If the water content in the steam is 5% by mass, then the steam is said to be 95% dry with a dryness fraction 0.95.
Dryness fraction can be expressed:
ζ = ws / (ww + ws) (1)
ζ = dryness fraction
ww = mass of water (kg, lb)
ws = mass of steam (kg, lb)
Actual enthalpy of wet steam can be calculated with the dryness fraction - ζ - and the specific enthalpy - hs - of "dry" steam picked from steam tables. Wet steam will always have lower usable heat energy than "dry" steam.
ht = hs ζ + (1 - ζ ) hw (2)
ht = enthalpy of wet steam (kJ/kg, Btu/lb)
hs = enthalpy of "dry" steam (kJ/kg, Btu/lb)
hw = enthalpy of saturated water or condensate (kJ/kg, Btu/lb)
The droplets of water in wet steam occupies a negligible space in the steam and the specific volume of wet steam will be reduced by the dryness fraction.
vt = vs ζ (3)
vt = specific volume of wet steam (m3/kg, ft3/lb)
vs = specific volume of the dry steam (m3/kg, ft3/lb)
Steam with pressure 5 bar gauge (6 bar abs) has a dryness fraction of 0.95. From the steam table
hs = 2755.46 (kJ/kg)
hw = 670.43 (kJ/kg)
The total enthalpy can be calculated:
ht = (2755.46 kJ/kg) 0.95 + (1 - 0.95) (670.43 kJ/kg)
= 2651 kJ/kg
Specific volume can be calculated:
v = (0.315 m3/kg) 0.95
= 0.299 m3/kg
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