Given a discharge Q and maximum allowable velocity Vmax, how is the minimum pipe diameter determined?

• A. D = Q / Vmax.
• B. A = Q / Vmax; D = 2 sqrt(A/π); choose D ≥ this value.
• C. A = Q Vmax; D = 2 sqrt(A/π).
• D. A = Q Vmax; D = sqrt(A/π).

Answer: (B)

To keep flow velocity within the limit, use Q = A × V, so the velocity V = Q / A. Requiring V ≤ Vmax gives A ≥ Q / Vmax, so the minimum cross-sectional area is A_min = Q / Vmax. For a circular pipe, A = πD^2/4, so solving for D with A = A_min yields D_min = 2 sqrt(A_min/π) = 2 sqrt( Q / (π Vmax) ). Therefore the minimum diameter must satisfy D ≥ 2 sqrt( Q / (π Vmax) ).