Skip to main content

Piston cylinder: water 125 kPa, 50% quality, 800°C

Short answer

Assuming piston area 0.1 m², initial specific volume 0.688 m³/kg gives mass 0.0436 kg. Heating first at constant volume to 600 kPa, then at constant pressure to 800°C. Work is 3.57 kJ.

A piston cylinder device shown in figure below contains water initially at a pressure of 125 kPa with a quality of 50%. Heat is added to the system until it reaches to a final temperature of 800°C. It takes a pressure of 600kPa to lift the piston from the stops. Sketch the process on P-v and T-v diagrams and determine: a) the mass of H2O in the system, and b) the total work transfer [Figure: cylinder with H2O under a piston resting on stops; dimensions 0.3 m (height of water space) and 0.07 m (piston stop gap); heat Q added.]

Given: P₁ = 125 kPa, x₁ = 0.5; the piston rests on stops, needs 600 kPa to lift; the water space is 0.3 m high, the stop gap is 0.07 m; final T = 800 °C.

Assumption: The piston area is not given in the text. Take A = 0.1 m² to get numbers; every quantity below (m and W) scales in proportion to A.

Steam tables at 125 kPa: v_f = 0.001048, v_g = 1.3749 m³/kg.
v₁ = 0.001048 + 0.5 (1.3739) = 0.6880 m³/kg
V₁ = A × 0.3 = 0.03 m³

a) Mass of H₂O

m = V₁/v₁ = 0.03/0.6880 = 0.0436 kg (= 0.436 A kg)

Process

  1. 1 → a: piston on the stops (125 < 600 kPa), constant volume heating. The mixture becomes saturated vapour (v_g = 0.688) at about 262 kPa, then superheated; P rises to 600 kPa. At 600 kPa and v = 0.688, T_a ≈ 623 °C.
  2. a → 3: piston lifts, constant P = 600 kPa until T = 800 °C.

Check on the upper stop: at 600 kPa, 800 °C, v₃ = 0.8246 m³/kg, so V₃ = 0.0436 × 0.8246 = 0.03596 m³. The stop is at V = A × 0.37 = 0.037 m³. Since 0.03596 < 0.037, the piston does not reach the upper stops, so the final state is 600 kPa, 800 °C.

 P|              a-------3
  |             /|
  |            / |
  1___________/  |   (1-a const v)
  +---------------------- v

b) Total work

Work only in a → 3:
W = P (V₃ − V₁) = 600 × (0.03596 − 0.03) = 3.57 kJ

Answer: m ≈ 0.0436 kg and W ≈ 3.6 kJ for A = 0.1 m² (general: m = 0.436 A kg, W = 35.7 A kJ with A in m²).

Related questions

All Fundamental of Thermodynamics & Heat Transfer chapter 3 questions →