---
title: "Piston cylinder: water 125 kPa, 50% quality, 800°C"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 3
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steam-piston-125kpa-quality-50-800c"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-piston-125kpa-quality-50-800c
site: BCT Engineering
---

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

Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances

- Asked in: 2073 Magh, 8 marks
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-piston-125kpa-quality-50-800c
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3

## Question

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.]

## 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.

## Answer

**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.

```text
 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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the mass of H2O found?
   A. m = V1/vg
   B. m = V1/v1 with v1 = vf + 0.5 vfg at 125 kPa = 0.688 m3/kg
   C. m = V1 x v1
   D. m = V1/vf
   Answer: B. With x = 0.5 and V1 = A x 0.3, m is about 0.0436 kg for A = 0.1 m2. (Key point: Mass)

2. Why is 1 to a at constant volume?
   A. The piston is held by friction
   B. The gas is ideal
   C. 125 kPa is below the 600 kPa needed to lift the piston
   D. Heat is removed
   Answer: C. The piston stays on the stops until pressure reaches 600 kPa. (Key point: Stage 1)

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