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. The lab reports the Concentrations (C) of the air sampling in ug/liter. The first step is to convert this value to lbs/cf (pounds per cubic foot) . 1 ug/1 x 0.000001g/ug x 0 .002205 /g x 28 .321/cf = O .QO.OQo06A1(lbycf The volume of air produced in one day, equals the flow rate (Q)x 1 the time of flow. I V = Q x T = of/day = cf/min x 1440min/day The volume must be corrected to standard temperature ander pressure(STP) . P = Pressure = 14.7 lb/int @ STP t V = volume cf T = Temperature in degrees above absolute Zero = 491 . 58-R @ STP. Using the Ideal Gas Law P1V1/T1 = P2V2/T2 Solving for V2 =P1V1T2/P2T1 Assuming P1 = P2 = 14 . 7 lb/int , P cancels from the equation leaving V2 = V1T2/Tl . V1 = Q cf/m x 1440 min/day T2 = 491 .58-R T1 = 459. 58 + T-F at sight . V2 = Q cf/miry x 1440 min/day x 491 .58-R/ ( 459 .58- + T-F) X lb/day = C ug/l x 0.0000000621 lb 1/ug cf x Q cf/min x 1440 min/day x 491 . 58-R/ ( 459 .58- + T-F) Q for the Influent sample = The well flow rate + the air flow. Q for the Effluent = The well flow + the air flow + the fuel flow rate. The fuel flow rate must be corrected to account for the change in volume from combustion. Since the gas volume depends on the number of molecules present the reaction of propane with oxygen produces an increase in volume. 1C3H8 + 502 -> 3CO2 + 4H2O , which gives an increase from 6 to 7 molecules per molecule of propane . This effectively means that after combustion the volume of flow resulting from the propane is double the propane flow rate. The reaction of -ethane (Natural Gas) and oxygen does not produce a net change in volume , CH4 + 2 02 -> CO2 + 2 H2O; therefore a correction for combustion does not need to be made when methane is burned. See table I for the results Of RT'MOX0—ft"�t a r _ ..... ..r.caca�.ivaa and 2 LOr results influent/effluent monitoring of the system. 6