---
title: "Slope break: jump depth and jump length by step"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:slope-break-jump-downstream-depth-and-length"
source: https://www.bctengineer.com/q/semester_4/hydraulics/slope-break-jump-downstream-depth-and-length
site: BCT Engineering
---

# Slope break: jump depth and jump length by step

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

- Asked in: 2075 Baisakh, 5+5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/slope-break-jump-downstream-depth-and-length
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

A rectangular channel section has a change in slope as shown in figure below. The channel section is 4 m wide having Manning's n = 0.0165. The bed slope $S_{02}$ = 0.0024 and the flowing discharge is 16 m³/sec. a) Calculate the depth that must exist in the downstream channel for a hydraulic jump to terminate at uniform flow condition. b) If upstream depth $Y_{01}$ = 0.4 m, calculate the length of hydraulic jump using at least three increments of depth in a step calculation. [Figure: steep upstream reach with uniform depth $Y_{01}$ = 0.4 m on slope $S_{01}$ changing to a milder downstream reach with slope $S_{02}$ and uniform depth $Y_{02}$]

## Short answer

Downstream uniform depth from Manning is Y02=1.495 m, so the jump toe occurs where the M3 curve depth is 0.908 m. Step calculation over three increments gives the M3 length about 92 m; jump length is about 3.5 m.

## Answer

**Given:** $b = 4$ m, $Q = 16\ \text{m}^3/\text{s}$ ($q = 4\ \text{m}^2/\text{s}$), $n = 0.0165$, downstream slope $S_{02} = 0.0024$, upstream uniform depth $Y_{01} = 0.4$ m (steep reach).

$$
y_c = \left(\frac{4^2}{9.81}\right)^{1/3} = 1.177\ \text{m}
$$

#### a) Depth in the downstream channel

For the jump to end in uniform flow, the depth after the jump must equal the normal depth of the downstream channel. From Manning's equation with $R = A/P$ (trial):

$$
Y_{02} = y_n = 1.495\ \text{m}\quad (> y_c,\ \text{so the slope is mild; } V = 2.68\ \text{m/s})
$$

For comparison, the sequent depth of $Y_{01} = 0.4$ m is:

$$
V_{01} = 10\ \text{m/s},\ Fr = 5.05,\quad y_2 = \frac{0.4}{2}\left(\sqrt{1 + 8(5.05)^2} - 1\right) = 2.66\ \text{m}
$$

Since $2.66 > 1.495$ m, the tailwater is too low for the jump to form at the break. The supercritical flow continues on the mild slope as an **M3** curve until its depth $y_j$ has a sequent depth equal to 1.495 m:

$$
y_j:\ \frac{y_j}{2}\left(\sqrt{1 + 8\frac{q^2}{g\,y_j^3}} - 1\right) = 1.495 \;\Rightarrow\; y_j = 0.908\ \text{m}
$$

#### b) Length of the M3 profile up to the jump (3 steps)

Depth from 0.40 m to 0.908 m, with $\Delta y = 0.169$ m:

$$
E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2V^2}{R^{4/3}},\quad \Delta x = \frac{E_2 - E_1}{S_{02} - \bar S_f}
$$

| $y$ (m) | $R$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ | $\bar S_f$ | $\Delta x$ (m) |
|---|---|---|---|---|---|---|
| 0.4000 | 0.3333 | 10.000 | 5.4968 | 0.11780 | | |
| 0.5693 | 0.4432 | 7.026 | 3.0851 | 0.03977 | 0.07878 | 31.6 |
| 0.7387 | 0.5395 | 5.415 | 2.2332 | 0.01818 | 0.02897 | 32.1 |
| 0.9080 | 0.6245 | 4.405 | 1.8971 | 0.00990 | 0.01404 | 28.9 |

$$
L = 31.6 + 32.1 + 28.9 \approx 92.5\ \text{m}
$$

The jump itself then raises the depth from 0.908 m to 1.495 m over a length of about $6(y_2 - y_1) = 6(1.495 - 0.908) \approx 3.5$ m.

**Answer: (a) $Y_{02} = 1.495$ m; (b) the toe of the jump is about 92 m downstream of the slope change (M3 profile), and the jump is about 3.5 m long.**

## Check yourself

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

1. What depth must exist downstream for the jump to end in uniform flow?
   A. Critical depth, 1.177 m
   B. The sequent depth of 0.4 m, 2.66 m
   C. The normal depth of the downstream channel, 1.495 m
   D. The upstream depth, 0.4 m
   Answer: C. The depth after the jump must equal the normal depth Y02 from Manning's equation. (Key point: Downstream depth)

2. Why does an M3 curve form before the jump?
   A. The sequent depth 2.66 m is greater than the normal depth 1.495 m, so the tailwater is too low to form the jump at the break
   B. The sequent depth is less than the normal depth
   C. The downstream slope is steep
   D. The upstream flow is subcritical
   Answer: A. Supercritical flow continues on the mild slope until its depth has a sequent depth equal to 1.495 m (0.908 m). (Key point: Why M3 curve)

3. Which formula is used for each step length in the direct step method?
   A. dx = (E2 - E1)/(S0 - Sf mean)
   B. dx = dy x S0
   C. dx = (E2 + E1)/Sf
   D. dx = dy/Sf
   Answer: A. The energy difference is divided by the bed slope minus the mean friction slope. (Key point: Step method)

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