Separate sewer, 0.6 full, 250000 persons

Design flow

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

$$Q_{max}=\frac{250000\times200\times0.8\times2.25}{86400\times1000}=1.0417\ \text{m}^3/\text{s}$$

Data: $S=1/600$, $n=0.013$ (constant), depth ratio $d/D=0.6$.

For $d/D=0.6$: $\theta = 2\cos^{-1}(1-2d/D)=203.07^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.6265$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1106$, so $v/v_f=(R/R_f)^{2/3}=1.0724$ and $Q/Q_f=(A/A_f)(v/v_f)=0.6718$ (n taken constant).

Size of the sewer

Discharge when running full:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{1.0417}{0.6718}=1.5505\ \text{m}^3/\text{s}$$

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 $n=0.013$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{1.5505\times0.013}{0.3117\times0.04082}\right)^{3/8}=1.188\ \text{m}$$

Adopt the next standard size, D = 1200 mm.

Check of velocity and capacity

For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.408$ m/s and $Q_f=1.5917$ m³/s.

$$\begin{aligned} v&=v_f\times(v/v_f)=1.408\times1.0724=1.51\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=1.5917\times0.6718=1.0694\ \text{m}^3/\text{s}\ (\ge 1.0417) \end{aligned}$$

The velocity at maximum flow, 1.51 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 = 1200 mm at slope 1 in 600; it runs 0.6 full with v = 1.51 m/s at Q = 1.0417 m³/s (computed diameter 1.188 m). Such a large flow may also be carried by two parallel sewers.