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Mass-Volume Calculation of Dissolved MTBE - 12 December 2001 to 03 January 2002 <br /> ` FLAG CITY CHEVRON - 6421 Capitol Avenue, Lodi, CA <br /> Assumptions <br /> • A total volume of 538,902 gallons was extracted from wells EW-4, EW-5, EW-9 and EW-10 <br /> between 12 December 2001 and 03 January 2002 (based on effluent flow meter <br /> observations) <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=(538,902 gallons)(0 1337 ft/gal) = 72,051 W 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 /> M,a,e�= (538,902 gallons)(8 337 lbs/gal) =4,492,826 lbs. processed water <br /> Multiplying the mass of the processed water by the average MTBE concentration(12-13-01=20µg/L <br /> and 01-03-02 = 41 ug/L, avg = 30 5 gg/L) yields the approximate mass of MTBE removed from <br /> processed water , <br /> MMTBE =(4,492,826 lbs)(0 0000000305) = 0.1.3 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 13 lbs MTBE (0 4536 kg/lb) = 59 g of MTBE <br /> Advanced GeoEnvironmental,Inc <br />