Chemical Reaction Engineering

Course Code: CHE 331

Course Cr. Hrs.: 4(3,1)
Course Instructor:
Dr. Muhammad Haris Hamayun
Assistant Professor,
Department of Chemical Engineering,
COMSATS University Islamabad, Lahore Campus.
Contact Email: mhhamayun@cuilahore.edu.pk
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 Course Contents

Kinetics of homogeneous reactions: rate of reaction, variables affecting the rate of

reaction, order of reaction, rate constant; searching for a mechanism of reaction,
activation energy and temperature dependency. Interpretation of batch reactor data for
single and multiple reactions. Integral method and differential method of analysis for
constant volume and variable volume batch reactors, search for a rate equation. Design
of homogeneous reactors, Batch, Mixed flow, Plug flow reactors, Comparison of single
reactor, multiple reactor systems in parallel/series. Temperature and pressure effects.
Adiabatic and non-adiabatic operations. Surface phenomenon and catalysis,
Heterogeneous reaction systems, rate equations for heterogeneous reactions, fluid
particle reactions, determination of rate controlling steps. Catalysis desorption
isotherms, kinetics of solid catalyzed reactions. Catalyst deactivation and regeneration.
Design of fluid-solid catalytic reactors.

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 Recommended Books

1) H. Scott Fogler, Elements of Chemical Reaction Engineering, 5th edition, Prentice

Hall, 2016.

2) Octave Levenspiel, Chemical Reaction Engineering, 3rd edition, Wiley India Pvt.
Limited, 2006.

3) Elsie Perkins, Chemical Reaction Engineering, WILLFORD Press, 2022.

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 CLOs and Mapping with PLOs

Understand • Describe the fundamentals of chemical reaction

(C2, PLO1) engineering.

Apply • Apply the fundamentals of chemical reaction

(C3, PLO1) engineering.

Analysis • Analyze the kinetic data using different methods of

(C4, PLO2) data analysis.

Design • Design isothermal and nonisothermal reactors (e.g.,

(C6, PLO3) Batch, CSTR, PFR, PBR etc.)

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 OBE in a Nutshell

◼ What do you want the students to have or able

to do? ◼ Knowledge, Skill, Affective

◼ How can you best help students achieve it?

◼ Student Centred Delivery

◼ How will you know what they have achieved it?

◼ Assessment

◼ How do you close the loop

◼ Plan, Do, Check, Act (PDCA)
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 Today’s Course Coverage (CLO # 2 and 4)

• Chapter # 5: Isothermal Design

❖ Design Structure for Isothermal Reactors

❖ Algorithm for Isothermal Reactors

❖ Batch Reactors
▪ Batch Reaction Time

▪ Example

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 Design
Structure
for
Isothermal
Reactors

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Algorithm
for
Isothermal
Reactors

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So, what we need to do in this Chapter?

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Scale Up of Processes

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 Batch Reactors

• In modeling a batch reactor, we assume there is no

inflow or outflow of material, and that the reactor is
well mixed.
• For most liquid-phase reactions, the density change with
reaction is usually small and can be neglected (i.e., V =
V0).
• In addition, for gas-phase reactions in which the batch
reactor volume remains constant, we also have V = V0.

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 Batch Reaction Times
2A ↔ B + C
dX 2
NA0 = −rA V0 −rA = kCA0 (1 − X)2
dt
Mole Balance: Combine:
dX −rA 2
dX kCA0 (1 − X)2
= =
dt CA0 dt CA0

dX
= k2CA0 dt
Rate Law: −rA = kCA2 (1 − X)2
Evaluate:
t X
dX
k 2 CA0 න dt = න 2
0 0 (1 − X)
Stoichiometry: CA = CA0 (1 − X) 1 X
tR =
k 2 CA0 1 − X 12
 Algorithm to Estimate tR

2A ↔ B + C

dX −rA
Mole Balance =
dtR CA0

Rate Law First Order −rA = kCA Second Order −rA = kCA2

Stoichiometry NAൗ
CA = V0 = CA0 (1 − X)
2
dX dX k2CA0 (1 − X)2
Combine First Order = k1(1 − X) Second Order =
dtR dtR CA0
1 1 1 X
Evaluate First Order t R = ln Second Order tR =
k1 1−X k 2 CA0 1 − X
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 Batch Reaction Times
2A ↔ B + C
First Order Reaction, X = 0.9, k1 = 10-4 s-1 Second Order Reaction, X = 0.9, k2CA0= 10-3 s-1

1 1 1 X
t R = ln tR =
k1 1−X k 2 CA0 1 − X

1 1 1 0.9
t R = −4 ln t R = −3
10 1 − 0.9 10 1 − 0.9

t R = 23,000 sec = 6.4 hours t R = 9,000 sec = 2.5 hours

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 Batch Reaction Times
• The total cycle time in any batch operation is considerably longer than the reaction time, tR, as
one must account for the time necessary to fill (tf) and heat (te) the reactor together with the
time necessary to clean the reactor between batches, tc.

tt = tf + te + tR + tc

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Example # 5.1

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Example # 5.1

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 Building Blocks # 1 and 2
A+B→C
Mole Balance

1 dNA dCA
= rA V = V0 − = −rA
V dt dt

Rate Law

−rA = kCA

Because water is present in such excess, the concentration of water at any time t is
virtually the same as the initial concentration, and the rate law is independent of the
concentration of H2O. CB = CB0
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 Building Block # 3
Initial Change Concentration
Specie Symbol Remaining (mol)
(mol) (mol) (mol/dm3)
CH2CH2O A NA0 −NA0 X NA = NA0 (1 − X) CA = CA0 (1 − X)
CB = CA0 (θB − X)
H2O B NB0 = θB NA0 −NA0 X NB = NA0 (θB − X)
CB ≡ CB0
CC = CA0 X
(CH2OH)2 C 0 +NA0 X NC = NA0 X
CC = CA0 − CA
NT0 NT = NT0 − NA0 X

NB0 55 NA = NA0 − NA0 X

θB = = = 55 CA0 X = CA0 − CA
NA0 1
The maximum value of X is 1, and θB ≫ 1 ,
therefore, CB is virtually constant. NA0 X = NA0 − NA 19
 Combine and Evaluate
Combine

dCA dCA
− = kCA − = kdt
dt CA

Evaluate

CA t
dCA CA0
−න = k න dt ln = kt CA = CA0 exp(−kt)
CA0 CA 0 CA

CA0 − CC
CC = CA0 (1 − exp −kt ) ln = −kt
CA0

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MS Excel Calculations

−rA = 0.3145 (min−1 )CA

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 Today’s Course Coverage (CLO # 2 and 4)

• Chapter # 5: Isothermal Design

❖ Continuous Stirred Tank Reactor (CSTR)

▪ A Single CSTR

✓ First and Second Order Reactions

✓ Damkohler Number

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 Continuous Stirred Tank Reactor

Mole Balance

FA0 X
V=
(−rA )exit

𝑣0 CA0 X
V=
(−rA )exit

V CA0 X
𝜏= =
𝑣0 (−rA )exit

This equation applies to a single CSTR or to the first reactor of CSTRs connected in series.
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 Single CSTR – 1st Order Reaction
Rate Law

−rA = kCA

Stoichiometry

CA = CA0 (1 − X)

Combine

CA0 X CA0 X 1 X
−rA = kCA0 (1 − X) 𝜏= = 𝜏=
(−rA )exit kCA0 (1 − X) k 1−X

𝜏k
X=
1 + 𝜏k 24
 Single CSTR – 1st Order Reaction

𝜏k
CA = CA0 1 − X = CA0 1−
1 + 𝜏k

CA0
CA =
1 + 𝜏k

We can increase τk by either increasing the temperature to increase k or increasing the space time τ by
increasing the volume V or decreasing the volumetric flow rate v0.
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 Single CSTR – 2nd Order Reaction
Rate Law

−rA = kCA2

Stoichiometry

CA = CA0 (1 − X)

Combine

2
CA0 X CA0 X
−rA = kCA0 (1 − X)2 𝜏= = 2
(−rA )exit kCA0 (1 − X)2

X
𝜏=
kCA0 (1 − X)2
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 Single CSTR – 2nd Order Reaction
Solving for X

1 + 2𝜏kCA0 − 1 + 2𝜏kCA0 2 − 2𝜏kCA0 2

X=
2𝜏kCA0

1 + 2𝜏kCA0 − 1 + 4𝜏kCA0
X=
2𝜏kCA0

1 + 2Da2 − 1 + 4Da2
X=
2Da2
This observation is a consequence of the fact that the CSTR operates
under the condition of the lowest reactant concentration (i.e., the exit
Da2 = 𝜏kCA0 concentration), and consequently the smallest value of the rate of reaction.
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 Damkohler Number
• It is a dimensionless number that can give us a quick estimate of the degree of
conversion that can be achieved in continuous flow reactors.

• It is the ratio of the rate of reaction of A to the rate of convective transport of A

evaluated at the entrance to the reactor.

−rA0 V Rate of reaction at entrance "A reaction rate"

Da = = =
FA0 Entering flowrate of A "A convection rate"

1st Order Reaction 2nd Order Reaction

−rA0 V k1 CA0 V 2
−rA0 V k 2 CA0 V
Da = = = 𝜏k1 Da = = = 𝜏k 2 CA0
FA0 𝑣0 CA0 FA0 𝑣0 CA0
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 Damkohler Number
• It is important to know what values of the Damköhler number, Da, give high and
low conversion in continuous-flow reactors.

• For irreversible reactions, a value of Da = 0.1 or less will usually give less than
10% conversion, and a value of Da = 10.0 or greater will usually give greater than
90% conversion; that is, the rule of thumb is

If Da < 0.1, then X < 0.1

If Da > 10, then X > 0.9

For 1st order liquid phase Da1

X=
reactions: 1 + Da1
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