How this calculation works
Normalize flow and total pressure rise, multiply for fluid output power, then divide by the entered efficiency. Integrate a constant input over the supplied hours for kWh.
Hydraulic W = total pressure rise Pa × flow m³/s; input W = hydraulic W / efficiencyA worked example
A total rise of 1 bar with 60 L/min gives 100 W of hydraulic output. At an entered 80% overall efficiency, required input is 125 W and modeled loss 25 W. Two hours uses 0.25 kWh at that constant input.
Good to know
Steady incompressible-flow arithmetic only. No pump curve, pipe losses, NPSH, startup, actual idle draw or equipment selection. Pressure and efficiency boundaries must match your scenario.
Reference: Reference: model, definitions and assumptions (opens in a new tab) ↗
Flow and pressure rise together determine useful fluid power
The default 60 L/min is 0.001 m³/s. A total pressure rise of 1 bar is 100,000 Pa. Multiplying gives 100 W of useful hydraulic power at that operating point. A pressure reading without a corresponding flow rate does not establish that power.
The supplied pressure rise must represent the energy boundary you intend to model. If inlet/outlet elevation or velocity changes matter, a difference between static gauge readings may omit those contributions. This page does not reconstruct the total pressure rise from a pipe layout.
Hydrostatic pressure describes a stationary-fluid column. Pump power needs volume flow and a matching total pressure rise and efficiency boundary.
Efficiency converts useful output into required input
At an entered 80% hydraulic-output/input ratio, producing 100 W of fluid power needs 100/0.8 = 125 W at the chosen input boundary. The difference is 25 W. Multiplying by 0.8 instead would produce 80 W, which reverses the intended input/output relationship.
At 50% efficiency the same useful output needs 200 W, while at 100% it needs 100 W. The comparison holds flow and total pressure rise fixed. It is not a measured pump curve and does not assume that one real pump can maintain the operating point at every entered efficiency.
A power result becomes energy only after a duration is supplied
A constant 125 W input for two hours uses 250 Wh, or 0.25 kWh. The calculation does not infer a start/stop schedule, a demand tariff or a separate motor efficiency. Include the intended chain in the overall efficiency input rather than applying an unstated extra factor.
The hydrostatic pressure calculator can describe a separate stationary-fluid pressure rise. A real pumping system can also have flow losses and elevation/velocity changes. The torque RPM and horsepower calculator concerns a shaft point, and connecting that mechanical input to fluid output would require a matching efficiency boundary.
A few questions, answered
Should I multiply or divide by efficiency to find input?
Divide hydraulic output by efficiency expressed as a fraction. With output fixed, lower efficiency requires more input.
Is the default efficiency a manufacturer rating?
No. It is an editable example assumption. No pump curve, motor specification or live equipment data is loaded.
Does zero flow prove the pump uses no electricity?
No. Zero flow gives zero useful fluid power in this ideal equation. Actual idle or shutoff electrical consumption is outside the model.
Does the result size a real pump?
No. It does not check system curves, NPSH, cavitation, startup requirements, pipe losses or equipment operating limits.