Evaluate the specific enthalpy  thermodynamics:
Steam is being generated in a boiler at a pressure of 15.25 bar. Calculate the specific enthalpy when
(i) Steam is dry saturated
(ii) Steam is wet saturated having 0.92 as the dryness fraction, and
(iii) Steam is already superheated, the temperature of steam being 270°C.
Sol.: Note. Specific enthalpy means enthalpy per unit mass. From steam table which is given below, we get the following data:
Absolute pressure (P)
bar

Saturation temperature (t) ºC

Specific enthalpy kJ/kg

Water (h_{f})

Latent heat (h_{fg})

15
15.55

198.3
200.0

844.6
852.4

1947
1941

Now 15.55  15.25 = 0.30 bar
15.5515 = 0.55bar
For difference of pressure of 0.55 bar, difference in t (or ts) = 200  198.0 = 2.0°C For difference of pressure of 0.30 bar, difference of
t = (2/0.55) × 0.30 =1.091°C. Corresponding to 15.25 bar, perfect value of
t = 200  1.091 = 198.909°C
For difference of pressure of 0.55 bar, difference of hf (which is heat of the liquid) = 852.4  844.6 = 7.8 kJ/kg For difference of pressure of 0.30 bar, difference of h_{f } = (7.8/0.55) × 0.30 = 4.255 kJ/kg. Corresponds to 15.25 bar, exact value of
h_{f } = 852.4  4.255 = 848.145 kJ/kg.
Again, for difference of pressure of 0.55 bar, difference of h_{fg} (which is latent heat of evaporation)
= 1947  1941 = 6 kJ/kg.
For difference of 0.30 bar, difference of
h_{f}_{g} = (6/0.55) × 0.30 = 3.273 kJ/kg. Corresponding to 15.25 bar, perfect value of
h_{f}_{g} = 1941 + 3.273 = 1944.273 kJ/kg
[Greater the pressure of steam generation less is latent heat of evaporation.]
The data computed above are written in a tabular form as below:
Absolute pressure (P)
bar

Saturation temperature (t) ºC

Specific enthalpy kJ/kg

Water (hf)

Latent heat (hfg)

15.25

198.909

848.145

1944.273

(i) When steam is dry saturated, its enthalpy can be given by
H_{Dry} = h_{f} + h_{f}_{g} kJ/kg
= 848.145 + 1944.273 = 2792.418 kJ/kg .......ANS
(ii) When steam is wet saturated, its enthalpy can be given by
H_{wet }= h_{f} + x_{h}_{f}_{g} kJ/kg
= 848.145 + 0.92 × 1944.273 = 2636.876 kJ/kg .......ANS
(iii) When steam is superheated, its enthalpy can be given by
H_{sup} = h_{f} + hfg + Cp(tsup  ts) kJ/kg