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Rigid vessel at critical state cooled to 2000 kPa

Short answer

Mass m = V / v_c = 0.02 / 0.003106 = 6.44 kg. At 2000 kPa, v = v_c, quality x₂ = (0.003106 − 0.001177)/(0.09958 − 0.001177) = 0.0196; liquid mass = 6.31 kg, vapour mass = 0.126 kg.

A rigid vessel having a volume of 0.02m³, initially contains water at its critical state. The vessel is cooled until its pressure drops to 2000kPa. Sketch the process on P-V and T-V diagrams and determine: a) The mass of H2O present in the vessel b) The quality at final state c) The mass of saturated liquid water and saturated water vapour at the final state. [Refer attached table for the properties of steam].

Given: V = 0.02 m³ (rigid), initial state: critical point, cooled to P₂ = 2000 kPa.

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

a) Mass

m = V/v_c = 0.02/0.003106 = 6.44 kg

b) Quality at final state

v₂ = v_c = 0.003106 (constant volume). At 2000 kPa: T_sat = 212.4 °C, v_f = 0.001177, v_g = 0.09958 m³/kg. Since v_f < v₂ < v_g, it is a wet mixture.
x₂ = (0.003106 − 0.001177)/(0.09958 − 0.001177) = 0.0196

c) Masses of saturated liquid and vapour

m_g = x₂ m = 0.0196 × 6.44 = 0.126 kg
m_f = m − m_g = 6.44 − 0.126 = 6.314 kg

 P|
  |  CP  (22.06 MPa)
  |  |
  |  |   v = vc = const
  |  |
  |  2   (2000 kPa)
  +------------- v

 T|
  |  CP  (373.9 C)
  |  |
  |  |   v = vc = const
  |  2   (212.4 C)
  +------------- v

On both diagrams the process is a vertical line from the critical point down into the dome.

Answer: m = 6.44 kg; x₂ = 0.0196; m_f = 6.31 kg, m_g = 0.126 kg.

✓ Answer checked by Claude Opus 5.5 (high effort)

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