MOTION OF PARTICLES
TROUGH FLUIDS
PART 2
By: Dr Akmal Hadi Bin Ma’ Radzi
Particle settling
Free settling


When a particle is at sufficient distance from the wall of the container and
from other particle, so that its fall is not affected by them, the process is
called free settling.
Terminal velocity is also known as free settling velocity.
Hindered settling



When the particles are crowded, they settle at a lower rate and the process
is called hindered settling.
The particles will interfere with the motion of individual particles
The velocity gradient of each particle are affected by the close presence of
other particles.
Hindered settling

The velocity for hindered setting can be computed by this
equation:
Stokes Law

Correction
factor
Where, ε is volume fraction of the slurry mixture and Ψp is
empirical correction factor.

Bulk density of mixture –

Empirical correction factor -
Hindered settling
The Reynolds number is then based on the velocity relative to
the fluid is
Where the viscosity of the mixture µm is given by;
Exercise

Calculate the settling velocity of glass spheres having a
diameter of 1.554 x10-4 m in water at 20ºC. The slurry
contains 60 wt % solids. The density of the glass sphere and
water is 2467kg/m3 and 998 kg/m3 respectively.

Calculate the Reynold number for this settling.
Solution
Density of the water  = 998 kg/m3, and viscosity of
water μ = 1.005 x 10-3 Pa.s. To calculate the volume
fraction ɛ of the liquid,
The bulk density of the slurry m is
Then,
The terminal velocity is:
The Reynolds number can be calculated as follows:
NRe
Fluidization



Fluidization is a process whereby a granular
material is converted from a static solid-like state to a
dynamic fluid-like state.
This process occurs when a fluid (liquid or gas) is passed
up through the granular material.
The most common reason for fluidizing a bed is to
obtain vigorous agitation of the solids in contact with
the fluid, leading to an enhanced transport mechanism
(diffusion, convection, and mass/energy transfer).
Fluidized Bed Reactor

When a gas flow is introduced through the bottom of a bed
of solid particles, it will move upwards through the bed via
the empty spaces between the particles.

At low gas velocities, aerodynamic drag on each particle is
also low, and thus the bed remains in a fixed state.

Increasing the velocity, the aerodynamic drag forces will
begin to counteract the gravitational forces, causing the bed
to expand in volume as the particles move away from each
other

Further increasing the velocity, it will reach a critical value at
which the upward drag forces will exactly equal the
downward gravitational forces, causing the particles to
become suspended within the fluid. At this critical value, the
bed is said to be fluidized and will exhibit fluidic behavior.

By further increasing gas velocity, the bulk density of the bed
will continue to decrease, and its fluidization becomes more
violent, until the particles no longer form a bed and are
“conveyed” upwards by the gas flow.

When fluidized, a bed of solid particles will behave as a fluid,
like a liquid or gas.

Objects with a lower density than the bed density will float
on its surface, bobbing up and down if pushed downwards,
while objects with a higher density sink to the bottom of the
bed

The fluidic behavior allows the particles to be transported
like a fluid, channeled through pipes, not requiring
mechanical transport
Applications of Fluidization
Conditions for Fluidization
Based on the Figure:




If the superficial velocity, VO is gradually increased, the
pressure drop will increases, but the particles do not move and
the height (L) remains the same.
At a certain velocity, the pressure drop across the bed
counterbalances the forces of gravity on the particles or the
weight of the bed
At point A = Any further increase in velocity, causes the
particles to move
At point B = Further increase in velocity, the particles become
separate enough to move about in the bed and true
fluidization begins.


From point B to point C = Once bed is fluidized, the
pressure drop across the bed stays constant , but the
bed heights continues to increase with increasing
velocity.
From point C to B = If the velocity is gradually
reduced, the pressure drop remains constant and the
bed height decreases.
*The pressure drop required for the liquid or the gas to
flow through the column at a specific flow rate
Minimum Fluidization Velocity



Minimum velocity of fluidization took place at incipient
(beginning) fluidization.
During this stage, the ratio of pressure drop to the
vessel height (L) is given by;
Where
is the minimum porosity


The minimum fluidization velocity
obtained by this equation;
can be
For roughly spherical particles,
is generally
between 0.4 and 0.45 (commonly taken as 0.45)
which increasing slightly with decreasing particle
diameter.

If the Reynolds number is used, then,
The equation becomes:
Exercise 2

A bed of ion exchange beads 8 ft deep is to be
backwashed with water to remove dirt. The
particles have a density of 1.24 g/cm3 and an
average size of 1.1 mm. The beads are assumed to
be spherical (Φ =1) and
is taken as 0.4. What
is the minimum fluidization velocity using water at
20ºC with density of 980 kg/m3?
Solution 2
)
𝟏𝟓𝟎(𝟏. 𝟎𝟎𝟓 𝐱 𝟏𝟎−𝟑 𝑽𝑶𝑴 𝟏 − 𝟎. 𝟒
𝟏𝟐 𝟏. 𝟏𝐱 𝟏𝟎−𝟑 𝟐
𝟎. 𝟒𝟑
𝟏. 𝟕𝟓(𝟗𝟖𝟎)𝑽𝟐 𝑶𝑴 𝟏
+
𝟏 𝟏. 𝟏 𝐱 𝟏𝟎−𝟑
𝟎. 𝟒𝟑
= 𝟗. 𝟖𝟏 𝟏𝟐𝟒𝟎 − 𝟗𝟖𝟎
𝑽𝑶𝑴 = 𝟎. 𝟎𝟎𝟐𝟏 𝒎/𝒔
Exercise 3
Solid particles having a size of 0.12 mm, a shape
factor Фs of 0.88, and a density of 1000 kg/m3 are to
be fluidized using air at 2.0 atm abs and 25 oC. The
voidage at minimum fluidizing condition is 0.42.
(a) If the cross section of the empty bed is 0.30 m2
and the bed contains 300 kg of solid, calculate the
minimum height of the fluidized bed.
(b) Calculate the pressure drop at minimum fluidizing
conditions
(c) Calculate the minimum velocity of fluidization
For part (a), the volume of solids = 300 kg/(1000
kg/m3) = 0.300 m3. The height of the solids would
occupy in the bed if ɛ1 = 0 is L1 = 0.300m3/(0.30m2
cross section) = 1.00 m.
Solving Lmf = 1.724 m
The physical properties of air at 2.0 atm and 25 oC
are μ = 1.845 x 10-5 Pa.s,  = 1.187 x 2 = 2.374 kg/m3,
and p = 2.0265 x 105 Pa. For the particle, Dp =
0.00012 m, p= 1000 kg/m3, s = 0.88 and ɛmf = 0.42.
For part (b),
To calculate Vmf ’ for part (c),
Solving,