DETERMINATION OF SIZES AND AREA OF ZONE OF

DETERMINATION OF SIZES AND AREA OF ZONE OF CHEMICAL CONTAMINATION. TASK 1 On the object has been destroyed ///, which keeps 10 tons of amiak. To Determine the sizes and area of zone of chemical pollution at night time, the area is opened. Initial data: meteorological conditions bright, Vwind= 3 m/s

SOLUTION 1) Determine the level of vertical stability of the air according to graph on pic. 1: Result: Invercia

2) 2. 1) From the table 11, 1 or 11, 2 of Annex it is necessary to find G of // for 10 tons of /// in terms of Vwind= 1 m/s. G= 4, 5 km for damaging concentration 2. 2) For Vwind= 3 m/s it is necessary to find correlation coefficient from table 12: K= 0, 45 2. 3) /// G=4, 5*0, 45=2, 02 km 3) Width of zone of chemical contamination is: Width= 0, 03 G – if inversion, Width= 0, 15 G – if izotermia, Width= 0, 008 G – if convectia. Width of zone of chemical contamination in condition of inversion is: Width= 0, 03 G = 0, 03 * 2, 02= 0, 06 km 4) Determination of the area of zone of chemical contamination: S= ½ G * Width= (2, 02 * 0, 06) / 2 = 0, 06 km 2

CONTAMINATED AIR TIME REACHING TO THE DEFINED MEASURE. TASK 2 As a result of accident on the object, placed on the distance 9 km from inhabited locality(settlement), were destroyed communications with ///. Meteorological conditions: izotermia, Vwind= 5 m/s. To determine the time of coming the cloud of contaminated air to the inhabited locality (settlement).

SOLUTION 1) From the table 14 for izotermia Vwind= 5 m/s it is necessary to find the average speed of moving cloud contaminated air: W = 7, 5 m/s 2) The time of coming cloud with contaminated air to inhabited locality(settlement) is: t = R : W = 9000 : (7, 5 * 60) = 20 min, where R – is distance to inhabited locality(settlement) W - average speed of moving cloud contaminated air Answer: 20 minutes

DATA FROM ANNEX TO TASK № 2 (TABLE 14)

DETERMINATION OF DAMAGING INFLUENCE (ACTION) OF HAZARDOUS CHEMICALS (NHR) TASK 3 On the object as a result of exposure was destroyed ///. Speed of wind is 3 m/s. To determine the time of damaging influence (action) of //. Solution: 1) From table 15 determine that t damag influence (action) of // (time of ) if Vwind= 1 m/s = 20 hours. 2) From table 16 determine correlation coefficient Vwind= 3 m/s = 0, 55. 3) Time of damage influence of t damag = 20 * 0, 55= 11 hours ( if Vwind= 3 m/s ) Answer: t damag = 11 hours

DATA FROM ANNEX TO TASK № 3 (TABLE 15; 16)