A circular sewer of diameter 0.5 m runs full at velocity 0.6 m/s. Discharge is nearly

Why is this the answer?

Discharge in a circular sewer running full: Q = A × V = (π D² / 4) × V. D = 0.5 m, V = 0.6 m/s. Q = π × 0.5² / 4 × 0.6 = 0.118 m³/s m³/s. For a sewer running full, the cross-sectional area equals π D²/4. Self-cleansing velocity is typically ≥ 0.6 m/s to prevent silt deposition.

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