Question

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

Answer 1

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