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APPENDIX F
<br /> DISSOLVED TPH-g VOLUME ASSUMPTIONS AND CALCULATIONS
<br /> PAGE 3 OF 7
<br /> 3) For the estimated 1,000 µg/1 ellipse cylinder, a= 62 ft,b=41 ft, and thickness c= 20 ft
<br /> Utilizing data collected from MW-3,the 10,000µg/l contour line and the 1,000µg/l contour line(Figure 9),
<br /> the average TPH-g concentration is estimated to be 4,233 ug/l (equivalent to 0.000004233 g/ml). A ml is
<br /> very nearly equivalent to a cubic centimeter of water, which by definition equals one gram, this
<br /> concentration is essentially unitless.
<br /> The area of the ellipse is given by: A1000=7c(62)(41)=7,986 ft' A10000=4,285 ft2
<br /> The volume of the ellipse cylinder is given by: V1000=A1000 x c=4,285 ft'x 20 ft = 85,700 f
<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: V1000=(0.40)(85,700 ft')=34,280 ft'
<br /> One ft' is equal to 7.48 gallons, so the volume of the water in the ellipse is given by:
<br /> V1000= (34,280 ft)(7.48 gal/ft') =256,414 gallons
<br /> One gallon of water weighs 8.337 lbs/gal, so the mass of the water in the ellipse is given by:
<br /> M1000=(256,414 gallons)(8.337 lb/gal)=2,137,724 lbs
<br /> Multiplying M1000 by the hydrocarbon concentration yields the approximate mass of dissolved hydrocarbons:
<br /> M1000 ( 2,137,724 lbs)(0.000004233) =9 lbs of dissolved TPH-g
<br /> Dividing M10000 by the weight of one gallon of gasoline,6.25 lbs/gallon,will yield the approximate volume
<br /> of TPH-g dissolved into the ground water of the inner ellipsoid:
<br /> V1000=(Mi000)/6.25lb/gal) =9 lbs/6.25 lbs/gal= 1.5 gallons of dissolved TPH-g
<br /> 4) For the estimated 50 µg/l ellipse cylinder, a=78 ft, b=48 ft, and thickness c=20 ft
<br /> Utilizing data collected from MW-6, the 1,000 µg/1 contour line and the 50 µg/l (ND) contour line
<br /> (Figure 9),the average TPH-g concentration is estimated to be 510 ug/1(equivalent to 0.000000510 g/ml).
<br /> A ml is very nearly equivalent to a cubic centimeter of water, which by definition equals one gram, this
<br /> concentration is essentially unitless.
<br /> The area of the ellipse is given by: A50=n(78)(48) = 11,762 fie - A1000=3,776 ft2
<br /> The volume of the ellipse cylinder is given by: V50=A1000 x c=3,776 ft2 x 20 ft =75,522 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: V50= (0.40)(85,700 ft) = 34,280 W
<br /> One ft' is equal to 7.48 gallons, so the volume of the water in the ellipse is given by:
<br /> V50=(34,280 fr')(7.48 gal/ft) =256,414 gallons
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