This condition is analyzed by equating eq 1037a to

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online