Skip to main content

Rigid container of steam cooled to 90°C

Short answer

Cooling at constant specific volume 1.3162 m³/kg. Condensation starts at 107.4°C. Final state at 90°C is a wet mixture with pressure 70.2 kPa and liquid mass fraction 0.442.

A rigid container with a volume of 0.170 m³ is initially filled with steam at 200 kPa, 300°C. It is cooled to 90°C. (Steam tables are attached here with) a) At what temperature does a phase change start to occur? b) What is the final pressure? c) What mass fraction of the water is liquid in the final state?

Given: V = 0.170 m³ (rigid, so v is constant), state 1: 200 kPa, 300 °C, cooled to 90 °C.

State 1 (superheated steam tables): at 200 kPa, 300 °C: v₁ = 1.3162 m³/kg.
Mass m = V/v₁ = 0.170/1.3162 = 0.1292 kg

The process is at constant volume, v = 1.3162 m³/kg.

a) Temperature at which the phase change starts

Condensation begins when v = v_g(T). From the saturated steam (temperature) table:

  • 105 °C: v_g = 1.4185 m³/kg
  • 110 °C: v_g = 1.2094 m³/kg

Interpolating for v_g = 1.3162:
T = 105 + 5 × (1.4185 − 1.3162)/(1.4185 − 1.2094) = ≈ 107.4 °C (P_sat ≈ 0.13 MPa)

b) Final pressure

At 90 °C, v = 1.3162 is less than v_g = 2.3591 and more than v_f = 0.001036, so it is a wet mixture; P = P_sat(90 °C) = 70.2 kPa (≈ 0.07 MPa).

c) Mass fraction of liquid

x = (v − v_f)/(v_g − v_f) = (1.3162 − 0.001036)/(2.3591 − 0.001036) = 0.5577
Liquid fraction = 1 − x = 0.442 (m_liquid ≈ 0.0571 kg)

 P|
  |  1   (200 kPa, 300 C)
  |  |
  |  |   const v = 1.316
  |  |
  |  2   (70.2 kPa, wet)
  +----------- v

 T|
  |  1   (300 C)
  |  |
  |  *   (≈ 107.4 C, meets sat. vapour line)
  |  |
  |  2   (90 C, wet)
  +----------- v

Answer: (a) ≈ 107 °C (b) 70.2 kPa (c) 44.2 % liquid by mass.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Fundamental of Thermodynamics & Heat Transfer chapter 3 questions →