---
title: "Jump upstream of a low dam"
subject: "Hydraulics"
chapter: 9
key: "semester_4/hydraulics:jump-low-dam-uniform-depth-0-5m"
source: https://www.bctengineer.com/q/semester_4/hydraulics/jump-low-dam-uniform-depth-0-5m
site: BCT Engineering
---

# Jump upstream of a low dam

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

- Asked in: 2071 Magh, 6 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/hydraulics/jump-low-dam-uniform-depth-0-5m
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9

## Question

The depth of uniform flow in a rectangular channel is 5 m wide (n = 0.02, $S_0$ = 0.04) is 0.5 m. A low dam raises the water depth to 2 m. Find whether a hydraulic jump takes place and if so at what distance upstream of the dam.

## Short answer

Discharge 13.95 m3/s, supercritical Fr=2.52. Sequent of 0.5 m is 1.55 m. Since dam depth 2 m exceeds 1.55 m, an S1 curve forms; jump begins where depth is 1.55 m, about 10 m upstream of the dam.

## Answer

**Given:** $b = 5$ m, $n = 0.02$, $S_0 = 0.04$, uniform depth $y_n = 0.5$ m; a low dam raises the depth to 2 m at the dam.

**Discharge**

$$
A = 2.5\ \text{m}^2,\ P = 6\ \text{m},\ R = 0.4167\ \text{m}
$$

$$
Q = \frac{1}{n}AR^{2/3}S_0^{1/2} = \frac{1}{0.02}(2.5)(0.4167)^{2/3}(0.2) = 13.95\ \text{m}^3/\text{s}
$$

$$
q = 2.789\ \text{m}^2/\text{s},\quad V = 5.58\ \text{m/s},\quad Fr = 2.52,\quad y_c = \left(\frac{2.789^2}{9.81}\right)^{1/3} = 0.926\ \text{m}
$$

Uniform flow is supercritical ($y_n < y_c$; steep slope). Upstream of the dam the water is deep and subcritical, so a jump is possible.

#### Does a jump occur?

Sequent depth of 0.5 m:

$$
y_2 = \frac{0.5}{2}\left(\sqrt{1 + 8(2.52)^2} - 1\right) = 1.549\ \text{m}
$$

The depth just upstream of the dam (2 m) is greater than 1.549 m, so the dam backwater is deeper than the jump needs. **A hydraulic jump does occur**, forming upstream of the dam where the S1 backwater curve has depth 1.549 m.

#### Distance upstream of the dam

Direct step method along the S1 curve from $y = 2.0$ m (dam) to 1.549 m (5 steps, $S_0 = 0.04$):

| $y$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ ($\times 10^{-4}$) | $\bar S_f$ ($\times 10^{-4}$) | $\Delta x$ (m) |
|---|---|---|---|---|---|
| 2.000 | 1.395 | 2.0991 | 6.8 | | |
| 1.910 | 1.461 | 2.0185 | 7.7 | 7.2 | -2.05 |
| 1.819 | 1.533 | 1.9392 | 8.8 | 8.2 | -2.02 |
| 1.729 | 1.613 | 1.8618 | 10.1 | 9.4 | -1.98 |
| 1.639 | 1.702 | 1.7865 | 11.7 | 10.9 | -1.93 |
| 1.549 | 1.801 | 1.7139 | 13.8 | 12.8 | -1.87 |

$$
L = 2.05 + 2.02 + 1.98 + 1.93 + 1.87 \approx 9.9\ \text{m}
$$

The steep bed slope ($S_0 = 0.04$) dominates the friction slope, so the S1 curve is short.

**Answer: A jump occurs (0.5 m to 1.55 m). Its downstream end is about 10 m upstream of the dam; the jump itself (length about $6(y_2 - y_1) \approx 6.3$ m) lies just upstream of this point.**

## Check yourself

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

1. How is Q found from the uniform depth of 0.5 m?
   A. Manning's equation Q = (1/n) A R^(2/3) S0^(1/2)
   B. Q = b yc sqrt(g yc)
   C. Momentum equation
   D. Q = A sqrt(g y)
   Answer: A. The uniform flow depth gives Q = 13.95 m^3/s. (Key point: Discharge)

2. Why does a jump occur?
   A. The dam depth (2 m) is less than the sequent depth
   B. The dam depth of 2 m exceeds the sequent depth 1.549 m of the supercritical flow
   C. The flow is subcritical
   D. Fr is 1
   Answer: B. The backwater is deeper than needed, so a jump forms upstream of the dam. (Key point: Does jump occur)

3. How is the distance upstream found?
   A. Using the jump length formula
   B. Using critical depth
   C. Direct step on the S1 curve from 2.0 m to 1.549 m, about 9.9 m
   D. Using Manning's equation
   Answer: C. The jump begins where the S1 depth reaches the sequent depth 1.549 m. (Key point: Distance)

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