Inputs
Results are estimates. Efficiencies are typical; real pump efficiency depends on make, pressure and viscosity. Verify motor and pump selection with the manufacturer catalog.
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When you print, usage data (calculated values and the information you entered) is sent to Sistemay; your e-mail is not used for contact unless you give consent. Choose “Save as PDF” in the print dialog.Formulas and method
Pump displacement is the required flow divided by motor speed and volumetric efficiency. Motor power is hydraulic power divided by overall efficiency, rounded up to the next standard motor rating.
The oil tank is usually 3–5 times the pump flow; systems that run hot need a larger tank or a cooler.
V [cc/rev] = Q [L/min] × 1000 ÷ (n × η_vol)
P_hyd [kW] = p [bar] × Q [L/min] ÷ 600
P_motor [kW] = P_hyd ÷ η_total × (1 + margin)
Tank [L] = k × Q [L/min] (k = 3…5)
Worked example
Example: 160 bar, 40 L/min, gear pump (η_v 0.90; η_t 0.80), 1500 rpm, 10% margin → pump ≈ 29.6 cc/rev, motor ≈ 14.7 kW → 15 kW, tank 4×40 = 160 L.
Practical tips
- Round displacement up to the next standard size and recheck flow at the real speed.
- Keep oil at 50–60 °C in continuous duty; add a cooler if the heat loss is high.
- Select the return filter for at least twice the pump flow.
- For intermittent duty an accumulator lets you use a smaller pump and motor.
Frequently asked questions
How is hydraulic motor power calculated?
Pressure × flow ÷ 600 gives hydraulic kW; divide by overall efficiency, add a margin and round to a standard motor.
Why do I need pump displacement?
Pumps are sold by displacement per revolution. It follows from flow and motor speed.
How big should the tank be?
A common rule is 3–5 times pump flow. A larger tank helps with heat and air release.
Where does pump loss go?
It turns into heat in the oil; this value drives the cooler requirement.
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