×
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

Compressor work

Compressor work

Compressor work

(OP)
I need some reality check on whether my classical calculation is correct or not.

I used some classical calculation that I got from a thermodynamics text book to calculate the work required to compress air.

Scenario 1
40g/s of air being compressed at 3 bar

Scenario 2
10g/s of air being compressed at 25 bar


Somehow, scenario 1 consumed much higher energy. This makes me conclude that flow rate is more significant than charge pressure when it comes to energy consumption. Can someone verify my conclusion

RE: Compressor work

(OP)
adiabatic

RE: Compressor work

(OP)
100% isentropic efficiency

RE: Compressor work

I agree with your intuition, the compression of 10 g/s to 25 bar needs more power than the other case  

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Compressor work

(OP)
Greg,

It's the other way around actually. My classical calculation result shows that scenario 1 consumes more power than scenario 2. That's why I need others to verify this as well

RE: Compressor work

Hmm, well to shoot my own credibility down, yes I agree now I've got a computer to do it.

Roughly a factor of two.


 

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

RE: Compressor work

(OP)
I used to think that the higher the charge pressure, the more power it will consume. I was thinking that the higher the pressure during the compression, the hotter the charge is going to be and this is where lots of power is needed.

However, when my spreadsheet tells otherwise, I have started to think that perhaps, the more charge mass involves and the reciprocating piston will have to induct and compress more.

RE: Compressor work

Assuming you're talking about an open system, for adiabatic compression the expression for power (dW/dt) is simply,

Power = mdot*cp*Tin[(1 + p_b/p_in)^(1 – 1/γ) - 1]/AE

where mdot = mass-air flow rate, cp = specific heat for constant pressure, Tin = absolute inlet temperature, p_b and p_in are boost and inlet pressure, respectively, γ = cp/cv, and AE = adiabatic efficiency. For air, at typical inlet temperatures, cp ≈ 1 kJ/(kg*°C) and γ ≈ 1.4.  So I find your scenario #2 should take just over 3 times the power that scenario #1 takes, assuming the same adiabatic efficiency and inlet temperature.  So I agree w/ Greg's intuition.
 

RE: Compressor work

In my haste, I must have typed in the numbers for the two cases wrong into my calculator.  Scenario #1 is indeed higher.

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