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

# Efficient rectangular section and triangular channel depth

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

- Asked in: 2079 Chaitra, 2+6 marks
- Video: [Hydraulically Efficient Rectangular Channel Section](https://www.youtube.com/watch?v=hSnQR1dZoYM) — WATERCORE
- 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-rect
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

What is the condition required for the most efficient rectangular 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

For the most efficient rectangular section, the bottom width must be twice the flow depth (b=2y). For the triangular channel with side slope 2H:1V, slope 0.0005, Q=0.5 m³/s, and n=0.04, Manning's equation gives normal depth y ≈ 0.90 m.

## Answer

#### Condition for the most efficient rectangular section
For a rectangular channel of width $b$ and depth $y$: $A = by$, $P = b + 2y = \dfrac{A}{y} + 2y$. For minimum $P$ at constant $A$:
$$
\frac{dP}{dy} = -\frac{A}{y^2} + 2 = 0 \;\Rightarrow\; A = 2y^2 \;\Rightarrow\; b = 2y
$$
So the **width is twice the depth** ($b = 2y$), and $R = \dfrac{2y^2}{4y} = \dfrac{y}{2}$.

#### 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. Condition for the most efficient rectangular section:
   A. b = y
   B. b = 2y
   C. b = 4y
   D. b = y/2
   Answer: B. Minimising P at constant A gives A = 2y^2, so width is twice the depth. (Key point: Rectangular best section)

2. For the best rectangular section R equals:
   A. y
   B. y/4
   C. y/2
   D. 2y
   Answer: C. R = 2y^2/4y = y/2. (Key point: Hydraulic radius)

3. For side slope 2H:1V the area and R are:
   A. A = 2y^2, R = 0.4472 y
   B. A = y^2, R = 0.35 y
   C. A = 2y^2, R = y/2
   D. A = 4y^2, R = 0.2236 y
   Answer: A. A = m y^2 = 2y^2, P = 2y sqrt5 = 4.472y, so R = 0.4472y. (Key point: Triangular geometry)

4. Solving Manning's equation gives a normal depth of about:
   A. 0.50 m
   B. 0.90 m
   C. 1.30 m
   D. 2.00 m
   Answer: B. 0.5 = 0.6538 y^(8/3) gives y^(8/3) = 0.7648, y = 0.904 m. (Key point: Normal depth)

## Close variants set in other exams

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

## Related in Hydro Power

- [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×
- [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×
- [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
- [Most economical rectangular channel section](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Amost-economical-rectangular-section) — Hydro Power, chapter 1, asked 2×
- [Most economical triangular channel section](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aeconomical-triangular-section) — Hydro Power, chapter 1
- [Most efficient trapezoidal channel (3H:4V)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aefficient-trapezoidal-3h4v) — 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
- [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
- [Normal depth in compound channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Acompound-channel-normal-depth-berms) — Hydraulics, chapter 6
- [Bed slope for critical normal depth](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Abed-slope-for-critical-flow-at-normal-depth) — Hydraulics, chapter 7