# Thermodynamic Relation

1. Sep 5, 2010

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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?

2. Sep 6, 2010

### CompuChip

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:

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

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 $\Delta p > 0$ (the final pressure is higher than the initial one) so $\Delta h > 0$ (the enthalpy increases).

Feel better now?

3. Sep 6, 2010