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Piston with linear spring: water 100 kPa, x = 0.8

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.

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.]

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

 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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