Lecture 20
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.
Last Lecture
Energy Balance Fundamentals
F
i0
E i0 
FE
i
 Q  W
i

dE sys
dt
Substituting for W
Hi0
F
i0
U i 0  P0V i 0  
F
i0
H i0 
Q  W S 
2
Hi
F
FH

i
i
Fi 0 H
i
U i  P V i   Q  W S 
 Q  WS 
i0
dE sys
  Fi H
dt
i
 0
dE sys
dt
Web Lecture 20
Class Lecture 16-Thursday 3/14/2013
 Reactors with Heat Exchange
 User friendly Energy Balance Derivations
 Adiabatic
 Heat Exchange Constant Ta
 Heat Exchange Variable Ta Co-current
 Heat Exchange Variable Ta Counter Current
3
Adiabatic Operation CSTR
FA0
FI

A
B
Elementary liquid phase reaction carried out in a CSTR
The feed consists of both - Inerts I and Species A with
the ratio of inerts I to the species A being 2 to 1.
4
Adiabatic Operation CSTR
 Assuming the reaction is irreversible for CSTR,
A  B, (KC = 0) what reactor volume is necessary to
achieve 80% conversion?
 If the exiting temperature to the reactor is 360K,
what is the corresponding reactor volume?
 Make a Levenspiel Plot and then determine the PFR
reactor volume for 60% conversion and 95%
conversion. Compare with the CSTR volumes at
these conversions.
 Now assume the reaction is reversible, make a plot
of the equilibrium conversion as a function of
temperature between 290K and 400K.
5
CSTR: Adiabatic Example
𝐹𝐴0
𝑚𝑜𝑙
=5
𝑚𝑖𝑛
FA0
Δ𝐻𝑅𝑥𝑛 =
𝑐𝑎𝑙
−20000
𝑚𝑜𝑙 𝐴
𝑇0 = 300𝑚𝑜𝑙
𝐾FI
𝐹𝐼 = 10
𝑚𝑖𝑛

A
1) Mole Balances:
6
𝑇 =?
𝑋 =?
B
V 
(exothermic)
FA 0 X
 rA
exit
CSTR: Adiabatic Example

CB 
 rA  k  C A 

K
C 

2) Rate Laws:
k  k 1e
KC
  H Rx
 K C 1 exp 
 R
3) Stoichiometry:
7
E 1 1 

 
R  T1 T 
 1
1 


 T  T 

 2
C A  C A 0 1  X 
C B  C A0X
CSTR: Adiabatic Example
4) Energy Balance
Adiabatic, ∆Cp=0
T  T0 
   H Rx X
  i C Pi
 T0 
   H Rx X
C PA   I C PI
    20 , 000  
20 , 000
T  300  
X
 X  300 
164  36
 164   2 18  
T  300  100 X
8
CSTR: Adiabatic Example
Irreversible for Parts (a) through (c)
 rA  kC A 0 1  X  (i.e., K C   )
(a) Given X = 0.8, find T and V
Calc
Calc
Calc
Calc
Given X  T   k    rA  V
Calc
KC
(if reversible)
9
CSTR: Adiabatic Example
Given X, Calculate T and V
V 
FA 0 X
 rA

exit
FA 0 X
kC A 0 1  X 
T  300  100 0 . 8   380 K
10 , 000  1
1 
k  0 . 1 exp

 3 . 81


1 . 989  298 380 
V 
10
FA 0 X
 rA

5 0 . 8 
 2 . 82
3 . 81  2 1  0 . 8 
dm
3
CSTR: Adiabatic Example
Given T, Calculate X and V
(b)
Given X  
 T  
 k  
  rA  
 V
Calc
Calc
Calc
Calc K C
(if reversible)
 r A  kC
A0
1 
X  (Irreversi ble)
T  360 K
X 
T  300
 0 .6
100
k  1 . 83 min
11
V 
1
5 0 . 6 

1 . 83  2 0 . 4 
2 . 05 dm
3
Calc
CSTR: Adiabatic Example
(c) Levenspiel Plot
FA 0
 rA

FA 0
kC A 0 1  X 
T  300  100 X
Calc
Calc
Calc
Calc
Choose X  T   k    rA  
12
FA 0
 rA
CSTR: Adiabatic Example
(c) Levenspiel Plot
13
CSTR: Adiabatic Example
CSTR
X = 0.6
T = 360 K
30
25
-F a 0 /R a
20
15
10
5
CS TR 6 0%
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
X
CSTR
30
X = 0.95
T = 395 K
25
-F a 0 /R a
20
15
10
14
5
C S T R 9 5%
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
CSTR: Adiabatic Example
PFR
30
X = 0.6
25
-Fa0/Ra
20
15
10
PFR 60%
5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
X
PFR
30
X = 0.95
25
-F a 0 /R a
20
15
10
PFR 9 5%
5
15
0
0
0.1
0 .2
0.3
0.4
0.5
0.6
0 .7
0.8
0.9
1
CSTR: Adiabatic Example - Summary
16
CSTR
X = 0.6
T = 360
V = 2.05 dm3
PFR
X = 0.6
Texit = 360
V = 5.28 dm3
CSTR
X = 0.95
T = 395
V = 7.59 dm3
PFR
X = 0.95
Texit = 395
V = 6.62 dm3
Energy Balance in terms of Enthalpy

Fi H i
V

 d  Fi H i
dV
 d  Fi H i
dV
17

Fi H i
V  V
 Ua T a  T
 Ua T a  T  V  0

0
dH i

    Fi

dV


Hi
dF i 
dV 
PFR Heat Effects
dF i
dV
 ri   i   r A 
H i  H i  C Pi T  T R 
0
dH i
dV
 C Pi
dT
dV
 d  Fi H i
dV

18
i
dT

    Fi C Pi

dV

H i  H R x


H i i   r A 

PFR Heat Effects
dT


   C Pi Fi
  H R   r A   Ua T a  T   0
dV


 FC
i
dT
dV
19
dT
Pi

dV
  H R r A  Ua T  T a 
  H R   rA   Ua T
 FC
i
Pi
Need to determine Ta
 Ta 
Heat Exchange:
dT
dV
 Fi C Pi
dT
 rA   H Rx   Ua T  Ta 
dV
20

 rA   H Rx   Ua T  Ta 

FA 0   i C Pi
Need to determine Ta
(16 B )
Heat Exchange Example:
Case 1 - Adiabatic
Energy Balance:
Adiabatic (Ua=0) and ΔCP=0
T  T0 
21
  H Rx X
  i C Pi
(16 A )
User Friendly Equations
A. Constant Ta
e.g., Ta = 300K
B. Variable Ta Co-Current
dT a

Ua T  T a

m C Pco o l
dV
,V  0
T a  T ao
(17 C )
C. Variable Ta Counter Current
dT a
dV
22

Ua T a  T
m C Pcool

V 0
T a  ? Guess
Guess Ta at V = 0 to match Ta0 = Ta0 at exit, i.e., V = V
Heat Exchanger Energy Balance
Variable Ta Co-current
Coolant Balance:
In - Out + Heat Added = 0
m C H C
 m C H C
V
dH
 m C
C
dV
V  V
 Ua  V T  T a   0
 Ua T  T a   0
H C  H C  C PC T a  T r 
0
dH
C
dV
dT a
23
dV
 C PC

dT a
dV
Ua T  T a 
m C C PC
, V  0 Ta  Ta0
Heat Exchanger Energy Balance
Variable Ta Co-current
In - Out + Heat Added = 0
m C H C
m C
dH
dT a
dV
24
dV

V  V
C
 m C H C
V
 Ua  V T  T a   0
 Ua T  T a   0
Ua T a  T
m C C PC

Heat Exchanger – Example
Case 1 – Constant Ta
Elementary liquid phase reaction carried out in a PFR
c
m
FA0
FI
Ta
T
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.
25
Heat Exchanger – Example
Case 1 – Constant Ta
1) Mole Balance:
(1)
dX
dV
2) Rate Laws:
  rA FA 0

CB 
( 2 ) rA   k  C A 

KC 

E
( 3 ) k  k 1 exp 
R
(4) K C  K C 2
26
 1
1 


 T  T 
 1

  H Rx
exp 
 R
 1
1 


 T  T 
 2

Heat Exchanger – Example
Case 1 – Constant Ta
3) Stoichiometry: C A  C A 0 1  X 
6 
C B  C A0X
4) Heat Effects:
dT

  H R   rA   Ua T
X eq 
27
 0
kC
1 kC
C
i
 Ta 
F A 0   i C Pi
dV
 C P
5 
Pi
8 
 C PA   I C PI
9 
7 
Heat Exchanger – Example
Case 1 – Constant Ta
Parameters:
 H R , E , R , T1 , T 2 ,
k 1 , k C 2 , Ua , T a , F A 0 ,
C A 0 , C PA , C PI ,  I ,
rate   r A
28
PFR Heat Effects
Heat
Heat
generated removed
dT

dV
Qg  Qr
 FC
i
 FC
i
dT
dV
29
Pi

Pi
  FA 0  i   i X C Pi  FA 0
  H R  rA   Ua T
FA0
  C
i
Pi
 Ta 
 C P X

  C
i
Pi
  C Pi X

Heat Exchanger – Example
Case 2 – Adiabatic
Mole Balance:
dX
dV

 rA
FA 0
Energy Balance:
Adiabatic and ΔCP=0
Ua=0

T  T0 
  H Rx X
  i C Pi
(16 A )
Additional Parameters
(17A) & (17B)
30
T 0 ,   i C Pi  C P A   I C P I
Adiabatic PFR
31
Example: Adiabatic
Find conversion, Xeq and T as a function of reactor volume
Xeq
X
rate
T
X
V
32
V
V
Heat Exchange
dT
dV
 Fi C Pi
dT
 rA   H Rx   Ua T  Ta 
dV
33

 rA   H Rx   Ua T  Ta 

FA 0   i C Pi
Need to determine Ta
(16 B )
User Friendly Equations
A. Constant Ta (17B) Ta = 300K
Additional Parameters (18B – (20B):
Ta ,   i C Pi , Ua
B. Variable Ta Co-Current
dT a

dV
Ua T  T a 
m C Pcool
V 0
T a  T ao
(17 C )
C. Variable Ta Countercurrent
dT a
dV
34

Ua T a  T
m C Pcool

V 0
Ta  ?
Guess Ta at V = 0 to match Ta0 = Ta0 at exit, i.e., V = Vf
Heat Exchange Energy Balance
Variable Ta Co-current
Coolant balance:
In - Out + Heat Added = 0
m C H C
 m C H C
V
dH
 m C
V  V
 Ua  V T  T a   0
 Ua T  T a   0
C
dV
H C  H  C PC T a  T r 
0
C
dH
C
dV
dT a
35
dV
 C PC

dT a
dV
Ua T  T a 
m C C PC
, V  0 Ta  Ta 0
All equations can be
used from before
except Ta parameter,
use differential Ta
instead, adding mC
and CPC
Heat Exchange Energy Balance
Variable Ta Co-current
In - Out + Heat Added = 0
m C H C
m C
dH
dV
V  V
C
 m C H C
V
 Ua  V T  T a   0
 Ua T  T a   0
dT a
dV

Ua T a  T

m C C PC
All equations can be used from before except dTa/dV
which must be changed to a negative. To arrive at the
correct integration we must guess the Ta value at V=0,
integrate and see if Ta0 matches; if not, re-guess the
value for Ta at V=0
36
Derive the user-friendly Energy Balance
for a PBR
W
Ua
  T
0
a
 T dW 
F
i0
H i 0   Fi H i  0
B
Differentiating with respect to W:
Ua
B
37
T a  T   0  
dF i
dW
H i   Fi
dH i
dW
0
Derive the user-friendly Energy Balance
for a PBR
Mole Balance on species i:
dF i
 ri    i   rA  


dW
Enthalpy for species i:
H i  H i T R  

T
C
TR
38
Pi
dT
Derive the user-friendly Energy Balance
for a PBR
Differentiating with respect to W:
dH i
dW
Ua
B
39
 0  C Pi
dT
dW
dT

Ta  T   rA   i H i   Fi C Pi
0
dW
Derive the user-friendly Energy Balance
for a PBR
Ua
B
dT

Ta  T   rA   i H i   Fi C Pi
0
dW
H
i
i
  H R T 
Fi  FA 0  i   i X 
Final Form of the Differential Equations in Terms of Conversion:
A:
40
Derive the user-friendly Energy Balance
for a PBR
Final form of terms of Molar Flow Rate:
Ua
dT

B
dW
B:
 T   rA   H
Fi C Pi
dX
dW
41
T a

 rA 
FA 0
 g X , T 
Reversible Reactions
AB CD
The rate law for this reaction will follow an elementary rate law.

C CC D
 rA  k  C A C B 
KC





Where Ke is the concentration equilibrium constant. We know from Le
Chaltlier’s law that if the reaction is exothermic, Ke will decrease as
the temperature is increased and the reaction will be shifted back to
the left. If the reaction is endothermic and the temperature is
increased, Ke will increase and the reaction will shift to the right.
42
Reversible Reactions
KC 
KP
 RT 

Van’t Hoff Equation:
d ln K P
dT
43

 H R T 
RT
2
 H R T R    Cˆ P T  T R 


RT
2
Reversible Reactions
For the special case of ΔCP=0
Integrating the Van’t Hoff Equation gives:
  H  R T R   1
1 


K P T 2   K P T1  exp 



R
T
T
2 
 1

44
Reversible Reactions
Xe
KP
endothermic
reaction
endothermic
reaction
exothermic
reaction
exothermic
reaction
T
45
T
End of Lecture 20
46