filteration

FILTRATION Removal of solid particles from a fluid by passing the

fluid through a filtering medium, or septum.

MECHANSIMS OF FILTRATION

(a) Clarifiers

* Also known as “deep-bed filters”.

* The particles of solid are trapped inside the filter medium.

* A typical cartridge filter:

MECHANSIMS OF FILTRATION (2/3)

(b) Cake filters

* The filter medium is relatively thin, compared with that of a clarifying filter.

* After the initial period, the cake of solids does the filtration, not the septum.

* A visible cake of appreciable thickness builds up on the surface and must be periodically removed.

MECHANSIMS OF FILTRATION (3/3)

EQUIPMENT FOR CONVENTIONAL FILTRATION

(1) Plate and Frame Filter Press

* The most common type, but less common for bioseparations.

* Used where a relatively dry cake discharge is desired.

* Cake removal: open the whole assembly

Should not be used where there are toxic fumes or biohazards.

(2) Horizontal Plate Filter:

* Filtration occurs from the top of each plate.

* Cake removal: removed with a sluicing nozzle or discharged by rapidly rotating the leaves.

EQUIPMENT FOR CONVENTIONAL FILTRATION (2/7)

(3) Vertical Leaf Filter and Candle Type Vertical Tank Filter:


(3) Vertical Leaf Filter and Candle Type Vertical Tank Filter (2/2):

* Have a relatively high filtration area per volume.

Require only a small floor area.

* Filter cake is formed on the external surface of the tubes.

* The tubes are cleaned by backwashing.


(4) Rotary Vacuum Filter:

* Rotate at a low speed during the operation.

* Pressure inside the drum is a partial vacuum.

Liquid is sucked through the filter cloth and solids are retained on the surface of the drum.

* Three chief steps of the filtration cycle:

(1) cake formation

(2) cake washing (to remove either valuable or unwanted solutes)

(3) cake discharge


* The workhorse of bioseparations.

* Common for large-scale operations whenever the solids are difficult to filter.

* Being automated.

Have a lower labor cost.


PRETREATMENT OF FILTRATION

“Filtration is a straightforward procedure for well-defined crystals.”

* Fermentation beers and other biological solutions are notoriously hard to filter, because of: (1) high, non-newtonian viscosity, and (2) highly compressible filter cakes. Conventional filtration is often too slow to be practical.

The filtration requires pretreatment: heating, coagulation and flocculation, or adsorption on filter aid.

A. Heating

* To improve the feed’s handling characteristics. (Thinking of filtering a dilution solution of egg white.)

* The simplest pretreatment (and the least expensive).

* Chief constraint: thermal stability of the product.

PRETREATMENT OF FILTRATION (2/12)

B. Coagulation and Flocculation

* Through the addition of electrolytes.

* Types of coagulants:

(1) Simple electrolytes (such as ferric chloride, alum, or acids and bases)

(2) Synthetic polyelectrolytes


coagulation flocculation

* Action of simple electrolytes: reduce the electrostatic repulsion existing between colloidal particles.

* Action of synthetic polyelectrolytes:

(1) Reduce electrostatic repulsion

(2) Adsorb on adjacent particles

* Commercially available polyelectrolytes (can be anionic, cationic, or nonionic): polyacrylamides, polyethylenimines, and polyamine derivatives.


* The effect of pH on filtrate volume for Streptomyces griseus:


C. Adsorption on Filter Aids

* Why filter-aid filtration?

Two major problems can be reduced:

(1) High compressibility of the accumulated biomass

(2) Penetration of small particles into the filter medium

Lengthen the filtration cycle; improve the quality of the filtered liquor.


* The effect of filter aid on filtrate volume for Streptomyces griseus:


* The effect of pH and filter aid on filtrate volume for Streptomyces griseus:


* How does the filter-aid help?

(1) Give porosity to the filter cake.

Solids to be filtered Porosity

Hard spheres of the same size ≈ 0.45

General cases 0.2−0.3

Compressible solids ≈ 0

Diatomaceous silica ( 藻土矽 ) ≈ 0.9

(2) Create a very large surface to trap the gelatinous precipitate.

Allow much more filtrate to be obtained before eventually clogging up.


- Protect the filter medium from fouling.

- Provide a finer matrix to exclude particles from the filtrate.

* How to use filter-aid?

(1) Precoat—a thin layer (0.1 to 0.2 lb/ft2) of filter aid is deposited on the filter medium prior to introducing the filter feed to the system

(2) Body feed—add the filter aid to the filter feed


-------------

____________________


* The use of filter-aid is mainly for removing small amounts of unwanted particulate material.

It cannot deal with large quantities of precipitate successfully.

* Types of filter-aid (the most effective):

(1) Diatomaceous earths such as Celite (consisting mainly of SiO2)

(2) Perlites (volcanic rock processed to yield an expanded form)

Note: some products like the aminoglycoside antibiotics may irreversibly bind to diatomaceous earth.


GENERAL THEORY FOR FILTRATION

Darcy’s law—relate the flow rate through a porous bed of solids to the pressure drop causing that flow.

µPk

v∆=

v = velocity of the liquid

∆P = pressure drop across the bed of thickness ℓ

∆P/ ℓ = pressure gradient

µ = viscosity of the liquid

k = permeability of the bed, a proportionality constant (dimension: L2)

* Like Ohm’s law, ℓ/k is the resistance of filtration.

Strictly speaking, Darcy’s law holds only when

5)1(

<−εµ

ρvd

where d is the particle size of the filter cake, ρ is the liquid density, and ε is the void fraction in the cake.

* Biological separations almost always obey this inequality.

For a batch filtration,

dt

dV

Av

1=µPk

dt

dV

A

∆=1

where V is the total volume of filtrate, A is the filter area, and t is the time.

GENERAL THEORY FOR FILTRATION (2/5)

µPk

v∆=Darcy’s law:

Two contributions to the filtration resistance:

CM RRk

+=

where RM is the resistance of the filter medium (constant), and RC is the resistance of the cake (varies with V).

The basic differential equation for filtration at constant pressure drop can thus be obtained as:

)(

1

CM RR

P

dt

dV

A +∆=

µµPk

dt

dV

A

∆=1


Incompressible Cakes

α = specific cake resistance, cm/g

ρ0 = mass of cake solids per volume of filtrate

)(

1

CM RR

P

dt

dV

A +∆=

µ

])/([

1

0 MRAV

P

dt

dV

A +∆=

αρµ (I.C.: t = 0, V = 0)

=

A

VRC 0αρ

BA

VK

P

R

A

V

PV

At M +

=

∆+

∆=

µµαρ2

0


Plot

V

Atversus

A

V Slope =

PK

∆=

20µαρ

Known µ , ρ 0, ∆P α can be determined.

* Often, the medium resistance RM is insignificant, B = 0.

20

2

∆=

A

V

Pt

µαρ

BA

VK

P

R

A

V

PV

At M +

=

∆+

∆= µµαρ

20


[Example] A suspension containing 225 g of carbonyl iron powder, Grade E, per liter of a solution of 0.01 N NaOH is to be filtered, using a leaf filter. Estimate the size (area) of the filter needed to obtain 100 lb of dry cake in 1 h of filtration at a constant pressure drop of 20 psi. The cake is incompressible. The specific cake resistance is 1011 ft/lb. The resistance of the medium is taken as 0.1 in−1.

P

R

A

V

PV

At M

∆+

∆=

µµαρ2

0

Solution:

330 lb/ft 0.14

g 453.6

lb

ft

L32.28

L

g 225

filtrate of volume

solid cake of mass =

==ρ

33

ft 1.7lb/ft 14.0

lb 100filtrate of volume ===V

(To be continued)

[Example] A suspension containing 225 g of carbonyl iron powder, Grade E, per liter of a solution of 0.01 N NaOH is to be filtered, using a leaf filter. Estimate the size (area) of the filter needed to obtain 100 lb of dry cake in 1 h of filtration at a constant pressure drop of 20 psi. The cake is incompressible. The specific cake resistance is 1011 ft/lb. The resistance of the medium is taken as 0.1 in-1.

P

R

A

V

PV

At M

∆+

∆=

µµαρ2

0Solution (cont’d):

t = filtration time = 1 h

×=∆2

22

2f

2f

3

h

s )3600(

s-lb

ft-lb2.32

psi 14.7

/ftlb 10116.2 psi 20P

= 1.2 × 1012 lb/ft-h2

α = specific cake resistance = 1011 ft/lb

RM = resistance of the medium = 0.1 in−1 = 1.2 ft−1

µ = viscosity of the liquid = 1 cp = 2.42 lb/ft-h (assumed)

(To be continued)

Solution (cont’d):

P

R

A

V

PV

At M

∆+

∆=

µµαρ2

0

1212

11

102.1

)2.1)(42.2(1.7

)102.1(2

)0.14)(10)(42.2(

1.7

)1(

×+

×=

A

A

A2 − 1.7 × 10-11A − 71.2 = 0

A = 8.4 ft2

#

[Example] A suspension containing 225 g of carbonyl iron powder, Grade E, per liter of a solution of 0.01 N NaOH is to be filtered, using a leaf filter. Estimate the size (area) of the filter needed to obtain 100 lb of dry cake in 1 h of filtration at a constant pressure drop of 20 psi. The cake is incompressible. The specific cake resistance is 1011 ft/lb. The resistance of the medium is taken as 0.1 in-1.

[Example] Streptomyces Filtration from an Erythromycin Broth. Using a test filter, we find the following data for a broth containing the antibiotic erythromycin and added filter aid:

The filter leaf has a total area of 0.1 ft2 and the filtrate has a viscosity of 1.1 cp. The pressure drop is 20 in. of mercury and the feed contains 0.015 kg dry cake per liter. Determine the specific cake resistance α and the medium resistance RM.

P

R

A

V

PV

At M

∆+

∆=

µµαρ2

0Solution:

(To be continued)

Example: Streptomyces Filtration from an Erythromycin Broth (cont’d)

Example: Streptomyces Filtration from an Erythromycin Broth (cont’d)

#

P

R

A

V

PV

At M

∆+

∆=

µµαρ2

0

[Example] We have filtered a slurry of sitosterol at constant pressure through a filtration medium consisting of a screen support mounted across the end of a Pyrex pipe. We find that the resistance of the filtration medium is negligible. We also find the following data in a laboratory test:

On the basis of this laboratory test, predict the number of frames (30 in × 30 in × 1 in thick) needed for a plate-and-frame press. Estimate the time required for filtering a 63 kg batch of steroid. In these calculations, assume that the feed pump will deliver 10 psi and that the filtrate from the press must be raised against the equivalent of 15 ft head.

(To be continued)

Example: filtering a slurry of sitosterol


(a) Predict the number of frames needed

33

g/cm 245.0cm 3.253

g 62 density Cake ==

Cake volume of 63 kg steroid = 35

3

3

cm 1057.2g/cm 0.245

g 1063 ×=×

Number of frames needed = 4.17cm 2.54

in

in 13030

cm 1057.23

3

35

=

×××

18 frames are needed.

(To be continued)

(b) Time required for filtering a 63 kg batch of steroid


For incompressible cake with a negligible filter medium resistance,

2

0

0

2

0 1

2or

2

∆=

∆=

A

V

Pt

A

V

Pt

ρρ

µαµαρ

In the laboratory test:

2

20 cm) 08.5(4

g 62

psi) 15( 2 min 163

=πρ

µα2

4

0 g

cm-psi-min 261

2=

ρµα


(To be continued)

(b) Time required for filtering a 63 kg batch of steroid (cont’d)

Solution:

In the large-scale operation:

252

2 cm 1009.2in

cm 54.2 in )3030(218 ×=

×××=A

psi 5.3(water) headft 33.9

psi 7.14 headft 15 psi 10 =

−=∆P

min 8.61009.2

000,63

5.3

1261

1

2

2

5

2

0

0

=

×××=

∆=

A

V

Pt

ρρ

µα


#

In the laboratory test: 2

4

0 g

cm-psi-min 261

2=

ρµα

Compressible Cakes

“Almost all cakes formed of biological materials are compressible. As these cakes compress, filtration

rates drop.”

To estimate the effects of compressibility, we assume that the cake resistance α is a function of the pressure drop.

sP)(' ∆= αα

where α’ = a constant related largely to the size and shape of the particles forming the cake

s = the cake compressibility

GENERAL THEORY FOR FILTRATION: Compressible Cakes (1/3)

=A

VRC 0αρRecall:

sP)(' ∆= αα Ps ∆+= log'loglog αα

Plot logα versus log∆P, slope = s, intercept = logα’.


sP)(' ∆= αα

• For a rigid, incompressible cake, s = 0.

• For a highly compressible cake, s ≈ 1.

• In practice, s ranges from 0.1−0.8.

• When values of s are high, one should consider pretreating the feed with filter aids.


=A

VRC 0αρRecall:

[Example] Filtration of Beer Containing Protease. We have a suspension of Bacillus subtilis fermented to produce the enzyme protease. To separate the biomass, we have added 1.3 times the biomass of a Celatom filter aid, yielding a beer containing 3.6 wt% solid, with a viscosity of 6.6 cp. With a Buchner funnel 5 cm in diameter attached to an aspirator, we have found that we can filter 100 cm3 of this beer in 24 min. However, previous studies with this type of beer have had a compressible cake with s equal to 2/3.

We now need to filter 3000 L of this material in a pilot plant’s plate-and-frame press. This press has 15 frames, each of area 3520 cm2. The spacing between these frames can be made large, so that we can filter all the beer in one single run. The resistance of the filter medium is much smaller than the filter cake, and the total pressure drop that can be used is 65 psi. How long will it take to filter this beer at 50 psi?

(To be continued)

Example: Filtration of Beer Containing Protease


Negligible RM 2

0

2

∆=

A

V

Pt

µαρ

Compressible cake, sP)(' ∆= αα2

1

0

2

'

∆= − A

V

Pt

s

ρµα

Laboratory test:

∆P = 14.7 psi (a Buchner funnel attached to an aspirator)

A = ;V = 100 cm3; t = 24 min; s = 2/3 2)cm 5(4

π

2

2

3

3/1

0

)cm 5(4

cm 100

)psi 7.14(2

' min 24

=π

ρµα

µα’ρ0 = 4.53 min psi1/3 cm−2

(To be continued)

2

1

0

2

'

∆= − A

V

Pt

s

ρµα µα’ρ 0 = 4.53 min psi1/3 cm−2

Pilot-plant operation:

V = 3000 L = 3 × 106 cm3

A = 15 × 2 × 3520 cm2 (Filtration occurs on both sides of the frame.)

26

3/1

2

10

3520215

103

)50(2

53.4

2

'

××

×=

∆= − A

V

Pt

s

ρµα

;

= 496 min = 8.3 h

Example: Filtration of Beer Containing Protease


#

ANALYSIS OF CONTINUOUS ROTARY VACUUM FILTERS

There are three stages involved in the operation:

(1) cake formation

(2) cake washing

(3) cake discharge(not affecting the filter size and the cycle time)

Cake Formation

For compressible cake and negligible medium resistance,2

10

2

10

2

'or

2

'

∆

=

∆= −− A

V

Pt

A

V

Pt f

sfs

ρµαρµα

where tf = cake formation time

Vf = volume of filtrate collected during the period of tf

A = filtration area (submerged area of filter)

ANALYSIS OF CONTINUOUS ROTARY VACUUM FILTERS (2/8)

Cake Formation (cont’d)

2

10

2

'

∆

= − A

V

Pt f

sf

ρµα

Let tf = β tc and A = βAT 2

1

0

2

'

∆

= −T

f

sc A

V

Pt

βρµα

β

where tc = cycle time

AT = total filter area

β = fraction of the drum submerged


Cake Washing

Two factors involved in the stage of cake washing:

(1) The fraction of soluble material remained after the wash

Governing the volume of wash liquid required.

(2) The rate of wash liquid passes through the cake

Controlling the fraction of cycle time for cake washing.


An empirical equation for the fraction of soluble material remained:

nr )1( ε−=

where r = ratio of soluble material remained after the wash to that originally present in the caken = volume of wash liquid divided by the volume of retained liquidε = washing efficiency of the cake


(1) The fraction of soluble material remained after the wash

Governing the volume of wash liquid required.


The wash liquid contains no additional solids.

(1) The cake thickness is constant.


(2) The rate of wash liquid passes through the cake

Controlling the fraction of cycle time for cake washing.

(2) Wash rate

= filtration rate at the end of cake formation

The flow of wash liquid is constant.


w

ww

At

V

dt

dV

A== 1

rate Wash

where Vw = volume of wash water required, and tw = time required for washing.

Filtration rate at the end of cake formation = ftt

dt

dV

A =

1

2/1

0

12

10

'

)(2or

2

'

∆=

∆=

−

− ρµαρµα tP

A

V

A

V

Pt

s

s

2/1

0

1

'2

)(1 rate Wash

∆=

==

−

== f

s

tttt t

P

A

V

dt

d

dt

dV

Aff

ρµα

2/1

0

1

'2

)(

∆=∴−

f

s

w

w

t

P

At

V

ρµα


A useful expression:

2/1

0

12/1

0

1

'

)(2 and

'2

)(

∆=

∆=−−

f

s

f

f

f

s

w

w

t

P

At

V

t

P

At

V

ρµαρµα

nfV

V

V

V

V

V

t

t

f

r

r

w

f

w

f

w 222 ===

2/1

1

0

2/1

1

0

)(2

' and

)(

'2

∆

=

∆

= −− s

fffs

fww P

t

A

Vt

P

t

A

Vt

ρµαρµα

where Vr = volume of liquid retained

f = ratio of the volume of retained liquid (Vr) to the volume of filtrate (Vf)


[Example] It is desired to filter a cell broth at a rate of 2000 L/h on a rotary vacuum filter at a vacuum pressure of 70 kPa. The cycle time for the drum is 60 s, and the cake formation time is 15 s. The broth to be filtered has a viscosity of 2.0 cp and a cake solid per volume of filtrate of 10 g/L. From laboratory tests, the specific cake resistance has been determined to be 9 × 1010 cm/g. Determine the area of the filter that is required.

Solution:

For incompressible cake,

Pt

VA

A

V

Pt

f

fff ∆

=

∆

=2

or 2

2

02

2

0µαρµαρ

(To be continued)

Example: Determine the area of a rotary vacuum filter


s-cm

g 0.02 cp 2 ==µ

g

cm 109 10×=α 3

30 cm

g 1010

L

g 10 −×==ρ; ;

33

3 cm 8333s 3600

h s) 15(

h

cm 102000 =

×

×=fV

25

223

s-cm

g 100.7

cm 100

m

kg

g 1000

s-N

m-kg

m

N 1070 kPa 70 ×=

×==∆P

475

2310202 cm 1095.5

)100.7)(15(2

)8333)(1010)(109)(02.0(

2×=

×××=

∆=

−

Pt

VA

f

fµαρ

A = 7715 cm2 = 0.7715 m2

2T m 09.3

15

607715.0 =×=×=

f

c

t

tAA

#

[Example] We want to filter 15,000 L/h of a beer containing erythromycin using a rotary vacuum filter originally purchased for another product. Our filter has a cycle time of 50 s and an area of 37.2 m2. It operates under a vacuum of 20 in Hg. The pretreated broth forms an incompressible cake with the resistance:

20 s/cm 292

=∆P

µαρ

We want to wash the cake until only 1% of the retained solubles is left, and we expect that the washing efficiency will be 70% and that 1% of the filtrate is retained. (a) Calculate the filtration time per cycle. (b) Find the washing time.

Solution:

For incompressible cake,

2

0

2

∆

==T

fcf A

V

Ptt

βµαρ

β

(To be continued)

Example: Filtration of erythromycin using rotary vacuum filter


For incompressible cake, 2

0

2

∆

==T

fcf A

V

Ptt

βµαρ

β

tc = 50 s

AT = 37.2 m2 = 37.2 × 104 cm2

20 s/cm 292

=∆P

µαρ

33 cm 10208 L208s 3600

h)s 50() L/h000,15( ×==

××= βββfV

s 1.9102.37

1020829

2 4

32

0 =

××

×=

∆

=β

ββ

µαρT

ff A

V

Pt

(To be continued)


n

f

w rnft

t)1( and 2 ε−==

Fraction of retained solubles, r = 0.01

Washing efficiency, ε = 0.7

Fraction of filtrate retained, f = 0.01

r = 0.01 = (1 − 0.7)n n = 3.82

tw = 2nf × tf = 2 × 3.82 × 0.01 × 9.1 = 0.7 s

Example: Filtration of erythromycin using rotary vacuum filter

(b) Find the washing time.

#

Application of Rotary Vacuum Filter

* It is commonly used to recover yeast and mycelia.

* Filtration of bacterial fermentation broth will usually require a precoat of filter aid.

* The separation of cell debris is performed by adding filter aid to the feed liquor.

CENTRIFUGAL FILTRATION

* A combination of a centrifuge and a filter.

* Accumulated solids can be washed.

CENTRIFUGAL FILTRATION (2/8)

Hydrostatic Equilibrium in a Centrifugal Field

In a rotating centrifuge, a layer of liquid is thrown outward from the axis of rotation and is held against the wall of the bowl by centrifugal force.


Consider a volume element of thickness dr at a radius r,

dmrdF 2ω= drrhdm )2( πρ=;

dF = centrifugal force

dm = mass of liquid in the element

ω = angular velocity

ρ = density of the liquid

h = height of the ring

rdrrh

dFdPdrrhdF 222

2 and 2 ρω

πωπρ ==−=∴

Integration )(2

1 21

22

221 rrPPP −=∆−=− ρω


Principles of Centrifugal Filtration

R1 = radius of the surface of feed solution

Rc = radius of the cake’s interface

Darcy’s law:

vk

PPkv µ

µ1

or =∆∆=

Set 0

1 αρ=k

vP

0µαρ=∆

For centrifugal filtration, the pressure drop varies with the radius, thus

vdr

dP0µαρ=−


vdr

dP0µαρ=−

The total volumetric flow rate, Q = (2πrh)v; or rh

Qv

π2=

(Note: v varies with r.)

=−∴

rh

Q

dr

dP

πµαρ

2 0

Integration cR

R

h

QP 0

0 ln2

=∆−

πµαρ

The pressure drop (−∆P) is due to the centrifugal force on the liquid.

)(2

1 21

20

2 RRP −=∆− ρω)/ln(

)(

0

21

20

0

2

cRR

RRhQ

−=µαρ

ρωπ

* Note: Rc is a function of time, and so is Q; however, Q is not a function of r.


)/ln(

)(

0

21

20

0

2

cRR

RRhQ

−=µαρ

ρωπ

Mass balance for the solids:

hRRV )( 2c

20c0 −= πρρ (where ρ c = cake density)

)/ln(

)()2(

0

21

20

0

2

0 c

cc

c

RR

RRh

dt

dRR

h

dt

dVQ

−=−==

µαρπρω

ρπρ

)/ln(

1

2

)(

0

21

20

2

ccc

c

RRR

RR

dt

dR

µαρρω −

=−

I. C.: t = 0, Rc = R0


)/ln(

1

2

)(

0

21

20

2

ccc

c

RRR

RR

dt

dR

µαρρω −

=−

I. C.: t = 0, Rc = R0

The integrated expression is complex, and can be approximated as:

−−

−

=cc

cc

R

R

R

R

RR

Rt 0

2

02

120

2

2

ln21)(2ρω

µαρ

This is the desired result to find the time needed for obtaining a cake of thickness (R0 − Rc).

* Recalling that for a flat cake,

2

0

2

∆=

A

V

Pt

µαρ


[Example] We can filter 250 cm3 of a slurry, containing 0.016 g progesterone ( 體激素黃 ) per cm3, in 32 min. Our filter has a surface area of 8.3 cm2, a pressure drop of 1 atm, and a filter medium of negligible resistance. The solids in the cake have a density of 1.09 g/cm3, and the slurry density is that of water.

We want to use this experiment to estimate the time to filter 1,600 liters of this slurry through a centrifugal filter. The filter has a basket of 51 cm radius and 45 cm height. It rotates at 530 rpm. When it is spinning, the liquid and cake together are 5.5 cm thick. How long will this filtration take?

Solution:

−−

−

=cc

cc

R

R

R

R

RR

Rt 0

2

02

120

2

2

ln21)(2ρω

µαρ

Need data of µα and Rc.

(To be continued)

[Example] We can filter 250 cm3 of a slurry, containing 0.016 g progesterone ( 體激素黃 ) per cm3, in 32 min. Our filter has a surface area of 8.3 cm2, a pressure drop of 1 atm, and a filter medium of negligible resistance. The solids in the cake have a density of 1.09 g/cm3, and the slurry density is that of water.

Example: filtration of progesterone (2/3)


In the laboratory test,

2

0

2

∆=

A

V

Pt

µαρ

t = 32 min = 1920 s; ρ 0 = 0.016 g/cm3

26

226

s-cm

g 1001.1

dyne

cm/s-g

atm

dyne/cm 1001.1 atm 1 ×=

×=∆P

V = 250 cm3; A = 8.3 cm2

2

6 3.8

250

)10(1.012

) 016.0( 1920

×= µα

µα = 2.67 × 108 s-1

(To be continued)

Using centrifugal filtration,

−−

−

=cc

cc

R

R

R

R

RR

Rt 0

2

02

120

2

2

ln21)(2ρω

µαρ

µα = 2.67 × 108 s-1 ; ρ c = 1.09 g/cm3 ; ρ = 1.0 g/cm3

ω = 530 rpm = 55.47 s-1 ; R0 = 51 cm ;

R1 = 51 − 5.5 = 45.5 cm

Mass balance for solids: hRRV cc )( 2200 −= πρρ

(0.016)(1,600 × 103) = (1.09)π[(51)2 − Rc2](45)

Rc = 49.3 cm

s 4663.49

51ln21

3.49

51

)5.4551()47.55)(0.1(2

)3.49)(09.1)(1067.2(

2

222

28

=

−−

−×=∴ t


#

Example: filtration of progesterone (3/3)

filteration

Engineering

conventional filtration

mechansims of filtration

filtration cycle

filter medium

filter aid

filter cloth

clarifying filter

high filtration area