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<br />GJ~..i;j'. <br /> <br /> <br />----------J-----Z6~~------- <br /> <br /> <br />~ <br />~ <br /> <br /> <br /> <br />Area = 148.75 ft.2 <br /> <br />9'- o' <br /> <br />AssUJIlptions: <br /> <br />Concrete lined section - I = .014 (coef. of roughness) <br />Hyd. r = 4.5 <br />J10w velocity = 4.05 ft./sec. =2!.76 M.P.H. <br />Slope = .0002 = 1. 056 ft ./mUe <br />Air temp. = _ 300y and lIl8XilllUJll vater temp. = + 390F <br />(Assumed wind velocity = 15 ft./sec. = approx. 10 M.P.H.) <br /> <br />Formula derived for determining heat losses: <br /> <br />(E +e) 'l'C = B.t.u./lb./hr. <br />AW <br /> <br />R = heat 10ss/ft.2/br./OF. temperature differential <br /> <br />between air and vater = 2 B.t.n. <br /> <br />, <br />e ~ heat lO8B/ft.2!hr.!ft. sec. wind velocity/oJ'. = .25 x <br />wind velocity in ft./sec. \ <br /> <br />~ = Temp. difference between air and vater.in OF. <br /> <br />C = Canal vater surface width in feet <br /> <br />-A = VolUJlle of 1 linear ft. of canal vater in ft.3 <br />(this volume is nWllerically the spe ae craBB <br />sectional area) <br /> <br />W = Density of vater - 62.4 Ibs./ft.3 <br /> <br />-For reservoirs use average volume or ~ter under 1 rt.2 <br />of surface. <br /> <br />10 <br />