che572 chapter 3 fluidization.pdf

8/19/2019 CHE572 Chapter 3 Fluidization.pdf

1/31

26

CHAPTER 3FLUIDIZATION

3.1 Introduction

• A fluidized bed is formed by passing a fluid usually agas upwards through a bed of particles supported ona distributor.

• As a fluid is passed upward through a bed of

particles, pressure loss due to frictional resistanceincreases as fluid flow increases.

• At a point, upward drag force exerted by the fluid onthe particle equal to apparent weight of particles inthe bed.

W

F F F = drag forceW = apparent weight


2/31

27

Figure 3.1: Elements of a Fluidized Bed

Gasin

Windbox

Gasdistributor

Fluidbed

Disengagementspace

Solid

feed

Soliddischarge

Dustout

Gasout

Dust separator


3/31

28

3.2 Characteristics of Gas Fluidized Bed

These can be roughly divided into two categories;

3.2.1 Primary Characteristics

• Bed behaves like liquid of the same bulk density –can add or remove particles, pressure-depthrelationship, wave motion, heavy objects sink, andlight ones float.

• Rapid particle motion – good solid mixing

• Very large surface area available – 1m3 of 100 µmparticles has a surface area of about 30,000 m2, and

1 m3 of 50 µm particles – 60,000 m2.

3.2.2 Secondary Characteristics

• Good heat transfer from surface to bed, and gas toparticles.

• Isothermal conditions radially and axially.

• Pressure drop through bed depends only on beddepth and particle density – does not increase with

gas velocity.

• Particles motions usually streamline – some erosionof surface or attrition of particles where gas velocitiesare high.


4/31

29

3.3 Advantages of Fluidized Bed

• High mobilityo

Gives superb heat transfer, which usuallyalways a problem to powders.o Heavily used for drying eg: pharmaceutical

industry.o Excellent reactors

• Good temperature controlo A perfect gas/liquid mixing equipment.

• Very flexibleo Can carry out many processes in a single

vessel.o Mix, dry, granule, separate etc. in one vessel.

• Less number of moving partso Easy to handle

3.4 Disadvantages of Fluidized Bed.

• Costlyo Blowing air into the system.o Trap air to make it fluidized.o Cleaning processo Some powders – costly in operation than others.

• Not all particles fluidizedo Cohesive and large particles are difficult to

fluidize.

• Difficult distributor designo Maldistribution of fluidizing gas


5/31

30

o ∆P across distributor = 30% of bed ∆P.

3.5 Pressure Drop Flow Relationship

• The force balance;

Pressure = Weight of particles – upthrust on particlesdrop

Bed cross-sectional area

• For a bed of particle density, ρ p , fluidized by a fluidwith ρ f to form a bed of depth, H and voidage, ε in avessel of cross sectional area, A;

( )

A

g HAP

f p ρ ρ ε −−=∆

1 (3.1)

or

( ) g H P f p ρ ε −−=∆ 1 (3.2)

• For a flow of fluid through a packed bed, two distincttypes of flow involved. They are laminar andturbulent flow.

• The pressure drop across a fluidized bed is the onlyparameter which can be accurately predicted:

A

MgPF =∆ N/m2 (3.3)

or


6/31

31

∆PF A

M 1.0= (kg/m2) (3.4)

where M in kg and A in m2.

( )( )g H

Pg pmf

F ρ ρ ε −−=∆

1 (3.5)

• ε mf is the bed voidage at U mf and a closeapproximation to it can be obtained by measuringthe aerated or most loosely packed bulk density,

ρ bLP .

• Equations 3.3 to 3.5 usually are used to predict thetheoretical pressure drop comparing toexperimental one.

3.5.1 Laminar Flow

• Through the work of Darcy and Poiseuille, it hasbeen known for more than 120 years that theaverage velocity through a packed bed, or througha pipe, is proportional to the pressure gradient.

• Pressure gradient ∝ fluid velocity

or

( )U

H

P∝

∆− (3.6)


7/31

32

• Based on Carmen-Kozeny (1927, 1933 and 1937),

( ) ( )32

21180

ε

ε µ

pd

U

H

P −=

∆−

(3.7)

→ Carmen-Kozeny equation for laminar flow.

3.5.2 Turbulent Flow

( ) ( )3

2

175.1ε

ε ρ p

g

d U

H P −=∆− (3.8)

→ Burke – Plumme equation for turbulent flowthrough a randomly packed bed of monosizedspheres of diameter, d p .

3.5.3 General equation for turbulent and laminarflow.

• Based on experimental data covering a wide range ofsize and shape of particles, Ergun (1952) suggestedthe following general equation for any flowconditions;

( ) ( ) ( )3

2

32

2 175.11150

ε

ε ρ

ε

ε µ

p

g

p d

U

d

U

H

P −+

−=

∆− (3.9)

Laminarcomponent

Turbulentcomponent


8/31

33

• Reynold number, ( )ε µ

ρ

−=

1Re*

U d g p (3.10)

For Re* < 10, → laminar flow

For Re* > 2000, → turbulent flow

• Ergun also expressed flow through a packed bed interms of friction factor;

Friction factor,

( )

( )ε

ε

ρ −

∆−=

1*

3

2U

d

H

P f

g

p

(3.11)

Compare this friction factor with Fanning frictionfactor.

• Equation (3.4) then becomes;

75.1Re*

150* += f (3.12)

with Re*

150* = f

for Re* < 10

and 75.1* = f for Re* > 2000


9/31

34

• For non-spherical particles; d p is replaced by d sv, then,

( ) ( ) ( )3

2

32

2 175.11150

ε

ε ρ

ε

ε µ

sv

g

sv d

U

d

U

H

P −+

−=

∆−

(3.13)

• The surface/volume size, d sv is used: if only sievesizes are available, depending on the particle shape,an approximation can be used for non-sphericalparticles;

Recalling, psv d d 87.0≈

where d p is the mean sieve size.

• Note also that: pv d d 13.1=

• And for Carmen – Kozeny equation for laminar flow;

( ) ( )32

21180

ε

ε µ

svd

U

H

P −=

∆− (3.14)


10/31

35

3.6 Minimum Fluidization Velocity, U mf .

• A plot of pressure drop across the bed vs. fluidvelocity as below.

Figure 3.2: Plot of ∆P vs. U o for fluidized bed system

• Line OA → packed bed region→ Solid particles do not move relative to

one another and their separation isconstant.

• ∆P vs. Uo relationships in region OA: use Carmen-Kozeny equation for laminar flow and Ergun equationin general.

ABed pressure

drop, ∆p

Gas velocity, U

B

O

C

Umf

∆p∆p∆p


11/31

36

• Region BC: fluidized bed region. In here, equation3.1, equation 3.2 and also Ergun equation in generalapplies.

• Point A: ∆P higher than predicted value fromequation 3.1 and 3.2.

• This is due to powders, which have been compactedto some extent before the fluidization process takesplace.

• Higher ∆P is associated with the extra force requiredto overcome inter particle attractive forces.

• Minimum fluidization velocity, U mf : superficial fluidvelocity at packed bed becomes a fluidized bed (asmarked on graph above).

• Also known as incipient fluidization velocity.

• U mf increases with particle size and particle densityand affected by fluid properties.

• Recalling Ergun (1952) for any flow condition;

( ) ( ) ( )3

2

32

2

175.11150ε

ε ρ ε

ε µ sv

g

sv d U

d U

H P −+−=∆− (3.15)

and ( ) g H P f p ρ ρ ε −−=∆ 1 (3.2)

substituting (3.15) into (3.2),


12/31

37

( )( ) ( ) ( )

3

2

32

2175.11150

1ε

ε ρ

ε

ε µ ρ ρ ε

sv

mf g

sv

mf

f pd

U

d

U g

−+

−=−− (3.16)

Rearranging,

( )( ) ( )

( )

−+

−=−−

2

222

3

2

3

3

2

3

2

..175.1

..1150

1

µ

ρ

ρ

µ

ε

ε

µ

ρ

ρ

µ

ε

ε ρ ρ ε

f svmf

sv f

f svmf

sv f

f p

d U

d

d U

d g

(3.17)

( )( ) ( )

( ) 2,3

,3

2

2

3

.175.1

.1150

1

mf e

mf e

sv f

f p

R

Rd

g

ε

ε

ε

ε

µ

ρ ρ ρ ε

−+

−=

−−

(3.18)

or

( ) ( ) 2,3,3

2

.175.1

.1150

mf emf e R R Ar ε

ε

ε

ε −+

−= (3.19)

where,

( )2

3

µ

ρ ρ ρ sv f p f gd Ar

−=

- Archimedes no. (3.20)

µ

ρ svmf f d U =Re - Reynolds no. (3.21)


13/31

38

• Wen and Yu (1966) correlation for U mf .

687.1

,, 1591060 mf emf e R R Ar += (3.22)

or

( ) 11059.317.33 5.05, −×+= −

Ar R mf e (3.23)

- for spheres ranging 0.01 < Re,mf < 1000

- used for particles larger than 100 µm- use d v instead of d sv for Wen and Yu

NB: Please check the Wen & Yu correlation in determiningU mf from Data Booklet.

• Baeyens and Geldart

- for particles, d p < 100 µm;

( )066 .0

f

87 .0

f

8.1

p

934.0934.0

f p

mf 1110

d gU

ρ µ

ρ ρ −=

(3.24)

Example

A bed of angular sand of mean sieve size 778 µm isfluidized by air. The particle density is 2540 kg/m3, µ g

(air) = 18.4 × 10-6 kg/ms, ρ g = 1.2 kg/m3 and 24.75 kg of

the sand are charged to the bed 0.216 m in diameter. Thebed height at incipient fluidization is 0.447 m. Find;

a) ε mf b) The pressure drop across the bubbling bed in cm

water gauge.

c) The incipient fluidization velocity, U mf .


14/31

43

Figure 3.3: Particles classification according to Geldart

100

1000

10000

10 100 1000

Particle size, (µµµµm)

ρ ρρ ρ p

- ρ ρρ ρ g

( k g / m 3 )

C

A

B D


15/31

44

3.7 Classification of Powders

• Geldart (1973) (Figure 3.3) classified powders into four

groups according to their fluidization properties at ambientcondition.

• There are 4 stages of particles: Aerated (A), Bubble (B),Cohesive (C) and Dense (D).

3.7.1 Group B

• Bubbling at U mf , thus U mb ≈ U mf

• Bubbles continue to grow, never achieving a maximumsize.

• This makes poor fluidization quality associated to largepressure fluctuation.

• However, lots of bubbles produced results in less ∆P togenerate, thus less entrainment.

• Example: construction sand.

3.7.2 Group D

• Large particles – able to produce deep spout bed.

• Need very large U mf and ∆P to fluidize.

• It is a costly operation since lots of air is needed for

blowing.


16/31

45

• Quite similar to group B particles, i.e. U mb ≈ U mf .

• Fluidization of group D and larger group B particles: jet

circulation/spout bed – technique used to get circulation.

• Example of operation: paddy drying.

• For B and D particles:o No inter particle involve.o Bed collapses instantly when gas supply interrupted.o Short residence time in bed.

• Example: paddy, beans, soy etc.

3.7.3 Group A

• For smaller particles structures where cohesivity becomes

significant.

• Lies between group C and free flowing particles (B).

• Existence of forces that holds particles together – whengas is supplied, bed expands but does not bubble.

• Non-bubbling fluidization at beginning of Umf, followed bybubbling fluidization as Uo increases (a.k.a. aeratablestate).

• Aeratable state = transformation from cohesive to free-flowing particles type.

• The freeboard has to be increased to allow for bedexpansion.


17/31

46

• Danger – if the powder is left in a drum → high voidageand it could cause blow-up.

• U mb > U mf , bubbles are constantly splitting and coalescing,and maximum stale bubble size is achieved.

• Take long time to de-aerate after gas supply is cut-off.

• Inter particle forces?? – yes, but significantly smaller thanhydrodynamic forces.

• Good quality and smooth fluidization.

• Gas bubbles are in limited size, break down at highvelocity and it gives good gas/solid contact

• Example: Fluid bed catalytic cracking (FCC) catalyst.

3.7.4 Group C

• Very cohesive particles and do not fluidized at all.

• Inter particle forces are large compared with the inertialforces on the particles.

• Structures are so strong:o At a given ∆P, not expanding and resist aeration.o Upon fluidization, cracks and rat hole form.o Slugging blows powder out.o Difficult to fluidize: inter particle forces >

hydrodynamic forces exerted on the particles by thefluidizing gas.


18/31

47

• Pressure loss across the bed is always less than apparentweight of the bed cross sectional area due to the particlesnot fully supported by fluidizing gas.

• However, group C fluidization can be improved:o Mechanical help: vibration, mixero Binary mixtures: act as flow conditioner

• Many industrial processes use fine powders, e.g.pharmaceutical, cosmetics, paint industries, foodindustries etc.

• Thus, many researches going on to improve and predictthe behaviour of group C particles.

• Example: the application of vibrations to the fluidized bedcolumn.

• With the aid of vibration, the bed is found to fluidize welland the pressure drop across the bed is close to thetheoretical pressure drop during fluidization.

• Theoretically, when vertical vibration is applied to afluidized bed column, the effect of forces between the bedand the distributor cause the break-up of interparticleforces and this cause the particles to fluidize well.

• According to Janssen et al. (1998), at a specific vibrationfrequency, the ratio between distributor’s plate and thebed displacement increases with an increase in vibrationintensity.

• This phenomenon caused the resultant force becomes

bigger and hence used to break the interparticle forcesbetween the particles.


19/31

48

d b

+

rBubblevolume, V b

Cusp

• Hence, these results in better fluidization quality andsmaller U mf values obtained compared to fluidizationwithout vibration.

• Vibration also is predicted to be able to reduce thedistance between particles and this reduces the voidage inthe bed.

• This is due to small compaction during negativedisplacement or due to the downward movement duringhalf cycle of vibration.

• However, equilibrium created between two mechanisms,i.e. the effect of pressure on the bed during vibration anddownward movement which produced the compaction andhence led to a stable fluidization.

3.8 Bubbles

• The shape of bubble is a hemispherical capped bubble.


20/31

49

• The upper surface of the bubble is approximatelyspherical, and it’s radius of curvature is denoted by r .

• Since r is not readily determinable, it is usually moreconvenient to express the bubble size as its ‘volume-equivalent diameter’, i.e. the diameter of the sphere whosevolume is equal to the bubble.

31

beq

V 6 d

=

π (3.25)

• Bubbling fluidization also known as lean phase.

• Condition at where the powder stops behaving like solidsbut they behave like liquid – two phase system.

• Bubbles are extremely important in supplying circulationas they are major circulating mechanism – hence, lead to

mixing.

• As bubbles rise, it grows and expand

• If the bed is deep enough and diameter of the column issmall,

o Then slugging could occuro This means problem because slugging will push the

powder up and possibly out of the vessel.

• Through bubbles, particles are transported out of the bed.

• Approximately, when U o , superficial gas velocity equals toparticle terminal velocity, V t , then carry over/entrainment

could occur.


21/31

50

• Refer figure 7.3(pg 173), figure 7.5 (pg 175), figure 7.7and 7.9 (pg 177 & 180) and figure 7.8 (pg 178) for

examples of bubbles formed for different groups.

3.9 Bubbling and Non-Bubbling Fluidization

• At U o above the U mf , fluidization may be generally eitherbubbling or non-bubbling.

• Most liquid fluidized bed system, except those involvingvery dense particles, does not bubble.

• Gas fluidized bed system give either only bubblingfluidization or non-bubbling fluidization beginning at U mf ,followed by bubbling fluidization as U o increases.

• Non-bubbling fluidization is also known as particulate or

homogenous fluidization is often referred to asaggregative or heterogeneous fluidization.

3.9 Expansion of non-bubbling bed

• Richardson and Zaki (1954) found the function f( ε ) whichapplied to both hindered settling and to non-bubblingfluidization.

• Thus, in general;

n

T o V U ε = (3.26)

• Khan and Richardson (1989), suggested the correlation inEquation (3.27) which permits the determination of the

exponent n at intermediate values of Re.


22/31

51

−=

−

−27 .0

p57 .0

D

d 4.21 Ar 043.0

4.2n

n8.4 (3.27)

• If the packed bed depth (H 1) and voidage (ε 1) are known,then if the mass remains constant, the depth at anyvoidage can be determined:

( )

( ) 1

2

1

2 H

1

1 H

ε

ε

−

−=

(3.28)

3.10 Entrainment

• Ejection of particles from the surface of bubbling bed.

• Also term as ‘carry over’ and ‘elutriation’.• Amongst the factors influencing rate of entrainment are:o gas velocityo particle densityo particle sizeo fines fractiono vessel diametero Increasing gas temperature

o Increasing gas pressure

Discuss these factors …

• Ejection of particles from fluidized bed depends on thecharacteristics of the bed: i.e. bubble size and velocity atsurface.

• If terminal velocity, Vt > Uo – entrained

Increasing drag


23/31

52

• If Vt < Uo – particle will fall back to the bed.

• Region above the fluidized bed surface:o Freeboardo Splash zoneo Disengagement zoneo Dilute-phase transport zone

(Refer to page 112 – from text book)

• Generally: fine particles – entrainedCoarse particles – stay in the bed.

• Practically: fine particles could stay in the bed and coarseparticles being entrained.

• TDH = Transport Disengagement heighto Height from bed surface to the top of the

disengagement height.o Entrainment flux and concentration of particles are

constant.

• Empirical estimation of entrainment rates from fluidizedbed:

Instantaneous

rate of loss ofsolid of size d pi

∝

Bed

area ∝

Fraction of bed

with size d pi attime, t .

( ) Bi*

ih Bi Bi AxK x M dt

d R =−= (3.29)


24/31

53

where

*

ihK = Elutriation rate constant (kg/m s)

M B = Total mass of solids in the bed (kg) A = Area of bed surface (m ) x Bi = Fraction of the bed mass with size

dpi at time, t.

• *

ihK = the entrainment flux at height, h above the bed

surface for the solid size, d pi when x Bi = 1.

• For continuous operation, x Bi and M B are constant, and so,

Bi

*

ihi AxK R = (3.30)

and total rate of entrainment,

∑ ∑== Bi*ihiT AxK R R (3.31)

• Total solids loading leaving the freeboard,

∑ ∑== AU / R oiiT ρ ρ (3.32)

• The elutriation rate constant,*

ihK : predicted value basedon experiment.

• Correlations are usually in terms of the carry over rate

above TDH,*

∞iK


25/31

54

• Examples of some widely accepted correlations are asbelow:

(i) Geldart et al (1979); for particles > 100 µm and Uo > 1.2m/s.

−=∞

o

ti

og

*

i

U

V 4.5exp7 .23

U

K

ρ

(ii) Zenz and Weil (1958) – for particles < 100 µm and Uo <1.2 m/s.

88.1

2

27

*

1026.1

×=∞

p pi

o

og

i

gd

U

U

K

ρ ρ when4

2

2

103 −×<

p pi

o

gd

U

ρ

and

18.1

2

24

*

1031.4

×=∞

p pi

o

og

i

gd

U

U

K

ρ ρ when4

2

2

103 −

×>

p pi

o

gd

U

ρ


26/31

55

3.10.1 Calculation of carryover rate

For continuous operation

• General case:

• Assumption: R E = R R = 0 and F and Q ≠ 0.

• Mass balance on the size fraction d pi gives:

T PiQiFi R xQ xF x += (3.33)

• Overall mass balance:

F = RT + Q (3.34)

Bi

*

ihT Piih AxK R x A E == (3.35)

• Recalling ∑ ∑== Bi*ihiT AxK R R (3.36)

• In a well mixed bed; xQi = x Bi (3.37)

R T , x Pi

F, x Fi

Q, x Qi

x Bi

R E , x Ei

R C , xR i

R R , x Ri R R , x Ri


27/31

56

• Substituting and rearranging from equation (3.33);

T *ih

Fi Bi

RF AK

F x x

−+=

(3.38)

• This equation cannot be solved directly because fromequation (3.36), R T depends on the value of xBi for eachsize fraction.

• In practice, a converging trial and error loop can be set up,

with R T = 0 for the first trial.

For batch operation

• For batch operation, the rates of entrainment of each sizerange, the total entrainment rate and the particle sizedistribution of bed change with time.

• Thus, the formula,

( ) t AxK M x Bi*

ih B Bi ∆∆ =− (3.39)

where ( ) B Bi M x∆ is the mass of solids in sizerange, i entrained in time increment, ∆t .

• By assuming that the mass of bed, M Bi does not changesignificantly with time, ∆t thus:

−= ∞

B

*

i

Bio Bi M

At K exp x x (3.40)


28/31

57

3.10.2 Total entrainment flux (overall carryover flux), E ih .

• Large, Martini and Bergougnau (1976) picture the total

entrainment flux, E ih, for a given size material, d pi consist oftwo partial fluxes:

o Continuous flux flowing upwards from bed to outlet, E i∞ .

o Flux of agglomerates ejected by bursting bubbles,which decreases exponentially as a function of

freeboard height.

• Expressed algebraically;

ha

ioiihie E E E

−

∞ += (3.41)

where E io is the component ejection flux = E o x Bi and

Bi

*

ii xK E ∞∞ = (3.42)

and

Bi

*

ihih xK E =

(3.43)

• The total solids carryover flux when gas offtake is at anyheight, h above the bed surface:

( )ah E E E oh −+= ∞ exp (3.44)


29/31

58

• Wen and Chen (1982) developed the idea further andproposed:

( ) ( )ahexp E E E E oh −−+= ∞∞ (3.45)

3.10.3 Terminal velocity, V t determination

(i) For spherical particles

• Laminar region (Ret < 0.2)

( ) µ

ρ ρ

18

2

,

vg p

ST t

gd V

−= , dp < 33 µm (3.46)

t

DC Re

24= (3.47)

• Turbulent region, (Ret >1000)

( )

g

vg p

t

gd N V

ρ

ρ ρ

43.0.

3

4,

−= , dp > 1500 µm (3.48)

CD ≈ 0.43

• Transition region, 0.2 < Ret < 1000

( ) 323

4Re vg pgt D gd C ρ ρ ρ −= (3.49)

or


30/31

59

3

t

g

2

g

g p

t

D

V

g

3

4

Re

C µ

ρ

ρ ρ −=

(3.50)

• Generally, 2.

3

4

tg

vgp

DV

gdC

ρ

ρ−ρ=

(3.51)

(ii) Non-spherical particles:

• For laminar region, Ret < 0.2

( ) µ

ρ ρ

18

gd K V

2

vg pST

ST ,t

−= (3.52)

where 065.0log843.0K ST = (3.53)

• Turbulent region, Ret > 1000

g N

vg p

N ,t K

gd .

3

4V

ρ

ρ ρ −=

(3.54)

where 88.431.5K N −= (3.55)


31/31

• Transition region, 0.2 < Ret < 1000

N

t

N ST TR K

43.0

2.01000

Re1000.

K

43.0K K +

−

−

−≈

(3.56)

Vt = KTR.VT(Sphere) (3.57)

che572 chapter 3 fluidization.pdf

Documents