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.

2
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

3
Review Last Lecture
CSTR with Heat Effects

Review Last Lecture
Energy Balance for CSTRs dT dt

F
A 0
 N i
C
P i
     
  
  
 H
Rx


   C
P
S

1    
T  T
C

 
UA
F
A 0
C
P 0
T
C

T
0

1 
 T a

4

 

Review Last Lecture
Energy Balance for CSTRs
   C
P
S

1    
T  T
C

Increasing T
0
5
T
Variation of heat removal line with inlet temperature.

Review Last Lecture
Energy Balance for CSTRs
κ=∞
   C
P
S

1    
T  T
C

κ=0
R(T) 
Increase κ
6
T a
T
0
T
Variation of heat removal line with κ (κ=UA/C
P0
F
A0
)

Review Last Lecture
Multiple Steady States (MSS)
7
Variation of heat generation curve with space-time.

Review Last Lecture
Multiple Steady States (MSS)
8
Finding Multiple Steady States with T
0 varied

Review Last Lecture
Multiple Steady States (MSS)
9
Temperature ignition-extinction curve

Review Last Lecture
Multiple Steady States (MSS)
Bunsen Burner Effect (Blowout)

Review Last Lecture
Multiple Steady States (MSS)
Bunsen Burner Effect (Blowout)

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Reversible Reaction
Gas Flow in a PBR with Heat Effects
A ↔ B
V


0
C
A 0
X kC
A 0


 1

X

X
K e



, X
  
H

 
H
 k
Rx G
R
Rx
X
 

C
P

1
 

1



 1

T

T
C


 k
1

K e
T
C
R


T
0
 
T a

310
1
 
400

T

310




 k
1
  k


 1

1
K e




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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
K
C

C
Be
C
Ae

C
A 0
C

A 0
X e
1  X y T
0 e
 y T
T
0
 
X e

K
C
1  K
C
T

Reversible Reaction
Gas Flow in a PBR with Heat Effects
A ↔ B
Exothermic Case:
K
C
X e
T
15
T
Endothermic Case:
K
C
X e
T T
~1

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Adiabatic Equilibrium Conversion
Conversion on Temperature
Exothermic ΔH is negative
Adiabatic Equilibrium temperature (T adia
) and conversion (X e,adia
)
X a
X eadi
T

T
0


 
H
Rx

X
C
PA
X e

1
K
C

K
C
T adia
T

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Gas Phase Heat Effects
Trends:
Adiabatic:
X exothermic X endothermic
Adiabatic T and X e
T
0
T
T

T
0


C
PA

H
 
Rx
I
X
C
PI
T
0
T

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Gas Phase Heat Effects
X
X e

K
C
1  K
C

X eadia
T
T  T
0

 H
Rx
 i
C
P i


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Gas Phase Heat Effects
Effect of adding inerts on adiabatic equilibrium conversion
Adiabatic:
X

I
 
Adiabatic Equilibrium

Conversion

0
I
T
T
0
X 

T  T
0
 
C
P
A
 H
Rx
 
I
C
P
I

, T  T
0


C
P
A
 H
 
Rx
I

C
P
I


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F
A0
T
0
F
A1
X
1
Q
1
T
0
F
A2
X
2
Q
2
T
0
F
A3
X
3

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X
X
3
X
2
X
1
T
0
X e
X
EB
X

  i
C
Pi

T

T
0

 
H
RX
T

Adiabatic Exothermic Reactions
A  B  H
Rx
  15 kcal mol
The heat of reaction for endothermic reaction is positive, i.e.,
 :
T  T
0

C

P
A
H
Rx
X
 
I
C
P
I
and X 

C
P
A
 C
P
I

 H
I
Rx
 
T
0
 T


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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

 r
A
F
A0
Combining the mole balance , rate law and stoichiometry
 dX dV
 kC
A0


0
C
1  X
A0

 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
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For Liquids, volumetric flow rates are additive.

0
 
A0
 
I0
 
A0

1  
I



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
ISO
T

V
  y
I




1
1
 k

I
I



 k

I

\
V
 k 1  y


 k
1  
I



ISO

1
1  
I



I
 
I

I OPT

I
   


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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  T
0

C

P
A
H
Rx
 
X
I
C
P
I


A.1.a. Case 2. Constant 
A,
Variable 
0 dX dV

  X


0


A
  X


1  
I

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A.2. Gas Phase
Without Inerts With Inerts and A

C
TA

F
A0

A
 C
A0

Taking the ratio of C
TA to C
TI
P
A
RT
A
Solving for

I
C
C
TI
TA


F
TI


I
F
TA
A

C

I
 
A
F
TI
F
TA
P
A
P
I
TI
P
I
RT
I
P
A
RT
A
T
I
T
A

F
TI

I

F
A0

 F
I0
I

P
I
RT
I
28 only A is fed.


Nomenclature note: Sub I with Inerts I and reactant A fed
Sub A with only reactant A fed
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F
TI
= Total inlet molar flow rate of inert, I, plus reactant A, F
TI
= F
A0
+ F
I0
F
TA
= Total inlet molar flow rate when no Inerts are fed, i.e., F
TA
= F
A0
P
I
, T
I
= Inlet temperature and pressure for the case when both Inerts (I) and A are fed
P
A
, T
A
= Inlet temperature and pressure when only A is fed
C
A0
= Concentration of A entering when no inerts are presents
C
TA
= Total concentration when no inerts are present
C
TI
C
A0I
C
A0

F
A0

A

P
A
RT
A

P
I
RT
I

F
A0

I

I


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
F
TI
F
TA

F
A0
 F
I0
F
A0
 
1  
I
 
1

 F
I0
F
A0
 F
A0



1 y
A0 y
A0


1
1  
I


I
 
A


1  
I

P
A
P
I
T
I
T
A



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
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.


1  
I

P
A
P
I
T
I
T
0


1
For example if

I
= 2 then

I entering pressures P
 I and P
A will be the same as

A
, but we need the to be in the relationship P
I
= 3P
A with T
A
= T
I

I
 
A



1  2
 
P
A
3P
A
T
A
T
A


A

3 
1 
3  A
 
0

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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

  X


0


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A.2.b. Case 2: Variable 
0 i.e., P
I
= P
A
, T

I
 
A
I
= T
A
F
TI  
F
TA
A

F
A0
 F
I0
F
A0

 
Constant T, P
A

1  
I


I
 
A

1  
I
 dX dV

1

A k
1  

1  X



B. Gas Phase Second Order Reaction
Pure A Inerts Plus A
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
C
A0

P
A
RT
A

F
A0

A
C
TI

F
A0

1  
I


I dX dV

 r
A
F
A0
 kC
2
A0I

1  X
 2

F
A0


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
B. Gas Phase Second Order Reaction

I
 
A

1  
I

P
A
P
I
T
I
T
A
C
2
A0I
F
A0


F
A0
F
A0

I
 2

F

A0
I
2


A
 
A
F
A0

1  
I
 2


P
A
P
I


2 

T
I
T
A


2


A

C
A0
1  
I
 2


P
I
P
A
T
A
T
I


2 dX dV
 k

1  
I
 2
C

A0
A


P
I
P
A
T
A
T
I


2

1  X
 2

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B. Gas Phase Second Order Reaction
For the same temperature and pressures for the cases with and without inerts, i.e., P
I
= P
A and T
I
= T
A
, then dX dV
 k

1  
I
 2
C

A0
A

1  X
 2


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Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance

38
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance

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End of Web Lecture 24
Class Lecture 20