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where <br /> Pv=velocity pressure in inches of water <br /> D=air density in pounds/cu ft = 1325 x P/T [EQ 4] <br /> EXAMPLE <br /> Using the data set from November 13, 1995 at 11 00 we have for the influent flow <br /> T=73 F, P= 18 w c vacuum, Pv= 0 18 orches of water, C=6,000 ppmv(TPHg) <br /> and converting units gives <br /> T=73 +460 = 533 R, P=(407- 18)/407 x 29 9=28 6 inches of mercury <br /> Using EQ 4 and EQ 3 we have <br /> D = 1 325 x 28 6/533 = 0 071 pounds/cu ft <br /> V= 1096 2 x 4(0 18/0 071) = 1744 ft/min <br /> Knowing that we have a two-inch internal diameter pipe and using EQ 2 for the flow rate yields <br /> Q = 1744 x 0218 x 528/533 x 28 6/29 9 = 36 scfm <br /> Applying EQ 1 we have <br /> M= 36 x 6,000 x 0 00032 = 69 pounds/day <br /> i <br />