The depth of flow of water, at a certain section of a rectangular channel of 5 m wide is 0.6 m. The discharge through the channel is 15 m³/s. If a hydraulic jump takes place on the downstream side, find the loss of energy per kg of water due to hydraulic jump.

Solution

1. Calculate Conditions Before and After the Jump

First, we find the velocities and the depth after the jump (\(d_2\)).

$$ \text{Initial Velocity, } V_1 = \frac{Q}{B \times d_1} = \frac{15}{5 \times 0.6} = 5.0 \, \text{m/s} $$ $$ \text{Froude Number, } Fr_1 = \frac{V_1}{\sqrt{g d_1}} = \frac{5.0}{\sqrt{9.81 \times 0.6}} \approx 2.06 $$ $$ \text{Depth after jump, } d_2 = \frac{d_1}{2} \left[ \sqrt{1 + 8 Fr_1^2} - 1 \right] = \frac{0.6}{2} [ \sqrt{1 + 8(2.06)^2} - 1 ] \approx 1.474 \, \text{m} $$ $$ \text{Final Velocity, } V_2 = \frac{Q}{B \times d_2} = \frac{15}{5 \times 1.474} \approx 2.035 \, \text{m/s} $$

2. Calculate Specific Energy Before and After the Jump

The specific energy (E) is the total energy head relative to the channel bed.

$$ E_1 = d_1 + \frac{V_1^2}{2g} = 0.6 + \frac{5^2}{2 \times 9.81} \approx 0.6 + 1.274 = 1.874 \, \text{m} $$ $$ E_2 = d_2 + \frac{V_2^2}{2g} = 1.474 + \frac{(2.035)^2}{2 \times 9.81} \approx 1.474 + 0.211 = 1.685 \, \text{m} $$

3. Calculate the Loss of Energy

The loss of energy head (\(h_L\)) is the difference in specific energy across the jump. This represents energy loss per unit weight of water.

$$ h_L = E_1 - E_2 = 1.874 - 1.685 = 0.189 \, \text{m} $$

Explanation of Energy Loss in a Hydraulic Jump

A hydraulic jump is a highly turbulent and chaotic event where the flow transitions from a high-velocity, shallow state (supercritical) to a low-velocity, deep state (subcritical). While momentum is conserved across the jump, mechanical energy is not.

The intense mixing and eddy formation within the jump convert a significant amount of the flow's kinetic energy into thermal energy (heat). This is considered an "energy loss" from a hydraulic perspective, as the useful mechanical energy of the flow decreases. The calculation finds this loss by comparing the specific energy (depth + kinetic energy) before the jump to the specific energy after it. The result is expressed as a "head loss" in meters.