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APPENDIX D DISSOLVED TPH -g VOLUME ASSUMPTIONS AND CALCULATIONS PAGE 7 OF 8 Utilizing boring CPT4 (5 , 900 pg/1 ) , CPT7 (1 , 800 pg/I) the 10, 000 pg/I contour line and the 1 ,000 pg/I , the average TPH-g concentration is estimated to be 4 ,675 ug/I (equivalent to 0 . 000004675 kg/1). Area : A1000 = 33434 ft2 The volume is given by: Vl000 = V1,0041gpo x In = 41 ,092 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: V1,00 _ (0.30 )(41 ,092 ft) = 12 ,328 ft3 One ft3 is equal to 7 .48 gallons , so the volume of the water in the ellipse is given by: V11000 = ( 127328 ft3)(7.48 gal/ft) = 92 ,210 gallons One gallon of water is equivalent to 3. 79 liters (/), so the volume water is given by: V1p00 = (92 ,210 gallons )(3 .79 //gal) = 349 ,4771 water Multiplying the volume of water by the average dissolved TPH-g concentration (4, 675 pg/1) yields the approximate mass of TPH-g in groundwater water: V111000 = (349 ,477 1) (0 .000004675 kg//) = 1 .6 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_9 = ( 1 . 6 kg TPH-g)(2 . 205 Ib/kg ) = 3.6 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 TPWg dissolved into the groundwater of the middle contour interval : V1 ,000 = (M1,000)/6 .25lb/gal) = 3 .6 lbs ./6 .25 lbs/gal = 0.6 gallons of dissolved TPH -g 4) For the estimated 100 pg/I middle contour interval , A100 = 6, 310 ftp , h = 30 ft (average thickness) Utilizing TPH-g data from well MW-6 (200 pg/1), the 1 , 000 pg/I contour line and the 100 pg/I (Figure 13), the average TPWg concentration is estimated to be 433 ug/I (equivalent to 0 . 000000433 kg/1) . Area : A100 = 63310ft2 The volume is given by: V100 = V100 N1,000 x In = 148 ,208 ft3