Energy Conservation(cont.)Example: Superheated water vapor is entering the steam turbine with a mass flow rate of 1 kg/s and exhausting as saturated steamas shown. Heat loss from the turbine is 10 kW under the following operating condition. Determine the power output of the turbine.P=1.4 MpaT=350° CV=80 m/sz=10 mP=0.5 Mpa100% saturated steamV=50 m/sz=5 m10 kwFrom superheated vapor table: hin=3149.5 kJ/kg From saturated steam table: hout=2748.7 kJ/kgdQdt+ + + = + + += − + −+−+−= − + + +=&( )&( )( ) ( )[( . . )( )( . )( )]. . .. ( )m hVgz m hVgzdWdtdWdtkWin out222 22 210 1 3149 5 2748 780 502 10009 8 10 5100010 400 8 195 0 049392 8• These properties are all dependent: specify one to determine all(because they are in a saturation state)• Liquid and vapor phases coexist, the total mass of the mixture, m, is the sum of the liquid mass and the vapor mass: m=mf+mg, The ratio of the mass of vapor to the total mass is called the quality of the mixture: x=mg/mf-liquid phase g-vapor phaseSaturated SteamTotal volume is the sum of liquid volume and vapor volume: V = Vf+ Vg= mfvf+ mgvg, where v is the specific volume or 1/ρ. [V = m(1/ρ) = mv]V/m = v = Vf/m + Vg/m = (mf/m)vf+ (mg/m)vg= [(m-mg)/m]vf+ (mg/m)vg= (1-x)vf+xvg= vf+ x(vg-vf) = vf+ xvfg, where vfg= vg-vfSimilarly, all other saturated thermodynamic properties can beexpressed in the same manner:Ex: internal energy: u = (1-x)uf+xug= uf+ x(ug-uf) = uf+ xufgsince U = Uf+ Ug= mfuf+ mgugSaturated Steam Tablehg(p=0.5 Mpa) = 2746.4 + (2758.1-2746.4)/(0.6178-0.4758)*(0.5-0.4758)=2748.4 kJ/kg for 100% quality saturated vaporp (Mpa)hg(kJ/Kg)Example: If the quality is 50% and the temperature is 150° Chf= 632.2, hfg= 2114.2, hg = 2746.4h = (1-x) hf + x hg = (1-0.5)(632.2) + 0.5(2746.4)= 1689.3 (kJ/kg)hf(kJ/kg)hfg(kJ/kg)h(p=1.5MPa, T=350°C)=3147.4 kJ/kgh(p=1MPa, T=350°C)=3157.7 kJ/kgSuperheated Steamh(p=1.4MPa, T=350°C)=3157.7+(3147.7-3157.7)*(0.4/0.5)=3149.7 (kJ/kg)Compressed Liquid• Similar to the format of the superheated vapor table• In general, properties are not sensitive to pressure, therefore, can treat thecompressed liquid as saturated liquid at the given TEMPERATURE.• Given: P and T: • But not h, since h=u+pv, and it depends more strongly on p. It can be approximated asTfTfTfssuuvv@@@,,≅≅≅h hvppf T f
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