M１Introduction Fc : Required cylinder power = 1,000 kN１20.1013 1.9615.1013１20.1013 1.9015.1013ＳＶ22● Volume calculation exampleIn advance, using the amount of hydraulic fluid that is charged in the accumulator, calculate the accumulator volume required for activating the cylinder.Specification conditionsDi : Cylinder bore = 300 mm (cross-sectional area (A) = 706.5 cm2)S : Cylinder stroke = 380 mmV : Cylinder speed = 0.75 m/secP : Pressure loss in piping etc. = 0.84 MPa(Pay attention to the pressure loss P between the accumulator and actuator)Q : Oil discharge volume from pump = 90 L/minWorking temperature = 20 to 80ºC Service fluid = Petroleum hydraulic oil *In calculation, convert all assigned pressure to the absolute pressure (MPa · abs).1) Find the required oil volume to be discharged from accumulator Vw (required cylinder oil amount).2) Considering the change in temperature during operation, find the gas charging pressure (P1) in the following steps.i) For Max. P1 at the maximum working temperature (80ºC), set the gas charging pressure ratio to 85%. (The gas charging pressure ratio can be up to 90% in consideration of the temperature change.)ii) Find Min. P1 at the minimum working temperature (20ºC) by the "Formula for gas charging pressure actual change due to temperature change".3) Find the gas charging pressure ratio (e) at the minimum working temperature.Relief ValveＰ3＋Ｐ2２ＶＷＶ1＝ ＝ｅ・η・Ｆ160˚CFP4) Find the polytropic exponent (m, n).· Find the oil charge time from Vw (the amount charged in the accumulator) and the pump flow rate.· The cylinder operation time becomes the accumulator oil discharge time.· From the nitrogen gas polytropic exponent list on page 205) Find the oil discharge coefficient (F) .6) Find the accumulator gas volume (V1). P3 : Maximum working pressure = 20 MPaP2 : Minimum working pressure = Fc / A x 10 + P = 15 MPaＰ１ｅ＝ ＝Ｐ２Average working pressureOil Charge TimeOil Discharge Timeπ・Ｄi２４Ｖ＝ ・Ｓ・１０－６π・３００２４＝ ×３８０×１０－６≒２６.９ＬMax. P1 = 0.85 · P2= 0.85 x 15.1013 MPa · abs= 12.84 MPa · absMin. P1 = 10.11 MPa · abs10.11（１５＋０.１０１３）≒０.６7(Ｐａ )＝ ＝（Ｔｍ）＝ ＝（Ｔｎ）＝ １０-３＝ ×１０-３ｍ＝１.９０ ｎ＝１.９６ａ －１ｎ１ａｍＦ＝ ＝ＶＷＱ２０.１０１３＋１５.１０１３２≒ １７.６ＭＰａ・ａｂｓ２６.９９０/６０≒ １７.９sec３８００.７５≒０.５sec－１≒0.135２６.９０.６7×０.９５×０.１３５≒３13Ｌ
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