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Piston cylinder 400 L to 600 L: work and heat

Short answer

Work occurs only during constant-pressure lift: W = 300 kPa × (0.6 – 0.4) m³ = 60 kJ. From first law, Q = m(u₄ – u₁) + W = 1.178 kg × (2549.6 – 835.0) kJ/kg + 60 kJ ≈ 2079 kJ added.

Consider the piston/cylinder arrangement as shown figure below. When the piston rests on the lower stops, the enclosed volume is 400L. When the piston reaches the upper stops, the volume is 600L. the cylinder initially contains water at 100kPa, 20% quality. It is heated until the water eventually exists as saturated vapor. It takes a pressure of 300kPa to lift the piston. Sketch P-v and T-v diagrams and determine the work transfer and heat transfer for the overall process. [Figure: cylinder containing H2O with a piston, lower stops and upper stops.]

Given: V_lower = 0.4 m³ (400 L), V_upper = 0.6 m³; state 1: 100 kPa, x₁ = 0.2, piston on the lower stops; 300 kPa is needed to lift the piston; heated until saturated vapour.

State 1 (100 kPa): v_f = 0.001043, v_g = 1.6940 m³/kg; u_f = 417.3, u_fg = 2088.2 kJ/kg.
v₁ = 0.001043 + 0.2 × 1.6930 = 0.3396 m³/kg
m = V₁/v₁ = 0.4/0.3396 = 1.178 kg
u₁ = 417.3 + 0.2 × 2088.2 = 835.0 kJ/kg

Process path:

  1. 1 → 2: constant volume (V = 0.4 m³) until P = 300 kPa (the piston just lifts).
  2. 2 → 3: constant P = 300 kPa as the piston rises. At 300 kPa, v_g = 0.6058, which would need V = 1.178 × 0.6058 = 0.714 m³ > 0.6 m³, so the piston reaches the upper stops first, at V = 0.6 m³ (v = 0.5095 m³/kg), while the steam is still wet.
  3. 3 → 4: constant volume (V = 0.6 m³) until saturated vapour: v_g = 0.5095 gives P₄ ≈ 361 kPa, T₄ ≈ 139.9 °C, u₄ = u_g = 2549.6 kJ/kg.
 P (kPa)
 361 |           4
     |           |
 300 |  2 ------ 3      2-3: const P (piston rising)
     |  |               1-2, 3-4: const V
 100 |  1
     +------------------- v
      0.340    0.510

 T (C)
 139.9 |           4   (on sat. vapour line)
       |           |
 133.5 |  2 ------ 3   (wet region, T = Tsat at 300 kPa)
       |  |
  99.6 |  1
       +------------------- v
        0.340    0.510

Work (only in 2 → 3 at 300 kPa):
W = 300 × (0.6 − 0.4) = 60 kJ

Heat transfer (first law, ΔKE = ΔPE = 0):
Q = m (u₄ − u₁) + W = 1.178 × (2549.6 − 835.0) + 60 = 2019.3 + 60 = 2079 kJ

Answer: W = 60 kJ; Q ≈ 2079 kJ (added).

✓ Answer checked by Claude Opus 5.5 (high effort)

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