Audited 31 Jul 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Hydraulic Jump Calculator

Find the depth of water after a hydraulic jump from the upstream depth in metres and upstream Froude number, using the Bélanger equation.

Hydraulic Jump Calculator

m
Conjugate downstream depth
1.886
The depth of the deep, slow flow just after the jump, in metres, that balances the momentum of the incoming flow.
Model scope
Ideal momentum relation for a horizontal rectangular channel; it does not locate the jump, calculate tailwater control, account for slope, width change, air entrainment, sediment, roughness, or energy-dissipator design.

Background.

Water rushing out from under a sluice gate or down a spillway chute is shallow and fast. Downstream, where the channel is deeper and the flow is slow, it cannot stay that way. The change happens abruptly, in a turbulent standing wave called a hydraulic jump: the surface leaps upward, the water churns and fills with air, and a large share of the flow's energy is turned into heat and noise. The same thing happens on a small scale in a kitchen sink, where the thin fast sheet spreading from the tap stream suddenly thickens into a ring.

Engineers build jumps on purpose. A stilling basin at the foot of a dam spillway is designed to hold a jump in place so that the destructive energy of the falling water is spent on a concrete apron rather than on the riverbed. Getting it right starts with one number: the depth the water will reach after the jump, called the conjugate or sequent depth. If the tailwater downstream is lower than that, the jump is swept downstream; if it is much higher, the jump drowns.

The conjugate depth comes from a momentum balance across the jump, first worked out by Bélanger in the nineteenth century. It needs only the incoming depth and the upstream Froude number, the ratio of the flow speed to the speed of a shallow-water wave. A jump can only form when that number is above 1, meaning the flow is supercritical. The relation is the ideal one for a horizontal rectangular channel. It assumes friction across the short jump is negligible and says nothing about where the jump will sit, how long it is, or how a sloping or widening channel changes it.

What is hydraulic jump calculator?

A hydraulic jump is the sudden transition from supercritical flow, shallow and fast with a Froude number above 1, to subcritical flow, deep and slow with a Froude number below 1. The conjugate depth y₂ is the downstream depth that balances the momentum of the incoming flow. For a horizontal rectangular channel the ratio y₂/y₁ depends only on the upstream Froude number, so the same Froude number always gives the same depth ratio whatever the size of the channel.

How to use this calculator.

  1. Enter the upstream depth y1 in metres: the depth of the fast, shallow flow just before the jump, for example at the vena contracta below a sluice gate.
  2. Enter the upstream Froude number. If you know the velocity V instead, compute it as V ÷ √(g × y1) with g = 9.81 m/s²; a value of 1 or less is refused because no jump can form.
  3. Read the conjugate downstream depth in metres and compare it with the tailwater depth your channel actually provides.
  4. Check the model-scope note: the result assumes a horizontal rectangular channel and does not size a stilling basin or locate the jump.

The formula.

y₂ = y₁[ sqrt(1 + 8Fr₁²) - 1 ] / 2

y₂ = y₁ × [√(1 + 8Fr₁²) − 1] ÷ 2. Here y₁ is the upstream depth, Fr₁ the upstream Froude number V₁ ÷ √(g y₁), and y₂ the conjugate depth. The equation comes from equating the momentum flux plus hydrostatic pressure force on each side of the jump, per unit width of a rectangular channel, and solving the resulting quadratic for the positive root. Energy is not conserved across a jump, so energy cannot be used in its place; it is momentum that carries through. The constant 8 arises from that algebra, not from measurement. The Froude number dominates the answer: for large Fr₁ the ratio y₂/y₁ grows almost in proportion to it, approaching √2 × Fr₁, while y₁ simply scales the result, so doubling the incoming depth at the same Froude number doubles y₂.

A worked example.

Example

Water leaves a sluice gate 0.5 m deep with an upstream Froude number of 3, which corresponds to a velocity of about 6.644 m/s. First, 8 × 3² = 72, and √(1 + 72) = √73 ≈ 8.544. Subtracting 1 gives 7.544, and multiplying by y₁ ÷ 2 = 0.25 gives a conjugate depth of 1.886 m, about 3.772 times the incoming depth. The flow per metre of width is 6.644 × 0.5 ≈ 3.322 m²/s, so after the jump the velocity falls to about 1.761 m/s and the Froude number to about 0.41. The specific energy drops from 2.75 m to about 2.044 m, so roughly 0.706 m of head, a quarter of the incoming energy, is dissipated in the jump. At a Froude number of 5 the same 0.5 m flow would jump to about 3.294 m.

upstream Froude Number3
upstream Depth M0.5

Frequently asked questions.

Why must the upstream Froude number be greater than 1?
A jump is a change from supercritical to subcritical flow, so the incoming flow must be supercritical. At a Froude number of exactly 1 the equation returns y₂ = y₁, meaning no jump at all, and below 1 the flow is already subcritical and has nothing to jump to.
What happens if the tailwater is not deep enough?
If the downstream channel cannot hold a depth close to the conjugate depth, the jump moves downstream until the balance is met, often off the protective apron and onto erodible ground. Stilling basins use features such as a lowered floor, chute blocks and end sills to hold the jump in place.
How much energy does a hydraulic jump dissipate?
For a rectangular channel the head loss is (y₂ − y₁)³ ÷ (4y₁y₂). In the 0.5 m, Froude 3 example that is about 0.706 m, roughly a quarter of the upstream specific energy, and the fraction lost grows as the Froude number rises.
What types of hydraulic jump are there?
The classification used in USBR stilling-basin work and standard open-channel texts runs by upstream Froude number: undular from 1 to 1.7, weak from 1.7 to 2.5, oscillating from 2.5 to 4.5, steady from 4.5 to 9, and strong above 9. Oscillating jumps can send irregular waves a long way downstream.
Can I use this for a trapezoidal or circular channel?
No. The equation assumes a rectangular cross-section, where depth and flow area are proportional. Trapezoidal and circular channels need the full momentum function solved for that shape, which usually means iteration rather than a single formula.

How this page was produced

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Quanta Calculator
Primary sources
3 cited below
Method
y₂ = y₁[ sqrt(1 + 8Fr₁²) - 1 ] / 2
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