Separate sewer for 0.5 m³/s, 0.7 depth

Given

Maximum flow $Q=0.5$ m³/s, slope $S=1/1000=0.001$, $n=0.012$, depth of flow $d/D=0.7$ at maximum flow.

Hydraulic elements

For $d/D=0.7$: $\theta = 2\cos^{-1}(1-2d/D)=227.16^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.7477$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1849$, so $v/v_f=(R/R_f)^{2/3}=1.1198$ and $Q/Q_f=(A/A_f)(v/v_f)=0.8372$ (n taken constant).

Size of the sewer

Discharge when running full:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{0.5000}{0.8372}=0.5972\ \text{m}^3/\text{s}$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00100$ and $n=0.012$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.5972\times0.012}{0.3117\times0.03162}\right)^{3/8}=0.887\ \text{m}$$

Adopt the next standard size, D = 900 mm.

Check of velocity and capacity

For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=0.975$ m/s and $Q_f=0.6202$ m³/s.

$$\begin{aligned} v&=v_f\times(v/v_f)=0.975\times1.1198=1.09\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.6202\times0.8372=0.5193\ \text{m}^3/\text{s}\ (\ge 0.5000) \end{aligned}$$

The velocity at maximum flow, 1.09 m/s, lies between the minimum (0.6 m/s) and the maximum non-scouring limit (3.0 m/s), so it is satisfactory.

Answer: Provide a circular sewer of diameter 900 mm at slope 1 in 1000; it runs 0.7 full with velocity 1.09 m/s at 0.5 m³/s.