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Rigid container of steam 1500 kPa cooled to 100°C

Short answer

For rigid container volume 0.170 m³, initial superheated steam at 1500 kPa, 200°C, v1=0.1324 m³/kg, mass m = V/v1 = 1.284 kg. Cooled to 100°C at constant volume, state becomes wet at saturation pressure 101.4 kPa; with v2=v1, quality x=0.0786, so liquid mass fraction = 1-x = 0.921.

A rigid container with a volume of 0.170 m³ is initially filled with steam at 1500 kPa, 200°C. It is cooled to 100°C. Determine a) Total mass of the system b) Final pressure c) Mass fraction of the liquid in the final state Also sketch the process on P-v and T-v diagrams. [Refer the attached table for properties of steam]

Given: V = 0.170 m³ (rigid); state 1: 1500 kPa, 200 °C; cooled to 100 °C.

State 1: T_sat(1500 kPa) = 198.3 °C, so steam at 200 °C is slightly superheated. Superheated table: v₁ = 0.1324 m³/kg.

a) Total mass

m = V/v₁ = 0.170/0.1324 = 1.284 kg

b) Final pressure

v₂ = v₁ = 0.1324 (constant volume). At 100 °C: v_f = 0.001043, v_g = 1.6719 m³/kg; since v_f < v₂ < v_g, it is a wet mixture.
P₂ = P_sat(100 °C) = 101.4 kPa (≈ 0.1 MPa)

c) Mass fraction of liquid

x₂ = (0.1324 − 0.001043)/(1.6719 − 0.001043) = 0.0786
Liquid fraction = 1 − x₂ = 0.921 (m_f = 1.183 kg; m_g = 0.101 kg)

 P|
  |  1   (1500 kPa, 200 C)
  |  |
  |  |   const v = 0.1324
  |  |
  |  2   (101.4 kPa, wet)
  +----------- v

 T|
  |  1   (200 C)
  |  *   (≈ 198 C, meets sat. vapour line)
  |  |
  |  2   (100 C, wet)
  +----------- v

On both diagrams, process 1-2 is a vertical line going down from state 1 (superheated vapour) into the dome, crossing the saturated vapour line at about 198 °C and ending at 2 inside the dome.

Answer: m = 1.284 kg; P₂ = 101.4 kPa; liquid mass fraction = 0.921.

✓ Answer checked by Claude Opus 5.5 (high effort)

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