---
title: "Bump followed by hydraulic jump, find bump height"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:flow-over-bump-then-jump-v1-v3-y4-h"
source: https://www.bctengineer.com/q/semester_4/hydraulics/flow-over-bump-then-jump-v1-v3-y4-h
site: BCT Engineering
---

# Bump followed by hydraulic jump, find bump height

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

- Asked in: 2072 Asoj, 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-v3-y4-h
- 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, $y_3$ = 30 cm, estimate $v_1$, $v_3$, $y_4$ and bump height h. Neglect friction. [Figure: section (1) upstream with depth $y_1$; section (2) on the bump crest of height h; section (3) after the bump with supercritical depth $y_3$; a hydraulic jump leading to section (4)]

## Short answer

Apply continuity and energy between sections 1 and 3 to get v1 = 1.17 m/s, v3 = 3.89 m/s. Sequent depth of supercritical y3 yields y4 = 0.82 m. Critical depth at crest gives bump height h ≈ 0.29 m.

## Answer

**Given:** horizontal channel, $y_1 = 1$ m (upstream, subcritical), $y_3 = 0.3$ m (supercritical, after the bump), then a hydraulic jump to $y_4$. Friction is neglected. Section 1 and section 3 are at the same bed level.

#### $v_1$ and $v_3$

Continuity: $v_1y_1 = v_3y_3$, so $v_3 = v_1(1/0.3) = 3.333\,v_1$.

Energy (smooth flow over the bump, no loss between 1 and 3):

$$
y_1 + \frac{v_1^2}{2g} = y_3 + \frac{v_3^2}{2g}
$$

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

$$
v_1 = 1.165\ \text{m/s},\qquad v_3 = 3.885\ \text{m/s}
$$

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

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

$$
y_4 = \frac{0.3}{2}\left(\sqrt{1 + 8(2.265)^2} - 1\right) = 0.822\ \text{m}
$$

#### Bump height $h$

The flow changes from subcritical to supercritical smoothly over the bump, so it passes through critical depth at the crest (section 2). With $q = v_1y_1 = 1.165\ \text{m}^2/\text{s}$:

$$
y_c = \left(\frac{1.165^2}{9.81}\right)^{1/3} = 0.517\ \text{m},\qquad E_c = 1.5\,y_c = 0.776\ \text{m}
$$

$$
E_1 = 1 + \frac{1.165^2}{2g} = 1.069\ \text{m}
$$

$$
h = E_1 - E_c = 1.069 - 0.776 = 0.293\ \text{m}
$$

**Answer: $v_1 = 1.17$ m/s; $v_3 = 3.89$ m/s; $y_4 = 0.82$ m; $h \approx 0.29$ m** (critical flow assumed at the crest).

## Check yourself

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

1. Which equations give v1 and v3?
   A. Momentum and continuity
   B. Manning and continuity
   C. Energy equation between 1 and 3 (no loss) plus continuity v1 y1 = v3 y3
   D. Belanger and continuity
   Answer: C. No loss over the smooth bump, so energy is conserved; with continuity v1 = 1.165 m/s. (Key point: Finding v1)

2. What is the flow depth at the bump crest?
   A. Critical depth
   B. The depth y1
   C. The depth y3
   D. The depth y4
   Answer: A. Flow passes smoothly from subcritical to supercritical, so it is critical at the crest. (Key point: Depth at crest)

3. How is h found?
   A. h = y1 - y3
   B. h = E1 - Ec, with Ec = 1.5 yc
   C. h = Ec - E1
   D. h = y4 - y1
   Answer: B. The crest energy is Ec = 1.5 yc, so h = 1.069 - 0.776 = 0.29 m. (Key point: Bump height)

4. How is y4 found?
   A. Energy equation from 3 to 4
   B. Belanger equation using Fr3 = 2.265 and y3 = 0.3 m
   C. y4 = y1
   D. Critical depth
   Answer: B. Across the jump momentum applies; y4 = 0.822 m. (Key point: Depth y4)

## Close variants set in other exams

- [Bump followed by hydraulic jump, find v1, v2, y4](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aflow-over-bump-then-jump-v1-v2-y4) — Hydraulics, chapter 9, asked 2×