Hidroteknik Calculators

Hydraulic Power Unit Sizing

A hydraulic power unit consists of a pump, an electric motor and a tank. From the target flow and pressure you determine the motor power, pump displacement and tank volume.

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Sizing Formulas

Electric motor power

P (kW) = p (bar) × Q (L/min) / (600 × η)

η: overall efficiency ≈ 0.85

Pump displacement

V_g = Q × 1000 / n

Tank volume

V_tank ≈ 3–5 × Q

Units

Quantity	Symbol	Unit
Flow	Q	L/min
Pressure	p	bar
Speed	n	rpm
Power	P	kW
Tank	Vtank	L

Worked Example

Given: Flow Q = 30 L/min, pressure p = 180 bar, motor speed n = 1450 rpm, efficiency η = 0.85.

1) Motor power: P = 180 × 30 / (600 × 0.85) = 10.6 kW → 11 kW standard motor

2) Pump displacement: Vg = 30 × 1000 / 1450 = ≈ 20.7 cc/rev

3) Tank volume: 3 × 30 = ≈ 90 L

Calculate with your own values →

Frequently Asked Questions

Why is the tank 3–5× the flow?

The oil needs time to cool, release air/moisture and settle. A volume of 3–5× the flow gives enough dwell time for heat balance and foam separation.

Why divide motor power by efficiency?

Hydraulic power is the theoretical value; because of pump losses the electric motor must supply more. η≈0.85 accounts for this.

Should I pick the next standard motor up?

Yes. Choose the nearest standard motor above the calculated power (e.g. 10.6 kW → 11 kW) to leave a safety margin for continuous full-load operation.