A rigid container of volume 1 m3 contains saturated water at 1Mpa.
If the quality is 0.10, what is the volume occupied by the
vapor?
Let X= volume of liquid
1Mpa correspnd to 180 deg.c at saturated conditions where specific volume of liquid = 1.128 cm3/g=1.128*0.001m3/kg=0.00128 m3/kg
and specific volume of vapor =193.8 cm3/g=198*0.001 m3/kg=0.198 m3/kg
Steam quality= 0.1
Volume of vapor =0.198*0.1 =0.0198m3
volume of liquid = 0.00128*0.9=0.001152 m3
total volume of 1 kg of steam = 0.0198+0.0011652=0.020952m3
0.020952 m3 coreespond to 1 kg
1m3 correspond to 1/0.020952=47.73 kg
volume of vapor =47.73Kg*0.0198 m3/kg =0.945 m3
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