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APPENDIX D DISSOLVED TPH -g VOLUME ASSUMPTIONS AND CALCULATIONS PAGE 5 OF 8 A conversion factor of 2 .205 Ib/kg . can be used to convert kilograms of TPH-g to pounds of TPH-g : MTPH_9 = (0 . 06 kg TPH -g )(2 . 205 Ib/kg ) = 0 . 1 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 : Vt00 = (M100)/6 . 25lb/gal ) = 0 . 1 lbs ./6 . 25 lbs/gal = 0 . 02 gallons of dissolved TPH -g 50 to 90 feet bsq groundwater plume mass calculation : 1 ) For the estimated 100 , 000 pg/l inner contour interval ; A100,000 = 129 ft2 and h = 23 ft (average thickness) Utilizing TPH-g data from well MW-8 ( 130 , 000 pg/1 ) and the 100 , 000 Ng/l contour line ( Figure 13), the average dissolved TPH-g concentration is estimated to be 115 , 000 ug/I (equivalent to 0 . 000115 kilograms per liter) . The volume is given by: V100.000 = A100,000 x h = 129 ftz x 23 ft = 2 , 967 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 is approximated by : V1000000 = (0 . 30 ) (21967 ft) = 890 ft3 One ft3 is equal to 7 .48 gallons , so the volume of the water is given by: V100,000 = (890 ft3)(7 .48 gal/ft3) = 6 ,657 gallons One gallon of water is equivalent to 3 . 79 liters (1) , so the volume water is given by: V100.000 (6 ,657 gallons )(3 .79 //gal) = 25 ,234 / water Multiplying the volume of water by the average dissolved TPH-g concentration (263 , 000 pg/1 ) yields the approximate mass of TPH-g in ground water: V100,000 = (25 , 234 /) (0 . 000115 kg//) = 2 . 9 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_0 = (2 . 9 kg TPH-g )(2 . 205 Ib/kg ) = 6.4 lbs . TPH -g