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Mass-V6lurrie-Calculation.of Dissolved (continued) <br /> -The volume of the outer elliptical column can be calculat d 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 coiumn.,is given by:. <br /> A,,,—nab = n(120ft)(60ft) 22,619"ft2 <br /> The volume of the outer elliptical columnis-given by: _ <br /> VOe ((A. ' c} =V1e= (22;619 ftz x 20 ft)--.175,92.9 ft' =276,451 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 outer ellipsoid is approximated by: <br /> Voe = (0.35)(276,451 ft') =96,757 ft3_ <br /> One cubic foot of water.contains 7.48.gallons,'so-the-v lume of the"water in the outer ellipsoid is ; <br /> given by: ;• . <br /> VOe = (96,.757 ft3)(7.48 gal/ft') =,723,74 gallons <br /> 'One gallon of water weighs 8.337.lbs/gal at 60°F,.so'th mass of the water in the-outer ellipsoid is - <br /> given by: <br /> Mce– (723,748 gallons)(8.337 lb/gal)" ,033,893 lbs <br /> Multiplying MOe by the hydrocarbon concentration-yield the approximateRrt'ass of dissolved MTBE" <br /> in the saturated-portion of the"outer ellipsoid:" <br /> MOeg– (6,033,893 lbs)(0.0040001) = 0.60 lbs MTBE" <br /> Dividing Mo,' by the weight'of one gallon of"gasoli e;6.17 lbs/gallon,:will yield the volume of <br /> gasoline dissolved into the ground.water of the outer el ipsoi&.. <br /> VieMTBE__(M' MTBE)/6."171b/gal)=0.6 lbs/6.17 lbs/gal="0.10.gallons of MTBE. <br /> Combining the volumes of MTBE iri the"inner and out r elliptical columns yields a grand total of <br /> 1,184,330 gallons of impacted ground water and an- stimated 0.72 gallons of.MTBE dissolved <br /> in ground water." <br /> .Advanced GeoEnviron]nental,Inc. " <br />