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Ve = CK sqr (P) Q = AVe <br /> Where : <br /> Ve= velocity in feet per minute (fpm) <br /> C = Orifice Coefficient = 0 65 (for orifice used) <br /> K = Constant = 794 6 mm water <br /> P = Pressure differential across the orifice <br /> Q = Flow rate in cubic feet per minute (CFM) <br /> A = Area orifice in square feet 1" = 0 . 00545 ft2 <br /> Q = A X 0 . 65 X 794 6 X sgr (P) 3 " = 0 . 04909 ft2 <br /> CALCULATIONS <br /> To calculate the pounds (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) lug/l x 0 0000019/ug x 0 . 0022051b/g x <br /> 28 321/cf = 0 . 00000006211b/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 pressure <br /> (STP) <br /> P = Pressure = 14 . 7 ib/xn2 Q STP <br /> V = Volume cf <br /> T = Temperature in degrees above absolute Zero = 491 58oR Q STP. <br /> Using the Ideal Gas Law P1V1/T1 = P2V2/T2 <br /> Solving for V2 =P1V1T2/P2TI <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 + TOF at sight . <br /> V2 = Q cf/min x 1440 min/day x 491 580R/ (459 . 580 + TOF) <br /> X lb/day = C ug/1 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 + the air flow. <br /> Q for the Effluent = The well flow + the air flow + the fuel flow <br /> rate . <br /> The fuel flow rate must be corrected to account for the change in <br /> volume from combustion <br /> 5 Regal 60 <br /> g 3 , 06/06/95 <br />