Sewer diameter for a 12 sq km area

The sewer is taken as a combined sewer. Missing data are assumed: slope 1 in 500, peak factor 3, sewer running half full. The 235 lpcd given is already the sewage flow, so no further reduction is applied.

Step 1: Sewage flow

Area = 12 km² = 1200 ha, population $=250\times1200=300000$.

$$Q_{sew}=\frac{300000\times235\times3}{86400\times1000}=2.4479\ \text{m}^3/\text{s}$$

Step 2: Storm water

$C=0.5(0.3)+0.5(0.75)=0.525$. Time of concentration = 25 min; storm duration = 20 min. The intensity for the given 20 min storm is taken from the assumed formula $i=\dfrac{760}{t+10}$ mm/hr:

$$\begin{aligned} i&=\frac{760}{20+10}=25.33\ \text{mm/hr}\\ Q_{storm}&=\frac{0.525\times25.33\times1200}{360}=44.333\ \text{m}^3/\text{s} \end{aligned}$$

(Because the 20 min storm is shorter than $t_c$ = 25 min, the farthest part of the area has not yet contributed. Using the full area is on the safe side.)

Step 3: Total discharge

$$Q=2.4479+44.333=46.781\ \text{m}^3/\text{s}$$

Step 4: Sewer design ($S=1/500$, $n=0.013$, half full)

For $d/D=0.5$: $\theta = 2\cos^{-1}(1-2d/D)=180.00^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.5000$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.0000$, so $v/v_f=(R/R_f)^{2/3}=1.0000$ and $Q/Q_f=(A/A_f)(v/v_f)=0.5000$ (n taken constant).

Size of the sewer

Discharge when running full:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{46.781}{0.5000}=93.562\ \text{m}^3/\text{s}$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00200$ and $n=0.013$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{93.562\times0.013}{0.3117\times0.04472}\right)^{3/8}=5.343\ \text{m}$$

Adopt the next standard size, D = 5400 mm.

Check of velocity and capacity

For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=4.203$ m/s and $Q_f=96.248$ m³/s.

$$\begin{aligned} v&=v_f\times(v/v_f)=4.203\times1.0000=4.20\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=96.248\times0.5000=48.124\ \text{m}^3/\text{s}\ (\ge 46.781) \end{aligned}$$

The velocity at maximum flow, 4.20 m/s, is above the maximum non-scouring limit of 3.0 m/s, so this design is not safe against scouring.

Step 5: Revised slope

Since the slope was only assumed, a flatter slope is chosen. With $S=1/1500$:

$$D=\left(\frac{93.562\times0.013}{0.3117\times0.02582}\right)^{3/8}=6.565\ \text{m}$$

Adopt $D=6.6$ m. Then $v_f=2.77$ m/s, and the half-full capacity is $47.45$ m³/s ($\ge 46.781$). The velocity of 2.77 m/s lies between 0.6 and 3.0 m/s, so this is satisfactory.

Answer: Design discharge = 46.78 m³/s. At 1 in 500 the required diameter is 5.343 m (5400 mm), but the velocity of 4.20 m/s is too high; at 1 in 1500 provide about 6.6 m diameter (v = 2.77 m/s). For such a large flow, two or more sewers in parallel or a box/arch section would be used in practice.