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

Power Dissipated by Hydraulic Jump

Formula Used:

The power dissipated is given by:

\[ \Delta P = \rho g Q \Delta E \]

• \(\Delta E\) = Change in specific energy
• \(\rho\) = Density of water (\(1000 \, kg/m^3\))
• \(g\) = Acceleration due to gravity (\(9.81 \, m/s^2\))
• \(Q\) = Discharge (\(15 \, m^3/s\))

Step 1: Calculate Change in Specific Energy

From the energy equation:

\[ \Delta E = \frac{(y_2 - y_1)^3}{4 y_1 y_2} \]

Step 2: Compute Froude Number

The Froude number at upstream is given by:

\[ Fr_1^2 = \frac{V_1^2}{g y_1} \]

Substituting values:

\[ Fr_1^2 = \frac{3^2}{9.81 \times 0.5} = 7.34 \]

Step 3: Compute \( y_2 \)

The conjugate depth after the hydraulic jump is:

\[ y_2 = y_1 \left( -1 + \sqrt{1 + 8 Fr_1^2} \right) \]

Substituting values:

\[ y_2 = 0.5 \left( -1 + \sqrt{1 + 8 \times 7.34} \right) = 1.68 \, {m} \]

Step 4: Calculate Change in Specific Energy

Substituting values into the energy equation:

\[ \Delta E = \frac{(1.68 - 0.5)^3}{4 \times 1.68 \times 0.5} = 0.49 \, {m} \]

Step 5: Compute Power Dissipated

Now, using the power equation:

\[ \Delta P = 1000 \times 9.81 \times 15 \times 0.49 \]

Solving:

\[ \Delta P = 72.10 \, {kW} \]

Final Answer:

Correct Answer: \( \mathbf{72} \) kW (rounded to the nearest integer).