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81/2,1 iy <br /> APPENDIX F <br /> TOTAL PETROLEUM HYDROCARBON-GASOLINE IMPACTED GROUND WATER <br /> VOLUME ASSUMPTIONS AND CALCULATIONS <br /> Sierra Lumber(Case No. 1) <br /> 375 West Hazelton Avenue, Stockton, California <br /> Volume of TPH-g Impacted Ground Water Pre-Remediation: <br /> Assumptions: <br /> • The distribution of maximum dissolved hydrocarbon concentrations on the site can be approximated <br /> by an elliptical cylinder(Figure 12);the distribution can be separated into four contoured areas(i.e. <br /> 100,000 µg/1, 10,000 µg/1, 1,000 µg/1 and 50 µg/1). <br /> • The effective porosity of soil at the site is estimated to be 40%(i.e. silt, silty sand and fine sand). <br /> • Data collected from the ground water sampling events are representative of the maximum dissolved <br /> TPH-g site conditions prior to remediation activities. <br /> • The thickness of dissolved plume is estimated to be between the average depth to ground water at <br /> the site and the bottom of the most impacted monitoring well screen intervals(i.e. 20 feet). <br /> The area of an ellipse cylinder can be used to illustrate the shape of the dissolved TPH-g plume, and <br /> calculated by the formula: <br /> A,_(long radius, a)(short radius, b) (a) <br /> 1) For the estimated 100,000 µg/1 ellipse cylinder, a=20 ft,b= 16 ft, and thickness c=20 ft <br /> Utilizing TPH-g data from wells MW-2,MW-4A and the 100,000µg/1 contour line(Figure 9),the average <br /> pre-remediation TPH-g concentration is estimated to be 136,667 ug/l(equivalent to 0.000136667 grams per <br /> milliliter). A ml is very nearly equivalent to a cubic centimeter of water, which by definition equals one <br /> gram,this concentration is nearly a unitless number. <br /> The area of the ellipse is given by: Ai00000=n(20)(16)= 1,005.3 ft <br /> The volume of the ellipse cylinder is given by: V,00aoo=A10000 x c= 1,005.3 ft? x 20 ft =20,106 ft' <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,00000=(0.40)(20,106 ft)= 8,042 ft <br /> One W is equal to 7.48 gallons, so the volume of the water in the ellipse is given by: <br /> V100000=(8,042 ft)(7.48 gal/ft)= 60,154 gallons <br /> One gallon of water weighs 8.337 lbs/gal, so the mass of the water in the ellipse is given by: <br /> Mi00000=(60,154 gallons)(8.337 lb/gal)= 501,505 lbs <br />