GATE 2025 CE (CE1) – Question 60
A hydraulic jump is formed in a 5 m wide rectangular channel, which has a horizontal bed and is carrying a discharge of 15 m$^3$/s. The depth of water upstream of the jump is 0.5 m. The power dissipated by the jump (in kW) is ___________ (*rounded off to the nearest integer*).
Note:
Acceleration due to gravity = 9.81 m/s$^2$
Density of water = 1000 kg/m$^3$
Kinetic energy correction factor = 1.0
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 72
Explanation
The discharge per metre is $q = 3$ m$^2$/s, so $V_1 = \frac{3}{0.5} = 6$ m/s and $Fr_1 = \frac{6}{\sqrt{9.81 \times 0.5}} = 2.709$. The conjugate depth is $y_2 = \frac{y_1}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right) = 0.25 \times (7.73 - 1) = 1.682$ m. The head loss is $\Delta E = \frac{(y_2 - y_1)^3}{4y_1y_2} = \frac{1.182^3}{4 \times 0.5 \times 1.682} = 0.4908$ m. The power dissipated is $\rho gQ\Delta E = 1000 \times 9.81 \times 15 \times 0.4908 = 72.2$ kW, which is 72 kW.