Filtration Report.

FILTRATION OBJECTIVE: • • •

To study the theory of filtration To determine the average cake resistance for the given slurry and the resistance of the filter medium used in the filtration process. To determine the time for filtration and hence the performance of the equipment based on the volume of the filtrate collected.

THEORETICAL STUDY: INTRODUCTION: The separation of particles is done on the basis of the state of the particles to be separated. There are a group of separation techniques where the separation is accomplished by the differences in the mechanical- physical forces in the system and not the molecular/ chemical forces. These mechanical- physical forces act on the particles, liquids, or the mixtures of particles and the liquids and not necessarily on the individual molecules.

Filtration is the removal of solid particles from a fluid by passing the fluid through a filtering medium, or septum, on which the solids are deposited. Filtration is the most common application of the flow of fluids through packed beds. It is analogous to the filtration carried out using a filter paper on a funnel or using a Buckner funnel in a laboratory. The objective is to separate the solid from the fluid in which it is present. The separation is carried out by forcing the fluid through the porous membrane. The solid particles are trapped within the pores of the membrane and a build up as a layer is seen on the surface of the membrane. Industrial filtrations range from simple straining to highly complex separations. The fluid may be a liquid or a gas; the valuable stream from the filter may be the fluid, or the solids or both. Sometimes it is neither, as when waste solids may be separated from the waste liquid prior to disposal. In industrial filtrations, the solid content ranges from a trace to a very high percentage. Often the feed is subjected to a pretreatment process to increase the filtration rate, as by heating recrystallizing, or by adding filter aids.

1

CLASSIFICATIONS:

Process Filtration Centrifugation Sedimentation cyclone separator Electrostatic precipitator Magnetic separator Screening

Phase Liquid – Solid Liquid – Solid & Liquid – Liquid Liquid – Solid Gas – Solid & Gas – Liquid Gas – Solid Solid – Solid & Solid – Liquid Solid – Solid

Separation Pressure reduction Centrifugal force Gravity Flow Electric field Magnetic field Size of particles

Industrial Filters: The classification of different equipments according to the different driving forces are listed below •

Pressure filters
Plate and frame filter press
Leaf filter
Sparkler filter
Candle filter
Line, Cartridge filter

•

Vacuum filters
Nutsche filter
Rotary Drum filters

•

Centrifugal filters
Top driven centrifuge
Bottom driven centrifuge

2

Mechanisms of filtration:

a) Cake filter

b) Clarifying filter

c) Cross- flow filter

•

Cake filters separate relatively large amounts of solids as a cake or crystals or sludge. Often they include provisions for washing the cake and removing some of the liquids form the solids before discharge.

•

Clarifying filters remove small amounts of solids to produce a clean gas or sparkling clear liquids, like beverages. The solid particles are trapped in the filter medium as shown. Clarifying filters differ from screens in that the pores in the filter medium are much larger in diameter than the particles to be removed.

•

In a Cross-flow filter the feed suspension flows under pressure at a very high velocity across the filter medium. A thin layer of solids may form on the surface of the medium but the high liquid velocity keeps the layer from building up. The filter medium may be ceramic, metal or a polymer membrane with pores small enough to exclude most of the suspended particles. Some liquid passes as a clear filtrate and leaves behind a suspension of solids.

3

CAKE FILTRATION: In cake filtration, the filter cloth often acts as no more than a substrate for building up the first thin layers of the filter cake, which itself then acts a filter to trap more and more particles of smaller and smaller size. The first particles form bridges over the pores of the medium. After this, smaller particles pass through the filter cloth, leading to a turbid liquid (filtrate). As soon as the first layer of the particles has accumulated on the filter medium, this cake will then act as the actual filter medium. The first solid particles enter the pores of the medium and are immobilized, but soon others begin to collect on the septum surface. After the brief initial period, a cake of apparent thickness builds up on the surface and must be periodically removed. Naturally, the pressure drop through the filter cake increases and flow decreases as the filter cake gets thicker. If the slurry contains too many fines, all pores of the cloth may become blocked.

DERIVATION: PRESURE DROP OF FLUID THROUGH THE FILTER CAKE: The solid-fluid suspension to be filtered is passed under pressure through a medium which allows the flow of the suspending fluid but retains the suspended particles to form a cake of the upstream side of the medium.

At time ‘t’ sec, thickness of cake = L m filter cross-sectional area = A m2 linear velocity for flow = v m/s for A m2

Section through a filter medium at definite time ‘t’ sec from the start of filtration For laminar flow, in a packed bed of particles, the Carman-Kozeny relation has been shown to apply to filtration. It is given by, …1

4

where,

∆pc

->

Pressure drop across the cake in N/m2 (lbf/ft2)

ν

->

Open tube velocity in m/s (ft/s)

D

->

Diameter in m (ft)

L

->

Length in m (ft)

µ

->

Viscosity in Pa.s or kg/m.s (lbm/ft.s)

k1

->

constant = 4.17 for random particles of definite size and shape

ε

->

Void fraction or porosity of cake

S0

->

Specific surface area of particle in m2 per volume of solid particle

The linear velocity depends on cross sectional are according to the relation, …2

where,

A

->

filter area in m2 (ft 2)

V

->

total m3 (ft3) volume of filtrate collected upto time ‘t’ sec

The cake thickness ‘L’ is related to volume by Material Balance …3

where,

Cs

->

kg solids /m3 (lbm/ft3) of filtrate

ρp

->

density of solid particles in the cake in kg/m3 (lbm/ft3) solid

The final term of this equation is the volume of filtrate held in the cake. This is usually small and neglected.

Substituting Eq. 2 in Eq. 1, using Eq. 3 to eliminate L, we get,

5

where,

α

->

Specific cake resistance in m/kg (ft/lbm), given by

For filter medium resistance,

where,

Rm

->

Resistance of the filter medium to filtrate flow ( m-1)

Rm is an empirical constant that includes the resistance to flow of the piping leads to and from the filter and the filter medium resistance. As α and Rm are in series,

where,

The volume of the filtrate V can be related to W, the kg of accumulated dry cake solids, as

where,

Cx m ρ

-> -> ->

mass fraction of the solids in the slurry mass ratio of dry cake to wet cake density of filtrate in kg/m3

Specific Cake Resistance: Specific cake resistance is a function of void fraction ε and S0. It is also a function of pressure as pressure can affect ε. By conducting various experiments at various pressure drops, the variation of α with ∆p can be calculated. If α is independent of -∆p, then the sludge is incompressible. Usually, α increases with –∆p, since most cakes are somewhat compressible.

6

An empirical equation is often used as,

where αo and s are empirical constants. The compressibility constant s is zero, for incompressible cakes. Usually, it varies between 0.1 and 0.8 .

Constant Pressure Conditions: For a Batch process,

where,

Kp is in s/m6 and B is in s/m3

For constant pressure, constant α, and incompressible cake, V and t are the only variables. Integrating to obtain the time of filtration in t s,

Dividing by V,

The plot of t/V Vs V gives a straight line with an intercept. The slope gives Kp and the intercept gives B.

7

Constant Rate Conditions: These conditions are attained if the slurry is fed to the filter by a positive displacement pump. Here, the flow rate of the slurry through the filter medium is kept a constant.

where,

where

Kv is in N/m5 , C is in N/m2.

Assuming that the cake is incompressible, Kv and C are constants characteristic of the slurry, cake, rate of the filtrate collected. A plot of pressure –∆p Vs V, the total volume of the filtrate collected, gives a straight line for a constant rate dV/dt. The slope of the line is Kv and the intercept is C. The pressure increases as the cake thickness increases and as the volume of the filtrate collected increases. The equations can also be rearranged in terms of –∆p and time ‘t’ as variables. At any moment during the filtration, the total volume V is related to the rate and total time t as,

Substituting in the first equation,

For a case when the specific cake resistance is not a constant, but varies as

this can be substituted as the value for α to obtain a final equation.

8

CENTRIFUGAL SEPARATION PROCESS: Centrifugal settling or sedimentation: Sometimes gravity settling may be too slow because of the closeness of the densities of the particles and the fluid, or because of forces holding the particles together as in the case of emulsions. In such cases, centrifugal separation is employed. Forces developed in centrifugal separation: Centrifugal separators make use of the principle that an object whirled about an axis or centre point at a constant radial distance from the point is acted on by a force. The object being whirled about an axis is constantly changing direction and is thus accelerating, even though the rotational speed is constant. This centripetal force acts in a direction toward the centre of rotation. If the object the object being rotated is a cylindrical container, the contents of fluid and solids exert an equal and opposite force, called centrifugal force, outward to the to the walls of the container. This is the force that causes settling or sedimentation of particles through a layer of liquid through a bed of filter cake held inside a perforated rotating chamber.

Here, a cylindrical bowl is shown rotating with a slurry feed of solid particles and liquid being admitted at the centre. The feed enters and is immediately thrown outward to the walls of the container, as in section (b). The liquid and solids are now acted upon by the vertical gravitational force and the horizontal centrifugal force. The centrifugal force is usually so large that the force of gravity may be neglected. The liquid layer then assumes the equilibrium position with the surface almost vertical. The particles settle horizontally outward and press against the vertical bowl wall. In section (c) two liquids having different densities are being separated by the centrifuge. The more dense fluid will occupy the outer periphery, since the centrifugal force is greater on the more dense fluid. 9

Equations for Centrifugal forces: In circular motion, the acceleration from the centrifugal force is

where,

ac

->

acceleration due to the centrifugal force in m/s2

r

->

radial distance from centre of rotation in m

ω

->

angular velocity in rad/s

The centrifugal force Fc in N, acting on the particle is given by

ince ω = ν/r, where ν is the tangential velocity of the particle in m/s

Often rotational speeds are given as N rev/in and

and Substituting

The gravitational force on a particle is given by, where, g is the acceleration of gravity = 9.80665 m/s2 In terms of gravity force, the centrifugal force is given as

Hence, the force developed is r ω2/g times gravity force. This is expressed as equivalent to so many g forces.

10

Equations for rates of settling in centrifuges: General equation for settling: If a centrifuge is used for sedimentation (removal of particles by settling), a particle of a given size can be removed from the liquid in the bow if sufficient residence time of the particle in the bowl is available for the particle to reach the wall. For a particle moving radially at its terminal setting velocity, the diameter of the smallest particle removed can be calculated. The feed enters at the bottom and it is assumed all the liquid moves upward at a uniform velocity, carrying solid particles with it. The particle is assumed to be moving radially at its terminal settling velocity vt. The trajectory or path of the particle is shown in the above figure. A particle of a given size is removed from the liquid if sufficient residence time is available for the particle to reach the wall of the bowl, where it is held. The length of the bowl is b m.

At the end of the residence time of the particle in fluid, the particle is at a distance rb m from the axis of rotation. If rb < r2, then the particle leaves the bowl with the fluid. If rb = r2 , it is deposited on the wall of the bowl and effectively removed from the liquid. For settling in the stokes’ law range, the terminal settling velocity at a radius r is obtained by,

where,

Vt

->

Settling velocity in the radial direction in m/s

Dp

->

Particle diameter in m

ρp

->

Particle density in kg/m3

ρ

->

Liquid density in kg/m3

µ

->

Liquid viscosity in Pa.s

If hindered settling occurs, the RHS of the above equation is multiplied by the factor (ε2Ψp).

11

Since,

Integrating between limits r=r1 at t=0 and r=r2 at t=tr

The residence time tr is equat to the volume of liquid V m3 in th bowl divided by the feed volumetric flow rate q in m3/s. The volume V is given by,

Substituting and solving for q,

Particles having diameters smaller than that calculated, will not reach the wall of the bowl and will go out with the exit liquid. Larger particles will reach the wall and be removed from the liquid. The critical diameter Dpc is be defined as the diameter of a particle that reaches ½ the distance between r1 and r2. This particle moves a distance of half the liquid layer or (r2-r1)/2 during the time this particle is in the centrifuge. The integration is then between r = (r1+r2)/2 at t = 0 and r = r2 at t = tr. Then we obtain,

At this flow rate qc, particles with a diagram greater than Dpc will predominantly settle to the wall and most smaller particles will remain in the liquid. 12

Sigma values and scale-up of centrifuges: A useful characteristic of a tubular-bowl centrifuge can be expressed as,

where νt is the terminal settling velocity of the particle in a gravitational field

Σ is physical characteristic of the centrifuge and not of the fluid- particle syste, being separated. For a special case of settling of a thin layer,

The value of Σ is really the area in m2 of a gravitational settler that will have the same sedimentation characteristics as the centrifuge for the same feed rate. To scale up from a laboratory test of q1 and Σ1 to q2 ( for νi 1 = νi 2 ),

This scale-up procedure is dependable for similar type and geometry centrifuges and if the centrifugal forces are within a factor of 2 from each other. If different configurations are used, efficiency factors E should be used where q1/Σ1E1 = q2Σ2E2 . These efficiencies are determined experimentally and values for different types of centrifuges are available.

Co- relation with the cake filtration theory: The theory of constant pressure filtration can be modified and used where the centrifugal force causes the flow instead of the impressed pressure difference. The equation will be derived for the case where a cake has already been deposited. The inside radius of the basket is r2, ri is the inner radius of the face of the cake, and r1 is the inner radius of the liquid surface. We will assume that the cake is nearly incompressible so that an average value of α can be used for the cake. Also the flow is laminar. If we assume a thin cake in a large-diameter centrifuge, then the area A for flow is approximately constant. The velocity of the liquid is,

13

where

q is the filtrate flow rate in m3/s and v is the velocity.

From the above we get,

where mc = csV , mass of cake in kg deposited on the filter. For a hydraulic head of dz m, the pressure drop is

In a centrifugal field, g is replaced by rω2 and dz by dr. Then,

Integrating between r1 and r2,

Combining the equations and solving for q,

For the case where the flow area A varies considerably with the radius, the following has been derived

Where A2 =2πr2b (area of filter medium), Āl=2πb(r2-ri)/ln(r2/ri) (logarithmic cake area), and Āa=(ri+r2)πb (arithmetic mean cake area). This equation holds for a cake of a given mass at a given time.

14

FILTER MEDIUM: It is defined as a material that is permeable to one or more components of a mixture, solution or suspension, and is impermeable to remaining components. Its principal role is to cause a clear separation of particulates from the fluid with minimum consumption of energy.

Properties of filter media: Filtration Properties Particle retention Flow resistance Dirt holding capacity Tendency to blind Cake discharge characteristics

Filter oriented properties Ability to be fabricated Resistance to stretch Vibrational stability Rigidity Strength

Types of filter media used: 1. Woven fabrics  Natural like cotton, wool  Synthetic includes polymers, metal carbons 2. Non woven media  Felts and needlefelts  Paper (cellulose and glass)  Filter sheets 3. Metal sheets  Perforated  Woven wire 4. Rigid porous media  Ceramics and stoneware  Carbon  Sintered metals 5. Plastic sheets  Woven monofilaments  Fibrillated film  Porous sheets

15

Feed oriented properties Chemical/thermal stability Biological stability Ad/absorptive characteristics Disposability Re-use

6. Membranes  Polymeric  Ceramic  Metal 7. Cartridges  Sheet fabrications  Bonded beds  Yarn wound

FILTER AIDS: It is frequently necessary to modify the slurry in order to provide an acceptable filtration rate, washing rate, or final cake moisture content. The most common treatment is the addition of flocculating agents. Filter aids are non-compressible particulate substances that are either mixed with the product slurry to improve filtration rate and reduce cloth blinding, or laid down to precoat the filter before the slurry is introduced to improve retention of fine particles. They are granular or fibrous solids capable of forming a highly permeable filter cake in which very fine solids are trapped. Application of filter aids allows the use of a much more permeable filter medium than would be provided by depth filtration.

Properties of Filter Aids: 

Filter aids should have low bulk density to minimize settling and aid good distribution on the filter-medium surface.  They should be porous and capable of forming a porous cake to minimize flow resistance.  They must be chemically inert to the filtrate. Examples: Diatomaceous earth (also called diatomite), which is an almost pure silica. Expanded perlite, particles of “puffed” lava that are principally aluminum alkali silicate. • Cellulosic fibers (ground wood pulp) are sometimes used when siliceous materials cannot be used. • Carbon, Gypsum, salt, Fine sand, Starch, and precipitated CaCO3.

• •

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LAB – SCALE EXPERIMENT: (CONSTANT PRESSURE PROCESS) OBJECTIVE: • To determine the average cake resistance for the given slurry and the resistance of the filter medium used in the constant pressure filtration process. • To determine the rating of the equipment, given the volume of the filtrate to be collected. APPARATUS REQUIRED: Sample to be filtered, pressure filtration set- up – Buchner funnel, filter medium of appropriate pore size, purified water for cleaning purposes THEORY:

In filtration, suspended solids in a fluid of liquid or gas are removed by using a porous medium that retains the particles as a separate phase or cake and passes the clear filtrate. The laboratory filtration is performed using a Buchner funnel. The liquid is caused to flow though the filter cloth by a vacuum at the exit end. The slurry consists of the liquid and the suspended particles. The passage of the particles is blocked by the small openings in the exit of the filter cloth. A support with relatively large holes is used to support the filter cloth. Solid particles build up in the form of a filter cake. The cake itself acts as a filter. The cake thickness increases with time. As the cakes builds up, the resistance to flow also increases.

17

From the derivation of the Pressure drop through the filter cake,

This equation is in the form of a straight line, y = mx + c, where Kp is the slope of the straight line and B is the intercept of the line in the y - axis.

A plot of t/V vs V thus, gives a straight line, from which the values of Kp and B can be calculated. Substitution of these values in the above expressions gives the values of α and Rm. ASSUMPTIONS MADE: •

There is negligible solid motion, that is, the solid phase velocity is considered zero.

•

There is a constant and time invariant wet to dry cake mass ratio in the slurry, that is, the concentration of the solids in the slurry is a constant.

•

Absolute vacuum conditions are assumed through the entire filtration process

•

Rm is a constant throughout the filtration process

•

The properties of the newly formed cake layer (concentration and specific resistance) are similar to the average filter cake properties.

EXPERIMENTAL PROCEDURE: 1) Settling Test: Allow about 1 liter of slurry to settle on the lab bench in a beaker. The material should settle and produce a clear liquor phase in less than 30 minutes. If the slurry remains cloudy for 30 minutes or more, the product will probably be difficult to isolate. The crystallization conditions may have to be altered to increase particle size or reduce fines.

18

2) Cake Permeability Test: • • •

Filter the slurry sample through a Buchner funnel with vacuum to obtain a filter cake about 2" thick. Measure the rate at which the clear ML is filtered through the cake. If the liquors filter at a rate of 1 gpm/ft2 of filter area or greater, then it is a good candidate for centrifugation or other large-scale filtration. If the liquor filtration rate is less than 0.5 gpm/ft2, it means that the slurry contains too many fines, or that the product is amorphous (noncrystalline) in nature and too easily compressed to allow liquid to drain through. Again, a change in crystallization conditions may be required.

3) Measuring Filtration data: • • • • •

Determine the solid concentration in the slurry sample before carrying the filtration tests. Pour the sample into the funnel and apply vacuum to the container Measure the volume of filtrate collected at various time intervals. The intervals may be increased progressively to compensate for the drop in the filtrate flow rate. Remove the cake from the filter and dry it and weigh it. This can also be used for the calculation of the concentration of the dry solids in the sample. The plot of time and filtrate data gives the desired parameters characteristic of the slurry.

PRECAUTIONS:  The cake should not crack at any point. The filtration should be stopped if a crack is observed in the filter cake.  Do not allow the cake to completely drain before adding more slurry.  The slurry should be kept as homogeneous as possible during the test.  The initial tests should be performed to check the amount of fines in the slurry.  The concentration of the solids in the slurry should be calculated initially before the experiment.  The rate at which the slurry is added into the funnel should not be altered too much so a smooth graph can be obtained.  If the filtration pressure is increased too quickly, the filtration rate will increase too fast. This should be avoided.  The selection of the filter medium should be done with the knowledge of the suspension characteristics of the slurry.

19

GRAPHS AND CALCULATIONS: COMPOUND 2 TIME t (sec) 8 12 22 25 40 56 75 92 115 147 175 203 236 283 345 385

t/V 0.160 0.120 0.147 0.125 0.160 0.187 0.214 0.230 0.256 0.294 0.318 0.338 0.363 0.404 0.460 0.497

VOLUME V (Ml) 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 775

t/V

COMPOUND-2 0.540 0.510 0.480 0.450 0.420 0.390 0.360 0.330 0.300 0.270 0.240 0.210 0.180 0.150 0.120 0.090 0.060 0.030 0.000 50

100

150

200

250

300

350

400

450

500

Volum e

From the graph, Kp =

1.06 * 10-5

B = 0.02

20

550

600

650

700

750

775

where µ

->

Viscosity in Pa.s or kg/m.s

= 1 centi Poise = 1 * 10-3 Pa.sec

A

->

filter area in m2

= ( 3.14) r2

∆pc

->

Pressure drop across the cake in N/m2

= 95.99 kPa

Cs

->

kg solids /m3 of filtrate

α

->

Specific cake resistance in m/kg

Rm

->

Resistance of the filter medium to filtrate flow (m-1)

Cs = 7.86 % w / w Assume 100 kg of slurry 7.86 kg of solid corresponds to 7.86 kg of solid corresponds to Density of ML Volume of ML

= = =

= 0. 0095 m2

100 kg of slurry 92.14 kg of ML

850 kg/m3 92.14/ 850 0.1084 m3

7.86 kg of wet cake corresponds to 0.1084 m3 of ML 175 kg of wet cake corresponds to 2.4134 m3 of ML which equals 170 kg of dry cake Cs

= =

170 kg of dry cake / 2.4134 m3 of ML 70.44 kg/ m3 of ML

Substituting in the equations of Kp and B,

α Rm

= =

1.3036 * 10-3 m/kg 1.8238 * 103 m-1

21

COMPOUND 4 Total time VOLUME V (Ml) 100 200 250 300 400 450 500 550 600 700 800 850

TIME t (sec) 7 10 14 19 22 26 28 34 36 46 55 70

111 secs t/V 0.070 0.050 0.056 0.063 0.055 0.058 0.056 0.062 0.060 0.066 0.069 0.082

COMPOUND-4

0.090 0.080 0.070

t/V

0.060 0.050 0.040 0.030 0.020 0.010 0.000 100

250

400

Volume 500

600

From the graph, Kp = 6.71 * 10-5

B = 0.041

22

800

where µ

->

Viscosity in Pa.s or kg/m.s

= 1 centi Poise = 10-3 Pa.sec

A

->

filter area in m2

= ( 3.14) r2

∆pc

->

Pressure drop across the cake in N/m2

= 95.99 kPa

Cs

->

kg solids /m3 of filtrate

α

->

Specific cake resistance in m/kg

Rm

->

Resistance of the filter medium to filtrate flow (m-1)

= 10 % w / w Cs Assume 100 kg of slurry 10 kg of solid corresponds to 10 kg of solid corresponds to Density of ML Volume of ML

= = =

= 0. 0095 m2

100 kg of slurry 90 kg of ML

900 kg/m3 90 / 900 0.10 m3

10 kg of wet cake corresponds to 0.10 m3 of ML 175 kg of wet cake corresponds to 1.75 m3 of ML which equals 155 kg of dry cake Cs

= =

155 kg of dry cake / 1.75 m3 of ML 88.5714 kg/ m3 of ML

Substituting in the equations of Kp and B,

α Rm

= =

6.563 * 10-3 m/kg 3.7388 * 103 m-1

23

COMPUOND 5 TIME t (sec) 7 10 13 20 25 28 34 36 42 48 53 70 80

VOLUME V (Ml) 100 150 200 300 350 400 450 500 550 600 650 700 750

t/V 0.070 0.067 0.065 0.067 0.071 0.070 0.076 0.072 0.076 0.080 0.082 0.100 0.107

COMPOUND -5 0.120 0.100

t/V

0.080 0.060 0.040 0.020 0.000 100

200

350

450

V

550

From the graph, Kp = 2.074 * 10 -4

B = 0.040

24

650

750

where µ

->

Viscosity in Pa.s or kg/m.s

= 0.6 centi Poise = .6 * 10-3 Pa.sec

A

->

filter area in m2

= ( 3.14) r2

∆pc

->

Pressure drop across the cake in N/m2

= 95.99k Pa

Cs

->

kg solids /m3 of filtrate

α

->

Specific cake resistance in m/kg

Rm

->

Resistance of the filter medium to filtrate flow (m-1)

Cs

= 11.60 % w / w

Assume 100 kg of slurry 11.60 kg of solid corresponds to 11.60 kg of solid corresponds to Density of ML Volume of ML

= = =

= 0.0095 m2

100 kg of slurry 88.4 kg of ML

800 kg/m3 88.4 / 800 0.1105 m3

11.60 kg of wet cake corresponds to 0.1105 m3 of ML 175 kg of wet cake corresponds to 1.667 m3 of ML which equals 140 kg of dry cake Cs

= =

140 kg of dry cake / 1.667 m3 of ML 83.98 kg/ m3 of ML

Substituting in the equations of Kp and B,

α Rm

= =

3.567 * 10-3 m/kg 3.647 * 103 m-1

25

COMPOUND 6

TIME t (sec) 7 10 13 17 20 24 27 31 35 41 46 51 61 70 80 103

VOLUME V (Ml) 100 250 300 400 450 500 550 600 650 700 750 800 850 900 950 1000

t/V 0.070 0.040 0.043 0.043 0.044 0.048 0.049 0.052 0.054 0.059 0.061 0.064 0.072 0.078 0.084 0.103

COMPOUND-6

t/V

0.110 0.100 0.090 0.080 0.070 0.060 0.050 0.040 0.030 0.020 0.010 0.000 100

250

300

400

450

500

550

600

650

700

750

800

V

From the graph, Kp = 1.05 * 10-4

B = 0.028 26

850

900

950 1000

where µ

->

Viscosity in Pa.s or kg/m.s

= 0.37 cPoise = 0.37 * 10-3 Pa.sec

A

->

filter area in m2

= ( 3.14) r2

∆pc

->

Pressure drop across the cake in N/m2

= 95.99 kPa

Cs

->

kg solids /m3 of filtrate

α

->

Specific cake resistance in m/kg

Rm

->

Resistance of the filter medium to filtrate flow (m-1)

Cs

= 9.25 % w / w

Assume 100 kg of slurry 9.25 kg of solid corresponds to 9.25 kg of solid corresponds to Density of ML Volume of ML

= = =

= 0. 0095 m2

100 kg of slurry 90.75 kg of ML

800 kg/m3 90.75 / 800 0.1134 m3

9.25 kg of wet cake corresponds to 0.1134 m3 of ML 300 kg of wet cake corresponds to 3.68 m3 of ML which equals 180 kg of dry cake Cs

= =

180 kg of dry cake / 3.68 m3 of ML 48.91 kg/ m3 of ML

Substituting in the equations of Kp and B,

α Rm

= =

5.0265 * 10-2 m/kg 6.9 * 104 m-1

27

COMPOUND 7

TIME t (sec) 10 14 18 25 34 80 86 90 108 117 150

[2 FILTER CLOTHS]

VOLUME V (Ml) 100 200 250 300 350 600 650 700 750 800 825

t/V 0.100 0.070 0.072 0.083 0.097 0.133 0.132 0.129 0.144 0.146 0.182

COMPOUND-7 0.200 0.180 0.160 0.140

t /V

0.120 0.100 0.080 0.060 0.040 0.020 0.000 100

200

250

300

350

600

650

700

750

V From the graph, Kp = 3.5 * 10 -4

B = 0.03

where

28

800

825

µ

->

Viscosity in Pa.s or kg/m.s

= 1 centi Poise = 1 * 10-3 Pa.sec

A

->

filter area in m2

= ( 3.14) r2

∆pc

->

Pressure drop across the cake in N/ m2

= 95.99 kPa

Cs

->

kg solids /m3 of filtrate

α

->

Specific cake resistance in m/kg

Rm

->

Resistance of the filter medium to filtrate flow (m-1)

Cs

= 1.55 % w / w

Assume 100 kg of slurry 1.55 kg of solid corresponds to 1.55 kg of solid corresponds to Density of ML Volume of ML

= = =

= 0. 0095 m2

100 kg of slurry 98.45 kg of ML

800 kg/m3 98.45 / 800 0.123 m3 of ML

1.55 kg of wet cake corresponds to 0.123 m3 of ML 80 kg of wet cake corresponds to 6.348 m3 of ML which equals 65 kg of dry cake Cs

= =

65 kg of dry cake / 6.348 m3 of ML 10.2386 kg/ m3 of ML

Substituting in the equations of Kp and B,

α Rm

= =

49.35 m/kg 4.559 * 104 m-1

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COMPOUND 8

TIME t (sec) 13 20 24 31 40 52 62 68 75 88 95 110 118 125 150

[2 FILTER CLOTHS]

VOLUME V (Ml) 100 150 200 250 300 350 400 450 500 550 600 650 675 700 750

t/V 0.130 0.133 0.120 0.124 0.133 0.149 0.155 0.151 0.150 0.160 0.158 0.169 0.175 0.179 0.200

COMPOUND-8

0.220 0.200 0.180 0.160 0.140

t/V

0.120 0.100 0.080 0.060 0.040 0.020 0.000 100

200

300

400

V

500

600

From the graph, Kp = 2.16 * 10 -4

B = 0.105

30

675

750

where µ

->

Viscosity in Pa.s or kg/m.s

= 1 centi Poise = 1 * 10-3 Pa.sec

A

->

filter area in m2

= ( 3.14) r2

∆pc

->

Pressure drop across the cake in N/ m2

= 95.99 kPa

Cs

->

kg solids /m3 of filtrate

α

->

Specific cake resistance in m/kg

Rm

->

Resistance of the filter medium to filtrate flow (m-1)

Cs

= 11.21 % w / w

Assume 100 kg of slurry 11.21 kg of solid corresponds to 11.21 kg of solid corresponds to Density of ML Volume of ML

= = =

100 kg of slurry 88.79 kg of ML

900 kg/m3 88.79 / 900 0.09865 m3

11.21 kg of wet cake corresponds to 0.09865 m3 of ML 240 kg of wet cake corresponds to 2.112 m3 of ML Cs

= =

240 kg / 3.68 m3 of ML 113.636 kg/ m3 of ML

Substituting in the equations of Kp and B,

α Rm

= =

1.6466 * 10-4 m/kg 9.575 * 104 m-1

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= 0. 0095 m2

COMPOUND 9

TIME t (sec) 9 12 16 22 27 31 36 40 47 52 59 68 79 87 93 97 105 113 122 146

[2 FILTER CLOTHS]

VOLUME V (Ml) 100 150 200 250 300 350 400 450 500 550 600 650 700 750 775 800 825 850 875 900

t/V 0.090 0.080 0.080 0.088 0.090 0.089 0.090 0.089 0.094 0.095 0.098 0.105 0.113 0.116 0.120 0.121 0.127 0.133 0.139 0.162

COMPOUND-9 0.180 0.160

0.140 0.120

t/V

0.100 0.080

0.060 0.040

0.020 0.000 100

200

300

400

500

600

V

32

700

775

825

875

From the graph, Kp = 2.84 * 10 -4

B = 0.0090

where µ

->

Viscosity in Pa.s or kg/m.s

= 0.6 centi Poise = 0.6 * 10-3 Pa.sec

A

->

filter area in m2

= ( 3.14) r2

∆pc

->

Pressure drop across the cake in N/ m2

= 95.99 kPa

Cs

->

kg solids /m3 of filtrate

α

->

Specific cake resistance in m/kg

Rm

->

Resistance of the filter medium to filtrate flow (m-1)

Cs

= 6.35 % w / w

Assume 100 kg of slurry 6.35 kg of solid corresponds to 6.35 kg of solid corresponds to Density of ML Volume of ML

= = =

= 0. 0095 m2

100 kg of slurry 93.65 kg of ML

800 kg/m3 93.65 / 800 0.117 m3

6.35 kg of wet cake corresponds to 0.117 m3 of ML 130 kg of wet cake corresponds to 2.395 m3 of ML which equals 100 kg of dry cake Cs

= =

100 kg of dry cake / 2.395 m3 of ML 41.75 kg/ m3 of ML

Substituting in the equations of Kp and B,

α Rm

= =

9.821 * 10-4 m/kg 13.678 * 102 m-1

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CONCLUSIONS: o Filtration was studied theoretically and on a lab scale. o Using the available equipment (Buchner funnel set- up), the resistance of the filter medium and the specific cake resistance of different cakes were calculated. o It was observed that as the cake thickness increases, the pressure difference across the cake also increases. Thus, the resistance to the fluid flow by the cake also increases. o A graph was plotted between t/V Vs Volume of the filtrate collected over the filter medium. The slope and the intercept of this graph gave values characteristic of the slurry. o The conclusions of the experiments conducted are applicable to Pressure filtration equipments such as Sparkler, Line, Candle, Cartridge filters and Vacuum filters such as Nutsche filters and ANFDs. o Scaling up of a centrifuge requires Pilot plant centrifuge devices for conducting lab- scale experiments. It requires the application of other equations determined by scale up factors after the calculation of the specific cake resistance and resistance of the filter medium. •

Theoretically, specific cake resistance values must lie in the range of 109 to 1012 and the resistance of the medium should lie in the range of 104

•

The DEVIATIONS in the values calculated from the graphs might be because  The Pressure assumed in the experimental conditions might not have been absolute vacuum.  The concentration of the sample was assumed to be the same as the batch concentration. This might not have been correct.  The graphical calculation of the slope and the intercept of the graph plotted might not have been precise.

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ACKNOWLEDGEMENTS

I thank Mr.Chellapandi, G.M., Process Engineering, for trusting my ability and knowledge and allowing me to work on a project. I extend my heartfelt gratitude to Mr. S. M. Krishnakumar and Mr. T. Sivaramakrishna for their wonderful support, guidance and assistance through my project. I should not fail to thank them for their immense encouragement for the same. I thank the Process Development Lab technicians for their assistance and guidance through the project. I am forever indebted to all the Phase-2 employees for allowing me to observe and study the equipments and letting me learn the ways of work in the industry. I thank the H.R. Department for their timely assistance. I thank Mr.Jacob Jayakumar for his help with the transportation, without which things wouldn’t have been so easy. I also thank Mr.Jai Ganesh for the assistance at the canteen. Lastly, I thank Orchid Chemicals & Pharmaceuticals for this wonderful opportunity provided to young, aspiring students to increase their aptitude and enrich their knowledge by working in the company for a brief period of time and providing them with the required exposure to be successful individuals.

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