Circular sewer just full, 1.6 m³/s

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.