Sewer hydraulics — minimum slope for a 2 m diameter circular sewer flowing nearly full A circular sewer of diameter 2.0 m must carry Q = 2 m^3/s when running nearly full. Assuming Manning’s n = 0.015 and using full-flow hydraulics as an approximation, what is the minimum bed slope S required to initiate this discharge?

Introduction / Context:
Gravity sewers are sized using uniform-flow relations such as Manning’s equation. For large circular conduits running full or nearly full, the hydraulic radius and area are well defined, allowing a direct computation of the slope needed to convey a given discharge.

Given Data / Assumptions:

• Diameter D = 2.0 m, discharge Q = 2 m^3/s.
• Manning’s n = 0.015.
• Full-flow approximation: A = πD^2/4 and R = A/P for a full circle.
• “Nearly full” ≈ “full” for area and hydraulic radius estimation.

Concept / Approach:
Manning’s equation: Q = (1/n) * A * R^(2/3) * S^(1/2). For a full circular section: A = πD^2/4 = π m^2 and R = A/P = (πD^2/4)/(πD) = D/4 = 0.5 m. Solve for slope S given Q.

Step-by-Step Solution:
Compute geometry: A = π ≈ 3.1416 m^2, R = 0.5 m.Compute coefficient: C = (1/n) * A * R^(2/3) ≈ (1/0.015) * 3.1416 * (0.5)^(2/3) ≈ 131.94.Manning form: Q = C * S^(1/2) ⇒ S = (Q/C)^2 = (2 / 131.94)^2 ≈ 2.30 × 10^-4.Therefore S ≈ 0.00023.

Verification / Alternative check:
Chezy’s relation with equivalent roughness gives a similar slope for the same Q. Using “nearly full” instead of exact depth introduces negligible change for this sizing level.

Why Other Options Are Wrong:
0.000036, 0.000014: Too flat; would not carry 2 m^3/s. 0.000091: Still too flat. 0.00045: Excessively steep for the given parameters.

Common Pitfalls:

• Using hydraulic radius R = D/8 (half-full value) instead of R = D/4 for full flow.
• Confusing S with S^(1/2) when rearranging Manning’s equation.

Final Answer:
0.00023