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Closed vessel at critical state cooled to 1 MPa

Short answer

From critical point vc=0.003106 m³/kg, mass m=V/vc=0.322 kg. Final state at 1 MPa has same specific volume; wet at Tsat=179.9°C, dryness fraction x=0.0102.

A one liter closed vessel contains water at its critical conditions. This vessel is cooled until its pressure drops to 1 MPa. Calculate the mass of water in the vessel, the final dryness fraction and final temperature. Also show the process on P-v and T-v diagrams.

Given: V = 1 L = 0.001 m³ rigid vessel (constant volume), initially at the critical point of water, cooled to P₂ = 1 MPa.

Critical state of water (steam tables): P_c = 22.06 MPa, T_c = 373.9 °C, v_c = 0.003106 m³/kg.

Mass of water

m = V/v_c = 0.001/0.003106 = 0.322 kg

Final state at 1 MPa

v₂ = v_c = 0.003106 m³/kg. At 1 MPa: T_sat = 179.9 °C, v_f = 0.001127, v_g = 0.19435 m³/kg. Since v_f < v₂ < v_g, it is a wet mixture.

Final temperature = T_sat(1 MPa) = 179.9 °C

Final dryness fraction
x₂ = (v₂ − v_f)/(v_g − v_f) = (0.003106 − 0.001127)/(0.19435 − 0.001127) = 0.0102

(So the vessel holds about 0.0033 kg of vapour and 0.319 kg of liquid.)

 P|
  |  CP  (22.06 MPa)
  |  |
  |  |   v = vc = const
  |  |
  |  2   (1 MPa)
  +----------- v

 T|
  |  CP  (373.9 C)
  |  |
  |  |   v = vc = const
  |  2   (179.9 C)
  +----------- v

On both diagrams the process is a vertical (constant-volume) line that starts at the critical point, goes down into the dome and ends at 2 (1 MPa).

Answer: m = 0.322 kg; x₂ = 0.0102; T₂ = 179.9 °C.

✓ Answer checked by Claude Opus 5.5 (high effort)

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