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CULVERT 2 Sta. 12+10 <br />1) QP = q AQ <br />• A = 9 Ac = .014 mi.2 Q = .46 in. runoff <br />2) Determine Tc. Consider all drainage is channel flow. From Ref, 1, <br />(Nomograph): <br />G H = 7150' - 7062' = 88' <br />-~ = 1000' channel flow. <br />Tc = .075 hrs. Use .1 hrs. <br />3) q @ Tc = .1 hrs = 1000 csm/in2 runoff. (Graph Fig. 5-2, Ref. 3) <br />4) Qp=qAQ - <br />_ (1000) (.014) (.46) <br />= 6.4 cfs. <br />5) -Check culvert capacity <br />HN = 1.0 Type 3 entrance Inlet control governs. (Ref. 4) <br />p _- - <br />6) Existing C.M.P. 12" ~. Use 21" p C.M.P. <br />CULVERT 3 Sta. 20+25 <br />• Reservoir outlet. No calculation. <br />CULUER7 4 Sta. 24+50 <br />1) QP = qAQ <br />A = 5 Ac = .O1 mi.2 <br />Q = .46 in. runoff <br />2) Determine Tc. All overland flow. <br />.>- = hydraulic length = 750' <br />s = slope 12~ <br />v = estimated velocity = 2.5 ft/sec (Fig. 3-1, Ref. 3) <br />Tt = Tc = '-v' = 750 = 300 secs. _ .1 hrs. <br />2.5 <br />3) q @ Tc = .1 hrs. = 1000 CSM/in.2 runoff. (Graph Fig. 5-2, Ref. 3) <br />. 4) QP = qAQ <br />_ (1000)(.01)(.46) <br />= 3.6 cfs. <br />_5_ <br />