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
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
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
RE: Compressor work
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
RE: Compressor work
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
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
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
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
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