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APPENDIX C
<br /> DISSOLVED TPH-g MASS-VOLUME ASSUMPTIONS AND CALCULATIONS
<br /> PAGE 2OF4
<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 is approximated by:
<br /> V10,000= (0.30)(10,980 ft)= 3,294 ft3
<br /> One ft3 is equal to 7.48 gallons, so the volume of the water is given by:
<br /> V100,= (3,294 ft3)(7.48 gal/ft)=24,639 gallons
<br /> One gallon of water is equivalent to 3.79 liters (/), so the volume water is given by:
<br /> V100DO= (24,639 gallons)(3.79 Ugal) = 93,382/water
<br /> Multiplying the volume of water by the average dissolved TPH-g concentration (12,666 pg/1)yields
<br /> the approximate mass of TPH-g in ground water:
<br /> V1o,000= (93,382 /)(0.000012666 kg//) = 1.18 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-g:
<br /> MTPH-g = (1.18 kg TPH-g)(2.205 Ib/kg) = 2.60 lbs. TPH-g
<br /> Dividing M10,000 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 /> V10,000 = (M10,000)/6.25 Ib/gal) = 2.60 lbs./6.25 lbs/gal = 0.42 gallons of dissolved TPH-g
<br /> 2) For the estimated 1,000 pg/1 middle contour interval, A1,000 = 3,430 ft2, h = 20 ft (average
<br /> screen interval)
<br /> Utilizing TPH-g data from well MW-2 (4,700 pg/1), MW-3 (9,400 pg/1), MW-6 (8,700 pg/1) the
<br /> 10,000 pg/l contour line and the 1,000 pg/I (Figure 6), the average TPH-g concentration is
<br /> estimated to be 6,760 pg/1 (equivalent to 0.000006760 kg/1).
<br /> Area: V1,000=A1,000 x h = 3,430 ft2 x 20 ft = 68,600 ft2
<br /> The volume is given by: V1,000=V1,000-V10.0o0 = 68,600 ft3- 10,980 ft3= 57,620 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 /> V1,000= (0.30)(57,620 ft)= 17,286 ft3
<br /> One ft3 is equal to 7.48 gallons, so the volume of the water in the ellipse is given by:
<br /> V1000= (17,286 ft3)(7.48 gal/ft)= 129,299 gallons
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