Neglect losses and find the depth of water on the raised section of the rectangular channel shown in Fig. P4.104. Assume uniform velocity profiles.

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Accepted Answer

Depth on raised section = [latex]y_2[/latex] . Continuity: 3×3 = [latex]V_2 y_2[/latex] .

Energy (see Eq. 4.5.21): [latex] \frac{3^2}{2g} +3 =\frac{V^2_2}{2g} +(0.4+y_2) [/latex].

∴ [latex]3.059 =\frac{9^2}{2gy_2^2}+y_2 , or y_2^3-3.059y_2^2 +4.128 =0[/latex]

Trial-and-error:

[latex] \left. \begin{matrix} y_2 =2.0 : -0.11 \underset{= }{?}0 \ y_2 =1.8 : +0.05 \underset{= }{?} 0 \end{matrix} ight\} ∴ y_2 =\underline{1.85 m}[/latex] .

[latex]\left. \begin{matrix} y_2 =2.1 : -0.1 \underset{= }{?}0 \ y_2 =2.3 : +0.1 \underset{= }{?} 0 \end{matrix} ight\} ∴ y_2 =\underline{2.22 m} [/latex] .

The depth that actually occurs depends on the downstream conditions. We cannot select a “correct” answer between the two.

Question 4.104: Neglect losses and find the depth of water on the raised sec...

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