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CENTRIFUGATION
•••• Sedimentation and centrifugation
Sedimentation
���� When a suspension is allowed to stand, the denser
solids slowly settle under the influence of gravity.
Centrifugation
���� A settling process that is accelerated with a centrifugal
field.
•••• Comparison between filtration and centrifugation:
Feature Filtration Centrifugation
Separation principal
Employment
Product obtained
Expense of equipment
Particle size
Removal of
insolubles
which are
dilute, large
and rigid
Dry cake
Less
Density
Used when
filtration is
ineffective
A paste or a more
concentrated
suspension
More
Introduction (2/8)
•••• Separation cost for recovering whole cells or cell debris:
Introduction (3/8)
Ultrafiltration
more
economical
Centrifugation
more
economical
Ultrafiltration
Centrifugation
•••• Schematic presentation of a laboratory centrifuge:
Introduction (4/8)
•••• Care of centrifuges:
(1) Avoid imbalance in the rotor, which may be caused by:
a. Tube cracking during the run
* Conventional glass (Pyrex) centrifuge tubes
withstand only 3−−−−4000 g.
���� Use centrifuge tubes made from
polypropylene or polycarbonate.
b. Misbalance of the tubes in the first place
���� Small tubes—balanced by volume by eye; large
tubes (> 200 mL)—should be weighed.
(2) Any spillage should be immediately rinsed away.
���� Avoid corrosion of centrifuge rotors.
(3) Do not use the machine at top speed constantly.
Introduction (5/8)
__________________ __________________ __________________ __________________
Introduction (6/8)
* Relative Centrifugal Force, RCF = g
r2ω
1s 1047.0s 60
min
min
2 rpm 1 -=
=π
g = 980 cm/s2
r: in cm
)((rpm) 10119.1cm/s 980
cm) (rpm
s 1047.0rpm)(
RCF 25
2
212
r
r−
−
×=
=
���� Often an average RCF is determined using a value for r
midway between the top and bottom of the sample
container.
Introduction (7/8)
ravg = 7 cm
20,000 rpm } RCF = 31,000
(centrifugal force = 31,000 ×××× g) ����
Introduction (8/8)
FORCES DEVELOPED IN CENTRIFUGAL
SEPARATION
The acceleration from a centrifugal force: a = ωωωω 2r
where ωωωω = angular velocity, rad/s
r = radial distance from center of rotation
Settling by gravity force: gd
v sg )(18
2
ρρµ
−=
Settling in centrifuges: rd
v s2
2
)(18
ωρρµω −=
•••• Gravitational sedimentation is too slow to be practical for bacteria, and conventional centrifugation is too slow
for protein macromolecules.
FORCES DEVELOPED IN CENTRIFUGAL SEPARATION (2/3)
__________
[Example] A laboratory bottle centrifuge is used to collect
yeast cells after fermentation. The centrifuge consists of a
number of cylinders rotated perpendicularly 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 µµµµm and a density of 1.05 g/cm3. The fluid has
physical properties close to those of water. The centrifuge is
to be operated at 500 rpm. How long does it take to have a
complete separation?
Solution:
rd
dt
drv
s
22
)(18
ωρρµω −== t
d
r
rs
22
1
2 )(18
ln ωρρµ
−=or
(To be continued)
Example: laboratory bottle centrifuge
Solution (cont’d):
td
r
rs
22
1
2 )(18
ln ωρρµ
−=
t = 0, r = 3 cm; t = ?, r = 10 cm
Data: d = 8.0 µµµµm = 8.0 ×××× 10-4 cm; µµµµ = 1 cP = 0.01 g/cm-s; ρρρρs = 1.05 g/cm3; ρρρρ = 1.0 g/cm3;
rad/s 3.5260
2500rpm 500 =
×==
πω
t××−×××
=−
224
)3.52()105.1(01.018
)100.8(
3
10ln
���� t = 2467 s = 41.3 min #
* Sedimentation Coefficient, s
““““The velocity of a particle through a viscous medium is usually proportional to the accelerating field.”
rd
v s
22
)(18
ωρρµω −= )(
18
2
ρρµ
−= s
ds����
Unit of s: svedberg (S; 1 S = 10−−−−13 second)
���� Svedberg: the inventor of ultracentrifuge
rsv2ωω =
FORCES DEVELOPED IN CENTRIFUGAL SEPARATION (3/3)
[Example] Estimate the time it would take to completely
clarify a suspension of 70 S ribosomes in a high speed
centrifuge operating at 10,000 rpm. During centrifugation,
the distance between the surface of liquid and the axis of
rotation is 4 cm, and the distance of travel of particles
radially outward is 1 cm.
Solution:
∫∫ =5
4
2
0
1
r
dr
sdt
t
ω����
( )h 8.1 s 29080
s 60
min 1
rev
2
min
rev 10000 s 1070
223.0
4
5ln
12
13
2==
×××
==− πωs
t
#
rsdt
drv 2ωω ==
TUBULAR BOWL CENTRIFUGE
•••• Suspension is usually fed through the bottom, and clarified liquid is removed
from the top.
•••• Solid deposits on the bowl’s wall as a thick paste.
•••• The suspension can be fed until solid loss in the effluent becomes prohibitive.
•••• An intermittent operation.
Assume that a particle is located at a distance z from the
bottom of the centrifuge and at a position r from the axis of
rotation. This particle is moving in both the z and r
directions.
TUBULAR BOWL CENTRIFUGE (2/6)
The movement of the particle in the z
direction (due to the convection of the
feed flow):
)( 2
1
2
0 RR
Q
dt
dz
−=
π
where Q = the volumetric flow rate
R1 = the distance of liquid interface
from the axis of rotation
The movement of the particle in the r direction:
rd
dt
drs
22
)(18
ωρρµ
−=
gd
vsg
)(18
2
ρρµ
−=
=
g
rv
dt
drg
2ω����
TUBULAR BOWL CENTRIFUGE (3/6)
)(2
1
2
0 RR
Q
dt
dz
−=
π
=
g
rv
dt
drg
2ω;
The trajectory of a particle in the
centrifuge:
Q
RR
g
rv
dtdz
dtdr
dz
drg
)(
/
/2
1
2
02 −
==
πω
Consider a particle enters the centrifuge at R1 (that is,
at z = 0, r = R1) and do not reach R0 until at z = l
Q
RR
g
v
R
R g l)(ln
2
1
2
0
2
1
0−
=πω
)/ln(
)(
10
22
1
2
0
RRg
vRRQ
gωπ −=
l
or
TUBULAR BOWL CENTRIFUGE (4/6)
For R0 and R1 being approximately equal,
)/ln(
)(
10
22
1
2
0
RRg
vRRQ
gωπ −=
l
2
101
110
1010
110
1010
10
2
1
2
0 2)(/)(
))((
]/)(1ln[
))((
)/ln(RRRR
RRR
RRRR
RRR
RRRR
RR
RR=+=
−
−+=
−+
−+=
−
⋅⋅⋅+−+−=+ 432
4
1
3
1
2
1)1ln( xxxxxNote:
)(2
22
Σ=
=∴ gg v
g
RvQ
ωπl
Note: vg is a function only of the particles themselves, and ΣΣΣΣ is a function only of the particular centrifuge.
TUBULAR BOWL CENTRIFUGE (5/6)
* Continuous tubular bowl
centrifuge for separation of
two liquids:
An internal baffle provides a
separate passage adjacent to
the bowl wall to conduct the
heavier-phase liquid to a
different discharge elevation.
TUBULAR BOWL CENTRIFUGE (6/6)
[Example] A bowl centrifuge is used to concentrate a
suspension of Escherichia coli prior to cell disruption. The
bowl of this unit has an inside radius of 12.7 cm and a
length of 73.0 cm. The speed of the bowl is 16,000 rpm and
the volumetric capacity is 200 L/h. Under these conditions,
this centrifuge works well. (a) Calculate the settling
velocity vg for the cells. (b) After disruption, the diameter of
debris is about one-half of that of cell and the viscosity is
increased four times. Estimate the volumetric capacity of
this same centrifuge operating under these new conditions.
Solution:
22
22
2or
2
ωπωπ
R
Qgv
g
RvQ gg
l
l=
=
(To be continued)
[Example] Analysis of bowl centrifuge
222
ωπ RQg
vgl
=Solution:
Data: R = 12.7 cm; = 73 cm; ωωωω = 16,000 rpm = 1674.7 rad/s; Q = 200 L/h = 55.56 cm3/s; g = 980 cm/s2
l
���� vg = 2.63 ×××× 10-7 cm/s
Using the same centrifuge, 1
2
1
2
1
2
g
g
g
g
v
v
v
v
Q
Q=
Σ
Σ=
gd
v sg )(18
2
ρρµ
−=16
1
4
)2/1(
/
/
/
/ 2
12
2
1
2
2
1
2
1
2
2
2
1
2 ====µµµ
µ dd
d
d
Q
Q����
(a) Calculate the settling velocity vg for the cells.
(b) Estimate the volumetric capacity of this same centrifuge for cell
debris.
#
[Example] Beer with a specific gravity of 1.042 and a
viscosity of 1.4 ×××× 10-3 N-s/m2 contains 1.5% solids, which
have a density of 1160 kg/m3. It is clarified at a rate of 240
L/h in a bowl centrifuge, which has an operating volume of
0.09 m3 and a speed of 10,000 rev/min. The bowl has a
radius 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 15,000
rev/min and the minimum particle size that can be removed
at the higher speed.
Solution:
All conditions except the bowl speed remain the same.
2
1
2
2
1
2
ωω
=Q
Q
)]([)]/[ln(18
)(
)/ln(
)(2
1
2
0
10
22
10
22
1
2
0RR
RR
d
RRg
vRRQ sg −
−=
−= l
lπ
µρρωωπ
(To be continued)
Calculate: when ωωωω = 15,000 rev/min, Q = ? d = ?
Solution (cont’d):
)]([)]/[ln(18
)( 21
20
10
22
RRRR
dQ s −
−= lπ
µρρω
2
1
2
2
1
2
ωω
=Q
Q���� 2
2
2
)10000(
)15000(
240=
Q
/sm 105.1s 3600
h
L1000
m L/h540 34
3
2
−×=
=Q����
[Example] Beer with a specific gravity of 1.042 and a viscosity of 1.4 ×××× 10-3 N-s/m2 contains 1.5% solids, which have a density of 1160 kg/m3. It
is clarified at a rate of 240 L/h in a bowl centrifuge, which has an
operating volume of 0.09 m3 and a speed of 10,000 rev/min. The bowl
has a radius of 5.5 cm and is fitted with a 4-cm outlet.
(To be continued)
)]([)]/[ln(18
)( 2
1
2
0
10
22
RRRR
dQ s −
−= lπ
µρρω
1s 157060
215000 -=×
=π
ω
Operating volume 32
1
2
0 m 09.0)]([ =−= RRlπ
���� ]09.0[)4/5.5ln()104.1(18
)10421160()1570(105.1
3
224
−−
×
−=×
d
���� d = 2.14 ×××× 10−−−−7 m
Calculate: when ωωωω = 15,000 rev/min, Q = ? d = ?
Solution (cont’d):
[Example] Beer with a specific gravity of 1.042 and a viscosity of 1.4 ×××× 10-3 N-s/m2 contains 1.5% solids, which have a density of 1160 kg/m3. It
is clarified at a rate of 240 L/h in a bowl centrifuge, which has an
operating volume of 0.09 m3 and a speed of 10,000 rev/min. The bowl
has a radius of 5.5 cm and is fitted with a 4-cm outlet.
#
SEPARATION OF LIQUIDS BY CENTRIFUGATION
���� A common operation in the food and other industries.
* Example: the dairy industry, in which the emulsion of
milk is separated into skim milk and cream.
The differential force across a
thickness dr is:
dF = rωωωω 2dm
ll
r
dFdPrdrdm
ππρ
2 and ])2[( ==
rdrr
rdrrdP 2
2
2
])2[(ρω
ππρω
==−l
l
Integration between r1 and r2:
( )2
1
2
2
2
212
rrPP −=−ρω
SEPARATION OF LIQUIDS BY CENTRIFUGATION (2/3)
( )2
1
2
2
2
212
rrPP −=−ρω
At the liquid-liquid interface,
Pressure exerted by the light
phase of thickness (r2 −−−− r1)
= Pressure exerted by the heavy
phase of thickness (r2 −−−− r4)
( ) ( )2
1
2
2
22
4
2
2
2
22rrrr LH −=−
ωρωρ����
LH
LH rrr
ρρρρ
−
−=
2
1
2
42
2����
* The interface at r2 must be located at a radius smaller
than r3.
SEPARATION OF LIQUIDS BY CENTRIFUGATION (3/3)
[Example] 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 919.5 kg/m3 and that of the aqueous
phase is 980.3 kg/m3. The radius for overflow of the light
liquid has been set at 10.160 mm and the outlet for the
heavy liquid at 10.414 mm. Calculate the location of the
interface in the centrifuge.
Solution:
LH
LH rrr
ρρρρ
−
−=
2
1
2
42
2
���� 5.9193.980
)160.10(5.919)414.10(3.98022
2
2 −−
=r
���� r2 = 13.75 mm
#
DISK CENTRIFUGE
•••• A short, wide bowl 8 to 20 in. in diameter turns on a vertical
axis. Inside the bowl and
rotating with it are closely
spaced “disks”, which are
actually cones of sheet metal
set one above the other.
•••• In operation, feed liquid enters the bowl at the bottom,
flows into the channels, and
upward past the disks.
DISK CENTRIFUGE (2/14)
•••• The operation can be made continuous.
DISK CENTRIFUGE (3/14)
___
____
___
Collection of solid:
•••• A properly operated disc centrifuge should separate 99% of the solids from the liquid stream and produce
an 80−−−−90% wet solids concentrate.
•••• The smaller the particle diameter, the lower the flow rate, and the longer the interval between discharges.
* Flow rate is proportional to the square of the
diameter of the particle.
gd
vvQ sgg )(18
; 2
ρρµ
−=Σ=
* Cell debris (particle size ≈≈≈≈ 0.5 µµµµm) can be separated with flow rates of 300−−−−500 L/h.
DISK CENTRIFUGE (6/14)
•••• In actual operation, the desired separation is achieved by empirically determining:
(a) The flow rate of feed that yields a clarified
supernatant liquid
(b) The time interval between solid discharges that will
minimize liquid loss while still allowing the solids to
flow
���� Discharge periods are on the order of 0.1 s.
DISK CENTRIFUGE (7/14)
Consider a particle located
at position (x, y), where x is
the distance from the edge
of the outer disks along the
gap between the disk, and
y is the distance normal to
the lower disk. Liquid is
fed into the centrifuge so
that it flows upward
through the gap between
the disks, entering at R0
and leaving at R1.
DISK CENTRIFUGE (8/14)
The velocity of the particle in the x direction is:
θω sin0 vvdt
dx−=
where v0 is the convective liquid velocity, and vωωωω is the
particle’s velocity under centrifugation.
DISK CENTRIFUGE (9/14)
There are three important characteristics of v0:
(1) Under most conditions, v0 >> vωωωωsinθθθθ.
(2) v0 is a function of radius.
(3) v0 is a function of y.
)()2(
0 yfrn
Qv
dt
dx
==
lπ
where Q = the total volumetric flow rate
n = number of disks
r = the distance from the axis of rotation
= the distance between disks
f(y) = some function giving the velocity variation
across the distance between disks
l
DISK CENTRIFUGE (10/14)
)()2(
0 yfrn
Qv
dt
dx
==
lπ
Note: Q = (total cross sectional area) ×××× (average velocity)
dyrn
yQfrndyvrnQ ∫∫ ×=×=
ll
lll
ll
00
0)2(
)(1)2(
1)2(
πππ
1)(1
0
=∫l
ldyyf����
DISK CENTRIFUGE (11/14)
The velocity of the particle in the y direction is:
θω
θω coscos2
==
g
rvv
dt
dyg
The trajectory of a particle between the disks of this
centrifuge is:
θωπ
cos )(
2
/
/ 2
2
ryQgf
vn
dtdx
dtdy
dx
dy g
==
l
)()2(
0 yfrn
Qv
dt
dx
==
lπ
DISK CENTRIFUGE (12/14)
θωπ
cos )(
2
/
/ 2
2
ryQgf
vn
dtdx
dtdy
dx
dy g
==
l
θsin0xRr −=
θθωπ
cos)sin( )(
22
0
2
xRyQgf
vn
dx
dy g −
=
l����
θsin0 xRr −=
DISK CENTRIFUGE (13/14)
θsin0 xRr −=
θθωπ
cos)sin( )(
22
0
2
xRyQgf
vn
dx
dy g −
=
l
Integration for those particles that are most difficult to
capture, that is,
At x = 0, y = 0 (The most unfavorable entering position.)
At x = (R0 −−−− R1)/sinθθθθ, y = (They are captured at the wall.) l
)(cot)(3
2 3
1
3
0
2
Σ=
−= gg vRR
g
nvQ θ
ωπ����
1)(1
0
=∫l
ldyyf
DISK CENTRIFUGE (14/14)
[Example] Chlorella cells are being cultivated in an open
pond. We plan to harvest this 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
1.07 ×××× 10-4 cm/s. The centrifuge has 80 discs with an angle of 40°°°°, an outer radius of 15.7 cm, and an inner radius of 6 cm. We plan to operate the centrifuge at 6000 rpm. Estimate the
volumetric capacity Q for this centrifuge.
Solution:
−= θ
ωπcot)(
3
2 3
1
3
0
2
RRg
nvQ g
Data: vg = 1.07 ×××× 10-4 cm/s; n = 80; R0 = 15.7 cm; R1 = 6 cm; θθθθ = 40°°°°; g = 980 cm/s2
rad/s 62860
26000=
×=
πω ���� Q = 3.14 ×××× 104 cm3/s = 31.4 L/s
#
SCALEUP OF CENTRIFUGATION
���� Use laboratory data to predict performance of
commercially available centrifuges.
•••• Commercially available centrifuges are designed on a mechanical basis and cannot be modified easily.
•••• Laboratory bottle centrifuges, being batch operation, give a clear liquid and a concentrated solid or paste.
���� An idealized separation, never reached in a
continuous flow centrifuge.
•••• There are two approaches of scaleup of centrifugation:
(1) use of the equivalent time Gt
(2) sigma analysis
•••• Scaleup of centrifugation based on the equivalent time Gt
Gt: a measurement of the difficulty of a given separation
tg
RGt
2ω=
where R = a characteristic radius, often the maximum
in the centrifuge
t = the time needed for a particle to reach R
* Once the value for Gt is determined, a large-scale
centrifuge that has a similar Gt should be considered.
* This approach must be regarded as only a crude
approximation.
SCALEUP OF CENTRIFUGATION (2/5)
Values of Gt for various solids:
SCALEUP OF CENTRIFUGATION (3/5)
[Example] It has been shown that bacterial cell debris
has Gt = 54 ×××× 106 s. For a centrifuge bowl of 10 cm in diameter, find the centrifuge speed if a full
sedimentation in 2 h is required.
Solution:
tg
RGt
2ω= )36002(
980
)5(1054
26 ××=×
ω����
���� rpm 580,11min
s 60
rad 2
rev 1 rad/s 1212 =
=
πω
#
•••• Scaleup of centrifugation using the ΣΣΣΣ factor (Q = vgΣΣΣΣ)
���� Scaleup involves choosing a centrifuge that has
the required ΣΣΣΣ value to meet the process requirements of vg and Q.
* The value of ΣΣΣΣ is really the area of a gravitational settler that will have the same sedimentation
characteristics as the centrifuge for the same feed
rate.
SCALEUP OF CENTRIFUGATION (4/5)
•••• Scaleup of centrifugation using the ΣΣΣΣ factor (Q = vgΣΣΣΣ)
* Scaleup from a laboratory test of Q1 and ΣΣΣΣ1 to Q2
using similar type and geometry centrifuges:
2
2
1
1
Σ=
Σ
* Scaleup if different centrifuges are used:
22
2
11
1
Σ=
Σ E
Q
E
Q
E is the efficiency of a centrifuge, which is determined
experimentally.
SCALEUP OF CENTRIFUGATION (5/5)
[Example] The old process for recovering starch particles
from a slurry of starch and gluten involved a gravitational
settling procedure in which the slurry was fed to one end of
a table where the starch particles settled and remained in
the table and starch-free liquid was discharged from the
opposite end of the table. We have been asked to evaluate a
process improvement involving the use of continuous
centrifuges. It has been reported that a starch table with
the dimensions of 2 ft wide and 120 ft long can handle a
slurry feed rate of 2 gal/min. The slurry has a viscosity of
10-3 kg/m-s and a density difference of 100 kg/m3. The
centrifuge has a ΣΣΣΣ value of 31,500 m2.
(a) Calculate the effective diameter of the starch particles.
(b) Estimate the centrifuge throughput, assuming that you
can operate at 50% of the theoretical maximum.
(To be continued)
[Example] Recovery of starch particles.
µµµµ = 10-3 kg/m-s; ρρρρs – ρρρρ = 100 kg/m3 (a) Calculate the effective diameter of the starch particles.
Solution:
m/s 1066.5s 60
min
ft 3.28
m
gal 7.48
ft
ft 1202
gal/min 2 63
2
−×=
×
=gv
A starch table with the dimensions of 2 ft wide and 120 ft long can
handle a slurry feed rate of 2 gal/min.
����
gd
v sg )(18
2
ρρµ
−= )8.9)(100()10(18
1066.53
26
−
−
×=×
d����
���� d = 1.02 ×××× 10-5 m (To be continued)
[Example] Recovery of starch particles. The centrifuge has a ΣΣΣΣ value of 31,500 m2. (b) Estimate the centrifuge throughput, assuming that you
can operate at 50% of the theoretical maximum.
Solution (cont’d):
Q at 50% of the theoretical maximum
= vg(0.5ΣΣΣΣ) = (5.66 ×××× 10-6) ×××× (0.5 ×××× 31500)
gal/min 1410min
s 60
L28.32
gal 7.48
m
L1000
s
m 089.0
3
3
=
=
#
m/s 1066.5 6−×=gv
[Example] A new recombinant protein is produced in
yeast. The company scientists, also known as “the boys in
the lab,” separate the cells in a laboratory bottle centrifuge
to give a thick paste that will be subsequently disrupted to
release the protein. This separation is accomplished by
centrifuging small quantities of the broth for 30 min at
2000 rpm. In the lab centrifuge, the inner radius of the
solution is 5 cm and the bottle tip radius is 15 cm. The cell
suspension contains only 7% by volume of cells. We are
asked to recommend the size and type of centrifuge for
separating 10 m3 of this suspension per day.
Solution:
=
g
rv
dt
drg
2ω∫∫ =t
gR
R
dtg
v
r
dr
0
20
1
ω����
t
RRgv
g
tv
R
Rg
g
2
10
2
1
0 )/ln(or ln
ω
ω==���� (To be continued)
Q = vgΣΣΣΣ
[Example] Recommend the size and type of centrifuge for separating
10 m3 of a yeast suspension per day.
t
RRgvg 2
10 )/ln(
ω=
Solution (cont’d):
Data: R1 = 5 cm; R0 = 15 cm; t = 30 min; ωωωω = 2000 rpm
���� vg = 1.36 ×××× 10−−−−5 cm/s
2
5
3
m 851cm/s 101.36
day/m 10=
×==Σ
−gv
Q
* In general, a safety factor of 2 is introduced for disc
centrifuges, while no safety factor is needed for tubular
bowl centrifuges.
#
[Example] We want to centrifuge chlorella cells using an
available disc bowl centrifuge operated at 6000 rpm. The
centrifuge has 80 discs with an angle of 40°°°°, an outer radius of 15.7 cm, and an inner radius of 6 cm. The cell suspension
has a viscosity of 1 cp and a density difference of 0.1 g/cm3.
The effective diameter of chlorella cells is 4.3 ×××× 10−−−−4 cm.
Assume the efficiency of the disc centrifuge is 0.5; estimate
the throughput.
Solution:
cm/s 1001.1)980)(1.0()01.0(18
)103.4()(
18
4242
−−
×=×
=−= gd
vsg
ρρµ
1s 628
s 60
min
rev
2 rev/min 6000
-=
=π
ω
×−= ERR
g
nvQ g θ
ωπcot)(
3
2 3
1
3
0
2
= 14,820 cm3/s
#
s-cm
g :poise
SCROLL TYPE OF DECANTING CENTRIFUGE
Horizontal Type
* An internal scroll conveyor is used to move the decanted
solid out of the machine.
* Centrifugal force: 500−−−−6,000 g
����
* Scroll Decanting
Centrifuge: Vertical
Type (2/2)
ULTRACENTRIFUGE
���� The term “ultracentrifuge” was originally applied by T.
Svedberg to any centrifuge that permitted observation of
the contents of the container during the act of
centrifuging.
���� It is now more commonly applied to any ultrahigh-force
centrifuge (up to 75,000 rpm, with RCF values up to
500,000 ×××× g).
ULTRACENTRIFUGE (2/3)
ULTRACENTRIFUGE (3/3)
SELECTION OF EQUIPMENT FOR LIQUID-
SOLID SEPARATIONS
Major function:
(1) Recover solids
(2) Clarify liquid
Operation mode:
(1) Continuous
(2) Batch, automatic
(3) Batch
Major function Operation Classification
Classification Equipment Subclassification
Major function Operation Classification
Classification Equipment Subclassification
CENTRIFUGAL EXTRACTOR
CENTRIFUGAL EXTRACTOR (2/2)
