Question 12.12: A control sluice spanning, the entry to a 3.5 m wide rectang...
A control sluice spanning, the entry to a 3.5 m wide rectangular channel, admits 5.5 \mathrm{m}^{3} / \mathrm{s} of water with a uniform velocity of 4.14 m/s. Explain under what conditions a hydraulic jump will be formed and, assuming that these conditions exist, calculate (i) the height of the jump, and (ii) power dissipated in the jump.
The upstream depth of flow is
h_{1}=\frac{5.5}{3.5 \times 4.14}=0.379 \mathrm{~m}The upstream Froude number, \mathrm{Fr}_{1}=\frac{4.14}{(9.81 \times 0.379)^{1 / 2}}=2.15
For the hydraulic jump to occur, the downstream flow must be tranquil and the depth of flow at downstream must satisfy the Eq. (12.46)
\frac{h_{2}}{h_{1}}=\frac{1}{2}\left[\left(1+8 \mathrm{Fr}_{1}^{2}\right)^{1 / 2}-1\right] (12.46)
Therefore,
h_{2}=\frac{0.379}{2}\left[\left\{1+8(2.15)^{2}\right\}^{1 / 2}-1\right]=0.978 \mathrm{~m}(i) Therefore, the height of the jump
\Delta h=\left(h_{2}-h_{1}\right)=(0.978-0.379)=0.6 \mathrm{~m}(ii) The loss of head in the jump is found out from Eq. (12.48) as
h_{j}=\frac{\left(h_{2}-h_{1}\right)^{3}}{4 h_{1} h_{2}} (12.48)
h_{j}=\frac{(0.6)^{3}}{4 \times 0.978 \times 0.379}=0.146 \mathrm{~m}The rate of dissipation of energy =\rho g Q h_{j}
=10^{3} \times 9.81 \times 5.5 \times 0.146 \mathrm{~W}=7.66 \mathrm{~kW}