Thermodynamics - Saturated Vapor Quality Question

[MechE2015]

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.

 

[collinsmark]

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. (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.

 

[MechE2015]

@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.

 

[collinsmark]

@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.

 

[Nickie]

I would like to clarify that both methods are correct, but they are used for different scenarios. The first method, using the mass of the liquid and vapor, is used when the total mass of the system is known. In this case, we can find the quality by dividing the mass of the vapor by the total mass of the system.

The second method, using specific volumes, is used when the total volume of the system is known. In this case, we can find the quality by calculating the difference between the specific volumes of the vapor and liquid, and then dividing it by the specific volume of the vapor.

It is important to understand the difference between these two methods and when to use them in order to accurately find the quality of the saturated vapor. It is also important to note that the specific volumes of the vapor and liquid may vary depending on the temperature and pressure, so it is important to use the correct values for the specific scenario given.