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Lecture 1
Chemical Reaction Engineering (CRE) is the
field that studies the rates and mechanisms of
chemical reactions and the design of the reactors in
which they take place.
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Today’s lecture
 Introduction
 Definitions
 General Mole Balance Equation




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Batch (BR)
Continuously Stirred Tank Reactor (CSTR)
Plug Flow Reactor (PFR)
Packed Bed Reactor (PBR)
Chemical Reaction Engineering
Chemical reaction engineering is at the heart of virtually every
chemical process. It separates the chemical engineer from other
engineers.
Industries that Draw Heavily on Chemical Reaction
Engineering (CRE) are:
CPI (Chemical Process Industries)
Examples like Dow, DuPont, Amoco, Chevron
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Smog (Ch. 1)
Wetlands (Ch. 7 DVD-ROM)
Hippo Digestion (Ch. 2)
Oil Recovery
(Ch. 7)
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Chemical Plant for Ethylene Glycol (Ch. 5)
Lubricant Design
(Ch. 9)
Cobra Bites
(Ch. 6 DVD-ROM)
Plant Safety
(Ch. 11,12,13)
Materials on the Web and CD-ROM
http://www.umich.edu/~essen/
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Let’s Begin CRE
 Chemical Reaction Engineering (CRE) is the field that
studies the rates and mechanisms of chemical reactions and
the design of the reactors in which they take place.
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Chemical Identity
 A chemical species is said to have reacted when it has lost its
chemical identity.
 The identity of a chemical species is determined by the kind,
number, and configuration of that species’ atoms.
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Chemical Identity
 A chemical species is said to have reacted when it has lost its
chemical identity. There are three ways for a species to loose
its identity:
1. Decomposition
2. Combination
3. Isomerization
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CH3CH3  H2 + H2C=CH2
N2 + O2  2 NO
C2H5CH=CH2  CH2=C(CH3)2
Reaction Rate
 The reaction rate is the rate at which a species looses its
chemical identity per unit volume.
 The rate of a reaction (mol/dm3/s) can be expressed as
either:
 The rate of Disappearance of reactant:
-rA
or as
 The rate of Formation (Generation) of product: rP
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Reaction Rate
Consider the isomerization
AB
rA = the rate of formation of species A per unit
volume
-rA = the rate of a disappearance of species A per unit
volume
rB = the rate of formation of species B per unit
volume
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Reaction Rate
EXAMPLE: AB
If Species B is being formed at a rate of
0.2 moles per decimeter cubed per second, ie,
rB = 0.2 mole/dm3/s
Then A is disappearing at the same rate:
-rA= 0.2 mole/dm3/s
The rate of formation (generation of A) is
rA= -0.2 mole/dm3/s
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Reaction Rate
 For a catalytic reaction, we refer to -rA', which is the rate of
disappearance of species A on a per mass of catalyst basis.
(mol/gcat/s)
NOTE: dCA/dt is not the rate of reaction
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Reaction Rate
Consider species j:
1. rj is the rate of formation of species j per unit volume
[e.g. mol/dm3s]
2. rj is a function of concentration, temperature, pressure,
and the type of catalyst (if any)
3. rj is independent of the type of reaction system (batch,
plug flow, etc.)
4. rj is an algebraic equation, not a differential equation
(e.g. = -rA = kCA or -rA = kCA2)
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General Mole Balance
System
Volume, V
Fj0
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Gj
Fj
 Molar Flow  Molar Flow   Molar Rate   Molar Rate 
 Rate of
   Rate of
  Generation    Accum ulation

 
 
 

 Species j in   Species j out of Species j  of Species j 
dN j
Fj 0

Fj

Gj

dt
 m ole
 m ole
 m ole
 m ole











 tim e 
 tim e 
 tim e 
 tim e 
General Mole Balance
If spatially uniform
G j  r jV
If NOT spatially uniform

V1
rj1
G j1  rj1V1
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

V2
rj 2
G j 2  rj 2 V2
General Mole Balance
W
G j   rjiVi
i1
Take limit
n
Gj  
rjiVi
i1 lim V  0 n  
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
 r dV
j
General Mole Balance
System
Volume, V
FA0
GA
FA
General Mole Balance on System VolumeV
 Out  Generation  Accumulation
dN A
FA 0  FA
  rA dV

dt
In
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Batch Reactor Mole Balance
Batch
FA 0  FA 

dN A
rA dV 
dt
FA 0  FA  0
Well Mixed

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r
A
dV  rAV
dNA
 rAV
dt
Batch Reactor Mole Balance
dN A
dt 
rAV
Integrating
when t = 0 NA=NA0
t = t NA=NA

t
NA

N A0
dN A
 rAV
Time necessary to reduce number of moles of A from NA0 to NA.
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Batch Reactor Mole Balance
t
NA

N A0
NA
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t
dN A
 rAV
CSTR Mole Balance
CSTR
dNA
FA 0  FA   rA dV 
dt
Steady State

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dNA
0
dt
CSTR Mole Balance
Well Mixed
 r dV  r V
A
A
FA 0  FA  rAV  0

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FA 0  FA
V 
rA
CSTR volume necessary to reduce the molar flow rate from
FA0 to FA.
Plug Flow Reactor
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Plug Flow Reactor Mole Balance
V
FA
FA


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V
V  V
 In  Out
 Generation

0
at V   



 at V  V  in V

FA V  FA V  V  rA V
0
Plug Flow Reactor Mole Balance
Rearrange and take limit as ΔV0
lim
V  0

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FA V V  FA V
V
 rA
dFA
 rA
dV
This is the volume necessary to reduce the entering molar flow
rate (mol/s) from FA0 to the exit molar flow rate of FA.
Alternative Derivation –
Plug Flow Reactor Mole Balance
PFR
dN A
FA0  FA   rA dV 
dt
Steady State
dN A
0
dt
FA0  FA   rA dV  0
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Alternative Derivation –
Plug Flow Reactor Mole Balance
Differientiate with respect to V
dFA
0
 rA
dV

The integral form is:
dFA
 rA
dV
V 
FA

FA 0
dFA
rA

This is the volume necessary to reduce the entering molar flow
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rate (mol/s) from FA0 to the exit molar flow rate of FA.
Packed Bed Reactor Mole Balance
PBR
dN A
FA W   FA W  W   rA W 
dt
dN A
Steady State
0
dt
lim
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W 0
FA W  W  FA W
W
 rA
Packed Bed Reactor Mole Balance
Rearrange:
dFA
 rA
dW
The integral form to find the catalyst weight is:

W
FA

FA 0
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dFA
rA
PBR catalyst weight necessary to reduce the entering molar
flow rate FA0 to molar flow rate FA.
Reactor Mole Balance Summary
Reactor
Batch
Differential
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Integral
t
dN A
 rAV
dt
NA

N A0
dN A
rAV
NA
t
FA 0  FA
V 
rA
CSTR
PFR
Algebraic
dFA
 
rA
dV
V
FA

FA 0
FA
dFA
drA
V
32
Fast Forward 10 weeks from now:
Reactors with Heat Effects
 EXAMPLE: Production of Propylene Glycol in an Adiabatic
CSTR
 Propylene glycol is produced by the hydrolysis of propylene
oxide:
H 2 SO4
CH2  CH  CH3  H2O 
CH2  CH  CH3
O
OH


OH
v0
Propylene Glycol
What are the exit conversion X and exit temperature T?
Solution
Let the reaction be represented by
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A+BC
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Evaluate energy balance terms
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Analysis
We have applied our CRE algorithm to calculate the
Conversion (X=0.84) and Temperature (T=614 °R) in a 300
gallon CSTR operated adiabatically.
T=535 °R
A+BC
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X=0.84
T=614 °R
KEEPING UP
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Separations
Filtration
Distillation
These topics do not build upon one another
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Adsorption
Reaction Engineering
Mole Balance
Rate Laws
These topics build upon one another
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Stoichiometry
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
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Mole Balance
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Rate Laws
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
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End of Lecture 1
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Supplemental Slides
Additional Applications of CRE
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Additional Applications of CRE
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Compartments for perfusion
Alcohol
Stomach
VG = 2.4 l
Gastrointestinal
VG = 2.4 l
tG = 2.67 min
Liver
VL = 2.4 l
tL = 2.4 min
Perfusion interactions between
compartments are shown by arrows.
VG, VL, VC, and VM are -tissue water
volumes for the gastrointestinal,
liver, central and muscle
compartments, respectively.
VS is the stomach contents volume.
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Central
VC = 15.3 l
tC = 0.9 min
Muscle & Fat
VM = 22.0 l
tM = 27 min
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