Separate sewer, 0.70 full, 100000 persons

Design flow

Assume 80% of the water supplied reaches the sewer. Peak factor = 2.25.

$$Q_{max}=\frac{100000\times200\times0.8\times2.25}{86400\times1000}=0.4167\ \text{m}^3/\text{s}$$

Data: $S=1/600$, $n=0.013$ (constant at all depths), depth ratio $d/D=0.70$.

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:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{0.4167}{0.8372}=0.4977\ \text{m}^3/\text{s}$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00167$ and $n=0.013$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.4977\times0.013}{0.3117\times0.04082}\right)^{3/8}=0.776\ \text{m}$$

Adopt the next standard size, D = 825 mm.

Check of velocity and capacity

For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.096$ m/s and $Q_f=0.5861$ m³/s.

$$\begin{aligned} v_{0.7}&=v_f\times(v/v_f)=1.096\times1.1198=1.23\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.5861\times0.8372=0.4907\ \text{m}^3/\text{s}\ (\ge 0.4167) \end{aligned}$$

At the design flow, $Q/Q_f=0.4167/0.5861=0.711$, which gives $d/D\approx0.62$ and $v/v_f\approx1.086$, so $v\approx1.19$ m/s.

The velocity at maximum flow, about 1.2 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 D = 825 mm at slope 1 in 600 (computed diameter 0.776 m runs exactly 0.7 full). The adopted pipe can carry 0.4907 m³/s at 0.7 depth; at Q = 0.4167 m³/s it runs about 0.62 full with v ≈ 1.19 m/s.