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- 2067 Mangsir · 7 marks
Rigid container of steam 200 kPa, 350°C cooled
Short answer
Phase change starts at about 104.7°C, when steam becomes saturated at constant specific volume 1.433 m³/kg. Final pressure equals saturation pressure at 90°C, which is 70.2 kPa. The final liquid mass fraction is 39.3% (quality 0.607).
A rigid container with a volume of 0.170 m³ is initially filled with steam at 200 kPa and 350°C. It is cooled to 90°C. a) At what temperature does a phase change starts to occur? b) What is the final pressure? c) What mass fraction of the water is liquid in the final state? Also sketch the process on P-v and T-v diagrams. [Refer the attached table for properties of steam]
Given: V = 0.170 m³ (rigid, so v is constant); state 1: 200 kPa, 350 °C; cooled to 90 °C.
State 1 (superheated table): v₁ = 1.4330 m³/kg; mass m = V/v₁ = 0.170/1.4330 = 0.1186 kg.
a) Temperature at which condensation starts
The phase change begins when v = v_g(T). Saturated table: v_g = 1.6719 m³/kg at 100 °C and 1.4185 m³/kg at 105 °C. Interpolating for v_g = 1.4330:
T = 100 + 5 × (1.6719 − 1.4330)/(1.6719 − 1.4185) = ≈ 104.7 °C
b) Final pressure
At 90 °C: v_f = 0.001036, v_g = 2.3591 m³/kg, so v₁ = 1.4330 lies between them: the final state is a wet mixture.
P₂ = P_sat(90 °C) = 70.2 kPa
c) Liquid mass fraction
x₂ = (1.4330 − 0.001036)/(2.3591 − 0.001036) = 0.607
Liquid fraction = 1 − x₂ = 0.393 (m_f = 0.0466 kg)
P (kPa) T (°C)
200 | 1 350 | 1
| | | |
| | v = const | | v = const
| | (enters dome) | | (enters dome at 104.7)
70.2| 2 90 | 2
+---------- v +---------- v
v = 1.433 v = 1.433
Since v is constant, process 1-2 is a vertical line on both diagrams: on P-v it falls from 200 kPa to 70.2 kPa, entering the saturation dome at about 105 °C; on T-v it falls from 350 °C into the dome and ends at 90 °C.
Answer: (a) 104.7 °C (b) 70.2 kPa (c) 39.3 % liquid by mass.
✓ Answer checked by Claude Opus 5.5 (high effort)
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