---
title: "Efficient trapezoidal section and triangular channel depth"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:triangular-channel-normal-depth-trap"
source: https://www.bctengineer.com/q/semester_6/hydro_power/triangular-channel-normal-depth-trap
site: BCT Engineering
---

# Efficient trapezoidal section and triangular channel depth

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

- Asked in: 2078 Chaitra, 2+6 marks
- Video: [Most economical trapezoid channel section //open channel flow //](https://www.youtube.com/watch?v=9PJhN517foA) — Civil Engineer FBH
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/triangular-channel-normal-depth-trap
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

What is the condition required for the most efficient trapezoidal section? Determine the normal depth for a triangular channel with side slope 2H: 1V and longitudinal slope 0.0005 to carry discharge of 0.5 m³/s if manning's coefficient n is 0.04.

## Short answer

The most efficient trapezoidal section occurs when half the top width equals the length of the sloping side, giving hydraulic radius R=y/2. For the triangular channel with side slope 2H:1V, slope 0.0005, Q=0.5 m³/s, and n=0.04, the normal depth is y ≈ 0.90 m.

## Answer

#### Condition for the most efficient trapezoidal section
For bottom width $b$, depth $y$, side slope $n$H : 1V, minimising $P = \dfrac{A}{y} - ny + 2y\sqrt{1+n^2}$ at constant $A$ gives:
- $\dfrac{b + 2ny}{2} = y\sqrt{1+n^2}$: **half the top width equals one sloping side**.
- Hydraulic radius $R = \dfrac{y}{2}$.
- The sides are tangent to a semicircle drawn with centre on the water surface. If the side slope is also free, the best is $60^\circ$ with the horizontal (half a regular hexagon).

#### Normal depth of triangular channel
Data: side slope $m = 2$ (2H : 1V), $S = 0.0005$, $Q = 0.5$ m³/s, $n = 0.04$.

$$
\begin{aligned}
A &= my^2 = 2y^2 \\
P &= 2y\sqrt{1+m^2} = 2y\sqrt{5} = 4.472y \\
R &= \frac{2y^2}{4.472y} = 0.4472y
\end{aligned}
$$

Manning's equation:
$$
\begin{aligned}
Q &= \frac{1}{n}AR^{2/3}S^{1/2} \\
0.5 &= \frac{1}{0.04}(2y^2)(0.4472y)^{2/3}(0.0005)^{1/2} \\
0.5 &= 25 \times 2 \times 0.5848 \times 0.02236\, y^{8/3} = 0.6538\,y^{8/3} \\
y^{8/3} &= 0.7648 \;\Rightarrow\; y = 0.904\ \text{m}
\end{aligned}
$$

Check: $A = 2(0.904)^2 = 1.636$ m², $R = 0.404$ m, $Q = 25 \times 1.636 \times 0.404^{2/3} \times 0.02236 = 0.50$ m³/s.

**Answer:** Normal depth $y_n \approx 0.90$ m.

## Check yourself

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

1. For the most efficient trapezoid:
   A. Half the top width equals one sloping side
   B. Bottom width equals depth always
   C. Top width equals bottom width
   D. Sloping side equals depth
   Answer: A. (b + 2ny)/2 = y sqrt(1+n^2). (Key point: Trapezoidal condition)

2. The hydraulic radius of the most efficient trapezoid is:
   A. y
   B. y/2
   C. 2y
   D. y/4
   Answer: B. R = y/2 for the best trapezoidal section. (Key point: Hydraulic radius)

3. If the side slope is also free, the best angle with the horizontal is:
   A. 30 degrees
   B. 45 degrees
   C. 60 degrees
   D. 90 degrees
   Answer: C. The best trapezoid is half a regular hexagon, with sides at 60 degrees to the horizontal. (Key point: Best side angle)

4. For the 2H:1V triangular channel with Q = 0.5, n = 0.04, S = 0.0005, the normal depth is about:
   A. 0.40 m
   B. 0.90 m
   C. 1.50 m
   D. 2.10 m
   Answer: B. 0.5 = 0.6538 y^(8/3) gives y = 0.904 m. (Key point: Normal depth)

## Close variants set in other exams

- [Efficient rectangular section and triangular channel depth](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-rect) — Hydro Power, chapter 1, asked 2×

## Related in Hydro Power

- [Most efficient trapezoidal channel (3H:4V)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aefficient-trapezoidal-3h4v) — Hydro Power, chapter 1
- [Normal depth in triangular channel (1:1)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-normal-depth-0-0004) — Hydro Power, chapter 1, asked 2×
- [Economical rectangular condition and trapezoidal bed slope](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atrapezoidal-channel-bed-slope) — Hydro Power, chapter 1
- [Flow depth in triangular channel](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atriangular-channel-depth-1-in-2000) — Hydro Power, chapter 1, asked 2×
- [Most economical trapezoidal channel of area 50 m²](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aeconomical-trapezoidal-area-50) — Hydro Power, chapter 1
- [Most economical triangular channel section](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aeconomical-triangular-section) — Hydro Power, chapter 1

## Related in other subjects

- [Efficient trapezoid with unequal side slopes](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aefficient-trapezoidal-two-side-slopes-critical-depth) — Hydraulics, chapter 6
- [Normal depth in compound channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acompound-channel-normal-depth-berms) — Hydraulics, chapter 6
- [Program for normal depth in trapezoidal channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aalgorithm-normal-depth-trapezoidal) — Hydraulics, chapter 6
- [Depth for maximum discharge, triangular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Atriangular-channel-depth-for-max-discharge) — Hydraulics, chapter 6