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Lecture 24
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.
Web Lecture 24
Class Lecture 20-Thursday 3/28/2013
Review of Multiple Steady States (MSS)
Reactor Safety (Chapter 13)
 Blowout Velocity
 CSTR Explosion
 Batch Reactor Explosion
2
Review Last Lecture
CSTR with Heat Effects
3
Review Last Lecture
Energy Balance for CSTRs
dT
dt

FA 0
 N i C Pi
G T   R T 
G T   rA V  H Rx 
R T   C PS 1   T  TC 

4
 
UA
FA 0C P 0
TC 
T0   Ta
1
Review Last Lecture
Energy Balance for CSTRs
R T   C PS 1   T  TC 
R(T)

Increasing T0
T
5
Variation of heat removal line with inlet temperature.
Review Last Lecture
Energy Balance for CSTRs
κ=∞
R T   C PS 1   T  TC 
κ=0
R(T)

Increase κ
Ta
6
T0
T
Variation of heat removal line with κ (κ=UA/CP0FA0)
Review Last Lecture
Multiple Steady States (MSS)
Variation of heat generation curve with space-time.
7
Review Last Lecture
Multiple Steady States (MSS)
Finding Multiple Steady States with T0 varied
8
Review Last Lecture
Multiple Steady States (MSS)
Temperature ignition-extinction curve
9
Review Last Lecture
Multiple Steady States (MSS)
Bunsen Burner Effect (Blowout)
Review Last Lecture
Multiple Steady States (MSS)
Bunsen Burner Effect (Blowout)
Reversible Reaction
Gas Flow in a PBR with Heat Effects
A↔B
0C A0X
V 
kC

X 


1

X

A0 
K e 

G    H Rx X 
, X 

1 

1   k  1 

K
e 

T0  Ta
1 
 310
R T   400 T  310 
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
1 

1   k  1 
K e 

  H Rx  k
R T   C P 1   T  T C 
TC 
k
Reversible Reaction
Gas Flow in a PBR with Heat Effects
A↔B
UA = 73,520
UA = 0
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Reversible Reaction
Gas Flow in a PBR with Heat Effects
A↔B
KC 
8 
14
C Be
C Ae
Xe 

C A 0 X e y T0 T
C A 0 1  X e y T0 T
KC
1 KC
Reversible Reaction
Gas Flow in a PBR with Heat Effects
A↔B
Exothermic Case:
KC
Xe
T
T
Endothermic Case:
KC
15
~1
Xe
T
T
Adiabatic Equilibrium Conversion
Conversion on Temperature
Exothermic ΔH is negative
Adiabatic Equilibrium temperature (Tadia) and conversion (Xe,adia)
X
T  T0 
Xeadi
  H Rx X
a
Xe 
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Tadia
T
C PA
KC
1 KC
Gas Phase Heat Effects
Trends:
Adiabatic:
X exothermic
X
endothermic
Adiabatic T
and Xe
T0
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T
T  T0 
  H Rx X
C PA   I C PI
T0
T
Gas Phase Heat Effects
X
Xe 
KC
1 KC

Xeadia
T
T  T0 

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 H Rx
  iC P
i
Gas Phase Heat Effects
Effect of adding inerts on adiabatic equilibrium conversion
Adiabatic:
X
I  
Adiabatic Equilibrium
Conversion
I  0
T
T0
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X 
T

 T0  C P   I C P
A
 H Rx
I
,
T  T0 
 H Rx 
CP  ICP
A
I
Q1
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FA0
FA1
T0
X1
Q2
FA2
T0
X2
FA3
T0
X3
X
Xe
X3
XEB
X 
X2
X1
T0
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T
  C T  T 
i
Pi
  H RX
0
Adiabatic Exothermic Reactions
 H Rx   15
A 
B
kcal
mol
The heat of reaction for endothermic reaction is positive, i.e.,
Energy Balance :

T  T0 

 H Rx X
C P   IC P
A
and
I
CP

X 
A
 CP I
I
T0  T 
 H Rx

We want to learn the effects of adding inerts on conversion. How the
conversion varies with the amount, i.e., I, depends on what you vary
and what you hold constant as you change I.
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A. First Order Reaction
dX
dV

rA
FA 0
Combining the mole balance, rate law and stoichiometry


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dX
dV

kC A 0 1  X 
 0C A 0

k
0
1  X 
Two cases will be considered
Case 1 Constant 0, volumetric flow rate
Case 2: Variable 0, volumetric flow rate
A.1. Liquid Phase Reaction
For Liquids, volumetric flow rates are additive.
 0   A 0   I 0   A 0 1   I 

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Effect of Adding Inerts to an Endothermic Adiabatic Reaction
What happens when we add Inerts, i.e., vary Theta I??? It all depends
what you change and what you hold constant!!!
I
I
ISO
ISO
k
T
1   I 


V
I

k 1  yXI 

 kk
 1  
 
I

 k 


1   I 




1 I 

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\
V
k 1  y I 
1
 I OPT

I
 I OPT
I
A.1.a. Case 1. Constant 0
To keep 0 constant if we increase the amount of Inerts,
i.e., increase I we will need to decrease the amount of
A entering, i.e., A0. So I  then A0 
T  T0 
 H Rx X
C P   IC P
A

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I
A.1.a. Case 2. Constant A, Variable 0
dX
dV
27

k 1  X 
0

k 1  X 
 A 1   I 
A.2. Gas Phase
Without Inerts
C TA 
FA 0
A
 CA0 
With Inerts and A
PA
C TI 
RT A
Taking the ratio of CTA to CTI
F TI
I

FA 0  FI 0
I

PI
RT I
PI
C TI
I
RT I


PA
C TA F TA

A
Solving for I
I  A

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F TI
RT A
F TI P A T I
F TA P I T A
We want to compare what happens when Inerts and A are fed to the case when
only A is fed.
Nomenclature note: Sub I with Inerts I and reactant A fed
Sub A with only reactant A fed
FTI = Total inlet molar flow rate of inert, I, plus reactant A, FTI = FA0 + FI0
FTA = Total inlet molar flow rate when no Inerts are fed, i.e., FTA = FA0
PI, TI = Inlet temperature and pressure for the case when both Inerts (I) and A are fed
PA, TA = Inlet temperature and pressure when only A is fed
CA0 = Concentration of A entering when no inerts are presents
CTA = Total concentration when no inerts are present 
CA0 
PA
RT A
CTI = Total concentration when both I and A are present 
PI
RT I
CA0I = Concentration of A entering when inerts
 A are entering

FA 0
I
I = Entering volumetric flow rate with both Inerts
 (I) and reactant (A)
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
FA 0
A
F TI
F TA

FA 0  FI 0
FA 0
 1   I  
yA0 
1

 F
A0


F

F
 I 0
A 0 
1
1   I 

P A T I 
 I   A 1   I 

T
P

A 
I

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
1
yA0
A.2.a. Case 1
Maintain constant volumetric flow, 0, rate as inerts are added. I.e., 0 =
I = A. Not a very reasonable situation, but does represent one extreme.
Achieve constant 0 varying P, T to adjust conditions so term in
brackets, [ ], is one.

PA TI 
1   I 
 1
PI T0 

For example if I = 2 then I will be the same as A, but we need the
entering pressures PI and PA to be in the relationship PI = 3PA with TA = TI


 1 
PA TA 
 I   A 1  2  
  A 3    A   0
 3 
3PA TA 

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
A.2.a. Case 1
That is the term in brackets, [ ], would be 1 which would
keep 0 constant with I = A = 0. Returning to our
combined mole balance, rate law and stoichiometry
dX
dV

32

k 1  X 
0
A.2.b. Case 2: Variable 0 Constant T, P
i.e., PI = PA, TI = TA
I  A
F TI
F TA
 A
F A 0  F I 0 
FA 0
  A 1   I 
 I   A 1   I 
dX
dV

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
1
k
A 1 
1  X 
B. Gas Phase Second Order Reaction
Pure A
CA0 
PA
RT A

Inerts Plus A
FA 0
A
dX
dV

34

C TI 

rA
FA 0
2

kC A 0 I 1  X 

FA 0
2
F A 0 1   I 
I
B. Gas Phase Second Order Reaction
 I   A 1   I 
2
C A0I
FA 0

F A 0
I
2
FA 0

dX
dV
35



FA 0
2
I
2
A
 A 1   I 
1   I 
CA0
2
PI TA
FA 0

A
2
2
2 P A   T I 
  A 1   I    

 P I  T A 
2
CA0
k
PA TI
 P T 
I
A


P A T I 
2
 P T 
2
I
A
1

X




P
T
 A I 
B. Gas Phase Second Order Reaction
For the same temperature and pressures for the cases
with and without inerts, i.e., PI = PA and TI = TA, then
dX
dV

36

k
1   I 
CA0
2
A
1  X 
2
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
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Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
38
End of
39
Web Lecture 24
Class Lecture 20

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