Combined circular sewer, 35 ha

Step 1: Sewage flow

Population $=100\times35=3500$. Assume 80% of the supply becomes sewage.

$$Q_{sew}=\frac{3500\times100\times0.8\times2.7}{86400\times1000}=0.00875\ \text{m}^3/\text{s}$$

Step 2: Storm water

$$Q_{storm}=\frac{C\,i\,A}{360}=\frac{0.4\times15\times35}{360}=0.5833\ \text{m}^3/\text{s}$$

Step 3: Combined design discharge

$$Q=0.00875+0.5833=0.5921\ \text{m}^3/\text{s}$$

Step 4: Diameter ($d/D=0.6$, $S=1/750$, $n=0.011$)

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:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{0.5921}{0.6718}=0.8813\ \text{m}^3/\text{s}$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00133$ and $n=0.011$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.8813\times0.011}{0.3117\times0.03651}\right)^{3/8}=0.942\ \text{m}$$

Adopt the next standard size, D = 1050 mm.

Check of velocity and capacity

For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.361$ m/s and $Q_f=1.1785$ m³/s.

$$\begin{aligned} v&=v_f\times(v/v_f)=1.361\times1.0724=1.46\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=1.1785\times0.6718=0.7917\ \text{m}^3/\text{s}\ (\ge 0.5921) \end{aligned}$$

The velocity at maximum flow, 1.46 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: Diameter of the combined circular sewer = 0.942 m; provide 1050 mm.