---
title: "Hydraulic jump: derivation and numerical"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_4/fluid_mechanics_mechanical:hydraulic-jump-conjugate-depth-loss"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/hydraulic-jump-conjugate-depth-loss
site: BCT Engineering
---

# Hydraulic jump: derivation and numerical

Fluid Mechanics · Chapter 4: Basic Equations for Fluid Flow

- Asked in: year not given, 4+4 marks
- Video: [Hydraulic Jump - The Basic Idea and Equations](https://www.youtube.com/watch?v=oEDxT99TM1U) — CE340Hydraulics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/hydraulic-jump-conjugate-depth-loss
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-4

## Question

(a) Define hydraulic jump. Derive the relation between the conjugate depths $y_1$ and $y_2$ of a hydraulic jump in a horizontal rectangular channel, and the expression for the head loss.
(b) Water flows at 6 m/s with a depth of 0.3 m in a wide horizontal rectangular channel and a hydraulic jump forms. Find the downstream depth and velocity, the downstream Froude number, the head loss, and the power dissipated per metre width.

## Short answer

Downstream depth is 1.34 m, velocity 1.34 m/s, Froude number 0.37, head loss 0.70 m, and power dissipated 12.4 kW per metre width.

## Answer

#### (a) Hydraulic jump
A **hydraulic jump** is the sudden rise in water surface when a rapid (supercritical, $Fr > 1$) flow changes to a slow (subcritical, $Fr < 1$) flow. It is a turbulent phenomenon with large energy dissipation.

**Derivation** (horizontal channel, unit width, neglect friction on the bed): apply the momentum equation to a control volume enclosing the jump.

Hydrostatic forces: $F_1 = \tfrac12\rho g y_1^2$ and $F_2 = \tfrac12\rho g y_2^2$. Discharge per unit width: $q = V_1y_1 = V_2y_2$.

$$
\tfrac12\rho g y_1^2 - \tfrac12\rho g y_2^2 = \rho q (V_2 - V_1)
$$

Using $V_1 = q/y_1$, $V_2 = q/y_2$:

$$
\frac{g}{2}(y_1^2 - y_2^2) = q^2\left(\frac{1}{y_2} - \frac{1}{y_1}\right) = \frac{q^2 (y_1 - y_2)}{y_1y_2}
$$

Dividing by $(y_1 - y_2)$ and rearranging:

$$
\frac{q^2}{g} = \frac{y_1y_2(y_1 + y_2)}{2}
$$

With $q = V_1y_1$ and $Fr_1 = V_1/\sqrt{gy_1}$, this gives the quadratic in $y_2/y_1$:

$$
\left(\frac{y_2}{y_1}\right)^2 + \frac{y_2}{y_1} - 2Fr_1^2 = 0 \quad \Rightarrow \quad \frac{y_2}{y_1} = \frac12\left(\sqrt{1 + 8Fr_1^2} - 1\right)
$$

**Head loss:** $h_L = E_1 - E_2$, where $E = y + \dfrac{V^2}{2g}$. Substituting from above:

$$
h_L = \frac{(y_2 - y_1)^3}{4y_1y_2}
$$

#### (b) Numerical
Given $y_1 = 0.3\ \text{m}$, $V_1 = 6\ \text{m/s}$.

$$
\begin{aligned}
q &= 0.3 \times 6 = 1.8\ \text{m}^2/\text{s} \\
Fr_1 &= \frac{6}{\sqrt{9.81 \times 0.3}} = 3.497 \\
y_2 &= \frac{0.3}{2}\left(\sqrt{1 + 8(3.497)^2} - 1\right) = 1.341\ \text{m} \\
V_2 &= \frac{q}{y_2} = \frac{1.8}{1.341} = 1.342\ \text{m/s} \\
Fr_2 &= \frac{1.342}{\sqrt{9.81 \times 1.341}} = 0.370
\end{aligned}
$$

Head loss and power:

$$
\begin{aligned}
h_L &= \frac{(1.341 - 0.3)^3}{4 \times 0.3 \times 1.341} = 0.702\ \text{m} \\
P &= \rho g q h_L = 1000 \times 9.81 \times 1.8 \times 0.702 = 12\,390\ \text{W/m} = 12.4\ \text{kW per m width}
\end{aligned}
$$

Since $Fr_2 < 1$, the downstream flow is subcritical, as expected.

**Answer: $y_2 = 1.34\ \text{m}$, $V_2 = 1.34\ \text{m/s}$, $Fr_2 = 0.37$, $h_L = 0.70\ \text{m}$, $P = 12.4\ \text{kW per metre width}$**

## Check yourself

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

1. A hydraulic jump is:
   A. a sudden rise of water surface when supercritical flow (Fr > 1) changes to subcritical flow (Fr < 1)
   B. a drop in surface when subcritical becomes supercritical
   C. a gradual change of depth in uniform flow
   D. flow over a weir
   Answer: A. It is a turbulent transition from rapid to slow flow with large energy dissipation. (Key point: Hydraulic jump definition)

2. The conjugate depth relation is derived by applying which equation to a control volume enclosing the jump?
   A. Energy equation
   B. Momentum equation with hydrostatic forces 1/2 rho g y^2 at each end
   C. Bernoulli's equation
   D. Continuity only
   Answer: B. Energy is lost in the jump, so momentum is used: 1/2 rho g (y1^2 - y2^2) = rho q (V2 - V1). (Key point: Derivation basis)

3. For y1 = 0.3 m and Fr1 = 3.497, y2/y1 = 0.5(sqrt(1 + 8 Fr1^2) - 1) gives y2 about:
   A. 0.70 m
   B. 1.34 m
   C. 2.10 m
   D. 3.50 m
   Answer: B. sqrt(1 + 8 x 12.23) = 9.93, so y2/y1 = 4.47 and y2 = 1.34 m. (Key point: Conjugate depth result)

4. The head loss in the jump is h_L = (y2 - y1)^3 / (4 y1 y2). The power dissipated per metre width is:
   A. rho g q h_L = 12.4 kW/m
   B. rho g y2 h_L
   C. 0.5 rho q V1^2
   D. rho q h_L / g
   Answer: A. P = rho g q h_L = 1000 x 9.81 x 1.8 x 0.702 = 12.4 kW per metre width. (Key point: Head loss and power)

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