Overall efficiency accounts for the loss of mechanical energy due to friction, and the loss of hydraulic energy due to the leakage flow. The clearances between moving parts can be thought of as an orifice. Fluid on one side of the orifice has a high pressure (up to the maximum pressure that exists at the pump outlet), and fluid on the other side of the orifice has a pressure equal to the pressure in the pump housing (generally less than 50 psi). This ?P multiplied by the leakage flow gives an estimate of the hydraulic power loss.

Overall efficiency for the Hydreco Model 1919 gerotor pump is plotted in Fig. 4.22a (1200 rpm) and Fig. 4.22b (1800 rpm). Efficiency is near a maximum at 1000 psi, so the comparison will be made at this pressure. Efficiency is 87.5% at 1200 rpm and 88% at 1800 rpm. The trend is the same as that found for volumetric efficiency; however, the increase in overall efficiency at the higher rpm is smaller.  The energy loss (fluid energy converted to heat) due to fluid moving past a stationary part, or a part moving in a fluid, is proportional to velocity squared . We expect the energy loss due to friction to increase as operating speed increases. This friction loss, however, is only part of the total power loss. Leakage flow represents a loss of hydraulic power. This flow is raised to some pressure, but it is not part of the output flow, and thus it is not included in the output power from the pump. Realizing that those who have done an in-depth study of fluid machinery will question the simplification, total power loss will now be discussed as the sum of two losses,

Total power loss = Friction power loss + Leakage power loss

A simple experiment might help the reader better understand the partitioning of total loss in this manner. A schematic for this experiment is shown in Fig. 4.23. Suppose a variable displacement pump is set for zero displacement and operated at a given speed. Fluid at the outlet has an unimpeded pathway back to the inlet, so developed pressure (?P across the pump) will be very small. Outlet flow will be small, because the pump displacement is set at zero. Delivered hydraulic power (small pressure × small flow) is very small and can be taken to be zero. The measured input power is the friction power loss.

Input power from the electric motor is given by Since no hydraulic power is delivered by the pump, this input power is the total power required to replace losses. (Speed is constant, so inertia is not a factor.). Friction power loss is defined by It is instructive to calculate the overall efficiency for a gerotor pump based on leakage power loss only (neglecting the friction loss component) and compare this calculated efficiency with the overall efficiency reported by the manufacturer. This comparison reveals the magnitude of the two components for the gerotor design. Data from the 1200 rpm, 1500 psi test will be used. As previously calculated, Ql = 3.53 GPM. Because of the gerotor pump design, it is difficult to measure pressure inside the housing at the point where the fluid is leaking from the outlet back to the inlet. We simply assume that the leakage ?P equals the pressure rise across the pump. Leakage power loss is The overall efficiency based on the input horsepower curves given by the manufacturer is 85.4%. This overall efficiency includes both leakage and friction losses. When we calculated leakage loss, we assumed that leakage ?P equaled the ?P across the pump. Actual leakage ?P is smaller, thus our estimate of leakage loss was too high. It is true that the leakage power loss is by far the dominant term for the gerotor design.

Key comparisons between volumetric and overall efficiencies for the gerotor pump are given in Table 4.1. The fact that the overall and volumetric efficiencies are approximately equal for each speed and pressure reinforces our conclusion that leakage loss is by far the dominant term for the gerotor design.

Overall efficiency at 2500 psi increases from 77% at 1200 rpm to 81.5% at 1800 rpm for the gerotor pump. Because leakage losses are such a significant term for this type of gear pump design, it is generally good policy to use a smaller displacement pump and operate at higher rpm to get a given flow.

If a gerotor motor with similar efficiency, around 80%, is used to convert the hydraulic power back to mechanical power, the overall efficiency (pump and motor combination) is  Less than two-thirds of the input power is delivered to the load. Mechanical transmissions can have efficiencies of 95% or better, so it is readily apparent that a decision to transmit mechanical energy with fluids must carefully weigh the advantages and disadvantages.