Lecture 19
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.
Today’s lecture
 Gas Phase Reactions
 Trends and Optimums
2
Last Lecture
User Friendly Equations Relate T and X or Fi
1. Adiabatic CSTR, PFR, Batch, PBR achieve this:
  Ĉ  0
W
S
P
X EB
 Ĉ T  T 


X
T  T0
3
i
Pi
0
 H Rx
~
CPA T  T0 
 H Rx

 H Rx X

 C
i
Pi
User Friendly Equations Relate T and X or Fi
2. CSTR with heat exchanger, UA(Ta-T) and a large coolant flow
rate
 UA

~

T T a    i C Pi T T 0 
FA 0


X EB 
 H Rx
C
m
Ta
4
T
User Friendly Equations Relate T and 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
FA 0  i CPi  C p X


User Friendly Equations Relate T and X or Fi
3B. In terms of molar flow rates, Fi
Ua
Ta  T   rAH Rx ij T 
dT B

dW
 FiCPi
4. For multiple reactions
Ua
Ta  T    rijH Rx
dT B

dV
 FiCP
5. Coolant Balance
i
6
dTA Ua T  Ta 

 c C Pc
dV
m
ij
Reversible Reactions
Xe
KP
endothermic
reaction
endothermic
reaction
exothermic
reaction
exothermic
reaction
T
7
T
Heat Exchange
Example: Elementary liquid phase reaction carried out in a PFR
c
m
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.
8
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.
9
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 Counter 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 = 20 dm3. The coolant
enters at 300 K. How do these curves differ from those
in the adiabatic case and part (a) and (b)?
Reversible Reactions
Gas Phase Heat Effects
Example: PBR A ↔ B
5) Parameters….
• For adiabatic:
• Constant Ta:
11
UA  0
dTa
0
dW
• Co-current:
Equations as is
• Counter-current:
dT
 (1) (or flip T - Ta to Ta - T)
dW
Reversible Reactions
Mole Balance:
dX
  rA FA 0
dW
(1)
W  b V
dX
rA  B
rA


dV
FA 0
FA 0
12
Reversible Reactions
Rate Law:

CB 
rA  k CA 

KC 

(2)
 E  1 1 
k  k1 exp    
 R  T1 T 
 H Rx
K C  K C 2 exp 
 R
13
(3)
 1 1 
  
 T2 T 
(4)
Reversible Reactions
Stoichiometry:
5 CA  CA 0 1  X y T0 T 
6 CB  CA 0 XyT0 T 


dy  FT T 
 T 

    
dW
y FT0 T0 
2y T0 
W  V
dy
b T 

 
dV
2y T0 
14
Reversible Reactions
Parameters:
FA 0 , k1 , E, R , T1 , K C 2 ,
H Rx , T2 , C A 0 , T0 , , b
15
(7)  (15)
Reversible Reactions
Gas Phase Heat Effects
Example: PBR A ↔ B
3) Stoich (gas): v  v 0 1 X  P0 T
P T0
16
5
FA 0 1  X  P T0 CA 0 1  X  T0
CA 

y
v 0 1 X  P0 T
1 X  T
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
17
KC
Xe 
1 KC
Reversible Reactions
Gas Phase Heat Effects
Example: PBR A ↔ B
Exothermic Case:
KC
Xe
T
T
Endothermic Case:
KC
18
~1
Xe
T
T
Adibatic 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
19
Tadia
T
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
Example A ↔ B
Gas Phase Heat Effects
dT rA H Rx   UaT  Ta 

dV
 FiCPi

FiCPi  FA 0 iCPi  CP X

Case 1: Adiabtic and ΔCP=0


T  T0
H Rx X


iCPi
(16A)
Additional Parameters (17A) & (17B)
22
T0,
iCPi  CPA  I CPI
Example A ↔ B
Gas Phase Heat Effects
Case 2: Heat Exchange – Constant Ta
ra H R    T  Ta 
Ua
Heat effects:

23
dT

dW
b
FA 0  iCPi
9
Example A ↔ B
Case 3. Variable Ta Co-Current
dTa Ua T  Ta 

,V0
 C Pcool
dV
m
Ta  Tao
(17C)
Case 4. Variable Ta Counter Current
dTa Ua Ta  T 

 C Pcool
dV
m
V0
Ta  ? Guess
GuessTa at V = 0 to match Ta0 = Ta0 at exit, i.e., V = Vf
24
25
26
27
28
29
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 
30
H Rx X
CPA   I CPI
Effects of Inerts In The Feed
31
Endothermic
First Order Irreversible
What happens when we vary
 k 


1



I 

32
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.
33
Gas Phase Heat Effects
Trends:
-Adiabatic:
X exothermic
X
endothermic
Adiabatic T
and Xe
T0
34
T
 H R X
T  T0 
C PA   I C PI
T0
T
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
X
X
T
T0
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.
35
Gas Phase Heat Effects
Effect of adding Inerts
Adiabatic:
X
X V1
Xe
V2
I  
I  0
T
T0
36
X
T  T0 C pA   I C pI 
H Rx
X
T
Exothermic Adiabatic
k
θI
As θI increase, T decrease and
dX
k

dV  0 H I 
37
38
39
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
Xe
X
V
V
V
Endothermic
V
41
Xe
KC
T
V
Xe
X
V
V
Derive the Steady State Energy Balance
(w/o Work)
W
Ua
0 B Ta  TdW   Fi0Hi0  Fi Hi  0
Differentiating with respect to W:
Ua
dFi
dH i
Ta  T   0   H i   Fi
0
B
dW
dW
42
Derive the Steady State Energy Balance
(w/o Work)
Mole Balance on species i:
dFi
 ri  i   rA 


dW
Enthalpy for species i:
T
H i  H i TR    C PidT

TR
43
Derive the Steady State Energy Balance
(w/o Work)
Differentiating with respect to W:
dH i
dT
 0  C Pi
dW
dW
Ua
dT

Ta  T   rA  i H i   Fi C Pi
0
B
dW
44
Ua
dT

Ta  T   rA  i H i   Fi C Pi  0
B
dW
 H
i
i
 HR T
Fi  FA 0 i  i X 
Final Form of the Differential Equations in Terms of Conversion:
A:
45
Final Form of terms of Molar Flow Rate:
Ua
 rAH
dT  B

dW
Fi C Pi
B:
46
dX  rA

 gX, T 
dW FA 0
(d) Gas Phase Counter Current He at Exchange V f = 20 dm
3
 Matches
47
End of Lecture 19
48
Gas Phase Heat Effects
Example: PBR A ↔ B
W  b V
1) Mole balance:
2) Rates:
dX  r

dW FA 0
'
A
1
rA'  kC A  C B k C  2
 E  1 1 
k  k1 exp      3
 R  T1 T  
49
 H R
k C  k C 2 exp 
 R
 1 1 
    4

 T2 T  