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Mass-Volume Calculation of Dissolved MTBE - 02 February to 08 March 2001 <br /> FLAG CITY CHEVRON - 6421 Capitol Avenue, Lodi, CA <br /> Assumptions <br />' • A total volume of 1.,005,519 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 February 2001 and 08 March 2001 (based on <br /> effluent flow meter observations between 02 February and 08 March 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 48allons conversely, gallon of water is equivalent to <br /> g Y, one g q <br /> 0 1337 cubic feet, so the volume of processed water is given by <br /> V=(1,005,519 gallons)(0 1337 ft/gal) = 134,438 ft'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,yater= (1,005,519 gallons)(8 337 lbs /gal) =8,383,012 lbs. processed water <br /> Multiplying the mass of the processed water by the average MTBE concentration (02-02-01= <br /> 150,ug/L and 03-08-01 = 101 ,ug/L, avg = 126 µg/L) yields the approximate mass of MTBE <br /> removed from processed water <br /> MMrsE= (8,383,012 lbs)(0 000000126) = 1 06 lbs MTBE <br /> A converstlon factor of 0 4536 kg/lb can be used to convert pounds of MTBE to kilograms of <br /> MTBE <br /> MSE = 106 lbs MTBE (0 4536 kg/lb) =481 g of MTBE <br /> r <br /> Advanced GeoEnvironmentaI,Inc <br /> I <br />