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V APPENDIX E DISSOLVED TPH VOLUME ASSUMPTIONS AND CALCULATIONS PAGE 2OF3 Water occupies the porosity in,the soil, which is estimated to be 35 % of the soil volume, so the total volume of water in the saturated portion of the cylinder is,approximated by: V1;000= (0.35)(8,960:0-2,688 ft3 One ft3 is equal to 7.48 gallons,ao the volume of the water in the cylinder is given by: V1.000=(2,688 ft3)(7.48 gal/ft3) =20,106 gallons One gallon of water,weighs 8.337 lbs/gala so the mass-of the water in the cylinder is given by: M1.000=(20„106 gallons)(8.337 Ib/gal) = 167,625 lbs. Multiplying M1,000 by the hydrocarbon concentration yields the approximate mass of dissolved hydrocarbons in the saturated-portion of the cylinder: M1,000- (167;625Ibs)(0.00000.105)= 0.17 lbs of dissolved TPH-g Dividing M,000 by the weight of onegallon of gasoline, 6.25 lbs/gallon, will yield the approximate volume of TPH-g dissolved into the groundwater of the inner contour interval Vi,000`= (M,.000)/6.25lb/gal)= O.17/6.25 lbs/gal =0.:027 gallons of.dissolved TPH-g 2) For the estimated 100 llg/I'.contour interval, A,.,= 1;198 ftZ, h =20 ft(height estimated and averaged from cross sections) Utilizing the 1,000 dig/1 contour line and the 100 pg/f,contour line (Figure XX), the average TPH=g concentration is estimated to be,550 ug/1 (equivalent to 0.00000055 g/ml). The area is given by: The volume is given by: V100=A100 x.h-V1,000=(1,198 ft2 x.20 ft)=8,960 ft3= 15,000 ft3 Water occupies the porosity in the soil, which is estimated,to be 35% of the.soil volume, so the total volume of water in'the saturated portion of the cylinder is approximated by: V10o= (0.35)(15;000.ft3) 5,250 ft3 One ft3 is equal to 7.48 gallons, so the volume of the water in the cylinder is given by: V1,0 (5,250 ft3)(7.48 gal/ft3) =39,210 gallons One gallon of water weighs 8:337 lbs/gal, so the mass of the water in the cylinder is given by: M100 (39;270 gallons)(8.337 lb/gal),= 327,393 lbs Multiplying M100 by the hydrocarbon concentration yields the approximate mass of dissolved hydrocarbons: M100=(327,393 Ibs)(0.00000055) =0.18 lbs of dissolved TPH-g. v