L1681_CH09

L1681_book.fm Page 207 Tuesday, October 5, 2004 10:51 AM

CHAPTER 9

Particulate Emission Control

Fresh air is good if you do not take too much of it; most of the achievements and pleasures of lifeare in bad air.

Oliver Wendell Holmes

9.1 PARTICULATE EMISSION CONTROL BASICS

Particle or particulate matter is defined as tiny particles or liquid droplets suspended in the air; theycan contain a variety of chemical components. Larger particles are visible as smoke or dust andsettle out relatively rapidly. The tiniest particles can be suspended in the air for long periods oftime and are the most harmful to human health because they can penetrate deep into the lungs.Some particles are directly emitted into the air from pollution sources.

Constituting a major class of air pollutants, particulates have a variety of shapes and sizes; aseither liquid droplet or dry dust, they have a wide range of physical and chemical characteristics.Dry particulates are emitted from a variety of different sources in industry, mining, constructionactivities, and incinerators, as well as from internal combustion engines — from cars, trucks, buses,factories, construction sites, tilled fields, unpaved roads, stone crushing, and wood burning. Dryparticulates also come from natural sources such as volcanoes, forest fires, pollen, and windstorms.Other particles are formed in the atmosphere by chemical reactions.

When a flowing fluid (engineering and science applications consider liquid and gaseous statesas fluid) approaches a stationary object (a metal plate, fabric thread, or large water droplet, forexample), the fluid flow will diverge around that object. Particles in the fluid (because of inertia)will not follow stream flow exactly, but tend to continue in their original directions. If the particleshave enough inertia and are located close enough to the stationary object, they collide with theobject and can be collected by it. This important phenomenon is depicted in Figure 9.1.

9.1.1 Interaction of Particles with Gas

To understand the interaction of particles with the surrounding gas, knowledge of certain aspectsof the kinetic theory of gases is necessary. This kinetic theory explains temperature, pressure, meanfree path, viscosity, and diffusion in the motion of gas molecules (Hinds, 1986). The theory assumesgases — along with molecules as rigid spheres that travel in straight lines — contain a large numberof molecules that are small enough so that the relevant distances between them are discontinuous.

Air molecules travel at an average of 1519 ft/sec (463 m/sec) at standard conditions. Speeddecreases with increased molecule weight. As the square root of absolute temperature increases,molecular velocity increases. Thus, temperature is an indication of the kinetic energy of gas

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208 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK


molecules. When molecular impact on a surface occurs, pressure develops and is directly relatedto concentration. Gas viscosity represents the transfer of momentum by randomly moving moleculesfrom a faster moving layer of gas to an adjacent slower moving layer of gas. Viscosity of a gas isindependent of pressure but will increase as temperature increases. Finally, diffusion is the transferof molecular mass without any fluid flow (Hinds, 1986). Diffusion transfer of gas molecules isfrom a higher to a lower concentration. Movement of gas molecules by diffusion is directlyproportional to the concentration gradient, inversely proportional to concentration, and proportionalto the square root of absolute temperature.

The mean free path, kinetic theory’s most critical quantity, is the average distance a moleculetravels in a gas between collisions with other molecules. The mean free path increases withincreasing temperature and decreases with increasing pressure (Hinds, 1986).

The Reynolds number characterizes gas flow, a dimensionless index that describes the flowregime. The Reynolds number for gas is determined by the following equation:

(9.1)

whereRe = Reynolds numberp = gas density, pounds per cubic foot (kilograms per cubic meter)Ug = gas velocity, feet per second (meters per second)D = characteristic length, feet (meters)η = gas viscosity, lbm/ft·sec (kg/m·sec)

The Reynolds number helps to determine the flow regime, the application of certain equations,and geometric similarity (Baron and Willeke, 1993). Flow is laminar at low Reynolds numbers andviscous forces predominate. Inertial forces dominate the flow at high Reynolds numbers, whenmixing causes the streamlines to disappear.

9.1.2 Particulate Collection

Particles are collected by gravity, centrifugal force, and electrostatic force, as well as by impaction,interception, and diffusion. Impaction occurs when the center of mass of a particle diverging fromthe fluid strikes a stationary object. Interception occurs when the particle’s center of mass closelymisses the object but, because of its size, the particle strikes the object. Diffusion occurs whensmall particulates happen to “diffuse” toward the object while passing near it. Particles that strike

Figure 9.1 Particle collection of a stationary object. (Adapted from USEPA-84/03, p. 1-5.)

Re = pU DG

η

PARTICULATE EMISSION CONTROL 209


the object by any of these means are collected if short-range forces (chemical, electrostatic, andso forth) are strong enough to hold them to the surface (Copper and Alley, 1990).

Different classes of particulate control equipment include gravity settlers, cyclones, electrostaticprecipitators, wet (Venturi) scrubbers, and baghouses (fabric filters). In the following sections wediscuss many of the calculations used in particulate emission control operations. Many of thecalculations presented are excerpted from USEPA-81/10.

9.2 PARTICULATE SIZE CHARACTERISTICS AND GENERAL CHARACTERISTICS

As we have said, particulate air pollution consists of solid and/or liquid matter in air or gas. Airborneparticles come in a range of sizes. From near molecular size, the size of particulate matter rangesupward and is expressed in micrometers (µm — one millionth of a meter). For control purposes,the lower practical limit is about 0.01 µm. Because of the increased difficulty in controlling theiremission, particles of 3 µm or smaller are defined as fine particles. Unless otherwise specified,concentrations of particulate matter are by mass.

Liquid particulate matter and particulates formed from liquids (very small particles) are likelyto be spherical in shape. To express the size of a nonspherical (irregular) particle as a diameter,several relationships are important. These include:

• Aerodynamic diameter• Equivalent diameter• Sedimentation diameter• Cut diameter• Dynamic shape factor

9.2.1 Aerodynamic Diameter

Aerodynamic diameter, da, is the diameter of a unit density sphere (density = 1.00 g/cm3) thatwould have the same settling velocity as the particle or aerosol in question. Note that becauseUSEPA is interested in how deeply a particle penetrates into the lung, the agency is more interestedin nominal aerodynamic diameter than in the other methods of assessing size of nonsphericalparticles. Nevertheless, a particle’s nominal aerodynamic diameter is generally similar to its con-ventional, nominal physical diameter.

9.2.2 Equivalent Diameter

Equivalent diameter, de, is the diameter of a sphere that has the same value of a physical propertyas that of the nonspherical particle and is given by

(9.2)

where V is volume of the particle.

9.2.3 Sedimentation Diameter

Sedimentation diameter, or Stokes diameter, ds, is the diameter of a sphere that has the same terminalsettling velocity and density as the particle. In density particles, it is called the reduced sedimentationdiameter, making it the same as aerodynamic diameter. The dynamic shape factor accounts for anonspherical particle settling more slowly than a sphere of the same volume.

dV

e =

61 3

π

/



9.2.4 Cut Diameter

Cut diameter, dc, is the diameter of particles collected with 50% efficiency, i.e., individual efficiencyεI = 0.5, and half penetrate through the collector — penetration Pt = 0.5.

(9.3)

9.2.5 Dynamic Shape Factor

Dynamic shape factor, χ, is a dimensionless proportionality constant relating the equivalent andsedimentation diameters:

(9.4)

The de equals ds for spherical particles, so χ for spheres is 1.0.

9.3 FLOW REGIME OF PARTICLE MOTION

Air pollution control devices collect solid or liquid particles via the movement of a particle in thegas (fluid) stream. For a particle to be captured, the particle must be subjected to external forceslarge enough to separate it from the gas stream. According to USEPA-81/10, p. 3-1, forces actingon a particle include three major forces as well as other forces:

• Gravitational force• Buoyant force• Drag force• Other forces (magnetic, inertial, electrostatic, and thermal force, for example)

The consequence of acting forces on a particle results in the settling velocity — the speed atwhich a particle settles. The settling velocity (also known as the terminal velocity) is a constantvalue of velocity reached when all forces (gravity, drag, buoyancy, etc.) acting on a body arebalanced — that is, when the sum of all the forces is equal to zero (no acceleration). To solve foran unknown particle settling velocity, we must determine the flow regime of particle motion. Oncethe flow regime has been determined, we can calculate the settling velocity of a particle.

The flow regime can be calculated using the following equation (USEPA-81/10, p. 3-10):

(9.5)

whereK = a dimensionless constant that determines the range of the fluid-particle dynamic lawsdp = particle diameter, centimeters or feetg = gravity force, cm/sec2 or ft/sec2

pp = particle density, grams per cubic centimeter or pounds per cubic footpa = fluid (gas) density, grams per cubic centimeter or pounds per cubic footµ = fluid (gas) viscosity, grams per centimeter-second or pounds per foot-second

Pt 1 –I I= ε

χ =

dd

e

s

2

K d (gp p /µ )p p a2 0.33=



USEPA-81/10, p. 3-10, lists the K values corresponding to different flow regimes as:

• Laminar regime (also known as Stokes’ law range): K < 3.3• Transition regime (also known as intermediate law range): 3.3 < K, 43.6• Turbulent regime (also known as Newton’s law range): K > 43.6

According to USEPA-81/10, p. 3-10, the K value determines the appropriate range of the fluid-particle dynamic laws that apply.

• For a laminar regime (Stokes’ law range), the terminal velocity is

(9.6)

• For a transition regime (intermediate law range), the terminal velocity is:

(9.7)

• For a turbulent regime (Newton’s law range), the terminal velocity is:

(9.8)

When particles approach sizes comparable with the mean free path of fluid molecules (alsoknown as the Knudsen number, Kn), the medium can no longer be regarded as continuous becauseparticles can fall between the molecules at a faster rate than that predicted by aerodynamic theory.Cunningham’s correction factor, which includes thermal and momentum accommodation factorsbased on the Millikan oil-drop studies and which is empirically adjusted to fit a wide range of Knvalues, is introduced into Stoke’s law to allow for this slip rate (Hesketh, 1991; USEPA-84/09, p. 58):

(9.9)

whereCf = Cunningham correction factor = 1 + (2Aλ /dp) A = 1.257 + 0.40e–1.10dp/2λλ = free path of the fluid molecules (6.53 × 10–6 cm for ambient air)

Example 9.1

Problem:Calculate the settling velocity of a particle moving in a gas stream. Assume the following

information (USEPA-81/10, p. 3-11):

Given:dp = particle diameter = 45 µm (45 microns)g = gravity forces = 980 cm/sec2

pp = particle density = 0.899 g/cm3

pa = fluid (gas) density = 0.012 g/cm3

µ = fluid (gas) viscosity = 1.82 × 10–4 g/cm-secCf = 1.0 (if applicable)

v gp (d ) / ( µp p= 2 18 )

v . g (d ) (p ) / [µ (p ).p

.p

. .a

.= 0 153 0 71 1 14 0 71 0 43 0 229]

v 1.74(gd p /p )p p a0.5=

v gp (d ) C / ( µp p f= 2 18 )



Solution:

Step 1. Use Equation 9.5 to calculate the K parameter to determine the proper flow regime:

The result demonstrates that the flow regime is laminar.

Step 2. Use Equation 9.9 to determine the settling velocity:

Example 9.2

Problem:Three differently sized fly ash particles settle through the air. Calculate the particle terminal

velocity (assume the particles are spherical) and determine how far each will fall in 30 sec.

Given:Fly ash particle diameters = 0.4, 40, 400 µmAir temperature and pressure = 238°F, 1 atmSpecific gravity of fly ash = 2.31

Because the Cunningham correction factor is usually applied to particles equal to or smaller than 1 µm,check how it affects the terminal settling velocity for the 0.4-µm particle.

Solution:

Step 1. Determine the value for K for each fly ash particle size settling in air. Calculate the particledensity using the specific gravity given:

Calculate the density of air:

K d (gp p / µ )p p a.= 2 330

= × ×45 10 [980(0.899)(0.012)/1.82 10–4 –4)2]] .0 33

= 3.07

v gp (d ) C / ( µp p f= 2 18 )

= × ×980(0.899)(45 10 ) (1)/[18(1.82 10–4 2 –4))]

= 5.38 cm/sec

Pp particle density (specific gravity o= = ff fly ash)(density of water)

= 2.31(62.4)

= 144.14 lb/ft3



(USEPA-84/09, p. 167)

Determine the flow regime (K):

For dp = 0.4 µm:

where 1 ft = 25,400(12) µm (USEPA-84/09, p. 183)

For dp = 40 µm:

For dp = 400 µm:

Step 2. Select the appropriate law, determined by the numerical value of K:K < 3.3; Stokes’ law range3.3 < K < 43.6; intermediate law range43.6 < K < 2360; Newton’s law rangeFor dp = 0.4 µm, the flow regime is laminar (USEPA-81/10, p. 3-10)For dp = 40 µm, the flow regime is also laminarFor dp = 400 µm, the flow regime is the transition regime

For dp = 0.4 µm:

For dp = 40 µm:

p = =air density PM/RT

= + =(1)(29)/(0.7302)(238 460) 0.0569 lb/ftt3

µ = = = ×air viscosity 0.021 cp 1.41 10–5 llb/ft-sec

K d (gp p / µ )p p a.= 2 0 33

K = [(0.4)/(25,400)(12)][32.2(144.14)(0.05699)/(1.41 10 ) ] 0.0144–5 2 0.33× =

K = [(40)/(25,400)(12)][32.3(144.14)(0.0569))/(1.41 10 ) ] 1.44–5 2 0.33× =

K = [(400)/(25,400)(12)][32.2(144.14)(0.05699)/(1.41 10 ) ] 14.4–5 2 0.33× =

v gp (d ) µp p= 2 18/( )

(32.2)[(0.4)/25,400)(12)] (144.14)/(18)(2= 11.41 10 )–5×

= ×3.15 10 ft/sec–5

v gp (d ) µp p= 2 /( )18

(32.2)[(40)/25,400)(12)] (144.14)(18)(1.2= 441 10 )–5×

0.315 ft/sec=



For dp = 400 µm (use transition regime equation):

Step 4. Calculate distance.For dp = 40 µm, distance = (time)(velocity):

For dp = 400 µm, distance = (time)(velocity):

For dp = 0.4 µm, without Cunningham correction factor, distance = (time)(velocity):

For dp = 0.4 µm with Cunningham correction factor, the velocity term must be corrected. For ourpurposes, assume particle diameter = 0.5 µm and temperature = 212°F to find the Cf value. Thus,Cf is approximately equal to 1.446.

Example 9.3

Problem:Determine the minimum distance downstream from a cement dust-emitting source that will be

free of cement deposit. The source is equipped with a cyclone (USEPA-84/09, p. 59).

Given:Particle size range of cement dust = 2.5 to 50.0 µmSpecific gravity of the cement dust = 1.96Wind speed = 3.0 mi/h

The cyclone is located 150 ft above ground level. Assume ambient conditions are at 60°F and 1 atm. Disregard meteorological aspects.

µ = air viscosity at 60°F = 1.22 × 10–5 lb/ft-sec (USEPA-84/09, p. 167)µm (1 µm = 10–6) = 3.048 × 105 ft (USEPA-84/09, p. 183)

v = 0.153g (d ) (p ) /(µ p0.71p

1.14p

0.71 0.43 0.29))

0.153(32.2) [(400)/(25,400)(12)] (0.71 1.14= 1144.14) /[(1.41 10 ) (0.0569)0.71 –5 0.43 0.29× ]]

= 8.90 ft/sec

Distance 30(0.315) 9.45 ft= =

Distance 30(8.90) 267 ft= =

Distance 30(3.15 10 ) 94.5 10 ft–5 –5= × = ×

The corrected velocity 3.15 10 (–5= = ×vCf 11.446) 4.55 10 ft/sec–5= ×

Distance 30(4.55 10 ) 1.365 10 ft–5 –3= × = ×



Solution:

Step 1. A particle diameter of 2.5 µm is used to calculate the minimum distance downstream free ofdust because the smallest particle will travel the greatest horizontal distance.

Step 2. Determine the value of K for the appropriate size of the dust. Calculate the particle density(pp) using the specific gravity given:

Calculate the air density (p). Use modified ideal gas equation, PV = nRuT = (m/M)RuT

Determine the flow regime (K):

For dp = 2.5 µm:

where 1 ft = 25,400(12) µm = 304,800 µm (USEPA-84/09, p. 183)

Step 3. Determine which fluid-particle dynamic law applies for the preceding value of K. Comparethe K value of 0.104 with the following range:

K < 3.3; Stokes’ law range3.3 ≤ K < 43.6; intermediate law range43.6 < K < 2360; Newton’s law range

The flow is in the Stokes’ law range; thus it is laminar.

Step 4. Calculate the terminal settling velocity in feet per second. For Stokes’ law range, the velocity is

pp = (specific gravity of fly ash)(density oof water)

= 1.96(62.4)

= 122.3 lb/ft3

P = (mass)(volume)

= PM / R Tu

= + =(1)(29)/[0.73(60 460)] 0.0764 lb/ft3

K d (gp p )p p.= a /µ2 0 33

K = [(2.5)/(25,400)(12)][32.2)(122.3)(0.07644)/(1.22 10 ) ] 0.104–5 2 0.33× =

v gp (d ) / µp p= 2 18( )

(32.2)[2.5/(25,400)(12)] (122.3)/(18)(1.2= 222 10 )–5×

1.21 10 ft/sec–3= ×



Step 5. Calculate the time for settling:

Step 6. Calculate the horizontal distance traveled:

9.4 PARTICULATE EMISSION CONTROL EQUIPMENT CALCULATIONS

Different classes of particulate control equipment include gravity settlers; cyclones; electrostaticprecipitators; wet (Venturi) scrubbers (discussed in Chapter 10); and baghouses (fabric filters). Inthe following section, we discuss calculations used for each of the major types of particulate controlequipment.

9.4.1 Gravity Settlers

Gravity settlers have long been used by industry for removing solid and liquid waste materialsfrom gaseous streams. Simply constructed (see Figure 9.2 and Figure 9.3), a gravity settler isactually nothing more than a large chamber in which the horizontal gas velocity is slowed, allowingparticles to settle out by gravity. Gravity settlers have the advantage of having low initial cost andare relatively inexpensive to operate because not much can go wrong. Although simple in design,

Figure 9.2 Gravitational settling chamber. (From USEPA, Control Techniques for Gases and Particulates, 1971.)

t (outlet height)/(terminal velocity)=

150/1.21 10–3= ×

1.24 10 sec 34.4 h5= × =

Distance (time for descent)(wind speed)=

= ×(1.24 10 )(3.0/3600)5

= 103.3 miles



gravity settlers require a large space for installation and have relatively low efficiency, especiallyfor removal of small particles (<50 µm).

9.4.2 Gravity Settling Chamber Theoretical Collection Efficiency

The theoretical collection efficiency of the gravity-settling chamber (USEPA-81/10, p. 5-4) is givenby the expression:

(9.10)

whereη = fractional efficiency of particle size dp (one size)vy = vertical settling velocityvx = horizontal gas velocityL = chamber lengthH = chamber height (greatest distance a particle must fall to be collected)

As mentioned, the settling velocity can be calculated from Stokes’ law. As a rule of thumb, Stokes’law applies when the particle size dp is less than 100 µm in size. The settling velocity is

(9.11)

wherevt = settling velocity in Stokes’ law range, meters per second (feet per second)g = acceleration due to gravity, 9.8 m/sec2 (32.1 ft/sec2)dp = particle diameter, micronspp = particle density, kilograms per cubic meter (pounds per cubic foot)pa = gas density, kilograms per cubic meter (pounds per cubic foot)

Figure 9.3 Baffled gravitational settling chamber. (From USEPA, Control Techniques for Gases and Particu-lates, 1971.)

η (v L)(v H)y x=

v [g(d ) (p p ) µ)t p p a= −2 18]/(



µ = gas viscosity, pascal-seconds (pound-feet-second)pa = N/m2

N = kilogram-meters per sec2

Equation 9.11 can be rearranged to determine the minimum particle size that can be collectedin the unit with 100% efficiency. The minimum particle size (dp)∗ in microns is given as:

(9.12)

Because the density of the particle pp is usually much greater than the density of gas pa, the quantitypp – pa reduces to pp. The velocity can be written as:

(9.13)

whereQ = volumetric flowB = chamber widthL = chamber length

Equation 9.12 is reduced to:

(9.14)

The efficiency equation can also be expressed as:

(9.15)

where Nc = number of parallel chambers: 1, for simple settling chamber and N trays +1, for aHoward settling chamber.

Equation 9.15 can be written as:

(9.16)

and the overall efficiency can be calculated using

(9.17)

whereηTOT = overall collection efficiencyηi = fractional efficiency of specific size particlewi = weight fraction of specific size particle

When flow is turbulent, Equation 9.18 is used.

(9.18)

(d )* {v ( µ) / [g(p p )]}p t p a.= −18 0 5

V Q / BL=

(d )* ( µQ) / [gp BL]}p p.= 18 0 5

η [(gp BLN ) / ( µQ)](d )p c p= 18 2

η [(gp BLN ) / ( µQ)](d )p c p= 0 5 18 2.

η ηTOT i iw= Σ

η [ (Lv / Hv )]y x= exp –



In using Equation 9.10 through Equation 9.16, note that Stokes’ law does not work for particlesgreater than 100 µm.

9.4.3 Minimum Particle Size

Most gravity settlers are precleaners that remove the relatively large particles (>60 µm) before thegas stream enters a more efficient particulate control device such as a cyclone, baghouse, electro-static precipitator (ESP), or scrubber.

Example 9.4

Problem:A hydrochloric acid mist in air at 25°C is collected in a gravity settler. Calculate the smallest

mist droplet (spherical in shape) collected by the settler. Stokes’ law applies; assume the acidconcentration is uniform through the inlet cross-section of the unit (USEPA-84/09, p. 61).

Given:Dimensions of gravity settler = 30 ft wide, 20 ft high, 50 ft longActual volumetric flow rate of acid gas in air = 50 ft3/secSpecific gravity of acid = 1.6Viscosity of air = 0.0185 cp = 1.243 × 10–5 lb/ft-secDensity of air = 0.076 lb/ft3

Solution:

Step 1. Calculate the density of the acid mist using the specific gravity given:

Step 2. Calculate the minimum particle diameter in feet and microns, assuming that Stokes’ law applies.For Stokes’ law range:

pp particle density (specific gravity o= = ff fly ash)(density of water)

1.6(62.4) 99.84 lb/ft3=

Minimum (18µQ/gp BL)p0.5dp =

Minimum [(18)(1.243 10 )(50)/(32.2)–5dp = × ((99.84)(30)(50)]0.5

= ×4.82 10 ft–5

= × ×(4.82 10 ft)(3.048 10 µm/ft)–5 5

= 14.7 µm



Example 9.5

Problem:A settling chamber that uses a traveling grate stoker is installed in a small heat plant. Determine

the overall collection efficiency of the settling chamber, given the operating conditions, chamberdimensions, and particle size distribution data (USEPA-84/09, p. 62).

Given:Chamber width = 10.8 ftChamber height = 2.46 ftChamber length = 15.0 ftVolumetric flow rate of contaminated air stream = 70.6 scfsFlue gas temperature = 446°FFlue gas pressure = 1 atmParticle concentration = 0.23 gr/scfParticle specific gravity = 2.65Standard conditions = 32°F, 1 atm

Particle size distribution data of the inlet dust from the traveling grate stoker are shown in Table9.1. Assume that the actual terminal settling velocity is one-half of the Stokes’ law velocity.

Solution:

Step 1. Plot the size efficiency curve for the settling chamber. The size efficiency curve is needed tocalculate the outlet concentration for each particle size (range). These outlet concentrations are thenused to calculate the overall collection efficiency of the settling chamber. The collection efficiencyfor a settling chamber can be expressed in terms of the terminal velocity, volumetric flow rate ofcontaminated stream, and chamber dimensions:

whereη = fractional collection efficiencyv = terminal settling velocityB = chamber widthL = chamber lengthQ = volumetric flow rate of the stream

Table 9.1 Particle Size Distribution Data

Particle size range, µm

Average particle diameter, µm

InletGrains/scf wt%

0–20 10 0.0062 2.720–30 25 0.0159 6.930–40 35 0.0216 9.440–50 45 0.0242 10.550–60 55 0.0242 10.560–70 65 0.0218 9.570–80 75 0.0161 780–94 85 0.0218 9.5

94 94 0.0782 34Total 0.23 100

η vB [gp (d ) µ ](B )p p= =L/Q /( L/Q2 18 )



Step 2. Express the collection efficiency in terms of the particle diameter dp. Replace the terminalsettling velocity in the preceding equation with Stokes’ law. Because the actual terminal settlingvelocity is assumed to be one half of the Stokes’ law velocity (according to the given problemstatement), the velocity equation becomes:

Determine the viscosity of the air in pounds per foot-second:

Determine the particle density in pounds per cubic foot:

Determine the actual flow rate in actual cubic feet per second. To calculate the collection effi-ciency of the system at the operating conditions, the standard volumetric flow rate of contami-nated air of 70.6 scfs is converted to actual volumetric flow of 130 acfs:

Express the collection efficiency in terms of dp, with dp in feet. Also express the collection effi-ciency in terms of dp, with dp in microns.

Use the following equation; substitute values for pp, g, B, L, µ, and Q in consistent units. Use theconversion factor for feet to microns. To convert dp from square feet to square microns, dp is di-vided by (304,800)2.

where dp is in microns.

Calculate the collection efficiency for each particle size. For a particle diameter of 10 µm:

v g(d ) p / µp p= 2 36

η [gp (d ) /36µ)(BL/Q)p p2=

Viscosity of air at 446 F 1.75 10 lb/–5° = × fft-sec

Pp 2.65(62.4) 165.4 lb/ft3= =

Q Q (T / T )a s a s=

= + +70.6(446 460)/(32 460)

= 130 acfs

η [gp (d ) /36µ)(BL/Q)p p2=

(32.2)(165.4)(10.8)(15)(d ) /[(36)(1.75p2= ×× 10 )(130)(304,800) ]–5 2

= ×1.134 10 (d )–4p

2

η (1.134 10 )(d ) (1.134 10 )(10–4p

2 –4= × = × )) 1.1 10 1.1%2 –2= × =



Table 9.2 provides the collection efficiency for each particle size. The size efficiency curve for thesettling chamber is shown in Figure 9.4; read off the collection efficiency of each particle size fromthis figure.

Calculate the overall collection efficiency (Table 9.3).

Figure 9.4 Size efficiency curve for settling chamber. (Adapted from USEPA-84/09, 136.)

Table 9.2 Collection Efficiency for Each Particle Size

dp, µm �, %

94 10090 9280 7360 4140 18.220 4.610 1.11

100

90

80

70

Effi

cien

cy, %

60

50

40

30

20

10

00 20 40 60 80 100 120

Particle size, microns

η ηwi i= Σ

= + + +(0.027)(1.1) (0.069)(7.1) (0.094)(14.0) (0..105)(23.0) (0.105)

(34.0) (0.095)(48.0) ((

+

+ + 00.070)(64.0) (0.095)(83.0) (0.340)(100.0)+ +

= 59.0%



9.4.4 Cyclones

Cyclones — the most common dust removal devices used within industry (Strauss, 1975) — removeparticles by causing the entire gas stream to flow in a spiral pattern inside a tube. They are thecollectors of choice for removing particles greater than 10 µm in diameter. By centrifugal force,the larger particles move outward and collide with the wall of the tube. The particles slide downthe wall and fall to the bottom of the cone, where they are removed. The cleaned gas flows out thetop of the cyclone (see Figure 9.5).

Cyclones have low construction costs and relatively small space requirements for installation;they are much more efficient in particulate removal than settling chambers. However, note that thecyclone’s overall particulate collection efficiency is low, especially on particles below 10 µm insize, and they do not handle sticky materials well. The most serious problems encountered withcyclones are with airflow equalization and with their tendency to plug (Spellman, 1999). They areoften installed as precleaners before more efficient devices such as electrostatic precipitators andbaghouses (USEPA-81/10, p. 6-1) are used. Cyclones have been used successfully at feed and grainmills; cement plants; fertilizer plants; petroleum refineries; and other applications involving largequantities of gas containing relatively large particles.

9.4.4.1 Factors Affecting Cyclone Performance

The factors that affect cyclone performance include centrifugal force, cut diameter, pressure drop,collection efficiency, and summary of performance characteristics. Of these parameters, the cutdiameter is the most convenient way of defining efficiency for a control device because it gives anidea of the effectiveness for a particle size range. As mentioned earlier, the cut diameter is definedas the size (diameter) of particles collected with 50% efficiency. Note that the cut diameter, [dp]cut,is a characteristic of the control device and should not be confused with the geometric mean particlediameter of the size distribution. A frequently used expression for cut diameter where collectionefficiency is a function of the ratio of particle diameter to cut diameter is known as the Lapple cutdiameter equation (method) (Copper and Alley, 1990):

(9.19)

where[dp]cut = cut diameter, feet (microns)µ = viscosity, pounds per foot-second (pascal-seconds) or (kilograms per meter-second)B = inlet width, feet (meters)

Table 9.3 Data for Calculation of Overall Collection Efficiency

dp, µm Weight fraction, wi �i

10 0.027 1.125 0.069 7.135 0.094 1445 0.105 2355 0.105 3465 0.095 4875 0.07 6485 0.095 8394 0.34 100

Total 1

[ ] {9 B/[2 ( )]}cutd n p pp t p g= −µ π



Figure 9.5 Convection reverse-flow cyclone. (From USEPA, Control Techniques for Gases and Particulates, 1971.)

Zone of inletinterference

Innervortex

Outervortex

Gas outlet

Body

Innercylinder(tubularguard)

Gasinlet

Top view

Side view

Outervortex

Innervortex

Dust outlet

Core



nt = effective number of turns (5 to 10 for common cyclones)vi = inlet gas velocity, feet per second (meters per second)pp = particle density, pounds per cubic foot (kilograms per cubic meter)pg = gas density, pounds per cubic foot (kilograms per cubic meter)

Example 9.6

Problem:Determine the cut size diameter and overall collection efficiency of a cyclone, given the particle

size distribution of a dust from a cement kiln (USEPA-84/09, p. 66).

Given:Gas viscosity µ = 0.02 centipoises (cP) = 0.02(6.72 × 10–4) lb/ft-secSpecific gravity of the particle = 2.9Inlet gas velocity to cyclone = 50 ft/secEffective number of turns within cyclone = 5Cyclone diameter = 10 ftCyclone inlet width = 2.5 ftParticle size distribution data are shown in Table 9.4.

Solution:

Step 1. Calculate the cut diameter [dp]cut, which is the particle collected at 50% efficiency. For cyclones:

whereµ = gas viscosity, pounds per foot-secondBc = cyclone inlet width, feetnt = number of turnsvi = inlet gas velocity, feet per secondpp = particle density, pounds per cubic footp = gas density, pounds per cubic foot

Determine the value of pp – p. Because the particle density is much greater than the gas density,pp – p can be assumed to be pp:

Table 9.4 Particle Size Distribution Data

Average particle size in range dp, µm % wt

1 35 20

10 1520 2030 1640 1050 660 3

>60 7

[ ] /[2 ( )]}0.5cut cd B n p pp ti p g= −{9µ π



Calculate the cut diameter:

Step 2. Complete the size efficiency table (Table 9.5) using Lapple’s method (Lapple, 1951). Asmentioned, this method provides the collection efficiency as a function of the ratio of particlediameter to cut diameter. Use the equation

Step 3. Determine overall collection efficiency:

Example 9.7

Problem:An air pollution control officer has been asked to evaluate a permit application to operate a

cyclone as the only device on the ABC Stoneworks plant’s gravel drier (USEPA-84/09, p. 68).

Given (design and operating data from permit application):Average particle diameter = 7.5 µmTotal inlet loading to cyclone = 0.5 gr/ft3 (grains per cubic foot)Cyclone diameter = 2.0 ftInlet velocity = 50 ft/secSpecific gravity of the particle = 2.75Number of turns = 4.5 turns

Table 9.5 Size Efficiency Table

dp, µm wi dp/[dp]cut �i,% wi�i%

1 0.03 0.1 0 05 0.2 0.5 20 4

10 0.15 1 50 7.520 0.2 2 80 1630 0.16 3 90 14.440 0.1 4 93 9.350 0.06 5 95 5.760 0.03 6 98 2.94

>60 0.07 — 100 7

p p pp p− = = =2.9(62.4) 180.96 lb/ft3

[d ]cp ut = ×[(9)(0.02)(6.72 10 )(2.5)/(2 )–4 π 55(50)(180.96)]0.5

= ×3.26 10 ft–5

= 9.94 µm

η 1 (1.0)/[1.0 ( ) ]2= + d / [d ]p p cut

Σw ni i(%) 0 4 7.5 16 14.4 9.3= + + + + + + 55.7 2.94 7+ +

= 66.84%



Operating temperature = 70°FViscosity of air at operating temperature = 1.21 × 10–5lb/ft-secThe cyclone is a conventional one.

Air pollution control agency criteria:Maximum total outlet loading = 0.1 gr/ft3

Cyclone efficiency as a function of particle size ratio is provided in Figure 9.6 (Lapple’s curve).

Solution:

Step 1. Determine the collection efficiency of the cyclone. Use Lapple’s method, which providescollection efficiency and values from a graph relating efficiency to the ratio of average particlediameter to the cut diameter. Again, the cut diameter is the particle diameter collected at 50%efficiency (see Figure 9.6). Calculate the cut diameter using the Lapple method (Equation 9.19):

Determine the inlet width of the cyclone, Bc. The permit application has established this cycloneas conventional. The inlet width of a conventional cyclone is one fourth of the cyclone diameter:

Determine the value of pp – p. Because the particle density is much greater than the gas density,pp – p can be assumed to be pp:

Calculate the cut diameter:

Figure 9.6 Lapple’s curve: cyclone efficiency vs. particle size ratio. (Adapted from USEPA-84/09, p 70.)

100

9080

70

60

50

40

Col

lect

ion

effic

ienc

y, %

30

20

10(0.1) 2 3 4 5 6 7 8 9 (1.0) 2 3 4 5 6 7 8 9 (10)

Particle size ratio, dp/(dp) cut

[ ] /[2 ( )]}0.5cut cd B n p pp tvi p g= −{9µ π

Bc = = =cyclone diameter/4 2.0/4 0.5 ft

p p pp p− = = =2.75(62.4) 171.6 lb/ft3



Calculate the ratio of average particle diameter to the cut diameter:

Determine the collection efficiency utilizing Lapple’s curve (see Figure 9.6):

Step 2. Calculate the required collection efficiency for the approval of the permit:

Step 3. Should the permit be approved? Because the collection efficiency of the cyclone is lower thanthe collection efficiency required by the agency, the permit should not be approved.

9.4.6 Electrostatic Precipitator (ESP)

Electrostatic precipitators (ESPs) have been used as effective particulate-control devices for manyyears. Usually used to remove small particles from moving gas streams at high collection efficien-cies, ESPs are extensively used when dust emissions are less than 10 to 20 µm in size, with apredominant portion in the submicron range (USEPA-81/10, p. 7-1). Widely used in power plantsfor removing fly ash from the gases prior to discharge, electrostatic precipitators apply electricalforce to separate particles from the gas stream. A high voltage drop is established between elec-trodes, and particles passing through the resulting electrical field acquire a charge. The chargedparticles are attracted to and collected on an oppositely charged plate, and the cleaned gas flowsthrough the device. Periodically, the plates are cleaned by rapping to shake off the layer of dustthat accumulates; the dust is collected in hoppers at the bottom of the device (see Figure 9.7). ESPshave the advantages of low operating costs; capability for operation in high-temperature applications(to 1300°F); low-pressure drop; and extremely high particulate (coarse and fine) collection effi-ciencies; however, they have the disadvantages of high capital costs and space requirements.

9.4.6.1 Collection Efficiency

According to USEPA-81/10, p. 7-9, ESP collection efficiency can be expressed by the followingtwo equations:

• Migration velocity equation• Deutsch–Anderson equation

[ ] { /[2 ( )]}cut cd B n p pp tvi p g= −9µ π

= ×[(9)(1.21 10 )(0.5)/(2 )4.5(50)(171.6)–5 π ]]0.5

= 4.57 microns

d /[d ] 7.5/4.57 1.64p p cut = =

η 72%=

µ [(inlet loading – outlet loading)/(inle= tt loading)](100)

= [(0.5 – 0.1)/(0.5)](100)

= 80%



Particle migration velocity. The migration velocity w (sometimes referred to as the drift velocity)represents the parameter at which a group of dust particles in a specific process can be collectedin a precipitator; it is usually based on empirical data. The migration velocity can be expressed interms of:

(9.20)

wherew = migration velocitydp = diameter of the particle, micronsEo = strength of field in which particles are charged, volts per meter (represented by peak voltage)Ep = strength of field in which particles are collected, volts per meter (normally, the field close to

the collecting plates)µ = viscosity of gas, pascal-seconds

Figure 9.7 Electrostatic precipitators. (From USEPA, Control Techniques for Gases and Particulates, 1971.)

to plates

Cleangas out

Dirtygas in

Dirtygas in

Dischargeelectrode (wire)

Cleangas out

Collectionplate

Highvoltage to wires

(a)

Weights

Cleangas out

Spray

Cleangas out

Dirtygas in

Dirtygas in

Dischargeelectrode

(wire)

(b)

Weights

w d E E µ)p o= p /(4π



Migration velocity is quite sensitive to the voltage, as the presence of the electric field twicein Equation 9.20 demonstrates. Therefore, the precipitator must be designed using the maximumelectric field for maximum collection efficiency. The migration velocity is also dependent on particlesize; larger particles are collected more easily than smaller ones.

Particle migration velocity can also be determined by the following equation:

(9.21)

wherew = migration velocityq = particle charge (charges)Ep = strength of field in which particles are collected, volts per meter (normally, the field close to

the collecting plates)µ = viscosity of gas, pascal-secondsr = radius of the particle, microns

Deutsch–Anderson equation. This equation or derivatives thereof are used extensively through-out the precipitator industry. Specifically, it has been used to determine the collection efficiency ofthe precipitator under ideal conditions. Though scientifically valid, a number of operating param-eters can cause the results to be in error by a factor of two of more. Therefore, this equation shouldbe used only for making preliminary estimates of precipitation collection efficiency. The simplestform of the equation is:

(9.22)

whereη = fractional collection efficiencyA = collection surface area of the platesQ = volumetric flow ratew = drift velocity

9.4.6.2 Precipitator Example Calculations

Example 9.8

Problem:A horizontal parallel-plate electrostatic precipitator consists of a single duct, 24 ft high and 20

ft deep, with an 11-in. plate-to-plate spacing. Given collection efficiency at a gas flow rate of 4200actual cubic feet per minute (acfm), determine the bulk velocity of the gas, outlet loading, and driftvelocity of this electrostatic precipitator. Also calculate revised collection efficiency if the flow rateand the plate spacing are changed (USEPA-84/09, p. 71).

Given:Inlet loading = 2.82 gr/ft3 (grains per cubic foot)Collection efficiency at 4200 acfm = 88.2%Increased (new) flow rate = 5400 acfmNew plate spacing = 9 in.

w qE / ( µr)p= 4π

η = −1 exp(–wA/Q)



Solution:

Step 1. Calculate the bulk flow (throughout) velocity v. The equation for calculating throughput velocityis

whereQ = volumetric flow rateS = cross-sectional area through which the gas passes

Step 2. Calculate outlet loading. Remember that

Therefore,

Step 3. Calculate the drift velocity, which is the velocity at which the particle migrates toward thecollection electrode with the electrostatic precipitator. Recall that Equation 9.22, the Deut-sch–Anderson equation, describes the collection efficiency of an electrostatic precipitator:

whereη = fractional collection efficiencyA = collection surface area of the platesQ = volumetric flow ratew = drift velocity

Calculate the collection surface area A. Remember that the particles will be collected on bothsides of the plate.

V Q / S=

V Q / S=

(4200)/[(11/12)(24)]=

191 ft/min=

3.2 ft/sec=

η (fractional) (inlet loading – outlet= = lloading)/(inlet loading)

outlet loading inlet loading)(1 )= −( η

(2.82)(1 – 0.882)=

0.333 gr/ft3=

η = 1 exp(–wA/Q)

A (2)(24)(20) 960 ft2= =



Calculate the drift velocity w. Because the collection efficiency, gas flow rate, and collection sur-face area are now known, the drift velocity can easily be found from the Deutsch–Andersonequation:

Solving for w:

Step 4. Calculate the revised collection efficiency when the gas volumetric flow rate is increased to5400 cfm. Assume the drift velocity remains the same:

Step 5. Does the collection efficiency change with changed plate spacing? No. Note that the Deut-sch–Anderson equation does not contain a plate-spacing term.

Example 9.9

Problem:Calculate the collection efficiency of an electrostatic precipitator containing three ducts with

plates of a given size, assuming a uniform distribution of particles. Also, determine the collectionefficiency if one duct is fed 50% of the gas and the other passages are fed 25% each (USEPA-84/09, p. 73).

Given:Volumetric flow rate of contaminated gas = 4000 acfmOperating temperature and pressure = 200°C and 1 atmDrift velocity = 0.40 ft/secSize of the plate = 12 ft long and 12 ft highPlate-to-plate spacing = 8 in.

Solution:

Step 1. What is the collection efficiency of the electrostatic precipitator with a uniform volumetricflow rate to each duct? Use the Deutsch–Anderson equation to determine the collection efficiencyof the electrostatic precipitator.

η = 1 – exp(–wA/Q)

0.882 1 – exp[–(960)(w)/(4200)]=

w 9.36 ft/min=

η = 1 – exp(–wA/Q)

= 1 – exp(–(960)(9.36)/(5400)]

= 0.812

= 81.2%

η 1 exp(–wA/Q)= −



Calculate the collection efficiency of the electrostatic precipitator using the Deutsch–Andersonequation. The volumetric flow rate (Q) through a passage is one third of the total volumetric flowrate:

Step 2. What is the collection efficiency of the electrostatic precipitator, if one duct is fed 50% of gasand the others 25% each? The collection surface area per duct remains the same. What is thecollection efficiency of the duct with 50% of gas, η1? Calculate the volumetric flow rate of gasthrough the duct in actual cubic feet per second:

Calculate the collection efficiency of the duct with 50% of gas:

What is the collection efficiency (η2) of the duct with 25% of gas flow in each? Calculate the volumetricflow rate of gas through the duct in actual cubic feet per second:

Calculate the collection efficiency (η2) of the duct with 25% of gas:

Calculate the new overall collection efficiency. The equation becomes:

Q (4000)/(3)(60)=

= 22.22 acfs

η 1 – exp(–wA/Q)=

= 1 – exp[–(288)(0.4)/(22.22)]

= 0.9944

= 99.44%

Q (4000)/(2)(60) 33.33 acfs= =

η1 1 exp[–(288)(0.4)/(33.33)]= −

= 0.9684

= 96.84%

Q (4000)/(4)(60) 16.67 acfs= =

η2 1 exp[–(288)(0.4)/(16.67)]= −

= 0.9990

= 99.90%



Example 9.10

Problem:A vendor has compiled fractional efficiency curves describing the performance of a specific

model of an electrostatic precipitator. Although these curves are not available, the cut diameter isknown. The vendor claims that this particular model will perform with a given efficiency underparticular operating conditions. Verify this claim and make certain the effluent loading does notexceed the standard set by USEPA (USEPA-84/09, p. 75).

Given:Plate-to-plate spacing = 10 in.Cut diameter = 0.9 µm Collection efficiency claimed by the vendor = 98%Inlet loading = 14 gr/ft3

USEPA standard for the outlet loading = 0.2 gr/ft3 (maximum)

The particle size distribution is given in Table 9.6.A Deutsch–Anderson type of equation describing the collection efficiency of an electrostatic

precipitator is:

whereη = fractional collection efficiencyK = empirical constantdp = particle diameter

Solution:

Step 1. Is the overall efficiency of the electrostatic precipitator equal to or greater than 98%? Becausethe weight fractions are given, collection efficiencies of each particle size are needed to calculatethe overall collection efficiency.

Determine the value of K by using the given cut diameter. Because the cut diameter is known,we can solve the Deutsch–Anderson type equation directly for K.

Table 9.6 Particle Size Distribution

Weight rangeAverage particle

size dp, µm

0–20% 3.520–40% 840–60% 1360–80% 1980–100% 45

η η ηt 1 2(0.5)( ) (2)(0.25)( )= +

= +(0.5)(96.84) (2)(0.25)(99.90)

= 98.37%

η 1 exp(– )= − Kdp



Solving for K,

Calculate the collection efficiency using the Deutsch–Anderson equation where dp = 3.5:

Table 9.7 shows the collection efficiency for each particle size.Calculate the overall collection efficiency.

whereη = overall collection efficiencywi = weight fraction of the ith particle sizeηi = collection efficiency of the ith particle size

Is the overall collection efficiency greater than 98%? Yes

Step 2. Does the outlet loading meet USEPA’s standard? Calculate the outlet loading in grains percubic foot:

Table 9.7 Collection Efficiency for Each Particle Size

Weight fraction wi

Average particle size dp, µm �i

0.2 3.5 0.93250.2 8 0.99790.2 13 0.99990.2 19 0.99990.2 45 0.9999

η 1 – exp(–kd )p=

0.5 1 – exp[–K(0.9)]=

K 0.77=

η 1 – exp[(–0.77)(3.5)]=

= 0.9325

η ηwi i= Σ

= + +(0.2)(0.9325) (0.2)(0.9979) (0.2)(0.99999) (0.2)(0.9999) (0.2)(0.9999)+ +

= 0.9861

= 98.61%

Outlet loading (1.0 )(inlet loading)= − η



where η is the fractional efficiency for the preceding equation.

Is the outlet loading less than 0.2 gr/ft3? Yes.

Step 3. Is the vendor’s claim verified? Yes.

9.4.7 Baghouse (Fabric) Filters

“The term baghouse encompasses an entire family of collectors with several types of filter bagshapes, cleaning mechanisms, and body shapes” (Heumann, 1997). Baghouse filters (fabric filtersor, more properly, tube filters) are the most commonly used air pollution control filtration systems.In much the same manner as the common vacuum cleaner, fabric filter material capable of removingmost particles as small as 0.5 µm and substantial quantities of particles as small as 0.1 µm isformed into cylindrical or envelope bags and suspended in the baghouse (see Figure 9.8). Theparticulate-laden gas stream is forced through the fabric filter, and as the air passes through thefabric, particulates accumulate on the cloth, providing a cleaned airstream. As particulates buildup on the inside surfaces of the bags, the pressure drop increases. Before the pressure drop becomes

Figure 9.8 Typical simple fabric filter baghouse design. (From USEPA, Control Techniques for Gases andParticulates, 1971.)

Outlet loading [(1.0 0.9861)(14)]= −

= 0.195 gr/ft3

Mechanism for shaking,rapping, or vibrating bags

Cleanair out

Cleanairout

Tubularfilterbags

Dirtyair in

Dirtyair in

Collecteddust out



too severe, the bags must be relieved of some of the particulate layer. The particulates are period-ically removed from the cloth by shaking or by reversing the airflow.

Fabric filters are relatively simple to operate, provide high overall collection efficiencies (up to99%+), and are very effective in controlling submicrometer particles, but they do have limitations.These include relatively high capital costs; high maintenance requirements (bag replacement, etc.);high space requirements; and flammability hazards for some dusts.

9.4.7.1 Air-to-Filter (Media) Ratio

The air-to-filter (cloth) ratio is a measurement of the velocity (filtration velocity) of the air passingthrough the filter media. The ratio definition is the volume of air (expressed in cubic feet per minuteor cubic meters per hour) divided by the filter media area (expressed in square feet or squaremeters). Generally, the smaller a particle diameter is, the more difficult it is to filter, thereby requiringa lower A/C value. The formula used to express air-to-filter (cloth) ratio (A/C ratio), filtrationvelocity, or face velocity (terms can be used interchangeably) is:

(9.23)

wherevf = filtration velocity, feet per minute (centimeters per second)Q = volumetric air flow rate, cubic feet per minute (cubic centimeters per second)Ac = area of cloth filter, square feet (square centimeters)

9.4.7.2 Baghouse Example Calculations

Example 9.11

Problem:The manufacturer does baghouse sizing. A simple check or estimate of the amount of baghouse

cloth needed for a given process flow rate can be computed by using the A/C ratio equation (Equation9.23) (USEPA-81/10, p. 8-34):

For example, if the process gas exhaust rate is given as 4.72 × 106cm3/sec (10,000 ft3/min) and thefiltration velocity is 4 cm/sec (A/C is 4:1 (cm3/sec)/cm2), the cloth area would be

Solution:To determine the number of bags required in the baghouse, use the formula:

where

v Q / Af c=

v Q / A or A Q / vf c c f= =

Ac 4.72 10 /46= ×

= 118 m (cloth required)2

A dhb = π



Ab = area of bag, meters (feet)d = bag diameter, meters (feet)h = bag height, meters (feet)

If the bag diameter is 0.203 m (8 in.) and the bag height is 3.66 m (12 ft), the area of each bag is

The calculated number of bags in the baghouse is:

Example 9.12

Problem:A proposal to install a pulse jet fabric filter system for cleaning an air stream containing

particulate matter must be evaluated. Select the most appropriate filter bag, considering performanceand cost (USEPA-84/09, p. 84).

Given:Volumetric flow rate of polluted air stream = 10,000 scfm (60°F, 1 atm)Operating temperature = 250°FConcentration of pollutants = 4 gr/ft3

Average air-to-cloth ratio (A/C ratio) = 2.5 cfm/ft2 clothCollection efficiency requirement = 99%

Table 9.8 lists information given by filter bag manufacturers. Assume no bag has an advantagefrom the standpoint of durability under the operating conditions for which the bag is to be designed.

Solution:

Step 1. Eliminate from consideration bags that, on the basis of given characteristics, are unsatisfactory.Considering the operating temperature and the bag tensile strength required for a pulse jet system:• Bag D is eliminated because its recommended maximum temperature (220˚F) is below the

operating temperature of 250°F.• Bag C is also eliminated because a pulse jet fabric filter system requires the tensile strength

of the bag to be at least above average.

Table 9.8 Filter Bag Properties

Property Filter bag A Filter bag B Filter bag C Filter bag D

Tensile strength Excellent Above average Fair ExcellentRecommended maximum temp, °F 260 275 260 220Resistance factor 0.9 1 0.5 0.9Cost per bag, $ 26.00 38.00 10.00 20.00Standard size 8 in. × 16 ft 10 in. × 16 ft 1 ft × 16 ft 1 ft × 20 ft

Ab = 3.14(0.203)(3.66)

= 2.33 m2

Number of bags 118/2.33=

= 51 bags



Step 2. Determine comparative costs of the remaining bags. Total cost for each bag type is the numberof bags times the cost per bag. No single individual bag type is more durable than the other.

Establish the cost per bag. From the information given in Table 9.8, the cost per bag is $26.00for Bag A and $38.00 for Bag B.

Determine number of bags, N, for each type. The number of bags required, N, is the total filteringarea required, divided by the filtering area per bag.

Calculate the total filter area At. Calculate given flow rate to actual cubic feet per minute, Qa.

Establish filtering capacity vf. This is given. The A/C ratio, expressed in cubic feet per minuteper square foot, is the same as the filtering velocity, which is given previously as 2.5 cfm/ft2

cloth. From the information given in Table 9.8, the filtering velocity is:

Calculate the total filtering cloth area, Ac, from the actual cubic feet per minute and filtering ve-locity determined before:

Calculate the filtering area per bag. Bags are assumed to be cylindrical; the bag area is A = πDh,where D = bag diameter and h = bag length:

Determine the number of bags required, N. N = (filtering cloth area of each bag Ac)/(bag area A):

Determine the total cost for each bag:

Step 3. Select the most appropriate filter bag, considering the performance and cost. Because the totalcost for bag A is less than for bag B, select bag A.

Q (10,000)(250 460)/(60 460) 13,65a = + + = 44 acfm

v 2.5 cfm/ftf2=

= 2.5 ft/min

A Q /v 13,654/2.5 5461.6 ftc a f2= = =

For bag A: (8/12)(16) 33.5 ft2A Dh= = =π π

For bag B: (10/12)(16) 41.9 ft2A Dh= = =π π

For bag A: 5461.6/33.5 163N A / Ac= = =

For bag B: 5461.6/41.9 130N = =

For bag A: total cost (N)(cost per bag)= = (163)(26.00) $4238=

For bag B: total cost (130)(38.00) $494= = 00



Example 9.13

Problem:Determine the number of filtering bags required and cleaning frequency for a plant equipped

with a fabric filter system. Operating and design data are given below (USEPA-84/09, p. 86).

Given:Volumetric flow rate of the gas stream = 50,000 acfmDust concentration = 5.0 gr/ft3

Efficiency of the fabric filter system = 98.0%Filtration velocity = 10 ft/minDiameter of filtering bag = 1.0 ftLength of filtering bag = 15 ft

The system is designed to begin cleaning when the pressure drop reaches 8.9 in. of water. Thepressure drop is given by:

where∆p = pressure drop, inches H2Ovf = filtration velocity, feet per minutec = dust concntration, pounds per cubic foott = time since the bags were cleaned, minutes

Solution:

Step 1. What is the number of bags N needed? To calculate N, we need the total required surface areaof the bags and the surface area of each bag.

Calculate the total required surface area of the bags Ac in square feet

whereAc = total surface area of the bagsq = volumetric flow ratevf = filtering velocity

Calculate the surface area of each bag A, in square feet:

∆p . v c(v ) tf f= +0 2 5 2

A q/vc f=

A q/vc f=

= 50,000/10

= 5000 ft2

A = Dhπ



whereA = surface area of a bagD = diameter of the bagh = length of the bag

Calculate the required number of bags N:

Step 2. Calculate the required cleaning frequency:

Because ∆p is given as 8.0 in. H2O, the time since the bags were cleaned is calculated by solving thepreceding equation:

Solving for t,

Example 9.14

Problem:An installed baghouse is presently treating a contaminated gas stream. Suddenly, some of the

bags break. We must now estimate this baghouse system’s new outlet loading (USEPA-84/09, p. 88).

Given:Operation conditions of the system = 60°F, 1 atmInlet loading = 4.0 gr/acfOutlet loading before bag failure = 0.02 gr/acfVolumetric flow rate of contaminated gas = 50,000 acfm

A = πDh

= (1.0)(15)π

= 47.12 ft2

N A / Ac=

5000/47.12=

= 106

∆p . v c(vf) tf= +0 2 5 2

5.0 gr/ft 0.0007143 lb/ft and p 0.2v3 3= =∆ ff t25c(v ) t+

8.0 (0.2)(10) (5)(0.0007143)(10) t2= +

t 16.8 min=



Number of compartments = 6Number of bags per compartment = 100Bag diameter = 6 in.Pressure drop across the system = 6 in. H2O Number of broken bags = 2 bags

Assume that all the contaminated gas emitted through the broken bags is the same as that passingthrough the tube sheet thimble.

Solution:

Step 1. Calculate the collection efficiency and penetration before the bag failures. Collection efficiencyis a measure of a control device’s degree of performance; it specifically refers to degree of removalof pollutants. Loading refers to the concentration of pollutants, usually in grains of pollutants percubic foot of contaminated gas streams. Mathematically, the collection efficiency is defined as:

From the preceding equation, the collected amount of pollutants by a control unit is the product ofcollection efficiency η and inlet loading. The inlet loading minus the amount collected gives theamount discharged to the atmosphere.

Another term used to describe the performance or collection efficiency of control devices is penetrationPt. USEPA-84/09, p. 3.3 gives it as:

(fractional basis)

(percent basis)

The following equation describes the effect of bag failure on baghouse efficiency:

wherePt1 = penetration after bag failurePt2 = penetration before bag failurePtc = penetration correction term, contribution of broken bags to Pt1

∆p = pressure drop, inches H2Oφ = dimensionless parameterQ = volumetric flow rate of contaminated gas, actual cubic feet per minuteL = number of broken bagsD = bag diameter, inchesT = temperature, degrees Fahrenheit

Collection efficiency η is:

η [(inlet loading outlet loading)/(inlet= − loading)](100)

Pt = −1 100η /

Pt = −100 η

P P Pt1 t2 tc= +

Ptc 0.582( p) /0.5= ∆ φ

φ Q/(LD (Y 460) )2 0.5= +



Penetration is:

Step 2. Calculate the bag failure parameter φ, a dimensionless number:

Step 3. Calculate the penetration correction Ptc; this determines penetration from bag failure:

Step 4. Calculate the penetration and efficiency after the two bags failed. Use the results of steps 1and 3 to calculate Pt1:

η (inlet loading outlet loading)/(inlet= − loading)

= (4.0 – 0.02)/(4.0)

= 0.005

= 99.5%

P 1.0t = − η

0.005=

φ Q/(LD (T 460) )2 0.5= +

= +50,000/(2)(6) (60 460)2 0.5

= 30.45

P 0.582( p) /tc0.5= ∆ φ

= (0.582)(6) /30.450.5

= 0.0468

P P Pt1 t2 tc= +

= +0.005 0.0468

= 0.0518

η* 1 – 0.0518=

= 0.948



Step 5. Calculate the new outlet loading after the bag failures. Relate inlet loading and new outletloading to the revised efficiency or penetration:

Example 9.15

Problem:A plant emits 50,000 acfm of gas containing a dust loading of 2.0 gr/ft3. A particulate control

device is employed for particle capture; the dust captured from the unit is worth $0.01/lb. Determinethe collection efficiency for which the cost of power equals the value of the recovered material.Also determine the pressure drop in inches of H2O at this condition (USEPA-84/09, p. 122).

Given:Overall fan efficiency = 55%Electric power cost = $0.06/kWh

For this control device, assume that the collection efficiency is related to the system pressuredrop ∆p through the equation:

where∆p = pressure drop, pounds per square footη = fractional collection efficiency

Solution:

Step 1. Express the value of the dust collected in terms of collection efficiency η:

Note that the collected dust contains 7000 grains per pound.

Step 2. Express the value of the dust collected in terms of pressure drop ∆p. Recall that η = ∆p/(∆p + 5.0).

New outlet loading (inlet loading)Pt1=

= (4.0)(0.0518)

= 0.207 gr/acf

η p/( p 5.0)= +∆ ∆

Amount of dust collected (Q)(inlet loadin= gg)( )η

The value of dust collected 50,000(ft /mi3= nn)2(gr/ft )(1/7000)(lb/gr) 0.01($/lb)3 × η

= 0.143 $/minη

The value of dust collected 0.143[ p/( p= ∆ ∆ ++ 5.0)]$/min



Step 3. Express the cost of power in terms of pressure drop ∆p:

whereη′ = fan efficiency∆p = pressure drop, poundsf per square footQ = volumetric flow rate

Step 4. Set the cost of power equal to the value of dust collected and solve for ∆p in poundsf persquare foot. This represents breakeven operation. Then, convert this pressure drop to inches of H2O.To convert from poundsf per square foot to inches of H2O, divide by 5.2.

Solving for ∆p:

Step 5. Calculate the collection efficiency using the calculated value of ∆p.

Example 9.16

Problem:Determine capital, operating, and maintenance costs on an annualized basis for a textile dye

and finishing plant (with two coal-fired stoker boilers), where a baghouse is employed for particulatecontrol. Use the given operating, design, and economic factors (USEPA-84/09, p. 123).

Given:Exhaust volumetric flow from two boilers = 70,000 acfmOverall fan efficiency = 60%Operating time = 6240 h/yrSurface area of each bag = 12.0 ft2

Bag type = Teflon® feltAir-to-cloth ratio = 5.81 acfm/ft2

Total pressure drop across the system = 17.16 lbf/ft2

Cost of each bag = $75.00Installed capital costs = $2.536/acfm

Bhp = ′ =Q p/ brake horsepower∆ η

Cost of power p(lb /ft )(50,000)[(ft /mif2 3= ∆ nn)(1/44,200)(kW-min/ft-lb )(1/0.55)f

×(0.06)($/kWh)(1/60)(h/min)]

0.002 p $/min= ∆

(0.143) p/( p 5) 0.002 p∆ ∆ ∆+ =

∆p 66.5 lb /ft 12.8 in. H Of2

2= =

η 66.5/(66.5 5) 0.93= + =

= 93.0%



Cost of electrical energy = $0.03/kWhYearly maintenance cost = $5000 plus yearly cost to replace 25% of the bagsSalvage value = 0Interest rate (i) = 8%Lifetime of baghouse (m) = 15 yrAnnual installed capital cost (AICC) = (installed capital cost) {i(1 + i)m/[(1 + i)m – 1]}

Solution:

Step 1. What is the annual maintenance cost? Calculate the number of bags N:

whereQ = total exhaust volumetric flow rateA = surface area of a bag

Calculate the annual maintenance cost in dollars per year.

Step 2. What is the annualized installed cost (AICC)? Calculate the installed capital cost in dollars:

Calculate the AICC using the equation given previously:

N Q/(air-to-cloth ratio)(A)=

N Q/(air-to-cloth ratio)(A)=

= (70,000)/(5.81)(12)

= 1004 bags

Annual maintenance cost $5000/year cost= + of replacing 25% of the bags each year

= +$5000 (0.25)(1004)(75.00)

= $23,825/year

Installed capital cost (Q)($2.536/acfm)=

(70,000)(2.536)=

= $177,520

AICC (installed capital cost){i(1 i) /[m= + ((1 i) 1]}m+ −

= + +(177,520){0.08(1 0.08) /[(1 0.08)15 15 −− 1]}

= $20,740/yr



Step 3. Calculate the operating cost in dollars per year:

Because 1 ft-lb/sec = 0.0013558 kW,

Step 4. Calculate the total annualized cost in dollars per year:

REFERENCES

Baron, P.A. and Willeke, K. (1993). Gas and particle motion, in K. Willeke and P.A. Baron (Eds.), AerosolMeasurement: Principles, Techniques and Applications. New York: Van Nostrand Reinhold.

Cooper, C.D. and Alley, F.C. (1990). Air Pollution Control. Philadelphia: Waveland Press.Hesketh, H.E. (1991). Air Pollution Control: Traditional and Hazardous Pollutants. Lancaster, PA: Technomic

Publishing Company, Inc.Heumann, W.L. (1997). Industrial Air Pollution Control Systems. New York: McGraw-Hill.Hinds, W.C. (1986). Aerosol Technology: Properties, Behavior, and Measurement of Airborne Particulates.

New York: John Wiley & Sons.Lapple, C.E. (1951). Fluid and Particle Mechanics. Newark, DE: University of Delaware.Spellman, F.R. (1999). The Science of Air: Concepts & Applications. Lancaster, PA: Technomic Publishing

Company, Inc.Strauss, W. (1975). Industrial Gas Cleaning, 2nd ed. Oxford: Pergamon Press.USEPA (1969). Control Techniques for Particulate Air Pollutants. USEPA (1971). Control Techniques for Gases and Particulates.USEPA-81/10 (1984). Control of particulate emissions, course 413, USEPA Air Pollution Training Institute,

EPA450-2-80-066.USEPA-84/09 (1984). Control of gaseous and particulate emissions, Course SI:412D, USEPA Air Pollution

Training Institute (APTI), EPA450-2-84-007.USEPA-84/03 (1984). Wet scrubber plan review, Course SI:412C, USEPA Air Pollution Training Institute

(APTI), EPA-450-2-82-020.

Operating cost Q p(operating time)(0.03/k= ∆ WWh/E)

Operating cost (70,000/60)(17.16)(6240)(0= ..03)(0.0012558)/0.6

$8470/yr=

Total annualized cost (maintenance cost)= ++ +AICC (operating cost)

= + +23,825 20,740 8470

= $53,035/yr

Documents

L1681_CH09