Depth for maximum velocity in circular channel

Geometry of a partly full circular channel

Let $\theta$ be the angle (radians) subtended at the centre by the wetted arc, $d$ the diameter and $y$ the depth.

$$A=\frac{d^2}{8}(\theta-\sin\theta),\qquad P=\frac{d\,\theta}{2},\qquad R=\frac{A}{P}=\frac d4\left(1-\frac{\sin\theta}{\theta}\right),\qquad y=\frac d2\left(1-\cos\frac\theta2\right)$$

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  /  |  \      y = depth
 |   |th |     th = angle at centre
  \  | _/      subtended by wetted arc
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Depth for maximum velocity

Chezy: $V=C\sqrt{RS_0}$. With $C$ and $S_0$ constant, $V$ is maximum when $R$ is maximum:

$$\frac{dR}{d\theta}=\frac d4\left[-\frac{\theta\cos\theta-\sin\theta}{\theta^2}\right]=0\;\Rightarrow\;\theta\cos\theta=\sin\theta\;\Rightarrow\;\tan\theta=\theta$$

Solving by trial, $\theta=4.4934$ rad $=257.45^\circ$ (the root between $\pi$ and $2\pi$).

$$y=\frac d2\left(1-\cos\frac{\theta}{2}\right)=\frac d2\left(1-\cos 128.73^\circ\right)=0.8128\,d$$

So the depth for maximum velocity is $y\approx0.81\,d$.

Discharge comparison with full flow

At maximum velocity: $A=\dfrac{d^2}{8}(\theta-\sin\theta)=0.6837\,d^2$ and $R=0.3043\,d$.

Running full: $A_f=\dfrac{\pi}{4}d^2=0.7854\,d^2$, $R_f=\dfrac d4=0.25\,d$.

$$\frac{V_{max}}{V_f}=\sqrt{\frac{R}{R_f}}=\sqrt{\frac{0.3043}{0.25}}=1.1033$$ $$\frac{Q_{V_{max}}}{Q_f}=\frac{A\sqrt R}{A_f\sqrt{R_f}}=\frac{0.6837\sqrt{0.3043}}{0.7854\sqrt{0.25}}=0.9604$$

Answer: depth for maximum velocity $=0.81\,d$; velocity is 1.103 times the full-flow velocity, and the discharge is 0.960 times (about 96.0% of) the full-pipe discharge. So the maximum velocity occurs at a discharge slightly less than the full-flow value.