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a <br /> tCALCULATIONS � <br /> To calculate the pounds (lb) per day the concentration is <br /> multiplied by the volume of air produced in one day. 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 <br /> (pounds per cubic foot) . <br /> 1 ug/l x 0 . 0000018/ug x 0 . 0022051g/g x 28 321/cf = <br /> 0 0000000621lb/cf <br /> The volume of air produced in one day, equals the flow rate (Q) x <br /> the time of flow. <br /> V = Q x T = cf/day = cf/min x 1440min/day <br /> 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 . 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 <br /> leaving V2 = V1T2/T1 <br /> V1 = Q cf/m x 1440 min/day <br /> T2 = 491 58OR T1 = 459 58 + T°F at sight . <br /> V2 = Q cf/min x 1440 min/day x 491 58°R/ (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 + T°F) <br /> Q for the Influent and Effluent samples = The well flow rate + <br /> the air flow <br /> page 5 of R604 , 04/30/93 <br /> S <br />