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Piston cylinder: 2 kg water 500 kPa, 20% quality

Short answer

The piston rises at constant pressure 500 kPa until volume reaches 0.4 m³, then heating continues at constant volume to saturated vapor. Final pressure is 970 kPa. Work done is 500 kPa × (0.4 − 0.1517) m³ = 124.2 kJ.

A piston cylinder device shown in figure below contains 2 kg of water initially at a pressure of 500 KPa with a quality of 20%. The water is heated until it becomes a saturated vapor. The volume of the system when the piston is at the upper stops is 0.4 m³. Sketch the process on P-v and T-v diagrams and determine: a) the final pressure, and b) the total work transfer [Figure: cylinder with H2O under a piston with upper stops, heat Q added.]

Given: m = 2 kg; P₁ = 500 kPa, x₁ = 0.2; heated until saturated vapour; V at the upper stops = 0.4 m³.

Steam table values at 500 kPa: T_sat = 151.8 °C, v_f = 0.001093, v_g = 0.3748 m³/kg.

State 1: v₁ = 0.001093 + 0.2 (0.3748 − 0.001093) = 0.07583 m³/kg
V₁ = m v₁ = 2 × 0.07583 = 0.1517 m³

Process path:

  1. 1 → 2: the piston is free and floats, so heating is at constant P = 500 kPa until the piston reaches the stops at V₂ = 0.4 m³ (v₂ = 0.2 m³/kg). Check: v_g (500 kPa) = 0.3748 > 0.2, so at this point the steam is still wet (x₂ = 0.53).
  2. 2 → 3: with the piston on the stops, heating continues at constant volume, v₃ = 0.2 m³/kg, until the steam is saturated vapour (v_g = 0.2).
 P|
  |            3   (on sat. vapour line, ≈ 970 kPa)
  |            |
  |            |   2-3: const v = 0.2
  | 1----------2   1-2: const P = 500 kPa
  +---------------- v
  0.0758      0.2

 T|
  |            3   (≈ 178.6 C, sat. vapour)
  |            |
  | 1----------2   (151.8 C, wet region)
  +---------------- v

a) Final pressure

At state 3, v_g = 0.2 m³/kg. From the saturated tables: v_g = 0.2149 at 900 kPa and 0.1943 at 1000 kPa. Interpolating:
P₃ = 900 + 100 × (0.2149 − 0.2)/(0.2149 − 0.1943) ≈ 970 kPa (T₃ ≈ 178.6 °C)

b) Total work

Work is done only in 1 → 2 (constant pressure):
W = P₁ (V₂ − V₁) = 500 × (0.4 − 0.1517) = 124.2 kJ

Answer: P₃ ≈ 970 kPa; W ≈ 124 kJ.

✓ Answer checked by Claude Opus 5.5 (high effort)

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