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Page 3 of 3 <br />1 <br /> FIGURE 2.05.3(3)(c)(i) (con't) <br />' <br /> t = 12 hours produces the greatest intensity <br /> <br /> Use C = .90 (Hydrology and Sedimentology of Surface Ltined Lands, <br /> page 91) <br />' i = 2.288 <br /> A = 0.28 <br />1 i <br />1 <br />8 <br /> Q = A) <br />(C <br />.008 = (.90)(2.28 <br />)(0.28 )(1.008) = 0.58 cfs. <br />' 6fin diameter for C.7P culvert is: <br /> s lope = x•0556 ft./f t. <br />' n <br />4 = 0.022 <br />fs <br />58 <br />= 0 <br /> c <br />. <br /> 4 = 1n486 AR2/3S 1/2 = 0.58 = <br />AR2/3(0.0556)1/2 <br />2 <br />' 0.0 <br />2 <br />' cho ose AR2/3 > 0.0115 <br /> 9" CMP A='R R2 = 0.44 <br /> .44 <br />.44 <br />A <br /> __ <br />= <br />= <br />R C 25rR 2.356 = 0.167 <br /> 4 (•44)(.187)2/3(0.0556)1/2 ~ 2.3 cfs <br />' 0.022 <br />' A 9" CMP at this location will convey 2.3 cfs. A 25 year design <br />storm will only produce 0.58 cfs. This size culvert is more then <br />adequate. <br />1 <br /> <br />II <br />II <br />i~ <br />REVISF,D May 15, 1981 <br />