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Lecture 21
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 21
Class Lecture 17 – Tuesday 3/19/2013
 Gas Phase Reactions
 Trends and Optimums
2
Review Last Lecture
User Friendly Equations relate T, X, or Fi
1. Adiabatic CSTR, PFR, Batch, PBR achieve this:
W S  Cˆ P  0
X EB
 Cˆ T  T 


X
T  T0
3
i
Pi
0
 H Rx
~
C PA T  T0 
 H Rx

 H Rx X

 C
i
Pi
User Friendly Equations relate T, X, or Fi
2. CSTR with heat exchanger, UA(Ta-T) and a
large coolant flow rate:
X EB
m C
T
Ta
4
 UA

~

T T a    iCPi T T 0 
FA0



 H Rx
User Friendly Equations relate T, X, or Fi
3. PFR/PBR with heat exchange:
FA0
T0
Coolant
Ta
3A. In terms of conversion, X
5
Ua
Ta  T   rAH Rx T 
dT
B

~
dW
FA0  i C Pi  C p X


User Friendly Equations relate T, X, or Fi
3B. In terms of molar flow rates, Fi
Ua
Ta  T   rAH Rxij T 
dT
B

dW
 FiCPi
4. For multiple reactions
Ua
Ta  T    rij H Rxij
dT  B

dV
 FiCPi
5. Coolant Balance
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dTA UaT  Ta 

dV
m c C Pc
Reversible Reactions
Xe
KP
endothermic
reaction
endothermic
reaction
exothermic
reaction
exothermic
reaction
T
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T
Heat Exchange
Example: Elementary liquid phase reaction carried out in a PFR
m c
FA0
FI
Ta
Heat Exchange
Fluid
A B
The feed consists of both inerts I and Species A
with the ratio of inerts to the species A being 2 to 1.
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Heat Exchange
a) Adiabatic. Plot X, Xe, T and the rate of
disappearance as a function of V up to V = 40 dm3.
b) Constant Ta. Plot X, Xe, T, Ta and rate of
disappearance of A when there is a heat loss to the
coolant and the coolant temperature is constant at
300 K for V = 40 dm3. How do these curves differ
from the adiabatic case.
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Heat Exchange
c) Variable Ta Co-Current. Plot X, Xe, T, Ta and
rate of disappearance of A when there is a heat
loss to the coolant and the coolant temperature
varies along the length of the reactor for V = 40
dm3. The coolant enters at 300 K. How do these
curves differ from those in the adiabatic case
and part (a) and (b)?
10
d) Variable Ta Countercurrent. Plot X, Xe, T, Ta
and rate of disappearance of A when there is a
heat loss to the coolant and the coolant
temperature varies along the length of the
reactor for V = 20 dm3. The coolant enters at 300
K. How do these curves differ from those in the
adiabatic case and part (a) and (b)?
Heat Exchange
Example: PBR A ↔ B
5) Parameters
• For adiabatic:
• Constant Ta:
• Co-current:
• Counter-current:
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Ua  0
dTa
0
dW
Equations as is
dT
 (1) (or flip T - Ta to Ta - T)
dW
Reversible Reactions
1) Mole Balances
dX
  rA FA 0
dW
(1)
W  b V
dX
rA  B
rA


dV
FA 0
FA 0
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Reversible Reactions

CB 
2) Rate Laws rA   k C A 

KC 

( 2)
 E  1 1 
k  k1 exp    
 R  T1 T 
 H Rx
K C  K C 2 exp 
 R
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(3)
 1 1 
  
 T2 T 
( 4)
Reversible Reactions
3) Stoichiometry
5 CA  CA 0 1  X y T0 T 
6 CB  CA 0 XyT0 T 


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FT  FT 0
dy  FT  T 
 T 
     

dW y FT0  T0 
2 y  T0 
W  V
 b
dy

dV
2y
T 
 
 T0 
Reversible Reactions
Parameters
FA0 , k1 , E , R, T1 , K C 2 ,
H Rx , T2 , C A0 , T0 ,  ,  b
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(7)  (15)
Reversible Reactions
Gas Phase Heat Effects
Example: PBR A ↔ B
P0 T
3) Stoichiometry: v  v 0 1 X 
P T0
Gas Phase
FA 0 1  X  P T0 CA 0 1  X  T0

y
5 CA 
v 0 1 X  P0 T
1 X  T
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6
CA 0 X T0
CB 
y
1 X  T
7
dy
 FT T 
T


1 X 
dW 2y FT 0 T0 2y
T0
Reversible Reactions
Gas Phase Heat Effects
Example: PBR A ↔ B
CBe
CA 0 X e y T0 T
KC 

CAe CA 0 1  X e y T0 T
8
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KC
Xe 
1 KC
Reversible Reactions
Gas Phase Heat Effects
Example: PBR A ↔ B
Exothermic Case:
KC
Xe
T
T
Endothermic Case:
KC
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~1
Xe
T
T
Reversible Reactions
Gas Phase Heat Effects
dT rA H Rx   UaT  Ta 

dV
 FiCPi

FiCPi  FA 0 iCPi  CP X
Case 1: Adiabatic and ΔCP=0
H Rx X

T  T0 

iCPi

(16A)
Additional Parameters (17A) & (17B)
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
T0,
iCPi  CPA  I CPI
Reversible Reactions
Gas Phase Heat Effects
Case 2: Heat Exchange – Constant Ta
Heat effects: dT 
dW
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 rA  H Rx  
Ua
b
FA0   i C Pi
T  Ta 
9
Reversible Reactions
Gas Phase Heat Effects
Case 3. Variable Ta Co-Current
dTa Ua T  Ta 

,V0
 C Pcool
dV
m
Ta  Tao
(17C)
Case 4. Variable Ta Countercurrent
dTa UaTa  T 

dV
m C Pcool
V 0
Ta  ?
Guess Ta at V = 0 to match Ta0 = Ta0 at exit, i.e., V = Vf
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23
24
25
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Endothermic

PFR A  B
 
1  
k1 1
X
dX
KC
  KC  

, Xe 
dV
0
1 KC

Xe
X 
XEB
X EB
 i CPi T  T0  C PA   I CPI T T0 


H Rx
H Rx
T0

T  T0 
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H Rx X
CPA   I CPI
Adiabatic Equilibrium
Conversion on temperature
Exothermic ΔH is negative
Adiabatic Equilibrium temperature (Tadia) and conversion (Xeadia)
X
T  T0
Xeadia

 H Rx X

C PA
KC
Xe 
1 KC
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Tadia
T
Q1
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FA0
FA1
T0
X1
Q2
FA2
T0
X2
FA3
T0
X3
X
Xe
X3
X EB
X2
X1
T0
30
T
 C T  T 


i
Pi
 H Rx
0
31
Gas Flow Heat Effects
Trends:
Adiabatic
X exothermic
X
endothermic
Adiabatic
T and Xe
T0
32
T
 H Rx X
T  T0 
C PA   I C PI
T0
T
Effects of Inerts in the Feed
33
 34
Endothermic
First Order Irreversible
 k 


1



I 
As inert flow increases the
conversion will increase.
However as inerts increase,
reactant concentration
decreases, slowing down the
reaction. Therefore there is an
optimal inert flow rate to
maximize X.
Gas Phase Heat Effects
Adiabatic:
As T0 decreases the conversion X will increase, however the
reaction will progress slower to equilibrium conversion and may not
make it in the volume of reactor that you have.
X
Xe
T
35
X
T0
X
T
T
Therefore, for exothermic reactions there is an optimum inlet
temperature, where X reaches Xeq right at the end of V. However,
for endothermic reactions there is no temperature maximum and
the X will continue to increase as T increases.
Gas Phase Heat Effects
Effect of adding inerts
Adiabatic:
X
X V1 V2
Xe
I  
I  0
T0
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X
T
T  T0 C pA   I C pI 
HRx
X
T
Exothermic Adiabatic
k
θI
As θI increase, T decrease and
dX
k

dV  0 H I 
37
38
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Adiabatic
Exothermic
V
k
Xe
KC
T
V
Xe
X
V
V
V
Endothermic
k
Xe
KC
T
Xe
Frozen
Xe
OR X
X
V
40
V
V
V
V
V
Heat Exchange
Exothermic
Xe
KC
T
V
V
Xe
X
V
V
Endothermic
V
41
Xe
KC
T
V
Xe
X
V
V
End of Web Lecture 21
End of Class Lecture 17
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