Usable volume, isothermal
dV = V0 x (p0/p1 - p0/p2) with absolute pressures: 10 l, precharge 90 bar(g), working 100 to 200 bar(g) gives 10 x (91.01325 / 101.01325 - 91.01325 / 201.01325) = 4.48 l.
Type a line (10 l precharge 90 bar 100 to 200 bar; need 2.5 l between 120 and 180 bar precharge 108 bar) or fill in the fields: the line and the fields stay in step. The figures in the examples are examples, not typical values.
Enter the gas volume of a hydraulic accumulator, its nitrogen precharge and the minimum and maximum working pressure, and the calculator gives the usable oil volume between the two pressures, for an isothermal and for an adiabatic change of the gas, with the gas volume at each pressure. In reverse, it gives the gas volume needed for a stated usable volume. Gauge pressures are made absolute, the working is shown with your numbers, a drawing sets out the gas volumes to scale, and the result is a line you can copy or add to a request.
Ideal-gas arithmetic for nitrogen only: the calculator does not select an accumulator and makes no real-gas or temperature correction. The maker data of the accumulator govern.
dV = V0 x (p0/p1 - p0/p2) with absolute pressures: 10 l, precharge 90 bar(g), working 100 to 200 bar(g) gives 10 x (91.01325 / 101.01325 - 91.01325 / 201.01325) = 4.48 l.
dV = V0 x [(p0/p1)^(1/n) - (p0/p2)^(1/n)] with n = 1.4 for nitrogen: for the same accumulator 10 x (0.928243 - 0.567806) = 3.60 l.
V0 = dV / [(p0/p1)^(1/n) - (p0/p2)^(1/n)]: 2.5 l between 120 and 180 bar(g) with a precharge of 108 bar(g) needs 8.37 l (isothermal) or 10.78 l (adiabatic). The precharge must be below p1.
| Quantity | Equals | Basis |
|---|---|---|
| Isothermal change | n = 1: p x V = const | slow charging and discharging |
| Adiabatic change, nitrogen | n = 1.4: p x V^n = const | short cycles |
| Standard atmosphere | 1.01325 bar(a) = 101.325 kPa | the default atmospheric pressure; p(a) = p(g) + p_atm |
| 1 psi | 6.894757293168 kPa = 0.06894757293168 bar | a force of 1 pound on 1 square inch |
| 1 kgf/cm2 (at) | 98.0665 kPa = 0.980665 bar | standard gravity, 9.80665 m/s2 |
| 1 MPa; 1 kPa | 10 bar; 0.01 bar | the bar: 1 bar = 100 kPa |
| 1 US gallon | 231 in3 = 3.785411784 l | 1 in = 25.4 mm |
| 1 in3 | 16.387064 cm3 = 0.016387064 l | 1 in = 25.4 mm |
Gauge and absolute pressures differ by the atmospheric pressure, which no unit factor takes care of: the gas law needs absolute pressures.
The arithmetic is the gas law for nitrogen as an ideal gas: p0 x V0^n = p1 x V1^n = p2 x V2^n with absolute pressures, n = 1 for an isothermal change and n = 1.4 for an adiabatic change of nitrogen. The usable volume formula agrees in the technical pages of a hydraulic accumulator supplier and in the accumulator model reference of a hydraulic simulation program; n = 1.4 for nitrogen agrees in those supplier pages and in an article of a fluid power trade journal. No precharge figure for a given accumulator and duty is given here: the sources differ, and the maker data govern. The standard atmosphere is 101325 Pa; 1 psi = 6.894757293168 kPa, 1 kgf/cm2 = 98.0665 kPa, 1 US gallon = 231 in3 = 3.785411784 l and 1 in3 = 16.387064 cm3 are exact by definition. The maker data of the accumulator always come first.
From the gas law for the nitrogen inside, with absolute pressures. The gas fills the whole gas volume V0 at the precharge p0 and is squeezed to V1 at the minimum working pressure p1 and to V2 at the maximum p2; the oil taken in and given back between p1 and p2 is V1 - V2. Isothermal: dV = V0 x (p0/p1 - p0/p2). Adiabatic: dV = V0 x [(p0/p1)^(1/n) - (p0/p2)^(1/n)] with n = 1.4 for nitrogen. An example: 10 l, precharge 90 bar(g), working 100 to 200 bar(g), that is 91.01325, 101.01325 and 201.01325 bar(a): isothermal 10 x (0.901003 - 0.452772) = 4.48 l, adiabatic 10 x (0.928243 - 0.567806) = 3.60 l.
The gas law holds for absolute pressures: a gas at 0 bar(g) still has the atmospheric pressure. A gauge shows gauge pressure, so the calculator adds the atmospheric pressure, 1.01325 bar(a) by default: 90 bar(g) is 91.01325 bar(a). The difference is small at high pressures and large at low ones, so every step of the working shows the absolute figures. Write bar(a), psia or a for a pressure that is already absolute; a pressure without a mark is read as gauge, and the card says so.
The isothermal figure (n = 1) suits slow charging and discharging, when the gas keeps the temperature of its surroundings. The adiabatic figure (n = 1.4 for nitrogen) suits short cycles, when the gas has no time to give off or take up heat. Without a word the calculator shows both. Write isothermal or adiabatic for one of them, or n with a figure from 1 to 2 (n 1.3) for another exponent from the maker data.
Turn the formula round: V0 = dV / [(p0/p1)^(1/n) - (p0/p2)^(1/n)]. For 2.5 l of usable oil between 120 and 180 bar(g) with a precharge of 108 bar(g): isothermal 2.5 / (0.900837 - 0.602239) = 8.37 l, adiabatic 2.5 / (0.928121 - 0.696133) = 10.78 l of gas volume. The sizes an accumulator comes in are in the maker range; the calculator does not pick one.
The precharge must be below the minimum working pressure p1: with p0 at or above p1 the accumulator holds no oil yet at p1, the formula gives no usable volume, and the calculator says so. How far below p1 depends on the accumulator type and the duty, and the maker data of your accumulator govern. The card shows the ratio p0 / p1 so it can be set against those data.
Volumes in l, ml, m3, US gal (231 in3 = 3.785411784 l exactly) or in3 (16.387064 cm3 exactly). Pressures in bar, mbar, kPa, MPa, psi (1 psi = 6.894757293168 kPa) or kgf/cm2 (98.0665 kPa exactly), each gauge or absolute. The atmospheric pressure in bar(a), and the exponent n as a plain figure.
The calculator works out volumes and pressures only. When a hydraulic accumulator, its bladder or diaphragm, a seal kit or a charging set is needed, send the line from the calculator with a photograph of the accumulator nameplate, and Vessel Core checks the details before an offer goes back.