•Applied where intimate contact is required between solid
particles and a gas stream (fluid phase).
•Developed mainly for the chemical and petroleum
industries for processes where high heat transfer
coefficients and degree of temperature uniformity is
required.
•E.g.:
–Drying of finely divided solids is carried out in a fluidized
system.
–Applied also in pyrolysis and gasification processes
–Removal of suspended dusts and mists from gases
Applications of Fluidized Bed Techniques
•Applicationinmineralprocessingandmetallurgical
engineeringintherecoveryofmetalsfromores.
–Agoodexampleisthegasphaseextractionofmetals,giving
highyieldsofrecovery.Avolatileorganicreagent(eg.
acetylacetate)passesthroughthefeedmaterialandreacts
selectivelywithmaterialtobeextracted.Theproductisavolatile
metalcomplexwhichisremovedfromtheresiduegasbycarrier
gas.
•Oneofthemostimportantpropertiesofafluidizedbedis
itsgoodheattransfercharacteristics.
Applications of Fluidized Bed Techniques
Applications of Fluidized Bed Techniques
Type Example Reasonfor using
fluidized bed
Homogenous catalytic
gas-phase reactions
Ethylene hydrogenationRapid heating of
entering gas,Uniform
controllable temperature.
Heterogeneous non-
catalytic reactions
Sulphideore roasting,
combustion
Ease of solids handling,
Temperature uniformity,
Good heat transfer.
Heterogeneous catalytic
reactions
Hydrocarboncracking,
Phthalic Anhydride,
Acrylonitrile
Ease of solids handling,
Temperature uniformity,
Good heat transfer.
The Froude numberdistinguishes the two types of fluidization.
Froude number =
Where;
–minimum fluidization velocity
d –diameter of particles
g –acceleration due to gravity
Froude Number<1 –Particulate fluidization
Froude Number >1 –Aggregate Fluidization
Types of Fluidization gd
u
mf
2 2
mf
u
Rice et al. (1958) and Romeo et al (1962) proposed dimensionless value Nconsisting
of four dimensionless values to characterize thetypeoffluidization occurring.
??????=??????�??????�
��
′
??????
�−??????
??????
�
??????
Where;
Fr–Froude number
Re
mf–Reynold’s number
ρ
sand ρ –Density of the particle and the fluid
l–Bed length
D –Diameter of the column
N < 100 –Particulate fluidization
N > 100 –Aggregate Fluidization
Froude Number
Effect of Fluidized Bed on Pressure Gradient
•As the superficial velocity
approaches u
mf the bed starts to
expand, at higher velocities it
becomes fluidized.
•At higher velocities, voidage is
higher, pressure gradient decreases
because weight of particles per unit
bed is smaller
•With even higher velocities transport
of the solid particles occur and
pressure gradient increases.
If the pressure gradient (−∆P/l) is plotted against the superficial velocity (u
c) using
logarithmic co-ordinates, a straight line of unit slope is obtained.
•Alinearrelationshipoccursupto
pointAwhereexpansionofthebed
starts.
•Thepressuredropreachesa
maximumatpointB.
•ItthendecreasestopointC.
•BeyondpointC,thepressuredropis
independentofthesuperficial
velocity
•Ifthefluidvelocityisreduced
again,thebedcontractsuntilit
reachestheconditionwherethe
particlesarejustrestingonone
another,pointE.
•Thepressuredrop(EF)acrossthis
reformedfixedbedatanyfluid
velocityisthenlessthanthatbefore
fluidisation.
Effect of Fluidized Bed on Pressure Drop
When pressure drop across the whole bed is plotted against velocity using
logarithmic coordinates.
Pressure Drop
•Whenthegasvelocityishighenoughthatthefrictionaldragforce
ontheparticlesequalstheweightoftheparticles(m×g)thebed
becomesfluidized.
•-∆Pisequalthebedweightperunitcross-sectionalarea.
•Thus,inabedofunitcross-sectionalarea,depthl,andporositye,
thepressuredropacrossthebedcausedbythelayoutweightof
particlesisgivenby:
Where
l–is length of bed
e –porosity of the bed
-density of solid
-density of fluid
g –acceleration due to gravityp
−∆??????=1−�??????
??????−?????????????????? (1)
Pressure Drop
Increasing superficial velocity will lead to the point of
incipient fluidization, where particles are just supported in
the fluid, eqn. (4) applies with voidage at being
??????
????????????=0.0055
�
????????????
3
1−�
????????????
??????
??????−????????????�
2
??????
(4)
The value of e
mf depends on shape, size distribution and
surface properties of particles.
A typical value is
Then eqn. 4 becomes mf
u mf
e 4.0
mf
e ( )
()5
00059.0
2
)4.0(
gd
u
s
e
mf
mf
−
=
=
Minimum Fluidizing Velocity
•Equations (4) and (5) apply only at streamline flow and
they are therefore restricted to fine particles only.
•For large particles, where streamline flow cannot be
achieved at incipient fluidization, the Ergun equation is
applied
Substituting for e=e
mfat incipient fluidization and -∆P from
equation (1)
Minimum Fluidizing Velocity( )
()7
1
75.1
)1(
150))(1(
2
323
2
d
u
e
e
d
u
e
e
ge
mf
mf
mfmf
mf
mf
smf
−
+
−
=−−
Multiplying eqn. (7) by yields )1(
2
3
mf
e
d
−
()8
75.1)1(
150
)(
2
33
2
2
3
+
−
=
−
du
e
du
e
egd
mf
mf
mf
mf
mfp
Minimum Fluidizing Velocity()8
75.1)1(
150
)(
2
33
2
2
3
+
−
=
−
du
e
du
e
egd
mf
mf
mf
mf
mfp Ga
gd
p
=
−
2
3
)(
Ga is the Galileo number mf
mf
du
Re=
Re
mf is the Reynolds number at the
minimum fluidizing velocity
Minimum Fluidizing VelocityGa
gd
p
=
−
2
3
)(
The Galileo number, Ga, is used in viscous flow and thermal expansion calculations—for
example, to describe fluid film flow over walls. These flows apply to condensers or
chemical columns.
Where
d – diameter of particles (m)
ρ
s – density of particles (kg/m³)
ρ – density of fluid (kg/m³)
g – acceleration due to gravity (m/s²)
The Reynolds number, Re, is a dimensionless quantity used in fluid mechanics to predict
flow patterns in different fluid flow situations.
Where
d – diameter of particles (m)
u – velocity of the fluid relative (m/s)
ρ – density of the fluid (kg/m³)
μ – dynamic viscosity of the fluid (Ns/m
2
)
????????????�
??????
=??????�
Equation (8) can be written as
Minimum Fluidizing Velocity ()9Re
75.1
Re
)1(150
2
33
mf
mf
mf
mf
mf
ee
e
Ga +
−
=
Equation 9 is a quadratic equation!!!
For a typical value of e
mf = 0.4, eqn. (9) becomes)11(]1)1053.51([7.25Re
5
−+=
−
Ga
mf
Solving Re
mf()10Re3.27Re1406
2
mf
mf
Ga +=
Similarly, for e
mf= 0.45
Minimum Fluidizing Velocity )12(]1)1039.91([6.23Re
5
−+=
−
Ga
mf
Therefore by definition, if , then the minimum
fluidization velocity is obtained from:)13(Re
mfmf
d
u
= mf
mf
du
Re=
Abedconsistsofuniformsphericalparticlesofdiameter
3mmanddensity4200kg/m
3
,whatwillbetheminimum
fluidizingvelocityinaliquidofviscosity3x10
-3
Pasm
-2
anddensityof1100kg/m
3
,Assumee
mf=0.4
= 1.0035 x 10
5
Example 1 2
3
)(
gd
Ga
p
−
= 23
33
)103(
81.9)11004200(1100)103(
−
−
−
=
x
x
Ga
Assuminge
mf= 0.4 from equation (11)
Example 1 78.8115]1)100035.1)(1053.51([7.25Re
55
=−+=
−
mf sm
x
x
u
d
mfmf
/377.7
1100103
)103(78.8115
Re
3
3
=
==
−
−
Inthecaseofnon-sphericalparticles,tousetheaboveequations,weaccountfornon
sphericityasfollows,fromeqn.(8)
Where
-mean linear dimension of the particles
-particle shape factor
When either and / or are unknown, the following
approximations by Wen and Yu (1966) can be used
Minimum Fluidizing Velocity ()14
75.111
150
)(
2
222
3232
3
+
−
=
−
pmf
mf
mfp
mf
mfp
du
e
ud
e
egd pd mfe 14
1
3
mf
e )15(11
1
32
−
mf
mf
e
e
and
Thevoidfractioneisdefinedas
The specific area of the particle is given by
Shape Factors v
a )17(
p
p
v
V
S
a=
Where
surface area of the particle
volume of the particlepS pV ( )
)16(
solidsvoidsbedofvolumetotal
bedinvoidsofvolume
e
+
=
Shape Factors
For non spherical particles, the effective diameter D
p is
defined as
Since (1- e) is the volume fraction of particles in the bed,
Where a, is the ratio of the total surface area in the bed to
the total volume of the bed (void volume + particle
volume)() () )19(1
6
1 e
D
eaa
p
v
−
=−= )18(
6
v
p
a
D=
Shape Factors
The sphericity shape factor ϕ of a particle is the ratio of the
surface area of the sphere having the same volume as
the particle to the actual surface area of the particle.
The surface area of a sphere,
And the volume;
2
pp
DS= 6
3
p
p
D
V
=
Shape Factors
For any particle,the shape factor is given by
Where
actual surface of the particle
is the equivalent diameter of the sphere having the same
volume as the particle.
Thereforethe specific area a
vof an irregular particle is: p
p
S
D
2
= pS pD )20(
6
6
3
2
pp
p
p
p
v
DD
D
V
S
a
=
==
Shape Factors
Therefore, for irregular shapes
(20)
For a sphere ϕ = 1
For a Cylinder where diameter = length, ϕ= 0.874
For a cube ϕ = 0.806
pp
p
v
DV
S
a
6
==
Bed Height & Porosity Relationship
The porosity of a packed bed is given by
Where
-density of the solid particle
-bulk density of the bed
The relationship between the height of packed bed and fluidizedbed
is then
Height of bed at minimum fluidizingconditionsp
bp
e
−
= b
p p
f
f
p
e
e
L
L
−
−
=
1
1 p
mf
mf
p
e
e
L
L
−
−
=
1
1
Heat Transfer
Fluidized beds have extremely good heat transfer properties
Nu is a dimensionless parameter used in calculations of heat transfer
between a moving fluid and a solid body.
Where for the particle =
for the particle =
- the superficial velocityK
hd
pc
du c
u '
Re
c '
uN
Pr for the particle =
??????????????????
??????
Heat Transfer
Pr =
??????????????????
??????
The Nusselt number (Nu) is a dimensionless parameter used in fluid mechanics and
heat transfer to characterize the convective heat transfer coefficient.
Where:
Nu – Nusselt number,
h – Convective heat transfer coefficient (W/(m²·K))
L – Characteristic length scale (such as the length of a surface
or the diameter of a pipe) (m)
K – Thermal conductivity of the fluid (W/(m·K))
Where:
Pr – Prandtl number,
μ – dynamic viscosity of the fluid (Pa·s or N·s/m²)
Cp – Specific heat capacity of the fluid at constant pressure (J/(kg·K))
K – thermal conductivity of the fluid (W/(m·K))
The Prandtl number (Pr) is a dimensionless parameter used in fluid mechanics and
heat transfer to characterize the relative importance of momentum diffusivity
(viscosity) to thermal diffusivity in a fluid.
Heat Transfer
Equation (29) is valued for
(Reynolds Number)
(Prantl Number)
(porosity Number)
For gas – solid systems 3'1
10Re10
−
c 14000Pr22 9.04.0 e ()
)30(
1
55.0
80.0
25.0
17.065.0
−
=
tc
p
ttt
du
epc
e
d
d
l
d
k
hd
Example 2
Spherical catalyst pellets 3 mm in diameter are to be fluidized with
nitrogen at 101.3 kPa at 60
0
C. The density of the catalyst particles are
980 kg/m
3
. The molecular weight of nitrogen is 29 kg/kmol. If it is
assumed that the point of incipient fluidization is reached
at e
mf = 0.43, Calculate the minimum fluidization velocity in the
vessel. Nitrogen is considered here to be an ideal gas with µ=
0.0000207 Ns/m
2
.
Ergun Equation
R (gas constant) = 8.314 m
3
∙Pa/(mol.K), g =9.81m/s
2
Example 2 Solution
The pressure drop across the bed caused by the layout weight of
particles is given by
Combining equation (1) and the Ergun equation
This equation can be simplified to
−∆??????=1−�??????
??????−?????????????????? (1)
1.75�????????????
????????????
2
+1501−�
????????????????????????
????????????−??????
??????−????????????�
2
�
????????????
3
=0( )
()7
1
75.1
)1(
150))(1(
2
323
2
d
u
e
e
d
u
e
e
ge
mf
mf
mfmf
mf
mf
smf
−
+
−
=−−
Example 2 Solution ( )
3
/061092365.1
/100060273314.8
29101300
mkg
kmolmolx
x
RT
PM
r
a
=
+
== ()( )( ) ( )( )
( )()()( )0003.043.081.9061092365.1980
0000207.043.01150061092365.1003.075.1
23
2
=−−
−+
mfmf
uu 01087183105.61076985.120055707349.0
332
=−+
−−
xuxu
mfmf 0
mfu
Ignore negative root
ρ
nitrogen is unknown
Example 3
Non spherical catalyst pellets, 4mm in diameter, shape factor of
1.1, are to be fluidized with air at 101.3 kPa at 70
0
C. The density
of the catalyst particles are 1100 kg/m
3
. Take the molecular weight
of air as 26.9 kg/kmol. If it is assumed that the point of incipient
fluidization is reached at , calculate the minimum
fluidizing velocity in the vessel. Air is considered here to be an
ideal gas with Ns/m
2
.43.0=
mf
0000207.0=
air