Minimum sewer diameter, 80000 population

Design flow

$$Q_{max}=\frac{80000\times145\times0.8\times3}{86400\times1000}=0.3222\ \text{m}^3/\text{s}$$

Data: $S=1/400$, $n=0.013$ (constant), depth ratio $d/D=0.75$.

For $d/D=0.75$: $\theta = 2\cos^{-1}(1-2d/D)=240.00^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.8045$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.2067$, so $v/v_f=(R/R_f)^{2/3}=1.1335$ and $Q/Q_f=(A/A_f)(v/v_f)=0.9119$ (n taken constant).

Size of the sewer

Discharge when running full:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{0.3222}{0.9119}=0.3534\ \text{m}^3/\text{s}$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00250$ and $n=0.013$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.3534\times0.013}{0.3117\times0.05000}\right)^{3/8}=0.632\ \text{m}$$

Adopt the next standard size, D = 675 mm.

Check of velocity and capacity

For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.175$ m/s and $Q_f=0.4203$ m³/s.

$$\begin{aligned} v&=v_f\times(v/v_f)=1.175\times1.1335=1.33\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.4203\times0.9119=0.3833\ \text{m}^3/\text{s}\ (\ge 0.3222) \end{aligned}$$

The velocity at maximum flow, 1.33 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: Minimum diameter required = 0.632 m; provide 675 mm.