Laserfiche WebLink
1 <br /> MASS-VOLUME CALCULATIONS OF EXTRACTED GASOLINE HYDROCARBONS <br /> 5 <br />' Assumptions <br /> • Utilizing analytical data collected from air samples (Table 3)and from field measurements collected during <br />' the SVE pilot test(Table 4),the volume of extracted gasoline hydrocarbons at the site can be approximated <br /> The hydrocarbon mass removed during the operating period can be calculated using the following equation <br />' M=C• Q - t <br /> where M=cumulative mass recovered(kg) <br /> C=vapor concentration(kg/m3) <br />' Q=extraction flow rate(m3/hr) <br /> t=operational period <br />' Mass-volume calculations were separated into the following three intervals <br /> 1) 15 February to 28 February 2000 <br />' 2) 29 February to 08 March 2000 <br /> 3) 08 March to 29 March 2000 <br />' The average inlet TPH-g concentration and average flow rates during each time interval was utilized for calculating the <br /> approximate mass/volume of extracted gasoline hydrocarbons <br /> 15 February to 28 February 2000 <br /> I 1) The average analytical TPH-g concentration for this tune interval was determined to be 490,ug/1 The average <br /> flow rate was calculated at 0 19 inches of water or approximately 24 cubic feet per minute, the operational <br /> period was equal to 336 hours of running time <br /> C = 490,Lcg/l=0 000490 kg/m3 <br /> Q = 0 19 inches water = 24 ft3/min <br /> t=336 hours <br /> Converting Q to m3/hour yields <br /> Q =(24 ft3/min) -(60 min/hour) -(0 0283168m3/ft3)=40 78 M3/hour <br /> IM=C - Q - t=(0 000490 kg/m3) -(40 78 m3/hour)- (336 hours) =6 7 kg of gasoline <br /> IConverting kg of gasoline to pounds of gasoline yields <br /> 6 7 kg of gas - 2 2046 lbs of gas/kg of gas= 14 8 lbs of gasoline <br /> Converting pounds of gasoline to gallons of gasoline yields <br /> 14 8 lbs of gas - 0 16 gal of gas/lbs of gas = 2.4 gallons of gasoline <br /> I <br />