Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here


Accumulator volume problem

Accumulator volume problem

Accumulator volume problem

I came across an accumulator volume problem that needs help.
I charged the an EPE AS55 accumulator from 0 to 4500 psi and turned off inlet valve. The pressure droped a little and stablized at 4000 psi. Then I opened outlet valve to collect water till pressure droped to 1600 psi.

In this case P1=1600 psi, P2=4500 psi, so accumulator precharge pressure P0=1300 psi. According to the chart, I expected the volume to be 24.5 litre, but only managed to get 17.5 litre.
Can anybody recomment on the shortage of volume. Thanks in advance.

RE: 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.

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


eBook - Efficient and Effective Production Support with 3D Printed Jigs and Fixtures
Jigs and fixtures offer manufacturers a reliable process for delivering accurate, high-quality outcomes, whether for a specific part or feature, or for consistency across multiples of parts. Although the methodologies and materials for producing jigs and fixtures have evolved beyond the conventional metal tooling of years past, their position as a manufacturing staple remains constant due to the benefits they offer. Download Now
Overcoming Cutting Tool Challenges in Aerospace Machining
Aerospace manufacturing has always been on the cutting edge, from materials to production techniques. However, these two aspects of aerospace machining can conflict, as manufacturers strive to maintain machining efficiency with new materials by using new methods and cutting tools. Download Now

Close Box

Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

Register now while it's still free!

Already a member? Close this window and log in.

Join Us             Close