Accumulator volume problem

The chart you are using assumed the changes to the accumulator are isothermal. That is, the P and V changes are occurring slowly and heat is being removed or added to maintain a constant T. However your results are adiabatic. That is, your changes are occurring faster than ambient heat can maintain the accumulator gas T constant.

For example, the AS55 has a 50 liter gas volume bladder and you have define P0 = 1300, P1=1600 and P2=4000.

Calculating the isothermal volume changes would be as per the equations
P0XV0=P2XV2 gives V3=(P0/P2)XV0 or (1300/4000)X50, which gives V2=16.3
Simairly P2XV2=P1XV1 gives V1=40.6. The expected fluid would be V1-V2 = 24.3
This volume agrees with your chart.


Assuming the accumulator is charges with N2 (ideal gas k=1.4). Allowing the pressure to stabilize after filling will be isothermal, thus V2 would be the same. The adiabatic expansion from P2 to P1 would be as per the equations
P2XV2^1.4=P1XV1^1.4, gives V1=V2X(P2/P1)^(1/1.4) or 16.3X(4000/1600)^0.71, which gives V1= 31.4. The expected fluid would be V1-V2=15.1 While calculated volume is less than your results, since some heat would have been absorbed by the accumulator, it agrees with what you observed.