---
title: "Economical rectangular condition and trapezoidal bed slope"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:trapezoidal-channel-bed-slope"
source: https://www.bctengineer.com/q/semester_6/hydro_power/trapezoidal-channel-bed-slope
site: BCT Engineering
---

# Economical rectangular condition and trapezoidal bed slope

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

- Asked in: 2080 Asoj, 2+6 marks
- Video: [Calculating Maximum Permissible Slope and Discharge in Trapezoidal Channels](https://www.youtube.com/watch?v=VTA-CegPPZo) — Hydropower
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/trapezoidal-channel-bed-slope
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

What is the condition for the most economical Rectangular section? A trapezoidal channel having bottom width 6 m and side slope 2:1, carries a discharge of 15 m³/s. Find the value of the bed slope of the channel if the depth of the flow is 1.2 m and Manning's 'n' = 0.013.

## Short answer

Most economical rectangular section has b = 2y and R = y/2. For the given trapezoidal channel (b=6 m, side slope 2:1, y=1.2 m, n=0.013, Q=15 m³/s), Manning's equation yields bed slope S ≈ 0.000439 (1 in 2277).

## Answer

#### Condition for the most economical rectangular section
For width $b$ and depth $y$, $P = A/y + 2y$. Setting $dP/dy = 0$ at constant $A$ gives $A = 2y^2$, i.e. **$b = 2y$** (width twice the depth) and **$R = y/2$**. This section carries maximum discharge for a given area.

#### Bed slope of trapezoidal channel
Data: $b = 6$ m, side slope 2H : 1V ($m = 2$), $Q = 15$ m³/s, $y = 1.2$ m, $n = 0.013$.

$$
\begin{aligned}
A &= (b + my)y = (6 + 2 \times 1.2) \times 1.2 = 10.08\ \text{m}^2 \\
P &= b + 2y\sqrt{1+m^2} = 6 + 2 \times 1.2\sqrt{5} = 11.367\ \text{m} \\
R &= \frac{10.08}{11.367} = 0.8868\ \text{m} \\
V &= \frac{15}{10.08} = 1.488\ \text{m/s}
\end{aligned}
$$

Manning's equation, solved for $S$:
$$
\begin{aligned}
Q &= \frac{1}{n}AR^{2/3}S^{1/2} \\
S &= \left(\frac{Qn}{AR^{2/3}}\right)^2 = \left(\frac{15 \times 0.013}{10.08 \times 0.8868^{2/3}}\right)^2 \\
&= \left(\frac{0.195}{9.304}\right)^2 = 4.39 \times 10^{-4}
\end{aligned}
$$

**Answer:** Bed slope $S \approx 0.000439$, i.e. about **1 in 2277**.

## Check yourself

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

1. The condition for the most economical rectangular section is:
   A. b = y
   B. b = 2y
   C. b = y/2
   D. b = 3y
   Answer: B. Minimum P at constant A gives b = 2y and R = y/2. (Key point: Rectangular condition)

2. For b = 6 m, y = 1.2 m, side slope 2:1, the area is:
   A. 7.2 m^2
   B. 10.08 m^2
   C. 12.0 m^2
   D. 8.64 m^2
   Answer: B. A = (b + my)y = (6 + 2.4) x 1.2 = 10.08 m^2, with P = 11.367 m. (Key point: Area and perimeter)

3. How is S obtained from Manning's equation?
   A. S = Qn / (A R^(2/3))
   B. S = (Qn / (A R^(2/3)))^2
   C. S = Q / (A R)
   D. S = sqrt(Qn/AR)
   Answer: B. Q = (1/n) A R^(2/3) S^(1/2) so S^(1/2) = Qn/(A R^(2/3)); square it. (Key point: Solving for slope)

4. The bed slope is about:
   A. 1 in 227
   B. 1 in 2277
   C. 1 in 22770
   D. 1 in 500
   Answer: B. S = 4.39 x 10^-4, about 1 in 2277. (Key point: Result)

## Related in Hydro Power

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## Related in other subjects

- [Trapezoidal to rectangular transition with bed lowering](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Atrapezoid-to-rectangular-transition-bed-lowering) — Hydraulics, chapter 7
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