Thermodynamics Rigid tank problem

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SUMMARY

The discussion focuses on a thermodynamics problem involving a closed, rigid tank filled with water, initially at 20 bar and a quality of 80%, which is cooled to a pressure of 4 bar. The user calculated the change in internal energy using the equations ΔU=Q-W and specific volume equations, arriving at a heat transfer value of -611.65 kJ. However, this result does not match the professor's provided answer of -8282 kJ, indicating a potential error in the calculations or assumptions made during the process.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the first law of thermodynamics.
  • Familiarity with properties of water at different pressures, including specific internal energy and specific volume.
  • Knowledge of quality in thermodynamics and its calculation.
  • Ability to interpret T-v diagrams for phase changes in substances.
NEXT STEPS
  • Review the calculation of specific internal energies at varying pressures using steam tables.
  • Study the concept of quality in thermodynamics and its impact on phase changes.
  • Learn how to accurately draw and interpret T-v diagrams for various thermodynamic processes.
  • Investigate common pitfalls in thermodynamic calculations, particularly in closed systems.
USEFUL FOR

Students studying thermodynamics, particularly those tackling problems involving phase changes in closed systems, as well as educators looking for examples of common calculation errors in heat transfer scenarios.

jorsh
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Homework Statement


A closed, rigid tank filled with water, initially at 20 bar, a quality of 80%, and a volume of .5 m^3, is cooled until the pressure reaches 4 bar. Evaluate the heat transfer, in kJ, and draw the process in a T-v diagram


Homework Equations


ΔU=Q-W (but W goes to zero here because of the constant volume)
u= ugX + (1-x)uf
v= vgX + (1-x)vf

The Attempt at a Solution


Ok, so. I know that the volume is constant. My plan was to use the quality to calculate specific internal energies at 20 bar and 4 bar, thus finding the change in internal energy.
I ended up with u1(20 bar) = 2261.528 kJ/kg and u2(4 bar) = 2163.742 kJ/kg using the above equation using quality.

However, it wants the answer in kJ, so I had to find the specific volume to find the mass to cancel the kJ/kg. So, i used vg and vf at 20 bar (using the equation v=vgx + (1-x)vf) and I got .07993934 m^3/kg. I divided the given .5 m^3 by the specific volume I got, and received the answer of 6.255 kg. I then multiplied this by the change in internal energy, which I found to be -97.786 (cooling process, makes sense).

The final answer I calculated was -611.65 kJ. My professor gave the answers in class so we could check, but did not go through the problems specifically. This answer does not agree with his answer of -8282 kJ. Maybe -82.82? I have it written down as -8282.

Thanks for any help
 
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