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• The lab reports the Concentrations (C) of the air sampling in ug/liter. The fust step is to convert th3-s value to lbs/cf (pounds per cubic toot) . 1 ug/l x 0.000001g/ug x 0.0022051g/g x 28 .321/cf = 0 .0000006211b/cf The voles of air produced in one day, equals the flow rate (Q)x the time of flow. V = Q x T - cf/dap - cf/min x 1440min/day The volume must be corrected to standard temperature and pres sure(STP). P = Pressure = 14.7 lb/int @ STP 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 ---PiV1T2/P2T1 Assuming PI - P2 = 24 . 7 lb/3.n2-, P cancels from the equation leaving V2 = V1T2/Tl. V1 = Q cf/a x 1444 sin/day T2 = 491.58-R T1 = 459.58 + T-F at sight. V2 = Q cf/min x 1440 min/day x 491 . 58-R/( 459 - 58- + T-F) • X lb/dap = C ug/1 x 0. 0000000621 lb 1/ug cf x Q cf/min x 1444 an/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 f or the change in volume fram coubustion. Since the gas volume depends on the number of molecules pr4sent the reaction of propane with oxygen produces an increase in volume. 1C3H8 + 502 -> 3CO2 + 4H20 , which gives an increase from 6 to 7 molecules per molecule of propane. This effectively means that after bastion the volume of flow resulting from the propane is double the propane flow rate. The reaction of methane (Natural Gas) and oxygen does not produce a net change in volume , C$4 + 2 02 -> CO2 + 2 H2O; therefore a correction for combustion does not need to be rade when methane is burned. See table 1 for the results of REMOX operation and table 2 for results influent/effluent monitoring of the system. • 6