Electrostatic Precipitators ESP

ESP.l DESIGN MODELS

An electrostatic precipitator (ESP) is being used to clean fly ash from a gas. The precipitator contains 30 ducts, with plates 12 ft high and 12 ft long. The spacing between the plates is 8 in. The gas is evenly distributed through all of the ducts. The following information can be used:

Gas volumetric flowrate = 40,000 acfm Particle drift velocity = 0.40 ft/s

Use the Deutsch-Anderson equation to calculate the efficiency of the precipitator. Also, use the modified Deutsch equation with a range of exponents varying from 0.4 to 0.7 (in increments of 0.05) to calculate the efficiency of the electrostatic precipitator.

Solution

The process of electrostatic precipitation consists of corona formation around a hightension wire, with particle charging by ionized gas molecules formed in the localized region of electrical breakdown surrounding the high-tension wire. This is followed by migration of the charged particles to the collecting electrodes. Finally, the particles collected from the collecting electrode are removed by rapping into hoppers. The approach taken by industry to size ESPs for various applications makes use of the Deutsch-Anderson equation referred to in the problem statement:

With the term w representing the effective migration or drift velocity. (Note: wA/q is often denoted as <f>.) The numerical value of w is selected on the basis of experience with a particular dust, a particular set of operating conditions, and a particular design. Precipitator manufacturers usually have a specific experience file from which the precipitation rate parameter can be selected for various applications and conditions. Average values of precipitation rate parameters for various applications, and the range of values that might be expected, are presented below.

Application

Precipitation Rate or Drift Velocity (ft/s)

Utility fly ash

0.13-0.67

Pulp and paper mills

0.21-0.31

Sulfuric acid mist

0.19-0.25

Cement (wet process)

0.33-0.37

Cement (dry process)

0.19-0.23

Gypsum

0.52-0.64

Smelter

0.06

Open-hearth furnace

0.16-0.19

Blast furnace

0.20-0.46

Hot phosphorous

0.09

Flash roaster

0.25

Multiple hearth roaster

0.26

Catalyst dust

0.25

Cupola

0.10-0.12

Since the desired collection efficiency and gas flowrate are usually specified, the required collecting area can be determined from the Deutsch-Anderson equation once an appropriate precipitation rate parameter has been chosen.

An attempt to account for the sensitivity of w on process variables, especially for small particle size distributions, appeared in 1957 and was later revised by Allander, Matts, and Ohnfeldt, who derived the expression

The second exponent (in this, the so-called modified Deutsch equation) provides a more accurate prediction of performance at high-efficiency levels but can become too pessimistic in certain situations. Typical values of m range between 0.4 and 0.7, with 0.5 as the norm. This and the previous equation are employed in the solution that follows.

The collection surface area per duct, A, is

The volumetric flowrate through each duct in acfs is q = 40,000/[(30)(60)] = 22.22 acfs

The collection efficiency using the Deutsch-Anderson model can now be calculated. E=\- e~(wAl<i) = 1 _ g-((288)(0.4)/22.22)

Using the modified Deutsch-Anderson (DA) equation to obtain an expression for the efficiency in terms of the exponent m leads to

E=l - e~(wA'q)m = 1 - e-«288)(0-4)/22.22)'" = j _ e-5.184™ The following table provides E for various values of m:

m

E(%)

0.40

85.51

0.45

87.72

0.50

89.74

0.55

91.56

0.60

93.17

0.65

94.58

0.70

95.78

While the DA equation predicts an efficiency of 99.44%, it can be seen that the efficiency is probably somewhat lower than that. At a value for the exponent of 0.5, it appears that the ESP operates at 89.7% efficiency. If expressed in terms of penetration, the DA equation gives a value of 0.0056 while the modified DA gives a value of 0.1026. This means that 18.3 times more fly ash is passing through (not collected) the precipitator than that predicted by the DA equation. Because of this, it can be seen that the design of an ESP can be somewhat tricky.

ESP.2 MODIFIED DEUTSCH-ANDERSON EQUATION

A horizontal parallel-plate ESP consists of a single duct 24 ft high and 20 ft deep with an 11-in. plate-to-plate spacing. A collection efficiency and 88.2% is obtained with a flowrate of4200 acfm. The inlet loading is 2.82 gr/ft3. Calculate the following using a modified form of the Deutsch-Anderson equation, with the exponent m = 0.5 (i.e., exponent on ({>):

1. The outlet loading.

2. The drift velocity for this system.

3. A revised collection efficiency if the flowrate is increased to 5400 acfm.

4. A revised collection efficiency if the plate spacing is decreased to 9 in.

Solution

The outlet loading (OL) is

The drift velocity, w, for this system is found from the modified form of the Deutsch-Anderson equation:

= 0.333 ft/s If the flowrate is increased to 5400 acfm,

Since q, w, and A are all constant, the modified Deutsch-Anderson equation predicts that the efficiency does not change if the plate spacing is decreased to 9 in.

ESP.3 FRACTIONAL EFFICIENCY CURVES

Fractional efficiency curves describing the performance of a specific model of an electrostatic precipitator have been compiled by a vendor. Although you do not possess these curves, you are told that the cut diameter (50% efficiency) for a precipitator with a 10-in. plate spacing is 0.9 |im. The vendor claims that this particular model will perform with an efficiency of 98% under your operating conditions. You are asked to verify this claim and to make certain that the effluent loading does not exceed 0.2 gr/ft3.

The inlet loading is 14 gr/ft3 and the aerosol has the following particle size distribution:

Weight Range (%)

Average Particle Size (|xm)

0-20

3.5

20^10

8.0

40-60

13.0

60-80

19.0

80-100

45.0

310 ELECTROSTATIC PRECIPITATORS (ESP) Assume a Deutsch-Anderson equation of the form

to apply. Solution

Using the Deutsch-Anderson equation,

The following table can now be generated:

Average Particle

Weight Fraction w.

Diameter {d¡) (pm)

E,

W-fii

0.2

3.5

0.93250

0.18650

0.2

8

0.99789

0.19958

0.2

13

0.99996

0.19999

0.2

19

0.99999

0.20000

0.2

45

0.99999

0.20000

£

= 0.98607

Thus, the overall efficiency, E0, is 98.607%. The outlet loading (OL) is

OL = (14)(1 - 0.98607) = 0.195 gr/ft3 < 0.2gr/ft3 Therefore, the OL standard is met.

ESP.4 THREE FIELDS IN SERIES

An electrostatic precipitator is to be used to treat 100,000 acfm of a gas stream containing particulates from a hazardous waste incinerator. The proposed precipitator consists of three bus sections (fields) arranged in series, each with the same collection surface. The inlet loading has been measured as 40 gr/ft3 and a maximum outlet loading of 0.18 gr/ft3 is allowed by local Environmental Protection Agency (EPA) regulations. The drift velocity for the particulates has been experimentally determined in a similar incinerator installation with the following results:

ESP.4 THREE FIELDS IN SERIES

First section (inlet): 0.37 ft/s Second section (middle): 0.35 ft/s Third section (outlet): 0.33 ft/s

1. Calculate the total collecting surface required based on the average drift velocity and the required total efficiency.

2. Find the total mass flowrate (lb/min) of particulates captured by each section using the above drift velocities.

Solution

Calculate the required total collection efficiency based on the given inlet and outlet loading.

Outlet loading

Inlet loading 0.18

= 0.9955 = 99.55% Calculate the average drift velocity, w:

Calculate the total surface area required using the Deutsch-Anderson equation.

Calculate the collection efficiency of each section. Assume that each section has the same surface area but employs individual section drift velocities:

£ ] _ e~(Awtßq) = j _ e-«25,732)(0.37)/[(3)(1666.7)])

£2 _ J _ e-(Aw2ßq) _ j _ e-«25,732)(0.35)/[(3)(l666.7)])

£3 _ J _ e-{Aw,ßq) _ j _ e-«25,732)(0.33)/[(3)(1666.7)])

Calculate the mass flowrate of particulates captured by each section using the collection efficiencies calculated above:

Wx = (£,)(Inlet loadingXi/) = 3.404 x 106 gr/min = 486.3 lb/min W2 = (1 - E| XFjXlnlet loading)^) = 4.977 x 105 gr/min = 71.1 lb/min W3=(l- Ex)( 1 - £2)(£3)( Inlet loading)^) = 8.034 x 104 gr/min = 11.48 lb/min

ESP.5 FOUR CHANNELS IN PARALLEL

A single-stage duct-type electrostatic precipitator contains five plates that are 10 ft high, 20 ft long, and spaced 9 in. apart. Air contaminated with gypsum dust enters the unit with an inlet loading of 53 gr/ft3 and a velocity through the unit of 5 ft/s. The solids bulk density is 47 lb/ft3.

1. Estimate the particle drift velocity, w, if the collection efficiency is 99%.

2. What is the outlet loading?

3. How many cubic feet of dust are collected per hour?

4. Determine the rapping frequency (intervals) minutes if the maximum allowable dust thickness on the plates is g in. (Assume this layer is uniform over the entire plate.)

5. Due to an anticipated increase in plant capacity, a larger volumetric flowrate would result in a gas velocity of 7.5 ft/s. Determine the new efficiency.

Solution

Collecting area, A, and gas flow rate, q, are given by A = (8 surfaces)(10 ft)(20 ft) = 1600 ft2 q = (4 channels)(5 ft/s)(9/12 ft)( 10 ft) = 150 ft3/s Using the Deutsch Anderson equation, ln(l -E)

The outlet loading (OL) is

The volumetric flowrate of particulates captured per hour, Fp, is

= 86.12 ft3/h The rapping cycle (RC) time is

The revised efficiency (for a 50% increase in flow) becomes

ESP.6 EFFECT OF INLET DISTRIBUTION

You have been requested to 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 collection efficiency if one duct is fed 50% of the gas and the other passages 25% each. Operating and design data include:

Volumetric flowrate of contaminated gas = 4000 acfm

Operating temperature and pressure = 20° C and 1 atm, respectively

Drift velocity = 0.40 ft/s

Size of the plate = 12 ft long and 12 ft high

Plate-to-plate spacing = 8 in.

Solution

Considering both sides of the plate,

Remembering the volumetric flowrate through a passage is one third of the total volumetric flowrate,

Calculate the collection efficiency using the Deutsch-Anderson equation:

This efficiency calculation assumes the gas is uniformly distributed at the inlet of the precipitator. A revised efficiency can be calculated if the flow is distributed as specified in the problem statement. First, calculate q in acfs through the middle section:

Calculate the collection efficiency, remembering the collection surface area per duct remains the same:

= 0.9684 = 96.84% Calculate q in acfs through an outer section:

The collection efficiency in the outside section is

ESP.7 EFFECT OF PARTICLE SIZE DISTRIBUTION 315 Calculate the new overall collection efficiency:

Note that the penetration (100 — E) has increased by a factor of 3. The reader, as an optional exercise should outline the calculational procedure to follow if the particle size distribution varies with each inlet duct.

ESP.7 EFFECT OF PARTICLE SIZE DISTRIBUTION

The following data are available for the proposed design of an ESP to operate at an efficiency of 97.5%:

Air volumetric flow = 100,000 acfm

Uniform flow distribution through 10 ducts

Duct height = 30 ft with 12 in. plate-to-plate spacing

Plate length = 36 ft

Inlet loading = 14 gr/ft3

Outlet loading = 0.35 gr/ft3:

In addition, the following particle size distribution and drift velocity data have been provided in terms of the average particle size (clp) of weight fraction (x,) in a given size range and the corresponding drift velocity (w).

dv (pm)

w (ft/s)

Mass Fraction x.

0.1

0.27

0.01

0.25

0.15

0.01

0.5

0.12

0.01

1.0

0.11

0.01

1.5

0.15

0.16

2.0

0.20

0.16

2.5

0.26

0.16

5.0

0.50

0.16

10.0

0.60

0.16

25.0

0.70

0.16

Determine if the proposed design will meet the desired efficiency. Also, prepare a graph of particle size vs. efficiency for the system and comment on the results.

Solution

Check if the average throughput velocity is in an acceptable range. The average velocity should be 2-8 ft/s.

This is in the acceptable range.

The volumetric flowrate through each (one) duct is q = (100,000)/10 = 10,000 acfm = 166.67 acfs

Calculate the plate area in square feet for each duct. Note that both plates contribute to the collection area:

If the Deutsch-Anderson equation applies,

£ = l - e-(wA/l) — 1 _ g-(w(2160)(60)/10,000) _ J _ e-12.96 w

Calculate the efficiency for the 0.1 |xm particle size:

Calculate the collection efficiencies for all of the other particle size ranges. Results are presented in the following table:

(pm)

E

0.1

0.9698

0.25

0.8569

0.5

0.7889

1.0

0.7596

1.5

0.8569

2.0

0.9251

2.5

0.9656

5.0

0.9985

10.0

0.9996

25.0

0.9999

ESP.7 EFFECT OF PARTICLE SIZE DISTRIBUTION Calculate the overall efficiency E of the unit:

dp (pm)

E\

xiEi

0.1

0.01

0.9698

0.009698

0.25

0.01

0.8569

0.008569

0.5

0.01

0.7889

0.007889

1.0

0.01

0.7596

0.007596

1.5

0.16

0.8569

0.137104

2.0

0.16

0.9251

0.148016

2.5

0.16

0.9656

0.154496

5.0

0.16

0.9985

0.159760

10.0

0.16

0.9996

0.159936

25.0

0.16

0.9999

0.159984

E = J2xiEi= 0.953048

E = J2xiEi= 0.953048

Thus, the efficiency is 95.30%

Since the desired efficiency is 97.5%, the proposed design is insufficient. As is typical with particulate control, collection of large particles is highly efficient. A decline in efficiency is seen as particle size decreases. However, when the particles become very small, diffusion effects occur that actually raise the collection efficiency for these particles. See the graph in Figure 69.

Particle diameter, microns

Figure 69. Effect of very small particles on collection efficiency.

Particle diameter, microns

Figure 69. Effect of very small particles on collection efficiency.

ESP.8 BUS SECTION FAILURE

A precipitator consists of two bus sections, each with five plates (four passages) in a field (see Figure 70). The corona wires between any two plates are independently controlled so that the remainder of the unit can be operated in the event of a wire failure. The following operating conditions exist:

Gas flowrate: 10,000 acfm

Plate dimensions: 10 ft x 15 ft; four rows per field

Drift velocity: 19.0 ft/min; section 1 16.3 ft/min; section 2

1. Determine the normal operating efficiency.

2. During operation, a wire breaks in section 1. As a result, all of the wires in that row are shorted and ineffective, but the others function normally. Calculate the collection efficiency under these conditions. Assume the gas stream leaving section 1 is uniformly redistributed on entering section 2, i.e., each of the four rows is fed the same volume of gas.

3. Redo part 2 assuming the flow in (through) each of the four rows (or passages) acts in a "railroad" manner, i.e., there is no redistribution after section 1.

4. Calculate a revised efficiency if a second wire fails in a different row. Assume "railroad" flow again.

Solution

The need for series sectionalization in a precipitator arises mainly because power input needs differ at various locations in a precipitator. In the inlet sections of a precipitator, concentrations of particulate matter will be relatively heavy. This requires a great deal of power input to generate the corona discharge required for optimal particle charging: heavy concentrations of dust particles tend to suppress corona current. On the other hand, in the downstream sections of a precipitator, dust

TOP VIEW

TOP VIEW

Section (Field) 1 Section (Field) 2

Figure 70. ESP with two bus sections and five plates.

Section (Field) 1 Section (Field) 2

Figure 70. ESP with two bus sections and five plates.

concentrations will be lighter. As a consequence, corona current will flow more freely and particle charging will tend to be limited by sparking rather than current suppression. Hence, excessive sparking is more likely to occur at lower voltage in downstream sections; if the precipitator has only a single power set, then this sparking, under spark rate-limited control, will limit power input to the entire precipitator, including the inlet sections. This will result in insufficient power being supplied to the discharge electrodes in the inlet sections, with a consequential fall in precipitator collection efficiency in the inlet sections of the precipitator.

A remedy for this situation is to divide the precipitator into a series of independently energized electric bus sections. Each bus section has its own transformer rectifier, voltage stabilization controls, and high-voltage conductors that energize the discharge electrodes within that section. This would allow greater power input, and increased particle charging of dust and particulate precipitation than in the previously described underpowered inlet sections.

For part 1, write the equation describing the overall or total efficiency, ET, in terms of the individual section efficiencies, Ex and E2:

The equation describing the total penetration, PT, in terms of the individual section penetrations, f, and P2, is

PT = PXP2 Calculate the efficiency of section 1, Ex:

Calculate the efficiency of section 2, E2:

The total efficiency ET is therefore

ET = 1 - (1 - 0.89772)(1 - 0.85858) = 0.98554 = 98.554%

Calculate a revised total efficiency for part 2:

= 1 - [1 - (0.75)(0.89772)][1 - 0.85858] = 0.95380 = 95.380%

Calculate a revised total efficiency for part 3:

ET = (1.0)[(0.75)(0.98554) + (0.25)(0.85858)] = (1.0)[0.73916 +0.21464] = 0.95380 = 95.380%

The calculation for part 4 is affected by where the second wire failure occurs. Determine a revised efficiency if the wire failure is located in section 1 :

ET = (1.0)[(0.5)(0.98554) + (0.5)(0.85858)] = 0.49277 + 0.42929 = 0.92206 = 92.206%

Determine a revised efficiency if the wire failure in part 4 occurs in a different row in section 2.

Et = (1.0)[(0.5)(0.98554) + (0.25)(0.85858) + (0.25)(0.89772)] = 0.49277 + 0.43895 = 0.93172 = 93.172%

Determine a revised efficiency if the wire failure in part 4 occurs in a same row in section 2:

Et = (1.0)[(0.75)(0.98544) + (0.25)(0.0)] = 0.73908 + 0.0 = 0.73908 = 73.908%

The reader should note the effect on efficiency of bus section failure and the location of the failure.

Parallel sectionalization provides the means for coping with different power input needs due to uneven dust and gas distributions that usually occur across the inlet face of a precipitator. Nevertheless, the gains in collection efficiency from parallel sectionalization are smaller than series sectionalization.

Bus section failure is one of the more important design operating and maintenance variables for ESPs. Detailed calculational procedures for estimating this effect are available in the literature. Two studies addressing this issue are:

Theodore, L., and Reynolds, J., "The Effect of Bus Section Failure on Electrostatic Precipitator Performance," JAPCA, 33: 1202-1205, 1983. Theodore, L., Reynolds, J., Taylor, F., Filippi, A., and Errico, S., "Electrostatic Precipitator Bus Section Failure: Operation and Maintenance," Proceedings of the Fifth USEPA Symposium on the Transfer and Utilization of Particulate Control Technology, Kansas City, 1984.

ESP.9 RESISTIVITY

Comment on the resistivity "problem" with ESPs. Solution

Resistivity is a term used to describe the resistance of a medium to the flow of an electrical current. By definition, the resistivity is the electrical resistance of a dust sample 1 cm2 in cross-sectional area and 1 cm thick. For ease of precipitation capture, dust resistivity values can be classified roughly into three groups:

3. Above 2.0 x 1010Q ■ cm (this value is frequently referred to as the critical resistivity).

Particulates in group 1 are difficult to collect. They are easily charged and precipitated; upon contacting the collection electrode, however, they lose their discharge electrode polarity and acquire the polarity of the collection electrode. The particulates are then repelled into the gas stream to either escape from the precipitator or become recharged by the corona field. Examples are unburned carbon in fly ash and carbon black. If the conductive particles are coarse, they may be removed upstream from the precipitator with another collection device, e.g., a cyclone. Baffles are often designed on the collection walls to limit this precipitation-repulsion phenomenon. Particulates with resistivities in group 3 cause back-ioniza-tion or back-corona, which is a localized discharge at the collection electrode due to the surface being coated by a layer of nonconductive material. A weak back-corona will merely lower the sparkover voltage, but a strong back-corona produces a positive ion discharge at the electrode. Back-corona phenomena becomes severe with a bulk particle layer resistivity greater than 10I2Q- cm. Particulates with resistivities in group 2 have been shown by experiment and experience to be the most acceptable for electrostatic precipitation. The particulates do not rapidly lose their charge on contact with the collection electrode or cause back-corona. Back-corona phenomena can be decreased by treatment of the gas stream, such as altering the temperature, moisture content, or chemical composition.

Particle resistivity decreases as gas temperature increases due to enhanced volume conductivity. Resistivity may also decrease as gas temperature decreases if surface conditioning agents such as moisture or acid gases are present in the gas stream. Adsorption of these on the particle surface is favored at lower temperatures and provides a conductive path on the particle surface.

ESP.10 DESIGN PROCEDURE

Provide a design procedure for electrostatic precipitators.

Solution

No critically reviewed design procedure exists for ESPs. However, one suggested "general" design procedure (L. Theodore, personal notes) is provided below.

1. Determine or obtain a complete description of the process, including the volumetric flowrate, inlet loading, particle size distribution, maximum allowable discharge, and process conditions.

2. Calculate or set the overall collection efficiency.

3. Select a migration velocity (based on experience).

4. Calculate the ESP size (capture area).

5. Select the field height (experience).

6. Select the plate spacing (experience).

7. Select a gas throughput velocity (experience).

8. Calculate the number of gas passages in parallel.

9. Select (decide) on bus sections, fields, energizing sets, specific current, capacity of energizing set for each bus section, etc.

10. Design and select hoppers, rappers, etc.

11. Perform a capital cost analysis, including materials, erection, and startup costs.

12. Perform an operating cost analysis, including power, maintenance, inspection, capital and replacement, interest on capital, dust disposal, etc.

13. Conduct a perturbation study to optimize economics.

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Responses

  • fulvus chubb-baggins
    How to calculate electrostatic precipitator efficiency?
    7 years ago
  • bertoldo
    How to calculate electrostatic precipitator capacity and current?
    3 years ago
  • Azzeza Sayid
    How to calculate of drift velocity of esp in m per sec?
    2 years ago
  • Novella
    How to calculate ESP dust loading?
    2 years ago

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