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<br /> APPENDIX F
<br /> DISSOLVED TPH-g VOLUME ASSUMPTIONS AND CALCULATIONS
<br /> PAGE 3 OF 7
<br /> 3) For the estimated 1,000 µg/l ellipse cylinder, a=62 ft,b=41 ft, and thickness c=20 ft
<br /> Utilizing data collected from MW 5,the 10,000µg/1 contour line and the 1,000µg/l contour line(Fi e 9),
<br /> the average TPH-g concentration is estimated to be 4,233'ug/l (equivalent to 0.000004233 g/mA�ml is
<br /> very nearly equivalent to a cubic centimeter of water, which by definition equals one gram, this
<br /> concentration is essentially unitless. *1 :I soy,3
<br /> The area of the ellipse is given by: AI000=a(62)(41)=7,986 ft-Ago 00 0=4,285 fiz
<br /> The volume of the ellipse cylinder is given by: V1000=A1000 x c=4,285 ft x 20 ft = 85,700 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: 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) =91bs/6.25 lbs/gal= 1.5 gallons of dissolved TPH-g
<br /> 4) For the estimated 50 µg/1 ellipse cylinder, a=78 ft,b=48 ft, and thickness c=20 ft
<br /> 480
<br /> Utilizing data collected from MW-6, the 1,000 µg/1 contour lined the 50 µg/1 (ND) contour e
<br /> (Figure 9),the average TPH-g concentration is estimated to be 510 ug/t(equivalent to 0.000000510 g/ml).
<br /> hiscoA ml is very nearly equivalent to a cubic centimeter of water,which by definition equals one gram, this,—
<br /> concentration
<br /> ncentration is essentially unitless. p 1^"4
<br /> p61�. 6
<br /> The area of the ellipse is given by: A50=n(78)(48)= 11,762 fl? -A100o=3,776 ft?"
<br /> The volume of the ellipse cylinder is given by: V50=A1000 x c=3,776 ft' 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 5
<br /> of water in the saturated portion of the ellipse is approximated by: V50 34 ft'
<br /> 175,S711 �0 '),00
<br /> One W is equal to 7.48 gallons, so the volume of the water in the ellipse is given by:
<br /> V50=(34,280 tf)(7.48 gal/ft')=256,414 gallons
<br /> 361ao a(�.� 8 - a25 ,A4V1b
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