Chapt10

# This condition is analyzed by equating eq 1037a to

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Unformatted text preview: .38) is valid for any channel shape. For a wide rectangular channel, b0 yc, the formula reduces to Wide rectangular channel: n2g 2 1/3 yc Sc f 8 This is a special case, a reference point. In most channel flows yn turbulent flow, the critical slope varies between 0.002 and 0.008. yc. For fully rough EXAMPLE 10.5 Part (a) y cot 50° Solution This is an easy cross section because all geometric quantities can be written directly in terms of depth y y csc 50° y 16 m3/s. Compute (a) yc, (b) Vc, The 50° triangular channel in Fig. E10.5 has a flow rate Q and (c) Sc if n 0.018. 2y csc 50° Rh E10.5 A y2 cot 50° 1 2 P 50° b0 2y cot 50° y cos 50° (1) The critical-flow condition satisfies Eq. (10.37a) 3 gAc 2 g(yc cot 50°)3 or 2Q2 g cot2 50° yc Part (b) v v (2yc cot 50°)Q2 2(16)2 9.81(0.839)2 1/5 With yc known, from Eqs. (1) we compute Pc 6.18 m, Rhc 3.97 m. The critical velocity from Eq. (10.37b) is Vc | 1/5 b0Q2 | e-Text Main Menu | Q A 16 m3/s 4.70 m2 2.37 m 0.760 m, Ac Ans. (a) 4.70 m2, a...
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