Filtration

FILTRATION Removal of solid particles from a fluid by passing the fluid through a filtering f iltering medium, or septum.

MECHANSIMS OF FILTRATION (a) Clarifiers

* Also known as “deep“deep-bed filters”. * The particles of solid are are trapped trapped inside inside the filter filt er medium.

M ECHA NSI NSI M S OF F I LTRATI LTRATI ON (2 (2/3) /3)

* A typical typical cartridge filter:

M ECHA NSI NSI M S OF F I LTRATI LTRATI ON (3 (3/3) /3)

(b) Cake filters

* The filter medium is relatively thin, thin, compared with that of a clarifying filter. * After the initial period, the cake of solids does the filtration, not filtration, not the septum. septum. * A visible cake of appreciable thickness builds up on the surface and surface and must be periodically removed.

EQUIPMENT FOR CONVENTIONAL FILTRATION (1) Plate and Frame Filter Press Press

* The most common type, type, but less common for bioseparations. * Used where a relatively dry cake discharge is discharge is desired. * Cake removal: open the whole assembly Should not be used where there are are toxic fumes or or biohazards.

EQUI PMEN T FOR CONVENTI CONVENTI ONAL F I LTRATION (2/7 (2/7))

(2) Horizontal Plate Filter:

* Filtration occurs from the top of each plate. plate. * Cake removal: removal: removed with a sluicing nozzle or discharged by rapidly rotating the leaves.


(3) Vertica Verticall Leaf Lea f Filter F ilter and a nd Candle Type Vertica Verticall Tank Tank Filter:


(3) Vertical Vertical Leaf Filter and an d Candle Can dle Type Type Vertical Vertical Tank Tank Filter (2/2):

* Have a relatively high filtration area per volume. volume. Require only a small floor area. * Filter cake is formed on the external surface of surface of the tubes. * The tubes are a re cleaned by backwashing. backwashing.

(4) Rotary Vacuum Filter:

* Rotate at a low speed during speed during the operation. * Pressure inside the drum is a partial vacuum. vacuum. Liquid is sucked through the filter cloth and cloth and solids are retained on the surface of the drum.

EQUI PME NT F OR CONVENTI CONVENTI ONAL F I LTRATI LTRATI ON (6 (6/7) /7)

* Three chief chief steps of the filtration cycle: (1) cake formation (2) cake washing (to washing (to remove either valuable or unwanted solutes) (3) cake discharge


* Being automated. Have a lower labor cost. * The workhorse workhorse of bioseparations. bioseparations. * Common for large-scale operations whenever the solids are difficult to filter.

PRETREATMENT OF FILTRATION Filtration is a straightforward procedur proceduree for well-defined crystals.” crystals.” * Fermentation beers and other biological solutions are are notoriously hard to filter, filter, because of: (1) high, nonnewtonian viscosity, and viscosity, and (2) highly compressibl compressiblee filter cakes. Conventional filtration is often too slow to be practical. practical. The filtration requires requires pretreatment pretreatment:: heating, coagulation and flocculation, or adsorption on filter aid. aid.

PRETREATM PRETREATM ENT OF F I LT RATI RATI ON (2/ (2/12 12))

A. Heating * To improve imp rove the feed’s handling characteristics. characteristics. (Thinking of filtering a dilution solution of egg white.) * The simplest pretr pretreatment eatment (and (and the least expensive). * Chief constraint: thermal stability of stability of the product.


B. Coagulation and Flocculation * Through the addition of electrolytes. electrolytes. * Types of coagulants: (1) Simple electrolytes (such electrolytes (such as ferric chloride, alum alum,, or acids and bases) (2) Synthetic polyelectrolytes

coagulation

flocculation


* Action of simple electrolytes: electrolytes: reduce the electrostatic repulsion existing between colloidal particles. particles . * Action of synthetic polyelectrolytes: polyelectrolytes: (1) Reduce electrostatic repulsion repulsion (2) Adsorb on adjacent particles * Commercially available polyelectrolytes (can polyelectrolytes (can be anionic, cationic, or nonionic): polyacrylamides, polyethylenimines, and polyamine derivatives.

PRETREATM PRETREATM ENT OF F I LTRATI ON (5 (5/12 /12)

* The effect of pH on filtrate volume for 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 pH and filter aid on filtrate volume for Str eptomyce ptom yces s gr gr i seu s :


* How does the filter-aid help? (1) Give porosity to the filter cake. Solids to be filtered

Porosity

Hard spheres of the same size General cases

0.2 0.3

Compressible Compressible solids Diatomaceous silica (

0.45

0 )

0.9

(2) Create a very large surface to trap the t he gelatinous precipitate. Allow much more filtrate to be obtained before eventually clogging up. up.

PRETREATM PRETREATM ENT OF F I LT RATI RATI ON (10 (10/12 /12)

* How to use filter-aid? filter-aid? (1) Precoat — — a thin layer (0.1 to 0.2 lb/ft2) of filter aid is deposited on the filter filt er medium prior to introducing to introducing the filter feed to the system - Protect the filter medium from fouling. - Provide a finer matrix to exclude particles from the filtrate. (2) Body feed — add add the filter aid to the filter feed


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

____________________


* The use of filter-aid filter-aid is mainly mainly for removing small amounts of unwanted particulate material. material. It cannot deal with large quantities of precipitate successfully. * Types Types of filter-aid (the most effective): (1) Diatomaceous earths such as Celite Celite (consisting (consisting mainly of SiO2) (2) Perlites Perlites (volcanic (volcanic rock processed to yield an expanded form)

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

GENERAL THEORY FOR FILTRATION Darcy’s law — relate relate the flow rate through a porous bed of solids to solids to the pressure drop causing that flow. v

k  P  

v = velocity of the liquid = P = pressure drop across = across the bed of thickness ℓ P / ℓ = = pressure gradient

= viscosity of the liquid k = permeability of the bed, a proportionality =

constant (dimension: L2) * Like Ohm’s law, law, ℓ /k is is the resistance of filtration. filtration.

GENERAL TH EORY EORY F OR F I LT RATI RATI ON (2/5 (2/5) )

Darcy’s law: v 

k  P  

Strictly speaking, Darcy’s law holds only when  vd vd  (1 -  )



5

where d is the particle size of the filter is fi lter cake, is the liquid density, and is the void fraction in the cake. * Biological separations almost always obey always obey this inequality. For a batch filtration, filtration, v

1 dV A dt



1 dV A dt



k  P

 

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


Two contributions to the t he filtration resistance: 

k



R M



RC

where R medium (constant), M is the resistance of the filter medium (constant), and R cake (varies with V ). ). C is the resistance of the cake (varies The basic differential equation for equation for filtration at constant pressure pressure drop can thus be obtained as: 1 dV A dt



k  P

1 dV

 

A dt



 P  ( R M  RC )


Incompressible Cakes  V   A  

RC  a  0 

= specific cake resistance, cm/g 0 =

mass of cake solids per volume of filtrate

 P  A dt  ( R M  RC ) 1 dV

 P  (I.C.: t = = 0, V = = 0) A dt  [a  0 (V / A)  RM ] 1 dV





 V    K    B    P 2 P  A  V  A 

At A t

 a  0  V 

 RM


At At



V

 a 0  V   R M  V   K    B   2 P  A   P  A 

 At   versus Plot   V  Known ,

0,

P

 V     A 

Slope = K 



 a  0 2 P

can be determined.

* Often, the medium resistance R = 0. = M is insignificant, B  a  0  V  t    2 P  A 

2

[Example] A suspension suspensio n containing contai ning 225 g of g of carbonyl iron powder, powder, Grade Gr ade E, per liter of liter of a solution of 0.01 N NaOH NaOH is to be filtered, using a leaf filter. Estimate the size (area) of the filter needed to needed to obtain 100 lb of lb of dry cake in 1 h of h of filtration at a constant pressure pressure drop of 20 psi. psi. The cake is incompressible. incompressible. The specific specific cake resistance is 1011 ft/lb ft/lb.. The resistance of the medium is taken as 0.1 in 1. Solution: At A t V  0 



 a  0  V 

 RM

  2 P  A   P

mass of cake solid volume of filtrate



V  volume of filtrate 

225 225 g  28.32 L   lb  3  14 . 0 lb/ft L  ft 3   453.6 g  100 100 lb

14.0 lb/ft 3

 7.1 ft 3 (To be be conti conti nu ed)

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

Solution (cont’d): (cont’d):

At At



V

a a  0  V   RM    P 2 P  A 

t = filtration time = 1 h =

 2.116 103 lb f /ft /ft 2   lb - ft   (3600) 2 s 2   P  20 psi  32.2   2 2 14.7 psi l b s h    f 

= 1.2 1012 lb/ft-h2 = specific cake resistance = 10 11 ft/lb 1 R M = resistance of the medium = 0.1 in = 1.2 ft

1

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

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

Solution (cont’d): (cont’d): At A t



V A(1)

7.1



 a 0  V 

 RM

2 P  A 

 P





(2.42)(1011 )(14.0) 7.1 2(1.2  1012 )

A 2 1.7 10-11A

A



(2.42)(1.2) 1.2  1012

71.2 = 0

A = 8.4 ft2

#

[Example] Streptomyces Filtration from 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. cp. The pressure drop is 20 in. of mercury mercury and and the feed contains 0.015 kg dry cake per liter. liter. Determine the specific cake resistance and the medium resistance R M. Solution:

 a  0  V   RM     P V 2 P  A 

At A t

(To be be conti conti nu ed)

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

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

At At V



a a  0  V 

 RM

  2 P  A   P

#

[Example] We have filtered a slurry of sitosterol sitosterol at constant pressure pressure through a filtration medium consisting of a screen support mounted across the end of a Pyrex pipe. We find that the resistance resistance of the filtration medium 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-andframe 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 psi and that the filtrate from the press must be raised against the equivalent of 15 ft head. ft head. (To be be conti conti nu ed)

Ex ample: ample: f il ter ter in g a sl sl ur ry of sitoste itosterr ol

Solution (cont’d): (cont’d): (a) Predict the number of frames needed Cake density 

62 g 253 253.3 cm3

Cake volume of 63 kg steroid =

245 g/cm3  0.245

63 103 g 0.245 g/cm3

 2.57 105 cm3 3

2.57 105 cm3  in   17.4 Number of frames needed = 3   30  30 1 in  2.54 cm 

18 frames are needed. (To be be conti conti nu ed)


Solution (cont’d): (cont’d): (b) Time required required for filtering a 63 kg batch of steroid For incompressible incompressible cake with a negligible filter medium resistance, a a  0  V  t    2 P  A 

2

a 1   0V  or t    2 P  0  A 

2

In the laboratory test:

     a 62 g 163 163 min    2 (15 psi ps i)  0   2  (5.08 cm)  4 

2

 a 2  0

 261 261

min - psi ps i - cm 4 g2 (To be be conti conti nu ed)


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

In the laboratory test:

2  0

 261 261

min - psi ps i - cm 4 g2

In the large-scale operation: A  18  2  (30  30) in

2

 2.54 cm  

in

2

5 2 2 . 09 10 cm   

  14.7 psi ps i ps i - 15 ft head  ps i  P  10 psi  3.5 psi   33.9 ft head (water)  2

2

a 1   0V  1  63,000 000  261 26 1     6.8 min t     5  2  0  P  A  3.5  2.09 10 

#

GEN ERA L T H EORY F OR FI L TRATI ON: Compr Compr essi ble Cakes Cakes (1/3) (1/3)

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 is a function of the pressur pressuree drop drop..

a  a ' ( P )

s

where

 V   A  

Recall: RC  a  0 

’ = = a constant related largely to the size and shape of the particles forming the cake s = the cake compressibility =


a  a ' ( P ) s

log a  log a ' s log P

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


a  a ' ( P )

s

 V   A  

Recall: RC  a  0 

For a rigid, rigid, incompressible incompressible cake, s = 0. = For a highly compressible compressible cake, s 1. In practice, s ranges from 0.1 0.8. ranges When values of s are high, one should consider pretreating are the feed with filter filt er aids.

[Example] Filtration of Beer Containing Protease. We have a suspension of Baci Baci l l us subtil i s 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, aspirator, we have found that we can filter 100 cm3 of this beer in 24 min. min. However, However, previous studies with this type of beer have had a compressible cake with s equal to 2/3. equal 2/3. We now need to filter to filter 3000 L of L of this material in a pilot plant’s plate-and-frame plate-and-frame press. This press has 15 frames, each of area 3520 cm2. The spacing spacing between 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 pressure drop that can be used is 65 psi. How long will it take to filter this beer at 50 psi? (To be be conti conti nu ed)

Ex ample: ample: F il trati on of B eer Contai Contai ni ng Proteas Protease e

Solution (cont’d) (cont’d):: a a  0  V  t    2 P  A 

Negligible R M

Compressible Compress ible cake,

2

a  a ' ( P ) s

t 

 a '  0  V    1- s 2 P  A 

2

Laboratory test: P = = 14.7 psi (a psi (a Buchner funnel attached to an aspirator) A =

 4

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

24 min 

 a '  0 2(14.7 psi ps i)1 / 3

   100 100 cm 3      (5 cm) 2   4 

’ 0 = 4.53 min psi 1/3 cm

2

2

(To be conti nu ed)

Ex ample: ample: F il trati on of B eer Contai Contai ni ng Proteas Protease e

Solution (cont’d) (cont’d)::  a '  0  V  t    1- s 2 P  A 

2

;

’ 0 = 4.53 min psi 1/3 cm

2

Pilot-plant operation: V = = 3000 L = 3 106 cm3 A = 15

2 3520 cm2 (Filtration occurs on both sides of the

frame.) 2

  a '  0  V  4.53  3 10   t    1- s  1/ 3  2 P  A  2(50)  15  2  3520  6

2

= 496 min = 8.3 h #

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)

AN AL YSI YSI S OF CONTI NU OUS ROTARY ROTARY VACUUM F I L TERS (2/8 (2/8))

Cake Formation

For compressible cake and negligible medium resistance, resistance, 2

2

 a '  0  V   a '  0  V f  or  t  t    f 1- s  1- s  2 P  A  2 P  A 

where t f = cake formation time V f = volume of filtrate collected during the period of t f A = filtration area (submerged area of filter)


Cake Formation (cont’d)

 a '  0  V f  t f    1- s  2 P  A 

2

Let t f = t c and A = AT b t c 

 a '  0  V f  1- s

2 P

2

   b A   T 

where t = cycle time c AT = total tot al filter area

= fraction of the drum submerged


Cake Washing 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.


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. required.

An empirical equation for the fraction of soluble material remained:

r  (1 -  ) n 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 volume of retained liquid = washing efficiency of the cake


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.

The wash liquid contains no additional solids. (1) The cake thickness is constant. The flow of wash liquid is constant. (2) Wash rate = filtration rate at the end of cake formation


Wash rate 

1 dV w A dt



V w At w

where V w = volume of wash water required, and t w = time required for washing. Filtration rate at the end of cake cak e formation = a '  0  V  t    2 P 1- s  A 

Wash rate 

2

1 dV A dt

t  t f

1/ 2

 2(  P )1- s t  or   A   a '  0  V

1 dV A dt

 t t f

d  V 

  dt  A  t t

 (  P )1- s  V w    At w  2 a '  0 t f 

f

1/ 2

1/ 2

 ( P )    t 2 a '   0 f   1- s


A useful expression: 1/ 2

 (  P )1- s  V w   At w  2  a '  0t f 

1/ 2

V f  2(  P )1- s   and and  At f   a '  0t f  1/ 2

V w  2a '  0t f  t w   1 s  A  (  P ) -  t w t f

2

V w V f

2

1/ 2

and and t f 

V w V r V r V f

V f   a '  0t f 



A  2(  P )

1- s

 

 2nf

where V r = volume of liquid retained f = ratio of the volume of retained liquid (V r ) to the volume of filtrate (V f )

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

For incompressible cake, t f 

a 0

 V f 

  2 P  A 

2

2

or A2 

a 0V f

2t f  P

(To be be conti conti nu ed)

Ex ample: ample: D eter ter mi ne the area of a rotar y vacuum vacuum f i l ter ter

Solution (cont’d): (cont’d):   2 cp  0.02

g

a  9 10 ; cm - s

10

cm g

;  0  10

g L

 10 10

-3

g cm 3

3   h  3 cm  3   (15 s)   V f   2000 10 8333 cm h   3600 s  

 P  70 kPa  70 103

A  2

a a  0V f 2 2t f  P



N  kg - m  1000 g   m  g 5   7 . 0 10 m 2  N - s 2   kg  100 100 cm  cm - s 2

(0.02)(9 1010 )(10 10 -3 )(8333) 2 2(15)(7.0 105 )

 5.95 10 7 cm4

A = 7715 cm2 = 0.7715 m2 AT  A 

t c t f

 0.7715 

60 15

 3.09 m 2

#

[Example] We want to filter 15,000 L/h of 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 s and an area of 37.2 m2. It operates under a vacuum of 20 in Hg. The pr pretreated etreated broth broth forms an incompressible incompressible cake with the resistance:  a  0

2 P

 29 s/cm2

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

For incompressible cake,

t f  b t c 

 a  0  V f 

2

   2 P  b AT  (To be conti nu ed)

Example: Example: F il tration of erythromycin rythromycin u sin g rotary rotary vacuum vacuum fi lter lter

Solution (cont’d) (cont’d):: For incompressible cake,

t f  b t c 



h

 a  0  V f 

2

   2 P  b AT 



000 L/h L/h)  (50  b s)  208 b L  208 208b 10 cm V f  (15,000  208   3600 s  3

3

t = 50 s c AT = 37.2 m2 = 37.2 104 cm2

 a  0

2 P

 29 s/cm2 2

 208 a a  0  V f  208 b 103     29   9.1 s t f  4  2 P  b AT   b  37.2 10  (To be be conti conti nu ed)

Example: Example: F il tration of erythromycin rythromycin u sin g rotary rotary vacuum vacuum fi lter lter

(b) Find the washing time.

Solution (cont’d): (cont’d): t w t f

and r  (1 -  ) n  2nf and

Fraction of retained solubles, r = 0.01 Washing efficiency, = 0.7 Fraction of filtrate f iltrate retained, f = 0.01 r = 0.01 = (1 t = 2n f w

0.7)n

n = 3.82 =

t f = 2 3.82 0.01 9.1 = 0.7 s

#

Application of Rotary Vacuum Filter * It is commonly commonly used to recover yeast and mycelia. mycelia. * Filtration of bacterial fermentation broth will broth will usually require a precoat of filter aid. aid. * The separation of cell debris is 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.

CENTRI CENTRI F UGAL F I LTRATI LTRATI ON (2 (2/8) /8)


Hydrostatic Equilibrium 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 wall of the bowl by centrifugal force.


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

dF  r  2 dm ; dm   (2 rh)dr dF = centrifugal force dm = mass of liquid in the element = = angular velocity

= density of the liquid h = height of the ring =

and - dP   dF  2 h 2 r 2 dr and

Integration

P 1 - P 2  - P 

dF

2 rh

1 2

  2 rdr

 2 (r 22 - r 12 )


Principles of Centrifugal Filtration Darcy’s law: v

Set

1

k

k  P

 

or

 a  0

 P 1   v k



 P 

  a  0 v

For centrifugal filtration, the pressure drop varies with the radius, thus radius, thus R 1 = radius of the surface of feed

solution R c = radius of the cake’s interface

-

dP dr

  a  0 v


-

dP dr

  a  0 v

(Note: v varies with r .) varies

The total volumetric flow rate, rate, Q = = (2 r h )v ; or v  -

Q

2 rh

 Q    a 0   dr r h 2   

dP

 Q  R0 -  P   a 0   ln  2 h  Rc

Integration

* Note: R ; however, Q is is c is a function of time, and so is Q not a function of r . The pressure drop ( the liquid.

-  P 

1 2

P ) is due to the centrifugal force on

 ( R - R ) 2

2 0

2 1

 h  2 ( R02 - R12 ) Q  a  0 ln( R0 / Rc )


 h  2 ( R02 - R12 ) Q  a  0 ln( R0 / Rc )

Mass balance for the solids:

 0V   c ( R02 - Rc2 )h Q

dV



 c h

dt

-

 0

dRc dt

(-2 Rc )



(where

dRc



c =

cake density)

2 2  2 h ( R0 - R1 )

dt

 a  0 ln( R0 / Rc )

 2 ( R02 - R12 )

1

2  c  a

Rc ln( R0 / Rc )

I. C.: t = 0, R = c = R 0


-

dRc dt



 2 ( R02 - R12 )

1

2  c  a

Rc ln( R0 / Rc )

I. C.: t = = 0, R c = R 0

The integrated expression is complex, and can be approximated as: 2   R0   R0   a c Rc   - 1 - 2 ln   t  2 2 2 2  ( R0 - R1 )  Rc   Rc   2

This is the desired result to result to find the time needed for obtaining a cake of thickness ( R R 0 c). 2

* Recalling that for a flat cake, cake, t 

 a 0  V 

  2 P  A 

[Example] [Exampl e] We can filter 250 cm3 of a slurry, slurry, containing 0.016 g progester progesterone one ( min. Our filter ) per cm3, in 32 min. has a surface area of 8.3 cm 2, a pressure drop of 1 atm, and a filter medium of negligible resistance. 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 liters of this slurry through a centrifugal filter. The filter has a basket of 51 cm radius and radius and 45 cm height. height. It rotates at 530 rpm. rpm. When it is spinning, the liquid and cake together are 5.5 cm thick . How long will this filtration take? Solution:

 R  2  R0  0   - 1 - 2 ln   t  2 2 2 2  ( R0 - R1 )  Rc   Rc    a  c Rc2

Need data of

and R c. (To be be conti conti nu ed)

Ex ample: ample: f i l tr ation of pr oges ogester ter one (2/3)

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

Solution (cont’d): (cont’d): In the laboratory test, t  t = = 32 min = 1920 s;

a a  0  V 

2

  2 P  A 

3 = 0.016 g/cm 0

V = = 250 cm3; A = 8.3 cm2

1.01106 dyne/cm2   g - cm/s2  g 6 atm   P  1 atm  1.0110    2 atm at m dyne cm s    016 )  250  a (0.016 250  1920    6 2(1.01  10 )  8.3 

2

= 2.67 108 s-1

Ex ample: ample: f i l tr ation of pr oges ogester ter one (3/3)

Solution (cont’d): (cont’d):  R  2  R0  0   - 1 - 2 ln   Using centrifugal filtration, t  2 2 2 2  ( R0 - R1 )  Rc   Rc    a c Rc2

= 2.67 108 s-1 ;

c =

1.09 g/cm3 ;

= 1.0 g/cm3

= 530 rpm = 55.47 s -1 ; R 0 = 51 cm ; R 1 = 51

5.5 = 45.5 cm

2 2 Mass balance for solids:  0V   c ( R0 - Rc )h

(0.016)(1,600 103) = (1.09) [(51)2

2](45) R c

R c = 49.3 cm 2 (2.67 108 )(1.09)(49.3) 2  51   51   t  466 s  - 1 - 2 ln    466 2 2 2  2(1.0)(55.47) (51 - 45.5 )  49.3   49.3 

#

Filtration

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