The isentropic compressor work for an ideal gas is given by: γR(ΔT)/γ -1 . This eq is usually reduced to Cp(ΔT) ; since Cp= γR/γ - 1. So, even though there is pressure rise in compressor, we are calculating work by using specific heat at constant pressure, which is fine if we only look at this as a slight mathematical manipulation. But another way to view this situation would be that since work for steady flow devices such as compressors = ΔH (change in enthalpy) & since enthalpy is a state function, therefore it is independent of the path chosen and depends only on end-states. Therefore, the use of the constant pressure heating path to calculate work of compressor may be justified, but the problem is that in the constant pressure heating path, the end states would not be the same as that during isentropic compression. Sure, we might achieve the same temperature values T1 and T2 in both paths, but in isentropic compression path, the initial pressure value would be P1 and finally P2, whereas in constant pressure heating path we would have some pressure p both in the beginning and at the end. So, how can we justify the use of the eq CpΔT to calculate compressor work from the state function perspective. The source of my confusion arises from the definition of state. How do we define it?

  • $\begingroup$ The enthalpy of an ideal gas if a function of temperature only, and not of pressure. $\endgroup$ – Chet Miller Jul 12 '18 at 14:13

You're right that the isentropic compressor work is the difference of enthalpy between the initial and final states (which are defined by the particular values of the state variables P, V, T).

And indeed, ΔH = Cp ΔT (assuming Cp does not depend much on T).

However, this formula is always valid for a perfect gas and does not assume that the pressure is constant. Ideed, Cp is called the heat capacity at constant pressure, but see it simply as a constant with a particular value. You do not necessarily need the transformation to be isobaric to use this constant at some point in your reasoning.

The "constant pressure" terminology just comes from the fact that for a perfect gas, for a transformation at constant pressure, the heat Q is equal to ΔH = Cp ΔT.

Of course, in the isentropic transformation, pressure is not constant, but the formula above is still correct.

  • $\begingroup$ This Cp thing is a mess. Anyway, thanks for clarification. Can you provide me with your sources, so that i can read about it in detail. $\endgroup$ – Mohammad Nayef Jul 12 '18 at 12:44
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    $\begingroup$ The enthalpy of an ideal gas is a function only of temperature, and is independent of pressure. $\endgroup$ – Chet Miller Jul 12 '18 at 13:18
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    $\begingroup$ Many good books define enthalpy. Maybe you can have a look at a detailed yet simple explanation here for instance. $\endgroup$ – D. Le Borgne Jul 12 '18 at 13:21

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