---
title: "Most efficient trapezoidal channel (3H:4V)"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:efficient-trapezoidal-3h4v"
source: https://www.bctengineer.com/q/semester_6/hydro_power/efficient-trapezoidal-3h4v
site: BCT Engineering
---

# Most efficient trapezoidal channel (3H:4V)

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2070 Magh, 6 marks
- Video: [Hydraulically Efficient Trapezoidal Channel Section](https://www.youtube.com/watch?v=Z1RrWWY45zA) — WATERCORE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/efficient-trapezoidal-3h4v
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Calculate dimensions of a most efficient trapezoidal channel section having side slopes 3H:4V with n = 0.012 consider bed slope 1 in 2000 which carries water at a velocity of 0.50 m/s.

## Short answer

From Manning's equation with V=0.5 m/s, n=0.012, S=1/2000, R=0.139 m, so depth y=2R=0.278 m. For most efficient trapezoid with 3H:4V slopes, bottom width b equals y, so b=0.278 m, carrying Q≈0.068 m³/s.

## Answer

For the most efficient trapezoidal section, $R = y/2$ and half the top width equals one sloping side. Velocity is known, so Manning's equation gives $R$ directly.

#### Data
Side slope 3H : 4V, so $n = 3/4 = 0.75$ horizontal per vertical; $\sqrt{1+n^2} = 1.25$. Manning's $n_m = 0.012$, $S = 1/2000$, $V = 0.5$ m/s.

#### Hydraulic radius from Manning
$$
\begin{aligned}
V &= \frac{1}{n_m}R^{2/3}S^{1/2} \\
R^{2/3} &= \frac{0.5 \times 0.012}{(1/2000)^{1/2}} = \frac{0.006}{0.02236} = 0.2683 \\
R &= 0.2683^{1.5} = 0.1390\ \text{m}
\end{aligned}
$$

#### Depth and bottom width
$$
\begin{aligned}
y &= 2R = 0.278\ \text{m} \\
b &= 2y\left(\sqrt{1+n^2} - n\right) = 2 \times 0.278 \times (1.25 - 0.75) = 0.278\ \text{m}
\end{aligned}
$$
(With 3H : 4V slopes, $b = y$.)

#### Check
$A = (0.278 + 0.75 \times 0.278) \times 0.278 = 0.1352$ m², $P = 0.278 + 2 \times 0.278 \times 1.25 = 0.973$ m, $R = 0.139$ m. Discharge $Q = AV = 0.1352 \times 0.5 = 0.0676$ m³/s.

**Answer:** Depth ≈ 0.278 m, bottom width ≈ 0.278 m (top width ≈ 0.695 m); it carries about 0.068 m³/s at 0.5 m/s. The section is small because the velocity is low and the channel is smooth.

## Check yourself

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

1. A side slope of 3H:4V gives n and sqrt(1+n^2) of:
   A. n = 1.33, 1.67
   B. n = 0.75, 1.25
   C. n = 0.75, 1.0
   D. n = 1.25, 0.75
   Answer: B. n = 3/4 = 0.75 and sqrt(1 + 0.5625) = 1.25. (Key point: Side slope)

2. Since V is given, how is R found?
   A. From Manning: R^(2/3) = V n / S^(1/2), giving R = 0.139 m
   B. From R = y/2 first
   C. From R = A/P using b
   D. From Chezy only
   Answer: A. V = (1/n)R^(2/3)S^(1/2) rearranged gives R^(2/3) = 0.2683, R = 0.139 m. (Key point: Finding R)

3. How is the depth then obtained?
   A. y = R
   B. y = 2R = 0.278 m
   C. y = R/2
   D. y = 4R
   Answer: B. For the most efficient trapezoid R = y/2, so y = 2R. (Key point: Depth)

4. The bottom width b = 2y(sqrt(1+n^2) - n) equals:
   A. 0.139 m
   B. 0.278 m (= y)
   C. 0.556 m
   D. 0.695 m
   Answer: B. b = 2y(1.25 - 0.75) = y = 0.278 m for this slope. (Key point: Bottom width)

## Related in Hydro Power

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## Related in other subjects

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- [Most efficient sections for area 6 m²](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aefficient-sections-area-6-triangle-rect-trap) — Hydraulics, chapter 6
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- [Efficient section needing minimum lining](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aefficient-section-minimum-lining-one-sloped-side) — Hydraulics, chapter 6