Thermo Properites Table Hw Question

In summary, the conversation discusses the temperature and quality of R134a in a compressed fluid can at 120psia and 0.38ft3/lbm, as well as the change in pressure after half of the mass is vented off and the can is allowed to exchange heat with the surroundings at 100°F. The table provided by DuPont is used to find the temperature and specific volume of the substance, and the ideal gas law is used to solve for pressure. The final pressure is found to be 70psia at a temperature of 100°F and a specific volume of 0.76ft^3/lb.
  • #1
Jstuff
15
0

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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  • #2
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.
 
  • #3
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
 
  • #4
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
 
  • #5
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
 
  • #6
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
 
  • #7
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
 
  • #8
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
 
  • #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
 

1. What is a Thermo Properties Table?

A Thermo Properties Table is a chart or database that lists the thermodynamic properties of various substances under different conditions, such as temperature and pressure. These properties include specific heat, enthalpy, entropy, and other important parameters that are used in thermodynamic calculations.

2. How is a Thermo Properties Table used?

A Thermo Properties Table is used to find the specific thermodynamic properties of a substance at a given set of conditions. It is often used in engineering and scientific calculations, such as in the design of heat exchangers, power plants, and refrigeration systems.

3. What types of substances are typically included in a Thermo Properties Table?

A Thermo Properties Table can include a wide range of substances, such as gases, liquids, and solids. These substances can be pure elements, compounds, or mixtures. Some tables may also include properties for solutions and mixtures of gases.

4. How accurate are the values in a Thermo Properties Table?

The accuracy of the values in a Thermo Properties Table depends on the source of the data and the methods used to measure or calculate the properties. In general, values from established sources, such as the National Institute of Standards and Technology (NIST), are considered highly accurate.

5. Are Thermo Properties Tables available for all substances?

No, Thermo Properties Tables may not be available for all substances. The availability of tables depends on the amount of research and data available for a particular substance. Some substances may have limited data or may not have been studied extensively, resulting in a lack of reliable thermodynamic data.

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