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
V
CSTR
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PFR
dX
t  N A0 
 rA V
0
dX
  r AV
dt
FA 0
dX
 rA
dV
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:
2AB
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    1501  
 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    1501  
 3  

 1
.75
G 


dz g c D p    
D
p 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    1501  
 3  
 1.75G 
Dp
   
 P T0 FT 0
G
0 
0 g c D p

 1    1501  
 3  
 1.75G 
Dp
   

9
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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
AB
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
dy
 T
1  X 

dW
2 y T0
For   0

dy

2y
dW
W hen W  0
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
12


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
18
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

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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
20
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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
dm 6
k  1.5
mol kg min
Case 1:
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Case 2:
W  100 kg
DP  2D P1
  0.0099kg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  kCA CB
3)
CA  CA0 1  Xy
4)
CB  CA0 1  Xy
W0
y 1
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Example 1: Gas Phase Reaction in PBR for δ = 0
5)
dy


dW
2y
2ydy  dW
y 2  1  W
y  1  W
12
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 rA  kC2A0 1  X y2  kC2A0 1  X 1  W
2
2
dX kC2A 0 1  X  1  W 

dW
FA 0
2
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Example 1: Gas Phase Reaction in PBR for δ = 0
kC2A 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 wit h pressure drop
X  0.75 wit hout 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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