ChE422 Topic 8

Introduction to Packed Bed Reactors (With Fixed Bed as Catalyst)

Topic 8

Ch.E. 422

Reference: ECRE by Fogler (pp. 170-187)

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.

©UST Ch.E. Department

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).

Fluidized Bed Reactor

Mole Balance for a steady state PBR for a Product A


W = weight of catalyst

r ' in Mole A/weight catalyst-timeA

Mole Balance for a steady state PBR for a Product A


For A as reactant:

Mole Balance for a steady state PBR for a Product A in terms of XA


(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.

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

Volumetric Flow Relations in a PBR and PFR


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

Volumetric Flow Relations in a PBR and PFR (Gas System)


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

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

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

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

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)

TURBULENT

LAMINAR

p

3

pc

G75.1D

11501

Dg

G

dz

dPErgun Equation:


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

0 3

0

0

150 111.75

Unit of : Pressure/length (e.g. kPa/m)

c p p

GG

g D D

Let


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

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

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


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

0

12

A A

dy TX

dW y T

12

A A

dyX

dW y

Isothermal case:


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

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

Exercises:


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.

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

Polymath Solution with y = 1


Polymath Solution with y = 1


Polymath Solution with y


Polymath Solution with y


Exercises:


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%.

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

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

Polymath Solution


Polymath Solution


Exercises:


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%.

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

Polymath Solution


Polymath Solution


Engineering Analysis



ChE422 Topic 8

Documents