L1-1
CHBE 424:
Chemical Reaction Engineering
Introduction & Lecture 1
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-2
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.
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-3
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!
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-4
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-5
What type of reactor(s) to use?
in
Continuously Stirred
Tank Reactor (CSTR)
out
Well-mixed batch reactor
Plug flow reactor (PFR)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-6
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-7
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-8
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-9
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-10
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)
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-11
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
Rate is rj2
within this
volume
m→∞ i1
DV→0
z
1
111
then G j     rj x, y, z  dx dy dz
000
1
1
y
Plug in rj and integrate over x, y, and z
x
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-12
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-13
Review of Frequently Encountered
Math Concepts
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-14
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
L1-15
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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.
Solve for c:
L1-16
 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
Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois, Urbana-Champaign.