Sizing a sewer main from flow and velocity: For a peak discharge of 0.0157 m^3/s conveyed at 0.9 m/s, what is the approximate internal diameter of the sewer main (assume full circular section)?

Introduction / Context:
Quick diameter estimates are routine in sewer design when target velocities are specified for self-cleansing and hydraulic efficiency. The continuity relationship Q = A * V directly links discharge, area, and velocity.

Given Data / Assumptions:

• Peak discharge Q = 0.0157 m^3/s.
• Mean velocity V = 0.9 m/s.
• Full circular flow assumed for sizing (conservative first-pass estimate).

Concept / Approach:
Use Q = A * V to find area A, then back-calculate diameter from A = (π/4) * d^2. Convert to centimeters for selection among the discrete options.

Step-by-Step Solution:
Compute area: A = Q / V = 0.0157 / 0.9 ≈ 0.017444 m^2.Relate to diameter: A = (π/4) * d^2 ⇒ d = sqrt(4A / π).Calculate: 4A ≈ 0.069776; divide by π ≈ 0.02222.d ≈ sqrt(0.02222) ≈ 0.149 m ≈ 0.15 m.Therefore, diameter ≈ 15 cm.

Verification / Alternative check:
Reverse check: A = π*(0.15^2)/4 ≈ 0.01767 m^2; Q = A * V ≈ 0.01767 * 0.9 ≈ 0.0159 m^3/s, close to the given 0.0157 m^3/s.

Why Other Options Are Wrong:
10–12 cm: Too small; would require higher velocity than 0.9 m/s for the same flow.18–20 cm: Oversized; would give lower velocity than desired for self-cleansing at this flow.

Common Pitfalls:

• Using nominal diameters without checking velocity limits for self-cleansing.
• Confusing internal diameters with external (nominal) diameters; always use internal for hydraulics.

Final Answer:
15 cm