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Air Pollution Examples for CEL 212-Environmental Engineering (Second Semester 2012-13) Air Quality and Meteorology Dr. Arun Kumar Civil Engineering (IIT Delhi) [email protected] Courtesy: Dr. Irene Xagoraraki (U.S.A.) April 27, 2013 [email protected] Dry Adiabatic Lapse Rate Altitude, z (km) Stability For this parcel of air the change in temperature with altitude was: Adiabatic lapse rate • = (T2-T1)/(z2-z1) z1 1 T2 Temperature, April 27, 2013 • When any parcel of air moves up or down, it’s temperature will change according to the adiabatic lapse rate T1 Dry adiabatic lapse rate: temperature decreases with increased altitude Γ=− = (10-20)oC/(2000-1000)m = -1 oC/100m 2 z2 2 dT = −1.00 °C/100m = -5.4 °F / 1000ft dz Atmospheric (actual) lapse rate < Г (temperature falls faster) unstable (super-adiabatic) > Г (temperature falls slower) stable (sub-adiabatic) = Г (same rate) neutral T (oC) [email protected] 3 Example 1 Z(m) April 27, 2013 [email protected] Unstable Conditions 4 Rapid vertical mixing takes place. T(ºC) 10 5.11 202 1.09 ∆T T2 − T1 1.09 − 5.11 = = = −0.0209 °C/m ∆z z 2 − z1 202 − 10 = −2.09 °C/100 m Since lapse rate is more negative than Г, (-1.00 ºC/100 m)=> atmosphere is unstable -1.25 oC/100 m < -1 oC/100m actual temperature falls faster than Г April 27, 2013 [email protected] 5 April 27, 2013 Unstable air encourages the dispersion and dilution of pollutants. [email protected] 6 1 Stable Conditions Air at a certain altitude remains at the same elevation. -0.5 oC/100 m > -1 oC/100m actual temperature falls slower than Г April 27, 2013 Stable air discourages the dispersion and dilution of pollutants. [email protected] Air at a certain altitude remains at the same elevation. Neutral Conditions -1 oC/100 m = -1 oC/100m 7 April 27, 2013 Neutrally stable air discourages the dispersion and dilution of pollutants. [email protected] 8 Why are these plumes so different? neutral Prediction for Pollutant Concentration under inversion layer Above inversion April 27, 2013 [email protected] 9 April 27, 2013 Point-Source Gaussian Plume Model April 27, 2013 [email protected] [email protected] 10 Point-Source Gaussian Plume Model 11 April 27, 2013 [email protected] 12 2 Point-Source Gaussian Plume Model Effective Stack Height • Model Structure and Assumptions – pollutants released from a “virtual point source” – advective transport by wind – dispersive transport (spreading) follows normal (Gaussian) distribution away from trajectory – constant emission rate – wind speed constant with time and elevation – pollutant is conservative (no reaction) – terrain is flat and unobstructed – uniform atmospheric stability April 27, 2013 [email protected] H = h + ∆H Where: H = Effective stack height (m) h = height of physical stack (m) ∆H = plume rise (m) 13 Effective Stack Height (Holland’s formula) for neutral conditions ∆H = vs u T −T −2 1.5 + 2.68 × 10 (P ) s a Ta April 27, 2013 [email protected] 14 Plume rise equation for neutral conditions d where v s = stack velocity (m/s) d = stack diameter (m) u = wind speed (m) P = pressure (kPa) Ts = stack temperature (ºK) Ta = air temperature (ºK) April 27, 2013 [email protected] 15 April 27, 2013 [email protected] 16 Atmospheric Stability Categories • How much will be % error in C(x,0,0) if one uses Heffective(unstable) for stability class? Think qualitatively. April 27, 2013 [email protected] 17 April 27, 2013 [email protected] 18 3 Vertical Dispersion Horizontal Dispersion April 27, 2013 [email protected] 19 April 27, 2013 Wind Speed Correction April 27, 2013 • A stack in an urban area is emitting 80 g/s of NO. It has an effective stack height of 100 m. The wind speed is 4 m/s at 10 m. It is a clear summer day with the sun nearly overhead. • Estimate the ground level concentration at: a) 2 km downwind on the centerline and b) 2 km downwind, 0.1 km off the centerline. p Where: ux = wind speed at elevation zx p = empirical constant [email protected] 21 April 27, 2013 [email protected] Example 2 1. 20 Example 2 • Unless the wind speed at the virtual stack height is known, it must be estimated from the ground wind speed z u2 = u1 2 z1 [email protected] 22 Example 2 2. Determine σy and σz σy = 290, σz = 220 Determine stability class Assume wind speed is 4 km at ground surface. Description suggests strong solar radiation. Stability class B 220 290 April 27, 2013 [email protected] 23 April 27, 2013 [email protected] 24 4 Example 2 3. Example 2 Estimate the wind speed at the effective stack height Note: effective stack height given – no need to calculate using Holland’s formula 4. Determine concentration a. x = 2000, y = 0 C (2000,0) = p z 100 u2 = u1 2 = 4 10 z1 April 27, 2013 0.15 C ( 2000 , 0 ) = 6 . 43 × 10 − 5 g/m 3 = 64 . 3 µg/m = 5.65 m/s [email protected] 1 0 2 1 100 2 80 exp − exp − π (290)(220)(5.6) 2 290 2 220 25 April 27, 2013 3 [email protected] 26 Example 2 Example 3 b. x = 2000, y = 0.1 km = 100 m C (2000,100) = 1 100 2 1 100 2 80 exp − exp − π (290)(220)(5.6) 2 290 2 220 C ( 2000 ,0 ) = 6 . 06 × 10 − 5 g/m 3 = 60 . 6 µg/m April 27, 2013 • If in example #2, there is another stack (downwind distance from 1st stack =500m) with physical height (203m). Now, calculate overall ground level concentration at 2 km downwind on the center line? This 2nd stack is also emitting NO at same 80 g/s rate (all other conditions remain constant) (for stack #2: inside diameter =1.07m; air temp:13degC; barometric pressure =1000 milibars; stack gas velocity=9.14m/s; stack gas temp: 149degC) 3 [email protected] 27 April 27, 2013 Example 3 hints [email protected] 28 Example 4 • Question: Suppose an anemometer at a height of 10 m above ground measure wind velocity =2.5m/s. estimate the wind speed at an elevation of 300 m in rough terrain if atmosphere is unstable (i.e., k=0.2)? • From stack #1, we know conc (C1) • For stack #2, first calculate effective stack height using Holland’s formula then calculate conc. at given distance using approach given in Example 2 (apply correction for x= distance of receptor from stack #2)say we get conc. C2 • Now total conc. at receptor =Ctotal=C1+C2 • Now see if this is less than Callowable • If not, then we need to control stack heights or source strength April 27, 2013 [email protected] • Answer: z u2 = u1 2 • U300/u10=(300/10)(0.2) z1 • Wind velocity at 300m=(2.5)*(30)(0.2)=4.9m/s 29 April 27, 2013 [email protected] p 30 5 CPCB minimum guideline for stack based on SO2 emission • CPCB minimum stack height =30m • So Choose maximum (30m; hSO2) April 27, 2013 [email protected] 31 Example 5 • A 40% efficient 1000MW coal fired power plant emitts SO2 at rate =6.47*108 microgram/s. the stack has effective height =20m (CPCB recommended minimum height =30m). An anemometer on a 10-m pole measures 2.5m/s of wind and atmospheric class is C. • Predict the ground-level concentration of SO2 4 km directly downwind? • What would be this concentration if stack height is changed to 30 m? • What is the recommended stack height based on SO2 emission rate? • Which stack height would you choose? April 27, 2013 [email protected] 32 Example 6 • Repeat Example 5 for stability classes : B,C and D for calculating C(x,0,0) where X=0-100m with 4 m gap. Now plot C(x,0,0) versus distance or for different stability classes. Use effective height obtained from Example 6. April 27, 2013 [email protected] 33 6