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f <br /> Mass-Volume Calculations <br /> AGE-NC Project No 98-0521 <br /> ' Page 2 of 3 <br /> ' 06 August to 27 September <br /> One gallon of water is equivalent to 3 79 liters (L), so the volume of processed water is given by <br /> V= (65,100 gallons)(3 79 L/gal)=246,729 L processed water <br /> 1 <br /> Dissolved MTBE <br /> ' f the processed water b the average of the 06 August and 15 September <br /> Multiplying the volume o p y g g p <br /> MTBE concentrations (119 �tg/1) yields the approximate mass of MTBE removed from processed <br /> water <br /> MMTBE_ (246,729 L)(0 000000119 kg/L) = 0.0294 kg MTBE <br /> A conversion factor of 2 205 lbs/kg can be used to convert kilograms of MTBE to pounds of MTBE <br /> MMTBH = (O 0294 kg MTBE)(2 205 lbs/kg) = 0.0647 lbs MTBE <br /> 1 <br /> Dissolved TPH-g <br /> ' <br /> Multiplyingthe volume of the processed water b the average of the 06 August and 15 September <br /> p Y g g p <br /> ' TPH-g concentrations (325 gg/1) yields the approximate mass of TPH-g removed from processed <br /> water <br /> ' MTPH s= (246,729 L)(0 000000325 kg/L)= 0.0802 kg TPH-g <br /> A conversion factor of 2 205 lbs/kg can be used to convert kilograms of TPH-g to pounds of TPH-g <br /> MTM _ (0 0802 kg TPH-g)(2 205 lbs/kg) =0.1768 lbs TPH-g <br /> i <br /> ' Advanced GeoEnvuronmental,Inc <br />