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Calculate the depth of cut-off for scour holes for an irrigation channel of F.S.Q equal to 354 cumecs and H.F.Q of natural drainage equal to 600 cumecs. Assume the value of friction factor as 1.

Correct Answer: 10.5 m
Lacey’s normal depth of scour = R = 0.47 (Q/f)1/3 where Q = 354 cumecs and f = 1. R = 0.47 (354/1)1/3 = 7 m The depth of cut-off provided for scour holes = 1.5 R (on both sides) = 1.5 x 7 = 10.5 m.

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