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 m

P=0.5 Mpa100% saturated steamV=50 m/sz=5 m

10 kw

From superheated vapor table: hin=3149.5 kJ/kg

From saturated steam table: hout=2748.7 kJ/kg

dQ

dt

( ) ( )

( ) ( )[( . . )

( )

( . )( )]

. . .

. ( )

m hV

gz m hV

gzdW

dtdW

dt

kW

in out

2 2

2 2

2 2

10 1 3149 5 2748 7

80 50

2 1000

9 8 10 5

1000

10 400 8 1 95 0 049

392 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 / mt

f-liquid phase g-vapor phase

Saturated Steam

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 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-vf

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)uf+xug= uf + x(ug-uf) = uf + xufg

since U = Uf + Ug = mfuf + mgug

Saturated Steam Table

hg(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) 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/kg

h(p=1MPa, T=350C)=3157.7 kJ/kg

Superheated Steam

h(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:

TfTfTf ssuuvv @@@ ,,

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 1.7%