--- title: "Water in piston cylinder with stops, 200 kPa" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 3 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:water-piston-stops-200kpa-50c" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/water-piston-stops-200kpa-50c site: BCT Engineering --- # Water in piston cylinder with stops, 200 kPa Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances - Asked in: 2075 Baisakh, 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/water-piston-stops-200kpa-50c - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3 ## Question A piston cylinder arrangement shown in figure below contains 2 kg of water initially at a pressure of 200 kPa and a temperature of 50°C. Heat is added until the piston reaches the upper stops where the total volume is 1.5 m³. It takes a pressure of 600 kPa to lift the piston. Sketch the process on P-v and T-v diagrams and determine the final temperature and the total work transfer. [Figure: cylinder with H2O, piston resting at the bottom of the stops region, heat Q added from below.] ## Short answer Heating water initially at 200 kPa, 50°C: constant-volume until 600 kPa, then constant pressure at 600 kPa to 1.5 m³. Using steam tables, final superheated temperature ≈ 704 °C, total work = 898.8 kJ. ## Answer **Given:** m = 2 kg; State 1: 200 kPa, 50 °C; the piston needs 600 kPa to lift; heat is added until the piston reaches the upper stops, V₂ = 1.5 m³. **State 1:** T_sat at 200 kPa = 120.2 °C, so water at 50 °C is compressed liquid. Use v₁ ≈ v_f(50 °C) = 0.001012 m³/kg (steam tables), so V₁ = 2 × 0.001012 = 0.00202 m³. **Process:** - 1 → a: the piston rests on the bottom stops (200 kPa < 600 kPa), so heating is at constant volume until P rises to 600 kPa. - a → 2: the piston lifts and the pressure stays at 600 kPa (constant P) until it reaches the upper stops with V₂ = 1.5 m³. ```text P (kPa) 600 | a -------------------------- 2 | | | | 1-a: const v (piston on stops) 200 | 1 +---------------------------------- v v1 = 0.00101 v2 = 0.75 T (C) 703.6| 2 | / 158.8| f ----------------- g / a-f-g-2: 600 kPa isobar | / 50 | 1,a +---------------------------------- v ``` On the T-v diagram, 1 and a almost coincide, because heating a compressed liquid at constant volume raises its pressure sharply with almost no rise in temperature. Along the 600 kPa isobar the liquid warms to 158.8 °C (f), evaporates at constant temperature (f to g), and is then superheated to state 2. **State 2:** v₂ = V₂/m = 1.5/2 = 0.75 m³/kg at P = 600 kPa. At 600 kPa, v_g = 0.3157 m³/kg < 0.75, so the steam is superheated. From the superheated table at 600 kPa: - 700 °C: v = 0.7472 m³/kg - 800 °C: v = 0.8246 m³/kg T₂ = 700 + 100 × (0.75 − 0.7472)/(0.8246 − 0.7472) = **≈ 703.6 °C** **Work:** work occurs only in a → 2 (constant P); the constant-volume heating 1 → a gives no work. W = P (V₂ − V₁) = 600 × (1.5 − 0.00202) = **898.8 kJ** **Answer: Final temperature ≈ 704 °C; total work transfer ≈ 899 kJ.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is water at 200 kPa, 50 C taken as compressed liquid with v about vf(50 C)? A. 50 C is below Tsat = 120.2 C at 200 kPa B. 50 C is above Tsat at 200 kPa C. Its quality is 0.5 D. The piston is resting on the stops Answer: A. Since T is below Tsat for its pressure, it is compressed liquid and v is approximated by vf at 50 C. (Key point: State 1 phase) 2. What happens in 1 to a while the piston sits on the bottom stops? A. Constant-pressure heating at 200 kPa B. Constant-volume heating until P reaches 600 kPa C. Isothermal expansion D. Adiabatic compression Answer: B. The piston needs 600 kPa to lift, so until then the volume is fixed and pressure rises. (Key point: First heating stage) 3. Why is state 2 (600 kPa, v = 0.75 m3/kg) superheated? A. v = 0.75 is less than vf B. Pressure is above the critical pressure C. v = 0.75 exceeds vg = 0.3157 at 600 kPa D. Quality there is 0.75 Answer: C. When v is greater than vg at that pressure the steam is superheated; interpolation gives about 704 C. (Key point: State 2 phase) 4. Where is work done and what is its value? A. Over the whole process using 200 kPa: about 300 kJ B. Only in 1 to a: about 899 kJ C. Zero, because the piston stops fix the volume D. Only in a to 2 at 600 kPa: about 899 kJ Answer: D. The constant-volume stage gives no work; W = 600 x (1.5 - 0.00202) = 898.8 kJ. 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