posted in Flowfit Hydraulics by Flowfit on 12:10 Oct 10th, 2019<< Back to Flowfit Hydraulics
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The role of a hydraulic pump is to provide sufficient pressure to overcome the operating pressure of a system to move hydraulic fluid at a required flow rate. To achieve the required flow and to ensure you have a suitable pump, you would need to calculate what operating pressure the system will need to be.
There are three categories of efficiency to consider when looking at hydraulic pumps and their flow rates;
This is determined by dividing the actual flow that is delivered to by the pump at a given pressure by the theoretical flow. The theoretical flow is calculated by multiplying a pump’s displacement per revolution by its driven speed.
For example; if a hydraulic pump has a displacement of 100cc/rev and is driven at 1000 RPM its theoretical flow will be 100 litres/minute.
This can be determined by dividing theoretical torque required to drive it by actual torque required to drive it. If there is an efficiency of 100%, this means if the pump was delivering flow at zero pressure, no force or torque would be required to drive it.
The product of volumetric and mechanical/hydraulic efficiency.
This is used to calculate the drive power required by a pump at a given flow and pressure.
Ideally the hydraulic power needed to drive a pump depends on these three things:
- The mass flow rate
- Liquid density
- Differential height
Key Formulas to use:
KW = L/MIN X BAR ÷ 510
BAR = KW X 510 ÷ L/MIN
L/MIN = KW X 510 ÷ BAR
HP (Horse Power) = KW ÷ 0.75
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