---
title: "Bump followed by hydraulic jump, find v1, v2, y4"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:flow-over-bump-then-jump-v1-v2-y4"
source: https://www.bctengineer.com/q/semester_4/hydraulics/flow-over-bump-then-jump-v1-v2-y4
site: BCT Engineering
---

# Bump followed by hydraulic jump, find v1, v2, y4

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

- Asked in: 2070 Magh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/flow-over-bump-then-jump-v1-v2-y4
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

Water in a horizontal channel accelerates smoothly over a bump and then undergoes a hydraulic jump, as in figure below. If $y_1$ = 1 m and $y_2$ = 30 cm, estimate $v_1$, $v_2$ and $y_4$. Neglect friction. [Figure: section (1) upstream; section (2) after the bump of height h with supercritical depth $y_2$; section (3) at the start of the jump; a hydraulic jump leading to section (4)]

## Short answer

Continuity and energy give v1 = 1.17 m/s, v2 = 3.89 m/s. Hydraulic jump from y2 = 0.3 m (Fr=2.27) gives sequent y4 = 0.82 m.

## Answer

**Given:** horizontal channel, $y_1 = 1$ m (upstream, subcritical), $y_2 = 0.3$ m (supercritical, after the bump). The jump takes the flow from section 3 (start of jump, still at $y_2 = 0.3$ m because friction is neglected) to section 4. Friction is neglected, and sections 1 and 2 are at the same bed level.

#### $v_1$ and $v_2$

Continuity: $v_1y_1 = v_2y_2$, so $v_2 = v_1/0.3$.

Energy (no loss between 1 and 2):

$$
y_1 + \frac{v_1^2}{2g} = y_2 + \frac{v_2^2}{2g}
$$

$$
1 - 0.3 = \frac{v_1^2}{2g}\left[\left(\frac{1}{0.3}\right)^2 - 1\right] \;\Rightarrow\; v_1^2 = \frac{2(9.81)(0.7)}{10.111} = 1.358
$$

$$
v_1 = 1.165\ \text{m/s},\qquad v_2 = 3.885\ \text{m/s}
$$

#### $y_4$ (depth after the jump)

The supercritical depth at the jump toe is $y_3 = y_2 = 0.3$ m (the same flow, no losses before the jump):

$$
Fr_3 = \frac{3.885}{\sqrt{9.81 \times 0.3}} = 2.265
$$

$$
y_4 = \frac{y_3}{2}\left(\sqrt{1 + 8Fr_3^2} - 1\right) = 0.15\left(\sqrt{1 + 41.04} - 1\right) = 0.822\ \text{m}
$$

Check by specific force (per unit width): $\dfrac{q^2}{g\,y_3} + \dfrac{y_3^2}{2} = \dfrac{1.358}{9.81 \times 0.3} + 0.045 = 0.506\ \text{m}^2$ and $\dfrac{1.358}{9.81 \times 0.822} + \dfrac{0.822^2}{2} = 0.168 + 0.338 = 0.506\ \text{m}^2$. The two are equal.

**Answer: $v_1 = 1.17$ m/s; $v_2 = 3.89$ m/s; $y_4 = 0.82$ m.**

## Check yourself

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

1. Why can the energy equation be used between sections 1 and 2?
   A. There is a jump between them
   B. There is no loss between them (friction neglected, smooth flow)
   C. The beds are at different levels
   D. Momentum is unknown
   Answer: B. The flow is smooth and frictionless there, so E1 = E2, giving v1 = 1.165 m/s. (Key point: Energy between 1 and 2)

2. What is the depth y3 at the start of the jump?
   A. Equal to y1 = 1 m
   B. Equal to y4
   C. Equal to y2 = 0.3 m, since there is no loss before the jump
   D. Critical depth
   Answer: C. With no losses the supercritical depth stays 0.3 m until the jump. (Key point: Depth at jump toe)

3. How can y4 = 0.822 m be checked?
   A. Equal specific energy at 3 and 4
   B. Equal specific force at 3 and 4 (0.506 m^2)
   C. Equal velocity
   D. Equal Froude number
   Answer: B. Momentum (specific force) is conserved across the jump. (Key point: Check of y4)

## Close variants set in other exams

- [Bump followed by hydraulic jump, find bump height](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aflow-over-bump-then-jump-v1-v3-y4-h) — Hydraulics, chapter 9, asked 2×