--- title: "Rigid vessel at critical state cooled to 2000 kPa" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 3 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:rigid-vessel-critical-state-2000kpa" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/rigid-vessel-critical-state-2000kpa site: BCT Engineering --- # Rigid vessel at critical state cooled to 2000 kPa Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances - Asked in: 2068 Bhadra, 6 marks - Video: [2.4 Analysis of Steam Cooling in a Rigid Vessel Determination of Dryness Fraction and Heat Rejection](https://www.youtube.com/watch?v=Ue-vV26M8mI) — Education Shop - 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-vessel-critical-state-2000kpa - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3 ## Question A rigid vessel having a volume of 0.02m³, initially contains water at its critical state. The vessel is cooled until its pressure drops to 2000kPa. Sketch the process on P-V and T-V diagrams and determine: a) The mass of H2O present in the vessel b) The quality at final state c) The mass of saturated liquid water and saturated water vapour at the final state. [Refer attached table for the properties of steam]. ## Short answer Mass m = V / v_c = 0.02 / 0.003106 = 6.44 kg. At 2000 kPa, v = v_c, quality x₂ = (0.003106 − 0.001177)/(0.09958 − 0.001177) = 0.0196; liquid mass = 6.31 kg, vapour mass = 0.126 kg. ## Answer **Given:** V = 0.02 m³ (rigid), initial state: critical point, cooled to P₂ = 2000 kPa. **Critical state of water:** v_c = 0.003106 m³/kg (P_c = 22.06 MPa, T_c = 373.9 °C). #### a) Mass m = V/v_c = 0.02/0.003106 = **6.44 kg** #### b) Quality at final state v₂ = v_c = 0.003106 (constant volume). At 2000 kPa: T_sat = 212.4 °C, v_f = 0.001177, v_g = 0.09958 m³/kg. Since v_f < v₂ < v_g, it is a wet mixture. x₂ = (0.003106 − 0.001177)/(0.09958 − 0.001177) = **0.0196** #### c) Masses of saturated liquid and vapour m_g = x₂ m = 0.0196 × 6.44 = **0.126 kg** m_f = m − m_g = 6.44 − 0.126 = **6.314 kg** ```text P| | CP (22.06 MPa) | | | | v = vc = const | | | 2 (2000 kPa) +------------- v T| | CP (373.9 C) | | | | v = vc = const | 2 (212.4 C) +------------- v ``` On both diagrams the process is a vertical line from the critical point down into the dome. **Answer: m = 6.44 kg; x₂ = 0.0196; m_f = 6.31 kg, m_g = 0.126 kg.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The mass in the vessel is found from: A. m = V/v_c = 0.02/0.003106 = 6.44 kg B. m = V × v_c = 6.2e-5 kg C. m = V/v_g at 2000 kPa D. m = P_c V/(R T_c) Answer: A. Rigid vessel at the critical point: use the critical specific volume. (Key point: Mass at critical state) 2. During cooling in the rigid vessel, the final specific volume is: A. Equal to v_g at 2000 kPa B. Equal to v_c, since mass and volume are fixed C. Equal to v_f at 2000 kPa D. Lower than v_c because cooling contracts it Answer: B. v2 = v1 = v_c = 0.003106 m³/kg. (Key point: Constant volume) 3. At 2000 kPa v_f = 0.001177 < 0.003106 < v_g = 0.09958, so the final state is: A. Superheated vapour B. Compressed liquid C. A wet saturated mixture at 212.4 °C with x ≈ 0.0196 D. Saturated vapour with x = 1 Answer: C. v lies between v_f and v_g, hence two-phase at T_sat. (Key point: Final phase) 4. With x2 = 0.0196, the masses are about: A. m_g = 6.31 kg, m_f = 0.126 kg B. m_g = m_f = 3.22 kg C. m_g = 0.0196 kg, m_f = 6.42 kg D. m_g = 0.126 kg, m_f = 6.31 kg Answer: D. m_g = x m = 0.0196 × 6.44; m_f = m - m_g. 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