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APPENDIX F <br /> DISSOLVED TPH-g VOLUME ASSUMPTIONS AND CALCULATIONS <br /> PAGE 6 OF 7 <br /> Multiplying M,000 Post by the hydrocarbon concentration yields the approximate mass of dissolved <br /> hydrocarbons in the saturated portion of the ellipse cylinder: <br /> Mt000po,t(39,409 lbs)(0.000001950)=0.08 lbs TPH-g <br /> Dividing M,000 Post by the weight of one gallon of gasoline,6.25 lbs/gallon,will yield the volume of TPH-g <br /> dissolved into the ground water of the ellipse cylinder: <br /> V t000post=(M1000P0s)/6.25lb/gal)=0.08 lbs/6.25 lbs/gal=0.013 gallons of dissolved TPH-g <br /> 2) For the estimated 50 (ND) µg/l ellipse cylinder volume: a= 10 ft, b= 9 ft, and thickness c=20 ft <br /> Utilizing the 1,000 µg/1 contour line and the 50 µg/1 contour line(Figure 10), the average post excavation <br /> TPH-g concentration for the ellipse is estimated to be 525 ug/l. This concentration is equivalent to <br /> 0.000000525 grams per ml. A ml is very nearly equivalent to a cubic centimeter of water, which by <br /> definition equals one gram, this concentration is nearly a unitless number. <br /> The area of the ellipse is given by: Aso Pos,=rc(10 ft)(9 ft)- At000 P.st=204 ftZ <br /> The volume of the ellipse is given by: V50 post=A50 Pa,,x c=204 ftZ x 20 ft =4,080 ft' <br /> Water occupies the porosity in the soil,which is estimated at this site to be 40%of the soil volume, so the <br /> total volume of water in the saturated portion of the ellipse cylinder is approximated by: <br /> V50 punt=(0.40)(4,080 ft') = 1,632 ft' <br /> One cubic foot of water contains 7.48 gallons,which weighs 8.337 lbs/gal,so the volume of the water in the <br /> ellipse cylinder is given by: <br /> V50post=(1,632 W)(7.48 gal/fe)= 12,207 gallons <br /> One gallon of water weighs 8.337 lbs/gal, so the mass of the water in the ellipse cylinder is given by: <br /> M50pW=(12,207 gallons)(8.337 lb/gal)= 101,770 lbs <br />