FILTRATION UNIT 2.pptx

FILTRATION
• a pressure drop has to be applied across the medium (screen/cloth)
• Separation of solids from liquids by passing a suspension through a
permeable medium which retains the particles
• fluid flow through small holes of a screen/cloth
• retains the large solid particles as a separate phase (porous cake)
• passes the clear filtrate
• porous filter cake acts as a filter for the suspended particles
• flow resistance increases as filter medium becomes clogged or cake builds
up
Slurry
flow
Filtrate
Filter medium
Filter cake

FILTRATION
- the clear filtrate from the filtration or
• The valuable product may be:
- the solid cake ( solid particles build up)
• essentially a mechanical operation
• less demanding in energy than evaporation or drying
• Used to remove solid particles from or to :
1) Clarify juices
2) Extracts
3) Vegetable and fish oils.
4) Fermented beverages.
5) Recirculated cooking oil
6) Flume water, milk, and soy milk.
7) Separate potato starch from potato fruit water,
8) High-melting fats from vegetable oils in fractionation processes.
9) Crystals from mother liquors.
10) Chemically precipitated impurities.

• The cake gradually builds up on the medium and the resistance
to flow progressively increases.
• During the initial period of flow, particles are deposited in the
surface layers of the cloth to form the true filtering medium.
PRINCIPLE OF FILTRATION

• The most important factors that influence the
rate of filtration are:
(a) The drop in pressure from the feed to the far side of the
filter medium.
(b) The area of the filtering surface.
(c) The viscosity of the filtrate.
(d) The resistance of the filter cake.
(e) The resistance of the filter medium
and initial layers of cake.

FILTRATION
As time passes during filtration, either
- the filtrate flow rate diminishes or
- pressure drop rises
Constant-pressure filtration
- pressure drop is held constant
- flow rate allowed to fall with time
Constant -rate filtration (less common)
- pressure drop is progressively increased
Liquid passes through 2 resistance in series:
- cake resistance (zero at start & increases with time)
- filter medium resistance (impt. during early stages of filtration)
during washing, both resistances are constant, and filter medium resistance is
usually negligible

BASIC THEORY OF FILTRATION
where
L = thickness of cake (m)
A = filter cross section area (m2)
 = specific cake resistance (m/kg)

dV
dt
 pA
 cS
V
A
Rm










 = viscosity of filtrate ( Pa.s)
cs = dry mass of cake deposited per unit volume of filtrate (kg
solids/m3 filtrate)
V = volume of filtrate (m3)
Rm = resistance of filter medium to filtrate flow (m-1)
p = total pressure drop = pcake + pfilter medium (N/m2)
Rate of filtration = driving force/resistance

CONSTANT PRESSURE FILTRATION (BATCH)
Time of filtration:
Slope = Kp/2
Intercept = B
Filtration volume , V (m3)
t/V (s/m3)
Determination of constants in a constant-pressure filtration run

dt
dV
 cs
2
A p


 


V  
A p


 


Rm
KpV B

t 
Kp
V2
2
BV
A = filter cross section area (m2)
 = specific cake resistance (m/kg)
 = viscocity of filtrate ( Pa.s)
cs = dry mass of cake deposited per unit volume of filtrate (kg solids/m3 filtrate)
V = volume of filtrate (m3)
Rm = resistance of filter medium to filtrate flow (m-1)
p = total pressure drop = pcake + pfilter (N/m2)

Kp  cs
2
A p


 



B 
A p


 


Rm
where
t/V = (Kp/2)V + B  plot as a graph

Example 1 ( lab scale)
Data for the laboratory filtration of CaCO3 slurry in water at 298.2 K are reported as
follows at a constant pressure (-∆p) of 338 kN/m2. The filter area of the plate and frame
press was A=0.0439 m2 and the slurry concentration was cs=23.47 kg/m3. Calculate the
constant α and Rm from the experimental data given, where t is time in s and V is filtrate
volume collected in m3.

EXAMPLE 1 CONSTANT PRESSURE FILTRATION

t 
K pV 2
2
 BV
t
V

K p
2
V  B
A = 0.0439 m2
 = ? m/kg
Kp/2 = 3.00 x 106 s/m6
cs = 23.47 kg solids/m3 filtrate
-ΔP = 338 kN/m2
Rm = ?m-1
CaCO3 slurry in water at 298.2K (25oC)
Given: Solution:
B =6400 s/m3

A = 0.0439 m2
 = ? m/kg
Kp/2 = 3.00 x 106 s/m6
-ΔP = 338 kN/m2
Rm = ?m-1
Solution:
Kp = 6.00 x 106 s/m6

 cs
2
A p
 
B =6400 s/m3

 
A p
 
Rm
From Appendix A.2-4:  = 8.937 x 10-4 Pa.s
Substituting all the known values,
Rm = 10.63 x 1010m-1
α = 1.863 x 1011 m/kg

Example 2 (larger scale)
The same slurry used in Example 1 is to be filtered in a plate-and-frame press
having 20 frames and 0.873 m2 area per frame. The same pressure will be used in
constant pressure filtration. Assuming the same filter-cake properties and filter
cloth, calculate the time to recover 3.37 m3 filtrate

Plate and frame press (batch)

A = 0.0439 m2
Plate-and-frame press having 20 frames (0.873 m2 per frame)
-ΔP = 338 kN/m2
Time to recover 3.37 m3 filtrate = ?
Solution:
Kp = 6.00 x 106 s/m6

 cs
2
A p
 
B =6400 s/m3

 
A p
 
Rm
From Appendix A.2-4:  = 8.937 x 10-4 Pa.s
Substituting all the known values plus the new area ( 20 x 0.873 = 17.46 m2),
Kp = 37.93 s/m6
B = 16.10 s/m3
t 
K pV2
2
 BV 
37.93(3.37)2
2
16.1(3.37)269.7s

Washing of filter cake:
Vf= total volume of filtrate for entire period at the end of filtration (m3)
where
= rate of washing (m3/s)

dV
dt








f
B
V
K
dt
dV
f
p
f 






 1
Leaf filter:

dV
dt








f
 1
4
1
Kp
Vf
B
Plate-&-frame filter:
Time of washing:

t  washing liquid
rate of washing
Total cycle filter time = filtration time + washing time + cleaning time
Cleaning time - remove the cake, clean the filter, and reassemble the filter

Example 3
At the end of the filtration cycle in previous example, a total filtrate
volume of 3.37 m3 is collected in a total time of 269.7 s. The cake
is to be washed by through-washing in the plate-and-frame press
using a volume of wash water equal to 10% of the filtrate volume.
Calculate the time of washing and the total filter cycle time if
cleaning the filter takes 20 min

Example 3
At the end of the filtration cycle in previous example, a total filtrate
volume of 3.37 m3 is collected in a total time of 269.7 s. The cake
is to be washed by through-washing in the plate-and-frame press
using a volume of wash water equal to 10% of the filtrate volume.
Calculate the time of washing and the total filter cycle time if
cleaning the filter takes 20 min
Time of washing:

t  washing liquid
rate of washing

Washing liquid = 10% filtrate volume m3 = 0.337 m3
V = 3.37 m3 filtrate
Time washing = ?
Solution:
Substituting all the known values ,
Kp = 37.93 s/m6
B = 16.10 s/m3
Total filter cycle = ?
Cleaning time = 20 min

dV
dt






f
 1
4
1
KpVf  B
rate of washing (m3/s)=

dV
dt






f
 1
4
1
37.93(3.37)16.1








1.737x103m3/s
Time of washing,

t washing liquid
rate of washing
 0.337
1.737x103
194s
Total cycle filter time = 269.7s + 194s + (20 x 60)s = 1663.7 s =27.73 min

filtration time 269.7s

FILTRATION EQUIPMENTS
Cake accumulation and removal in batch mode
Filter funnel
Filter press
Leaf pressure filter
Vacuum leaf filter
Continuous cake accumulation and removal
Horizontal continuous filter
Rotary drum filter

CONSTANT PRESSURE FILTRATION( CONTINUOUS)
Filter medium resistance = negligible i.e. B = 0

tKp
V2
2
Time required for formation of cake:
Feed, filtrate & cake move at steady constant rates

V
tc
W
cX
cS
W = mass flowrate of slurry (kg/s)
cX= slurry concentration in mass fraction
Flowrate of filtrate:
where
tC= total cycle time
cs = dry mass of cake deposited per unit volume of filtrate


cX
1mcX
 = density of filtrate (kg/m3)
m = mass ratio of wet cake to dry cake (kg wet cake/kg dry cake)

CONSTANT PRESSURE FILTRATION( CONTINUOUS)
Filter medium resistance = negligible i.e. B = 0

V
Atc

2f p
 
tc
cs










1/2
f = fraction of the cycle used for cake formation
where
f = the fraction of submergence of the drum surface in the slurry
Short cycle times and/or the filter medium resistance  large:
t ftC
 Kp
V2
2
BV
Flowrate of filtrate per filter area:
In the rotary drum:

t  ftC
 f
n n = drum speed
where
V
Atc

Rm
/tc
 Rm
2
/tc
2
 2cs p
 f / tc

 








1/2
cs

EXAMPLE 4
EXAMPLE 4
A rotary drum filter having a 33 % submergence of the drum in the slurry is to be used
to filter aqueous CaCO3 slurry as given in Example 1 using a pressure drop of 67 kPa.
The solids concentration in the slurry is 0.191 kg solid/kg slurry and the filter cake is
such that the kg wet cake/kg dry cake = m = 2. The density and viscosity of the filtrate
can be assumed as those of water at 298.2 K.
Calculate the filter area needed to filter 0.778 kg slurry/s. The filter cycle time is 250 s.
The specific cake resistance can be represented
–ΔP is in Pa and α is m/kg

  (4.37x109
)P0.3

Early stages of filtration - resistance of cake = negligible
CONSTANT RATE FILTRATION
For a constant rate (dV/dt) in m3/s:
KV is in N/m5
where
C is in N/m2
Cake is incompressible:
Pressure  as cake thickness  & volume of filtrate 
p
cs
2
A
dV
dt












V 
Rm
A
dV
dt










KV
V C Slope = KV
Intercept = C
Filtration volume , V (m3)
-p(N/m2)
Determination of constants in a constant-rate filtration run
Total volume V:

V t dV
dt
Hence: p
cs
2
A
dV
dt








2












t
Rm
A
dV
dt










Slurry fed to the filter by a positive-displacement pump

CONSTANT RATE FILTRATION
Example 7
The filtration equation for filtration at a constant pressure of 38.7
psia ( 266.8 kPa) is
where t is in s, -ΔP in psia, and V in liters. The specific resistance of
the cake is independent of pressure. If the filtration is run at
constant rate of 10 liters/s, how long will it take to reach 50 psia?
t/V = 6.10 x 10-5 V + 0.01

FILTRATION UNIT 2.pptx

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FILTRATION UNIT 2.pptx