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Appendix C - Calculation of Dissolved MTBE Volume-Mass ' Assumptions: • The distribution of dissolved MTBE concentrations on the site can be approximated by an ' ellipse (See Figure 1); and The effective porosity of soil at the site is estimated to be 30%. ' • Data collected from the June 2002 ground water sampling event are representative of existing site conditions. � The area of C ellipse can�be card by the formula: � 3 D 3' (-j Ar = (long radius, a)(short radius, b)( } -' ' For the estimated ellipse, a= 180 ft, b= 80 ft, and thickness c = 30 ft The average MTBE concentration for the ellipse, between MW-1 and MW-3, bounded by wells MW-2, MW-6 and MW-5, is estimated to be 20.86 µg/l, this concentration is equivalent to 0.0000002 g/ml. A ml is very nearly equivalent to a cubic centimeter of water, which by definition ' equals one gram, this concentration is nearly a unitless number. '• The area of the ellipse is given by: Ae= 180(80)(0.7854) = 12,016 ftZ tThe volume of the ellipse is given by: ' Ve =Ae x c =12,016 ft2 x 30 ft = 360,498 ft' Water occupies the porosity in the soil, which is estimated to be 30%of the soil volume, so the total ' volume of water in the saturated portion of the ellipse is approximated by: Ve= (.30)(360,498 ft) = 108,149 ft3 One cubic foot ofwater contains 7.48 gallons,which weighs 8.3371bs/gal,so the volume of the water in the ellipse is given by: ' Ve = (108,149 ft)(7.48 gal/ft') = 808,958 gallons One gallon of water equals 3.78 5 liter of water, so the volume of the water expressed as liters in the ellipsoid is given by; V= (808,958 gallons) (3.785 liter/gallon) = 3,061,906 liters of water 1