--- title: "Frictionless piston 20 kg with H2O, five parts" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 3 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:frictionless-piston-20kg-h2o" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/frictionless-piston-20kg-h2o site: BCT Engineering --- # Frictionless piston 20 kg with H2O, five parts Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances - Asked in: 2074 Bhadra, 8 marks - 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/frictionless-piston-20kg-h2o - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3 ## Question The frictionless piston shown in figure below has a mass of 20 kg and a cross sectional area of 78.48 cm². Heat is added until the temperature reaches 400°C. If the quality of the H2O at the initial state is 0.2, determine: i) The initial pressure ii) The mass of H2O iii) The quality of the system when the piston hits the tops iv) The final pressure and v) The total work transfer [Take Patm = 100 kPa, g = 9.81 m/s²] [Figure: cylinder with H2O closed by a frictionless piston; dimensions marked 50 mm (height of H2O space below the piston) and 30 mm (gap between the piston and the top stops); heat Q added from below.] ## Short answer Initial pressure = 125 kPa (atmospheric plus 20 kg piston). Mass = 1.42×10⁻³ kg from x=0.2 and initial volume. At stops, quality = 0.320. Final pressure at 400°C ≈ 698 kPa. Work = 0.0294 kJ. ## Answer **Given:** m_p = 20 kg, A = 78.48 cm² = 7.848 × 10⁻³ m², P_atm = 100 kPa, g = 9.81 m/s²; x₁ = 0.2; the H₂O space is 50 mm high initially; the piston hits the top stops after a further 30 mm; final T = 400 °C. **Process:** 1 → 2: heating at constant pressure (the piston floats) until it touches the stops; 2 → 3: constant-volume heating to 400 °C. #### i) Initial pressure P₁ = P_atm + m_p g/A = 100 + (20 × 9.81)/(7.848 × 10⁻³ × 1000) = 100 + 25.0 = **125 kPa** #### ii) Mass of H₂O At 125 kPa (steam tables): v_f = 0.001048, v_g = 1.3749 m³/kg, v_fg = 1.3739 m³/kg. v₁ = v_f + x v_fg = 0.001048 + 0.2 × 1.3739 = 0.27582 m³/kg V₁ = A × 0.050 = 7.848 × 10⁻³ × 0.05 = 3.924 × 10⁻⁴ m³ m = V₁/v₁ = 3.924 × 10⁻⁴ / 0.27582 = **1.423 × 10⁻³ kg (about 1.42 g)** #### iii) Quality when the piston hits the stops V₂ = A × (0.05 + 0.03) = 6.278 × 10⁻⁴ m³ v₂ = V₂/m = 0.4413 m³/kg x₂ = (v₂ − v_f)/v_fg = (0.4413 − 0.001048)/1.3739 = **0.320** #### iv) Final pressure From 2 → 3 at constant volume, v₃ = 0.4413 m³/kg at T₃ = 400 °C (superheated). From the superheated tables at 400 °C: v = 0.5137 m³/kg at 600 kPa and v = 0.4398 m³/kg at 700 kPa. Interpolating for v = 0.4413 gives P ≈ 698 kPa. **P₃ ≈ 0.70 MPa (about 698 kPa)** #### v) Total work Work is done only in 1 → 2 at constant P = 125 kPa: W = P (V₂ − V₁) = 125 × (6.278 × 10⁻⁴ − 3.924 × 10⁻⁴) = **0.0294 kJ (≈ 29.4 J)** ```text P| | 3 (400 C, ≈ 698 kPa) | | | | 2-3: const v (piston on stops) | 1------2 1-2: const P = 125 kPa +------------- v ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the initial pressure of the water found? A. P1 = Patm + mp g/A = 125 kPa B. P1 = mp g/A only = 25 kPa C. P1 = Patm - mp g/A = 75 kPa D. P1 = Patm = 100 kPa Answer: A. The piston weight adds to atmospheric pressure: 100 + 25 = 125 kPa. (Key point: Initial pressure) 2. How is the mass of water obtained? A. m = V1 x vg B. m = V1/v1, where v1 = vf + x1 vfg at 125 kPa C. m = V1/vf D. m = V1/vg Answer: B. The initial specific volume is that of a wet mixture with x = 0.2, giving m about 1.42 g. (Key point: Mass of H2O) 3. What are the process stages? A. Constant-volume heating first, then constant-pressure B. Isothermal expansion then adiabatic C. Constant-pressure heating until the piston hits the stops, then constant-volume heating to 400 C D. Constant pressure throughout to 400 C Answer: C. The piston floats at 125 kPa until it reaches the stops, after which volume is fixed and pressure rises. (Key point: Process stages) 4. How is the quality when the piston hits the stops found? A. x2 is unchanged at 0.2 B. x2 = 1 since it hits the stops C. x2 = v2/vg at 700 kPa D. x2 = (v2 - vf)/vfg with v2 = V2/m, giving about 0.32 Answer: D. Volume increases at constant mass, so v2 = 0.4413 and x rises to 0.32. (Key point: Quality at stops) 5. Where is work done and how much? A. Only in 1-2 at 125 kPa: about 29.4 J B. Only in 2-3: about 0.7 kJ C. In both stages: about 100 kJ D. None, since the piston is frictionless Answer: A. W = P (V2 - V1) = 125 x 2.354e-4 kJ, about 0.0294 kJ. 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