How Does Adiabatic Compression Affect Steam and Water Mixtures?

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Homework Help Overview

The discussion revolves around the adiabatic and reversible compression of a mixture of saturated water and steam, specifically focusing on calculating the fraction of steam at the compressor entrance, the exit entropy, and the exit pressure. Participants are exploring the implications of specific volumes and the properties of the mixture at given pressures.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculation of the fraction of steam and express confusion regarding the specific volume of the mixture. There are inquiries about the relationship between the inlet and exit conditions, particularly concerning entropy and pressure. Some participants suggest checking steam tables and considering the quality of steam.

Discussion Status

Some participants have made progress in calculating certain properties, such as the entrance quality of steam and exit entropy. However, challenges remain regarding the exit pressure and the connection between various thermodynamic properties. Guidance has been offered regarding the use of steam tables and the trial-and-error method for determining final states.

Contextual Notes

Participants note constraints related to the specific volumes of saturated water and steam, as well as the need for interpolation in steam tables. There is an acknowledgment of homework deadlines influencing the pace of the 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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