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Hydraulics
29.5
Hydraulic accumulator
203 Articles  2254 Elements

Hydraulic accumulator

29.5

In addition to low-loss energy conversion, the quality parameters of hydraulic systems can be increased by storing energy in the separate hydraulic accumulator unit.

Hydraulic accumulators therefore fulfill the main task of enabling more economical energy utilization in the hydraulic system. In addition, hydraulic accumulators improve the smoothness of energy transfer by performing the following tasks:87

  • Reduce flow pulsation
  • absorb dynamic pressure changes (pressure surges) in the line system (pulsation damping)
  • absorb mechanical shocks from the load system (shock absorption)
  • increase the uniformity of the movement sequence (running quality).
  • cover large current requirements for a short time, for example by high start-up acceleration
  • to be realized (covering power peaks)
  • satisfy small leakage current requirements in the long term, around
  • maintain biasing force (leakage current compensation)

General conditions for energy storage: p 0 = 0.9 p 1

The following relationship applies to bladder accumulators: p 2 ≤ 4 p 0

The maximum operating pressure of the hydraulic system p3 results from the pressure for which the hydraulic system is secured against excess pressure. This is the maximum pressure setting pressure relief valve to trigger opening tank. (response pressure).

The usual condition for the maximum operating pressure p 2 of the hydraulic accumulator is: p 2 ≤ 0.9 · p 3

Eqn. 1
\require{color}\definecolor{myred}{RGB}{255,0,0} \Delta V=V_{\color{myred}0}\cdot\left[\left(\frac{p_{\color{myred}0}}{p_{\color{myred}1}}\right)^\frac1\kappa-\left(\frac{p_{\color{myred}0}}{p_{\color{myred}2}}\right)^\frac1\kappa\right]
Useful volumeΔV=19.81dm3 
Memory sizeV0 = 100dm3
Gas filling pressurep0 = 135bar
Minimum working pressurep1 = 150bar
Max working pressurep2 = 210bar
Relationshipκ = 1.4
100
Calc 1

Ratio κ = c p / c v = 1.4 for nitrogen N260

Eqn. 2
\require{color}\definecolor{myred}{RGB}{255,0,0} V_{\color{myred}0}=\frac{\Delta V}{\left[\left(\frac{p_{\color{myred}0}}{p_{\color{myred}1}}\right)^\frac1\kappa-\left(\frac{p_{\color{myred}0}}{p_{\color{myred}2}}\right)^\frac1\kappa\right]}
Memory sizeV0=100.93dm3 
Useful volumeΔV = 20dm3
Gas filling pressurep0 = 135bar
Minimum working pressurep1 = 150bar
Max working pressurep2 = 210bar
Relationshipκ = 1.4
20
Calc 2

Ratio κ = c p / c v = 1.4 for nitrogen N260

Eqn. 3
\require{color}\definecolor{myred}{RGB}{255,0,0} p_{\color{myred}2}=\frac{p_{\color{myred}0}}{\left[\left(\frac{p_{\color{myred}0}}{p_{\color{myred}1}}\right)^\frac1\kappa-\frac{\Delta V}{V_{\color{myred}0}}\right]^\kappa}
Max working pressurep2=179bar 
Gas filling pressurep0 = 135bar
Minimum working pressurep1 = 150bar
Relationshipκ = 1.4
Useful volumeΔV = 11dm3
Memory sizeV0 = 100dm3
135
Calc 3
60
Böge, A. / Arbeitshilfen und Formeln für das technische Studium / Vieweg / Braunschweig / 1985 / 6. Auflage
87
Findeisen, D.; Helduser, S. / Ölhydraulik / Springer Vieweg / Berlin / 2015 / 6. Auflage
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