MECHANICAL SEPARATION METHODS
Mechanical

physical separation processes employed in food industries include; filtration, settling
and sedimentation, centrifugation, expression and size reduction.
CENTRIFUGATION
: Centrifugation
is defined as a unit operation that involves the
application of a centrifugal force for separating distinct phases

solids and liquids

in suspension
or solution. Commercial centrifuges are either of
two types: sedimenting or filtering, each of
which can
be classified as batch or continuous feed. Further sub

division in each category is
based on the mechanism of separating, advancing and discharging separated solids or liquid
phases.
In centrifugal settling or sedimentation, the centrifugal force is so lar
ge that the force of gravity
may be neglected. The liquid layer then assumes the equilibrium position with the surface almost
vertical.
The particles settle horizontally outward and press against the vertical bowl wall.
Centrifugal settling is employed in
many food industries such as breweries, vegetable

oil
processing, fish

protein

concentrate processing, fruit juice processing, dairy industry etc.
Centrifugal filtration is used in place of pressure filtration where a centrifugal force causes the
flow of
slurry in a filter where a cake of solids builds up on a screen. The cake of granular solids
from the slurry is deposited on a filter medium held in a rotating basket, washed and then spun
dry. Centrifugal filters are used for separating
insoluble solids f
rom liquids.
Centrifugation like filtration or expression changes the nutritional and sensory qualities due to
the separation or concentration of food components but the processing conditions do not involve
heat and cause little damage to food.
slurry
feed
slurry feed
liquid

liquid feed
ooooo
oo
ooooo
oo
oooooo
oo
oo
oo
oo
oo
oo
oo
liquid solid
heavy liquid light
liquid
(a)
(b)
(c)
o

0

o

o


o

o

o

o
/
\
/
\
/
\
/
\
Sketch of centrif
ugal separation
:
(a) initial slurry feed entering (b) settling of solids from
a liquid (c) separation of two liquid fractions.
Principles of Centrifugation
Centrifugal
force is generated when materials are rotated; the size of the force depends on
the radius, speed of ro
tation and the mass (density for a unit volume)
of the centrifugal
material.
I.e. centrifugal force, F = ma = mω²r
ω =
, N
= number
of revs per
minute,
a = rω²
The above equation can also be expressed as;
F = mr (
)² = 0.01097
mrN²
Separation of Immiscible Liquids
In separation of immiscible liquids, the denser liquids (A) moves to the wall of the bowl
and the lighter liquid (B) is displaced to an inner annulus. Provision can be made for a
separation of the two layers.
Considering the annular ring of the liquid of the thickness r
2

r
1
, the centrifugal force
acting on the volume element of thickness
dr
and of mass
dm
at radium
r
is
dF= w
2
r
dm

(i)
mass of element, dm=
dv
elemental
volume, dv= 2
brdr,…b= height or length of cylinder
dm=
2
b
r
d
r

(ii)
dF = 2
w
2
π
b
r²dr………(iii)
Thus the pressure drop over the element,
dP=
= ω²
rdr……………..(
iv)
the pressure drop over a ring of liquid (
) thick
is
(
)
= ω²
∫
=
(
)
(
)
=
(
)
………………(v)
Hence the pressure drop over the two liquid layers are;
(
)
=
(
)
………………(vi)
(
)
=
(
)
……………..(vii)
where r
n
is the boundary radius between the two layers.
The thickness of the layers (annular rings) depends on the density of the liquids, the
pressure difference across the layers and the speed of rotation. The boundary region
between the liquids at a giv
en centrifuge speed forms at a radius r
n
where the hydrostatic
pressure of the two layers is equal. This is termed the neutral zone and is important in
equipment design to determine the position of feed and discharge pipe.
If the neutral zone is to remain
stable then,
(
)
=
(
)
Solving for
gives
=
=
[
(
)
(
)
]
If the difference in density is very small the neutral zone becomes unstable. Hence, the
difference between
A
and
B
should not be less than 3%. If the duty is to remove light
liquids from a mass of heavier liquid (for example in cream separation from milk), the
residence time in the outer layer should exceed that in the inner layer, by reducing the
radius of the outer
layer and hence the radius of the neutral zone.
Conversely, if a dense liquid is to be separated from a mass of lighter liquids (e.g water
from oil), the radius of the outer layers (and the neutral zone) is increased.
Question 1.
A centrifuge having a radi
us of the bowl of 0.1016m is rotating at 1000rpm. Calculate the
centrifugal force developed in terms of gravity forces.
Solution
Centrifugal force (Fc) in terms of gravity forces (Fg) is expressed as
But Fc=mw
2
r, Fg=mg
=
=
But w
=
, N = 1000rpm , r = 0.1016m , g = 9.807m/s
²
=
=
(
)
²
=
(
)
²
(
)
=
Question 2
A bowl centrifuge is used to break an oil

in

water emulsion. Determine the radius of the neutral
zone in order to position the feed pipe correctly. Assume that the density of the continuous phase
is 1000kg/m
3
and the density of the oil is 870kgm

3
. The inlet and outlet radii from the centrifuge
are 3cm and 45.5cm.
Solution
=
(
)
m
Substituting into the equation
=
(
(
)
(
)
)
=
√
(
)
= 0.097m
Question 3
In a vegetable

oil

refining process,
an aqueous phase is being separated from the oil phase in a
centrifuge. The density of the oil is 919.5kg/m
3
and that of the aqueous phase is 980.3kg/m
3
. The
radius r
1
for overflow of the light liquid has been set at 10.160mm and the outlet for the heavy
l
iquids at 10.414mm. Calculate the location of the interface in the centrifuge.
[
L
= 919.5kgm

3
and
H
= 980.3kgm

3
]
Separ
ation
of
Insoluble Solids
from
L
iquids
If a centrifuge is used for sedimentation, a particle of a given size can be removed from the
liquid in the bowl if there is sufficient residence time of the particle in the bowl for the particle to
reach the wall. For a particle moving radially at its set
tling velocity, the diameter of the smallest
particle which can be removed can be calculated.
As liquids containing insoluble solids with greater density than the liquid is fed to a rotating
cylindrical bowl, the solids
will move towards the
bowl wall. if one outlet is provided for the
liquid near the center of rotation, then those particles of solids which reach the bowl wall will
remain in the bowl. Those particles which do not reach the bowl wall will be carried out in the
liquid. The fract
ion remaining in the bowl and the fraction passing out in the liquid will be
controlled by the rate of feed.
b
r
1
r
b
r
2
NB: at the end of the residence time of the particle in the fluid, the particle is at a distance rb
from the axis of rotation. If r
b
<
r
2
, then the particle leaves the bowl with the fluid. If r
b
= r
2
, it is
deposited on the wall of the bowl and effectively
removed from the liquid.
The force balance on a particle in the centrifuge
Fnet = Fcentrifuge
–
Fdrag
–
F buoyant……………………………………..
(i
)
F net =
dt
mdv
, F centrifugal = mr
w
2
=
2
6
3
rw
d
p
p
F drag = 3π
d
p
V
r
, F buoyant =
2
6
rw
d
p
Substituting the terms into (i) gives
2
2
3
3
6
3
6
6
rw
d
V
d
rw
d
dt
dv
d
p
r
p
p
p
p
Since the acceleration phase for the moving particles is fairly,
the velocity can be treated as
constant
.
0 =
r
p
p
p
V
d
rw
rw
d
3
6
2
2
3
Making the radial
velocity Vr the subject
Vr =
p
p
p
d
rw
d
3
6
2
3
Vr =
18
2
2
p
p
rw
d
d
p
= diameter of particle
r = radius of centrifuge
p
= density of particle
= density of liquid
= viscosity of liquid
Expressing the above equation in terms of residence time
and volumetric flowrate
But Vr =
dt
dr
18
2
2
p
p
d
rw
dt
dr
dt
d
w
r
dr
p
p
2
2
tres
p
p
rb
ra
dt
d
w
r
dr
0
2
2
18
18
ln
2
2
tres
d
w
r
p
p
rb
ra
ln
ra
r
b
18
2
2
tres
d
w
p
p
r
a
= point from the axis where a particle begins to settle
r
b
= point it has reached after a time interval t res
t
res
= residence time = average time an element of fluid remains in the centrifuge.
Making t
res
the subject,
t
res
=
(
)
(
)
Solving for volumetric flow rate
t
res
=
=
NB :V = volume of liquid in the centrifuge.
=
(
)
(
)
Q =
(
)
(
)
OR
Q =
=
(
)
=
(
)
(
)
(
)
Where
= inner radius of the fluid layer
= outer radius of the fluid layer (which is usually the wall)
Cut point or critical diameter
This is the diameter of a particle which reaches half the distance between r
1
and r
2
. To be
removed, a particle need to be at the wall, so r
b
= r
2
. A parti
cle of cut point diameter must
therefore have begun to settle at the halfway point,
=
, then
=
=
=
(
)
(
)
(
)
If the flow through the centrifuge is greater than
, almost all particles larger than the cut
point diameter will be removed while the smaller particles will remain.
Question 4
Beer with a specific gravity of 1.042 and a viscosity of 1.40
Ns
m

2
contains 1.5% solids
which have a density of 1160kgm

3
. It
is clarified at a rate of 240Lh

1
in a bowl centrifuge which
an operating volume of 0.09m
3
and a speed of 10000rpm. The bowl has a diameter of 5.5cm and
is fitted with a 4cm outlet. Calculate t
he effect on feed rate of an increase in bowl speed to
15000rpm
and minimum particle size that can be removed at the higher speed.
Solution
Initial
flow rate
=
(
)
(
)
=
(
)
(
)
(
)
New flow rate Q
2
=
(
)
(
)
(
)
All conditions except the bowl speed remain the same
=
(
)
(
)
=
=
(
)
(
)
= 540Lh

1
or
= 0.15LS

1
From the above
equation
=
(
(
)
)
(
)
(
)
=
(
)
(
)
(
)
(
)
(
)
√
= 0.68
m
Question 5
A viscous solution containing particles with a density
= 1461kgm

3
is to be clarified by
centrifugation. The solution density
= 801
kgm

3
and its viscosity is 100cp. The centrifuge has
a bowl with
= 0.02225m and
= 0.00716m and height b= 0.1970m. Calculate the critical
particle diameter of the largest particles in the exit stream if N = 23000rpm and the flow rate q =
0.002832m
3
h
Solution
ω =
=
(
)
=
2410 rad/s
the bowl volume V is
V =
πb
(
)
= π
(
)
{
(
)
(
)
}
= 2.747
m
3
Viscosity
= 100
= 0.100pa.s = 0.100kg
/m.s
The flow rate,
Q =
= 7.87
m
3
/s
Q =
(
)
(
)
(
)
(
)
(
)
(
)
(
)
d
pc
=
0.746
m
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