FLOW IN FLUIDIZED BEDS
• Fluidization refers to those gas-solids and liquidsolids system in which the solid phase is
subjected to behave more or less like a fluid by
the upwelling current of gas or liquid stream
moving through the bed of solid particles.
• When a fluid flows upward thro a packed bed of
particles at low velocities, the particles remain
stationary.
• As the fluid velocity is increased, ∆P increases
according to Ergun equation.ppt
• Fluidization starts at a point when the bed pressure drop
exactly balances the net downward forces (gravity minus
buoyancy forces) on the bed packing.
• Then the particles just begin to move, and this is the
onset of fluidization or minimum fluidization.
• The fluid velocity at which fluidization begins is the min.
fluidization velocity (v’mf) based on empty cross section
of the tower.
•
The porosity of the bed when true fludization occurs is
the min. porosity for fluidiztoin and is εmf
• The bed expands to this voidage or porosity before
particle motion appears.
• As stated earlier, the ∆P increases as the
gas velocity is increased until the onset of
min. fluidization.
• Then as the velocity is further increased,
∆P decreases very slightly and then
remains practically unchanged as the bed
continues to expand or increase in
porosity with increase in velocity
Relationship bet. Bed height &
porosity
• Since the volume LA(1- ε) is equal to the
total volume of solids……..
L1 A(1   1 )  L2 A(1   2 )
L1 1   2

L2 1   1
Min. fluidization velocity
• Fluidization starts at a point when the force
obtained from the pressure drop times the cross
sectional area must equal the gravitational force
exerted by the mass of the particles minus the
buoyant force of the displaced fluid.
PA  Lmf A(1   mf )( p   ) g
P

 (1   mf )( p   ) g
Lmf
Fg  m g  vg
 Fg  LA p (1   ) g
Fb  LA (1   ) g
1
• From Ergun eqn…..and at min. fluidization…
P

Lmf
150v mf'
(1   mf ) 2
 s2 D p2
 mf
3

1.75 v mf2 (1   mf )
s D p
 mf
2
3
• By comparing (1) & (2)
150v mf'
 s2 D p2
(1   mf )
 mf 3

1.75  v mf2
 s D p  mf 3
 g ( p   )
3
N Re,mf 
D p vmf' 

• For very small particles, only the laminar-flow
(NRe,mf < 20) term of the Ergun eqn is significant
v mf' 
g (  p   )  mf3
150 
1   mf
 s2 D p2
• For large particles, only the turbulent-flow
(NRe,mf >1000) term of the Ergun eqn is significant
v
'
mf
 s DP g (  p   )

1.75

3
mf



1/ 2
we know
150v mf' (1   mf )
 D
2
s
 mf
2
p
3

1.75  v mf2
 s D p  mf
3
 g ( p   )
3
 DP3

m ultiply both sides by 
.......
2  and rearrange



'
2
 DP3
150v mf (1   mf )
 DP3
1.75  v mf
 DP3






2
2
2  g ( p   )
3
3
2
2
  s D p
   s D p  mf
 
 mf



150 (1   mf )
  mf
2
s
3
N Re,mf 
1.75
 s  mf
3
N

2
Re,mf
 DP3


2  g ( p   )
 

• If the terms v’mf and/or Φs are not known, Wen and Yu found
for a variety of systems that
s
N Re,mf
3
mf
1   mf
1

and 2 3  11
14
 s  mf


 D

2
 33.7  0.0408 
2  g (  p   )
 



3
P
1/ 2
 33.7
• This eqn. holds good for a Reynolds number range of 0.001
to 4000
Prob. Minimum velocity for
fluidization
• Solid particles having a size of 0.12mm, a shape factor
of 0.88 and a density of 1000kg/m3 are to be fluidized
using air at 2 atm abs and 25ºC. The voidage at min.
fluidizing is 0.42
• a). If the cross section of the empty bed is 0.3m2 and the
bed contains 300kg of solid, calculate the min.height of
the fluidized bed.
• b). Calculate the pressure drop at min. fluidizing condns.
• c). Calculate the min. velocity for fluidization
Air properties at 2 atm abs and 25ºC are
Viscosity=1.845x10-5 Pa.s ; density = 2.374 kg/m3
• a). vol. of solids (300kg / 1000 kg/m3) = 0.3 m3
• Initial bed ht…(ht of solids if ε1 = 0), L1=0.3m3/0.3m2
• We know
L1 1   2

L2 1   1
• By sub. values…Lmf=1.724m
• b).we know
P

 (1   mf )( p   ) g
Lmf
== ∆P=9786 N/m2
• c). sub. the values in the eqn…….
150 (1   mf )
  mf
2
s
3
N Re,mf 
1.75
 s  mf
3
N
• NRe,mf = 0.07764
=== v’mf =0.004618 m/s

2
Re,mf
 DP3


2  g ( p   )
 

Prob……..
• A tower having a diameter of 0.1524m is
being fluidized with water at 20.2ºC. The
uniform spherical beads in the tower bed
have a dia of 4.42mm and a density of
1603 kg/m3. Estimate the min. fluidization
velocity and compare with the
experimental value of 0.02307 m/s of
Wilhelm and Kwauk
• sub. the values in the eqn…
N Re,mf

 33.7 2  0.0408


 D


2  g (  p   )
 


3
P
1/ 2
 33.7
==NRe,mf = 114.5445
v’mf = 0.02592 m/s.
which is app. equal to the
expt. value 0.02307 m/s
Motion of particle through fluid
• Whenever a particle is moving thro a fluid, a
number of forces will be acting on the particle.
• For a rigid particle moving in a fluid, there are 3
forces acting on the body.
»Gravity force acting downward (Fg)
»Buoyant force acting upward (Fb)
»Resistance or Drag force acting in
opposite direction to the particle
motion (FD)
• The gravity force , Fg = mg
• Buoyant force, Fb = Vpρg = mρg / ρp
• The drag force FD = CD (v2/2) ρ Ap
• where CD is the drag coefficient which is similar
to friction factor in flow of fluids in pipes
FD
AP
CD 
 v2
2
• The resultant force on the body is then Fg-Fb-FD.
• This resultant force must equal the force due to acceleration.
dv
m
 Fg  Fb  FD
dt
dv
mg C D v 2 A
 m  mg 

 (1)
dt
p
2
• If we start from the moment the body is released from its
position of rest, the falling of the body consists of two periods:
– The period of accelerated fall (very short)
– The period of constant velocity fall which is MAX
(which is called as free settling velocity or terminal
settling velocity ‘vt’ )
• To solve for terminal velocity by sub. dv/dt = 0 in
the eqn (1) and re-arranging
vt 
2 g (  p   )m
Ap  p C D 
for a spherical particle
m

6
Ap 
D 3p  p

4
D p2
 vt 
4g ( p   )D p
3C D 
for N Re, p  1.0
24
CD 
N Re, p
 vt 
gD (  p   )
2
p
18
for 1000 N Re, p  200,000
C D  0.44
 vt  1.75
gDp (  p   )

Types of settling
• When the particle is at sufficient distance
from the boundaries of the container and
from other particles, so that its fall is not
affected by them, the process is called
free settling.
• If the motion of the particle is impeded by
other particles, which will happen when
the particles are near each other even
though they may not actually be colliding,
the process is called hindered settling.
Hindered settling
• In hindered settling, for a uniform suspension,
the settling velocity ‘us’ can be estimated from
the terminal velocity for an isolated particle
using the empirical equation of Maude and
Whitmore:
us  ut ( )
n
‘n’ is known as Richardson zaki index which
varies from 4.6 in the Stokes’ law to
2.5 in the Newton’s law region
Terminal settling velocity of dust
particles
• Calculate the terminal settling velocity of
dust particles having a dia of 60μm in air
at 294K and 101.32kPa. The dust particles
can be considered spherical with a density
of 1280 kg/m3.
Air properties: density = 1.137 kg/m3
viscosity = 1.90x10-5 Pa-s
• Ans: 0.1372m/s
Hindered settling
• Particles of sphalerite (S.G 4) are settling under
the force of gravity in CCl4 (S.G 1.594) @ 20ºC.
The dia of sphalerite particles is 0.004”. The
volume fraction of sphalerite in CCl4 is 0.2. What
is the settling velocity of sphalerite? Assume
Stoke’s law is valid. Richardson zaki index is =
4.1. Take viscosity is 1.03 cP
• ut=0.0131m/s
• us=5.2mm/s