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Manning Equation. <br />q - 1,49 ARz~3 Sl/z <br />zs - <br />n <br />Hydraulic Data: <br />~ = 3,34 cFs <br />D = 0.54 FT, <br />A = 1,66 FTZ <br />R = 0,38 Fr, <br />n = 0,03 <br />V = Z,O1 FPS <br />Assume reservoir high waterline to discharge the <br />design storm is equal to the spillway invert elevation <br />plus the depth of floor in the outlet channel: <br />N,IJ.L,zs yr = 5434,5' + 0,54' = 5435,14' <br />• To provide a minimum 1 FT of freeboard the top of <br />embankment shall be at an elev. of 5436,14', <br />c) Waste Rock Disposal Site = Sedimentation Reservoir Interceptor <br />Channel Design - Section E - E <br />Problem: Determine hydraulic data of a trapezoidal shaped inter- <br />ceptor channel, depth of flow at the design storm and <br />maximum anticipated velocity. <br />Solution: <br />Maximum allowable velocity, AMAX = 3FP5 <br />Minimum allowable velocity, MIN = ZFPS <br />q10 - 2'19 CFS (From Peak Discharge Calcs.) <br />n = Q,03 (Manning coefficient for riprap) <br />• sideslope = 2,5;1 (horizontal to vertical) <br />slopes (s) _ ~,Q2 FT~FT <br /> <br />