TEKNOLOGI PENGENDALIAN PENCEMAR UDARA. Fabric filter. Rachmat Boedisantoso Jurusan Teknik Lingkungan FTSP ITS Kampus Sukolilo, Surabaya 60111

Fabric filter

Rachmat Boedisantoso

Jurusan Teknik Lingkungan FTSP – ITS Kampus Sukolilo, Surabaya – 60111

TEKNOLOGI PENGENDALIAN PENCEMAR UDARA

Advantages of Fabric Filters

 _{Very high collection efficiency}

 _{They can operate over a wide range of volumetric}

flow rates

 _{The pressure drops are reasonably low.}

 _{Fabric Filter houses are modular in design, and can}

Fabric Filters (contd.)

 Disadvantages of Fabric Filters

 _{Fabric Filters require a large floor area.}

 _{The fabric is damaged at high temperature.}

 _{Ordinary fabrics cannot handle corrosive gases.}

 _{Fabric Filters cannot handle moist gas streams}

Fabric Filters

 Principle

 _{The filters retain particles larger than the}

mesh size

 _{Air and most of the smaller particles flow}

through. Some of the smaller particles are retained due to interception and diffusion.

 _{The retained particles cause a reduction in}

the mesh size.

 _{The primary collection is on the layer of}

 Fabric Filtration merupakan alat kontrol udara yang paling umum dipergunakan

 Fabric filter menggunakan filter yang terbuat dari nilon atau wol

 Partikulat yang telah disisihkan/ terkumpul kemudian dibersihkan dengan mekanisme pembersihan tertentu.

 Fabric filter juga disebut :

Baghouse

Fabric filter collectors,

Bag filters,

Fabric dust collectors,

Filter collectors,

Dust collectors,

Cloth collectors

keuntungan dan kerugian Fabric Filter :

 Keuntungan penggunaan fabric filter adalah :

 Efisiensi sangat tinggi, bahkan untuk partikel yang halus

 Dapat dipakai untuk berbagai macam debu

 Dapat untuk volume gas yang besar

 Dapat beroperasi pada pressure drop yang rendah

 Sedangkan kerugiannya adalah :

 Memerlukan tempat luas

 Bahan filter dapat rusak pada temperatur tinggi atau bahan

asam

 Tidak dapat beroperasi pada lingkungan yang lembab

 Fabric filter terdiri atas :

inlet,

outlet,

filter bag,

Mekanisme pengumpulan Fabric Filter pada umumnya ada tiga cara utama, yaitu :

 Impaction, partikel memiliki gaya inersia yang terlalu besar untuk mengikuti aliran garis pada filter fiber sehingga tertumbuk pada permukaan filter

 Interception, partikel mempunyai inersia yang sangat kecil (partikel yg lebih kecil). Partikel akan berada

pada aliran viscous, bergerak melambat dan menyentuh barrier dan berhenti

 Diffusion, partikel lebih kecil dari 1 mikron berada pada kisaran gerak Brown, sehingga terjadi gerakan random yang akhirnya terintersepsi dengan dust

Proses Filtrasi Fabric Filter terdapat dua jenis

desain yang dapat digunakan yaitu:

 Interior Filtration, partikulat dikumpulkan pada

bagian dalam dari bag filter. Gas yang

mengandung partikulat memasuki fabric filter melalui bagian bawah dari kolektor dan

diarahkan ke dalam bag dengan

menggunakan diffuser vanes atau baffle dan juga cell plate.

 Exterior Filtration, partikulat dikumpulkan

pada bagian luar dari bag fliter. Proses

penyaringan berlangsung dari luar bag filter ke dalam bag filter.

Jenis Proses filtrasi ( kiri : Interior Filtrasi, kanan : Eksterior Filtrasi)

Terdapat beberapa cara pembersihan yang dapat

dipergunakan untuk menyisihkan partikulat yang menempel pada permukaan bag filter, tiga cara yang paling sering

digunakan adalah :

 (i). Shaking

 Mechanical shaking menggunakan motor yang dihubungkan dengan bag

 Energi yang diperlukan rendah

 Gerakan dan kecepatan tergantung endapan debu

 Arah gerakan Horisontal atau vertical

 Gerakan di bagian atas frame tempat bag diletakkan

 Aliran gas berhenti saat dilakukan proses pembersihan

 Shaker baghouse umumnya menggunakan interior filtration  Diameter 15,2 – 45,7 cm ( 6-18 inch )  Panjang sampai 12,2 m ( 40 ft )  Lama pembersihan 30 dt – beberapa menit

 Terdiri dari beberapa kompartemen

 (ii). Reverse Air

 Mekanisme yang paling sederhana

 Aliran udara kotor dihentikan

 Mengalirkan backwash air (udara bersih yang berlawanan arah)

 Aliran udara bertekanan rendah

 Debu akan jatuh ke hopper

 Lama pembersihan 30 menit – beberapa jam  Durasi pembersihan 10 – 30 detik

 Terdapat ring dengan jarak 10 – 46 cm

 Reverse air baghouse berdiameter 20 – 46 cm, dan

panjang 6,1 – 12,2 cm

 Terdiri dari beberapa kompartemen

 (iii). Pulse Jet

 Disebut juga pressure jet cleaning

 40 – 50% baghouse baru di Amerika

 Menggunakan high pressure jet dari udara

 _{Sistem exterior filtration}

 Menimbulkan shock wave

 Pulse jet baghouse berdiameter 10,2 – 16,2

cm

 Panjang umumnya 2,4 – 3,7, tapi dapat

Bahan bag house

 _{Woven, terbuat dari benang, dipakai untuk}

pembersihan energi rendah

 Felted filter terdiri dari fiber yang dikompres

ke dalam mar dan dilekatkan pada woven

 Ada yang terbuat dari bahan alam seperti katun atau wol

 Temperatur lebih kecil dari 212oF atau 100oC  Katun temperatur lebih kecil

 Fiber sintesis

 Nilon, orlon dan polyester tahan temperatur lebih tinggi

dan tahan terhadap bahan kimia

 Nilon memiliki abrasive resistant yang paling tinggi

 Polyester atau Dacron baik untuk menahan asam,

alkali dan abrasi dan relatif murah

 Nomex buatan Dupont

 Membran material terbuat dari berbagai jenis fiber yang disusun membentuk

membrane diantaranya adalah Gore-Tex membrane, jenis ini dapat

mereduksi emisi dengan baik, pressure drop yang relatif rendah, umur bag yang meningkat dan ratio air-cloth yang lebih tinggi

Design of Fabric Filters

The equation for fabric filters is based on

Darcy’s law for flow through porous media.

Fabric filtration can be represented by the

following equation:

S = K_e + K_sw

Where,

S = filter drag, N-min/m3

K_e = extrapolated clean filter drag, N-min/m3

K_s = slope constant. Varies with the dust, gas and fabric, N-min/kg-m W= Areal dust density = LVt, where

L = dust loading (g/m3), V = velocity (m/s)

Both K_e and K_s are determined empirically

Fabric Filters

Δ P Total pressure drop

Δ P_f Pressure drop due to the fabric

Δ P_p Pressure drop due to the particulate layer

Darcy’s equation

ΔP_f Pressure drop N/m2

ΔP_p Pressure drop N/m2

D_f Depth of filter in the direction of flow (m)

D_p Depth of particulate layer in the direction of flow (m) μ Gas viscosity kg/m-s

V superficial filtering velocity m/min

K_f, K_p Permeability (filter & particulate layer m2₎

60 Conversion factor δ/min V = Q/A

Q volumetric gas flow rate m3_/min

Dust Layer

L Dust loading kg/m3

t time of operation min

ρ_L Bulk density of the particulate layer kg/m3

ΔP = ΔP_f + ΔPp

Filter Drag S = ΔP/V

Areal dust density W = LVt S= k₁+k₂W

Parameter yang penting dan perlu

dipertimbangkan dalam merancang Fabric Filter diantaranya adalah :

 (i). Pressure Drop

 Dinyatakan sebagai pressure drop per unit

area sebgai fungsi dari karakteristik medium filter

 Biasanya berkisar 2-4 inch

 Dihitung dengan cara :

t V C k p V k p p p p f c f f c f 1 2 1         

 dimana :

 pf = pressure drop sepanjang FF

 k1 = resistensi FF (inch air /menit atau cm/menit),

merupakan fungsi dari karakteristik viscositas gas dan filter seperti ketebalan dan porositas (permeabilitas)

 Vf = kecepatan filtrasi (ft/menit atau m/menit)

 pf = pressure drop sepanjang cake dalam inch (cm)

 k2 = resistensi dari cake (inch air /menit atau

cm/menit

 C1 = dust loading (lb/ft3 atau kg/m3) ditentukan

secara experimen. Koefisien ini tergantung dari viscositas gas, densitas partikel dan porositas.

t V C k p V k p p p p f c f f c f 1 2 1         

 (ii). Kecepatan Penyaringan

 Kecepatan penyaringan dinyatakan sebagai :

 kecepatan merupakan kecepatan superficial filtering

A

Q



 (iii). Performance Factor

 Salah satu variable yang penting dalam

mendesain baghouse adalah ratio air to cloth (A/C) atau ratio udara terhadap bahan filter

 _{A/C menggambarkan berapa banyak gas}

kotor yang melewati permukaan filter dengan luas tertentu selama waktu tertentu.

 Ratio yang tinggi berarti sejumlah besar

udara yang melewati fabric

 _{Satuan cm}3_/detik/cm2 _{atau ft}3_/menit/ft2

 Tergantung dari mekanisme pembersihan,

Problem

 Estimate the net cloth area for a shaker

bag house that must filter 40,000 cfm of air with 10 grams of flour dust per cubic foot of air. Also specify the number of

components to be used and calculate the total number of bags required if each bag is 8 feet long and 0.5 feet in diameter.

The maximum filtering velocity for flour dust is 2.5 ft/min.

Solution

 Step 1: Calculate total area and number of

components required: A = Q/V

 Step 2:

Calculate the area of each bag: A = Π(d)l

 Step 3:

Calculate the total number of bags required. Number of bags required = Total area / Area

30

2. Dividing Collection Devices

 Filters and scrubbers do not drive the particles to a wall, but rather divide the flow into smaller parts where they can collect the particles.

2.1 Surface Filters

 A surface filter is a membrane (sheet steel,

cloth, wire mesh, or filter paper) with holes smaller than the dimensions of the

31

 However, one only needs to ponder the mechanical problem of drilling holes of

0.1-µ diameter or of weaving a fabric with threads separated by 0.1µ to see that

such filters are not easy to produce.

 They are much too expensive and fragile for use as high-volume industrial air

32

 Although industrial air filths rarely have holes smaller than the smallest particles captured, they often act as if they did.

 The reason is that, as fine particles are

caught on the sides of the holes of a filter, they tend to bridge over the holes and

make them smaller.

 Thus as the amount of collected particles increases, the cake of collected material becomes the filter.

33

 The particles collect on the front surface of the growing cake.

 For that reason this is called a surface filter.

 The flow through a simple filter is shown schematically in Fig. 9.12.

35

 If we follow the gas stream in Fig. 9.12

from point 1 to point 3 we see that the flow is horizontal and has a small change in

velocity because the pressure drop,

causing the gas to expand, and because the gas is leaving behind its contained particles.

36

 In most industrial filters, both for gases and liquids, the flow velocity in the individual pores is so low that the flow is laminar.

 Therefore, we may use the well-known relations for laminar flow of a fluid in a porous medium, which indicate

Here, k is the permeability, a property of the bed. (19) s Q p k A x    _{ } _   _{   }    

37

 For a steady fluid flow through a filter cake supported by a filter medium, there are two resistances to flow in series, but the flow rate is the same through each of them.

 We find:

 Solving for P₂, we get

2 3 1 2 cake filter -P = s P P P k k x x      _{ }  _{ }  _{     }           2 1 3 cake filter = s s x x P P P k k           _{ }  _{ }    

38

 Then solving for v_s:

 This equation describes the instantaneous flow rate through a filter; it is analogous to Ohm’s law for two resistors in series.

 The Δx/k terms are called the cake resistance and

the cloth resistance.

1 3

( )

(20)

[( / ) ( / ) ]

s

cake filter filter

p p Q v x k x k A       

39

 The resistance of the filter medium is usually

assumed to be a constant that is independent of time, so (Δx/k)_filter is replaced with a constant α.

 If the filter cake is uniform, then its resistance is proportional to its thickness:

1 1 cake cake cake mass of cake x area

volume of gas mass of solids removed area volume of gas

        _{  }  _{ }  _{ }        _{ }  

40

 Customarily we define:

Here W is the volume of cake per volume of gas processed, which corresponds to a collection efficiency, η, of 1.0.

 For most surface filters η=1.0, so the η is normally

dropped in the equation. Thus

Here V is the volume of gas cleaned.

1

cake

mass of solids removed volume of cake

W

volume of gas volume of gas processed

      _{  }    ( ) and cake (21) cake s d x V x W W A dt       _{ }   

41

 _{Substituting Eq. (21) for the cake thickness in Eq. (20), we}

find

 For most industrial gas filtrations the filter is supplied by a centrifugal blower at practically constant pressure, so (P₁ -P₃) is a constant, and Eq. (22) may be rearranged and integrated to 1 3 (P -P ) Q 1 dV = = = (22) A A dt [(V /kA+ ] s W          2 1 3 ( ) (23) 2 V W V P P t A k A  _      _{  }            

42

 For many filtrations the resistance α of the

filter medium is negligible compared with the cake resistance, so the second term of Eq. (23) may be dropped.

 The two most widely used designs of

industrial surface filters are shown in Figs. 9.13 and 9.14 (next and second slides).

45

 For the baghouse in Fig. 9.13 there must be some way of removing the cake of particles that accumulates on the filters.

 Normally this is not done during gas-cleaning operations.

 A weak flow of gas in the reverse direction may also be added to help dislodge the cake, thus deflating the bags.

 Often metal rings are sewn into filter bags at regular intervals so that they will only partly collapse when the flow in reversed.

46

 Because it cannot filter gas while it is being cleaned, a shake-deflate baghouse cannot serve as the sole

pollution control device for a source that produces a continuous flow of dirty gas.

 Typically, for a major continuous source like a power plant, about five baghouses will be used in parallel, with four operating as gas cleaners during the time that the other one is being shaken and cleaned.

 Each baghouse might operate for two hours and then be cleaned for 10 minutes.

47

 The other widely used baghouse design, called a pulse-jet filter, is shown in Fig. 9.14.

 In it the flow during filtration is inward

through the bags, which are similar to the bags in Fig. 9.13 except their ends open at the top.

 The bags are supported by internal wire cages to prevent their collapse.

48

 The bags are cleaned by intermittent jets of compressed air that flow into the inside of the bag to blow the cake off.

 Often these baghouses are cleaned while they are in service.

49

Example 15

 The shake-deflate baghouse on a power

station has six compartments, each with 112 bags that are 8 in. in diameter and 22 ft long, for an active area of 46 ft2 per bag.

 The gas being cleaned has a flow rate of

86,240 ft3/min.

 The pressure drop through a freshly

cleaned baghouse is estimated to be 0.5 in. H₂O.

50

 The bags are operated until the pressure drop is 3 in. H₂O, at which time they are taken out of service and cleaned.

 The cleaning frequency is once per hour.

 The incoming gas has a particle loading of 13 grams/ft3.

51

 The collection efficiency is 99%, and the filter cake is estimated to be 50% solids, with the balance being voids.

 Estimate how thick the cake is when the gags are taken out of service for cleaning. What is the permeability, k, of the cake?

52

 Solution:

The average velocity coming to the filter surface:

The 5 is used here because one of the six

compartment is always out of service for cleaning.

 v_s is commonly referred to as the air-to-cloth ratio, face velocity, or superficial velocity.

3 2 86240 / 3.35 1.02 (5)(112)(46 ) s Q ft min ft m A ft min min     

53

 If the filter remains in service for 1 hour before

cleaning and v_s is constant, the 1 square foot of bag will collect the following mass of particles:

  3 2 2 60 13 3.35 0.99 (1 ) 7000 0.369 1.80 s m gr lbm ft min c t h A ft gr min h lbm kg ft m          _{      }         

54

 The thickness of the cake collected in 1 hour is

 Taking 2 3 3 3 3 / 0.369 / (2 / )(0.5)(62.4 / / ) 5.9 10 0.071 . 1.8 m A lbm ft g cm lbm cm ft g ft in mm        

0, we can solve Eq. (20) for k,

filter x k     _{ }     5 2 2 2 2 12 2 13 2 (3.35 / min)(0.071 /12 .)(0.018 )(2.09 10 / / )( / 60 ) ( ) (3 . )(5.202 / / . ) 7.96 10 7.40 10 s x ft ft in cp lbf s ft cp min s k p in H O lbf ft in H O ft m               

55

 Compare this with values found in ground water flow:

 The calculated permeability of this material is roughly the same as that of a highly permeable sandstone. #

12 2 11 2 (7.96 10 ) 0.75 1.06 10 darcy k ft darcies ft       _{ }    

56

 This calculation shows that the collected is about 1.8 mm thick.

 If the cleaning were perfect, this would be the cake thickness.

 However, it is hard to clean the bags

completely, and in power plant operation it is common for the average cake thickness on the bags to be up to 10 times this

57

 One of the advantages of the pulse-jet design is that it cleans the bags more thoroughly, allowing a higher v_s, at the cost of a somewhat shortened bag life.

 Fig. 9.15 (next slide) is a set of typical results from tests of collection efficiency for this kind of filter.

59

 If the superficial velocity increases, the

efficiency falls; for a superficial velocity of 3.35 m/min the outlet concentration is

about 20 percent of the inlet concentration.

 The particles that pass through such a filter do not pass through the cake but

through pinholes, which are regions where the cake did not establish properly (Fig.

61

 The pinholes are apparently about 100 µ in diameter, much too large for a single particle to block because there are rarely 100-µ particles in the streams being

treated.

 When the superficial velocity is high, more pinholes form.

62

 _{Example 16}

 Estimate the velocity through a pinhole in a filter with a pressure drop of 3 in. of

water.

 Assuming that this is the pressure drop

corresponding to the curve for 0.39 m/min on Fig. 9.15, that the steady-state

penetration at that velocity is 0.001, and

that the pinholes have a diameter of 100 µ, estimate how many pinholes per unit area there are in the cake.

63

 _Solution:

The flow through the pinhole is best described by Bernoulli’s equation, from which we find the average velocity:

 Here the (area ∙velocity) of the pinholes must be 0.001 times the

(area ∙velocity) of the rest of the cake. Hence

1/ 2 1/ 2 2 3 2 2 2 2(3 ) 249 0.61 (1.20 / ) 21.5 / pinholes P in H O Pa kg C kg m in H O Pa m s m s    _{ }         _{ }  _{   }  _   _   _   _  2 7 2 0.001 0.001 (0.39 / ) 3.0 10 (21.5 / )(60 / ) pinholes _s cake pinholes A _{m min m} A m s s min m       

64

 Each pinhole has an area of A =

(100  10-6 _m)2 (π/4) = 7.85  10-9 _m2_{, so there}

must be

 The calculated velocity through a pinhole is (21.6 x 60) /0.39 = 3300 times the velocity through the cake.

7 2 9 2 3.0 10 38 pinholes/m # 7.85 10 m     

65

2.2 Depth Filter

 Filters do not form a coherent cake on the surface, but instead collect particles

throughout the entire filter body are called depth filters.

 The examples with which the student is probably familiar are the filters on filter-tipped cigarettes and the lint filters on many home furnaces.

66

 In both of these a mass of randomly

oriented fibers (not woven to form a single surface) collects particles as the gas

passes through it.

 In Fig. 9.17 (next slide), we see a particle-laden gas flowing toward a target, which we may think of as a cylindrical fiber in a filter.

68

 To determine whether a particle bumps into the target or flows around it, we can compute the relative velocity between particle and gas using the appropriate equivalents of Stokes’ law.

 That task was first undertaken by

Langmuir and Blodgett, and Fig. 9.18 (next slide) conveniently summarizes the

mathematical solutions for the small

70

 To see how they obtained Fig. 9.18,

consider a single particle in a turning part of the gas stream as shown in Fig. 9.19 (next slide).

72

 In Fig. 9.19, the appropriate velocity to use in Stokes’ law is not the overall stream velocity but rather the difference in y-directed velocity between the particle and the gas stream.

 Generally, the gas stream will have a larger velocity in this direction than the particle, so

3

(

)

y drag y gas y particle

73

 The resisting (inertial) force of the particle is

 If there are no electrostatic, magnetic, or other forces acting, then these two forces are equal and opposite, so 3 6 y particle y inertial d F ma D dt        2 18 ( )

y particle y gas y particle

d

dt D

   



74

 Rearrange,

 Δt is the time available for the y-directed forces to

move the particle around the garget, which must be proportional to the time it takes the main gas flow to go past the target.

2 18 (24) ( ) y particle y gas y particle d _t D  _         



75

 This time = D_b/v, where D_b is the diameter of the barrier, so we may substitute in Eq. (24), finding

 N_s is the separation number, which appears on the horizontal axis in Fig. 9.18.

 It is equal to the diameter of the barrier divided by the stokes stopping distance.

2 1 2 18 1 (25) ( ) y _{y particle} b y y gas y particle s d _D D N  _          



76

 Some authors call N_s the impaction parameter or inertial parameter.

 In Eq. (25) we can see that if the integral on the left is a large number (a low value of N_s), then there is plenty of time and

force for the flow to move the particle

around the target and the target efficiency (η_t) will be low.

77

Example 17

 A single, cylindrical fiber 10 µ in diameter is placed perpendicular to a gas stream that is moving at 1 m/s.

 The gas stream contains particles that are

1 µ in diameter and the particle concentration is 1 mg/m3.

 What is the rate of collection of particles on the fiber?

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 Solution:

If all the particles that start moving directly toward the fiber hit it, then the collection rate would be equal to the volumetric flow rate approaching the fiber times the concentration of particles:

 If we catch them all, we will collect 10-8 _{g/s for every}

meter of fiber length.

5 3 3

8

maximum possible rate = (1 / )(10 )(10 / ) 10 / / b D c m s m g m g m s      

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 The actual amount caught will be this number times the target efficiency.

 The separation number is

 From Fig. 9.18, we see that for cylinders this value of N_s corresponds to a target efficiency of about 0.42, so we would expect to collect about 0.42  10-8

g/m/s. # 3 6 2 5 5 (2000 / )(10 ) (1 / ) 0.617 (18)(1.8 10 / / )(10 ) s kg m m m s N kg m s m      

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Example 18

 A filter consists of a row of parallel fibers

across a flow, as described in Example 17, with the center-to-center spacing of the

fibers equal to five fiber diameters.

 What collection efficiency will the filter have for the particles?

 Assume that the fibers are far enough

apart that each one behaves as if it were in an infinite fluid, uninfluenced by the

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 _Solution:

From Example 17, η_t = 42% for a single fiber.

If the fibers are spaced five fiber diameters apart, then the open area is 80%, and the blocked area is 20%.

 So,

Collection efficiency = (target efficiency) (percentage blocked) = 42%20% = 8.4% #

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Example 19

 A filter consists of 100 rows of parallel fibers as described in Example 18,

arranged in series.

 They are spaced far enough apart that the

flow field becomes completely uniform between one row and the next.

 What is the collection efficiency of the entire filter.

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 _Solution:

η_overall = 1 - p_overall = 1 - (p_individual)n

= 1 - (1-0.084)100 = 0.9998 #

 These three examples show, in idealized form, what goes on within depth filters.

 Most such filters do not have an orderly array of parallel fibers; the filter medium consists of a tangled jumble of fibers in a random orientation, making up a thick mat.

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 In depth filters that diffusion leads to small particle collection in addition to that

computed above by impaction.

 Friedlander developed a theoretical

equation, with constants determined by experiment, for the case of diffusion

collection of particles from a gas steam flowing past a cylinder under

circumstances where impaction was negligible.

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 Most of the published data could be represented by

where all the terms are as defined previously, D_i is the diffusivity and ν is the kinematic viscosity.

 The first term on the right is for diffusion collection, where as the second is for collection by noninertial contact (interception) 2 / 3 _{2 1/ 2} 1/ 2 3/ 2 1/ 6 1/ 2 1/ 2 6 3 (26) i t b b D D D D       

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Example 20

 Repeat Example 17 for particles having a diameter of 0.1 µ. Take into account

impactions, diffusion, and interception.

 _Solution:

In this case N_s is (0.1)2 = 0.01 times the previous value, or 0.0062, for which, from Fig. 9.18 we can read η_t ≈ 0.

 From Fig. 8.1, we can read that the

diffusivity is about 6 10-6 _cm2_{/s = 6 10} -10 _m2_/s.

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 So

 The diffusion term is (0.0086 / 0.00025) = 34.4 times the interception term.

 Table 9.3 (next slide) shows the effect of changes in velocity and particle diameter on the collection

mechanisms. 10 2 2 / 3 7 2 1/ 2 5 2 5 1/ 2 1/ 2 5 2 1/ 2 5 3/ 2 6(6 10 / ) 3(10 ) (1 / ) (1.49 10 / )(10 ) (1 / ) (1.49 10 / ) (10 ) 0.0086 0.00025 0.0088 0.9% # t m s m m s m s m m s m s m   _   _  _ _      

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 There is some particle size at which there is a minimum collection efficiency.

 Typically, this size is in the range 0.1 to 1 µ, which is the size most likely to be

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2.3 Filter Media

 For shake-deflate baghouses, the filter bags are made of tightly woven fibers

(surface filter), much like those in a pair of jeans.

 Pulse-jet baghouses use high strength felted fabrics, so that they act partly as depth filters and partly as surface filters.

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 Filter Fabrics are made of cotton, wool, glass fibers, and a variety of synthetic fibers.

 Cotton and wool cannot be used above

180 and 200oF, respectively, without rapid deterioration, whereas glass can be used to 500oF (and short-time excursions to

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 In addition the fibers must be resistant to acids or alkalis if these are present in the gas stream or the particles as well as to flexing wear caused by the repeated

cleaning.

Sumber:

 Control of Primary Particulates by Hsin

Chu, Professor, Dept. of Environmental Engineering, National Cheng Kung

University, Taiwan

 Technology for Air Pollution Control by