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r <br /> • The lab reports the Concentrations (C) of the air sampling in <br /> ug/liter The first step is to convert this value to lbs/cf (pounds <br /> per cubic foot) lug/l x 0 . 0000018/ug x 0 . 00220519/9 x 28 . 321/cf ` <br /> 0 . 0000000621lb/cf <br /> The volume of air produced in one day, equals the flow rate (Q) x the <br /> time of flow. <br /> V = Q x T = cf/day = cf/min x 1.440min/day <br /> The volume must be corrected to standard temperature and pressure <br /> (STP) . <br /> P = Pressure = 14 7 lb/int @ STP <br /> V = Volume cf <br /> T = Temperature in degrees above absolute Zero = 491 58OR @ STP. <br /> Using the Ideal Gas Law P1V1/T1 = P2V2/T2 <br /> Solving for V2 =P1V1T2/P2T1 <br /> Assuming P1 = P2 = 14 7 lb/int , P cancels from the equation leaving <br /> V2 = V1T2/T1 . <br /> V1 = Q cf/m x 1440 min/day <br /> T2 = 491 58oR T1 = 459 58 + TOF at sight <br /> V2 = Q cf/min x 1440 min/day x 491 . 580R/ (459 . 580 + TOF) <br /> X lb/day = C ug/1 x 0 0000000621 lb 1/ug cf x Q cf/min x 1440 <br /> min/day x 491 580R/ (459 580 + TOF) <br /> Q for the Influent sample = The well flow rate + the air flow. <br /> Q for the Effluent = The well flow + the air flow + the fuel flow <br /> rate . <br /> The fuel flow rate must be corrected to account for the change in <br /> volume from combustion <br /> Since the gas volume depends on the number of molecules present the <br /> reaction of propane with oxygen produces an increase in volume . The <br /> reaction, 1C3H8 + 502 -> 3CO2 + 4H2O, increases the number of <br /> molecules from 6 to 7 per molecule of propane This effectively <br /> means chat after combustion the volume of flow resulting from the <br /> propane is double the propane flow rate . The reaction o£ methane <br /> (Natural Gas) and oxygen does not produce a net change in volume, <br /> CH4 + 2 02 -> CO2 + 2 H2O, therefore a correction for combustion <br /> does not need to be made when methane is burned. <br /> CO2 EMISSIONS <br /> Analysis of the vapor stream has revealed that a mayor constituent <br /> of the extracted vapor is carbon dioxide (CO2) The CO2 is found in <br /> concentrations well above its normal concentration in air and <br /> therefore is probably the product of the oxidation of hydrocarbons <br /> by biological and/or chemical means . <br /> page 4 of R603 , 12/28/92 <br />