The product of power factor and pump efficiency is used to determine which efficiency?

• A. Pump efficiency times motor efficiency
• B. Power factor and pump efficiency
• C. Discharge head times pump efficiency
• D. Power factor and motor efficiency

Answer: (B)

When evaluating a pump that’s powered by electricity, think about how much of the electrical power you actually get as useful hydraulic power. Two pieces control that: power factor, which tells you how much of the drawn electrical power is real power (usable energy), and pump efficiency, which tells you how effectively the pump converts that real power into hydraulic energy in the fluid. Multiply these two factors together to get the fraction of the apparent electrical power input that ends up as hydraulic power. That’s why the product of power factor and pump efficiency is used to gauge the pump system’s efficiency relative to the electrical supply.

For example, if the power factor is 0.80 and the pump efficiency is 0.70, the product is 0.56. This means about 56% of the apparent power input is effectively converted into hydraulic power, with the remainder lost to other inefficiencies and reactive power.

The other options either mix in motor efficiency, which relates to converting electrical input to shaft power (not hydraulic energy directly), or involve discharge head in a way that calculates power rather than efficiency, so they don’t capture this specific relationship.