Compressor work

[Azmio]

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

 

[GregLocock]

Isothermal or adiabatic?

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Azmio]

adiabatic

 

[GregLocock]

Did you use 100% isentropic efficiency?


Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Azmio]

100% isentropic efficiency

 

[GregLocock]

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

Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Azmio]

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

 

[GregLocock]

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

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Azmio]

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.

 

[black2003cobra]

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/?) - 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, ? = cp/cv, and AE = adiabatic efficiency. For air, at typical inlet temperatures, cp ? 1 kJ/(kg*°C) and ? ? 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.

 

[black2003cobra]

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