L1-1
Chemical Reaction Engineering
Introduction & Lecture 1
L1-2
Required Texts
• Fogler, H. S. Elements of Chemical Reaction Engineering. 4th ed.
Upper Saddle River, NJ: Prentice-Hall PTR, 2006. ISBN:
9780130473943.
Recommended Texts
• Levenspiel, O. Chemical Reaction Engineering. 3rd ed. New York,
NY: Wiley, 1999. ISBN: 9780471254249.
• Smith, J. Chemical Engineering Kinetics. 3rd ed. New York, NY:
McGraw-Hill, 1981. ISBN: 9780070587106.
• Steinfeld, J. I., J. S. Francisco, and W. L. Hase. Chemical Kinetics
and Dynamics. 2nd ed. Upper Saddle River, NJ: Prentice Hall, 1999.
ISBN: 9780137371235.
• Bailey, J. E., and D. F. Ollis. Biochemical Engineering Fundamentals.
2nd ed. New York, NY: McGraw-Hill, 1986. ISBN: 9780070032125.
• Stephanopoulos, G., A. Aristidou, and J. Nielsen. Metabolic
Engineering: Principles and Methodologies. San Diego, CA:
Academic Press, 1998. ISBN: 9780126662603.
L1-3
Grading
ACTIVITIES
Homework
Quiz
Midterm exam 1
Midterm exam 2
Final exam
PERCENTAGES
L1-4
What is Chemical Reaction Engineering
(CRE) ?
Understanding how chemical reactors work lies at the heart of
almost every chemical processing operation.
Raw
material
Separation
Process
Chemical
process
Separation
Process
Products
By-products
Design of the reactor is no routine matter, and many
alternatives can be proposed for a process. Reactor design
uses information, knowledge and experience from a variety of
areas - thermodynamics, chemical kinetics, fluid mechanics,
heat and mass transfer, and economics.
CRE is the synthesis of all these factors with the aim of
properly designing and understanding the chemical reactor.
L1-5
How do we design a chemical reactor?
Type & size
Maximize the space-time yield of the desired product
(productivity lb/hr/ft3)
Stoichiometry
Kinetics
Basic molar balances
Fluid dynamics
Reactor volume
Use a lab-scale reactor to determine the kinetics!
L1-6
Reactor Design
Reaction
Stoichiometry
Kinetics: elementary vs non-elementary
Single vs multiple reactions
Reactor
Isothermal vs non-isothermal
Ideal vs nonideal
Steady-state vs nonsteady-state
L1-7
What type of reactor(s) to use?
in
Continuously Stirred
Tank Reactor (CSTR)
out
Well-mixed batch reactor
Plug flow reactor (PFR)
L1-8
What size reactor(s) to use?
Answers to this questions are based on the desired
conversion, selectivity and kinetics
Reactor type
&
size
Kinetics
Material &
energy
balances
Conversion
&
selectivity
L1-9
Chemical Reaction
• A detectable number of molecules have lost their identity
and assumed a new form by a change in the kind or number
of atoms in the compound and/or by a change in the atoms’
configuration
• Decomposition
• Combination
• Isomerization
• Rate of reaction
– How fast a number of moles of one chemical species are
being consumed to form another chemical species
L1-10
Elementary and Nonelementary rxns
◉ elementary rxn: 양론계수와 화학반응 차수가 같은 반응
예) A + B→ R의 반응 속도가 −𝑟𝐴 = 𝑘𝐶𝐴 𝐶𝐵 인 경우.
(기초 반응은 주로 반응속도가 분자간의 충돌 속도에 의존하는 경우임)
◉ nonelementary rxn: 양론계수와 화학반응 차수 사이에
아무런 대응 관계가 없는 반응
예) 𝐻2 + 𝐵𝑟2 → 2𝐻𝐵𝑟 의 반응 속도(실험적으로):
𝑟𝐻𝐵𝑟 =
𝑘1 [𝐻2 ][𝐵𝑟2 ]1/2
𝑘2 + 𝐻𝐵𝑟 /[𝐵𝑟2 ]
비기초 반응은 단일 반응으로 관찰되는 것이 실제로는 2개 이상의 기초
반응이 연속적으로 일어나는 것이며, 이것은 중간 생성물의 양이 아주
작거나 검출하지 못하기 때문임.
L1-11
Rate Law for rj
• rA: the rate of formation of species A per unit volume [e.g., mol/m3•s]
• -rA: the rate of a consumption of species A per unit volume
A  B  products
r A  kC A CB
1st order in A, 1st order in B, 2nd order overall
r A  kCAn
 rA 
nth order in A
k1CA
Michaelis-Menton: common in enzymatic reactions
1  k 2CA
rj depends on concentration and temperature:
-rA 
 Ea 
 RT 
C
A e
A
Arrhenius dependence on temperature
A: pre-exponential factor
R : ideal gas constant
E A : activation energy
T:temperature
L1-12
Basic Molar Balance (BMB)
Fj0
Fj
Gj
System volume
Rate of
flow of j into
system
Rate of
Rate of
Rate of
flow of j + generation of j Rate of
decomposition =
out of
by chemical
accumulation
of j
system
rxn
combine
Fj
0

 mol 


 s 
in
Fj

 mol 


 s 
-
out
Gj
 mol 


 s 
+
generation
Nj: moles j in
system at time t

dN j
dt
d
mol 
dt
= accumulation
L1-13
Basic Molar Balance (BMB)
Rate of
flow of j into
system
Fj
0
 mol 


 s 

Rate of
Rate of
Rate of
flow of j + generation of j Rate of
decomposition =
out of
by chemical
accumulation
of j
system
rxn
Fj
 mol 


 s 

Gj
dN j

dt
 mol 


 s 
d
mol 
dt
If the system is uniform throughout its entire volume, then:
G j  rj V
Moles
Moles j
generated per
generated
= unit time and
per unit time
volume
(mol/s)
(mol/s•m3)
Volume
(m3)
L1-14
Non-Uniform Generation
system
If rj varies with position (because
the temperature or concentration
varies) then rj1 at location 1 is
surrounded by a small subvolume
DV within which the rate is uniform
DV
V
m
G j  lim
DV
Rate is rj1
within this
volume
 rjDV   rjdV
m→∞ i1
DV→0
z
1
111
then G j     rj x, y, z  dx dy dz
000
1
x
1
y
Plug in rj and integrate over x, y, and z
Rate is rj2
within this
volume
L1-15
Basic Molar Balance Equations
Fj0
Fj
Gj
System volume
In - Out + Generation = Accumulation
dN j
Fj0  F j G j 
dt
dN j
Fj0  F j rj V 
dt
V
dN j
Fj0  F j   rjdV 
dt
uniform rate in V
nonuniform rate in V
Next time: Apply BME to ideal batch, CSTR, & PFR reactors
L1-16
Review of Frequently Encountered
Math Concepts
L1-17
Basic Math Review
1
x
 
ln x y  y ln  x 
n
 x n
p
x
q
ln a
e   a
q p
 x
x
ln  x   ln  y   ln  
y
ln  x   ln  y   ln  xy 
Example: Problems that Contain Natural Logs
Solve for X:
a
ln  bt  1  ln  x   ln  y 
b


a
x
a b
ln  bt  1  ln  x   ln  y   ln  bt  1
 ln  
b
y

ln  bt 1a b
e

x
ln 
e y
  bt  1
a b
x

y
 y  bt  1
a b
x
L1-18
Review of Basic Integration
1
x
b
1
b
n
p
 x n
x

 n1
For n≠1:  n dx   x dx   x
ax
a
b
n
q

q p
 x
b
b n1 a n1


 n  1 
a
n  1 n  1
1
b
b


dx

ln
x

ln

ln
b

ln
a


For n=1:  n




 

a
a
x
a
Solve
for t:
5
c t   1    1   c  t  0 
 1
ct
 1


 2    dt        t  0
5
d


d
d0
 x 1
1x
5 dx
c
 0.2  1   t
d
c
 0.8   t
d
d
 0.8  t
c
Solve for c:
L1-19
 k 
 k 
dc
dc

c



dt




1

k
t
dt
1

k
t
c

d 

d 
t
c dc
1
 k 
dt  
0 1 kdt
c c
Do NOT move t or c outside of the integral
0
x
x
x dx
From
1

 ln 1   x  
Appendix A:  1   x  
0
0
ε is a
constant
t
1

c
k  ln 1  k d t    ln  c   c
0
kd
0
1

1
 k  ln 1  k d t   ln 1  k d 0    ln  c   ln  c 0 
kd
0
kd


k
ln  k d t  1  ln  c   ln  c 0 
kd
 k

ln
k
t

1




d
kd


e

  c 
ln  
c
e  0 

 c 
k

ln  k d t  1  ln  
kd
 c0 
 k

 ln k dt 1 
k

e d

c
c0
 c0
 k

ln
k
t

1




d
kd


e
c