Thermodynamics - Saturated Vapor Quality Question

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Homework Help Overview

The discussion revolves around calculating the quality of saturated vapor in a rigid vessel containing both liquid water and water vapor at a specified pressure. The problem involves thermodynamics concepts, specifically relating to the properties of saturated liquids and vapors.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss two methods for calculating quality: one based on mass calculations and another using specific volumes. There is confusion regarding the validity of these methods and the results they yield.

Discussion Status

Some participants have provided calculations and expressed confidence in their results, while others have raised questions about the reasoning behind certain values, particularly the specific volume used in calculations. There is an ongoing exploration of potential errors in computations and the implications for the quality determination.

Contextual Notes

Participants are working under the constraints of specific volume values at a given temperature and pressure, and there is uncertainty regarding the source of some specific volume figures mentioned in the discussion.

MechE2015
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Homework Statement



A 4 cubic meter rigid vessel contains 0.04 cubic meters of liquid water and 3.96 cubic meters of water vapor at 101.325 kPa. What is the quality of the saturated vapor?

Homework Equations



Specific volume of saturated liquid at 100 C and 101.325 kPa: 0.001044 m^3/kg
Specific volume of saturated vapor at 100 C and 101.325 kPa: 1.67290 m^/kg
Mass of Vapor = (Volume vapor) / (Specific Volume of sat. vapor)
Mass of water = (Volume water) / (Specific Volume of sat. liquid)
x = (Mass of Vapor) / (Total Mass)


The Attempt at a Solution



Mass of liquid: 0.04/0.001044 = 38.314 kg
Mass of vapor: 3.96/1.67290 = 2.367 kg

x = 2.367/(38.314+2.367) = 0.05818

Also, for some reason, I thought the quality could be found by specific volumes, which would give me: x = (1.5313-0.001044)/(1.6729) = 0.9147

Could anyone tell me which method is correct? We have used specific volume to find the quality a few times, which confuses me as to why this would be different.
 
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MechE2015 said:

Homework Statement



A 4 cubic meter rigid vessel contains 0.04 cubic meters of liquid water and 3.96 cubic meters of water vapor at 101.325 kPa. What is the quality of the saturated vapor?

Homework Equations



Specific volume of saturated liquid at 100 C and 101.325 kPa: 0.001044 m^3/kg
Specific volume of saturated vapor at 100 C and 101.325 kPa: 1.67290 m^/kg
Mass of Vapor = (Volume vapor) / (Specific Volume of sat. vapor)
Mass of water = (Volume water) / (Specific Volume of sat. liquid)
x = (Mass of Vapor) / (Total Mass)

The Attempt at a Solution



Mass of liquid: 0.04/0.001044 = 38.314 kg
Mass of vapor: 3.96/1.67290 = 2.367 kg

x = 2.367/(38.314+2.367) = 0.05818
That looks right to me. :approve: (ignoring any trivial rounding differences, if any)
Also, for some reason, I thought the quality could be found by specific volumes, which would give me: x = (1.5313-0.001044)/(1.6729) = 0.9147
Forgive me, but I have no idea what you're doing there. You'll have to explain your reasoning.
 
@collinsmark

I think I had an error computing the specific volume, and if I correct it with the specific volume of 0.09833 m^3/kg, the I get a quality = (0.09833 - 0.001044) / (1.6729 - 0.001044) = 0.05818

Sorry for the confusion but I think I found my mistake with the incorrect computation for specific volume.
 
MechE2015 said:
@collinsmark

I think I had an error computing the specific volume, and if I correct it with the specific volume of 0.09833 m^3/kg, the I get a quality = (0.09833 - 0.001044) / (1.6729 - 0.001044) = 0.05818

Sorry for the confusion but I think I found my mistake with the incorrect computation for specific volume.
Okay. But where does the the 0.09833 m3/kg come from? It's the specific volume of what exactly? Again, forgive me, but I'm just not following.
 

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