Question #136167

An amount of water vapor with x = 0.7, occupies a volume of 0.197 m3 at a pressure of 1.4 MPa. If the pressure is kept constant, determine the heat that enters the system to make the steam exactly in a dry saturation state (x = 1, meaning it's all gaseous) and the heat we put into this system is seen as work. If now the volume is kept constant while heat is removed until the pressure drops to 1.25 MPa, then determine the dryness fraction (x) and the heat extracted from the water vapor.

Expert's answer

Dryness fraction of water vapor=0.7,Occupies volume =0.197 "m^3" ,"P_1=1.4" MPa,

"V_{fg}=xV_{g}"

"V_g=\\frac {0.197}{0.7}=0.2814 m^3"

"W=pdV=1.4 \\times 10^6\\times (0.2814-0.197)=0.118 \\times 10^6 J=1.18 \\times 10^6 J"

As per thermodynamic law

"dQ=dU+dW,dU=0\ndQ=1.18 \\times 10^6"

Now on removing we get

1.25=x1.4

x=0.893

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