Lecture 24 (Ch. 13)
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 22-Thursday 3/24/2011
 Review of Multiple Steady States
 Blowout Velocity
 CSTR Explosion
 Batch Reactor Explosion
2
3
dT
FA 0

GT   RT 

dt  N iCPi
GT   rAV H Rx 
RT   CPS 1  T  TC 

4
UA

FA 0CP 0
T0  Ta
TC 
1 
RT  CPS 1  T  TC 
R(T)
Increasing T0

T
5
Variation of heat removal line with inlet temperature.
κ=∞
RT  CPS 1  T  TC 
κ=0
R(T)

Increase κ
Ta
6
T0
T
Variation of heat removal line with κ (κ=UA/CP0FA0)
Variation of heat generation curve with space-time.
7
Finding Multiple Steady States with T0 varied
8
Temperature ignition-extinction curve
9
A <-> B
0 C A0 X
k
, X


X 
1 


kCA 0 1  X 
1  k1 
K
K
e 
e 


 H Rx k
G  H Rx X 

1 

1  k1 
 Ke 
V
R T   C P 1   T  TC 
T0  Ta
 310
1 
R T   400T  310
TC 
12
UA = 73,520
UA = 0
13
Example: PBR A ↔ B
CBe
CA 0 X e y T0 T
KC 

CAe CA 0 1  X e y T0 T
8
14
KC
Xe 
1 KC
Example: PBR A ↔ B
Exothermic Case:
KC
Xe
T
T
Endothermic Case:
KC
15
~1
Xe
T
T
Conversion on Temperature
Exothermic ΔH is negative
Adiabatic Equilibrium temperature (Tadia) and conversion (Xeadia)
X
T  T0
Xeadia

 H Rx X

CPA
KC
Xe 
1 KC
16
Tadia
T
Trends:
-Adiabatic:
X exothermic
X
endothermic
Adiabatic T
and Xe
T0
17
T
 H R X
T  T0 
C PA   I C PI
T0
T
X
Xe 
KC
1 K C

Xeadia
T
T  T0 

18
H Rx
 iCPi
Effect of adding inerts on adiabatic equilibrium
conversion
Adiabatic:
X
I  
Adiabatic Equilibrium Conversion
I  0
T
T0
19
X
T  T0 CPA  I CPI 
H Rx
, T  T0 
H Rx 
CPA   I CPI
Q1
20
FA0
FA1
T0
X1
Q2
FA2
T0
X2
FA3
T0
X3
X
Xe
X3
XEB
 C T  T 

X
i
 H RX
X2
X1
T0
21
Pi
T
0
A 
 B
HRx
The heat of reaction for endothermic reaction is positive, i.e.,
 X
H
Rx
T  T0 
C PA   I C P I
Energy Balance
and
kcal
 15
mol
C P  C P  I T0  T

X

A
I
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.
22
dX rA

dV FA0
A. First Order Reaction
Combining the mole balance, rate law and stoichiometry

dX kCA0 1 X k


1 X
dV
 0CA0
0

23
Two cases will be considered
Case 1 Constant 0, volumetric flow rate
Case 2: Variable 0, volumetric flow rate
For liquids the volumetric flow rates are additive
0  A0  I0  A0 1 I 

24
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


V

25
\
V
k 1  y I 
k 1  yXI 
k 

k
1  


 1  II 
 k 


1  I 
 I OPT

1
1 I 
I
I


 I OPT
I
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 

26
H Rx X
CPA   I CP I
k1 X
dX k1 X


dV
0
 A 1 I 

27
Without Inerts
CTA 
With Inerts and A
FA0
P
 CA0  A
A
RTA
CTI 
FTI FA0  FI0
P

 I
I
I
RTI
Taking the ratio of CTA to CTI
FTI
PI
CTI  I
RTI


PA
CTA FTA
 A RTA

Solving for I
I  A

28
FTI PA TI
FTA PI TA
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 C A0 
PA
RTA
  PI
CTI = Total concentration when both I and A are present
RTI
FA0
A
CTA = Total concentration when no inerts are present 
 A are entering  FA0
CA0I = Concentration of A entering when inerts
I
 (I) and reactant (A)
I = Entering volumetric flow rate with both Inerts
29

1
1
FTI FA0  FI0

 1  I  

 FA0  y A0
FA0
FTA


F

F
 I0
A0 
y A0 
1
1  I 

PA TI 
 I   A 1  I 

T
P

I A 

30
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 


31
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 k1 X

dV
0

32
I  A
FTI
F  F 
  A A0 I 0   A 1  I 
FTA
FA0
 I   A 1  I 
dX
1 k

1 X
dV  A 1 

33
Pure A
C A0 
Inerts Plus A
PA
F
 A0
RTA  A
CTI 
2
2
dX rA kCA0I 1 X


dV FA0
FA0



34
FA0 1 I 
I
PA TI
 I   A 1  I 
PI TA
2
C 2A0I FA0  I 
F

 A0

2
FA0
FA0
I
FA0
2
2




T
2 P
 A   A 1  I   A   I 
PI  TA 
2


C A0
PI TA

2 P T 
 A 1  I   A I 
dX
k

dV 1  I 2
35

2


C A0 PI TA
2
1
X




 A PA TI 
For the same temperature and pressures for the cases with and without inerts,
i.e., PI = PA and TI = TA, then
dX
k
CA0
2

1
X
 
dV 1 I 2  A

36
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
37
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
38
39