A hydraulic jump occurs, in a triangular (V-shaped) channel with side slopes 1:1 (vertical to horizontal). The sequent depths are 0.5 m and 1.5 m. The flow rate (in m3/s, round off to two decimal places) in the channel is _________.Correct answer is '1.73'. Can you explain this answer?

Question

Answer

Given data:

- Side slopes of the triangular channel: 1:1 (vertical to horizontal)
- Sequent depths: 0.5 m and 1.5 m

Understanding a hydraulic jump:

- A hydraulic jump occurs when the flow changes from supercritical to subcritical.
- It is a sudden change in flow velocity and water depth.
- In a hydraulic jump, the flow energy is converted into turbulence and heat energy.

Equation for flow rate:

- The flow rate (Q) in a channel can be calculated using the equation Q = A * V, where A is the cross-sectional area and V is the velocity of flow.

Calculating the flow rate:

1. Calculate the cross-sectional areas of the two depths:
- Cross-sectional area at the first depth (A1) = (1/2) * b1 * y1, where b1 is the base width and y1 is the depth.
- Cross-sectional area at the second depth (A2) = (1/2) * b2 * y2, where b2 is the base width and y2 is the depth.

2. Calculate the base width:
- Since the side slopes are 1:1, the base width (b) can be calculated using the equation b = 2 * y, where y is the depth.

3. Calculate the velocities at the two depths:
- The velocity at the first depth (V1) can be calculated using the equation V1 = Q / A1.
- The velocity at the second depth (V2) can be calculated using the equation V2 = Q / A2.

4. Apply the energy equation:
- The energy equation for a hydraulic jump in a triangular channel is given by:
(V1^2 / (2g)) + (y1 - y2) = (V2^2 / (2g))
where g is the acceleration due to gravity (9.81 m/s^2).

5. Substitute the given values and solve for Q:
- (V1^2 / (2g)) + (y1 - y2) = (V2^2 / (2g))
- (Q^2 / (2g * A1^2)) + (y1 - y2) = (Q^2 / (2g * A2^2))
- (Q^2 / (2 * 9.81 * (0.5^2))) + (0.5 - 1.5) = (Q^2 / (2 * 9.81 * (1.5^2)))
- Simplifying the equation, we get: Q^2 = 1.73 * 9.81 * (0.5^2) * (1.5^2)
- Taking the square root of both sides, we get: Q ≈ 1.73 m^3/s

Therefore, the flow rate in the channel is approximately 1.73 m^3/s.

Similar GATE Doubts

The pre-jump Froude Number for a particular flow in a horizontal rectangular channel is 10. The ratio of sequent depths (i.e., post-jump depth to pre-jump depth) is ___________ Correct answer is '13.65'. Can you explain this answer?

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