Open-channel hydraulics — circular channel condition for maxima For a circular open channel, when the hydraulic mean depth Dm equals 0.30 D (D = diameter), this condition corresponds to the maximum of which quantity?

Introduction / Context:Design of circular open channels (partially full pipes, sewers) often uses optimal depth conditions for maximum discharge or maximum velocity. Hydraulic mean depth Dm = A/P (area over wetted perimeter) is a convenient parameter. This question targets the specific condition Dm = 0.30 D.

Given Data / Assumptions:

• Circular channel partially full (open-channel flow).
• D = full pipe diameter; Dm = A/P.
• Known hydraulic optima exist for max discharge and max velocity.

Concept / Approach:For a circular channel: maximum discharge (flow rate) occurs at a depth close to 0.94–0.96 D, for which hydraulic mean depth is about 0.29–0.30 D. Maximum mean velocity occurs at a shallower depth (roughly 0.80–0.82 D), corresponding to Dm about 0.24–0.25 D. Therefore, Dm = 0.30 D signals the maximum discharge condition, not the maximum velocity condition.

Step-by-Step Solution:Recall optimums: Dm ≈ 0.30 D → max discharge; Dm ≈ 0.25 D → max velocity.Match the given Dm to the appropriate maximum.Select “flow rate.”

Verification / Alternative check:Derivations from Manning or Chezy formulations with geometric relations for circular segments confirm the distinct depths for maxima.

Why Other Options Are Wrong:

• Mean velocity: associated with Dm ≈ 0.25 D, not 0.30 D.
• Both/neither: contradict standard geometric-hydraulic results.

Common Pitfalls:

• Assuming maxima for velocity and discharge occur at the same depth—they do not.

Final Answer:flow rate