Combined sewer, half full, 1.5 m³/s

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.