Learn: Hydraulic-jump energy loss
A hydraulic jump dissipates mechanical energy even though a momentum balance can relate the depths. For a rectangular horizontal channel, the loss can be written directly using the conjugate depths.
What the formula is saying
Subtract y₁ from y₂, cube the depth increase, and divide by 4y₁y₂. The cubic numerator makes a stronger jump lose disproportionately more head.
Read the symbols in plain language
- y₁
- Upstream depth
Upstream depth. The downstream depth exceeds the upstream depth for the modeled hydraulic jump.
mMetres measure length; 1 m = 1000 mm.
- y₂
- Downstream depth
Downstream depth. The downstream depth exceeds the upstream depth for the modeled hydraulic jump.
mMetres measure length; 1 m = 1000 mm.
- ΔE
- Result to find
Hydraulic-jump energy loss. The cubed rise divided by the depth product leaves metres of energy loss.
m
Sort out the units first
Both depths use m. The numerator has unit m³ and the denominator m², leaving a head loss in m. This is not a flow rate or a percentage of total energy.
Assumptions before calculating
Use positive conjugate depths with y₂ > y₁ from the same rectangular-channel jump, constant discharge, horizontal bed and negligible external work across the jump.
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 ΔE and explain the result in the stated output unit.
- y₁ · Upstream depth
- 0.5 m
- y₂ · Downstream depth
- 2 m
Find the rise through the jump
The downstream depth exceeds the upstream depth for the modeled hydraulic jump.
(2)-(0.5) = 1.5 mForm the depth-product denominator
Use both positive conjugate depths and the numerical factor four.
4 × (0.5) × (2) = 4 m²Calculate dissipated head
The cubed rise divided by the depth product leaves metres of energy loss.
(1.5)^3 ÷ (4) = 0.84375 m
Does this worked answer make sense?
As the two conjugate depths approach equality, modeled loss tends to zero. Scaling both depths by the same factor scales this head loss by that factor.
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
- y₂ · Downstream depth
- 1.6 m
Find the rise through the jump
The downstream depth exceeds the upstream depth for the modeled hydraulic jump.
(1.6)-(0.4) = 1.2 mForm the depth-product denominator
Use both positive conjugate depths and the numerical factor four.
4 × (0.4) × (1.6) = 2.56 m²Calculate dissipated head
The cubed rise divided by the depth product leaves metres of energy loss.
(1.2)^3 ÷ (2.56) = 0.675 m
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
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
- y₂ · Downstream depth
- 2.4 m
Find: Learn: Hydraulic-jump energy loss
A hint, not the answer
Subtract y₁ from y₂, cube the depth increase, and divide by 4y₁y₂. The cubic numerator makes a stronger jump lose disproportionately more head.
Both depths use m. The numerator has unit m³ and the denominator m², leaving a head loss in m. This is not a flow rate or a percentage of total energy.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the rise through the jump
The downstream depth exceeds the upstream depth for the modeled hydraulic jump.
(2.4)-(0.6) = 1.8 mForm the depth-product denominator
Use both positive conjugate depths and the numerical factor four.
4 × (0.6) × (2.4) = 5.76 m²Calculate dissipated head
The cubed rise divided by the depth product leaves metres of energy loss.
(1.8)^3 ÷ (5.76) = 1.0125 m
Avoid the common trap
Do not replace the cube by a square or take the absolute value to conceal reversed upstream and downstream labels. Energy loss is positive in the chosen jump direction.
When this method applies — and when it does not
The calculator checks depth ordering, but two arbitrary depths are not automatically a physically compatible conjugate pair. Confirm momentum compatibility using flow data. This expression does not describe a pipe expansion or a sloping nonrectangular jump unchanged.
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 energy loss. Water-measurement principles; for discharge devices, read the orifice/weir chapters and the installation and head-measurement conditions, not only the coefficient formula.
- U.S. Bureau of Reclamation — Water Measurement Manual, 3rd edition (1997; revised reprint 2001)
- Dawei Han, University of Bristol — Concise Hydraulics (2008, Ventus Publishing)
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.