Design and Construction of Sewers

Part 1: Size of the sewer

At half depth ($d/D=0.5$) the area and hydraulic radius are half and equal to the full values, respectively: $A=A_f/2$, $R=R_f$, so $v=v_f$ and, with $n$ constant, $Q=Q_f/2$.

$$Q_f=2\times1.5=3.0\ \text{m}^3/\text{s}$$

Full-pipe Manning's formula: $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$ with $S=\dfrac{1}{2000}=0.0005$, $n=0.013$.

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{3.0\times0.013}{0.3117\times0.02236}\right)^{3/8}=1.907\ \text{m}$$

Adopt a standard size $D=2.10$ m. Check: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.120$ m/s (this is also the velocity at half depth), $Q_f=3.877$ m³/s, so half-full capacity $=1.939$ m³/s, greater than 1.5 m³/s. Velocity at design flow is above 0.6 m/s.

Part 2: Dry-season flow of 0.5 m³/s

Using the adopted $D=2.10$ m, $Q_f=3.877$ m³/s and $v_f=1.120$ m/s:

$$\frac{Q}{Q_f}=\frac{0.5}{3.877}=0.1290$$

For circular sections with constant $n$, this ratio corresponds to $d/D=0.243$ (solving $Q/Q_f=\frac{A}{A_f}\left(\frac{R}{R_f}\right)^{2/3}$). Then $\dfrac{v}{v_f}=0.689$ and

$$v=0.689\times1.120=0.77\ \text{m/s}$$

Since 0.77 m/s is greater than 0.60 m/s, the flow does maintain the self-cleansing velocity in the dry season.

Answer: $D\approx2.10$ m (computed 1.907 m). In the dry season the velocity is about 0.77 m/s, so the self-cleansing velocity is maintained.

Step 1: Peak runoff (rational method)

$t_c=55$ min. No rainfall formula is given, so $i=\dfrac{760}{t+10}$ mm/hr is assumed:

$$\begin{aligned} i&=\frac{760}{55+10}=11.69\ \text{mm/hr}\\ Q&=\frac{CiA}{360}=\frac{0.45\times11.69\times200}{360}=2.923\ \text{m}^3/\text{s} \end{aligned}$$

Step 2: Hydraulic elements at $d/D=0.6$

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).

Step 3: Diameter

Here the velocity at peak flow is fixed ($v=1.5$ m/s), so the flow area is

$$A_{p}=\frac{Q}{v}=\frac{2.923}{1.5}=1.949\ \text{m}^2$$ $$A_f=\frac{A_p}{A/A_f}=\frac{1.949}{0.6265}=3.111\ \text{m}^2\ \Rightarrow\ D=\sqrt{\frac{4A_f}{\pi}}=1.990\ \text{m}$$

Adopt $D=2.10$ m.

Step 4: Slope

$R=\dfrac{R}{R_f}\times\dfrac{D}{4}=1.1106\times\dfrac{1.990}{4}=0.5525$ m. From Manning's formula $v=\dfrac{1}{n}R^{2/3}S^{1/2}$:

$$S=\left(\frac{v\,n}{R^{2/3}}\right)^2=\left(\frac{1.5\times0.013}{0.5525^{2/3}}\right)^2=0.00084\approx\frac{1}{1192}$$

(For the adopted $D=2.10$ m, the slope needed for the same 1.5 m/s is $1$ in 1281.)

Answer: Diameter about 1.990 m (adopt 2.10 m), slope about 1 in 1192 (0.6 depth, v = 1.5 m/s, Q = 2.923 m³/s).

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.

Design flow

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

$$Q_{max}=\frac{100000\times200\times0.8\times2.25}{86400\times1000}=0.4167\ \text{m}^3/\text{s}$$

Data: $S=1/600$, $n=0.013$ (constant at all depths), depth ratio $d/D=0.70$.

For $d/D=0.7$: $\theta = 2\cos^{-1}(1-2d/D)=227.16^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.7477$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1849$, so $v/v_f=(R/R_f)^{2/3}=1.1198$ and $Q/Q_f=(A/A_f)(v/v_f)=0.8372$ (n taken constant).

Size of the sewer

Discharge when running full:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{0.4167}{0.8372}=0.4977\ \text{m}^3/\text{s}$$

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 $n=0.013$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.4977\times0.013}{0.3117\times0.04082}\right)^{3/8}=0.776\ \text{m}$$

Adopt the next standard size, D = 900 mm.

Check of velocity and capacity

For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.162$ m/s and $Q_f=0.7391$ m³/s.

$$\begin{aligned} v&=v_f\times(v/v_f)=1.162\times1.1198=1.30\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.7391\times0.8372=0.6188\ \text{m}^3/\text{s}\ (\ge 0.4167) \end{aligned}$$

The velocity at maximum flow, 1.30 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 = 900 mm at slope 1 in 600; it runs 0.7 full with v = 1.30 m/s at Q = 0.4167 m³/s (computed diameter 0.776 m).

Given

Maximum flow $Q=0.5$ m³/s, slope $S=1/1000=0.001$, $n=0.012$, depth of flow $d/D=0.7$ at maximum flow.

Hydraulic elements

For $d/D=0.7$: $\theta = 2\cos^{-1}(1-2d/D)=227.16^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.7477$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1849$, so $v/v_f=(R/R_f)^{2/3}=1.1198$ and $Q/Q_f=(A/A_f)(v/v_f)=0.8372$ (n taken constant).

Size of the sewer

Discharge when running full:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{0.5000}{0.8372}=0.5972\ \text{m}^3/\text{s}$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00100$ and $n=0.012$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.5972\times0.012}{0.3117\times0.03162}\right)^{3/8}=0.887\ \text{m}$$

Adopt the next standard size, D = 900 mm.

Check of velocity and capacity

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

$$\begin{aligned} v&=v_f\times(v/v_f)=0.975\times1.1198=1.09\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.6202\times0.8372=0.5193\ \text{m}^3/\text{s}\ (\ge 0.5000) \end{aligned}$$

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.

Answer: Provide a circular sewer of diameter 900 mm at slope 1 in 1000; it runs 0.7 full with velocity 1.09 m/s at 0.5 m³/s.

Given

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

Design for peak flow

For $d/D=0.7$: $\theta = 2\cos^{-1}(1-2d/D)=227.16^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.7477$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1849$, so $v/v_f=(R/R_f)^{2/3}=1.1198$ and $Q/Q_f=(A/A_f)(v/v_f)=0.8372$ (n taken constant).

Size of the sewer

Discharge when running full:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{0.5000}{0.8372}=0.5972\ \text{m}^3/\text{s}$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00100$ and $n=0.012$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.5972\times0.012}{0.3117\times0.03162}\right)^{3/8}=0.887\ \text{m}$$

Adopt the next standard size, D = 900 mm.

Check of velocity and capacity

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

$$\begin{aligned} v&=v_f\times(v/v_f)=0.975\times1.1198=1.09\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.6202\times0.8372=0.5193\ \text{m}^3/\text{s}\ (\ge 0.5000) \end{aligned}$$

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 $=\dfrac{Q_{max}}{\text{peak factor}}=\dfrac{0.5}{3}=0.1667$ m³/s. For the adopted pipe ($D=0.900$ m, $Q_f=0.6202$ m³/s, $v_f=0.975$ m/s):

$$\frac{Q}{Q_f}=\frac{0.1667}{0.6202}=0.2687\ \Rightarrow\ \frac{d}{D}=0.354,\ \ \frac{v}{v_f}=0.848$$ $$v=0.848\times0.975=0.83\ \text{m/s}$$

At the minimum flow (about one-third of average, $0.0556$ m³/s), $d/D=0.202$ and $v=0.60$ m/s.

Since the self-cleansing velocity is 0.6 m/s (0.75 m/s is preferred for sewage), the velocity at average dry weather flow, 0.83 m/s, is above 0.6 m/s; at minimum flow the velocity is above 0.6 m/s.

Answer: Provide 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 self-cleansing velocity is maintained at average flow.

Part 1: Diameter when running just full

For a circular pipe flowing full, $Q=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=1/150=0.006667$ and $n=0.013$:

$$D=\left(\frac{Q\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{1.6\times0.013}{0.3117\times0.08165}\right)^{3/8}=0.927\ \text{m}$$

Full velocity: $v_f=\dfrac{Q}{A}=\dfrac{1.6}{\frac{\pi}{4}(0.927)^2}=2.37$ m/s (check: $\frac{0.3969}{n}D^{2/3}S^{1/2}=2.37$ m/s).

Adopt a standard diameter of 1.05 m (the calculated size 0.927 m is the minimum required).

Part 2: Discharge when half full

At $d/D=0.5$: $A=A_f/2$ and $R=R_f$, so $v=v_f$ and with $n$ constant,

$$Q_{half}=\frac{Q_f}{2}=\frac{1.6}{2}=0.80\ \text{m}^3/\text{s}$$

(for the calculated diameter; for the adopted $D=1.05$ m the half-full capacity is $1.115$ m³/s).

Answer: D = 0.927 m (adopt 1.05 m); half-full discharge = 0.80 m³/s with the same velocity 2.37 m/s.

Given

$D=1.8$ m, $S=1/600$, $n=0.013$, depth of flow $d=0.75D$.

Full-flow values

$$\begin{aligned} v_f&=\frac{0.3969}{n}D^{2/3}S^{1/2}=\frac{0.3969}{0.013}(1.8)^{2/3}\left(\frac{1}{600}\right)^{1/2}=1.844\ \text{m/s}\\ Q_f&=v_f\times\frac{\pi}{4}D^2=1.844\times2.545=4.693\ \text{m}^3/\text{s} \end{aligned}$$

Proportionate values at $d/D=0.75$

For $d/D=0.75$: $\theta = 2\cos^{-1}(1-2d/D)=240.00^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.8045$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.2067$, so $v/v_f=(R/R_f)^{2/3}=1.1335$ and $Q/Q_f=(A/A_f)(v/v_f)=0.9119$ (n taken constant).

$$\begin{aligned} v&=1.1335\times1.844=2.09\ \text{m/s}\\ Q&=0.9119\times4.693=4.28\ \text{m}^3/\text{s} \end{aligned}$$

Answer: Velocity = 2.09 m/s; discharge = 4.28 m³/s.

Given

Peak discharge $Q=0.70$ m³/s, $S=1/1000$, $n=0.012$, depth of flow $d/D=2/3=0.667$.

For $d/D=0.6667$: $\theta = 2\cos^{-1}(1-2d/D)=218.94^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.7082$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.1645$, so $v/v_f=(R/R_f)^{2/3}=1.1069$ and $Q/Q_f=(A/A_f)(v/v_f)=0.7839$ (n taken constant).

Size of the sewer

Discharge when running full:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{0.7000}{0.7839}=0.8930\ \text{m}^3/\text{s}$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00100$ and $n=0.012$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.8930\times0.012}{0.3117\times0.03162}\right)^{3/8}=1.032\ \text{m}$$

Adopt the next standard size, D = 1050 mm.

Check of velocity and capacity

For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.081$ m/s and $Q_f=0.9355$ m³/s.

$$\begin{aligned} v&=v_f\times(v/v_f)=1.081\times1.1069=1.20\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.9355\times0.7839=0.7333\ \text{m}^3/\text{s}\ (\ge 0.7000) \end{aligned}$$

The velocity at maximum flow, 1.20 m/s, lies between the minimum (0.6 m/s) and the maximum non-scouring limit (3.0 m/s), so it is satisfactory.

For non-scouring, the velocity must not exceed the maximum permissible value (about 3 m/s for concrete sewers), and for self-cleansing it should be at least 0.6 m/s. The velocity is 1.20 m/s at peak flow, which is within 0.6-3.0 m/s, so the sewer is safe against scouring and also self-cleansing at peak flow. (At lower flows the velocity falls, so flows below about 0.058 m³/s give less than 0.6 m/s, and the minimum flow should be checked separately.)

Answer: D = 1050 mm (computed 1.032 m); v = 1.20 m/s < 3 m/s, so safe.

Part 1: Diameter and velocity (just full)

$S=1/400=0.0025$, $n=0.012$, $Q=1$ m³/s. For a full circular pipe:

$$D=\left(\frac{Q\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{1\times0.012}{0.3117\times0.05}\right)^{3/8}=0.907\ \text{m}$$ $$v=\frac{Q}{\frac{\pi}{4}D^2}=\frac{1}{\frac{\pi}{4}(0.907)^2}=1.55\ \text{m/s}$$

Adopt $D=1.05$ m. For it: $Q_f=1.479$ m³/s and $v_f=1.708$ m/s.

Part 2: Check at 0.6 m³/s

$$\frac{Q}{Q_f}=\frac{0.6}{1.479}=0.4056$$

From the proportionate-flow relations of a circular section (constant $n$), this gives $d/D=0.443$ and $v/v_f=0.948$, so

$$v=0.948\times1.708=1.62\ \text{m/s}$$

Since 1.62 m/s is greater than the self-cleansing velocity of 0.6 m/s, the sewer does maintain self-cleansing velocity when the flow drops to 0.6 m³/s.

Answer: D = 0.907 m (adopt 1.05 m), full velocity = 1.55 m/s; at 0.6 m³/s the velocity is 1.62 m/s, so self-cleansing is maintained.

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 $=400\times250=100000$. Assume 80% of the supply becomes sewage; peak factor = 2.7.

$$Q_{sew}=\frac{100000\times110\times0.8\times2.7}{86400\times1000}=0.2750\ \text{m}^3/\text{s}$$

Step 2: Storm water

Weighted coefficient: $C=0.5(0.5)+0.5(0.6)=0.55$. No intensity formula is given, so $i=\dfrac{760}{t+10}$ 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).

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

Step 3: Total design discharge

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

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

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:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{9.6388}{0.5000}=19.2776\ \text{m}^3/\text{s}$$

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 $n=0.013$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{19.2776\times0.013}{0.3117\times0.04472}\right)^{3/8}=2.955\ \text{m}$$

Adopt the next standard size, D = 3000 mm.

Check of velocity and capacity

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

$$\begin{aligned} v&=v_f\times(v/v_f)=2.840\times1.0000=2.84\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=20.0739\times0.5000=10.0370\ \text{m}^3/\text{s}\ (\ge 9.6388) \end{aligned}$$

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).

The sewer is taken as a combined sewer. Missing data are assumed: slope 1 in 500, peak factor 3, sewage = 80% of the 235 lpcd given, sewer running half full.

Step 1: Sewage flow

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

$$Q_{sew}=\frac{300000\times235\times0.8\times3}{86400\times1000}=1.9583\ \text{m}^3/\text{s}$$

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 storm of 20 min 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

$$Q=1.9583+44.333=46.292\ \text{m}^3/\text{s}$$

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:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{46.2917}{0.5000}=92.5833\ \text{m}^3/\text{s}$$

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 $n=0.013$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{92.5833\times0.013}{0.3117\times0.04472}\right)^{3/8}=5.322\ \text{m}$$

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 $Q_f=96.2406$ m³/s.

$$\begin{aligned} v&=v_f\times(v/v_f)=4.203\times1.0000=4.20\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=96.2406\times0.5000=48.1203\ \text{m}^3/\text{s}\ (\ge 46.2917) \end{aligned}$$

The velocity at maximum flow, 4.20 m/s, lies between the minimum (0.6 m/s) and the maximum non-scouring limit (3.0 m/s), so it is NOT within limits.

Answer: Diameter = 5.322 m; provide about 5400 mm (for such a large flow two or more sewers in parallel or a box/arch section may be used).

Assumptions

Sewage 135 lpcd of which 80% reaches the sewer, peak factor 3, $n=0.013$ for concrete, circular section running full at peak flow, intensity from the assumed formula $i=\dfrac{760}{t+10}$ mm/hr with storm duration 26 min.

Step 1: Discharge

$$\begin{aligned} Q_{sew}&=\frac{1500\times135\times0.8\times3}{86400\times1000}=0.0056\ \text{m}^3/\text{s}\\ i&=\frac{760}{26+10}=21.11\ \text{mm/hr}\\ Q_{storm}&=\frac{0.45\times21.11\times85}{360}=2.243\ \text{m}^3/\text{s}\\ Q&=Q_{sew}+Q_{storm}=2.249\ \text{m}^3/\text{s} \end{aligned}$$

Step 2: Slope

$$S=\frac{\text{fall}}{\text{length}}=\frac{22}{5000}=0.00440\approx\frac{1}{227}$$

Step 3: Size

$$D=\left(\frac{Q\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{2.249\times0.013}{0.3117\times0.0663}\right)^{3/8}=1.139\ \text{m}$$

Adopt $D=1.200$ m. Then $v_f=\dfrac{0.3969}{0.013}(1.200)^{2/3}S^{1/2}=2.29$ m/s and $Q_f=2.586$ m³/s.

Step 4: Check

The full-flow velocity 2.29 m/s is greater than the required self-cleansing velocity 0.98 m/s and below the maximum of 3 m/s, so the section is satisfactory. At the very small dry weather flow (about 0.0056 m³/s) the velocity in this big pipe will be lower, so a small low-flow channel (cunette) in the invert or an egg-shaped section is advisable.

Answer: A circular combined sewer of about 1.200 mm diameter at slope 22/5000 (1 in 227), carrying 2.249 m³/s at velocity 2.29 m/s.

Design flow

$$Q_{max}=\frac{80000\times145\times0.8\times3}{86400\times1000}=0.3222\ \text{m}^3/\text{s}$$

Data: $S=1/400$, $n=0.013$ (constant), depth ratio $d/D=0.75$.

For $d/D=0.75$: $\theta = 2\cos^{-1}(1-2d/D)=240.00^\circ$, $A/A_f=\dfrac{\theta-\sin\theta}{2\pi}=0.8045$, $R/R_f=1-\dfrac{\sin\theta}{\theta}=1.2067$, so $v/v_f=(R/R_f)^{2/3}=1.1335$ and $Q/Q_f=(A/A_f)(v/v_f)=0.9119$ (n taken constant).

Size of the sewer

Discharge when running full:

$$Q_f=\frac{Q}{Q/Q_f}=\frac{0.3222}{0.9119}=0.3534\ \text{m}^3/\text{s}$$

Manning's formula for a full circular pipe, $Q_f=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$, with $S=0.00250$ and $n=0.013$:

$$D=\left(\frac{Q_f\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.3534\times0.013}{0.3117\times0.05000}\right)^{3/8}=0.632\ \text{m}$$

Adopt the next standard size, D = 675 mm.

Check of velocity and capacity

For the adopted pipe: $v_f=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=1.175$ m/s and $Q_f=0.4203$ m³/s.

$$\begin{aligned} v&=v_f\times(v/v_f)=1.175\times1.1335=1.33\ \text{m/s}\\ Q_{capacity}&=Q_f\times(Q/Q_f)=0.4203\times0.9119=0.3833\ \text{m}^3/\text{s}\ (\ge 0.3222) \end{aligned}$$

The velocity at maximum flow, 1.33 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: Minimum diameter required = 0.632 m; provide 675 mm.

Part 1: Dry weather flow (sewage) [2 marks]

Assume 80% of 235 lpcd becomes sewage and the peak factor is 3.

$$Q_{sew}=\frac{65000\times235\times0.8\times3}{86400\times1000}=0.4243\ \text{m}^3/\text{s}$$

Part 2: Design of the combined sewer [6 marks]

Storm water. $t_c=t_e+t_f=5+18=23$ min. With the assumed formula $i=\dfrac{760}{t+10}$ mm/hr:

$$\begin{aligned} i&=\frac{760}{23+10}=23.03\ \text{mm/hr}\\ Q_{storm}&=\frac{0.55\times23.03\times60}{360}=2.111\ \text{m}^3/\text{s}\\ Q&=Q_{sew}+Q_{storm}=2.535\ \text{m}^3/\text{s} \end{aligned}$$

Size. The sewer is designed to run full at peak flow. Take a design velocity of 3.0 m/s (below the maximum permissible 3.2 m/s):

$$A=\frac{Q}{v}=\frac{2.535}{3.0}=0.845\ \text{m}^2,\qquad D=\sqrt{\frac{4A}{\pi}}=1.037\ \text{m}$$

Adopt $D=1.05$ m. Then $v=\dfrac{Q}{\frac{\pi}{4}D^2}=2.93$ m/s (less than 3.2 m/s).

Slope. From Manning's formula with $R=D/4=0.263$ m, $n=0.013$:

$$S=\left(\frac{v\,n}{R^{2/3}}\right)^2=\left(\frac{2.93\times0.013}{0.263^{2/3}}\right)^2=0.00862\approx\frac{1}{116}$$

Check at average dry weather flow $=Q_{sew}/3=0.1414$ m³/s: $Q/Q_f=0.0558$, $d/D=0.160$, $v=1.58$ m/s. This is greater than 0.6 m/s, so the self-cleansing condition is satisfied.

Answer: Circular combined sewer, D = 1.05 m, slope about 1 in 116, carrying 2.535 m³/s at 2.93 m/s (running full).

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.

Sewers are made of materials that must be strong, smooth, watertight, durable against sewage gases and acids, and economical. The common materials are given below.

1. Asbestos cement (AC) pipes

• Merits: light, easy to cut and lay, smooth inside (low friction), available in long lengths with few joints, resistant to corrosion.
• Demerits: weak against impact and heavy loads; attacked by acids and sulphates; health concern about asbestos fibres, so use is declining.

2. Cast iron (CI) pipes

• Merits: strong, durable, withstand high internal and external pressure and heavy traffic; long lengths; good for river crossings, inverted siphons, pumping mains.
• Demerits: costly, heavy; corrodes with sewage and acids unless lined; rough surface after rusting.

3. Concrete pipes (plain and reinforced, PCC and RCC)

• Merits: cheap, can be made at site in large sizes (up to 3 m), strong when reinforced, easy to repair, and widely available in Nepal.
• Demerits: attacked by acid wastes and hydrogen sulphide (crown corrosion), porous, heavy to transport, many joints in small lengths.

4. Vitrified clay / stoneware pipes

• Merits: resistant to acids and corrosion, smooth, impervious, durable, cheap in small sizes.
• Demerits: brittle and weak in tension, break in handling; available only in small sizes (up to 600 mm) and short lengths, so many joints.

5. Plastic pipes (PVC, HDPE, UPVC)

• Merits: very light, smooth (lower slope needed), corrosion free, flexible joints, long lengths, easy installation; popular for house connections and small sewers.
• Demerits: low strength under heavy loads and shallow cover; damaged by heat and sunlight; costly in large sizes.

6. Brick and stone masonry sewers

• Merits: can be built in any shape and large size at site using local material.
• Demerits: joints leak, rough surface, attacked by sewage, slow to construct, so used only for large old sewers with lining.

7. Steel pipes

• Merits: strong, light compared to CI, can have long lengths, welded watertight joints; used for pressure mains and siphons.
• Demerits: corrode easily (need lining and coating), costly.

For a modern urban sewer in Nepal, RCC pipes are used for large diameters, and HDPE/UPVC or stoneware for small ones.

Sewers are constructed in the following steps (for pipe sewers in an urban area).

 Ground level
 |<---- trench width ---->|
 |   \                  /   |   Sheeting & strutting
 |    \   backfill      /    |
 |     \______________/      |
 |      |  bedding   |       |
 |      |  ( O pipe )|       |
 |      |____________|       |
        gravel / concrete bed

1. Survey, setting out and alignment

The centre line and the levels (invert levels) of the sewer are marked on the ground from the design drawings with a theodolite and levelling instrument. Manhole locations are fixed. Sight rails (boning rods) are set up at intervals to control the gradient.

2. Excavation of trench

• The trench is excavated by hand or machine (excavator) to the required depth and width. Width = pipe diameter + 0.3-0.6 m working space on each side.
• The excavated soil is placed on one side of the trench at least 0.6 m from the edge, and the other side is left free for pipes.
• The depth must give a minimum cover of about 1 m to 1.2 m over the pipe (to protect from traffic).

3. Timbering (sheeting and shoring)

In loose soil or deep trenches (more than 1.5 m), the sides are supported by timber planks and struts or steel trench boxes to prevent collapse and accidents.

4. Dewatering

Water in the trench is removed by sump and pump, or well points, so that the pipe can be laid in dry condition.

5. Preparation of bed (bedding)

A bed of sand, gravel or 10-15 cm of concrete (1:3:6) is laid and levelled exactly to the slope. Holes are dug for the socket of the pipes. Rock is removed so the pipe rests uniformly.

6. Handling and laying of pipes

• Pipes are inspected for cracks and lowered into the trench by ropes, a chain pulley or crane.
• Laying starts from the downstream end, with the socket facing upstream, and each length is checked for level and line with a boning rod, so that the gradient is exact.

7. Jointing

• Spigot and socket joints: the annular gap is filled with tarred gasket (yarn), then cement mortar 1:1 or lead. A fillet is formed at the mouth.
• For concrete and PVC: collar joints, rubber ring joints or solvent joints. Joints must be watertight.

8. Testing of the sewer

Pipe lines are tested (see testing) before backfilling.

9. Backfilling

Selected soil, free from stones, is filled around and 30 cm above the pipe by hand, and compacted in layers (15-20 cm), then the rest of the trench is filled and road restored.

10. Manholes and house connections

Manholes are built at specified points, and the house connections are made with Y-junctions.

Steps in construction of sewers in an urban area

1. Setting out: survey; mark the centre line and levels of the sewer and manhole positions from the drawings. Place sight rails over the line.
2. Excavation: dig the trench by machine or hand, wide enough for pipes (diameter + 0.6 to 1.0 m) and deep enough for the invert level, with a minimum cover of about 1 m. Keep excavated soil away from the edge.
3. Timbering and dewatering: support trench sides with sheeting and struts, and remove water by pumping.
4. Bedding: lay a bed of sand, gravel or concrete, levelled to the design gradient.
5. Laying of pipes: lower the pipes carefully and lay from downstream to upstream, with sockets facing upstream, checking the line and level with boning rods.
6. Jointing: make water-tight joints with cement mortar, tarred yarn, rubber rings or solvent cement.
7. Testing: test the joints before backfilling.
8. Backfilling and compaction in layers, then restoration of the road surface.
9. Construction of manholes and house connections.

Testing of sewer line

Sewers are tested to ensure they are watertight and straight.

• Water (hydrostatic) test: a length of sewer between two manholes is plugged at the lower end, and filled with water to a head of about 1.5 m above the crown at the upper end. After allowing absorption, the drop in water level in a standpipe is measured over 30 min. The loss of water (leakage) must not exceed the limit given in the specification; otherwise the joints are repaired and the test repeated.
• Air test: the pipe is sealed and air at a pressure of about 100 mm of water is pumped in; the pressure should not fall by more than 25 mm in 5 minutes.
• Smoke test: smoke is blown in and any leak shows as smoke coming out (used for house drains).
• Mirror and lamp test: a mirror is held at one manhole and a lamp at the next; the circle of light shows the line and the clear bore of the pipe is checked for obstruction or sagging.
• Ball test: a ball of diameter about 13 mm less than the pipe is rolled through to check that there is no obstruction or projecting joint material.

Testing of sewers is done after laying and jointing, before backfilling, to check that the sewer is watertight, straight, true to gradient and free from obstruction. A defective sewer causes exfiltration (pollution of groundwater) or infiltration (extra flow).

Tests

1. Water (hydrostatic) test: the length between two manholes is plugged at the lower end and filled with water to give a head of about 1.5 m above the crown at the upper end. After the pipe has absorbed water, the fall of level in a standpipe over a set time (often 30 min) is noted. Leakage must be within the specified limit.
2. Air test: the sealed pipe is pressurised with air to about 100 mm of water gauge; the drop of pressure over 5 minutes should not exceed about 25 mm. It is quick and does not require water.
3. Smoke test: smoke is forced into the sewer; escape at any point shows a leak (mostly for house drains and old sewers).
4. Mirror and lamp test (alignment test): a mirror at one manhole reflects light from a lamp at the next manhole. A full bright circle shows that the sewer is straight and clear.
5. Ball (obstruction) test: a ball slightly smaller than the pipe is rolled along; obstructions or projecting joint material stop it.
6. Infiltration test: in a sewer below the water table, the flow entering is measured at the lower manhole.

Defects found are repaired and the test is repeated until the line is accepted.