Sewer at 2/3 depth, 0.70 m³/s

Given

Peak discharge $Q=0.70$ m³/s, $S=1/1000$, $n=0.012$, depth of flow $d/D=2/3=0.667$.

For $d/D=0.6667$: $\theta = 2\cos^{-1}(1-2d/D)=218.94^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.7082$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1645$, so $v/v_f=(R/R_f)^{2/3}=1.1069$ and $Q/Q_f=(A/A_f)(v/v_f)=0.7839$ (n taken constant).

Size of the sewer

Discharge when running full:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{0.7000}{0.7839}=0.8930\ \text{m}^3/\text{s}$$

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 $n=0.012$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.8930\times0.012}{0.3117\times0.03162}\right)^{3/8}=1.032\ \text{m}$$

Adopt the next standard size, D = 1050 mm.

Check of velocity and capacity

For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.081$ m/s and $Q_f=0.9355$ m³/s.

$$\begin{aligned} v&=v_f\times(v/v_f)=1.081\times1.1069=1.20\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.9355\times0.7839=0.7333\ \text{m}^3/\text{s}\ (\ge 0.7000) \end{aligned}$$

The velocity at maximum flow, 1.20 m/s, lies between the minimum (0.6 m/s) and the maximum non-scouring limit (3.0 m/s), so it is satisfactory.

For non-scouring, the velocity must not exceed the maximum permissible value (about 3 m/s for concrete sewers), and for self-cleansing it should be at least 0.6 m/s. The velocity is 1.20 m/s at peak flow, which is within 0.6-3.0 m/s, so the sewer is safe against scouring and also self-cleansing at peak flow. (At lower flows the velocity falls, so flows below about 0.058 m³/s give less than 0.6 m/s, and the minimum flow should be checked separately.)

Answer: D = 1050 mm (computed 1.032 m); v = 1.20 m/s < 3 m/s, so safe.