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APPENDIX F <br /> DISSOLVED TPH-g VOLUME ASSUMPTIONS AND CALCULATIONS <br /> PAGE 4 OF 7 <br /> One gallon of water weighs 8.337 lbs/gal, so the mass of the water in the ellipse is given by: <br /> M50=_ (2*,4414 gallons)(8.3371b/gal)=2,137,724 lbs �l H 3 r 843, <br /> � % <br /> Multiplying M50 by the hydrocarbon concentration yids the approximate mass of dissolved hydrocarbons: <br /> M50(2 4-�))4 lbss ', � <br /> 00000510)1 1 lbs dissolved TPH-g <br /> I' <br /> Dividing M50 by the weight o?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 /> V50=(M50)/6.25lb/gal)= 1.1 lbs/6.251bs/gal=0.2 gallons of dissolved TPH-g <br /> — 0 411 <br /> Combining the total volume of dissolved TPH-g at the different ggncentrations, impacted PRE- <br /> REMEDIAITON ground water yields a total of approximately 235,240 ft' containing an estimated 25.5 <br /> gallons of TPH-g, equivalent to 155.8 pounds of TPH-g. <br /> 70y <br /> � � � � ty <br />