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# 0015 what is the normal depth yn v v e text main menu

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Unformatted text preview: f Contents | Study Guide 668 Chapter 10 Open-Channel Flow b0 yn 50° W 6 ft E10.2 Solution From Table 10.1, for asphalt, n is unknown b0 6 ft 0.016. The area and hydraulic radius are functions of yn, which 2yn cot 50° P 6 1 2 A 2W (6 6 From Manning’s formula (10.19) with a known Q 1.49 (6yn 0.016 300 or (6yn y2 cot 50°) n y2 cot 50°)5/3 n b0)yn y2 cot 50° n 6yn 2yn csc 50° 300 ft3/s, we have 6yn y2 cot 50° n 6 2yn csc 50° 83.2(6 2/3 (0.0015)1/2 2yn csc 50°)2/3 One can iterate this formula laboriously and eventually find yn 4.6 ft. However, it is a perfect candidate for EES. Instead of manipulating and programming the final formula, one might simply evaluate each separate part of the Chézy equation (in English units, with angles in degrees): P 6 2*yn/sin(50) A 6*yn yn^2/tan(50) Rh A/P 300 1.49/0.016*A*Rh^(2/3)*0.0015^0.5 Hit Solve from the menu bar and EES complains of “negative numbers to a power”. Go back to Variable Information on the menu bar and make sure that yn is posit...
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