fixed bed and fluidized bed

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

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 Area2

Area where flow occurs = 4

D 2

4

DForce P

Resistance : due to Shear Find Contact Area Find shear stress

Contact areaForce

Until now, we haven’t 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 p

R

p

VA

V

= 1 pR

p

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



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 p

R

p

VA

V

4H

Flow AreaD

ContactPerimeter

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 pavg

p

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

0

2

2 1 p

p

AVV

2

4

DForce P

02

2

2 1 1

4

p

p pR

p

AV V AD

P VV

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

AV

PV

D DL



Pressure drop

2

2

0 2

2

2

4

1

4

2

p

p

AV V

P LD D

2

2

0

3

2 1

p

p

ALV V

P

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

pDLVP

2

02 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 V

P

In terms of specific surface area

2

02 3

1

150

p

LVP

D

In terms of average particle diameter


Fixed bed - pressure drop calculation (Turbulent Flow)

Pressure drop and shear stress equations2

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=constant 2 avgV

202

=V

K

0avg

VV

However



We have already developed an expression for contact area

Wetted Area= ppN A = 1 pR

p

AV

V

=1

pR

p

VA

V

2

02

1 pR

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

30 1 p

p

AVP K L

V



2

03

16

p

VP L

DK

In terms of average particle diameter

2

03

1 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

0Laminar 2 3

150 1

p

LVP

D

20

3

1

7

4Turbulentp

LVP

D

2 20 02 3 3

150 1 7 1

4totalp 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 20 02 3 3

150 1 7 1

4totalp p

LV LVP

D D

Ergun Equation for

packed bed

0 20

2 2

2

2 1724

1

2

p

p

avg

AV V V

fV

Some texts provide equation for friction factor

212 avg

fV

laminar

212

turbulent

avg

fV



0

2

2

20

2

2

0

2 1

1

2

p

p

AV

K

fV

VV

0

4 17

12

p

p

V

AV

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

AV

Vf

0

25 17

3 12

p

p

AV

V



0

25 17

3 12

p

p

AV

fV

Multiply by 3 on both sides (why?)

0

25 17

3 1

6

2p

V

D

0

150 1 73

4p

fD 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 Ren

A Bf C

1p oD V


Fixed Bed Sphericity vs Void Fraction

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

2

2( , , , )p p o p

o column

D D V DPf

V L D


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.

At fluidization, Weight of the particles == pressure drop (area) Remember to include buoyancy

2

14 s f R

DP V

2

14s f

DL


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


Fluidized bed: Operation

Range of operation depends on Ar

Ar

100 104 108

R

0

80

40


Fluidized bed: Operation 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

fixed bed and fluidized bed

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