Frictionless piston 20 kg with H2O, five parts

Given: m_p = 20 kg, A = 78.48 cm² = 7.848 × 10⁻³ m², P_atm = 100 kPa, g = 9.81 m/s²; x₁ = 0.2; the H₂O space is 50 mm high initially; the piston hits the top stops after a further 30 mm; final T = 400 °C.

Process: 1 → 2: heating at constant pressure (the piston floats) until it touches the stops; 2 → 3: constant-volume heating to 400 °C.

i) Initial pressure

P₁ = P_atm + m_p g/A = 100 + (20 × 9.81)/(7.848 × 10⁻³ × 1000) = 100 + 25.0 = 125 kPa

ii) Mass of H₂O

At 125 kPa (steam tables): v_f = 0.001048, v_g = 1.3749 m³/kg, v_fg = 1.3739 m³/kg.
v₁ = v_f + x v_fg = 0.001048 + 0.2 × 1.3739 = 0.27582 m³/kg
V₁ = A × 0.050 = 7.848 × 10⁻³ × 0.05 = 3.924 × 10⁻⁴ m³
m = V₁/v₁ = 3.924 × 10⁻⁴ / 0.27582 = 1.423 × 10⁻³ kg (about 1.42 g)

iii) Quality when the piston hits the stops

V₂ = A × (0.05 + 0.03) = 6.278 × 10⁻⁴ m³
v₂ = V₂/m = 0.4413 m³/kg
x₂ = (v₂ − v_f)/v_fg = (0.4413 − 0.001048)/1.3739 = 0.320

iv) Final pressure

From 2 → 3 at constant volume, v₃ = 0.4413 m³/kg at T₃ = 400 °C (superheated). From the superheated tables at 400 °C: v = 0.5137 m³/kg at 600 kPa and v = 0.4398 m³/kg at 700 kPa. Interpolating for v = 0.4413 gives P ≈ 698 kPa.
P₃ ≈ 0.70 MPa (about 698 kPa)

v) Total work

Work is done only in 1 → 2 at constant P = 125 kPa:
W = P (V₂ − V₁) = 125 × (6.278 × 10⁻⁴ − 3.924 × 10⁻⁴) = 0.0294 kJ (≈ 29.4 J)

 P|
  |        3   (400 C, ≈ 698 kPa)
  |        |
  |        |   2-3: const v (piston on stops)
  | 1------2   1-2: const P = 125 kPa
  +------------- v