Transcript Lecture 6
Lecture 6
Kjemisk reaksjonsteknikk
Chemical Reaction Engineering
Review of previous lectures
Pressure drop in fixed bed reactor
PFR reactor design with pressure drop (ε=0)
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Reaction Engineering
Stoichiometry
Rate Laws
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Mole Balance
Isothermal reactor design
These topics build upon one another
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Reactor Mole Balances in terms of conversion
Reactor
Differential
Algebraic
Integral
X
X
Batch
N A0
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PFR
0
V
CSTR
FA 0
dX
rA
dV
dX
rA V
t N A0
dX
r AV
dt
t
FA 0 X
rA
X
V FA 0
0
dX
rA
X
PBR
FA 0
dX
rA
dW
X
W FA 0
0
dX
rA
W
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Concentration Flow System:
Gas Phase Flow System:
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FA
CA
0 1 X
T P0
T0 P
FA 0 1 X
C A 0 1 X T0 P
FA
CA
P
T
1 X T P0
0
0 1 X
T0 P
b
b
FA 0 B X C A 0 B X
FB
a
a T0 P
CB
P
T
1 X
1 X
T P0
0
0
T0 P
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Pressure Drop in Packed Bed Reactors
Note: Pressure drop does NOT affect liquid phase reactions
Sample Question:
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Analyze the following second order gas phase reaction that occurs
isothermally in a PBR:
2AB
Mole Balance:
Must use the differential form of the mole balance to separate variables:
dX
FA 0
rA
dW
Rate Law:
2
r
kC
Second order in A and irreversible:
A
A
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Pressure Drop in Packed Bed Reactors
1 X P T0
CA
CA 0
1 X P0 T
FA
Stoichiometry:
1 X P
CA CA 0
1 X P0
Isothermal, T=T0
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Combine:
2
A0
dX kC
dW
FA 0
1 X P
1 X 2 P0
2
2
Need to find (P/P0) as a function of W (or V if you have a PFR)
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Pressure Drop in Packed Bed Reactors
Ergun Equation:
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dP
G 1 1501
3
1
.75
G
dz g c D p
D
p TURBULENT
LAMINAR
P pressure,
Φ porosity (volume of void/total bed volume)
1- Φ (volume of solid/total bed volume)
gc conversion factor. 1.0 for metric system
Dp diameter of particle in bed
μ viscosity of gas passing through the bed
Z length down the packed bed
u, superficial velocity
ρ gas density
G= ρu superficial mass velocity
kPa
m
kg/m.s
m
m/s
kg/m3
kg/m2,s
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Pressure Drop in Packed Bed Reactors
dP
G 1 1501
3
1
.75
G
dz g c D p
Dp
TURBULENT
LAMINAR
Ergun Equation:
Constant mass flow:
m
0
m
0 0
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0
0
FT P0 T
0
FT 0 P T0
P0 T
0 (1 X)
P T0
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Pressure Drop in Packed Bed Reactors
Variable Density
dP
G
dz 0 g c D p
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Let
P T0 FT 0
0
P0 T FT
P0 T FT
1 1501
3
1.75G
Dp
P T0 FT 0
G
0
0 g c D p
1 1501
3
1.75G
Dp
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Pressure Drop in Packed Bed Reactors
Catalyst Weight
W zA zA 1
c
Where
b
c
c
b bulk density
c solid catalyst density
porosity (a.k.a., void fraction)
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Let
Ac, cross section area
0
P0 T FT
dP
dW A c 1 c P T0 FT 0
2 0
1
A c 1 c P0
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Pressure Drop in Packed Bed Reactors
dy
T FT
dW
2 y T0 FT 0
P
y
P0
We will use this form for single reactions:
d P P0
1
T
1 X
dW
2 P P0 T0
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dy
T
1 X
dW
2 y T0
dy
1 X
dW
2y
Isothermal
case
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Pressure Drop in Packed Bed Reactors
dX kC2A 0 1 X 2
y
2
dW FA 0 1 X
2
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dX
dP
dy
f X, P and
f X, P or
f y, X
dW
dW
dW
The two expressions are coupled ordinary differential equations. We
can only solve them simultaneously using an ODE solver such as
Polymath. For the special case of isothermal operation and epsilon =
0, we can obtain an analytical solution.
Polymath will combine the mole balance, rate law and
stoichiometry.
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PBR
AB
1) Mole Balance:
dX rA
dW FA 0
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1 X P
1 X
CA 0
CA CA 0
y
1 X P0
1 X
1 X 2
1 X y
2
2) Rate Law:
rA kC
2
A0
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PBR
For
dy
T
1 X
dW
2 y T0
0
dy
dW 2 y
When
W0
y 1
Initial condition
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dy 2 dW
y 2 (1 W )
y (1 W )
1/ 2
2 0
1
A c 1 c P0
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1
P
y 1 W
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W
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2
CA
P
CA CA 0 1 X
P0
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No P
P
W
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3
-rA
rA kCA2
No P
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P
W
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4
X
No P
P
W
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P0 T
0 (1 X)
P T0
P0
T T0 , y
P
0
1
f
(1 X) y
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5
P
1.0
No P
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W
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Example 1: Gas Phase Reaction in PBR for δ = 0
Gas Phase Reaction in PBR with δ = 0 (Polymath Solution)
A + B 2C
Repeat the previous one with equil molar feed of A and B and
kA = 1.5dm9/mol2/kg/min
α = 0.0099 kg-1
Find X at 100 kg
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Example 1: Gas Phase Reaction in PBR for δ = 0
A + B 2C
dm6
k 1.5
mol kg min
Case 1:
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Case 2:
W 100 kg
DP 2DP1
0.0099 kg 1
X?
1
P02 P01
2
P?
X?
P?
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Example 1: Gas Phase Reaction in PBR for δ = 0
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1) Mole Balance:
dX r 'A
dW
FA 0
2) Rate Law:
r'A kCACB
3)
CA CA 0 1 X y
4)
CB CA 0 1 X y
W0
y 1
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Example 1: Gas Phase Reaction in PBR for δ = 0
5)
dy
dW
2y
2 ydy dW
y 2 1 W
y 1 W
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rA kC2A0 1 X y2 kC2A0 1 X 1 W
2
2
dX kC 2A 0 1 X 1 W
dW
FA 0
2
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Example 1: Gas Phase Reaction in PBR for δ = 0
kC 2A 0
dX
1 W dW
2
FA 0
1 X
kC2A 0
X
1 X
FA 0
W 2
W
2
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W 0, X 0, W W, X X
X 0.6 with pressure drop
X 0.75 without pressure drop, i.e. 0
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Example A + B → 2C
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Example A + B → 2C
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