How Does Adiabatic Compression Affect Steam and Water Mixtures?

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SUMMARY

The discussion focuses on the adiabatic and reversible compression of a saturated water and steam mixture at an initial pressure of 1.02 bar and specific volume of 1.25 m³/kg, compressing it to a specific volume of 0.473 m³/kg. Participants calculated the entrance quality of steam and exit entropy but faced challenges determining the exit pressure. Key insights include the importance of using steam tables to find the specific volumes and entropies of saturated steam and water, and the necessity of ensuring consistency between calculated fractions and specific volumes.

PREREQUISITES
  • Understanding of thermodynamic principles, particularly adiabatic processes.
  • Familiarity with steam tables and their application in thermodynamics.
  • Knowledge of specific volume and entropy calculations for mixtures.
  • Ability to calculate quality (dryness fraction) of steam in a saturated mixture.
NEXT STEPS
  • Review steam tables for specific volumes and entropies of saturated steam and water at various pressures.
  • Learn how to calculate the exit pressure using the final entropy and specific volume of the mixture.
  • Explore the concept of quality in steam mixtures and its impact on thermodynamic properties.
  • Investigate graphical methods for estimating specific volumes and temperatures in thermodynamic processes.
USEFUL FOR

Mechanical engineers, thermodynamics students, and professionals involved in steam system design and analysis will benefit from this discussion.

zircons
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A compressor adiabatically and reversibly compresses a mixture of saturated water and steam from a pressure 1.02 bar and specific volume of 1.25 m^3/kg to 0.473 m^3/kg.

Calculate the fraction of steam at the compressor entrance, the exit entropy, and the exit pressure.

For the fraction of steam at the entrance, I thought it would be the saturated volume of saturated water and steam/the given volume of 1.25. However, the saturated volume of water and steam at 1.02 bar is higher than the given volume of 1.25. That's impossible. I'm confused; what am I doing wrong?
 
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dryness fraction is = (weight of steam) / (weight of steam + weight of water)

its late now so i can't do it, (plus i have an assignment due on wednesday) but either tomorrow or wednesday i will help out some more :)

Anthony
 
zircons said:
A compressor adiabatically and reversibly compresses a mixture of saturated water and steam from a pressure 1.02 bar and specific volume of 1.25 m^3/kg to 0.473 m^3/kg.

Calculate the fraction of steam at the compressor entrance, the exit entropy, and the exit pressure.

For the fraction of steam at the entrance, I thought it would be the saturated volume of saturated water and steam/the given volume of 1.25. However, the saturated volume of water and steam at 1.02 bar is higher than the given volume of 1.25. That's impossible. I'm confused; what am I doing wrong?
One may wish to check the specific volume of water and steam for saturated conditions at 1.02 bar. Also, consider the quality of the steam, the mass fraction x that is steam and the fraction (1-x) that is liquid. Then consider the significance of "adiabatically and reversibly" and how that relates to the exit conditions.
 
Isn't the specific volume of liquid water, even at saturation, about 0.001 m3/kg? That doesn't sound higher than 1.25.
 
Thank you for your guidance!

I have now calculated the entrance quality of steam and exit entropy. However, I'm having trouble with the exit pressure. I know the inlet pressure, volume, quality, and entropy. I know the exit volume and entropy. The exit quality of steam and temperature must be higher (right?). I'm at a loss at how to connect it to the exit pressure though.
 
Did you reach the conclusion that the exit entropy is the same as the inlet entropy? Assume a final temperature. Can you look up in the steam tables the entropy of saturated steam, the specific volume of saturated steam, the entropy of saturated liquid water, and the specific volume of saturated liquid water at that temperature? From the final entropy of the mixture, calculate the fraction of saturated steam and saturated water to make good on that final entropy. Then check to see if these fractions are also consistent with the final specific volume of the mixture. If they are not consistent, try another temperature.

Depending on what your steam tables are like, you may be able to find the final state without this trial-and-error approach.
 
Oh man, that's the only solution I could think of, but I was hoping I wouldn't need to do the trial-and-error way. My steam tables will require it, along with interpolation :/ Regardless, thank you for your help!
 
It shouldn't be too bad. Make a graph of the mismatch in overall specific volume as a function of the assumed temperature. After plotting a few points, you will see where the graph is heading, and you will make much better guesses of the temperature. Three or four temperature guesses ought to be enough to get you there.
 

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