Question #152152
find the enthalpy of entropy of steam when the pressure is 20 Bar and specific volume is 0.8
1
Expert's answer
2020-12-22T05:07:44-0500

The Dryness fraction(x) of the wet steam=0.8


The pressure of wet steam(P)=20 bar=2 MPa J/kg


The saturation properties of wet steam at 2 MPa is given


in the steam table as

V=0.0012m3/kgVtg(0.09960.0012)=0.0984m3/kgV= 0.0012 m3/kg Vtg (0.0996-0.0012)=0.0984 m3/kg

h=908.5kJ/kghig=1889.8kJ/kgh=908.5 kJ/kg hig=1889.8kJ/kg

U=906.15kJ/kgUfq=(2599.1906.15)=1692.95kJ/KgS=2.4468kJ/kgKs=3.8923KJ/kgKU = 906.15 kJ/kg Ufq= (2599.1-906.15) =1692.95 kJ/Kg S= 2.4468 kJ/kg-K s=3.8923 KJ/kg-K

Specific volume at x=0.8=v=vf+x.vfg=0.07992m3/kgx=0.8 = v=vf+x.vfg=0.07992 m3/kg

Specific enthalpy at x=0.8=>h=hf+x.hfg=2420.34kJ/kgx=0.8 => h=hf+x.hfg=2420.34 kJ/kg

Specific internal energy at x=0.8=>u=uf+x.ufg=2260.51kx=0.8=>u=uf+x.ufg=2260.51 k

Specific entropy at x=0.8=>s=sf+xsfg=55606kJ/kgKx=0.8 => s=sf+x sfg=5 5606 kJ/kg-K


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