Thermodynamic Relation: Gas Entropy in Compressor

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SUMMARY

The discussion centers on the thermodynamic behavior of gas entropy in a compressor during isentropic compression. It is established that as the gas is compressed, the specific enthalpy (h) increases due to the positive relationship between specific volume (v) and pressure (p), as described by the equation \(\left(\frac{\partial{h}}{\partial{p}} \right)_s= v\). The participants express concerns regarding the integration of the relationship between h and p, emphasizing the need to consider how specific volume varies with pressure for a comprehensive understanding of the enthalpy change.

PREREQUISITES
  • Understanding of isentropic processes in thermodynamics
  • Familiarity with the concept of specific enthalpy and specific volume
  • Knowledge of differential calculus and integration techniques
  • Proficiency in applying the fundamental theorem of calculus in thermodynamic equations
NEXT STEPS
  • Study the derivation and application of the first law of thermodynamics in isentropic processes
  • Learn about the relationship between specific volume and pressure in real gases
  • Explore advanced thermodynamic properties and their implications in compressor design
  • Investigate the impact of varying specific volume on enthalpy changes during compression
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Students and professionals in mechanical engineering, particularly those specializing in thermodynamics and fluid mechanics, as well as engineers involved in compressor design and analysis.

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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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