Hydraulically equivalent sections — for a square sewer (side = 1000 mm), find the diameter of a circular section with the same discharge capacity under equal slope and roughness

Introduction / Context:
Two closed conduits are “hydraulically equivalent” if, under the same slope and roughness, they convey the same discharge. For full-flow sewers, discharge (for turbulent regime) scales approximately with A * R^(2/3), where A is area and R is hydraulic radius (A/P). Converting between shapes assists in design substitutions and catalog selection.

Given Data / Assumptions:

• Square section: side a = 1000 mm = 1.0 m.
• Target: equivalent circular diameter d.
• Equal slope S and Manning/Chezy roughness for both sections.
• Use Q ∝ A * R^(2/3) for comparison (same proportionality constant).

Concept / Approach:

For a square running full: A_s = a^2, P_s = 4a, R_s = A_s / P_s = a/4. For a circle running full: A_c = (π d^2)/4, P_c = π d, R_c = d/4. Equate A_s * R_s^(2/3) = A_c * R_c^(2/3) and solve for d in terms of a.

Step-by-Step Solution:

Verification / Alternative check (if short method exists):

Using hydraulic mean depth equality alone would give d = a; however, equal discharge requires the A * R^(2/3) criterion, yielding the 1.095 factor.

Why Other Options Are Wrong:

1045, 1065, and 1075 mm underestimate the required diameter; 1000 mm ignores the difference in wetted perimeter/area interplay between shapes.

Common Pitfalls (misconceptions, mistakes):

Equating only hydraulic radius or only area; forgetting that both A and R influence discharge capacity.

Final Answer:

1095 mm