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If you have a saturated steam (quality=1.0) volume that is compressed adiabtically will the steam condense to follow the saturation curve or superheat?

I found in an old textbook: The Principles of Thermodynamics (1899) on Google the following passage:

For compression, adiabatic:

  1. If we start with pure saturated steam, without admixture of water, it will be superheated by compression.
  2. If the initial steam weight is greater than that of the water, steam is generated by the compression.
  3. If there is more water than steam, steam is condensed during compression.

I am still at a bit of a loss for this makes physical sense and how this is properly captured in an energy balance. (i.e., $\frac{\text{d}U}{\text{d}t} = -\frac{p\text{d}V}{\text{d}t}$)

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  • $\begingroup$ One of the nice things about classical thermodynamics is that nothing's changed since that book was published (and rather a while before that :-) ) $\endgroup$ – Carl Witthoft Sep 23 '16 at 14:26
  • $\begingroup$ So true. Also, it's quite amusing as you read different textbooks often you find many similar examples and verbage that were used decades ago. $\endgroup$ – Scott G Sep 23 '16 at 14:28
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Take a look on this T-V water diagram:

enter image description here

  1. If we start with pure saturated steam, without admixture of water, it will be superheated by compression.

Suppose we have a steam located at point C or G or K where dryness factor = 1 (no water), further compressing/pressurizing of the steam will make the substance end up in the superheated vapor region (outside the saturation dome).

  1. If the initial steam weight is greater than that of the water, steam is generated by the compression.

If we have a point between point B and C at P = 0.1 MPa with a dryness factor of 0.8 (near point C) for example meaning that we have greater amount of vapor than water, further compressing (say to FG line, 1 Mpa) will lead us to a point that could be very close to saturated vapor line (higher dryness factor than 0.8) or in the superheated region, which means steam (more precisely vapor) is generated in both cases.

  1. If there is more water than steam, steam is condensed during compression.

This is the opposite situation of case 2, if we have a point near B, say having dryness factor of 0.2, further compressing (say to FG line, 1 Mpa) will lead us to a point on the FG line with dryness factor that is lower than 0.2, meaning less vapor content (condensation).

I hope all is clear now.

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  • $\begingroup$ The question pertains to an adiabatic process; your TV diagram does not make lines of adiabatic volume change explicit; but we do know they wont be lines of constant temperature, for sure, so it tells us nothing about what would happen to the saturation. $\endgroup$ – Eelco Hoogendoorn Dec 16 '18 at 13:38
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An ideal adiabatic compression of steam is a reversible process. This means that on the state of the fluid will follow lines of constant entropy. An alternative view is this is the opposite of an ideal turbine where the initial assumption is ds = 0 (adiabatic expansion in reverse).

For the T-s diagram constant entropy will put you at steam or liquid depending on the condition of the steam-water.

  1. If x > 0.5 then compression will lead to steam
  2. If x < 0.5 then compression will lead to liquid
  3. If x ~ 0.5 then compression will stay at roughly 50/50 mixture until supercritical water conditions are met (the maximum of the saturation curve).

Or for the Mollier P-h diagram. If you start at "B", compressing the fluid will follow the red line and go to superheated steam.

Temperature-Entropy Diagram for Water Mollier Pressure-Enthalpy Diagram of Water

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