---
title: "Equivalent roughness of compound channel"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:compound-channel-equivalent-roughness-horton-einstein"
source: https://www.bctengineer.com/q/semester_4/hydraulics/compound-channel-equivalent-roughness-horton-einstein
site: BCT Engineering
---

# Equivalent roughness of compound channel

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2080 Chaitra, 4 marks
- Video: [Open Channel Flow - 13 [How to calculate roughness coefficient of a section in uniform flow]](https://www.youtube.com/watch?v=EUZdYRgZZdI) — Abhijit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/compound-channel-equivalent-roughness-horton-einstein
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

A symmetrical compound channel section has main channel geometry of trapezoidal shape with base width 20 m and flood plain both sides as shown in figure below. For a depth of 5 m in symmetrical compound section, determine equivalent roughness using Horton-Einstein Method. Assume Manning's n for the main channel and for the flood plain are 0.021 and 0.039 respectively. [Figure: main channel trapezoid with bed width 20 m, depth 4 m and side slopes 1 vertical to 4 horizontal; flood plains 5 m wide on each side with outer side slope 1 vertical to 2 horizontal]

## Short answer

Apply Horton-Einstein formula: n_e = (∑ P_i n_i^{1.5}/P)^{2/3}. Main channel wetted perimeter 52.99 m (n=0.021), flood plains 14.47 m (n=0.039), total P=67.46 m; yields n_e = 0.0254.

## Answer

**Horton-Einstein method:** each part of the section has the same mean velocity as the whole section, and the equivalent roughness is
$$
n_e=\left[\frac{\sum P_in_i^{3/2}}{P}\right]^{2/3}
$$
where $P_i$ is the wetted perimeter of the part having roughness $n_i$, and $P=\sum P_i$ is the total wetted perimeter.

#### Assumptions from the figure
Main channel: bed width 20 m, depth 4 m, side slopes 1V:4H. Flood plain on each side: horizontal part 5 m wide at the level of the top of the main channel, with an outer bank 1V:2H. Total depth $=5$ m, so the water depth on the flood plain is $5-4=1$ m. The vertical interface lines between the main channel and flood plain are not counted as wetted perimeter.

```text
 flood plain   ~~~~~~~~~~~~~~~~~~~~~  flood plain
  \ 5 m    (1 m deep)      5 m /
   \_______      ________/
           \    /  4 m deep
            \__/  20 m
```

#### Wetted perimeters
- Main channel (n = 0.021): slope length each side $=4\sqrt{1+4^2}=4\times4.1231=16.492$ m
$$
P_m=20+2\times16.492=52.985\ \text{m}
$$
- Each flood plain (n = 0.039): horizontal part $5$ m + outer slope length $1\times\sqrt{1+2^2}=2.236$ m
$$
P_f=5+2.236=7.236\ \text{m};\quad\text{both}=2P_f=14.472\ \text{m}
$$
- Total: $P=52.985+14.472=67.457$ m.

#### Equivalent roughness
$$
n_e=\left[\frac{52.985\,(0.021)^{1.5}+14.472\,(0.039)^{1.5}}{67.457}\right]^{2/3}
$$
$$
(0.021)^{1.5}=0.003043,\qquad (0.039)^{1.5}=0.007702
$$
$$
n_e=\left[\frac{0.2727}{67.457}\right]^{2/3}=(0.004043)^{2/3}=0.0254
$$
**Answer: equivalent Manning's roughness $n_e\approx0.0254$** (between 0.021 and 0.039, weighted towards the larger perimeter of the main channel).

## Check yourself

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

1. Which formula gives the equivalent roughness?
   A. n_e = [sum(P_i n_i^1.5)/P]^(2/3)
   B. n_e = sum(P_i n_i)/P
   C. n_e = [sum(P_i n_i^2)/P]^(1/2)
   D. n_e = sum(A_i n_i)/A
   Answer: A. Horton-Einstein assumes equal mean velocity in each part: n_e = [sum P_i n_i^(3/2) / P]^(2/3). (Key point: Horton-Einstein formula)

2. What does the Horton-Einstein method assume?
   A. Each part has the same discharge
   B. Each part has the same mean velocity as the whole section
   C. Each part has the same area
   D. Each part has the same shear
   Answer: B. It assumes the mean velocity in each subsection equals that of the whole section. (Key point: Basic assumption)

3. How are the vertical interface lines treated?
   A. Counted as wetted perimeter with n = 0.039
   B. Counted with n = 0.021
   C. Not counted in the wetted perimeter
   D. Counted with an average n
   Answer: C. The imaginary interface lines carry no boundary friction and are excluded. (Key point: Interface)

4. What is the equivalent n for this section?
   A. 0.030
   B. 0.039
   C. 0.021
   D. 0.0254
   Answer: D. Total P = 67.457 m and the sum gives n_e = (0.004043)^(2/3) = 0.0254, between 0.021 and 0.039. (Key point: Result)

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