CENTRIFUGE
CENTRIFUGATION

A process used to separate or concentrate
materials suspended in a liquid medium.

Centrifugation separates on the basis of the
particle size and density difference between
the liquid and solid phases.
THEORY OF CENTRIFUGATION
the effect of gravity on particles (including
macromolecules) in suspension.
 Two particles of different masses will settle in
a tube at different rates in response to gravity.
 Centrifugal force is used to increase this
settling rate in an instrument called a
centrifuge.


Centrifuges are devices used in a variety of
scientific and technical applications

The centrifugal force generated is
proportional to the rotation rate of the rotor
(in rpm) and the distance between the rotor
center and the centrifuge tube.
CENTRIFUGE SEPARATION PROCESS
Settling: acceleration from gravity (Fg)
 Centrifuge:

 acceleration
from centrifugal force (Fc)
 circular
motion and acceleration occurred
from centrifugal force
ac
 r
2
ac = acceleration from centrifugal force (m/s2)
r = radial distance (m)
ω = angular velocity (rad/s)
CENTRIFUGAL FORCE (FC)

The centrifugal force, Fc acting on an object of
mass m, rotating in a circular path of radius R, at
an angular velocity of ω is :
Fc  mR
2
(1)
and
2N N


60
30
(2)
where N = rotational speed (rpm) ω= an angular
velocity (rad s-1)
G FORCE (GRAVITIES OR G’S)

The steady-state velocity of particles moving in
a streamline flow under the action of an
accelerating force
from vt 
g (  s  l ) Dp2
18
r (  s  l ) D
2
vt 
2
p
18
Where vt=terminal velicity of partical; ρs and ρl
= density of solid and liquid ; r = distance of the
particle from center of rotation;µ = viscosity of
liquid.
CENTRIFUGATION TIME

Time taken by the particle to move though the
liquid layer is called residence time (tr).
dr
Vt 
dt
 ( p   )D r
vt 
18
2
2
p
dr D r (  p   )

dt
18
2
p
2
D  ( p   )
1
dt

r r dr 

18
0
i
ro
2
2
p
t
ro D  (  p   )
 tr
ln 
18
ri
2
p
2
ro
18 ln
ri
tr 1  2 2
D p (  p   )
CALCULATION OF FLOW RATE FOR
CONTINUOUS CENTRIFUGE

flow rate (Q)
 (ro2  ri 2 )b
V
Q 
ro
tr
18 ln
ri
2 2
D p (  p   )
Q
 (ro2  ri 2 )b  D p2 2 (  p   )
ro
18 ln
ri
 ri
= inside radius (m)
 ro = outside radius (m)
 b = height of centrifuge(m)
 µ = viscosity (Pa.s)
 ω = an angular velocity (rad s-1)
ρp = density of solid (kg/m3)
 ρ = density of liquid (kg/m3)
 Dp= diameter of particle(m)

LIQUID SEPARATION
Figure 1 Liquid centrifuge
(a) Pressure difference
Consider a thin differential cylinder, of
thickness dr and height b as shown in Fig.
1(a): the differential centrifugal force across
the thickness dr is given by equation (1):
dFc = (dm)r2
 where dFc is the differential force across the
cylinder wall, dm is the mass of the
differential cylinder,  is the angular velocity
of the cylinder and r is the radius of the
cylinder.

dm = 2πρrbdr
where  is the density of the liquid and b is
the height of the cylinder. The area over
which the force dFc acts is 2πrb , so that:
dFc /2πrb = dP =ρ 2rdr
where dP is the differential pressure across
the wall of the differential cylinder.

To find the differential pressure in a
centrifuge, between radius r1 and r2, the
equation for dP can be integrated, letting the
pressure at radius r1 be P1 and that at r2 be
P2, and so
P2 - P1 = ρω2 (r22 - r12)/2
(3)
Equation (3) shows the radial variation in
pressure across the centrifuge.
Figure 1 Liquid centrifuge
(b)neutral zone
ρAω2 (rn2 - r12)/2 = ρB ω2(rn2– r22)/2
rn2 = (ρAr12 - ρBr22) / (ρA - ρB)
(4)
where ρA is the density of the heavier liquid
ρB is the density of the lighter liquid

Equation (4) shows that as the discharge radius for
the heavier liquid is made smaller, then the radius
of the neutral zone must also decrease
CENTRIFUGE EQUIPMENT
LIQUID/LIQUID SEPARATION
CENTRIFUGES
FIG. 2 Liquid centrifuges: (a) conical bowl

In liquid/liquid separation centrifuges,
conical plates are arranged as illustrated in
Fig. 2(a) and these give smoother flow and
better separation.

Whereas liquid phases can easily be
removed from a centrifuge, solids present
much more of a problem.
LIQUID/SOLID SEPARATION CENTRIFUGES
FIG. 3 Liquid/solid centrifuges (a) telescoping
bowl, (b) horizontal bowl, scroll discharge
LIQUID/SOLID SEPARATION CENTRIFUGES
(c)
FIG. 3 Liquid/solid centrifuges (c) nozzle

One method of handling solids from
continuous feed is to employ telescoping
action in the bowl, sections of the bowl
moving over one another and conveying the
solids that have accumulated towards the
outlet, as illustrated in Fig. 3(a).

The horizontal bowl with scroll discharge,
centrifuge, as illustrated in Fig.3(b) can
discharge continuously. In this machine, the
horizontal collection scroll (or screw) rotates
inside the conical-ended bowl of the
machine and conveys the solids with it,
whilst the liquid discharges over an overflow
towards the centre of the machine and at
the opposite end to the solid discharge.

Another method of handling solids is to
provide nozzles on the circumference of the
centrifuge bowl as illustrated in Fig. 3(c).
These nozzles may be opened at intervals to
discharge accumulated solids together with
some of the heavy liquid.
EXAMPLE 1
Find centrifugation time tr of a
particle d=1mm. In a
centrifuge
Given
N  995RPM
Ri
  8.110 Pa.s
4
 P  1100kg / m
3
 f  1000kg / m
3
Ri  0.20m.
Ro  0.25m.
Ro
Find ω
2N

60
2    995

60
  104.20rad / s
Find time
18 ln(ro / ri )
tr  2 2
d   p   f 
4
18 8.110  ln(0.25 / 0.20)
tr 
2
2
0.001  104.20  1100 1000
3
t r  3.2510 sec
tr of particle d=1mm. in centrifuge≥3.25x10-3sec
EXAMPLE2

A bowl centrifuge is used to break an oil-inwater 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 1000 kg/m3 and the
density of the oil is 870 kg/m3. the outlet
radius from the centrifuge are 3 cm and 4.5
cm).

Solution
1000(0.045)  870(0.03)
rn 
1000 870
2.025 0.783
rn 
130
rn  0.098m
2
2
2
EXAMPLE 3
Beer with a specific gravity of 1.042 and a viscosity
of 1.04x10-3 N s/m2 contains 1.5% solids which
have a density of 1160kg/m3. It is clarified at a rate
of 240 l/h in a bowl centrifuge which has and
operating volume of 0.09 m3 and a speed of 10000
rev/min. The bowl has a diameter of 5.5 cm and is
fitted with a 4 cm outlet. Calculate the effect on
feed rate of an increase in bowl speed to 15000
rev/min and the minimum particle size that can be
removed at the higher speed.

Solution
Initial flow rate
V  (2N1 / 60) D  p   f 
2
Q1 
2
18  ln(ro / ri )
new flow rate
Q2 
V  (2N 2 / 60) 2 D 2  p   f 
18  ln(ro / ri )
As all conditions except the bowl speed
remain the same,
Q2 (2N 2 / 60) 2

Q1 (2N1 / 60) 2
Q2
(2  3.142 15000/ 60) 2

(240/ 3600) (2  3.142 10000/ 60) 2
Therefore,
Q2 = 0.15 l/s
TO FIND THE MINIMUM PARTICLE SIZE
Q2 [18 ln(ro / ri )]
D 
(2N 2 / 60) 2 (  p   f )V
2
0.15[181.40103  ln(0.0275/ 0.02)]

(2  3.14215000/ 60) 2 (1160 1042)0.09
3
1.2010
D 
 6.8m
7
2.6210