A saturated mixture is heated, volume is doubled and pressure remains the same.

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SUMMARY

The discussion focuses on a thermodynamics problem involving a saturated water mixture heated from an initial state with a pressure of 200 kPa and a volume of 0.3 m³ to a final volume of 0.6 m³ while maintaining constant pressure. The initial temperature T1 was not provided, but the final temperature T2 was calculated to be 120.12°C using pressure tables and interpolation. Additionally, the internal energy can be determined by calculating the specific volume at state 2 and referencing the internal energy values from the tables.

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  • Understanding of saturated water mixtures and phase changes
  • Familiarity with thermodynamic tables, specifically pressure and temperature tables
  • Knowledge of specific volume calculations
  • Ability to perform interpolation for thermodynamic properties
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allly
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Homework Statement



A saturated water mixture is heated.

P1=200kPa
m=0.5kg
V1=0.3m^3
V2=0.6m^3

find
T1
T2
Internal Energy

2. The attempt at a solution

Using the pressure tables and interpolation I found T2=120.12

stuck on T2 and internal energy.
 
Last edited:
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allly said:
2. The attempt at a solution

Using the pressure tables and interpolation I found T2=120.12

stuck on T2 and internal energy.

Since you have the mass, for state 2, you can get the specific volume v2 by using v2=V2/m and read that off from the table.

Your table will also have values for specific internal energy, which when multiplied by mass, gives the internal energy.
 

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