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Mass-Volume Calculation of Dissolved TPH-g (continued) <br /> SIERRA LUMBER- 375 West Hazelton Avenue, Stockton, CA <br />! <br /> One gallon of water weighs 8 337 lbs/gal at 60°F, so the mass of the water in the Inner ellipsoid is <br />! given by <br /> M,, = (77,111 gallons)(8 337 lb/gal) = 642,875 lbs <br /> Multiplying M,, by the hydrocarbon concentrationyl elds the approximate mass of dissolved <br /> hydrocarbons In the saturated portion of the Inner ellipsoid <br /> M,eg= (642,875 lbs)(0 000040) — <br /> 25 7 lbs gasoline <br /> Dividing M., by the weight of one gallon of gasoline, 6 17 lbs/gallon, will yield the volume of <br /> gasoline dissolved Into the ground water of the Inner ellipsoid <br /> Vleg = (M1,,d/6 17lb/gal) = 25 7 lbs/6 17 lbs/gal =4.2 gallons of gasoline <br /> The volume of the outer elliptical column can be calculated by the method shown above The volume <br /> of the Inner elliptical column should be subtracted from the overall volume of the outer column <br /> The area of the outer elliptical column Is given by <br /> IAOe----tab = 7c(45ft)(27MI) = 3,888 ft' <br /> The volume of the outer elliptical column is given by <br /> I <br /> Vce = (A,,, c) - V1, = (3,888 ft' x 20 ft) - (2/3)29,454 ft' = 58,119 ft' <br /> Water occupies the porosity In the soil, which Is estimated to be 35% of the soil volume, so the total <br /> volume of water In the saturated portion of the Inner ellipsoid Is approximated by <br /> Vpe = ( 35)(58,119 ft') =20,342 ft' <br /> I One cubic foot of water contains 7 48 gallons, so the volume of the water in the Inner ellipsoid is <br /> given by <br /> VO, = (20,342 ft3)(7 48 gal/ft3) = 152,157 gallons <br /> One gallon of water weighs 8 337 lbs/gal at 60'F, so the mass of the water in the Inner ellipsoid Is <br /> given by <br /> IM0, = (152,157 gallons)(8 337 lb/gal) = 1,268,533 lbs <br /> Advanced GmEnvironmental,Inc <br /> I <br />