Hydraulic Jump Calculations Step by step Answers with Explanation

Your question:

A hydraulic jump occurs in a 0.2 m wide rectangular channel at the point where the depth of water flow is 0.15 m and the Froude number is 2. Make calculations for the specific energy, critical and sequent depths, loss of head, and the energy dissipated.

Hydraulic Jump Calculations Answers and Explanation

Calculations for Hydraulic Jump Properties

We need to find:

• Specific energy (E1, E2)

Specific Energy (E)

Specific energy (E) in a hydraulic jump represents the total energy per unit weight of the water and is the sum of potential energy (due to depth) and kinetic energy (due to velocity).

• g = acceleration due to gravity (9.81 m/s²)

However, we are not directly given the velocity (V) in this case. We can utilize the Froude number (Fr) for calculations.

V₁ ≈ 2.77 m/s

Now we can calculate the specific energy before the jump (E₁):

The critical depth (hc) is the depth at which the specific energy has a minimum value. There are various methods to find hc, but we can use the following empirical relationship for rectangular channels:

hc ≈ y₁ / Fr₁²

Since we know E₁ and can assume a negligible velocity after the jump (due to subcritical flow), we can simplify the equation:

E₁ ≈ y

Δh = 0.47 m - 0.15 m = 0.32 m

Energy Dissipated (Ed)

• Specific energy before jump (E₁): 0.47 m (assumed constant)

• Critical depth (hc): 0.0375 m