Question

Pasquill-Gifford represents them by ơ,-1000 x tan(T)12.15 and or-ad in Table 7.4. Cy= 1000 x tan 2.15 where x is downwind distance ínkm and T is one-half Pasquill's θ in degrees. T, as a function ofx, is determined for each stability from the following: Stability Equation for T T24.167 2.5334 Inx) T = 18.333-1.8096 ln(x) T = 12.5-1.0857 ln(x) T8.3333 0.72382 In(x) T 6.25-0.54287 In(x) T 4.1667-0.36191 Ln(x) Table 7.4a Pasquill-Gifford horizontal dispersion parameters [Turner] Stability Distance (km) at upper boundary 3.11 0.5 3.11 0.40.5 0.3 0.4 0.250.3 0.20.25 0.15 0.2 0. 0.15 453.85 2.1166 346.75 1.7283 104.7 258.89 14094 71.2 217.41 1.2644 47.4 179,52 1.1262 170.22 1.0932 29.3 158.08 10542 122.8 0.9447 14.0 37.7 21.4 B 35. 0.435 0.2 04 109.30 1.0971 98.483 0.98332 40.0 90.673 0.93198 20.2 all x 61.141 0.91465 44.053 0.51179 36.650 0.56589 251.2 33.504 0.60486 134.9 32.093 0.64403 65.1 D 30. 30. 10. <0.3 34.459 0.86974 12.1 E 40. 47.618 0.29592 35.420 0.37615 141.9 26.970 0.46713 109.3 24.703 0.50527 79.1 22.534 0.57154 49.8 21.628 0.63077 33.5 21.628 0.75660 21.6 23.331 0.81956 8.7 20.40. 10.20. 0.31 0.10.3 <0.1 24.260 0.83660 3.5 F 60. 60. 30 34.219 0.21716 27.074 0.27436 83.3 15 22.651 0.32681 68.8 17.836 0.4150 54.9 16.187 0.4649 40.0 14.823 0.54503 27.0 0.7 0.2 $0.2 13.953 0.63227 21.6 13.953 0.68465 14.0 14.457 0.78407 10.9 0.7 15.209 0.81558 4.1 Table 7.4b Pasquill-Gifford vertical dispersion parameter, oz where x is in km Turner ахь 30

Answer 1

source err mision Rate: 100 Eleckive 2 15 Stabi lity (A) T2y 167- 2.5334 n K a= 453.85 → From Giveh lable Accoredng o pasguit! -GiffoRd 15 113.03 m 2-1166 104- 65 CUSsia ume Moce

C-Con cen tsahion emicsiat an er 6 x 113.03 x oy S x2T C- 3.348 x 10 )/m3 H- 80 (50) C= 100 (113.03 6 x 113.03 x 104 .6 s x 2 T (So) メ(104.65 C- 303.5 47/m3