In order that the hydraulic jump below a spillway will not be swept downstream, establish the relationship between the variables indicated in Fig. 10-16. (Prof. E. A. Elevatorski suggested the dimensionless parameters that follow.)

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The energy equation is applied between a section upstream of the dam where h is measured and section
1, neglecting the velocity head of approach, i.e.,

[latex](h+d)+0+ ext { negl. }- ext { losses (neglected) }=0+0+V_{1}^{2} / 2 g[/latex].

[latex] ext { or } V_{1}=\sqrt{2 g(h+d)}[/latex].

[latex] ext { Since } q=y_{1} V_{1}[/latex].

[latex]y_{1}=\frac{q}{V_{1}}=\frac{q}{\sqrt{2 g(d+h)}}[/latex].

or [latex]y_{1}=\frac{q}{\sqrt{2 g}(d / h+1)^{1 / 2} h^{1 / 2}}[/latex] (A).

From Problem 10.54, the hydraulic jump relationship is

[latex]\frac{y_{2}^{2}-y_{1}^{2}}{2}=\frac{q V_{1}}{g}\left(\frac{y_{2}-y_{1}}{y_{2}} ight) \quad ext { or } \quad g y_{2}^{2}+g y_{1} y_{2}=2 q V_{1}[/latex].

Solving,

[latex]y_{2}=\frac{-y_{1} \pm \sqrt{y_{1}^{2}+8 q V_{1} / g}}{2}[/latex].

[latex] ext { Dividing by } y_{1} ext { produces a dimensionless form }[/latex]

[latex]\frac{y_{2}}{y_{1}}=-\frac{1}{2} \pm \frac{1}{2} \sqrt{1+8 q V_{1} / y_{1}^{2} g}=\frac{1}{2}\left[\sqrt{1+8 q^{2} / g y_{1}^{3}}-1 ight][/latex]. (B).

[latex] ext { Since } y_{2}=(d-D), y_{2} / y_{1}=(d-D) / y_{1} ext { is substituted in }(B) ext { together with the value of } y_{1} ext { from }(A)[/latex] to yield

[latex]\frac{d-D}{y_{1}}=\frac{1}{2}\left[\sqrt{1+8 q^{2} / g y_{1}^{3}}-1 ight][/latex].

[latex]\frac{2(d-D) \sqrt{2 g}(d / h+1)^{1 / 2} h^{1 / 2}}{q}+1=\sqrt{1+\frac{8\left(2^{3 / 2} ight)\left(g^{3 / 2} ight)(d / h+1)^{3 / 2} h^{3 / 2}}{q g}}[/latex].

The equation is put in dimensionless form by multiplying the left-hand side by h/h, dividing through by [latex]\sqrt{8}[/latex], and collecting terms:

[latex]\left(\frac{h^{3 / 2} g^{1 / 2}}{q} ight)\left(\frac{d-D}{h} ight)\left(\frac{d}{h}+1 ight)^{1 / 2}+0.354=\sqrt{\frac{1}{8}+2.828\left(\frac{g^{1 / 2} h^{3 / 2}}{q} ight)\left(\frac{d}{h}+1 ight)^{3 / 2}}[/latex] (C).

The dimensionless terms in (C) may be written

[latex]\pi_{1}=\frac{h^{3 / 2} g^{1 / 2}}{q}, \quad \pi_{2}=\frac{D}{h}, \quad \pi_{3}=\frac{d}{h}[/latex]

Equation (C) then becomes

[latex]\pi_{1}\left(\pi_{3}-\pi_{2} ight)\left(\pi_{3}+1 ight)^{1 / 2}+0.354=\sqrt{\frac{1}{8}+2.828 \pi_{1}\left(\pi_{3}+1 ight)^{3 / 2}} [/latex] (D).

Prof. Elevatorski has prepared a graph of Equation (D) to allow for ready solution. For calculated values of [latex]\pi_{1} ext { and } \pi_{2} ext {, the graph gives the value of } \pi_{3} [/latex].
Prof. Elevatorski, in commenting on the omission of the energy loss on the spillway face, states "by neglecting the loss due to friction, a slight excess of tailwater will be produced in the stilling basin. A slightly-drowned jump produces a better all-around energy dissipator as compared to one designed for the [latex]y_{2}[/latex] depth."

Question 10.59: In order that the hydraulic jump below a spillway will not b...

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