Unformatted text preview: viz., S 1 / 2 c = nQ χ by c X b + 2 y c by c ~ 2 / 3 = ⇒ S c = X nQ χ by c ~ 2 X b + 2 y c by c ~ 4 / 3 Then, since n = 0 . 025 , we have S c = . 025 p 600 ft 3 / sec Q p 1 . 49 ft 1 / 3 / sec Q (10 ft)(4 . 82 ft) 2 ^ 10 ft + 2(4 . 82 ft) (10 ft)(4 . 82 ft) ± 4 / 3 = 0 . 0132 Finally, the critical angle, θ c , is θ c = tan − 1 S c = 0 . 75 o...
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 Spring '06
 Phares
 Trigraph, yc, ChezyManning equation

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