A container contains water at 100 kPa. The specific volume of the water inside the container...

A container contains water at 100 kPa. The specific volume of the water inside the container is 1.00 m3/kg. What is the quality (x) of the water in the container, as a percentage and what is the phase of the water in the container?

Solutions

Expert Solution

At 100 kPa, from saturated steam tables,

Specific volume of saturated liquid water = 1.043 *10^-3 m³/Kg

Specific volume of saturated water vapour = 1.694 m³/Kg

But given that the specific volume of water in the container = 1 m³/Kg

This volume lies between saturated vapour and saturated liquid water. Thus the container has water in both, liquid and vapour, phase.

Basis : 1 Kg of water in the container. Let 'x' be the mass of liquid and '1-x' becomes mas of vapour.

Volume of total water in the container = Mass * Speccific volume = 1*1 = 1 m³

Volume of total water in the container = Volume contributed by liquid + Volume contributed by vapour

1 = specific volume of liquid * mass of liquid + specific volume of vapour * mass of vapour

1 = 1.043*10^-3*x + 1.694*(1-x)

x = 0.4099 Kg

Mass of liquid water = 0.4099 Kg

Mass of water vapour = 1 - 0.4099 = 0.5901 Kg

Mass percent of liquid water (Quality) = (Mass of liquid water / Total mass of water)*100 = (0.4099/1)*100 = 40.99%

Amanda K answered 1 year ago

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