# Wet Steam Quality and the Dryness Fraction

## An introduction and definition of vapor or steam quality and dryness fraction. Includes formulas for calculating wet steams enthalpy and specific volume

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To produce *100%* dry steam in an boiler, and keep the steam dry throughout the piping system, is in general not possible. Droplets of water will escape from the boiler surface. Because of turbulence and splashing when bubbles of steam break through the water surface the steam space will contain a mixture of water droplets and steam.

In addition heat loss in the pipes will condensate steam to droplets of water.

Steam - produced in a boiler where the heat is supplied to the water and where the steam are in contact with the water surface of the boiler - will contain approximately *5%* water by mass.

### Dryness fraction of Wet Steam

If the water content of the steam is *5% *by mass, then the steam is said to be *95%* dry and has a dryness fraction of *0.95*.

Dryness fraction can be expressed as:

ζ = w_{s}/ (w_{w}+ w_{s})(1)

where

ζ= dryness fraction

w_{w}= mass of water (kg, lb)

w_{s}= mass of steam (kg, lb)

### Enthalpy of Wet Steam

The actual enthalpy of evaporation of wet steam is the product of the dryness fraction - *ζ -* and the specific enthalpy - *h*_{s} - from the steam tables. Wet steam have lower usable heat energy than dry saturated steam.

h_{t}= h_{s}ζ + (1 - ζ ) h_{w}(2)

where

h_{t}= enthalpy of wet steam (kJ/kg, Btu/lb)

h_{s}= enthalpy of steam (kJ/kg, Btu/lb)

h_{w}= enthalpy of saturated water or condensate (kJ/kg, Btu/lb)

### Specific Volume of Wet Steam

The droplets of water in wet steam will occupy negligible space in the steam and the specific volume of wet steam will be reduced according the dryness fraction.

v = v_{s}ζ(3)

where

v= specific volume of wet steam (m^{3}/kg, ft^{3}/lb)

v_{s}= specific volume of the dry steam (m^{3}/kg, ft^{3}/lb)

### Example - Enthalpy and Specific Volume of Wet Steam

Steam at pressure *5 bar gauge (6 bar abs) *has a dryness fraction of *0.95.* From the steam table

*h*_{s}* = 2755.46 (kJ/kg)*

*h _{w}*

*= 670.43 (kJ/kg)*

Total enthalpy can be expressed as:

h_{t}= (2755.46 kJ/kg) 0.95 + (1 - 0.95) (670.43 kJ/kg)

= 2651 kJ/kg

Specific volume can be expressed as:

v= (0.315 m^{3}/kg) 0.95

= 0.299 m^{3}/kg

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