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Ij <br /> CALCULATIONS ` <br /> To calculate the 'Npounds (lb) per day the concentration is <br /> multiplied by the volume of air produced in one day.` <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 /> I 1 ug/l x 0 . 000001g/ug x 0 .-0022-051g/g x 28 .321/cf <br />` 0 . 00000006211b/cf - ` <br /> M i <br /> The volume of air produced in one ;day, equals the flow rate (Q)x ' <br /> the time of flow. I <br /> c V = Q x. T = cf/day 'E"-E cf/min x 1440min/day <br /> II <br /> The volume must be ' corrected , to standard temperature and; <br />' pressure (STP) !� <br /> P = Pressure = 14 . 7Ilb/in2 @ STP # <br /> V = Volume cf iE <br /> k T = Temperature in degrees above absolute Zero = 491 . 58oR @ STP. 4 <br /> F <br /> Using the Ideal GasllLaw P1V1/T1 = P2V2/T2 <br /> Solving for V2 =P1V1T2/P2T1 <br /> Assuming P1 = P2 14 . 7 lb/in2, i° P cancels from ° .the equation) <br /> leaving V2 = V1T2/T1 . i <br />�i i� <br /> V1 = Q cf/m x 1440 min/day ` <br /> i3 <br /> T2 = 491 . 580R T1 = 459 . 58 + TOF at sight . <br /> V2 = Q cf/min x 1440 min/day x 451 . 580R/ (459 . 580 + ,TOF) k <br /> X lb/day = C ug/1x 0 . 0000000621 lb 1/ug cf x Q cf/min x 14401, <br /> min/day x 491 . 580R% (459 . 580 + ToF) <br /> Q for the Influent and Effluent samples = The well : flow rate +,f <br /> the air flow. j <br /> 4 <br /> .w iE i� I <br /> ij <br /> 4 5 <br /> Ili page 5 of R60.4, 03/03/93 '. <br /> i <br />