Geometric properties of a circular channel

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.

     ____  <- 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$^{8/3}$.