3
0.0457129
4
CB
5.1
0.8457129
5.1
0.8457129
5
CS
3.
2.999997
3.
2.999997
6
Cv1
3360.
3360.
3360.
3360.
7
Cv2
5.36E+04
5.36E+04
5.36E+04
5.36E+04
8
DHRx1A -4.54E+04
-4.54E+04
-4.54E+04
-4.54E+04
9
DHRx2S
-3.2E+05
-3.2E+05
-3.2E+05
-3.2E+05
10 E1A
1.28E+05
1.28E+05
1.28E+05
1.28E+05
11 E2S
8.0E+05
8.0E+05
8.0E+05
8.0E+05
12 FD
2467.445
61.27745
8874.034
61.27745
13 Fvent
2467.445
61.27745
8874.034
61.27745
14 k1A
0.0562573
0.0562573
1.82787
0.7925113
15 k2S
8.428E-16
8.428E-16
2.367E-06
1.276E-08
16 P
4.4
4.4
4.4
4.4
17 r1A
-1.233723
-4.437017
-0.0306385
-0.0306385
18 r2S
-2.529E-15
-7.102E-06
-2.529E-15
-3.829E-08
19 SumNCp 1.26E+07
1.26E+07
1.26E+07
1.26E+07
20 SW1
1.
1.
1.
1.
21 Sw2
0
0
0
0
22 T
422.
422.
466.4882
454.9731
23 t
0
0
4.
4.
24 UA
0
0
2.77E+06
0
25 V0
4000.
4000.
4000.
4000.
26 VH
5000.
5000.
5000.
5000.
Differential equations
1 d(CA)/d(t) = SW1*r1A
mol/dm3/hr
2 d(CB)/d(t) = SW1*r1A
change in concentration of cyclomethylpentadiene
3 d(CS)/d(t) = SW1*r2S
change in concentration of diglyme
4 d(P)/d(t) = SW1*((FD-Fvent)*0.082*T/VH)
5 d(T)/d(t) = SW1*((V0*(r1A*DHRx1A+r2S*DHRx2S)-SW1*UA*(T-373.15))/SumNCp) + SW1*Sw2*4/60
13-17
Explicit equations
1
V0 = 4000
dm3
2
VH = 5000
dm3
3
DHRx1A = -45400
J/mol Na
4
DHRx2S = -3.2E5
J/mol of Diglyme
5
SumNCp = 1.26E7
J/K
6
A1A = 4E14
per hour
7
E1A = 128000
J/kmol/K
8
k1A = A1A*exp(-E1A/(8.31*T))
rate constant reaction 1
9
A2S = 1E84
per hour
10 E2S = 800000
J/kmol/K
11 k2S = A2S*exp(-E2S/(8.31*T))
rate constant reaction 2
12 SW1 = if (T>600 or P>45) then (0) else (1)
13 r1A = -k1A*CA*CB
mol/dm3/hour (first order in sodium and cyclomethylpentadiene)
14 r2S = -k2S*CS
mol/dm3/hour (first order in diglyme)
15 FD = (-0.5*r1A-3*r2S)*V0
16 Cv2 = 53600
17 Cv1 = 3360
18 Fvent = if (FD<11400) then (FD) else(if (P<28.2) then ((P-1)*Cv1) else ( (P-1)*(Cv1 +Cv2)))
19 UA = if T> 455 then (2.77e6) else (0)
no cooling
20 Sw2 = if ((T>300) and (T<422)) then (1) else (0)
13-18
(x)
Solving the equations with Ua = 0 for the entirety of the process, we get that the temperature
reaches at 0.94 hours while the system reaches runaway at 3.649 hours. This implies the
maximum time in minutes that the cooling can be lost is 163 minutes.
(xi)
Individualized solution
P13-1 (g)
The following table shows the value of conversion for different entering temperature, calculated at t=4 h
To
68
72
76
80
X
0.158
0.729
0.805
0.844
13-19
In polymath code, change coolant rate and run Polymath code to obtain the following table
Coolant rate (lbmol/hr)
400
700
950
1000
2000
3000
6000
8000
10000
Conversion
0.89
0.796
0.66
0.194
0.159
0.15
0.143
0.141
0.140
Temp (deg F)
155.6
139.2
124.2
83.6
79.2
78.04
76.97
76.72
76.56
13-20
P13-1 (h) Using the base code from PRS Example R13-3a, given as follows
Polymath Code
d(Ca)/d(t) = 1/tau*(Ca0-Ca)+ra #
d(Cb)/d(t) = 1/tau*(Cb0-Cb)+ra #
d(Cc)/d(t) = 1/tau*(0-Cc)-ra #
d(Cm)/d(t) = 1/tau*(Cm0-Cm) #
d(T)/d(t) = (-Q-Fa0*ThetaCp*(T-T0)+(-36000)*ra*V)/NCp #
d(I)/d(t) = T-Tsp #
Fa0 = 80 #
T0 = 70 #
V = (1/7.484)*500 #
Tsp = 138 #
UA = 16000 #
Ta1 = 60 #
kc = 8.5 #
k = 16.96e12*exp(-32400/1.987/(T+460)) #
Fb0 = 1000 #
Fm0 = 100 #
mc0 = 1000 #
ra = -k*Ca #
NCp = Ca*V*35+Cb*V*18+Cc*V*46+Cm*V*19.5 #
ThetaCp = 35+Fb0/Fa0*18+Fm0/Fa0*19.5 #
v0 = Fa0/0.923+Fb0/3.45+Fm0/1.54 #
Ca0 = Fa0/v0 #
Cb0 = Fb0/v0 #
Cm0 = Fm0/v0 #
tau = V/v0 #
X = (Ca0-Ca)/Ca0 #
mc = mc0+kc/tau*I #
Ta2 = T-(T-Ta1)*exp(-UA/(18*mc)) #
Q = mc*18*(Ta2-Ta1) #
t(0)=0
Ca(0)=0.03789
Cb(0)=2.12
Cc(0)=0.143
Cm(0)=0.2265
T(0)=138.53
I(0)=0
t(f)=4
we can determine the value of kc for which the reactor will fall to the lower steady-state and when it
becomes unstable. The following two graphs show those points when kc = 24 and 0.2 respectively. The
third graph shows what happens when T0 = 65. It becomes unstable at a much lower temperature.
13-21
P13-1 (i) Using the code from PRS Example R13-3b,
Polymath Code
d(Ca)/d(t) = 1/tau*(Ca0-Ca)+ra #
d(Cb)/d(t) = 1/tau*(Cb0-Cb)+ra #
d(Cc)/d(t) = 1/tau*(0-Cc)-ra #
d(Cm)/d(t) = 1/tau*(Cm0-Cm) #
d(T)/d(t) = (-Q-Fa0*ThetaCp*(T-T0)+(-36000)*ra*V)/NCp #
d(I)/d(t) = T-Tsp #
Fa0 = 80 #
T0 = 70 #
V = (1/7.484)*500 #
Tsp = 138 #
UA = 16000 #
Ta1 = 60 #
kc = 200 #
k = 16.96e12*exp(-32400/1.987/(T+460)) #
Fb0 = 1000 #
Fm0 = 100 #
mc0 = 1000 #
ra = -k*Ca #
tauI=0.1
NCp = Ca*V*35+Cb*V*18+Cc*V*46+Cm*V*19.5 #
ThetaCp = 35+Fb0/Fa0*18+Fm0/Fa0*19.5 #
v0 = Fa0/0.923+Fb0/3.45+Fm0/1.54 #
Ca0 = Fa0/v0 #
Cb0 = Fb0/v0 #
Cm0 = Fm0/v0 #
tau = V/v0 #
X = (Ca0-Ca)/Ca0 #
mc = mc0+kc*(T-Tsp)+kc/tauI*I #
Ta2 = T-(T-Ta1)*exp(-UA/(18*mc)) #
Q = mc*18*(Ta2-Ta1) #
t(0)=0
Ca(0)=0.03789
Cb(0)=2.12
Cc(0)=0.143
Cm(0)=0.2265
T(0)=138.53
I(0)=0
t(f)=4
we can change the value of kc and t1 and find values that produce the lowest oscillations and the
quickest return to steady-state. For example, at kc = 200 and t1 = 0.1 the following graph results
(increasing kc further will generate even lower oscillations):
13-22
P13-1 (j) Individualized solution
P13-2
After the feed was shut off,
See the following Polymath program
Polymath Code:
d(T)/d(t) = -dH * k / Cpa
dH = -336
Cpa = .38
k = (0.307/60)* exp(44498 * (1 / 970 - 1 / T))
T(0)=980
t(0) = 0
t(f) = 4
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Cpa
0.38
0.38
0.38
0.38
2 dH
-336.
-336.
-336.
-336.
3 k
0.0081711
0.0081711
4.285E+17
4.285E+17
4 T
980.
980.
3.302E+20
3.302E+20
5 t
0
0
4.
4.
Differential equations
1 d(T)/d(t) = -dH * k / Cpa
Explicit equations
1 dH = -336
2 Cpa = .38
3 k = (0.307/60)* exp(44498 * (1 / 970 - 1 / T))
13-23
As can be seen from the above plots, temperature shoots at time t = 3.16 mins
Since NA doesn’t come in the temperature equation, there will not be any effect of feed present in the
reactor.
If T0 = 100 °F, Temperature will remain constant for next 4 minutes, which means no explosion.
If T0 = 500 °F, there will not be any explosion and the temperature profile will look like
13-24
P13-3 (a)
Polymath Code
d(Na)/d(t) = ra*V
d(Nb)/d(t) = ra*V+Fbo
d(Nc)/d(t) = -ra*V
d(T)/d(t) = ((6000*(-ra*V))-(Fbo*15*(T-323)))/(15*Na+15*Nb+30*Nc)
d(X)/d(t) = -ra*V/Nao
Fbo = if(t<50)then(10)else(0)
Nao = 500
Cbo = 10
13-25
k = .01*exp((10000/1.987)*(1/300-1/T))
vo = Fbo/Cbo
V = if(t<50)then(50+(vo*t))else(100)
ra = -k*Na*Nb/(V^2)
Qg = -6000*ra*V
Qr = 0
Na(0)=500
Nb(0)=0
Nc(0)=0
T(0)=298
t(0) = 0
t(f) = 120
X(0)=0
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Cbo
10.
10.
10.
10.
2
Fbo
10.
0
10.
0
3
k
0.0089352
0.0089352
10.08511
10.08511
4
Na
500.
0.1396707
500.
0.1396707
5
Nao
500.
500.
500.
500.
6
Nb
0
0
42.43357
0.1397745
7
Nc
0
0
499.8603
499.8603
8
Qg
0
0
1.24E+05
11.81314
9
Qr
0
0
0
0
10 ra
0
-0.3460808
0
-1.969E-05
11 T
298.
298.
510.4441
510.4441
12 t
0
0
120.
120.
13 V
50.
50.
100.
100.
14 vo
1.
0
1.
0
15 X
0
0
0.9997207
0.9997207
Differential equations
1 d(Na)/d(t) = ra*V
2 d(Nb)/d(t) = ra*V+Fbo
3 d(Nc)/d(t) = -ra*V
4 d(T)/d(t) = ((6000*(-ra*V))-(Fbo*15*(T-323)))/(15*Na+15*Nb+30*Nc)
5 d(X)/d(t) = -ra*V/Nao
Explicit equations
1 Fbo = if(t<50)then(10)else(0)
2 Nao = 500
3 Cbo = 10
4 k = .01*exp((10000/1.987)*(1/300-1/T))
5 vo = Fbo/Cbo
6 V = if(t<50)then(50+(vo*t))else(100)
7 ra = -k*Na*Nb/(V^2)
8 Qg = -6000*ra*V
9 Qr = 0
13-26
P13-3 (a) Continued
13-27
P13-3 (b)
This is the same as part (a) except the energy balance.
Energy balance:
Polymath Code
d(Na)/d(t) = ra*V
d(Nb)/d(t) = ra*V+Fbo
d(Nc)/d(t) = -ra*V
d(T)/d(t) = ((6000*(-ra*V))-(Fbo*15*(T-323))-Qr)/(15*Na+15*Nb+30*Nc)
d(X)/d(t) = -ra*V/Nao
UA=100
Ta=323
Fbo = if(t<50)then(10)else(0)
Nao = 500
Cbo = 10
k = .01*exp((10000/1.987)*(1/300-1/T))
vo = Fbo/Cbo
V = if(t<50)then(50+(vo*t))else(100)
ra = -k*Na*Nb/(V^2)
Qg = -6000*ra*V
Qr = UA*(T-Ta)
Na(0)=500
Nb(0)=0
Nc(0)=0
T(0)=298
t(0) = 0
t(f) = 180
X(0)=0
Polymath Output
13-28
P13-3 (c)
This is the same as part (b) except the reaction is now reversible.
See the following Polymath program
Polymath Code
d(Na)/d(t) = ra*V
d(Nb)/d(t) = ra*V+Fbo
d(Nc)/d(t) = -ra*V
d(T)/d(t) = ((6000*(-ra*V))-(Fbo*15*(T-323))-Qr)/(15*Na+15*Nb+30*Nc)
d(X)/d(t) = -ra*V/Nao
UA=100
Ta=323
Fbo = if(t<50)then(10)else(0)
Nao = 500
Cbo = 10
13-29
k = .01*exp((10000/1.987)*(1/300-1/T))
k2 = .1*exp((16000/1.987)*(1/300-1/T))
vo = Fbo/Cbo
V = if(t<50)then(50+(vo*t))else(100)
ra = -k*Na*Nb/(V^2)+k2*Nc/V
Qg = -6000*ra*V
Qr = UA*(T-Ta)
Na(0)=500
Nb(0)=0
Nc(0)=0
T(0)=298
t(0) = 0
t(f) = 180
X(0)=0
Polymath output
13-30
P13-4 (a)
13-31
P13-4 (b)
P13-5
13-32
See the following Polymath program
Polymath code
d(X)/d(t) = -ra/.1
d(T)/d(t) = ((40000+(10*(T-298)))*(-ra)*(1/.1))/(56.25-(10*X))
k1 = .002*exp((100000/8.314)*(1/373-1/T))
Ca = .1*(1-X)
Cb = .1*(1.25-X)
k2 = .00003*exp((150000/8.314)*(1/373-1/T))
Cc = .1*X
ra = -((k1*(Ca^.5)*(Cb^.5))-(k2*Cc))
X(0)=0
T(0)=373
t(0) = 0
t(f) = 10
Polymath Output
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
0.1
0.0749517
0.1
0.0749517
2 Cb
0.125
0.0999517
0.125
0.0999517
3 Cc
0
0
0.0250483
0.0250483
4 k1
0.002
0.002
106.1361
106.1361
5 k2
3.0E-05
3.0E-05
366.7509
366.7508
6 ra
-0.0002236
-0.162921
1.14E-06
1.3E-07
7 t
0
0
10.
10.
8 T
373.
373.
562.918
562.918
9 X
0
0
0.2504829
0.2504829
Differential equations
1 d(X)/d(t) = -ra/.1
2 d(T)/d(t) = ((40000+(10*(T-298)))*(-ra)*(1/.1))/(56.25-(10*X))
13-33
Explicit equations
1 k1 = .002*exp((100000/8.314)*(1/373-1/T))
2 Ca = .1*(1-X)
3 Cb = .1*(1.25-X)
4 k2 = .00003*exp((150000/8.314)*(1/373-1/T))
5 Cc = .1*X
6 ra = -((k1*(Ca^.5)*(Cb^.5))-(k2*Cc))
P13-6 (a)
A + 2B ® C
13-34
Where Q=UA (𝑇& − 𝑇)=2500 (290-T)
See the following Polymath program
Polymath Code
d(Na)/d(t) = ra*V
d(Nb)/d(t) = 2*ra*V+Fbo
d(Nc)/d(t) = -ra*V
d(X)/d(t) = -ra*V/50
d(T)/d(t) = ((2500*(290-T))-(80*vb*(T-325))+(-55000*(-ra*V)))/(35*Na+20*Nb+75*Nc)
vb = 0.9
k = .0005*exp((8000/1.987)*(1/300-1/T))
V = 10+(vb*t)
Fbo = 4*vb
Ca = Na/V
Cb = Nb/V
Cc = Nc/V
ra = -k*Ca*Cb^2
Na(0)=50
Nb(0)=0
Nc(0)=0
X(0)=0
T(0)=300
t(0) = 0
t(f) = 450
Polymath Output
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
5.
0.0238595
5.
0.0238595
2 Cb
0
0
3.71037
3.71037
3 Cc
0
0
0.1192843
0.0966225
4 Fbo
3.6
3.6
3.6
3.6
5 k
0.0005
0.0002882
0.0005
0.000315
6 Na
50.
9.90168
50.
9.90168
7 Nb
0
0
1539.803
1539.803
8 Nc
0
0
40.09832
40.09832
9 ra
0
-0.0033037
0
-0.0001035
10 t
0
0
450.
450.
11 T
300.
288.1691
300.
290.0131
12 V
10.
10.
415.
415.
13 vb
0.9
0.9
0.9
0.9
14 X
0
0
0.8019664
0.8019664
13-35
Differential equations
1 d(Na)/d(t) = ra*V
2 d(Nb)/d(t) = 2*ra*V+Fbo
3 d(Nc)/d(t) = -ra*V
4 d(X)/d(t) = -ra*V/50
5 d(T)/d(t) = ((2500*(290-T))-(80*vb*(T-325))+(-55000*(-ra*V)))/(35*Na+20*Nb+75*Nc)
Explicit equations
1 vb = 0.9
2 k = .0005*exp((8000/1.987)*(1/300-1/T))
3 V = 10+(vb*t)
4 Fbo = 4*vb
5 Ca = Na/V
6 Cb = Nb/V
7 Cc = Nc/V
8 ra = -k*Ca*Cb^2
We have iterated by changing vb to achieve X=0.8, T<403 K at all time and Nc in 23.5 hrs to be
approximately 120 mol, we find that for vb= 0.9 dm3/min, X= 0.8, Nc=40.1, T=290 K.
In 24 hrs, allowing 23.5 hrs for reaction, moles of Nc formed = (40.1/450)*(23.5*60) =126 moles C.
P13-6 (b)
If the maximum coolant rate falls to 200 mol/min, then it may not be prudent to assume that the coolant
leaves at the entering ambient temperature, Ta. It should be assumed that the coolant temperature varies
spatially along the heat exchanger pipes and the required term for the heat exchange would be:
Q = mc Cpcool (Ta1 – Ta2)
Ta2 = T – Ta1).expI−
where Ta1 = ambient T coolant entering
Ta2 = ambient T coolant leaving
J/
O
KL A'LMMN
The reduced flow rate and hence heat exchange, may increase the reactor temperature to approaching
130°C, the upper limit, at conversions approaching 80%m, and so more caution is required. The
incorporation of temperature control would be prudent.
P13-7
P13-7 (a)
See Polymath program
Polymath Code (Base case)
A+B®C
d(Ca)/d(t) = ra
d(Cb)/d(t) = rb
d(Cc)/d(t) = rc
Na=Ca*V
Nb=Cb*V
Nc=Cc*V
sumNcp=Na*15+Nb*15+Nc*30
dh=-6000
d(T)/d(t) = ((UA*(450-T))+(dh*(ra*V)))/(sumNcp)
UA = 0
V = 10
k=0.01*exp((10000/1.987)*(1/300-1/T))
ra = -k*Ca*Cb
rb = ra
13-36
rc=-ra
Ca(0)=0.1
Cb(0)=0.1
Cc(0)=0
T(0)=300
t(0) = 0
t(f) = 250
Polymath Output
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
0.1
0.0013239
0.1
0.0013239
2 Cb
0.1
0.0013239
0.1
0.0013239
3 Cc
0
0
0.0986761
0.0986761
4 dh
-6000.
-6000.
-6000.
-6000.
5 k
0.01
0.01
7.779782
7.779782
6 Na
1.
0.0132387
1.
0.0132387
7 Nb
1.
0.0132387
1.
0.0132387
8 Nc
0
0
0.9867613
0.9867613
9 ra
-0.0001
-0.001954
-1.364E-05
-1.364E-05
10 rb
-0.0001
-0.001954
-1.364E-05
-1.364E-05
11 rc
0.0001
1.364E-05
0.001954
1.364E-05
12 sumNcp 30.
30.
30.
30.
13 T
300.
300.
497.3523
497.3523
14 t
0
0
250.
250.
15 UA
0
0
0
0
16 V
10.
10.
10.
10.
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
4 d(T)/d(t) = ((UA*(450-T))+(dh*(ra*V)))/(sumNcp)
Explicit equations
1 V = 10
2 Na = Ca*V
3 Nb = Cb*V
4 Nc = Cc*V
5 dh = -6000
6 UA = 0
7 sumNcp = Na*15+Nb*15+Nc*30
8 k = 0.01*exp((10000/1.987)*(1/300-1/T))
9 ra = -k*Ca*Cb
10 rb = ra
11 rc = -ra
13-37
P13-7 (b) Modify the Base case Polymath code to change UA and V as follows
UA = 10000 J/min.K
V = 10 dm3
UA= 10000/(4.18*60)=39.87 cal/s.K
UA = 40000 J/min.K
V = 10 dm3
UA= 40000/(4.18*60)= 159.5 cal/s.K
13-38
UA = 10000 J/min.K
UA= 398.7 cal/s.k
V = 10 dm3
P13-7 (c) Change UA and To in base case Polymath code to following values:
Ua=159.5 cal/s.K and To=380 K
Use the base case polymath code to change the value of UA and To to obtain results
P13-8
See the following Polymath code with values taken from P12-16
Polymath Code
d(Ca)/d(t) = 1/tau*(Ca0-Ca)+ra #
d(Cb)/d(t) = 1/tau*(0-Cb)+rb #
d(T)/d(t) = (Qg-Qr)/NCp #
Fa0 = 0.6 # kmol/hr
T0 = 310 # K
V = 0.01 # m3
UA = 216 # kcal/hr.K
dh=-80000 # kcal/kmol
Ta1 = 310 # K
k = 60*exp((20000)*(1/400-1/T)) # hr-1
Kc=100*exp((-80000/1.987)*(1/400-1/T))
Fb0 = 0 #
mc = 453.6# kmol/hr
ra = -k*(Ca-Cb/Kc) # kmol/m3.h
rb = -ra #
Na = Ca*V # kmol
Nb = Cb*V # kmol
ThetaCp = 40 # kcal/kmol.K
13-39
v0 = Fa0/14.8 # m3/h
Ta2 = T-(T-Ta1)*exp(-UA/(18*mc)) #
Ca0 = Fa0/v0 # kmol/m3
Qr2 = mc*18*(Ta2-Ta1) #
tau = V/v0 #
NCp = Na*40+Nb*40 #
Qr1=Fa0*ThetaCp*(T-T0)#
Qr=Qr1+Qr2
Qg=ra*V*dh
t(0)=0
Ca(0)=0.001 # kmol/m3
Cb(0)=0
T(0)=298
t(f)=4
Polymath Output
P13-9 (a) Yes. As more of C is produced, the reaction rate of reaction 3 may increase. Since reaction 3 is
high exothermic, there is still possibility that the reactor will run away if Qr > Qg at t = 0.
P13-9 (b) Q
r t=0
= UA(T0 −Ta ) = (100)(450 − 400) = 5000'cal/s
13-40
mol
s dm3
3
3
3 1
mol
r2 C 0 = r2 A0 = - k2 AC A0CC 0 = - ( ´ 10-3 ) (1.0 )( 0.2 ) = -0.0001
2
2
2 3
s dm3
1
1
1
mol
r3 B 0 = r3C 0 = - k3C CB 0CC 0 = - (0.6 ´ 10-3 ) ( 0.5 )( 0.2 ) = -0.00002
3
3
3
s dm3
Qg t =0 = r1B 0V DH Rx1B + r2C 0V DH Rx 2 C + r3 B 0V DH Rx 3 B
P13-9 (c) r1B 0 = 2r1 A0 = -2k1 AC A0CB 0 = -2(0.001) (1.0 )( 0.5 ) = -0.0005
2
2
= (-0.0005)(2000)(-5000) + (-0.0001)(2000)(10000) + (-0.00002)(2000)(-50000)
= 5000 - 2000 + 2000
= 5000 cal/s
P13-9 (d)
Qg - Qr
dT
=
=0
dt t =0 N A0CP + N B 0CP + N C 0CP
A
B
C
P13-9 (e) Qr t =0 (Ta = 350K ) = UA(T0 - Ta ) = (100)(450 - 350) = 10000 cal/s
Qg - Qr
dT
(Ta = 350 K ) =
dt t =0
N A 0 C P + N B 0 C P + N C 0C P
A
=
B
C
5000 - 10000
= -0.1 K / s
2000[(1)(10) + (0.5)(10) + (0.2)(50)]
P13-9 (f)
Qg - Qr
dT
=
dt N A0CP + N B 0CP + N C 0CP + N I CP
A
B
C
I
As inert is added, temp will increase slower and therefore runaway will be less likely.
13-41
P13-10
(i) For the original recipe, the reaction begins at a set temperature, and due to the initial charge of
reactants in the reactor, the reaction rate is immediately large. This initial large rate of exothermic
reaction causes an almost immediate temperature spike. During this initial reaction spike, the rate
of cooling is unable to keep up with the rate of generation. Once all the immediate reactants have
reacted, the reaction rate is dependent on the volumetric flow of reactants into the system, which
results in a smaller amount of reactants in the reactor at a given time, so the rate of cooling can
match the rate of heat generation and bring the reaction to an equilibrium temperature at 300 K.
(ii) The critical initial volume (Vo) for this situation is 1.2 m3. Above this initial volume, the temperature
peak will be high enough where sufficient cooling is not achieved by 500 seconds. Note this time is a
rough estimate because of the more complicated factors such as pressure increase and the failure of
the pressure relief valve which occurred in the actual Synthron explosion.
(iii) Increasing vo increases the final temperature within the reactor greatly, which has a large effect on
the explosion and the reactor’s likelihood of staying above 350 K.
Decreasing UA has a similarly large effect in in increasing the final temperature within the reactor, so
this would still affect the explosion. Decreasing Vo decreases the initial temperature peak, as it
decreases the amount of time the reactor stays above its equilibrium temperature. This is because
the coolant has a smaller temperature range to cool over because of the lower peak.
(iv) Variables associated with the initial charge of reactants have an effect on the initial temperature peak.
This initial temperature peak will determine how long the temperature takes to equilibrate. The
variables associated with the rate of cooling and generation after the initial charge will determine
what the equilibrated temperature of the reaction is. The large initial charge of reactants in the
modified recipe caused a temperature spike that was higher than the original recipe, which means
there was a longer amount of time the temperature was not at equilibrium. This caused a buildup of
pressure, and this higher temperature for a longer period of time resulted in the vapor cloud release
and eventual explosion.
P13-11
(i) Governing ODEs for the system
(1)
(2)
PQ
PR
PU
PR
= −𝑘1 𝑆 − 𝑘, 𝐼𝑆
= +𝑘1 𝑆 + 𝑘, 𝐼𝑆 − 𝑘W 𝐼 − 𝑘X 𝐼
P*
(3) PR = 𝑘W 𝐼
PY
(4) PR = 𝑘X 𝐼
This set of equations can be solved numerically to get S, I, D and R as functions of time.
(ii) Using Wolfram, parameter values required for fitting the given data are:
Parameter
value
Initial condition
Value
k1
6*10-7
S(0)
1.3864*109
k2
10.58*10-9
I(0)
770
k3
14.5645
R(0)
0
k4
2*10-3
D(0)
0
13-42
(iii) Varying the k4 sliders we can see that, at k4 = 3.4×10-3 and I0 = 540, the death rate becomes critical.
(iv) With k1 = 0 and I (0) = 770, keeping other parameters fixed, the graphs obtained are as follows:
Even on adjusting k2, k3 and k4, it is observed that there is a phase lag between the values predicted by
the model and the actual data. There is a tradeoff between the maximum value and this phase lag when
k1 is set to zero. Hence it can be concluded that having a finite value of k1, i.e., simulating the spread of
infection by contacting a contaminated surface has an important role in removing this trade-off, and thus
modelling the phenomenon more accurately. This analysis also points out the need for keeping surfaces
clean!
13-43
With k1 = 2*10–7day–1 and I (0) = 0
It can be observed that there is a rightwards shift (phase lag) in the simulated curves w.r.t data as
compared with the first case.
(v) When I (0) is decreased, all three curves shift rightward, with a slight decrease in its maximum value.
When I(0) is increased, the curves shift leftwards and maximum value increases slightly.
When s(0) is decreased, the curves shift rightward with a decrease in amplitude. The opposite is
observed on increasing S(0), ie., leftward shift and increase in amplitude.
By varying the sliders, we can see that the more sensitive parameter is S(0).
It takes only one susceptible person to get infected by a surface, it doesn't depend on the presence
of infected people. As for the order, this represents a first order kinetics.
Further, in the graph of the number of healthy cases, there is no local maxima. Only a minima is
observed.
(vi) Applying PSSH on the concentration of infected people concentration,
PU
PR
= +𝑘1 𝑆 + 𝑘, 𝐼𝑆 − 𝑘W 𝐼 − 𝑘X 𝐼 = 0
(𝑘1 + 𝑘, 𝐼)𝑆 = (𝑘W + 𝑘X )𝐼
Solving will give,
QD
]
𝐼 = D _D aD
Q
^
`
E
Substituting in (1),
PQ
Q ED D
PR
= −𝑘1 𝑆 − D _D ]aDE Q
PQ
QD] (D^_D`)
PR
^
`
E
= − D _D aD Q
^
`
E
13-44
Integrating this equation will give an intrinsic equation of S wrt time,
Q
(𝑘W + 𝑘X )𝑙𝑛(Q ) − 𝑘, (𝑆 − 𝑆c ) = −𝑘1 (𝑘W + 𝑘X )𝑡
M
Plotting this curve and comparing it with the numerical solution looks like,
As can be seen, the two curves are very different and clearly PSSH doesn't work.
PSSH doesn't seem to be valid, which can also be explained due to the nature of the numerical I vs t graph.
We can see from the below plot that the Pseudo steady-state assumption is far from being valid.
(vii) Individualized solution
13-45
Page intentionally blank
13-46
Synopsis for Chapter 14 – Mass Transfer Limitations in
Reacting Systems
General: This chapter is a review of transport phenomena that focuses on external mass transfer to the
catalyst surface.
Questions
l Q14-1A (21 seconds) Questions Before Reading (QBR).
AA Q14-3A (5 min) Oops! Should say discussed in Example 14-4 not 14-2. Otherwise reasonably good
discussion question.
O Q14-4A (8 min) Oops!! Should say it is referencing to Example 14-4 not 14-3.
AA Q14-5A (8 min) Oops!! Should say See Example 14-4.
AA Q14-6B (8 min) Dust explosions are very important and one should view the video on the dust
explosion in the Chemical Reaction Engineering sub page of (http://umich.edu/~safeche/).
AA Q14-8B (6 min) Go to the Safety Website (http://umich.edu/~safeche/) and view the video on the
dust explosion.
Computer Simulations and Experiments
l P14-1B (a) – (c) & (f) (5-10 min per simulation) Wolfram simulations are always assigned simulations.
5-10 minutes per simulation including writing conclusions.
Problems
O P14-2B (40 min) Very interesting and motivational problem suggested by Bob Kabel of Penn State
University.
S P14-3B (40 min) California Professional Engineers Registration Exam Problem.
O P14-4B (15 min max) quick calculation of pellet burning time.
O P14-5C (30 min) this problem is similar to the web problem on transdermal drug delivery in the
expanded material on the web. Similar to P14-10B and Expanded Material on the web.
AA P14-6B (30 min) PBR in which reaction is controlled by external mass transfer.
I P14-7C (40 min) Shows the transition from reaction rate limited at low temperatures to mass
transfer controlled at high temperatures in CVD (Chapter 10).
l P14-8B (25 min) Old Exam Question (OEQ). Similar to problem P14-6B, mass transfer limited reaction
in a PBR.
AA P14-9B (45 min) Old Exam Question (OEQ). Reversible reaction in mass transfer limited PBR. More
difficult than P14-8B.
AA P14-10B (50 min) An alternative to problems P14-6B and P14-8B.
S P14-11B (50 min) If the student looks on the web under Expanded Material for Chapter 14 they will
find this problem is solved in the expanded material on the web.
S P14-12D (75 min) Requires knowledge of error function solution.
S14-1
AA P14-13B (30 min) Old Exam Question (OEQ). Good question, reasonably straight forward derivative
of a particle dissolving in a liquid.
S P14-14C (45 min) Derivation of dissolution equation in spherical coordinates to find both CA(r,t) and
WA(r,t).
O P14-15B (35 min) Straight forward application of algorithm for mass transfer as applied on a
dissolving particle.
I P14-16B (45 min) More advanced problem than the single pill dissolution.
O P14-17B (30 min) Fairly straight forward calculation of the dissolution of a particle
S14-2
Solutions for Chapter 14 – Mass Transfer Limitations in
Reacting Systems
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-14
http://umich.edu/~elements/5e/14chap/obj.html#/
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q14-1 Individualized solution.
Q14-2 Individualized solution.
Q14-3 Individualized solution.
Q14-4
T1 = 300K
T2 = 350K
23
16
12
12
k c2 æ DAB2 ö æ u1 ö æ U 2 ö æ d P1 ö
=ç
÷ ç ÷ ç ÷ ç
÷
k c1 è DAB1 ø è u 2 ø è U1 ø è d P 2 ø
(11-70)
As a first approximation assume
D AB2 µ1
=
D AB1 µ 2
then
56
12
12
k c2 æ u1 ö æ d P1 ö æ U 2 ö
=ç ÷ ç
÷ ç ÷
k c1 è u 2 ø è d P 2 ø è U1 ø
At T1 = 300K µ1 ≈ 0.883cP
At T2 = 350K µ2 ≈ 0.380cP
Assume density doesn’t change that much, u =
u1 µ1
=
= 2.32
u2 µ2
µ
r
U 2 1 d P1 1
= ,
=
U1 2 d P 2 2
14-1
k1 = 4.61´ 10-6 m s
12
5 6 é1ù
k2 = 4.61´10-6 m s [2.32 ] ê ú
ë2û
12
é1ù
êë 2 úû
= 4.65 ´10-6 m s
(
)(
WA = −rA"" = kc2C Ab = 4.61×10−6 m s 103 mol m3
)
-rA¢¢ = 0.00465 mol m 2 s
Q14 -5
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
" d P1 %1 2
" 1 %1 2
k c2 = k c1$
=
2.9m
s
$ '
'
#5&
# d P2 &
= 1.3m s
)
,
1063
X = 1− exp+−1.3•
0.05.
*
15
= 1− exp(−4.6)
X = 1− 0.01 = 0.99
again, virtually complete conversion.
€
Q14-6 Individualized solution.
Q14-7 Individualized solution.
Q14-8 Individualized solution.
P14-1
P14-1 (a)
(i)
Flux WAr is most sensitive to diffusivity DAB
(ii) By varying the slider for liquid viscosity, we see that theflux decreases very slowly as the viscosity
increases. As diffusivity is increased, molar flux also increases. So, when both viscosity and
diffusivity are increased simultaneously, the overall effect is an increase in molar flow rate.
(iii) As velocity increases, flux increases, but its effect is small compared to the other parameters.
(iv) Among all the parameters, changing DAB has the greatest effect on molar flux. Liquid diffusivity
and velocity tend to increase the molar flux, whereas viscosity tends to decrease it, though its
effect is negligible.
14-2
P14-1 (b)
(i)
Most sensitive parameter is DAB
(ii) Both DAB and U tend to increase the flux. Hence, increasing them simultaneously increases kc and
thus molar flux. Decreasing them decreases the molar flux.
(iii) DAB is the most sensitive parameter and increasing this tends to increase the molar flux.
DAB and U tend to increase molar flux, whereas viscosity tends to decrease it.
Increasing bulk concentration increases flux, and decreasing it decreases the flux. This relation is
the opposite for surface concentration.
P14-1 (c)
(i)
For 100 % solid, tc is 274 ms. If solid is 10 %, then tc=~40 ms (from sliders)
As percentage of solid increases, tc increases.
(ii) For CA∞=8.58, when particle diameter =132.78 µm, tc=500 ms, other parameters remaining same
for DAB=0.00005, tc=500 ms, other parameters remaining same.
ρc=4*106g/m3, tc=500 ms, other parameters remaining same.
(iii) Increasing density, particle diameter and volume fraction of solid increases the buning time tc ,
and increasing DAB decreases it.
P14-1 (d) Example 14-4
Using parameter values in Table E14-7.1,
𝑑% = (1.5 ∗ 0.002+ ∗ 0.005)-// = 0.0031 𝑚
For liquid phase,
𝜌 ∗ 𝑣 ∗ 𝑑 1000 ∗ 0.1 ∗ 0.0031
=
= 310
𝜇
10𝑒 − 3
𝜈
10@A
𝑆; =
= @B = 100
𝐷>? 10
𝑅4 =
𝑅4C =
𝑅𝑒
310
=
= 517
(1 − 𝜙)𝛾 (1 − 0.4) ∗ 1
-/+
I
I
𝑆HC = 𝑅4C
∗ 𝑆;J = 517K.L ∗ 100 J = 105.5
𝑆HC ∗ 𝐷>? (1 − 𝜙)𝛾 105.5 ∗ 10@B ∗ 0.6
𝑘; =
∗
=
= 0.00051 𝑚/𝑠
𝑑%
𝜙
0.0031 ∗ 0.4
Similarly for vapor phase
𝜌 ∗ 𝑣 ∗ 𝑑 1 ∗ 0.1 ∗ 0.0031
𝑅4 =
=
= 31
𝜇
10𝑒 − 5
𝜈
10@L
𝑆; =
= @L = 1
𝐷>? 10
𝑅4C =
𝑅𝑒
31
=
= 52
(
(1 − 𝜙)𝛾
1 − 0.4) ∗ 1
-/+
I
I
𝑆HC = 𝑅4C
∗ 𝑆;J = 52K.L ∗ 1J = 7.19
𝑆HC ∗ 𝐷>? (1 − 𝜙)𝛾 719 ∗ 10@L ∗ 0.6
𝑘; =
∗
=
= 0.035 𝑚/𝑠
𝑑%
𝜙
0.0031 ∗ 0.4
14-3
P14-1 (e) Example 14-5
Assuming T1= 500 C (773 K) and T2 as 600 C (873 K)
𝜈+
1
1
= 𝑒𝑥𝑝 S−4000 ∗ T − VW
𝜈𝑇1 𝑇2
== 𝑒 @K.LX = 0.55
Assume
DA 2 ! µ1 "
1
~# $=
DA1 % µ2 & 0.55
kc 2 " DA 2 #
=$
%
kc1 & DA1 '
23
16
" υ1 #
$ %
& υ2 '
"µ #
=$ 1 %
& µ2 '
46
16
" µ1 #
$ %
& µ2 '
"µ #
=$ 1 %
& µ2 '
56
𝑘4+
= (1/0.55)L/A = 1.65
𝑘4𝑙𝑛
1
=2
1 − 𝑋-
So,
-
𝑙𝑛 -@\ = 2 ∗ 1.65= 3.3
]
𝑋+ = 0.96
P14-1 (f)
(i)
As kr is increased, the X increases rapidly and the X vs. W curve remains constant afterwards.
Changing kr has no effect on molar flux WA as a function of W. The rate constant k increases very
slightly on increasing kr.
Increasing U decreases conversion X, increases molar flux ,and rate constant k increases. The slope
of k as a function of W remains same.
(ii) On increasing DAB, k increases. On increasing dp, k decreases. However, the rate constant k
remains almost independent of W over the whole range of parameters.
ρc=4*106g/m3, tc=500 ms, other parameters remaining same.
(iii) As the ratio DAB/ν is increased, the conversion as a function of W increases and approaches one.
(iv) Particle diameter dp, porosity ɸ and cross sectional area Ac are some of the most important
parameters.
Least sensitive parameters include fluid viscosity, shape factor, CA0 and diffusivity DAB
(v) Conversion increases with DAB,kr,ρ,ν and decreases with: U, dp,µ,ɸ, AC,ρc
The rate constant remains almost independent of W over the whole range of parameters.
As W increases, conversion increases and approaches a value of 1. Molar flux decreases with W.
14-4
P14-2
Given,
• Minimum respiration rate of chipmunk,
FAL = 1.5 μmol O2/min
• Breathing rate of Chipmunk,
v0 = 0.05 dm3 gas/min
• Diameter of hole, D = 3cm
Assuming,
• A represents oxygen
• B represents nitrogen
• Chipmunk has a constant breathing rate of 0.05 dm3 of
gas/min
Minimum flow rate of oxygen to the bottom, FAL
=Minimum respiration rate of chipmunk
= 1.5 μmol of O2/min
Flow%rate%of%A%down%the%hole
= %Flux%of%A × %Cross3sectional%Area
C −C
∴FAL = DAB A0 AL × AC
L
C A0 −C AL
or %%L = DAB
× AC %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%3333%(1)
FAL
DAB = 0.18%cm2 / s = 0.18×10−4 m2 / s
π D2 π ×(0.03m)2
=
= 7.069×10−4 m2
4
4
−6
FAL = 1.50 ×10 mol / min = 2.5×10−8 mol / s
A C=
P14-2 (a) At Ann Arbor, Michigan
P
C A0 = T × y A """"""(Ideal"gas"law)
RT
where,
PT = 1.013×105 N / m2 (at(Ann(Arbor(,(Michigan(((situated(at(sea(level()
R = 8.314(J / mol.K
T = 250 C = 298K
y A = 0.21((((((((((((((Mole(fraction(of(oxygen(in(air)
C A0 =
1.013×105 N / m2
"
%
J
×298K '
$ 8.314*
mol.K
#
&
× 0.21
∴*C A0 = 8.59*mol / m3
Now,
Flow rate of A = (Concentration of A at the bottom) x (Volumetric intake of gas)
14-5
FAL = C AL ×v0
v0 = 0.05'dm3 / min = 0.05×10+3 m3 (gas/min
F
2.5×10−8 mol / s
∴C AL = AL =
= 0.03-mol / m3
'
3
v0 $
& 0.05×10+3 m × 1min - )
&
min 60s )(
%
Solving for the length from (1),
C −C
L = DAB A0 AL × AC
FAL
#
&
(8.59+ − 0.030)+mol / m3 (
L = 0.18×10−4 m2 / s × %
×7.069×10−4 m2
%
(
−8
2.5×10 mol / s
$
'
∴L = 4.36m
P14-2 (b) At Boulder, Colorado
Boulder, Colorado is 5430 feet above sea-level. The corresponding atmospheric pressure is 0.829 x 10-5
N/m2.
PT = 0.829×105 N / m2 )at)Boulder,)Colorado
P
C A0 = T × y A )
RT
!C A0 =
0.829×105 N / m2
"
%
J
×298K '
$ 8.314!
mol.K
#
&
× 0.21!
∴!C A0 = 7.03!mol / m3
Solving for the length from (1),
C −C
!L = DAB A0 AL × AC
FAL
#
&
(7.03! − 0.03)!mol / m3 (
!L = 0.18×10−4 m2 / s × %
×7.069×10−4 m2
%
(
−8
$ 2.5×10 mol / s '
∴L = 3.56m
P14-2 (c)
During winter at Ann Arbor, Michigan
T = 0°F = −17.78°C = 255.37K
P
C A0 = T × y A +
RT
!C A0 =
1.013×105 N / m2
"
%
J
×255.37K '
$ 8.314!
mol.K
#
&
× 0.21!
14-6
∴C A0 = 8.20mol m3
Solving for the length from (1),
C −C
!L = D AB A0 AL × AC
FAL
# 255 &1.75 # (10.02! − 0.03)!mol / m3 &
( ×7.069×10−4 m2
L = 0.18×10 m / s × %
×%
(
%
(
−8
298
$
'
$ 2.5×10 mol / s
'
−4
2
∴L = 3.87m
During winter at Boulder, Colorado
T = 0°F = −17.78°C = 255.37K
P
C A0 = T × y A +
RT
C A0 =
0.829×105 N m2
"
%
J
×255.37K '
$ 8.314
mol.K
#
&
× 0.21
∴C A0 = 8.20mol m3
Solving for the length from (1),
C −C
!L = DAB A0 AL × AC
FAL
# 255 &1.75 # (8.20! − 0.03)!mol / m3 &
( ×7.069×10−4 m2
!L = 0.18×10 m / s × %
( %
%
−8
$ 298 ' $ 2.5×10 mol / s ('
−4
2
∴L = 3.17m
P14-2 (d) Individualized solution
P14-3
14-7
14-8
P14-4
(a) Diffusion controlled:
Ks=(365*10-10)*0.1883*106/(188)=3.656*10-5
t=dp02/Ks=(100*10-6)2/(3.656*10-5)=0.274 ms
P14-4 (b) Reaction controled:
t= ρcdp0/(2krCAs)= 188*(10-4)/(2*0.01*8.58)=109.55 ms
P14-4 (c) Combined diffusion and reaction controlled:
W = k (C — C ) = –r² = k C
Ar
c
A
As
(Diffusion)
As
r
As
(Surface Reaction)
For small particles, Sh~2
kc=2De/D , where D is diameter of dissolving particle
D*=2De/kr
where 2De/kr is the diameter at which the resistances to mass transfer
and reaction rate are equal.
A mole balance on the solid particle yields
In - Out +Generation= Accumulation
where ρ is molar density of carbon
At time t= 0, the initial diameter is D=Di
So, time to complete burning of particle =
α=2*0.01*8.58/188=9.13*10-4
D =10 m
i
-4
D*=2De/kr=2*10-4/0.01=0.02
tc=1095.29*(10-4+(10-4)2/(2*0.02))=0.1098 s=109.8 ms
14-9
P14-5
14-10
14-11
14-12
P14-6
14-13
14-14
P14-6 (a & b)
See the following Polymath code:
Polymath Code:
d(x)/d(z)=2.22*(1-x)/(1+0.15*x)^.5
z(0)=0
z(f)=5
x(0)=0
Output:
The results show that only 3.3 cm of the tube is required for
converison of 99.9%, much less than the full 20 m.
14-15
P14-6 (a) From the above table, required tube length is only 3.3 cm for 99.9 % conversion. So number of
pipe necessary is 1
P14-6 (b) Refer above graph of X vs z
P14-6 (c)
14-16
P14-6 (d)
P14-6 (e)
P14-7
As flow rate increases at a constant temperature, deposition rate increases.
As temperature increases, for a fixed flow rate, deposition rate increases.
Plotting deposition rate vs. temperature , it can be seen that the deposition rate increases linearly with
temperature.
From the graph of dispersion rate vs. flow rate, we can infer that deposition rate increases with increase
in flow rate, but not linearly. Deposition rate increases and then saturates with increase in flow rate.
2.5
2
1.5
500 SCCM
600 SCCM
1
750 SCCM
0.5
1000 SCCM
0
Deposition rate (mg/cm 2 . h)
Deposition rate (mg/cm 2 . h)
2.5
2
1.5
670 C
1
750 C
800 C
0.5
0
650
700
750
800
850
0
Temperature (0 C)
200
400
600
800
Flow Rate (SCCM)
14-17
1000
1200
P14-8
Mass transfer limited, rapid reaction, packed bed.
Proposed guidelines for optimum process conditions:
Use high T, so that the removal of chlorine is only mass transfer limited (i.e., rapid reaction rate of
diffusion).
Increasing u (directly proportional to velocity u), enhances the mass transfer coefficient but also acts to
reduce converison X. the latter dependency is ore influential on X (raised to a high power), so reduction
u may be preferable (for constant packed bed properties). If converison is already high enough, then a
larger velocity may be preferable for a mass transfer limited reaction.
The mass transfer rate can be increased by increasing the concentration gradient achieved by increasing
the bulk CA.
14-18
P14-9
Mass transfer limited, packed bed, reversible reaction
Assume a Langmuir-Hinshelwood mechanism surface reaction
14-19
Hence the required catalyst weight W2 will be 1.41 Wt, the original catalyst weight, if the strategy is to
increase bed length rather than diameter.
let
E increases, eE increses, 1/eE decreases, X increases
14-20
P14-10 (a)
14-21
Polymath Code:
d(x)/d(w)=-ra/fa0
k=0.01
fa0=10
kc=4242
ca0=0.001
ca=ca0*(1-x)
ra=-k*kc*ca/(k+kc)
w(0)=0
w(f)=1e6
x(0)=0
Output:
14-22
P14-10 (b)
14-23
See the following Polymath code:
Polymath Code:
d(x)/d(w)=-ra/fa0
k=0.01
fa0=10
kc=4242
ca0=0.001
T0=300
T=(10*x*10000/1000)+T0
ca=ca0*(1-x)*(T0/T)
ra=-k*kc*ca/(k+kc)
w(0)=0
w(f)=1.2e6
x(0)=0
Output:
14-24
For a conversion of X = 60%, Wcat = 1020 kg
P14-11
See Web Expanded material for Chapter 14.
To find the rate of drug delivery, we simply multiply the area of the patch times the flux of drug A at the
pat—skin interface.
P14-11 (a)
To find the flux WA we can either carryout a differential mole balance or use Equation (14-2) we will use
Equation (14-2).
Step 1. Diffusion of A through the Epidermis film general equation
WA z − WA z+Δz = 0
Dividing by Dz and taking the limit as Δz → 0 we obtain the a differential equation for diffusion
through a stagnant film at dilute concentration which is
dWAz
=0
dz
14-25
(W14-1.1)
P14-11 (b)
Step 2. Use Fick’s law to relate the flux WAz to the concentration gradient for dilute concentrations. In
the epidermis layer (0 – d1)
WA1 = −D A1
dC A
=0
dz
(W14-1.2)
dC A
dz
(W14-1.3)
In the dermis layer
WA2 = −D A2
Step 3. State the boundary condition. In the epidermis layer
z=0
CA = CA0
z = δ1
CA = CA1
z = δ1
C A = C A1
z = δ2
CA = 0
(W14-1.4)
In the dermis layer
(W14-1.5)
Step 4. Next we use Fick’s law to substitute for WAz and then divide out the diffusivity.
d 2CA
=0
dz
(W14-1.6)
C A = K1z + K1
(W14-1.7)
Integrating twice
Step 5. Solve for the concentration profile in the epidermis layer (0 < z < d1) we have
Substituting (W14-1.4) into (W14-1.7)
C A = K1
C A1 = K 2δ1 + K1 = K 2δ2 + C A0
C A0 − C A
z
=
C A0 − C A1 δ1
WAz1 = −D A1
dC A
(C − CA1 )
= D A1 A0
dz
δ1
(W14-1.8)
(W14-1.9)
P14-11 (c)
Step 6. In the dermis layer we substitute (W14-1.5) into (14-1.7)
C A1 = K1δ1 + K 2δ2
0 = K1δ2 + K 2
CA
δ −z
= 2
C A1 δ2 − δ1
WAz2 = −D A2
dC A
C A1
= D A2
dz
δ2 − δ1
14-26
(W14-1.10)
(W14-1.11)
P14-11 (d)
Step 7. Evaluate
The flux at d1 are equal
(W14-1.12)
WAz1 = WAz2
D A1
(CA0 − CA1 ) = D
δ1
A2
C A1
δ2 − δ1
C A0
δ1
C A1 =
D A2
D
+ A1
δ2 − δ1 δ1
D A1
(W14-1.13)
P14-11 (e)
Step 8. Evaluate concentration profiles
In the dermis layer rearranging Equation (W14-1.8)
C A = C A1
CA =
(δ2 − z)
δ2 − δ1
D A1C A0
(δ2 − z)
δ1D A2 + D A1 (δ2 − δ1 )
(W14-1.14)
In the epidermis layer
C A = C A0 − (C A0 − C A1 )
z
δ1
(W14-1.15)
Substitute for CA1
#
D A1C A0 &
%
(z
δ1
(
C A = C A0 − %C A0 −
D A2
D A1 ( δ1
%
+
%$
δ2 − δ1 δ1 ('
Rearranging
#
&
D A2
% δ −δ
(z
2
1
(
C A = C A0 − C A0 %
% D A2 + D A ( δ1
%$ δ2 − δ1 δ1 ('
(W14-1.16)
#
&
C
D A1 A0 (
%
C
D
δ1
(
WAz = D A2 A1 = A2 %
δ2 − δ1 δ2 − δ1 % D A2 + D A1 (
%$ δ2 − δ1 δ1 ('
(W14-1.17)
P14-11 (f)
Flux in the dermis layer
14-27
WAz =
1
δ −δ
CA0
+ 2 1 CA0 =
δ1
DA2
R1 + R2
DA1
(W14-1.18)
where R1 and R2 are the resistance to diffusion. The molar flow of Species A into the skin is area
of the patch, Ap, times the flux into the skin, WA
P14-11 (g)
(W14-1.19)
If we consider there is a resistance to the drug release in the parth RP, then the total resistance is
(W14-1.20)
Then
(W14-1.21)
P14-12
14-28
From the above graph, it can be seen that at z=0.4 m, CA= 0.01 CA0, hence 𝛿= 18-0.4 m =17.6 m
The time since the diffusion started (i.e. at the end of last glacial)
-`.A +
-
-
t= _-.B+ a ∗ b∗+.Ab∗-KcId ∗ /AKK∗+b∗/AL = 2808 𝑦𝑒𝑎𝑟𝑠
P14-13
=volume dissolved / initial volume
14-29
P14-14
Balance on drug particles in a spherical shell of radius r
and thickness dr gives:
Wdr.4πr2|r- Wdr.4πr2|r+dr+rd. 4πr2=
Wdr=-De
\
dC
dr
,
dC
.4pr 2 Dr
dt
rd=-kC
De d 2 dC
dC
(r
) - kC =
2
r dr
dr
dt
C(r,t)=Cr(r).Ct(t)
If the solubility is small, it is possible to assume a quasi-steady state exists for thediffusion
equation.
So, C(r,t)=Cr(r)
\
De
dC t (t )
d 2 dC r (r )
(r
)-k =
=0
2
C r (r )r dr
dr
C t (t )dt
\
De
d 2 dC r (r )
(r
)-k =0
2
C r (r )r dr
dr
k 2
r 2C!! r (r ) + 2rC! r (r ) r C r (r ) = 0
De
2
k
C!! r (r ) + C! r (r ) C r (r ) = 0
r
De
rCr (r ) = A cosh(F
kR
F= i
De
r
r
) + B sinh(F
)
R( t )
R( t )
2
14-30
Boundary conditions:
R(t=0)=Ri
C(r,0)=0 , r Î ( Ri , ¥) --------------------(i)
C(∞,t)=0, t Î (0, ¥) -----------------------(ii)
C(0,t) = finite (until the drug dissolves completely)--------------(iii)
Þ
dC
|r = R ( t ) = 0 Þ C(R(t),t)=f(t) ----------(iv)
dr
Using boundary condition (iii) gives
r
C(r) = B sinh(F
) /r
R(t )
Overall balance on solid particle gives:
In-Out+Generation=Accumulation
d ( rpD 3 / 6)
rds’’πD =
dt
2
2kr C
dD
-a
,
=
,a =
*
dt 1 + D / D
r
Di - D +
1
( D 2 i - D 2 ) = at
*
2D
R(t)=D(t)/2
Qualitative plots:
14-31
P14-15 (a)
P14-15 (b)
CC
Surface reaction limited −𝑟>i
= 𝑘j– 𝐶>K
Mass transfer effects are not important when the surface reaction rate is limiting.
CC
CC
For 1 mole A dissolving 1 mole B then −𝑟>i
= −𝑟?i
CC )
𝑑𝐷
2(−𝑟>i
2
o = − 𝑘j 𝐶>K
= −n
𝑑𝑡
𝜌
5
14-32
Boundary conditions, t = 0, D = Di
Time for dissolution, tc, at D = 0, assuming particle density, r = 2000 kg/m3
tc=
rDi
2k r C A 0
=
2000 *10 -5
= 5 *1012 s
-10
2 *10 * 2000
P14-15 (c)
14-33
= 10 sec
14
P14-15 (d)
To reduce tc, increase CA and / or decrease Di dissolves faster
To increase tc, decrease CA and / or increase Di dissolves slower
P14-16 (a)
14-34
P14-16 (b)
14-35
14-36
using polymath code in part c, following graph is obtained
P14-16 (c)
14-37
See the following polymath code:
Polymath Code:
d(D1)/d(t)=if(t>0.24) then (if(D1>0.00001) then (-4*Dab*Sinn /(D1*rho)) else (0)) else (0)
d(D2)/d(t)=if(t>0.52) then (if(D2>0.00001) then (-4*Dab*Sinn /(D2*rho)) else (0)) else (0)
d(D3)/d(t)=if(t>1.18) then (if(D3>0.00001) then (-4*Dab*Sinn /(D3*rho)) else (0)) else (0)
d(Cb)/d(t)=ka*Ca*V/Wbody
d(Ca1)/d(t)=if(t>0.24) then(2*Dab*Sinn*3.14*D1/V) else (0)
d(Ca2)/d(t)=if(t>0.52) then(2*Dab*Sinn*3.14*D2/V) else (0)
d(Ca3)/d(t)=if(t>1.18) then(2*Dab*Sinn*3.14*D3/V) else (0)
Dab=0.0006
Sinn=400
rho=35.4
ka=0.166667
V=1200
Wbody=75000
Ca=Ca1+Ca2+Ca3-(Cb*Wbody/V)
t(0)=0
t(f)=45
D1(0)=0.3
D2(0)=0.3
D3(0)=0.3
Cb(0)=0
Ca1(0)=0
Ca2(0)=0
Ca3(0)=0
Output:
14-38
P14-16 (d)
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 distribution of outer
layers is needed, with thin layers for initial response and thicker layers for delayed response. This
distribution would be back calculated given the necessary stomach concentration 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
required to maintain this level. The range of pill sizes depends on the number of pills which can be
reasonably consumed in one siting, the period for effect and the limits of practical pill size.
P14-17
The diameter is plotted as a function of time:
The data suggests that initially, the decrease in diameter with time is non-linear, but from ~ 50 time units, the
rate is almost linear. On fitting a linear equation to the datafrom 50 time units to 110 time units, the
equation of the line is as follows:
y=-0.0735x+12.155
If we neglect the initial non-linearity, then the timerequired for complete decomposition of the POHC is ~165
time units.
14-39
Page intentionally blank
14-40
Synopsis for Chapter 15 – Diffusion and Reaction
General: This chapter focuses on internal diffusion limitation. Many of the problems are conceptual.
Questions
l Q15-1A (19 seconds) Questions Before Reading (QBR).
AA Q15-3A (3 min) Good discussion problem.
G Q15-4 (30 min) Usually assigned in the graduate course and requires reading slurry reactor material
on CRE website for Chapter 15.
O Q15-7A (20 min) Good screencasts on diffusion and reactions.
AA Q15-8 (10 min) Always assign. Get the students thinking about Critical Thinking.
Computer Simulations and Experiments
S P15-1B (a) (3 min) Seldom assigned, but shows effective diffusivity can be a fraction of the binary
diffusivity.
l P15-1B (b) (20-35 min) Example 15-3. This Wolfram problem is a Stop and Smell the Roses
problem. The students should spend at least 20 minutes exploring this problem to get an
intuitive understanding of diffusion limitations in a packed porous catalyst bed.
Problems
l P15-2B (20 min) Old Exam Question (OEQ). Excellent conceptual problem with minimal calculations.
l P15-3B (20 min) Old Exam Question (OEQ). Excellent conceptual problem with minimal calculations.
l P15-4A (15 min) Old Exam Question (OEQ). Conceptual problem with minimal calculations.
AA P15-5B (35 min) Oops Should say Figure 15-4 not 15-3. Old Exam Question (OEQ). Straight forward
problem of spherical catalyst concentration profiles.
AA P15-6B (35 min) Very interesting problem on the swimming of sperm. Very straight forward.
O P15-7B (30 min) Derivation of effectiveness factor on the walls of a single cylindrical pellet.
O P15-8B (35 min) Interesting problem where part of the pore is poisoned. Students need to derive an
equation for the concentration profile as they cannot use any formation in the text.
I P15-9A (35 min) Derivation of effectiveness factors.
AA P15-10B (40 min) Very often assigned as it is a rigorous derivation of the effectiveness factor for a
zero order reaction.
O P15-11C (45 min) PBR that is internal diffusion limited. Fairly difficult problem.
S P15-12C (60 min) Fairly difficult as it involves Bessel function solutions.
l P15-13C (45 min) Very important problem. Reasonably straight forward application of applying
falsified kinetics.
S15-1
AA P15-14B (40 min) Derivation of equation used in falsified kinetics. Rather challenging problem.
G P15-15C (45-55 min) Reasonably straight forward. The students will need to read the Additional
Material on slurry reactors on the Web USED IN GRADUATE COURSE.
G P15-16C 45-55 min) Reasonably straight forward. Another slurry reactor problem in the Additional
Material for Chapter 15 on the web. Excellent Problem.
S P15-17C (45 min) If the student happens to look at the Chapter 15 Expanded Material on the web
site they will see the problem worked out for them.
S15-2
Solutions for Chapter 15 – Diffusion and Reaction
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Chapter 15:
http://www.umich.edu/~elements/5e/15chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram codes):
http://www.umich.edu/~elements/5e/tutorials/Wolfram_LEP_tutorial.pdf
Q15-1 Individualized solution.
Q15-2 Individualized solution
Q15-3 Heat is released in an exothermic reaction as a result the temperature inside the pores of the
catalyst pellet will be higher than the temperature at the surface. Therefore, the rate constant will be
higher inside the pores and hence higher reaction rate which means ƞ > 1.
Page 6 of the following reference develops on this idea furthur:
http://www.umich.edu/~elements/5e/15chap/Ch15_Web-Additional%20Material.pdf
Q15-4 Increase in temperature will result in reduced diffusivity and hence the resistance due to internal
and external diffusion will decrease.
Q15-5 Individualized answer.
Q15-6 Using wrong E, wrong n one may estimate the reaction to occur in a particular regime disguised
as another regime. The particle size can be reduced hoping that the internal diffusion limitations
become neglible. But this could lead make the reaction temperature sensitive, possibly leading to a
runaway reaction.
Q15-7
(a) The equation for the amount reacting in 1 cm is:
𝐹𝑙𝑢𝑥 × 𝑎𝑟𝑒𝑎 = 4𝜋𝐷. 𝐿𝐶12
(b) Effectiveness factor in terms of thiele modulus:
4
ƞ = 5 6 (𝛷𝑐𝑜𝑡ℎ𝛷 − 1)
Q15-8 Individualized answer
P15-1 (a) (i) Individual solution
15-1
P15-1 (b)
Wolfram
(i)
For fixed surface area (= 500m2/g),
R1 = 0.001 and R2 = 5 R1
Internal effectiveness factor changes by a factor of 1/8.
For fixed radius of pellet (=0.002m),
Sa1 = 100 and Sa2 = 5 Sa1
Internal effectiveness factor changes by a factor of 5/9.
Therefore, radius has a greater impact in reducing internal diffusion limitations.
(ii) 90% Conversion can be achieved by increasing effective diffusivity, internal surface area and rate
constant
(iii) Combination of high viscosity, low rate constant and low internal surface area can given WP ~1.
(iv) In the base case, radius of pellet R = 0.003 m
Ƞ = 0.15
Ω = 0.07
WP = 50
MR = 0.9
Since Ƞ is very small and WP>>1, so reaction is internal diffusion limited. Since Ω is small and MR>
0.15, so reaction is also limited by external diffusion.
So percentage of resistance for both internal and external diffusion is significant while percentage
of resistance for surface reaction is negligible.
(v) With increase in temperature, diffusivity increases. Therefore, the resistance due to internal and
external diffusivities will decrease.
(vi) Internal surface area tends to decrease MR. The curve of MR with respect to radius of pellet
always tends to saturate. Kinematic viscocity tends to increase MR.
(vii) Rate constant and surface area seem to affect the internal effectiveness factor the most.
(viii) Rate constant and surface area seem to affect the overall effectiveness factor the most.
(ix) Effective diffusivity seems to affect MR the most.
(x) Effective diffusivity seems to affect WP the most.
(xi) For a catalyst particle of radius 0.005 m, the WP ~ 80. That means, the internal diffusion is a major
contributor to the resistance.
(xii) We can see that the the rate constant and the internal surface area have a major effect on the
internal and external diffusion in the reaction. Increasing the effective diffusivity can give higher
conversions.
Polymath
(i)
If the reaction were zero order, then the rate doesnot depend on the concentration of reactants
at any point in the pellet. Therefore, the rate will be the same at the surface and at any point
inside the pore.
(ii) Refer part (iv) and (v) of the Wolfram questions.
(iii) For example 15-3, we see that WP>>1 and MR<0.15, therefore, internal diffusion severely effects
the reaction and the external diffusion has negligible effect on the reaction.
(iv) Decrease in internal surface area with time will result in lower reaction rates and hence the
internal effectiveness factor will decrease with time.
(v) Individual solution
P15-2 (a) Yes
P15-2 (b) All temperatures, FT0 = 10 mol/hr. The rate of reaction changes with flow rate in this range and
rate of reaction changes linearly with temperature throughout. At higher flow rates (5000 mol/hr) we
can see that rate does not change much with flow rate.
15-2
P15-2 (c) Yes
P15-2 (d)
T < 367 K, FT0 = 1000 mol/hr, 5000 mol/hr.
T < 362 K, FT0 = 100 mol/hr.
P15-2 (e) Yes
P15-2 (f)
T > 367 K, FT0 = 1000 mol/hr, 5000 mol/hr.
T > 362 K, FT0 = 100 mol/hr.
P15-2 (g)
Ω=
(
A
Ω=
)
−rA at 362K ,FT 0 = 10mol / hr
actualrateofreaction
=
idealrateofreaction −r at 362K ,F = 5000mol / hr
(
T0
)
0.26
= 0.37
0.70
P15-2 (h)
At FT0 = 5000 mol/hr, there is no external diffusion limitation, so the external effectiveness factor is 1.
η=
η=
(
actualrateofreaction at,362K,,FT0 = 5000mol / hr
(
)
extrapolatedrateofreaction at,362K,,FT0 = 5000mol / hr
)
1.2
= 0.86
1.4
P15-2 (i)
3"#φ coshφ −1$%
η=
= 0.86
2
φ
by iterative solution f = 1.60
ϕ=
CA
C AS
=
( )
1 sinh φλ
λ sinhφ
P15-3 (a) External mass transfer limited at 400 K and dP = 0.8 cm. Alos at all FT0 < 2000 mol/s
P15-3 (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.
P15-3 (c) Internal diffusion limited at T = 400 K and 0.1 < dP < 0.8
P15-3 (d) η =
rate with dP = 0.8
10
= = 0.625
rate with dP = 0.03 16
15-3
P15-4
Curve A: At low temperatures (high 1/T) the reaction is rate limited as evidenced by the high activation
energy. At high temperatures (low 1/T) the reaction is diffusion limited as evidenced by the weak
dependence on temperature,
Curve B: Weak dependence on temperature suggests diffusion limitations
Curve C: Strong dependence on temperature suggests reaction limited.
P15-5 (a)
! 1 $ sinh φ1λ
Then ϕ = # &
" λ % sinh φ1
3
Effectiveness factor: η = 2 [φ1 cosh φ1 −1] first order reaction.
φ1
𝜆=
7 ∗ 10DE
= 0.7
1 ∗ 10D4
! sinh (φ 2) $
1
1
& ⇒ φ1 = 6.0
At λ = , ϕ = 0.1 : 0.1 = 2 #
2
#" sinh (φ1 ) &%
∴ϕ =
1 # sinh ( 6 × 0.7) &
%
(
0.7 $ sinh 6 '
P15-5 (b)
3
η = 2 [φ1 cosh φ1 −1] = 0.80 ⇒ φ1 = 2.04
φ1
Use the following equation to calculate R
Since SA, 𝜌H , 𝑘, De is constant we can find R2 as
K
6 O
0.001 O
N =K
N
2.04
𝑅
Or, R=(2.04/6)*0.001 = 3.4*10-4
Thus, dp= 2*R =6.8*10-4 cm
15-4
P15-5 (c) Since most of the reaction is taking place on the external surface and reactant concentration is
negligible in catalyst interior surface, we can suggest to avoid plating the catayst interior with Platinum as
Platinum is costly.
P15-6 (a)
Start with a mole balance on A in the tail:
Divide by ADz and take the limit as z ® 0.
From the flux equation:
Combining the two equations we get:
Dividing by –De we get:
We need boundary conditions
15-5
P15-6 (b)
P15-6 (c)
2𝐷𝐶1R
2 ∗ (3.6 ∗ 10DT ) ∗ (4.36 ∗ 10DU )
𝐿=Q
=Q
𝑘
(23 ∗ 10DVW )/(3 ∗ 10DVR ∗ 𝐿)
P15-6 (d)
The answer in Part © is equal to the average tail length. h = 1 in this problem. If h = 1, then it contradicts
the assumption of diffusion being rate-limiting.
P15-7 (a)
First-order irreversible reaction:
A®B
Let ψ = CA CAS
15-6
Concentration distribution:
𝐶1
Z𝐶 =
1Y
𝑐𝑜𝑠ℎ(1 − 𝑧/𝐿)
𝑐𝑜𝑠ℎ𝜙
"
( " L %+ %
$ cosh *φ $1− '- '
1
) # L &, '
At z = L, CA = CAS = CAS $
$
'
10
cosh φ
$
'
#
&
1
1
=
10 cosh φ
φ = 2.9932
P15-7 (b)
At z = 1/2 L:
"
( " 0.5L %+ %
$ cosh *φ $1−
'- '
L &, '
) #
$
CA = CAS
$
'
cosh φ
$
'
#
&
% cosh !2.9932 ( 0.5)# (
"
$*
CA = ( 0.001) '
' cosh ( 2.9932) *
&
)
CA = 2.345 ×10−4 gmol l
P15-7 (c)
φ = 2L
2
⇒ φα ( 2L )
Dd
ηold =
tanh φold tanh ( 2.9932)
=
= 0.3324
φold
2.9932
ηnew = 0.8 =
tanh φ new
= φ new = 0.8880
φ new
%φ
(
% 0.8880 (
−4
2L new = 2L old ' new * = 2 ×10−3 '
* = 5.993×10 cm
& 2.9932 )
& φold )
(
)
15-7
P15-7 (d)
At z = L, C A =
C AS
0.001
=
= 7.038 ×10−4 gmol l
cosh φ new cosh ( 0.8880)
Thus mimimum C is now 70.38% of C . Therefore the suggestion is plating entire surface of the inside of
the pore.
A
AS
P15-8
15-8
P15-8 (b)
$ 2k − '1 2
tanh φ1
)
where η = φ1 = L &&
Before poisoning, η =
)
φ1
rD
% Ab (
After poisoning, the differential equation and boundary conditions are the same for the unpoisoning
^_
region of the pore, z1 < z < L if z is replaced by L – z; and if we let CA = CA1 at z¢ = 0 and ^a` = 0 at z = L – z1.
Then, for the unpoisoned section of the catalyst pore h applies if f is replaced by:
+ # z &.
tanh -φ1 %1− (0
# 2 k"" &
# z&
, $ L '/
( = %1− ( φ1 , i.e., η =
( L − z) %
# z&
$ L'
$ rD Ab '
φ1 %1− (
$ L'
The effectiveness factor for the unpoisoned section of catalyst pore is defined as:
12
This can be related to the overall effectivness factor for the entire pore by
15-9
P15-9 (a)
Therefore, concentration profile can be written as:
$" cosh (αz) $&
C A = C AS #
'
%$ cosh (αL ) ($
* sinh αz
* dC
( ) -/
WA A P = −De , A
/A P = −A P DeC ASα ,
, cosh (αL )
/
+ dz z=L .
+
z=L .
WA A P = −A P DeαC AS tanh (αL )
15-10
−WA A P ηrA# aA P L = −ηkC ASaA P L
By the sign convention:
∴ηkC ASaA P L = A P DeαC AS tanh (αL )
$ ka '
$ αD '
ka $ De ' 1
η = & e ) tanh (αL ) =
L ))
& ) tanh &&
% kaL (
De % ka ( L
% De (
$ ka 1 '
$ ka '
)) tanh (αL ) &&
η = &&
L ))
% De L (
% De (
−WA A P = k C A P (C A0 − C AS ) = αA P DeC AS tanh (αL )
" αD
%
e tanh αL
C A0 = C AS $1+
( )'
kC
#
&
"
%
'
C AS $$
1
'
=
C A0 $1+ αDe tanh αL '
( )'
$#
kC
&
* ka * D 1e
L //
,
/tanh ,,
η
+ ka L .
+ De .
Ω=
=
*
αDe
1+
tanh (αL ) 1+ kaDe tanh , ka L /
, D /
kC
kC
+
e .
P15-9 (b)
AàB
WAπ rL −WAπ rL
r
r+Δr
+ rAπ rΔrL = 0
1 d
rW + r = 0
r dr Ar A
EMCD therefore, WA = −De
1
D
r e
d
dC A
dr
rdC A
" 2
%
dr = D $ d C A − 1 dC A '
e$
2
dr
r dr '
# dr
&
rA = −kC A
1 d
rW + r = 0
r dr Ar A
" 2
%
d C A 1 dC A '
De $
−
− kC A = 0
$ dr 2 r dr '
#
&
ψ=
CA
C A0
λ=
r
R
15-11
d 2ψ
dλ 2
d 2ψ
dλ
2
−
1 dψ kR2ψ
−
=0
λ dλ
De
−
1 dψ
−D ψ = 0
λ dλ a
At λ = 1, ψ = 1 and at λ = 1,
Bessel Function Solution
dψ
=0
dλ
P15-10 (a)
EMCD WA = −D
dC A
, rA = −k
dz
In − Out + Gen = 0
€
€
WA Ac z − WA Ac z+Δz + rAΔzA c = 0
dWA
+ rA = 0 ,
dz
DAd˜ C A
−k =0
dz 2
CA
C As
d 2ψ C A0 d 2C A
=
dλ2 L2
dz 2
−
ψ=
, λ=
Z
,
L
d 2ψ
kL2
−
=0
dλ2 DAC As
€
(1)
dψ
kL2
=
λ + C1
dλ DAC As
€
Using the symmetry B. C.
λ=0
€
dψ
=0
dλ
C1 = 0
Integrating equation (1)
€
ψ=
€
kL2 λ2
+ C2
2DAC As
at λ = 1 ψ = 1
€
€
15-12
[
]
ψ = 1+ φ 20 λ2 −1
P15-10 (b)
Now let’s find what value of l that y = 0 for different φ 0 .
€
[
]
For φ 20 = 1 : 0 = 1+ 1 λ2 −1 = 1+ λ2 −1
2
λ =0
€
Therefore, the concentration is zero (i.e., y = 0) at
€
€
λ=0
[
]
For φ 20 = 16 : 0 = 1+ λ2 −1 = 1+ 16λ2 −16
λ2 =
€
€
€
15
= 0.938
16
Seems okay, but let’s look further and calculate the concentration ratio y at l = 0 for φ 0 = 4.
[
]
2
ψ = 1+ 16 (0.2) −1 = 1+ 16[0.04 −1] = −14.9
Negative concentration.
€
P15-10 (c)
Let’s try again with φ 0 = 10
€
2
[
]
ψ = 1+ (10) 0.12 −1 = 1−10 2 (0.99)
ψ = 1− 99 = −99 not possible
€
y will be negative for any value of φ 0 greater than one.
€
€
P15-10 (d)
We now need to resolve the problem with the fact that there is a critical value of l, lc, for which both
dψ
=0
y = 0 and
dλ
d 2ψ
− 2φ 20 = 0
2
dλ
€ dψ − 2φ 2 λ + C
dλ2
€
0
€
1
ψ = φ 20 λ2 + C1λ + C 2
At λ = 1 , ψ = 1
1 = φ 20 + C1 + C 2
€
€
At λ = λ C , ψ = 0
0 = φ 20 λ2C + C1λ C + C 2
€
15-13
€
€
Subtracting
1 = φ 20 − φ 20 λ2C + C1 (1− λ C )
Solving for C1
C1 =
€
(
1− φ 20 1− λ2C
)
1− λ C
Solving for C2
C 2 = 1− φ 20 −
€
(1− φ (1− λ ))
2
0
2
C
1− λ C
(1− φ (1− λ )) λ + 1− φ − (1− φ (1− λ ))
ψ=φ λ +
2
0
2 2
0
2
C
2
0
1− λ C
€
2
0
(
1− λ C
(
1− φ 20 1− λ2C
dψ
2
= 0 = 2φ 0 λ C −
At λ = λ C then
dλ
1− λ C
€
2
C
))
(
0 = φ 20 λ2C − 2φ 20 λ C − 1− φ 20
€
€
λ C = 2φ 20 −
λ C = 1−
( )( (
4φ 40 − 4 φ 20 − 1− φ 20
2φ 20
)
)) = 1− 4φ 40 + 4φ 20 − 4φ 40 = 1− 4φ 20
2φ 20
2φ 20
1
φ0
€
€
Sketch of concentration profile for different values of f0
f0 = 1 then lc = 0
f = 2 then lc = 0.5
That is for f0 = 2, the concentration of A is zero halfway (l = 0.5) through the slab.
1
1− λ C =
φ0
[
]
[
]
= φ 20 λ2 + [ φ 0 − φ 20 − φ 20 + φ 0 ] λ + 1− φ 20 − 2φ 0 + 2φ 20
ψ = φ 20 λ2 + φ 0 − φ 20 (1+ λ C ) + 1− φ 20 − φ 0 − φ 20 (1+ λ C )
€
ψ = φ 20 λ2 + 2φ 0 (1− φ 0 ) λ + 1− 2φ 0 + φ 20
For l > lc
15-14
€
P15-10 (e)
zc
η=
η=
∫ 0 −rAAcdz + ∫
0


L
−rA Acdz
z
c
−rA Acdz
=
−rA Acz c + 0
L
∫ 0 −rAAcdz
=
zc
L
zc
1
= λ c = 1−
L
φ0
€
−rA = 0 for z c < z < L
€
P15-10 (f)
€
P15-10 (g) Use spherical coordinates equation to solve this part
P15-10 (h) Individualized solution
P15-11
Given: Second-order decomposition reaction: A ® B + 2C
k = 50 m4/sec. gmol ; dp = 0.4 cm ; U = 3 m/s ; T = 250°C = 523°K;
P = 500 kPa = 4.936 atm ; X = 0.80 ; De = 2.66 ´ 10–8 m2/s ; Eb = 0.4 ;
rb = 2 ´ 106 g/m3 ; Sa = 400 m2/g.
Neglecting axial diffusion with respect ot forced axial convection, we have:
15-15
𝑔
𝑔𝑚𝑜𝑙 104 𝑙
𝑚E
𝑚O
K50
N K400
N g2 ∗ 10T 4 h K0.115
NK 4 N
𝑔𝑚𝑜𝑙.
𝑠
𝑔
𝑙
𝑚
𝑚
ΦO = 0.2 ∗ 10DO 𝑚d
O
𝑚
2.66 ∗ 10DW 𝑠
Reactor length:
3𝑚
1
𝑠 g1 − 0.8 − 1h
𝐿=
𝑔
𝑔𝑚𝑜𝑙
𝑚E
𝑚O
(9.313 ∗ 10DW )2 ∗ 10T 4 K50
N K400 N K115
N
𝑠. 𝑔𝑚𝑜𝑙
𝑔
𝑚
𝑚4
𝐿 = 2.80 × 10DU 𝑚
P15-12 (a)
15-16
, where Io and Ko are Bessel functions.
P15-12 (b) No solution will be given.
P15-13 (a)
15-17
15-18
The unit of k is
k l.m
knop.m.kqr.stuv
which is for k’
P15-13 (b)
P15-13 (c)
R.U
𝑘 w 𝜌x 𝐶1R
ΦO = 𝑅Q
𝐷.
(1 ∗ 10DO )O ∗ (2.035 ∗ 10E ) ∗ (2.3 ∗ 10T ) ∗ (0.196 ∗ 1000)R.U ∗ 60
ΦO = Q
0.23 ∗ 10DE
ΦO = 13.1 ∗ 10T
𝜂=K
2 V/O
3
N ∗K N
𝑛+1
Φr
𝜂=K
V/O
2
3
N ∗K
N = 0.2 ∗ 10DT
1.5 + 1
13.1 ∗ 10T
P15-13 (d)
To make the catalyst more effective, we should use a smaller diameter.
P15-13 (e)
15-19
P15-14
d2y
− φ n yn = 0
2
dλ
dy
Multiply by 2 y
dλ
€
2y
dy d # dy &
2 n dy
% ( = φn y 2
dλ dλ $ dλ '
dλ
€
Manipulating
the L.H.S.
2
d # dy &
dy d 2 y
% ( =2
dλ $ dλ '
dλ dλ2
€
2
d # dy &
n
% ( = φ n y 2dy
dλ $ dλ '
# dy & 2
y n +1
+ C1
% ( = 2φ n
$ dλ '
n +1
C
dψ
y = ψλ = A , λ = 0 y = 0
= 0 therefore C1 = 0
C A0
dλ
€
Taking the derivative of y and evaluating at l =1
€
dy
2φ 2n y n +1
=
dλ λ=1
n +1
=
λ=1
2φ 2n
n +1
The effectiveness factor is
$
dC '
πR 2 &DA A )
%
dr ( r=k
η=
4
kC nAs πR 3
3
€
In dimensionless form
€
η=
% dψ (
3' *
& dλ ) λ=1
φ 2n
€
λ = ψλ , differentiating gives
dy
dψ
=λ
+ψ
dλ
dλ
€
at λ = 1
€
15-20
€
dy
dψ
=
+1
dλ λ=1 dλ
dψ
2φ 2n
=
−1
dλ
n +1
€
%
(
2
' 2φ n −1* 3 2
n +1 − 3
η = 3' n +21 * =
'
*
φn
φn
φ 2n
'
*
&
)
For larger φ n
€
η=3
2 1
n + 1 φn
€
€
P15-15 (a) Compare the slopes of the lines given by S/-rL vs 1/m.
Slope in case of Catalyst A = 330 min.g/dm3
Slope in case of Catalyst B = 50 min.g/dm3
Clearly, Catalyst A has higher diffusion resistance and hence will have a smaller effectiveness factor.
P15-15 (b) A more efficient gas sparger would mean increases hydrogen solubility. As catalyst A has
smaller effectiveness factor, the system is diffusion limited and hence increased hydrogen solubility
won’t give any significant increase in rates.
P15-5 (c) For Catalyst B, the overall resistance is given by,
Y
V
= 50 g h + 1
D|}
k
For the combined diffusional resistances to be less than 50% of the total resistance,
~
•
URg h
~
•
URg h€V
< 0.5
Therefore, m > 50 g/dm3.
P15-16 (a) S/-r and 1/m for the three runs are:
𝑆
−𝑟
0.5
3
1
1
𝑚
0.333
2
0.667
Size of catalyst
12
50
50
The three points given by (S/-r, 1/m) of the three runs are collinear. Therefore, overall resistance due to
diffusion doesnot seem to depend on the radius of pellet in this range.
This line is given by,
Y
D|}
V
= 1.5 gkh
This equation implies, resistance due to gas absorbtion is zero and that resistance is mainly due to
diffusion-reaction effects. But as the diffusion-reaction resistance is not depending on radius of pellets,
this implies reaction limited rate.
15-21
Therefore, out of the given recommendations by installing a more efficient sparger overall rate can be
increased.
P15-16 (b) From Run 1 and 2 we can see that the rate is independent of catalyst size. This implies the
overall rate is reaction limited. Therefore, effectiveness factor = 1
P15-16 (c) In case of Run 4,
Y
D|
V
= 3.004, k = 0.5
Overall resistance = 3.004/0.5 = 6
Ratio of resistances in case of 50 µm and 750 µm = 4
Ratio of particle sizes = 15
Implying 𝑟t 𝛼 𝑑O . Therefore reaction is external diffusion limited in this range of particle size.
We know,
V
𝑟t = … u
† †
T
T
𝑎t = ‡ ‰ = ŠUR‹k∗Os/tk4 = 0.4𝑑𝑚 O /𝑔
ˆ ˆ
𝑘t = 9.32 × 10-4 mol/dm.min
As resistance due to gas absorbtion is zero, percentage of external resistance in overall resistance is
100%.
P15-17 See solution on Web in Chapter 15 Expanded Material.
Now consider the case of a first order reaction
(1) First Order Reaction Kinetics
De
d2C A
2
dz
d2 ψ
dλ2
−kC A = 0*,*ψ =
− Daψ = 0*****Da =
CA
C A0
*,*λ =
z
b
kb2
De
ψ = A*cosh Daλ +B*sinh Daλ
dψ
= A Da sinh Daλ +B Da coshλ
dλ
Symmetry
B = −A
dψ
= 0#@#λ = 1#∴
dλ
sinh Da
cosh Da
ψ = 1"@"λ = 0
= −A)tanh Da
1 = A#cosh Da
A=
ψ=
1
cosh Da
#,##B =
cosh Da#λ
cosh Da
−
tanh Da
cosh Da
tanh Da
cosh Da
sinh Da#λ
15-22
(2) Now consider Monod Kinetics
d2C A
µ C C
− max A C = 0
KS + C A
dz2
De
dCC
µ C C
= max A C = 0
KS + C A
dz
2
Use
Quasi
Steady
State
Analysis
No further solution to Monod Kinetics will be given.
(3) Now consider Variable Diffusion Coefficient
dFw
= υc WO2
A
z=0 c
dt
D C dψ
dC
WA = −De A = − e A0
dz
L dλ λ=0
D C dψ
= e A0
υ A V
dt
L dλ λ=0 c c
dFw
Mole balance
! dC $
d#De A &
dz &%
#"
−k = 0
dz
for hindered diffusion
DAB
De =
1+ α2 Fw2 1−Fw
(
)
As a first approximation, assume no variation in De with l
d2 ψ
kL2
=0
dλ2 DeC A0
as before
−
φ0 =
kL2
2DeC A0
d2 ψ
−2φ0 = 0
dλ2
Solution the same as before Equation (E12-2.13)
ψ = φ0 λ λ −2 +1
(
WA = −De
)
dC A
dz
z=0
D C dψ
= − e A0
L dλ λ=0
dψ
2kL2
= 2φ0 λ −2φ0
= −2φ0 = −
dλ
2DeC A0
λ=0
(
)
% D C (% kL2 (
* = kL
WO = WA = '' − e A0 **' −
' DC *
2
L
&
)& e A0 )
The flux of O2 in does not depend upon De, which is not uprising since this reaction is zero order.
15-23
For the buildup of material that hinders diffusion
dFW
= υc AckL = AcLk = kV
dt
FW = kVt
From a quasi-steady state approximation as time goes on FW increases De decreases and φ 0 increases.
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 l = 0 and l = 1 (see solution to P12-10(c) in which
the solution manual) we will find
€
1
λc =
φ0
We see that as t increases lc decreases, that is the point at which the oxygen concentration is zero
moves toward the top of the gel.
15-24
Synopsis for Chapter 16 –Residence Time Distributions of
Chemical Reactors
General: The purpose of the problems in this chapter is for the reader to become familiar with using the
RTD to characterize reactors.
Questions
l Q16-1A (14 seconds) Questions Before Reading (QBR).
Q16-3B (4 min total) Part (a) is a quick answer and Part (b) can be sketched directly on the figure in
the example.
O Q16-4B (6-12 min) View screencasts supported on the Extra Help on the Web.
O Q16-5B (10 min) The students should practice thinking about actions they would take to prevent
unwanted reactor behavior. The Monsanto Explosion, Example 13-2, is a very good choice to apply
AWFOS-S16 to.
Computer Simulations and Experiments
l P16-1A (5-8 min) Wolfram LEP to compare differences between Xseg and Xmm.
Problems
l P16-2B (a) (30 min) This part strongly recommended. Students use their imagination to suggest a
diagnosis (e.g., Figure 16-12 and or Figure 16-14) for each malfunctioning reactor to match RTD
curves with reactor operation. I almost always assign all of parts (a). (b) (45 min max) often assigned.
l P16-3C (35 min) Old Exam Question (OEQ). Somewhat difficult problem to find t, s2, tm and E(t).
AA P16-4B (30 min) Straight forward problem using tracer input to find tm and s2. Similar to P16-6B.
l P16-5B (30 min) Old Exam Question (OEQ). Easier problem than P16-3C to find t, s2, E(t) and tm.
AA P16-6B (30 min) Similar to problem P16-5B.
S P16-7 (60 min) Rather difficult derivation of E(t).
O P16-8 (30 min) Old Exam Question (OEQ). Somewhat difficult. Students need to realize they have to
find the parameters A and B first.
O P16-9A (10 min) Easy problem to carry out the integration.
S P16-10B (35 min) Novel problem using RTD to wait in line at gasoline stations.
AA P16-11B (40 min) Old Exam Question (OEQ). Straight forward problem but with a lot of parts. Covers
all necessary calculations of RTD. Similar to problem P16-12B.
I P16-12B (50 min) Similar to Problem P16-11B.
AA P16-13C (35 min) Good problem with real data, t, can be calculated from tank dimensions.
S16-1
Page intentionally blank
S16-2
Solutions for Chapter 16 – Residence Time Distributions of
Chemical Reactors
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-16
http://umich.edu/~elements/5e/16chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q16-1 Individualized solution.
Q16-2 Individualized solution.
Q16-3B (a)
The area of a triangle (h=0.044, b=5) can approximate the area of the tail :0.11
~0.11
Q16-3B (b)
For a PFR/CSTR Series
$0 t < τ
p
&
&
E t = % −(t−τp )/τ s
&e
&'
τs
()
t ≥ τp
16-1
Q16-4 Individualized solution.
Q16-5 Individualized solution.
P16-1
(i)
One set of parameters can be: 𝜏" = 1 min, k = 1 m3/kmol.min, 𝜏# = 5 min
(ii) 𝜏" = 12 min, k = 1 m3/kmol.min, 𝜏# = 1 min
(iii) 1. Early mixing and late mixing give virtually same conversion for large 𝜏# .
2. PFR here is more efective than CSTR.
3. Conversion increases with mean residence time and k.
P16-2 (a)
(1) PFR with dead volume
(2) PFR with bypassing
(3) Ideal CSTR in parallel
(4) PFR/CSTR in series with bypassing
(5) PFR/CSTR in series with bypassing
(6) 4 PFR in series or PFR with recycle
(7) LFR with dead zones and channelling
(8) Two LFR in parallel
(9) Two PFR in parallel
(10) PFR and CSTR in parallel
P16-2 (b)
16-2
LFR
P16-3 (a)
Mean Residence Time
∞
∫
By definition
E(t)dt = 1 . The area of the semicircle representing the E(t) is given by A =
0
τ=
2
2
= 0.8min .
. For constant volumetric flow tm = τ =
π
π
P16-3 (b)
Variance
2
∞
2
∞
σ = ∫ t − τ E(t)dt = ∫ t 2E t dt − τ 2
0
∞
( )
t 2E t dt =
∫ ()
0
σ2 =
0
2τ
∫
0
()
t2 τ 2 − t − τ
2
( )
0
5π
"
%
dt = −τ 4 $cos2 x +2cos x +1'sin2 x dx = τ 4
#
&
8
∫
()
()
()
π
5π 4 2 1
τ −τ =
= 0.159
8
2π
P16-4 (a)
The cumulative distribution function F(t) is given:
The real reactor can be modelled as two parallel PFRs:
PFR
PFR
16-3
PFR
πτ 2
= 1 and
2
"1
3
The relative E(t) = # δ t − τ 1 + δ t − τ 2
4
$% 4
(
)
(
)
Mean Residence Time
1
tm = t dF = (10min*1)+(20min*0.75) = 25min
∫
0
or
∞
$1
' 1
3
3
tm = tE(t)dt = t & δ t − τ 1 + δ t − τ 2 ) = τ 1 + τ 2 = 25min
4
4
%4
( 4
0
∫
(
∫
)
(
)
P16-4 (b)
Variance
∞
'
2
2$1
2 3
2
3
1
σ 2 = ∫ t − τ E(t)dt = ∫ t −tm & δ t − τ 1 + δ t − τ 2 )dt = τ 1 −tm + τ 2 −tm =75min2
4
4
4
%4
(
0
( )
(
)
(
)
(
)
(
P16-5 (a)
)
$ 𝐸(𝑡) = 1
*
For reactor (a), (0.5)(0.2)(t1-0) = 1, so t1 = 10 min.
For reactor (b), (0.5)(0.2)(t1-5) = 1, so t1 = 15 min.
P16-5 (b)
Mean residence time:
ì0.04t , if t < 5
ï
(a) E (t ) = í- 0.04t + 0.4, if 5 £ t £ 10
ï0,otherwise
î
∞
∫ ()
tm = tE t dt =
0
5
10
∫ 0.04t dt + ∫ (−0.04t2 + 0.4t)dt =5 min
0
2
5
ì0, if t < 5
ï
(b) E (t ) = í- 0.02t + 0.3, if 5 £ t £ 15
ï0,if t > 15
î
∞
15
tm = tE t dt = (−0.02t 2 + 0.3t)dt =
∫ ()
0
∫
5
25
min
3
16-4
)
(
)
Variance:
∞
2
∞
2
2
σ = ∫ t −tm E(t)dt = ∫ t 2E t dt −tm
0
(
)
()
0
∞
7
∫ t2E (t )dt = 6 tm2
(a)
0
2
7 2 2 tm 25
−tm =
= = 4.167min2
So σ 2 = tm
6
6
6
∞
15
t 2E t dt = (−0.02t 3 + 0.3t 2 )dt =75
∫ ()
(b)
∫
0
5
2
So σ 2 = 75− tm
= 75−(
25 25 50
)( ) =
= 5.56min2
3 3
9
P16-5 (c)
Fraction of fluid spends longer than 7 minutes:
∞
10
7
7
∫ E (t )dt = ∫ (−0.04t + 0.4)dt =0.18
(a)
)
67
(b) ∫/ 𝐸 (𝑡)𝑑𝑡 = ∫/ (−0.02𝑡 + 0.3)𝑑𝑡 = 0.64
P16-6 (a)
"0,if t < 1
$$
E(t) = #1,if 1 ≤ t ≤ 2
$
$%0,if t > 2
2
∞
∫ ()
tm = tE t dt = t dt =1.5min
∫
0
∞
2
1
2
(
)
∞
σ = ∫ t −tm E(t)dt = ∫
0
0
2
2
t E t dt −tm
=
()
2
1
P16-6 (b) Fraction of fluid spends longer than 1.5 minutes:
2
∞
∫ ()
E t dt =
1.5
∫ 1dt =0.5
1.5
P16-6 (c) Fraction of fluid spends 2 minutes or less:
2
2
0
0
7
1
∫ t2 dt −tm2 = 3 −1.52 = 12 = 0.0833min2
ò E (t )dt = ò1dt =1
16-5
P16-6 (d) Fraction of fluid spends between 1.5 and 2 minutes:
2
2
1.5
1.5
∫ E (t )dt = ∫ 1dt =0.5
P16-7
16-6
P16-7 continued
16-7
P16-7 continued
t
F(t) =
t
∫ ()
∫
E t dt =
49
τ
60
0
∞
tm =
∞
∫ tE (t )dt = ∫
49
τ
60
0
∞
3.392
τ8
t
9
[1− (
49 τ 7
τ 1 1 49 τ 7
) ]dt = 1− 3.392( )8 [ − (
) ]
60 t
t 8 15 60 t
τ8
49 τ 7
t(3.392 [1− (
) ])dt =
9
60 t
t
∞
2
∞
∞
49
τ
60
∞
=
∫
3.392
49
τ
60
Where τ =
τ8
t7
[1− (
49
τ
60
τ8
t9
[1− (
π R2 L
in above equations.
v0
P16-8
P16-8 (a)
#
% A− B(t 0 − t)3 ,if 0 ≤ t ≤ 2 t 0
E(t) = $
%&0,if t > 2 t 0
$
t
& At+ 0.25B(t 0 − t)4 − 0.25Bt 04 ,if 0 ≤ t ≤ 2 t 0
F(t) = E t dt = %
&'1,if t > 2 t 0
0
16-8
π R2 L
0
49 τ 7
) ]dt − τ 2 = 0.133τ 2
60 t
∫ ()
49 τ
49
τ
60
2
σ 2 = ∫ (t − tm ) E(t) dt = ∫ t 2 E (t ) dt − tm
= ∫ t 2 (3.392
0
τ8
∫ 3.392 t8 [1− ( 60 t )7 ]dt = τ = v
49 τ 7
2
) ])dt − tm
60 t
P16-8 (b)
2 t0
∞
tm =
2
∫ tE (t )dt = ∫ t[A− B(t0 − t)3 ]dt = 2 At02 + 5 Bt50
0
0
∞
2
2
∞
σ = ∫ (t − tm ) E(t) dt = ∫
0
2
2
t E t dt − tm
=
0
()
2 t0
∫ t 2[A− B(t0 − t)3 ]dt − tm2
0
8
4
2
8
4
8
4
= At 30 + Bt 60 −[2 At 02 + Bt 50 ]2 = At 30 − 4 At 04 + Bt 60 − ABt 70 − B2 t10
0
3
5
5
3
5
5
25
P16-8 (c)
∞
∫ E (t )dt = 1
0
2 t0
So
∫ A− B(t0 − t)3dt = 1
0
A(2 t 0 ) + 0.25B(t 0 − 2 t 0 )4 − 0.25Bt 04 = 1
At 0 =
1
2
Also E(t) ≥ 0,if 0 ≤ t ≤ ∞
Hence A− B(t 0 − t)3 ≥ 0,for0 ≤ t ≤ 2t0
B(t− t 0 )3 + A ≥ 0,for0 ≤ t ≤ 2t0
A− Bt 30 ≥ 0
P16-9
First Moment about the mean: by definition is always equal to zero.
m1 =
∞
∞
∞
0
0
0
∫ (t − τ )E (t ) dt = ∫ tE (t ) dt − τ ∫ E (t ) dt = τ − τ = 0
m1CSTR = m1PFR = m1LFR = 0
P16-10 (a)
Assuming each station has 1 person only
Average time required=
*∗*<=∗><?∗=<@∗7<6A∗B<67∗A<6B∗6<A6∗*
A=
= 219D23 = 9.52 𝑚𝑖𝑛𝑠
P16-10 (b)
Hint: We can model this problem as a CSTR. The time spend is given by total waiting time.
For a CSTR, internal age distribution function can be used to characterize the time person has been (and
still is) on the gas station
16-9
P16-10 (b) continued
1 N
𝐼 (𝛼 ) = − 𝑒 M ⁄O
𝜏
Q
Q
6
N
Average waiting time = ∫* 𝛼 𝐼 (𝛼 )𝑑𝛼 = − O ∫* 𝛼𝑒 M ⁄O 𝑑𝛼
P16-10 (c) No solution will be given
P16-11 (a)
External age distribution E(t)
By plotting Cx105 as a function of time, the curve shown is obtained
C(t)
C(t)
900
800
700
600
C
500
400
300
200
100
0
0
10
20
30
40
50
60
t[min]
To obtain the E(t) curve from the C(t) curve, we just divide C(t) by the integral
6
∞
10
50
60
∫ 0 C(t)dt = ∫ 0 C(t)dt + ∫ 6 C(t)dt + ∫10 C(t)dt + ∫ 50 C(t)dt
3
6
∫ 0 C(t)dt = 8 (1)$%1(0)+ 3(622)+ 3(812)+2(831)+ 3(785)+ 3(720)+(650)"#10−5 = 4173.4⋅10−5
10
1
50
1
∫ 6 C(t)dt = 3 (2)"#(650)+ 4(523) +(418)$%10−5 = 2106.7⋅10−5
∫10 C(t)dt = 3 (5)$%418 + 4(238)+2(136)+ 4(77)+2(44)+ 4(25)+2(14)+ 4(8)+5"#10−5 = 3671.7⋅10−5
60
1
∫ 50 C(t)dt = 2 (10)"#1(5)+1(1)$%10−5 = 30⋅10−5
∞
∫ 0 C(t)dt = 9981.7⋅10−5 ≅ 0.1
16-10
70
We now calculate:
C(t)
E(t) =
=
∞
∫ 0 C(t)dt
C(t)
0.1
E(t)
0.09
0.08
0.07
E(t) [1/min]
0.06
0.05
0.04
0.03
0.02
0.01
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
t[min]
Using Excel we fit E(t) to a polynomial:
$
−3 4
−2 3
−2 2
& for 0 ≤ t ≤ 3 E1 (t) = −1.1675⋅10 t +1.1355⋅10 t − 4.7492⋅10 t
&
−2
&+9.9505⋅10 t
& for 3 ≤ t ≤ 20 E (t) = −1⋅8950⋅10−6 t 4 + 8.7202⋅10−5 t 3 −1.1739⋅10−3 t 2
2
&
&
−4
E(t) = %−1.7979⋅10 t + 0.092343
&
−8 4
−6 3
−4 2
& for 20 ≤ t ≤ 60 E3 (t) = 1.2618⋅10 t −2.4995⋅10 t +1.8715⋅10 t −
&
−3
&6.3512⋅10 t + 0.083717
& for t > 60 0
&'
P16-11 (b)
External age cumulative distribution F(t)
F(t) =
t
∫ 0 E(t)dt
:
Integrating the E(t), we obtain the F(t):
16-11
48
50
52
54
56
58
60
62
64
1.2
1
F(t)
0.8
0.6
0.4
0.2
0
0
10
20
30
40
t[min]
P16-11 (c)
Mean residence time and variance
tm =
∞
∫ 0 E(t)t dt
The area under the curve of a plot tE(t) as a function of t will yield tm.
tm=t=10min
16-12
50
60
t
0
0.4
1
2
3
4
5
6
8
10
15
20
25
30
35
40
45
50
60
tm =
C(t)
0
329
622
812
831
785
720
650
523
418
238
136
77
44
25
14
8
5
1
E(t)
0
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
6
∞
tE(t)
0
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
10
t-tm
-10
-9.6
-9
-8
-7
-6
-5
-4
-2
0
5
10
15
20
25
30
35
40
50
50
(t-tm)2E(t)
0
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
60
∫ 0 E(t)t dt = ∫ 0 E(t)t dt + ∫ 6 E(t)t dt + ∫10 E(t)t dt + ∫ 50 E(t)t dt
tm = τ = 9.88min ≅ 10min
We can calculate the variance by calculating the area under the curve of a plot of: (t-tm)2E(t)
s2=74min2
σ2 = ∫
50
2
∞
2
6
10
2
t −tm ) E(t)t dt = ∫ (t −tm ) E(t)dt + ∫ (t −tm ) E(t)dt +
0 (
0
6
2
60
2
∫10 (t −tm ) E(t)dt + ∫ 50 (t −tm ) E(t)dt
σ 2 = 73.81min2 ≅ 74min2
16-13
P16-11 (d)
Fraction of the material that spends between 2 and 4min in the reactor
4
1
∫ 2 E(t)dt = shaded area = 3 "#1(0.0812)+ 4(0.0831)+1(0.0785)$% = 0.16
P16-11 (e)
Fraction of the material that spends longer than 6min
16-14
10
∞
50
60
∫ 6 E(t)dt = shaded area = ∫ 6 E(t)dt + ∫10 E(t)dt + ∫ 50 E(t)dt
10
2
50
5
∫ 6 E(t)dt = 3 (1(0.065)+ 4(0.0523)+1(0.0418)) = 0.210
∫10 E(t)dt = 3 (1(0.0418)+ 4(0.0238)+2(0.0136)+ 4(0.0077)+2(0.0044)+ 4(0.0025)
)
+2(0.0014)+ 4(0.0008)+1(0.0005) = 0.367
10
60
∫ 50 E(t)dt = tail = 2 (0.0005+ 0.0001) = 0.003
∞
∫ 6 E(t)dt = shaded area = 0.581
P16-11 (f)
Fraction of the material that spends less than 3min
3
3
∫ 0 E(t)dt = shaded area = 8 "#1(0)+ 3(0.0622)+ 3(0.0812)+1(0.0831)$% = 0.192
P16-11 (g)
Normalized distributions
Normalized RTD
Θ=
t
τ
( )
()
E Θ = τE t
16-15
0.9
0.8
0.7
E(Q)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
Q
Normalized cumulative RTD
Θ
Θ
( ) ∫ 0 E (Θ) dΘ = ∫ 0 E (t ) dt
F Θ =
1.2
1
F(Q)
0.8
0.6
0.4
0.2
0
0
1
2
3
Q
P16-11 (h)
Reactor Volume
F=10 dm3/min
V = F ⋅ τ = 100dm3
P16-11 (i)
Internal age distribution
1
I t = "#1− F t $%
τ
()
()
16-16
4
5
6
0.12
0.1
I(t)
0.08
0.06
0.04
0.02
0
0
10
20
30
40
50
60
70
t[min]
P16-11 (j)
Mean internal age
∞
6
10
0
0
6
αm = ∫ I(t)t dt = ∫ I(t)t dt + ∫
I(t)t dt +
50
60
∫10 I(t)t dt + ∫ 50 I(t)t dt = 1min
P16-11 (k) Refer Ch-17 and Ch-18 solution manual where the solution to this problem is continued.
P16-12 (a)
C-t curve is given by
The area under C(t) curve is calculated to be
E(t) curve is given by
E(t)= C(t)/ 0.973
16-17
P16-12 (b)
P16-12 (c)
P16-12 (d)
The fraction spending less than 250 s is 0.422
P16-12 (e)
P16-12 (f)
16-18
P16-12 (g)
Θ = 𝑡/𝜏
E(Θ) = τ E(𝑡)
Where τ = 273 s
t (s)
0
150
175
200
225
250
275
300
325
350
375
400
425
450
C (g/dm^3)
0
0
1
3
7.4
9.4
9.7
8.2
5
2.5
1.2
0.5
0.2
0
E(t)
0.00
0.00
1.03
3.08
7.61
9.66
9.97
8.43
5.14
2.57
1.23
0.51
0.21
0.00
𝚯
0.00
0.55
0.64
0.73
0.82
0.92
1.01
1.10
1.19
1.28
1.37
1.47
1.56
1.65
E(𝚯)
0.00
0.00
280.6
841.7
2076.3
2637.4
2721.6
2300.7
1402.9
701.4
336.7
140.3
56.1
0.00
Using data in above table, we can plot E(𝚯) vs 𝚯
Y
F(Θ) = ∫* 𝐸 (Θ)𝑑Θ
Applying Simpson’s formula
For Θ = 0.55 to 0.92
h= (0.92-0.55)/4 = 0.092
*.*@A
A1 = = (0 + 4 ∗ 280.6 + 2 ∗ 841.7 + 4 ∗ 2076.3 + 2637.4) = 419.6𝐸 − 3
For Θ = 0.92 to 1.01
A2= 0.09* (2637.4+2721.6)*0.5 = 241E-3
For Θ = 1.01 to 1.37
h= (1.37-1.01)/4 = 0.092
*.*@A
A3 = = (2721.6 + 4 ∗ 2300.7 + 2 ∗ 1402.9 + 4 ∗ 701.4 + 336.7) = 545.5𝐸 − 3
16-19
P16-12 (g) continued
For Θ = 1.37 to 1.65
h= (1.65-1.37)/3 = 0.092
=∗*.*@A
(336.7 + 3 ∗ 140.3 + 3 ∗ 56.1 + 0) = 31.8𝐸 − 3
A4 =
B
Sum of Area= A1+A2+A3+A4 = (419.6+241+545.5+31.8 )*10^-3 =1.24
There is some inaccuracy in calculation as Simpsons rule is not perfect and there may be some error in
E(t) calculation
𝚯
0.55
0.92
1.01
1.37
1.65
F (𝚯)
0
0.419
0.66
0.12
0.124
Using above data, following graph is obtained
P16-13
16-20
16-21
Page intentionally blank
16-22
Synopsis for Chapter 17 – Predicting Conversion Directly From
the Residence Time Distribution
General: Problems in this chapter use the parameters s2, tm, and E(t) that we showed how to calculate in
the last chapter to predict conversion in zero and one parameter RTD models.
Questions
l Q17-1A (22 seconds) Questions Before Reading (QBR).
O Q17-3 (6-12 min) See Extra Help on the Web site for suggested screencast.
AA Q17-4 (7 min) Encourages students to think about the questions they ask.
Computer Simulations and Experiments
l P17-1B (10-12 min each) Parts (a) through (f). I always assign all Wolfram problems.
Problems
AA P17-2B (35 min) Old Exam Question (OEQ). Fairly straight forward calculation after finding the
Damköhler number tk from the data given for 86.5% conversion.
O,G P17-3C (50 min) Rather tricky derivation.
l P17-4C (40 min) Continuation of Chapter 16 problem P16-3C in order to predict conversion.
AA P17-5B (40 min) Old Exam Question (OEQ). Continuation of Chapter 16 problem P16-4B in order to
predict conversion.
l P17-6B (35 min) Continuation of Chapter 16 problem P16-5B in order to predict conversion.
Sometimes I specify Figure (a) and (b) but not both.
O P17-7B (10 min max) Compares conversion obtained at different values of rate constant.
AA P17-8B (25 min) Straight forward calculation to find conversion but need to calculate new k1 at 10°C
higher temperature.
AA P17-9B (40 min) Old Exam Question (OEQ). Continuation of Chapter 16 problem P16-6B in order to
predict conversion. Sometimes I only assign Part (a).
AA P17-10A (30 min) Straight forward calculation to find conversion in PFR and LFR.
O P17-11A (20 min) Straight forward integration.
AA P17-12B (10 min) Easy calculation once the students see rate law at high and low CA extremes.
AA P17-13B (30 min) Only one or two of the parts are assigned. Choosing two problems from part (a)
through (d) are recommended. Straight forward calculations.
O P17-14C (50 min) Not entirely straight forward. Sometimes I only assign part (a) or part (b).
O,G P17-15C (55 min) Relatively easy, but not entirely straight forward.
I,G P17-16C (55 min) Straight forward, similar to P17-13B.
O,G P17-17B (55 min) Catalyst decay continuation of Chapter 16 problem P16-11B to calculate
conversion.
S17-1
AA,G P17-18B (40 min) Old Exam Question (OEQ). Good problem to assign as it uses real data from an
industrial reactor.
S17-2
Solutions for Chapter 17 – Predicting Conversion Directly from
the Residence Time Distribution
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-17
http://umich.edu/~elements/5e/17chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q17-1 Conversion predicted by segregation model is less than conversion for PFR, and is more than
conversion for LFR and CSTR.
Q17-2 Individualized solution.
Q17-3 Individualized solution.
Q17-4 Individualized solution.
P17-1 (a)
(i)
Increase in k increases X for both segregation model as well as LFR, increase in n decreases X for
both segregation model as well as LFR whereas change in Cao has no effect on both.
(ii) 1. Observe that the mean residence time does not change as we vary n, k or Cao. Same is the
case with F-curve as well. This is clear as we see differential equations governing them.
Both of them are just dependent on E while E is just a function of time.
2. Conversion X obtained for ideal reactor is always higher than X for both segregation model as
well as for LFR.
Polymath (i)
17-1
P17-1 (b)
(i)
For k=0.0004, X and Xbar are farthest apart and for k=0.01 these both are closest to each other.
(ii) As we increase n, conversion obtained by segregation model decreases.
(iii) E curve is independent of n, CA0, and k. As τ is increased, the E curve shifts rightward, and at a
particular time, E(t) is more than that with smaller τ.
(iv) As k increases both X and Xseg increases. As CA0 increases, X and Xseg increases. As n increases, X
and Xseg decreases. As tau increases, Xseg increases but no effect on X.
E is independent of k, Ca0 and n
(v) E(t) is dependent only on τ.
X>Xseg always
X is independent of τ.
Polymath (i)
(1)
Plot XPFR and XLFR vs. k:
XPFR and XLFR come close together when k is small or large and become farther apart at medium values of k.
(2) Use the E(t) equation given in example 16-1 and 16-2 to get the results
P17-1 (c)
(i)
E(t) is dependent only on τ.
(ii) As n increases, Xseg increases
(iii) E curve is independent of k, CA0 and n
(iv) Increasing k, CA0, n increases both X and Xseg. 𝜏 has no effect so it is not present on slider
(v) (a) Increasing k increases rate of reaction and hence X and Xseg increases
(b) Increasing n increases rate of reaction and hence X and Xseg increases
17-2
Polymath (i)
(1)
Xseg and Xmm come close together when kCa0 is small and become farther apart when kCa0 is large.
(2)
For 2nd order reaction, CSTR equation gives
"
= 𝑘𝜏𝐶+, =3.20
($%")'
𝑋./01 =0.576
(3) For T=350 K
d(X) / d(t) = -ra/Cao
X(0) = 0
d(Xseg) / d(t) = X*E
Xseg(0) = 0
t(0) = 0
t(f) = 200
ra = -k*Ca^2
E = if(t<=70) then(E1) else (E2)
E1 = 4.44658e-10*t^4-1.1802e-7*t^3+1.35358e-5*t^2 -0.000865652*t+0.028004
E2 = -2.64e-9*t^3+1.3618e-6*t^2-0.00024069*t+0.015011
k = 0.01*exp(10000/1.9872*(1/320-1/350))
Ca = Cao*(1-X)
Cao = 8
17-3
Polymath code
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
8.
0.1278158
8.
0.1278158
2 Cao
8.
8.
8.
8.
3 E
0.028004
0.000225
0.028004
0.000225
4 E1
0.028004
0.0028731
0.1635984
0.1635984
5 E2
0.015011
0.000225
0.015011
0.000225
6 k
0.0384938
0.0384938
0.0384938
0.0384938
7 ra
-2.463604
-2.463604
-0.0006289
-0.0006289
8 t
0
0
200.
200.
9 X
0
0
0.984023
0.984023
10 Xseg
0
0
0.8045648
0.8045648
Differential equations
1 d(X)/d(t) = -ra/Cao
2 d(Xseg)/d(t) = X*E
Explicit equations
1 Cao = 8
2 E2 = -2.64e-9*t^3+1.3618e-6*t^2-0.00024069*t+0.015011
3 E1 = 4.44658e-10*t^4-1.1802e-7*t^3+1.35358e-5*t^2 -0.000865652*t+0.028004
4 E = if(t<=70) then(E1) else (E2)
5 k = 0.01*exp(10000/1.9872*(1/320-1/350))
6 Ca = Cao*(1-X)
7 ra = -k*Ca^2
For this case, conversion increases to 0.80
For a pseudo first order reaction:
Polymath code
d(X) / d(t) = -ra/Cao
X(0) = 0
d(Xseg) / d(t) = X*E
Xseg(0) = 0
t(0) = 0
t(f) = 200
Ca = Cao*(1-X)
ra = -k*Ca
E = if(t<=70) then(E1) else (E2)
E1 = 4.44658e-10*t^4-1.1802e-7*t^3+1.35358e-5*t^2 -0.000865652*t+0.028004
E2 = -2.64e-9*t^3+1.3618e-6*t^2-0.00024069*t+0.015011
k = 0.01
Cao=0.4
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 Ca
0.4
0.0541341
0.4
0.0541341
2 Cao
0.4
0.4
0.4
0.4
3 E
0.028004
0.000225
0.028004
0.000225
4 E1
0.028004
0.0028737
0.1635984
0.1635984
5 E2
0.015011
0.000225
0.015011
0.000225
17-4
6 k
0.01
0.01
0.01
0.01
7 ra
-0.004
-0.004
-0.0005413
-0.0005413
8 t
0
0
200.
200.
9 X
0
0
0.8646647
0.8646647
10 Xseg
0
0
0.271287
0.271287
Differential equations
1 d(X)/d(t) = -ra/Cao
2 d(Xseg)/d(t) = X*E
Explicit equations
1 Cao = 0.4
2 Ca = Cao*(1-X)
3 E2 = -2.64e-9*t^3+1.3618e-6*t^2-0.00024069*t+0.015011
4 E1 = 4.44658e-10*t^4-1.1802e-7*t^3+1.35358e-5*t^2 -0.000865652*t+0.028004
5 E = if(t<=70) then(E1) else (E2)
6 k = 0.01
7 ra = -k*Ca
Thus, conversion decreases to 0.27 for pseudo first order reaction
P17-1 (d)
(i)
Increasing k increases conversion. Increasing CA0 increases conversion and increasing n also
increases conversion
(ii) Increasing k increases reaction rate. Increasing CA0 increases reaction rate and increasing n also
increases reaction rate
(iii) (a) As CA0 is increased, the exit concentration of A increases. (b) As n is increased, concentration
of A decreases sharply
Polymath
(i) (1) The solution is same as that given in P17-1B (c) Polymath problem part (1)
(2) The solution is same as that given in P17-1B (c) Polymath problem part (2)
(3) See the following solution
Liquid phase, first order, Maximum Mixedness model
Rate Law:
where
17-5
See the following Polymath program:
Polymath Code:
d(x)/d(z) = -(ra/cao+E/(1-F)*x)
cao = 8
k = 0.08
lam = 200-z
ca = cao*(1-x)
E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004
E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011
F1 = 4.44658e-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-.000865652/2*lam^2+.028004*lam
F2 = -(-9.30769e-8*lam^3+5.02846e-5*lam^2-.00941*lam+.618231-1)
ra = -k*ca
E = if(lam<=70)then(E1)else(E2)
F = if(lam<=70)then(F1)else(F2)
EF = E/(1-F)
z(0)=0
x(0)=0
z(f)=200
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
z
0
0
200
200
x
0
0
0.7829342
0.7463946
cao
8
8
8
8
k
0.08
0.08
0.08
0.08
lam
200
0
200
0
ca
8
1.7365447
8
2.0288435
E1
0.1635984
0.0028731
0.1635984
0.028004
E2
2.25E-04
2.25E-04
0.015011
0.015011
F1
5.6333387
0
5.6333387
0
F2
0.9970002
0.381769
0.9970002
0.381769
ra
-0.64
-0.64
-0.1389236
-0.1623075
E
2.25E-04
2.25E-04
0.028004
0.028004
F
0.9970002
0
0.9970002
0
EF
0.075005
0.0220691
0.075005
0.028004
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] k = 0.08
[3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-10*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-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-.000865652/2*lam^2+.028004*lam
[8] F2 = -(-9.30769e-8*lam^3+5.02846e-5*lam^2-.00941*lam+.618231-1)
[9] 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)
17-6
At z = 200, i.e. λ = 0 (exit), conversion X = 75 %.
The decrease in reaction order from 2nd to 1st 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.
(4) Liquid phase, third order, Maximum Mixedness model
Rate Law:
Where
See the following Polymath program:
Polymath Code:
d(x)/d(z) = -(-k*(1-x)^3+E/(1-F)*x)
cao = 8
k = 0.08
lam = 200-z
ca = cao*(1-x)
E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004
E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011
F1 = 4.44658e-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-.000865652/2*lam^2+.028004*lam
F2 = -(-9.30769e-8*lam^3+5.02846e-5*lam^2-.00941*lam+.618231-1)
E = if(lam<=70)then(E1)else(E2)
F = if(lam<=70)then(F1)else(F2)
EF = E/(1-F)
z(0)=0
x(0)=0
z(f)=200
17-7
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
z
0
0
200
200
x
0
0
0.4867311
0.4614308
cao
8
8
8
8
k
0.08
0.08
0.08
0.08
lam
200
0
200
0
ca
8
4.1061501
8
4.3085534
E1
0.1635984
0.0028733
0.1635984
0.028004
E2
2.25E-04
2.25E-04
0.015011
0.015011
F1
5.6333387
0
5.6333387
0
F2
0.9970002
0.381769
0.9970002
0.381769
E
2.25E-04
2.25E-04
0.028004
0.028004
F
0.9970002
0
0.9970002
0
EF
0.075005
0.0220689
0.075005
0.028004
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)^3+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[2] k = 0.08
[3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-10*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-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-.000865652/2*lam^2+.028004*lam
[8] F2 = -(-9.30769e-8*lam^3+5.02846e-5*lam^2-.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)
At z = 200, i.e. λ = 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 mol/dm3 then the rate falls rapidly and CA is not consumed so
quickly, resulting in a smaller conversion.
(5) Liquid phase, half order, Maximum Mixedness model
Rate Law:
Where
See the following Polymath program:
Polymath Code:
d(x)/d(z) = -(-k*(1-x)^(.5)+E/(1-F)*x)
cao = 8
k = 0.08
lam = 200-z
ca = cao*(1-x)
17-8
E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004
E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011
F1 = 4.44658e-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-.000865652/2*lam^2+.028004*lam
F2 = -(-9.30769e-8*lam^3+5.02846e-5*lam^2-.00941*lam+.618231-1)
ra = -k*ca
E = if(lam<=70)then(E1)else(E2)
F = if(lam<=70)then(F1)else(F2)
EF = E/(1-F)
z(0)=0
x(0)=0
z(f)=200
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
z
0
0
200
200
x
0
0
0.9334778
0.9038179
cao
8
8
8
8
k
0.08
0.08
0.08
0.08
lam
200
0
200
0
ca
8
0.532148
8
0.7694568
E1
0.1635984
0.002873
0.1635984
0.028004
E2
2.25E-04
2.25E-04
0.015011
0.015011
F1
5.6333387
0
5.6333387
0
F2
0.9970002
0.381769
0.9970002
0.381769
E
2.25E-04
2.25E-04
0.028004
0.028004
F
0.9970002
0
0.9970002
0
EF
0.075005
0.0220677
0.075005
0.028004
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)^(.5)+E/(1-F)*x)
Explicit equations as entered by the user
[1] cao = 8
[2] k = 0.08
[3] lam = 200-z
[4] ca = cao*(1-x)
[5] E1 = 4.44658e-10*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-10/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-.000865652/2*lam^2+.028004*lam
[8] F2 = -(-9.30769e-8*lam^3+5.02846e-5*lam^2-.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)
17-9
At z = 200, i.e. λ = 0 (exit), conversion X = 90 %.
The decrease in reaction order from 2nd to ½ has the effect of increasing the exit conversion by 34%. The
smaller the dependency of the rate on CA means that when CA falls below 1 mol/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.
P17-1 (e)
(i) Increasing k1 increases conversion. Increasing k2 increases conversion, but Increasing k3 decreases
conversion
(ii)
Effect of rate constants on concentration
(a) on increasing k1, concentration of C increases but concentration of all other species decreases
(b) on increasing k2, concentration of A and C decreases but concentration of B, D, and E decreases
(c) on increasing k2, concentration of A and E increases but concentration of B, C, and D decreases
Effect of rate constants on selectivity
(a) on Increasing k1 both SC/D and SD/E increases
(b) on Increasing k2, SC/D decreases but SD/E increases
(c) on increasing k3, SC/D increases but SD/E decreases
(iii) Effect of rate constants on reaction rate
(a) on increasing k1, reaction rate of A, B, C increases but reaction rate of D and E decreases
(b) on increasing k2, reaction rate of A, D, E increases, reaction rate of C decreases and there is no
change in reaction rate of B
(c) on increasing k3, reaction rate of B and E increases, reaction rate of D decreases and there is no
change in reaction rate of A and C
(iv) (a) By varying reaction rate constant, one can alter the selectivity and maximize the production of
desired product
(b) the difference between X and Xbar can be narrowed by appropriately selecting values of rate
constants.
Polymath (i)
(1) Asymmetric RTD:
See the following Polymath program for base case simulation at T=350 K
Polymath Code:
d(Ca)/d(t) = ra
d(Cb)/d(t) = rb
d(Cc)/d(t) = rc
d(Cabar)/d(t) = Ca*E
d(Cbbar)/d(t) = Cb*E
d(Ccbar)/d(t) = Cc*E
d(Cd)/d(t) = rd
d(Ce)/d(t) = re
d(Cdbar)/d(t) = Cd*E
d(Cebar)/d(t) = Ce*E
T = 350
k1 = exp((5000/1.987)*(1/350-1/T))
k2 = exp((1000/1.987)*(1/350-1/T))
E1 = -2.104*t^4+4.167*t^3-1.596*t^2+0.353*t-0.004
E2 = -2.104*t^4+17.037*t^3-50.247*t^2+62.964*t-27.402
rc = k1*Ca*Cb
k3 = exp((9000/1.987)*(1/350-1/T))
ra = -k1*Ca*Cb-k2*Ca
re = k3*Cb*Cd
E = if(t<=1.26)then(E1)else(E2)
rb = -k1*Ca*Cb-k3*Cb*Cd
17-10
Scd = Cc/(Cd+.000000001)
Sde = Cd/(Ce+.000000000001)
rd = k2*Ca-k3*Cb*Cd
Ca(0)=1
Cb(0)=1
Cc(0)=0
Cd(0)=0
Ce(0)=0
Cabar(0)=0
Cbbar(0)=0
Ccbar(0)=0
Cdbar(0)=0
Cebar(0)=0
t(0)=0
t(f)=2.52
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
1.
0.0228578
1.
0.0228578
2
Cabar
0
0
0.1513598
0.1513306
3
Cb
1.
0.2840909
1.
0.2840909
4
Cbbar
0
0
0.4543234
0.4539723
5
Cc
0
0
0.3992785
0.3992785
6
Ccbar
0
0
0.3570959
0.3566073
7
Cd
0
0
0.3178411
0.2612331
8
Cdbar
0
0
0.3029636
0.3026417
9
Ce
0
0
0.3166306
0.3166306
10 Cebar
0
0
0.1782569
0.1778722
11 E
-0.004
-0.0272502
0.9590814
-0.0272502
12 E1
-0.004
-27.41437
0.9590814
-27.41437
13 E2
-27.402
-27.402
0.9568354
-0.0272502
14 k1
1.
1.
1.
1.
15 k2
1.
1.
1.
1.
16 k3
1.
1.
1.
1.
17 ra
-2.
-2.
-0.0293515
-0.0293515
18 rb
-1.
-1.
-0.0807076
-0.0807076
19 rc
1.
0.0064937
1.
0.0064937
20 rd
1.
-0.0522662
1.
-0.0513561
21 re
0
0
0.1762699
0.0742139
22 Scd
0
0
1.528438
1.528438
23 Sde
0
0
38.92844
0.8250406
24 T
350.
350.
350.
350.
25 t
0
0
2.52
2.52
Differential equations
1 d(Ca)/d(t) = ra
2
d(Cb)/d(t) = rb
3
d(Cc)/d(t) = rc
17-11
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
d(Cdbar)/d(t) = Cd*E
10 d(Cebar)/d(t) = Ce*E
Explicit equations
1
T = 350
2
k1 = exp((5000/1.987)*(1/350-1/T))
3
k2 = exp((1000/1.987)*(1/350-1/T))
4
E1 = -2.104*t^4+4.167*t^3-1.596*t^2+0.353*t-0.004
5
E2 = -2.104*t^4+17.037*t^3-50.247*t^2+62.964*t-27.402
6
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(t<=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
Now change the temperature in above code. For example, if T is raised to 380 K then,
Cabar=0.103, Scd= 3.7 and Sde=0.34.
Therefore, ff the temperature is raised, the conversion of A increases. The selectivity Sc/d increases with
temperature and Sd/e decreases with increasing temperature
(2) Bimodal RTD
See the following Polymath program for base case simulation at T=350 K
Polymath Code:
d(ca)/d(z) = -(-ra+(ca-cao)*EF)
d(cb)/d(z) = -(-rb+(cb-cbo)*EF)
d(cc)/d(z) = -(-rc+(cc-cco)*EF)
d(cd)/d(z) = -(-rd+(cd-cdo)*EF)
d(ce)/d(z) = -(-re+(ce-ceo)*EF)
d(F)/d(z) = -E
cbo = 1
cao = 1
cco = 0
cdo = 0
ceo = 0
lam = 6-z
T = 350
k1 = exp((5000/1.987)*(1/350-1/T))
k2 = exp((1000/1.987)*(1/350-1/T))
k3 = exp((9000/1.987)*(1/350-1/T))
E1 = 0.47219*lam^4-1.30733*lam^3+0.31723*lam^2+0.85688*lam+0.20909
E2 = 3.83999*lam^6-58.16185*lam^5+366.2097*lam^4-1224.66963*lam^3+2289.84857*lam^2-2265.62125*lam+925.46463
E3 = 0.00410*lam^4-0.07593*lam^3+0.52276*lam^2-1.59457*lam+1.84445
ra = -k1*ca*cb-k2*ca
rb = -k1*ca*cb-k3*cb*cd
17-12
rc = k1*ca*cb
rd = k2*ca-k3*cb*cd
re = k3*cb*cd
E = if(lam<=1.82)then(E1)else(if(lam<=2.8)then(E2)else(E3))
EF = E/(1-F)
Scd = cc/(cd+.0000000001)
Sde = cd/(ce+.0000000001)
ca(0)=1
cb(0)=1
cc(0)=0
cd(0)=0
ce(0)=0
z(0)=0
z(f)=6
F(0)=0.999999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
ca
1.
0.2681151
1.
0.2681151
2
cao
1.
1.
1.
1.
3
cb
1.
0.5376702
1.
0.5376702
4
cbo
1.
1.
1.
1.
5
cc
0
0
0.2764056
0.2764056
6
cco
0
0
0
0
7
cd
0
0
0.2695551
0.2695551
8
cdo
0
0
0
0
9
ce
0
0
0.1859242
0.1859242
10 ceo
0
0
0
0
11 E
0.00911
0.0061155
0.6294825
0.20909
12 E1
346.3456
0.1018044
346.3456
0.20909
13 E2
6737.445
0.0741533
6737.445
925.4646
14 E3
0.00911
0.0061155
1.84445
1.84445
15 EF
9110.
0.2104826
9110.
0.2104826
16 F
0.999999
0.0066163
0.999999
0.0066163
17 k1
1.
1.
1.
1.
18 k2
1.
1.
1.
1.
19 k3
1.
1.
1.
1.
20 lam
6.
0
6.
0
21 ra
-2.
-2.
-0.4122726
-0.4122726
22 rb
-1.
-1.
-0.2890893
-0.2890893
23 rc
1.
0.1441575
1.
0.1441575
24 rd
1.
0.1231833
1.
0.1231833
25 re
0
0
0.1557555
0.1449318
26 Scd
0
0
1.04344
1.025414
27 Sde
0
0
23.62022
1.449812
28 T
350.
350.
350.
350.
29 z
0
0
6.
6.
17-13
Differential equations
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 d(ce)/d(z) = -(-re+(ce-ceo)*EF)
6 d(F)/d(z) = -E
Explicit equations
1
cbo = 1
2
cao = 1
3
cco = 0
4
cdo = 0
5
ceo = 0
6
lam = 6-z
7
T = 350
8
k1 = exp((5000/1.987)*(1/350-1/T))
9
k2 = exp((1000/1.987)*(1/350-1/T))
10 k3 = exp((9000/1.987)*(1/350-1/T))
11 E1 = 0.47219*lam^4-1.30733*lam^3+0.31723*lam^2+0.85688*lam+0.20909
12
E2 = 3.83999*lam^6-58.16185*lam^5+366.2097*lam^4-1224.66963*lam^3+2289.84857*lam^22265.62125*lam+925.46463
13 E3 = 0.00410*lam^4-0.07593*lam^3+0.52276*lam^2-1.59457*lam+1.84445
14 ra = -k1*ca*cb-k2*ca
15 rb = -k1*ca*cb-k3*cb*cd
16 rc = k1*ca*cb
17 rd = k2*ca-k3*cb*cd
18 re = k3*cb*cd
19 E = if(lam<=1.82)then(E1)else(if(lam<=2.8)then(E2)else(E3))
20 EF = E/(1-F)
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
Sd/e decreases with increasing temperature
(2) Comparing Table E17-6.2 and E17-6.4
For Asymmetric Distribution:
(a) The conversion predicted by maximum mixedness model is slightly more than that predicted by
segregation model
(b) There is significant difference in prediction between these two models for Concentration of E
For Bimodal Distribution:
(a) The conversion predicted by maximum mixedness model is less than that estimated by segregation model
(b) There is significant difference in estimation between these two models for selectivity S C/D and S D/E and
some of the concentration values.
Generalization: For Asymmetric RTD, the results of segregation model and maximum mixedness model
matches closely to each other while there is some difference in estimation for Bimodal distribution
17-14
17-1 (f)
(i)
Conversion increases with an increase in k1. On increasing k2, conversion increases and the
increase is faster than that on increasing k1. On increasing k3, conversion decreases very slightly,
this decrease is almost negligible.
(ii) On increasing k1, concentration of C increases, and concentration of A, B, D and E decreases. On
increasing k2, concentration of A, C decreases, concentration D, E increases, and concentration of
B remains unchanged.
On increasing k3, selectivity of B decreases, A increases, C decreases, D decreases, E increases.
On increasing k1, SC/D and SD/E increase. On increasing k2, SCID decreases, and SDIE increases.
On increasing k3, SC/D increases, SD/E decreases.
(iii) As k1 is increased, rA decreases, and two minima are observed in the profile. Similar behavior is
observed for rB. rC increases on increasing k1, and two maxima are observed. rD and rE decreases.
On increasing k2, rA ,rB and rC decrease, and rD and rE increase.
On increasing k3, rB decreases, rA increases slightly (almost constant), rC decreases slightly, rD
decreases, rE increases
(iv) Desired selectivity can be achieved by varying rate constants
P17-2
Irreversible, first order, long tubular reactor, constant volume, isothermal
For a PFR
For
For laminar flow with negligible diffusion (LFR), the mean conversion is given by:
E(t) for laminar flow =
where
Therefore
We can apply the approximated solution due to Hilder:
where Da=kτ=2
17-15
P17-3
Equivalency Maximum Mixedness and Segregation model for first order reaction:
Maximum Mixedness Model:
Rearranging:
Using the integration factor:
by definition
gives:
changing the variables from λ to t in the RHS integral:
(1)
Exit concentration is when λ=0, F(0)=0 hence equation (1) becomes:
This is the same expression as for the exit concentration for the Segregation model.
P17-4 (a)
Mean Residence Time
∞
By definition
∫
E(t)dt = 1 . The area of the semicircle representing the E(t) is given by A =
0
τ=
2
2
= 0.8min .
. For constant volumetric flow tm = τ =
π
π
17-16
πτ 2
= 1 and
2
P17-4 (b)
Variance
∞
σ 2 = ∫ t −τ
2
∞
2τ
0
t 2E t dt =
∫ ()
0
σ2 =
∞
( ) E(t)dt = ∫ t2E (t ) dt − τ 2
0
∫
t2 τ 2 − t − τ
0
2
( )
0
5π 4
"
%
dt = −τ 4 $cos2 x +2cos x +1'sin2 x dx =
τ
#
&
8
∫
()
()
()
π
5π 4 2 1
τ −τ =
= 0.159
8
2π
Using Polymath to calculate Variance:
See the following Polymath program:
Polymath Code:
d(sigma)/d(t) = (t-tau)^2*E
tau = (2/3.14)^0.5
t1 = 2*tau
E2 = (t*(2*tau-t))^(1/2)
E = if(t<t1)then(E2)else(0)
t(0)=0
t(f)=1.596
sigma(0)=0
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
t
0
0
1.596
1.596
sigma
0
0
0.1593161
0.1593161
tau
0.7980869
0.7980869
0.7980869
0.7980869
t1
1.5961738
1.5961738
1.5961738
1.5961738
E2
0
0
0.7980614
0.0166534
E
0
0
0.7980614
0.0166534
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(sigma)/d(t) = (t-tau)^2*E
Explicit equations as entered by the user
[1] tau = (2/3.14)^0.5
[2] t1 = 2*tau
[3] E2 = (t*(2*tau-t))^(1/2)
[4] E = if (t<t1) then (E2) else (0)
17-17
P17-4 (c)
Conversion predicted by the Segregation model
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t) = X*E
tau = (2/3.14)^0.5
t1 = 2*tau
E2 = (t*(2*tau-t))^(1/2)
E = if(t<t1)then(E2)else(0)
k = .8
X = 1-exp(-k*t)
t(0)=0
t(f)=1.596
Xbar(0)=0
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
t
0
0
1.596
1.596
Xbar
0
0
0.4447565
0.4447565
tau
0.7980869
0.7980869
0.7980869
0.7980869
t1
1.5961738
1.5961738
1.5961738
1.5961738
E2
0
0
0.7980671
0.0166534
E
0
0
0.7980671
0.0166534
k
0.8
0.8
0.8
0.8
X
0
0
0.7210716
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] tau = (2/3.14)^0.5
[2] t1 = 2*tau
[3] E2 = (t*(2*tau-t))^(1/2)
[4] E = if (t<t1) then (E2) else (0)
[5] k = .8
[6] X = 1-exp(-k*t)
17-18
P17-4 (d)
Conversion predicted by the Maximum Mixedness model
See the following Polymath program:
Polymath Code:
d(x)/d(z) = -(-k*(1-x)+E/(1-F)*x)
d(F)/d(z) = -E
k = .8
lam = 1.596-z
tau = (2/3.14)^0.5
E1 = (tau^2-(lam-tau)^2)^0.5
E = if(lam<=2*tau)then(E1)else(0)
z(0)=0
z(f)=1.596
x(0)=0
F(0)=0.9999
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
z
0
0
1.596
1.596
x
0
0
0.4445289
0.4445289
F
1
-5.053E-04
1
-5.053E-04
k
0.8
0.8
0.8
0.8
lam
1.596
0
1.596
0
tau
0.7980869
0.7980869
0.7980869
0.7980869
E1
0.0166534
0
0.7980666
0
E
0.0166534
0
0.7980666
0
17-19
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(x)/d(z) = -(-k*(1-x)+E/(1-F)*x)
[2] d(F)/d(z) = -E
Explicit equations as entered by the user
[1] k = .8
[2] lam = 1.596-z
[3] tau = (2/3.14)^0.5
[4] E1 = (tau^2-(lam-tau)^2)^0.5
[5] E = if (lam<=2*tau) then (E1) else (0)
P17-4 (e)
X = 44.5% as for the Segregation Model, but we knew this because for first order reactions Xseg = XMM
P17-5 (a)
From P16-4:
The cumulative distribution function F(t) is given:
The real reactor can be modeled as two parallel PFRs:
"1
3
The relative E(t) = # δ t − τ 1 + δ t − τ 2
%4
4
$
(
)
(
)
Mean Residence Time
PFR
PFR
1
tm =
PFR
∫ t dF = (10min*1)+(20min*0.75) = 25min
0
17-20
or
∞
tm =
∫
tE(t)dt =
0
∫
$1
' 1
3
3
t & δ t − τ 1 + δ t − τ 2 ) = τ 1 + τ 2 = 25min
4
4
%4
( 4
(
)
(
)
P17-5 (b)
Variance
∞
'
2
2$1
2 3
2
3
1
σ = ∫ t − τ E(t)dt = ∫ t −tm & δ t − τ 1 + δ t − τ 2 )dt = τ 1 −tm + τ 2 −tm =75min2
4
4
4
%4
(
0
2
( )
(
)
(
)
(
)
(
)
(
)
P17-5 (c)
For a PFR, second order, liquid phase, irreversible reaction with k = 0.1 dm3 /mol×min-1, τ = 25 min and
CAo = 1.25 mol/dm3
kτ C Ao
X=
= 0.758
1+ kτ C Ao
For a CSTR, second order, liquid phase, irreversible reaction with k = 0.1 dm3 /mol×min-1, τ = 25 min and
CAo = 1.25 mol/dm3
X
= kτ C Ao → X = 0.572
2
1− X
(
)
P17-5 (d)
(1)-Conversion predicted by the Segregation Model
∞
X=
∫
0
()
X(t)E t dt =
∞
kC t & 1
0
Ao
3
)
∫ 1+kCAo t (' 4 δ (t − τ1 ) + 4 δ (t − τ2 )+*dt =
1 kC Ao τ1
3 kC Ao τ2
+
= 0.731
4 1+kC Ao τ1 4 1+kC Ao τ2
(2)-Conversion predicted by the Maximum Mixedness model
()
()
E λ
dX rA
=
+
X
dλ C Ao 1−F λ
rA = −kC2A = −kC2Ao1− X
(
2
)
()
()
E λ
2
dX
= −kC Ao 1− X +
X
dλ
1−F λ
(
)
17-21
We need to change the variable such the integration proceeds forward:
(
)
E T −z
2
dX
= kC Ao 1− X −
X
dz
1−F T − z
(
)
(
)
See the following Polymath program:
Polymath Code:
d(x)/d(z) = -(ra/cao+E/(1-F)*x)
d(F)/d(z) = -E
cao = 1.25
k = .1
lam = 40-z
ca = cao*(1-x)
t1 = 10
t2 = 30
E3 = 0
ra = -k*ca^2
E2 = 0.75/(t2*2*(1-0.99))
E1 = 0.25/(t1*2*(1-0.99))
E = if((lam>=0.99*t1)and(lam<1.01*t1))then(E1)else(if((lam>=0.99*t2)and(lam<1.01*t2))then(E2)else(E3))
EF = E/(1-F)
z(0)=0
z(f)=40
x(0)=0
F(0)=0.9999
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
z
0
0
40
40
x
0
0
0.7125177
0.7061611
F
0.9999
-1.081E-04
0.9999
-1.081E-04
cao
1.25
1.25
1.25
1.25
k
0.1
0.1
0.1
0.1
lam
40
0
40
0
ca
1.25
0.3614311
1.25
0.3672986
t1
10
10
10
10
t2
30
30
30
30
E3
0
0
0
0
ra
-0.15625
-0.15625
-0.0130632
-0.0134908
E2
1.25
1.25
1.25
1.25
E1
1.25
1.25
1.25
1.25
E
0
0
1.25
0
EF
0
0
2.8604931
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
Explicit equations as entered by the user
[1] cao = 1.25
[2] k = .1
[3] lam = 40-z
[4] ca = cao*(1-x)
[5] t1 = 10
[6] t2 = 30
[7] E3 = 0
[8] ra = -k*ca^2
[9] E2 = 0.75/(t2*2*(1-0.99))
[10] E1 = 0.25/(t1*2*(1-0.99))
[11] E = if ((lam>=0.99*t1)and(lam<1.01*t1)) then (E1) else( if ((lam>=0.99*t2)and(lam<1.01*t2)) then (E2) else (E3))
[12] EF = E/(1-F)
17-22
(3) For T-I-S model,
𝜏 3 25 ∗ 25
𝑛= 3=
= 8.33
𝜎
75
𝜏< = 𝜏⁄𝑛 = 25>8.33 = 3.0
Follow the additional material on Ch-17 website to calculate CA1 to CA9.
%$@√$@B∗C∗,.$∗$.3D
CA1=
3∗C∗,.$
=0.968
Similarly calculate CA2 to CA9
CA2=0.784, CA3=0.655, CA4=0.56, CA5= 0.49, CA6= 0.432, CA7= 0.387, CA8= 0.35, CA9 =0.32
For n= 8
𝐶+, − 𝐶+F 1.25 − 0.35
𝑋=
=
= 72 %
𝐶+,
1.25
For n=9
𝐶+, − 𝐶+J 1.25 − 0.32
𝑋=
=
= 74.4 %
𝐶+,
1.25
Since n is little higher than 8, so X would be between than 72 % and 74 %
P17-5 (e)
For 2nd order reaction
P17-5 (f)
For T-I-S model,
𝜏 3 25 ∗ 25
𝑛= 3=
= 8.33
𝜎
75
𝜏< = 𝜏⁄𝑛 = 25>8.33 = 3.0
𝑋 =1−
1
(1 + 3.0 ∗ 0.4)F.CC
X=0.999
P17-6 (a)
From P16-5 (a) – (c):
P
M 𝐸(𝑡) = 1
,
For reactor (a), (0.5)(0.2)(t1-0) = 1, so t1 = 10 min.
For reactor (b), (0.5)(0.2)(t1-5) = 1, so t1 = 15 min.
P17-6 (b)
Mean residence time:
#0.04t ,if t < 5
%
(a) E(t) = $−0.04t + 0.4,if 5 ≤ t ≤ 10
%0,otherwise
&
∞
tm =
∫ ()
tE t dt =
0
5
∫
0
10
0.04t 2 dt + (−0.04t 2 + 0.4t)dt =5
∫
5
17-23
#0,if t < 5
%%
E(t)
=
$−0.02t + 0.3,if 5 ≤ t ≤ 15
(b)
%
%&0,if t > 15
∞
15
0
5
tm =
25
∫ tE (t )dt = ∫ (−0.02t2 + 0.3t)dt = 3
Variance:
∞
∞
2
2
σ 2 = ∫ t −tm E(t)dt = ∫ t 2E t dt −tm
0
(a)
(
)
0
∞
()
7
∫ t2E (t )dt = 6 tm2
0
2
7 2 2 tm 25
So σ = tm −tm =
= = 4.167min2
6
6
6
2
∞
(b)
15
2
∫ t E (t )dt = ∫ (−0.02t 3 + 0.3t2 )dt =75
0
5
2
So σ 2 = 75−tm
= 75−(
25 25 50
)( ) =
= 5.56min2
3 3
9
P17-6 (c)
Fraction of fluid spends longer than 7 minutes:
(a)
∞
∫ ()
E t dt =
7
10
∫ (−0.04t + 0.4)dt =0.18
7
P
$D
(b) ∫S 𝐸 (𝑡)𝑑𝑡 = ∫S (−0.02𝑡 + 0.3)𝑑𝑡 = 0.64
P17-6 (d) Segregation Model
(a) Polymath code:
d(Xseg) / d(t) = X*E
Xseg(0) = 0
t(0) = 0
t(f) = 20
k1 = 0.2
X = k1*t/(1+k1*t)
tau = 5
t1 = 2*tau
E1 = 0.04*t
E2 = -0.04*t+0.4
E = if(t<tau)then(E1)else(if(t<t1)then(E2)else(0))
(b) Polymath code:
d(Xseg) / d(t) = X*E
Xseg(0) = 0
t(0) = 0
t(f) = 20
k1 = 0.2
X = k1*t/(1+k1*t)
E1 = -0.02*t + 0.3
E = if(t<5)then(0)else(if(t<=15)then(E1)else(0))
17-24
P17-6 (e) Maximum Mixedness Model
(a) Polymath code:
d(Xmm) / d(z) = -(-k*(1-Xmm)^2 + E/(1-F)*Xmm)
Xmm(0) = 0
d(F) / d(z) = -E
F(0) = 0.999
z(0) = 0
z(f) = 20
k = 0.2
tau = 5
t1 = 2*tau
lam = 20 - z
E1 = 0.04*lam
E2 = -0.04*lam+0.4
E = if(lam<tau)then(E1)else(if(lam<=t1)then(E2)else(0))
(b) Polymath code:
d(Xmm) / d(z) = -(-k*(1-Xmm)^2 + E/(1-F)*Xmm)
Xmm(0) = 0
d(F) / d(z) = -E
F(0) = 0.999
z(0) = 0
z(f) = 20
k = 0.2
lam = 20 - z
E1 = -0.02*lam + 0.3
E = if(lam<5)then(0)else(if(lam<=15)then(E1)else(0))
P17-6 (f) For T-I-S model, for Reactor A
𝜏3
5∗5
𝑛= 3=
=6
𝜎
4.167
𝜏< = 𝜏⁄𝑛 = 5>6 = 0.834
For CA0=1, k=0.2
%$@√$@B∗,.FCB∗,.3∗$
CA1=
3∗,.FCB∗,.3
=0.87
CA2=0.77, CA3=0.69, CA4= 0.63, CA5= 0.57, CA6= 0.53
X= (1-0.53)/1 = 47 %
Thus, Conversion is 47 %
For Reactor B
3D
𝜏 3 (25/3) ∗ ( C )
𝑛= 3=
= 12.5
𝜎
5.56
𝜏< = 𝜏⁄𝑛 = 25>(3 ∗ 12.5) = 0.67
For CA0=1, k=0.2
%$@√$@B∗,.VS∗,.3∗$
CA1=
3∗,.VS∗,.3
=0.89
17-25
CA2=0.81, CA3=0.73, CA4= 0.67, CA5= 0.62, CA6= 0.58, CA7=0.54, CA8=0.50, CA9=0.47, CA10=0.45,
CA11=0.42, CA12=0.4, CA13=0.38
X12= (1-0.4)/1 = 60 %
X13= (1-0.38)/1= 58 %
Thus, Conversion at n=12.5 is 59 %
P17-6 (g) See part (d), (e) and (f)
P17-6 (h) For reactor A,
For 1st order reaction, conversion in T-I-S model is
𝑋 = 1−
1
(1 + 0.834 ∗ 0.2)V
Solving above equation gives X= 60 %
For Segregation model
d(Xseg) / d(t) = X*E
Xseg(0) = 0
t(0) = 0
t(f) = 20
k1 = 0.2
X = 1-exp(-k1*t)
tau = 5
t1 = 2*tau
E1 = 0.04*t
E2 = -0.04*t+0.4
E = if(t<tau)then(E1)else(if(t<t1)then(E2)else(0))
The above code gives Xseg=60 %
For maximum mixedness model
d(Xmm) / d(z) = -(-k*(1-Xmm) + E/(1-F)*Xmm)
Xmm(0) = 0
d(F) / d(z) = -E
F(0) = 0.999
z(0) = 0
z(f) = 20
k = 0.2
tau = 5
t1 = 2*tau
lam = 20 - z
E1 = 0.04*lam
E2 = -0.04*lam+0.4
E = if(lam<tau)then(E1)else(if(lam<=t1)then(E2)else(0))
The above code gives Xmm= 60 %
Thus for 1st order reaction, Xseg=Xmm= XTIS = 60 %
P17-7
X=1-1/(1+τk/n)n
(a)k=0.4 min -1
for reactor A:
τ=5 min, σ2=29.167-25=4.167 min2
n=τ2/σ2=25/4.167=6
X=1-1/(1+τk/n)n=1-1/(1+5*0.4/6)6=0.822
17-26
for reactor B:
τ=8.33min, σ2=75-8.332=5.611 min2
n=τ2/σ2=8.332/4.167=12.4~12
X=1-1/(1+τk/n)n=1-1/(1+8.33*0.4/12)12=0.947
(b)k=0.04 min -1
for reactor A:
τ=5 min, σ2=29.167-25=4.167 min2
n=τ2/σ2=25/4.167=6
X=1-1/(1+τk/n)n=1-1/(1+5*0.04/6)6=0.179
for reactor B:
τ=8.33min, σ2=75-8.332=5.611 min2
n=τ2/σ2=8.332/4.167=12.4~12
X=1-1/(1+τk/n)n=1-1/(1+8.33*0.04/12)12=0.28
(c)k=4 min -1
for reactor A:
τ=5 min, σ2=4.167 min2
n=τ2/σ2=25/4.167=6
X=1-1/(1+τk/n)n=1-1/(1+5*4/6)6=0.99984
for reactor B:
τ=8.33min, σ2=5.611 min2
n=τ2/σ2=8.332/4.167=12.4~12
X=1-1/(1+τk/n)n=1-1/(1+8.33*4/12)12=0.99999
P17-8
For a first order reaction, Xmm=XTIS=Xseg
Xseg is given by
From the conversion it is possible to determine k at 300K:
17-27
The conversion at T=310 K is given by:
E
E
=Xseg
P17-9 (a)
The E(t) is a square pulse
Third order liquid-phase reaction: -rA = kCA3 with CAo = 2mol/dm3 and k = 0.3 dm6/mol2/min (Isothermal
Operation)
(1)-Conversions
For CSTR,
For PFR,
17-28
For LFR,
Polymath code:
d(Xbar)/d(t) = X*E
Xbar(0) = 0
t(0) = 0
t(f) = 2e4
k = 0.3
Ca0 = 2
X = 1-(1/(1+2*k*Ca0^2*t))^0.5
tau = 1.5
t1 = tau/2
E2 = tau^2/2/(t^3+.000001)
E = if(t<t1)then(0)else(E2)
𝑋W = 0.494
For T-I-S Model
First, we need to determine sigma2. See the following Polymath code
d(Sigma2) / d(t) = (t-tmf)^2*E
Sigma2(0) = 0
tmf=1.5
E2 = 1
t1=1
E = if(t>=t1)then(E2)else(0)
t(0) = 0
t(f) = 2
From above code result, we get sigma2= 0.0833
𝜏3
1.53
=
= 27
𝜎 3 0.0833
𝜏< = 𝜏⁄𝑛 = 1.5>(27) = 0.055
𝑛=
Now, follow procedure given in P17-6 (f) to calculate exit conversion
Segregation Model
0.53
17-29
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t) = X*E
k = .3
Cao = 2
t1 = 1
E2 = 1
E = if(t>=t1)then(E2)else(0)
X = 1-1/(1+2*k*Cao^2*t)^(1/2)
Xbar(0)=0
t(0)=0
t(f)=2
Output:
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
t
0
0
2
2
Xbar
0
0
0.5296583
0.5296583
k
0.3
0.3
0.3
0.3
Cao
2
2
2
2
t1
1
1
1
1
E2
1
1
1
1
E
0
0
1
1
X
0
0
0.5847726
0.5847726
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 = .3
[2] Cao = 2
[3] t1 = 1
[4] E2 = 1
[5] E = if (t>=t1) then (E2) else (0)
[6] X = 1-1/(1+2*k*Cao^2*t)^(1/2)
(2)-Maximum Mixedness Model
See the following Polymath program
POLYMATH Results
Calculated values of the DEQ variables
Variableinitial valueminimal valuemaximal valuefinal value
z
0
0
2
2
x
0
0
0.5215389
0.5215389
F
0.9999
-9.999E-05
0.9999
-9.999E-05
cao
2
2
2
2
k
0.3
0.3
0.3
0.3
lam
2
0
2
0
ca
2
0.9569223
2
0.9569223
E1
1
1
1
1
17-30
ra
t2
t1
E
EF
-2.4
2
1
1
1.0E+04
-2.4
2
1
0
0
-0.2628762
2
1
1
1.0E+04
-0.2628762
2
1
0
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
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(0)
[10] EF = E/(1-F)
The conversion shows an inflection point in correspondence of z = 1, where start the pulse
P17-9(b)
(1) Segregation Model
d(ca)/d(t) =ra
d(cb)/d(t) =rb
d(cc)/d(t) =rc
d(cabar)/d(t) =ca*E
d(cbbar)/d(t) =cb*E
d(ccbar)/d(t) =cc*E
cabar(0) = 0
cbbar(0) = 0
ccbar(0) = 0
ca(0) = 2
cb(0) = 0
cc(0) = 0
t(0) = 0
t(f) = 2e4
k1= 0.3
k2=1
ra=-k1*ca
rb=k1*ca-k2*cb
rc=k2*cb
tau = 1.5
t1 = tau/2
E2 = tau^2/2/(t^3+.000001)
E = if(t<t1)then(0)else(E2)
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 ca
2.
-3.269E-153
2.
0
2 cabar
0
0
1.351847
1.351847
3 cb
0
-3.646E-10
6.545E-10
1.117E-11
4 cbbar
0
0
0.314049
0.314049
17-31
.
5 cc
0
0
2.
2.
6 ccbar
0
0
0.3341022
0.3341022
7 E
0
0
1.756E-08
1.406E-13
8 E2
1.125E+06
1.406E-13
1.125E+06
1.406E-13
9 k1
0.3
0.3
0.3
0.3
10 k2
1.
1.
1.
1.
11 ra
-0.6
-0.6
9.808E-154
0
12 rb
0.6
-6.545E-10
0.6
-1.117E-11
13 rc
0
-3.646E-10
6.545E-10
1.117E-11
14 t
0
0
2.0E+04
2.0E+04
15 t1
0.75
0.75
0.75
0.75
16 tau
1.5
1.5
1.5
1.5
The exit concentration using Segregation model is
Cabar= 1.35
Cbbar= 0.31
Ccbar= 0.33
(2) Maximum Mixedness Model
d(ca)/d(z) = -(-ra+(ca-cao)*EF) #
d(cb)/d(z) = -(-rb+(cb-cbo)*EF) #
d(cc)/d(z) = -(-rc+(cc-cco)*EF) #
d(F)/d(z) = -E #
cbo = 0 #
cao = 2 #
cco = 0 #
lam = 2e4-z
k1= 0.3
k2=1
ra=-k1*ca
rb=k1*ca-k2*cb
rc=k2*cb
EF = E/(1-F) #
tau = 1.5
t1 = tau/2
E2 = tau^2/2/(lam^3+.000001)
E = if(lam<t1)then(0)else(E2)
z(0)=0
ca(0)=2
cb(0)=0
cc(0)=0
F(0)=0.99
z(f)=2e4
Polymath Output
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 ca
2.
9.957E-11
2.
1.338463
2 cao
2.
2.
2.
2.
3 cb
0
0
0.3109399
0.3109399
4 cbo
0
0
0
0
5 cc
0
0
2.
0.3505966
6 cco
0
0
0
0
17-32
7 cdo
0
0
0
0
8 ceo
0
0
0
0
9 E
1.406E-13
0
1.419E-07
0
10 E2
1.406E-13
1.406E-13
1.125E+06
1.125E+06
11 EF
1.406E-11
0
1.417E-05
0
12 F
0.99
-0.0099986
0.99
-0.0099986
13 k1
0.3
0.3
0.3
0.3
14 k2
1.
1.
1.
1.
15 lam
2.0E+04
0
2.0E+04
0
16 ra
-0.6
-0.6
-2.987E-11
-0.401539
17 rb
0.6
-2.11E-14
0.6
0.0905991
18 rc
0
0
0.3109399
0.3109399
19 t1
0.75
0.75
0.75
0.75
20 tau
1.5
1.5
1.5
1.5
21 z
0
0
2.0E+04
2.0E+04
The exit concentration using maximum mixedness model is
Cabar= 1.33
Cbbar= 0.31
Ccbar= 0.35
P17-10 (a)
3rd order, k=175 dm6/(mol2 min), CBo=0.0313 dm3/min
(LFR, PFR, CSTR with τ=100s)
PFR
Design equation
Using the quadratic solution
The conversion for a PFR X=83.2%
17-33
CSTR
Design equation
The conversion for a CSTR X=66.2 %
LFR (completely segregated)
E(t)=0
for t<τ/2
=τ2 /2t3
τ=V/v=1000/10=100s
for t>= τ/2
Where
See the following Polymath program:
Polymath Code:
d(xbar)/d(t) = x*E
cbo = 0.0313
k = 175
tau = 100
E1 = tau^2/(2*t^3)
x = 1-(1/(1+2*k*cbo^2*t))^0.5
E = if(t>=tau/2)then(E1)else(0)
t(0)=1e-5
t(f)=100
xbar(0)=0
Output:
Variable Initial value Minimal value Maximal value Final value
1 cbo
0.0313
0.0313
0.0313
0.0313
2 E
0
0
0.0363537
0.005
3 E1
5.0E+18
0.005
5.0E+18
0.005
4 k
175.
175.
175.
175.
5 t
1.0E-05
1.0E-05
100.
100.
6 tau
100.
100.
100.
100.
7 x
1.714E-06
1.714E-06
0.8316631
0.8316631
8 xbar
0
0
0.5944894
0.5944894
17-34
Differential equations
1 d(xbar)/d(t) = x*E
Explicit equations
1 cbo = 0.0313
2 k = 175
3 tau = 100
4 E1 = tau^2/(2*t^3)
5 x = 1-(1/(1+2*k*cbo^2*t))^0.5
6 E = if(t>=tau/2)then(E1)else(0)
The integral
gives mean conversion=59.5%
P17-10 (b)
(Segregation Model , Maximum Mixedness Model and T-I-S)
Batch reactor
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t) = E*X
cbo = 0.0313
k = 175
X = 1-(1/(1+2*k*cbo^2*t))^0.5
E = if(t<=70) then(E1) else (E2)
E1 = 4.44658e-10*t^4-1.1802e-7*t^3+1.35358e-5*t^2 -0.000865652*t+0.028004
E2 = -2.64e-9*t^3+1.3618e-6*t^2-0.00024069*t+0.015011
t(0)=0
t(f)=100
Xbar(0)=0
17-35
Output:
Variable Initial value Minimal value Maximal value Final value
1 cbo
0.0313
0.0313
0.0313
0.0313
2 E
0.028004
0.00192
0.028004
0.00192
3 E1
0.028004
0.0028731
0.028004
0.0032426
4 E2
0.015011
0.00192
0.015011
0.00192
5 k
175.
175.
175.
175.
6 t
0
0
100.
100.
7 X
0
0
0.8316631
0.8316631
8 Xbar
0
0
0.554339
0.554339
Differential equations
1 d(Xbar)/d(t) = E*X
Explicit equations
1 cbo = 0.0313
2 k = 175
3 X = 1-(1/(1+2*k*cbo^2*t))^0.5
4 E2 = -2.64e-9*t^3+1.3618e-6*t^2-0.00024069*t+0.015011
5 E1 = 4.44658e-10*t^4-1.1802e-7*t^3+1.35358e-5*t^2 -0.000865652*t+0.028004
6 E = if(t<=70) then(E1) else (E2)
Conversion predicted by segregation model is 55.44 %
Maximum Mixedness
Rate Law :
where k=175 dm6/mol2 min
See the following Polymath program:
Polymath Code:
d(X)/d(z) = -(ra/Cao+E/(1-F)*X) #
lam = 100-z #
Cbo = .0313
k = 175
Cao = .0313
Ca = Cao*(1-X)
Cb = Cbo*(1-X)
ra = -k*Ca*Cb^2
E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004 #
E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011 #
F1 = (4.44658e-10)/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-.000865652/2*lam^2+.028004*lam
F2 = -(-9.30769e-8*lam^3+5.02846e-5*lam^2-.00941*lam+.618231-1) #
E = if (lam<=70) then (E1) else (E2) #
F = if (lam<=70) then (F1) else (F2) #
17-36
EF = E/(1-F) #
z(0)=0
X(0)=0
z(f)=100
Output:
Variable Initial value Minimal value Maximal value Final value
1 Ca
0.0313
0.0133884
0.0313
0.0139412
2 Cao
0.0313
0.0313
0.0313
0.0313
3 Cb
0.0313
0.0133884
0.0313
0.0139412
4 Cbo
0.0313
0.0313
0.0313
0.0313
5 E
0.00192
0.00192
0.028004
0.028004
6 E1
0.0032426
0.0028731
0.028004
0.028004
7 E2
0.00192
0.00192
0.015011
0.015011
8 EF
0.0220689
0.0220689
0.028004
0.028004
9 F
0.9129999
0
0.9129999
0
10 F1
0.9228893
0
0.9228893
0
11 F2
0.9129999
0.381769
0.9129999
0.381769
12 k
175.
175.
175.
175.
13 lam
100.
0
100.
0
14 ra
-0.0053663
-0.0053663
-0.00042
-0.0004742
15 X
0
0
0.5722557
0.5545934
16 z
0
0
100.
100.
Differential equations
1 d(X)/d(z) = -(ra/Cao+E/(1-F)*X)
Explicit equations
1 lam = 100-z
2 Cbo = .0313
3 k = 175
4 Cao = .0313
5 Ca = Cao*(1-X)
6 Cb = Cbo*(1-X)
7 ra = -k*Ca*Cb^2
8 E1 = 4.44658e-10*lam^4-1.1802e-7*lam^3+1.35358e-5*lam^2-.000865652*lam+.028004
9 E2 = -2.64e-9*lam^3+1.3618e-6*lam^2-.00024069*lam+.015011
10 F1 = (4.44658e-10)/5*lam^5-1.1802e-7/4*lam^4+1.35358e-5/3*lam^3-.000865652/2*lam^2+.028004*lam
11 F2 = -(-9.30769e-8*lam^3+5.02846e-5*lam^2-.00941*lam+.618231-1)
12 E = if (lam<=70) then (E1) else (E2)
13 F = if (lam<=70) then (F1) else (F2)
14 EF = E/(1-F)
Xmaximum mixedness= 55.46 %
For T-I-S model, follow the procedure as given in earlier examples
17-37
P17-11 (a)
∞
m1 =
∫(
) ()
t − τ E t dt =
0
∞
∞
∫ ()
tE t dt − τ
0
∫ E (t ) dt = τ − τ = 0
0
m1CSTR = m1PFR = m1LFR = 0
P17-11 (b)
Second-order liquid-phase reaction Da= τkCAo=1.0,τ=2min and kCAo=0.5min-1.
CSTR
FAo − FA = −rAV
FAo − FA = FAo X
V=
FAo X
(−rA )exit
Liquid-phase τ =
V C Ao −C A
=
vo
−r
( A)
C −C
Second-order −rA = kC 2A and τ = Ao A =
kC 2A
kC
Solved
X=
F Ao
2
Ao
(1− X )
(1+2Da) − 1+ 4Da = 0.382
2Da
PFR
−dFA
dV
X
= −rA
dX
= - rA
dV
X
V = FAo
dX
∫ −r
0
A
Second-order
X
V = FAo
∫
dX
2
0 kC A
where C A = C Ao
(1− X )
(1+ ε X)
Liquid-phase ε = 0 and integrating
Da
1 " X %
= 0.5
τ=
$
' or X =
1+ Da
kC Ao # 1− X &
LFR
In the ring globule of radius r
dC A
= rA Where −rA = kC 2A
dt
ktC Ao
(2order batch)
X=
1+ ktC Ao
17-38
#
%0 for t < 1min
(E(t) LFR )
E(t) = $
3
−1
%
&4 / (2t )min for t ≥ 1min
'
)
kC
τ
dt
τ
1
X = XE(t)dt = kC Ao
= kC Ao & +
− Ao )dt =
2
2
2
2
t )
1+ ktC Ao
τ
τ /2 1+ ktC Ao t
τ /2& t
%
(
∞
∫
2
2
∞
∫
2
(
)
$
(
∞ &
∫
kC Ao
(
2
)
)
∞
* 1+ ktC -'
$ Da * 1+ Da / 2 -'
τ 2kC Ao $ 1
Ao /)
&− + kC ln,
&1− ln,
=
=
Da
/)
Ao ,
/)
2 &% t
t
2 + Da / 2 .)(
&%
+
.(τ /2
Evaluate for Da=1, X = 0.451
CSTR
0.382
PFR
0.5
LFR
0.451
P17-11(c) Choose a reaction order and then using E(t) equation for an ideal reactor, calculate the
conversion
P17-12
The criteria
Xseg>XMM
Xseg=XMM
Xseg<XMM
The following figure shows the reaction rate as function of the concentration.
The second derivative is initially negative (Xseg<XMM), then positive (Xseg>XMM). The flex point is for
CA=8mol/dm3 (Xseg=XMM).
17-39
In the limit of low concentration
(First order) and Xseg=XMM
in the limit of high concentration
(Reaction order=-1) and Xseg>XMM
P17-13
(a)
(b)
(c)
(d)
(e) (a)
(e) (b)
(e) (c)
(e) (d)
(f)(a)
(f)(b)
(f)(c)
(f)(d)
(g)(a)
(g)(b)
(g)(c)
(g)(d)
PFR
0.399
0.399
0.338
0.296
0.0769
0.0769
0.0741
0.0715
0.918
0.918
0.714
0.59
0.393
0.393
0.333
0.293
CSTR
0.338
0.338
0.28
0.244
0.0741
0.0741
0.0696
0.0659
0.6959
0.6959
0.56
0.48
0.333
0.333
0.277
0.242
LFR
0.355
0.355
0.302
0.266
0.0734
0.0734
0.06955
0.0665
0.83
0.83
0.65
0.54
0.35
0.35
0.298
0.263
P17-13(a)
dN A
= rAV
dt
N A = N A0 (1 - X A )
dX A
= -rAV = kC A0 (1 - X A )V
dt
\ X A = 1 - e -kt
\ N A0
¥
X = ò X A (t ) E (t )dt
0
(a)PFR:
E(t)=δ(t-τ)
τ=tm=5.1 min
X = 1-e-kτ=1=1-e-0.1*5.1=0.399
17-40
TIS
0.381
0.381
0.32
0.281
0.0762
0.0762
0.073
0.0699
0.89
0.89
0.696
0.58
0.381
0.381
0.322
0.282
Segregation
Model
0.382
0.382
0.321
0.281
0.0765
0.0765
0.0732
0.0702
0.87
0.87
0.6875
0.572
0.380
0.380
0.320
0.280
Maximum
Mixedness
0.382
0.382
0.316
0.274
0.0761
0.0761
0.0728
0.0697
0.87
0.87
0.66
0.549
0.380
0.380
0.317
0.276
(b)CSTR:
E(t)=
1
t
-
t
e t
0.3377
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=5.1
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.1960784
5.982E-10
0.1960784
5.982E-10
2 k
0.1
0.1
0.1
0.1
3 t
0
0
100.
100.
4 tau
5.1
5.1
5.1
5.1
5 X
0
0
0.9999546
0.9999546
6 Xbar
0
0
0.3377483
0.3377483
Differential equations
1 d(Xbar)/d(t) = X*E
Explicit equations
1 k = 0.1
2 tau = 5.1
3 X = 1-exp(-k*t)
4 E = (1/tau)*exp(-t/tau)
(c)LFR:
E(t)= 0 if t<τ/2
=τ2/2t3 if t>=τ/2
0.355
See the following Polymath program:
Polymath Code:
k = 0.1
t0 = 0.000001
tau = 5.1
E1 = tau*tau/(2*t^3)
X = 1-exp(-k*t)
E = if(t< tau/2) then (0) else (E1)
d(Xbar)/d(t) = X*E
Xbar(0)=0
t(0) = 1E-07
t(f) = 100
17-41
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.3973402
1.301E-05
2 E1
1.301E+22
1.301E-05
1.301E+22
1.301E-05
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
5.1
5.1
5.1
5.1
7 X
1.0E-08
1.0E-08
0.9999546
0.9999546
8 Xbar
0
0
0.3551329
0.3551329
Differential equations
1 d(Xbar)/d(t) = X*E
Explicit equations
1 k = 0.1
2 t0 = 0.000001
3 tau = 5.1
4 E1 = tau*tau/(2*t^3)
5 X = 1-exp(-k*t)
6 E = if(t< tau/2) then (0) else (E1)
(d)Tanks in Series Model
σ=2.49 min
Y'
D.$'
n=Z' =3.BJ' =4.195 ~ 4
0.381
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)
taui = tau/4
E = t^(3)*exp(-t/taui)/(6*taui^4)
k=0.1
tau=5.1
Xbar(0)=0
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
6.307E-23
5.464E-30
0.17183
5.464E-30
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
17-42
4 tau
5.1
5.1
5.1
5.1
5 taui
1.275
1.275
1.275
1.275
6 X
0
0
0.9999546
0.9999546
7 Xbar
0
0
0.3812235
0.3812235
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^1
Explicit equations
1 tau = 5.1
2 taui = tau/4
3 k = 0.1
4 E = t^(3)*exp(-t/taui)/(6*taui^4)
(e)Segregation Model
0.382
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)
C1=0.0038746+0.2739782*t+1.574621*t^2-0.2550041*t^3
C2=-33.43818+37.18972*t-11.58838*t^2+1.695303*t^3-0.1298667*t^4+0.005028*t^5-(7.743*10^(-5))*t^6
C=if(t<=4andt>=0)then(C1)elseif(t>4andt<=14)then(C2)else0
E=C/51
t(0)=0
t(f)=100
Xbar(0)=0
X(0)=0
k=0.1
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 C
0.0038746
0
9.937666
0
2 C1
0.0038746
-2.392E+05
9.994775
-2.392E+05
3 C2
-33.43818
-3.855E+07
10.48137
-3.855E+07
4 E
7.597E-05
0
0.1948562
0
5 k
0.1
0.1
0.1
0.1
6 t
0
0
100.
100.
7 X
0
0
0.9999546
0.9999546
8 Xbar
0
0
0.3822647
0.3822647
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)
17-43
Explicit equations
1 C1 = 0.0038746+0.2739782*t+1.574621*t^2-0.2550041*t^3
2
C2 = -33.43818+37.18972*t-11.58838*t^2+1.695303*t^3-0.1298667*t^4+0.005028*t^5-(7.743*10^(5))*t^6
3 C = if(t<=4 and t>=0) then (C1) else if(t>4 and t<=14) then (C2) else 0
4 E = C/51
5 k = 0.1
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)
Explicit equations
1 C1 = 0.0039+0.274*t+1.57*t^2-0.255*t^3
2 C2 = -33.4+37.2*t-11.6*t^2+1.7*t^3-0.13*t^4+0.005*t^5-(7.7*10^(-5))*t^6
3 C = if(t<4 and t>0) then (C1) else if(t>=4 and t<14) then (C2) else 0
4 E = C/51
5 k = 0.1
(f)Maximum Mixedness Model:
r
dX
E (l )
= A +
X
dl C A0 1 - F (l )
0.3818
See the following Polymath program:
Polymath Code:
d(X)/d(z)=-((rA/1)+(E*X/(1-F)))
d(F)/d(z)=-E
rA=-k*(1-X)
C1=0.0038746+0.2739782*lam+1.574621*lam^2-0.2550041*lam^3
C2=-33.43818+37.18972*lam-11.58838*lam^2+1.695303*lam^3-0.1298667*lam^4+0.005028*lam^5-(7.743*10^(5))*lam^6
C=if(lam<=4andlam>=0)then(C1)elseif(lam>4andlam<=14)then(C2)else0
E=C/51
lam=14-z
k=0.1
z(0)=0
z(f)=14
X(0)=0
F(0)=0.9999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
C
0.0148043
0.0038746
9.958392
0.0038746
2
C1
-387.266
-387.266
10.03479
0.0038746
3
C2
0.0148043
-33.43818
10.48832
-33.43818
4
E
0.0002903
7.597E-05
0.1952626
7.597E-05
5
F
0.9999
-0.0013411
0.9999
-0.0013411
6
k
0.1
0.1
0.1
0.1
7
lam
14.
0
14.
0
17-44
8
rA
-0.1
-0.1
-0.0618154
-0.0618154
9
X
0
0
0.3818459
0.3818459
10 z
0
0
14.
14.
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 lam = 14-z
3 rA = -k*(1-X)^1
4 C1 = 0.0038746+0.2739782*lam+1.574621*lam^2-0.2550041*lam^3
5
C2 = -33.43818+37.18972*lam-11.58838*lam^2+1.695303*lam^3-0.1298667*lam^4+0.005028*lam^5(7.743*10^(-5))*lam^6
6 C = if(lam<=4 and lam>=0) then (C1) else if(lam>4 and lam<=14) then (C2) else 0
7 E = C/51
P17-13(b)
This is also a first order reaction in A, hence conversions for PFR, CSTR, and LFR are the same as in (a)
(a) PFR:
E(t)=δ(t-τ)
τ=tm=5.1 min
1-e-kτ=1=1-e-0.1*5.1=0.399
(b)CSTR:
E(t)=
1
t
e
-
t
t
0.34
(c)LFR:
E(t)= 0 if t<τ/2
=τ2/2t3 if t>=τ/2
0.36
(d)Tanks in Series Model
σ=2.49 min
Y'
D.$'
n=Z' =3.BJ' =4.195 ~ 4
0.381
(e)Segregation Model
0.382
(f)Maximum Mixedness Model:
r
dX
E (l )
= A +
X
dl C A0 1 - F (l )
0.382
17-45
P17-13 (c)
(a) PFR:
dN A
= rAV
dt
N A = N A0 (1 - X A )
dX A
= - rAV = k (C A0 (1 - X A ))2V
dt
kt
\XA =
1 + kt
E (t ) = d (t - t )
\ N A0
¥
X = ò X A (t ) E (t )dt
0
0.338
\
(b) CSTR
0.28
\
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=5.1
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.1960784
5.982E-10
0.1960784
5.982E-10
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
5 tau
5.1
5.1
5.1
5.1
6 X
1.0E-08
1.0E-08
0.9090909
0.9090909
7 Xbar
0
0
0.2806789
0.2806789
Differential equations
1 d(Xbar)/d(t) = X*E
Explicit equations
1 k = 0.1
2 t0 = 0.000001
3 tau = 5.1
4 X = k*t/(1+k*t)
5 E = (1/tau)*exp(-t/tau)
17-46
(c) LFR
0.302
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
t0=0.000001
E1=tau*tau/(2*t^3)
E=if(t<tau/2)then(0)else(E1)
Xbar(0)=0
k=0.1
tau=5.1
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.3899176
1.301E-05
2 E1
1.301E+22
1.301E-05
1.301E+22
1.301E-05
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
5.1
5.1
5.1
5.1
7 X
0
0
0.9090909
0.9090909
8 Xbar
0
0
0.3021389
0.3021389
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 t0 = 0.000001
2 tau = 5.1
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if(t< (E1) else (0) then 2)>
(d) Tanks in series Model
σ=2.49 min
Y'
D.$'
n=Z' =3.BJ' =4.195 ~ 4
0.32
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
taui = tau/4
E = t^(3)*exp(-t/taui)/(6*taui^4)
k=0.1
17-47
tau=5.1
Xbar(0)=0
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
6.307E-23
5.464E-30
0.1745929
5.464E-30
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
5.1
5.1
5.1
5.1
5 taui
1.275
1.275
1.275
1.275
6 X
0
0
0.9090909
0.9090909
7 Xbar
0
0
0.3203103
0.3203103
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 tau = 5.1
2 taui = tau/4
3 k = 0.1
4 E = t^(3)*exp(-t/taui)/(6*taui^4)
(e) Segregation Model
0.321
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
C1=0.0038746+0.2739782*t+1.574621*t^2-0.2550041*t^3
C2=-33.43818+37.18972*t-11.58838*t^2+1.695303*t^3-0.1298667*t^4+0.005028*t^5-(7.743*10^(-5))*t^6
C=if(t<=4andt>=0)then(C1)elseif(t>4andt<=14)then(C2)else0
E=C/51
t(0)=0
t(f)=100
Xbar(0)=0
X(0)=0
k=0.1
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 C
0.0038746
0
9.967042
0
2 C1
0.0038746
-2.392E+05
9.983794
-2.392E+05
3 C2
-33.43818
-3.855E+07
10.48861
-3.855E+07
17-48
4 E
7.597E-05
0
0.1954322
0
5 k
0.1
0.1
0.1
0.1
6 t
0
0
100.
100.
7 X
0
0
0.9090909
0.9090909
8 Xbar
0
0
0.3211241
0.3211241
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 C1 = 0.0038746+0.2739782*t+1.574621*t^2-0.2550041*t^3
2
C2 = -33.43818+37.18972*t-11.58838*t^2+1.695303*t^3-0.1298667*t^4+0.005028*t^5-(7.743*10^(5))*t^6
3 C = if(t<=4 and t>=0) then (C1) else if(t>4 and t<=14) then (C2) else 0
4 E = C/51
5 k = 0.1
(f) Maximum mixedness model:
r
dX
E (l )
= A +
X
dl C A0 1 - F (l )
0.316
See the following Polymath program:
Polymath Code:
d(X)/d(z)=-((rA/1)+(E*X/(1-F)))
d(F)/d(z)=-E
rA=-k*(1-X)^2
C1=0.0038746+0.2739782*lam+1.574621*lam^2-0.2550041*lam^3
C2=-33.43818+37.18972*lam-11.58838*lam^2+1.695303*lam^3-0.1298667*lam^4+0.005028*lam^5-(7.743*10^(5))*lam^6
C=if(lam<=4andlam>=0)then(C1)elseif(lam>4andlam<=14)then(C2)else0
E=C/51
lam=14-z
k=0.1
z(0)=0
z(f)=14
X(0)=0
F(0)=0.9999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
C
0.0148043
0.0038746
9.90457
0.0038746
2
C1
-387.266
-387.266
10.0249
0.0038746
3
C2
0.0148043
-33.43818
10.48867
-33.43818
4
E
0.0002903
7.597E-05
0.1942073
7.597E-05
5
F
0.9999
-0.0013431
0.9999
-0.0013431
6
k
0.1
0.1
0.1
0.1
7
lam
14.
0
14.
0
17-49
8
rA
-0.1
-0.1
-0.0467677
-0.0467677
9
X
0
0
0.3161305
0.3161305
10 z
0
0
14.
14.
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 lam = 14-z
3 rA = -k*(1-X)^2
4 C1 = 0.0038746+0.2739782*lam+1.574621*lam^2-0.2550041*lam^3
5
C2 = -33.43818+37.18972*lam-11.58838*lam^2+1.695303*lam^3-0.1298667*lam^4+0.005028*lam^5(7.743*10^(-5))*lam^6
6 C = if(lam<=4 and lam>=0) then (C1) else if(lam>4 and lam<=14) then (C2) else 0
7 E = C/51
P17-13(d)
(a)PFR:
∴
0.296
(b)CSTR:
0.244
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=5.1
t(0) = 0.0000001
t(f) = 100
X(0)=0
17-50
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.1960784
5.982E-10
0.1960784
5.982E-10
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
5.1
5.1
5.1
5.1
5 X
0
0
0.7817821
0.7817821
6 Xbar
0
0
0.2448442
0.2448442
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 tau = 5.1
2 k = 0.1
3 E = (1/tau)*exp(-t/tau)
(c)LFR:
0.266
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
t0=0.000001
E1=tau*tau/(2*t^3)
E=if(t<tau/2)then(0)else(E1)
Xbar(0)=0
k=0.1
tau=5.1
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.3887385
1.301E-05
2 E1
1.301E+22
1.301E-05
1.301E+22
1.301E-05
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
5.1
5.1
5.1
5.1
7 X
0
0
0.7817821
0.7817821
8 Xbar
0
0
0.266644
0.266644
17-51
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 t0 = 0.000001
2 tau = 5.1
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if(t< (E1) else (0) then 2)>
(d)Tanks in series Model
σ=2.49 min
Y'
D.$'
n=Z' =3.BJ' =4.195 ~4
0.281
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
taui = tau/4
E = t^(3)*exp(-t/taui)/(6*taui^4)
k=0.1
tau=5.1
Xbar(0)=0
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
6.307E-23
5.464E-30
0.1706799
5.464E-30
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
5.1
5.1
5.1
5.1
5 taui
1.275
1.275
1.275
1.275
6 X
0
0
0.7817821
0.7817821
7 Xbar
0
0
0.2807419
0.2807419
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 tau = 5.1
2 taui = tau/4
3 k = 0.1
4 E = t^(3)*exp(-t/taui)/(6*taui^4)
17-52
P17-13(d) continued
(e)Segregation Model
0.281
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
C1=0.0038746+0.2739782*t+1.574621*t^2-0.2550041*t^3
C2=-33.43818+37.18972*t-11.58838*t^2+1.695303*t^3-0.1298667*t^4+0.005028*t^5-(7.743*10^(-5))*t^6
C=if(t<=4andt>=0)then(C1)elseif(t>4andt<=14)then(C2)else0
E=C/51
t(0)=0
t(f)=100
Xbar(0)=0
X(0)=0
k=0.1
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 C
0.0038746
0
9.990736
0
2 C1
0.0038746
-2.392E+05
9.973866
-2.392E+05
3 C2
-33.43818
-3.855E+07
10.48753
-3.855E+07
4 E
7.597E-05
0
0.1958968
0
5 k
0.1
0.1
0.1
0.1
6 t
0
0
100.
100.
7 X
0
0
0.7817821
0.7817821
8 Xbar
0
0
0.2814134
0.2814134
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 C1 = 0.0038746+0.2739782*t+1.574621*t^2-0.2550041*t^3
2
C2 = -33.43818+37.18972*t-11.58838*t^2+1.695303*t^3-0.1298667*t^4+0.005028*t^5-(7.743*10^(5))*t^6
3 C = if(t<=4 and t>=0) then (C1) else if(t>4 and t<=14) then (C2) else 0
4 E = C/51
5 k = 0.1
(f) Maximum mixedness model:
X=0.274
See the following Polymath program:
Polymath Code:
d(X)/d(z)=-((rA/1)+(E*X/(1-F)))
d(F)/d(z)=-E
rA=-k*(1-X)^3
C1=0.0038746+0.2739782*lam+1.574621*lam^2-0.2550041*lam^3
C2=-33.43818+37.18972*lam-11.58838*lam^2+1.695303*lam^3-0.1298667*lam^4+0.005028*lam^5-(7.743*10^(5))*lam^6
17-53
C=if(lam<=4andlam>=0)then(C1)elseif(lam>4andlam<=14)then(C2)else0
E=C/51
lam=14-z
k=0.1
z(0)=0
z(f)=14
X(0)=0
F(0)=0.9999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
C
0.0148043
0.0038746
9.915535
0.0038746
2
C1
-387.266
-387.266
10.03581
0.0038746
3
C2
0.0148043
-33.43818
10.4734
-33.43818
4
E
0.0002903
7.597E-05
0.1944223
7.597E-05
5
F
0.9999
-0.0013426
0.9999
-0.0013426
6
k
0.1
0.1
0.1
0.1
7
lam
14.
0
14.
0
8
rA
-0.1
-0.1
-0.0382678
-0.0382678
9
X
0
0
0.2739869
0.2739869
10 z
0
0
14.
14.
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 lam = 14-z
3 rA = -k*(1-X)^3
4 C1 = 0.0038746+0.2739782*lam+1.574621*lam^2-0.2550041*lam^3
5
C2 = -33.43818+37.18972*lam-11.58838*lam^2+1.695303*lam^3-0.1298667*lam^4+0.005028*lam^5(7.743*10^(-5))*lam^6
6 C = if(lam<=4 and lam>=0) then (C1) else if(lam>4 and lam<=14) then (C2) else 0
7 E = C/51
P17-13 (e)
For reaction in part (a):
(e)(a) PFR
17-54
(e)(a)PFR:
E(t)=δ(t-τ)
τ=tm=0.8 min
1-e-kτ=1=1-e-0.1*0.8=0.0769
(e)(a)CSTR:
E(t)=
1
t
e
-
t
t
0.0741
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^1
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=0.8
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
1.25
6.458E-55
1.25
6.458E-55
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
0.8
0.8
0.8
0.8
5 X
0
0
0.9999546
0.9999546
6 Xbar
0
0
0.0740741
0.0740741
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^1
Explicit equations
1 tau = 0.8
2 k = 0.1
3 E = (1/tau)*exp(-t/tau)
(e)(a)LFR:
0.0734
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)
t0=0.000001
E1=tau*tau/(2*t^3)
E=if(t<tau/2)then(0)else(E1)
Xbar(0)=0
17-55
k=0.1
tau=0.8
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.032064
3.2E-07
2 E1
3.2E+20
3.2E-07
3.2E+20
3.2E-07
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
0.8
0.8
0.8
0.8
7 X
0
0
0.9999546
0.9999546
8 Xbar
0
0
0.0733361
0.0733361
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^1
Explicit equations
1 t0 = 0.000001
2 tau = 0.8
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if (t< (E1) else (0) then 2)>
(e)(a)Tanks in series:
0.0762
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)
taui = tau/4
E = t^(3)*exp(-t/taui)/(6*taui^4)
k=0.1
tau = 0.8
Xbar(0)=0
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
1.042E-19
7.421E-210
0.0136383
7.421E-210
2 k
0.1
0.1
0.1
0.1
17-56
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
0.8
0.8
0.8
0.8
5 taui
0.2
0.2
0.2
0.2
6 X
0
0
0.9999546
0.9999546
7 Xbar
0
0
0.0761546
0.0761546
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^1
Explicit equations
1 tau = 0.8
2 taui = tau/4
3 k = 0.1
4 E = t^(3)*exp(-t/taui)/(6*taui^4)
(e)(a)Segregation Model:
0.0765
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)
E=if(t>0andt<2*tau)then(sqrt(tau^2-(t-tau)^2))else(0)
tau=0.8
k=0.1
t(0)=0
t(f)=100
Xbar(0)=0
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0
0
2 k
0.1
0.1
0.1
0.1
3 t
0
0
100.
100.
4 tau
0.8
0.8
0.8
0.8
5 X
0
0
0.9999546
0.9999546
6 Xbar
0
0
0.0765497
0.0765497
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^1
Explicit equations
1 tau = 0.8
2 k = 0.1
3 E = if(t>0 and t<2*tau) then (sqrt(tau^2-(t-tau)^2)) else (0)
17-57
P17-13 (e) continued
(e)(a)Maximum Mixedness Model:
0.0761
See the following Polymath program:
Polymath Code:
d(X)/d(z)=-((rA/1)+(E*X/(1-F)))
d(F)/d(z)=-E
rA=-k*(1-X)
lam=1.6-z
E=if(lam>0andlam<2*tau)then(sqrt(tau^2-(lam-tau)^2))else(0)
k=0.1
tau=0.8
z(0)=0
z(f)=1.6
X(0)=0
F(0)=0.9999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.7999086
0
2 F
0.9999
-0.0054097
0.9999
-0.0054097
3 k
0.1
0.1
0.1
0.1
4 lam
1.6
0
1.6
0
5 rA
-0.1
-0.1
-0.0923848
-0.0923848
6 tau
0.8
0.8
0.8
0.8
7 X
0
0
0.0761521
0.0761521
8 z
0
0
1.6
1.6
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 lam = 1.6-z
3 rA = -k*(1-X)
4 tau = 0.8
5 E = if(lam>0 and lam<2*tau) then (sqrt(tau^2-(lam-tau)^2)) else (0)
For reaction in part (b):
Conversion is same as above (reaction in part (a))
For reaction in part (c):
(e)(c) PFR:
17-58
\ X = 0.0741
(e)(c)CSTR:
0.0696
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=0.8
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
1.25
6.458E-55
1.25
6.458E-55
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
0.8
0.8
0.8
0.8
5 X
0
0
0.9090909
0.9090909
6 Xbar
0
0
0.069559
0.069559
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 tau = 0.8
2 k = 0.1
3 E = (1/tau)*exp(-t/tau)
(e)(c)LFR:
0.06955
17-59
P17-13 (e) continued
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
t0=0.000001
E1=tau*tau/(2*t^3)
E=if(t<tau/2)then(0)else(E1)
Xbar(0)=0
k=0.1
tau=0.8
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.0322837
3.2E-07
2 E1
3.2E+20
3.2E-07
3.2E+20
3.2E-07
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
0.8
0.8
0.8
0.8
7 X
0
0
0.9090909
0.9090909
8 Xbar
0
0
0.0695591
0.0695591
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 t0 = 0.000001
2 tau = 0.8
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if (t< (E1) else (0) then 2)>
(e)(c)Tanks in Series Model:
0.073
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
taui = tau/4
E = t^(3)*exp(-t/taui)/(6*taui^4)
k=0.1
tau = 0.8
Xbar(0)=0
t(0) = 0.0000001
t(f) = 100
X(0)=0
17-60
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
1.042E-19
7.421E-210
0.0105451
7.421E-210
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
0.8
0.8
0.8
0.8
5 taui
0.2
0.2
0.2
0.2
6 X
0
0
0.9090909
0.9090909
7 Xbar
0
0
0.0728439
0.0728439
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 tau = 0.8
2 taui = tau/4
3 k = 0.1
4 E = t^(3)*exp(-t/taui)/(6*taui^4)
(e)(c)Segregation Model:
0.0732
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
E=if(t>0andt<2*tau)then(sqrt(tau^2-(t-tau)^2))else(0)
tau=0.8
k=0.1
t(0)=0
t(f)=100
Xbar(0)=0
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0
0
2 k
0.1
0.1
0.1
0.1
3 t
0
0
100.
100.
4 tau
0.8
0.8
0.8
0.8
5 X
0
0
0.9090909
0.9090909
6 Xbar
0
0
0.0731865
0.0731865
17-61
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 tau = 0.8
2 k = 0.1
3 E = if(t>0 and t<2*tau) then (sqrt(tau^2-(t-tau)^2)) else (0)
(e)(c)Maximum Mixedness Model:
0.0728
See the following Polymath program:
Polymath Code:
d(X)/d(z)=-((rA/1)+(E*X/(1-F)))
d(F)/d(z)=-E
rA=-k*(1-X)^2
lam=1.6-z
E=if(lam>0andlam<2*tau)then(sqrt(tau^2-(lam-tau)^2))else(0)
k=0.1
tau=0.8
z(0)=0
z(f)=1.6
X(0)=0
F(0)=0.9999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.7999208
0
2 F
0.9999
-0.0054097
0.9999
-0.0054097
3 k
0.1
0.1
0.1
0.1
4 lam
1.6
0
1.6
0
5 rA
-0.1
-0.1
-0.0859787
-0.0859787
6 tau
0.8
0.8
0.8
0.8
7 X
0
0
0.0727528
0.0727528
8 z
0
0
1.6
1.6
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 lam = 1.6-z
3 rA = -k*(1-X)^2
4 tau = 0.8
5 E = if(lam>0 and lam<2*tau) then (sqrt(tau^2-(lam-tau)^2)) else (0)
17-62
P17-13 (e) continued
For reaction in part (d):
(e)(d) PFR:
τ=0.8
σ=0.399
dN A
= rAV
dt
N A = N A0 (1 - X A )
dX A
= -rAV = k (C A0 (1 - X A ))3V
dt
1
\ X A = 11 + 2kt
E (t ) = d (t - t )
\ N A0
¥
X = ò X A (t ) E (t )dt
0
0.0715
(e)(d)CSTR:
0.0659
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=0.8
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
1.25
6.458E-55
1.25
6.458E-55
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
0.8
0.8
0.8
0.8
5 X
0
0
0.7817821
0.7817821
6 Xbar
0
0
0.0658886
0.0658886
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
17-63
Explicit equations
1 tau = 0.8
2 k = 0.1
3 E = (1/tau)*exp(-t/tau)
(e)(d)LFR:
0.0665
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
t0=0.000001
E1=tau*tau/(2*t^3)
E=if(t<tau/2)then(0)else(E1)
Xbar(0)=0
k=0.1
tau=0.8
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.037176
3.2E-07
2 E1
3.2E+20
3.2E-07
3.2E+20
3.2E-07
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
0.8
0.8
0.8
0.8
7 X
0
0
0.7817821
0.7817821
8 Xbar
0
0
0.0664998
0.0664998
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 t0 = 0.000001
2 tau = 0.8
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if (t< (E1) else (0) then 2)>
(e)(d)Tanks in series:
τ=0.8
σ=0.399
n=4.02~4
0.0699
17-64
P17-13 (e) continued
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
taui = tau/4
E = t^(3)*exp(-t/taui)/(6*taui^4)
k=0.1
tau = 0.8
Xbar(0)=0
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
1.042E-19
7.421E-210
0.0217544
7.421E-210
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
0.8
0.8
0.8
0.8
5 taui
0.2
0.2
0.2
0.2
6 X
0
0
0.7817821
0.7817821
7 Xbar
0
0
0.0699403
0.0699403
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 tau = 0.8
2 taui = tau/4
3 k = 0.1
4 E = t^(3)*exp(-t/taui)/(6*taui^4)
(e)(d)Segregation Model:
0.0702
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
E=if(t>0andt<2*tau)then(sqrt(tau^2-(t-tau)^2))else(0)
tau=0.8
k=0.1
t(0)=0
t(f)=100
Xbar(0)=0
X(0)=0
17-65
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0
0
2 k
0.1
0.1
0.1
0.1
3 t
0
0
100.
100.
4 tau
0.8
0.8
0.8
0.8
5 X
0
0
0.7817821
0.7817821
6 Xbar
0
0
0.070226
0.070226
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 tau = 0.8
2 k = 0.1
3 E = if(t>0 and t<2*tau) then (sqrt(tau^2-(t-tau)^2)) else (0)
(e)(d)Maximum Mixedness Model:
0.0697
See the following Polymath program:
Polymath Code:
d(X)/d(z)=-((rA/1)+(E*X/(1-F)))
d(F)/d(z)=-E
rA=-k*(1-X)^3
lam=1.6-z
E=if(lam>0andlam<2*tau)then(sqrt(tau^2-(lam-tau)^2))else(0)
k=0.1
tau=0.8
z(0)=0
z(f)=1.6
X(0)=0
F(0)=0.9999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.7999316
0
2 F
0.9999
-0.0054097
0.9999
-0.0054097
3 k
0.1
0.1
0.1
0.1
4 lam
1.6
0
1.6
0
5 rA
-0.1
-0.1
-0.0805068
-0.0805068
6 tau
0.8
0.8
0.8
0.8
7 X
0
0
0.0697261
0.0697261
8 z
0
0
1.6
1.6
17-66
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 lam = 1.6-z
3 rA = -k*(1-X)^3
4 tau = 0.8
5 E = if(lam>0 and lam<2*tau) then (sqrt(tau^2-(lam-tau)^2)) else (0)
P17-13 (f)
For reaction in part (a):
(a)PFR:
E(t)=δ(t-τ)
τ=25 min
1-e-0.1*25=0.918
(f)(a) CSTR:
0.6959
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=25
t(0) = 0.0000001
t(f) = 100
X(0)=0
17-67
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.04
0.0007326
0.04
0.0007326
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
25.
25.
25.
25.
5 X
0
0
0.9999546
0.9999546
6 Xbar
0
0
0.6959703
0.6959703
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^1
Explicit equations
1 tau = 25
2 k = 0.1
3 E = (1/tau)*exp(-t/tau)
(f)(a)LFR:
0.83
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)
t0=0.000001
E1=tau*tau/(2*t^3)
E=if(t<tau/2)then(0)else(E1)
Xbar(0)=0
k=0.1
tau=25
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.1254587
0.0003125
2 E1
3.125E+23
0.0003125
3.125E+23
0.0003125
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
25.
25.
25.
25.
7 X
0
0
0.9999546
0.9999546
8 Xbar
0
0
0.8272304
0.8272304
17-68
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^1
Explicit equations
1 t0 = 0.000001
2 tau = 25
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if (t< (E1) else (0) then 2)>
(f)(a) TIS:
0.89
n=252/75=8.333~8
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t) = X*E
d(X)/d(t) = k*(1-X)^1
tau = 25
taui = tau/8
k = 0.1
E = t^(7)*exp(-t/taui)/(5040*taui^8)
t(0)=0
t(f)=100
Xbar(0)=0
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
2.182E-57
2.182E-57
0.0476485
2.763E-08
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
25.
25.
25.
25.
5 taui
3.125
3.125
3.125
3.125
6 X
0
0
0.9999546
0.9999546
7 Xbar
0
0
0.8864451
0.8864451
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^1
Explicit equations
1 tau = 25
2 taui = tau/8
3 k = 0.1
4 E = t^(7)*exp(-t/taui)/(5040*taui^8)
17-69
P17-13 (f) continued
(f)(a) Segregation Model:
1
4
3
4
E(t)= d (t - 10) + d (t - 30)
dN A
= rAV
dt
N A = N A0 (1 - X A )
dX A
= -rAV = kC A0 (1 - X A )V
dt
\ X A = 1 - e -kt
\ N A0
¥
X = ò X A (t ) E (t )dt =
0
1
3
1
3
X A (10) + X A (30) = (1 - e -10 k ) + (1 - e -30 k ) = 0.87
4
4
4
4
(f)(a) Maximum mixedness Model:
0.87
See the following Polymath program:
Polymath Code:
d(X)/d(z)=-((rA/1)+(E*X/(1-F)))
d(F)/d(z)=-E
rA=-(k*(1-X)^1)
k=0.1
t1=10
t2=30
lam = 40-z
E2 = 0.75/(t2*2*(1-0.99))
E1 = 0.25/(t1*2*(1-0.99))
E3=0
E = if((lam>=0.99*t1)and(lam<1.01*t1))then(E1)else(if((lam>=0.99*t2)and(lam<1.01*t2))then(E2)else(E3))
z(0)=0
z(f)=40
X(0)=0
F(0)=0.999999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
E
0
0
1.25
0
2
E1
1.25
1.25
1.25
1.25
3
E2
1.25
1.25
1.25
1.25
4
E3
0
0
0
0
5
F
0.999999
-3.956E-06
0.999999
-3.956E-06
6
k
0.1
0.1
0.1
0.1
7
lam
40.
0
40.
0
8
rA
-0.1
-0.1
-0.0129321
-0.0129321
9
t1
10.
10.
10.
10.
10 t2
30.
30.
30.
30.
11 X
0
0
0.8706792
0.8706792
12 z
0
0
40.
40.
17-70
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 rA = -(k*(1-X))
3 t1 = 10
4 t2 = 30
5 lam = 40-z
6 E2 = 0.75/(t2*2*(1-0.99))
7 E1 = 0.25/(t1*2*(1-0.99))
8 E3 = 0
9
E = if ((lam>=0.99*t1)and(lam<1.01*t1)) then (E1) else( if ((lam>=0.99*t2)and(lam<1.01*t2)) then (E2) else
(E3))
For reaction in part (b):
Same as above
For reaction in part (c):
(f)(c) PFR:
\ X = 0.714
(f)(c) CSTR:
0.56
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=25
t(0) = 0.0000001
t(f) = 100
X(0)=0
17-71
P17-13 (f) continued
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.04
0.0007326
0.04
0.0007326
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
25.
25.
25.
25.
5 X
0
0
0.9090909
0.9090909
6 Xbar
0
0
0.5639471
0.5639471
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 tau = 25
2 k = 0.1
3 E = (1/tau)*exp(-t/tau)
(f)(c) LFR:
0.65
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
t0=0.000001
E1=tau*tau/(2*t^3)
E=if(t<tau/2)then(0)else(E1)
Xbar(0)=0
k=0.1
tau=25
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.1265542
0.0003125
2 E1
3.125E+23
0.0003125
3.125E+23
0.0003125
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
25.
25.
25.
25.
7 X
0
0
0.9090909
0.9090909
8 Xbar
0
0
0.648511
0.648511
17-72
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 t0 = 0.000001
2 tau = 25
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if (t< (E1) else (0) then 2)>
(f)(c) TIS:
0.696
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t) = X*E
d(X)/d(t) = k*(1-X)^2
tau = 25
taui = tau/8
k = 0.1
E = t^(7)*exp(-t/taui)/(5040*taui^8)
t(0)=0
t(f)=100
Xbar(0)=0
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
2.182E-57
2.182E-57
0.0476582
2.763E-08
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
25.
25.
25.
25.
5 taui
3.125
3.125
3.125
3.125
6 X
0
0
0.9090909
0.9090909
7 Xbar
0
0
0.6959218
0.6959218
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 tau = 25
2 taui = tau/8
3 k = 0.1
4 E = t^(7)*exp(-t/taui)/(5040*taui^8)
17-73
P17-13 (f) continued
(f)(c) Segregation:
1
4
3
4
E(t)= d (t - 10) + d (t - 30)
dN A
= rAV
dt
N A = N A0 (1 - X A )
dX A
= -rAV = k (C A0 (1 - X A )) 2 V
dt
kt
\XA =
1 + kt
¥
1
3
0.1*10
0.1* 30
X = ò X A (t ) E (t )dt = X A (10) + X A (30) = 0.25 *
+ 0.75 *
= 0.6875
4
4
1 + 0.1*10
1 + 0.1* 30
0
\ N A0
(f)(c) Maximum Mixedness Model:
0.66
See the following Polymath program:
Polymath Code:
d(X)/d(z)=-((rA/1)+(E*X/(1-F)))
d(F)/d(z)=-E
rA=-(k*(1-X)^2)
k=0.1
t1=10
t2=30
lam = 40-z
E2 = 0.75/(t2*2*(1-0.99))
E1 = 0.25/(t1*2*(1-0.99))
E3=0
E = if((lam>=0.99*t1)and(lam<1.01*t1))then(E1)else(if((lam>=0.99*t2)and(lam<1.01*t2))then(E2)else(E3))
z(0)=0
z(f)=40
X(0)=0
F(0)=0.999999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
E
0
0
1.25
0
2
E1
1.25
1.25
1.25
1.25
3
E2
1.25
1.25
1.25
1.25
17-74
4
E3
0
0
0
0
5
F
0.999999
-0.0035796
0.999999
-0.0035796
6
k
0.1
0.1
0.1
0.1
7
lam
40.
0
40.
0
8
rA
-0.1
-0.1
-0.011168
-0.011168
9
t1
10.
10.
10.
10.
10 t2
30.
30.
30.
30.
11 X
0
0
0.6658143
0.6658143
12 z
0
0
40.
40.
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 rA = -(k*(1-X)^2)
3 t1 = 10
4 t2 = 30
5 lam = 40-z
6 E2 = 0.75/(t2*2*(1-0.99))
7 E1 = 0.25/(t1*2*(1-0.99))
8 E3 = 0
9 E = if ((lam>=0.99*t1)and(lam<1.01*t1)) then (E1) else( if ((lam>=0.99*t2)and(lam<1.01*t2)) then (E2) else (E3))
For reaction in part (d):
(f)(d) PFR:
dN A
= rAV
dt
N A = N A0 (1 - X A )
dX A
= -rAV = k (C A0 (1 - X A ))3V
dt
1
\ X A = 11 + 2kt
E (t ) = d (t - t )
\ N A0
¥
X = ò X A (t ) E (t )dt
0
X =0.59
CSTR:
0.48
17-75
P17-13 (f) continued
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=25
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.04
0.0007326
0.04
0.0007326
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
25.
25.
25.
25.
5 X
0
0
0.7817821
0.7817821
6 Xbar
0
0
0.4750084
0.4750084
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 tau = 25
2 k = 0.1
3 E = (1/tau)*exp(-t/tau)
(f)(d)
LFR:
0.54
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=25
t(0) = 0.0000001
t(f) = 100
X(0)=0
17-76
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.1267909
0.0003125
2 E1
3.125E+23
0.0003125
3.125E+23
0.0003125
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
25.
25.
25.
25.
7 X
0
0
0.7817821
0.7817821
8 Xbar
0
0
0.5401314
0.5401314
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 t0 = 0.000001
2 tau = 25
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if (t< (E1) else (0) then 2)>
(f)(d)TIS:
0.58
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t) = X*E
d(X)/d(t) = k*(1-X)^3
tau = 25
taui = tau/8
k = 0.1
E = t^(7)*exp(-t/taui)/(5040*taui^8)
t(0)=0
t(f)=100
Xbar(0)=0
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
2.182E-57
2.182E-57
0.0474927
2.763E-08
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
25.
25.
25.
25.
5 taui
3.125
3.125
3.125
3.125
6 X
0
0
0.7817821
0.7817821
7 Xbar
0
0
0.5782912
0.5782912
17-77
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 tau = 25
2 taui = tau/8
3 k = 0.1
4 E = t^(7)*exp(-t/taui)/(5040*taui^8)
(f)(d)Segregation Model:
dN A
= rAV
dt
N A = N A0 (1 - X A )
dX A
= -rAV = k (C A0 (1 - X A ))3V
dt
1
\ X A = 11 + 2kt
\ N A0
¥
X = ò X A (t ) E (t )dt =
0
1
3
1
1
X A (10) + X A (30) = 0.25 * (1 ) + 0.75 * (1 )
4
4
1 + 2 * 0.1*10
1 + 2 * 0.1* 30
= 0.572
(f)(d)Maximum Mixedness Model
Conversion=0.549
See the following Polymath program:
Polymath Code:
d(X)/d(z)=-((rA/1)+(E*X/(1-F)))
rA=-(k*(1-X)^3)
d(F)/d(z)=-E
k=0.1
z(0)=0
z(f)=40
X(0)=0
F(0)=0.9999
t1=10
t2=30
lam = 40-z
E2 = 0.75/(t2*2*(1-0.99))
E1 = 0.25/(t1*2*(1-0.99))
E3=0
E = if((lam>=0.99*t1)and(lam<1.01*t1))then(E1)else(if((lam>=0.99*t2)and(lam<1.01*t2))then(E2)else(E3))
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
E
0
0
1.25
0
2
E1
1.25
1.25
1.25
1.25
3
E2
1.25
1.25
1.25
1.25
17-78
4
E3
0
0
0
0
5
F
0.9999
-9.736E-05
0.9999
-9.736E-05
6
k
0.1
0.1
0.1
0.1
7
lam
40.
0
40.
0
8
rA
-0.1
-0.1
-0.0090389
-0.0091717
9
t1
10.
10.
10.
10.
10 t2
30.
30.
30.
30.
11 X
0
0
0.5512155
0.549028
12 z
0
0
40.
40.
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 rA = -(k*(1-X)^3)
3 t1 = 10
4 t2 = 30
5 lam = 40-z
6 E2 = 0.75/(t2*2*(1-0.99))
7 E1 = 0.25/(t1*2*(1-0.99))
8 E3 = 0
9 E = if ((lam>=0.99*t1)and(lam<1.01*t1)) then (E1) else( if ((lam>=0.99*t2)and(lam<1.01*t2)) then (E2) else (E3))
P17-13 (g)
For reaction in part (a):
(g)(a)PFR:
E(t)=δ(t-τ)
τ=5 min
X = 1-e-0.1*5=0.393
(g)(a)CSTR:
0.333
17-79
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^1
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=5
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.2
4.122E-10
0.2
4.122E-10
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
5.
5.
5.
5.
5 X
0
0
0.9999546
0.9999546
6 Xbar
0
0
0.3333333
0.3333333
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)
Explicit equations
1 tau = 5
2 k = 0.1
3 E = (1/tau)*exp(-t/tau)
(g)(a)LFR:
0.35
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
t0=0.000001
E1=tau*tau/(2*t^3)
E=if(t<tau/2)then(0)else(E1)
Xbar(0)=0
k=0.1
tau=5
t(0) = 0.0000001
t(f) = 100
X(0)=0
17-80
P17-13 (g) continued
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.4008871
1.25E-05
2 E1
1.25E+22
1.25E-05
1.25E+22
1.25E-05
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
5.
5.
5.
5.
7 X
0
0
0.9999546
0.9999546
8 Xbar
0
0
0.3500068
0.3500068
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)
Explicit equations
1 t0 = 0.000001
2 tau = 5
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if (t< (E1) else (0) then 2)>
(g)(a)TIS:
n=τ2/σ2=52/4.167=5.99 ~6
0.381
See the following Polymath program:
Polymath Code:
taui=tau/6
E=t^(5)*exp(-t/taui)/(120*taui^6)
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)
Xbar(0)=0
k=0.1
tau=5
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
2.488E-37
1.908E-44
0.2099462
1.908E-44
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
5.
5.
5.
5.
17-81
5 taui
0.8333333
0.8333333
0.8333333
0.8333333
6 X
0
0
0.9999546
0.9999546
7 Xbar
0
0
0.3813751
0.3813751
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)
Explicit equations
1 tau = 5
2 taui = tau/6
3 k = 0.1
4 E = t^(5)*exp(-t/taui)/(120*taui^6)
(g)(a) Segregation Model:
0.380
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.1992851
0
2 E1
0
0
0.6
0.6
3 E2
0.4
-0.2
0.4
-0.2
4 k
0.1
0.1
0.1
0.1
5 t
0
0
15.
15.
6 t1
10.
10.
10.
10.
7 tau
5.
5.
5.
5.
8 X
0
0
0.7768698
0.7768698
9 Xbar
0
0
0.3807271
0.3807271
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)
Explicit equations
1 k = 0.1
2 tau = 5
3 E1 = t/tau^2
4 t1 = 2*tau
5 E2 = -(t-t1)/tau^2
6 E = if (t< else(if(t<="t1)then(E2)else(0))" (E1) then>
(g)(a) Maximum Mixedness Model:
0.380
17-82
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
E
0
0
0.1975427
0
2
E1
0.8
0
0.8
0
3
E2
-0.4
-0.4
0.4
0.4
4
E3
0
0
0
0
5
F
0.999999
-1.4E-06
0.999999
-1.4E-06
6
k
0.1
0.1
0.1
0.1
7
lam
20.
0
20.
0
8
rA
-0.1
-0.1
-0.0368778
-0.0619272
9
t1
10.
10.
10.
10.
10 tau
5.
5.
5.
5.
11 X
0
0
0.6312225
0.380728
12 z
0
0
20.
20.
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 rA = -(k*(1-X))
3 lam = 20-z
4 tau = 5
5 t1 = 2*tau
6 E3 = 0
7 E1 = lam/tau^2
8 E2 = -(lam-t1)/tau^2
9 E = if (lam< (0)) then(E2)else (lam<="t1)" else(if (E1) then>
For reaction in part (b):
Same as above
For reaction in part (c):
(g)(c)PFR:
17-83
P17-13 (g) continued
\ X = 0.333
(g)(c)CSTR:
0.277
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=5
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.2
4.122E-10
0.2
4.122E-10
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
5.
5.
5.
5.
5 X
0
0
0.9090909
0.9090909
6 Xbar
0
0
0.2773428
0.2773428
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 tau = 5
2 k = 0.1
3 E = (1/tau)*exp(-t/tau)
(g)(c) LFR:
0.298
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
t0=0.000001
E1=tau*tau/(2*t^3)
E=if(t<tau/2)then(0)else(E1)
17-84
Xbar(0)=0
k=0.1
tau=5
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.3978846
1.25E-05
2 E1
1.25E+22
1.25E-05
1.25E+22
1.25E-05
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
5.
5.
5.
5.
7 X
0
0
0.9090909
0.9090909
8 Xbar
0
0
0.2982341
0.2982341
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 t0 = 0.000001
2 tau = 5
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if (t< (E1) else (0) then 2)>
(g)(c) TIS:
0.322
See the following Polymath program:
Polymath Code:
taui=tau/6
E=t^(5)*exp(-t/taui)/(120*taui^6)
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^2
Xbar(0)=0
k=0.1
tau=5
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
2.488E-37
1.908E-44
0.2105158
17-85
1.908E-44
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
5.
5.
5.
5.
5 taui
0.8333333
0.8333333
0.8333333
0.8333333
6 X
0
0
0.9090909
0.9090909
7 Xbar
0
0
0.3216361
0.3216361
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 tau = 5
2 taui = tau/6
3 k = 0.1
4 E = t^(5)*exp(-t/taui)/(120*taui^6)
(g)(c) Segregation Model:
0.320
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.1986949
0
2 E1
0
0
0.6
0.6
3 E2
0.4
-0.2
0.4
-0.2
4 k
0.1
0.1
0.1
0.1
5 t
0
0
15.
15.
6 t1
10.
10.
10.
10.
7 tau
5.
5.
5.
5.
8 X
0
0
0.6
0.6
9 Xbar
0
0
0.3203968
0.3203968
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^2
Explicit equations
1 k = 0.1
2 tau = 5
3 E1 = t/tau^2
4 t1 = 2*tau
5 E2 = -(t-t1)/tau^2
6 E = if (t< else(if(t<="t1)then(E2)else(0))" (E1) then>
(g)(c) Maximum Mixedness Model:
0.317
17-86
P17-13 (g) continued
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
E
0
0
0.1977739
0
2
E1
0.8
0
0.8
0
3
E2
-0.4
-0.4
0.4
0.4
4
E3
0
0
0
0
5
F
0.999999
-5.069E-06
0.999999
-5.069E-06
6
k
0.1
0.1
0.1
0.1
7
lam
20.
0
20.
0
8
rA
-0.1
-0.1
-0.0250298
-0.0466224
9
t1
10.
10.
10.
10.
10 tau
5.
5.
5.
5.
11 X
0
0
0.4997024
0.3171939
12 z
0
0
20.
20.
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 rA = -(k*(1-X)^2)
3 lam = 20-z
4 tau = 5
5 t1 = 2*tau
6 E3 = 0
7 E1 = lam/tau^2
8 E2 = -(lam-t1)/tau^2
9 E = if (lam< (0)) then(E2)else (lam<="t1)" else(if (E1) then>
For reaction in part (d):
(g)(d) PFR:
0.293
17-87
(g)(d) CSTR:
0.242
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.1
tau=5
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.2
4.122E-10
0.2
4.122E-10
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
5.
5.
5.
5.
5 X
0
0
0.7817821
0.7817821
6 Xbar
0
0
0.2421278
0.2421278
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 tau = 5
2 k = 0.1
3 E = (1/tau)*exp(-t/tau)
(g)(d) LFR:
0.263
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
t0=0.000001
E1=tau*tau/(2*t^3)
E=if(t<tau/2)then(0)else(E1)
Xbar(0)=0
k=0.1
tau=5
t(0) = 0.0000001
t(f) = 100
X(0)=0
17-88
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.3983975
1.25E-05
2 E1
1.25E+22
1.25E-05
1.25E+22
1.25E-05
3 k
0.1
0.1
0.1
0.1
4 t
1.0E-07
1.0E-07
100.
100.
5 t0
1.0E-06
1.0E-06
1.0E-06
1.0E-06
6 tau
5.
5.
5.
5.
7 X
0
0
0.7817821
0.7817821
8 Xbar
0
0
0.2634677
0.2634677
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 t0 = 0.000001
2 tau = 5
3 E1 = tau*tau/(2*t^3)
4 k = 0.1
5 E = if (t< (E1) else (0) then 2)>
(g)(d) TIS:
0.58
See the following Polymath program:
Polymath Code:
taui=tau/6
E=t^(5)*exp(-t/taui)/(120*taui^6)
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)^3
Xbar(0)=0
k=0.1
tau=5
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
2.488E-37
1.908E-44
0.2103056
1.908E-44
2 k
0.1
0.1
0.1
0.1
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
5.
5.
5.
5.
5 taui
0.8333333
0.8333333
0.8333333
0.8333333
6 X
0
0
0.7817821
0.7817821
7 Xbar
0
0
0.2823623
0.2823623
17-89
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 tau = 5
2 taui = tau/6
3 k = 0.1
4 E = t^(5)*exp(-t/taui)/(120*taui^6)
(g)(d) Segregation Model:
0.280
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.1994053
0
2 E1
0
0
0.6
0.6
3 E2
0.4
-0.2
0.4
-0.2
4 k
0.1
0.1
0.1
0.1
5 t
0
0
15.
15.
6 t1
10.
10.
10.
10.
7 tau
5.
5.
5.
5.
8 X
0
0
0.5
0.5
9 Xbar
0
0
0.2809295
0.2809295
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)^3
Explicit equations
1 k = 0.1
2 tau = 5
3 E1 = t/tau^2
4 t1 = 2*tau
5 E2 = -(t-t1)/tau^2
6 E = if (t< else(if(t<="t1)then(E2)else(0))" (E1) then>
(g)(d) Maximum Mixedness Model:
0.276
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
E
0
0
0.1995877
0
2
E1
0.8
0
0.8
0
3
E2
-0.4
-0.4
0.4
0.4
4
E3
0
0
0
0
17-90
5
F
0.999999
-2.204E-06
0.999999
-2.204E-06
6
k
0.1
0.1
0.1
0.1
7
lam
20.
0
20.
0
8
rA
-0.1
-0.1
-0.0192471
-0.0379882
9
t1
10.
10.
10.
10.
10 tau
5.
5.
5.
5.
11 X
0
0
0.4226286
0.2757593
12 z
0
0
20.
20.
Differential equations
1 d(X)/d(z) = -((rA/1)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 k = 0.1
2 rA = -(k*(1-X)^3)
3 lam = 20-z
4 tau = 5
5 t1 = 2*tau
6 E3 = 0
7 E1 = lam/tau^2
8 E2 = -(lam-t1)/tau^2
9 E = if (lam< (0)) then(E2)else (lam<="t1)" else(if (E1) then>
P17-14 (a)
Segregation Model:
Liquid phase, Segregation Model, second order, non-ideal CSTR, adiabatic:
E(t)=IF (t<=1) THEN (t) ELSE ( IF (t>=2) THEN (0) ELSE (2-t))
For a batch globule:
Where CAo=2 mol/dm3
Where
17-91
T=300+150X
See the following Polymath program:
Polymath Code:
d(xbar)/d(t)=E*x
d(x)/d(t)=k*ca0*((1-x)^2)
ca0=2
T=300+150*x
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)-(1/T)))
xbar(0)=0
x(0)=0
t(0)=0
t(f)=3
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
ca0
2.
2.
2.
2.
2
E
0
0
0.9946149
0
3
E1
0
0
3.
3.
4
E2
2.
-1.
2.
-1.
5
Ea
1.25E+04
1.25E+04
1.25E+04
1.25E+04
6
k
0.5
0.5
2.424701
2.424701
7
T
300.
300.
437.9808
437.9808
8
t
0
0
3.
3.
9
x
0
0
0.9198721
0.9198721
10 xbar
0
0
0.6673438
0.6673438
Differential equations
1 d(xbar)/d(t) = E*x
2 d(x)/d(t) = k*ca0*((1-x)^2)
Explicit equations
1 ca0 = 2
2 T = 300+150*x
3 E1 = t
4 E2 = 2-t
5 E = if (t<=1) then (E1) else (if (t>=2) then (0) else (E2))
6 Ea = 12500
7 k = 0.5*exp(Ea/8.314*((1/300)-(1/T)))
=0.667
Maximum Mixedness Model:
See the following Polymath program:
17-92
Polymath Code:
d(X)/d(z)=-((rA/Ca0)+(E*X/(1-F)))
rA=-k*(Ca0^2)*(1-X)^2
Ca0=2
T=300+150*X
E1 = lam
E2 = 2-lam
E = if(lam<=1)then(E1)else(if(lam>=2)then(0)else(E2))
Ea = 12500
k = 0.5*exp(Ea/8.314*((1/300)-(1/T)))
X(0)=0
d(F)/d(z)=-E
lam=3-z
z(0)=0
z(f)=3
F(0)=0.9999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca0
2.
2.
2.
2.
2
E
0
0
0.9898387
0
3
E1
3.
0
3.
0
4
E2
-1.
-1.
2.
2.
5
Ea
1.25E+04
1.25E+04
1.25E+04
1.25E+04
6
F
0.9999
-0.0001165
0.9999
-0.0001165
7
k
0.5
0.5
1.862411
1.801067
8
lam
3.
0
3.
0
9
rA
-2.
-2.056925
-0.6201651
-0.7052221
10 T
300.
300.
406.721
403.0691
11 X
0
0
0.7114734
0.687127
12 z
0
0
3.
3.
Differential equations
1 d(X)/d(z) = -((rA/Ca0)+(E*X/(1-F)))
2 d(F)/d(z) = -E
Explicit equations
1 Ea = 12500
2 Ca0 = 2
3 T = 300+150*X
4 lam = 3-z
5 E1 = lam
6 E2 = 2-lam
7 k = 0.5*exp(Ea/8.314*((1/300)-(1/T)))
8 rA = -k*(Ca0^2)*(1-X)^2
9 E = if (lam<=1) then (E1) else (if (lam>=2) then (0) else (E2))
0.687
17-93
P17-14 (b)
Parallel reactions, isothermal:
Segregation model:
Batch globules
Exit concentrations
E(t)=IF (t<=1) THEN (t) ELSE ( IF (t>=2) THEN (0) ELSE (2-t))
See the following Polymath program:
Polymath Code:
d(Ca)/d(t)=-k1a*Ca^2-k2c*Ca*Cb
d(Cb)/d(t)=k1a*Ca^2-k2c*Ca*Cb
d(Cc)/d(t)=k2c*Ca*Cb
d(Cabar)/d(t)=Ca*E
d(Cbbar)/d(t)=Cb*E
d(Ccbar)/d(t)=Cc*E
E1 = t
E2 = 2-t
E = if(t<=1)then(E1)else(if(t>=2)then(0)else(E2))
k2c=0.12
k1a=0.5
S=Cbbar/(Ccbar+0.00001)
Ca0=2
t(0)=0
t(f)=2
Ca(0)=2
Cb(0)=0
Cc(0)=0
Cabar(0)=0
Cbbar(0)=0
Ccbar(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
2.
0.5799685
2.
0.5799685
2
Ca0
2.
2.
2.
2.
3
Cabar
0
0
0.9952971
0.9952971
4
Cb
0
0
1.075943
1.075943
5
Cbbar
0
0
0.838566
0.838566
17-94
6
Cc
0
0
0.1720441
0.1720441
7
Ccbar
0
0
0.083065
0.083065
8
E
0
0
0.9900897
0
9
E1
0
0
2.
2.
10 E2
2.
0
2.
0
11 k1a
0.5
0.5
0.5
0.5
12 k2c
0.12
0.12
0.12
0.12
13 S
0
0
55.10606
10.09408
14 t
0
0
2.
2.
Differential equations
1 d(Ca)/d(t) = -k1a*Ca^2-k2c*Ca*Cb
2 d(Cb)/d(t) = k1a*Ca^2-k2c*Ca*Cb
3 d(Cc)/d(t) = k2c*Ca*Cb
4 d(Cabar)/d(t) = Ca*E
5 d(Cbbar)/d(t) = Cb*E
6 d(Ccbar)/d(t) = Cc*E
Explicit equations
1 E1 = t
2 E2 = 2-t
3 E = if (t<=1) then (E1) else (if (t>=2) then (0) else (E2))
4 k2c = 0.12
5 k1a = 0.5
6 S = Cbbar/(Ccbar+0.00001)
7 Ca0 = 2
Selectivity=10.094
Maximum Mixedness model:
See the following Polymath program:
Polymath Code:
d(Ca)/d(z)=-(-(ra/1)+(E*(Ca-Ca0)/(1-F)))
d(Cb)/d(z)=-(-(rb/1)+(E*(Cb-Cb0)/(1-F)))
d(Cc)/d(z)=-(-(rc/1)+(E*(Cc-Cc0)/(1-F)))
d(F)/d(z)=-E
E1 = lam
E2 = 2-lam
E = if(lam<=1)then(E1)else(if(lam>=2)then(0)else(E2))
lam=2-z
ra=-k1a*Ca^2-k2c*Ca*Cb
rb=k1a*Ca^2-k2c*Ca*Cb
rc=k2c*Ca*Cb
k1a=0.5
k2c=0.12
S=Cb/(Cc+0.00001)
Ca0=2
Cb0=0
Cc0=0
z(0)=0
z(f)=2
Ca(0)=2
Cb(0)=0
Cc(0)=0
F(0)=0.9999999
17-95
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
2.
1.014957
2.
1.014957
2
Ca0
2.
2.
2.
2.
3
Cb
0
0
0.8072371
0.8072371
4
Cb0
0
0
0
0
5
Cc
0
0
0.0889027
0.0889027
6
Cc0
0
0
0
0
7
E
0
0
0.9867406
0
8
E1
2.
0
2.
0
9
E2
0
0
2.
2.
10 F
0.9999999
2.783E-06
0.9999999
2.783E-06
11 k1a
0.5
0.5
0.5
0.5
12 k2c
0.12
0.12
0.12
0.12
13 lam
2.
0
2.
0
14 ra
-2.
-2.
-0.6133867
-0.6133867
15 rb
2.
0.416752
2.
0.416752
16 rc
0
0
0.0988635
0.0983174
17 S
0
0
350.2289
9.078987
18 z
0
0
2.
2.
Differential equations
1 d(Ca)/d(z) = -(-(ra/1)+(E*(Ca-Ca0)/(1-F)))
2 d(Cb)/d(z) = -(-(rb/1)+(E*(Cb-Cb0)/(1-F)))
3 d(Cc)/d(z) = -(-(rc/1)+(E*(Cc-Cc0)/(1-F)))
4 d(F)/d(z) = -E
Explicit equations
1
lam = 2-z
2
E1 = lam
3
E2 = 2-lam
4
E = if (lam<=1) then (E1) else (if (lam>=2) then (0) else (E2))
5
k2c = 0.12
6
k1a = 0.5
7
ra = -k1a*Ca^2-k2c*Ca*Cb
8
rb = k1a*Ca^2-k2c*Ca*Cb
9
rc = k2c*Ca*Cb
10 S = Cb/(Cc+0.00001)
11 Ca0 = 2
12 Cb0 = 0
13 Cc0 = 0
Selectivity=9.079
17-96
P17-15
T=613 0R
k=16.96*1012*e(-32400/RT) h-1=16.96*1012*e(-32400/(1.987*613)) *(1/60) min-1=0.7923min-1
P17-15 (a)
rA = -kC A
CA0=0.132
ΘB=18.65
⸫CB0=2.4618
Segregation model:
Exit Selectivity=CB bar/CC bar=19.08997
See the following Polymath program:
Polymath Code:
d(Ca)/d(t)=rA
d(Cb)/d(t)=rB
d(Cc)/d(t)=rC
d(Cabar)/d(t)=Ca*E
d(Cbbar)/d(t)=Cb*E
d(Ccbar)/d(t)=Cc*E
C1=0.0039+0.274*t+1.57*t^2-0.255*t^3
C2=-33.4+37.2*t-11.6*t^2+1.7*t^3-0.13*t^4+0.005*t^5-(7.7*10^(-5))*t^6
C=if(t<=4andt>=0)then(C1)elseif(t>4andt=<14)then(C2)else0
E=C/51
t(0)=0
t(f)=100
k=0.7923
rA=-k*Ca
rB=rA
rC=-rA
S=Cbbar/(Ccbar+0.00001)
Cabar(0)=0
Cbbar(0)=0
Ccbar(0)=0
Ca(0)=0.132
Cb(0)=2.4618
Cc(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
C
0.0039
-3.813273
10.1031
0
2
C1
0.0039
-2.393E+05
9.908888
-2.393E+05
3
C2
-33.4
-3.841E+07
10.47255
-3.841E+07
4
Ca
0.132
5.099E-36
0.132
5.099E-36
5
Cabar
0
0
0.0081258
0.0081252
6
Cb
2.4618
2.3298
2.4618
2.3298
7
Cbbar
0
0
2.345803
2.006578
8
Cc
0
0
0.132
0.132
9
Ccbar
0
0
0.1243205
0.1051017
10 E
7.647E-05
-0.0747701
0.1981
0
11 k
0.7923
0.7923
0.7923
0.7923
17-97
12 rA
-0.1045836
-0.1045836
-4.04E-36
-4.04E-36
13 rB
-0.1045836
-0.1045836
-4.04E-36
-4.04E-36
14 rC
0.1045836
4.04E-36
0.1045836
4.04E-36
15 S
0
0
26.77842
19.08997
16 t
0
0
100.
100.
Differential equations
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
1 C1 = 0.0039+0.274*t+1.57*t^2-0.255*t^3
2 C2 = -33.4+37.2*t-11.6*t^2+1.7*t^3-0.13*t^4+0.005*t^5-(7.7*10^(-5))*t^6
3 C = if(t<=4 and t>=0) then (C1) else if(t>=4 and t<14) then (C2) else 0
4 E = C/51
5 k = 0.7923
6 rA = -k*Ca
7 rB = rA
8 rC = -rA
9 S = Cbbar/(Ccbar+0.00001)
P17-15 (b)
Maximum Mixedness Model:
Exit Selectivity= FB/FC=CB /CC =2.3298/0.132=19.09027
See the following Polymath program:
Polymath Code:
d(Ca)/d(z)=-(-(rA/1)+(E*(Ca-Ca0)/(1-F)))
d(Cb)/d(z)=-(-(rB/1)+(E*(Cb-Cb0)/(1-F)))
d(Cc)/d(z)=-(-(rC/1)+(E*(Cc-Cc0)/(1-F)))
d(F)/d(z)=-E
lam=14-z
C1=0.0039+0.274*lam+1.57*lam^2-0.255*lam^3
C2=-33.4+37.2*lam-11.6*lam^2+1.7*lam^3-0.13*lam^4+0.005*lam^5-(7.7*10^(-5))*lam^6
C=if(lam<=4andlam>=0)then(C1)elseif(lam>4andlam<=14)then(C2)else0
E=C/51
rA=-k*Ca
rB=rA
rC=-rA
S=Cb/(Cc+0.00001)
k=0.7923
z(0)=0
z(f)=14
F(0)=0.999999
Ca0=0.132
Cb0=2.4618
Cc0=0
Ca(0)=0.132
Cb(0)=2.4618
Cc(0)=0
17-98
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
C
-6.134272
-6.134272
10.07874
0.0039
2
C1
-388.1601
-388.1601
9.95707
0.0039
3
C2
-6.134272
-33.4
10.58096
-33.4
4
Ca
0.132
-1.440473
1.867199
0.0094726
5
Ca0
0.132
0.132
0.132
0.132
6
Cb
2.4618
0.8893268
4.196999
2.339273
7
Cb0
2.4618
2.4618
2.4618
2.4618
8
Cc
0
-1.735199
1.572473
0.1225274
9
Cc0
0
0
0
0
10 E
-0.1202798
-0.1202798
0.1976223
7.647E-05
11 F
0.999999
0.1422428
1.146239
0.1422428
12 k
0.7923
0.7923
0.7923
0.7923
13 lam
14.
0
14.
0
14 rA
-0.1045836
-1.479382
1.141287
-0.0075051
15 rB
-0.1045836
-1.479382
1.141287
-0.0075051
16 rC
0.1045836
-1.141287
1.479382
0.0075051
17 S
2.462E+05
-174.3352
2.462E+05
19.09027
18 z
0
0
14.
14.
Differential equations
1 d(F)/d(z) = -E
2 d(Ca)/d(z) = -(-(rA/1)+(E*(Ca-Ca0)/(1-F)))
3 d(Cb)/d(z) = -(-(rB/1)+(E*(Cb-Cb0)/(1-F)))
4 d(Cc)/d(z) = -(-(rC/1)+(E*(Cc-Cc0)/(1-F)))
Explicit equations
1
lam = 14-z
2
C1 = 0.0039+0.274*lam+1.57*lam^2-0.255*lam^3
3
C2 = -33.4+37.2*lam-11.6*lam^2+1.7*lam^3-0.13*lam^4+0.005*lam^5-(7.7*10^(-5))*lam^6
4
C = if(lam<=4 and lam>=0) then (C1) else if(lam>4 and lam<=14) then (C2) else 0
5
E = C/51
6
k = 0.7923
7
rA = -k*Ca
8
rB = rA
9
rC = -rA
10 Ca0 = 0.132
11 Cb0 = 2.4618
12 Cc0 = 0
13 S = Cb/(Cc+0.00001)
17-99
P17-15 (c)
PFR:
X(t)=1-e-kt
¥
X = ò X A (t ) E (t )dt =1- e-0.7923*5.1 ~ 0.982
0
CB=CA0*(ΘB-XA)=0.132*(18.65-0.982)=2.332
CC= CA0*XA=0.1297
Selectivity= CB /CC=17.98
CSTR:
0.802
CB=CA0*(ΘB-XA)=0.132*(18.65-0.802)=2.356
CC= CA0*XA=0.106
Selectivity= CB /CC=22.255
See the following Polymath program:
Polymath Code:
d(Xbar)/d(t)=X*E
d(X)/d(t)=k*(1-X)
E=(1/tau)*exp(-t/tau)
Xbar(0)=0
k=0.7923
tau=5.1
t(0) = 0.0000001
t(f) = 100
X(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.1960784
5.982E-10
0.1960784
5.982E-10
2 k
0.7923
0.7923
0.7923
0.7923
3 t
1.0E-07
1.0E-07
100.
100.
4 tau
5.1
5.1
5.1
5.1
5 X
0
0
1.
1.
6 Xbar
0
0
0.801616
0.801616
Differential equations
1 d(Xbar)/d(t) = X*E
2 d(X)/d(t) = k*(1-X)
Explicit equations
1 tau = 5.1
2 k = 0.7923
3 E = (1/tau)*exp(-t/tau)
17-100
P17-16 (a)
T=1000C=1273.15 K
8000 1
1
(
)) =1140614.2
1.987 300 1273.15
12000 1
1
k2C= 2 exp(
(
)) =9630485.5
1.987 300 1273.15
k1A= 40 exp(
Segregation Model:
Exit Selectivity=0.558
See the following Polymath program:
Polymath Code:
d(Ca)/d(t)=ra
d(Cb)/d(t)=rb
d(Cc)/d(t)=rc
d(Cd)/d(t)=rd
d(Cabar)/d(t)=Ca*E
d(Cbbar)/d(t)=Cb*E
d(Ccbar)/d(t)=Cc*E
d(Cdbar)/d(t)=Cd*E
E2=-(2.64*10^(-9))*t^3+(1.3618*10^(-6))*t^2-0.00024069*t+0.015011
E1=(4.44658*10^(-10))*t^4-(1.1802*10^(-7))*t^3+(1.35358*10^(-5))*t^2-0.000865652*t+0.028004
E=if(t<=70)then(E1)else(E2)
ra=-k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
rb=-2*k1a*Ca*Cb^2
rc=k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
rd=(1/3)*k2c*(Ca^2)*(Cc^3)
k1a=1140614.2
k2c=9630485.5
S=Ccbar/(Cdbar+0.0000001)
Cabar(0)=0
Cbbar(0)=0
Ccbar(0)=0
Cdbar(0)=0
Ca(0)=0.05
Cb(0)=0.05
Cc(0)=0
Cd(0)=0
t(0)=0
t(f)=200
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.05
0.0093921
0.05
0.0093921
2
Cabar
0
0
0.0111656
0.0111656
3
Cb
0.05
2.175E-07
0.05
2.175E-07
4
Cbbar
0
0
6.362E-06
6.362E-06
5
Cc
0
0
0.0070758
0.0015878
6
Ccbar
0
0
0.0043527
0.0043527
7
Cd
0
0
0.007804
0.007804
8
Cdbar
0
0
0.0068065
0.0068065
9
E
0.028004
0.000225
0.028004
0.000225
17-101
10 E1
0.028004
0.0028744
0.1635984
0.1635984
11 E2
0.015011
0.000225
0.015011
0.000225
12 k1a
1.141E+06
1.141E+06
1.141E+06
1.141E+06
13 k2c
9.63E+06
9.63E+06
9.63E+06
9.63E+06
14 ra
-142.5768
-142.5768
-2.268E-06
-2.268E-06
15 rb
-285.1536
-285.1536
-1.013E-09
-1.013E-09
16 rc
142.5768
-0.0005809
142.5768
-3.4E-06
17 rd
0
0
0.0001939
1.134E-06
18 S
0
0
2.040885
0.6394888
19 t
0
0
200.
200.
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
4 d(Cd)/d(t) = rd
5 d(Cabar)/d(t) = Ca*E
6 d(Cbbar)/d(t) = Cb*E
7 d(Ccbar)/d(t) = Cc*E
8 d(Cdbar)/d(t) = Cd*E
Explicit equations
1
E2 = -(2.64*10^(-9))*t^3+(1.3618*10^(-6))*t^2-0.00024069*t+0.015011
2
E1 = (4.44658*10^(-10))*t^4-(1.1802*10^(-7))*t^3+(1.35358*10^(-5))*t^2-0.000865652*t+0.028004
3
E = if(t<=70) then (E1) else (E2)
4
k2c = 9630485.5
5
k1a = 1140614.2
6
rb = -2*k1a*Ca*Cb^2
7
ra = -k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
8
rc = k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
9
rd = (1/3)*k2c*(Ca^2)*(Cc^3)
10 S = Ccbar/(Cdbar+0.0000001)
P17-16 (b)
Maximum Mixedness Model:
Exit selectivity =1.12
See the following Polymath program:
Polymath Code:
d(Ca)/d(z)=-(-(ra/1)+(E*(Ca-Ca0)/(1-F)))
d(Cb)/d(z)=-(-(rb/1)+(E*(Cb-Cb0)/(1-F)))
d(Cc)/d(z)=-(-(rc/1)+(E*(Cc-Cc0)/(1-F)))
d(Cd)/d(z)=-(-(rd/1)+(E*(Cd-Cd0)/(1-F)))
d(F)/d(z)=-E
E1=(4.44658*10^(-10))*lam^4-(1.1802*10^(-7))*lam^3+(1.35358*10^(-5))*lam^2-0.000865652*lam+0.028004
E2=-(2.64*10^(-9))*lam^3+(1.3618*10^(-6))*lam^2-0.00024069*lam+0.015011
E=if(lam<=70)then(E1)else(E2)
lam=200-z
ra=-k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
rb=-2*k1a*Ca*Cb^2
rc=k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
17-102
rd=(1/3)*k2c*(Ca^2)*(Cc^3)
k1a=1140614.2
k2c=9630485.5
S = Cc/(Cd+0.000001)
Ca0=0.05
Cb0=0.05
Cc0=0
Cd0=0
z(0)=0
z(f)=200
Ca(0)=0.05
Cb(0)=0.05
Cc(0)=0
Cd(0)=0
F(0)=0.999999
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.05
0.0128578
0.05
0.0130281
2
Ca0
0.05
0.05
0.05
0.05
3
Cb
0.05
0.0002024
0.05
0.0002175
4
Cb0
0.05
0.05
0.05
0.05
5
Cc
0
0
0.011628
0.0067703
6
Cc0
0
0
0
0
7
Cd
0
0
0.0061218
0.0060403
8
Cd0
0
0
0
0
9
E
0.000225
0.000225
0.028004
0.028004
10 E1
0.1635984
0.0028728
0.1635984
0.028004
11 E2
0.000225
0.000225
0.015011
0.015011
12 F
0.999999
0.0089863
0.999999
0.0089863
13 k1a
1.141E+06
1.141E+06
1.141E+06
1.141E+06
14 k2c
9.63E+06
9.63E+06
9.63E+06
9.63E+06
15 lam
200.
0
200.
0
16 ra
-142.5768
-142.5768
-0.0008971
-0.0010414
17 rb
-285.1536
-285.1536
-0.0012013
-0.0014065
18 rc
142.5768
0.0001557
142.5768
0.000196
19 rd
0
0
0.001388
0.0001691
20 S
0
0
2.667759
1.120667
21 z
0
0
200.
200.
Differential equations
1 d(Ca)/d(z) = -(-(ra/1)+(E*(Ca-Ca0)/(1-F)))
2 d(Cb)/d(z) = -(-(rb/1)+(E*(Cb-Cb0)/(1-F)))
3 d(Cc)/d(z) = -(-(rc/1)+(E*(Cc-Cc0)/(1-F)))
4 d(Cd)/d(z) = -(-(rd/1)+(E*(Cd-Cd0)/(1-F)))
5 d(F)/d(z) = -E
17-103
Explicit equations
1
lam = 200-z
2
E1 = (4.44658*10^(-10))*lam^4-(1.1802*10^(-7))*lam^3+(1.35358*10^(-5))*lam^2-0.000865652*lam+0.028004
3
E2 = -(2.64*10^(-9))*lam^3+(1.3618*10^(-6))*lam^2-0.00024069*lam+0.015011
4
E = if(lam<=70) then (E1) else (E2)
5
k2c = 9630485.5
6
k1a = 1140614.2
7
rb = -2*k1a*Ca*Cb^2
8
ra = -k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
9
rc = k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
10 rd = (1/3)*k2c*(Ca^2)*(Cc^3)
11 S = Cc/(Cd+0.000001)
12 Ca0 = 0.05
13 Cb0 = 0.05
14 Cc0 = 0
15 Cd0 = 0
P17-16 (c)
PFR:
dC A
= rA
dt
dC B
= rB
dt
dCC
= rC
dt
dC D
= rD
dt
17-104
See the following Polymath program or calculating t:
Polymath Code:
d(tm)/d(t) = E*t
d(F)/d(t) = E
E2 = -(2.64*10^(-9))*t^3+(1.3618*10^(-6))*t^2-0.00024069*t+0.015011
E1 = (4.44658*10^(-10))*t^4-(1.1802*10^(-7))*t^3+(1.35358*10^(-5))*t^2-0.000865652*t+0.028004
E = if(t<=70)then(E1)elseif(t<=200)then(E2)else(0)
t(0)=0
t(f)=1000
F(0)=0
tm(0)=0
Output:
POLYMATH Report (For calculating t)
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0.028004
0
0.028004
0
2 E1
0.028004
0.0028736
339.3362
339.3362
3 E2
0.015011
-1.503879
0.015011
-1.503879
4 F
0
0
0.9910104
0.9910104
5 t
0
0
1000.
1000.
6 tm
0
0
37.52774
37.52774
Differential equations
1 d(tm)/d(t) = E*t
2 d(F)/d(t) = E
Explicit equations
1 E2 = -(2.64*10^(-9))*t^3+(1.3618*10^(-6))*t^2-0.00024069*t+0.015011
2 E1 = (4.44658*10^(-10))*t^4-(1.1802*10^(-7))*t^3+(1.35358*10^(-5))*t^2-0.000865652*t+0.028004
3 E = if(t<=70) then (E1) else if(t<=200)then (E2) else (0)
τ=37.53 min
See the following Polymath program:
Polymath Code:
d(Ca)/d(t)=ra
d(Cb)/d(t)=rb
d(Cc)/d(t)=rc
d(Cd)/d(t)=rd
tau=37.53
ra=-k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
rb=-2*k1a*Ca*Cb^2
rc=k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
rd=(1/3)*k2c*(Ca^2)*(Cc^3)
k1a=1140614.2
k2c=9630485.5
S=Cc/(Cd+0.0000001)
Ca(0)=0.05
Cb(0)=0.05
Cc(0)=0
Cd(0)=0
t(0) = 0.0000001
t(f) =37.53
17-105
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.05
0.0104381
0.05
0.0104381
2
Cb
0.05
1.007E-06
0.05
1.007E-06
3
Cc
0
0
0.0121458
0.0031559
4
Cd
0
0
0.0072812
0.0072812
5
k1a
1.141E+06
1.141E+06
1.141E+06
1.141E+06
6
k2c
9.63E+06
9.63E+06
9.63E+06
9.63E+06
7
ra
-142.5768
-142.5768
-2.2E-05
-2.2E-05
8
rb
-285.1536
-285.1536
-2.415E-08
-2.415E-08
9
rc
142.5768
-0.0046551
142.5768
-3.297E-05
10 rd
0
0
0.0015576
1.099E-05
11 S
0
0
2.837981
0.4334263
12 t
1.0E-07
1.0E-07
37.53
37.53
13 tau
37.53
37.53
37.53
37.53
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
4 d(Cd)/d(t) = rd
Explicit equations
1 k2c = 9630485.5
2 k1a = 1140614.2
3 rb = -2*k1a*Ca*Cb^2
4 ra = -k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
5 rc = k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
6 rd = (1/3)*k2c*(Ca^2)*(Cc^3)
7 tau = 37.53
8 S = Cc/(Cd+0.0000001)
Selectivity=0.4334
CSTR:
See the following Polymath program:
Polymath Code:
d(Ca)/d(t)=ra
d(Cb)/d(t)=rb
d(Cc)/d(t)=rc
d(Cd)/d(t)=rd
tau=37.53
ra=-k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
rb=-2*k1a*Ca*Cb^2
rc=k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
rd=(1/3)*k2c*(Ca^2)*(Cc^3)
k1a=1140614.2
k2c=9630485.5
17-106
S=Cc/(Cd+0.0000001)
Ca(0)=0.05
Cb(0)=0.05
Cc(0)=0
Cd(0)=0
t(0) = 0.0000001
t(f) =37.53
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.05
0.00871
0.05
0.00871
2
Cabar
0
0
0.0112397
0.0112397
3
Cb
0.05
2.435E-08
0.05
2.435E-08
4
Cbbar
0
0
6.15E-06
6.15E-06
5
Cc
0
0
0.0029787
0.000565
6
Ccbar
0
0
0.0043518
0.0043518
7
Cd
0
0
0.008145
0.008145
8
Cdbar
0
0
0.0068817
0.0068817
9
E
0.0266454
1.913E-25
0.0266454
1.913E-25
10 k1a
1.141E+06
1.141E+06
1.141E+06
1.141E+06
11 k2c
9.63E+06
9.63E+06
9.63E+06
9.63E+06
12 ra
-142.5768
-142.5768
-8.783E-08
-8.783E-08
13 rb
-285.1536
-285.1536
-1.178E-11
-1.178E-11
14 rc
142.5768
-2.71E-05
142.5768
-1.317E-07
15 rd
0
0
9.035E-06
4.391E-08
16 S
0
0
0.7944665
0.6322882
17 t
1.0E-07
1.0E-07
2000.
2000.
18 tau
37.53
37.53
37.53
37.53
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
4 d(Cd)/d(t) = rd
5 d(Cabar)/d(t) = Ca*E
6 d(Cbbar)/d(t) = Cb*E
7 d(Ccbar)/d(t) = Cc*E
8 d(Cdbar)/d(t) = Cd*E
Explicit equations
1 tau = 37.53
2 k2c = 9630485.5
3 k1a = 1140614.2
4 rb = -2*k1a*Ca*Cb^2
5 ra = -k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
6 rc = k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
17-107
7 rd = (1/3)*k2c*(Ca^2)*(Cc^3)
8 S = Ccbar/(Cdbar+0.000001)
9 E = (1/tau)*exp(-t/tau)
Selectivity=0.6323
P17-16 (d)
Segregation model:
See the following Polymath program:
Polymath Code:
d(Ca)/d(t)=ra
d(Cb)/d(t)=rb
d(Cc)/d(t)=rc
d(Cd)/d(t)=rd
d(Cabar)/d(t)=Ca*E
d(Cbbar)/d(t)=Cb*E
d(Ccbar)/d(t)=Cc*E
d(Cdbar)/d(t)=Cd*E
E=if(t<=10)then(5*t)elseif(t>=20)then(0)else(100-5*t)
ra=-k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
rb=-2*k1a*Ca*Cb^2
rc=k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
rd=(1/3)*k2c*(Ca^2)*(Cc^3)
k1a=1140614.2
k2c=9630485.5
S=Ccbar/(Cdbar+0.00001)
Cabar(0)=0
Cbbar(0)=0
Ccbar(0)=0
Cdbar(0)=0
Ca(0)=0.05
Cb(0)=0.05
Cc(0)=0
Cd(0)=0
t(0)=0
t(f)=200
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.05
0.0093921
0.05
0.0093921
2
Cabar
0
0
6.001059
6.001059
3
Cb
0.05
2.175E-07
0.05
2.175E-07
4
Cbbar
0
0
0.0020931
0.0020931
5
Cc
0
0
0.0070322
0.0015878
6
Ccbar
0
0
2.749143
2.749143
7
Cd
0
0
0.007804
0.007804
8
Cdbar
0
0
3.249823
3.249823
9
E
0
0
49.7432
0
10 k1a
1.141E+06
1.141E+06
1.141E+06
1.141E+06
11 k2c
9.63E+06
9.63E+06
9.63E+06
9.63E+06
12 ra
-142.5768
-142.5768
-2.268E-06
-2.268E-06
13 rb
-285.1536
-285.1536
-1.013E-09
-1.013E-09
17-108
14 rc
142.5768
-0.0005676
142.5768
-3.4E-06
15 rd
0
0
0.0001895
1.134E-06
16 S
0
0
1.516931
0.8459334
17 t
0
0
200.
200.
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
4 d(Cd)/d(t) = rd
5 d(Cabar)/d(t) = Ca*E
6 d(Cbbar)/d(t) = Cb*E
7 d(Ccbar)/d(t) = Cc*E
8 d(Cdbar)/d(t) = Cd*E
Explicit equations
1 E = if(t<=10) then (5*t) else if (t>=20) then (0) else ( 100-5*t)
2 k2c = 9630485.5
3 k1a = 1140614.2
4 rb = -2*k1a*Ca*Cb^2
5 ra = -k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
6 rc = k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
7 rd = (1/3)*k2c*(Ca^2)*(Cc^3)
8 S = Ccbar/(Cdbar+0.00001)
Selectivity=0.846
Maximum Mixedness Model:
See the following Polymath program:
Polymath Code:
d(Ca)/d(z)=-(-(ra/1)+(E*(Ca-Ca0)/(1-F)))
d(Cb)/d(z)=-(-(rb/1)+(E*(Cb-Cb0)/(1-F)))
d(Cc)/d(z)=-(-(rc/1)+(E*(Cc-Cc0)/(1-F)))
d(Cd)/d(z)=-(-(rd/1)+(E*(Cd-Cd0)/(1-F)))
d(F)/d(z)=-E
E=if(lam<=10)then(5*lam)elseif(lam>=20)then(0)else(100-5*lam)
lam=20-z
ra=-k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
rb=-2*k1a*Ca*Cb^2
rc=k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
rd=(1/3)*k2c*(Ca^2)*(Cc^3)
k1a=1140614.2
k2c=9630485.5
S=Cc/(Cd+0.00001)
Ca(0)=0.05
Cb(0)=0.05
Cc(0)=0
Cd(0)=0
z(0)=0
z(f)=20
Ca0=0.05
Cb0=0.05
Cc0=0
Cd0=0
F(0)=0.999999
17-109
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.05
0.012881
0.05
0.012881
2
Ca0
0.05
0.05
0.05
0.05
3
Cb
0.05
6.046E-05
0.05
6.046E-05
4
Cb0
0.05
0.05
0.05
0.05
5
Cc
0
0
0.0187435
0.006746
6
Cc0
0
0
0
0
7
Cd
0
0
0.0060746
0.0060746
8
Cd0
0
0
0
0
9
E
0
0
49.98469
0
10 F
0.999999
-499.
0.999999
-499.
11 k1a
1.141E+06
1.141E+06
1.141E+06
1.141E+06
12 k2c
9.63E+06
9.63E+06
9.63E+06
9.63E+06
13 lam
20.
0
20.
0
14 ra
-142.5768
-142.5768
-0.0003807
-0.0003807
15 rb
-285.1536
-285.1536
-0.0001074
-0.0001074
16 rc
142.5768
-0.0004368
142.5768
-0.0004368
17 rd
0
0
0.0108236
0.0001635
18 S
0
0
10.79879
1.108696
19 z
0
0
20.
20.
Differential equations
1 d(Ca)/d(z) = -(-(ra/1)+(E*(Ca-Ca0)/(1-F)))
2 d(Cb)/d(z) = -(-(rb/1)+(E*(Cb-Cb0)/(1-F)))
3 d(Cc)/d(z) = -(-(rc/1)+(E*(Cc-Cc0)/(1-F)))
4 d(Cd)/d(z) = -(-(rd/1)+(E*(Cd-Cd0)/(1-F)))
5 d(F)/d(z) = -E
Explicit equations
1
lam = 20-z
2
E = if(lam<=10) then (5*lam) else if (lam>=20) then (0) else ( 100-5*lam)
3
k2c = 9630485.5
4
k1a = 1140614.2
5
rb = -2*k1a*Ca*Cb^2
6
ra = -k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
7
rc = k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
8
rd = (1/3)*k2c*(Ca^2)*(Cc^3)
9
S = Cc/(Cd+0.00001)
10 Ca0 = 0.05
11 Cb0 = 0.05
12 Cc0 = 0
13 Cd0 = 0
Selectivity=1.109
17-110
PFR:
Calculating τ:
See the following Polymath program for calculating t:
Polymath Code:
E = if(t<=10)then(5*t/500)elseif(t>=20)then(0)else((100-5*t)/500)
d(tm)/d(t)=E*t
d(F)/d(t)=E
t(0)=0
t(f)=1000
F(0)=0
tm(0)=0
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.0996511
0
2 F
0
0
0.9999986
0.9999986
3 t
0
0
20.
20.
4 tm
0
0
9.999986
9.999986
Differential equations
1 d(tm)/d(t) = E*t
2 d(F)/d(t) = E
Explicit equations
1 E = if(t<=10) then (5*t/500) else if (t>=20) then (0) else ( (100-5*t)/500)
Thus τ=10
See the following Polymath program for calculating selectivity:
Polymath Code:
d(Ca)/d(t) = ra
d(Cb)/d(t) = rb
d(Cc)/d(t) = rc
d(Cd)/d(t) = rd
k2c = 9630485.5
k1a = 1140614.2
rb = -2*k1a*Ca*Cb^2
ra = -k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
rc = k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
rd = (1/3)*k2c*(Ca^2)*(Cc^3)
tau = 10
S = Cc/(Cd+0.0000001)
t(0)=0
t(f)=10
Ca(0)=0.05
Cb(0)=0.05
Cc(0)=0
Cd(0)=0
17-111
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.05
0.0117996
0.05
0.0117996
2
Cb
0.05
3.26E-06
0.05
3.26E-06
3
Cc
0
0
0.0168424
0.0051954
4
Cd
0
0
0.006601
0.006601
5
k1a
1.141E+06
1.141E+06
1.141E+06
1.141E+06
6
k2c
9.63E+06
9.63E+06
9.63E+06
9.63E+06
7
ra
-142.5768
-142.5768
-0.0001255
-0.0001255
8
rb
-285.1536
-285.1536
-2.861E-07
-2.861E-07
9
rc
142.5768
-0.0175509
142.5768
-0.0001879
10 rd
0
0
0.0059155
6.268E-05
11 S
0
0
6.229288
0.7870518
12 t
1.0E-07
1.0E-07
10.
10.
13 tau
10.
10.
10.
10.
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
4 d(Cd)/d(t) = rd
Explicit equations
1 k2c = 9630485.5
2 k1a = 1140614.2
3 rb = -2*k1a*Ca*Cb^2
4 ra = -k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
5 rc = k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
6 rd = (1/3)*k2c*(Ca^2)*(Cc^3)
7 tau = 10
8 S = Cc/(Cd+0.0000001)
Selectivity=0.787
CSTR:
See the following Polymath program for calculation of selectivity:
Polymath Code:
d(Ca)/d(t)=ra
d(Cb)/d(t)=rb
d(Cc)/d(t)=rc
d(Cd)/d(t)=rd
d(Cabar)/d(t)=Ca*E
d(Cbbar)/d(t)=Cb*E
d(Ccbar)/d(t)=Cc*E
d(Cdbar)/d(t)=Cd*E
E=(1/tau)*exp(-t/tau)
ra=-k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
rb=-2*k1a*Ca*Cb^2
17-112
rc=k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
rd=(1/3)*k2c*(Ca^2)*(Cc^3)
k1a=1140614.2
k2c=9630485.5
Cabar(0)=0
Cbbar(0)=0
Ccbar(0)=0
Cdbar(0)=0
Ca(0)=0.05
Cb(0)=0.05
Cc(0)=0
Cd(0)=0
S=Ccbar/(Cdbar+0.000001)
tau=10
t(0) = 0.0000001
t(f) =2000
Output:
POLYMATH Report
Ordinary Differential Equations
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
0.05
0.0088521
0.05
0.0088521
2
Cabar
0
0
0.0129044
0.0129044
3
Cb
0.05
4.746E-08
0.05
4.746E-08
4
Cbbar
0
0
1.871E-05
1.871E-05
5
Cc
0
0
0.0039762
0.000778
6
Ccbar
0
0
0.0068333
0.0068333
7
Cd
0
0
0.008074
0.008074
8
Cdbar
0
0
0.0060525
0.0060525
9
E
0.1
3.72E-45
0.1
3.72E-45
10 k1a
1.141E+06
1.141E+06
1.141E+06
1.141E+06
11 k2c
9.63E+06
9.63E+06
9.63E+06
9.63E+06
12 ra
-142.5768
-142.5768
-2.37E-07
-2.37E-07
13 rb
-285.1536
-285.1536
-4.549E-11
-4.549E-11
14 rc
142.5768
-7.303E-05
142.5768
-3.554E-07
15 rd
0
0
2.436E-05
1.185E-07
16 S
0
0
1.242945
1.128818
17 t
0
0
1000.
1000.
18 tau
10.
10.
10.
10.
Differential equations
1 d(Ca)/d(t) = ra
2 d(Cb)/d(t) = rb
3 d(Cc)/d(t) = rc
4 d(Cd)/d(t) = rd
5 d(Cabar)/d(t) = Ca*E
6 d(Cbbar)/d(t) = Cb*E
7 d(Ccbar)/d(t) = Cc*E
8 d(Cdbar)/d(t) = Cd*E
17-113
Explicit equations
1 tau = 10
2 k2c = 9630485.5
3 k1a = 1140614.2
4 rb = -2*k1a*Ca*Cb^2
5 ra = -k1a*Ca*Cb^2-(2/3)*k2c*(Ca^2)*(Cc)^3
6 rc = k1a*Ca*Cb^2-k2c*(Ca^2)*(Cc)^3
7 rd = (1/3)*k2c*(Ca^2)*(Cc^3)
8 E = (1/tau)*exp(-t/tau)
9 S = Ccbar/(Cdbar+0.000001)
Selectivity=1.1288
P17-17(a)
P17-17 (b)
17-114
P17-17 (c)
P17-17 (d)
17-115
P17-17 (e)
P17-17 (f)
P17-17 (g)
17-116
See the following Polymath Code:
d(Xseg)/d(t)=X*E
d(X)/d(t)=k*Ca0*(1-X)^2
Ca0=1
k=0.1
Ca=Ca0*(1-X)
E1=-0.0011721*t^4+0.0113999*t^3-0.0476813*t^2+0.0999015*t
E2=2.79737e-8*t^4-4.94398e-6*t^3+0.000326287*t^2-0.00968055*t+0.111693
E=if(t<=3)then(E1)else(E2)
t(0)=0
t(f)=60
X(0)=0
Xseg(0)=0
Output
Variable Initial value Minimal value Maximal value Final value
1 Ca
1.
0.1428571
1.
0.1428571
2 Ca0
1.
1.
1.
1.
3 E
0
-0.000316
0.0853663
0.0001327
4 E1
0
-1.289E+04
0.0841716
-1.289E+04
5 E2
0.111693
-0.000316
0.111693
0.0001327
6 k
0.1
0.1
0.1
0.1
7 t
0
0
60.
60.
8 X
0
0
0.8571429
0.8571429
9 Xseg
0
0
0.4227603
0.4214206
Differential equations
1 d(Xseg)/d(t) = X*E
2 d(X)/d(t) = k*Ca0*(1-X)^2
Explicit equations
1 Ca0 = 1
2 k = 0.1
3 Ca = Ca0*(1-X)
4 E1 = -0.0011721*t^4+0.0113999*t^3-0.0476813*t^2+0.0999015*t
5 E2 = 2.79737e-8*t^4-4.94398e-6*t^3+0.000326287*t^2-0.00968055*t+0.111693
6 E = if(t<=3)then(E1)else(E2)
Xseg= 42.1 %
P17-17 (h)
See the following Polymath program:
Polymath Code:
d(X)/d(z)=-ra/Ca0-E*X/(1-F)
d(F)/d(z)=-E
17-117
Ca0=1
k=0.1
lam=60-z
Ca=Ca0*(1-X)
E1=-0.0011721*lam^4+0.0113999*lam^3-0.0476813*lam^2+0.0999015*lam
E2=2.79737e-8*lam^4-4.94398e-6*lam^3+0.000326287*lam^2-0.00968055*lam+0.111693
ra=-k*Ca^2
E=if(lam<=3)then(E1)else(E2)
z(0)=0
z(f)=60
X(0)=0
F(0)=0.99
Output
Variable Initial value Minimal value Maximal value Final value
1 Ca
1.
0.441799
1.
0.5975925
2 Ca0
1.
1.
1.
1.
3 E
0.0001327
-0.0003165
0.0840493
0
4 E1
-1.289E+04
-1.289E+04
0.0835502
0
5 E2
0.0001327
-0.0003165
0.111693
0.111693
6 F
0.995
-0.0061825
0.9965758
-0.0061825
7 k
0.1
0.1
0.1
0.1
8 lam
60.
0
60.
0
9 ra
-0.1
-0.1
-0.0195186
-0.0357117
10 X
0
0
0.558201
0.4024075
11 z
0
0
60.
60.
Differential equations
1 d(X)/d(z) = -(ra/Ca0+E*X/(1-F))
2 d(F)/d(z) = -E
Explicit equations
1 Ca0 = 1
2 k = 0.1
3 lam = 60-z
4 Ca = Ca0*(1-X)
5 E1 = -0.0011721*lam^4+0.0113999*lam^3-0.0476813*lam^2+0.0999015*lam
6 E2 = 2.79737e-8*lam^4-4.94398e-6*lam^3+0.000326287*lam^2-0.00968055*lam+0.111693
7 ra = -k*Ca^2
8 E = if(lam<=3)then(E1)else(E2)
Thus Xmm= 40.2 %
P17-18 (a) (b)
The tracer experiment for the poor operation indicate that initially there is channeling in the bed,
corresponding to the rapid peak in concentration. After the peak the concentration falls as the tracer flow
is hindered by compacted packing in the other parts of the bed.
P
The mean conversion for RTD analysis is 𝑋W = ∫, \1 − 𝑒 %^._ ` . 𝐸(𝑡). 𝑑𝑡
𝐸 (𝑡) =
𝐶 (𝑡)
P
∫, 𝐶 (𝑡)
17-118
Using E(t) data from the flow conditions we get the following mean conversion after evaluating the above
integral.
XA values for different operations
Flow condition
Poor operation
Good operation
k1 = 0.1 hr–1
0.1
0.4
k2 = 2 hr–1
0.364
0.518
P17-18 (c)
Results point to the importance of flow condition in achieving high conversion.
17-119
Page intentionally blank
17-120
Synopsis for Chapter 18 – Models for Nonideal Reactors
General: This chapter uses parameters calculated from the RTD to predict conversion for one and two
parameter models.
Questions
l Q18-1A (21 seconds) Questions Before Reading (QBR).
G Q18-2B (12 min) Good problem to find tm and sigma squared and E(t) and F(t).
S Q18-3A (7-8 min) Quick calculation to get value for the dispersion coefficient.
I Q18-4 (10 min) Good question as it leads the reader through how to make a simplification to obtain
an analytical solution and avoid writing a computer program to determine the solution.
AA Q18-5 (6-12 min) See Extra Help on the Web site for recommendations for screencasts.
Computer Simulations and Experiments
l P18-1B (a), (b), (e) & (f) (8 min per simulation). Use Wolfram to better understand the effects of the
system parameters. I always assign all Wolfram problems.
O,G P18-1B (55 min) Parts (c) COMSOL LEP. Always assigned for the graduate CRE course.
O,G P18-1B (35 min) Parts (d) COMSOL often assigned in the graduate CRE course.
Problems
AA P18-2B (55 min) Old Exam Questions (OEQ). Challenging problem.
l P18-3B (30 min) Old Exam Questions (OEQ). Very interesting problem with simple calculations using
Big Ten sports team colors.
AA P18-4B (45 min) Old Exam Questions (OEQ). Challenging problem as students need to find the
parameter values from chapter correlations or figures.
AA P18-5A (55 min) Fairly straight forward problem.
AA P18-6A (40 min) Old Exam Questions (OEQ). Good problem, straight forward.
l P18-7A (25 min) Old Exam Questions (OEQ). Short problem, straight forward.
O P18-8B (a) (25 min) Is very straight forward. (b) (20 min) Does not relate to part (a).
l P18-9B (30 min) Good problem, short and straight forward.
AA,G P18-10B (45 min) Requires creativity to develop a model. Calculations are straight forward.
AA,G P18-11B (30 min) COMSOL conceptual problem.
AA,G P18-12D (90 min) Comprehensive problem requiring some thinking as reaction is second order.
Relatively difficult. Professor Rane Curl (1929-2019) of the University of Michigan, Chemical
Engineering, developed this rather difficult problem.
O P18-13B (90 min) Creativity needed to develop model and evaluate model parameters.
O P18-14B (40 min) Requires some creativity to match reaction to P16-2B(a). Oops, should have read
P16-2B(a), not (b).
S18-1
l P18-15B (30 min) Old Exam Questions (OEQ).Short problem to test understanding of concepts.
AA,G P18-16C (70 min) Straight forward once you find a and b.
AA P18-17B (35 min) Old Exam Questions (OEQ). Straight forward.
AA P18-18B (45 min) Old Exam Questions (OEQ).Straight forward.
AA P18-19B (30 min) Old Exam Questions (OEQ). Relatively straight forward and made very easy if one
remembers Xseg = Xmm = XTIS for a first order reactor.
AA,G P18-20C (70 min) Evolves expanding terms in analytical solution.
Key to Nomenclature
l = Always assigned
AA = Always assign one from the group of
alternates
O = Often assigned
I = Infrequently assigned
S = Seldom assigned
G = Graduate level
N = Never assigned
E.g., means problem l P1-3B will be assigned every time I teach the course, problem AA P1-8 means that
this problem or one of the other problems with the prefix AA is always assigned for this chapter, Problem
l P1-2 will be infrequently assigned, Problem O P1-6B will often be assigned, and Problem S P3-16B is
seldom assigned.
Alternates: In problems that have a dot in conjunction with AA means that one of the problems, either
the problem with a dot or any one of the alternates are always assigned.
Time: Approximate time in minutes it would take a B student to solve the problem.
S18-2
Solutions for Chapter 18 – Models for Nonideal Reactors
Useful Links:
1. Click on the link given below to download Wolfram/python codes for Ch-18
http://umich.edu/~elements/5e/18chap/live.html
2. Click on the link given below to view Wolfram tutorial (for running Wolfram Codes)
http://umich.edu/~elements/5e/software/Wolfram_LEP_tutorial.pdf
3. Click on the link given below to view Polymath tutorial (for running Polymath Codes)
http://umich.edu/~elements/5e/tutorials/Polymath_LEP_tutorial.pdf
Q18-1 Individualized solution.
Q18-2 Individualized solution.
Q18-3 Re = Udp/v = 0.33
For Sc = 0.35, D𝜑/Udp = 1 (Figure 18-10)
So, D = 20 cm2/s (Note: This solution is for a PBR)
Q18-4 This can be a good approximation if CA doesn’t vary much in which case we don’t need to divide
CA0 by something. Linearizing other non-first order reactions may give inaccurate results also. This can
be tested by comparing with conversion found using other models like TIS.
Q18-5 Individualized solution.
Q18-6 Individualized solution.
P18-1 (a)
(i)
As Peclet number is varied from its largest to smallest value, outlet concentration for “reaction
with dispersion” deviates more from “Ideal PFR” case.
(ii) At space time = 16min, conversion reaches 95% for reaction with dispersion.
(iii) At Peclet no = 16.5, conversion reaches 90%.
(iv) 1. For large values of Peclet no., there is virtually no dispersion and the concentration and
conversion profile approaches that of a PFR.
2. Conversion increases with an increase in reaction rate constant and space time constant.
P18-1 (b)
(i)
Increase the average velocity (U) so that conversion by dispersion model approaches to that of
ideal PFR model.
(ii) The minimum no of tanks required so that conversion obtained by tank in series is just greater
than dispersion model is 6.
18-1
(iii)
1. Reaction rate constant is the only parameter that affects the conversion in all cases.
2. Conversion with dispersion decreases as Peclet no decreases. But we cannot decrease that
conversion below as compared to conversion with a CSTR.
3. Space time and k have greatest effects on conversion with dispersion when we observe the
graph of conversion along reactor length.
P18-1 (c)
Students can go to COMSOL link LEP 18-3 on
http://www.umich.edu/~elements/5e/comsol/comsol_access_instructions.html
Use the parameter values as in example 18-3 with 2nd order reaction
(i)
For plug flow reactor, if velocity is increased, the concentration at a fixed axial position also
increases since there would be less residence time. The concentration is uniform in radial direction
and increases with velocity. For laminar flow reactor, if velocity is increased, concentration increases
in both radial and axial direction and radial concentration profile is not uniform.
(ii) For plug flow reactor, reactor radius doesn’t have impact on profiles. For laminar flow reactor, if
radius is increased, then outlet axial concentration decreases.
(iii) For plug flow reactor, on increasing diffusivity, exit concentration increases. Also the
concentration increases in radial concentration profile at a fixed axial distance. For laminar flow
reactor, on increasing diffusivity exit concentration increases and radial concentration profile
becomes flat
(iv) For plug flow reactor, since the concentration is uniform in radial direction, the values obtained in
radial concentration profile and radially averaged concentration profile is same. For laminar flow
reactor, if DAB or velocity is increased, then radially averaged profile and radial concentration
profile (at outlet) matches with each other.
(v) If DAB is increased then velocity needs to be reduced to obtain same graph
(vi) Parameters Dispersion Model
Closed-Closed dispersion model
X = 1−
( )
2
2
(1+ q) exp(Per q / 2)− (1− q) exp(−Per q / 2)
4qexp Per / 2
Where q = 1+ 4Da / Per
Da = kτ = k
Per =
l
= Damköhler number
U
Ul
= Peclet number
Da
Per q = Pe22 + 4DaPer
Where DaPer =
kl 2 Rateof consumptionof Aby reaction
=
Da
Rateof transport by diffusion
Numerical example
L
= const = 5.15min
U
L = τ *U = 5.15min*0.1cm / s = 0.309m
τ=
Sc =
µ
= 1000 (Liquids region in Fig. 14.11)
DAB
Da = kτ = 1.288
18-2
dt
1 cm
1 dm
1m
Re
10
100
1000
Per
172
0.077
1.03·10-4
ReSc
104
105
106
L/dt
30.9
3.09
0.309
Q
1.015
8.226
223.609
X
0.721
0.567
0.563
D/(U*dt)
0.18 From Fig14.10
40 From Fig. 14.11
3000 From Fig. 14.11
D (cm2/s)
0.018
40
30000
Yes, there will be a radius that maximizes the conversion.
(vii) The exit concentration is higher in plug flow velocity profile
(viii) Two parameters that reduces radial variation are increasing diffusivity and increasing velocity
(ix) Radius has least effect on radial variation in concentration
(x) Increasing diffusivity increases outlet concentration
(xi) Effect of Dispersion on outlet concentration is more important at higher reaction orders
(xii) (a) Dispersion reduces the conversion and thus exit concentration increases. (b) Increasing flow
velocity reduces conversion
P18-1 (d)
(i)
If UA is increased, then temp near reactor wall decreases and so the conversion also decreases near
the wall. At very high value of UA, the temp near the wall would be closer to coolant temp and there
would be very little conversion. Since the temp is already near coolant temp, any further increase
in UA will not help in decreasing temp as LMTD would be too low.
(ii) Increasing diffusivity increases outlet conversion. So, diffusivity should be increased
(iii) The required parameter is activation energy. The value of activation energy is 74,700 J/mol at which
maximum of radial conversion becomes 1.
(iv) The conversion is higher in plug flow reactor with U=Uo. The temp is higher in Plug flow profile
(v) With diffusion effect, the conversion increases at location close to wall
P18-1 (e)
(i)
Outlet concentration increases as we increase bypass volume fraction, β, approaching to ideal
CSTR conc when beta goes to 0.
(ii) Outlet concentration decreases as we increase alpha and at alpha =1, both conc curves meet.
(iii) Keeping alpha=1 and Beta=0.5, we get outlet conc= 1.1 kmol/m3. However, if Beta=0 and alpha
=0.5, out conc= 0.48 kmol/m3 which is closer to ideal case. So Beta has more effect
(iv) 1. Concentration and conversion for ideal CSTR is just affected by rate constant k. X increases
while Ca decreases as we increase rate constant k.
2. The only case when both of the concentration and conversion curves meet are when there is
no dead volume (alpha=1) and no bypass volume (beta=0).
P18-1 (f)
(i)
For a fixed value of alpha, increases in beta increases conversion. This is because now there is
more material transfer in between two CSTR resulting in higher conversion.
(ii) As we increase alpha, the models turns to that of an ideal CSTR. Hence with an increase in alpha,
the volume of CSTR with higher agitation increases and hence conversion increases, thereby
decreasing final concentration.
(iii) With an increase in rate constant k, exit conc of both ideal CSTR as well as CSTR with an
interchange decreases.
(iv) 1. Alpha has greater effect on both concentration as well as conversion profiles as compared to
beta. Both alpha and beta have a positive effect on conversion profile.
18-3
2. Concentration and conversion for ideal CSTR is just affected by rate constant k. X increases
while Ca decreases as we increase rate constant k.
3. 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.
P18-2 (a)
+
𝑡# = ∫, 𝑡 ∗ 𝐸(𝑡)𝑑𝑡
+
𝑡# = 10 ∫, 𝑡 ∗ 𝑒 01,2 𝑑𝑡 = 0.1 𝑚𝑖𝑛
𝑡# = 𝜏 = 0.1 𝑚𝑖𝑛
𝑉 = 𝜏 ∗ 𝑣, = 0.1 ∗ 2 = 0.2 𝑑𝑚 <
P18-2 (b)
The given E(t) looks like that of a CSTR: 𝐸 (𝑡) =
When 𝑣, = 0.2 𝑑𝑚E /𝑚𝑖𝑛
𝜏=
= >?/A
B
=
=
C
>?/( )
CD
1/1,
𝑉 0.2
=
=1
𝑣, 0.2
Assuming 1st Order reaction, For a CSTR
𝜏𝑘 =
𝑋
1−𝑋
For X=0.5, k=1 min-1
For 𝑣, = 0.02 𝑑𝑚E /𝑚𝑖𝑛
𝜏=
𝑉
0.2
=
= 10
𝑣, 0.02
For X=0.91, k=1 min-1
For both the cases, k= 1 min-1, so the reaction is indeed first order
When 𝑣, = 10 𝑑𝑚E /𝑚𝑖𝑛
𝜏=
𝑉 0.2
=
= 0.02 𝑚𝑖𝑛
𝑣, 10
𝑋=1−
1
1
= 1−
= 0.02
1 + 𝜏𝑘
1 + 0.02
X= 20 %
P18-2 (c)
The reaction is first order as shown in part (b). For 1st Order PFR reaction,
ln(1 − 𝑋) = −𝑘𝜏
𝜏=
𝑉 1
= =1
𝑣, 1
ln(1 − 𝑋) = −1
𝑋 = 63.2 %
18-4
P18-2 (d)
At U=1 m/s, L= 6 m, dp =10 m, Re= 106, from fig 18.9
D/Udt =1/3
Or Da= 10/3
𝑈𝐿 18
=
= 1.8
𝐷𝑎 10
𝐿
6
1
𝐷V1 = 𝜏𝑘 = 1 ∗ =
=
𝑈 60 10
𝑃=P =
𝑞 = X1 +
4 ∗ 𝐷V1
4
= X1 +
= 1.1
𝑃=P
10 ∗ 1.8
4𝑞 exp (𝑃=P /2)
𝑃
𝑞
𝑃 𝑞
<
^
]− =P ^
(1 + 𝑞)< exp ] =P
2 − (1 − 𝑞) exp
2
4 ∗ 1.1 ∗ 2.46
10.82
𝑋=1−
= 1−
= 0.087
4.41 ∗ 2.69 − 0.01 ∗ 0.37
11.86
𝑋=1−
𝑋 = 8.7 %
P18-3 (a)
Money for buying reactors
Using the tank in series model:
2Da +1− 4Da +1
where Da=kτCao
2Da
C −C
C A = C Ao 1− X
X = Ao A
C Ao
Second order reaction X =
(
)
τ
Assume that τ = τ t and that in reactors modeled as more than one tank, that τ = t . Number of tanks
n
n=
τ2
rounded to the nearest integer.
σ2
Reactor
Σ(min) τ(min) n
Maze & blue
2
2
1
Green & white
4
4
1
Scarlet & grey
3.05
4
2
Orange & blue
2.31
4
3
Purple & white
5.17
4
1
Silver & black
2.5
4
3
Crimson & white 2.5
2
1
X
0.50
0.61
0.69
0.72
0.61
0.72
0.5
Where
Scarlet & grey: X1 = 0.5, CA1=0.5 ® X2 = 0.38, CA2=0.31 ® X = 0.69
Orange & blue: X1 = 0.43, CA1=0.57 ® X2 = 0.34, CA2=0.38 ® X3 = 0.27,
CA3=0.28 ® X=0.72
Using the combination of maze & blue followed by crimson & white reactor (same overall conversion
either way)
X1 = 0.5, CA1=0.5 ® X2 = 0.17, CA2 = 0.41 ® X=0.59
18-5
The orange & blue or silver & black reactors which both approximate to 3 tanks in series give the greatest
conversion.
P18-3 (b)
More money for buying reactors
Try:
Green & white and Maze & blue: X1 = 0.61, CA1=0.39 ® X2 = 0.34, CA2=0.26 ® X = 0.74
Scarlet & grey and Maze & blue: X1 = 0.69, CA1=0.31 ® X2 = 0.42, CA2=0.18 ® X = 0.82
Orange & blue and Maze & blue: X1 = 0.72, CA1=0.28 ® X2 = 0.40, CA2=0.17 ® X = 0.83
The highest conversion is now obtained from the Orange & blue reactor combined with the Maze & blue
reactor.
P18-3 (c)
Ann Arbor, MI
East Lansing, MI
Columbus, OH
Urbana, IL
Evanston, IL
West Lafayette, IN
Madison, WI
P18-4
Packed bed reactor with dispersion
1st order, k1=0.0167/s, ε=0.5, dp=0.1 cm,
µ
ν = = 0.01cm2 / s L=10 cm, U= 1 cm/s
ρ
Re =
ρUd p
µ
= 10 and Sc =
ν
no data concerning DAB
DAB
From packed bed correlation for Da , and liquid phase region of graph,
Gives
Per =
ab c
def
= 2 (𝑎𝑝𝑝𝑟𝑜𝑥) ® Da =
2Ud p
ε
=
2*1*0.1
= 0.4cm2 / s
0.5
UL 1*10
=
= 25
Da
0.4
X = 1−
)
( −Pe q +%
"
% "
2
2
r
-'
$(1+ q) exp (Per q 2)' − $(1− q) exp**
2 -,'&
$
#
& #
)
q = 1+
(
4qexp Per
2
4Da
Per
Da=τκ and τ =
L 10
= = 10s ® Da=0.167 and q=1.013
U 1
X = 0.15
Conversion X=15%.
18-6
P18-5 (a)
Number of tanks in series
Assuming the Peclet-Bodenstein relation: n =
Where Bo =
Bo
+1
2
UL
Da
d u 5*2
ν
0.01
=
=2
To estimate Bo, Re = t =
= 1000 and Sc =
DAB 0.005
v
0.01
From gas phase dispersion correlation chart,
Da
udt
=8
2
Gives Da = 80cm / s
Bo =
2*200
=5
80
5
n = +1 = 3.5
2
P18-5 (b)
Conversion
Using individual reactor material balances:
Reactor 1:
rV
Mole balance: X = − A
FAo
2
Rate law: −rA = kC A
(
)
Stoichiometry: C A = C Ao 1− X1 , d=0 and e=0 hence no volume change
)<
𝑘 ∗ 𝐶l, ∗ (1 − 𝑋1 ∗ 𝑉 25 ∗ 0.01 ∗ (1 − 𝑋1 )< ∗ 𝑉
𝑋1 =
=
𝑣,
𝑣,
𝑉
𝐿 200
𝜏=
= =
= 100𝑠
𝑣, 𝑈
2
𝜏 100
𝜏o = =
= 28.6 𝑠
𝑛 3.5
𝑋1 = 7.15 ∗ (1 − 𝑋1 )<
𝑋1 = 0.689
𝐶l1 = 𝐶l, (1 − 𝑋1 ) = 0.0031 𝑚𝑜𝑙/𝑑𝑚E
Reactor 2:
𝑘 ∗ 𝐶l1 ∗ (1 − 𝑋< )< ∗ 𝑉 25 ∗ 0.0031 ∗ (1 − 𝑋< )< ∗ 𝑉
𝑋< =
=
𝑣,
𝑣,
<
𝑋< = 2.2165 ∗ (1 − 𝑋< )
𝑋< = 0.517
𝐶l< = 𝐶l1 (1 − 𝑋< ) = 0.0015 𝑚𝑜𝑙/𝑑𝑚E
18-7
Reactor 3:
𝑘 ∗ 𝐶l< ∗ (1 − 𝑋E )< ∗ 𝑉 25 ∗ 0.0015 ∗ (1 − 𝑋E )< ∗ 𝑉
𝑋E =
=
𝑣,
𝑣,
<
𝑋E = 1.0725 ∗ (1 − 𝑋E )
𝑋E = 0.394
𝐶lE = 𝐶l< (1 − 𝑋E ) = 0.000909 𝑚𝑜𝑙/𝑑𝑚 E
Reactor 4:
𝑘 ∗ 𝐶lE ∗ (1 − 𝑋q )< ∗ 𝑉 25 ∗ 0.000909 ∗ (1 − 𝑋q )< ∗ 𝑉
𝑋q =
=
𝑣,
𝑣,
<
𝑋q = 0.65 ∗ (1 − 𝑋q )
𝑋q = 0.3097
𝐶lq = 𝐶lE (1 − 𝑋q ) = 0.000627 𝑚𝑜𝑙/𝑑𝑚 E
Bounds on conversion:
r
0rst
3 tanks 𝑋 = sDr
r
sD
0rsv
4 tanks 𝑋 = sDr
sD
=
=
,.,10,.,,,u,u
,.,1
,.,10,.,,,w<x
,.,1
= 0.909
= 0.937
P18-5 (c)
Change of the fluid velocity
Let U=0.1cm/s
Re=50 and Sc=2
From gas phase dispersion correlation chart,
Da
udt
= 0.5
2
Gives Da = 0.5udt = 0.5*0.1*5 = 0.25cm / s
Bo =
UL 0.1*200
=
= 80
Da
0.25
Bo
+1 = 41
2
The conversion is close to the one PFR 2nd order reaction:
k=25dm3/(mol·s)
τ=l/U=200/0.1=2000s
Da=kτCAo=500
n=
Da
= 0.998
1+Da
Let U=100cm/s
Re=50000 and Sc=2
X=
From gas phase dispersion correlation chart,
Da
udt
= 0.21
2
gives Da = 0.21udt = 0.21*100*5 = 0.25cm / s
18-8
Bo =
UL 100*200
=
= 190.5
Da
105
Bo
190.5
+1 =
+1 = 96
2
2
The conversion is close to the one PFR 2nd order reaction:
k=25dm3/(mol·s)
τ=l/U=2/1=2s
Da=kτCAo=0.5
Da
X=
= 0.333
1+Da
n=
P18-5 (d)
Change of the superficial velocity
d u 0.2*4
Re = t =
= 80
v
0.01
From packed bed dispersion correlation chart,
D a=
Bo =
0.55ud p
ε
=
Daε
ud p
= 0.55
0.55*4*0.2
= 1.1
0.4
UL 4*200
=
= 727
Da
1.1
Bo
727
+1 =
+1 = 364.5
2
2
The conversion is close to the one PFR 2nd order reaction:
Check
k=25dm3/(mol·s)
τ=l/U=2/1=2s
Da=kτCAo=0.5
n=
X=
Da
= 0.333
1+Da
P18-5 (e)
Liquid instead of gas
d u 5*4*0.001
Part (a) Re = t =
= 0.2
v
0.1
Sc =
ν
100
=
= 2*107
DAB 5e −6
This is off the scale of the graph for liquid phase dispersion, hence Da cannot be evaluated.
P18-6 (a)
Peclet numbers
From Example 16.2, s2=6.21min2 and tm=5.15min
18-9
Closed:
σ2
2
tm
=
2
2 "
−Pe %
−
$1− e r ' → Per = 7.414
&
Per Pe2 #
r
Open:
σ2
2
tm
=
2
8
+
→ Per = 11.68
Per Pe2
r
P18-6 (b)
Space–time and dead volume
tm
τ=
= 4.40
1+2 Per
Vs = τ *vo = 263.8dm3
VD =V −Vs = 156.2dm3
%deadvolume =
156.2
= 37.2%
420
P18-6 (c)
Conversions for 1st order isomerization
Dispersion model
Da=kt=0.927
q = 1+
4Da
= 1221
Per
(
)
4qexp Per / 2
X = 1−
2
2
= 0.570
(1+ q) exp(Per q / 2) − (1− q) exp(−Per q / 2)
Tanks-in-series
n=
τ2
σ2
= 4.35
1
X = 1−
(1+ τ iK )
n
= 0.568
P18-6 (d)
Conversions PFR and CSTR
PFR: X = 1− e−kτ X=0.604
CSTR: X = 1−
1
X=0.481
1+ τ k
XDisp
0.570
XT-I-S
0.568
XPFR
0.604
XCSTR
0.481
18-10
P18-7
Tubular Reactor
1st order, irreversible, pulse tracer test ® σ 2 = 65 s 2 and tm = 10 s
For a 1st order reaction, PFR: X =1- e-kt = 0.98
We need t and k. There being no data for diffusivity (Schmidt number) Da and hence Per cannot be
obtained using tubular flow correlations.
Calculate
V= 3 m * 25 dm2 * (0.1 m/dm)2 = 0.75 m3
t = 0.75 m3 / (3*10-2 m3/s) = 25 s
This is greater than tm so channeling is occurring
Calculate, k
1
ln
k = 1 − X = 3.91/25 s = 0.156 s-1
τ
Therefore, assume closed vessel dispersion model:
tm = t = 10 s space-time
Calculate, Pe
σ2
2
tm
=
2
2 "
Pe % 65
−
= 0.65
$1− e r ' =
& 102
Per Pe2 #
r
Iterating ® Per = 1.5
Calculate, X using the measured V and v 0 giving t = 25 s
" Pe %
4qexp$$ r ''
# 2 &
X = 1−
(
" Pe *q %+ (
" −Pe *q %+
2
2
r
* 1+ q exp$ r '- − * 1− q exp$
'$ 2 '- *
$ 2 '*)
#
&, )
#
&,
( )
(
)
Da = t k = 25 * 0.156 = 3.9
q = 1+
4Da
=
Per
1+
4*3.9
= 3.376
1.5
" 1.5 %
4*3.376exp$ '
# 2 &
X = 1−
= 0.88
(
+
(
" 1.5*3.376 %
" −1.5*3.376 %+
2
2
* 1+ 3.376 exp$
'- − * 1− 3.376 exp$
'2
2
*)
#
&-, *)
#
&-,
(
)
(
)
Conversion for the real reactor assuming the closed dispersion (X=0.88) model is less than for the ideal
PFR (X=0.98)
tideal =
V
= 25 s
v0
V = VD + VS
VS = α * V
18-11
Note,
tRTD = 10 s = a * t , a=10/25 =0.4
Dispersion Volume
Dead Volume
Dead Volume = (1- a) V= 0.6 V, Vs = 0.4V
Use tm = 10s, Da = tm * k = 10*0.156 = 1.56
q = 1+
4*1.56
= 2.27
1.5
" 1.5 %
4*2.27exp$ '
# 2 &
XD = 1−
(
" 1.5*2.27 %+ (
" −1.5*2.27 %+
2
2
* 1+2.27 exp$
'- − * 1−2.27 exp$
'2
2
*)
#
&-, *)
#
&-,
(
)
(
)
XD = 1- (19.22/58.38) = 0.67
P18-8 (a)
E(t):
Reaction Volume =*V
Conversion T-I-S and Maximum Mixedness Model
XT-I-S=0.5
For a first order reaction Xmm=XT-I-S= 1−
∞
2
4
0
0
2
1
" τ %n
$1+ k '
# n &
τ = tm = ∫ E(t)t dt = ∫ 0.25⋅t 2dt + ∫ 1− 0.25⋅t tdt = 2min
∞
(
)
∞
2
σ = ∫ E(t)(t −tm ) dt = ∫ E(t)t 2 dt − τ 2 = min2
3
2
0
2
0
18-12
n=
τ2
=6
σ2
From the conversion it is possible to determine k at 300K:
" 1
%
$$
−1''
# n 1− X
&
k=
= 0.367min−1
τ
n
The conversion at T=310 K is given by:
# E
E &
k310 = k300 exp%
−
( = 1.422min−1
$ R300K R ⋅310K '
R = 1.986
cal
molK
Xmm=XT-I-S= 1−
1
" τ %n
$1+ k '
# n &
= 0.903
P18-8 (b)
Complex Reactions
dC A
dτ
dCB
dτ
dCC
dτ
= −k1C A − k3C A
= k1C A − k2CB
= k2CB
dCD
= k3C A
dτ
Where k1= k2= k3=0.1 min-1
dC A
dλ
()
()
E λ
∑riA + (C A −C Ao ) 1− F λ
=−
E (λ )
)
dλ
1− F ( λ )
E (λ )
dCB
= − (k1C A − k2CB ) + (CB −CBo )
dλ
1− F ( λ )
E (λ )
dCC
= − (k2CB ) + (CC −CCo )
dλ
1− F ( λ )
E (λ )
dCD
= − (k3C A ) + (CD −CDo )
dλ
1− F ( λ )
dC A
(
) (
= − −k1C A − k3C A + C A −C Ao
18-13
Where CAo=1mol/dm3 and CBo= CCo= CDo=0.
Changing variable: l=z-4
Polymath Code:
d(ca)/d(z) = -(-ra+(ca-cao)*EF)
d(cb)/d(z) = -(-rb+(cb-cbo)*EF)
d(cc)/d(z) = -(-rc+(cc-cco)*EF)
d(cd)/d(z) = -(-rd+(cd-cdo)*EF)
d(F)/d(z) = -E
k1 = 0.1
k2 = 0.1
k3 = 0.1
rc = k2*cb
ra = -k1*ca-k3*ca
rb = k1*ca-k2*cb
t2 = 4
rd = k3*ca
cao = 1
cbo = 0
cco = 0
cdo = 0
t1 = 2
lam = 4-z
E1 = 0.25*lam
E2 = 1-0.25*lam
E = if (lam<t1) then (E1) else ( if (lam<=t2) then (E2) else (0))
EF = E/(1-F)
z(0) = 0
z(f) =4
ca(0)=1
cb(0)=0
cc(0)=0
cd(0)=0
F(0)=0.99
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
ca
1.
0.6770285
1.
0.6770285
2
cao
1.
1.
1.
1.
3
cb
0
0
0.1429386
0.1429386
4
cbo
0
0
0
0
5
cc
0
0
0.0185472
0.0185472
6
cco
0
0
0
0
7
cd
0
0
0.1614858
0.1614858
8
cdo
0
0
0
0
9
E
0
0
0.4985019
0
10 E1
1.
0
1.
0
11 E2
0
0
1.
1.
12 EF
0
0
3.535202
0
13 F
0.99
-0.0099979
0.99
-0.0099979
14 k1
0.1
0.1
0.1
0.1
15 k2
0.1
0.1
0.1
0.1
16 k3
0.1
0.1
0.1
0.1
17 lam
4.
0
4.
0
18-14
18 ra
-0.2
-0.2
-0.1354057
-0.1354057
19 rb
0.1
0.053409
0.1
0.053409
20 rc
0
0
0.0142939
0.0142939
21 rd
0.1
0.0677028
0.1
0.0677028
22 t1
2.
2.
2.
2.
23 t2
4.
4.
4.
4.
24 z
0
0
4.
4.
Differential equations
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 d(F)/d(z) = -E
Explicit equations
1
k1 = 0.1
2
k2 = 0.1
3
k3 = 0.1
4
rc = k2*cb
5
ra = -k1*ca-k3*ca
6
rb = k1*ca-k2*cb
7
t2 = 4
8
rd = k3*ca
9
cao = 1
10 cbo = 0
11 cco = 0
12 cdo = 0
13 t1 = 2
14 lam = 4-z
15 E1 = 0.25*lam
16 E2 = 1-0.25*lam
17 E = if (lam< (0)) else (E2) then (lam<="t2)" if ( (E1)>
18 EF = E/(1-F)
18-15
P18-8 (c) No solution will be given
P18-9 (a)
From P16-3c:
1
2
= 0.159 min2 , tau=0.8 min and k= 0.8 min-1
min, σ 2 =
2π
π
Tanks-in-series
For a first order reaction, X T - I - S = X seg
tm = τ =
Refer solution of P17-4 where Xseg was calculated to be 0.447
So, X T - I - S = X seg = 0.447
P18-9 (b)
Closed-closed vessel dispersion model
For a first order reaction the conversion is given:
X =10.25 =
4q exp(Pe r / 2)
(1 + q ) exp( Per q / 2) - (1 - q )2 exp(- Per q / 2)
2
2
2 "
−Pe %
−
$1− e r '
&
Per Pe2 #
Per=6.839
r
Da = τ k = 0.638
𝑞 = X1 +
4𝐷𝑎
4 ∗ 0.638
= X1 +
= 1.172
𝑃𝑒P
6.84
X=0.448
XT-I-S
0.447
XDispersion
0.448
P18-10 No solution will be given
P18-11
(Link for accessing COMSOL and COMSOL tutorials:
http://www.umich.edu/~elements/6e/comsol/comsol.html)
(a) A decrease in the thermal conductivity of the reaction mixture causes an increase of the bulk
fluid temperature and can result a minimum in concentration.
(b) An increase in heat transfer coefficient results in less temp and uniform temp and hence no
minimum.
(c) A decrease in heat transfer coefficient results in more temp and no uniform temp and hence
minimum can exist.
(d) The external heat transfer coefficient increases, this implies a lower wall temperature and a
resulting lower temperature in the reacting mixture and no minimum.
(e) This implies an increase in the wall temperature and a resulting higher temperature in the reacting
mixture that can cause a minimum in the concentration.
18-16
P18-12
From P16-11 (k):
t=10 min and s2=74min2
$
−3 4
−2 3
−2 2
& for 0 ≤ t ≤ 3 E1 (t) = −1.1675⋅10 t +1.1355⋅10 t − 4.7492⋅10 t
&
−2
&+9.9505⋅10 t
& for 3 ≤ t ≤ 20 E (t) = −1⋅8950⋅10−6 t 4 + 8.7202⋅10−5 t 3 −1.1739⋅10−3 t 2
2
&
&
−4
E(t) = %−1.7979⋅10 t + 0.092343
&
−8 4
−6 3
−4 2
& for 20 ≤ t ≤ 60 E3 (t) = 1.2618⋅10 t −2.4995⋅10 t +1.8715⋅10 t −
&
−3
&6.3512⋅10 t + 0.083717
& for t > 60 0
&'
P18-12 (a)
2nd order, kCA0=0.1min-1, CA0=1mol/dm3, Segregation Model
Segregation Model
∞
X=
∫ X (t )E (t ) dt
0
kC t
( ) 1+ kCAo t
Where X t =
Ao
Polymath code:
d(Xbar)/d(t) = E*X
kCao = 0.1
E1 = -0.0011675*t^4+0.011355*t^3-0.047492*t^2+0.0995005*t
E2 = -1.8950*10^(-6)*t^4+8.7202*10^(-5)*t^3-1.1739*10^(-3)*t^2-1.7979*10^(-4)*t+0.092343
E3 = 1.2618*10^(-8)*t^4-2.4995*10^(-6)*t^3+1.8715*10^(-4)*t^2-6.3512*10^(-3)*t+0.083717
E4 = 0
E = if(t<=3)then(E1)else(if(t<=20)then(E2)else(if(t<=60)then(E3)else(E4)))
X = kCao*t/(1+kCao*t)
t(0) = 0
t(f) = 70
Xbar(0) = 0
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1 E
0
0
0.0834735
0
2 E1
0
-2.436E+04
0.0834735
-2.436E+04
3 E2
0.092343
-21.26102
0.092343
-21.26102
4 E3
0.083717
1.774E-05
0.083717
0.0017977
5 E4
0
0
0
0
6 kCao
0.1
0.1
0.1
0.1
7 t
0
0
70.
70.
8 X
0
0
0.875
0.875
9 Xbar
0
0
0.4224876
0.4224876
18-17
Differential equations
1 d(Xbar)/d(t) = E*X
Explicit equations
1 kCao = 0.1
2 E1 = -0.0011675*t^4+0.011355*t^3-0.047492*t^2+0.0995005*t
3 E2 = -1.8950*10^(-6)*t^4+8.7202*10^(-5)*t^3-1.1739*10^(-3)*t^2-1.7979*10^(-4)*t+0.092343
4 E3 = 1.2618*10^(-8)*t^4-2.4995*10^(-6)*t^3+1.8715*10^(-4)*t^2-6.3512*10^(-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)
P18-12 (b)
2nd order, kCA0=0.1min-1, CA0=1mol/dm3, Maximum Mixedness Model
()
()
E λ
dX rA
=
+
X
d λ C Ao 1− F λ
Rate Law : −rA = kC 2A
(
C A = C Ao 1− X
)
rA = -kC Ao (1 - X ) where k=0.1 dm3 /mol min
2
2
rA
2
= -kC Ao (1 - X )
C Ao
dF
= −E z
dz
()
where z = 60 – λ
Polymath code:
d(X)/d(z) = -(ra/Cao+E/(1-F)*X)
d(F)/d(z) = -E
Cao = 1
lam = 60-z
Ca = Cao*(1-X)
k = 0.1
ra = -k*Ca^2
E1 = -0.0011675*lam^4+0.011355*lam^3-0.047492*lam^2+0.0995005*lam
E2 = -1.8950*10^(-6)*lam^4+8.7202*10^(-5)*lam^3-1.1739*10^(-3)*lam^2-1.7979*10^(-4)*lam+0.092343
E3 = 1.2618*10^(-8)*lam^4-2.4995*10^(-6)*lam^3+1.8715*10^(-4)*lam^2-6.3512*10^(-3)*lam+0.083717
E4 = 0
E = if(lam<=3)then(E1)else(if(lam<=20)then(E2)else(if(lam<60)then(E3)else(E4)))
z(0) = 0
z(f) = 60
X(0) = 0
F(0) = 0.99
Polymath Output:
Calculated values of DEQ variables
Variable Initial value Minimal value Maximal value Final value
1
Ca
1.
0.5226948
1.
0.5952897
2
Cao
1.
1.
1.
1.
3
E
0
0
0.0834812
0
18-18
4
E1
-1.284E+04
-1.284E+04
0.0834812
0
5
E2
-9.868052
-9.868052
0.092343
0.092343
6
E3
2.228E-05
1.998E-05
0.083717
0.083717
7
E4
0
0
0
0
8
F
0.99
-0.010344
0.99
-0.010344
9
k
0.1
0.1
0.1
0.1
10 lam
60.
0
60.
0
11 ra
-0.1
-0.1
-0.027321
-0.035437
12 X
0
0
0.4773052
0.4047103
13 z
0
0
60.
60.
Differential equations
1 d(X)/d(z) = -(ra/Cao+E/(1-F)*X)
2 d(F)/d(z) = -E
Explicit equations
1
Cao = 1
2
lam = 60-z
3
Ca = Cao*(1-X)
4
k = 0.1
5
ra = -k*Ca^2
6
E1 = -0.0011675*lam^4+0.011355*lam^3-0.047492*lam^2+0.0995005*lam
7
E2 = -1.8950*10^(-6)*lam^4+8.7202*10^(-5)*lam^3-1.1739*10^(-3)*lam^2-1.7979*10^(4)*lam+0.092343
8
E3 = 1.2618*10^(-8)*lam^4-2.4995*10^(-6)*lam^3+1.8715*10^(-4)*lam^2-6.3512*10^(-3)*lam+0.083717
9
E4 = 0
10 E = if(lam<=3)then(E1)else(if(lam<=20)then(E2)else(if(lam<60)then(E3)else(E4)))
(Check F(t))
P18-12 (c)
Tanks in series and 1st order reaction with k=0.1min-1
n=
τ2
σ2
X = 1−
= 1.35
1
" τ %n
$1+ k '
# n &
= 0.527
P18-12 (d)
Dispersion models and 1st order reaction with k=0.1min-1
Peclet number open system
σ2
2
tm
=
2
8
+
→ Pe = 4.906
Pe Pe2
Peclet number closed system
18-19
σ2
2
tm
=
2
2
−
1− e−Pe → Pe = 0.98
Pe Pe2
X = 1−
(
)
( )
2
2
(1+ q) exp(Per q / 2)− (1− q) exp(−Per q / 2)
4qexp Per / 2
q = 1+
4Da
Per
Da = τ k
X= 0.53
P18-12 (e)
Dispersion models and 2nd order reaction with k=0.1dm3/mol×s and CA0=1mol/dm3
Conversion
Linearizing the 2st order reaction rate :
C
−rA = kC 2A ≅ k A0 C A = k'C A
2
it is possible to obtain an approximated solution;
C
k' = k A0 = 0.05s−1
2
Da = τ k' = 30
Per=0.98
q = 1+
X = 1−
(from Part(d))
4Da
4*30
= 1+
= 11.111
Per
0.98
( )
2
2
(1+ q) exp(Per q / 2)− (1− q) exp(−Per q / 2)
4qexp Per / 2
X=0.998
Full solution can be obtained with FEMLAB.
P18-12 (f)
𝑣1
𝑣,
V1
V2
𝑣1
For 1st Reactor,
𝑉1 e2yC = 𝑣1 𝐶z< − 𝑣1 𝐶z1
er
(1)
𝑉1 e2yC = 𝑣1 (𝐶z< − 𝐶z1 )
er
(2)
18-20
𝑣,
For 2nd Reactor,
ery{
= 𝑣1 𝐶z1 − 𝑣1 𝐶z< − 𝑣, 𝐶z<
(3)
𝑉< e2 = 𝑣1 (𝐶z1 − 𝐶z< ) − 𝑣, 𝐶z<
(4)
𝑉<
e2
ery{
Defining:
V
v
V
α = 1 , β = 1 and τ =
V
v0
v0
Equation 2 and 3 becomes
er
𝛼𝜏 e2yC = 𝛽(𝐶z< − 𝐶z1 )
(5)
(1 − 𝛼)𝜏 e2 = 𝛽(𝐶z1 − 𝐶z< ) − 𝐶z<
(6)
ery{
The above two equation can be solved using data of effluent tracer concentration as a function of time
as given in P16-11. Solving the equation will give model parameters 𝛼 𝑎𝑛𝑑 𝛽
P18-12 (g)
Two parameters model models and 2nd order reaction with k=0.1dm3/mol×s and CA0=1mol/dm3
Using CSTR reaction mol balance for 1st Reactor:
<
𝑣, 𝐶l, + 𝑣1 𝐶l< − 𝑣1 𝐶l< = −𝑘𝐶l1
𝑉1
C A0 + βC A2 − βC A1 = −kατ C 2A1
(1)
Using CSTR reaction mol balance for 2nd Reactor:
<
𝑣1 𝐶l1 − 𝑣1 𝐶l< − 𝑣, 𝐶l< = −𝑘𝐶l<
𝑉<
βC A1 − βC A2 −C A2 = −kC 2A2 1− α τ
(
)
(2)
Equation (1) and (2) can be solved to obtain 𝐶l1 and 𝐶l< . Conversion can be obtained by below
equation
C −C
X = A0 A2
C A0
P18-12 (h)
Table of the conversions
XT_I_S
0.527
XMM
0.405
Xseg
0.422
X Dispersion
0.53
P18-13 (a)Two parameters model
(
)
0min ≤ t < 10min C(t) = 10⋅ 1− exp(−0.1t)
(
)
10min ≤ t C(t) = 5+10⋅ 1− exp(−0.1t)
2
0min ≤ t < 10min F1 (t) = ⋅ 1− exp(−0.1t)
3
1 2
10min ≤ t F2 (t) = + ⋅ 1− exp(−0.1t)
3 3
(
(
)
)
18-21
Xtwoparameters
?
.
1
β=1/3
F(t)
0.5
0
0
5
10
15
20
t(min)
The F (t) can be representative of the ideal PFR and ideal CSTR in parallel model:
vPFR=βvo
V1=αV
node1
V2
a= Fractional volume
b= Fraction of flow
1
1
β = , α = , τ = 10min
3
3
τ PFR =
V1
vPFR
=
(
(
)
)
1− α
α
τ = 10min τ CSTR =
τ = 10min
β
1− β
P18-13 (b)
Conversion
2nd order, vo=1 dm3/min, k=0.1dm3/mol×min, CA0=1.25mol/dm3
Balance around node 1
(
)
βvoCPFR + 1− β v0CCSTR = v0C A
For the PFR:
Second-order
C −C
DaPFR
XPFR = A0 PFR =
= 0.556
C A0
1+DaPFR
α
Where DaPFR = k τ C Ao = 1.25
β
CPFR = 0.556mol / dm3
18-22
For the CSTR:
(
)
1+2DaCSTR − 1+ 4DaCSTR
C −C
XCSTR = A0 CSTR =
= 0.42
C A0
2DaCSTR
Where DaCSTR = k
1− α
τ C = 1.25
1− β Ao
CCSTR = 0.725mol / dm3
Where C A = βCPFR + 1− β CCSTR = 0.668mol / dm3
(
)
C −C
X = A0 A = 0.465
C A0
P18-13 (c)
Two parameters model
0min ≤ t < 10min F1 (t) = 0
1 2
10min ≤ t F2 (t) = + ⋅ 1− exp− "#0.2(t −10)$%
.
3 3
{
}
1
F(t)
0.5
0
0
5
10
15
20
t
The F (t) can be representative of the ideal PFR and ideal CSTR in series with a bypass between them:
vb=βvo
V1=αV
node1
V2
a= Fractional volume
b= Fraction of flow
1
3
40
β = , α = , τ = min
3
4
3
(
(
)
)
1− α
V
τ PFR = 1 = ατ = 10min τ CSTR =
τ = 5min
vo
1− β
18-23
Conversion
2nd order, vo=1 dm3/min, k=0.1dm3/mol×min, CA0=1.25mol/dm3
Balance around node 1
(
)
(
)
βvoCb + 1− β v0CCSTR = βvoCPFR + 1− β v0CCSTR = v0C A
For the PFR:
Second-order
C −C
DaPFR
XPFR = A0 PFR =
= 0.556
C A0
1+DaPFR
Where DaPFR = kατ C Ao = 1.25
CPFR = 0.556mol / dm3
For the CSTR:
(
)
1+2DaCSTR − 1+ 4DaCSTR
C −C
XCSTR = PFR CSTR =
= 0.185
CPFR
2DaCSTR
Where DaCSTR = k
1− α
τ C = 0.278
1− β PFR
CCSTR = 0.453mol / dm3
C −C
X = A0 A = 0.61
C A0
Where C A = βCPFR + 1− β CCSTR = 0.487mol / dm3
(
)
P18-14
18-24
LFR
P18-15 (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.
P18-15 (b)
Higher dispersion coefficient
Curve 2 has a higher dispersion coefficient: a higher Peclet number corresponds to a lower dispersion
coefficient.
P18-15 (c)
Large number of T-I-S
Curve 1: Increasing the number of T-I-S corresponds to increase the Peclet Number (i.e. Bo=2(n-1)).
18-25
P18-16 (a)
vb
node1
V1
V2
v
V
F(t) and E(t) curves for α = 1 = 0.4 and β = b = 0.2
V
v0
The analytical expression for E (t) is given by:
# βδ (t) t < τ
1
%%
−
t−
τ 1 ) /τ 2
(
E(t) = $
e
% 1− β
t ≥ τ1
τ2
%&
(
)
Where:
τ 1 = τ PFR =
V1
τ 2 = τ CSTR =
=
αV
1
= τ
1− β v0 2
(v0 −vb ) (
V2
)
(1− α )V = 3 τ
(v0 −vb ) (1− β )v0 4
=
18-26
vb=βvo
V1=αV
Integrating we can obtain the analytical expression for F (t):
)
τ
+β t <
2
+
F(t) = *
" τ % " 3τ %
−$t− '/$ '
+
τ
2
4
+1− 1− β e # & # & t ≥
,
2
(
)
P18-16 (b)
Conversion
2nd order, kCAo=0.5min-1, τ=2min
Function balance around node 1
(1− β )voCCSTR + βv0Cb = v0C A
For the PFR:
Second-order
C −C
DaPFR
XPFR = A0 PFR =
= 0.333
C A0
1+DaPFR
Where DaPFR = kC Ao
α
τ = 0.5
1− β
(
)
kCPFR = 0.333min−1
For the CSTR:
(
)
1+2DaCSTR − 1+ 4DaCSTR
C −C
XCSTR = PFR CSTR =
= 0.268
CPFR
2DaCSTR
Where DaCSTR = k
1− α
τ C = 0.5
1− β PFR
kCCSTR = 0.366min−1
kC A = 1− β kCCSTR + β kCb = 1− β kCCSTR + β kC Ao = 0.393min−1
(
X=
)
kC A0 − kC A
kC A0
(
)
= 0.214
In absence of bypass (β=0) the conversion would be X=0.245
18-27
P18-17
Two parameters model
A possible two-parameter model is the PFR with Bypass (vb) and Dead Volume (VD)
-1
Where vo=1mv3min
, V=2m3, α=V1/V=0.5, b=vb/vo=0.5 and τ=V/vo=2min
b
To evaluate the conversion, we write a balance around node 1on species A:
(1− β )voCPFR + βv0Cb = v0C A
PFR, 2nd order, liquid phase
V1
rA = −kC ACB
node1
3
Where k=1.5 m3/(kmol×min) and C Ao = CBo = 2mol / dm
C −C
DaPFR V2
XPFR = A0 PFR =
= 0.857
C A0
1+DaPFR
Where DaPFR = kC Ao
VD
α
τ =6
1− β
(
)
CPFR = C Ao − XPFRC Ao = 0.286mol / dm3
C A = 1− β CPFR + βCb = 1− β CPFR + βC Ao = 1.143mol / dm3
(
)
(
)
C −C
X = A0 A = 0.429
C A0
P18-18
∞
∫ E(t)dt = 1
By definition:
0
∞
∫ E(t)dt =
0
0.1*t1
2
= 1 → t1 = 20min
2
E(0)
The analytical expression for E(t) is given by:
For a triangular E(t): t1 =
18-28
vb=βvo
V1=αV
For 0 ≤ t ≤ t1 E(t) = −
0.1
t + 0.1
t1
For t > t1 E(t) = 0
P18-18 (a)
Mean residence time
tm =
t1
2
2
*
3
2'
0.1
−0.1t
0.1t
0.1 2 0.1t1 0.1t1
,
/
&−
)
& t t + 0.1)tdt = , 3t + 2 / = − 3 t1 + 2 = 6 = 6.67min
% 1
(
1
+
.
∞
t1 $
0
0
∫ E (t )tdt = ∫
0
t
(For a triangular E(t): tm = 1 )
3
Variance
∞
2
σ 2 = ∫ E t t −tm dt =
0
( )(
)
t1=20min
*
4
3'
0.1
−0.1t
0.1t
2
2
2
,
/
&−
)
−tm
& t t + 0.1)t dt −tm = , 4t + 3 /
% 1
(
1
+
.
t1 $
∫
0
0
3
0.1 3 0.1t1 2 0.1⋅203
t +
−tm =
−6.672 = 22.2min
4 1
3
12
Assuming closed-closed system: τ = t m
=−
P18-18 (b)
Using the below equation for closed-closed system
𝜎<
2
2
0•€• )
(
< = 𝑃𝑒 − 𝑃𝑒𝑟 < 1 − 𝑒
𝑡#
P
22.2
2
2
(1 − 𝑒 0•€• )
= 0.5 =
−
<
6.67
𝑃𝑒P 𝑃𝑒𝑟 <
Solving, we get 𝑃𝑒P = 2.56
P18-18 (c)
Tanks in series
𝜏 < 6.67<
𝑛= <=
=2
𝜎
22.2
Thus, required no of tank in series is 2
P18-19
+
Since ∫, 𝐸(𝑡)𝑑𝑡 = 1, so max value of E(t) in the curve is 1.
So, following is the equation for E(t) curve
𝐸(𝑡) = 0, 𝑖𝑓 𝑡 < 10
E(t)= -0.02t+0.4 if 10≤ 𝑡 ≤ 20
𝐸(𝑡) = 0, 𝑖𝑓 𝑡 > 20
+
<,
𝑡# = 𝜏 = ∫, 𝑡 𝐸(𝑡)𝑑𝑡 = ∫1, (−0.02𝑡 < + 0.4𝑡) 𝑑𝑡 =13.4 min
𝑘𝜏 = 0.1 ∗ 13.4 = 1.34
18-29
(a) For an ideal PFR
𝑋‡ = 1 − 𝑒 0ˆ2 = 1 − exp(−1.34) = 0.738
X=73.8 %
(b) For an ideal CSTR
𝜏𝑘
1.34
𝑋‡ =
=
= 0.573
1 + 𝜏𝑘 1 + 1.34
Or X=57.3 %
(c) For an ideal laminar flow
1
1
𝑋‡ = 1 −
=1−
= 0.66
(1 + 0.25𝜏𝑘 )𝑒 ,.‰ˆB + 0.25𝜏𝑘
2.61 + 0.335
Or X= 66 %
(d) For segregation model
For a first order reaction
𝑋(𝑡) = 1 − 𝑒 0ˆ2
<,
𝑋‡ = Š (1 − 𝑒 0,.12 ) ∗ (−0.02𝑡 + 0.4)𝑑𝑡 = 0.73
1,
Or X=73 %
(e) For Maximum mixedness model
See following Polymath code
d(X) / d(t) = k*(1-X)
X(0) = 0
d(Xseg) / d(t) = X*E
Xseg(0) = 0
k=0.1
E1=0
E2= -0.02*t+0.4
E=if (t<=10) then (E1) else (E2)
t(0) = 0
t(f) = 20
Polymath Output
Calculated values of DEQ variables
Initial value
Minimal value
Maximal value
Final value
1 E
Variable
0
0
0.2
0
2 E1
0
0
0
0
3 E2
0.4
0
0.4
0
4 k
0.1
0.1
0.1
0.1
5 t
0
0
20.
20.
6 X
0
0
0.8646647
0.8646647
7 Xseg
0
0
0.7293294
0.7293294
(f) Dispersion Model
+
𝜎 < = Š (𝑡 − 𝜏)< 𝐸 (𝑡)𝑑𝑡
,
<,
𝜎 < = Š (𝑡 − 13.4)< ∗ (−0.02𝑡 + 0.4)𝑑𝑡 = 5.56
1,
Using equation 18-29 to calculate Per
𝜎<
2
2
(1 − 𝑒 0•=• )
< = 𝑃𝑒 −
𝑡#
𝑃𝑒P<
P
18-30
On solving we get Per=62.5
𝐷𝑎1 = 𝑘𝜏 = 0.1 ∗ 13.4 = 1.34
4𝐷𝑎1
= 1.042
𝑃𝑒P
4𝑞 exp (𝑃=P /2)
𝑋=1−
𝑃 𝑞
𝑃 𝑞
<
^
]− =P ^
(1 + 𝑞)< exp ] =P
2 − (1 − 𝑞) exp
2
𝑞 = X1 +
𝑋=1−
4 ∗ 1.042 ∗ 3.7 ∗ 101E
15.42
= 1−
= 0.737
1q
01‰
4.17 ∗ 1.385 ∗ 10 − 0.0017 ∗ 7.21 ∗ 10
57.75
𝑋 = 73.7 %
(g) T-I-S Model
+
𝜎 < = Š (𝑡 − 𝜏)< 𝐸 (𝑡)𝑑𝑡
,
<,
𝜎 < = Š (𝑡 − 13.4)< ∗ (−0.02𝑡 + 0.4)𝑑𝑡 = 5.56
1,
𝜏 < 13.4 ∗ 13.4
𝑛= <=
= 32.3
𝜎
5.56
𝜏 13.4
𝜏o = =
= 0.415
𝑛 32.3
1
𝑋=1−
= 0.73
(1 + 0.415 ∗ 0.1)E<.E
Or X= 73 %
Ideal PFR
Ideal CSTR
73.8 %
57.3 %
Ideal Laminar
flow Reactor
66 %
Segregation
Model
73 %
Maximum
Mixedness
73%
Dispersion
Tank in Series
73.7 %
73 %
P18-20 (a)
(Refer to Section 18.3 of the book
Use equations (18-17) and (18-18))
For small deviations from plug flow, Per is large
So, for large Per, q tends to 1 and (1-q)2exp(-Perq/2) tend to zero
Therefore, X = 1 – 4exp(Per/2)/22exp(Perq/2)
= 1-exp[Per/2(1-q)]
(Bˆ){
= 1-exp(-𝜏𝑘 + •= )
•
(Use Da1 = 𝜏𝑘, q = (1 + x)n = 1+ nx +n(n-1)x2/2! + ……….)
P18-20 (b)
Assume no volume expansion
Same conversion implies same outlet concentration (Ca)
For a PFR: ln(Ca/Ca0) = -k𝜏p , where 𝜏p = VP/𝑣 ---(1)
(Bˆ){
For the real reactor, from part (a) above, ln(Ca/Ca0) = exp(-𝜏𝑘 + •= ) , where 𝜏 = V/𝑣 ---(2)
•
Equating equation (1) and (2) and inserting values of 𝜏 and 𝜏p,
18-31
We get VP/V = 1- (𝜏k)/ 𝑃𝑒P ---(3)
Also, 𝑃𝑒P = UL/Da and 𝜏 = L/U
Using these equations in (3) gives VP/V = 1- (𝜏k)/ 𝑃𝑒P = 1- kDa/U2
P18-20 (c) Use Figure 13.19 of
O. Levenspiel, Chemical Reaction Engineering, 3rd ed. Wiley,1999.
P18-20 (d)
(Go to http://www.umich.edu/~elements/6e/live/chapter18/LEP-18-20.cdf to download the Wolfram
code for this part)
By using numerical techniques, this equation (P18-20.3) can be validated. The equation can be written
as:
Ž01
CA = 𝐶lf‹Œ• (1 + (n𝜏𝑘𝐶l,
)ln(CA0/𝐶lf‹Œ• )/ Per)
Ž01
We can plot CA and 𝐶lf‹Œ• (1 + (n𝜏𝑘𝐶l,
)ln(CA0/𝐶lf‹Œ• )/ Per) on the same graph as a function of reactor
length and observe that, for any n, they both should converge to the same value at the exit if the
equation is to be true.
(Note: In the below graphs, the red curve represents C A and the blue curve represents 𝐶lf‹Œ• (1 +
Ž01
(n𝜏𝑘𝐶l,
)ln(CA0/𝐶lf‹Œ• )/ Per))
For example, for n = 2 we can see, from the below graph, that this suggestion holds.
Similar result holds for n = 1
18-32
For n = 0
(You may not be able to see both the curves because they are overlapping)
Therefore, by the above graphs, it can be presumed that this equation holds for higher orders as well.
(Note: For n other than 0 and 1, equation (18-9) becomes non-linear and can’t be solved analytically
(unless some appropriate assumptions are made)).
P18-20 (e)
Dispersion doesn’t affect zeroth order reaction.
18-33
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18-34
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