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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 practicalfor 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��


)(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


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 ΣΣΣΣis a function only of the particular centrifuge.


* 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.


[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 = ?


)]([)]/[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−−−−7m

Calculate: when ωωωω = 15,000 rev/min, Q = ? 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.


___

____

___

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.


•••• 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.


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 R0and leaving at R1.


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.


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


)()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��


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


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 −=


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


[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:


[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 =

=π

ω

#

•••• 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 using the ΣΣΣΣ factor (Q = vgΣΣΣΣ)

* Scaleup from a laboratory test of Q1 and ΣΣΣΣ1 to Q2using similar type and geometry centrifuges:

2

2

1

1

Σ=

Σ

QQ

* 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.


[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.


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=


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)

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