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<br /> f•
<br /> 1 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 ellipse are approximated by
<br /> 1 Vel =245,040 ftp ( 35) =85,764 ft'
<br /> Ve2=442,960 ft3 ( 35) = 155,036 ft'
<br /> Ve3 = 524,640 ft; ( 35) = 183,624 fl
<br /> Ve4=69,115 113 (35)=24,190 fe
<br /> 1 One cubic foot of water contains 7 48 gallons, so the volume of the water in the ellipses are given
<br /> by
<br /> Vel = 85,764 ft' (7 48 gallW) = 641,514 gallons
<br /> Ve2= 15 5,03 6 ft3 (7 48 gal/ff) = 1,159,669 gallons
<br /> 1 Ve3= 183,624 ft3 (7 48 gal/ft3) = 1,373,507 gallons
<br /> Ve4=24,190 W (7 48 gal/W)= 180,941 gallons
<br />' One gallon of contains 3 785 liter so the volume of the water to the ellipses are given by
<br /> Vel = 641,514 gallons (3 785 liter/gal) =2,428,130 liters
<br /> Ve2= 1,159,669 gallons (3 785 liter/gal) =4,389,347 liters
<br /> Ve3 = 1,373,507 gallons(3 785 liter/gal) =5,198,723 liters
<br /> Ve4 = 180,941 gallons (3 785 liter/gal) =684,861 liters
<br /> li
<br /> 1 Multiplying the volumes of the ellipse by the current(approximated)hydrocarbon concentrations
<br /> yields the approximate mass of dissolved hydrocarbons in the saturated portion of the ellipse
<br /> 1 Mel =2,428,130 liters (1 0 µg/1)=3,428,130 µg MTBE
<br /> MMTBE as grams 2.42 g MTBE
<br /> MMTBE as kilograms 0.0024 kg MTBE
<br /> 1 Met=4,389,347 liters (5 0 gg/l)=21,946,735 pg MTBE
<br /> MMTBE as grams 21.9 g MTBE
<br /> MM.,.BE as kilograms 0.0219 kg MTBE
<br /> 1 Me3 =5,198,723 liters (10 µg/1)= 51,987,230 µg MTBE
<br /> MMTBE as grams 5198 g MTBE
<br /> MMTBE as kilograms 0.05198 kg MTBE
<br /> 1 Me4=684,861 liters(100 gg/l) =68,486,100 µg MTBE
<br /> MMTBE as grams 68 g MTBE
<br /> M,,,,,,, as Uograms 0.068 kg MTBE
<br /> 1 The total of the dissolved mass equals 0 14 kg of MTBE or 0 14 kg(2 205 lb/kg)
<br /> 0.3 lbs of MTBE
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<br /> .\ Advanced GeoEnviranmental,Inc
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