BS 2 Filtration

7/30/2019 BS 2 Filtration

1/70

FILTRATION

Removal of solid particles from a fluid by passing thefluid through a filtering medium, or septum.


2/70

MECHANSIMS OF FILTRATION

(a) Clarifiers

* Also known as deep-bed filters.

* The particles of solid are trapped inside the filter

medium.


3/70

* A typical cartridge filter:

MECHANSIMS OF FI LTRATION (2/3)


4/70

(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 FI LTRATION (3/3)


5/70

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 orbiohazards.


6/70

(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 CONVENTI ONAL FILTRATION (2/7)


7/70

(3) Vertical Leaf Filter and Candle Type Vertical Tank

Filter:



8/70

(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.



9/70

(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.


10/70

* Three chief steps of the filtration cycle:

(1) cake formation

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

solutes)

(3) cake discharge

EQUIPMENT FOR CONVENTI ONAL FI LTRATION (6/7)


11/70

* The workhorse of bioseparations.

* Common for large-scale operations whenever the solids

are difficult to filter.

* Being automated.

Have a lower labor cost.



12/70

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 onfilter aid.


13/70

A. Heating

* To improve the feeds 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 F I LTRATION (2/12)


14/70

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


15/70

* 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.



16/70

* The effect of

pH onfiltrate

volume for

Streptomyces

griseus:



17/70

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; improvethe quality of the filtered liquor.



18/70

* The effect of

filter aid on

filtrate volume

for Streptomyces

griseus:



19/70

* The effect ofpH andfilter aid on filtrate volume for

Streptomyces griseus:



20/70

* How does the filter-aid help?

(1) Give porosity to the filter cake.

Solids to be filtered Porosity

Hard spheres of the same size 0.45General cases 0.2-0.3Compressible solids 0Diatomaceous silica () 0.9

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

Allow much more filtrate to be obtained beforeeventually clogging up.



21/70

- Protect the filter medium from fouling.

- Provide a finer matrix to exclude particles

from the filtrate.

* How to use filter-aid?

(1) Precoata thin layer (0.1 to 0.2 lb/ft2

) of filter aid isdeposited on the filter medium prior to introducing

the filter feed to the system

(2) Bodyfeedadd the filter aid to the filter feed



22/70

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

____________________

PRETREATMENT OF F ILTRATION (11/12)


23/70

* The use of filter-aid is mainly for removing small

amounts of unwanted particulate material.

It cannot deal with large quantities of precipitatesuccessfully.

* 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 antibioticsmay irreversibly bind to diatomaceous earth.



24/70

GENERAL THEORY FOR FILTRATION

Darcys lawrelate the flow rate through a porous bed of

solids to the pressure drop causing that flow.

Pkv

v= velocity of the liquidP= pressure drop across the bed of thickness P/ = pressure gradient = viscosity of the liquidk= permeability of the bed, a proportionality

constant (dimension: L2)

* Like Ohms law, /kis the resistance of filtration.


25/70

Strictly speaking, Darcys law holds only when

5)1(

-

vd

where dis the particle size of the filter cake, is theliquid density, and is the void fraction in the cake.

* Biological separations almost always obey this inequality.

For a batch filtration,

dt

dV

Av

1

Pk

dt

dV

A

1

where Vis the total volume of filtrate, A is the filter

area, and tis the time.

GENERAL THEORY FOR FI LTRATION (2/5)

Pkv

Darcys law:


26/70

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

Pk

dt

dV

A

1



27/70

Incompressible Cakes

a = specific cake resistance, cm/g0 = mass of cake solids per volume of filtrate

)(

1

CM RR

P

dt

dV

A

])/([

1

0 MRAV

P

dt

dV

A

a (I.C.: t= 0, V= 0)

AVRC 0a

BA

VK

P

R

A

V

PV

At M

a

2

0



28/70

Plot

V

Atversus

A

V Slope =

PK

2

0a

Known , 0, P a can be determined.* Often, the medium resistance RM is insignificant, B= 0.

2

0

2

AV

Pt a

B

A

VK

P

R

A

V

PV

At M

a

2

0



29/70

[Example] A suspension containing 225 g of carbonyl iron

powder, Grade E, per liter of a solution of 0.01 NNaOH 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 offiltration at a constant pressure drop of20 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

a

2

0

Solution:

3

30 lb/ft0.14g453.6

lb

ft

L32.28

L

g225

filtrateofvolume

solidcakeofmass

3

3ft1.7

lb/ft14.0

lb100filtrateofvolume V

(To be continued)


30/70

[Example] A suspension containing 225 g of carbonyl iron powder, Grade E, per

liter of a solution of 0.01 NNaOH 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 of20 psi. The cake is incompressible. The specific cake

resistance is 10

11

ft/lb. The resistance of the medium is taken as 0.1 in

-1

.

P

R

A

V

PV

At M

a

2

0Solution (contd):

t= filtration time = 1 h

2

22

2

f

2

f

3

h

s)3600(

s-lb

ft-lb2.32

psi14.7

/ftlb10116.2psi20P

= 1.2 1012 lb/ft-h2a = specific cake resistance = 1011 ft/lbRM = resistance of the medium = 0.1 in

-1 = 1.2 ft-1

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


31/70

Solution (contd):

P

R

A

V

PV

At M

a

2

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 = 0A = 8.4 ft2

#

[Example] A suspension containing 225 g of carbonyl iron powder, Grade E, per

liter of a solution of 0.01 NNaOH 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 of20 psi. The cake is incompressible. The specific cake

resistance is 10

11

ft/lb. The resistance of the medium is taken as 0.1 in

-1

.


32/70

[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 of0.1 ft2 and the filtrate has a

viscosity of1.1 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 a and the medium resistance RM.

P

R

A

V

PV

At M

a

2

0Solution:

(To be continued)


33/70

Example: Streptomyces Filtration from an Erythromycin Broth (contd)

E l S Fil i f E h i B h ( d)


34/70

Example: Streptomyces Filtration from an Erythromycin Broth (contd)

#

P

R

A

V

PV

At M

a

2

0


35/70

[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. Wealso 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 kgbatch 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 of15 ft head.

(To be continued)


36/70

Example: f il ter ing a slur ry of sitosterol

Solution (contd):

(a) Predict the number of frames needed

3

3 g/cm245.0cm3.253

g62densityCake

Cake volume of 63 kg steroid =35

3

3

cm1057.2g/cm0.245

g1063

Number of frames needed = 4.17cm2.54

in

in13030

cm1057.2 3

3

35

18 frames are needed.

(To be continued)


37/70

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

Solution (contd):

For incompressible cake with a negligible filter

medium resistance,2

0

0

2

0 1

2or

2

A

V

Pt

A

V

Pt

aa

In the laboratory test:

2

20 cm)08.5(4

g62

psi)15(2min163

a

2

4

0 g

cm-psi-min261

2

a


(To be continued)


38/70

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

Solution:

In the large-scale operation:

25

2

2 cm1009.2in

cm54.2in)3030(218

A

psi5.3(water)headft33.9

psi7.14headft15psi10

-P

min8.61009.2

000,63

5.3

1261

1

2

2

5

2

0

0

A

V

Pt

a


#

In the laboratory test: 2

4

0 g

cm-psi-min261

2

a


39/70

Compressible Cakes

Almost all cakes formed of biological materials arecompressible. As these cakes compress, filtration

rates drop.

To estimate the effects of compressibility, we assume that

the cake resistance a is a function of the pressure drop.s

P)(' aa

where a= a constant related largely to the size and shapeof the particles forming the cakes= the cake compressibility

GENERAL THEORY FOR FI LTRATION: Compressible Cakes (1/3)

A

VRC 0aRecall:


40/70

sP)(' aa Ps log'loglog aa

Plot logaversus logP, slope = s,intercept = loga.



41/70

sP)(' aa

For a rigid, incompressible cake, s= 0. For a highly compressible cake, s 1. In practice, sranges from 0.1-0.8. When values ofsare high, one should consider pretreating

the feed with filter aids.


A

VR

C 0aRecall:


42/70

[Example] Filtration of Beer Containing Protease. We have a

suspension ofBacil lus subtil isfermented 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 3.6wt% 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

sequal to 2/3.

We now need to filter 3000 L of this material in a pilot

plants 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)


43/70

Example: F il tration of Beer Containing Protease

Solution (contd):

Negligible RM2

0

2

A

V

Pt

a

Compressible cake,s

P)(' aa2

1

0

2

'

- A

V

Pt

s

a

Laboratory test:P= 14.7 psi (a Buchner funnel attached to an aspirator)A = ;V= 100 cm3; t= 24 min; s= 2/32)cm5(

4

2

2

3

3/1

0

)cm5(4

cm100

)psi7.14(2

'min24

a

a0

= 4.53 min psi1/3 cm-2

(To be continued)


44/70

2

1

0

2'

- A

VP

ts

a a0= 4.53 min psi1/3 cm-2

Pilot-plant operation:

V= 3000 L = 3 106 cm3A = 15 2 3520 cm2 (Filtration occurs on both sides of theframe.)

2

6

3/1

2

1

0

3520215103

)50(253.4

2'

-A

V

Pt

s

a

;

= 496 min = 8.3 h

Example: F il tration of Beer Containing Protease

Solution (contd):

#


45/70

ANALYSIS OF CONTINUOUS ROTARY

VACUUM FILTERS

There are threestages involved in the

operation:

(1) cake formation

(2) cake washing

(3) cake discharge

(not affecting

the filter sizeand the cycle

time)

ANALYSIS OF CONTINUOUS ROTARY VACUUM FI LTERS (2/8)


46/70

Cake Formation

For compressible cake and negligible medium resistance,

2

1

0

2

1

0

2

'or

2

'

--A

V

Pt

A

V

Pt

f

sfs

aa

where tf= cake formation time

Vf= volume of filtrate collected during the period

oftf

A = filtration area (submerged area of filter)




47/70

Cake Formation (contd)

2

1

0

2

'

-

A

V

Pt

f

sf

a

Let tf= btcand A = bAT2

1

0

2

'

-

T

f

sc

A

V

P

t

b

ab

where tc= cycle time

AT= total filter area

b = fraction of the drum submerged




48/70

Cake WashingTwo 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.

( )



49/70

An empirical equation for the fraction of soluble material

remained:n

r )1( -

where r= ratio of soluble material remained after the

wash to that originally present in the cake

n= volume of wash liquid divided by the volumeof retained liquid

= washing efficiency of the cake

Two factors involved in the stage of cake washing:

(1) The fraction of soluble material remained after the washGoverning the volume of wash liquid required.

( )



50/70

The wash liquid contains no additional

solids.

(1) The cake thickness is constant.

Two factors involved in the stage of cake washing:

(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.

( )



51/70

w

ww

At

V

dt

dV

A

1rateWash

whereVw= volume of wash water required, andtw= time

required for washing.

Filtration rate at the end of cake formation =ftt

dt

dV

A

1

2/1

0

12

1

0

'

)(2or

2

'

-

- a

a tP

A

V

A

V

Pt

s

s

2/1

0

1

'2

)(1

rateWash

-

f

s

ttttt

P

A

V

dt

d

dt

dV

Aff

a

2/1

0

1

'2

)(

-

f

s

w

w

t

P

At

V

a

( )

ANALYSIS OF CONTINUOUS ROTARYVACUUM FI LTERS (8/8)


52/70

A useful expression:

2/1

0

1

2/1

0

1

')(2and

'2)(

--

f

s

f

f

f

s

w

w

tP

AtV

tP

AtV

aa

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

ff

fs

fw

w

P

t

A

Vt

P

t

A

Vt

aa

where Vr= volume of liquid retained

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

the volume of filtrate (Vf)



53/70

[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 aviscosity of2.0 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 thearea of the filter that is required.Solution:

For incompressible cake,

PtVA

AV

Pt

f

ff

f

2or

2

2

02

2

0 aa

(To be continued)


54/70

Example: Determine the area of a rotary vacuum f i l ter

Solution (contd):

s-cmg0.02cp2

gcm109 10a 3

30cm

g1010

L

g10 -; ;

33

3 cm8333s3600

hs)15(

h

cm102000

fV

2

5

22

3

s-cm

g100.7

cm100

m

kg

g1000

s-N

m-kg

m

N1070kPa70

P

47

5

23102

02 cm1095.5

)100.7)(15(2

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

2

-

Pt

VA

f

fa

A = 7715 cm2 = 0.7715 m22

T m09.315

607715.0

f

c

t

tAA

#

[Example] We want to filter 15,000 L/h of a beer containing


55/70

[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

of50 s and an area of37.2 m2. It operates under a vacuum

of 20 in Hg. The pretreated broth forms an incompressible

cake with the resistance:

20 s/cm292

P

a

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:


2

0

2

T

f

cfA

V

Ptt

b

ab

(To be continued)


56/70

Example: Fi ltr ation of erythromycin using rotary vacuum fi lter

Solution (contd):


2

0

2

T

fcf

AV

Ptt

bab

tc= 50 s

AT= 37.2 m2 = 37.2 104 cm2

20 s/cm292

P

a

33 cm10208L208s3600

h)s50()L/h000,15(

bbbfV

s1.9102.37

1020829

2 4

32

0

b

b

b

a

T

f

fA

V

Pt

(To be continued)


57/70

Solution (contd):

n

f

w rnft

t)1(and2 -

Fraction of retained solubles, r= 0.01

Washing efficiency, = 0.7Fraction of filtrate retained, f= 0.01

r= 0.01 = (1 - 0.7)n n= 3.82tw= 2nftf= 2 3.82 0.01 9.1 =0.7 s

Example: Fi ltr ation of erythromycin using rotary vacuum fi lter

(b) Find the washing time.

#


58/70

Application of Rotary Vacuum Filter

* It is commonly used to recover yeast and mycelia.

* Filtration ofbacterial fermentation broth will usuallyrequire a precoat of filter aid.

* The separation ofcell debris is performed by adding

filter aid to the feed liquor.


59/70

CENTRIFUGAL

FILTRATION

* A combination of acentrifuge and a filter.

* Accumulated solids

can be washed.

CENTRIFUGAL F I LTRATION (2/8)


60/70

( )



61/70

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.



62/70

Consider a volume element of

thickness drat a radius r,

dmrdF2 drrhdm )2( ;

dF= centrifugal force

dm= mass of liquid in the element

=angular velocity = density of the liquidh= height of the ring

rdrrh

dF

dPdrrhdF222

2and2 -

Integration )(2

1 21

2

2

2

21 rrPPP ---



63/70

Principles of Centrifugal Filtration

R1 = radius of the surface of feed

solution

Rc= radius of the cakes interface

Darcys law:

vk

PPkv

1or

Set 01

a

k

vP

0a

For centrifugal filtration, the

pressure drop varies with the

radius, thus

vdr

dP0a-



64/70

vdr

dP0a-

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

v 2

(Note: vvaries with r.)

-

rh

Q

dr

dP

a

2 0

Integration c

R

R

h

QP

0

0 ln2

- a

The pressure drop (-P) is due to the centrifugal force onthe liquid.

)(2

1 21

2

0

2RRP --

)/ln(

)(

0

2

1

2

0

0

2

c

RR

RRhQ

-

a

* Note: Rcis a function of time, and so is Q;however,Qis

not a function ofr.



65/70

)/ln()(

0

2

1

2

0

0

2

cRRRRhQ -

a

Mass balance for the solids:

hRRV )(2

c

2

0c0 - (where c = cake density)

)/ln(

)()2(

0

2

1

2

0

0

2

0 c

c

c

c

RR

RRh

dt

dRR

h

dt

dVQ

--

a

)/ln(

1

2

)(

0

2

1

2

0

2

ccc

c

RRR

RR

dt

dR

a

--

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



66/70

)/ln(

1

2

)(

0

2

1

2

0

2

ccc

c

RRR

RR

dt

dR

a

--

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

0

2

1

2

0

2

2

ln21)(2

a

This is the desired result to find the time needed forobtaininga cake of thickness (R0-Rc).

* Recalling that for a flat cake,2

0

2

A

V

P

ta

[Example] We can filter 250 cm3 of a slurry, containing


67/70

[Example] We can filter 250 cm of a slurry, containing

0.016 g progesterone()per cm3, in 32 min. Our filterhas a surface area of 8.3 cm2, a pressure drop of 1 atm, and

a filter medium of negligible resistance. The solids in thecake have a density of1.09 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 of51 cm radius and 45 cm height. It

rotates at 530 rpm. When it is spinning, the liquid and cake

together are5.5 cm thick. How long will this filtration take?

Solution:

--

-

cc

cc

R

R

R

R

RR

Rt 02

0

2

1

2

0

2

2ln21

)(2a

Need data ofa and Rc.(To be continued)

Example: f i l tration of progesterone (2/3)


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[Example] We can filter 250 cm3 of a slurry, containing 0.016 g

progesterone () per cm3, in 32 min. Our filter has a surfacearea 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 of1.09 g/cm3

,and the slurry density is that of water.

Solution (contd):

In the laboratory test,

2

0

2

A

V

P

ta

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

2

6226

s-cmg1001.1

dynecm/s-g

atmdyne/cm1001.1atm1

P

V= 250 cm3; A = 8.3 cm2

2

6 3.8

250

)10(1.012

)016.0(1920

a a = 2.67 108 s-1(To be continued)

Example: f i l tration of progesterone (3/3)


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Using centrifugal filtration,

--

-

cc

cc

R

R

R

R

RR

Rt 0

2

0

21202

2

ln21

)(2

a

a = 2.67 108 s-1 ; c = 1.09 g/cm3 ; = 1.0 g/cm3 = 530 rpm = 55.47 s-1 ; R0 = 51 cm ;R1 = 51 - 5.5 = 45.5 cm

Mass balance for solids: hRRV cc )(22

00 -

(0.016)(1,600 103) = (1.09)[(51)2-Rc2](45) Rc = 49.3 cm

s4663.49

51ln21

3.49

51

)5.4551()47.55)(0.1(2

)3.49)(09.1)(1067.2(

2

222

28

--

-

t

Solution (contd):

#


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Documents

BS 2 Filtration