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1 <br /> n <br /> G APPENDIX D <br /> DISSOLVED TPH-g VOLUME ASSUMPTIONS AND CALCULATIONS <br /> PAGE 8OF8 <br /> Water occupies the porosity in the soil , which is estimated to be 30% of the soil volume, so the <br /> total volume of water in the saturated portion of the ellipse is approximated by: <br /> Vtao = (0 .30 )(148,208 ft) = 44,462 ft3 <br /> One ft' is equal to 7 .48 gallons , so the volume of the water in the ellipse is given by: <br /> Vim _ (44,462 ft)(7 .48 gal/ft) = 332,578 gallons <br /> One gallon of water is equivalent to 3 . 79 liters (q, so the volume water is given by: <br /> Vim = (332,578 gallons)(3 .79 //gal) = 1,260,473 / water <br /> Multiplying the volume of water by the average dissolved TPH-g concentration (433 pg/1 ) yields <br /> the approximate mass of TPH-g in groundwater: <br /> V,1Qo = ( 1 ,260 ,473 1) (0 .000000433 kg//) = 0 .5 kg TPH-g <br /> A conversion factor of 2 .205 Ib/kg . can be used to convert kilograms of TPH-g to pounds of <br /> TPH-g : <br /> MTPH_9 = (0 . 5 kg TPH-g )(2 . 205 Ib/kg ) = 1 .2 lbs . TPH -g <br /> Dividing Mtoo by the weight of one gallon of TPH-g , 6 .25 lbs/gallon , will yield the approximate <br /> volume of TPH-g dissolved into the groundwaterof the middle contour interval : <br /> Vtoo = (M,00)/6. 25Ib/gal ) = 1 .2 lbs ./6.25 lbs/gal = 0.2 gallons of dissolved TPH-g <br /> Combining the total volume of dissolved TPH-g yields an estimated 49 pounds of TPH -g , <br /> equivalent to 7.8 gallons of TPH-g . <br />