For a rectangular channel, develop an expression for the relation between the depths before and after a hydraulic jump. Refer to Fig. 10-14.

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Accepted Answer

For the free body between sections 1 and 2, considering a unit width of channel and unit flow q,

[latex]P_{1}=\gamma \bar{h} A=\gamma\left(\frac{1}{2} y_{1} ight) y_{1}=\frac{1}{2} \gamma y_{1}^{2}, \quad ext { and similarly } \quad P_{2}=\frac{1}{2} \gamma y_{2}^{2}[/latex].

Table 10.1 Types of Varied Flow

Form of Profile	Type of Flow	Symbol	Type of Profile	[latex]\frac{d y}{d L}[/latex]	Depth Relations	Channel Slope
	Subcritical	[latex]M_{1}[/latex]	Backwater	+	[latex]y>y_{N}>y_{i}[/latex]	Mild [latex]0
	Subcritical	[latex]M_{2}[/latex]	Dropdown	-	[latex]y_{N}>y>y_{c}[/latex]
	Supercritical	[latex]M_{3}[/latex]	Backwater	+	[latex]y_{N}>y_{ C }>y[/latex]
	Subcritical	[latex]H_{2}[/latex]	Dropdown	-	[latex]y>y_{i}[/latex]	Horizontal S-=0 [latex]y_{N}=\infty[/latex]
	Supercritical	[latex]H_{3}[/latex]	Backwater	+	[latex]y_{C}>y[/latex]	Horizontal S-=0 [latex]y_{N}=\infty[/latex]
	Subcritical	[latex]C_{1}[/latex]	Backwater	+	[latex]y>y_{c}=y_{N}[/latex]	Critical [latex]\begin{aligned}&S_{N}=S_{c} \&y_{N}=y_{c}\end{aligned}[/latex]
	Uniform, critical	[latex]C_{2}[/latex]	Parallel to bed		[latex]y_{c}=y=y_{N}[/latex]
	Supercritical	[latex]C_{3}[/latex]	Backwater	+	[latex]y_{C}=y_{N}>y[/latex]
	Subcritical	[latex]S_{1}[/latex]	Backwater	+	[latex]y>y_{C}>y_{N}[/latex]	Steep [latex]S>S_{c}>0[/latex]
	Supercritical	[latex]S_{2}[/latex]	Dropdown	-	[latex]y_{c}>y>y_{N}[/latex]
	Supercritical	[latex]S_{3}[/latex]	Backwater	+	[latex]y_{C}>y_{N}>y[/latex]
	Subcritical	[latex]A_{2}[/latex]	Dropdown	-	[latex]y>y_{c}[/latex]	Adverse [latex]\begin{gathered}S<0 \y_{N}=\infty\end{gathered}[/latex]
	Supercritical	[latex]A_{3}[/latex]	Backwater	+	[latex]y_{c}>y[/latex]

From the principle of impulse and momentum,

[latex]\Delta P_{x} d t=\Delta ext { linear momentum }=\frac{W}{g}\left(\Delta V_{x} ight)[/latex].

[latex]\frac{1}{2} \gamma\left(y_{2}^{2}-y_{1}^{2} ight) d t=\frac{\gamma q d t}{g}\left(V_{1}-V_{2} ight)[/latex].

[latex] ext { Since } V_{2} y_{2}=V_{1} y_{1} ext { and } V_{1}=q / y_{1} ext {, the above equation becomes }[/latex]

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

[latex] ext { Since } q^{2} / g=y_{c}^{3} ext {, }[/latex] [latex]y_{c}^{3}=\frac{1}{2} y_{1} y_{2}\left(y_{1}+y_{2} ight)[/latex].

[latex] ext { The length of the jump has been found to vary between } 4.3 y_{2} ext { and } 5.2 y_{2} [/latex].

The hydraulic jump is an energy dissipator. In designing hydraulic jump stilling basins, knowledge of jump length and depth [latex]y_{2} ext { is important. Good energy dissipation occurs when } V_{1}^{2} / g y_{1}=20 ext { to } 80 [/latex].

Question 10.54: For a rectangular channel, develop an expression for the rel...

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