---
title: "Efficient section needing minimum lining"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:efficient-section-minimum-lining-one-sloped-side"
source: https://www.bctengineer.com/q/semester_4/hydraulics/efficient-section-minimum-lining-one-sloped-side
site: BCT Engineering
---

# Efficient section needing minimum lining

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2076 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/efficient-section-minimum-lining-one-sloped-side
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

If the channel in the figure is to deliver 10 m³/s when laid on a slope of 0.0001, calculate dimensions of the efficient section which require minimum lining. Take n = 0.015. [Figure: channel with horizontal bed width b, depth h, one vertical side and one side slope m = 2]

## Answer

Minimum lining means minimum wetted perimeter for the given flow, which is the most efficient hydraulic section. The channel has one vertical side ($z_1=0$) and one side slope $z_2=2$ (2H:1V).

#### Given data
$Q=10$ m³/s, $S_0=0.0001$, $n=0.015$, bed $b$, depth $h$.

#### Geometry
$$
A=bh+\frac{z_1+z_2}{2}h^2=bh+h^2
$$
$$
P=b+h\left(\sqrt{1+0^2}+\sqrt{1+2^2}\right)=b+h(1+2.2361)=b+3.2361\,h
$$

#### Condition for minimum $P$ at fixed $A$
$$
P=\frac{A}{h}-h+3.2361\,h\;\Rightarrow\;\frac{dP}{dh}=-\frac{A}{h^2}+2.2361=0\;\Rightarrow\;\frac{A}{h^2}=2.2361
$$
$$
b=\frac{A}{h}-h=2.2361h-h=1.2361\,h
$$
Then $A=2.2361h^2$, $P=4.4721h$ and $R=\dfrac{A}{P}=\dfrac h2$.

#### Manning's equation
$$
Q=\frac1nAR^{2/3}S_0^{1/2}\;\Rightarrow\;10=\frac{1}{0.015}(2.2361h^2)\left(\frac h2\right)^{2/3}(0.0001)^{1/2}
$$
$$
h^{8/3}=\frac{10\times0.015}{2.2361\times0.5^{2/3}\times0.01}=10.6486\;\Rightarrow\;h=2.428\ \text{m}
$$
$$
b=1.2361\times2.428=3.001\ \text{m}
$$
Check: $A=13.181$ m², $P=10.858$ m, $R=1.214$ m, $V=0.759$ m/s and $Q=\dfrac{1}{0.015}(13.181)(1.214)^{2/3}(0.01)=10.00$ m³/s.

**Answer: depth $h=2.43$ m and bed width $b=3.00$ m** (wetted perimeter $=10.86$ m is the minimum lining per metre).

## Check yourself

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

1. What does minimum lining mean in hydraulic terms?
   A. Minimum area
   B. Minimum wetted perimeter for the given flow (most efficient section)
   C. Maximum depth
   D. Minimum slope
   Answer: B. Lining covers the wetted perimeter, so minimum lining means minimum P. (Key point: Minimum lining)

2. For one vertical side and one side slope of 2, what is the perimeter?
   A. P = b + 2h
   B. P = b + 2 sqrt5 h
   C. P = b + h(1 + sqrt5) = b + 3.236h
   D. P = b + 4h
   Answer: C. The vertical side contributes h and the sloping side h sqrt(1+4) = 2.236h. (Key point: Geometry)

3. What does dP/dh = 0 at constant A give?
   A. A/h^2 = 2.236, b = 1.236h and R = h/2
   B. b = 2h
   C. b = h
   D. R = h
   Answer: A. A/h^2 = 2.236 gives b = 2.236h - h = 1.236h and R = h/2. (Key point: Optimum proportions)

4. What are the final dimensions?
   A. h = 1.5 m, b = 1.9 m
   B. h = 3.0 m, b = 2.43 m
   C. h = 2.43 m, b = 3.00 m
   D. h = 2.0 m, b = 2.5 m
   Answer: C. h^(8/3) = 10.65 gives h = 2.43 m and b = 1.2361 x 2.428 = 3.00 m. (Key point: Result)

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