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Water with linear spring cooled from 4 MPa

Short answer

With a linear spring P=300+37000V, mass m = V1/v1 = 0.1/0.08644 = 1.157 kg. At P2=1000 kPa, V2=0.01892 m³, v2=0.01635 m³/kg, wet at T2=179.9°C. Work W = 0.5(P1+P2)(V2-V1) = -202.7 kJ.

A piston cylinder device with a linear spring initially contains water at a pressure of 4 MPa and 500°C with an initial volume being 0.1 m³, as shown in figure. The system now cools until the pressure reaches 1000 kPa. If the piston is at the bottom, the system pressure is 300 kPa. Sketch the process on P-v diagram and determine: a) The mass of water b) The final temperature and volume and c) The total work transfer [Refer the attached tables for properties of steam] [Figure: cylinder containing H2O with a piston held by a linear spring above it.]

Given: state 1: 4 MPa (4000 kPa), 500 °C, V₁ = 0.1 m³; cooled until P₂ = 1000 kPa; with the piston at the bottom (V = 0) the pressure is 300 kPa; the spring is linear, so P is linear in V.

a) Mass

Superheated steam at 4 MPa, 500 °C: v₁ = 0.08644 m³/kg
m = V₁/v₁ = 0.1/0.08644 = 1.157 kg

b) Final temperature and volume

Linear relation: P = 300 + c V, with c = (4000 − 300)/0.1 = 37 000 kPa/m³.
At P₂ = 1000 kPa: V₂ = (1000 − 300)/37 000 = 0.01892 m³
v₂ = V₂/m = 0.01892/1.157 = 0.01635 m³/kg

At 1000 kPa: v_f = 0.001127, v_g = 0.1943 m³/kg. Since v_f < v₂ < v_g, it is a wet mixture, so
T₂ = T_sat(1 MPa) = 179.9 °C, x₂ = (0.01635 − 0.001127)/(0.1943 − 0.001127) = 0.079

c) Total work

Work is the area under the straight line on P-V:
W = ½ (P₁ + P₂)(V₂ − V₁) = ½ (4000 + 1000)(0.01892 − 0.1) = −202.7 kJ (negative: work is done on the water by the surroundings as the volume shrinks).

 P|1 (4000)
  |  \
  |   \
  |    2 (1000)
  |  300 at V=0 (extension)
  +------------- V
   0.019   0.1

Answer: m = 1.157 kg; T₂ = 179.9 °C, V₂ = 0.0189 m³; W = −202.7 kJ.

✓ Answer checked by Claude Opus 5.5 (high effort)

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