---
title: "Geometric properties of a circular channel"
subject: "Hydraulics"
chapter: 6
key: "semester_4/hydraulics:circular-channel-geometric-properties-y1125"
source: https://www.bctengineer.com/q/semester_4/hydraulics/circular-channel-geometric-properties-y1125
site: BCT Engineering
---

# Geometric properties of a circular channel

Hydraulics · Chapter 6: Uniform flow in open channel

- Asked in: 2080 Chaitra, 1+1+1+1 marks
- Video: [KTU S3 Civil Fluid Mechanics and Hydraulics Geometrical Properties of a Circular Channel Section](https://www.youtube.com/watch?v=nkOc2mqeDns) — Splash Environs
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/circular-channel-geometric-properties-y1125
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-6

## Question

Estimate following geometric properties of a circular channel shown in figure below: Angle ($\alpha$), Hydraulic Radius, Top width and Section factor for uniform flow. [Figure: circular channel of radius 0.75 m (diameter 1.5 m) flowing at depth y = 1.125 m; $\alpha$ is the angle subtended at the centre by the water surface]

## Short answer

Angle α subtended by water surface is 120°. Hydraulic radius R=0.453m, top width T=1.299m, and section factor for uniform flow AR²/³=0.838m⁸/³. These are computed using chord geometry for a circular channel of diameter 1.5m at flow depth 1.125m.

## Answer

#### Given data
$D=1.5$ m, $r=0.75$ m, flow depth $y=1.125$ m. The depth is more than the radius ($y/D=0.75$), so the water surface is above the centre.

```text
     ____  <- water surface (chord)
   /  a   \     a = angle at centre by surface
  |   *    |    * = centre
   \______/
```

#### (i) Angle $\alpha$
The centre is $y-r=1.125-0.75=0.375$ m below the water surface.
$$
\cos\frac{\alpha}{2}=\frac{y-r}{r}=\frac{0.375}{0.75}=0.5\;\Rightarrow\;\frac\alpha2=60^\circ,\quad \alpha=120^\circ
$$
Angle subtended at the centre by the wetted arc (the part below the surface) is $\theta=360^\circ-120^\circ=240^\circ=4.1888$ rad. This $\theta$ is used for area and perimeter.

#### (ii) Area and wetted perimeter
$$
A=\frac{r^2}{2}(\theta-\sin\theta)=\frac{0.75^2}{2}(4.1888-\sin240^\circ)=\frac{0.5625}{2}(4.1888+0.8660)=1.4217\ \text{m}^2
$$
$$
P=r\theta=0.75\times4.1888=3.1416\ \text{m}
$$

#### Hydraulic radius
$$
R=\frac{A}{P}=\frac{1.4217}{3.1416}=0.4525\ \text{m}
$$

#### Top width
$$
T=2r\sin\frac\alpha2=2\times0.75\times\sin60^\circ=1.2990\ \text{m}
$$

#### Section factor for uniform flow
$$
AR^{2/3}=1.4217\times(0.4525)^{2/3}=0.8380\ \text{m}^{8/3}
$$
(The section factor for critical flow, $Z=\sqrt{A^3/T}=1.4873$ m$^{2.5}$, can be computed with the same $A$ and $T$.)

**Answer: $\alpha=120^\circ$, $R=0.453$ m, $T=1.299$ m, $AR^{2/3}=0.838$ m$^{8/3}$.**

## Check yourself

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

1. With r = 0.75 m and y = 1.125 m, how is alpha found?
   A. cos(alpha/2) = (y - r)/r, giving alpha = 120 degrees
   B. sin(alpha) = y/D
   C. alpha = 360 degrees
   D. cos(alpha) = y/r
   Answer: A. The centre is 0.375 m below the surface, so cos(alpha/2) = 0.5, alpha/2 = 60 degrees. (Key point: Angle alpha)

2. Which angle is used for the area and wetted perimeter?
   A. alpha = 120 degrees
   B. theta = 360 - alpha = 240 degrees
   C. 90 degrees
   D. 60 degrees
   Answer: B. The wetted arc subtends theta = 240 degrees = 4.1888 rad at the centre. (Key point: Wetted arc angle)

3. How is the top width found?
   A. T = r theta
   B. T = D
   C. T = 2r sin(alpha/2) = 1.299 m
   D. T = A/R
   Answer: C. T = 2 x 0.75 x sin60 = 1.299 m. (Key point: Top width)

4. What is R for this flow?
   A. 0.375 m
   B. 0.75 m
   C. 0.25 m
   D. A/P = 1.4217/3.1416 = 0.453 m
   Answer: D. A = 1.4217 m^2 and P = r theta = 3.1416 m, so R = 0.4525 m. (Key point: Hydraulic radius)

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