Given the hydraulic power P_h and pump efficiency η_pump, which expression gives the input electrical power P_elec?

Given the hydraulic power P_h and pump efficiency η_pump, which expression gives the input electrical power P_elec?

• A. P_elec = η_pump × P_h.
• B. P_elec = P_h / η_pump.
• C. P_elec = P_h + η_pump.
• D. P_elec = P_h - η_pump.

Answer: (B)

Efficiency tells us how effectively electrical power is converted into hydraulic power. By definition, pump efficiency η_pump equals the hydraulic power delivered to the fluid divided by the electrical power drawn: η_pump = P_h / P_elec. Rearranging gives the input electrical power: P_elec = P_h / η_pump. This means for a given hydraulic output, the electrical input is larger when efficiency is less, and it matches the hydraulic power if efficiency were 100%. For example, with P_h = 50 kW and η_pump = 0.85, P_elec ≈ 58.8 kW. The other forms would incorrectly mix the quantities (they either multiply by efficiency or add/subtract a dimensionless efficiency term) and don’t reflect that the input power must account for efficiency losses.

Efficiency tells us how effectively electrical power is converted into hydraulic power. By definition, pump efficiency η_pump equals the hydraulic power delivered to the fluid divided by the electrical power drawn: η_pump = P_h / P_elec. Rearranging gives the input electrical power: P_elec = P_h / η_pump. This means for a given hydraulic output, the electrical input is larger when efficiency is less, and it matches the hydraulic power if efficiency were 100%. For example, with P_h = 50 kW and η_pump = 0.85, P_elec ≈ 58.8 kW. The other forms would incorrectly mix the quantities (they either multiply by efficiency or add/subtract a dimensionless efficiency term) and don’t reflect that the input power must account for efficiency losses.