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APPENDIX F <br /> DISSOLVED TPH-g VOLUME ASSUMPTIONS AND CALCULATIONS <br /> PAGE 2 OF 7 <br /> Multiplying Mi00000 by the hydrocarbon concentration yields the approximate mass of dissolved <br /> hydrocarbons in the saturated portion of the ellipse: <br /> Mi00000 (501,505 lbs)(0.000136667)= 68.5 lbs of dissolved TPH-g <br /> Dividing M,00000 by the weight of one gallon of gasoline,6.25 lbs/gallon,will yield the approximate volume <br /> of TPH-g dissolved into the ground water of the inner ellipsoid: <br /> Vi00000=(Mi00000)/6.25lb/ al)=68.5/6.25lbs/gal=11 gallons of dissolved TPH- <br /> g d t <br /> 2) For the estimated 10,000 µg/1 ellipse cylinder, a=38 ft,b =31 ft, and thickness c=20 ft <br /> tq,6-6-b <br /> Utilizing TPH-g data from well MW-1, the 100,000 µg/1 contour line and the 10,001 contour line <br /> (Figure 9),thej�yerage pre-remediation TPH-g concentration is estimated to be 59,667 ug/1(equivalent to <br /> 0.000059667 g/ml).A ml is very nearly equivalent to a cubic centimeter ofwater,which by definition equals,_ <br /> one gram,this concentration is essentially unitless. 1 <br /> The area of the ellipse is given by: Ai0000=a(38)(3 1) =3,701 ftZ A100000=2,695.7 ft' <br /> The volume of the ellipse cylinder is given by: Vi0000=A10000 x c=2695.7 ftZ x 20 ft =53,914 W = <br /> Water occupies the porosity in the soil,which is estimated to be 40%of the soil volume,so the total volume <br /> of water in the saturated portion of the ellipse is approximated by: V,0000=(0.40)(53,914 ft')=21,566 W <br /> One ft is equal to 7.48 gallons, so the volume of the water in the ellipse is given by: <br /> V10000=(21,566 W)(7.48 gal/ft')= 161,314 gallons <br /> One gallon of water weighs 8.337 lbs/gal, so the mass of the water in the ellipse is given by: <br /> Ml0000= (161,314 gallons)(8.337 lb/gal)= 1,344,875 lbs -� <br /> Multiplying M,0000 by the hydrocarbon concentration yields the approximate mass ofdissolvedhydrocarbons: <br /> Mi0000 ( 1,344,875 lbs)(0.000059667) =80.2 lbs of dissolved TPH-g <br /> Dividing Mi0000 by the weight of one gallon of gasoline,6.251bs/gallon,will yield the approximate volume <br /> of TPH-g dissolved into the ground water of the inner-middle ellipsoid: <br /> V10000=(nh0000)/6.251b/gal)=80.2lbs/6.25 lbs/gal=12.8 gallons of dissolved TPH-g <br />