For a peak discharge of 0.0157 cumec, with a velocity of 0.9 m/sec, the diameter of the sewer main, is A. 10 cm B. 12 cm C. 15 cm D. 18 cm

The correct answer is C. 15 cm.

The formula for calculating the diameter of a sewer main is:

$d = \sqrt{\frac{4Q}{v}}$

where:

• $d$ is the diameter of the sewer main in meters
• $Q$ is the peak discharge in cubic meters per second
• $v$ is the velocity of the flow in meters per second

Substituting the given values into the formula, we get:

$d = \sqrt{\frac{4(0.0157)}{0.9}} = 15$ cm

The other options