Thermo Properites Table Hw Question

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SUMMARY

The discussion revolves around a thermodynamics homework problem involving R134a at 120 psia with a specific volume of 0.38 ft³/lbm. Participants clarify that the initial temperature is approximately 90.5°F, not -40°F, and that the quality of the substance can be determined using the formula V1=Vf+x(Vgf). After venting half the mass, the new specific volume doubles, leading to further calculations using the ideal gas law, Pv=RT. The importance of using the correct thermodynamic tables for R134a properties is emphasized, particularly for determining pressure at thermal equilibrium.

PREREQUISITES
  • Understanding of thermodynamic properties and tables for refrigerants like R134a.
  • Familiarity with the ideal gas law, specifically Pv=RT.
  • Knowledge of quality calculations in thermodynamics.
  • Ability to convert units and interpret specific volumes in thermodynamic contexts.
NEXT STEPS
  • Study the properties of R134a using the DuPont thermodynamic properties table.
  • Learn how to calculate quality using the formula V1=Vf+x(Vgf).
  • Practice unit conversion techniques specific to thermodynamics.
  • Explore the implications of venting mass on specific volume and pressure calculations.
USEFUL FOR

Students studying thermodynamics, particularly those working with refrigerants, engineers involved in HVAC systems, and anyone seeking to understand the behavior of R134a under varying conditions.

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Homework Statement


R134a rests in a compressed fluid can at 120psia, and a specific volume of 0.38ft3/lbm. What is its temperature and quality? Half of the mass is now vented off to the atmosphere. The can is allowed to exchange heat with the surroundings at 100°F until they are in thermal equilibrium. What is the pressure in the can now?

Homework Equations


Properties Tables in the back of my thermo books.
Pv=RT

The Attempt at a Solution


So I looked at the properties table in the back of my book and saw that the temperature of R134a in a compressed fluid is -40F. However, this is incorrect. I also thought that the quality would be 0 since quality only applies to a substance that is saturated i.e. in between the liquid and gaseous state. However, this is also wrong. For the second part I have no idea how to go forward. Please help. I'm trying to understand this problem not just get the answer.
Thanks[/B]
 
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Is it possible that the solution is around 93.2F, I have a table of R134a in my book and it's in metric, but when I converted to close values, it shows that temperature at a saturated vapor state. It seems my book uses data from this paper. I could send a screenshot of the table if the result matches what you need.
 
See the following reference from the DuPont company, which gives tables of R134a thermodynamic properties: http://www2.dupont.com/Refrigerants/en_US/assets/downloads/h47751_hfc134a_thermo_prop_eng.pdf

The table shows that at 120 psia, the specific volume of the saturated liquid is 0.0136 ft^3/lb, and the specific volume of the saturated vapor is about 0.397 ft^3/lb (what does this tell you about the quality?). Check the table to see what temperature this occurs at. Gezibash was pretty close, but a little high.

For the second part, if you vent half the mass, what is the new specific volume?

Chet
 
Chet
I found on the table where it states the temperature, 90.5F, at the specified Pressure and volume. I am confused though because in the problem it says it is a compressed liquid however the tables are for a vapor. Why then do I look in the vapor table?
Thanks
 
Jstuff said:
Chet
I found on the table where it states the temperature, 90.5F, at the specified Pressure and volume. I am confused though because in the problem it says it is a compressed liquid however the tables are for a vapor. Why then do I look in the vapor table?
Thanks
The problem statement says "compressed fluid", not compressed liquid. The term fluid applies to both liquids and gases. Also, the tables are not for a vapor. They are for any combination of saturated liquid and vapor. The statement asks for the quality, which automatically implies that both saturated liquid and vapor are present. You can tell from the specific volume that most of it is vapor. You can use the data in the table to precisely calculate the quality.

Chet
 
Hey Chett, so I found the quality by using a formula I found in the book. V1=Vf+x(Vgf) where x is the formula.
And now I am working on the second part of the problem. So since the mass is half the specific volume will double since v=V/m.
Using that I an now using the ideal gas law, Pv=RT, and I can solve for pressure knowing the temperature and specific volume. However, I am having trouble with my units and what R should be. Any explanation on a good way to keep the units straight I always seem to have problems with them in thermodynamics. It never happened in Physics.
Thanks
 
Jstuff said:
Hey Chett, so I found the quality by using a formula I found in the book. V1=Vf+x(Vgf) where x is the formula.
And now I am working on the second part of the problem. So since the mass is half the specific volume will double since v=V/m.
Using that I an now using the ideal gas law, Pv=RT, and I can solve for pressure knowing the temperature and specific volume. However, I am having trouble with my units and what R should be. Any explanation on a good way to keep the units straight I always seem to have problems with them in thermodynamics. It never happened in Physics.
Thanks
You shouldn't have a problem if you remember to carry the units along in the calculations, and to properly cancel units. You learned how to do that in Physics, correct? As far as Pv=RT is concerned, v in this equation is the volume per mole, and R is the universal gas constant.

You don't need to use the ideal gas law to solve this problem. You can use the tables in the reference I sent you, and find the pressure at which the temperature is 100F and the specific volume is 0.76. You can use the ideal gas law to help you with a starting point for where to look in the tables.

Chet
 
Sorry for all the questions Chett, but at 100F the volume is different. I don't see any other place where it says 100F. Can you further explain how to find the answer using the table.
Thanks
 
Table 2, page 26.

Chet
 
  • #10
Chet, the table on page 26. At 100F none of the volumes are at .76ft^3/lb which is the new specific volume once half the mass is gone.
 
  • #11
At 100F and 70 psia, what does the Table 2 give for the specific volume?

Chet
 
  • #12
Aww okay in your previous post you wrote page 26 when you meant to write 24.
Thanks for all your help Chet!
 
  • #13
Jstuff said:
Aww okay in your previous post you wrote page 26 when you meant to write 24.
Thanks for all your help Chet!
I guess on my computer, it comes out to be on page 26. That's weird.

Glad to be of help.

Chet
 

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