# Thermodynamics - Saturated Vapor Quality Question

1. Mar 4, 2013

### MechE2015

1. The problem statement, all variables and given/known data

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?

2. Relevant 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)

3. 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.

2. Mar 4, 2013

### collinsmark

That looks right to me. (ignoring any trivial rounding differences, if any)
Forgive me, but I have no idea what you're doing there. You'll have to explain your reasoning.

3. Mar 4, 2013

### 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.

4. Mar 5, 2013

### collinsmark

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.