---
title: "Economical triangular section hydraulic radius"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:most-economical-triangular-section-hydraulic-radius"
source: https://www.bctengineer.com/q/semester_4/hydraulics/most-economical-triangular-section-hydraulic-radius
site: BCT Engineering
---

# Economical triangular section hydraulic radius

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2080 Chaitra, 4 marks
- Video: [Open channel Flow - 21 [Hydraulically efficient triangular section derivation]](https://www.youtube.com/watch?v=eQvLlh16j1Y) — Abhijit
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/most-economical-triangular-section-hydraulic-radius
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Prove that the most economical triangular channel section has hydraulic radius $\frac{1}{\sqrt{2}}$ times hydraulic radius of most economical rectangular section.

## Short answer

The most economical triangular section has side slope z=1 (45° sides), giving hydraulic radius Rₜ=y/(2√2). The most economical rectangular section has b=2y and Rₙ=y/2. Hence Rₜ=(1/√2)Rₙ, proving the required relation.

## Answer

#### Most economical triangular section
Let side slope be $z$ (H:V), depth $y$:
$$
A=zy^2,\qquad P=2y\sqrt{1+z^2}
$$
For a given area, $y^2=A/z$, so
$$
P^2=4y^2(1+z^2)=\frac{4A(1+z^2)}{z}
$$
Minimise $P$ with respect to $z$:
$$
\frac{d}{dz}\left(\frac{1+z^2}{z}\right)=\frac{2z\cdot z-(1+z^2)}{z^2}=0\;\Rightarrow\;z^2=1\;\Rightarrow\;z=1
$$
So the best triangular section has side slopes $45^\circ$ (right-angled vertex at the bottom):
$$
A=y^2,\qquad P=2\sqrt2\,y,\qquad R_t=\frac{y^2}{2\sqrt2\,y}=\frac{y}{2\sqrt2}
$$

#### Most economical rectangular section
$b=2y$: $A=2y^2$, $P=4y$, so
$$
R_r=\frac{2y^2}{4y}=\frac y2
$$

#### Comparison (at the same depth $y$)
$$
\frac{R_t}{R_r}=\frac{y/(2\sqrt2)}{y/2}=\frac{1}{\sqrt2}
$$
$$
\boxed{R_t=\frac{1}{\sqrt2}R_r}
$$
Hence the hydraulic radius of the most economical triangular section is $1/\sqrt2$ times that of the most economical rectangular section of the same depth.

## Check yourself

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

1. Minimising P for a given area in a triangular channel gives which side slope?
   A. z = 1 (45 degrees)
   B. z = 2
   C. z = 0.5
   D. z = sqrt3
   Answer: A. d/dz[(1+z^2)/z] = 0 gives z^2 = 1, so z = 1. (Key point: Best triangular slope)

2. What is R of the best triangular section?
   A. y/2
   B. y/4
   C. y/(2 sqrt2)
   D. y sqrt2
   Answer: C. A = y^2, P = 2 sqrt2 y, so R = y/(2 sqrt2). (Key point: Triangular R)

3. For the most economical rectangular section (b = 2y), R is:
   A. y
   B. y/2
   C. y/4
   D. 2y
   Answer: B. A = 2y^2 and P = 4y, so R = y/2. (Key point: Rectangular R)

4. What is the ratio Rt/Rr at the same depth?
   A. sqrt2
   B. 1/2
   C. 2
   D. 1/sqrt2
   Answer: D. (y/(2 sqrt2))/(y/2) = 1/sqrt2. (Key point: Ratio)

## Related in Hydraulics

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