Transcript lec8_anim
Lecture 8
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
Block 1: Mole Balances on PFRs and PBR
Must Use the Differential Form
Block 2: Rate Laws
Block 3: Stoichiometry
Pressure Drop:
Liquid Phase Reactions:
Pressure Drop does not affect the concentrations in liquid phase rx.
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Gas Phase Reactions:
Epsilon not Equal to Zero
d(P)/d(W)=..
Polymath will combine with d(X)/f(W)=..for you
Epsilon = 0 and Isothermal
P=f(W)
Combine then Separate Variables (X,W) and Integrate
Reactor Mole Balances in terms of
conversion
Reactor
Differential
Algebraic
Integral
X
Batch
N A0
dX
dt
r AV
0
V
CSTR
PFR
t N A0
FA 0
dX
dV
dX
rA V
t
FA 0 X
rA
X
rA
X
V FA 0
0
dX
rA
X
X
PBR
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FA 0
dX
dW
rA
W FA 0
0
dX
rA
W
Concentration Flow System:
CA
CB
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FA 0 1 X
0 1 X
T P0
0 1 X
Gas Phase Flow System:
FA
CA
FA
T P0
T0 P
C A 0 1 X T0 P
1 X
T P0
T0 P
b
b
FA 0 B X
C A0 B X
FB
a
a T0 P
P
T
1 X
T P0
0
0 1 X
T0 P
Pressure Drop in Packed Bed Reactors
Note: Pressure drop does NOT affect liquid phase reactions
Sample Question:
Analyze the following second order gas phase reaction that occurs
isothermally in a PBR:
AB
Mole Balance:
Must use the differential form of the mole balance to separate
variables:
dX
FA 0
5
dW
rA
Rate Law:
2
r
kC
Second order in A and irreversible:
A
A
Pressure Drop in Packed Bed Reactors
1 X P T0
CA
CA0
1 X P0 T
FA
Stoichiometry:
1 X P
CA CA0
1 X P0
Isothermal, T=T0
Combine:
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dX
dW
kC
2
A0
FA 0
1 X
1 X 2
2
P
P
0
2
Need to find (P/P0) as a function of W (or V if you have a PFR)
Pressure Drop in Packed Bed Reactors
Ergun Equation:
Constant mass flow:
dP
G 1 150 1
3
1
. 75
G
dz
g cD p
Dp
TURBULENT
LAMINAR
m
0
m
0 0
0
0
7
0
FT P0 T
FT 0 P T 0
0 (1 X )
P0 T
P T0
Pressure Drop in Packed Bed Reactors
Variable Density
0
P T 0 FT 0
P0 T FT
P T F
1 150 1
T
3
1 . 75 G 0
dz
0 g c D p
Dp
P T 0 FT 0
dP
Let
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G
1 150 1
3
0
1 . 75 G
0 g c D p
Dp
G
Pressure Drop in Packed Bed Reactors
Catalyst Weight W zA c b zA c 1 c
Where b bulk density
c solid catalyst
density
porosity ( a.k.a., void
Let
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dP
dW
0
fraction )
P0 T FT
A c 1 c P T0 FT 0
2 0
1
A c 1 c P0
Pressure Drop in Packed Bed Reactors
dy
dW
T FT
P
y
2 y T 0 FT 0
P0
We will use this form for single reactions:
d P P0
dW
dy
dW
dy
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dW
1
2 P P0 T 0
T
2 y T0
T
2y
1 X
1 X
1 X
Isothermal case
Pressure Drop in Packed Bed Reactors
dX
dW
dX
dW
kC A 0 1 X
2
FA 0 1 X
2
2
f X , P and
y
2
dP
dW
f X , P or
dy
dW
f y, X
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.
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Polymath will combine the mole balance, rate law and
stoichiometry.
PBR
AB
1) Mole Balance:
dX
dW
C A C A0
rA
FA 0
1 X P
1 X
CA0
y
1 X P0
1 X
1 X 2
y
1 X
2
2) Rate Law:
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rA kC
2
A0
PBR
For
0
dy
dW
2y
When
dy
2
W 0
dW
y (1 W )
2
y (1 W )
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1/ 2
y 1
1
P
y 1 W
1 2
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W
2
CA
C A C A 0 1 X
P
P0
No P
P
W
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3
-rA
2
rA kC A
No P
P
W
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4
X
No P
P
W
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0 (1 X )
T T0 ,
f
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0
P0 T
y
P T0
P0
P
1
(1 X ) y
5
P
1.0
No P
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
k 1 .5
dm
6
mol kg min
Case 1:
Case 2:
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W 100 kg
D P 2 D P1
0 . 0099 kg
P ?
X ?
P02
1
1
2
P01
X ?
P ?
Example 1: Gas Phase Reaction in PBR
for δ = 0
1) Mole Balance:
dX
dW
FA 0
2) Rate Law:
r ' A kC A C B
3)
C A C A 0 1 X y
4)
C B C A 0 1 X y
W 0
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r 'A
y 1
Example 1: Gas Phase Reaction in PBR
for δ = 0
5)
dy
dW
2 ydy dW
2y
y 1 W
2
y 1 W
1 2
rA kC A 0 1 X y kC A 0 1 X 1 W
2
dX
dW
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2
2
kC A 0 1 X 1 W
2
2
2
FA 0
2
Example 1: Gas Phase Reaction in PBR
for δ = 0
2
dX
1 X
2
kC A 0
FA 0
1 W dW
2
2
kC A 0
W
W
1 X
FA 0
2
X
W 0, X 0, W W , X X
X 0 . 6 with pressure
X 0 . 75 without
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drop
pressure
drop , i.e . 0
Example A + B → 2C
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Example A + B → 2C
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Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
Polymath Solution
A + 2B C
is carried out in a packed bed reactor in which there is pressure
drop.The fed is stoichiometric in A and B.
Plot the conversion and pressure ratio y = P/P0 as a function of
catalyst weight upto 100 kg.
Additional Information
kA = 6dm9/mol2/kg/min
α = 0.02 kg-1
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Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
A + 2B C
1) Mole Balance:
dX
dW
2) Rate Law:
r 'A
FA 0
r ' A kC A C B
2
3) Stoichiometry: Gas, Isothermal
V V 0 1 X
CA C A0
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P0
P
1 X
y
1 X
Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
q B - 2X )
(
4)
CB = CA0
y
(1+ e X )
5)
dy
dW
6) f =
7)
2y
1 X
u (1+ e X )
=
u0
y
2
3
C A 0 2 FA 0 2 k 6 0 . 02
Initial values: W=0, X=0, y=1 W=100
Combine with Polymath.
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If δ≠0, polymath must be used to solve.
Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
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Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
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T = T0
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Engineering Analysis
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Engineering Analysis
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Engineering Analysis
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Pressure Change – Molar Flow Rate
0
dP
dW
dW
dy
dW
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FT 0
FT 0 P T 0
A c 1 c
0
dy
FT
FT P0 T
FT T
FT 0 T 0
yP 0 A c 1 c
FT T
2 y FT 0 T 0
1 X
2 0
P0 A C 1 C
Use for heat effects, multiple rxns
Isothermal: T = T0
dX
dW
2y
1 X
Heat Effects
Isothermal Design
Stoichiometry
Rate Laws
Mole Balance
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Gas Phase Reaction with Pressure Drop
Concentration Flow System:
CA
CB
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FA 0 1 X
0 1 X
T P0
0 1 X
Gas Phase Flow System:
FA
CA
FA
T P0
T0 P
C A 0 1 X T0 P
1 X
T P0
T0 P
b
b
FA 0 B X
C A0 B X
F
a
a T0 P
B
T P0
1 X
T P0
0 1 X
T0 P
Example 2: Gas Phase Reaction in PBR
for δ ≠ 0
A + 2B C
1) Mole Balance:
dX -rA¢
=
dW FA0
2) Rate Law:
-rA¢ = kCAC2B
3) Stoichiometry: Gas, Isothermal
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(
)
P0
u = u 0 1+ eX
P
1- X)
(
CA = CA0
y
(1+ eX)
End of Lecture 8
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