Most economical circular channel section for maximum discharge in open-channel hydraulics: identify the correct combined conditions (depth–diameter ratio, hydraulic mean depth, and wetted perimeter relations).
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ADepth of water = 0.95 × diameter of the circular section
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BHydraulic mean depth (R) = 0.286 × diameter of the circular section
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CWetted perimeter = 2.6 × diameter of the circular section
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DWetted perimeter = 2.83 × depth of water
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EAll of the above
Answer
Correct Answer: All of the above
Explanation
Introduction / Context:
For circular channels (like sewers and tunnels) operating partially full in steady uniform flow, hydraulic efficiency is maximized when geometric proportions create the greatest discharge for a given roughness and slope. The “most economical section” minimizes wetted perimeter for a given area (or equivalently maximizes hydraulic radius R = A/P), which directly influences discharge via Manning or Chezy equations.
Given Data / Assumptions:
- Prismatic circular channel conveying water under gravity in open-channel flow.
- Steady, uniform flow; roughness and bed slope are fixed.
- Objective: maximum discharge (economical section conditions).
Concept / Approach:
For a circular section, the best hydraulic condition occurs at a specific depth ratio y/D (not necessarily running full). Classical derivations using calculus on A(θ) and P(θ) with θ as half central angle lead to standard proportionalities for the optimum: y/D ≈ 0.94–0.95, R ≈ 0.286 D, and characteristic wetted perimeters with respect to D and y. These relations are widely used as design thumb rules.
Step-by-Step Solution:
Optimal depth ratio: y ≈ 0.95 D (rounded from ~0.94).Hydraulic mean depth: R = A/P ≈ 0.286 D at the optimum.Wetted perimeter: P ≈ 2.6 D when the section is most economical.Alternative representation: P ≈ 2.83 y (since y ≈ 0.95 D, both forms are consistent).Verification / Alternative check:
Using Manning’s Q ∝ A R^(2/3) S^(1/2), maximizing Q for fixed S and n reduces to maximizing A R^(2/3). Differentiation with respect to θ for a circular segment yields the listed numerical factors, all consistent with standard hydraulics handbooks.
Why Other Options Are Wrong:
- Each single statement (options A–D) is correct individually; hence choosing just one under-represents the complete condition set.
- The only fully correct choice consolidating all truths is 'All of the above'.
Common Pitfalls:
- Assuming maximum discharge occurs when the pipe runs full (not true for roughness-controlled open-channel behavior).
- Confusing hydraulic radius R with geometric radius D/2.
Final Answer:
All of the above