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Riechers Spence & Associates Job #: 21 02 049.0 <br /> Consulting Civil Engineers Date: 10/21 /02 <br /> Po m,H = (Patm.HkN/m2) - (0.9994)(101.325kN/M2) =10.3289 m <br /> (An/m) (9.804kN/m') <br /> iv) Oxygen concentration assuming the percent oxygen <br /> concentration leaving the aeration tank is 19% <br /> / 1 P.,H +Pw,Effdepth Ot <br /> _ Cs.T.H = 1C,r H�2( Palm + 21 space in equation <br /> .N <br /> 11.27 g/m')1 10.3289m+(3.1394-0.3048)m 19 <br /> 2( 10.3289m + 21 <br /> = 12.2798 g/m3 <br /> b.) Standard oxygen transfer rate (SOTR) using a = 0.65, p = 0.95 and <br /> F = 0.90 1 <br /> SOTR = AOTR C�,20 J (1 .02420-T) <br /> IaF(� H <br /> sr <br /> _ {(1 .3849 kg/h)(9.08 g/m3)(1 .024)20-10} / <br /> / <br /> {(0.65)(0.9)[(0.95)(12.2798 g/m3)-2.0 g/m31) <br /> = 15.9406 / 5.6545 = 2.8191 kg/h (6.2150 Ib/h) <br /> c.) Air flowrate <br /> Air flowrate, m3/min = (SOTRkg/h) <br /> (E)(60 min/ h)(kgO, l m'air) <br /> The density of air at a temperatur of 10°C and an atmospheric pressure <br /> of 0.9994 * 101 .325 kPa = 101 .2642 Kpa is: <br /> _ (1.012642*105 N/m')(28.97kg/kg-mole) = 1 .2462 kg/m3 <br /> P (8314N•m/kg-mole•K)[(273.15+10)k] <br /> The correspondin amount of of oxygen by weight is 23.18% <br /> 02 = 1 .2462 kg/m3 *0.2318 = 0.2889 kg/m3 <br /> Air flowrate, m3/min = (2.8191kg/h) = 0.4647 m3/min <br /> (0.35)(60min/ h)(0.2889kgO2 lm'air) <br /> 20 <br />