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i <br /> Appendix B <br /> AGE-NC-Project No. 98-0520 <br /> Page 5 of 6 <br /> 3) For the estimated third (outer) zone (50 µg11 contour line): A= 15,818 ft2 and h = 10.6 ft <br /> The average TPH-g concentration in each zone was estimated by adding all ground water samples <br /> within the selected area and dividing by the total number of samples or adding the two boundary <br /> concentrations and dividing by two.The average zones(reported as micrograms per liter: µg/l)were <br /> 1,400 ug/l (0.0000014), 433 µg11 (0.00000043) and 75 µg11 (0.000000075). <br /> 1) Approximate volume of soil in the inner zone (V 1): <br /> V 1 =(6,219 ft2)(5.8 ft) -+ 36,070 ft3 <br /> The average porosity of the soil on-site is 45%; using this, the volume of ground water in the soil <br /> can be calculated: <br /> V,Gw= average soil porosity • V 1 = 0.45 • 36,070 ft3 -# 16,231 ft3 <br /> Converting to gallons of water: <br /> V,GW • average density of water -+ 16,231 ft3 • 7.48 gal/ft3 -4 121,408 gallons <br /> SS33 <br /> Using'1,400 µ .1 as the average concentration of gasoline in the inner zone, the gallons of gasoline <br /> are: LI Ir-<G i 5`.i <br /> Gallons of water • average concentration of contaminant 4 <br /> 121,408 gal • 0.0000014 -+ 0.17 gallon of gasoline <br /> Oa44:�6 „ j <br /> 2) Approximate volume of soil to the boundary of the second zone (V2): <br /> V2 =(I 3,232 ftZ)(8.3 ft) -4 109,826 ft3 <br /> V2Gw= average soil porosity• V2 = 0.45 • 109,826 ft3 4 49,422 ft3 <br /> Subtract the volume of water in the inner zone(V,Gw) to get the volume of water within the second <br /> zone of contamination (V2GW): <br /> . V2Gw= V2Gw - ViGw=49,422 ft3 - 16,231 ft3 -+ 33,191 ft3 <br /> Advanced GeoEnvironmental,Inc. <br />