Hydraulic Pump Power Formula Reference

Calculate pump input power

For flow q in litres per minute and differential pressure Δp in bar, the Parker pump formula is P = (q × Δp) / (600 × ηt), where P is kilowatts and ηt is overall efficiency as a decimal. This gives the pump input power required for the hydraulic output under the stated conditions.

The pressure term is the difference between pump inlet and outlet pressure. If efficiency is omitted, the result represents ideal hydraulic power rather than the higher input power required by a real pump.

Estimate flow from displacement and speed

The manual gives pump flow as q = (D × n × ηv) / 1000, where displacement D is cm³/rev, shaft speed n is rpm and ηv is volumetric efficiency. Use efficiency as a decimal fraction and use compatible units.

Overall efficiency is ηt = ηv × ηhm, with ηhm the mechanical efficiency. Use measured or manufacturer-supported efficiencies at the operating point; they can vary with pressure, speed, temperature and fluid.

Check the result before sizing equipment

Confirm that the calculation uses differential pressure, actual flow and the correct efficiency values. The 600 divisor converts L/min × bar to kW. A unit mismatch or assumed efficiency can materially change the result.

This reference is not a pump selection or heat-loss analysis. Check the selected pump’s operating envelope, fluid, speed, pressure rating, duty cycle, drive and thermal limits against current manufacturer data.

Worked input-power example

For q = 120 L/min, Δp = 250 bar and ηt = 0.90, P = (120 × 250) / (600 × 0.90) = 55.6 kW. This is an estimated shaft input at that operating point; losses and changing efficiency should be considered over the full duty cycle.

If 250 bar is only the outlet gauge pressure, determine inlet pressure too. The formula uses the pressure difference across the pump, not a single pressure reading.

Check flow and efficiency values

The displacement-flow relation uses volumetric efficiency to account for internal leakage. A value of 0.95 means 95 percent, entered as 0.95 in the formula. Overall efficiency combines volumetric and mechanical efficiency and should be derived consistently for the same operating point.

For example, with D = 45 cm³/rev, n = 1,500 rpm and ηv = 0.95, the manual’s relation gives q = 64.1 L/min. Use the pump’s performance data rather than assuming a fixed efficiency across the full pressure range.

Distinguish input and output power

Hydraulic output power is q × Δp / 600 in kW. The required pump input power is higher and is calculated by dividing by overall efficiency. A motor’s delivered output power uses a different efficiency placement, so check that the selected equation matches a pump.

These formulas estimate steady-state power. They do not account for transient pressure spikes, relief-valve flow, acceleration, thermal balance or drive losses outside the pump. Include those checks when sizing a motor or power unit.

State the operating point

Flow, pressure and efficiency should describe the same pump speed, fluid, temperature and load condition. Combining a peak pressure with a nominal flow or an unrelated efficiency can give a misleading power estimate.

For equipment selection, check both continuous and intermittent duty and compare calculated shaft power with the drive’s rated output and service factor. The manufacturer’s pump curve remains the authority for actual operation.

Source and scope

Source: Parker, Series F11 Hydraulic Pump & Motor Service Manual. Use the source edition and conditions stated in the table.