---
title: "Most economical triangular channel section"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:economical-triangular-section"
source: https://www.bctengineer.com/q/semester_6/hydro_power/economical-triangular-section
site: BCT Engineering
---

# Most economical triangular channel section

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

- Asked in: 2071 Bhadra, 2+5 marks
- Video: [Conditions for Most Economical Triangular Section | Most Economical Triangle Channel Derivation](https://www.youtube.com/watch?v=XpoHWZDdqqk) — Professor Nidhi Trivedi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/economical-triangular-section
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

What is meant by economical channel section? Prove that a triangular channel section will be most economical when each of its slopes is 1:1 and sloping sides makes an angle of 45° with vertical.

## Short answer

Economical section minimizes wetted perimeter for given area. For triangular channel, minimizing P = 2√A √((1+m²)/m) yields m=1, so side slope 1:1 (45° with vertical).

## Answer

#### Economical channel section
A channel section is **economical** (most efficient / best hydraulic) when for a given cross-sectional area, bed slope and roughness it gives the **maximum discharge**. This happens when the wetted perimeter is minimum, so the cost of excavation and lining is also least.

#### Proof for triangular section
Let depth $y$, side slope $m$ H : 1 V on both sides, and $\theta$ the angle of each side with the vertical, so $m = \tan\theta$.

```text
   |<------ 2my ------>|
   \                  /
    \       y        /
     \  theta|      /
      \      |     /
       \     |    /
        \    |   /
         \___|__/
             V (apex)
```

$$
\begin{aligned}
A &= \frac{1}{2}(2my)y = my^2 \;\Rightarrow\; y = \sqrt{\frac{A}{m}} \\
P &= 2y\sqrt{1+m^2} = 2\sqrt{\frac{A}{m}}\sqrt{1+m^2} = 2\sqrt{A}\sqrt{\frac{1+m^2}{m}}
\end{aligned}
$$

For minimum P at constant A, minimise $F = \dfrac{1+m^2}{m} = \dfrac{1}{m} + m$:
$$
\begin{aligned}
\frac{dF}{dm} &= -\frac{1}{m^2} + 1 = 0 \\
m^2 &= 1 \;\Rightarrow\; m = 1
\end{aligned}
$$
$d^2F/dm^2 = 2/m^3 > 0$, so it is a minimum.

So the side slope is **1 : 1** (1 H : 1 V), and $\tan\theta = m = 1$ gives $\theta = 45^\circ$ with the vertical. The two sides are then at right angles to each other.

Additional results for this section: $A = y^2$, $P = 2\sqrt{2}\,y$, $R = \dfrac{y}{2\sqrt{2}}$.

## Check yourself

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

1. For a given area, bed slope and roughness, an economical section:
   A. Has maximum wetted perimeter
   B. Gives maximum discharge with minimum wetted perimeter
   C. Has minimum area
   D. Has minimum hydraulic radius
   Answer: B. Minimum perimeter means maximum discharge and the least excavation and lining. (Key point: Economical section)

2. For fixed A, P of a triangular section is proportional to which function of m?
   A. m + m^2
   B. sqrt(m)
   C. (1+m^2)/m = 1/m + m under the root
   D. 1/m^2
   Answer: C. P = 2 sqrt(A) sqrt((1+m^2)/m), so minimise (1+m^2)/m = 1/m + m. (Key point: Perimeter in terms of m)

3. Setting d/dm (1/m + m) = 0 gives:
   A. m = 0.5
   B. m = 1
   C. m = 2
   D. m = sqrt(2)
   Answer: B. -1/m^2 + 1 = 0, so m = 1, a side slope of 1:1. (Key point: Optimum slope)

4. What angle do the sides make with the vertical at the best section?
   A. 30 degrees
   B. 60 degrees
   C. 45 degrees
   D. 90 degrees
   Answer: C. tan(theta) = m = 1 gives theta = 45 degrees, so the sides are at right angles to each other. (Key point: Angle)

## Related in Hydro Power

- [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 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
- [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×
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- [Most efficient trapezoidal channel (3H:4V)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aefficient-trapezoidal-3h4v) — Hydro Power, chapter 1
- [Economical trapezoidal channel side length](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aeconomical-trapezoidal-side-length) — Hydro Power, chapter 1

## Related in other subjects

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- [Derive most economical rectangular section](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Amost-economical-rectangular-section-derive) — Hydraulics, chapter 6
- [Most economical trapezoid of area 40 m²](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aeconomical-trapezoidal-section-area-40) — Hydraulics, chapter 6
- [Most economical span, double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-double-circuit) — Transmission and Distribution System Design, chapter 4, asked 6×