Question #127487

A frictionless piston-cylinder device contains 8 kg of superheated water vapor at 500 kPa and 300 C. Steam is now cooled at constant pressure until 70 percent of it, by mass, condenses

Expert's answer

The super heated steam at pressure=P1=500kpa and Tempwerature(T1)= 300 C

mass(m)=8 kg

therefore, The enthalpy at initial state=H1=mxh1=8x(3064.6)=24516.8 KJ

If it is cooled at same pressure , steam becomes saturated at temp (T2)=152 C.

At P2=500kPa, hf=640.09 kJ/Kg, hg=2748.01 kJ/Kg

The enthalpy of saturated steam (H2) =mxh2=m(hg)=8x2748.01=21984.08 kJ.

If 70% of steam is condensed at state-3 , dryness fraction at state-3 is X=0.3.

Therefore enthalpy of steam at state -3 is given by

H3=m(hf+X(hg-hf))=10179.4 Kj

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