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APPENDIX C
<br /> PRE-REMEDIATION DISSOLVED TPH-g VOLUME ASSUMPTIONS AND CALCULATIONS
<br /> PAGE 2OF5
<br /> One ft3 is equal to 7.48 gallons, so the volume of the water is given by:
<br /> V100,0,_ (516 ft3)(7.48 gal/ft3)= 3,859 gallons
<br /> One gallon of water is equivalent to 3.79 liters (/), so the volume water is given by:
<br /> V100000= (3,859 gallons)(3.79 Ugal) = 14,628/water
<br /> Multiplying the volume of water by the average dissolved TPH-g concentration (155,000 pg/1)
<br /> yields the approximate mass of TPH-g in ground water:
<br /> V100,000= (14,628/)(0.000155 kg//)=2.26 kg TPH-g
<br /> A conversion factor of 2.205 Ib/kg. can be used to convert kilograms of TPH-g to pounds of TPH-
<br /> g:
<br /> MTPH-g _ (2.26 kg TPH-g)(2.205 Ib/kg) = 4.9 lbs. TPH-g
<br /> Dividing M1,000 by the weight of one gallon of TPH-g, 6.25 lbs/gallon, will yield the approximate
<br /> volume of TPH-g dissolved into the groundwater of the inner contour interval:
<br /> V100,000 = (M100,000)/6.25 Ib/gal) = 4.9 lbs./6.25 lbs/gal = 0.78 gallons of dissolved TPH-g
<br /> 2) For the estimated 10,000 pg/I middle contour interval, A100 = 1,985 ft2, h = 20 ft (average
<br /> thickness)
<br /> Utilizing TPH-g data from well MW-1 (65,000 pg/1), MW-2 (50,000 pg/1), MW-9A(36,000 pg/1)the
<br /> 10,000 pg/I contour line and the 100,000 pg/I (Figure 5), the average TPH-g concentration is
<br /> estimated to be 52,200 pg/I (equivalent to 0.000053 kg/1).
<br /> Area: V10,000=A10,000 x h = 1,985 ft2 x 20 ft = 39,700 ft2
<br /> The volume is given by: V10,000=V10.000-V1oo,o00 = 39,700 ft3- 1,720 ft3= 37,980 ft3
<br /> Water occupies the porosity in the soil, which is estimated to be 30% of the soil volume, so the
<br /> total volume of water in the saturated portion of the ellipse is approximated by:
<br /> V10,000= (0.30)(37,980 ft3) = 11,394 ft3
<br /> One ft3 is equal to 7.48 gallons, so the volume of the water in the ellipse is given by:
<br /> V1o,000= (11,394 ft3)(7.48 gal/ft3)= 85,227 gallons
<br /> One gallon of water is equivalent to 3.79 liters (/), so the volume water is given by:
<br /> V10,0o0= (85,227 gallons)(3.79//gal) = 323,010 /water
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