---
title: "Concrete-lined canal: Q 350"
subject: "Irrigation and Drainage Engineering"
chapter: 4
key: "semester_6/irrigation_and_drainage_engineering:lined-canal-q350-slope-15000"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lined-canal-q350-slope-15000
site: BCT Engineering
---

# Concrete-lined canal: Q 350

Irrigation and Drainage Engineering · Chapter 4: Design of Canals

- Asked in: 2078 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/lined-canal-q350-slope-15000
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-4

## Question

Design a concrete-lined canal to carry a discharge of 350 m³/s at a slope of 1 in 5000. The side slopes of the canal may be taken as 1.5:1. The value of n for lining is 0.014. Assume limiting velocity in the canal is 2 m/s.

## Short answer

Most economical section exceeds velocity limit. With V fixed at 2 m/s, required depth 3.11 m, bed width 51.6 m, side slope 1.5H:1V, bed slope 1 in 5000.

## Answer

Given: $Q = 350\ \text{m}^3/\text{s}$, $S = 0.000200$ (1 in 5000), $n = 0.014$, side slope 1.5H:1V, limiting velocity 2 m/s. 

#### Most economical (best hydraulic) section

For a trapezoidal section of least perimeter: $R = D/2$ and $B = 2D(\sqrt{1+z^2} - z)$.

$$
B = 2D(\sqrt{1+1.5^2} - 1.5) = 0.6056D,\qquad A = D^2\left(2\sqrt{1+z^2} - z\right) = 2.1056D^2
$$

Using Manning's equation with $R = D/2$:

$$
Q = \frac{1}{n}\,2.1056D^2\left(\frac{D}{2}\right)^{2/3}S^{1/2} \Rightarrow D^{8/3} = \frac{Q n\,2^{2/3}}{2.1056\sqrt{S}}
$$

$$
D = 8.061\ \text{m},\quad B = 4.881\ \text{m},\quad A = 136.81\ \text{m}^2,\quad R = 4.030\ \text{m},\quad V = 2.558\ \text{m/s}
$$

#### Velocity check

$V = 2.56$ m/s is more than the limiting velocity 2 m/s, so the best section cannot be used. Fix $V = 2$ m/s and find the section from the given slope.

$$
\begin{aligned}
R &= \left(\frac{Vn}{\sqrt S}\right)^{3/2} = \left(\frac{2\times 0.014}{\sqrt{0.000200}}\right)^{3/2} = 2.786\ \text{m} \\
A &= \frac{Q}{V} = 175.00\ \text{m}^2,\qquad P = \frac{A}{R} = 62.82\ \text{m}
\end{aligned}
$$

From $A = (B + zD)D$ and $P = B + 2D\sqrt{1+z^2}$:

$$
\begin{aligned}
2.1056D^2 - 62.82D + 175.00 &= 0 \\
D &= 3.110\ \text{m},\qquad B = P - 3.6056D = 51.60\ \text{m}
\end{aligned}
$$

**Answer:** $D \approx 3.11$ m, $B \approx 51.6$ m, side slope 1.5H:1V, $S = 1$ in 5000, $V = 2$ m/s (the best section would give velocity above the limit).

## Check yourself

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

1. For the most economical trapezoidal section, which relations hold?
   A. R = D and B = D
   B. R = 2D and B = 2D
   C. R = D/2 and B = 2D(sqrt(1+z^2) - z)
   D. R = D/2 and B = D(sqrt(1+z^2) + z)
   Answer: C. These come from minimum wetted perimeter. (Key point: Best section)

2. Why is the best section rejected?
   A. Its depth is too small
   B. Its slope is too flat
   C. Its V is below 2 m/s
   D. Its V = 2.56 m/s exceeds the limiting 2 m/s
   Answer: D. Velocity must not exceed the limit, so V is fixed at 2 m/s. (Key point: Velocity check)

3. How is R found with V = 2 m/s?
   A. R = (V n/sqrt(S))^(3/2) = 2.786 m
   B. R = D/2 = 4.03 m
   C. R = Q/V
   D. R = V^2 n
   Answer: A. From Manning's equation with the known slope; then A = Q/V = 175 m^2 and P = A/R = 62.82 m. (Key point: Fix V)

## Related in Irrigation and Drainage Engineering

- [Concrete-lined canal: Q 45, 45° slopes](https://www.bctengineer.com/question.md?key=semester_6%2Firrigation_and_drainage_engineering%3Alined-canal-q45-slope-17000) — Irrigation and Drainage Engineering, chapter 4
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## Related in other subjects

- [Economical channel: unlined versus concrete lined](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aeconomical-trapezoidal-unlined-vs-lined) — Hydraulics, chapter 6, asked 2×
- [Efficient trapezoid with unequal side slopes](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aefficient-trapezoidal-two-side-slopes-critical-depth) — Hydraulics, chapter 6
- [Maximising discharge with same lining](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-channel-max-discharge) — Hydro Power, chapter 1
- [Rectangle with same lining for maximum discharge](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Arectangular-channel-same-lining-max-discharge) — Hydraulics, chapter 6