Thermodynamic Relation: Gas Entropy in Compressor

AI Thread Summary
In an isentropic compression process, the specific enthalpy of a gas increases as it moves from the inlet to the exit of a compressor. The relationship between changes in enthalpy and pressure is expressed through the equation involving specific volume, which is positive during compression. However, concerns arise regarding the assumption that specific volume remains constant, as it typically decreases with increasing pressure. A more thorough analysis would require integrating the relationship between enthalpy and pressure to account for variations in specific volume. Ultimately, while the conclusion that enthalpy increases is supported, the dependence of specific volume on pressure complicates the understanding of the overall thermodynamic behavior.
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Homework Statement



A gas enters a compressor and is compressed isentropically. Does the specific enthalpy (h) increase or decrease as the gas passes from inlet to exit?

Homework Equations



\left(\frac{\partial{h}}{\partial{p}} \right)_s= v\qquad(1)

The Attempt at a Solution



Since the specific volume v is a positive number we know that pressure increases (since it is being compressed), then the enthalpy must also increase.


This is the answer that was given in the book. I don't really like it. The left side of (1) is a differential change and hence the right hand side is a single value. When we extend this idea to a finite change, what happens to the right hand side?

Does anyone see what I mean by "I don't like it?" We are looking at values of h and p at two different states 1 and 2. But what the heck is v supposed to do?

I feel like to get the full story, we would need to integrate dh = v(p) dp. We know that v should decrease with an increase in p. Any thoughts?
 
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Personally I always find these thermodynamic problems tricky, but I suppose that you could take your equation (1) and integrate it from the initial to the final pressure:

<br /> \int_{p_1}^{p_2} \left(\frac{\partial{h}}{\partial{p}} \right)_s \, \mathrm{d}p = \int_{p_1}^{p_2} v \, \mathrm{d}p<br />

It follows (insert stuff about fundamental theorem of calculus here) that
h_2 - h_1 = v (p_2 - p_1)
or, more compactly,
\Delta h = v \Delta p

Then compression means that \Delta p &gt; 0 (the final pressure is higher than the initial one) so \Delta h &gt; 0 (the enthalpy increases).

Feel better now?
 
Hi CompuChip!

I actually don't just yet since we know that v varies as well as h and p. It might not matter though seeing as it is always positive, but I still feel like it should be more thorough. It will depend on the difference of the product v2p2 - v1p1 I think. And thus it will depend on how v varies with p. I am just trying to see if there is something general we can say without knowing explicitly how v varies with p.
 
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