×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

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

Jobs

Time to pressurise air vessel

Time to pressurise air vessel

Time to pressurise air vessel

(OP)
Hi All

Hopefully someone can help shed some light on a little problem - what is the best way for me to calculate how long it will take for an empty pressure vessel (at atmospheric) to reach the pressure of my compressed air main?  What do I need to know/derive?    

To me it appears this is like charging a capacitor - no?  Can I draw direct comparisons?

Thanks in advance for any help.

rmacp


RE: Time to pressurise air vessel

Until the vessel gets to about half (in absolute terms) the supply pressure, the flow is: (1) constant; and (2) sonic.  During that time, calculate sonic velocity and multiply it times flow area to get actual cubic ft/sec.  Calculate the receiver volume at 1/2 main pressure, then divide volume by volume flow rate to get time to half way.

Then it gets ugly.  There isn't a closed-form equation that will let you predict the last half because all of the flow equations (other than choked flow) start with a differential pressure and length - your differential pressure changes substantially from second to second.

You can approximate it with an equation like the AGA compressible-flow equation that you recalculate every 5 psi or so, but I've never been close doing that.

David Simpson, PE
MuleShoe Engineering
www.muleshoe-eng.com
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The Plural of "anecdote" is not "data"

RE: Time to pressurise air vessel

The differential equation describing the charging process of a capacitor is a linear type while the differential equations describing the charging process of a vessel with a pressurize gas are non-linear equations as explained by zdas04.

RE: Time to pressurise air vessel

Will your source (main) pressure and temp remain constant.

If so, consider that for a perfect gas, const specific heats and an adiabatic tank, the energy equation yields

E-Einitial =Hsource(stagnation)*(Mass-Mass initial)

where Mass is mass in tank.

Cv*(Mass*T)-Massinitial*Tinitial)=Hsource(mass-massinitial)
The pressure in the tank is then calculated from:
Ptank*Vtank/R=Minitial*Tinitial+GAMMA*Tstagsource(Mass-Massinitial)

RE: Time to pressurise air vessel

Hi
?f your pressure vessel is the only unit that consumes compressed air than you have to calculate the total volume (pipe line and the pressure vessel ) and than multiply with the determined pressure you will reach. That will be the total air you will need to reach the desired pressure. Of course temperature is a nother figure to consider but effect will be minimum.Than come to he compressor capacity. ?f your compressor is a screw or piston type compressor the  capacity will not change some much (not more than % 5) due the pressure output.Only the power demand will change. Than divide the air need to the compressor capacity in liter/ sec than you will find the time to reach  to that pressure.But if your compressor is a centrifugal compressor you have to consider that Centrifugal compressors capacity changes due to the pressure output. So the the compressing prosses of an emty wessel is dynamic not linear.

RE: Time to pressurise air vessel

(OP)
Thanks for the replies everyone - perhaps not as easy as I first thought!!!

I'll devote more thinking time to it - I may even have to resort to experimentation!

RE: Time to pressurise air vessel

Machinery's Handbook has formulas for HP required to compress air. One is for isothermal and the other is for adiabatic compression. They say that neither of these two theoretical extremes are obtainable, but power required is about the medium. From the HP you should be able to determine the time.

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!


Resources