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A hydraulic jump forms at the downstream end of the spillway carrying $17.93 m^3/s$ discharge if the depth before jump is 0.8m. Determine the depth after jump and the energy losses.
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written 2.6 years ago by | • modified 2.6 years ago |
Solution :
Given :
Discharge ,$ Q=17.93 \mathrm{~m}^{3} / \mathrm{s}$
Depth before jump, $ d_{1}=0.8 \mathrm{~m}$
Taking width, $b=1 \mathrm{~m}$,
we get Discharge per unit width,
$ q=\frac{17.93}{1}=17.93$
Let
$ d_{2}=$ Depth after jump and
$h_{L}=$ Loss of energy.
we get
$\begin {aligned} d_{2} &=-\frac{d_{1}}{2}+\sqrt{\frac{d_{1}^{2}}{4}+\frac{2 q^{2}}{g d_{1}}} \\ &=-\frac{0.8}{2}+\sqrt{\frac{0.8^{2}}{4}+\frac{2 \times 17.93^{2}}{9.81 \times 0.8}} \\ &=-0.4+\sqrt{0.16+81.927} \\ &=-0.4+9.06 \\ &=8.66 \mathrm{~m} \end{aligned} $
for loss of energy,
$\begin {aligned} h_{L} &=\frac{\left(d_{2}-d_{1}\right)^{3}}{4 d_{1} d_{2}} \\ &=\frac{(8.66-0.8)^{3}}{4 \times 0.8 \times 8.66} \\ &=17.52 \mathrm{~m} \end{aligned}$
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