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APPENDIX C DISSOLVED TPH-g MASS-VOLUME ASSUMPTIONS AND CALCULATIONS PAGE 3 OF 4 One gallon of water is equivalent to 3.79 liters (n, so the volume water is given by: V1,000= (129,299 gallons)(3.79 //gal)=490,044 /water Multiplying the volume of water by the average dissolved TPH-g concentration (6,760 pg/1)yields the approximate mass of TPH-g in groundwater: V,1,000= (490,044/)(0.000006760 kg//) = 3.31 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: MTPHy = (3.31 kg TPH-g)(2.205 Ib/kg) = 7.30 lbs. TPH-g 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) = 7.30 lbs./6.25 lbs/gal = 1.16 gallons of dissolved TPH-g 3) For the estimated 100 pg/1 outer contour interval, A100= 5,480 ft2,' h =20 ft (average screen interval) Utilizing TPH-g data from MW-8A (240 pg/1) the 1,000 pg/I contour line and the 100 pg/1 , the average TPH-g concentration is estimated to be 446 pg/I (equivalent to 0.000000446 pg/1). Area: A,00= 5,480 ftZ The volume is given by: V100= V100 N1,000 x h =41,000 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)(41,000 ft3) = 12,300 ft3 One ft3 is equal to 7.48 gallons, so the volume of the water in the ellipse is given by: V100= (12,300 ft3)(7.48 gal/ft3) =92,004 gallons One gallon of water is equivalent to 3.79 liters (/), so the volume water is given by: V100= (92,004 gallons)(3.79 //gal)= 348,695/water Multiplying the volume of water by the average dissolved TPH-g concentration (446 pg/1) yields the approximate mass of TPH-g in groundwater water: V100= (348,695 /) (0.000000446 kg/1)= 0.15 kg TPH-g