---
title: "Hydraulic jump definition and downstream Froude number"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:hydraulic-jump-what-is-and-froude-after-jump"
source: https://www.bctengineer.com/q/semester_4/hydraulics/hydraulic-jump-what-is-and-froude-after-jump
site: BCT Engineering
---

# Hydraulic jump definition and downstream Froude number

Hydraulics · Chapter 9: Non-uniform rapidly varied flow (RVF)

- Asked in: 2072 Magh, 1+1+4 marks
- Video: [hydraulic jump explanation](https://www.youtube.com/watch?v=C6LizKJdjOk) — Anand kudoli
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/hydraulic-jump-what-is-and-froude-after-jump
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

What is hydraulic jump? Why is energy principle not applied for the analysis of the jump? Water flows in a 5 m wide rectangular channel at Froude number 3.5; the depth of flow is 1.2 m. If water undergoes a hydraulic jump, what is the Froude number downstream of jump?

## Short answer

A hydraulic jump is the turbulent transition from supercritical to subcritical flow. Energy principle fails because dissipation in the roller is unknown; momentum is used instead. For Fr1=3.5 and y1=1.2 m, y2=5.37 m and Fr2=0.37.

## Answer

#### Hydraulic jump

A **hydraulic jump** is the sudden, turbulent rise of the water surface that occurs when a supercritical flow ($Fr > 1$) changes into subcritical flow ($Fr < 1$) in an open channel. It has a violent roller, strong mixing and large energy loss.

#### Why the energy principle is not used

The jump contains intense turbulence and eddies, and the energy dissipated (converted to heat) is unknown and cannot be written as a simple term. Hence the energy equation cannot be solved for the unknown depth. Instead the **momentum equation** is used, because the forces on the short control volume are known (the hydrostatic forces at the ends), while friction on the short, horizontal bed is negligible.

#### Numerical part

**Given:** $b = 5$ m, $Fr_1 = 3.5$, $y_1 = 1.2$ m.

**Sequent depth**

$$
y_2 = \frac{y_1}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right) = 0.6\left(\sqrt{1 + 8(12.25)} - 1\right) = 5.37\ \text{m}
$$

**Velocities**

$$
V_1 = Fr_1\sqrt{g\,y_1} = 3.5\sqrt{9.81 \times 1.2} = 12.01\ \text{m/s}
$$

$$
V_2 = V_1\frac{y_1}{y_2} = 12.01 \times \frac{1.2}{5.37} = 2.68\ \text{m/s}
$$

**Froude number after the jump**

$$
Fr_2 = \frac{V_2}{\sqrt{g\,y_2}} = \frac{2.68}{\sqrt{9.81 \times 5.37}} = 0.37
$$

**Answer: $Fr_2 \approx 0.37$** (subcritical; $y_2 = 5.37$ m).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is a hydraulic jump?
   A. A sudden turbulent rise of the surface as supercritical flow changes to subcritical flow
   B. A gradual rise from subcritical to supercritical flow
   C. A fall in the water surface over a hump
   D. A rise from subcritical to supercritical at a gate
   Answer: A. It is a turbulent roller changing flow from Fr > 1 to Fr < 1 with large energy loss. (Key point: Definition)

2. Why is the energy principle not used for the jump?
   A. Energy is always conserved
   B. The energy dissipated in turbulence is unknown, so momentum (with known hydrostatic forces) is used
   C. Friction on the bed is large
   D. The flow is subcritical
   Answer: B. Unknown loss makes the energy equation unsolvable, while momentum forces are known. (Key point: Why not energy)

3. What is the sequent depth for Fr1 = 3.5 and y1 = 1.2 m?
   A. 2.4 m
   B. 4.2 m
   C. 5.37 m
   D. 3.5 m
   Answer: C. y2 = 0.6(sqrt(1+8 x 12.25) - 1) = 5.37 m. (Key point: Sequent depth)

4. What is Fr2 after the jump?
   A. About 0.37, subcritical
   B. About 1.0
   C. About 3.5
   D. About 2.7
   Answer: A. V2 = V1 y1/y2 = 2.68 m/s, so Fr2 = 2.68/sqrt(9.81 x 5.37) = 0.37. (Key point: Downstream Froude)

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