Calculators · Hydraulics and pneumatics

Hydraulic pump and motor calculator

A hydraulic pump turns shaft rotation into flow, and a hydraulic motor turns flow back into rotation. Their size is given as displacement — the oil moved per revolution. These calculators give pump flow and the power needed to drive the pump, and the torque and speed a hydraulic motor delivers, allowing for the leakage and friction losses that real units have.

Pump flow from displacement and speed

Multiplies pump displacement by shaft speed and applies volumetric efficiency, which accounts for internal leakage. Gear pumps are typically 85–93% volumetrically efficient; piston pumps 92–97%.

Results
Pump flow—L/minActual flow delivered to the circuit.
Pump flow—US gal/minThe same flow in US gallons per minute.

Formula
Q [L/min] = displacement [cc] × rpm × η_v / 1,000

Hydraulic power and pump input power

Hydraulic power is pressure multiplied by flow. The engine or motor driving the pump must supply more than that, because of the pump’s overall efficiency. This is how drive motors are sized for hydraulic power units.

Results
Hydraulic power—Power carried by the oil.
Required drive power—Shaft power the engine or electric motor must provide.

Formula
P_hyd [kW] = p [bar] × Q [L/min] / 600
P_in = P_hyd / η_overall

Hydraulic motor torque

Output torque depends on motor displacement and the pressure difference across it, reduced by mechanical efficiency. Larger displacement gives more torque at lower speed from the same flow.

Results
Output torque—Torque at the motor shaft.

Formula
T [Nm] = Δp [bar] × displacement [cc] / (20π) × η_m

Hydraulic motor speed

Divides the flow reaching the motor by its displacement, allowing for internal leakage. It shows how fast a wheel motor, winch or auger will turn with a given pump.

Results
Motor speed—rpmOutput shaft speed.

Formula
rpm = Q [L/min] × 1,000 × η_v / displacement [cc]

Worked example: a tractor hydraulic system

  1. A 45 cc/rev pump at 1,500 rpm and 95% volumetric efficiency delivers 64.1 L/min.
  2. At 200 bar that carries 200 × 64.1 / 600 = 21.4 kW of hydraulic power.
  3. At 85% overall efficiency the pump needs 25.1 kW from the engine.
  4. Feeding a 100 cc/rev motor, the flow gives about 609 rpm and, at 200 bar, 286 Nm of torque.

Typical pump efficiencies

Pump typeVolumetric efficiencyOverall efficiencyTypical pressure
External gearabout 85 – 93%about 80 – 88%up to about 250 bar
Vaneabout 85 – 92%about 78 – 85%up to about 175 bar
Axial pistonabout 92 – 97%about 85 – 92%up to about 400 bar

Related

About these results

Results are theoretical estimates from standard engineering formulas. Real-world figures depend on conditions these formulas do not capture. Always follow the manufacturer's specifications for maintenance, loading and safety decisions.