A circular sewer flows at a depth equal to one-quarter of its diameter (y = D/4). What is the wetted perimeter of the flowing segment, expressed in terms of D?

Introduction / Context:
Hydraulic computations for partially full circular sewers require geometric elements such as wetted perimeter P and area A. From these, hydraulic radius R = A/P and velocity by Manning or Chezy can be obtained.

Given Data / Assumptions:

• Circular sewer of diameter D, radius R = D/2.
• Flow depth y = D/4.
• Rigid geometry; no deformation.

Concept / Approach:

For a circular segment: if 2θ is the central angle (in radians) subtended by the water surface at the pipe centre, the relation between depth and θ is y = R(1 − cos θ). The wetted perimeter equals the arc length P = 2θR.

Step-by-Step Solution:

Verification / Alternative check:

At half-full, P would be πR = πD/2. Our depth is less than half-full, so P = πD/3 is consistent (smaller than πD/2).

Why Other Options Are Wrong:

πD/6 underestimates the arc; πD/2 corresponds to half-full; πD and 2πD/3 are too large for this shallow depth.

Common Pitfalls:

Confusing θ with 2θ; using degrees in place of radians when multiplying by radius; substituting diameter for radius in arc-length formula.

Final Answer:

πD/3