Cap 09 Filtracion

7/25/2019 Cap 09 Filtracion

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2003 by CRC Press LLC

235

9Filtration

9.1 IntroductionFiltration is a unit operation whose purpose is the separation of an insolublesolid present in solidliquid suspension by passing the suspension througha porous membrane that retains the solid particles. The porous membraneis called the filtering medium, while the retained particles in the membraneform a layer known as cake and the liquid that passes through the porousmembrane and does not contain solids is called a filtrate.

The desired phase in filtration can occur in the filtrate, cake, or both. When

the desired solid part is obtained, the cake should be washed to eliminatethe impurities it may contain. Filtration can operate under simple gravity toobtain the flow of filtrate that crosses the filtrating medium, or by applyinga pressure higher than the atmospheric pressure to the front part of thefiltering medium or vacuum to the back side. These processes are known aspressure filtration and vacuum filtration, respectively.

Filtration is an operation widely used in the industry in general; in theparticular case of the food industry, three characteristic types of filtrationcan be distinguished. The first includes suspensions containing significant

amounts of insoluble solids that, when filtered, form a cake on the filtratingmedium; thus, obtaining the solid, filtrate, or both could be interesting.Another type of filtration includes suspensions with few insoluble solidsthat usually are undesired, in which case the filtration is called clarification.Finally, microfiltration should be mentioned. Microfiltration occurs when thesize of the solid particles to be separated is around 0.1 mm.

9.2 Fundamentals of Filtration

The study performed next applies to filters in which cakes are formed onthe filtering medium. Among these types of filters, the most used are theplate-and-frame press filters.


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236 Unit Operations in Food Engineering

Initially, the feed or slurry passes through the filtering medium, but, as itfiltrates, a cake is formed that increases its thickness, so the feed should passthrough not only the filtering medium but also the cake. Hence, the pressure

drop along the filter is supposed to be greater with time, resulting in thefiltrate flow being smaller with time. This indicates that filtration can bedone at constant pressure drop or at constant filtrate flow. In the first case,when maintaining a constant pressure drop, the filtrate flow will decreasealong the filtration time. On the other hand, when a constant filtrate flow isdesired, the pressure drop will be greater over time. As filtration proceeds,the thickness of the deposited solids increases, becoming a non-stationarycirculation of fluids through variable filling height case.

Two zones can be distinguished in the filter: the filtering cake and thefiltering medium. The former constitutes a filling that can change its char-acteristics (specific surface, porosity, etc.), while the filtering medium hasfixed characteristics.

In general, for a filter in which a cake is formed, the total pressure dropexperienced by the fluid (P) is the summation of the pressure drop expe-rienced through the filtering medium (Pm) plus the pressure drop experi-enced through the cake (Pc):

(9.1)

9.2.1 Resistance of the Filtering Cake

If a plate-and-frame press filter is considered, the filtering surface is flat andvertical, as shown in Figure 9.1.a. The fluid crosses through a cake of thick-ness z, in which the pressure drop is Pc. Applying the Bernoulli equation

to the cake section obtains:

(9.2)

According to the conditions of the filter:

The circulation of the fluid is laminar, so EVcan be obtained from Kozenysequation as:

(9.3)

( )= +( )= ( )+ ( ) P P P P Pc m c m

P vg EV

+ + + =2

2z 0

PEV + =

0

EP v K a

zVS= =

( ) ( )

12

0

2

3


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Filtration 237

in which vis the velocity in the free zone given by:

(9.4)

where Vis the filtrates volume andAis the cross-sectional area.

FIGURE 9.1.APress-and-frame filter. (Adapted from Ocn, J. and Tojo, G., Problemas de Ingenieria Quimica,Aguilar, Madrid, 1968.)

FIGURE 9.1.BCake formation scheme.

Fixed

head

Plate FrameMoving

head

Liquor

Smash

feeding

Filtering

medium

Cake

Suspension

Filtrate

Filteringmedium

z

zm

v

A

dV

d t=

1


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Combining the equations mentioned before results in:

(9.5)

The specific resistance of the cake is defined as:

(9.6)

where the units of are m/kg. physically represents the pressure dropneeded to obtain a unit superficial velocity times volume of filtrate per unitviscosity through the cake containing a unit mass of solid per unit area offilter.

Substituting the expression of the cakes specific resistance in Equation 9.5obtains:

(9.7)

In order to integrate this equation, the cakes thickness should be a functionof the filtrate volume. Hence, some variables will be defined that will allowthickness to be a function of the volume of filtrate.

The cake thicknesszis a function of the weight of the deposited dry cake(mDC):

(9.8)

To obtain the relationship between the deposited dry cake (mDC) and thevolume of filtrate (V),Mis defined as the relation between the wet cake anddry cake weight:

(9.9)

where the subindexes WCand DCindicate wet cake and dry cake, respec-tively, while mRLis the mass of liquid retained by the cake, whose value is:

(9.10)

1

1

3

2

0

2A

dV

dt K a zP

S

tc= ( ) ( ) ( )

=

( )( )K aSS

1 02

3

11A

dVdc

PczS

= ( )( )

m A zDC S= ( ) 1

M

m

m

m m

m

A z

A zWC

DC

DC RL

DC S= =

+

= + ( )1 1

m m MRL DC= ( )1


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Filtration 239

If Sis the mass fraction of solid in the filtered suspension:

(9.11)

Hence:

(9.12)

This last equation and Equation 9.8 allow one to obtain the relationbetween thickness of the cake and volume of filtrate:

(9.13)

Substituting and rearranging this expression in Equation 9.7 obtains:

(9.14)

In this equation it can be observed that the flow of filtrate is directlyproportional to the pressure drop experienced by the fluid when crossingthrough the cake and inversely proportional to the volume of filtrate. If the

pressure drop is kept constant, the filtrate flow decreases with filtration time,since the volume of filtrate increases with time.

9.2.2 Filtering Medium Resistance

The filtrate flow that crosses the filtering medium is the same flow thatcrosses the cake. The equation of Kozeny can be applied to this zone tocalculate the pressure drop Pm, but actually a resistance of the filtering

medium Rfis used, so the filtrates flow would be:

(9.15)

The units of Rfare m1, a resistance that includes various constants of the

filter. When compared with Kozenys equation, this constant is:

Sm

m M VDC

DC= = +

solid mass

total mass

m

V S

M SDC=

1

zV S

M S AS=

( )

11

1

dV

d t

A Pc

V S

M S

= ( )

2

1

dV

dt

A P

Rm

f

= ( )


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(9.16)

in whichzmis the thickness of the filtering medium and all the others are itscharacteristic variables. The subindex mindicates that these are characteris-tics of the filtering medium.

It is convenient to consider the resistance offered by the filtering mediumas equivalent to the resistance offered by a certain cake thicknesszfthat has

been deposited due to crossing a volume of filtrate Vf. It is obvious that Vfis a constant fictitious value. Similar to Equation 9.14 obtained for the cake,

the filtrate flow can be expressed as a function of this fictitious volume offiltrate according to the expression:

(9.17)

Comparing this last equation with Equation 9.15, it is observed that:

(9.18)

in which the resistance of the filtering medium is a function of the fictitiousvolume of filtrate.

9.2.3 Total Filtration Resistance

As seen before, the total pressure drop experienced by the fluid is the sum-mation of the losses caused when crossing through the cake and the filteringmedium.Therefore, the filtrate can be expressed according to the equations:

(9.19a)

(9.19b)

Once these equations are obtained, they should be integrated. Two casescan occur: one when the pressure drop experienced by the fluid is constantand another when operating at a constant volumetric flow.

RK a z

fm

m m S m

= ( )

( ) ( )

3

2

0

21

dV

d t

A P

V S

MS

m

f

= ( )

2

1

R VS

A MSf f=

( ) 1

dV

d t

A P

S

MSV Vf

= ( )

+( )

2

1

dV

d t

A P

S

M SV A Rf

= ( )

+

2

1


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Filtration 241

9.2.4 Compressible Cakes

When the specific resistance of the cake does not change with the cake

thickness and is independent of the pressure Pc

, the cake is called incom-pressible. This happens when the cake is formed by rigid solid particles thatkeep their shape, in which the values of cake porosity and specific surfaceof the particles are not affected by the compression applied on the bed.

In general, particles that form the cake are flexible and deformable. In thesetypes of cakes (called compressible), the resistance to flow depends on thepressure drop, which varies along the thickness of the cake. In this case thespecific resistance of the cake, , varies along the cake thickness, and anaverage value should be used if one wants to integrate Equations 9.19a and b.

Empirical formulas exist that allow one to calculate the specific resistanceof the cake . One of the most used of these equations is given by Almy andLewis:

(9.20)

where 0is the specific resistance of the cake when there is no pressure drop,and n is the compressibility factor, whose value varies between 0.1 and 1,

with higher values corresponding to the more compressible cakes. Althoughit has been proved that certain dependence exists, n is supposed to beindependent of pressure.

The values of 0and nshould be obtained experimentally from measure-ments of of a known pressure drop in the cake and by plotting into doublelogarithmic paper the corresponding pairs of values and Pc. The slope ofthe fitted straight line will be the value of the compressibility factor n, whilethe ordinate to the origin allows one to obtain the value of 0. Anotherexpression that gives the variation of the cakes specific resistance as a

function of the pressure drop is Ruths equation:

(9.21)

in which 0, , and nare parameters that should be obtained in an empiricalform.

9.3 Filtration at Constant Pressure Drop

When the filtration operation is carried out at a constant pressure drop, it isinteresting to know the variation of the volume of filtrate with time. Thiscan be done by integrating Equation 9.19b. This equation can be expressedas separated variables:

= ( )0 Pc

n

= 1+ ( )

0 Pcn


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(9.22)

If a new constant Cis defined:

(9.23)

then Equation 9.22 can be expressed as:

If this equation is integrated on the limit condition (t= 0, V= 0), then:

(9.24)

Rearranging:

where, if the new constants K1and K2are defined as:

(9.25)

(9.26)

it is obtained that:

(9.27)

corresponding to the equation of a straight line when plotting t/Vvs. thevolume of filtrate. The slope of this straight line is the constant K1, while thevalue of the constant K2is the ordinate to the origin.

S

M SV AR dV

A Pd tf1

2

+

= ( )

C

MS

S=

1

V

CA R dV

A Pd tf+

= ( )2

V

CAR V

A Ptf

2 2

2 + =

( )

t

V A P CV

R

A P

f=( )

+( )

2 2

KA P C1 22

=( )

KR

A P

f

2= ( )

t

VK V K= +1 2


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Filtration 243

Obtaining the constants K1and K2should be performed in experimentalform. If a filtration at constant pressure is carried out and the values of thevolume of filtrate obtained for different filtration times are recorded, a t/V

against Vgraphic can be built. When adjusting these values by the methodof least squares to the straight line given in Equation 9.27, the values of theconstants K1 and K2 can be obtained from the values of the slope and theordinate to the origin, respectively.

It is possible to obtain the value of the constant Cfrom K1and from C, toobtain the value of the specific resistance . The value of filtering mediumresistance Rfcan be obtained from the ordinate to the origin.

Once and Rfare known, from Equation 9.27, it is possible to determinethe filtration time needed to obtain a given volume of filtrate. Similarly, thevolume of filtrate can be calculated for a determined time. This can be donefrom Equation 9.24, which is a second-order equation with respect to thevolume of filtrate. When solving this equation, two roots are obtained, oneof which has no physical meaning since it is a negative volume. Choosingthe adequate root, the volume of filtrate is obtained from the equation:

(9.28)

This equation allows calculation of the volume of filtrate as a function oftime.

Equation 9.28 can be expressed as a function of the fictitious volume Vfgiven in Equation 9.18; hence, the following expression is obtained:

(9.29)

The flow rate that crosses the cake and the filtering medium is obtainedfrom the derivative of the last equation with respect to time:

(9.30)

The flow rate is maximum when the filtration time is null, thus:

(9.31)

V A C RC P

t C Rf f

= + ( )

2 22

1/2

V V C A P t Vf f=2 2

1 2

2+ ( )

/

qdV

d t

AC P

C RC P

tf

= = ( )

+ ( )

2 2

1 2

2/

qA P

Rfmax =

( )


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9.4 Filtration at Constant Volumetric Flow

Instant volumetric flow can be expressed as the variation of the filteredvolume with the filtration time:

If the volumetric flow remains constant, this will be equal to the filteredvolume collected during a determined time:

Substituting this expression in Equation 9.19b obtains:

(9.32)

This equation allows one to obtain the variation of the pressure drop inorder to keep a constant volumetric flow:

or:

(9.33)

If new constants K3and K4are defined as:

(9.34)

(9.35)


(9.36)

q dV dt=

q V t= = constant

qd V

d t

A P

S

MSV AR

A P

V

CARf f

= = ( )

+

= ( )

+

2 2

1

( )= +P

CAqV

R

Aq

f 2

( )= +PCA

q tR

Aq

f 2

2

Kq

C A3

2

2=

K

R q

Af

4=

( )= +P K t K 3 4


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Filtration 245

Analogous to the last section, using experimental results, if the pressuredrop is plotted against filtration time, a series of points is obtained that,when adjusted to a straight line (Equation 9.36), allows one to find the

constants K3and K4from the slope, and the ordinate to the origin, respec-tively. The values of and Rfcan then be obtained from the values of theconstants K3and K4.

The expression that relates the volume of filtrate collected with filtrationtime and pressure drop can be obtained from Equation 9.32 and written as:

When solving this second-order equation, taking the adequate root, thefiltered volume as a function of time can be obtained:

(9.37)

It can be observed that this expression is similar to the equation obtainedfor filtration at constant pressure (Equation 9.28).

9.5 Cake Washing

Once the filtration cake is formed, it is usually washed to eliminate the

undesirable solids that it may contain. This washing step is carried out onthe filter by passing a washing liquid through the cake. This operation isperformed at constant pressure drop and volumetric flow, and the washingliquid can follow the same path as the filtrate or a different one. Either way,the washing liquid will contain soluble solids once it has crossed the cakein such way that their concentration will decrease along time.

When the washing liquid follows the same path as the filtrate, the concen-tration of the washing solids eliminated is up to 90% at an initial period andthen abruptly decreases (Figure 9.2). When the washing liquid does notfollow the same path as the filtrate, the elimination of the soluble solids isgradual.

In most of the filters, the washing water follows the same path as thefiltrate. However, in press-and-frame press filters, the path is different.Figures 9.3and9.4show the filtration and washing operations, respectively,in a press-and-frame filter. It can be observed that the washing area is halfof the filtration area, while the resistance of the filtering medium is doublefor the washing area.

V

CAR V A P

tf

22 0+ ( ) =

V AR C P

tC Rf f

=C2 2

1 2

4 2+

( )

/


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FIGURE 9.2Concentration of soluble solids in washing water: (a) same path; (b) different path.

FIGURE 9.3Filtration operation in a plate-and-frame press filter.

Concent

rationinwashingwater

Washing time

b

a

Filteringmedium

Filtrate

Feed

Inlet

PlateFrame


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Filtration 247

It is important to highlight that the flow of washing liquid coincides withthe final flow of filtrate. Thus, if Equation 9.19.b is taken, then:

(9.38)

whereAwis the washing area.Where the washing area coincides with that of filtration (Aw = A), the

expression to calculate the washing flow in which Vis the volume of filteredliquid at the end of the filtration operation will be used.

For plate-and-frame press filters, the washing area is half of the filtrationarea. However, the washing liquid should cross the filtering medium twotimes. Hence, in Equation 9.38, 2Rfshould be used instead of Rf, and Aw=A/2. Taking all these remarks into account, the following expression isobtained:

(9.39a)

FIGURE 9.4Washing operation of the cake.

water

Washing

Plate Frame

Cake

q qA P

SMS V R

L FINAL

W

W f

= = ( ) ( )

+

2

1

A

qA P

S

M SAR

L

f

= ( )

+

1

4

1

2


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This equation, expressed as a function of K1 and K2, defined inEquations 9.25 and 9.26, allows one to express the washing flow as:

(9.39b)

The washing time (tL) is obtained by dividing the volume of liquid neededfor the washing operation by the flow:

(9.40)

where VLis the volume of liquid that will be used to wash the cake.

9.6 Filtration Capacity

The filtration capacity is defined as the quotient between the volume of

filtrate and the time of one filtration cycle:

(9.41)

The time of one cycle is the sum of filtration time (t) plus nonoperativetime (t), which includes the washing time (tL) and a complementary time(t*) needed for discharge, cleaning, and assembly and adjustment of the filterto begin a new filtration stage.

(9.42)

9.7 Optimal Filtration Conditions at Constant PressureIt is evident that during the filtration processes at constant pressure, thefiltrate flow decreases as filtration time goes on. For this reason, there will

be a time at which continuing the filtration process would not be feasible,so the optimum will be looked for. This optimum occurs at the time at whichthe filtration capacity is maximum.

Maximum filtration conditions will be obtained by maximizing the filtra-tion capacity function. Hence:

q

K V KL=

+1

8 41 2

t V qL L L=

F c

V

tCYCLE( )=

t t t t t tC L= + = + + *


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Filtration 249

which is the maximum condition. When the derivative of the filtration capac-ity with respect to the volume of filtrate is equal to zero, it is possible toobtain the optimum volume. In the same way, the optimum filtration timecan be obtained if the function is derived with respect to time. The differenceis based on the fact that, for the first case, the time should be expressed asa function of the volume of filtrate, while in the second case, the volumeshould be expressed as a function of filtration time.

For the particular case in which the nonoperative time, t , is a determined

value, the way to obtain the optimum is described next. The filtration timeis a function of the volume of filtrate and can be obtained from Equation 9.27:

(9.43)

In this way, the filtration capacity can be expressed as:

The optimum can be found from the derivative of this expression andequaling it to zero:

Thus, the following expression is obtained:

(9.44)

This equation indicates the nonoperative time required to obtain an opti-mum volume of filtrate. If a nonoperative time is given, the volume of filtrateto have an optimum filtration capacity will be:

(9.45)

The optimum time will be obtained by substituting this value of volumein Equation 9.42:

(9.46)

d F c

d V

( )= 0

t K V K V = +12

2

F cV

K V K V t( )=

+( )+ 1 2 2

d F c

dV

d

d t

V

K V K V t

( )=

+( )+

=

12

2

0

=t K V12

V tKOPTIMUM

= 1

t t Kt

KOPTIMUM= +

2

1


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The graphical method of Sbarbaugh allows one to obtain the values of Voptand topt from the curve in which the filtrates volume is a function of time(Figure 9.5). To obtain Vopt and topt, the tangent line to the filtration curvefrom the point (t , 0) is plotted. The tangent point has as coordinates thevalues of the optimum time and volume of filtrate.

9.8 Rotary Vacuum Disk Filter

Another type of filter widely used in the industry is the rotary filter thatoperates under vacuum. These filters consist of a cylindrical drum withdifferent sections, which rotates on its axis and in which vacuum is appliedto facilitate the crossing of the fluid. The solids are retained in the lateralcylindrical surface (Figure 9.6). As can be observed in this figure, part of thecylinder is submerged in the suspension to be filtered. The cake depositedon the surface is washed by showers of washing liquid. Vacuum is appliedto the interior of the drum at the filtration zone as well as in the washingzone, but once out of these zones, vacuum is eliminated to easily separatethe cake with a scraper. In this way, the filter can begin the filtration cycleagain. This kind of operation is carried out at constant pressure drop andvolumetric flow.

FIGURE 9.5Graphical method to determine optimum conditions of filtration.

Filtered

Volume

TimetOPTIMUM

VOPTIMUM

t


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Filtration 251

Mathematical treatment and calculations are performed assuming that thetotal filtering area (A) is formed by nsurfaces Ai, small enough to assumethat they behave as a flat filter:A= n Ai.

The volume filtered (Vi) by one of the sections Ai, since it is a flat filter,can be obtained from Equation 9.28:

where tsis the time during which the sectionAiis submerged, correspondingto the filtration time.

If the filtration drum rotates at a rotation velocity N, and its immersionangle is , then the time it will be submerged is:

(9.47)

in which the angle is given in radians.

FIGURE 9.6Rotary vacuum filter.

Washingwater

Cake

Feeding

Scraper

Air

Vacuum

Vacuum

V A RC P

t C Ri i f S f = + ( )

C2 21 2

2

/

t

NS=

2


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The total volume of filtrate is obtained by adding together the volumes ofeach filtration section considered:

(9.48)

The filtrate flow is a function of the volume of filtrate and is obtained bymultiplying the volume of filtrate times the spin velocity of the drum:

(9.49)

Sometimes it is interesting to calculate the thickness of the cake depositedon the filters surface. During one filtration cycle, when a point of the filtersurface penetrates into the suspension to be filtered, the insoluble solids

begin to deposit. As time goes on, more solids are deposited on this point

and, as a consequence, the thickness of the deposited cake increases, depend-ing on the amount of time that this point has been submerged. As was seenin Equation 9.47, such time depends on the immersion angle.

The mass of solids deposited on one of the areas Ai is obtained fromEquation 9.13:

in which the thickness of the deposited cake is a function of the filteredvolume Vi. Such volume is given by Equation 9.48, with Airepresenting thefiltration area; if the definition of the constant Cas a function of the resistanceof the cake is taken into account, then thickness can be obtained by meansof the following equation:

(9.50)

This equation allows one to obtain the cake thickness for an angle ofsubmerged filter of radians.

V nV A RC P

NC Ri f f= = +

( )

C2 21 2

2

2

/

q NV NA RC P

CRf f= = + ( )

CN

2 2

1 22

2

/

zV S

MS Ai

S i

=

( )

1

1

1

z RC P

C NR

Sf f= ( )

+ ( )

1

1

2

22

1 2

/


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Filtration 253

Problems

9.1

A plate-and-frame filter that operates under constant pressure requires 1 hto separate 600 l of filtrate of an aqueous suspension. Calculate the filtrationcapacity if the initial filtration velocity is 60 l/min, 8 l of water are neededto wash the deposited cake, and 35 min are needed to discharge, clean, andassemble the filter.

The filtrate flow is given by Equation 9.19b, which, with the definition ofconstant C(Equation 9.23), is transformed into:

For a filtration at constant pressure drop, (P) = constant, the integrationof this equation yields:

K1and K2are defined by Equations 9.25 and 9.26.

For t= 0: (dV/dt) = 0

from which it is obtained that K2= 1000 s/m3

For t= 60 min: V= 600 l = 0.6 m3, so:

obtaining:

It is important to point out that the washing flow coincides with the finalvolume of filtrate; for plate-and-frame filters, Equation 9.39b allows one tocalculate the filtrate flow:

dVd t

A PV

CARf

= ( )+

2

K V K V t

1

2

20+ =

dV

d t

A P

R Kt f

= = ( )

==0 2

601l

min

K K1

2

20 6 0 6 3600 0. .( ) + ( ) =

K1 8333 33= . s m6


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Since, at the end of filtration, 0.6 m3was obtained and the constants K1andK2were previously obtained, then the washing volumetric flow rate will be:

qL= 2.273 105m3/s

The washing time is obtained by dividing the washing volume by thevolumetric flow rate:

The filtration capacity is obtained from Equation 9.41, in which the timeof one cycle is given by Equation 9.42. Therefore:

in whichV= 0.6 m3

t = 3600 stL = 3520 st* = 35 60 s = 2100 s

So the filtration capacity is:

9.2

A plate-and-frame press filter is used to filtrate an aqueous suspension.Working under constant filtration velocity, 250 l of filtrate are obtained after45 min. During this period, the pressure drop increases from 0.40 kg/cm2to3.5 kg/cm2. If it were desired to work at a constant pressure of 3.5 kg/cm2,what amount of filtrate would be obtained after 45 min?

The expression that relates the variation of the volume of filtrate to timeis given by Equation 9.32:

Since the filtration is at constant volumetric flow, then q = (dV/dt) =constant:

qK V KL

=+

1

8 41 2

tL=

80 10 m2.273 10 m s =3520s

3 3

5 3

F cV

t t tL*( )= + +

F c( )= =

6 51 10 3 91

5

. .m s l min

3

qdV

d t

A P

V

CARf

= = ( )

+

2


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Filtration 255

It is not necessary to integrate the last equation because it is operatedunder a constant volumetric flow, but the pressure drop (P) can be directlyobtained as a function of filtration time (Equation 9.36):

where the constants K3and K4are given by Equations 9.34 and 9.35:

For t= 0:

For t= 45 min = 2700 s:

Filtration at constant pressure drop (P) = 3.5 kg/cm2

When integrating the differential equation, under the limit condition t= 0,V= 0, a second-order equation is obtained with respect to the volume offiltrate (Equation 9.27):

The following is obtained from the expression of K3:

q= = =

250

455 556 9 26 10 5

ll min m s3

min. .

( )= +P K t K 3 4

K

q

C AK

R q

Af

3

2

2 4= =

( )= =P 0 4 39 200. ,kg cm Pa2

K P4 39 200= ( )= , Pa

( )= =P 3 5 343 000. ,kg cm Pa2

K P Kt s3

4 343 000 39 2002700

112 52= ( ) = ( ) = , , .Pa Pa s

V C AR V

CA P tf

22

22

0+ ( ) =

CA q

K

2 2

3

52

119 26 10

112 527 62 10

= =

( )=

.

..

m s

Pa s

m

Pa s

3 6


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From the relation:


The different values obtained are substituted in the second-order equation:

When solving this equation, the negative root does not have a physical

meaning, so only the positive root is taken:V

= 0.345 m

3

.Therefore, after 45 min of filtration at a constant pressure drop of 3.5 kg/cm2,345 l of filtrate are obtained.

9.3

An aqueous suspension that contains 7% of insoluble solids is filtered at arate of 10 ton/h, using a plate-and-frame press filter that works underconstant pressure of 3 atm. It was found, experimentally, that the depositedcake contains 50% moisture, the density of the dry solids is 3.5 g/cm3, andthe equivalent diameter of the deposited particles is 0.002 mm. The washingof the cake begins after 10,000 kg/h of feeding have been filtered using 150l of water, while 30 min are used for the discharge, cleaning, and assemblyoperations. If the resistance of the filtering medium to filtration is considerednegligible, calculate: (a) the specific resistance of the cake; (b) volume ofwater filtered after one hour; (c) time needed to carry out washing; and(d) filtration capacity.

Data: K= 5 can be taken for Kozenys constant.Properties of water: Density = 1000 kg/m3; Viscosity = 1 mPa s

10,000 kg of the suspension are filtered in 1 h.

Dry solids deposited: W = (10,000) (0.07) = 700 kg

K

K

R q

A

CA

q

CA R

qf f

4

3

2

2

2

= = =

C AR qK

Kf= = ( )

( )( )

= 43

5 39 26 1039 200

112 5232 26 10.

,

..m s

Pa

Pa sm3 3

V V2 3 11 42 32 26 10 2 7 62 10 3 5 9 8 10 2700 0+ ( ) ( ) ( ) = . . . .

V V2 0 0645 0 1411 0+ =. .

M= = =weight of wet cake

weight of dry cake

100 kg wet cake

50 kg dry cake2


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Filtration 257

If VT= volume of deposited cake:

Substituting the variables with their values obtains that porosity is =0.778.

Particles specific surface:

The cake-specific resistance is obtained from Equation 9.6:

For a filtration under constant pressure drop, the volume of filtrate is givenby Equation 9.28. Since the resistance of the medium is negligible, Rf= 0, theequation can be rearranged as:

in which:

Also, from the definition of the constant C(Equation 9.23):

Volume of filtrate collected in 1 h:

M=

mass of dry solid+ mass of retained liquid

mass of dry solid

MV

VT

T P P

= +( )

= +( )

11

11

a dpS0

3 6 16 6 0 002 10 3 10= = ( )= . m m

= ( )( ) = ( ) ( )( )( )

=

K aS

S

1 5 1 0 778 3 10

0 778 35006 07 100

2 6 12

39

3 3

m

kg m

mkg

.

..

V A C P t= ( )

21 2

/

( )=

= P 3

9 8 10

12 94 10

45atm

Pa

atmPa

..

CMS

S=

=

( )( )( )=

1 1 2 0 07

6 07 10 10 0 072 024 10

9 3

12 2

.

. ..

m kg kg mm

3


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Thus, the filtration area is obtained from Equation 9.28, with Rf= 0

Hence, one filtration area:A= 4.155 m2

The washing flow coincides with that of the final filtrate, and since Rf= 0,Equation 9.39 becomes simpler:

since:

A = 4.155 m2

C = 2.024 1012m2V = 8.6 m3

= 103Pas(P) = 2.94 105Pa

then:

The washing time will be:

Calculation of the filtration capacity:

VW MS

S=

( )=

( ) ( )( )( )

=1 700 1 2 0 07

1000 0 078 6 3

kgd.s.

kg mm

3

.

..

8 62 2 024 10 2 94 10

1036003

12 2 5

3

1 2

.. .

/

mm Pa

Pas=

( ) ( )( )

( )

A s

qC A P

VL=

( )14

2

qL=

2 99 10

4

. m s

3

tV

qLL

L

= = ( )

( )=

0 15

2 99 10502

3

4

.

.

m

m ss

3

F cV

t t tL( )=

+ + =

+ +( ) =

*

..

8 6

3600 502 18001 457 10

33m

s m s3

FC= =1 457 10 87 423. .m s l min3


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Filtration 259

9.4

An aqueous solution that contains 10% suspended solids is filtered in a plate-

and-frame press filter. In a previous experiment it was determined that thewet cake to dry cake ratio is 2.2, using an incompressible cake with a specificresistance of 2.5 1010m/kg. During filtration at constant pressure of 3 atm,the variation on the amount of filtrate with time is recorded in the followingtable:

From these data, calculate: (a) total area of the filter and resistance of thefiltering medium; (b) if the nonoperative time of each filtrating cycle is26 min, calculate the volume of filtrate recovered at the end of 10 h if theoperation is carried out at the optimum filtration cycle; and (c) it is desiredto filtrate the same solution but working at constant volumetric flow. If, atthe end of 142 min, the pressure drop experienced by the fluid when crossingthe cake and the filtering medium is 4.5 atm, calculate the volume of filtrateobtained and the volumetric flow rate at which it circulates.

Data: Water properties: Density = 1000 kg/m3; Viscosity = 1.2 mPa s

Filtration at constant pressure drop:In this type of operation, the variation on the volume of filtrate and time

are correlated by a second-order equation (Equation 9.27):

in which the constants K1 and K2 are given by Equations 9.25 and 9.26.Equation 9.27 is a straight line if t/V is plotted against V, with a slope K1and intercept K2.

(a) The following table can be constructed with the data stated in theproblem:

The constants K1and K2can be obtained from the fitting of these data usingthe method of least squares.

Time (minutes) 8 18 31 49 70 95Filtrates mass (kg) 1600 2700 3720 4900 6000 7125

t(s) 480 1080 1860 2940 4200 5700V(m3) 1.6 2.7 3.72 4.9 6.0 7.125

t/V(s/m3) 300 400 500 600 700 800

K V K V t12

2 0+ =

K K1 290 40 157 58= =. .s m s m6 3


26/30



The value of constant Cis obtained from Equation 9.23:

Calculation of the filtration area:It is possible to obtain such area from the expression of constant K1

(Equation 9.25)

Then A = 8.506 m2.

Calculation of the resistance of the filtration medium:It is possible to obtain the value from the expression of the constant K

2(Equation 9.26):

(b) Filtration capacity: it is given by Equation 9.41 that, in this case:

in which t= 26 60 s = 1560 s.Hence, from Equation 9.27:

So the filtration capacity will be:

C MSS= =

( )( )( )( )( )= 1

1 2 2 0 1

2 5 10 1000 0 13 12 1010 13 2

. .

. ..

m kg kg mm3

A K C P2

1

3

13 2 4

2

2

1 2 10

2 90 4 3 12 10 3 9 8 10

72 357

= ( ) =

( ) ( ) ( )

=

.

. . .

.

Pas

s m m Pa

m

6

RK A P

f= ( )

=( ) ( ) ( )

( )

=

2

2 4

3

11 1

157 58 8 506 3 9 8 10

1 2 10

3 284 10

. . .

.

.

s m m Pa

Pa s

m

3

F c

V

t t( )=

+

t K V K V = +12

2

F cV

K V K V t( )=

+ + 12

2


27/30


Filtration 261

The optimum cycle is obtained from the derivation of this function withrespect to the volume of filtrate and equaling it to zero:

Since the numerator of this expression is equal to zero, then:

Thus, Vopt= (t/K1)1/2.Since t = 1560 s and K1= 90.40 s/m6, the optimum volume of filtrate is

Vopt= 4.154 m3.

Filtration time obtained from the expression:

t= K1V2+ K2V

t= 90.4 s/m6(4.154 m3)2+ 157.58 s/m3(4.154 m3)

t= 2215 s = 36 min 55 s

One cycles time:

Number of cycles:

Nine complete cycles have been completed with +0.536 cycles that corre-spond to 0.536 cycles 3775 s/cycle = 2025 s. The volume of filtrate for thistime can be obtained by means of a second-order equation: t= K1V

2+ K2Vo

The negative root of this equation has no physical meaning, so the positiveroot yields the filtrates volume obtained during the 2025 s: V= 3.941 m3.

The volume of filtrate recovered at the end of 10 h of operation will be:

d

dtF c

K V K V t V K V K

K V K V t( )( )=

+ + +( )+ + ( )

=12

2 1 2

12

2

2

20

K V t12 =

t t tcycle= + = + =2215 1560 3775s s s

n of cycles = 10 h 3600 s h

3775 scycles ( )( ) = 9 536.

2025 90 4 157 582= +. .V V

V= + =9 4 154 3 941 41 327cycles m cycle m m3 3 3. . .


28/30



(c) Filtration under a constant volumetric flow: The volume of filtrate isobtained from Equation 9.37:

The values of the different variables of this equation are:

A= 8.506 m2

C= 3.12 1013m2t= 8520 s

Rf= 3.284 1011m1

(P) = (4.5) (9.8 104) Pa

Hence, the volume of filtrate is V= 7.983 m3.

Filtrate volumetric flow rate:

9.5

It is desired to obtain 60 l/min of filtrate from an aqueous suspension thatcontains 0.25 kg of insoluble solids per each kg of water, using a rotary drumfilter. The pressure drop suffered by the fluid is 300 mm of Hg, obtaining acake that has a 50% moisture content and a filtrate that has a viscosity of1.2 mPa s and a density of 1000 kg/m3. The filters cycle time is 5 min, with30% of the total surface submerged. The particles that form the suspensioncan be considered as spheres with a 6 103mm diameter and a 900 kg/m3density. If it is supposed that the resistance offered by the filtering medium

to filtration is negligible and that a value of 5 can be taken for the Kozenyconstant, calculate (a) the specific resistance of the cake; (b) the filters area;and (c) thickness of the deposited cake.

Suspended solids:

V AR C P

tC Rf f= +

( )

C22

1 2

4 2

/

q

V

t= = =

7 983

1420 0562

3 3.

min.

min

m m

q l= = =0 0562 56 2 3 373. . .m min min m h3 3

S=+( )

= ( )0 25

1 0 250 2

.

..

kg solids

total kg 20% solids


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Filtration 263

Since:

it is possible to obtain the value of porosity using the values of = 1000kg/m3, P= 900 kg/m3, and = 0.4737.

Particles specific surface:

The cake specific resistance is obtained from Equation 9.6:

Constant Cis obtained from Equation 9.23:

Number of rps: N= 1/(5 min) = 0.2 rpm = 3.33 103rps

Drop pressure:

For a rotary filter, the filtrate volumetric flow rate is obtained fromEquation 9.49, which, for the present case, becomes simpler since the resis-tance of the filtering medium can be neglected (Rf= 0):

M =

mass of wet cake

mass of dry cake

1 kg wet cake

0.5 kg dry cake= = 2

MV

VC

C P

= +( )

11

adS P

0 66 16 6

6 1010= =

=

mm

= ( )( ) = ( )( )( )( )

= K aS

S

1 5 1 0 4737 10

900 0 47372 75 100

2 6 12

310

3 3

m

kg mm kg

.

..

CMS

S=

=

( )

( )( )( )

= 1 1 2 0 2

2 75 10 1000 0 21 09 10

10

13 2

.

. ..

m kg kg mm

3

( )=

=P

300

760

1 0333

1

9 8 10

139973

4

atmatm

atm

Pa

atmPa

. .

q NV NAC P

N= =

( )

2

2

1 2

/


30/30


The available data are

= 0.3 2radians;

= 1.2 103Pa sq= 60 l/min = 103m3/s;

(P) = 39,972.4 PaC= 1.09 1013m2;N= 3.33 103s1

It is possible to calculate the area of the filter by substituting these data in

the equation stated before, thus obtainingA= 11.73 m2

.The cake thickness is obtained from Equation 9.50, but since Rf = 0, theequation can be rearranged as:

in which (P), , , C, and Nhave the values indicated above, and also:P= 900 kg/m3; = 0.4737; and = 2.75 1010m/kg. These data allow oneto obtain the cake thickness:z= 0.018 m = 18 mm.

zC P

C NS

=1

1

2

2

1 2

( )

( )

/

Cap 09 Filtracion

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