Cylinder force and speed calc
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Transcript of Cylinder force and speed calc
477
Hydraulic CylinderForce and Speed Calculations
Calculation of Hydraulic Cylinder Force…EXAMPLE: A certain application requires a cylinder force of 25 tons. What should be the
cylinder bore diameter used and at what gauge pressure?
SOLUTION: The required force is 25 tons × 2000 = 50,000 pounds. Refer to the“HydraulicCylinderForce”tableonpages478 and 479 which shows several combinations ofpistondiameterandPSIpressurewhichwillproduce50,000poundsofforceormore.Forexample,a6inchpistonwillproduce56,550poundsat2000PSI;a7inchpistonwillproduce57,725lbsat1500PSI;an8inchpistonwillproduce50,265lbsat1000PSI,a10inchpistonwillproduce58,900 lbs.at750PSI,etc.So therearemanycombinationswhichcouldbeused, and the final choice is a matter of preference or of matching the pressure and flow capability of other components, particularly the pump.
In practice, choose a combination which will produce from 10% to 25% more thanactually required by the load alone. This will provide a safety allowance which will take care of pressure losses in valves and piping, and mechanical losses in the cylinder.
EXAMPLE:Howmanypoundsofforcewillbedevelopedontheextensionstrokeofa3Z\v˝borecylinderoperatingat1500PSI?Ifthiscylinderhasa1C\v˝diameterpistonrod,howmuchforce will be developed on the retraction stroke?
SOLUTION:Refertothe“HydraulicCylinderForce”tableonpages478 and 479. The chart shows12,444lbs.Asolutioncanalsobeobtainedbyusingthepistonarea(8.296squareinches)andmultiplyingbythepressure(1500PSI);8.296squareinches×1500PSI=12,444lbs.
Ontheretractionstroketheamountofforcedevelopedonthe2.41squareinchrodareamust be subtracted: 12,444 – 3608 = 8836 lbs.
EXAMPLE:WhatPSIgaugepressureisrequiredforretractionofa50,000lb.loadwithan 8 inch bore cylinder having a 4 inch diameter rod?
SOLUTION: The net piston area must be found which is the full piston area minus the rod area.50.27(pistonarea)–12.57(rodarea)=37.7squareinches.PSI=50,000÷37.7=1326PSI.Theactualpressurewillbeslightlygreaterduetofrictionofthepistoninthebarrel.
Calculation of Hydraulic Cylinder Speed…EXAMPLE: At what speed would the piston of a 4 inch bore cylinder extend on an oil
flowof12GPM?
SOLUTION:The tableof “HydraulicCylinderSpeeds”onpages480 and 481 may be used or the speed figured with the formula which says that “speed is equal to the incoming flow of oil in cubic inches per minute, divided by the square inch area of the piston”. The speed will be in inches per minute.
Aflowof12GPMis231×12=2772cubicinchesperminute.Thespeedis2772(flowrate)÷12.57(pistonarea)=220.5inchesperminute.Thischecksverycloselywiththevalueshown in the table on page 480.
EXAMPLE:FindtheGPMflownecessarytocausea5inchborecylindertotravelatarate of 175 inches per minute while extending.
Howfastwouldthiscylinderretractonthesameoilflowifithada2inchdiameterpistonrod?
SOLUTION: Flow is determined by multiplying the piston area in square inches times the travel rate in inches per minute. This gives flow in cubic inches per minute. Divide by 231 toconverttoGPM:19.64(pistonarea)×175=3437cubicinchesperminute.3437÷231=14.88GPM.Thischecksverycloselywith15GPMat174inchesperminuteshownonthechart on page 480.
Tofindtheretractionspeedon14.88GPM,thenetpistonareamustbefound.This isthe fullpistonareaminustherodarea:19.64(pistonarea)–6.5(rodarea)=16.5squareinches.Theflowrateis3437cubicinchesperminute(equivalentto14.88GPM)÷16.5(netarea)=208inchesperminute.Notethatthisisfasterthantheextensionspeedonthesameoil flow.