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The motor of a hydraulic press pump is chosen not by multiplying the press's highest pressure by its highest flow, but by the power the pump actually draws at each step of the cycle. At the moment of pressing, pressure is high but flow is low; during rapid approach, flow is high and pressure is low. For a correct choice, the power is first found step by step, then the highest value and the equivalent power representing average heating over the cycle are assessed together.
In this article we focus not on the mechanical details of the motor-pump connection but on the power calculation and the operating cycle.
The basic relationship for pump power
The power a hydraulic pump requires at the shaft is directly proportional to the flow and pressure of the oil it delivers. The practical formula is:
Shaft power (kW) = Pressure (bar) × Flow (L/min) / (600 × pump overall efficiency)
Overall efficiency includes both the volumetric and mechanical losses of the pump; for gear and vane pumps it is usually taken as 0.80–0.90, and somewhat higher for piston pumps. The correct value is on the pump manufacturer's data sheet.
Flow also depends on motor speed. The flow of a fixed-displacement pump is:
Flow (L/min) ≈ Displacement (cm³/rev) × Motor speed (rpm) / 1000 × volumetric efficiency
Four-pole motors are common on hydraulic power units; under load they turn at about 1,450–1,480 rpm. A 10 cm³/rev pump at 1,460 rpm delivers about 14.6 L/min, excluding losses. Most pumps are designed for this speed range; before running one with a two-pole motor, confirm the pump's maximum permissible speed and inlet conditions with the manufacturer.
The press cycle: each step is a different load
A typical hydraulic press cycle consists of four or five steps. Pressure and flow change at each step, and so does the load seen by the motor.
- Rapid approach: The cylinder is unloaded and descends quickly to the die. Flow is high; pressure is only enough to overcome friction and weight.
- Pressing (working stroke): The die contacts the part and pressure quickly rises to the set value. Speed falls and flow decreases.
- Pressure hold: The part is held under pressure for a set time. The pump only makes up for leakage; depending on the arrangement, power is low or moderate.
- Decompression and return: Pressure is released in a controlled way and the cylinder returns upward. Flow is high again, pressure moderate.
- Idle: While the part is changed, the pump circulates without pressure; the motor only covers no-load losses.
In a simple arrangement with a single fixed-displacement pump, the pump delivers the same flow at every step and the excess oil returns to the tank through the relief valve. During pressing this means full flow at high pressure, and the motor power becomes very large. For this reason, presses are mostly designed with a two-stage arrangement:
- Double pump (high-low): A high-flow, low-pressure pump is active during rapid approach; when pressure rises, an unloading valve opens it to the tank and pressing is done with a small-flow, high-pressure pump.
- Pressure-compensated variable pump: When pressure reaches the set value, the pump automatically reduces its displacement; power drops markedly during the pressing and hold steps.
Which arrangement is used sometimes changes the motor power by a factor of two or three. Before choosing the motor, always study the hydraulic circuit diagram or the pump nameplates.
Peak power and equivalent power
Once the step powers in the cycle have been found, two separate questions are asked:
- Can the motor cover the highest step power? The motor must not drop out of speed and stall at the hardest moment of the cycle. An induction motor has a short-term torque capacity above its rated torque; but relying on this capacity continuously is not right. Starting and breakdown torque values are in the 2026 technical and product catalogue.
- How much does the motor heat up over the cycle? Winding heating is related to the square of the current. So, to represent the cycle, the root-mean-square (equivalent) power is used, not the arithmetic average:
Equivalent power = √( Σ (Step power² × Step time) / Cycle time )
The motor's rated power must not be less than the equivalent power; the highest step power must also remain within the motor's safe short-term capacity. If the cycle is short and repeats continuously, the motor heats up close to S1 continuous duty conditions; if the press stands idle for hours and only presses occasionally, the situation is different. That is why it matters to state the cycle time and strokes per hour with the order.
Hypothetical calculation example: a double-pump press
The values below are hypothetical and chosen only to show the flow of the calculation. Pump overall efficiency is assumed to be 0.85 in all steps; the cycle time is 30 seconds.
| Step | Pressure | Flow | Time | Step power |
|---|---|---|---|---|
| Rapid approach (both pumps) | 40 bar | 60 L/min | 4 s | ≈ 4.7 kW |
| Pressing (high-pressure pump) | 250 bar | 15 L/min | 6 s | ≈ 7.4 kW |
| Pressure hold | 250 bar | leakage only | 5 s | ≈ 1.0 kW |
| Return | 60 bar | 60 L/min | 4 s | ≈ 7.1 kW |
| Idle circulation | low | — | 11 s | ≈ 0.8 kW |
In this cycle, the highest step power is about 7.4 kW and the equivalent power about 4.6 kW. Although 5.5 kW looks sufficient based on equivalent power, the 7.4 kW at the moment of pressing pushes the motor continually into the overtorque region. So a 7.5 kW, four-pole motor is chosen; the highest step stays within the rated power limit and the heating margin is preserved.
Had the same press been designed with a single fixed pump delivering 60 L/min at 250 bar, the shaft power would have exceeded 29 kW. Selection made only by "maximum pressure × maximum flow" without knowing the cycle can make the motor four times larger than necessary; conversely, looking only at the pressing step of a double-pump press and ignoring the return can lead to an undersized motor.
