# Hydraulic jump calculations answers and explanation

__Hydraulic
Jump Calculations__ **Step by step Answers with
Explanation**

__Hydraulic Jump Calculations__

## 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

__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