Enthalpy and humidity ratio calculator
Enter dry bulb and relative humidity (or wet bulb) to get moist air enthalpy, humidity ratio and dew point. Same ASHRAE 2017 equations as the psychrometric calculator.
Definitions
- Enthalpy h (kJ/kg dry air)
- Heat content of 1 kg of dry air plus its vapor, measured from dry air at 0 °C, so very cold air is negative. Coil load = mass flow × enthalpy difference.
- Humidity ratio W (g/kg dry air)
- Mass of water vapor per kg of dry air. In Korean HVAC usage "absolute humidity" means this ratio, not the vapor density in g/m³ used in meteorology; both are shown.
- Dew point (°C)
- Temperature where condensation starts when the air is cooled at constant moisture. Same humidity ratio, same dew point.
- Relative humidity and wet bulb
- Relative humidity is vapor pressure as a share of saturation pressure at the same temperature; wet bulb is the reading of a wetted thermometer. Either one plus dry bulb fixes the state.
Enthalpy and humidity ratio at typical conditions (101.325 kPa)
Common design and exam conditions, computed with the same engine as the calculator. kcal/kg = kJ/kg ÷ 4.1868.
| Condition | Dry bulb °C | RH % | Enthalpy kJ/kg(DA) | Enthalpy kcal/kg | Humidity ratio g/kg(DA) | Dew point °C | Wet bulb °C |
|---|---|---|---|---|---|---|---|
| Reference | 25 | 50 | 50.3 | 12.02 | 9.88 | 13.9 | 17.9 |
| Summer room | 26 | 50 | 52.9 | 12.64 | 10.50 | 14.8 | 18.7 |
| Summer room (cooler) | 24 | 50 | 47.8 | 11.42 | 9.30 | 12.9 | 17.1 |
| Summer outdoor (example) | 32 | 60 | 78.4 | 18.72 | 18.03 | 23.3 | 25.6 |
| Summer outdoor (exam example) | 33 | 60 | 82.2 | 19.63 | 19.11 | 24.2 | 26.4 |
| Cooling coil leaving air | 16 | 90 | 41.9 | 10.02 | 10.21 | 14.4 | 15.0 |
| Winter room | 22 | 40 | 38.8 | 9.27 | 6.56 | 7.8 | 13.9 |
| Winter outdoor (example) | −5 | 70 | −0.7 | −0.17 | 1.73 | −9.1 | −6.3 |
| Cold winter outdoor | −10 | 70 | −7.3 | −1.74 | 1.12 | −14.0 | −11.0 |
Equations (ASHRAE Handbook Fundamentals 2017, chapter 1)
- Saturation pressure pws(t): Hyland-Wexler (Eq. 5 and 6, over water and ice)
- Vapor pressure pw = RH/100 × pws(t), humidity ratio W = 0.621945 pw / (p − pw) (Eq. 20)
- From wet bulb: W = ((2501 − 2.326 t*) Ws* − 1.006 (t − t*)) / (2501 + 1.86 t − 4.186 t*) (Eq. 33, t* = wet bulb)
- Enthalpy h = 1.006 t + W (2501 + 1.86 t) kJ/kg dry air (Eq. 32)
- Dew point: solve pws(td) = pw iteratively (no approximation)
Coil load from an enthalpy difference
Cooling 1,000 m³/h of room air at 26 °C, 50 % (h 52.9 kJ/kg, specific volume 0.862 m³/kg) to 16 °C, 90 % (h 41.9 kJ/kg): mass flow 1,000 ÷ 0.862 = 1,160 kg/h, load 1,160 × (52.9 − 41.9) ÷ 3,600 = 3.5 kW. Put both points and the airflow into the psychrometric calculator to split it into sensible and latent.
Sources
- ASHRAE Handbook Fundamentals 2017 (SI), chapter 1 Psychrometrics: Eq. 5, 6, 20, 32, 33
- Hyland, R. W. and Wexler, A. (1983), ASHRAE Transactions 89(2A): saturation pressure formulations
- Standard atmosphere (altitude to pressure): ASHRAE Handbook Fundamentals 2017, chapter 1, Eq. 3
Frequently asked questions
How do I convert enthalpy from kJ/kg to kcal/kg?
Divide the kJ/kg value by 4.1868. For example, air at 25 °C and 50 % RH has 50.3 kJ/kg(DA), about 12.0 kcal/kg(DA). Useful when checking older textbooks that use kcal.
How do I get coil capacity from an enthalpy difference?
Total capacity (kW) = dry air mass flow (kg/s) × (inlet enthalpy − outlet enthalpy). 3,000 m³/h at 1.2 kg/m³ is about 1.0 kg/s, so a 20 kJ/kg difference gives about 20 kW. For exact values, use the inlet specific volume to get the dry air mass flow.
Is absolute humidity in g/m³ the same as humidity ratio in g/kg?
No. g/m³ is the mass of water vapour in 1 m³ of air (vapour density), while g/kg(DA) is per kilogram of dry air. In HVAC the humidity ratio is often called absolute humidity. At 25 °C and 50 % RH the humidity ratio is 9.88 g/kg(DA) and the vapour density is about 11.5 g/m³.
Last updated 2026-10-12.