Energy Dissipation | Civil Engineering Optional Notes for UPSC PDF Download

Introduction

• Water flowing over a spillway has very high kinetic energy due to conversion of potential energy to kinetic energy.
• If discharged directly into downstream channel, can cause serious scour of channel bed.
• To protect the channel bed, kinetic energy needs to be dissipated before discharge into the downstream channel.

Types of Energy Dissipating Devices

1. Devices using a hydraulic jump for energy dissipation
2. Devices using a bucket for energy dissipation

Choice depends on tailwater depth and characteristics of hydraulic jump formed.

Hydraulic Jump Characteristics

• A hydraulic jump is a sudden, turbulent rise of water occurring when flow changes from supercritical to subcritical state.
• Accompanied by extremely turbulent rollers and significant energy dissipation.

Momentum equation will be applied to the control volume taken at the hydraulic jump section for a unit width perpendicular to the control volume,

On deriving the above equation, we finally get the following result.

The ratio of flow depths after and before the hydraulic jump (y2/y1) is a function of the Froude number of the subcritical flow before hydraulic jump.

Types of Hydraulic Jumps

1. Undular Jump (F1 = 1.0 to 1.7): Undulating water surface, low energy dissipation (~5%)

2. Weak Jump (F1 = 1.7 to 2.5): Small rollers, smooth downstream surface, energy dissipation ~20%

3. Oscillating Jump (F1 = 2.5 to 4.5): Oscillating jet, energy dissipation 20-40%

4. Steady Jump (F1 = 4.5 to 9.0): Stable and balanced, good performance, energy dissipation 45-70%

5. Strong Jump (F1 > 9.0): Rough action, strong surface waves downstream, energy dissipation 70-85%

Hydraulic Jump as an Energy Dissipater

• Analytical equation for energy dissipated with hydraulic jump: ΔE = (y2 - y1)^3 / (4y1y2)
• Power lost by hydraulic jump: P = γw * Q * ΔE

Length of Hydraulic Jump

Empirical equations:

•      a. Safranez equation: L = 5.2y₂
•      b. Bakhmeteff equation: L = 5(y₂ - y₁)
•      c. Smetana equation: L = 6(y₂ - y₁)
•      d. Page equation: L = 5.6y₂

Solved Example: If the Froude number at the drop of a hydraulic jump pool is 6 and the water depth is 0.50 m, find out the length of the hydraulic jump. Calculate the power dissipated with the hydraulic jump if the discharge on the spillway is 1600 m^3/s.
Solution:
Given:
Froude number (F₁) = 6
Initial water depth (y₁) = 0.50 m
Discharge (Q) = 1600 m³/s
Calculate the depth after the hydraulic jump (y₂) using the equation:
(y₂/y₁) = (√(1 + 8F₁²) + 1)/2

Substituting the values:
y₂/y₁ = (√(1 + 8 * 6²) + 1)/2
= (√(1 + 288) + 1)/2
= (√289)/2
= 8.5
y₂ = 8.5 * y₁
= 8.5 * 0.50 m
= 4.0 m

Calculate the length of the hydraulic jump using different empirical equations:
a) Safranez equation: L = 5.2y2 = 5.2 * 4.0 m = 20.8 m
b) Bakhmeteff equation: L = 5(y2 - y1) = 5(4.0 - 0.5) m = 17.5 m
c) Smetana equation: L = 6(y2 - y1) = 6(4.0 - 0.5) m = 21.0 m
d) Page equation: L = 5.6y2 = 5.6 * 4.0 m = 22.4 m
For a conservative design, the longest result is chosen.
Therefore, the length of the hydraulic jump is taken as L = 22.4 m.

Calculate the energy dissipated as head (ΔE) using the equation:

ΔE = (y₂ - y₁)³ / (4y₁y₂)
Substituting the values:
ΔE = (4.0 - 0.5)³ / (4 * 0.5 * 4.0)
= 5.36 m
Calculate the power dissipated (P) using the equation:
P = γw * Q * ΔE
Substituting the values:
P = 9.81 kN/m³ * 1600 m^3/s * 5.36 m
= 84,131 kW
Therefore, the length of the hydraulic jump is 22.4 m, and the power dissipated with the hydraulic jump is 84,131 kW for the given conditions.