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Accumulator maximum pressure & minimum pressure 2

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RobertHasty

Mechanical
Jun 14, 2012
81
Hello,

I have been looking into accumulator formulas and would like your opinion on it.

Let's say I would like to use an accumulator to extract a cylinder at 100 bar.
What would be in that case the minimum working pressure (P1) and maximum working pressure (P2).

I'm guessing the minimum would be 100 bar and maximum would be defined by pressure relief valve?

Looking forward reading your opinions.
Thank you.
 
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You're using an accumlator for what exactly? "Extract a cylinder"??

A bit more description of your process , pump etc.

For me an accumulator is used where you have a small pump and a large load / volume in a short time.

Hence you spend say 30 minutes pumping up the accumulator to xx bar, then use it in say 30 seconds down to your min required pressure of yy bar.

Is your accumulator a fixed gas volume or is it air over oil?
Is all the volume you need coming from the accumulator or does the pump add some volume?

Lots of ways to do this but you need to understand the process first.



Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
It does not matter if the cylinder is extending or retracting, all that matters to the accumulator is the volume that is needed to move the cylinder to its full travel.

If the cylinder needs 100 BAR minimum (P2), the pre charge (P1) will be 90 BAR.

The maximum pressure in the accumulator (P3) will be a function of the volume of gas that is stored.

If you start with 90 BAR gas pressure and fill the accumulator with oil to the lowest oil pressure = 100 BAR, the gas volume will go down, but it's not possible to know what the new volume is until you select the accumulator volume.

If the maximum pressure is 200 BAR, the pressure ratio is 2:1 and Boyle's law tells us that in isothermal conditions, halving the volume of the gas will double the pressure.

P1 = 90 BAR
P2 = 100 BAR
P3 = Max Pressure ?
V1 = Gas Volume ?
V2 = Pre Charge volume ?
V3 = Compressed gas volume?

When you know how much oil needs to flow from the accumulator to the cylinder, you can calculate V3-V2. When you know this you can work out how much gas is required to go between P2 and P3.

The max pressure might be set by the volume of gas you can store, or it might be set by the system limits.

Example:

A cylinder needs to have minimum pressure of 100 BAR
The pre charge will be 90% of the minimum pressure to make sure that there is always some compression to push the oil out of the accumulator.
Let's say that the max pressure of the system is 210 BAR.

Let's also say that the cylinder will require 450cc to move full stroke.

P2 x V2 = P3 x V3

We want to know what volume of gas is required to ensure that the pressure is not more than 210 BAR, when the difference between V2 and V3 is 450cc.

P3/P2 = V3/V2
210/100 = 2.1 Therefore the volume ratio is also 2.1 : 1

Your gas volume needs to be 450cc x 2.1 to work between 100 and 210 BAR.

You need 945cc of gas volume differential to achieve these pressures.

However, the gas will also expand as it's compressed and contract as it decompresses. Therefore, the polytropic constant for nitrogen needs to be added to the equation.

With adiabatic compression and decompression the pre-charge volume will need to be 1500cc.

If you can live with higher max pressure, like 300 BAR, you can cope with less gas volume and you would need 1260cc.

Accumulator volumes are standardised to 1, 2, 4 and 10 Litres. You would need to use a 2 litre accumulator in this case.

If you can make space for a 4 litre accumulator, the max pressure will come down to 122 BAR.

If you use a 10 Litre accumulator, the max pressure will be 108 BAR.

Really, the max pressure is function of the gas volume that you can store.

 
Dear FluidPowerUser, thank you for the great reply!
It took a lot of effort to write it.

I have read it in detail and have some additional questions.

FluidPowerUser said:
Let's also say that the cylinder will require 450cc to move full stroke.

P2 x V2 = P3 x V3

We want to know what volume of gas is required to ensure that the pressure is not more than 210 BAR, when the difference between V2 and V3 is 450cc.

P3/P2 = V3/V2

I believe there is a typo here. I believe you meant P3/P2 = V2/V3?

FluidPowerUser said:
210/100 = 2.1 Therefore the volume ratio is also 2.1 : 1Th

Your gas volume needs to be 450cc x 2.1 to work between 100 and 210 BAR.

You need 945cc of gas volume differential to achieve these pressures.
I am having a hard time figuring this out (945 cc).
The volume difference is 450 cc. That would mean that V2 = V3 + 450 cc in the example above.
That would mean folliwing:
P3/P2 = V2/V3
P3/P2 = (V3+450)/V3
2,1*V3=V3+450
1,1V3=450
V3=409,09

And that would mean that V2=409,09+450=859,09 cc?
 
I think you're right about the P3/P2 = V2/V3

This impacts the statement that pressure ratio and volume ratio are the same. They are, but now it's the inverse.

So I agree with your working out to find min value of V2 at 100 bar.

Still assumes that you have no impact from any pump though?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
Ref the calculation for the pressure and volume, it is just an rearrangement to find the ratios. It's not really important in what we are trying to establish here. I don't think there is a typo though.

Ref your last calculation, it has to be assumed that the nitrogen is expanding adiabatically. It's cooling down as it expands and the volume is shrinking as it cools. The polytropic index for nitrogen is 1.4 and that has to be added into the equation. Hence, the volume of gas needs to be larger to compensate for the thermodynamic process.
 
It’s not necessarily the case that an accumulator is there to supplement pump flow.

Accumulators can be stored energy to supplement pump flow, provide emergency power when the pump is not working, or pulsation dampening.

If a pump is in the circuit, then it will need a non return valve to prevent the accumulator from motoring the pump. There will be no flow from the pump until the pressure differential over the check valve is low enough to allow the valve to open. Filling and emptying the accumulators is highly dependent on the system dynamics and the accumulators will always react more quickly than the pump.

At the point where the accumulator pressure is lower than the pressure at the cylinder, the pump will fill the accumulator and cylinder will slow down or stop.

None of this changes the storage volume of the accumulators, it just changes the point at which they start to fill or empty.
 
FPU - Very true and many accumulators are sized for at least one open / close of actuators and sometimes more.

The pump though is what normally charges the accumulator and therefore I would expect it to have an MOP higher than the max operating pressure of the accumulator.

But this seems to be a general "How do I do it?" question, not a specific one?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
LittleInch said:
But this seems to be a general "How do I do it?" question, not a specific one?
Yes, it's a general question
 
FluidPowerUser said:
Ref your last calculation, it has to be assumed that the nitrogen is expanding adiabatically. It's cooling down as it expands and the volume is shrinking as it cools. The polytropic index for nitrogen is 1.4 and that has to be added into the equation. Hence, the volume of gas needs to be larger to compensate for the thermodynamic process.
Thanks!
 
There are now simple formulas. I use a simulator. If necessary I can simulate the motion of two actuators running asynchronously but usually two are running in tandem so it is simple.
Hyd_sizing_parameters_e9t0vn.png
hyd_sizing_plots_hi99a4.png


Peter Nachtwey
Delta Motion
IFPS Hall of Fame Member
 
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