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N j APPENDIX D DISSOLVED TPH -g VOLUME ASSUMPTIONS AND CALCULATIONS PAGE 4 OF 8 V1,000 = (21 ,210 gallons )(3 .79 //gal) = 80,387 / water Multiplying the volume of water by the average dissolved TPH-g concentration (5, 500 pg/1 ) yields the approximate mass of TPH-g in groundwater water: V,1 ,000 = (80 ,387 /) (0 .0000055 kg/0 = 0 .4 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 .4 kg TPH-g )(2 .205 Ib/kg) = 1 .0 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 ) = 1 .0 lbs./6 . 25 lbs/gal = 0.2 gallons of dissolved TPH -g 4) For the estimated 100 pg/I middle contour interval , A100 = 1 , 346 ft' , h = 25 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 TPH-g concentration is estimated to be 433 ug/I (equivalent to 0. 000000433 kg/1 ) . Area : A1GG = 11346 ft2 The volume is given by: V10o = VIGO -V1,oaox h = 171458 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: VIGO _ (0 .30)(17 ,458 ft) = 53237 ft3 One ft3 is equal to 7 .48 gallons , so the volume of the water in the ellipse is given by: V100 = (5237 ft3)(7 .48 gal/ft3) = 39, 176 gallons One gallon of water is equivalent to 3. 79 liters (n, so the volume water is given by: V1oo = (39, 176 gallons )(3 .79 //gal) = 148 ,476 / water Multiplying the volume of water by the average dissolved TPH-g concentration (433 pg/1) yields the approximate mass of TPH-g in groundwater: V 100 = ( 148 ,476 � (0 .000000433 kg/1) = 0 .06 kg TPH -g