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i Ve = CK sqr (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) CALCULATIONS To calculate the pounds (lb) 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/l x 0 0000018/ug x 0 002205lb/g x 28 . 321/cf = 0 0000000621lb/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/zn2 @ STP V = volume cf T = Temperature in degrees above absolute Zero = 491 58oR @ 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 58oR T1 = 459 58 + ToF at sight . V2 = Q cf/min x 1440 min/day x 491 580R/ (459 580 + ToF) X lb/day = C ug/1 x 0 0000000621 lb 1/ug cf x Q cf/min x 1440 min/day x 491 58oR/ (459 580 + ToF) Q for the influent sa-ple = The -.,ell flo%, rate - the air flo- SUMMARY There has been no evidence of floating product or product sheen on the ground water in MW1 since April 14, 1992 and MW3 and MW10 since November 1.1, 1992 The last evidence of product, as a sheen in the condensate recovery system, was on March 1, 1993 Vapor recovery has reduced the hydrocarbons in ground water significantly since the May 25, 1990 sampling, see Table 1 PAGE 4