Most economical circular channel for maximum velocity (open-channel flow) For a circular channel carrying water (not running full), the section that gives maximum mean velocity satisfies which of the following conditions?
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Adepth of water = 0.810 × diameter
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Bhydraulic mean depth (R) = 0.304 × diameter
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Cwetted perimeter = 2.245 × diameter
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Dall the above
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Eonly (a) and (b)
Answer
Correct Answer: all the above
Explanation
Introduction / Context:For a given circular conduit running partially full as an open channel, there is a specific depth that maximizes the mean velocity (equivalently minimizes hydraulic losses per unit area). This is often called the most economical circular section for velocity in open-channel hydraulics.
Given Data / Assumptions:
- Circular conduit flowing with a free surface (not pressurized).
- Resistance described by Manning or Chezy; objective is to maximize V ∝ R^(2/3) for given slope and roughness.
- Geometry relates depth, area, and wetted perimeter via the central angle of the wetted segment.
Concept / Approach:
By differentiating V with respect to depth (or equivalently maximizing R = A/P), the optimal depth occurs at y ≈ 0.81 D. The corresponding geometric relations yield R ≈ 0.304 D and wetted perimeter P ≈ 2.245 D.
Step-by-Step Solution:
Set up A(θ) and P(θ) for a circular segment; express R(θ) = A/P.Differentiate R with respect to θ and set derivative to zero to find optimal θ.Back-substitute to find y/D ≈ 0.81, R/D ≈ 0.304, and P/D ≈ 2.245.Verification / Alternative check:
Standard hydraulics tables list the same ratios for the most economical circular section for maximum velocity, confirming the results.
Why Other Options Are Wrong:
Each item (a), (b), and (c) is individually correct; therefore, any single selection is incomplete. The comprehensive correct choice is “all the above.”
Common Pitfalls:
Confusing “maximum velocity” with “maximum discharge” depth (which has a different optimal ratio); applying full-pipe relations to free-surface flow.
Final Answer:
all the above