Hydraulically equivalent full-flow sewers: A circular pipe (diameter D) and a square conduit (side b) discharge the same on the same grade and roughness. Which relation holds good?

Introduction / Context:
“Hydraulically equivalent” full-flow conduits discharge the same flow for the same slope and roughness. For gravity sewers, Manning/Chézy formulations show discharge depends on cross-sectional area and hydraulic radius. Equating these appropriately yields a relationship between characteristic dimensions of different shapes.

Given Data / Assumptions:

• Both conduits run full, same slope S and roughness n.
• Discharge relation: Q ∝ A * R^(2/3) for Manning's form.
• Circle: A_c = π D^2 / 4, P_c = π D, R_c = A_c / P_c = D/4.
• Square: A_s = b^2, P_s = 4 b, R_s = b/4.

Concept / Approach:

Equate capacity terms A R^(2/3) for the two shapes. Because R = A/P, the capacity term becomes A * (A/P)^(2/3) = A^(5/3) / P^(2/3), which leads to a simple power relation between D and b.

Step-by-Step Solution:

Verification / Alternative check:

Dimensional consistency and monotonic relation: larger D corresponds to larger b, as expected for equal capacity.

Why Other Options Are Wrong:

Other exponents do not arise from the A^(5/3)/P^(2/3) equivalence; they would imply incorrect dependence on shape parameters.

Common Pitfalls:

Using R instead of A^(5/3)/P^(2/3) directly; mixing running full with part-full relations.

Final Answer:

π D^(8/3) = 4 b^(8/3)