---
title: "Economical trapezoidal channel side length"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:economical-trapezoidal-side-length"
source: https://www.bctengineer.com/q/semester_6/hydro_power/economical-trapezoidal-side-length
site: BCT Engineering
---

# Economical trapezoidal channel side length

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

- Asked in: 2075 Bhadra, 6 marks
- Video: [Condition for Most Economical Trapezoidal Channel | Most Economical Trapezoidal Section Derivation](https://www.youtube.com/watch?v=Fzj0G3jbxbk) — 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-trapezoidal-side-length
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Verify that for the most economical trapezoidal channel, the length of one channel side must be equal to half the length of the top-width of the section.

## Short answer

For most economical trapezoidal channel, minimizing wetted perimeter gives condition (b+2ny)/2 = y√(1+n²), meaning half top width equals sloping side length.

## Answer

A trapezoidal channel is **most economical** when its wetted perimeter is minimum for a given area, which gives maximum discharge for given $A$, $S$ and roughness.

Let bottom width $b$, depth $y$, side slope $n$ H : 1 V.

```text
   |<-------- T = b + 2ny ------->|
   \                             /
    \ sloping side = y*sqrt(1+n^2)
     \__________________________/
          |<----- b ----->|
```

#### Area and wetted perimeter
$$
\begin{aligned}
A &= (b + ny)y \;\Rightarrow\; b = \frac{A}{y} - ny \\
P &= b + 2y\sqrt{1+n^2} = \frac{A}{y} - ny + 2y\sqrt{1+n^2}
\end{aligned}
$$

#### Condition for minimum P (A constant)
$$
\begin{aligned}
\frac{dP}{dy} &= -\frac{A}{y^2} - n + 2\sqrt{1+n^2} = 0 \\
\frac{A}{y} &= 2y\sqrt{1+n^2} - ny
\end{aligned}
$$
Substitute $A/y = b + ny$:
$$
\begin{aligned}
b + ny &= 2y\sqrt{1+n^2} - ny \\
b + 2ny &= 2y\sqrt{1+n^2} \\
\frac{b + 2ny}{2} &= y\sqrt{1+n^2}
\end{aligned}
$$

#### Interpretation
- Top width $T = b + 2ny$, so the left side is **half the top width**.
- Length of one sloping side $= \sqrt{(ny)^2 + y^2} = y\sqrt{1+n^2}$, which is the right side.

Therefore, for the most economical trapezoidal section, **length of one sloping side = half the top width**. Hence verified.

Also $d^2P/dy^2 = 2A/y^3 > 0$, so this is indeed a minimum of $P$. A related result is $R = y/2$.

## Check yourself

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

1. A trapezoidal channel is most economical when:
   A. Its area is maximum for given perimeter ignoring discharge
   B. Its wetted perimeter is minimum for a given area
   C. Its top width is minimum
   D. Its depth is maximum
   Answer: B. Minimum P for given A gives maximum discharge for given A, S and roughness. (Key point: Economical meaning)

2. What is the wetted perimeter of a trapezoid with bottom width b, depth y and side slope n?
   A. b + 2ny
   B. b + 2y sqrt(1+n^2)
   C. b + 2y
   D. b + y sqrt(1+n^2)
   Answer: B. Each sloping side has length y sqrt(1+n^2), so P = b + 2y sqrt(1+n^2). (Key point: Perimeter expression)

3. Setting dP/dy = 0 and substituting A/y = b + ny gives:
   A. b = y
   B. (b + 2ny)/2 = y sqrt(1+n^2)
   C. b = 2y
   D. b + ny = y
   Answer: B. This says half the top width equals the sloping side length. (Key point: Condition derived)

4. Why does (b + 2ny)/2 = y sqrt(1+n^2) mean side = half top width?
   A. b + 2ny is the top width and y sqrt(1+n^2) is one sloping side
   B. b + 2ny is the area and y sqrt(1+n^2) is depth
   C. b + 2ny is the perimeter
   D. Both sides are the bed width
   Answer: A. Top width T = b + 2ny, and one sloping side = sqrt((ny)^2 + y^2) = y sqrt(1+n^2). (Key point: Interpretation)

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