Question

A spillway discharges flood at a rate of 9 m³/s per meter width. If the depth of flow on the horizontal apron at the toe of the spillway is 46 cm, the tailwater depth needed to form a hydraulic jump is approximately given by which of the following?
1. 2.54 m
2. 4.90 m
3. 5.77 m
4. 6.23 m

Answer 1

Correct Answer - Option 3 : 5.77 m

Concept:

Hydraulic Jump: 

In an open channel when a rapidly flowing stream abruptly changes into a slowly flowing stream, a distinct rise or jump in elevation of fluid surface happens, this phenomenon is known as Hydraulic Jump.

For a hydraulic jump, the height of the slow stream or subcritical stream is (y₂) = \(- \frac{y_1}{2} + \sqrt{\frac{y_1^2}{4}\ + \frac{q^2}{gy_1}}\)

y₁ = Height of the rapidly flowing stream or supercritical stream

g = Acceleration due to gravity

q = Discharge per metre width in m³/sec

Calculation:

Given:

y1 = height of horizontal apron = 46 cm = 0.46 m

q = 9 m3/sec

∴ y₂ = \(- \frac{0.46}{2} + \sqrt{\frac{(0.46)^2}{4}\ + \frac{9^2}{9.81\times 0.46}}\)

y2 = 5.77 m.