Pumping power calculation in wastewater systems: Find the brake horsepower (BHP) required for a pump delivering 0.075 m^3/s against a total head of 12 m, if overall motor efficiency is 60%.

Introduction / Context:
Determining the brake horsepower for pumps is a core calculation in water and wastewater design. It links hydraulic power requirements with motor efficiency to select an adequate drive. This problem reinforces unit handling and the relationship between flow, head, efficiency, and power.

Given Data / Assumptions:

• Discharge, Q = 0.075 m^3/s
• Total head, H = 12 m
• Overall efficiency (motor + pump), η = 0.60
• Density of water, rho_w = 1000 kg/m^3; g = 9.81 m/s^2

Concept / Approach:
Hydraulic power Ph is rho_w * g * Q * H (in watts). Shaft power Ps equals Ph / η. Brake horsepower (BHP) is the shaft power expressed in horsepower. Use 1 HP = 0.746 kW for conversion.

Step-by-Step Solution:
Ph = rho_w * g * Q * HPh = 1000 * 9.81 * 0.075 * 12Compute: 9.81 * 12 = 117.72; 117.72 * 0.075 = 8.829; 8.829 * 1000 = 8829 WPs = Ph / η = 8829 / 0.60 = 14715 W = 14.715 kWBHP = 14.715 / 0.746 ≈ 19.73 HP ≈ 20 HP

Verification / Alternative check:
Cross-check by back-converting 20 HP to kW (≈14.92 kW) and multiplying by 0.60 to ensure hydraulic power close to 8.83 kW. This is consistent with the computed Ph = 8.829 kW.

Why Other Options Are Wrong:
10 HP, 15 HP: Too low for the specified flow and head; would under-drive the pump.25 HP: Overestimates required power for the given duty and efficiency.18 HP: Still below the calculated requirement.

Common Pitfalls:

• Mistaking hydraulic power for motor power (forgetting to divide by efficiency).
• Using wrong HP–kW conversion or omitting rho_w and g.
• Rounding too early, causing noticeable error in final BHP.

Final Answer:
20 HP