The reservoir level upstream of a 30-m wide spillway for a flow of 800 m³/s is at El. 200 m. The downstream river level for this flow is at El. 100 m. Determine the invert level of a stilling basin having the same width as the spillway so that a hydraulic jump is formed in the basin. Assume the

Question

The reservoir level upstream of a 30-m wide spillway for a flow of 800 m³/s is at El. 200 m. The downstream river level for this flow is at El. 100 m. Determine the invert level of a stilling basin having the same width as the spillway so that a hydraulic jump is formed in the basin. Assume the losses in the spillway are negligible.

Given:

Q = 800 m³/s
B = 30 m
Upstream water level = El 200 m
Downstream water level = El 100 m.

Determine:

Stilling basin invert elevation to form the jump ?

Given:

Q = 800 m³/s
B = 30 m
Upstream water level = El 200 m
Downstream water level = El 100 m.

Determine:

Stilling basin invert elevation to form the jump ?

Accepted Answer

Let z be the invert elevation of the stilling basin. Then, referring to Fig. 2-12, [latex]y_{2}[/latex] = 100−z. Since the losses on the spillway face are negligible and assuming [latex]y_{1}[/latex] to be small,

[latex]V_{1} =\sqrt{2g (200-z)}[/latex]

Now, Q = B[latex]V_{1}y_{1}[/latex]. Hence

[latex]y_{1} = \frac{800}{30 imes \sqrt{19.62(200-z)} }[/latex]

[latex]=\frac{6.02}{\sqrt{200-z} }[/latex]

Substituting expressions for [latex]y_{1}[/latex] and [latex]V_{1}[/latex]

[latex]F^{2}_{r1} =\frac{V^{2}_{1} }{gy_{1} }[/latex]

[latex]=\frac{19.62(200-z)}{9.81 imes 6.02/\sqrt{200-z} }[/latex]

[latex]=0.332(200-z)^{1.5}[/latex]

Substitution of expressions for [latex]y_{1}[/latex], [latex]y_{2}[/latex], and [latex]F^{2}_{r1}[/latex] into Eq. 2-61 yields

[latex]\frac{y_{2} }{y_{1} } =\frac{1}{2} \left(-1+\sqrt{1+8F^{2}_{r1} } ight)[/latex] (2-61)

[latex]\frac{100-z}{6.02\sqrt{200-z} } =\frac{1}{2} \left(-1+\sqrt{1+8 imes 0.332(200-z)^{1.5} } ight)[/latex]

Simplifying this equation, we obtain

[latex]\left(100-z ight) \sqrt{200-z} =-3.01+3.01\sqrt{1+2.656\left(200-z ight) ^{1.5} }[/latex]

Solving this equation by trial and error

z = 84.18 m

Thus, the stilling basininvert should be at El. 84.18 to form the jump.

Question 2.2: The reservoir level upstream of a 30-m wide spillway for a f...

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