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APPENDIX C PRE-REMEDIATION DISSOLVED TPH-g VOLUME ASSUMPTIONS AND CALCULATIONS PAGE 4 OF 5 Dividing M1,000 by the weight of one gallon of TPH-g, 6.25 lbs/gallon, will yield the approximate volume of TPH-g dissolved into the groundwater of the middle contour interval: V1,000 = (M1,000)/6.25lb/gal) = 3.5 lbs./6.25 lbs/gal = 0.56 gallons of dissolved TPH-g 4) For the estimated 100 pg/I second outer contour interval, A100= 9,589 ft', h = 20 ft (average thickness) Utilizing TPH-g data from well MW-11 (220 pg/1) the 1,000 pg/I contour line and the 100 pg/l (Figure 5),the average TPH-g concentration is estimated to be 440 pg/I (equivalent to 0.00000044 kg/1). Area: V100=A100 x h = 9,589 ft2 x 20 ft = 191,780 ft2 The volume is given by: V100=V100 41,000 = 191,780 ft3-82,940 ft3= 108,840 ft3 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: V100= (0.30)(108,840 ft3)= 32,652 ft3 One ft3 is equal to 7.48 gallons, so the volume of the water in the ellipse is given by: V100= (32,652 ft3)(7.48 gal/ft3) = 244,237 gallons One gallon of water is equivalent to 3.79 liters (/), so the volume water is given by: V100= (244,237 gallons)(3.79//gal)=925,658/water Multiplying the volume of water by the average dissolved TPH-g concentration (440 pg/1) yields the approximate mass of TPH-g in groundwater: V100= (925,658/) (0.00000044 kg//)=0.41 kg TPH-g A conversion factor of 2.205 Ib/kg.can be used to convert kilograms of TPH-g to pounds of TPH-g: MTPH-g = (0.41 kg TPH-g)(2.205 Ib/kg) = 0.90 lbs. TPH-g Dividing M100 by the weight of one gallon of TPH-g, 6.25 lbs/gallon, will yield the approximate volume of TPH-g dissolved into the groundwater of the middle contour interval: V100 = (M1oo)/6.25lb/gal) = 0.90 lbs./6.25 lbs/gal = 0.14 gallons of dissolved TPH-g Combining the total volume of dissolved TPH-g yields an estimated 47.05 pounds of TPH-g, equivalent to 7.53 gallons of TPH-g.