Question

A hydraulic jump is formed in a 5 m wide rectangular channel, which has a horizontal bed and is carrying a discharge of 15 m³/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).

Hint:

When calculating the power dissipated by a hydraulic jump, use the Froude number and specific energy change to determine the change in energy. Multiply by the discharge and density of water for the final result.

Answer 1

Correct Answer: 69

The power dissipated by the jump is given by the formula: \[ \Delta P = \rho g Q \Delta E \] where: \(\Delta E\) is the change in specific energy, \(\rho\) is the density of water, \(g\) is the acceleration due to gravity, \(Q\) is the discharge. From the energy equation, \[ \Delta E = \frac{(y_2 - y_1)^3}{4 y_1 y_2} \] For Froude number at upstream: \[ Fr_1^2 = \frac{V_1^2}{g y_1} - \frac{q^2}{g y_1^3} \] \[ Fr_1^2 = \frac{3^2}{9.81 \times 0.5} = 7.34 \] Now, \[ y_2 = y_1 \left( -1 + \sqrt{1 + 8 Fr_1^2} \right) = 0.5 \left( -1 + \sqrt{1 + 8 \times 7.34} \right) = 1.68 \, {m} \] Next, calculate the change in specific energy: \[ \Delta E = \frac{(1.68 - 0.5)^3}{4 \times 1.68 \times 0.5} = 0.49 \, {m} \] Now, from the power equation: \[ \Delta P = 1000 \times 9.81 \times 15 \times 0.49 = 72.10 \, {kW} \quad {(rounded to the nearest integer)}. \] Thus, the power dissipated by the jump is \( \boxed{72} \, {kW} \) (rounded to the nearest integer).