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

[amp_mcq option1=”10 cm” option2=”12 cm” option3=”15 cm” option4=”18 cm” correct=”option3″]

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 are incorrect because they do not result in a diameter that is large enough to accommodate the given peak discharge.