Lecture 5 - University of Michigan
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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
2
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
4
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
8
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
9
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
10
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
13
X
14
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
16
B
Species
A
B
17
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
21
22
23
End of Lecture 5
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