Question

2. An environmental consultancy wishes to attain approval to start up a petroleum refinery operation. Before operation can begin, the consultancy needs to confirm that their projected airbome particulate output would be less than 0.3 mg/m3 at 25°C. The area of interest is a 12 km by 11.5 km plot of land with an average ceiling height of 0.6 km. If the daily airborne particulate production was 181 kg/km2, and the average wind velocity was 0.73 m/s, determine if the projected output meet the requirement. Assume that the influent air contains no particulate, the particulate is evenly distributed throughout the projected area and steady state conditions apply. Make sure to inelude a system diagram.

Answer 1

Schematic diagram for the projected area is shown below: Wind in → Wind out 0.6 km > L 11.5 km 12 km Daily airborne particulate production per km² is given as, 181 kg/km? Calculate the total land area as: A=LxB =12x11.5 =138 km? For 1 day, the airborne production for whole projected area will become: Cos kg X(138 km) El 181 - km day =181x138 = 24978 kg/day

Speed of the wind is, v=0.73 m/s. Wind flows across the shaded area in the diagram. Area of this shaded region is: 4. hzied =11.5x0.6 = 6.9 km? Volumetric flowrate of wind passing this shaded region will be: O= vnazied -(0.73 m., 167 km)*(6.9 km) 86400 s =5.037x10-km S day 4352 km day For this volumetric flowrate of wind over the projected area, the airborne particulate output available will be: Р Panatalie- 24978 kg/day 435.2 km day = 57.394 kg/km = 0.0574 mg/m At 25°C, this available airborne particulate output is less than 0.3 mg/m". Therefore, the projected output meets the desired requirements.

Answer 2

Given daten wird Dally airborne production = 185 kg kan? Average wind velocity = 0.73 m/s Average ceiling night Area of intrest & 12+ 45.5 = 138 km² x 120(km) & Total airboone production - 161 kg = 24978 kg day across the Shaded cerca wind is flourling Shaded area = 41.5x 0.6 -6.9km² Volume of wind passing this crossection/ second = velocity of wind & Ana (Shaded) = 6.9 km2 x 0.73 m x 103 km)

the Q = 5.0578103 Kod Totul quontity of wond flowing anoss Chaded cancer per day = Q x (Total seconds in a day) = 5.0378103 km3x (24 x 3600) - 435.1966 Kon't This 435.1966 km wind will take away 24976 kg of arbome The output = 24976 kg 24970 kg = 57.394 kg UT3.1966 m3 km or 57.394 4100 m2 = .057 mg 109 which is uss than song