Transcript Lecture 5 - University of Michigan

Lecture 5
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.
Lecture 5 - Thursday 1/20/2011
 Block 1:
 Block 2:
 Block 3:
 Stoichiometric Table: Flow
 Definitions of Concentration: Flow
 Gas Phase Volumetric Flow rate
 Calculate the Equilibrium Conversion Xe
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Reactor
Differential
Algebraic
Integral
X
Batch
N A0
dX
dt
  r AV
0
V 
CSTR
PFR
t  N A0 
FA0
dX
dV
dX
 r AV
t
FA0 X
 rA
X
  rA
X
V  FA0 
0
dX
 rA
X
PBR
3
FA0
dX
dW
X
  rA
W  FA0 
0
dX
 r A
W
How to find
Step 1: Rate Law

Step 2: Stoichiometry
 rA  f  X 
rA  g C i 
C i   h  X 
Step 3: Combine
 to get  rA  f  X 

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Mole Balance
Rate Laws
These topics build upon one another
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Stoichiometry
Species
Symbol
Reactor Feed
Change
Reactor Effluent
A
A
FA0
-FA0X
FA=FA0(1-X)
B
B
FB0=FA0ΘB
-b/aFA0X
FB=FA0(ΘB-b/aX)
C
C
FC0=FA0ΘC
+c/aFA0X
FC=FA0(ΘC+c/aX)
D
D
FD0=FA0ΘD
+d/aFA0X
FD=FA0(ΘD+d/aX)
Inert
I
FI0=FA0ΘI
----------
FI=FA0ΘI
FT0
Where:
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i 
Fi 0
FA0

C i 0 0
C A 0 0

C i0
C A0
Concentration – Flow System

yi0
and
y A0
CA 
FT=FT0+δFA0X
FA


d
a

c
a

b
a
1
Concentration Flow System: C A 
Liquid Phase Flow System:
CA 
FA
CB 
NB
etc.
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
V


F A 0 1  X
0

FA

  0
 C A 0 1  X
 Flow Liquid Phase
N A0 
b
b





X

C


X
 B


A0 
B
V0 
a
a



If the rate of reaction were
then we would have
This gives us
 r A  kC A C B
 rA  C A 0
 rA  f  X
2
b


1  X   B  X 
a



FA0
 rA
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X
Combining the compressibility factor equation of
state with Z = Z0
Stoichiometry:
CT 
CT 0 
P
ZRT
P0
Z 0 R 0T0
FT  C T 
FT 0  C T 0 0
We obtain:
  0
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FT P0 T
FT 0 P T 0
C A  FA
 P   T 0  FT 0


 



0
 FT   P0   T 

 0 

F
 0 
FA
C T 0  FT 0  0
CB
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 F B   P   T0 


 C T 0 




 FT   P0   T 
 P   T0 


 P  T 

 0 
The total molar flow rate is: F  F  F  X
T
T0
A0
 FT 0  F A 0  X
Substituting FT gives:    0 

FT 0


FA0
   0  1 
X
FT 0

 T P0

T P
 0
   0 1  y A 0  X
T P0
   0 1   X
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 T P0

T P
 0


T0 P
T P0
T0 P
Concentration Flow System: C A 
CB
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FA


F A 0 1  X
 0 1   X


T P0

   0 1   X
Gas Phase Flow System:
CA 
FA


T P0
T0 P
C A 0 1  X  T 0 P
1   X 
T P0
T0 P
b
b




FA0   B  X  C A0   B  X 
FB
a
a



 T0 P



P
T
1   X 

T P0
0
 0 1   X 
T0 P
If –rA=kCACB


b


 1  X    B  X   P T  2 
a


2
0

 
 rA  k A C A 0 
P T 




1


X
1


X

 0
 




This gives us
FA0/-rA
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X
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Xef
Consider the following elementary reaction with KC=20 dm3/mol
and CA0=0.2 mol/dm3.
Calculate Equilibrium Conversion or both a batch reactor (Xeb)
and a flow reactor (Xef).
2A  B
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 2 CB 
 rA  k A  C A 

KC 

2A  B
X eb  0 .703
X ef  ?
 Solution:


Rate law:
 2
CB 
 rA  k A  C A 

KC 

A 
1
2
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B
Species
A
B
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Fed
FA0
0
FT0=FA0
Change
-FA0X
+FA0X/2
Remaining
FA=FA0(1-X)
FB=FA0X/2
FT=FA0-FA0X/2
A
FA0
-FA0X
FA=FA0(1-X)
B
0
FA0X/2
FB=FA0X/2
Stoichiometry:
Gas isothermal T=T0, isobaric P=P0
V  V 0 1   X 
CA 
CB 
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F A 0 1  X 
V 0 1   X 
FA 0 X 2
V 0 1   X 


C A 0 1  X 
1   X 
C A 0 1  X 
2 1   X 
  C 1  X
 r A  k A   A 0
  1   X 

 

2 1   X  K C 



2
C A0 X
Pure A  yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0
1
1

  y A 0   1   1   
2
2

@ eq: -rA=0
2 K C C A0 
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X e 1   X e 
1 
Xe
2
2 K C C A0
3

dm  
mol 
  0 .2
 2  20
8
3

mol  
dm 

1

1
  y A 0  1  1  
2

2
8
X e  0.5 X e
1  2 X
e
2
 Xe
2

2
8 .5 X e  17 X e  8  0
20
Flow: X ef  0.757
Recall
Batch:
X eb  0 .70
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End of Lecture 5
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