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Moist-air psychrometric state

Find humidity ratio, enthalpy, dew/frost point and moist-air density from temperature, RH and absolute pressure.

Find humidity ratio, enthalpy, dew/frost point and moist-air density from temperature, RH and absolute pressure.

How this calculation works

Start with the measured dry-bulb temperature, relative humidity and local absolute pressure. Relative humidity specifies the fraction of saturation vapor pressure, so it cannot be substituted directly for humidity ratio. The model first finds water-vapor partial pressure using the ASHRAE saturation relation, then calculates water mass per kilogram of dry air. That humidity ratio supports the enthalpy and specific-volume calculations. Moist-air density includes both dry air and water vapor. The dew or frost point is found by solving for the saturation temperature at the same vapor partial pressure. Below freezing, the reported saturation point refers to ice. Use the result to describe one equilibrium air state; a foggy, supersaturated mixture needs a different model. The enthalpy reference is dry air and liquid water at zero degrees Celsius.

Inputs and units

  • Dry-bulb temperature (°C)
  • Relative humidity (%): Over ice below 0.01 °C; over water above it. Zero RH has no finite dew point.
  • Absolute air pressure (kPa)

Method and formula

pw = RH × pws(T)/100; W = 0.621945 pw/(p−pw); h = 1.006T + W(2501 + 1.86T); v = 287.042(T+273.15)(1+1.607858W)/p; density = (1+W)/v. W is kg/kg dry air, p and pw are Pa, T is °C. Dew/frost point solves pws(Td)=pw using the ASHRAE saturation fit.

Worked example

Example inputs

  • Dry-bulb temperature: 25 °C
  • Relative humidity: 50 %
  • Absolute air pressure: 101.325 kPa

Calculation steps

  1. At 25 °C, the saturation fit gives pws ≈ 3169.216470 Pa; 50% RH gives pw ≈ 1584.608235 Pa.
  2. Convert total pressure: 101.325 kPa × 1000 = 101325 Pa. W = 0.621945 × 1584.608235 / (101325 − 1584.608235) ≈ 0.009881043691 kg/kg, or 9.881044 g/kg dry air.
  3. h = 1.006 × 25 + 0.009881043691 × (2501 + 1.86 × 25) ≈ 50.321959 kJ/kg dry air.
  4. Solving pws(Td) = 1584.608235 Pa gives Td ≈ 13.863973 °C. The moist-air specific volume is about 0.8580432639 m³/kg dry air; density ≈ 1.176958 kg/m³.

Example results

  • Humidity ratio: 9.881043691 g/kg dry air
  • Specific enthalpy: 50.3219588 kJ/kg dry air
  • Dew/frost point: 13.86397327 °C
  • Moist-air density: 1.176958186 kg/m³

Assumptions

  • Ideal mixture of dry air and water vapor at equilibrium.
  • Enthalpy reference is dry air and liquid water at 0 °C.

Limitations

  • Supported dry bulb −20 to 50 °C, RH 1–100%, absolute pressure 60–110 kPa.
  • Does not model supersaturated fog, chemical vapors or high-pressure process gases.
  • Preliminary educational check; verify inputs and equipment data before design or operation.

Sources

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