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?
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Aπ D^(8/3) = 4 b^(8/3)
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Bπ D^(3/8) = 4 b^(3/8)
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Cπ D^(2/3) = 4 b^(2/3)
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Dπ D^(3/2) = 4 b^(3/2)
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ENone of these
Answer
Correct Answer: π D^(8/3) = 4 b^(8/3)
Explanation
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:
For circle: capacity term = (π D^2 / 4)^(5/3) / (π D)^(2/3) = [π D^(8/3)] / [4^(5/3)].For square: capacity term = (b^2)^(5/3) / (4b)^(2/3) = b^(8/3) / 4^(2/3).Equate and simplify: π D^(8/3) / 4^(5/3) = b^(8/3) / 4^(2/3) ⇒ π D^(8/3) = 4 b^(8/3).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)