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 (hf)
|
Latent heat (hfg)
|
|
15
15.55
|
198.3
200.0
|
844.6
852.4
|
1947
1941
|
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 hf = (7.8/0.55) × 0.30 = 4.255 kJ/kg. Corresponds to 15.25 bar, exact value of
hf = 852.4 - 4.255 = 848.145 kJ/kg.
Again, for difference of pressure of 0.55 bar, difference of hfg (which is latent heat of evaporation)
= 1947 - 1941 = 6 kJ/kg.
For difference of 0.30 bar, difference of
hfg = (6/0.55) × 0.30 = 3.273 kJ/kg. Corresponding to 15.25 bar, perfect value of
hfg = 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
HDry = hf + hfg kJ/kg
= 848.145 + 1944.273 = 2792.418 kJ/kg .......ANS
(ii) When steam is wet saturated, its enthalpy can be given by
Hwet = hf + xhfg kJ/kg
= 848.145 + 0.92 × 1944.273 = 2636.876 kJ/kg .......ANS
(iii) When steam is superheated, its enthalpy can be given by
Hsup = hf + hfg + Cp(tsup - ts) kJ/kg