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orifice is used to measure the total flow An orifice plate restricts the flow of air across it This restriction causes a pressure drop across the orifice By measuring the resulting pressure change across the orifice it is possible to calculate the air flow rate The flow rate is calculated by the pressure drop (millimeters (mm) water) across a square edge orifice plate Ve = CK s qr (P) Q = AVe Where - Ve= velocity in feet per minute (fpm) C = Orifice Coefficient = 0 65 (for orifice used) K = Constant = 794 6 mm water P = Pressure differential across the orifice Q = Flow rate in cubic feet per minute (CFM) A = Area orifice in square feet 1" = 0 00545 ft2 Q = A X 0 65 X 794 6 X sqr (P) 3 " = 0 04909 ft2 CALCULATIONS To calculate the pounds (1b) per day the concentration is multiplied by the volume of air produced in one day The lab reports the Concentrations (C) of the air sampling in ug/liter. The first step is to convert this value to lbs/cf (pounds per cubic foot) . lug/1 x 0 0000019/ug x 0 0022051b/g x 28 321/cf = 0 00000006211b/cf The volume of air produced in one day, equals the flow rate (Q) x the time of flow V = Q x T = cf/day = cf/min x 1440min/day The volume must be corrected to standard temperature and pressure (STP) P = Pressure = 14 7 lb/in2 @ STP V = Volume cf T = Temperature in degrees above absolute Zero = 491 580R @ STP Using the Ideal Gas Law P1V1/T1 = P2V2/T2 Solving for V2 =P1V1T2/P2T1 Assuming P1 = P2 = 14 7 lb/in2 , P cancels from the equation leaving V2 = V1T2/T1 V1 = Q cf/m x 1440 min/day T2 = 491 580R T1 = 459 . 58 + TOF at site V2 = Q cf/min x 1440 min/day x 491 . 58oR/ (459 580 + ToF) X lb/day = C ug/1 x 0 . 0000000621 1b 1/ug cf x Q cf/min x 1440 min/day x 491. . 58oR/ (459 580 + ToF) Q for the Influent sample = The well flow rate + the air flow. Q for the Effluent = The well flow + the air flow + the fuel flow rate 3 Regal 603 , 10/24/95