Thermal Dynamics
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Thermal dynamics help needed please?
An 85m^3 rigid vessel contains 10 kg of water (both liquid and vapor in thermal equilibrium at a pressure of 0.01 MPa). Calculate the volume and mass of both the liquid and vapor. Thanks
The volume within the tank taken up by unheated water is
( 10 kg ) / ( 1000 kg / m³ ) = 0.01 m³
thus a significant volume within the 85 m³ vessel can be taken up by steam
85 m³ - 0.01 m³ = 84.99 m³
From the saturated steam table at:
http://www.engineeringtoolbox.com/saturated-steam-properties-d_457.html
Pressure 0.01 MPa = 0.1 bar
we obtain...
Specific volume (steam) = 14.675 m³ / kg
( 84.99 m³ ) / ( 14.675 m³ / kg ) = 5.79 kg (steam)
10 kg - 5.79 kg (steam) = 4.21 kg (water)
So lets reiterate the calculation one more time
4.21 kg (water) / ( 1000 kg / m³ ) = 0.00421 m³
85 m³ - 0.00421 m³ = 84.99579 m³
( 84.99579 kg ) / ( 14.675 m³ / kg ) = 5.79187 kg (steam)
Volume
liquid = 0.00421 m³
steam = 84.99579 m³
Mass
liquid = 4.21 kg
steam = 5.79 kg
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