Determining hydraulic pump condition using volumetric efficiency
I was recently asked to give a second opinion on the condition of a variable displacement hydraulic pump. My client had been advised that its volumetric efficiency was down to 80%. Based on this advice, he was considering having this hydraulic pump overhauled.
What is volumetric efficiency?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. 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 stated when stating volumetric efficiency. For example, a hydraulic pump with a theoretical output of 100 GPM, and an actual output of 94 GPM at 5000 PSI and 120 SUS is said to have a volumetric efficiency of 94% at 5000 PSI and 120 SUS. In practice, fluid viscosity is established by noting the fluid temperature at which actual output is measured and reading the viscosity off the temperature/viscosity graph for the grade of fluid in the hydraulic system.
What is the significance of volumetric efficiency?
As a hydraulic 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.
Calculating the volumetric efficiency of variable hydraulic pumps
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 100 SUS. When I advised my client that the pump's volumetric efficiency was in fact 92%, he was alarmed by the conflicting results. To try and 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 at maximum rpm and no load (presumably the maximum capacity of his flow-tester). At 4,350 PSI the recorded output was 320 liters per minute. From these results, volumetric efficiency had been calculated to be 80% (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 dependant 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 were conducted with pump displacement set to 100 liters per minute at no load, pump output would be at 20 liters per minute at 4,350 PSI - all other things being equal. This means that this pump has a volumetric efficiency of 20% at 10% displacement, 80% at 40% displacement and 92% at 100% 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% displacement, which is within acceptable limits. If my client had based their decision on volumetric efficiency at 40% displacement, they would have paid thousands of dollars for unnecessary repairs.
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