Specific volume and density problem

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SUMMARY

The discussion focuses on a pressure vessel containing 1 kg of liquid water and 80 liters of steam, with a total volume of 100 liters. The density of the liquid water is calculated to be 50 kg/m³ using the formula ρ = m/V, where the volume of water is derived from the remaining volume of the vessel. The mass of the steam is determined using its specific volume of 0.02 m³/kg, leading to a mass of 4 kg. Finally, the overall density of the system is calculated by combining the masses of water and steam with the total volume of the vessel.

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Xyius
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1. A pressure vessel of volume 100 liters is filled with 1 kg of liquid water and 80 liters of steam with a specific volume of 0.02 m3/kg. Determine
a. The density of the liquid water in SI fundamental units
b. The mass of the steam in grams
c. The density and the specific volume of the system (liquid water + steam)

Homework Equations


[tex]Specific\ Volume = \nu = \frac{1}{\rho} = \frac{V}{m}[/tex]

The Attempt at a Solution


Basically, I am unsure about the assumptions that should be made for part A only. The vessel is 100 L and is filled with 1 kg of liquid water and 80L of steam. Is it correct of me to assume that the entire vessel is filled? So therefore.. (Since 20L = 0.02 cubic meters)
[tex]\rho_{water} = \frac{m}{V} = \frac{1kg}{0.02m^3} = 50\frac{kg}{m^3}[/tex]

Does this sound right to you guys? Part B would be solved in a similar manner using the specific volume of steam that is given, and part C I am assuming you would do something like this..
[tex]\rho_{system} = \frac{m_{water + steam}}{V_{system}}[/tex]
 
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Xyius said:

The Attempt at a Solution


Basically, I am unsure about the assumptions that should be made for part A only. The vessel is 100 L and is filled with 1 kg of liquid water and 80L of steam. Is it correct of me to assume that the entire vessel is filled? So therefore.. (Since 20L = 0.02 cubic meters)
[tex]\rho_{water} = \frac{m}{V} = \frac{1kg}{0.02m^3} = 50\frac{kg}{m^3}[/tex]

Does this sound right to you guys?

Yep that would be correct.

Xyius said:
Part B would be solved in a similar manner using the specific volume of steam that is given, and part C I am assuming you would do something like this..
[tex]\rho_{system} = \frac{m_{water + steam}}{V_{system}}[/tex]

part b would just be modifying your formula to get mass.

part c, that is correct as well (your method that is)
 
Cool thanks :)
 

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