--- title: "Piston with linear spring: water 100 kPa, x = 0.8" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 3 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:water-linear-spring-p100-x08-v001" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/water-linear-spring-p100-x08-v001 site: BCT Engineering --- # Piston with linear spring: water 100 kPa, x = 0.8 Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances - Asked in: 2068 Magh, 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-linear-spring-p100-x08-v001 - 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 P.9 contains water initially at P1 = 100 kPa, x1 = 0.8 and V1 = 0.01 m³. When the system is heated, it encounters a linear spring (k = 100 kN/m). At this state volume is 0.015 m³. The heating continues till its pressure is 200 kPa. If the diameter of the piston is 0.15 m, determine (a) the final temperature, and (b) the total work transfer. Also sketch the process on P-v diagram. [Refer attached table for properties of steam] [Figure P.9: cylinder containing water with a piston and a linear spring above it.] ## Short answer Mass is 0.00738 kg; final volume is 0.015312 m³. At 200 kPa, the steam is superheated with specific volume 2.075 m³/kg, giving final temperature about 627°C. Total work is 0.5 kJ (constant-pressure) + 0.047 kJ (linear spring) = 0.547 kJ. ## Answer **Given:** state 1: P₁ = 100 kPa, x₁ = 0.8, V₁ = 0.01 m³. The piston moves freely until it touches the spring at V₂ = 0.015 m³; then the spring is compressed until P₃ = 200 kPa. k = 100 kN/m; D = 0.15 m. **Step 1: Mass** At 100 kPa: v_f = 0.001043, v_g = 1.6940 m³/kg, so v_fg = 1.6930 m³/kg. v₁ = 0.001043 + 0.8 × 1.6930 = 1.3554 m³/kg m = V₁/v₁ = 0.01/1.3554 = 0.00738 kg **Step 2: Piston area** A = (π/4)(0.15)² = 0.017671 m² **Step 3: Process** - 1 → 2: constant P = 100 kPa from 0.01 to 0.015 m³ (piston free). - 2 → 3: the spring is compressed, and the pressure rises linearly with V. The pressure rise is ΔP = k Δx/A, with Δx = ΔV/A, so ΔP = k ΔV/A². ΔV = ΔP A²/k = (200 − 100) × (0.017671)² / 100 = 3.123 × 10⁻⁴ m³ V₃ = 0.015 + 0.000312 = 0.015312 m³ #### a) Final temperature v₃ = V₃/m = 0.015312/0.00738 = 2.075 m³/kg at P₃ = 200 kPa. At 200 kPa, v_g = 0.8857 m³/kg < 2.075, so the steam is superheated. From the superheated tables at 200 kPa: 500 °C → v = 1.7814; 600 °C → v = 2.0130; 700 °C → v = 2.2444 m³/kg. Interpolating between 600 and 700 °C gives **T₃ ≈ 627 °C** #### b) Total work W₁₂ = P₁ (V₂ − V₁) = 100 × (0.015 − 0.01) = 0.5 kJ W₂₃ = ½(100 + 200)(0.015312 − 0.015) = 0.0468 kJ **W = 0.5468 kJ ≈ 0.547 kJ** ```text P| 3 (200) | / | / 1-------2/ (1-2 const P = 100) +-------------------- V 0.01 0.015 0.0153 ``` **Answer: T₃ ≈ 627 °C; W ≈ 0.547 kJ.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The piston-spring process is: A. Constant pressure until the spring is touched at 0.015 m³, then pressure rises linearly with volume B. Constant volume then constant pressure C. Isothermal compression of the spring D. Linear pressure rise from the start Answer: A. Piston free first, spring compressed only after contact. (Key point: Process stages) 2. The pressure rise with the spring is ΔP = k ΔV/A² because: A. The spring force equals PV B. The spring force kΔx is spread over area A and Δx = ΔV/A C. Δx = ΔV × A D. The spring changes mass in the cylinder Answer: B. ΔP = kΔx/A with Δx = ΔV/A gives kΔV/A². (Key point: Spring relation) 3. At 200 kPa with v3 ≈ 2.075 m³/kg (v_g = 0.8857) the water is: A. A wet mixture at 120.2 °C B. Saturated vapour at 120.2 °C C. Superheated vapour, T ≈ 627 °C by interpolation D. Compressed liquid Answer: C. v3 > v_g so it is superheated; T is interpolated from the superheat table. (Key point: Final state phase) 4. The total work is the sum of: A. Only the 1-2 constant pressure work B. P3(V3 - V1) with final pressure C. Only the spring work D. P1(V2 - V1) for 1-2 plus the trapezoid area (average pressure × ΔV) for 2-3, ≈ 0.547 kJ Answer: D. 0.5 kJ + 0.0468 kJ ≈ 0.547 kJ. 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