Design discharge, separate and combined

Assumptions

• 80% of water supplied becomes sewage; peak factor = 3; infiltration neglected.
• No rainfall formula is given, so $i=\dfrac{760}{t+10}$ mm/hr is assumed for the intensity at $t_c$.

(a) Separate system (sewage only)

$$Q_{sew}=\frac{45000\times150\times0.8\times3}{86400\times1000}=0.1875\ \text{m}^3/\text{s}=187.5\ \text{l/s}$$

Design the sewer running full ($S=1/500$, $n=0.013$). From Manning's formula for a full circular pipe, $Q=\dfrac{0.3117}{n}D^{8/3}S^{1/2}$:

$$D=\left(\frac{Q\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{0.1875\times0.013}{0.3117\times0.04472}\right)^{3/8}=0.520\ \text{m}$$

Adopt $D=0.525$ m (standard size). Full velocity $v=\dfrac{0.3969}{n}D^{2/3}S^{1/2}=0.89$ m/s, which is above 0.6 m/s.

(b) Combined system

$$\begin{aligned} i&=\frac{760}{30+10}=19.0\ \text{mm/hr}\\ Q_{storm}&=\frac{0.45\times19.0\times20}{360}=0.475\ \text{m}^3/\text{s}\\ Q_{comb}&=Q_{sew}+Q_{storm}=0.1875+0.475=0.662\ \text{m}^3/\text{s} \end{aligned}$$ $$D=\left(\frac{0.662\times0.013}{0.3117\times0.04472}\right)^{3/8}=0.835\ \text{m}$$

Adopt $D=0.90$ m. Full velocity $=1.27$ m/s (below the 3 m/s maximum).

Answer: Separate system: $Q=0.1875$ m³/s, sewer about $D=0.525$ m. Combined system: $Q=0.662$ m³/s, sewer about $D=0.90$ m, both at slope 1 in 500.