Technology for Air Pollution Control – Part2

1. Technology for Air Pollution Control – Part2

2. 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

3. 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 • 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. • Disadvantages of Cyclones • They can be used for small particles • High pressure drops contribute to increased costs of operation.

7. 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.

8. 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%

9. Dimensions of a Standard Cyclone

11. 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

12. 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

13. Filter Cloth

14. Fabric Filters Fabric Filter

15. 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.

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 + 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 L = dust loading (g/m3), V = velocity (m/s) • Both Ke and Ks are determined empirically from pilot tests.

17. Empirical Determination of Ke and Ks

18. 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

19. 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

20. 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

21. Advantages of Fabric Filters • Very high collection efficiency • They can operate over a wide range of volumetric flow rates • The pressure drops are reasonably low. • Fabric Filter houses are modular in design, and can be pre-assembled at the factory

22. Fabric Filters (contd.) • Disadvantages of Fabric Filters • Fabric Filters require a large floor area. • The fabric is damaged at high temperature. • Ordinary fabrics cannot handle corrosive gases. • Fabric Filters cannot handle moist gas streams • A 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-s V superficial filtering velocity m/min Kf, Kp Permeability (filter & particulate layer m2) 60 Conversion factor δ/min V = Q/A Q volumetric gas flow rate m3/min A cloth area m2

24. Dust Layer L Dust loading kg/m3 t time of operation min ρL Bulk density of the particulate layer kg/m3 ΔP = ΔPf + ΔPp Filter Drag S = ΔP/V Areal dust density W = LVt S= k1+k2W

25. 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.

26. 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

27. Electrostatic Precipitator Electrostatic Charging of Dust Particles Cutaway of Electrostatic Precipitator

28. 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.

29. 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

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. 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).

33. 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

34. ELECTROSTATIC PRECIPITATOR (contd.) • Advantages of Electrostatic Precipitators • 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. • Disadvantages of Electrostatic Precipitators • 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.

35. Wet Scrubbers Aeromix Wet Scrubber

36. Wet Scrubbers Floating Bed Wet Scrubber

37. 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.

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 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)

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 Scrubbers • High potential for corrosive problems • Effluent 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 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.

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.

44. Venturi Scrubbers

45. 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.

46. 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

47. 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?

48. Solution • Step 1: Absolute pressure drop is given by Δp = 4.8

49. HYDROCARBON CONTROL

50. General Methods For Control Of Hydrocarbon Emissions • Incineration or after burning • Direct flame incineration • Thermal incineration • Catalytic incineration