---
title: "Water with linear spring cooled from 4 MPa"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 3
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:water-linear-spring-4mpa-500c"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/water-linear-spring-4mpa-500c
site: BCT Engineering
---

# Water with linear spring cooled from 4 MPa

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

- Asked in: 2070 Bhadra, 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-4mpa-500c
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3

## Question

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.]

## 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.

## Answer

**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).

```text
 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.**

## Check yourself

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

1. m =
   A. V1 x v1
   B. V1/v1 = 0.1/0.08644 = 1.157 kg
   C. 0.1 kg
   D. v1/V1
   Answer: B. v1 from superheated table 4 MPa, 500 C. (Key point: Mass)

2. Spring gives:
   A. P constant
   B. PV constant
   C. P = 300 + cV, c = 37000 kPa/m3
   D. P = cV
   Answer: C. Linear in V. (Key point: Linear P-V)

3. At 1000 kPa V2 = 0.01892; phase is:
   A. Superheated
   B. Compressed liquid
   C. Critical
   D. Wet, T = Tsat = 179.9 C
   Answer: D. v2 = 0.01635 between vf and vg. (Key point: Final state)

4. W =
   A. Area under line: -202.7 kJ
   B. +202.7 kJ
   C. -4500 kJ
   D. 0
   Answer: A. Negative because volume shrinks. (Key point: Work)

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