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Ji <br /> CALCULATIONS a <br /> To calculate the pounds (lb) per day the concentration isa! <br /> multiplied by the .volume of air produced in one da <br /> Y- <br /> The lab reports the Concentrations (C) of the air sampling inl <br /> ug/liter. The first 'step is to convert this value to � 1bs/cf <br /> (pounds percubic cubic foot);. �3 <br /> 1 ug/1 x 0 . 000001,g/ug x 0 . 0022051g/g x ; 28 .321/cf = '1e <br /> 0 . 0000000621lb/cf <br /> The volume of air 'produced in one day, equals the flow., rate (Q)x <br /> the time of flow. F. <br /> ,I <br /> d V <br /> t (i F <br /> V = Q. x T = cf/day = cf;/min x 1440min/day <br /> The volume must . be corrected: to standard '; <br /> temperature and s <br /> pressure (STP) y F M <br /> P = Pressure = 14 ..7 lb/ n2 @ STP i <br /> V = Volume cf �' 4 <br /> T = Temperature in degrees above absolute Zero = 491 .°589R @ STP. <br /> ,E <br /> Using the Ideal Gas LawP1V1/T1 = P2V2/T2 <br /> Solving for V2 =P1V1T2/P2T1 lj <br /> Assuming Pi = P2 = 14- 7 lb/in2, 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 + T9F at sight . <br /> V2 = Q cf/min x 1440 min/day x 491 . 58°R/ (459 . 580 + ;iT°F)± . <br /> :I I <br /> X lb/day = C ug/'i x 0.. 0000000621 lb 1/ug cf 'x Q cf/min x 1440 <br /> min/day x 491 . 58°R/ (459 .58° + T°F) <br /> f <br /> Q for the Influent andi Effluent samples W The well,if:flow rate + <br /> the air flow. ji r <br /> c <br /> .ii <br /> fj <br />�. it •Y .1 <br /> page 5 of R604, 04/07/93 <br /> l <br />