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IMASS CALCULATIONS <br />' The hydrocarbon mass removed during the operating period can be calculated using the following <br /> equation M = C•Q•t <br />' where M= cumulative mass recovered(kg) <br /> C vapor concentration (kg/n?) <br /> Q = extraction flow rate (m?/hr) <br /> t= operational period(hrs) <br /> 01 October 2000 to 09 October 2000 <br />' using C = 120 gg/1= 120 x 10-4 kg/m? <br /> Q = 207 cfm(approximate) = 340 m3/hr <br />' t = 216 hrs (total of known operation) <br /> 1 20x 10-4 kg/m? • 340 m?/hr • 216 hrs = 9 0 kg of gasoline <br />' • 2 2046 lbs/kg = 20 lbs of gasoline <br /> 90kg g <br /> tto convert lbs of gasoline to gallons of gasoline, use 0 16 gal/lbs <br />' lbs of gal • 0 16 gal/lbs = 3 2 gallons of gasoline <br />' 09 October 2000 to 30 November 2000 <br /> using C = 77 gg/1= 7 7 x 10.5 kg/In? <br /> Q = 207 cfm(approximate) = 340 rn r <br />' t = 1248 hrs (total of known operation) <br /> 7 7 x 10-5 kg/m? • 340 m?/hr • 1248 hrs = 33 kg of gasoline <br /> 33 kg • 2 2046 lbs/kg = 72 8 lbs of gasoline <br /> to convert lbs of gasoline to gallons of gasoline, use 0 16 gal/lbs <br />' lbs of gal • 0 16 gal/lbs = 12 gallons of gasoline <br /> 30 November 2000 to 11 December 2000 <br /> using C = 250 gg/l= 2 50 x 10-4 kg/m? <br />' Q = 207 cfrn(approximate) = 340 m3/hr <br /> t = 264 hrs {total of known operation} <br />' Advanced GeuEnv►ronmenta4 Inc <br />