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• Volume for 15-35 foot interval <br /> V,= 20' x 20' x 20' x 5 squares = 40.000'3 = 1,480 yds 3 <br /> 27 cu-ft/yd 27 <br /> 7udgnng from Figures 10 and 11, approximately half this volume was removed by overexcavation, <br /> leaving an estimated 750 yds3 m place <br /> Volume for the 35-45 foot interval <br /> Vx= 20 x 20 x 10 x 14 squares = 56,000 ft3= 2,075 yds3 <br /> 27 27 <br /> Volume for the 45-50 foot interval <br /> V3=20' x 20' x 10' x 13 squares= 52,000 ft3 = 1,925 yds3 <br /> 27 27 <br /> Volume for the 50-55 foot interval <br /> V3 =20' x 20' x 10' x 16 squares = 64,00._0 ft' = 2,370 yd2 <br /> 27 27 <br /> • Vtw = V, + V2 + V3 +V4= 750+2,075 +1,925+2,370= 7,120 yds3 <br /> 5.6 Estimated Volume of Hydrocarbon Contaminants <br /> A very rough approximation of the number of gallons of diesel remaining in the soil can be arrived at <br /> by estimating the volume of contaminated soil contained between each diesel isocontour and an <br /> average diesel concentration for that incremental volume The following equation was used in this <br /> calculation <br /> VP,.= X ft x Y ft x T ft x 7 48 gal/cu-ft x [TPH-d(ppm)] — 1,000,000 <br /> where X and Y are the dimensions of the area contained between two nsocontours, T is the <br /> thickness of the contour interval, [TPH-d] is the concentration of diesel in parts per million, and <br /> 7 48 is the conversion factor from cubic feet to gallons <br /> Summing the volume of diesel contained in the depth intervals mapped in Figures 11-15 yields a <br /> cumulative volume of diesel between the depths of 15 and 55 feet The relatively few gallons present <br /> below this depth are ignored in this calculation Using this method, we calculate that approximately <br /> 2500 gallons of diesel are present 1n the soil <br /> 14 <br />