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r � APPENDIX J DISSOLVED TPH VOLUME ASSUMPTIONS AND CALCULATIONS PAGE 3OF5 3) For the estimated 50 µg/1 ellipse cylinder, a--'50.5 ft,b =25 ft, and thickness c =.15 ft Utilizing the 1,000 gg/l contour line and the 50 gg/l contour line(Figure 12),the average TPH concentration is estimated to be 525 ug/l (equivalent to 0.000000525 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: A50= rc (50.5)(25) = 3,966 ft2 -A,0000= 1,856 ft2 The volume of the ellipse cylinder is given by: V50=Aso x c= 1,856 ft2 x 15 ft =27,840 ft' i 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: V50= (0.40)(27,840 ft3) = 11,136 ft3 -One ft' is equal to 7.48 gallons, so the volume of the water in the ellipse is given by: V50= (11,136 ft3)(7.48 gal/ft 3)=83,297 gallons One gallon of water weighs 8.337 lbs/gal,so the mass-of the water in the ellipse is given by: Mi000= (83,297 gallons)(8.337 lb/gal) = 694,447 lbs Multiplying M,000 by the hydrocarbon concentration yields the approximate mass of dissolved hydrocarbons: M50 ( 694,447 lbs)(0.000000525)=0.37 lbs of dissolved TPH Dividing M1000 by the weight of one gallon of gasoline,6.25 lbs/gallon,will yield the approximate volume of TPH dissolved into the ground water.of the inner ellipsoid: Vso= (Mi0000)/6.25lb/gal) = 0.37lbs/6.25 lbs/gal= 0.06 gallons of dissolved TPH Combining the total volume of dissolved TPH at the different concentrations, impacted PRE- EXCAVATION ground water yields a total of approximately 59,490 ft' containing an estimated 1.83 gallons of TPH, equivalent to 11.41 pounds of TPH. i i