Steam Quality from Enthalpy

Dryness fraction, quality, x. It is the mass of vapour divided by the total mass of the mixture, so x = 0.95 means 950 g of every kilogram is actually steam and the other 50 g is still water riding along as fog. The arithmetic is a lever: saturated water at that pressure sits at $h_f$, dry saturated steam sits at $h_f + h_{fg}$, and a mixture sits proportionally between them. At 1 MPa absolute the table gives $h_f = 762.5$ and $h_{fg} = 2{,}014.6$ kJ/kg, so a sample measured at 2,576 kJ/kg works out to (2,576 − 762.5)/2,014.6 = 0.900. Ninety percent dry, which in a plant is poor.

Two things matter more than the sum. The first is that both table values have to be read at the SAME pressure as the sample, because $h_f$ and $h_{fg}$ each move hundreds of kilojoules across a normal working range, and mixing rows is how a perfectly good measurement turns into a nonsense answer. The second is what the answer costs. Wet steam gives up only the fraction x of the latent heat it should, so a coil fed 90% steam is short about 10% of its rating, and the water it carries is doing damage on the way through: droplets at header velocity erode valve seats and turbine blades, and slugs of it are what causes water hammer. In practice quality gets measured with a throttling calorimeter, which expands a small sample to atmospheric pressure where it becomes superheated and its enthalpy can be read from a thermometer alone. And note the honest boundary on this whole page. Quality is only defined ON the saturation line. Below $h_f$ the water is subcooled and above $h_f + h_{fg}$ the steam is superheated, and in both cases the answer this page hands back is a message rather than a number worth using.