Geometry of a partially full circular sewer: If the central angle of the water-filled segment is α (in radians) for a pipe of radius r, what is the cross-sectional flow area A?
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AA = (r^2 / 2) * (α − sin α)
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BA = r^2 * (α + sin α)
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CA = (r^2 / 2) * (α + sin α)
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DA = r^2 * (α − sin α) / 4
Answer
Correct Answer: A = (r^2 / 2) * (α − sin α)
Explanation
Introduction / Context:Many hydraulic calculations for sewers flowing partially full require geometric properties of a circular segment: area, wetted perimeter, and hydraulic radius as functions of the central angle α. These relationships enable evaluation of conveyance using Manning or Chézy formulas.
Given Data / Assumptions:
- Circular pipe of radius r.
- Water-filled segment subtends angle α at the center (radians).
- Uniform cross-section; ignore crown roughness variations.
Concept / Approach:
The area of a circular segment equals the area of the sector minus the area of the isosceles triangle formed by the two radii and the chord. Sector area for angle α is (α * r^2) / 2. Triangle area with side 2r sin(α/2) and included angle α is (r^2 * sin α) / 2. Subtracting yields the standard formula.
Step-by-Step Solution:
Sector area = (α * r^2) / 2.Triangle area = (r^2 * sin α) / 2.Segment (flow) area A = sector − triangle = (r^2 / 2) * (α − sin α).Use α consistent with flow depth to compute A for hydraulics.Verification / Alternative check:
Check limits: if α = 0, A = 0; if α = π (half full), A = (r^2 / 2) * (π − 0) = (π r^2)/2; if α = 2π (full), expression gives full circle area (r^2/2)*(2π − 0) = π r^2, which is correct.
Why Other Options Are Wrong:
Options with plus signs overestimate area by adding the triangle; divisors not equal to 2 break the correct sector-triangle relationship.
Common Pitfalls:
Using α in degrees; forgetting to subtract the triangle area; mixing diameter D and radius r inconsistently.
Final Answer:
A = (r^2 / 2) * (α − sin α)