- 2071 Bhadra · 8 marks
Piston cylinder with stops: 2 kg H2O to 250°C
Short answer
Initial constant-volume heating until pressure reaches 400 kPa, then constant-pressure expansion to 250°C. Work is 400 kPa × 2 kg × (0.5952−0.025) m³/kg = 456.2 kJ.
A piston cylinder device shown in figure below contains 2 kg of H2O with an initial temperature and volume of 80°C and 0.05 m³ respectively. It requires a pressure of 400 kPa to lift the piston from the stops. The system is heated until its temperature reaches 250°C. Sketch the process on P-v and T-v diagrams and determine the total work transfer. [Refer attached table for the properties of steam] [Figure: cylinder with H2O under a piston resting on stops, heat Q added.]
Given: m = 2 kg; state 1: 80 °C, V₁ = 0.05 m³; 400 kPa is needed to lift the piston; heated to 250 °C.
State 1: v₁ = 0.05/2 = 0.025 m³/kg. At 80 °C: v_f = 0.001029, v_g = 3.405 m³/kg, P_sat = 47.4 kPa. Since v_f < v₁ < v_g, it is a two-phase mixture with x₁ = (0.025 − 0.001029)/(3.405 − 0.001029) = 0.007. P₁ = 47.4 kPa < 400 kPa, so the piston rests on the stops.
Process:
- 1 → 2: constant volume (v = 0.025) until P = 400 kPa. At 400 kPa, v_g = 0.4624 > 0.025, so it is still wet (T₂ = T_sat = 143.6 °C). No work.
- 2 → 3: the piston lifts and the pressure stays at 400 kPa until T = 250 °C. Superheated table at 400 kPa, 250 °C: v₃ = 0.5952 m³/kg.
P (kPa)
400 | 2 ----------------------- 3
| |
| | 1-2: const v (piston on stops)
47.4 | 1
+----------------------------- v
0.025 0.5952
T (C)
250 | 3
| /
143.6 | 2 ------------------ g / 2-g-3: 400 kPa isobar
| |
80 | 1
+----------------------------- v
On the T-v diagram, 1-2 is a vertical line inside the dome; along the 400 kPa isobar the mixture evaporates at 143.6 °C (2 to g) and is then superheated to 250 °C (g to 3).
Work: only in 2 → 3:
W = P m (v₃ − v₂) = 400 × 2 × (0.5952 − 0.025) = 456.2 kJ
Answer: Total work transfer ≈ 456 kJ.
✓ Answer checked by Claude Opus 5.5 (high effort)
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