PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Energy 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 out2 22 22 210 1 3149 5 2748 780 502 10009 8 10 5100010 400 8 1 95 0 049392 8• At saturation, these properties are all dependent: specify one to determine all • When Liquid and vapor phases coexist, the total mass of the mixture, m, is the sum of the liquid mass and the vapor mass: mt = mf+mg The ratio of the mass of vapor to the total mass is called the quality of the mixture denoted by x, where x = mg / mtf-liquid phase g-vapor phaseSaturated SteamNote that for a liquid-vapor phase mixture, the quality, x, of the mixture is an independent property.Using the quality, x, to determine mixture porpertiesI. Volume:Total volume is the sum of liquid volume and vapor volume: V = Vf + Vg = mfvf + mgvg, where v is the specific volume or 1/V = m(1/) = 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-vfII. Internal Energy:Similarly, all other saturated thermodynamic properties, such as internal energy, u can be expressed in the same manner:internal energy: u = (1-x)uf+xug= uf + x(ug-uf) = uf + xufg since 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 C hf = 632.2, hfg = 2114.2, hg = 2746.4 h = (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=350C)=3147.4 kJ/kgh(p=1MPa, T=350C)=3157.7 kJ/kgSuperheated Steamh(p=1.4MPa, T=350C)=3157.7+(3147.7-3157.7)*(0.4/0.5) =3149.7 (kJ/kg)Compressed Liquids• Similar to the format of the superheated vapor table• In general, thermodynamic properties are not sensitive to pressure, therefore, one can treat compressed liquids as a saturated liquid at the given TEMPERATURE.• Given: p and T: • However, the same does not usually apply to enthalpy, h. Since h=u+pv, it depends more strongly on p. It can be approximated as:TfTfTfssuuvv@@@,, h h v p pf T f sat @( )Compressed Liquid (contd)Example: Internal Energy, T = 100 Cu5 MPA = 417.52, u10 MPA = 416.52, uf = 418.94 where Psat = 0.1MPaA pressure change by a factor of 100, leads to a change in u of less than 0.5%Enthalpy, T = 100 Ch5 MPA = 422.72, h10 MPA = 426.12, hf = 419.04 where Psat = 0.1MPaA change of
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