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1 <br /> . Mass-Volume Calculation of Dissolved MTBE - 06 September 2001 to 01 October 2001 <br /> FLAG CITY CHEVRON- 6421 Capitol Avenue, Lodi, CA <br /> Assumptions• <br />' • A total volume of 730,219 gallons was extracted from wells EW-2, EW-4, EW-5, EW-6, <br /> EW-7, EW-8, EW-9 and EW-10 between 06 September and 01 October 2001 (based on <br />' effluent flow meter observations between 06 September and October 01 2001) <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= (730,219 gallons)(0 1337 ft/gal)=97,630 ft'processed water <br /> Oneallon of water wen ° <br /> g weighs 8 3 371bs/gal at 60 F, so the mass of processed water Is given by <br />' Mw.,,=(730,219 gallons)(8 337 lbs/gal)= 6,087,836 lbs. processed water <br />' Multiplying the mass of the processed water by the average MTBE concentration (09-06-01=7µg/1) <br /> yields the approximate mass of MTBE removed from processed water <br /> MmmE=(6,087,836 lbs)(0 000000007) = 0.04 lbs MTBE <br /> A converstion factor of 0 4536 kg/lb can be used to convert pounds of MTBE to grams of MTBE <br /> 1 MSE= 0 04 lbs MTBE (0 4536 kg/lb)= 19 g of MTBE <br /> 1 ' <br /> I <br /> IAdvanced GeoEnv:ronmenta4 Inc- <br />