Hydraulic Power Calculation
The power carried by a hydraulic line is found from the product of flow rate and pressure. The practical relation is this: multiply flow rate in liters per minute by pressure in bar and divide by 600 to get fluid power in kilowatts; the drive motor must be selected larger than this.
Let us work through an example: 40 l/min flow rate and 200 bar pressure means 40 x 200 / 600 = 13.3 kW of fluid power. This is the power carried by the line; the power the motor must deliver is greater, because the pump has its own efficiency.
- Total pump efficiency is typically in the 0.80-0.90 range; motor power is found by dividing fluid power by this value.
- You can produce the same power with high pressure and low flow rate or with low pressure and high flow rate; the first means a smaller cylinder, the second a thicker hose.
- Every liter of oil returning to tank through the relief valve is lost power converted directly into heat.
The calculation is most often used when selecting the drive motor and the cooler. All of the lost power produced in the system eventually heats the oil; that is why tank volume and cooler capacity must grow as power increases. Failing to estimate lost power correctly is the typical reason for systems that constantly overheat in summer.
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