UNDERSTAND IT. WORK IT OUT.

Learn: Hydraulic-jump sequent depth

A hydraulic jump changes shallow supercritical flow into deeper subcritical flow. In a horizontal rectangular channel, momentum balance relates the upstream depth and Froude number to the downstream conjugate depth.

Beginner-friendlyFree · No accountTwo worked examples + separate practice
01

What the formula is saying

Compute √(1 + 8Fr₁²), subtract 1, halve the result, then multiply by y₁. The intermediate factor is the ratio y₂/y₁, not the downstream depth itself.

y₂/y₁ = ½(√(1+8Fr₁²) − 1)

Read the symbols in plain language

y₁
Upstream depth

Upstream depth. Multiply the dimensionless ratio by the actual upstream water depth.

m

Metres measure length; 1 m = 1000 mm.

Fr₁
Upstream Froude number

Upstream Froude number. Square the upstream Froude number before applying the factor eight and adding one.

ratio / no unit

A dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.

y₂
Result to find

Hydraulic-jump sequent depth. Multiply the dimensionless ratio by the actual upstream water depth.

m

Sort out the units first

y₁ and y₂ use m; upstream Fr₁ is dimensionless and must exceed one for the jump modeled here. Fr₁ must be based on the upstream section, not the downstream section.

Assumptions before calculating

Assume a short jump in a horizontal rectangular channel with approximately hydrostatic sections before and after the jump, negligible wall-friction force across it, and unchanged discharge and width.

02

Let’s solve one together

Read the given values, follow each operation, then check what the result means.

Read the supplied values as one complete study case. Find y₂ and explain the result in the stated output unit.

y₁ · Upstream depth
0.5 m
Fr₁ · Upstream Froude number
4
  1. Evaluate the momentum-balance root

    Square the upstream Froude number before applying the factor eight and adding one.

    √(1 + 8 × (4)^2) ≈ 11.35781669
  2. Find the conjugate-depth ratio

    Subtracting one and halving gives downstream depth divided by upstream depth.

    ((11.35781669)-1) ÷ 2 ≈ 5.178908346
  3. Recover the downstream depth

    Multiply the dimensionless ratio by the actual upstream water depth.

    (0.5) × (5.178908346) ≈ 2.589454173 m
Answer2.589454173 m

Does this worked answer make sense?

For Fr₁ > 1, the predicted y₂ must exceed y₁. As Fr₁ approaches 1 from above, the depth ratio approaches 1 and the jump weakens.

A second worked example — different values

A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.

y₁ · Upstream depth
0.4 m
Fr₁ · Upstream Froude number
3
  1. Evaluate the momentum-balance root

    Square the upstream Froude number before applying the factor eight and adding one.

    √(1 + 8 × (3)^2) ≈ 8.544003745
  2. Find the conjugate-depth ratio

    Subtracting one and halving gives downstream depth divided by upstream depth.

    ((8.544003745)-1) ÷ 2 ≈ 3.772001873
  3. Recover the downstream depth

    Multiply the dimensionless ratio by the actual upstream water depth.

    (0.4) × (3.772001873) ≈ 1.508800749 m
Answer1.508800749 m
03

Now try your own values

Change a value or its unit. The same method will show your calculation, step by step.

Upstream depth. Multiply the dimensionless ratio by the actual upstream water depth.

Upstream Froude number. Square the upstream Froude number before applying the factor eight and adding one.

English, Arabic and Persian digits are supported. The steps convert inputs to the formula’s base units.

Results update only when you calculate. The lesson example above stays unchanged.

04

Your turn — check your understanding

Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.

y₁ · Upstream depth
0.6 m
Fr₁ · Upstream Froude number
2.5

Find: Learn: Hydraulic-jump sequent depth

For repeating decimals, use at least four significant figures. Accepted rounding tolerance: 0.05% of the expected value; zero uses an absolute tolerance of 10⁻¹².

A hint, not the answer

Compute √(1 + 8Fr₁²), subtract 1, halve the result, then multiply by y₁. The intermediate factor is the ratio y₂/y₁, not the downstream depth itself.

y₁ and y₂ use m; upstream Fr₁ is dimensionless and must exceed one for the jump modeled here. Fr₁ must be based on the upstream section, not the downstream section.

Show the full practice solution

Compare the steps with your work; revealing a solution does not mark the lesson complete.

  1. Evaluate the momentum-balance root

    Square the upstream Froude number before applying the factor eight and adding one.

    √(1 + 8 × (2.5)^2) ≈ 7.141428429
  2. Find the conjugate-depth ratio

    Subtracting one and halving gives downstream depth divided by upstream depth.

    ((7.141428429)-1) ÷ 2 ≈ 3.070714214
  3. Recover the downstream depth

    Multiply the dimensionless ratio by the actual upstream water depth.

    (0.6) × (3.070714214) ≈ 1.842428529 m
Answer1.842428529 m

Avoid the common trap

Do not use downstream Fr in place of upstream Fr or conserve energy through a dissipative jump. Multiplying by y₁ is essential: the square-root expression alone is a ratio.

When this method applies — and when it does not

The equation predicts conjugate depths, not the jump length, location, aeration or stilling-basin design. Tailwater must be considered to determine whether the jump can form where intended; energy is not conserved through the jump.

For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.

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Sources & further reading

References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.

Reading focus: Hydraulic-jump sequent depth. Water-measurement principles; for discharge devices, read the orifice/weir chapters and the installation and head-measurement conditions, not only the coefficient formula.

Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.

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