bs 2 filtration
TRANSCRIPT
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FILTRATION
Removal of solid particles from a fluid by passing thefluid through a filtering medium, or septum.
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MECHANSIMS OF FILTRATION
(a) Clarifiers
* Also known as deep-bed filters.
* The particles of solid are trapped inside the filter
medium.
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* A typical cartridge filter:
MECHANSIMS OF FI LTRATION (2/3)
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(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)
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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.
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(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)
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(3) Vertical Leaf Filter and Candle Type Vertical Tank
Filter:
EQUIPMENT FOR CONVENTI ONAL FILTRATION (3/7)
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(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.
EQUIPMENT FOR CONVENTI ONAL FILTRATION (4/7)
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(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.
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* 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)
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* The workhorse of bioseparations.
* Common for large-scale operations whenever the solids
are difficult to filter.
* Being automated.
Have a lower labor cost.
EQUIPMENT FOR CONVENTI ONAL FILTRATION (7/7)
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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.
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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)
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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
PRETREATMENT OF F I LTRATION (3/12)
coagulation flocculation
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* 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.
PRETREATMENT OF F I LTRATION (4/12)
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* The effect of
pH onfiltrate
volume for
Streptomyces
griseus:
PRETREATMENT OF F I LTRATION (5/12)
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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.
PRETREATMENT OF F I LTRATION (6/12)
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* The effect of
filter aid on
filtrate volume
for Streptomyces
griseus:
PRETREATMENT OF F I LTRATION (7/12)
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* The effect ofpH andfilter aid on filtrate volume for
Streptomyces griseus:
PRETREATMENT OF F I LTRATION (8/12)
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* 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.
PRETREATMENT OF F I LTRATION (9/12)
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- 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
PRETREATMENT OF F I LTRATION (10/12)
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-------------
____________________
PRETREATMENT OF F ILTRATION (11/12)
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* 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.
PRETREATMENT OF F I LTRATION (12/12)
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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.
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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:
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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
GENERAL THEORY FOR FI LTRATION (3/5)
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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
GENERAL THEORY FOR FI LTRATION (4/5)
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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
GENERAL THEORY FOR FI LTRATION (5/5)
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[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)
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[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)
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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
.
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[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)
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Example: Streptomyces Filtration from an Erythromycin Broth (contd)
E l S Fil i f E h i B h ( d)
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Example: Streptomyces Filtration from an Erythromycin Broth (contd)
#
P
R
A
V
PV
At M
a
2
0
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[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)
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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)
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(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
Example: f il ter ing a slur ry of sitosterol
(To be continued)
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(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
Example: f il ter ing a slur ry of sitosterol
#
In the laboratory test: 2
4
0 g
cm-psi-min261
2
a
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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:
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sP)(' aa Ps log'loglog aa
Plot logaversus logP, slope = s,intercept = loga.
GENERAL THEORY FOR FI LTRATION: Compressible Cakes (2/3)
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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.
GENERAL THEORY FOR FI LTRATION: Compressible Cakes (3/3)
A
VR
C 0aRecall:
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[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)
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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)
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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):
#
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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)
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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)
ANALYSIS OF CONTINUOUS ROTARY VACUUM FI LTERS (2/8)
ANALYSIS OF CONTINUOUS ROTARY VACUUM FI LTERS (3/8)
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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
ANALYSIS OF CONTINUOUS ROTARY VACUUM FI LTERS (3/8)
ANALYSIS OF CONTINUOUS ROTARY VACUUM FI LTERS (4/8)
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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.
( )
ANALYSIS OF CONTINUOUS ROTARY VACUUM FI LTERS (5/8)
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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.
( )
ANALYSIS OF CONTINUOUS ROTARY VACUUM FI LTERS (6/8)
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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.
( )
ANALYSIS OF CONTINUOUS ROTARY VACUUM FI LTERS (7/8)
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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)
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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)
ANALYSIS OF CONTINUOUS ROTARY VACUUM FI LTERS (8/8)
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[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)
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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
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[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:
For incompressible cake,
2
0
2
T
f
cfA
V
Ptt
b
ab
(To be continued)
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Example: Fi ltr ation of erythromycin using rotary vacuum fi lter
Solution (contd):
For incompressible cake,
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)
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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.
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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.
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CENTRIFUGAL
FILTRATION
* A combination of acentrifuge and a filter.
* Accumulated solids
can be washed.
CENTRIFUGAL F I LTRATION (2/8)
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( )
CENTRIFUGAL F I LTRATION (3/8)
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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.
CENTRIFUGAL F I LTRATION (4/8)
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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 ---
CENTRIFUGAL F I LTRATION (5/8)
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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-
CENTRIFUGAL F I LTRATION (6/8)
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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.
CENTRIFUGAL F I LTRATION (7/8)
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)/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
CENTRIFUGAL F I LTRATION (8/8)
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)/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
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[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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