How is hydraulic power calculated and how do you determine system efficiency?

• A. Hydraulic power P_h = ΔP × Q; overall efficiency = useful hydraulic power / electrical power input (accounting for pump and component losses).
• B. P_h = Q / ΔP; efficiency = hydraulic power divided by mechanical input.
• C. P_h = ΔP × Q × efficiency factor; efficiency = hydraulic power divided by input energy.
• D. P_h = ΔP / Q; efficiency = input energy minus losses.

Answer: (A)

Hydraulic power is the rate at which the fluid does work, which comes from the pressure driving the flow. The correct expression uses the pressure rise (ΔP) multiplied by the volumetric flow rate (Q): P_h = ΔP × Q. If ΔP is in pascals and Q is in cubic meters per second, you get watts. This reflects that more pressure and/or more flow both increase the power the fluid can deliver.

Efficiency is about how much of the input energy actually becomes useful hydraulic work. The standard overall efficiency compares the useful hydraulic power to the electrical power you put into the system (including losses from the pump and other components). So, overall efficiency = useful hydraulic power ÷ electrical power input. For example, if the system delivers 1000 W of hydraulic power but draws 1200 W from the motor, the efficiency is 1000/1200 ≈ 83%.

The other formulations aren’t correct because dividing ΔP by Q would not yield power, and multiplying by an arbitrary efficiency factor isn’t a standard way to define hydraulic power. Likewise, describing efficiency as hydraulic power divided by mechanical input or as input energy minus losses doesn’t form a proper ratio consistent with how efficiency is defined.