Velocity and discharge, 1.8 m sewer

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.