Filtration.pdf

EKC313 Separation Processes Lecture #1: Filtration

Semester 1, 2010

What is separation process? In chemistry and chemical engineering, a separation process is used to transform a mixture a mixture of substances into two or more distinct products. The separated products could differ in chemical properties or some physical property, such as size, or crystal modification or other separation into different components. [Wikipedia]

Separation process and chemical engineering The most conventional separation process used in Chemical Industry is distillation process. In fact distillation technique itself is much older than the Chem. Eng. discipline. Distillation was introduced to medieval to medieval Europe through Latin through Latin translations of Arabic chemical treatises in the 12th century [Wikipedia]. [Wikipedia]. Nevertheless, distillation play a pivotal role in establishing Chemical Engineering as an independent discipline itself, which, differentiate our job scope from mechanical engineers and chemist. Strictly speaking, our unique job function started from the time when new engineering skill sets were desperately needed to handle a newly founded industry: Refinery of crude oil. In other word, mastering separation technique really defines chemical engineer uniqueness during that time. Ironically, this saying is almost still true in this new millennium.

Figure 1. Larger distilling apparatus for getting cone o il from Abies Alba (European Silver Fir ) [http://chestofbooks.com/health/aromatherapy/The-Volatile-Oils-Vol2/54-Cone-Oil-From-Abies-Alba.html]

Early civilization developed techniques to (1) extract metals from ores, perfumes from flowers, dyes from plants, and potash from the ashes of burnt plants, (2) evaporate sea water to obtain salt, (3) refine rock asphalt, and (4) distill liquors. [Seader and Henley]

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Semester 1, 2010

Filtration In filtration, suspended solid particles in a fluid and liquid or gas are physically or mechanically removed by using a porous medium that retains the particles as a separate phase or cake and passes the clear filtrate. [Geankoplis]

The objectives for performing filtration usually fall into one of the following categories: 1. clarification for liquor purification, 2. separation for solids recovery, 3. separation for both liquid and solids recovery, and/or 4. separation aimed at facilitating or improving other plant operations.

Clarification involves the removal of relatively small amounts of suspended solids from suspension (typically below 0.15% concentration). A first approach to considering any clarification option is to define the required degree of purification. That is, the maximum allowable percentage of solids in the filtrate must be established. Compared with other filter devices, clarifying filters are of lesser importance to pure chemical process work. They are primarily employed in beverage manufacturing and water polishing operations, pharmaceutical filtration, fuel/ lubricating oil clarification, electroplating solution conditioning, and dry-

cleaning solvent recovery.

In filtration for solids recovery, the concentration of solids suspension must be high enough to allow the formation of a sufficiently thick cake for discharge in the form of a solid mass before the rate of flow is materially reduced. However, solids concentration alone is not the only criterion for adequate cake formation. For example, an 0.5% suspension of paper pulp may be readily cake-forming whereas a 10% concentration of certain chemicals may require thickening to produce a dischargeable cake. [Cheremisinoff]

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Semester 1, 2010

Industrial filtration equipments For chemical industry, the quality and quantity of feed or slurry solution handled for daily operation varying significantly due to different application purposes. Those streams may carry a heavy load of solid particles or just trace amount of them. Due to the wide diversity of these encountered filtration problems, various types of filtering unit/process have been developed to cope with different situations.

Bed filters

Figure 2. Typical bed filter. [http://kscst.iisc.ernet.in/rwh_files/images/Sand%20bed%20filter.jpg], [http://www.cee.vt.edu/ewr/environmental/teach/wtprimer/rapid/filter3.jpg]

•

Used to handle large amount of water with relatively small amount of solids

•

Coarse pieces of gravel at the bottom while fine sands at the top layer (Why?)

•

The processed water is drawn out at the bottom

• Normally used to remove precipitates which are not adhere strongly to the sand and can be easily removed by backwashing.

•

This kind of filter is very popular to treat municipal water. [Geankoplis]

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Semester 1, 2010

Plate-and-frame filter presses

Figure 3. Plate and frame filter presses [http://www.stpats.com/images2007/PF30-2.jpg], [http://www.mine-engineer.com/mining/Flow_Illus.jpg]

•

The plate are covered with a filter medium/cloth

•

Each plate is squeezed tightly together by a screw or a hydraulic ram.

•

Solids are deposited on the cloth-covered faces of the plates while liquid passes through the cloth, down grooves or corrugations in the plate faces and out of the press.

•

Operated under a pressure of 3 to 10 atm. [McCabe and Smith]

Figure 4. Plate and frame filter in one of the refinery plant. (Left) The technician is tightening up the filter cloth. (Right) Fully automated continuous filtration process.

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Semester 1, 2010

Continuous rotary-drum filter

Figure 5. Rotary drum filter [http://upload.wikimedia.org/wikipedia/commons/thumb/a/a2/Rotary_vacuum-drum_filter.svg/480pxRotary_vacuum-drum_filter.svg.png], [http://img.directindustry.com/images_di/photo-g/rotary-drum-vacuum-filter384475.jpg]

•

In most cases, operated at 0.1 to 2 rotations/min in an agitated slurry trough.

•

A filter medium covers the face of the drum and is partly submerged in the liquid

•

Vacuum and air are alternately applied through central duct as the drum rotates.

•

The filter cake is removed by scraping it off with horizontal knife. [McCabe and Smith]

Principles and Basic Theory of Filtration

Figure 6. Pictorial representation of cake filtration. (a) True case scenario, and (b) Schematic showing how mathematical analysis can be done on (a).

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Semester 1, 2010

Due to the combining effects of hydrodynamic and physicochemical factors, the study of cake structure and resistance is extremely complex, and any mathematical description based on theoretical considerations is at best only descriptive. [Liquid Filtration: http://www.knovel.com/knovel2/Toc.jsp?BookID=427&VerticalID=0] (This link contains a free liquid filtration textbook in pdf format)

The flow of filtrate through the packed bed of cake can be described by an equation similar to Poiseuille’s (Pwah-sui-ya) Law (Assumptions: flow is laminar, viscous, incompressible and the flow is through a constant circular cross-section with length >> diameter). Poiseuille’s equation for laminar flow in a straight tube:

− ∆ p =

32µν L D

2

(1.1) 2

Where ∆ p is pressure drop in N/m , length in m,

µ is

ν is

open-tube velocity in m/s, D is diameter in m and L is

viscosity in kg/m*s. For laminar flow in a packed bed of particles, the Eq. 1.1

can be rewrite according to Carman-Kozeny relation as:

k 1 µν (1 − ε ) S 0 L 2

− ∆ pc =

ε

3

2

(1.2)

Where k 1 is a constant and equals to 4.17 for random particles of definite size and shape, linear velocity based on filter area in m/s,

ε is

ν

is

void fraction or porosity of cake, L is thickness of 2

3

cake in m, S0 is specific surface area of particle m of particle area per m volume of solid particle, and ∆ pc = pa – p b is pressure drop in cake in N/m . 2

The linear velocity is based on the empty cross-sectional area and is defined as:

ν

=

dV / dt A

(1.3) 3

2

Where V is total volume of filtrate (m ) collected up to time t s and A is filter area in m . The cake thickness L may be related to the volume of filtrate V by a material balance. If cs is kg3

solid/m of filtrate, a material balance gives

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Semester 1, 2010

LA(1 − ε ) ρ p = c s (V + ε LA)

(1.4)

Substituting Eq. 1.2 and 1.3 into 1.4 to eliminated L, we obtain the final equation as:

dV Adt

=

− ∆ pc k 1 (1 − ε ) S 0

2

ρ p ε

Where

α is

3

µ csV

A

=

− ∆ pc α

µ c sV

(1.5)

A

the specific cake resistance in m/kg defined as:

k 1 (1 − ε )S 0

2

α

=

ρ p ε

3

(1.6)

For the filter-medium resistance, we can write, by analo gy with Eq. 1.5

dV Adt

=

− ∆ p f µ Rm

(1.7)

Where Rm is the resistance of the filter medium to filtrate flow in m and ∆ p f = pb – pc is the -1

pressure drop across the filter medium. When Rm is treated as an empirical constant, it includes the resistance to flow of the piping leads to and from the filter and the filter medium resistance.

Since the cake and the filter medium are positioned in series, hence dV

=

Adt

− ∆ p α c sV   µ  Rm +  A  

(1.8)

Where ∆ p = ∆ p f + ∆pc . Eq 1.8 can be further modified to accommodate the practical ability aspect of filtration as follow:

dV Adt

=

− ∆ p µα c s

A

(1.9)

(V + V e )

Where V e is a volume of filtrate necessary to build up a fictitious filter cake whose resistance is equal to Rm.

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Semester 1, 2010

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

W = c sV =

ρ c x

(1 − mc x )

V

(1.10)

Where c x is mass fraction of solids in the slurry, m is mass ratio of wet cake to dry cake, and ρ is 3

density of filtrate in kg/m .



Why we even bother to rearrange those equations as such from Eq. 1.8 to 1.10??

For constant-pressure filtration done in batch process, the basis equations are (rearranged from Eq. 1.8)

dt dV

µα c s

=

A (−∆ p ) 2

V +

µ

A( −∆ p )

Rm = K pV + B

6

(1.11a)

3

Where K p is in [s/m ] and B in [s/m ]

K p = B =

µα c s

A ( −∆ p ) 2

µ

A(−∆ p )

and

(1.11b)

Rm

(1.11c)

For constant pressure, constant α, and incompressible cake, V and t are the only variables in Eq. 1.11a:

t

∫ 0

V

∫

dt = ( K pV + B ) dV 

t =

K p

0

Dividing Eq. 1.12 by V

t V

=

2

V + BV

K p

2

2

V + B

(1.12)

(1.13)

3

Where V is the total volume of filtrate in m collected to t secs.

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Semester 1, 2010

Example 1.1 Evaluation of Filtration Constant for Constant Pressure Filtration [Geankoplis]

Data for the laboratory filtration of CaCO3 slurry in water at 298.2 K are reported as follows at 2

2

constant pressure of 338 kN/m . The filter area of the plate-and-frame press was A = 0.0439 m 3

and the slurry concentration was 23.47 kg/m . Calculate the constant

α

and Rm from the

experimental data given. 3

3

t (s) 4.4 9.5 16.3 24.6 34.7 46.1 59.0 73.5 89.4

V (m ) 0.000498 0.001000 0.001501 0.002000 0.002498 0.003002 0.003506 0.004004 0.004502

107.3

0.005009

Firstly the data are calculated as t/V

3

V (m ) 0.000000 0.000498 0.001000 0.001501 0.002000 0.002498 0.003002 0.003506 0.004004 0.004502

(t/V) (s/m )

0.005009

21421.44141

8835.341365 9500 10859.42705 12300 13891.11289 15356.42905 16828.29435 18346.65335 19857.84096

The data are then plotted as t/V versus V in following figure:

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Semester 1, 2010 3

6

From the figure, the intercept is determined as B = 6414 s/m and the slope as K p/2 = 3 x 10 6

6

6

s/m and hence K p = 6 x 10 s/m .

At 298.2 K the viscosity of water is roughl y 8.9 × 10 −4 Pa*s = 8.9 × 10 −4 kg/m*s. From Eq. 1.11b

K p =

µα c s

A ( −∆ p ) 2



6 ×10 = 6

(8.9 ×10 −4 )α ( 23.47) (0.0439) 2 (338 ×10 3 )

 α = 1.863 × 10

11

m/kg

From Eq. 1.11c B =

µ

A(−∆ p )

−

Rm 

6414 =



(8.9 × 10 4 ) Rm (0.0439)(338 × 103 )

10 Rm = 10.63 × 10 m

-1

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Semester 1, 2010

Range of void function:

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Semester 1, 2010

Example 1.2 Time required to perform a filtration [Geankoplis]

The same slurry used in example 1.1 is to be filtered in a plate-and-frame press havin g 20 frames 2

and 0.873 m area per frame. The same pressure will be used in constant-pressure filtration. 3

Assuming the same filter cake properties and filter cloth, calculate the time to recover 3.37m filtrate. 2

6

6

3

In the previous example the area A = 0.0439 m , K p/2 = 3 x 10 s/m , and B = 6414 s/m . Since the α and Rm will be the same as before, K p can be corrected. From Eq. 1.11b K p is proportional 2

2

to 1/A and hence the new K p with A = 0.873 m is:

6 2 K p = 6 × 10 (0.0439 / 17.46) = 37.93 s/m

6

The new B is proportional to 1/A from Eq. 1.11c:

B = 6400

0.0439 17.46

3

= 16.10 s/m

Substituting these new values into Eq. 1.12 t =

K p

2

V + BV = 2

37.92 2

(3.37) 2 + (16.10)(3.37) = 269.7 s

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Semester 1, 2010

Equations for washing of filter cakes and total c ycle time Filtration for both solids and liquid recovery differs from filtration for solids recovery alone in the cake building, washing and drying stages. If the filtrate is valuable liquor, maximum washing is necessary to prevent its loss; but if it is valueless, excess wash liquor can be applied without regard to quality [Cheremisinoff].

The washing of a cake after the filtration cycle is necessary by displacement of filtrate and by diffusion. The amount of washing liquid should be sufficient to give the desired washing effect. To calculate washing rates, it is assumed that the conditions during washing are the same as those that existed at the end of the filtration. It is assumed that the cake structure is not affected when wash liquid replaces the slurry in liquid in the cake.

For constant pressure filtration, using the same pressure in washing as in filtering, the final filtering rate is the reciprocal of Eq. 1.11a

1  dV    = dt K V B +   f p f

(1.14) 3

where (dV/dt)f = rate of washing in m /s and Vf is the total volume of filtrate for the entire period 3

at the end of filtration in m . For plate-and-frame filter presses, the wash liquid travels through a cake twice as thick and an area only half as large as in filtering, so the predicted washing rate is one-fourth of the final filtration rate:

1 1  dV    = dt 4 K V B +   f p f

(1.15)

[http://www.andritz.com/ep-filter_cake_washing.jpg]

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Semester 1, 2010

Another reason for dewatering the filter cake: A question posted by Mansperger on ChemE forum at [http://www.finishing.com/146/91.shtml] We're having problems with our sludge filter-press cake. It is not dewatering properly. Instead of a nice solid , we're getting slime. I'd like some information about a floc or filtering aid which would help resolve this problem.

TIA, Debra Mansperger (- Chandler, AZ)

Figure 7. Good filter cake [http://www.goodquarry.com/images/productiontechnology/PPT_photo25_large.jpg]

Filter cloths also come in many different styles of fabric weave. Mesh opening or pore size is also an important consideration when selecting the proper filter cloths for a particular application. The mesh opening or pore size is determined by the number of fibers, size of the fiber, and the type of weave. These factors in turn govern flow rates, particle retention, and the strength of the fabric itself. As shown below, these two fabrics, having the same type of open area, will have similar flow rates, but the fabric on the left will retain finer particles.

Figure 8. Filter cloth with different scale of permeability. http://www.water.siemens.com/SiteCollectionImages/products/sludge_biosolids_processing/filter_press/faq1_R.jpg

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Semester 1, 2010

Example 1.3 Rate of washing and total filter-cycle time [Geankoplis] 3

At the end of the filtration cycle in previous examples, a total filtrate volume of 3.37 m is collected in a total time of 269.7 s. The cake is to be washed by through-washing in the plateand-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 mins. 6

3

For this filter, Eq. 1.15 holds. Substituting K p= 37.93 s/m , and B = 16.10 s/m , and 3

Vf = 3.37 m , the washing rate is as follow:

1 1  dV  = 1.737 ×10 −3 m 3 / s   =  dt  f 4 (37.93)(3.37) + 16.10 3

3

The time of washing is then as follow for 10 % of filtrate volume 0.10(3.37 m ) = 0.337 m

t =

0.337 1.737 × 10 −3

= 194.0 s

The total filtration cycle is

269.7 60

+

194.0 60

+ 20 = 27.73 min

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Semester 1, 2010

Continuous filtration In a filter that is continuous in operation, such as a rotary-drum vacuum type, the feed, filtrate and cake move at steady, continuous rates. In a rotary drum the pressure drop is held constant for the filtration. The cake formation involves a continual change in conditions. In continuous filtration, the resistance of the filter medium is generally negligible compared with the cake resistance, hence, B ≈ 0 . Integrating Eq. 1.11a with B = 0

∫

t

0

d t = K p

V

∫



0

VdV

t = K p

(1.16)

V

2

2

(1.17)

where t is the time required for formation of the cake. In a rotary-drum filter, the filter time t is less than the total cycle time t c by

t = ft c

(1.18)

where f is the fraction of the cycle used for cake formation. In the rotary drum, f is the fraction submergence of the drum surface in the slurry. Substituting Eq. 1.11b and Eq. 1.18 into Eq. 1.17 and rearranging,

 2 f (−∆ p)  Flow rate = =  At c  t c µα cs  V

1/ 2

(1.19)

When short cycle times are used in continuous filtration and/or the filter medium resistance is relatively large, the filter resistance term B must be included, and Eq. 1.11a becomes

2

t = ft c = K p

V

2

+ BV

Hence, Flow rate =

V At c

(1.20)

=

 Rm2   2csα (−∆ p ) f  − Rm / t c +   2 + µ t c   t c    α c s

1/ 2

(1.21)

All the mathematical derivations so far are taken from [Geankopolis].

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Semester 1, 2010

Example 1.4 Filtration in a Continuous Rotary-Drum Filter

A rotary-vacuum-drum filter having a 33% submergence of the drum in the slurry is to be used to filter a CaCO3 slurry as given in Example 1.1 using a pressure drop of 67.0 kPa. The solids concentration in the slurry is c x = 0.191 kg-solid/kg-slurry and the filter cake is such that the kgwet-cake/kg-dry-cake, m = 2.0. 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 cake time is 250 s. The specific cake resistance can be represented by α = ( 4.37 ×10 9 )(−∆p ) 0.3 , where

− ∆ p is in Pa and α in m/kg. 3

-3

For water ρ = 996.9 kg/m , µ = 0.8937 x 10 Pa*s

cs =

ρ c x

1 − mc x

=

(996.9)(0.191) 1 − (2.0)(0.191)

= 308.1 kg-solid/m3-filtrate

Solving for α α

= ( 4.37 ×109 )(−∆p ) 0.3 = (4.37 × 109 )(67 × 103 ) 0.3 = 1.225 × 1011 m/kg. To calculate the flow rate

of the filtrate,

    1 V kg − slurry  kg − solid   c x   = 0.778 c =  0.778  0.191 kg − solid  s t c s kg − slurry     308.1 m 3 − filtrate    = 4.823 × 10 −4 m -filtrate/s 3

Substituting into Eq. 1.19

V At c

=

4.823 × 10 −4

  2(0.33)(67 × 103 =  −3 11  250(0.8937 × 10 )(1.225 × 10 )(308.1) 

A

1/ 2

2

Hence, A = 6.60 m

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Semester 1, 2010

References [Seader and Henley] J.D. Seader & E.J. Henley John, 'Separation Process Principles' ,2nd Ed. John Wiley & Sons, 2006

[Geankoplis] C.J. Geankoplis, 'Transport Processes and Unit Operations', 4th Ed. Prentice-Hall International, 2003.

[McCabe and Smith] Mc Cabe W.L., J.C.Smith, P.Harriott, ‘ Unit Operations in Chemical Engineering’,7th Ed. McGraw Hill, 2005.

[Cheremisinoff] Cheremisinoff N.P. ‘Liquid Filtration’ 1st Ed. Butterworth-Heinemann, 1998.

Good online-resources for ChemE Separation Techniques http://people.clarkson.edu/~wwilcox/Design/refsepn.htm

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Filtration.pdf

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