Our experienced technical engineers can help with the detailed number-crunching for your particular project and application, but here is a quick summary for your reference:
Hydraulic power is defined as flow multiplied by pressure. The hydraulic power supplied by a pump is:
For example, if a pump delivers 180 litres/minute and the pressure is 250 bar, then the power of the pump is:
Power = (250 x 180) ÷ 600 = 75 kW.
When calculating the power input to the pump, the total pump efficiency ηtotal must be included. This efficiency is the product of volumetric efficiency, ηvol and the hydromechanical efficiency, ηhm. Power input = Power output ÷ ηtotal. The average for axial piston pumps, ηtotal = 0.87.
Continuing with the example above, the power source (for example a diesel engine or an electric motor), must be capable of delivering at least 75 ÷ 0.87 = 86 [kW].
The hydraulic motors and cylinders that the pump supplies with hydraulic power also have efficiencies and the total system efficiency (without including the pressure drop in the hydraulic pipes and valves) will end up at approximately 0.75.
Cylinders normally have a total efficiency of around 0.95 while hydraulic axial piston motors have 0.87, the same as the pump. In general the power loss in a hydraulic energy transmission is thus around 25% or more at ideal viscosity range 25-35 [cSt].
To calculate the required maximum power output for the diesel engine…
First check the maximum power point, i.e. the point where pressure times flow reach their maximum value, then the basic estimation would be:
Qtot = the theoretical pump flow for the consumers, not including leakages, at maximum power point.
Pmax = actual pump pressure at maximum power point.
Note: η is the total efficiency = (output mechanical power ÷ input mechanical power). For rough estimations, η = 0.75., although you can typically add 10-20% depending on the application to this power value.
Calculate the required pump displacement from the required maximum sum of flow for the consumers in the worst case scenario and the diesel engine rpm at this point. The maximum flow can differ from the flow used for calculation of the diesel engine power.
Calculation of preliminary cooler capacity: Heat dissipation from hydraulic oil tanks, valves, pipes and hydraulic components is less than a few percent in standard mobile equipment and the cooler capacity must include some margins. Minimum cooler capacity, Ecooler = 0.25Ediesel
At least 25% of the input power must be dissipated by the cooler when peak power is utilized for long periods. In normal case however, the peak power is used for only short periods, thus the actual cooler capacity required might be considerably less. The oil volume in the hydraulic tank is also acting as a heat accumulator when peak power is used.
The system efficiency is very much dependent on the type of hydraulic work tool equipment, the hydraulic pumps and motors used and power input to the hydraulics may vary considerably. Each circuit must be evaluated and the load cycle estimated. New or modified systems must always be tested in practical work, covering all possible load cycles.
An easy way of measuring the actual average power loss in the system is to equip the machine with a test cooler and measure the oil temperature at the cooler inlet, the oil temperature at the cooler outlet and the oil flow through the cooler, when the machine is in normal operating mode. From these figures the test cooler power dissipation can be calculated and this is equal to the power loss when temperatures are stabilized. From this test the actual required cooler can be calculated to reach specified oil temperature in the oil tank. One problem can be to assemble the measuring equipment in-line, especially the oil flow meter.