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} APPENDIX J DISSOLVED TPH VOLUME ASSUMPTIONS AND CALCULATIONS PAGE 2 OF 5 Multiplying M10000 by the hydrocarbon concentration yields the approximate mass of dissolved hydrocarbons in the saturated portion of the ellipse: Mi0000 (117,552 lbs)(0.000062333)= 7.3 lbs of dissolved TPH Dividing M10000 by the weight of one gallon of gasoline,6.251bs/gallon,will yield the approximate volume of TPH dissolved into the ground water of the inner ellipsoid: V,0000= (MI0000)/6.25lb/gal)= 7.3 /6.25 lbs/gal= 1.17 gallons of dissolved TPH I 2) For the estimated 1,000 gg/l ellipse cylinder, a= 39.5 ft,b= 17 ft, and thickness c = 15 ft Utilizing TPH data from well MW-7, the 10,000 µg/l contour line and the 1,000 gg/1 contour line(Figure 12),the average pre-excavation TPH concentration is estimated to be 5,566 ug/l(equivalentto 0.000005566 g/ml).A ml is very nearly equivalent to a cubic centimeter of water,which by definition equals one gram, this concentration is essentially unitless. The area of the ellipse is given by: A1000 = Tc (39.5)(17)= 2110 ft' -A,O000= 1,796 ft' The volume of the ellipse cylinder is given by: V1000=A1000 x c= 1,796 ft' x 15 ft =26,940 ft3 Water occupies the porosity in the soil,which is estimated to be 40%of the soil volume,so the total volume of water in the saturated portion of the ellipse is approximated by: V1000 = (0.40)(26,940 ft)= 10,776 ft' One ft' is equal to 7.48 gallons, so the volume of the water in the ellipse is given by: V1000= (10,776 ft3)(7.48 gal/ft') = 80,605 gallons One gallon of water weighs 8.337 lbs/gal, so the mass of the water in the ellipse is given by: MiOoo= (80,605 gallons)(8.337 lb/gal) =672,000 lbs Multiplying M,000 by the hydrocarbon concentration yields the approximate mass of dissolved hydrocarbons: M,000 ( 672,000 lbs)(0.000005566)=3.74 lbs of dissolved TPH Dividing M1000o by the weight of one gallon of gasoline, 6.251bs/gallon,will yield the approximate volume of TPH dissolved into the ground water of the inner ellipsoid: V,000= (Mi0000)/6.25lb/gal)= 3.74lbs/6.25 lbs/gal= 0.6 gallons of dissolved TPH i