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Mass-Volume Calculation of Dissolved MTBE - 03 May to 04 June 2001 <br /> Assumptions Assump <br /> FLAG CITY CHEVRON - 6421 Capitol Avenue,Lodi, CA <br />' <br />' • A total volume of 866,753 gallons was extracted from wells EW-2, EW-4, EW-5, EW-6, <br /> EW-7, EW-8,EW-9 and EW-10 between 03 May 2001 and 04 June 2001 (based on effluent <br />' flow meter observations between 03 May and 04 June 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 /> I <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= (866,753 gallons)(0 1337 fl/gal) = 11.5,885 ft3 processed water <br />' One gallon of water weighs 8 337 lbs/gal at 60°F, so the mass of processed water is given by <br />' MW,,,=(866,753 gallons)(8 337 lbs/gal) = 7,226,120 lbs. processed water <br /> Multiplying the mass of the processed water by the average MTBE concentration (05-03-01= <br /> 140µg/L and 06-04-01 =41 ,ug/L, avg = 91 jzg/L)yields the approximate mass of MTBE removed <br /> from processed water <br /> MMTBE _ (7,226,1201bs)(0 00000009 1)= 0.66 lbs MTBE <br /> I A converstion factor of 0 4536 kg/lb can be used to convert pounds of MTBE to kilograms of <br /> MTBE <br />' MMTBE= 0 661bs MTBE (0 4536 kg/lb) =299 g of MTBE <br />' Advanced GeoEnnronmental,Inc <br />