Lecture 18
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
 Solution to in-class problem
 User friendly Energy Balance Derivations
 Adiabatic
 Heat Exchange Constant Ta
 Heat Exchange Variable Ta Co-current
 Heat Exchange Variable Ta Counter Current
2
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.
3
Adiabatic Operation for CSTR
a) Assuming the reaction is irreversible for CSTR, A  B,
(KC = 0) what reactor volume is necessary to achieve
80% conversion?
b) If the exiting temperature to the reactor is 360K, what is
the corresponding reactor volume?
c) 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.
d) Now assume the reaction is reversible, make a plot of the
equilibrium conversion as a function of temperature
between 290K and 400K.
4
CSTR Adiabatic Example
mol
min
T0  300K
mol
FI  10
min
FA0  5
H Rx  20,000cal mol A (exot hermic)

A
T
X
B

Mole Balance:
5

FA 0 X
V
 rA exit
CSTR Adiabatic Example

CB 
 rA  k C A 

K
C

Rate Law:
k  k1e
E 1 1 
  
R  T1 T 
 H Rx
K C  K C1 exp
 R
Stoichiometry:
6
 1 1 
  
 T2 T 
C A  C A 0 1  X 
CB  CA0X
CSTR Adiabatic Example
Energy Balance - Adiabatic, ∆Cp=0:
T  T0

 H Rx X

T
 i C Pi
0

 H Rx X
C PA   I C PI
   20,000 
20,000
T  300 
X  300
X

164 36
164 218
T  300  100 X
7
CSTR Adiabatic Example
Irreversible for Parts (a) through (c)
 rA  kCA0 1  X (i.e.,KC  )
(a) Given X = 0.8, find T and V
Given X Calc
T Calc
k Calc
rA Calc
V
Calc K C
(if reverible)
8
CSTR Adiabatic Example
Given X, Calculate T and V
FA 0 X
FA 0 X
V

 rA exit kCA 0 1  X 
T  300 1000.8  380K
10,000  1
1 
k  0.1 exp

 3.81


1.989  298 380

FA 0 X
50.8
V

 2.82 dm 3
3.8121  0.8
 rA
9
CSTR Adiabatic Example
Given T, Calculate X and V
(b)
Given X Calc

 T Calc

 k Calc

 rA Calc

 V
Calc K C
(if reverible)
 rA  kCA 0 1  X  (Irreversible)
T  360K
T  300
X
 0.6
100
k  1.83min1
10

50.6
V
 2.05 dm 3
1.8320.4
CSTR Adiabatic Example
(c) Levenspiel Plot
FA 0
FA 0

 rA kCA 0 1  X 
T  300  100 X
FA 0
Choose X T k rA 
rA
Calc
11
Calc
Calc
Calc
CSTR Adiabatic Example
(c) Levenspiel Plot
12
CSTR Adiabatic Example
CSTR
X = 0.6
T = 360
30
25
-Fa0/Ra
20
15
10
5
CSTR 60%
0
0
0.1
0.2
CSTR
0.3
0.4
0.5
0.6
X = 0.95
X
0.7
0.8
0.9
1
T = 395
30
25
-Fa0/Ra
20
15
10
5
CSTR 95%
13
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
0.8
0.9
1
X
PFR
X = 0.95
30
25
-Fa0/Ra
20
15
10
PFR 95%
14
5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
CSTR Adiabatic Example
Summary
CSTR
PFR
CSTR
PFR
15
X = 0.6
X = 0.6
X = 0.95
X = 0.95
T = 360
Texit = 360
T = 395
Texit = 395
V = 2.05 dm3
V = 5.28 dm3
V = 7.59 dm3
V = 6.62 dm3
Calculate Adiabatic Equilibrium Conversion
and Temperature:
(d) At Equilibrium
CBe
CA 0 X e
Xe
KC 


CAe CA 0 1  X e  1  X e
KC
Xe 
1 KC
 H R
K C  K C 2 exp
 R
Calc
16
 1 1 
  
 T2 T 
Calc
Choose T KC X e , repeat
Calculate Adiabatic Equilibrium Conversion
and Temperature:
(d) At Equilibrium
  20,000 1
1 
K C  1,000exp
 

 1.987  290 T 
T  290 K C  1,000
T  330 K C  14.9
X e  0.999
X e  0.937
T  350 K C  2.6
X e  0.72
T  370 K C  0.95 X e  0.355
T  390 K C  0.136 X e  0.12
17
Calculate Adiabatic Equilibrium Conversion
and Temperature:

X  CPA  I CPI

T  T0
200
T  300

 H Rx  20,000
X  0.1 T  300
18
(e) Te = 358
Xe = 0.59
PFR Heat Effects
mc , H C
Ta
T
FA, Fi
V
In - Out + Heat Added = 0
Q  UATa  T
Ta
T
Fi
19
V
Fi
V+ΔV
V+ΔV
WS  0 (no turbine)
A  DL
D 2
V 
L
4
a  4/D
PFR Heat Effects
FH
i
i V
  Fi H i
 d  Fi H i
dV
 d  Fi H i
dV
 U a Ta  T   0
dH i
dFi 

   Fi
  Hi

dV
dV


Q  U a Ta  T V
20
V  V
 U a Ta  T V  0
PFR Heat Effects
H i  H i0  C Pi T  TR 
dFi
 ri  i  ra 
dV
dH i
dT
 C Pi
dV
dV
 d Fi Hi
dV
21
dT


  Fi CPi
  Hi i  ra  i Hi  H R
dV


PFR Heat Effects
dT


  CPi Fi
 H R  ra   U a Ta  T   0
dV


dT
 Fi CPi dV  H R ra  U a T  Ta 
dT H R  ra   Ua T  Ta 

dV
 FiCPi
22
PFR Heat Effects
Heat
generated
Heat
removed
dT Qg  Qr

dV  Fi CPi
FC F    XC
i
Pi
A0
i
Pi
 FA0
dT H R ra   Ua T  Ta 

dV
FA 0  i CPi  CP X

23
i

 C
i
Pi

 CPi X
User Friendly Equations Relate T and X or Fi
3. PBR in terms of molar flow rates
dT

dW
Ua

rA H Rx T  
T  Ta 
b
 FiCP
i
4. For multiple reactions

Ua
dT B

dV
Ta  T   rij HRx
F C
5. Coolant Balance
24

dTA UaT  Ta 

dV
mÝcCPc
i
Pi
ij
Heat Exchange Example
Elementary liquid phase reaction carried out in a PFR
c
m
Ta
FA0
FI
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 Exchange Example
1) Mole Balance:
2) Rate Law:
dX
(1)
  rA FA 0
dV

CB 
(2) rA   k C A 

KC 

 E  1 1 
(3) k  k1 exp   
 R  T1 T 
 H Rx
(4) K C  K C 2 exp
 R
26
 1 1 
  
 T2 T 
Heat Exchange Example:
Case 1- Constant Ta
3) Stoichiometry:
C A  C A 0 1  X  5
6
CB  CA0 X
4) Heat Effects:
dT H R ra   Ua T  Ta 

dV
FA 0  iCPi
CP  0

27
kC
X eq 
1 kC
 C
i
Pi
8
 CPA  ICPI
9
7
Heat Exchange Example:
Case 1- Constant Ta
Parameters:
H R , E, R , T1 , T2 ,
k1 , k C 2 , U a , Ta , FA 0 ,
C A 0 , C PA , C PI ,  I ,
rat e   ra
28
PFR Heat Effects
Heat
generated
Heat
removed
dT Qg  Qr

dV  Fi CPi
FC F    XC
i
29
Pi
A0
i
i
Pi
 FA0
dT H R ra   UaT  Ta 

dV
FA 0  iCPi  CP X


 C
i
Pi

 CPi X
Heat Exchange Example:
Case 2 Adiabatic
Mole Balance:
Energy Balance:
dX rA

dV FA 0
Adiabtic and ΔCP=0
Ua=0
T  T0

 H Rx X

 i CPi
(16A)
Additional Parameters (17A) & (17B)
30
T0 ,  i CPi  CPA  ICPI
Adibatic PFR
31
Example: Adiabatic
Find conversion, Xeq and T as a function of reactor volume
X
Xeq
X
V
32
rate
T
V
V
Heat Exchange:
dT  rA  H Rx   UaT  Ta 

dV
 Fi CPi
dT  rA  H Rx   UaT  Ta 

dV
FA 0  i CPi
33
Need to determine Ta
(16B)
User Friendly Equations
A. Constant Ta (17B) Ta = 300K
Additional Parameters (18B – (20B):
Ta ,  i CPi , Ua
B. Variable Ta Co-Current
dTa UaT  Ta 

,V0
 CPcool
dV
m
Ta  Tao
C. Variable Ta Counter Current
34
(17C)
dTa UaTa  T 

V  0 Ta  ? Guess
 CPcool
dV
m
GuessTa at V = 0 to match Ta0 = Ta0 at exit, i.e., V = Vf
Heat Exchanger Energy Balance
Variable Ta Co-current
Coolant balance:
In - Out + Heat Added = 0
 CHC V  m
 C H C V  V  U a VT  Ta   0
m
C
m
dH C
 U a T  Ta   0
dV
H C  H  C PC Ta  Tr 
0
C
dH C
dTa
 C PC
dV
dV
35
dTa U a T  Ta 

, V  0 Ta  Ta 0
 CCPC
dV
m
All equations can be
used from before
except Ta parameter,
use differential Ta
instead, adding mC
and CPC
Heat Exchanger Energy Balance
Variable Ta Counter-current
In - Out + Heat Added = 0
 C H C V V  m
 C H C V  U a VT  Ta   0
m
dH C
C
m
 U a T  Ta   0
dV
dTa U a Ta  T 

 CCPC
dV
m
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
0 B Ta  T dW   Fi0Hi0  Fi Hi  0
Differentiating with respect to W:
Ua
dFi
dH i
Ta  T   0   Hi   Fi
0
B
dW
dW
37
Derive the User Friendly Energy Balance
for a PBR
Mole Balance on species i:
dFi
 ri  i   rA 


dW
Enthalpy for species i:
T
H i  H i TR    C Pi dT

TR
38
Derive the User Friendly Energy Balance
for a PBR
Differentiating with respect to W:
dH i
dT
 0  C Pi
dW
dW
Ua
dT

Ta  T   rA  i Hi   FiCPi
0
B
dW
39
Derive the User Friendly Energy Balance
for a PBR
Ua
dT

Ta  T   rA  i Hi   FiCPi
0
B
dW
 H
i
i
 HR T
Fi  FA0 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
Ta  T   rAH
dT  B

dW
Fi C Pi
B:
41
dX  rA

 g X, T 
dW FA 0
Reversible Reactions
AB CD
The rate law for this reaction will follow an elementary rate law.

CCCD 

 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
KP
KC 

RT 
Van’t Hoff Equation:

d ln K P H R T  H R TR   Cˆ P T  TR 


2
dT
RT
RT 2
43
Reversible Reactions
For the special case of ΔCP=0
Integrating the Van’t Hoff Equation gives:
 H  R TR   1 1 
  
K P T2   K P T1  exp
R
 T1 T2 

44
Reversible Reactions
Xe
KP
endothermic
reaction
endothermic
reaction
exothermic
reaction
exothermic
reaction
T
45
T
End of Lecture 18
46