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A <br /> Mass-Volume Calculation of Dissolved MTBE - 09 April to 03 May 2001 <br /> P Y <br /> IFLAG CITY CHEVRON- 6421 Capitol Avenue, Lodi, CA <br /> Assumptions <br /> • A total volume of 543,352 gallons was extracted from wells EW-2, EW-4, EW-5, EW-6, <br /> EW-7,EW-8,EW-9 and EW-10 between 09 April 2001 and 03 May 2001 (based on effluent <br /> flow meter observations between 09 April and 03 May 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 equlvalernt to <br /> I0 1337 cubic feet, so the volume of processed water is given by <br /> V=(543,352 gallons)(0 1337 fl/gal) =72,646 fe processed water <br /> IOne gallon of water weighs 8 337 lbs/gal at 60°F, so the mass of processed water is given by <br /> M,,,tpr=(543,352 gallons)(8 337 lbs/gal) =4,529,926 lbs. processed water <br /> Multiplying the mass of the processed water by the average MTBE concentration (04-09-01= <br /> 100yg/L and 05-03-01 = 140 yg/L, avg = 120 ,ug/L) yields the approximate mass of MTBE <br /> removed from processed water <br />' MMTBE _ (4,529,926 lbs)(0 000000120) =0.54 lbs MTBE <br /> A converstion factor of 0 4536 kg/lb can be used to convert pounds of MTBE to lalograms of <br /> MTBE <br />' MMraE= 0 54 lbs MTBE (0 4536 kg/lb) =245 g of MTBE <br /> I ' <br /> Advanced GeoEnwronmental,Inc. <br />