Hydraulic Jump: Formula, Types, Energy Loss and Uses
A hydraulic jump is a sudden rise in the water surface in an open channel, where fast and shallow flow suddenly becomes slow and deep. You can see one below a dam spillway or a sluice gate as a turbulent, foaming roller. It is one of the most useful energy dissipaters in hydraulic engineering.
When does a jump form?
A jump forms only when supercritical flow (Froude number F > 1) meets subcritical flow (F < 1) downstream. The Froude number is
F = V / √(g y)
where V is velocity and y is depth (for a rectangular channel). The depth before the jump is y1 and the depth after it, called the sequent or conjugate depth, is y2.
Key formulas (rectangular channel)
| Quantity | Formula |
|---|---|
| Sequent depth ratio | y2/y1 = ½ [ √(1 + 8F1²) − 1 ] |
| Energy (head) loss | ΔE = (y2 − y1)³ / (4 y1 y2) |
| Length of jump (approx.) | L ≈ 5 to 7 (y2 − y1); 6 (y2 − y1) is commonly used |
| Height of jump | h = y2 − y1 |
The sequent depth formula comes from the momentum equation: the specific force is the same just before and just after the jump. Energy is not conserved — that loss is the whole point of the jump.
Types of hydraulic jump (by F1)
| Froude number F1 | Type | Behaviour |
|---|---|---|
| 1.0 – 1.7 | Undular jump | Small surface waves, very little energy loss |
| 1.7 – 2.5 | Weak jump | Small rollers, smooth downstream flow, energy loss about 5–15% |
| 2.5 – 4.5 | Oscillating jump | Unstable jet, waves travel far downstream — avoid in design |
| 4.5 – 9.0 | Steady jump | Well-formed, stable; 45–70% energy dissipated — best for stilling basins |
| above 9.0 | Strong jump | Very rough, high energy loss, can damage the basin |
Worked example
Water flows in a wide rectangular channel at q = 2 m³/s per metre width with depth y1 = 0.25 m. Find the sequent depth, energy loss and jump length.
- V1 = q / y1 = 2 / 0.25 = 8.0 m/s
- F1 = 8.0 / √(9.81 × 0.25) = 8.0 / 1.566 = 5.11 → a steady jump
- y2/y1 = ½ [√(1 + 8 × 5.11²) − 1] = 6.74, so y2 = 6.74 × 0.25 = 1.69 m
- ΔE = (1.69 − 0.25)³ / (4 × 0.25 × 1.69) = 1.75 m
- Energy before the jump E1 = y1 + V1²/2g = 0.25 + 3.26 = 3.51 m, so about 50% of the energy is dissipated
- Length L ≈ 6 × (1.69 − 0.25) = 8.6 m
Uses of a hydraulic jump
- Energy dissipation below spillways, chutes and sluice gates, so the high-velocity water does not scour the river bed.
- Raising the water level on the downstream side for canal irrigation.
- Mixing chemicals (such as coagulants) in water treatment plants.
- Removing air pockets from water supply lines.
Exam tip: A jump always moves flow from supercritical to subcritical, never the reverse. If y1 and y2 are given, the energy loss formula (y2 − y1)³/(4y1y2) is the quickest route to the answer.
Frequently asked questions
What is a hydraulic jump?
A sudden rise of water surface when fast, shallow (supercritical) flow changes to slow, deep (subcritical) flow. It happens only when the upstream Froude number is greater than 1.
What is the formula for sequent depth?
For a rectangular channel, y2/y1 = ½[√(1 + 8F1²) − 1], where F1 = V1/√(g y1) is the Froude number before the jump.
How much energy is lost in a hydraulic jump?
The head loss is (y2 − y1)³ / (4 y1 y2). In a steady jump it is often 45–70% of the incoming energy.
Where are hydraulic jumps used?
Below spillways and sluice gates to dissipate energy and protect the river bed, to mix chemicals in water treatment, and to raise water level for irrigation canals.
Fluid mechanics and open channel flow are covered in detail in our notes, with worked examples and a handwritten summary.
See the Fluid Mechanics notes ›