che422 topic 8
DESCRIPTION
ChE422 Topic 8TRANSCRIPT
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Introduction to Packed Bed Reactors (With Fixed Bed as Catalyst)
Topic 8
Ch.E. 422
Reference: ECRE by Fogler (pp. 170-187)
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After this topic, you should be able to:
Describe the types and enumerate some applications of a Packed Bed Reactor (PBR);
Set up a mole balance for a steady state PBR with the fixed bed as catalyst;
Set up and apply pressure drop equations across the same PBR.
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Packed Bed Reactor
Reactants are continuously
fed to a packed bed which
could be a catalyst or
another reactant. Reaction
takes place across the bed
and products exit at the
other end. The bed may be
stationary (fixed) or moving
(fluidized).
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Fluidized Bed Reactor
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Mole Balance for a steady state PBR for a Product A
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W = weight of catalyst
r ' in Mole A/weight catalyst-timeA
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Mole Balance for a steady state PBR for a Product A
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For A as reactant:
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Mole Balance for a steady state PBR for a Product A in terms of XA
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(1 )A Ao AF F X A Ao AdF F dX
1
1
0'
AX
AAo
A
dXW F
r
Integral may be solved if rA’ may be expressed in
terms of XA or data of XA and rA’ is available.
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Summary of Reactor Mole Balances in terms of Conversion, XA (For Finals)
Reactor
Differential
Algebraic
Integral
0A A
A
F XV
r
CSTR
0A
A A
dXF r
dV 0
0
AfX
AA
A
dXV F
r
PFR
0A
A A
dXN r V
dt 0
0
AX
AA
A
dXt N
r V
Batch
XA
t
0A
A A
dXF r
dW
1
1 0
0
AX
AA
A
dXW F
r
PBR
XA
W 8
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Volumetric Flow Relations in a PBR and PFR
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0
0
0T
T0
T
T
P
P
F
F
0
0 0
0 0
= mass rate symbol
(mass rate = density x volumetric rate)
F (mass rate = molal rate x Molecular Weight)T T
m m
M F M
Applying Continuity Equation: mass flow rate = constant
0 00
0
P MPM
RT RT 0 0
0
0
P M T
P M T
00
0
0
T
T
M F
M F
For Liquids, assume incompressible or 𝜌=𝜌0; 𝜐 = 𝜐0
For Gases: assume ideal gas behavior
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Volumetric Flow Relations in a PBR and PFR (Gas System)
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0
0
0T
T0
T
T
P
P
F
F
Recall Batch Relation between Total Moles with Conversion, XA:
0(1 )T T A AN N X
For a Continuous Flow Reactor:
0(1 )T T A AF F X 0
1TA A
T
FX
F
00
0
(1 )A A
P TX
P T
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Gas Phase Flow System:
Concentration for Flow System:
0 0 0
0 00
0
1 1
11
A A A AAA
A AA A
F X C X TF PC
PT X T PX
T P
A
A
FC
00
0
1 A A
PTX
T P
0 0
0
0 00
0
11
A B A A B A
BB
A AA A
b bF X C X
TF Pa aC
PT X T PX
T P
For: aA + bB -> R
BoB
Ao
CM
C
Concentration Relations
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Decomposition Reaction: a AB
0A
A A
dXF r
dW
Mole Balance:
n
A Ar kC Rate Equation:
Pressure Drop in Packed Bed Reactors Conducting a Gaseous Decomposition
1Unit of k : (Volume /mol ) / kg catalyst-timen n
A
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00
0
1
1
AAA A
A A
X TF PC C
X P T
Stoichiometry:
0
0
1
1
A
A A
A A
X PC C
X P
Isothermal, T=T0
0
0 0
1
1
n
AA AA
A A A
XdX k PC
dW F X P
Combine:
Need to find (P/P0) as a function of W for Gas System
Pressure Drop in Packed Bed Reactors Conducting a Gaseous Decomposition
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TURBULENT
LAMINAR
p
3
pc
G75.1D
11501
Dg
G
dz
dPErgun Equation:
Pressure Drop in Packed Bed Reactors
P = pressure (psf) – varies with z
z = distance from reactor entrance
Φ = porosity = volume of void/total bed volume = constant
1-Φ = volume solids/total bed volume = constant
gc = 4.17 x 108 (lbm-ft/hr2/lbf)
Dp = particle diameter (ft) = constant
μ = gas viscosity (lbm/ft-hr) = f( T)
G = superficial mass velocity of gas (lbm/ft2-hr) = constant
ρ = gas density (variable with T, P, FT)
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TURBULENT
LAMINAR
p
3
pc
G75.1D
11501
Dg
G
dz
dPErgun Equation:
Pressure Drop in Packed Bed Reactors
0T
T
0
0
p
3
pc0 F
F
T
T
P
PG75.1
D
11501
Dg
G
dz
dP
00
0
0
0T
T0
T
T
P
P
F
F 0 0
0 0
T
T
PF T
F P T
0
0 0 0
1 1 T
T
PF T
F P T
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0 3
0
0
150 111.75
Unit of : Pressure/length (e.g. kPa/m)
c p p
GG
g D D
Let
Pressure Drop in Packed Bed Reactors
0T
T
0
0
p
3
pc0 F
F
T
T
P
PG75.1
D
11501
Dg
G
dz
dP
00
0 0
T
T
P FdP T
dz P T F
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ccbc 1zAzAW Catalyst Weight
Relation Between W and z
cross sectional area
= distance from reactor entrance
bulk density of the solid (with void volume)
= density of solid catalyst
= porosity or void volume fraction
1- = solids volume fraction
c
b
c
A
z
(1 )
(1 )
c c
c c
dW A dz
dWdz
A
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0T
T
0
0
cc
0
F
F
T
T
P
P
1AdW
dP
0
0
2 1 Unit of : 1/mass
1c cA P
Let
Relation Between P/Po and W
but: (1 )c c
dWdz
A
From the Ergun Equation:
00
0 0
T
T
P FdP T
dz P T F
2
0
0 02T
T
P FdP T
dW P T F
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Pressure Drop in Packed Bed Reactors
2
0
0 02T
T
P FdP T
dW P T F
0
0 0
0
1
2T
T
PdP FT
dW T FPP
0
0
But: (1 ) or 1TT T A A A A
T
FF F X X
F
Let y = 𝑃
𝑃0
0
0 0
11
2A A
d P P TX
dW P P T
0
12
A A
dy TX
dW y T
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0
12
A A
dy TX
dW y T
12
A A
dyX
dW y
Isothermal case:
Pressure Drop in Packed Bed Reactors
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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 ε= 0, we can obtain
an analytical solution.
Polymath will combine the mole balance, rate law and
stoichiometry.
Pressure Drop in Packed Bed Reactors Applied to a Gaseous Decomposition
2
1
0 0
0 0 0
1 1
1 1
n n
A AA A AA A
A A A A A A
X XdX k kPC C y
dW F X P F X
(1 )2
A A
dyX
dW y
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Special Case where εA = 0
1 0
2
1/2
0
2
2
0 1
2
1
(1 )
A
y W
For
dy
dW y
y dy dW
When W y
y dy dW
y W
y W
(1 )2
A A
dyX
dW y
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Exercises:
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1. Consider the elementary, isothermal gas phase decomposition:
2 A -> B +C taking place in a 20 m 1.5” Sch. 40 pipe packed with a
catalyst. The flow and packed bed conditions are as follows:
Po = 1013 kPa vo = 7.15 m3/hr
Solid catalyst density: 1923 kg/m3
Porosity = 0.45
Cross Sectional Area of Pipe: 0.0013 m2
Pressure Drop Parameter βo = 25.8 kPa/m
Entering Concentration: 0.1 kmol/m3
Rate constant (k) : 12 m6/kmol-kg-cat-(hr)
a) Calculate the Final Conversion neglecting the pressure drop
b) Calculate the Final Conversion considering the pressure drop
c) Determine how Answer in (b) will change if Dp is doubled.
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Exercise 1
1) Mole Balance:
0A
A
F
r
dW
dX
2) Rate Law:
2
2
2
0AA yX1
X1kCr
2AB +C
yX1
X1C
P
P
X1
X1CC 0A
0
0AA
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Polymath Solution with y = 1
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Polymath Solution with y = 1
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Polymath Solution with y
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Polymath Solution with y
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Exercises:
©UST Ch.E. Department
2 . Consider the elementary, isothermal gas phase decomposition:
A + B -> 2C taking place in a fixed bed catalyst reactor under the
following conditions:
CAo = CBo = 0.2 M
FAo = 2 mols/min
Rate constant (k) : 1.5 dm6/mol-kg-cat-min
α = 0.0099/kg
Weight of Catalyst = 100 kg
a) Calculate the Final Conversion and Final Pressure
b) Determine how Answer in (b) will change if Dp is doubled and
the entering pressure is reduced by 50%.
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Example 3: 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 equimolar feed of A and B and kA = 1.5dm6/mol/kg/min α = 0.0099 kg-1
Find X at 100 kg
3
0
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A + B 2C
min kg mol
dm5.1k
6
1kg 0099.0
kg 100W ?X ?P
1PP D2D 0102 P2
1P Case 2:
Exercise 2: Gas Phase Reaction in PBR for δ = 0
Case 1:
?X ?P
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Polymath Solution
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Polymath Solution
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Exercises:
©UST Ch.E. Department
3 . Consider the elementary, isothermal gas phase decomposition:
A + 2B -> 2C taking place in a fixed bed catalyst reactor under the
following conditions:
CAo = 0.2 M; CBo = 0.4 M
FAo = 2 mols/min
Po = 10 atm
Rate constant (k) : 6 dm9/mol2-kg-cat-min
α = 0.02/kg
Weight of Catalyst = 100 kg
a) Calculate the Final Conversion and Final Pressure
b) Determine how Answer in (b) will change if Dp is doubled and
the entering pressure is reduced by 50%.
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Exercise 3: 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 feed is stoichiometric in A and B.
Find the conversion and pressure ratio y = P/P0 for a catalyst weight of 100 kg.
Additional Information kA = 6dm9/mol2/kg/min α = 0.02 kg-1
3
5
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Polymath Solution
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Polymath Solution
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Engineering Analysis
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Engineering Analysis
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Engineering Analysis