Reading Water and Steam Saturation Results

Use Water Saturation Temperature when you know absolute pressure in pascals. Use Water Saturation Pressure when you know temperature in kelvin. Celsius input must first be converted by adding 273.15. A pressure gauge reading requires the local atmospheric pressure: absolute pressure = gauge pressure + atmospheric pressure. Do not assume every site is at 101325 Pa.

These tools implement the IAPWS supplementary saturation release for pure water between 273.16 and 647.096 K. They return equilibrium properties on the liquid–vapor boundary. The pressure and temperature are linked; they are not two independently selectable saturation inputs. This correlation is SR1-86(1992), not a full IF97 steam-state implementation.

Original calculated example: enter 101325 Pa in the pressure-input tool. The model gives 373.12430 K, or 99.97430 °C. The following table is generated from that same calculation, with energy results converted from joules to kilojoules for readability.

For equal masses at this saturation condition, the vapor occupies about 1604 times the liquid volume. Density and specific volume are reciprocals for each phase. Neither column is the density of a liquid–vapor mixture.

The vapor-minus-liquid enthalpy difference is 2256.665 kJ/kg. In an ideal steady vaporizer that converts 0.01 kg/s of saturated liquid to saturated vapor at this pressure, with no shaft work or kinetic/potential energy changes, the corresponding heat input is 22.567 kW. This example excludes feedwater heating, pressure losses, equipment losses and superheating. It is an energy balance, not a heater selection.

Across these examples, higher saturation pressure raises the temperature while the enthalpy gap shrinks. At the model’s critical endpoint, 647.096 K and 22.064 MPa, the liquid and vapor properties coincide and latent heat is zero. That endpoint does not describe a distinct boiling liquid and vapor pair.

Steam quality is the vapor mass fraction, not its volume fraction. Saturation pressure alone does not determine quality. A wet mixture needs another property to determine its state; do not average the two density columns. The DOE thermodynamics handbook explains quality and steam-table notation. Engivault’s saturation tools return the separate phases. To describe a mixture, use Wet Steam Mixture Properties with pressure and quality, or Steam Quality from Enthalpy with pressure and compatible enthalpy.

Original mixture example: at 101325 Pa and quality 0.9, 90% of the mass is vapor. Mass weighting gives h = 0.1 h_liquid + 0.9 h_vapor = 2450.053 kJ/kg and v = 1.50616 m³/kg. Density is 1/v = 0.66394 kg/m³. Vapor occupies 99.9931% of this equilibrium inventory’s volume. A flowing-pipe volume fraction cannot be inferred from mass flow fraction alone when the phase velocities differ.

The inverse relation is x = (h − h_liquid)/(h_vapor − h_liquid). Enter enthalpy in J/kg in the tool, using a compatible reference convention. An enthalpy outside the phase interval is rejected. At the critical point that interval collapses, so both mixture tools exclude it; near the critical point, small enthalpy errors can cause large quality errors.

Enthalpy and entropy use a reference convention; their zero is not an absence of energy. This implementation follows IAPWS’s triple-point liquid internal-energy and entropy reference. For an energy difference, use compatible states from the same convention. The calculator’s enthalpy unit is J/kg; this article displays kJ/kg.

Superheated vapor, subcooled liquid, solutions and below-triple-point ice equilibrium require other models. For the supported low-pressure liquid range, see Water Properties for Pipe Flow. Check that model’s temperature and pressure limits before using its viscosity or density.