Thermodynamics of an Insulated Box: Water-Steam Question

  • Thread starter Thread starter Master1022
  • Start date Start date
  • Tags Tags
    Box Thermodynamics
Click For Summary
SUMMARY

The discussion focuses on a thermodynamics problem involving a sealed, thermally insulated tank with a volume of 2 m³ and a safe working pressure of 4 bar. The scenario includes an initial state where 10% of the tank is filled with water at 20 degrees Celsius, with the remainder as water vapor. The participant outlines a method for calculating the heat that can be added without exceeding the pressure limit, but encounters an issue with a calculated dryness fraction exceeding 1. The suggestion to utilize steam tables for specific volumes of liquid and vapor is emphasized as a critical step in resolving the issue.

PREREQUISITES
  • Understanding of the 1st Law of Thermodynamics
  • Familiarity with steam tables for water and vapor properties
  • Knowledge of the ideal gas law
  • Ability to calculate specific internal energy
NEXT STEPS
  • Study the application of steam tables in thermodynamic calculations
  • Learn about the concept of dryness fraction in phase change processes
  • Review the ideal gas law and its implications in thermodynamic systems
  • Explore the calculation of specific internal energy for different phases of water
USEFUL FOR

Engineering students, thermodynamics practitioners, and anyone involved in the analysis of phase changes in thermodynamic systems will benefit from this discussion.

Master1022
Messages
590
Reaction score
116

Homework Statement


A sealed, thermally insulated tank of volume 2 m^3 has a safe working pressure of 4 bar. At 20 degrees celsius, 10% of the volume is occupied by water, the remainder by water vapour. Calculate how much heat can be added without exceeding the safe working pressure.

This question is from "Basic Engineering Thermodynamics' - P. B. Whalley - p. 63, q6.6

Homework Equations


1st law of Thermodynamics: Q - W = \Delta U
Other equations in the photo

The Attempt at a Solution


The attached photo is my attempt (apologies, was trying to find out how to make it just an attachment). Hopefully, it is legible, was initially working in draft...
IMG_6178.JPG


My method outline:
1. work out the initial masses of liquid and vapour
2. work out the dryness fraction
3. work out the initial (specific) internal energy
4. using ideal gas law to calculate specific volume that would give pressure of 4 bar
5. calculate final dryness fraction
6. calculate final (specific) internal energy
7. find the difference between the U values (after including mass in the calculation)

However, in step 5, I am getting a dryness fraction > 1, which doesn't make sense. I would appreciate any help regarding any flaws in my method or working.

Thanks in advance
 

Attachments

  • IMG_6178.JPG
    IMG_6178.JPG
    48.7 KB · Views: 726
Physics news on Phys.org
In the final state, you still have liquid and vapor present. Why didn't you use the steam tables to get the specific volumes of the liquid and vapor in the final state?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
Replies
5
Views
3K
Replies
49
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
16
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
2K
  • · Replies 44 ·
2
Replies
44
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
6K