--- title: "Rigid container of steam 200 kPa, 350°C cooled" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 3 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:rigid-container-200kpa-350c-cooled-90c" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/rigid-container-200kpa-350c-cooled-90c site: BCT Engineering --- # Rigid container of steam 200 kPa, 350°C cooled Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances - Asked in: 2067 Mangsir, 7 marks - Video: [Thermodynamics: First Law Rigid Tank. Steam Tables](https://www.youtube.com/watch?v=XQ1nlXe5ByI) — THE CLASS OF TORCHIA - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/rigid-container-200kpa-350c-cooled-90c - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3 ## Question 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] ## 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). ## Answer **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) ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the rigid container, v during cooling is: A. Constant at 1.4330 m³/kg, so the process is a vertical line on P-v and T-v diagrams B. Decreasing steadily C. Increasing because pressure drops D. Equal to v_g at 90 °C Answer: A. Fixed V and m give constant v. (Key point: Constant v) 2. Condensation begins when: A. The temperature reaches 100 °C exactly B. v equals v_g(T) on the saturated vapour line, at about 104.7 °C C. v equals v_f D. The pressure reaches 200 kPa Answer: B. Cooling at constant v hits the saturated vapour line where v_g(T) = 1.4330. (Key point: Start of condensation) 3. The final pressure at 90 °C is: A. 200 kPa, unchanged B. Found by ideal gas P1T2/T1 C. P_sat(90 °C) = 70.2 kPa, because the state is a wet mixture D. 101.3 kPa Answer: C. Since v_f < v < v_g at 90 °C, pressure equals saturation pressure. (Key point: Final pressure) 4. The liquid mass fraction is: A. x2 = 0.607 B. 0.001036/2.3591 C. 0.5 D. 1 - x2 = 0.393, with x2 = (v - v_f)/(v_g - v_f) = 0.607 Answer: D. Liquid fraction is 1 minus the quality. 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