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Separate sewer, check self-cleansing

Short answer

The sewer is designed as 900 mm at 1 in 1000. At average dry weather flow (0.1667 m³/s), the velocity is 0.83 m/s, which exceeds the self-cleansing velocity of 0.6 m/s, so self-cleansing is maintained.

Design a sewer for separate system to carry peak flow 0.5 m³/sec at a slope 10 in 10000. Sewer should run 0.7 times depth at peak flow. The value of n in Manning's formula is 0.012. Will the self cleansing velocity be maintained in the sewer during dry weather flow? Take peak factor = 3.

Given

Peak flow Qmax=0.5Q_{max}=0.5 m³/s, S=10/10000=0.001S=10/10000=0.001, n=0.012n=0.012, d/D=0.7d/D=0.7 at peak flow, peak factor = 3.

Design for peak flow

For d/D=0.7d/D=0.7: θ=2cos⁡−1(1−2d/D)=227.16∘\theta = 2\cos^{-1}(1-2d/D)=227.16^\circ, A/Af=θ−sin⁡θ2π=0.7477A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.7477, R/Rf=1−sin⁡θθ=1.1849R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1849, so v/vf=(R/Rf)2/3=1.1198v/v_f=(R/R_f)^{2/3}=1.1198 and Q/Qf=(A/Af)(v/vf)=0.8372Q/Q_f=(A/A_f)(v/v_f)=0.8372 (n taken constant).

Size of the sewer

Discharge when running full:

Qf=QQ/Qf=0.50000.8372=0.5972 m3/sQ_f=\frac{Q}{Q/Q_f}=\frac{0.5000}{0.8372}=0.5972\ \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=0.00100S=0.00100 and n=0.012n=0.012:

D=(Qf n0.3117 S1/2)3/8=(0.5972×0.0120.3117×0.03162)3/8=0.887 mD=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.5972\times0.012}{0.3117\times0.03162}\right)^{3/8}=0.887\ \text{m}

Adopt the next standard size, D = 900 mm.

Check of velocity and capacity

For the adopted pipe: vf=0.3969nD2/3S1/2=0.975v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=0.975 m/s and Qf=0.6202Q_f=0.6202 m³/s.

v=vf×(v/vf)=0.975×1.1198=1.09 m/sQcapacity=Qf×(Q/Qf)=0.6202×0.8372=0.5193 m3/s (≥0.5000)\begin{aligned} v&=v_f\times(v/v_f)=0.975\times1.1198=1.09\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.6202\times0.8372=0.5193\ \text{m}^3/\text{s}\ (\ge 0.5000) \end{aligned}

The velocity at maximum flow, 1.09 m/s, lies between the minimum (0.6 m/s) and the maximum non-scouring limit (3.0 m/s), so it is satisfactory.

Check in dry weather flow

Average (dry weather) flow =Qmaxpeak factor=0.53=0.1667=\dfrac{Q_{max}}{\text{peak factor}}=\dfrac{0.5}{3}=0.1667 m³/s. For the adopted pipe (D=0.900D=0.900 m, Qf=0.6202Q_f=0.6202 m³/s, vf=0.975v_f=0.975 m/s):

QQf=0.16670.6202=0.2687 ⇒ dD=0.354,  vvf=0.848\frac{Q}{Q_f}=\frac{0.1667}{0.6202}=0.2687\ \Rightarrow\ \frac{d}{D}=0.354,\ \ \frac{v}{v_f}=0.848 v=0.848×0.975=0.83 m/sv=0.848\times0.975=0.83\ \text{m/s}

At the minimum flow (about one-third of the average, 0.05560.0556 m³/s), d/D=0.202d/D=0.202 and v=0.60v=0.60 m/s.

The self-cleansing velocity is 0.6 m/s (0.75 m/s is preferred for sewage). The velocity at the average dry weather flow, 0.83 m/s, is well above 0.6 m/s, and even at the minimum flow the velocity is just about 0.6 m/s.

Answer: Provide a 900 mm sewer at 1 in 1000. Velocity at peak flow = 1.09 m/s; at dry weather (average) flow = 0.83 m/s, so the self-cleansing velocity is maintained.

✓ Answer checked by Claude Opus 5.5 (high effort)

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