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Technology for Air Pollution Control – Part2 - PowerPoint PPT Presentation

Technology for Air Pollution Control – Part2. Cyclones . Principle

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Technology for Air Pollution Control – Part2

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Technology for Air Pollution Control – Part2

Cyclones

• Principle

• The particles are removed by the application of a centrifugal force. The polluted gas stream is forced into a vortex. the motion of the gas exerts a centrifugal force on the particles, and they get deposited on the inner surface of the cyclones

Cyclones (contd.)

Construction and Operation

The gas enters through the inlet, and is forced into a spiral.

• At the bottom, the gas reverses direction and flows upwards.

• To prevent particles in the incoming stream from contaminating the clean gas, a vortex finder is provided to separate them. the cleaned gas flows out through the vortex finder.

Cyclones (contd.)

• Cyclones have a lost capital cost

• Reasonable high efficiency for specially designed cyclones.

• They can be used under almost any operating condition.

• Cyclones can be constructed of a wide variety of materials.

• There are no moving parts, so there are no maintenance requirements.

• They can be used for small particles

• High pressure drops contribute to increased costs of operation.

Problem

• A cyclone with a flow rate of 150 m3/min has an efficiency of 80%. Estimate the efficiency if the flow rate is doubled.

Solution

• Step 1

Q1 = 150 m3/min

Q2 = 300 m3/min

Pt1 = 100% - 80% = 20%

Pt2/Pt1 = (Q1/Q2)0.5

Step 2

Final Efficiency = 1- Pt2

= 86%

Problem

• Design of Cyclone

Body diameter = 0.75 m

Flow rate = 2.75 m3/s

ρp = 1600 kg/m3

ρg = 1.1 kg/m3

µ = 2.5*10-5 kg/m-s

High throughput

H = 0.8 * body diameter

W = 0.35 * body diameter

Lb = 1.7 * body diameter

Lc = 2.0 * body diameter

Calculate the diameter of particle with 50% efficiency

Solution

• Step 1:

Inlet Velocity = Q/A

= 17.46 m/s

• Step 2:

Number of effective turns

Ne = (Lb + Lc/2)/H

Ne = 3.375

• Step 3: Diameter of particle

dpc = 10 µm

Fabric Filter

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 previously deposited particles.

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 = Ke + Ksw

Where,

S = filter drag, N-min/m3 S = ∆P/V

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

Ks = slope constant. Varies with the dust, gas and fabric, N-min/kg-m

W= Areal dust density = L V t

• Both Ke and Ks are determined empirically from pilot tests.

Empirical Determination of Ke and Ks

Problem

• Estimate the values of Ke and Ks for the filter drag model:

Limestone dust loading L = 1.00 g/m3Fabric Area A = 1.00 m2Air flow rate Q = 0.80 m3/min

Solution

• Step 1:

Calculate the air velocity

Air velocity =  0.80 (m3/min)/1.00 m2

= 0.80 m/min

• Step 2:

• Step 3:

Determine Ke and Ks graphically

Ke = 470 N-min/m3 Ks = 0.563 N-min/g-m

Fabric Filters

Δ P Total pressure drop

Δ Pf Pressure drop due to the fabric

Δ Pp Pressure drop due to the particulate layer

Δ Ps Pressure drop due to the bag house structure

• 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 be pre-assembled at the factory

Fabric Filters (contd.)

• 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

• A fabric filtration unit is a potential fire hazard

Darcy’s equation

ΔPfPressure drop N/m2

ΔPpPressure drop N/m2

DfDepth of filter in the direction of flow (m)

DpDepth of particulate layer in the direction of flow (m)

μGas viscosity kg/m-s

Vsuperficial filtering velocity m/min

Kf, KpPermeability (filter & particulate layer m2)

60Conversion factor δ/min

V = Q/A

Qvolumetric gas flow rate m3/min

Acloth area m2

Dust Layer

ttime of operation min

ρLBulk density of the particulate layer kg/m3

ΔP = ΔPf + ΔPp

Filter Drag S = ΔP/V

Areal dust density W = LVt

S= k1+k2W

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 per bag

= 1270 bags

Number of compartments 4000 sq. ft. / compartment

Electrostatic Precipitator

Electrostatic Charging of Dust Particles

Cutaway of Electrostatic Precipitator

Electrostatic Precipitator

• Principle

• The particles in a polluted gas stream are charged by passing them through an electric field.

• The charged particles are led through collector plates

• The collector plates carry charges opposite to that on the particles

• The particles are attracted to these collector plates and are thus removed from the gas steam

• Construction and Operation of Electrostatic Precipitator

• Charging Electrodes in the form of thin wires are placed in the path of the influent gas.

• The charging electrodes generate a strong electric field, which charges the particles as they flow through it.

• The collector plates get deposited with the particles. the particles are occasionally removed either by rapping or by washing the collector plates.

Design of Electrostatic Precipitators

• The efficiency of removal of particles by an Electrostatic Precipitator is given by

η = fractional collection efficiency

w = drift velocity, m/min.

A = available collection area, m2

Q = volumetric flow rate m3/min

Migration velocity

Where,

q = charge (Columbus)

Ep = collection field intensity (volts/m)

μ = dynamic viscosity of gas (Pa-S)

c = Cunningham correction factor

• Cunningham correction factor

where,

T = absolute temperature (°k)

dp = diameter of particle (μm)

Problem

• An ESP is designed to treat 50,000 m3/min with 97 % efficiency. Assuming an effective drift velocity of 2.5 m/min, calculate the required plate area and the number of plates. The plate size is 10 m by 5 m (height by length).

Solution

• Step 1:

Efficiency of an Electrostatic Precipitator is given by

A =-[ (Q/w)*ln(1- η)]

A = 70,000 m2

• Step 2:

Number of plates = total area/plate area

= 1400

ELECTROSTATIC PRECIPITATOR (contd.)

• Electrostatic precipitators are capable very high efficiency, generally of the order of 99.5-99.9%.

• Since the electrostatic precipitators act on the particles and not on the air, they can handle higher loads with lower pressure drops.

• They can operate at higher temperatures.

• The operating costs are generally low.

• The initial capital costs are high.

• Although they can be designed for a variety of operating conditions, they are not very flexible to changes in the operating conditions, once installed.

• Particulate with high resistivity may go uncollected.

Wet Scrubbers

Aeromix Wet Scrubber

Wet Scrubbers

Floating Bed Wet Scrubber

Wet Scrubbers

• Principle

• Wet scrubbers are used for removal of particles which have a diameter of the order of 0.2 mm or higher.

• Wet scrubbers work by spraying a stream of fine liquid droplets on the incoming stream.

• The droplets capture the particles

• The liquid is subsequently removed for treatment.

• Construction and Operation

• A wet scrubber consists of a rectangular or circular chamber in which nozzles are mounted.

• The nozzles spray a stream of droplets on the incoming gas stream

• The droplets contact the particulate matter, and the particles get sorbed.

• The droplet size has to be optimized.

Wet Scrubbers (contd.)

• Construction and Operation (contd.)

• Smaller droplets provide better cleaning, but are more difficult to remove from the cleaned stream.

• The polluted spray is collected.

• Particles are settled out or otherwise removed from the liquid.

• The liquid is recycled.

• Wet scrubbers are also used for the removal of gases from the air streams.

Scrubber

• Efficiency

where,

k = Scrubber coefficient (m3 of gas/ m3 of liquid)

R = Liquid-to-gas flow rate (QL/QG)

ψ = internal impaction parameter

• Internal impaction parameter

where,

c = Cunningham correction factor

ρp = particle density (kg/m3)

Vg = speed of gas at throat (m/sec)

dp = diameter of particle (m)

dd = diameter of droplet (m)

μ = dynamic viscosity of gas, (Pa-S)

Wet Scrubbers (contd.)

• Wet Scrubbers can handle incoming streams at high temperature, thus removing the need for temperature control equipment.

• They can handle explosive gases with little risk.

• Gas adsorption and dust collection are handled in one unit.

• Corrosive gases and dusts are neutralized.

• High potential for corrosive problems

• Effluent scrubbing liquid poses a water pollution problem.

Cyclone Spray Chambers

• These scrubbers combine a cyclone with a spray nozzle.

• The added centrifugal force permits good separation of the droplets, hence a smaller droplet size can be used.

• Cyclone spray chambers provide up to 95% removal of particles > 5 micron.

Orifice Scrubbers

• The gas is impacted onto a layer of the scrubbing liquid.

• The gas passes through the liquid, thus removing almost all the particulate matter, and a large portion of the probable gases.

• After coming out of the liquid, the gas is passed through baffles to remove the liquid droplets.

Impingement Scrubbers

• In Impingement scrubbers, the gas impacts a layer of liquid/froth through a perforated tray.

• Passing through this layer removes the particulate matter.

• The wet gas stream is then passed through a mist collector.

Venturi Scrubbers

• The dirty gas is led in to the chamber at high inlet velocities.

• At the inlet throat, liquid at low pressure is added to the gas stream

• This increases the relative velocity between the gas and the droplets, thus increasing the efficiency of removal.

• Efficiencies of the range of 95% for particles larger than 0.2 mm have been obtained.

Venturi Scrubber

Absolute Pressure Drop

Δp = pressure drop ( cm of water)

ug = gas velocity (cm/s)

Qt = liquid volume flow rate

Qg= gas volume flow rate

Problem

• Water is introduced into the throat of a venturi scrubber. The air velocity through the scrubber is 550 fps and the liquid to gas ratio is 8.5 gal/1000 actual ft3. Determine the pressure drop?

Solution

• Step 1:

Absolute pressure drop is given by

Δp = 4.8

HYDROCARBON CONTROL

General Methods For Control Of Hydrocarbon Emissions

• Incineration or after burning

• Direct flame incineration

• Thermal incineration

• Catalytic incineration

Problem

• Calculate the removal efficiency of a burner if the concentration of HC was reduced from 1300 ppm to 100 ppm.

Solution

• Step 1:

Removal Efficiency of the burner

Efficiency = (initial concentration – final concentration)/ initial concentration

Efficiency of the burner = 92.3%

VOC Incinerators

• Principle

• VOC incinerators thermally oxidize the effluent stream, in the presence of excess air.

• The complete oxidation of the VOC results in the formation of carbon monoxide and water. The reaction proceeds as follows:

CxHy + ( x + y/4 ) O2 x CO2 + (y/2) H2O

• Operation

The most important parameters in the design and operation of an incineration system are what are called the

' three T's ' Temperature, Turbulence, and residence Time.

VOC Incinerators (contd.)

• Temperature

• The reaction kinetics are very sensitive to temperature

• The higher the temperature, the faster the reaction

• Timing

• A certain time has to be provided for the reaction to proceed

• Turbulence

• Turbulence promotes mixing between the VOC's and oxygen

• Proper mixing helps the reaction to proceed to completion in the given time.

VOC Incinerators (contd.)

• The various methods for incineration are:

• Elevated fires, for concentrated streams

• Direct thermal oxidation, for dilute streams

• Catalytic oxidation, for dilute streams.

Problem

• In a workshop a mixture of chemicals were released,

Benzene – 3000 ppm

Toulene – 1000 ppm

Methane – 2000 ppm

Calculate the lower exposure limit (LEL) of the mixture

LEL by volume for each chemical

Benzene – 1.4 %

Toulene – 1.27 %

Methane – 5.00%

Solution

LEL of a mixture:

Xi - Volume of i component in the mixture

Xm - Volume of mixture

LELi - LEL of i component

LELmix = [3000/(6000*1.4) + 1000/(6000*1.27) + 2000/(6000*5)]-1

= 1.79%

GASES

Air Pollution Control For Gases

• Thermal Incernation

• Catalytic Combustion

Air Correction Equipment for Gases and Vapors

• Principle

• Adsorption towers use adsorbents to remove the impurities from the gas stream.

• The impurities bind either physically or chemically to the adsorbing material.

• The impurities can be recovered by regenerating the adsorbent.

• Adsorption towers can remove low concentrations of impurities from the flue gas stream.

• Construction and Operation

• The adsorbent is supported on a heavy screen

• Since adsorption is temperature dependent, the flue gas is temperature conditioned.

• Vapor monitors are provided to detect for large concentrations in the effluent. Large concentrations of the pollutant in the effluent indicate that the adsorbent needs to be regenerated.