WORK PLANS - Laserfiche WebLink

Home Browse Search

APPENDIX D DISSOLVED TPH-g VOLUME ASSUMPTIONS AND CALCULATIONS PAGE 2 OF 8 Utilizing TPH-g data from boring B1 (426 , 000 pg/l) and the 100 , 000 pg/l contour line (Figure 12), the average dissolved TPH-g concentration is estimated to be 263 , 000 ug/1 (equivalent to 0 . 000263 kilograms per liter) . The volume is given by: V100,000 = A100,000 x h = 44 ft2 x 10 ft = 440 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: V100,00o = (0 .30)(440 ft3) = 132 ft3 One ft3 is equal to 7 .48 gallons , so the volume of the water is given by: V100,000 = ( 132 ft3)(7.48 gal/ft3) = 987 gallons One gallon of water is equivalent to 3 . 79 liters (1) , so the volume water is given by: V1001000 (987 gallons)(3 .79 1/gal) = 3,742 / water Multiplying the volume of water by the average dissolved TPH-g concentration (263,000 pg/l) yields the approximate mass of TPH-g in ground water: V100,000 = (3,742 0 (0 .000263 kg/1) = 1 .0 kg TPH-g A conversion factor of 2. 205 Iblkg . can be used to convert kilograms of TPH-g to pounds of TPH-g : MTPH_9 = ( 1 .0 kg TPH-g )(2 .205 Ib/kg) = 2.2 lbs. TPH-g Dividing M700,000 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 inner contour interval : V1002000 = (M , 0a,a00)/6.25 Ib/gal) = 2.2 lbs./6 .25 lbs/gal = 0.35 gallons of dissolved TPH-g 2) For the estimated 10, 000 pg/1 middle contour interval , A10,000 = 337 ft' , h = 20 ft (average thickness) Utilizing TPH-g data from well MW- 1A (36 , 000 pg/I ), the 100 , 000 lag/I contour line and the 10 , 000 pg/I (Figure 12), the average TPWg concentration is estimated to be 48 ,667 ug/I (equivalent to 0 . 000048667 kg/1) . Area: V10,000 = A ,,,aoo x h = 337 ft2 x 20 ft = 63740 ftz The volume is given by: V10,000 = V10,aoo -V1001000 = 61740 ft3 - 440 ft3 = 61300 ft3