# Wet Steam Quality and Dryness Fraction

## Introduction and definition of steam quality and dryness fraction including calculating wet steam enthalpy and specific volume

The steam dryness fraction is used to quantify the amount of water within steam.

- Dry steam - all water molecules are in the gaseous state
- Wet steam - a portion of the water molecules have lost their energy - latent heat - and condensed to tiny water droplets

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 due to turbulence and splashing when bubbles of steam break through the water surface. The steam leaving the boiler space will contain a mixture of water droplets and steam.

In addition heat loss in the pipe lines condensates parts of the steam to droplets of water.

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

### Dryness fraction of Wet Steam

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 = 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

Actual enthalpy of wet steam can be calculated with the dryness fraction - *ζ -* and the specific enthalpy - *h*_{s} - of "dry" steam picked from steam tables. Wet steam will always have lower usable heat energy than "dry" 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 "dry" 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 occupies a negligible space in the steam and the specific volume of wet steam will be reduced by the dryness fraction.

v_{t}= v_{s}ζ(3)

where

v_{t}= 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 with 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)*

The total enthalpy can be calculated:

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

= 2651 kJ/kg

Specific volume can be calculated:

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

= 0.299 m^{3}/kg