1. 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 Mpa T=350  C V=80 m/s z=10 m P=0.5 Mpa 100% saturated steam V=50 m/s z=5 m 10 kw From superheated vapor table: h in =3149.5 kJ/kg From saturated steam table: h out =2748.7 kJ/kg

2. 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: m t = m f +m g The ratio of the mass of vapor to the total mass is called the quality of the mixture denoted by x, where x = m g / m t f-liquid phase g-vapor phase Saturated Steam

3. Note that for a liquid-vapor phase mixture, the quality, x, of the mixture is an independent property. Using the quality, x, to determine mixture porperties I. Volume: Total volume is the sum of liquid volume and vapor volume: V = V f + V g = m f v f + m g v g, where v is the specific volume or 1/   V = m(1/  ) = mv] V/m = v = V f /m + V g /m = (m f /m)v f + (m g /m)v g = [(m-m g )/m]v f + (m g /m)v g = (1-x)v f +xv g = v f + x(v g -v f ) = v f + xv fg, where v fg = v g -v f II. 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)u f +xu g = u f + x(u g -u f ) = u f + xu fg since U = U f + U g = m f u f + m g u g

4. Saturated Steam Table h g (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 vapor p (Mpa) h g (kJ/Kg) Example: If the quality is 50% and the temperature is 150  C h f = 632.2, h fg = 2114.2, h g = 2746.4 h = (1-x) h f + x h g = (1-0.5)(632.2) + 0.5(2746.4) = 1689.3 (kJ/kg) h f (kJ/kg) h fg (kJ/kg)

5. h(p=1.5MPa, T=350  C)=3147.4 kJ/kg h(p=1MPa, T=350  C)=3157.7 kJ/kg Superheated Steam h(p=1.4MPa, T=350  C)=3157.7+(3147.7-3157.7)*(0.4/0.5) =3149.7 (kJ/kg)

6. 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:

7. Compressed Liquid (contd) Example: Internal Energy, T = 100  C u 5 MPA = 417.52, u 10 MPA = 416.52, u f = 418.94 where P sat = 0.1MPa A pressure change by a factor of 100, leads to a change in u of less than 0.5% Enthalpy, T = 100  C h 5 MPA = 422.72, h 10 MPA = 426.12, h f = 419.04 where P sat = 0.1MPa A change of 1.7%