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CALCULATIONS To calculate the pounds (lb) per day the concentration is multiplied by the volume of air produced in one day The lab reports the Concentrations (C) of the air sampling in ug/liter The first step is to convert this value to lbs /cf (po = per cubic foot) . lug/l x 0 . 000001 /u x 0 0022051g/9 0 0000000621lb/cf The volume of air produced in one day, equals the flow rate (Q) x the time of flow V = Q x T = cf/day = cf/min x 1440min/day The volume must be corrected to standard temperature and pressure (STP) P = Pressure = 14 7 lb/in2 @ STP V = volume cf T = Temperature rn degrees above absolute Zero = 491 58oR @ STP Using the Ideal Gas Law P1V1/T1 = P2V2/T2 Solving for V2 =PIV1T2/P2Tl Assuming P1 = P2 = 14 7 lb/in2 , P cancels from the equation leaving V2 = V1T2/Tl V1 = Q cf/m x 1.440 min/day T2 = 491 58oR T1 = 459 58 + TOF at sight . V2 = Q cf/min x 1440 min/day x 491 . 58oR/ (459 580 + TOF) X lb/day = C ug/l x 0 0000000621 lb 1/ug cf x Q cf/min x 1440 min/day x 491 58aR/ (459 580 + TOF) 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 preseat the reaction of propane with oxygen produces an increase in volume . The reaction, 1C3H8 + 502 -> 3CO2 + 4H2O, increases sThise of tuber if molecules from 6 to 7 per molecule of propane means that after combustion the�volume Nof flow Yresultjngoffrom the e propane is double the propane Lic�w �a� - --•�-Mv- �--- ----- (Natural Gas) and + 2gH2O, thereforedaccorrectionchange for in vo combustion CH4 + 2 02 -> CO2 + , does not need to be made when methane is burned page 5 of R603 , 01/26/93