# Calculating Hydraulic Cylinder Efficiency

Blog | September 23rd, 2016

Engineering mathematics is an intimidating topic, but it can be distilled down to a manageable set of constants and variables when a particular machine is studied. For instance, governed by the laws of fluid dynamics, the ability to calculate hydraulic cylinder efficiency relies primarily on the compressibility of the fluid medium as it relates to the mechanical power of a cylinder piston. Essentially, volumetric efficiency, on this occasion, is a function of constancy versus loss.

Losses versus Work Completed

Down on a filthy work site, a clean inner cylinder triggers its mechanism, pushes the rod longitudinally, and generates a near perfect work cycle. But wait, friction inside the cylinder blunts the efficiency rating. The result of this tug-o-war is lost energy and an overall drop in work efficiency. Other reasons for this lossy situation include but are not limited to leakage in the seals and heat losses during the compression of the hydraulic oil. Of course, one of the biggest culprits in this scenario is the actual structural build of the rod and its associated components, for engineering irregularities cause wear, and wear fights the lubricating agent to hamper smooth functionality. This is why the high-quality build of a hydraulic component is an essential part of any efficient hydraulic system.

Back to Calculating Hydraulic Cylinder Efficiency

If engineering tolerances and the inviolable laws of fluid dynamics hamper pure functionality, these are the main components of the sum. The engineering value, the design of the cylinder, is almost a constant, but finite changes will take place in this constant as the part wears. Otherwise, mechanical efficiency is expressed as a percentage of the measured losses in kilonewtons (kN) when it’s divided by the theoretically sustainable piston force. Now, two results are calculated during this application of the formula because two piston force values are plugged into the formula, one after the other. The first value is measured as the piston extends. The second figure, not surprisingly, is a result of the piston retracting.

Again, these figures are affected by system variables, but a measurable loss factor does produce a fairly accurate mathematical model of efficiency problems when it’s used to create a divisible quotient based on practically measured piston force, but that measurement has to be repeated so that both halves of the cycle (extension and contraction) can be used to work out overall hydraulic cylinder efficiency, which is a figure that’s directly influenced by the mechanical-to-hydraulic product, as rated during the extension stage.

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