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1. 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
1

2. 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.
2

3. 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.
3
Recommended Books

4. 4
CLOs and Mapping with PLOs
• Describe the fundamentals of chemical reaction
engineering.
Understand
(C2, PLO1)
• Apply the fundamentals of chemical reaction
engineering.
Apply
(C3, PLO1)
• Analyze the kinetic data using different methods of
data analysis.
Analysis
(C4, PLO2)
• Design isothermal and nonisothermal reactors (e.g.,
Batch, CSTR, PFR, PBR etc.)
Design
(C6, PLO3)

5. 5
OBE in a Nutshell
◼ What do you want the students to have or able
to do?
◼ How can you best help students achieve it?
◼ How will you know what they have achieved it?
◼ How do you close the loop
◼ Knowledge, Skill, Affective
◼ Plan, Do, Check, Act (PDCA)
◼ Student Centred Delivery
◼ Assessment

6. • Chapter # 5: Isothermal Design - Conversion
❖ Example # 5.4
❖ Analytical Solution for Reaction with Pressure Drop
❖ Example # 5.5
6
Lecture # 18 (CLO # 2, 3 and 4)

7. 7
Example # 5.4
p =
P
P0
= (1 −
2β0L
P0
) β0 =
G
ρ0gcDp
1 − φ
φ3
150 1 − φ μ
Dp
+ 1.75 G

8. 8
Calculation of Total Pressure Drop
G =
ሶ
m
Ac
=
104.4 (
lbm
h
)
0.01414 ft2
= 7383.3
lbm
h. ft2
For air at 260 °C, and 10 atm:
μ = 0.0673 lbm/ft. h
𝑝0 = 0.413 lbm/ft3
𝑣0 =
ሶ
m
p0
=
104.4
0.413
= 252.8 ft3/h
From Problem Statement:
Dp = ¼ inches = 0.0208 ft
φ = 0.45
gc = 4.17×108 lbm. ft/lbf. h2

9. 9
Calculation of Total Pressure Drop
β0 =
G
ρ0gcDp
1 − φ
φ3
150 1 − φ μ
Dp
+ 1.75 G
β0 =
7383.3
0.413 × 4.17 × 108 × 0.0208
1 − 0.45
0.453
×
150 1 − 0.45 (0.0673)
0.0208
+ 1.75 × 7383.3
β0 = 0.01244 × 266.9 + 12920.8 = 164.1
lbf
ft3
×
1 ft2
144 in.2 ×
1 atm
14.7
lbf
in.2
= 0.0775
atm
ft
= 25.8
kPa
m

10. 10
Calculation of Total Pressure Drop
p =
P
P0
= 1 −
2β0L
P0
p =
P
P0
= 1 −
2 × 0.0775 × 60
10
= 0.265
P = 0.265P0 = 2.65 atm (268 kPa)
∆P = P0 − P = 10 − 2.65 atm = 7.35 atm (744kPa)

11. 11
Plot for Pressure Drop
𝑣 = 𝑣0 1 + εX
P0
P
T
T0
• For ε = 0, and isothermal operation (T = T0):
𝑣 =
𝑣0
𝑝 p =
P
P0
= 1 −
2β0z
P0

12. 12
Plot for Pressure Drop
• For ρc = 120
lbm
ft3 :
α =
2β0
(1 − φ)AcρcP0
α =
2(0.0775)
(1 − 0.45)(0.01414)(120)(10)
α = 0.0165lbm
−1 = 0.037 kg−1

13. 13
Important Trends!!!
Mole Balance
Combine
Evaluate

14. 14
Second Order Reaction in PBR
A → B
1) Mole Balance:
FA0
dX
dW
= −rA
′
2) Rate Law:
−rA
′
= kCA
2
3) Stoichiometry:
For isothermal operation
and ε = 0;
CA =
FA
𝑣
=
FA0(1 − X)
𝑣0 1 + εX
P0
P
= CA0 1 − X
P
P0
= CA0 1 − X p

15. 15
Second Order Reaction in PBR
4) Combine:
p =
P
P0
= (1 − αW)
CA = CA0 1 − X p = CA0 1 − X (1 − αW)
dX
dW
=
kCA
2
FA0
=
kCA0
2
1 − X 2(1 − αW)
𝑣0CA0
=
kCA0 1 − X 2(1 − αW)
𝑣0

16. 16
Second Order Reaction in PBR
5) Evaluate: 𝑣0
kCA0
න
0
X
dX
(1 − X)2
= න
0
W
1 − αW dW
𝑣0
kCA0
X
1 − X
= W 1 −
αW
2
Integration:
Solving for X: X =
kCA0W/𝑣0 1 −
αW
2
1 + kCA0W/𝑣0 1 −
αW
2
Solving for W:
W =
1 − 1 −
2𝑣0α
kCA0
X
1 − X
1/2
α

17. 17
Example # 5.5

18. 18
Solution
X =
kCA0W/𝑣0 1 −
αW
2
1 + kCA0W/𝑣0 1 −
αW
2
ρb = ρc 1 − φ = 1923 1 − 0.45 = 1058
kg
m3
W = ACρc 1 − φ L = ACρbL = 0.0013 1058 20 = 27.5 kg
kCA0W
𝑣0
=
(12)(0.1)(27.5)
7.15
= 4.6

19. 19
A) Solution (for ΔP = 0)
X =
kCA0W/𝑣0 1 −
αW
2
1 + kCA0W/𝑣0 1 −
αW
2
α = 0
For no ΔP:
X =
kCA0W/𝑣0
1 + kCA0W/𝑣0
= 0.82

20. 20
B) Solution (for ΔP ≠ 0)
α =
2β0
P0ACρb
= 0.037kg−1
1 −
αW
2
= 1 −
0.037 × 27.5
2
= 0.49
X =
kCA0W/𝑣0 1 −
αW
2
1 + kCA0W/𝑣0 1 −
αW
2
=
4.6 ∗ 0.49
1 + (4.6 ∗ 0.49)
= 0.693
kCA0W
𝑣0
=
(12)(0.1)(27.5)
7.15
= 4.6
From Slide 13:
IMPORTANT:
We see the predicted
conversion dropped from
82.2% to 69.3% because
of pressure drop. It
would be not only
embarrassing but also an
economic disaster if we
had neglected pressure
drop and the actual
conversion had turned
out to be significantly
smaller.

21. 21
C) Solution (for Dp is doubled)
• The particle diameter has inverse relationship with both α & β.
α2 = α1
Dp1
Dp2
= 0.037 ∗
1
2
= 0.0185 kg−1
X =
kCA0W/𝑣0 1 −
αW
2
1 + kCA0W/𝑣0 1 −
αW
2
= 0.774

22. 22
Optimal Particle Diameter