Circular culvert diameter and slope

Given data

$Q_{full}=0.5$ m³/s, $n=0.012$. At depth $y=d/4$ the velocity must be at least $0.7$ m/s (so the limiting case is $V=0.7$ m/s at $y=d/4$).

Geometry at $y=d/4$

$$\cos\frac\theta2=1-\frac{2y}{d}=0.5\;\Rightarrow\;\theta=120^\circ=2.0944\ \text{rad}$$ $$R'=\frac d4\left(1-\frac{\sin\theta}{\theta}\right)=\frac d4\left(1-\frac{0.8660}{2.0944}\right)=0.14663\,d$$

Full flow: $R_f=\dfrac d4=0.25\,d$.

Relating velocities (same $n$ and $S_0$)

From Manning, $V\propto R^{2/3}$:

$$\frac{V_{full}}{V_{d/4}}=\left(\frac{R_f}{R'}\right)^{2/3}=\left(\frac{0.25}{0.14663}\right)^{2/3}=1.4272$$ $$V_{full}=1.4272\times0.7=0.9990\ \text{m/s}$$

Diameter

$$Q=V_{full}\frac{\pi d^2}{4}\;\Rightarrow\;d=\sqrt{\frac{4\times0.5}{\pi\times0.9990}}=0.798\ \text{m}\approx0.80\ \text{m}$$

Slope

$$V_{full}=\frac1n\left(\frac d4\right)^{2/3}S_0^{1/2}\;\Rightarrow\;S_0=\left[\frac{nV_{full}}{(d/4)^{2/3}}\right]^2=\left[\frac{0.012\times0.9990}{(0.1996)^{2/3}}\right]^2=0.001232$$

i.e. about 1 in 811.

Answer: diameter $d\approx0.80$ m (provide 0.80 m) and bed slope $S_0\approx0.00123$ (1 in 811). Check with $d=0.80$ m: the slope needed for $Q=0.5$ m³/s is $0.00122$ (1 in 821), and then $V$ at $y=d/4$ is $0.697$ m/s, about 0.7 m/s.