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:

Now  15.55 - 15.25 = 0.30 bar

15.55-15 = 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_fg  = (6/0.55) × 0.30 = 3.273 kJ/kg. Corresponding to 15.25 bar, perfect value of

h_fg = 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:

(i) When steam is dry saturated, its enthalpy can be given by

H_Dry = h_f  + h_fg  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_hfg   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