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

# Most economical rectangular channel section

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

- Asked in: 2081 Chaitra, 1+5 marks; 2077 Chaitra, 6 marks (asked 2 times)
- Video: [Most economical Rectangular channel section Derivation](https://www.youtube.com/watch?v=OzKP7zWgBmk) — Vinay Mudiraj
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/most-economical-rectangular-section
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

What do you mean by most-economical section of an open channel? Show that the hydraulic mean depth of a most-economical rectangular channel section is half of the channel flow depth.

## Short answer

A most economical section minimizes wetted perimeter for given area, maximizing discharge. For a rectangular channel, setting dP/dy=0 yields width b = 2y, and hydraulic mean depth R = A/P = y/2.

## Answer

A **most economical (best hydraulic) section** of an open channel is the section which, for a given area, bed slope and roughness, carries the maximum discharge. Since $Q = A C\sqrt{RS}$ (Chezy) or $Q = \frac{1}{n}AR^{2/3}S^{1/2}$ (Manning), $Q$ is maximum when $R = A/P$ is maximum, i.e. when the **wetted perimeter $P$ is minimum** for the given area. Such a section also needs the least lining and excavation.

#### Proof for a rectangular section
```text
   |<------- b ------->|
   |                   |  ^
   |      water        |  y
   |___________________|  v
```
Let width $= b$, depth $= y$.
$$
A = by \quad \Rightarrow \quad b = \frac{A}{y}
$$
$$
P = b + 2y = \frac{A}{y} + 2y
$$
For given $A$, $P$ is minimum when $dP/dy = 0$:
$$
\frac{dP}{dy} = -\frac{A}{y^2} + 2 = 0 \;\Rightarrow\; A = 2y^2
$$
Since $A = by$: $by = 2y^2 \Rightarrow b = 2y$. (Check: $d^2P/dy^2 = 2A/y^3 > 0$, so it is a minimum.)

Hydraulic mean depth:
$$
\begin{aligned}
R &= \frac{A}{P} = \frac{by}{b + 2y} \\
&= \frac{2y\cdot y}{2y + 2y} = \frac{2y^2}{4y} = \frac{y}{2}
\end{aligned}
$$

#### Result
For the most economical rectangular channel:
- width is twice the depth, $b = 2y$;
- hydraulic mean depth is half the depth, $R = y/2$.

Example: if $y = 1.5$ m, then $b = 3$ m and $R = 0.75$ m.

## Check yourself

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

1. A most economical section of an open channel is one that, for given area, slope and roughness:
   A. Has maximum wetted perimeter
   B. Carries maximum discharge, i.e. has minimum wetted perimeter
   C. Has minimum area
   D. Has maximum depth
   Answer: B. Q is max when R = A/P is max, which for a given A means P is minimum. (Key point: Most economical section)

2. For a most economical rectangular channel of width b and depth y, the relation is:
   A. b = y
   B. b = y/2
   C. b = 2y
   D. b = 3y
   Answer: C. Setting dP/dy = 0 with P = A/y + 2y gives A = 2y^2, so by = 2y^2 and b = 2y. (Key point: Condition for rectangle)

3. The hydraulic mean depth of the most economical rectangular section is:
   A. y
   B. y/2
   C. 2y
   D. y/4
   Answer: B. R = A/P = 2y^2/(2y+2y) = y/2. (Key point: Hydraulic mean depth)

4. If y = 1.5 m in a most economical rectangular channel, the width b and hydraulic mean depth R are:
   A. b = 1.5 m, R = 0.75 m
   B. b = 3 m, R = 1.5 m
   C. b = 3 m, R = 0.75 m
   D. b = 0.75 m, R = 3 m
   Answer: C. b = 2y = 3 m and R = y/2 = 0.75 m. (Key point: Numerical check)

## Related in Hydro Power

- [Most economical triangular channel section](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aeconomical-triangular-section) — 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×
- [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
- [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
- [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×
- [Uniform flow in rectangular channel (Chezy)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-channel-chezy-flow) — Hydro Power, chapter 1

## Related in other subjects

- [Derive most economical rectangular section](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Amost-economical-rectangular-section-derive) — Hydraulics, chapter 6
- [Economical triangular section hydraulic radius](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Amost-economical-triangular-section-hydraulic-radius) — Hydraulics, chapter 6
- [Hydraulic exponent for rectangular section 10/3](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahydraulic-exponent-rectangular-10-3) — Hydraulics, chapter 6, asked 2×
- [Program for uniform depth in rectangular channel](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aalgorithm-uniform-depth-rectangular) — Hydraulics, chapter 6