Sewer diameter and velocity, just full

Part 1: Diameter and velocity (just full)

$S=1/400=0.0025$, $n=0.012$, $Q=1$ m³/s. For a full circular pipe:

$$D=\left(\frac{Q\,n}{0.3117\,S^{1/2}}\right)^{3/8}=\left(\frac{1\times0.012}{0.3117\times0.05}\right)^{3/8}=0.907\ \text{m}$$ $$v=\frac{Q}{\frac{\pi}{4}D^2}=\frac{1}{\frac{\pi}{4}(0.907)^2}=1.55\ \text{m/s}$$

(In practice, a standard size such as 975 mm or 1.05 m would be adopted.)

Part 2: Check at 0.6 m³/s

For the sewer that runs just full at 1 m³/s ($D=0.907$ m, $Q_f=1$ m³/s, $v_f=1.55$ m/s):

$$\frac{Q}{Q_f}=\frac{0.6}{1.0}=0.60$$

From the proportionate-flow relations of a circular section (constant $n$), this gives $d/D=0.558$ and $v/v_f=1.045$, so

$$v=1.045\times1.55=1.62\ \text{m/s}$$

Since 1.62 m/s is greater than the self-cleansing velocity of 0.6 m/s, the sewer does maintain the self-cleansing velocity when the flow drops to 0.6 m³/s.

Answer: D = 0.907 m, full velocity = 1.55 m/s; at 0.6 m³/s the depth is about 0.56D and the velocity is 1.62 m/s, so self-cleansing is maintained.