After flowing over the concrete spillway of a dam, 9000 cfs then passes over a level concrete apron (n = 0.013). The velocity of the water at the bottom of the spillway is 42.0 ft/sec, and the width of the apron is 180 ft. Conditions will produce a hydraulic jump, the depth in the channel below

Question

After flowing over the concrete spillway of a dam, 9000 cfs then passes over a level concrete apron (n = 0.013). The velocity of the water at the bottom of the spillway is 42.0 ft/sec, and the width of the apron is 180 ft. Conditions will produce a hydraulic jump, the depth in the channel below the apron being 10.0 ft. (a) In order for the jump to be contained on the apron, how long should the apron be built? (b) How much energy is lost from the foot of the spillway to the downstream side of the jump?

Accepted Answer

[latex] ext { (a) Referring to Fig. 10-15, first calculate depth } y_{2} ext { at the upstream end of the jump. }[/latex]

[latex]q^{2} / g=\frac{1}{2} y_{2} y_{3}\left(y_{2}+y_{3} ight), \quad(9000 / 180)^{2} / 32.2=\frac{1}{2}\left(10 y_{2} ight)\left(y_{2}+10 ight) \quad y_{2}=1.37 ft[/latex].

Also,

[latex]y_{1}=q / V_{1}=(9000 / 180) / 42.0=1.19 ft[/latex].

[latex] ext { Now calculate length } L_{A B} ext { of the retarded flow. }[/latex]

[latex]V_{1}=42.0 ft / sec , \quad V_{1}^{2} / 2 g=27.4 ft , \quad R_{1}=(180 imes 1.19) / 182.38=1.174 ft[/latex].

[latex]V_{2}=\frac{q}{y_{2}}=\frac{50.0}{1.37}=36.5 ft / sec , \quad \frac{V_{2}^{2}}{2 g}=20.7 ft \quad R_{2}=\frac{(180 imes 1.37)}{182.74}=1.349 ft[/latex].

[latex] ext { Then } V_{ ext {mean }}=39.25 ft / sec , R_{ ext {mean }}=1.262 ft ext {, and }[/latex]

[latex]L_{A B}=\frac{\left(V_{2}^{2} / 2 g+y_{2} ight)-\left(V_{1}^{2} / 2 g+y_{1} ight)}{S_{0}-S}=\frac{(20.7+1.37)-(27.4+1.19)}{0-\left(\frac{0.013 imes 39.25}{(1.486)(1.262)^{2 / 3}} ight)^{2}}=75.4 ft[/latex].

[latex] ext { The length of the jump } L_{J} ext { from } B ext { to } C ext { is between } 4.3 y_{3} ext { and } 5.2 y_{3} ft ext {. Assuming the conservative }[/latex][latex] ext { value of } 5.0 y_{3}[/latex],

[latex]L_{J}=5.0 imes 10.0=50.0 ft[/latex].

Hence, total length ABC = 16 + 50 = 126 ft (approximately).

[latex] ext { (b) Energy at } A=y_{1}+V_{1}^{2} / 2 g=1.19+27.4=28.59 ft - lb / lb [/latex].

[latex] ext { Energy at } C=y_{3}+V_{3}^{2} / 2 g=10.0+(5.0)^{2} / 2 g=10.39 ft - lb / lb[/latex].

[latex] ext { Total energy lost }=\gamma Q H=(62.4)(9000)(28.59-10.39)=1.02 imes 10^{7} ft - lb / sec ext {, or } 18,600 hp [/latex].

Question 10.58: After flowing over the concrete spillway of a dam, 9000 cfs ...

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