Question

unwuy downwind (in m) for part (b) and what in Ppiny NEmergency response for the rupture of a chemical pipeline is being evaluated. The chemi- cal is estimated to discharge from the pipeline at a rate of 4.53 kg/s. The concentration of concern is 100 ppm-anyone downwind who is exposed to a concentration higher than this must be evacuated. Calculate the downwind evacuation distance. The pipeline is at ground level. The molecular weight of the released material is 17.03. The sunlight is mod- erate and the wind speed is 3.5 m/s. Assume 1 atm, 25°C, and rural conditions.

Answer 1

ALL the calculation is done with the help of calculator

Solution given date 4.53 byl u = 3.5 ms in = b =0X100 pprt = '10 1 M. = 17.03 (kg 2 P= latry - 101.325 kpa. using Ideal Gas Law PV = nRT occupied by a At gas STP x the vol 22.9 lit 101.325 x 22.9 X 203 8.314 x 298 n = 0.916 kmo) 7163 mo = kg N. My = 0.9168103 x 17.03 d = more my = 156 x 203 ky = sy. - 15.6x10 ²x 3.5 bol 4.53 8109 4.83 x 109 d = 120.53 m