# Centrifugation

```Centrifugation
Theory of centrifugation
Types of centrifuges
Applications
Centrifugal separations
• Centrifugal separation is a sedimentation
operation accelerated by centrifugal force.
• Prerequisite for the separation is a difference
in density between the phases.
• This applies to both
– solid–liquid separation
– liquid–liquid separation
Sedimentation by Gravity
• A particle suspended in a liquid medium of
lesser density tends to sediment downward due
to the force of gravity (Fg)
Fg  mg  m  980cm.s
2
• There are two forces that oppose the
gravitational force;
– the buoyancy force, Fb
– the frictional force, Ff
Buoyancy force
Fb  mM g  V p  M g
mM = mass of the fluid medium
displaced,
Vp = volume of the particle,
ρM = density of the displaced fluid
The net gravitational effect, taking into account the buoyancy
force is
Fg  net
4 3
4 3
2
  r   P   M  g   r   P   M   980cm.s
3
3
ρM is the density of the medium (g.cm-3);
ρP is the particle density (g.cm-3)
r is the particle radius (cm).
Frictional force
The movement of a particle through a fluid medium is hindered by the viscosity of
the medium, η as described for a spherical particle by Stokes’ equation
 dx 
Ff  6 r  
 dt 
η, is the viscosity of the medium in
poise, P (g cm-1s-1);
r is the radius of the particle (cm);
(dx/dt) is the velocity of the moving
particle (cm.s-1).
Terminal velocity
At low velocities and pressures, the frictional force is again negligible in a gas.
At higher velocities, even in gases, this force becomes substantial, combining
with the buoyancy force eventually to exactly oppose the gravitational force,
resulting in no further acceleration of the particle.
Mathematically, the conditions for attaining
terminal velocity are met when:
Fg  Fb  F f
Effect of Diffusion
• Random Brownian motion
results in the net movement of
solute or suspended particles
from regions of higher
concentration to regions of
lower concentration
• Diffusion works in opposition to
centrifugal sedimentation,
which tends to concentrate
particles.
• The rate of diffusion of a
particle is given by Fick’s law:
dP
 dP 
  DA 

dt
 dx 
D is the diffusion coefficient which
varies for each solute and particle;
A is the cross sectional area through
which the particle diffuses
dP/dx is the particle concentration
Sedimentation in a Centrifugal Field
A particle moving in a circular path continuously experiences a
centrifugal force, Fc.
This force acts in the plane described by the circular path and is
directed away from the axis of rotation. The centrifugal force may
be expressed as
Fc  ma  m x
2
m is the particle mass (g);
a is the acceleration (cm.s-2)
ω is the angular velocity (radians s-1 π.rpm/60)
x is the radial distance from the axis of rotation to the particle (cm)
Relative centrifugal force (RCF)
Ratio of acceleration of the centrifugal field to that of
acceleration owing to the earth’s gravity
Fc
RCF 

Fg
2
m

 x
 mg 

2

 x
g
Alternatively RCF is given by
RCF  1.119 10
5
 rpm 
2
x
m = particle mass (g);
a = acceleration (cm.s-2);
ω = angular velocity (radians s-1 2π.rpm/60);
x = radial distance from the axis of rotation to the particle (cm).
Forces acting on a particle in a centrifugal
field
When centrifugal force is equaled by buoyancy and frictional
forces,
Fc  Fb  Ff
 dx 
m x  VP  M  x  6 r  
 dt 
2
2
Assuming spherical particles the above equation
becomes
Fb = buoyancy force
Ff = frictional force
Fc = centrifugal force
Fg = gravitational force
4 3
4 3
 dx 
2
2
 r  P x   r  M  x  6 r  
3
3
 dt 
Solving for (dx/dt)
2
2

dx  2r   P   M   x 

dt
9
Forces acting on a particle in a
centrifugal field

In terms of particle diameter, d and
d 2  P  M   2 x
particle velocity, v
v

18
Upon integration the equation above yields the time required for
a particle to traverse a radial distance from x0 to x1

 x 
18
1

t 2
ln
 
2
  d   P   M      x0 
x0 is the initial position of the particle
x1 is the final position of the particle
Problem: The effect of the microorganisms’s size on
settling velocity
Compare the settling velocities of E.coli and yeast at a cell
concentration of 1.5%
Parameter
Yeast
E. coli
Size
Density
Apparent
viscosity
7 micron
1.03 g/ml
1.8 cP
1 micron
1.03 g/ml
1.8 cP
Assume isothermal conditions and the settling behaviour
same as that of a sphere
Sedimentation is governed by
v
2
2
d






 P M x
18
Settling velocity for E. coli = 0.0033 cm/sec
Settling velocity for yeast cells = 22.7 cm/sec
Parameters that govern settling velocity
• The sedimentation rate (i.e. limiting velocity)
of a particle in a centrifugal field
– increases as the square of the particle diameter
and rotor speed, i.e. doubling the speed or
particle diameter will lessen the run time by a
factor of four
– increases proportionally with distance from the
axis or rotation
– is inversely related to the viscosity of the carrier
medium
Sedimentation Coefficient
• In a homogeneous medium, the
following parameters are
For a given set of run conditions, the
constant for a given particle sedimentation coefficient, Sr, may be
–
–
–
–
The viscosity
Particle size
Particle density
Density of the medium
• The sedimentation rate is
proportional to ω2x,
calculated as
 dx 
  2r 2     
dt 
P
M

Sr  2 
9
 x 
– expressed in terms of the
sedimentation coefficient, S.
• Measure of the sedimentation
velocity per unit of centrifugal
force
Sedimentation Coefficient
• The sedimentation coefficient, S, has the dimensions of
seconds and is expressed in Svedberg units equal to 10-13 s
S20, w 
ST ,MT ,M   P  M 
20,W   P  T ,M 
• Sedimentation coefficient is dependent on
– the particle being separated
– the centrifugal force
– the properties of the sedimentation medium.
• Useful to compare sedimentation coefficients obtained
– under differing conditions
– sedimentation media by reference to the behaviour of the particle in
water at 20oC.
Rotor Efficiency (k-factor)
• The time required for a particle to traverse a rotor is
known as the pelleting efficiency or k-factor
• k-factor is calculated at the maximum rated rotor
speed, is a function of rotor design and is a constant
for a given rotor.
• k-factors provide a convenient means of
– determining the minimum residence time required to
pellet a particle in a given rotor
– useful for comparing sedimentation times for different
rotors
The k-factor is derived from the equation
ln  rmax  rmin  1013
k
2
3600
rmax and rmin are the maximum and minimum
distances from the centrifugal axis
k  2.531011  ln  rmax  rmin  / rpm2
If the sedimentation coefficient of a particle is known, then the
rotor k-factor can also be calculated from the relation:
k  TS
T is the time in hours required for pelleting
S is the sedimentation coefficient in Svedberg units
For runs conducted at less than the maximum rated rotor speed, the
k-factor may be adjusted according to
 rpmmax 
k

 rpmact 
2
k-Factors are also useful when switching from a
rotor with a known pelleting time, t1, to a second
rotor of differing geometry by solving for t2 in the
relation
Stress in the bowl wall limits centrifuge
speed
• Self stress (SS)
– Stress in the bowl wall is due to rotation
S S  4.1110
10
n Db m
2
2
• Stress due to the liquid in the bowl (Sl)
2
2
2
n
D

D

D
b
b
i
10
S  1.03 10



• Total stress (ST) ST = SS + Sl
2
2



D

D

b
i 
2

ST  4.1110  10n Db  m Db 


4


Problem: Separating cells growing on a support
Animal cells can be cultivated on the external surface of dextran
1.02 g/ml and a diameter o 150 μm.
A 50 litre stirred tank is used to cultivate cells grown on
microcarriers to produce a viral vaccine. After growth, the stirring
is stopped and are allowed to settle. The microcarrier-free fluid is
then withdrawn to isolate the vaccine.
The tank has a liquid height to diameter ratio of 1:5; the carrierfree fluid has density of 1 g/ml and a viscosity of 1.1 cP.
a) Estimate the settling time to reach the velocity
b) Estimate the time to reach this velocity
Solution
Using the equation for terminal velocity
d 

v
2
 M   x 
2
P
18
Substituting the values we get
v  0.022cm.s 1
Problem: Centrifugation of yeast cells
A laboratory bottle centrifuge consists of a number of cylinders
rotated perpendicular to the axis of rotation.
During centrifugation the distance between the surface of liquid
and the axis of rotation is 3 cm, and the distance from the bottom
of the cylinder to that axis is 10 cm.
The yeast cells can be assumed to be spherical, with a diameter of
8.0 μm and a density of 1.05 g/ml.
The fluid has physical properties close to those of pure water.
The centrifuge is to be operated at 500 rpm.
How long does it take to have complete separation?
From the equation
It was found that
d2
v 
( s   ) 2 r
18
dr
d2
 v 
( s   ) 2 r
r
18
We are interested in the yeast cell which takes longest to settle,
which is that starting near the liquid surface, t = 0; r = 3 cm.
Integrating the initial equation,
we find
2
r
d


2
ln 
(  s   ) t

 3cm  18
Substituting the values, we get
(8 104 cm) 2 
g   500  2 2 
 10cm 
ln 
0.05 3  


1
1 
cm   60sec 
 3cm  18(0.01g.cm .sec 
t  2500sec
Types of Centrifugal Separation
• According to the phase of the medium and the
phase of the material to be purified
– Gas-gas
– Liquid-liquid
– Liquid-solid
• According to the method by which purified
fractions are recovered
– Batch mode
– Semi-batch mode
– continuous mode
Types of centrifuges
• Tubular bowl centrifuges
– Simple yet can provide very high G
– Can be cooled
– Disadvantage: Requirement for intermittent dismantling
for cleaning
• Disc type centrifuges: Three types
– Solids-retaining
– Intermittent solids-ejecting
– Continuous solids-ejecting
– Used for centrifugal filtration
Tubular bowl centrifuge
Utilize a vertically mounted, imperforate cylindrical-bowl
design to process feed streams with a low solids content
Liquid(s) is discharged continuously and
solids are manually recovered after the
rotor capacity is reached
Industrial models are available with
Diameters up to 1.8 m
Holding capacities up to 12 kg
Throughput rates of 250 m3 h-1
Centifugal forces ranging up to 20000 g
Laboratory models are available with
Diameters of 4.5 cm
Throughput rates of 150 L.h-1
Centrifugal forces ranging up to 62000 g
Performance analysis of a tubular centrifuge
Analysis depends on finding the position of a
particle as a function of time
R0
R1
l
Assumptions
1. Particle located at a distance z from the bottom
of the centrifuge
2. It is also located at position r from the axis of
z
rotation
r
Liquid Interface
3. This position is between the liquid surface R1
Idealization of the
4. Feed freely flows in the bottom and out the top Tubular bowl
5. Solids are thrown out by centrifugal force and centrifuge
trapped against the wall, located at R0
6. The centrifugal force is so high that the liquid
interface R1 is constant
Performance analysis of a tubular centrifuge
The particle is moving in both the z and r directions.
Its movement in the z direction comes from the convection of the feed pumped
in the bottom of the centrifuge dz
Q
dt

 ( R0 2  R12 )
Q is the feed flow rate
The particle movement in the r direction is related to its radial position by
2
dr d
2




r

 s r
dt 18
 r 2 
dr
 vg 

dt
 g 
Which can be rewritten in terms the velocity of a
particle settling under the influence of gravity
To find the trajectory of the particle within the centrifuge
2
2

R

R

 r 2 
0
1 
dr dr / dt

 vg 

dz dz / dt
g


Q
For particles which are most difficult to capture, they enter the centrifuge
at r=R1 and do not reach r=R0 until the end of the unit at z=l
Integration and rearrangement of the equation for the particle trajectory gives the
maximum flow possible flow rate in the centrifuge as a function of both particle
properties and centrifuge characteristics
 l  R0  R
2
Q
2
1
v 
2
g
g ln  R0 / R1 
In most tubular centrifuges as R0 and R1 are approximately equal, we can simplify the
above equation
R0  R1  R0  R1 

( R0 2  R12 )

ln( R0 / R1 ) ln 1   R0  R1  / R1 
R0  R1  R0  R1 


 R1 ( R0  R1 )
 R0  R1  / R1  ....
 2R2
 2 lR 2 2 
Q  vg 
 vg []

g


The Generalized Σ Formula
The most-used quantity to characterize centrifuges, the Σ concept
Qtheor  vg .
Assumptions
•
•
•
•
•
•
Where,
2 lR 2 2

g
Viscous drag is determining the particle movement.
The flow in disk bowls between the disks is laminar and
symmetrical.
The liquid rotates at the same speed as the bowl
The particle concentration is low (no hindered settling).
The particle always moves at its final settling velocity.
This settling velocity (Vc) is proportional to the g force.
The equation for critical diameter becomes
 18. .Q
Se
theor
dc  
.
   p   f  V . 2 .r 2

 18. .Q

theor

dc  
 .   p   f  .g 


1
2




1
2
Solids-retaining Disc centrifuge
Appropriate for liquid-solid or liquid-liquid
separations where the solids content is less than
For liquid-solid separations, the solids that
accumulate on the bowl wall are recovered
when the rotor capacity is reached and the
centrifuge is stopped
some designs to facilitate solids removal
Recovery of two liquid streams can be
achieved by positioning exit
Ports at different radial distances as
dictated by the relative concentration of
the liquids
Intermittent solids ejecting disc centrifuge
Suitable for processing samples with solids contents
Solids or sludge that accumulate on the bowl
wall are intermittently discharged through a
hydraulically activated, peripheral opening
Laboratory models to 18 cm diameter and
industrial units to 60 cm
Industrial centrifuges capable of
throughputs in excess of 100 m3 h-1
Bowl section of a self-cleaning disc stack centrifuge indicating direction of
fluid flow and ejection of sedimented solids through passages controlled with
hydraulically operated pistons
Discharge is intermittent.
Feed
Discharge pump
Nozzle machines allow
for continuous
discharge of solids
through throttled
nozzles
Timing unit
Solid bowl machines
without solid discharge
mechanisms require
manual cleaning from
time to time depending
upon feedstock solidsAnnular piston
Photocell
Discharge
Discs
Sediment holding space
Solids ejection ports
Opening chamber
Closing chamber
Drain hole
Operating water valve
Continuous solids ejecting disc centrifuge
Solids contents ranging from 5 to 30% by volume
Solids are continuously discharged via backwardfacing orifices
Newer designs discharge to an internal chamber
where the discharge is pumped out as a product
stream
Industrial units are available to 200 m3 h-1
throughput capacity, elevated
temperature (&lt;200oC) or pressure (7 bar)
capability, and particle removal to 0.1 μm.
Performance analysis of disc type centrifuge
ω
Objective: To find the location of a given particle
Consider a particle at position (x, y)
θ
The velocity in the x direction is due to
convection and sedimentation
dx
 v0  v sin 
dt
Average
 Q 
v0  
f ( y)

 n(2 rl ) 
y
x
R1
convective
velocity
The volume of v0 averaged over y must
equal the convective velocity
R0
Characteristics of v0
Much larger than vωsinθ
Function of y
Performance analysis of disc type centrifuge
ω
 Q 
1
v0 dy  


l0
n
(2

rl
)


l
θ
Performing integration we get
l
1
f ( y )dy  1

l0
y
x
R1
Again considering that the convective velocity
is much greater than that of sedimentation
dx
 v0  v sin   v0
dt
 Q 

f ( y)

 n(2 rl ) 
R0
Performance analysis of disc type centrifuge
ω
The velocity in the y direction is due to
convection and sedimentation
θ
dy
 v cos 
dt
dy dy / dt

dx dx / dt
 2 nlvg  2 r  2

r cos 
 Qgf ( y ) 


In terms of R0
y
x
R1
R0
2

2

nlv

r
dy
g
2

(
R

x
sin

)
cos 

0
dx  Qgf ( y ) 
This describes the trajectory of the particle
between the discs
Performance analysis of disc type centrifuge
For particles which are most difficult to capture
ω
These particles enter at the outer edge of the discs where
x=0 and y=0
θ
They are captured at the inner edge of the discs at
y=l and x=(R0-R1)/sinθ
After capture they and other particles are
forced along the disc surface to the outer
edge, where they are discharged.
y
x
R1
 2 n 2

3
3
Q  vg 
( R0  R1 ) cot   vg []
 3g

R0
In both the cases the quantity in square brackets Σ has dimensions of (length)2
Essentially the term Σ is dependent on the geometry of the centrifuge
Horizontal continuous-conveyer centrifuge
Integrate an active mechanical solids discharge mechanism in an imperforate bowl
for the continuous processing of larger sample volumes
The solids-discharge mechanism: A helical screw turning at a slightly slower rate
than the rotor
Capable of very high throughput, up to 300 000 L h-1
Combination of a centrifuge and a filter with a rapidly rotating perforated
Suspension is fed along the axis of the bowl and solids accumulate on the wall
Liquid flows under centrifugal force through the cake which
accumulates on the basket wall and out through the perforations in the wall
Used to wash accumulated
cake solids in filtration
Drainage number

d (G )1/ 2

Higher drainage
numbers correspond to
more rapid drainage
The theoretical sizing of a centrifugal
separator
• Viscous drag is determining the particle movement.
• The flow in disk bowls between the disks is laminar
and symmetrical.
• The liquid rotates at the same speed as the bowl
• The particle concentration is low (no hindered
settling)
• The particle always moves at its final settling
velocity.
• This settling velocity (Vc) is proportional to the g
force.
Problem: Complete recovery of bacterial cells in a
tubular bowl centrifuge
For complete recovery of bacterial cells from a fermentation broth
with a pilot plant scale tubular centrifuge.
It has been already determined that the cells are approximately
spherical with a radius of 0.5 μm and have density of 1.1 g.cm-3
The speed of the centrifuge is 5000 rpm
The bowl diameter is 10 cm
The bowl length is 100 cm
The outlet opening of the bowl has a diameter of 4 cm
Estimate the maximum flow rate of the fermentation broth that can
be attained.
Problem: Tubular centrifugation of E.coli
A bowl centrifuge is used to concentrate a suspension of E.coli prior
to cell disruption.
The bowl of this unit has an inside radius of 12.7 cm and a length of
73 cm. the speed of the bowl is 16000 rpm.
The volumetric capacity is 200 litres/hr. Under these conditons the
centrifuge works well.
a) Calculate the settling velocity vg for the cells
b) After disruption the diameter of the debris is about one half of the
original cell diameter and the viscosity is increased four times.
Estimate the volumetric capacity of this centrifuge under these
new conditions
Problem: Disc centrifugation of chlorella
Chlorella cells are being cultivated in open ponds. We plan to
harvest the biomass by passing the dilute stream of cells through an
available disc bowl centrifuge.
The settling velocity vg for these cells has been measured as 1.04 x
10-4 cm.s-1.
The centrifuge has 80 discs with an angle of 40o
We plan to operate the centrifuge at 600 rpm.
Estimate the volumetric capacity Q for this centrifuge.
Equivalent time
• To assess the approximate properties of a particle
type to be separated
• Define a dimensionless acceleration G
• This dimensionless unit is measured in terms of ‘g’s
(multiples of earth’s acceleration)
• A rough approximation of the difficulty of a
separation by centrifugation is the product of
– the dimensionless acceleration and
– the time required for separation
• Determination by
– Centrifuging samples for various times until a constant PCV is
reached
– The equivalent time is calculated as the product of G and the
time required for reaching the constant PCV
• For scale up of centrifugation we can assume
constant equivalent time
G
Gt 
 R
2
g
 R
2
g
t
(Gt )1  (GT )2
Problem: Scale-up based on equivalent time
If bacterial cell debris has Gt = 54 x 106 s, how large must be the
centrifuge bowl and what centrifuge will be needed to effect a full
sedimentation in reasonable amount of time?
Assume the reasonable amount of time as 2 hours and the bowl
Characteristics of Separator Types
Coagulants and flocculants
• Metal salts
– especially of aluminium or ferric iron
• Natural flocculants
• Starch, Gums, Tannin, Alginic acid, Sugar/sugar acid polymers,
Polyglucosamine (chitosan)
• Synthetic flocculants
– Polyacrylamides, Polyamines/imines, Cellulose derivatives (e.g.
carboxymethyl cellulose), Polydiallydimethyl ammonium chloride
• Chilling temperatures below 20oC, particularly yeast cells
• pH adjustment in range 3-6
• Concentration
– increases particle concentration, increasing collision frequency
```