Circular combined sewer, 60 ha

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).