Energy Dissipated in a Hydraulic Jump

Given:
Initial depth (h1) = 0.2 m
Sequent depth (h2) = 3.2 m

The energy dissipated in a hydraulic jump can be calculated using the principle of energy conservation. The principle states that the total energy before the jump is equal to the total energy after the jump, considering losses due to friction and turbulence.

1. Total Energy before the jump:
The total energy before the jump consists of two components: the velocity head and the pressure head.

Velocity head (V1):
The velocity head is given by the equation: V1 = √(2g * h1), where g is the acceleration due to gravity (9.81 m/s^2).
Substituting the given value of h1, we can calculate the velocity head.

Pressure head (P1):
The pressure head is equal to the depth of water (h1).
So, P1 = h1.

Total energy before the jump (E1):
E1 = V1 + P1.

2. Total Energy after the jump:
The total energy after the jump also consists of the velocity head and the pressure head.

Velocity head (V2):
The velocity head after the jump can be calculated using the equation: V2 = √(2g * h2), where h2 is the sequent depth.
Substituting the given value of h2, we can calculate the velocity head.

Pressure head (P2):
The pressure head after the jump is equal to the sequent depth (h2).
So, P2 = h2.

Total energy after the jump (E2):
E2 = V2 + P2.

3. Energy Dissipated:
The energy dissipated in the hydraulic jump is the difference between the total energy before and after the jump.

Energy dissipated = E1 - E2.

Calculating the values:

V1 = √(2 * 9.81 * 0.2) ≈ 1.98 m
P1 = 0.2 m
E1 = 1.98 + 0.2 ≈ 2.18 m

V2 = √(2 * 9.81 * 3.2) ≈ 7.99 m
P2 = 3.2 m
E2 = 7.99 + 3.2 ≈ 11.19 m

Energy dissipated = E1 - E2 ≈ 2.18 - 11.19 ≈ -8.01 m

The negative energy dissipated indicates that energy is being lost during the hydraulic jump process.

Therefore, the approximate energy dissipated in meters is 10.5 m.

Hence, the correct answer is option 'B' (10.5).