---
title: "Economical trapezoid and power saving"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:economical-trapezoid-saving-in-power-vs-rectangular"
source: https://www.bctengineer.com/q/semester_4/hydraulics/economical-trapezoid-saving-in-power-vs-rectangular
site: BCT Engineering
---

# Economical trapezoid and power saving

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2072 Asoj, 4+2 marks
- Video: [Hydraulics || Unit:5 uniform Flow in open channel || Numerical on economical trapezoidal channel || ](https://www.youtube.com/watch?v=SR4_i75Cqrk) — Study with Jyoti
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/economical-trapezoid-saving-in-power-vs-rectangular
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Determine the most economical section of a trapezoidal channel with side slope of 2:1, carrying a discharge of 9 m³/s with a velocity of 0.75 m/s. Take Manning's n = 0.025. For conveying the same discharge, if a rectangular channel 1.2 m deep and 3 m wide is provided, what would be the saving in power per km length of channel?

## Short answer

Most economical trapezoidal section (2:1 side slope) for Q=9 m³/s has depth 2.20 m, bed width 1.04 m, slope 1 in 3236. A rectangular channel 1.2 m deep by 3 m wide requires a much steeper slope, resulting in a power saving of about 565 kW per km.

## Answer

**Reading:** side slope $2:1$ is taken as 2 horizontal to 1 vertical ($z=2$).

#### Given data
$Q=9$ m³/s, $V=0.75$ m/s, $n=0.025$.

#### Most economical trapezoid
$$
A=\frac QV=\frac{9}{0.75}=12\ \text{m}^2
$$
$$
A=y^2\left(2\sqrt{1+z^2}-z\right)=y^2(2\sqrt5-2)=2.4721\,y^2\;\Rightarrow\;y=\sqrt{\frac{12}{2.4721}}=2.203\ \text{m}
$$
$$
b=2y(\sqrt5-2)=2\times2.203\times0.2361=1.040\ \text{m},\qquad R=\frac y2=1.102\ \text{m}
$$
Bed slope from Manning's equation:
$$
S_t=\left(\frac{nV}{R^{2/3}}\right)^2=\left(\frac{0.025\times0.75}{1.102^{2/3}}\right)^2=3.090\times10^{-4}\quad(\text{1 in }3236)
$$

#### Rectangular channel 3 m wide, 1.2 m deep, same $Q=9$ m³/s
$$
A_r=3.6\ \text{m}^2,\quad V_r=\frac{9}{3.6}=2.5\ \text{m/s},\quad P_r=3+2(1.2)=5.4\ \text{m},\quad R_r=\frac{3.6}{5.4}=0.6667\ \text{m}
$$
$$
S_r=\left(\frac{nV_r}{R_r^{2/3}}\right)^2=\left(\frac{0.025\times2.5}{0.6667^{2/3}}\right)^2=6.707\times10^{-3}\quad(\text{1 in }149)
$$

#### Power
For the same discharge, the power lost per unit length is $\gamma QS_f$ (the head loss per unit length is $S_f$). Over 1 km:
$$
P_{trap}=\gamma Q S_t L=9.81\times9\times3.090\times10^{-4}\times1000=27.28\ \text{kW}
$$
$$
P_{rect}=9.81\times9\times6.707\times10^{-3}\times1000=592.19\ \text{kW}
$$
$$
\text{Saving}=P_{rect}-P_{trap}=564.9\ \text{kW per km}
$$
Equivalently, the rectangular channel needs 6.71 m of fall per km, against 0.31 m for the economical trapezoid.

**Answer: economical section $y=2.20$ m, $b=1.04$ m (bed slope 1 in 3236); saving in power $\approx565$ kW per km.**

## Check yourself

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

1. For Q = 9 m3/s and V = 0.75 m/s, how is the economical trapezoid (z = 2) sized?
   A. A = 12 m2 and y = sqrt(12/2.4721) = 2.20 m
   B. A = 6.75 m2 and y = 1.65 m
   C. A = 12 m2 and y = 12/3 = 4 m
   D. A = 12 m2 and y = 1.2 m
   Answer: A. A = Q/V = 12 m2 and A = y^2(2 sqrt5 - 2) = 2.4721 y^2 give y = 2.203 m. (Key point: Area and depth of the economical trapezoid)

2. Why does the 3 m x 1.2 m rectangle need a much steeper slope for the same Q?
   A. Because it has a larger area
   B. Because its Manning n is larger
   C. Because its velocity (2.5 m/s) is far higher than 0.75 m/s, so a greater slope is needed to drive it
   D. Because the rectangle has a larger hydraulic radius
   Answer: C. S = (nV/R^(2/3))^2, so the higher velocity needs a far larger slope (1 in 149 against 1 in 3236). (Key point: Why the slopes differ)

3. How is the power lost in a length L of channel computed?
   A. P = gamma Q S L (specific weight x discharge x slope x length)
   B. P = gamma A V L
   C. P = rho Q^2 L
   D. P = gamma Q y L
   Answer: A. Head loss per length equals the slope S, so power = gamma Q (S L). (Key point: Power lost per length)

4. The trapezoid loses about 27 kW/km and the rectangle about 592 kW/km. The saving is about:
   A. 27 kW
   B. 320 kW
   C. 620 kW
   D. 565 kW per km
   Answer: D. Saving = 592.19 - 27.28 = about 565 kW per km. (Key point: Saving per km)

## Related in Hydraulics

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