A rectangular channel 6.1 m wide carries 11.3 m^3/s and discharges onto a 6.1-m-wide apron with no slope at a mean velocity of 6.1 m/s. What is the height of the hydraulic jump? What energy is absorbed (lost) in the jump?

Question

A rectangular channel 6.1 m wide carries 11.3 m³/s and discharges onto a 6.1-m-wide apron with no slope at a mean velocity of 6.1 m/s. What is the height of the hydraulic jump? What energy is absorbed (lost) in the jump?

Accepted Answer

[latex] ext { (a) } \quad V_{1}=6.1 m / s , q=11.3 / 6.1=1.85\left( m ^{3} / s ight) / m ext { width, and } y_{1}=q / V_{1}=1.85 / 6.1=0.303 m ext {. Then }[/latex]

[latex] q^{2} / g=\frac{1}{2} y_{1} y_{2}\left(y_{1}+y_{2} ight), \quad(1.85)^{2} / 9.81=\frac{1}{2}(0.303) y_{2}\left(0.303+y_{2} ight), \quad 0.349=0.0459 y_{2}+0.152 y_{2}^{2}[/latex].

[latex] ext { from which } y_{2}=-1.67 m ,+1.37 m ext {. The negative root being extraneous, } y_{2}=1.37 m ext { and the height of the hydraulic jump is (1.37 — 0.303) = 1.07 m}[/latex].

[latex] ext { Note that } y_{c}=\sqrt[3]{(1.85)^{2} / 9.81} ext { or } \sqrt[3]{\frac{1}{2} y_{1} y_{2}\left(y_{1}+y_{2} ight)}=0.70 m[/latex].

Hence the flow at 0.303-m depth is supercritical, and the flow at 1.37-m depth is subcritical.

[latex] ext { (b) Before the jump, } E_{1}=V_{1}^{2} / 2 g+y_{1}=(6.1)^{2} / 2 g+0.303=2.20 m \cdot N / N [/latex].

[latex] ext { After the jump, } E_{2}=V_{2}^{2} / 2 g+y_{2}=[11.3 /(6.1 imes 1.37)]^{2} / 2 g+1.37=1.46 m \cdot N / N [/latex].

[latex] ext { Lost energy per second }=\gamma Q H=(9.79)(11.3)(2.20-1.46)=81.9 kW[/latex].

Question 10.55: A rectangular channel 6.1 m wide carries 11.3 m^3/s and disc...

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