---
title: "Bresse's method for 20 m wide channel"
subject: "Hydraulics"
chapter: 8
key: "semester_4/hydraulics:bresse-method-20m-channel-65-m3s"
source: https://www.bctengineer.com/q/semester_4/hydraulics/bresse-method-20m-channel-65-m3s
site: BCT Engineering
---

# Bresse's method for 20 m wide channel

Hydraulics · Chapter 8: Non-uniform gradually varied flow (GVF)

- Asked in: 2078 Chaitra, 3+6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/bresse-method-20m-channel-65-m3s
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8

## Question

A rectangular channel is 20 m wide carries a discharge of 65 m³/s. It is laid at a slope of 0.0001. At a certain section along the channel length the flow depth is 2 m. What is the type of surface profile? How far upstream or downstream will the depth be 2.6 m? Use direct integration (Bresse's) method considering increment of depth = 0.3 m. Take energy correction factor equals to 1.1 and manning's roughness coefficient n = 0.025.

## Short answer

The profile is M2 on a mild slope because normal depth 4.024 m exceeds critical depth 1.058 m and the section depth lies between them. Bresse integration gives the 2.6 m depth about 1.29 km upstream of the 2 m section.

## Answer

**Given:** $b = 20\ \text{m}$, $Q = 65\ \text{m}^3/\text{s}$ ($q = 3.25\ \text{m}^2/\text{s}$), $S_0 = 0.0001$, $n = 0.025$, $\alpha = 1.1$, $y = 2\ \text{m}$ at a section.

**Critical and normal depth**

$$
y_c = \left(\frac{\alpha q^2}{g}\right)^{1/3} = \left(\frac{1.1 \times 3.25^2}{9.81}\right)^{1/3} = 1.058\ \text{m}
$$

Manning's equation with $R = A/P$, solved by trial: $y_n = 4.024\ \text{m}$ ($A = 80.5\ \text{m}^2$, $R = 2.87$ m, $V = 0.81$ m/s).

#### Type of profile

$y_n > y > y_c$, and the slope is mild ($y_n > y_c$), so the profile is **M2** (drawdown). On an M2 curve the depth increases upstream, so the depth of 2.6 m lies **upstream** of the 2 m section.

#### Bresse's method

For a wide channel with constant Chezy $C$:

$$
x = \frac{y_n}{S_0}\left[u - \left(1 - \left(\frac{y_c}{y_n}\right)^3\right)\phi(u)\right],\qquad u = \frac{y}{y_n}
$$

$$
\phi(u) = \frac{1}{6}\ln\frac{u^2 + u + 1}{(u-1)^2} + \frac{1}{\sqrt3}\tan^{-1}\frac{2u + 1}{\sqrt3}
$$

Here $\dfrac{y_n}{S_0} = 40\,238\ \text{m}$ and $1 - (y_c/y_n)^3 = 1 - (0.2629)^3 = 0.9818$.

| $y$ (m) | $u$ | $\phi(u)$ | $u - 0.9818\,\phi$ | $x$ (m) |
|---|---|---|---|---|
| 2.0 | 0.4970 | 0.8158 | -0.3039 | -12 228 |
| 2.3 | 0.5716 | 0.9039 | -0.3159 | -12 709 |
| 2.6 | 0.6461 | 1.0004 | -0.3360 | -13 521 |

Distances relative to the 2 m section:

- 2.0 m to 2.3 m: $-12\,709 - (-12\,228) = -481$ m
- 2.3 m to 2.6 m: $-13\,521 - (-12\,709) = -812$ m

$$
x_{2.0 \to 2.6} = -1292\ \text{m}
$$

The negative sign means upstream. As a check, the direct step method with the actual rectangular geometry gives about 1220 m with the same two 0.3 m steps and about 1276 m with a fine integration, so the Bresse result is reasonable.

**Answer: M2 profile; the depth is 2.6 m about 1.29 km upstream of the section where the depth is 2 m.**

## Check yourself

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

1. Why is the profile M2?
   A. y < yc on a mild slope
   B. yn > y > yc on a mild slope (yc = 1.058, y = 2, yn = 4.024)
   C. y > yn on a mild slope
   D. The slope is steep
   Answer: B. With yn > y > yc on a mild slope the profile is the drawdown curve M2. (Key point: Profile type)

2. Where is the 2.6 m depth relative to the 2 m section?
   A. Downstream, since M2 depth rises downstream
   B. At the same section
   C. Upstream, since M2 depth rises upstream toward yn
   D. Cannot be determined
   Answer: C. For M2 depth increases going upstream towards yn, so 2.6 m lies upstream (about 1.29 km). (Key point: Direction)

3. How is critical depth found with energy coefficient 1.1?
   A. yc = (q^2/g)^(1/3) ignoring alpha
   B. yc = (alpha q^2 / g)^(1/3)
   C. yc = (q/alpha)^(1/3)
   D. yc = 2yn/3
   Answer: B. Critical flow requires alpha Q^2 T/(gA^3)=1, giving yc = (alpha q^2/g)^(1/3) = 1.058 m. (Key point: Critical depth with alpha)

4. How is the distance between two depths obtained in Bresse's method?
   A. Add the x values of both depths
   B. Take the difference of x = (yn/S0)[u - (1-(yc/yn)^3) phi(u)] at the two depths
   C. Divide energy change by slope
   D. Multiply depth change by S0
   Answer: B. x is computed for each u = y/yn and the difference gives the length between depths. (Key point: Bresse distance)

## Related in Hydraulics

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