Pump sizing – minimum horsepower: A well discharges 3.25 × 10^-2 m^3/s and must be lifted against a total head of 30 m. What is the minimum required pump horsepower (assume practical overall efficiency so the installed power meets duty)?

Introduction / Context:Selecting a pump–motor set requires estimating hydraulic power and adjusting for realistic efficiency. Oversights here can result in underpowered installations, poor service pressures, and shortened equipment life.

Given Data / Assumptions:

• Discharge Q = 3.25 × 10^-2 m^3/s.
• Total dynamic head H = 30 m.
• Water density ρ ≈ 1000 kg/m^3; g ≈ 9.81 m/s^2.
• Overall (pump + motor + transmission) efficiency taken as a practical value to estimate minimum installed H.P.

Concept / Approach:Hydraulic power Ph = ρ * g * Q * H. Brake power Pb = Ph / η_overall. Convert watts to horsepower by dividing by 746.

Step-by-Step Solution:Compute Ph: Ph = 1000 * 9.81 * 0.0325 * 30.Numerical value: 9.81 * 0.0325 = 0.318825; 0.318825 * 1000 * 30 ≈ 9564.75 W.Hydraulic horsepower ≈ 9564.75 / 746 ≈ 12.8 H.P.Assume realistic η_overall ≈ 0.6–0.7 for minimum installed rating.At 0.6 efficiency: required ≈ 12.8 / 0.6 ≈ 21.3 H.P.; nearest standard option ≈ 20 H.P.

Verification / Alternative check:If η_overall were 0.7, Pb ≈ 18.3 H.P.; selecting 20 H.P. still provides margin for wear, voltage variation, and minor head underestimation.

Why Other Options Are Wrong:

• 10 or 15 H.P.: insufficient once efficiency and safety margin are included.
• 18 H.P.: borderline if η is lower or headlosses are slightly higher.
• 25 H.P.: feasible but not the minimum required per given duty.

Common Pitfalls:

• Forgetting to include overall efficiency; choosing by hydraulic power alone underestimates motor rating.
• Ignoring extra minor losses and future performance drift.

Final Answer:20 H.P.