Skip to main content

Sewer diameter for a 2.5 sq km area

Short answer

Computed sanitary flow 0.275 m³/s, storm flow 9.364 m³/s, total 9.639 m³/s. For half-full sewer, Q_full=19.28 m³/s; using Manning's with n=0.013, S=1/500 yields diameter 2.955 m. Provide 3000 mm circular sewer (runs half full, v=2.84 m/s within limits).

Calculate the diameter of a sewer to serve an area of 2.5 sq. km. with a population density of 400 persons per hectare. The average rate of water supply is 110 lpcd. The coefficient of runoff for 50% area is 0.5 and the remaining area is 0.6. The time taken to reach by the storm water at the considered point of sewer from the farthest point of the catchment area is 18 minute. Consider the storm duration of 21 minutes, values of Manning's n = 0.013 at all the depths of flow. The permissible slope is 1 in 500. Consider peak factor of 2.7. The sewer should run half full.

The sewer is taken as a combined sewer (sewage + storm water), running half full at maximum flow.

Step 1: Sanitary sewage

Area = 2.5 km² = 250 ha, so population =400×250=100000=400\times250=100000. Assume 80% of the supply becomes sewage; peak factor = 2.7.

Qsew=100000×110×0.8×2.786400×1000=0.2750 m3/sQ_{sew}=\frac{100000\times110\times0.8\times2.7}{86400\times1000}=0.2750\ \text{m}^3/\text{s}

Step 2: Storm water

Weighted coefficient: C=0.5(0.5)+0.5(0.6)=0.55C=0.5(0.5)+0.5(0.6)=0.55. No intensity formula is given, so i=760t+10i=\dfrac{760}{t+10} mm/hr is assumed for the given storm duration of 21 min (the time of concentration is 18 min, so a 21 min storm covers the whole catchment).

i=76021+10=24.52 mm/hrQstorm=0.55×24.52×250360=9.364 m3/s\begin{aligned} i&=\frac{760}{21+10}=24.52\ \text{mm/hr}\\ Q_{storm}&=\frac{0.55\times24.52\times250}{360}=9.364\ \text{m}^3/\text{s} \end{aligned}

Step 3: Total design discharge

Q=0.2750+9.364=9.639 m3/sQ=0.2750+9.364=9.639\ \text{m}^3/\text{s}

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

For d/D=0.5d/D=0.5: θ=2cos⁡−1(1−2d/D)=180.00∘\theta = 2\cos^{-1}(1-2d/D)=180.00^\circ, A/Af=θ−sin⁡θ2π=0.5000A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.5000, R/Rf=1−sin⁡θθ=1.0000R/R_f=1-\dfrac{\sin\theta}{\theta}=1.0000, so v/vf=(R/Rf)2/3=1.0000v/v_f=(R/R_f)^{2/3}=1.0000 and Q/Qf=(A/Af)(v/vf)=0.5000Q/Q_f=(A/A_f)(v/v_f)=0.5000 (n taken constant).

Size of the sewer

Discharge when running full:

Qf=QQ/Qf=9.63880.5000=19.2776 m3/sQ_f=\frac{Q}{Q/Q_f}=\frac{9.6388}{0.5000}=19.2776\ \text{m}^3/\text{s}

Manning's formula for a full circular pipe, Qf=0.3117nD8/3S1/2Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}, with S=0.00200S=0.00200 and n=0.013n=0.013:

D=(Qf n0.3117 S1/2)3/8=(19.2776×0.0130.3117×0.04472)3/8=2.955 mD=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{19.2776\times0.013}{0.3117\times0.04472}\right)^{3/8}=2.955\ \text{m}

Adopt the next standard size, D = 3000 mm.

Check of velocity and capacity

For the adopted pipe: vf=0.3969nD2/3S1/2=2.840v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=2.840 m/s and Qf=20.0739Q_f=20.0739 m³/s.

v=vf×(v/vf)=2.840×1.0000=2.84 m/sQcapacity=Qf×(Q/Qf)=20.0739×0.5000=10.0370 m3/s (≥9.6388)\begin{aligned} v&=v_f\times(v/v_f)=2.840\times1.0000=2.84\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=20.0739\times0.5000=10.0370\ \text{m}^3/\text{s}\ (\ge 9.6388) \end{aligned}

The velocity at maximum flow, 2.84 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 required = 2.955 m; provide 3000 mm circular sewer (running half full, v = 2.84 m/s).

✓ Answer checked by Claude Opus 5.5 (high effort)

All Sanitary Engineering chapter 3 questions →