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Extensive workbook with more than 200 up-to-date solved problems on advanced chemical reactors for deeper understanding of chemical reactor design
Problem Solving in Chemical Reactor Design provides in-depth coverage of more than 200 solved complex reactor design problems extracted from core chemical engineering subject areas. The problems in this book cover the design of non-ideal, catalytic, multiphase, heterogeneous, and biochemical reactors rather than focusing on basic Chemical Reactor Engineering concepts.
Each complex problem is solved using simple procedures and mathematical tools, enabling readers to better understand the correct procedure for solving problems and solve them faster, more conveniently, and more accurately.
This book is inspired by more than two decades of the author's teaching experience in chemical reactor engineering. Accompanying electronic materials include spreadsheets and easily understandable Matlab® programs, which can both be downloaded from the Wiley website.
Some of the topics covered in Problem Solving in Chemical Reactor Design include:
- Optimization, operation, and complexities of reactor design in the face of non-idealities such as mixing issues and residence time distributions
- Utilization of the tanks-in-series model, dispersion model, and intricate combinations of ideal reactors to elucidate the impact on conversion rates
- Signal processing within the domain of chemical reactor engineering, specifically focusing on convolution and deconvolution methodologies
- Reaction kinetics, diffusion dynamics, and catalyst efficiency in catalytic reactor design, and design of gas-catalytic and gas-liquid-solid catalyst systems in multiphase reactors
Problem Solving in Chemical Reactor Design is an excellent learning resource for students and professionals in the fields of chemical engineering, pharmaceuticals, biotechnology, and fine chemistry.
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Veröffentlichungsjahr: 2024
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Table of Contents
Cover
Table of Contents
Title Page
Copyright
Dedication
Preface
Nomenclature
Part I: Non-ideal Flow Characterization and Chemical Reaction
1 Non-ideal Flow and Reactor Characterization
Summary of Residence Time Distribution Properties and Most Important Models
2 Chemical Reaction in Non-ideal Reactors
Summary of Most Important Models
3 Transfer Function in Chemical Reactor Design
Summary of the Equations and Concepts
Part II: Convolution and Unsteady State in Chemical Reactors
4 Convolution and Deconvolution of Signals in Chemical Reactor Engineering
Summary of Equations and Methods
5 Partial Differential Equations in Chemical Reactor Engineering
Summary
Part III: Catalytic and Multiphase Reactor Design
6 Reaction Rate in Catalytic Processes
Summary of Equations for the Catalytic Reactor Design
7 Catalytic Reactor Design
8 Multiphase Reactor Design
Summary of Rate Expressions
Part IV: Biochemical Reactor Design
9 Biochemical Reactor Design: Enzymatic Processes
Summary of Kinetic Expressions
10 Biochemical Reactor Design: Microbial Growth
Summary of Kinetic Expressions and Mass Balances in Bioreactors
Bibliography
Index
End User License Agreement
Guide
Cover
Table of Contents
Title Page
Copyright
Dedication
Preface
Nomenclature
Begin Reading
Bibliography
Index
End User License Agreement
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Problem Solving in Chemical Reactor Design
Juan A. Conesa
Author
Prof. Juan A. Conesa
Universidad de Alicante
Alicante 03690
Spain
Cover: © Shaiith/Shutterstock
All books published by WILEY-VCH are carefully produced. Nevertheless, authors, editors, and publisher do not warrant the information contained in these books, including this book, to be free of errors. Readers are advised to keep in mind that statements, data, illustrations, procedural details or other items may inadvertently be inaccurate.
Library of Congress Card No.: applied for
British Library Cataloguing-in-Publication Data: A catalogue record for this book is available from the British Library.
Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at <http://dnb.d-nb.de>.
© 2025 WILEY-VCH GmbH, Boschstraße 12, 69469 Weinheim, Germany
All rights reserved (including those of translation into other languages, text and data mining and training of artificial technologies or similar technologies). No part of this book may be reproduced in any form – by photoprinting, microfilm, or any other means – nor transmitted or translated into a machine language without written permission from the publishers. Registered names, trademarks, etc. used in this book, even when not specifically marked as such, are not to be considered unprotected by law.
Print ISBN: 978-3-527-35411-5
ePDF ISBN: 978-3-527-84802-7
ePub ISBN: 978-3-527-84801-0
oBook ISBN: 978-3-527-84803-4
To my beloved wife, Marga, and children, Margarita, Juan, Pablo, and Miguel.
Preface
This book has been written after more than two decades of teaching experience in chemical reactor engineering. During these years, I have been collecting problems posed to undergraduate and master’s students, which are now compiled here along with their solutions. This is not a basic chemical reactor engineering (CRE) book; instead, it covers the design of nonideal, catalytic, multiphase, and biochemical reactors.
Following the publication of my book Chemical Reactor Design: Mathematical Modeling and Applications in 2019, this book addresses aspects not covered in basic textbooks dedicated to reactor design. The concepts introduced in my earlier book are further developed here through solved examples. This approach aims to help the reader gain a deeper understanding of reactor design procedures, especially in the context of nonideal and heterogeneous reactors.
This collection of solved problems in advanced chemical reactor engineering provides in-depth coverage of complex reactor design issues, such as the design of catalytic or biochemical reactors. It is not a basic book covering simple concepts.
The book is accompanied by electronic materials, including spreadsheets and Matlab® programs, which can be downloaded from the Wiley website at http://www.wiley-vch.de/ISBN9783527354115.
The tools used in this book are not complex, and the solutions are presented in a simplified manner. I primarily use spreadsheets and occasionally incorporate Matlab. Importantly, the programs are designed to be highly understandable.
Elche (Spain)May 2024
Juan A. Conesa
Nomenclature
Suggested units are indicated for each variable.
A
m×n
matrix of convolution
C
concentration (usually mol/m
3
, or kg/m
3
)
C
As
concentration of A in the surface (mol/m
3
)
c
p
calorific capacity of the reacting flow (J/K · g)
C
p
concentration of product “P” (mol/m
3
)
C
T
total concentration (mol/m
3
)
d
p
particle diameter (m)
D
dilution rate (s
−1
)
D
A
diffusion coefficient of species A (m
2
/s)
D
AB
diffusion coefficient (diffusivity) of A in B (m
2
/s)
D
e
effective diffusion coefficient (m
2
/s)
E
(in the reaction rates context) → activation energy (J/mol)
E
(in the RTD context) → residence time distribution function, RTD (—)
F
integral form of the residence time distribution function (—)
g
A
generation term of a mole balance (mol/s)
h
heat transfer coefficient (J/K · s)
H
(
s
)
transfer function (—) (Laplace space)
h
(
t
)
time-dependent transfer function (—)
k
kinetic constant (units depend on the kinetic law)
k
0
pre-exponential factor of the kinetic constant (units depend on the reaction order)
k
L
mass transfer coefficient in the liquid phase (mol/s · m
2
/(mol/m
3
) = m/s)
K
M
Michaelis constant (g/l)
K
S
Monod constant (g/l)
L
characteristic length (m)
L
{
h
(
t
)}
Laplace transform of function
h
(
t
)
M
tracer mass (g or mol)
N
amount of substance (mol)
n
A
molar flow of component “A” (mol/s or mol/s · m
2
)
N
ad
dimensionless adiabatic temperature increase (—)
N
c
dimensionless cooling capacity (—)
n
t
number of tanks in the tanks-in-series model (—)
n
T
total molar flow (mol/s)
q
heat flux (J/s)
Q
volumetric flowrate (m
3
/s)
q
C
heat flux in cooling media (J/s)
R
radius (m)
reaction (or process) rate based on the volume of catalyst particles (mol/s · m
3
)
reaction (or process) rate based on the volume of reacting species (mol/s · m
3
)
reaction (or process) rate based on the weight of catalyst (mol/s · g)
r
A
reaction (or process) rate based on the external catalyst surface (mol/s · m
2
)
R
g
universal gas constant (J/mol · K)
s
(in the Laplace transform context) → main variable in the Laplace space
S
section (m
2
)
T
temperature (K)
t
time (s)
average residence time (=
V
/
Q
) (s)
T
C
temperature of the cooling media (K)
t
m
first moment of the RTD; mean value of time (s)
U
global heat transfer coefficient (J/K · s)
u
linear velocity (m/s)
V
volume (m
3
)
V
d
dead volume (m
3
)
V
p
volume with plug flow regime (m
3
)
W
weight of catalyst (g)
We
Weisz modulus (—)
X
(
s
)
stimulus function (—) (Laplace space)
x
(
t
)
time-dependent stimulus function (—)
X
A
molar conversion of reactant A (—)
Y
(
s
)
response to stimulus function (—) (Laplace space)
y
(
t
)
time-dependent response to stimulus function (—)
Greek Symbols
τ
tortuosity factor (—)
α
fraction of volume of a subsystem (—)
β
fraction of flow passing through a subsystem (—)
δ
dirac delta function (—)
μ
growth rate per unit of cell (1/s)
ε
s
solid load (m
3
solid/m
3
reactor)
ε
G
gas fraction (m
3
gas/m
3
reactor)
ε
L
liquid fraction (m
3
liquid/m
3
reactor)
ε
b
bed porosity (m
3
void/m
3
reactor)
σ
2
variance (s
2
)
σ
standard deviation (s)
η
effectiveness (—)
Δ
H
r
enthalpy of a reaction (kJ/mol)
Subscripts
i
actual position
i
+ 1
position of the following interval
i
− 1
position of the preceding interval
S
surface
0
inlet conditions
F
fluid
C
catalyst
R
reactor
Superscripts
t
actual time
t
+ 1
time of the following interval
t
− 1
time of the previous interval
Part INon-ideal Flow Characterization and Chemical Reaction
1Non-ideal Flow and Reactor Characterization
Summary of Residence Time Distribution Properties and Most Important Models
Residence Time Distribution
In a reactor, C(t) is obtained by injecting pulse of tracer. From that:
Total amount of tracer injected is:
In a step tracer run:
Mean residence time:
Variance of the residence time distribution (RTD):
The square root of the variance, σ, is called standard deviation.
Average residence time:
RTD in Ideal Reactors
For the plug flow reactor (PFR):
For the continuous stirred tank reactor (CSTR):
Tanks-in-series (TIS) Model
Dispersion Model
Bo < 0.01
Bo > 0.01, Closed–Closed Recipient
E(t) is only integrable by numerical methods.
Bo > 0.01, Open–Open Recipient
Designing E1 to the E(t) given by the first expression (valid for Bo < 0.01) and E2 to the one predicted by the open–open (o–o) assumption, the differences between these RTDs are small at low Bo, but at a high Bo number, E1 is not valid (for more details, consult spreadsheet “dispersion model Bo fitting.xls”):
Problem 1.1 A solution of potassium permanganate (KMnO4) was rapidly injected into a water stream flowing through a circular tube at a linear velocity of 35.70 cm/s. A photoelectric cell located 2.74 m downstream from the injection point was utilized to monitor the local concentration of KMnO4.
(a) By using the given effluent KMnO
4
concentrations, calculate the average residence time of the fluid as well as the variance,
E
(
t
),
F
(
t
), and
I
(
t
).
(b) Determine the number of ideal tanks (
n
t
), the variance, the dispersion number, and the Peclet number.
Time (s)
KMnO
4
concentration
0.0
0.0
2.0
11.0
4.0
53.0
6.0
64.0
8.0
58.0
10.0
48.0
12.0
39.0
14.0
29.0
16.0
22.0
18.0
16.0
20.0
11.0
22.0
9.0
24.0
7.0
26.0
5.0
28.0
4.0
30.0
2.0
32.0
2.0
34.0
2.0
36.0
1.0
38.0
1.0
40.0
1.0
42.0
1.0
Solution to Problem 1.1
For details refer the Wiley website at http://www.wiley-vch.de/ISBN9783527354115
(a) To solve this problem, we should use a spreadsheet. First, data given in the statement are introduced, and then we should do the following calculations:
t
(s)
C
(
t
)
C
(
t
)d
t
E
(
t
)
t
·
E
(
t
)
t
·
E
(
t
)·Δ
t
(
t
−
t
m
)
2
·
E
(
t
)
(
t
−
t
m
)
2
·
E
(
t
)·Δ
t
F
(
t
)
I
(
t
)
0
0
0
0.000
0.000
0.000
0.000
0.000
0.000
1.000
2
11
11
0.014
0.029
0.029
1.149
1.149
0.029
0.971
4
53
64
0.069
0.275
0.304
3.345
4.494
0.166
0.834
6
64
117
0.083
0.498
0.773
2.055
5.399
0.332
0.668
8
58
122
0.075
0.602
1.100
0.666
2.721
0.482
0.518
10
48
106
0.062
0.623
1.224
0.059
0.725
0.607
0.393
12
39
87
0.051
0.607
1.230
0.053
0.112
0.708
0.292
14
29
68
0.038
0.527
1.134
0.344
0.397
0.783
0.217
16
22
51
0.029
0.457
0.983
0.720
1.065
0.840
0.160
18
16
38
0.021
0.374
0.830
1.024
1.744
0.882
0.118
20
11
27
0.014
0.285
0.659
1.162
2.186
0.911
0.089
22
9
20
0.012
0.257
0.542
1.419
2.581
0.934
0.066
24
7
16
0.009
0.218
0.475
1.540
2.959
0.952
0.048
26
5
12
0.006
0.169
0.387
1.464
3.004
0.965
0.035
28
4
9
0.005
0.145
0.314
1.504
2.968
0.975
0.025
30
2
6
0.003
0.078
0.223
0.939
2.443
0.981
0.019
32
2
4
0.003
0.083
0.161
1.147
2.086
0.986
0.014
34
2
4
0.003
0.088
0.171
1.375
2.522
0.991
0.009
36
1
3
0.001
0.047
0.135
0.812
2.187
0.994
0.006
38
1
2
0.001
0.049
0.096
0.947
1.759
0.996
0.004
40
1
2
0.001
0.052
0.101
1.093
2.040
0.999
0.001
42
1
2
0.001
0.054
0.106
1.248
2.341
Σ(
C
(
t
)d
t
) = 771
t
m
= 10.98 s
σ
2
= 46.88 s
2
We can calculate time increment as the difference between time in two experimental data; in the third column, concentration and time increment are multiplied. For a better calculation, instead of calculating C(t)·Δt for each time directly, we can use the average value of C(t) between two consecutive data. This is called Simpson’s rule for evaluating areas graphically. The sum of the values in this column would give the area of the C(t) curve, so in the fourth column, the following relationship is applied to calculate E(t):
In the next columns, the average value of E(t) between two consecutive data is multiplied by time, by time increment, and the sum would represent mean time, as we have:
In a similar way, variance is calculated:
Finally, we can calculate F(t) and I(t) according to their definitions:
We can check the form of the graphs showing the distributions:
(b) From the calculated parameters, it is easy to find the number of tanks in the TIS model:
For the dispersion model to be applied, first we should assume Bo < 0.01, and then:
We obtain Bo = 11.24 ≫ 0.01, so this assumption is not valid.
Assuming now Bo > 0.01 and closed–closed recipient:
From that:
and we obtain Bo = 0.2413, and Pe = 1/Bo = 4.14.
On the other hand, if an o–o recipient is assumed:
Obtaining Bo = 0.127 and Pe = 7.85.
For checking which one of the conditions is fulfilled, data about average time in the recipient is needed in order to test if the equation is satisfied.
So, we have that:
Obtaining Bo = 0.215. This difference between tm and indicates that probably the recipient is open.
Problem 1.2 An experiment to characterize a tubular reactor was conducted using a technique in which a tracer is continuously fed into the system. At a specific time, the tracer supply is halted. From this point onward, the exit signal is recorded, resembling what can be termed a “decreasing step input” (or negative step). In this experiment, the following data were obtained:
t
(min)
0
1
2
4
6
7
8
10
C
(g/l)
0.5
0.5
0.5
0.4
0.1
0
0
0
(a) Calculate the RTD of this reactor.
(b) What fraction of the fluid spends more than 3 minutes in the reactor? Make a plot of the procedure.
(c) What fraction of the fluid spends between 3 and 4 minutes in the reactor?
Solution to Problem 1.2
For details refer the Wiley website at http://www.wiley-vch.de/ISBN9783527354115
(a) Following this procedure, the exit signal will be related to the cumulative RTD of the system,
F
(
t
), but the actual function obtained is the internal age function,
I
(
t
). In the spreadsheet, we can do the following calculation:
t
(min)
C
(g/l)
C
/
C
max
I
(
t
)
F
(
t
)
E
(
t
) = Δ
F
/Δ
t
0
0.5
1
1
0
0.00
1
0.5
1
1
0
0.00
2
0.5
1
1
0
0.10
4
0.4
0.8
0.8
0.2
0.30
6
0.1
0.2
0.2
0.8
0.20
7
0
0
0
1
0.00
8
0
0
0
1
0.00
10
0
0
0
1
In the table, I(t) is derived directly from the data, and:
For the calculation of E(t), a numerical derivative is done:
Obtaining:
(b) For calculating the fraction of fluid that spends a time in the reactor, we use the definition of
E
(
t
):
This integration can be done numerically, in increments:
where Δt is obviously 1 minute. Graphically, the approximation of the integral would be the following:
Obviously, this is a rough approximation. A better approximation would involve using smaller time increments, but this is not possible with the obtained data. More precision is obtained if average value of the E(t) function is used, so:
In this situation, we are doing the following approximation in the graph:
(c) If we look for the fraction of fluid passing between 3 and 4 minutes:
Problem 1.3 This problem involves a reactor with a flow rate of 10 l/min. Concentration measurements were taken at the outlet during a pulse test at various time intervals. The data obtained are as follows:
t
(min)
c
× 10
5
(g/l)
t
(min)
c
× 10
5
(g/l)
0
0
15
238
0.4
329
20
136
1.0
622
25
77
2.0
812
30
44
3
831
35
25
4
785
40
14
5
720
45
8
6
650
50
5
8
523
60
1
10
418
(a) Determine the volume of the reactor and plot the functions
E
(
t
),
F
(
t
), and
I
(
t
) as a function of time.
(b) Determine the number of tanks required to model the reactor using a TIS system.
(c) Calculate the Peclet number (Pe) if the reactor is modeled using a dispersion model.
Solution to Problem 1.3
For details refer the Wiley website at http://www.wiley-vch.de/ISBN9783527354115
(a) This problem is quite similar to
Problem 1.1
. In a spreadsheet, we calculate:
t
(min)
C
(
t
) (g/l)
C
(
t
)·Δ
t
E
(
t
)
t
·
E
(
t
)·Δ
t
(
t
−
t
m
)
2
·
E
(
t
)
(
t
−
t
m
)
2
·
E
(
t
)·Δ
t
0
0
0
0.0000
0
0
0
0.4
3.29·10
−3
1.32·10
−3
0.0368
0.0059
3.4842
1.3937
1
6.22·10
−3
3.73·10
−3
0.0697
0.0418
5.7993
3.4796
2
8.12·10
−3
8.12·10
−3
0.0909
0.1819
6.0023
6.0023
3
8.31·10
−3
8.31·10
−3
0.0931
0.2792
4.7237
4.7237
4
7.85·10
−3
7.85·10
−3
0.0879
0.3516
3.2975
3.2975
5
7.20·10
−3
7.20·10
−3
0.0806
0.4031
2.1175
2.1175
6
6.50·10
−3
6.50·10
−3
0.0728
0.4367
1.2383
1.2383
8
5.23·10
−3
1.05·10
−2
0.0586
0.9371
0.2644
0.5288
10
4.18·10
−3
8.36·10
−3
0.0468
0.9362
0.0007
0.0015
15
2.38·10
−3
1.19·10
−2
0.0267
1.9989
0.6335
3.1675
20
1.36·10
−3
6.80·10
−3
0.0152
1.5230
1.4853
7.4263
25
7.70·10
−4
3.85·10
−3
0.0086
1.0779
1.9080
9.5401
30
4.40·10
−4
2.20·10
−3
0.0049
0.7391
1.9464
9.7322
35
2.50·10
−4
1.25·10
−3
0.0028
0.4899
1.7324
8.6618
40
1.40·10
−4
7.00·10
−4
0.0016
0.3136
1.3993
6.9965
45
8.00·10
−5
4.00·10
−4
0.0009
0.2016
1.0896
5.4482
50
5.00·10
−5
2.50·10
−4
0.0006
0.1400
0.8903
4.4515
60
1.00·10
−5
1.00·10
−4
0.0001
0.0672
0.2786
2.7857
Σ(
C
(
t
)d
t
) = 8.93·10
−2
t
m
= 10.12 s
σ
2
= 80.99 s
2
We easily obtain:
For calculating the volume of the system, we have:
So the volume is 101.2 l.
(b)
(c) As in
Problem 1.1
, we do not know if the recipient is closed or open. Assuming closed–closed condition, we obtain Bo = 1.33 and Pe = 0.75.
Problem 1.4
In the given reactor configurations, where the total volume is represented as V = V1 + V2, and the ratio V1/V2 is imposed, we need to determine the ratio of the volumes V1/V2 in the second configuration in order to achieve the same exit conversion, XA2, for both a first-order reaction and a second-order reaction.
Solution to Problem 1.4
For details refer the Wiley website at http://www.wiley-vch.de/ISBN9783527354115
In the first case:
We can write:
and in the CSTR:
And solving for XA2:
In the second configuration:
we have, in the CSTR: Q0CA0 = (−rA)V2 + Q0CA1
So: . Then, in the PFR:
that is:
We can check that both configurations give the same final conversion.
In the case where a second-order reaction took place, the procedure is similar. In that case, for the first configuration, in the PFR:
and in the CSTR:
So finally:
This expression is a second-order equation. It is quite complicated to solve the final conversion, but finally, for the first configuration:
Obviously, the other solution with minus sign does not make sense.
In the second configuration, following the same procedure:
and finally, in the PFR:
We can check the form of the solution by assuming values of the constants. For example, k·CA0 = 0.1 mol/l·s, Q0 = 1 l/s, V = 100 l, and the following graphs can be calculated. As we can check, in any case, the first configuration (PFR + CSTR) gives a higher conversion than the second configuration (CSTR + PFR). In all cases, V1 refers to the volume of the PFR and V2 to that of CSTR.
Problem 1.5
(a) Obtain the cumulative probability function for the following data arising from stable operation with concentration of 0.122 M, followed by a step disturbance with
C
= 0.548 M.
t
(s)
10
20
30
45
60
90
120
200
400
1000
C
(M)
0.126
0.43
0.207
0.314
0.378
0.484
0.505
0.527
0.544
0.547
(b) Obtain a polynomial for its interpolation, and calculate the average residence time using numerical integration by the method of trapezoids.
(c) Calculate the average time using the polynomial.
Solution to Problem 1.5
For details refer the Wiley website at http://www.wiley-vch.de/ISBN9783527354115
(a) The resolution of this problem required a spreadsheet. Let us calculate the
E
(
t
) as the derivative of the
C
step
(
t
) function, i.e.,
Furthermore, . We can do the following calculation, similar to that presented in Problems 1.1 and 1.3:
t
(s)
C
step
(M)
C
(
t
)·d
t
E
(
t
)
t
·
E
(
t
)
t
·
E
(
t
)·d
t
(
t
−
t
m
)
2
·
E
(
t
)
(
t
−
t
m
)
2
·
E
(
t
)·d
t
10
0.004
0.0002
0.0024
65.133
20
0.021
1.35
0.0003
0.0055
0.04
71.085
681.09
30
0.085
1.75
0.0004
0.0120
0.09
98.874
849.79
45
0.192
3.91
0.0006
0.0274
0.30
141.053
1799.46
60
0.256
5.19
0.0007
0.0440
0.54
159.384
2253.28
90
0.362
12.93
0.0009
0.0845
1.93
178.663
5070.70
120
0.383
14.84
0.0010
0.1175
3.03
161.660
5104.84
200
0.405
41.28
0.0010
0.2044
12.88
108.806
10 818.61
400
0.422
107.10
0.0011
0.4220
62.64
16.824
12 562.99
1000
0.425
327.30
0.0011
1.0608
444.85
238.058
76 464.73
Sum
t
m
(s)
σ
2
515.6375
526.28
115 605.49
Note that the concentration measured has been subtracted from the 0.122 M of the stable operation.
And then:
(b) The polynomial of the fitting is shown in the previous figure. Using numerical integration of the data, we obtain
t
m
= 532.79 s.
(c) Using the polynomial shown in the figure, we can do:
as:
This function cannot be integrated because the integral does not converge, as the upper limit is infinity. If we change this limit to, for example, 1000 (high enough to bear in mind all the tracer leaving the system), this can be solved, and:
Problem 1.6 A continuous reactor normally operates with a feed of 50 l/s. Under these conditions, the dynamic data for different pulses presented below were obtained.
(a) To what type of reactor do these data surely correspond?
(b) Obtain the residence time distribution.
(c) Estimate the residence time and the volume of the reactor, and
(d) Calculate the total mass of the tracer used in the three runs.
C
tracer
(mg/l)
t
(s)
Run 1
Run 2
Run 3
0
0.011
0.012
0.001
1
0.553
0.534
0.538
2
0.899
0.907
0.887
3
0.923
0.953
0.934
5
0.860
0.856
0.876
8
0.681
0.723
0.758
12
0.550
0.586
0.585
17
0.413
0.435
0.442
27
0.320
0.331
0.350
38
0.251
0.236
0.276
58
0.162
0.158
0.176
86
0.085
0.103
0.097
130
0.023
0.008
0.015
Solution to Problem 1.6
For details refer the Wiley website at http://www.wiley-vch.de/ISBN9783527354115
(a) and (b) In a spreadsheet, we can calculate numerically:
Let us plot the given data:
This is quite similar to the RTD of a CSTR, so the reactor behaves as one CSTR.
(c) Mean residence time is calculated following the procedure explained in other problems, obtaining 32.97 (Run 1), 32.53 (Run 2), and 32.98 s (Run 3) (32.83 average value).
Volume of the reactor can be calculated:
resulting V = 1641.3 l
(d) Mass of tracer can be calculated by applying:
The calculation gives N = 1382 mg (Run 1), N = 1412 mg (Run 2), and N = 1480 mg (Run 3).
Problem 1.7 The following data correspond to a 2250 l reactor that is fed 150 l/s:
t
(s)
0
2
4
6
8
10
12
14
16
18
p
(
t
)
0.00
0.02
0.03
0.05
0.1
0.19
0.32
0.47
0.6
0.71
t
(s)
20
22
24
26
28
30
32
34
36
38
40
p
(
t
)
0.80
0.86
0.90
0.93
0.95
0.97
0.98
0.98
0.99
0.99
1.00
Estimate:
(a) The Peclet number for the dispersion model.
(b) The number of reactors for the model of stirred TIS.
Solution to Problem 1.7
For details refer the Wiley website at http://www.wiley-vch.de/ISBN9783527354115
The given function p(t) should correspond to F(t) data, i.e., to a step tracer run, as the value of this function is continuously growing from 0 to 1. These data have been plotted to verify that the trend is as expected. Before beginning to look for the solution that they ask us for, previous calculations must be made:
t
(s)
p
t
(
t
)
E
(
t
)
E
(
t
)·d
t
t
·
E
(
t
)
Integral
(
t
−
t
m
)
2
·
E
(
t
)
Integral
0
0
0
0.02
0
0
2
0.02
0.01
0.01
0.02
0.02
1.994
1.994
4
0.03
0.005
0.02
0.02
0.04
0.734
2.728
6
0.05
0.01
0.05
0.06
0.08
1.024
1.759
8
0.1
0.025
0.09
0.2
0.26
1.648
2.673
10
0.19
0.045
0.13
0.45
0.65
1.685
3.334
12
0.32
0.065
0.15
0.78
1.23
1.103
2.789
14
0.47
0.075
0.13
1.05
1.83
0.337
1.440
16
0.6
0.065
0.11
1.04
2.09
0.001
0.338
18
0.71
0.055
0.09
0.99
2.03
0.194
0.195
20
0.8
0.045
0.06
0.9
1.89
0.677
0.872
22
0.86
0.03
0.04
0.66
1.56
1.037
1.715
24
0.9
0.02
0.03
0.48
1.14
1.242
2.279
26
0.93
0.015
0.02
0.39
0.87
1.464
2.706
28
0.95
0.01
0.02
0.28
0.67
1.411
2.876
30
0.97
0.01
0.01
0.3
0.58
1.927
3.338
32
0.98
0.005
0
0.16
0.46
1.261
3.187
34
0.98
0
0.01
0
0.16
0.000
1.261
36
0.99
0.005
0
0.18
0.18
1.976
1.976
38
0.99
0
0.01
0
0.18
0.000
1.976
40
1
0.005
0.2
0.2
2.851
2.851
Sum =
1.00
16.12
42.29
The first thing to do is to calculate E(t) by using the data from F(t). This is done by numerical derivation, i.e.,
Once E(t) is known, both tm and σ2 (whose equations are shown in the header of the table) can be calculated.
When we already have the values of both tm and σ2, we can calculate the required values.
The first model that is studied is the TIS. The parameter of such a model is nt:
In order to be able to compare the E(t) obtained by the model with the E(t) obtained experimentally, it is also necessary to calculate . On the other hand, to determine the deviation between the points obtained experimentally and with the method, the sum of the squared differences between all the points has been calculated:
Since the deviation is of the order of 2.36·10−3, it can be concluded that the method can well predict the experimental E(t) values.
For the dispersion model, the first thing to calculate is the Bo number. In this case, the operation that has been done is: σ2 = 2Bo, so Bo = 21.14.
Since Bo is greater than 0.01, a closed–closed system or an o–o system must be chosen. In this exercise, we will use the o–o assumption. The first thing that is done is to calculate the Bo with the equations that govern this o–o dispersion model:
Obtaining Bo = 0.074. The next step is to calculate E(t) with the closed–closed dispersion model. To do this, the following equation is used:
In view of the results, it is clear that the E(t) obtained with the model differs greatly from the E(t) obtained experimentally. Therefore, it can be concluded that this model will not correctly predict this example. On the other hand, when Bo is greater than 0.01 (as in this case), the impulse response is wide, and this gives an asymmetric E(t).
Problem 1.8 Consider a continuous reactor fed with a flow Q0 with a composition CA0. Both operating variables are held constant. The (total) volume of the reactor is VR and the space time, τ, is defined based on this volume. The kinetics obeys the expression , and there are no secondary reactions. The reactor is modeled as three identical continuous reactors, perfectly stirred, but it is considered that there is also a certain current “mixed” between the three cameras, which is equal to α·Q0. To visualize this, analyze the following diagram:
Get the dynamic expressions for , and as functions of CA0, CA1, CA2, CA3, α, τ, a, and k, as required.
Solution to Problem 1.8
We will assume well-mixed behavior in all three reactors but consider the accumulation term in the mass balance. In this sense, in the first reactor:
Dividing by the flow rate Q0:
In the second reactor:
And in the third one:
Problem 1.9 Draw the E(t) curves for the following systems, with each system represented in a separate sketch:
(a) An ideal PFR with a volume of 2 l and a feed rate of 1 l/min.
(b) An ideal CSTR with a volume of 2 l and a feed rate of 1 l/min.
(c) An ideal PFR with a volume of 2 l, followed by an ideal CSTR with a volume of 2 l. The feed flows in at a rate of 1 l/min.
(d) An ideal CSTR with a volume of 2 l, followed by an ideal PFR with a volume of 2 l. The feed flows in at a rate of 1 l/min.
(e) An ideal CSTR with a volume of 2 l in parallel with an ideal PFR with a volume of 2 l. The feed, with a volume flow rate of 1 l/min, is evenly split between the two reactors: 0.5 l/min to the CSTR and 0.5 l/min to the PFR. After the reactors, the streams are mixed. Provide the
E
(
t
) curve for the entire system, not for each individual reactor.
Solution to Problem 1.9
(a) An ideal PFR (volume 2 l, feed 1 l/min):
(b) An ideal CSTR (volume 2 l, feed 1 l/min).
(c) An ideal PFR (volume 2 l) is followed by an ideal CSTR (volume 2 l). The feed has a volume flow rate of 1 l/min.
(d) An ideal CSTR (volume 2 l) is followed by an ideal PFR (volume 2 l). The feed has a volume flow rate of 1 l/min.
Same E(t) as in case (c).
(e) An ideal CSTR (volume 2 l) in parallel with an ideal PFR (volume 2 l). The feed (volume flow rate 1 l/min) is equally divided over the reactors: 0.5 l/min to the CSTR and 0.5 l/min to the PFR. After the two reactors, the streams are mixed together. Give the
E
(
t
) curve of the complete system, not of each individual reactor.
Problem 1.10 During the “Chemical Reactor Engineering” course, there was a problem that involved sketching the E(t) curve for a system of interconnected ideal reactors. Several answers were provided, and four of the sketches are as follows:
Answer (1) appears to be the correct answer.
(a) Sketch a possible system of interconnected ideal reactors, resulting in sketch 1.
(b) What systems of interconnected reactors correspond to the other sketches?
(c) Sketch the corresponding
F
-curve for each system.
Denote PFRs in your answer as P1, P2 … and CSTRs as T1, T2 …, with respective volumes VP1, VP2, …, VT1, VT2, …. Denote the various volume flow rates as Q1, Q2 …
(d) Sketch the figure yourself, and put relevant information in it (points of intersection, areas, etc.), expressed in
V
P1
,
V
P2
, …,
V
T1
,
V
T2
, …,
Q
1
,
Q
2
, …
If the first-order irreversible reaction A → products, with rate constant k, takes place in this system of reactors, what is the conversion at the outlet of the system? (again, expressed in VP1, VP2, …, VT1, VT2, …, Q1, Q2, …)?
Solution to Problem 1.10
(a) and (b)
(c)
(d)
(e) In the PFR:
Then, in the CSTR:
In the mix point, at the end of the system:
Problem 1.11 At time t = 0, a step change in tracer concentration occurs in the feed of a reactor with unknown flow behavior. At the reactor outlet, the following F-curve is measured:
(a) Please sketch the
E
-curve of this unknown reactor, including all relevant information in the sketch.
(b) Which combination of ideal reactors has the same
E
-curve as the unknown reactor?
(c) What is the residence time of the unknown reactor?
(d) If the reaction A → B, with a rate
(−
r
B
) = − 4
C
A
kmol/m
3
·min, occurs in the unknown reactor at the residence time determined in part (c), what will be the conversion of A?
Solution to Problem 1.11
(a)
(b)
(c) Residence time is:
t
= 0.33 · 0 + 0.66 · 0.5 = 0.33 min
(d) A → B, with
(−
r
B
) = − 4
C
A
kmol/m
3
·min, takes place in the unknown reactor. What will be the conversion of A?
In a PFR, first-order reaction:
As only 66% of the flow passes through the reactor, the correct conversion is:
Problem 1.12 Three ideal PFRs with volumes of 2, 4, and 6 m3 are interconnected, as shown in the figure below. The water feed, with a volume flow rate (Q0) of 3 m3/min, is divided between the parallel PFRs as indicated:
(a) At
t
= 0, a pulse of salt is injected into the feed at point A. Please sketch the salt concentration versus time curves at points B, C, D, and E. Make sure to indicate the scale divisions on the axes.
After conducting the experiment described above, the split ratio of the feed over the two PFRs is changed such that in the new situation Q1 = 1 m3/min and Q2 = 1 m3/min. Once again, at t = 0, a pulse of 2 mol of salt is injected into the feed at point A.
(b) Sketch the salt concentration versus time curves at points B, C, D, and E, considering the new split ratio. Indicate the scale divisions on the axes.
(c) Sketch the
F
-curve for the overall system in both situations. Please note that the volume of the tubing connecting the PFRs can be neglected.
Solution to Problem 1.12
(a)
(b)
(c) In the first situation:
And in the second:
Problem 1.13 During the oxidation of phenol to cyclohexanone, assume that the reaction rate is given by r = k·CA (“A” being phenol) when excess O2 is present. The following parameter values are known:
V
= 66 ml
Q
0
= 3 ml/min
C
A0
= 0.1 mol/l
C
A
(out of the reactor) = 0.01 mol/l
T
= 80 °C
P
= 3 MPa
(a) What is the residence time for this reactor? Solve the steady-state PFR material balance. Given the steady-state outlet concentration of A, what is the value of the rate constant
k
?
(b) To perform an RTD step test on the reactor, two inert solvents are used: toluene and cyclohexane. The reactor (PFR) begins with toluene flowing through at steady state. At time
t
= 0, a valve is turned introducing the cyclohexane flow and cutting off the toluene flow. The measured RTD is shown in the figure.
What is the mean residence time for the reactor?
(c) From the data, estimate the dimensionless dispersion number, Bo, describing this reactor.
Solution to Problem 1.13:
For details refer the Wiley website at http://www.wiley-vch.de/ISBN9783527354115
(a) Residence time is:
t
=
V
/
Q
0
= 66/3 = 22 min
For the PFR in steady state: dnA = rAdV, which for a first-order reaction gives:
In the present case:
Giving a value of k = 0.1 min−1
(b) We can read in the plot the time–
F
(
t
) data and do the corresponding calculations:
Time (s)
F
(
t
)
E
(
t
) (s
−1
)
E
(
t
)·Δ
t
t
·
E
(
t
)
t
·
E
(
t
)·Δ
t
(
t
−
t
m
)
2
·
E
(
t
)
(
t
−
t
m
)
2
·
E
(
t
)·Δ
t
1200
0
0.0010
0.05
1.20
61.25
101
5062.5
1250
0.05
0.0010
0.05
1.25
161.25
1
62.5
1300
0.1
0.0040
0.2
5.20
231.25
605
30 250
1350
0.3
0.0030
0.15
4.05
223.75
1654
82 687.5
1400
0.45
0.0035
0.35
4.90
395.00
8409
840 875
1500
0.8
0.0020
0.1
3.00
113.75
6503
325 125
1550
0.9
0.0010
0.05
1.55
58.75
4651
232 562.5
1600
0.95
0.0005
0.05
0.80
40.00
6301
630 125
1700
1
0.0006
Sum=
1
t
m
(s)=
1245
σ
2
(s
2
)
2 146 750
t
m
(min)=
20.75
σ
2
(min
2
)
596.32
The value of E(t) is calculated by using a numerical method for deriving the F(t) values read in the graph. In this way:
The value of tm is calculated by integrating (i.e. sum) of the values of (t·E(t)·Δt). As we can see, in this system . In fact, so we would expect to get the E(t) signal later. This is probably due to the fact that flow rate is not measured correctly, or that there is a dead volume not contributing to the total residence time in the reactor. In any case, we first try to find the Bo number for applying the dispersion model. Assuming that Bo < 0.01, we can do σ2 = 2 · Bo, and we find that Bo ≫ 0.01, so the assumption is not correct. For Bo > 0.01 and assuming o–o conditions:
finding that Bo = 0.3113 that is actually quite a high value (Bo > 0.2). We can assume that the reactor would be similar to a CSTR. Indeed, if we apply the TIS model, we find that:
Problem 1.14 A reaction system is described by the following model:
Do a mass balance in this system for a tracer pulse input and determine the residence time distribution function in the case where tank 1 is very small compared to tank 2 and the transfer speed between the reactors is very small.
Solution to Problem 1.14
If we propose a molar balance on the tracer, with a pulse injected at t = 0 for each of the tanks, we obtain:
where CT1 and CT2 are, respectively, tracer concentrations in both reactors. These two differential equations are coupled and should be solved simultaneously.
In this model, the two adjustable parameters are the flow rate exchanged (Q1) and the volume of the most agitated region (V1). Remember that the measured volume (V) is the sum of V1 and V2. We will call β the fraction of the total flow that is transferred between both reactors:
and α to the fraction of the total volume that corresponds to the most agitated area:
On the other hand, the average time () is given by the quotient V/Q0.
The initial conditions (t = 0) for this model are: (i) CT1 = (CT1)0, and (ii) (CT2)0 = 0.
Analytical solution is possible in the present case and is the following:
being:
Previous equation shows that, if tank 1 is small compared to 2 (α small), and the transfer speed between both reactors is small (β small), the second exponential term tends to 1 during the first part of the response to an impulse injection. During the second part, the first exponential term tends to 0. If we represent the logarithm of the tracer concentration versus time, the response curve will tend to a straight line at both ends of the curve, and the parameters will be obtained from the slopes (m1 for t → ∞ and m2 for t → 0) and the cut points of both lines (for t → ∞, the cut point is −{αm2 + β + 1}/α{m1 − m2}).
Problem 1.15 Figure shows a combination of ideal reactors used to model a real reactor.
(a) Qualitatively sketch the RTD (i.e.
E
(
t
),
F
(
t
)) that would result if two of the following combinations of ideal reactors were connected in series (i.e.
n
= 1).
(b) How would your results change if two were connected in series (i.e.
n
= 2)?
(c) How about
n
= 5 or 10?
Solution to Problem 1.15
For details refer the Wiley website at http://www.wiley-vch.de/ISBN9783527354115
For n = 1, we have a signal corresponding to one CSTR with interchange displaced by a value of time corresponding to the RFP residence time. If n = 2, the signal is displaced two times:
For values of “n” higher, the system would behave as a PFR, with residence time n times the residence time of the first reactor.
In Problem 3.2, we will see that the transfer function of this system of reactors is (n = 1):
being β the ratio Q1/Q0. For a system where n = 2:
The inverse of the Laplace transform is a very complicated function E(t).
We can do a simulation using Matlab®. For a system consisting of two CSTRs with flow rate Q1 between them, expression for the pulse run concentration is that shown in the previous problem, but now α is the ratio V2/(V2 + V3).
For example, for n = 1–4, with V1 = 0.9 V and Q1 = 0.1·Q0, and , we can do:
clear all
close all
alfa=0.9;
beta=0.9;
tm_pfr=10;
tm_cstrs=20;
m_1=[(1-alfa+beta)/(2*alfa*(1-alfa))]*[-1+((1-(4*alfa*beta*(1-alfa))/
(1-alfa+beta)∧2))∧0.5];
m_2=[(1-alfa+beta)/(2*alfa*(1-alfa))]*[-1-((1-(4*alfa*beta*(1-alfa))/
(1-alfa+beta)∧2))∧0.5];
t=0:150;
td=(t-tm_pfr)/tm_cstrs;
E1=((alfa*m_1+beta+1)*exp(m_2*td)-(alfa*m_2+beta+1)*exp(m_1*td))/
alfa/(m_1-m_2);
E1=(t>tm_pfr).*E1;
subplot(2,2,1), plot(t,E1)
area=sum(E1*1);
E1=E1/area;
tmdistrib=sum(E1.*t*1)
title('n=1')
xlabel('Time')
ylabel('E(t)')
%%%%
E2=conv(E1,E1);
t2=2*t(1):2*t(end);
area=sum(E2*1);
E2=E2/area;
tmdistrib=sum(E2.*t2*1)
subplot(2,2,2), plot(t2,E2)
title('n=2')
xlabel('Time')
ylabel('E(t)')
%%%%
E3=conv(E2,E1);
t3=3*t(1):3*t(end);
area=sum(E3*1);
E3=E3/area;
tmdistrib=sum(E3.*t3*1)
subplot(2,2,3), plot(t3,E3)
title('n=3')
xlabel('Time')
ylabel('E(t)')
%%%%
E4=conv(E3,E1);
t4=4*t(1):4*t(end);
area=sum(E4*1);
E4=E4/area;
tmdistrib=sum(E4.*t4*1)
subplot(2,2,4), plot(t4,E4)
title('n=4')
xlabel('Time')
ylabel('E(t)')
Getting:
During the calculation, we can check that, for n = 1, .
For n = 2,
For n = 3,
For n = 4,
And comparing all four distributions:
Problem 1.16 A tracer step input experiment is being performed, feeding pure water, switching to salt water, and analyzing the conductivity at the outlet. The data represented in the figure was obtained.
(a) What will the output signal be like if the tracer had been injected into pulse?
(b) Calculate the mean residence time of the reactor.
Solution to Problem 1.16
(a) What we have obtained is a cumulative RTD (i.e.
F
(
t
) curve), and it is easy to find the time that the tracer will exit with
C
max
:
(b) The mean residence time is:
And:
Mean residence time:
The E(t) function is:
We can check that:
Problem 1.17 A test is performed with a tracer pulse, and a curve is obtained that follows the expression C = (t − 2)2 for 0 ≤ t ≤ 2 and C = 0 at any other time. Calculate the mean residence time using the equation for E(t) and F(t).
Solution to Problem 1.17
In the reactor, we have a pulse that input gives:
The area under the curve is defined between 0 and 2 (time where C(t) equals to zero). And in this way:
And so:
Following the definition of F(t):
being “K” an integration constant. By definition, we know that:
so the constant “K” should be equal to zero.
Problem 1.18 A reaction system is described by the following model:
At a given moment, the system presents values of α = 0.1 and β = 0.9. A series of reforms are carried out in the system, and it is achieved that α = 0.9 and β = 0.1.
(a) Qualitatively draw the response of the system to a tracer impulse, comparing the two situations.
(b) Also, draw the corresponding
F
(
t
) curves for both situations.
(c) Indicate what reforms had to be made to achieve this change.
Solution to Problem 1.18
(a) In this system, the residence time compared to that expected is:
As we have that α = 0.1 and β = 0.9 at the beginning:
In the second case:
So it will be very difficult to observe these changes. Nevertheless, the bypass produces a very important peak at time equal to zero in both situations, but the amount of tracer passing through the bypass is much higher in the first case. We can have something similar to:
(b) The cumulative time function would have the following aspect:
(c) Usually, a better agitation is procured to eliminate dead volumes. Also, reduction of bypass flow is achieved with better agitation or by changing the reactor filling.
Problem 1.19 In a tracer input experiment, a perfect pulse is injected, and the following distribution is obtained:
(a) Calculate the mean residence time of the reactor.
