2 CyclonesPrincipleThe 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
4 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.
5 Cyclones (contd.) Advantages of Cyclones Disadvantages of Cyclones Cyclones have a lost capital costReasonable 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.Disadvantages of CyclonesThey can be used for small particlesHigh pressure drops contribute to increased costs of operation.
11 Problem Design of Cyclone Body diameter = 0.75 m Flow rate = 2.75 m3/s ρp = 1600 kg/m3ρg = 1.1 kg/m3µ = 2.5*10-5 kg/m-sHigh throughputH = 0.8 * body diameterW = 0.35 * body diameterLb = 1.7 * body diameterLc = 2.0 * body diameterCalculate the diameter of particle with 50% efficiency
12 Solution Step 1: Inlet Velocity = Q/A = 17.46 m/s Step 2: Number of effective turnsNe = (Lb + Lc/2)/HNe = 3.375Step 3: Diameter of particledpc = 10 µm
15 Fabric Filters Principle The filters retain particles larger than the mesh sizeAir 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.
16 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 + KswWhere,S = filter drag, N-min/m3 S = ∆P/VKe = extrapolated clean filter drag, N-min/m3Ks = slope constant. Varies with the dust, gas and fabric, N-min/kg-mW= Areal dust density = L V tL = dust loading (g/m3), V = velocity (m/s)Both Ke and Ks are determined empirically from pilot tests.
18 Problem Estimate the values of Ke and Ks for the filter drag model: Limestone dust loading L = 1.00 g/m3 Fabric Area A = 1.00 m2 Air flow rate Q = 0.80 m3/minTime (min)51015202530Filter ∆P (Pa)330490550600640700
19 Solution Step 1: Calculate the air velocity Air velocity = 0.80 (m3/min)/1.00 m2= 0.80 m/minStep 2:Step 3:Determine Ke and Ks graphicallyKe = 470 N-min/m3 Ks = N-min/g-mS = ∆P/V412.5612.5687.5750800875W = LVt4812162024
20 Fabric Filters Δ P Total pressure drop Δ Pf Pressure drop due to the fabricΔ Pp Pressure drop due to the particulate layerΔ Ps Pressure drop due to the bag house structure
21 Advantages of Fabric Filters Very high collection efficiencyThey can operate over a wide range of volumetric flow ratesThe pressure drops are reasonably low.Fabric Filter houses are modular in design, and can be pre-assembled at the factory
22 Fabric Filters (contd.) Disadvantages of Fabric FiltersFabric 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 streamsA fabric filtration unit is a potential fire hazard
23 Darcy’s equation ΔPf Pressure drop N/m2 ΔPp Pressure drop N/m2 Df Depth of filter in the direction of flow (m)Dp Depth of particulate layer in the direction of flow (m)μ Gas viscosity kg/m-sV superficial filtering velocity m/minKf, Kp Permeability (filter & particulate layer m2)60 Conversion factor δ/minV = Q/AQ volumetric gas flow rate m3/minA cloth area m2
24 Dust Layer ΔP = ΔPf + ΔPp Filter Drag S = ΔP/V L Dust loading kg/m3t time of operation minρL Bulk density of the particulate layer kg/m3ΔP = ΔPf + ΔPpFilter Drag S = ΔP/VAreal dust density W = LVtS= k1+k2W
25 ProblemEstimate 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.
26 SolutionStep 1: Calculate total area and number of components required. A = Q/VStep 2:Calculate the area of each bag.A = Π (d) lStep 3:Calculate the total number of bags required.Number of bags required = Total area / Area per bag= 1270 bagsNumber of compartments sq. ft. / compartment
27 Electrostatic Precipitator Electrostatic Charging of Dust ParticlesCutaway of Electrostatic Precipitator
28 Electrostatic Precipitator PrincipleThe particles in a polluted gas stream are charged by passing them through an electric field.The charged particles are led through collector platesThe collector plates carry charges opposite to that on the particlesThe particles are attracted to these collector plates and are thus removed from the gas steamConstruction and Operation of Electrostatic PrecipitatorCharging 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.
29 Design of Electrostatic Precipitators The efficiency of removal of particles by an Electrostatic Precipitator is given byη = fractional collection efficiencyw = drift velocity, m/min.A = available collection area, m2Q = volumetric flow rate m3/min
30 Migration velocity Where, q = charge (Columbus) Ep = collection field intensity (volts/m)r = particle radius (m)μ = dynamic viscosity of gas (Pa-S)c = Cunningham correction factor
31 Cunningham correction factor where,T = absolute temperature (°k)dp = diameter of particle (μm)
32 ProblemAn 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).
33 SolutionStep 1:Efficiency of an Electrostatic Precipitator is given byA =-[ (Q/w)*ln(1- η)]A = 70,000 m2Step 2:Number of plates = total area/plate area= 1400
34 ELECTROSTATIC PRECIPITATOR (contd.) Advantages of Electrostatic PrecipitatorsElectrostatic precipitators are capable very high efficiency, generally of the order of %.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.Disadvantages of Electrostatic PrecipitatorsThe 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.
37 Wet Scrubbers Principle Construction and Operation 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 particlesThe liquid is subsequently removed for treatment.Construction and OperationA 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 streamThe droplets contact the particulate matter, and the particles get sorbed.The droplet size has to be optimized.
38 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.
39 Scrubber Efficiency where, k = Scrubber coefficient (m3 of gas/ m3 of liquid)R = Liquid-to-gas flow rate (QL/QG)ψ = internal impaction parameterInternal impaction parameterc = Cunningham correction factorρp = 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)
40 Wet Scrubbers (contd.) Advantages of Wet Scrubbers Wet Scrubbers can handle incoming streams at high temperature, thus removing the need for temperature control equipment.Wet scrubbers can handle high particle loading.Loading fluctuations do not affect the removal efficiency.They can handle explosive gases with little risk.Gas adsorption and dust collection are handled in one unit.Corrosive gases and dusts are neutralized.Disadvantages of Wet ScrubbersHigh potential for corrosive problemsEffluent scrubbing liquid poses a water pollution problem.
41 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.
42 Orifice ScrubbersThe 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.
43 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.
45 Venturi ScrubbersThe 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 streamThis 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 mm have been obtained.
46 Venturi Scrubber Absolute Pressure Drop Δp = pressure drop ( cm of water)ug = gas velocity (cm/s)Qt = liquid volume flow rateQg = gas volume flow rate
47 ProblemWater 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?
48 SolutionStep 1:Absolute pressure drop is given byΔp = 4.8
58 CxHy + ( x + y/4 ) O2 x CO2 + (y/2) H2O VOC IncineratorsPrincipleVOC 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) H2OOperationThe 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.
59 VOC Incinerators (contd.) TemperatureThe reaction kinetics are very sensitive to temperatureThe higher the temperature, the faster the reactionTimingA certain time has to be provided for the reaction to proceedTurbulenceTurbulence promotes mixing between the VOC's and oxygenProper mixing helps the reaction to proceed to completion in the given time.
63 VOC Incinerators (contd.) The various methods for incineration are:Elevated fires, for concentrated streamsDirect thermal oxidation, for dilute streamsCatalytic oxidation, for dilute streams.
64 Problem In a workshop a mixture of chemicals were released, Benzene – 3000 ppmToulene – 1000 ppmMethane – 2000 ppmCalculate the lower exposure limit (LEL) of the mixtureLEL by volume for each chemicalBenzene – 1.4 %Toulene – 1.27 %Methane – 5.00%
65 Solution LEL of a mixture: Xi - Volume of i component in the mixtureXm - Volume of mixtureLELi - LEL of i componentLELmix = [3000/(6000*1.4) /(6000*1.27) /(6000*5)]-1= 1.79%
67 Air Pollution Control For Gases Adsorption TowersThermal IncernationCatalytic CombustionAir Correction Equipment for Gases and Vapors
68 Adsorption Towers 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.
69 Adsorption Towers (contd.) Construction and OperationAdsorption towers consist of cylinders packed with the adsorbent.The adsorbent is supported on a heavy screenSince 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.Advantages of Adsorption TowersVery low concentrations of pollutants can be removed.Energy consumption is low.Do not need much maintenance.Economically valuable material can be recovered during regeneration.