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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 pasteor a more
concentrated
suspension
More
I ntroduction (2/8)
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Separation costfor recovering whole cells or cell debris:I ntroduction (3/8)
Ultrafiltration
more
economical
Centrifugation
more
economical
Ultrafiltration
Centrifugation
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Schematic presentation ofa laboratory centrifuge:I ntroduction (4/8)
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Care of centrifuges:(1) Avoid imbalancein the rotor, which may be caused by:
a. Tube crackingduring the run
* Conventional glass (Pyrex) centrifuge tubes
withstand only 34000 g.Use centrifuge tubes made from
polypropylene or polycarbonate.
b. Misbalanceof the tubesin the first place
Small tubesbalanced by volume by eye; large
tubes (> 200 mL)should be weighed.
(2) Any spillage should be immediatelyrinsed away.
Avoid corrosion of centrifuge rotors.
(3) Do notuse the machine at top speed constantly.
I ntroduction (5/8)
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________________________________________________________________________
I ntroduction (6/8)
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* Relative Centrifugal Force, RCF=g
r2
1s1047.0
s60
min
min
2rpm1
-
g= 980 cm/s2
r: in cm
)((rpm)10119.1
cm/s980
cm)(rpm
s1047.0rpm)(
RCF 252
21
2
r
r-
-
Often an average RCF is determined using a value for rmidway between the top and bottom of the sample
container.
I ntroduction (7/8)
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ravg= 7 cm
20,000 rpm RCF = 31,000(centrifugal force = 31,000 g)
I ntroduction (8/8)
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FORCES DEVELOPED IN CENTRIFUGAL
SEPARATION
The acceleration from a centrifugal force: a= 2rwhere = angular velocity, rad/s
r= radial distance from center of rotation
Settling by gravity force: gd
v sg )(18
2
-
Settling in centrifuges: r
d
v s2
2
)(18 -
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Gravitational sedimentation istoo slow to be practicalfor bacteria, and conventional centrifugation is too slow
for protein macromolecules.
FORCES DEVELOPED IN CENTRIFUGAL SEPARATION (2/3)
__________
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[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 mand a density of 1.05 g/cm3. The fluid hasphysical 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
- td
r
rs
22
1
2 )(18
ln
-or
(To be conti nued)
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Example: laboratory bottle centr ifuge
Solution (contd):
td
r
rs
22
1
2 )(18
ln
-
t= 0, r= 3 cm; t= ?, r= 10 cm
Data: d= 8.0 m = 8.0
10-4
cm; = 1 cP = 0.01 g/cm-s; s= 1.05 g/cm3; = 1.0 g/cm3;rad/s3.52
60
2500rpm500
t-
-2
24
)3.52()105.1(01.018
)100.8(
3
10ln
t= 2467 s = 41.3 min#
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* Sedimentation Coefficient, s
The velocity of a particle through a viscous medium
is usuallyproportional to the accelerating field.
rd
v s2
2
)(18
- )(18
2
- sd
s
Unit of s:svedberg(S; 1 S = 10
13second)
Svedberg: the inventorof ultracentrifuge
rsv 2
FORCES DEVELOPED IN CENTRI FUGAL SEPARATION (3/3)
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[Example] Estimate the time it would take to completely
clarify a suspension of 70 Sribosomes 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
h8.1s29080
s60
min1
rev
2
min
rev10000s1070
223.045ln1 2
13
2
- st
#
rsdt
drv 2
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TUBULAR BOWL CENTRIFUGE (3/6)
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The movement of the particle in the z
direction(due to the convection of the
feed flow):
)( 212
0 RR
Q
dt
dz
-
where Q= the volumetric flow rateR1= the distance of liquid interface
from the axis of rotation
The movement of the particle in the rdirection:
rd
dt
drs
22
)(18
-
gd
v sg )(
18
2
-
g
rv
dt
drg
2
TUBULAR BOWL CENTRIFUGE (3/6)
TUBULAR BOWL CENTRIFUGE (4/6)
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)( 212
0 RR
Q
dt
dz
-
g
rvdt
drg
2
;
The trajectory of a particle in the
centrifuge:
Q
RR
g
rv
dtdz
dtdr
dz
drg
)(
/
/2
1
2
02 -
Consider a particle enters the centrifuge at R1(that is,
at z= 0, r= R1)anddo not reach R0until at z =
Q
RR
g
v
R
R g )(ln
2
1
2
0
2
1
0 -
)/ln(
)(
10
22
1
2
0
RRg
vRRQ
g -
or
TUBULAR BOWL CENTRIFUGE (4/6)
TUBULAR BOWL CENTRIFUGE (5/6)
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For R0and R1being approximately equal,
)/ln(
)(
10
22
1
2
0
RRg
vRRQ
g -
2101
110
1010
110
1010
10
2
1
2
0 2)(/)(
))((
]/)(1ln[
))((
)/ln(RRRR
RRR
RRRR
RRR
RRRR
RR
RR
-
--
-
-
-- 4324
1
3
1
2
1)1ln( xxxxxNote:
)(222
gg v
gRvQ
Note: vgis a function only of the particles themselves,and is a function only of the particular centrifuge.
TUBULAR BOWL CENTRIFUGE (5/6)
TUBULAR BOWL CENTRIFUGE (6/6)
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* Continuous tubular bowl
centrifugefor separation of
two liquids:
An internal baffleprovides a
separate passage adjacent to
the bowl wall to conduct the
heavier-phase liquid to a
different discharge elevation.
TUBULAR BOWL CENTRIFUGE (6/6)
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[Example] A bowl centrifuge is used to concentrate a
suspension of Escher ichia coli prior to cell disruption. The
bowl of this unit has an inside radius of 12.7 cmand alength of 73.0 cm. The speed of the bowl is 16,000 rpmand
the volumetric capacity is 200 L/h. Under these conditions,
this centrifuge works well. (a) Calculate the settling
velocity vgfor 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
R
Qgv
g
RvQ gg
(To be conti nued)
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[Example] Analysis of bowl centrifuge
222
R
Qgvg
Solution:
Data: R =12.7 cm; = 73 cm; = 16,000 rpm = 1674.7 rad/s;Q= 200 L/h = 55.56 cm3/s; g= 980 cm/s2
vg= 2.63 10-7cm/s
Using the same centrifuge,1
2
1
2
1
2
g
g
g
g
vv
vv
gd
v sg )(18
2
-16
1
4
)2/1(
/
/
/
/ 2
12
2
1
2
2
1
2
1
2
2
2
1
2
dd
d
d
Q
Q
(a) Calculate the settling velocity vgfor the cells.
(b) Estimate the volumetric capacity of this same centrifuge for celldebris.
#
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[Example] Beer with a specific gravity of 1.042and a
viscosity of 1.4 10-3N-s/m2contains 1.5% solids, whichhave a density of 1160 kg/m3. It is clarified at a rate of 240
L/hin a bowl centrifuge, which has an operating volume of0.09 m3and a speed of 10,000 rev/min. The bowl has a
radius of 5.5 cmand is fitted with a 4-cm outlet. Calculate
the effect on feed rate of an increase in bowl speed to 15,000
rev/min andthe minimum particle size that can be removed
at the higher speed.
Solution:
All conditions except the bowl speedremain the same.
2
1
2
2
1
2
Q
Q
)]([)]/[ln(18
)(
)/ln(
)(2
1
2
010
22
10
22
1
2
0RR
RR
d
RRg
vRRQ s
g --
-
(To be conti nued)
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Calculate: when = 15,000 rev/min, Q= ? d= ?Solution (contd):
)]([)]/[ln(18
)( 21
2
0
10
22
RRRR
dQ s -
-
2
1
2
2
1
2
Q
Q 2
22
)10000(
)15000(
240
Q
/sm105.1s3600
h
L1000
mL/h540 34
3
2
-
Q
[Example] Beer with a specific gravity of 1.042and a viscosity of 1.4 10-3N-s/m2contains 1.5% solids, which have a density of 1160 kg/m3. It
is clarified at a rate of 240 L/hin a bowl centrifuge, which has an
operating volume of 0.09 m3and a speed of 10,000 rev/min. The bowlhas a radius of 5.5 cmand is fitted with a 4-cm outlet.
(To be conti nued)
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)]([
)]/[ln(18
)( 21
2
0
10
22
RR
RR
dQ s -
-
1s157060
215000 -
Operating volume32
1
2
0 m09.0)]([ - RR
]09.0[)4/5.5ln()104.1(18
)10421160()1570(105.1
3
224
-
-
-
d
d= 2.14 107m
Calculate: when = 15,000 rev/min, Q= ? d= ?Solution (contd):
[Example] Beer with a specific gravity of 1.042and a viscosity of 1.4 10-3N-s/m2contains 1.5% solids, which have a density of 1160 kg/m3. It
is clarified at a rate of 240 L/hin a bowl centrifuge, which has an
operating volume of 0.09 m3and a speed of 10,000 rev/min. The bowl
has a radius of 5.5 cmand is fitted with a 4-cm outlet.
#
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SEPARATION OF L IQUIDS BY CENTRI FUGATION (2/3)
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The differential force across a
thickness dris:dF= r2dm
r
dFdPrdrdm
2and])2[(
rdrr
rdrrdP
22
2
])2[(
-
Integration between r1and r2:
21222
212
rrPP --
SEPARATION OF L IQUIDS BY CENTRI FUGATION (2/3)
SEPARATION OF L IQUIDS BY CENTRI FUGATION (3/3)
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21222
212
rrPP --
At the liquid-liquid interface,
Pressure exerted by the light
phase of thickness (r2 r1)
= Pressure exerted by the heavyphase of thickness (r2 r4)
21222
2
4
2
2
2
22rrrr LH --
LH
LH rrr
-
-
21
242
2
* The interface at r2must be located at a radius smaller
than r3.
SEPARATION OF L IQUIDS BY CENTRI FUGATION (3/3)
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[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/m3and that of the aqueous
phase is 980.3 kg/m3. The radius for overflow of the light
liquid has been set at 10.160 mmand the outlet for the
heavy liquid at 10.414 mm. Calculate the location of the
interface in the centrifuge.
Solution:
LH
LH rrr
-
-
2
1
2
42
2
5.9193.980
)160.10(5.919)414.10(3.980 2222
-
-
r
r2= 13.75 mm#
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DISK CENTRIFUGE
DI SK CENTRIFUGE (2/14)
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A short, wide bowl8 to 20 in.in diameter turns on a verticalaxis. 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 liquidenters the bowl at the bottom,
flows into the channels, and
upward past the disks.
DI SK CENTRIFUGE (2/14)
DI SK CENTRIFUGE (3/14)
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The operation can be made continuous.DI SK CENTRIFUGE (3/14)
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___
____
___
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Collection of solid:
DI SK CENTRIFUGE (6/14)
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A properly operated disc centrifuge should separate99% of the solids from the liquid streamand produce
an 8090% 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 thediameter of the particle.
gd
vvQ sgg )(18
;2
-
* Cell debris (particle size 0.5 m)can beseparated with flow rates of 300500 L/h.
( )
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DI SK CENTRIFUGE (8/14)
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Consider a particle located
at position (x, y), where xis
the distance from the edge
of the outer disks along the
gapbetween the disk, and
yis the distance normal to
the lower disk. Liquid isfed into the centrifuge so
that it flows upward
through the gap between
the disks, entering at R0and leaving at R1.
( )
DI SK CENTRIFUGE (9/14)
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The velocity of the particle in the xdirection is:
sin0 vvdtdx -
where v0is the convective liquid velocity, and vis theparticles velocity under centrifugation.
( )
DISK CENTRIFUGE (10/14)
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There are three important characteristics of v0:
(1) Under most conditions, v0>> vsin.(2) v0is a function of radius.
(3) v0is a function of y.
)()2(
0 yfrn
Qv
dt
dx
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
DI SK CENTRIFUGE (11/14)
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)()2(
0 yfrn
Qv
dt
dx
Note: Q= (total cross sectional area)(average velocity)
dy
rn
yQfrndyvrnQ
00
0
)2(
)(1)2(
1)2(
1)(1
0
dyyf
DISK CENTRIFUGE (12/14)
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The velocity of the particle in the ydirection is:
coscos2
g
rvv
dt
dyg
The trajectory of a particlebetween the disks of this
centrifuge is:
cos)(
2
/
/ 22
ryQgf
vn
dtdx
dtdy
dx
dy g
)()2(
0 yfrn
Qvdtdx
DISK CENTRIFUGE (13/14)
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cos)(
2
/
/ 22
ryQgf
vn
dtdx
dtdy
dx
dy g
sin0 xRr -
cos)sin()(
2 20
2
xRyQgf
vndxdy g -
sin0 xRr -
DISK CENTRIFUGE (14/14)
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sin0 xRr -
cos)sin()(
220
2
xRyQgf
vn
dx
dy g-
Integration for those particlesthat are most difficult to
capture,that is,
At x= 0, y= 0 (The most unfavorable entering position.)
At x= (R0 R1)/sin, y= (They are captured at the wall.)
)(cot)(3
2 31
3
0
2
- gg vRR
g
nvQ
1)(1
0
dyyf
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[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 vgfor these cells has been measured as1.07 10-4cm/s. The centrifuge has 80 discswith an angle of40, 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 capacityQ
for this centrifuge.
Solution:
-
cot)(
3
2 31
3
0
2
RRg
nvQ g
Data: vg= 1.07 10-4cm/s; n= 80; R0= 15.7 cm; R1= 6 cm; = 40; g=980 cm/s2
rad/s62860
26000
Q= 3.14 104cm3/s = 31.4 L/s
#
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SCALEUP OF CENTRIFUGATION
Use laboratory data to predict performance of
commercially available centrifuges.
Commercially available centrifuges are designed on amechanical basisand 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 (2/5)
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Scaleup of centrifugation based on the equivalent time GtGt: a measurement of the difficulty of a given separation
tg
RGt
2
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 Gtis determined, a large-scale
centrifuge that has a similar Gtshould be considered.
* This approach must be regarded as only a crude
approximation.
SCALEUP OF CENTRIFUGATION (3/5)
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Values of Gtfor various solids:
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[Example] It has been shown that bacterial cell debris
has Gt= 54 106s. For a centrifuge bowl of 10 cmindiameter, find the centrifuge speedif a fullsedimentation in 2 his required.
Solution:
tg
RGt
2
)36002(980
)5(
1054
26
rpm580,11min
s60
rad2
rev1rad/s1212
#
SCALEUP OF CENTRIFUGATION (4/5)
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Scaleup of centrifugation using the factor (Q= vg)Scaleup involves choosing a centrifuge that hasthe required valueto meet the processrequirements of vgand Q.
* The value of is really the area of a gravitationalsettlerthat will have the same sedimentation
characteristics as the centrifuge for the same feed
rate.
SCALEUP OF CENTRIFUGATION (5/5)
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Scaleup of centrifugation using the factor (Q= vg)
* Scaleup from a laboratory test of Q1and 1to Q2using similar type and geometry centrifuges:
2
2
1
1
* Scaleup if different centrifuges are used:
22
2
11
1
E
Q
E
Q
Eis the efficiency of a centrifuge, which is determined
experimentally.
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[Example] Recovery of starch particles.
= 10-3kg/m-s; s= 100 kg/m3(a) Calculate the effective diameter of the starch particles.
Solution:
m/s1066.5s60
min
ft3.28
m
gal7.48
ft
ft1202
gal/min2 63
2
-
gv
A starch table with the dimensions of 2 ft wide and120 ft longcan
handle a slurry feed rate of 2 gal/min.
gdv sg )(18
2
- )8.9)(100()10(18
1066.53
26
--
d
d= 1.02 10-5m(To be continued)
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[Example] Recovery of starch particles. The centrifuge has a value of31,500 m2. (b) Estimate the centrifuge throughput, assuming that you
can operate at 50% of the theoretical maximum.
Solution (contd):
Qat 50% of the theoretical maximum
= vg(0.5) = (5.66 10-6) (0.5 31500)
gal/min1410min
s60
L28.32
gal7.48
m
L1000
s
m089.0
3
3
#
m/s1066.5 6-gv
[Example] A new recombinant protein is produced in
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[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 minat
2000 rpm. In the lab centrifuge, the inner radius of the
solution is 5 cmand the bottle tip radius is 15 cm. The cell
suspension contains only 7% by volume of cells. We areasked to recommend the size and type of centrifuge for
separating 10 m3 of this suspension per day.
Solution:
g
rv
dt
drg
2 t
gR
R
dtg
v
r
dr
0
20
1
t
RRgv
g
tv
R
Rg
g
2
10
2
1
0 )/ln(orln
(To be continued)
Q= vg
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[Example] Recommend the size and type of centrifuge for separating
10 m3 of a yeast suspension per day.
t
RRgvg 2
10 )/ln(
Solution (contd):
Data: R1= 5 cm; R0= 15 cm; t= 30 min; = 2000 rpmvg= 1.36 105cm/s
2
5
3
m851
cm/s101.36
day/m10
-
gv
Q
* In general, a safety factor of 2is introduced for disc
centrifuges, while no safety factor is neededfor tubular
bowl centrifuges.
#
[Example] We want to centrifuge chlorella cells using an
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[Example] We want to centrifuge chlorella cells using an
available disc bowl centrifuge operated at 6000 rpm. The
centrifuge has 80 discswith an angle of 40, an outer radiusof 15.7 cm, and an inner radius of 6 cm. The cell suspensionhas a viscosity of 1 cpand a density difference of 0.1 g/cm3.
The effective diameter of chlorella cells is 4.3 104cm.Assume the efficiency of the disc centrifuge is 0.5; estimate
the throughput.
Solution:
cm/s1001.1)980)(1.0()01.0(18
)103.4()(
18
4242
--
- gd
v sg
1s628s60
min
rev
2rev/min6000 -
- ERR
g
nvQ g
cot)(
3
2 31
3
0
2
= 14,820 cm3/s
#
s-cm
g:poise
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SCROLL TYPE OF DECANTING CENTRIFUGE
Horizontal Type
* An internal scroll conveyoris used to move the decanted
solid out of the machine.
* Centrifugal force: 5006,000 g
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* Scroll Decanting
Centrifuge: Vertical
Type (2/2)
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ULTRACENTRIFUGE
The term ultracentrifuge was originally applied by T.Svedbergto 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-forcecentrifuge(up to 75,000 rpm, with RCF values up to
500,000 g).
ULTRACENTRIFUGE (2/3)
8/13/2019 BS 5 Centrifugation
60/69
ULTRACENTRIFUGE (3/3)
8/13/2019 BS 5 Centrifugation
61/69
8/13/2019 BS 5 Centrifugation
62/69
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
8/13/2019 BS 5 Centrifugation
63/69
Classification Equipment Subclassification
8/13/2019 BS 5 Centrifugation
64/69
Major function Operation Classification
8/13/2019 BS 5 Centrifugation
65/69
Classification Equipment Subclassification
8/13/2019 BS 5 Centrifugation
66/69
CENTRIFUGAL EXTRACTOR
8/13/2019 BS 5 Centrifugation
67/69
CENTRIFUGAL EXTRACTOR
CENTRIFUGAL EXTRACTOR (2/2)
8/13/2019 BS 5 Centrifugation
68/69
8/13/2019 BS 5 Centrifugation
69/69