Hydraulic jump
A hydraulic jump forms where fast, shallow (supercritical) flow suddenly changes to slow, deep (subcritical) flow. The water surface rises abruptly with violent turbulence and rolling. It happens below spillways, sluice gates and steep chutes, and engineers deliberately create it in stilling basins to dissipate the huge energy of water released from a dam before it reaches the erodible river bed.
The key parameter is the Froude number, Fr = V / √(g y), comparing flow velocity with the speed of a small surface wave. Fr above 1 means supercritical flow, below 1 subcritical. Applying momentum conservation across the jump, since energy is not conserved, gives the relation between the depths before and after, called conjugate or sequent depths. The energy lost in the jump follows from those depths. The jump's character depends on Fr₁: weak and undular near 1, oscillating and producing waves between about 2.5 and 4.5 (undesirable), steady and efficient between about 4.5 and 9, and strong but rough above that.
Stilling basins use chute blocks, baffle blocks and end sills to stabilise and shorten the jump, with designs chosen by Froude number. The tailwater depth must match the required sequent depth across the range of discharges; if it is too low the jump sweeps out of the basin, and if too high it drowns. Jumps also aid mixing of chemicals at treatment plants.
Froude: Fr = V / √(g y)Sequent depth: y₂ / y₁ = ½ [√(1 + 8 Fr₁²) − 1]Energy loss: ΔE = (y₂ − y₁)³ / (4 y₁ y₂)Jump length ≈ 6 y₂ (approximate, unbaffled)
Problem. Flow leaves a sluice at 0.4 m depth and 8 m/s in a wide rectangular channel. Find Fr₁, the sequent depth and energy loss.
- Fr₁ = 8 / √(9.81 × 0.4) = 8 / 1.981 = 4.04.
- y₂ = 0.2 × [√(1 + 8 × 16.32) − 1] = 0.2 × (11.47 − 1) = 2.09 m.
- ΔE = (2.09 − 0.4)³ / (4 × 0.4 × 2.09) = 4.83 / 3.34 = 1.44 m.
- Fr₁ about 4 lies in the oscillating range, so a basin with appurtenances is needed.
Answer. Fr₁ ≈ 4.0, y₂ ≈ 2.1 m, energy loss ≈ 1.4 m
Quick quiz
1What is the Froude number?
The ratio of flow velocity to the speed of a small surface wave, V/√(gy).
2Why use momentum, not energy, across a jump?
Energy is lost in the turbulence; momentum is conserved.
3What happens if tailwater is too low?
The jump sweeps out of the basin downstream.
Clauses and documents to read
- IS 4997:1968 (design of hydraulic jump type stilling basins)
- IS 6934:1998 (hydraulic design of high ogee overflow spillways)
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- P. N. Modi, S. M. Seth, Hydraulics and Fluid Mechanics including Hydraulic Machines, Standard Book House, latest edition.
- V. T. Chow, Open-Channel Hydraulics, McGraw Hill, 1959.
- K. Subramanya, Flow in Open Channels, McGraw Hill Education (India), latest edition.
Free and official sources
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