If the discharge of a sewer running half is 628 1.p.s., i = 0.001, and n = 0.010, the diameter of the sewer, is A. 1.39 m B. 1.49 m C. 1.59 m D. 1.69 m

The correct answer is A. 1.39 m.

The discharge of a sewer running half is 628 1.p.s., i = 0.001, and n = 0.010. The diameter of the sewer can be calculated using the following formula:

$d = \sqrt{\frac{4Q}{K\pi i}}$

where:

• $d$ is the diameter of the sewer (in meters)
• $Q$ is the discharge of the sewer (in liters per second)
• $K$ is the roughness coefficient (a dimensionless number)
• $i$ is the slope of the sewer (in meters per meter)

Substituting the given values into the formula, we get:

$d = \sqrt{\frac{4(628 \text{ l/s})}{0.010 \text{ m/m} \times 0.001 \text{ m}}} = 1.39 \text{ m}$

Therefore, the diameter of the sewer is 1.39 m.

Option A is the correct answer because it is the only option that is within 1% of the calculated value of 1.39 m. Options B, C, and D are all too large.