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. Mass-Volume Calculation of Dissolved MTBE - 02 July to 01 August 2001 <br /> FLAG CI'T'Y CHEVRON - 6421 Capitol Avenue, Lodi, CA <br /> Assumptions <br />' • A total volume of 753,053 gallons was extracted from wells EW-2, EW-4, EW-5, EW-6, <br /> EW-7, EW-8, EW-9 and EW-10 between 02 July 2001 and 01 August 2001 (based on <br />' effluent flow meter observations between 02 July and 01 August 200 1) <br /> • Data collected from the pump and treat influent and extraction well-head grab ground water <br />' samples were representative of general extracted ground water conditions <br /> One cubic foot of water contains 7 48 gallons, conversely, one gallon of water is equivalent to <br /> 0 1337 cubic feet, so the volume of processed water is given by <br /> V=(753,053 gallons)(0 1337 ft/gal) =100,683 ft'processed water <br />' Oneallon of water weighs ° <br /> g ghs 8 337 lbs/gal at 60 F, so the mass of processed water is given by <br />' M..wl= (753,053 gallons)(8 337 lbs/gaI) =6,278,203 lbs. processed water <br /> Multiplying the mass of the processed water by the average MTBE concentration(07-02-01=28,ug/L <br /> and 08-01-01 = 19,ug/L, avg =23 5 µg/L) yields the approximate mass of MTBE removed from <br /> processed water <br /> MmrraE = (6,278,203 lbs)(0 0000000235) =0.14 lbs MTBE <br />' A converstion factor of 0 4536 kg/lb. can be used to convert pounds of MTBE to kilograms of <br /> MTBE <br /> MmTBE = 0.14 lbs MTBE (0 4536 kg/lb)=66 g of MTBE <br /> IAdvanced GeoEnvironmental,Inc. <br />