Water, piston-cylinder problem [Thermodynamics]

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SUMMARY

The discussion centers on a thermodynamics problem involving 10 kg of water in a piston-cylinder arrangement at 100 kPa, with a quality of 50%. The water is heated until its volume triples, leading to a final pressure of 200 kPa. The final temperature is determined to be approximately 827-829 degrees Celsius, indicating that the water transitions into a superheated vapor state. The use of steam tables from "Fundamentals of Engineering Thermodynamics" by Moran is essential for accurate calculations, particularly for interpolating values between pressures of 1.5 bar and 3 bar.

PREREQUISITES
  • Understanding of thermodynamic properties, specifically saturated liquid and vapor states.
  • Familiarity with steam tables and their application in thermodynamic calculations.
  • Knowledge of specific volume calculations and phase quality in two-phase mixtures.
  • Basic principles of the ideal gas law and its limitations for non-ideal gases.
NEXT STEPS
  • Study the use of steam tables for superheated vapor calculations in thermodynamics.
  • Learn about the interpolation methods for determining thermodynamic properties between tabulated values.
  • Research the significance of the Z factor and its application in real gas behavior.
  • Explore the implications of using the ideal gas law for non-ideal gases in various thermodynamic scenarios.
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Students and professionals in thermodynamics, mechanical engineers, and anyone involved in heat transfer and fluid mechanics applications.

leafjerky
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Homework Statement


Okay, so these are usually pretty easy for me to understand, but this one doesn't make sense.

10 kg of water in a piston cylinder arrangement exists as saturated liquid/vapor at 100 kPa, with a quality of 50%. It is now heated so the volume triples. The mass of the piston is such that a cylinder pressure of 200 kPa will float it. Find the final temperature and the heat transfer in the process.

Homework Equations


v = V/m
v - specific volume
V - volume
m - mass

y = yf +xyfg

The Attempt at a Solution



State 1:[/B]
m1 = m2 = m = 10kg
x1 = .5 ---> 2-phase mixture
P1 = 100 kPa = 1 bar
v1 = .0010432 m3/kg + (.5)(1.694 - .0010432)
v1 = .8475 m3/kg

State 2:
v2 = 3*v1 = 2.5425 m3/kg
P2 = 200 kPa --> pressure is constant from this point on

Solutions I have found online say that the final temp is in the 800's. How is that possible? I went into my steam tables (Fundamentals of Engineering Thermodynamics, Moran) and I couldn't figure out the state. I assume it is a superheated vapour. Is this right? If so, I go to the tables and the temp could probably be interpolated between 1.5 bar and 3 bar, but even then it's not close to 800. I believe the answer is 827 or 829 degrees C. Thanks for any help.
 
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What temperature does the ideal gas law give you? (This is a pretty low pressure).
 
Thank you! I've always been under the assumption to never use ideal gas for non ideal gases though? This is a first for me. Is there a way to know? Thanks again
 
leafjerky said:
Thank you! I've always been under the assumption to never use ideal gas for non ideal gases though? This is a first for me. Is there a way to know? Thanks again
Calculate the reduced pressure and the reduced temperature. Then, see the chart in Moran et al for the Z factor. Or better yet, calculate the pseudo-reduced volume and the reduced pressure and use the Z chart in Moran.

Chet
 

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