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Water in piston cylinder with stops, 200 kPa

Short answer

Heating water initially at 200 kPa, 50°C: constant-volume until 600 kPa, then constant pressure at 600 kPa to 1.5 m³. Using steam tables, final superheated temperature ≈ 704 °C, total work = 898.8 kJ.

A piston cylinder arrangement shown in figure below contains 2 kg of water initially at a pressure of 200 kPa and a temperature of 50°C. Heat is added until the piston reaches the upper stops where the total volume is 1.5 m³. It takes a pressure of 600 kPa to lift the piston. Sketch the process on P-v and T-v diagrams and determine the final temperature and the total work transfer. [Figure: cylinder with H2O, piston resting at the bottom of the stops region, heat Q added from below.]

Given: m = 2 kg; State 1: 200 kPa, 50 °C; the piston needs 600 kPa to lift; heat is added until the piston reaches the upper stops, V₂ = 1.5 m³.

State 1: T_sat at 200 kPa = 120.2 °C, so water at 50 °C is compressed liquid. Use v₁ ≈ v_f(50 °C) = 0.001012 m³/kg (steam tables), so V₁ = 2 × 0.001012 = 0.00202 m³.

Process:

  • 1 → a: the piston rests on the bottom stops (200 kPa < 600 kPa), so heating is at constant volume until P rises to 600 kPa.
  • a → 2: the piston lifts and the pressure stays at 600 kPa (constant P) until it reaches the upper stops with V₂ = 1.5 m³.
 P (kPa)
 600 |  a -------------------------- 2
     |  |
     |  |  1-a: const v (piston on stops)
 200 |  1
     +---------------------------------- v
      v1 = 0.00101                v2 = 0.75

 T (C)
 703.6|                               2
      |                             /
 158.8|      f ----------------- g /    a-f-g-2: 600 kPa isobar
      |     /
   50 |  1,a
      +---------------------------------- v

On the T-v diagram, 1 and a almost coincide, because heating a compressed liquid at constant volume raises its pressure sharply with almost no rise in temperature. Along the 600 kPa isobar the liquid warms to 158.8 °C (f), evaporates at constant temperature (f to g), and is then superheated to state 2.

State 2: v₂ = V₂/m = 1.5/2 = 0.75 m³/kg at P = 600 kPa. At 600 kPa, v_g = 0.3157 m³/kg < 0.75, so the steam is superheated. From the superheated table at 600 kPa:

  • 700 °C: v = 0.7472 m³/kg
  • 800 °C: v = 0.8246 m³/kg

T₂ = 700 + 100 × (0.75 − 0.7472)/(0.8246 − 0.7472) = ≈ 703.6 °C

Work: work occurs only in a → 2 (constant P); the constant-volume heating 1 → a gives no work. W = P (V₂ − V₁) = 600 × (1.5 − 0.00202) = 898.8 kJ

Answer: Final temperature ≈ 704 °C; total work transfer ≈ 899 kJ.

✓ Answer checked by Claude Opus 5.5 (high effort)

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