Lecture 22

Report
Lecture 22
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
 Blowout Velocity
 CSTR Explosion
 Batch Reactor Explosion
2
Last Lecture
CSTRs with Heat Effects
3
Energy balance for CSTRs
 
Q  W
S
n
F
i0
i 1
dT
dt
4

n
d Eˆ sys
i 1
dt
H i 0   Fi H i 
 
Q  W
 Fi 0 C Pi T  Ti 0     H Rx T  rA V 
S
NC
i
Pi
Energy balance for CSTRs
dT
dt

FA 0
 N i C Pi
G T   R T 
G T   rA V  H Rx 
R T   C PS 1   T  TC 

5
 
UA
FA 0C P 0
TC 
T0   Ta
1
Steady State Energy Balance for CSTRs
At Steady State
dT
dt

dN
A
0
dt
 rA V  FA 0 X
G T   R T   0
   H Rx FA 0 X  FA 0   i C P T  T 0   UA T  T a   0
i
6
Energy balance for CSTRs
Solving for X
  i C Pi  T  T 0  
X 
 H
UA
FA 0
T  T a 
 X EB

Rx
Solving for T
T 
7
FA 0 X    H Rx   UAT
a
 FA 0   i C Pi T 0
UA  FA 0   i C Pi
Energy balance for CSTRs
R T   C PS 1   T  TC 
R(T)
Increasing T0
T
8
Variation of heat removal line with inlet temperature.
Energy balance for CSTRs
R T   C PS 1   T  TC 
κ=∞
κ=0
R(T)
Increase κ
Ta
9
T0
T
Variation of heat removal line with κ (κ=UA/CP0FA0)
V 
FA 0 X
 rA  X , T 
A  B
1) Mole Balance:
2) Rate Law:
10
V 
FA 0 X
 rA
 rA  kC
A
3) Stoichiometry:
4) Combine:
C A  C A 0 1  X 
V 
FA 0 X
kC A 0 1  X 
k 
11

C A00X
kC A 0 1  X 
X
1 X
V 
FA 0 X
kC A 0 1  X 
k 
X 
k
1  k
12
kC A 0 1  X 
X
1 X

G T   X    H Rx  

C A00X
 E RT
 Ae
1  Ae
 Ae
 E RT
 E RT
1  Ae
 E RT
   H Rx 
Variation of heat generation curve with space-time.
13
Finding Multiple Steady States with T0 varied
14
Finding Multiple Steady States with T0 varied
15
Temperature ignition-extinction curve
16
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
17
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
18
Example B: Liquid Phase CSTR
Same reactions, rate laws, and rate constants as example A
A  2 B  C (1)
2
 r1A  k1A C A C B
NOTE: The specific reaction rate k1A is defined with respect to
species A.

3C  2 A  D ( 2 )
3
C
 r2C  k 2C C C
2
A
NOTE: The specific reaction rate k2C is defined with respect to
species C.
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
Example B: Liquid Phase CSTR
The complex liquid phase reactions take place in a 2,500 dm3
CSTR. The feed is equal molar in A and B with FA0=200
mol/min, the volumetric flow rate is 100 dm3/min and the
reation volume is 50 dm3.
Find the concentrations of A, B, C and D existing in the reactor
along with the existing selectivity.
Plot FA, FB, FC, FD and SC/D as a function ofV
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Example B: Liquid Phase CSTR
Solution
Liquid CSTR
Mole Balances:
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(1)
f C A    0 C A 0   0 C A  rA V
(2)
f C B    0 C B 0   0 C B  rB V
(3)
f C C     0 C C  rC V
(4)
f C D     0 C D  rD V


Net Rates:
(5) 

rA  r1A  r2 A
Selectivity
If one were to write SC/D=FC/FD in the Polymath program,
Polymath would not execute because at V=0, FC=0 resulting in
an undefined volume (infinity) at V=0. To get around this
problem we start the calculation 10-4 dm3 from the reactor
entrance where FD will not be zero and use the following IF
statement.
(15)
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
S˜ C
D
 if
V  0.001 
 F 
C
then   else
F D 
0 
Selectivity
Stoichiometry:
(16)
C A  FA  0
(17)
C B  FB  0
(18)
C C  FC  0
(19)
C D  FD  0
Parameters:
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(20)
 0  100 dm
(21)
k 1 A  10 dm
(22)
k 2 C  15 dm
3
min

3
mol


3
mol

2
4
min
min
Example 1: Safety in Chemical Reactors
H 2 O

Gas
N
O
 2
T0  200 F
m A0  310 lb h
17 % H 2 O
P
200°F
Liquid

83 % NH 4 NO 3
510°F

Ta

24

X
NH 4 NO 3
M  500 lb
NH 4 NO 3  N 2O  2H 2O


Ta 0
Example 1: Safety in Chemical Reactors
NH 4 NO 3  2 H 2 0  N 2 O
A  2B  C
dN
A
 F A 0  F A  rA V
B
 F B 0  F B  rB V
t
dN
t
dN C
t
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 FC 0  FC  rC V
FA0
FI0
A
Example 1: Safety in Chemical Reactors
T
t

Qg  Qr
NC
i
(only A in vat, B, C are gases) 
Pi
Qg  Qr
N A C PA
Q g  ( rA V )(  H rxA )
Q r  FA 0 C PA ( T  T0 )   B ( H B  H B 0 )   UA ( T  Ta )
If the flow rate shut off, the temperature will rise (possibly to
point of explosion!)
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Rearranging:
Q
dT

Q
  g        
r      

 rA V   H Rx   FA 0   i  H i  H i 0   UA T  Ta 
 N i C Pi
dt

Qg  Qr
 N i C Pi
Additional information (approximate but close to the real case):
H

Rx
  336 Btu / lb ammonium
C P  0 . 38 Btu / lb ammonium
C P  0 . 47 Btu / lb ofsteam
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 rA V  kC A V  k
M
V

nitrate at 500 F ( cons tan t )
nitrate  F
nitrate  F
V  kM ( lb / h )
dT
Q  FA 0 C PA T  T 0   FA 0  B H B g   H B 0   rA V  H Rx  FA  H Avap  N A C PA
dt
Q
Q
Q
g
           r 1      
 r 2  
dT
 rA V   H Rx    FA 0 C PA T  T0    B  H B g   H B 0   FA  H Vap  UA T  Ta   N A C PA


dt




Complete conversion FA = 0
Q g  Q r 1  Q r 2   N A C PA
dT
dt
Batch Reactors with Heat Effects
Single Reactions
Q g  rA V H Rx 
Multiple Reactions Q g   r ij H Rxij V
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
Risk Rupture
Q r 2  mÝvap  H vap
dT
dt

Qg  Qr
 N i C Pi
Q r 1  UA T  T a 
Keeping MBAs Away From Chemical
Reactors
 The process worked for 19 years before they showed up!
 Why did they come?
 What did they want?
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Nitroaniline Synthesis Reaction
NO2
NO2
Cl
NH2
+
ONCB
30
+
2NH3
Ammonia 
+
Nitroanaline
+
NH4Cl
Ammonium
Chloride
Nitroaniline Synthesis Process
NH3 in H2O
ONCB
Autoclave
175 oC
~550 psi
O-Nitroaniline
Product Stream
NH3
Separation
Filter
Press
To Crystallizing Tanks
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“fast” Orange
Nitroaniline Synthesis Reactor
Old
3 kmol ONCB
43 kmol Ammonia
100 kmolWater
V = 3.25 m3
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Nitroaniline Synthesis Reaction
NO2
NO2
Cl
NH2
+
ONCB
+
2NH3
Ammonia 
+
Nitroanaline
+
Batch Reactor, 24 hour reaction time
33
Management said:TRIPLE PRODUCTION
NH4Cl
Ammonium
Chloride
MBA Style Nitroaniline Synthesis Reactor
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New
9 kmol ONCB
33 kmol Ammonia
100 kmolWater
V = 5 m3
Temperature oC
Temperature-time trajectory
400
Cooling Restored
175
Isothermal
Operation
9:55
t=0
35
10:40 10:50
fuse
midnight 12:18
Temperature-time trajectory
dT
dt

UA ( T 0  T )  ( rA V )( DH
)
N A 0 C pA  N B 0 C pB  N W C pW
dT
dt
36
rx

Qg  Qr
NC p
End of Lecture 22
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