A circular sewer of diameter 2.0 m runs partially full at depth y = 0.50 m. What is the wetted perimeter (in metres) of the flowing segment? (Use geometric relations for a circular segment.)

Civil Engineering Waste Water Engineering Difficulty: Medium
Choose an option
  • A
    π/3 m
  • B
    2π/3 m
  • C
    π m
  • D
    4π/3 m
  • E
    2 m

Answer

Correct Answer: 2π/3 m

Explanation

Introduction / Context:Hydraulic elements for partially full circular sewers (area, wetted perimeter, hydraulic radius) are needed to compute discharge and velocity using Manning or Chezy equations. This problem asks for the wetted perimeter from basic geometry.

Given Data / Assumptions:

  • Diameter D = 2.0 m → radius R = 1.0 m.
  • Flow depth y = 0.50 m.
  • Steady cross-sectional geometry; no deformation.

Concept / Approach:

For a circular segment of depth y in a circle of radius R, the half-central angle θ satisfies y = R(1 − cos θ). The wetted perimeter Pw equals arc length subtended by 2θ: Pw = R * 2θ (with θ in radians).

Step-by-Step Solution:

Compute cos θ = 1 − y/R = 1 − 0.50/1.0 = 0.5.Thus θ = arccos(0.5) = π/3 radians (60°).Wetted perimeter Pw = R * 2θ = 1.0 * 2 * (π/3) = 2π/3 m.

Verification / Alternative check:

At y = R (i.e., 1.0 m), Pw would be half the circumference = πR = π m; since 0.5 m is shallower than R, Pw should be less than π m; 2π/3 m satisfies this expectation.

Why Other Options Are Wrong:

(a) π/3 m is the half-angle, not the arc length. (c) π m corresponds to half-full, not y = 0.5 m. (d) 4π/3 m exceeds half-circumference for this shallow depth. (e) 2 m is not an arc length derived from geometry.

Common Pitfalls:

Mixing the central angle with arc length; using degrees instead of radians in arc computations; confusing depth y with rise from invert to water surface across the diameter.

Final Answer:

2π/3 m

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