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The volume must be corrected to standard temperature and <br /> pressure (STP) . <br /> P = Pressure = 14 7 lb/int @ STP <br /> V = Volume cf <br /> T = Temperature in degrees above absolute Zero = 491 . 580R @ <br /> 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 <br /> leaving V2 = V1T2/Tl <br /> V1 = Q cf/m x 1440 min/day <br /> T2 = 491 580R T1 = 459 58 + TOF at site. <br /> V2 = Q cf/min x 1440 min/day x 491 580R/ (459 580 + TOF) <br /> X lb/day = C ug/l 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 <br /> 6 R603 APPEND D <br />