Tags: hydraulics

**H**ydraulic pumps convert mechanical energy into hydraulic energy. A high-performance piston pump can convert mechanical energy into hydraulic energy with an efficiency of 92 percent. If the pump drives a piston motor, the motor is able to convert this hydraulic energy back into mechanical energy with an efficiency of 92 percent. The overall efficiency of this hydraulic drive, without considering flow losses, is 85 percent (0.92 x 0.92 x 100 = 85).

Table 1 shows what the typical efficiency would be if a gearbox accomplished this same drive transfer.

**Table 1. Typical Gearbox Efficiencies**

The inefficiencies or losses in a hydraulic drive can be divided into two categories: hydraulic-mechanical, which comprise flow and mechanical friction losses, and volumetric, which comprise leakage and compressibility losses (Figure 1).

**Figure 1. Losses in a Hydraulic Drive (Bosch Rexroth)**

The advantages of a hydraulic drive, which include high-power density (high-power output per unit mass), variable speed control, simple overload protection and both rotary and linear motion, are possible from a single system. As Table 1 shows, a key disadvantage of a hydraulic drive is that it is far less efficient than a mechanical drive. What’s worse, the wear process decreases a hydraulic drive’s volumetric efficiency (and therefore total efficiency) causing the drive to slow down and more energy to be given up to heat.

The hydraulic pump is usually the hardest working component of a hydraulic system. As the pump wears in service, internal leakage increases and therefore the percentage of theoretical flow available to do useful work (volumetric efficiency) decreases. If volumetric efficiency falls below a level considered acceptable for the application, the pump will need to be overhauled. In a condition-based maintenance environment, the decision to change-out the pump is often based on remaining bearing life or deterioration in volumetric efficiency, whichever occurs first.

Volumetric efficiency is the percentage of theoretical pump flow available to do useful work. In other words, it is a measure of a hydraulic pump’s volumetric losses through internal leakage and fluid compression. It is calculated by dividing the pump’s actual output in liters or gallons per minute by its theoretical output, expressed as a percentage. Actual output is determined using a flow-tester to load the pump and measure its flow rate.

Because internal leakage increases as operating pressure increases and fluid viscosity decreases, these variables should be included when stating volumetric efficiency. For example, a hydraulic pump with a theoretical output of 100 GPM, and an actual output of 94 GPM at 5,000 PSI and 46 cSt is said to have a volumetric efficiency of 94 percent at 5,000 PSI and 46 cSt. In practice, fluid viscosity is established by noting the fluid temperature at which actual pump output is measured and reading the viscosity off the temperature/viscosity graph for the grade of fluid in the hydraulic system.

**Determining Pump Efficiency **

When calculating the volumetric efficiency of a variable displacement pump, internal leakage must be expressed as a constant. Consider this example: I was recently asked to give a second opinion on the condition of a large, variable displacement pump. My client had been advised that its volumetric efficiency was down to 80 percent and based on this advice, he was considering having the pump overhauled.

The hydraulic pump in question had a theoretical output of 1,000 liters per minute at full displacement and maximum rpm. Its actual output was 920 liters per minute at 4,350 PSI and 46 cSt. When I advised my client that the pump’s volumetric efficiency was in fact 92 percent he was alarmed by the conflicting assessments. To explain the disparity, I asked to see the first technician’s test report.

After reviewing this test report, I realized that the results actually concurred with mine, but had been interpreted incorrectly. The test had been conducted to the same operating pressure and at a fluid temperature within one degree of my own test, but at reduced displacement. The technician had limited the pump’s displacement to give an output of 400 liters per minute (presumably the maximum capacity of his flow-tester) at maximum rpm and no load. At 4,350 PSI the recorded output was 320 liters per minute. From these results, volumetric efficiency had been calculated to be 80 percent (320/400 x 100 = 80).

To help understand why this interpretation is incorrect, think of the various leakage paths within a hydraulic pump as fixed orifices. The rate of flow through an orifice is dependent on the diameter (and shape) of the orifice, the pressure drop across it and fluid viscosity. This means that if these variables remain constant, the rate of internal leakage remains constant, independent of the pump’s displacement.

Note that in the above example, the internal leakage in both tests was 80 liters per minute. If the same test was conducted with pump displacement set to 100 liters per minute at no load, pump output would be 20 liters per minute at 4,350 PSI - all other things equal. This means that this pump has a volumetric efficiency of 20 percent at 10 percent displacement, 80 percent at 40 percent displacement and 92 percent at 100 percent displacement. As you can see, if actual pump output is measured at less-than-full displacement (or maximum rpm), an adjustment needs to be made when calculating volumetric efficiency.

**Time for an Overhaul? **

In considering whether it is necessary to have this hydraulic pump overhauled, the important number is volumetric efficiency at 100 percent displacement, which is within acceptable limits. If my client had based his decision on volumetric efficiency at 40 percent displacement, his company would have paid thousands of dollars for unnecessary repairs.