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Hi Jack,

If you multiply your answer (24.6 hp) times the ratio of specific heats you do indeed get 34.4 hp (the answer I gave).

Let's consider a cylinder with a piston that's compressing air. We put a closed control volume around the air in the cylinder and work is done on the air, compressing it. How does this process compare to an open control volume in which you have air entering at some mass flow rate and then exiting the control volume at higher pressure and temperature due to work done on it?

The first law for the closed control volume gives:

dU = W

The first law for the open control volume gives:

dU = W + dH

where dU = 0 since there is no air stored inside the control volume. Thus:

dH = W

You calculated dU = W. I calculated dH = W. Which is correct? For a compressor, you have air in and air out (enthalpy in and enthalpy out). If you only do an analysis on the closed control volume of air being compressed, you don't get the entire picture. There is additional work being done by the compressor to force the air out of the cylinder at the higher pressure. The work done by a compressor is dH = W. The correct answer looks at the open control volume with a mass flow in, mass flow out and work added.

See also:

http://www.grc.nasa.gov/WWW/K-12/airplane/compth.html

http://www.codecogs.com/reference/engineering/thermodynamics/air_compressors.php

You are correct. I've been looking deeper to see where I was wrong. I knew I saw what I demonstrated somewhere (I told you I've been doing this from memory) and found http://en.wikipedia.org/wiki/Adiabatic_process" [Broken] which gives the same answer as me. But they say it's the work done

*by*the system.

Then I found http://en.wikipedia.org/wiki/Enthalpy#Open_systems", which essentially say the same thing you say, which is you have to add the shaft work. So

*dW = dH*in a adiabatic process.

And then, from http://en.wikipedia.org/wiki/Isentropic_process#Derivation_of_the_isentropic_relations",

*dH = Vdp*in an isentropic process and by re-doing the whole derivation I did earlier it gives:

[tex]W= \frac{k}{k-1}P_1 V_1 \left( PR^{\frac{k-1}{k}} - 1 \right)[/tex]

Hence, the same one you gave, i.e. with C

_{p}. So my initial equation (and hopefully the final one!) becomes:

[tex]p=0.00436 \frac{PR^{0.286}-1}{0.286 \left( PR-1 \right)}\frac{\Delta P \ CFM}{\eta_s}[/tex]

Where the units are

*psi*,

*cfm*and

*hp*.

Thanks, that's why I like this forum.

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