Hydraulic cylinders convert the pressure and flow of a hydraulic fluid into proportional force and linear displacement of a piston in the cylinder. Unlike air, hydraulic fluid is incompressible, which means that whatever volume of fluid the hydraulic pump can push is the exact amount that the cylinder will have to increase its volume by pushing its piston out. Double-acting (push-pull) hydraulic cylinders typically retract faster than they elongate since the displacement volume of the side containing the connecting rod is smaller than that of the pushing side.
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Things you need
- Calculator or spreadsheet
- Stop watch
Determine the volumetric flow rate of the hydraulic pump powering the hydraulic system. Hydraulic pumps are rated to deliver a specified number of gallons-per-minute at a given pressure. Barring leakage or backpressure-relief within the system, whatever volume is sent into the line to the push side of the cylinder must be resolved by the cylinder increasing its volume. Convert a gallons-per-minute (gpm) flow into cubic-feet per-minute (cfm) by dividing the gallon rate by 7.48-gallons-per cubic foot. For example, assume a 14.96-gpm pump which converts to a 2-cfm pump rate.
Calculate the volumetric displacement of the cylinder in cubic feet per foot of linear piston travel. In this example, a cylinder has a 3-inch diameter piston and a 2-inch diameter piston rod. The area of the piston is the diameter of the piston, 3-inches squared times pi (3.1416) divided by 4 which yields 7.07-square-inches (sq.in.) on the push side. For each foot the cylinder extends itself, its push-side volume increases by 7.07-sq.in. times 12 yielding 84.84-cu-in displacement gain per foot of movement.
Divide the 84.84-cu-in. displacement gain per foot by 1728-cu-in. per cubic-foot, each foot of piston travel provides 0.0491 cubic feet of displacement. 2-cfm pump would therefore move the piston 40.73-feet in one minute, or about 8.14-inches per second cylinder elongation.
Calculate the speed of retraction for the same cylinder assuming the pump rate of 2-cfm remains constant. Since the pull-side of the cylinder contains the piston rod, with its own surface area contribution, the displacement change per foot is smaller than the push side. The area of the 2-inch piston rod is 2-inches squared times pi (3.1416) divided by 4 which yields 3.14-square-inches. Subtracting 3.14-sq-in from 7.07-sq-in yields an effective net area of 3.93-sq-inches. Dividing 7.07-sq-in by 3.93-sq-in results in a ratio of 1.8 which means the cylinder has to retract 1.8-times faster than it extends with the same pump volume of 2-cfm being delivered. So the cylinder retracts at a speed of 1.8 times 8.14-inches per second or 14.62-inches-per-second.
Tips and warnings
- Conservative hydraulic machine speeds are far more efficient than high speeds.
- Always wear protective clothing and safety glasses when working with hydraulics.
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