94% depth for maximum discharge in circular channel

Condition for maximum discharge

Manning: $Q=\dfrac1nAR^{2/3}S^{1/2}=\dfrac{S^{1/2}}{n}\dfrac{A^{5/3}}{P^{2/3}}$. For constant $d$, $n$ and $S$, $Q$ is maximum when $A^{5/3}P^{-2/3}$ is maximum:

$$\frac{dQ}{d\theta}=0\;\Rightarrow\;5P\frac{dA}{d\theta}=2A\frac{dP}{d\theta}$$

Geometry

Let $\theta$ be the angle (rad) subtended at the centre by the wetted arc:

$$A=\frac{d^2}{8}(\theta-\sin\theta),\quad P=\frac{d\theta}{2},\quad \frac{dA}{d\theta}=\frac{d^2}{8}(1-\cos\theta),\quad\frac{dP}{d\theta}=\frac d2$$

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  /  |  \     y = depth
 |   |th |    d = diameter
  \  |__/
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Solution

$$5\left(\frac{d\theta}{2}\right)\frac{d^2}{8}(1-\cos\theta)=2\,\frac{d^2}{8}(\theta-\sin\theta)\frac d2$$ $$5\theta(1-\cos\theta)=2(\theta-\sin\theta)$$ $$3\theta-5\theta\cos\theta+2\sin\theta=0$$

Solving by trial (or Newton's method): $\theta=5.2781$ rad $=302.41^\circ$.

Depth of flow

$$y=\frac d2\left(1-\cos\frac\theta2\right)=\frac d2\left(1-\cos151.21^\circ\right)=0.9382\,d$$

So the depth for maximum discharge is $y\approx0.94\,d$ (94% of the diameter).