Troubleshooter
Pump efficiency formula
The following formula is used to calculate overall pump efficiency: Pump efficiency = Volumetric eff. x Mech. eff. x Hyd. eff.
Volumetric Efficiency
A pump is subject during operation to oil leakage losses. Leakage losses will increase at higher pressures. If the leakage loss at maximum working pressure is 10% of the amount of the oil intake – 100% – then only 90% will be deliver to the hydraulic system by the pump. The energy lost here through leakage is transformed into unwanted heat energy.
The volumetric efficiencies for different pump categories are given in the table below.
Mechanical efficiency
A certain proportion of the power supplied is lost as a result of friction between the different moving parts, plungers, inlet and output valve, bearings and et cetera. If 10% of the power supplied is lost because of mechanical causes, the efficiency of the pump will be 90%.
Hydraulic efficiency
The individual molecules of the flowing liquid will rub together, resulting in the loss of a proportional of the supplied power. If the flows occur at a higher pressure, the losses will increase, and hydraulic efficiency will decrease.
Difference between pumps efficiencies
This picture demonstrates the substantial differences that can occur between different types of pumps.
Let’s compare the two shown. Both of them generate the same capacity, namely 5 kW. The top pump has a poor level of efficiency and for that demands an installed capacity of 6.7 kW in order to generate 5 kW. The energy lost here is therefore 1.7 kW. In contrast the pump underneath has an excellent level of efficiency and only needs a capacity of 5.3 kW. The energy lost for this pump is thus no more than 0.3 kW.
Therefore of the poorer pump the leakage loss is 25% of the amount of the oil intake -100% – then only 75% will be delivered to the system by the pump. For the excellent pump the leakage lost is 5% then 95% will be delivered to the system by the pump.
