Fluidized Bed

Fixed bed and fluidized bed

Why fixed (or fluidized) bed?

Expensive Catalyst

enzyme (immobilized)

Large Surface area

Used in reaction/adsorption/ elution (for example)

Goal: Expression for pressure drop, try some

examples

Ref: BSL, McCabe & Smith

Fixed bed

Filled with particles

Usually not spherical

To increase surface area

To increase void fraction

To decrease pressure drop

For analytical calculation, assume all particles are

identical

Usable, because final formula can be modified by a

constant factor (determined by experiment)

Fixed bed

What are important parameters?

(For example, for adsorption of a protein from a

broth)

rate of adsorption (faster is better)

saturation concentration (more is better)

From the product requirement (eg X kg per day),

density and product concentration in broth ==>

volumetric flow rate

Fixed bed

Sphericity

Volume of particle = Vp

Surface Area of particle = Ap

Surface Area of sphere of same volume (Vs =Vp) = As

Sphericity = As/Ap

May be around 0.3 for particles used in packed beds

lower sphericity ==> larger surface area

Assume quick adsorption (rate of adsorption is high)

Calculate the surface area of particles needed for

operation

As,

Vs

Ap,

Vp

Sphericity specific surface area average particle diameter

Fixed bed

Specific surface area

= Ap /Vp

Minimal value for sphere

Some books use S to denote area (instead of A)

Assume all the particles are identical

==> all particles have exactly same specific surface area

Tarus saddlePall Ring

Rings (Raschig,etc)

Fixed bed

What is the pressure drop we need, to force the fluid through

the column?

(i.e. what should be the pump spec)

We know the volumetric flow rate (from adsorption

equations, productivity requirements etc)

We know the area per particle (we assume all particles are

identical). And the total area for adsorption (or reaction in

case of catalytic reactor).

Hence we can calculate how many particles are needed

Given a particle type (eg Raschig ring) , the approximate

void fraction is also known (based on experimental results)

Fixed bed What is void fraction?

Volume of reactor = VR

Number of particles = Np

Volume of one particle = Vp

Volume of all the particles = Vp * Np = VALL-PARTICLES

R ALL PARTICLES

R

V V

V

VOIDS

R

VVoid fraction

V

R P P

R

V V N

V

1RP

P

VN

V

Knowing void fraction, we can find the reactor volume

needed

Alternatively, if we know the reactor volume and void

fraction and the Vp, we can find the number of particles

Fixed bed To find void fraction experimentally

Prepare the adsorption column (or reactor....) and fill it

with particles

Fill it with water

Drain and measure the quantity of water

(= void volume)

Calculate void fraction

Fixed bed Since we know Vp, Np, , we can find VR

Choose a diameter and calculate the length (i.e. Height)

of the column (for now)

In normal usage, both the terms height and length may

be used interchangeably (to mean the same thing)

Adsorption rate, equilibrium and other parameters will

also influence the determination of height & diameter

To calculate the pressure drop

Note: columns with large dia and shorter length (height)

will have lower pressure drop

What can be the disadvantage(s) of such design ?

(tutorial)

Fixed bed To calculate the pressure drop

You want to write it in terms of known quantities

Length of column, void fraction, diameter of particles, flow rate of

fluid, viscosity and density

Obtain equations for two regimes separately (turbulent and laminar)

Consider laminar flow

Pressure drop increases with

velocity

viscosity

inversely proportional to radius

Actually, not all the reactor area is available for flow. Particles

block most of the area. Flow path is not really like a simple tube

Hence, use hydraulic radius

Fixed bed - pressure drop calculation (Laminar flow)

To calculate the pressure drop, use Force balance

Force P Area

2

Area whereflow occurs = 4

D

2

4

DForce P

Resistance : due to Shear

Find Contact Area

Find shear stress

Contact areaForce

Until now, we havent said anything about laminar

flow. So the above equations are valid for both

laminar and turbulent flows

Fixed bed - pressure drop calculation (Laminar Flow)

Find contact area

Wetted Area= ppN A =1

pR

p

VA

V

= 1 pRp

AV

V

To calculate the shear stress, FOR LAMINAR FLOW

max42 avgVV

R R

r R

dV

dr

8 avgV

D

2

max 21

rV V

R

max 2 avgV V

Here V refers to velocity for flow in a tube

However, flow is through bed, NOT a simple tube

R P P

R

V V N

V

1RP

P

VN

V


Find effective diameter (i.e. Use Hydraulic radius), to

substitute in the formula

Also relate the velocity between particles to some quantity

we know

To find hydraulic radius ( and hence effective dia)

RFlowvolume V

Wetted Area= ppN A

=1

pR

p

VA

V

4HFlow Area

DContactPerimeter

Hydraulic diameter*

4*

Flow Area Column Height

ContactPerimeter Column Height

4Flowvolume

wetted area


4

1H

p

p

DA

V

8 avg

H

V

D

8 1

4

pavg

p

AV

V

2 1 pavgp

AV

V

Vavg is average velocity of fluid in the bed, between

particles

Normally, volumetric flow rate is easier to find


Can we relate volumetric flow rate to Vavg ?

Use a new term Superficial velocity (V0)

0

Volumetric flowrateV

Column Area 0 2

4

QV

D

I.e. Velocity in an empty column, that will provide the

same volumetric flow rate

Can we relate average velocity and superficial velocity?

0avg

VV


2 1gp

pav

AV

V

02

2 1 p

p

A

VV

2

4

DForce P

02

2

2 1

14

p

p p

R

p

AV

V ADP V

V

Force balance: Substitute for etc.

Contact areaForce

2

4R

DV L

Volume of reactor (say, height of bed = L)

2

0

2

2

2

2

4

1

4

2

p

p

A

VP

VD D

L


Pressure drop

2

2

0 2

2

2

4

1

4

2

p

p

AV

VP L

D D

2

2

0

3

2 1

p

p

ALV

VP

Specific surface area vs average diameter

p

p

A

V

Define average Dia of particle as

6p

p

p

DA

V

Some books (BSL) use Dp


Pressure drop

2

2

0

3

62 1

pD

LV

P

2

0

2 3

72 1

p

LV

D

However, using hydraulic radius etc are only

approximations

Experimental data shows, we need to multiply the

pressure requirement by ~ 2 (exactly 100/48)

2

2

0

3

25

6

1

p

p

ALV

VP

In terms of specific surface area

2

0

2 3

1

150

p

LVP

D

In terms of average particle diameter

Fixed bed - pressure drop calculation (Turbulent Flow)

Pressure drop and shear stress equations

2

4

DForce P

Contact areaForce

Only the expression for shear stress changes

f

Re

For high turbulence (high Re),

2=constant

12 avg

fV

21=constant2 avg

V

2

0

2=

VK

0avg

VV

However


We have already developed an expression for contact area

Wetted Area= ppN A = 1 pRp

AV

V =

1 p

R

p

VA

V

2

0

21

p

R

p

AVK V

V

2

Contact area4

DForce P

Hence, force balance

2

4R

DV L

Volume of reactor (say, height of bed = L)

2

3

0 1p

p

AVP K L

V


2

0

31

6

p

VP L

DK

In terms of average particle diameter

2

0

31

p

p

AVP K L

V

In terms of specific surface area

Value of K based on experiments ~ 7/24

What if turbulence is not high?

Use the combination of laminar + turbulent pressure drops:

valid for all regimes!

2

0

Laminar 2 3

150 1

p

LVP

D

203

1

7

4Turbulent

p

LVP

D

2 2

0 0

2 3 3

150 1 7 1

4total

p p

LV LVP

D D

Ergun Equation for

packed bed

Fixed bed - pressure drop calculation (Laminar OR Turbulent Flow)

If velocity is very low, turbulent part of pressure drop is

negligible

If velocity is very high, laminar part is negligible

2 2

0 0

2 3 3

150 1 7 1

4total

p p

LV LVP

D D

Ergun Equation for

packed bed

0 20

2 2

2

2 17

24

1

2

p

p

avg

AV

V V

fV

Some texts provide equation for friction factor

212 avg

fV

laminar

212

turbulent

avg

fV


02

2

2

0

2

2

0

2 1

1

2

p

p

A

VK

fV

VV

0

4 17

12

p

p

V

A

V

For pressure drop, we multiplied the laminar part by 2

(based on data) . For the turbulent part, the constant was

based on data anyway.

Similarly...

0

100

48

4 17

12

p

p

A

V

Vf

0

25 17

3 12

p

p

A

V

V


0

25 17

3 12

p

p

A

Vf

V

Multiply by 3 on both sides (why?)

0

25 17

3 1

6

2

p

V

D

0

150 1 73

4pf

D V

0

150 1 73

4pD Vf

Packed bed friction factor = 3 f

150 13 1

R.75

e ppf f

Eqn in McCabe and Smith

Reynolds number for packed bed

Example

Adsorption of Cephalosporin (antibiotic)

Particles are made of anionic resin(perhaps resin coatings on ceramic

particles)

void fraction 0.3, specific surface area = 50 m2/m3(assumed)

column dia 4 cm, length 1 m

feed concentration 2 mg/liter (not necessary to calculate pressure drop, but

needed for finding out volume of reactor, which, in this case, is given). Superficial

velocity about 2 m / hr

Viscosity = 0.002 Pa-s (assumed)

What is the pressure drop needed to operate this column?

Fixed Bed

What is the criteria for Laminar flow?

Modified Reynolds Number

Turbulent flow:- Inertial loss vs turbulent loss

Loss due to expansion and contraction

Packing uniformity

In theory, the bed has a uniform filling and a constant void fraction

Practically, near the walls, the void fraction is more

Edge Center Edge

0.2

0.4

0.8 Ergun Eqn commonly

used, however, other

empirical correlations are

also used

e.g. Chilton Colburn eqn

Re RenA B

f C

1p oD V

Fixed Bed

Sphericity vs Void Fraction

f

0 1

1

~0.4

Fixed Bed

Alternate method to arrive at Ergun equation (or similar correlations)

Use Dimensional analysis

dependent variableP

( subscript, means fluid density or )fwithout

, , , , , , (i.e. sphericity)p o columnD L V D f

2

2( , , , )

p p o p

o column

D D V DPf

V L D

f

Fluidized bed

When the fluid (moving from bottom of the column to the top) velocity is increased, the particles begin to move at (and above) a certain velocity.

Fluidized bed

At fluidization,

Weight of the particles == pressure drop (area)

Remember to include buoyancy

2

14

s f R

DP V

2

14

s f

DL

IIT-Madras, Momentum Transfer: July 2005-Dec 2005

Fluidized bed: Operation

Empirical correlation for porosity

n

t

V

V

Types of fluidization: Aggregate fluidization vs Particulate

fluidization

Larger particles, large density difference (SOLID - FLUID)

==> Aggregate fluidization (slugging, bubbles, etc)

==> Typically gas fluidization

Even with liquids, lead particles tend to undergo

aggregate fluidization

Archimedes number 3

2

f pg DAr

Fluidized bed: Operation Porosity increases

Bed height increases

Fluidization can be sustained until terminal velocity is reached

If the bed has a variety of particles (usually same material, but

different sizes)

calculate the terminal velocity for the smallest particle

Range of operability = R

Minimum fluidization velocity = incipient velocity (min range)

Maximum fluidization velocity = terminal velocity (max range)

Other parameters may limit the actual range further

e.g. Column may not withstand the pressure, may not be tall

enough etc

R = Vt/VOM

Theoretically R can range from 8.4 to 74


Range of operation

depends on Ar

Ar

100 104 108

R

0

80

40


Criteria for aggregate fluidization

Semi empirical

0.5

20.6 ( )

0.3 ( )

p

s

Dfor liquid

for gas

Particulate fluidization

Typically for low Ar numbers

More homogenous mixture

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Fluidized Bed

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