acalculator

What is the psychrometric state?

Type the dry-bulb temperature and one humidity value (relative humidity, wet bulb or dew point). The psychrometric calculator reads the point off the psychrometric chart by formula and gives the humidity ratio, dew point, wet bulb, enthalpy, specific volume and density.

Your numbers

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Humidity from
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Humidity ratio (g/kg)
9.88

The humidity ratio is 9.88 g/kg, with a dew point of 13.9 °C, a wet bulb of 17.9 °C and an enthalpy of 50.32 kJ/kg.

Humidity ratio (gr/lb)
69.2
Relative humidity (%)
50
Dew point
13.9 °C
Wet bulb
17.9 °C
Enthalpy (kJ/kg dry air)
50.32
Specific volume (m³/kg dry air)
0.8581
Specific volume (ft³/lb dry air)
13.745
Density (kg/m³)
1.1769
Vapour pressure (kPa)
1.5843
Saturation pressure (kPa)
3.1685
Degree of saturation (%)
49.2

Humidity ratio (g/kg): 9.88. The humidity ratio is 9.88 g/kg, with a dew point of 13.9 °C, a wet bulb of 17.9 °C and an enthalpy of 50.32 kJ/kg.

How to calculate

Psychrometric calculator: humidity ratio, relative humidity, dew point, wet bulb, enthalpy, specific volume and density of moist air from the dry bulb and the relative humidity, wet bulb or dew point.

Example with the default inputs (Dry-bulb temperature 76.9999999999999 °F, Humidity from Relative humidity, Relative humidity 50%, Air pressure 14.6959487755135 psi): The humidity ratio is 9.88 g/kg, with a dew point of 13.9 °C, a wet bulb of 17.9 °C and an enthalpy of 50.32 kJ/kg.

Method: p_sat(t) by Buck (1996); p_w = RH × p_sat(t), p_sat(dew point), or from the wet bulb t*: W = ((2501 − 2.326 t*) W_s* − 1.006 (t − t*)) ÷ (2501 + 1.86 t − 4.186 t*); W = 0.62198 p_w ÷ (p − p_w); h = 1.006 t + W (2501 + 1.86 t); v = 287.058 T ÷ (p − p_w).

  • Moist air is a mix of ideal gases; enthalpy is zero for dry air and liquid water at 0 °C.
  • Below 0 °C the saturation pressure is over ice (the dew point is a frost point); the wet-bulb balance uses 4.186 t* for the added water at any temperature.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Dry-bulb temperature 77 °F, Humidity from Relative humidity, Relative humidity 50, Air pressure 14.7 psi gives Humidity ratio (g/kg) 9.87943, Enthalpy (kJ/kg dry air) 50.317847, Wet-bulb temperature 64.2 °F.Source: Wikipedia, Arden Buck equation (Buck 1996: 6.1121 exp((18.678 − T/234.5)(T/(257.14 + T))) hPa over water, 6.1115 exp((23.036 − T/333.7)(T/(279.82 + T))) hPa over ice), https://en.wikipedia.org/wiki/Arden_Buck_equation (retrieved 2026-10-05); Engineering ToolBox, Humidity Ratio of Air (x = 0.62198 p_w ÷ (p_a − p_w)), https://www.engineeringtoolbox.com/humidity-ratio-air-d_686.html (retrieved 2026-10-05); Engineering ToolBox, Enthalpy of Moist Air (h = 1.006 t + x (1.86 t + 2501) kJ/kg), https://www.engineeringtoolbox.com/enthalpy-moist-air-d_683.html (retrieved 2026-10-05)
  2. Dry-bulb temperature 68 °F, Humidity from Dew point, Dew point 50 °F, Air pressure 14.7 psi gives Relative humidity 52.509908, Dew point 50 °F, Humidity ratio (g/kg) 7.629633.Source: Wikipedia, Arden Buck equation (Buck 1996: 6.1121 exp((18.678 − T/234.5)(T/(257.14 + T))) hPa over water, 6.1115 exp((23.036 − T/333.7)(T/(279.82 + T))) hPa over ice), https://en.wikipedia.org/wiki/Arden_Buck_equation (retrieved 2026-10-05); Engineering ToolBox, Humidity Ratio of Air (x = 0.62198 p_w ÷ (p_a − p_w)), https://www.engineeringtoolbox.com/humidity-ratio-air-d_686.html (retrieved 2026-10-05)
  3. Dry-bulb temperature 86 °F, Humidity from Wet bulb, Wet-bulb temperature 86 °F, Air pressure 14.7 psi gives Relative humidity 100, Dew point 86 °F.Source: Wikipedia, Arden Buck equation (Buck 1996: 6.1121 exp((18.678 − T/234.5)(T/(257.14 + T))) hPa over water, 6.1115 exp((23.036 − T/333.7)(T/(279.82 + T))) hPa over ice), https://en.wikipedia.org/wiki/Arden_Buck_equation (retrieved 2026-10-05)

How it works

Temperatures t are in °C, T = t + 273.15 in kelvins, and pressures in pascals.

  • Saturation pressure (Buck 1996): p_sat(t) = 611.21 × exp((18.678 − t ÷ 234.5) × (t ÷ (257.14 + t))) over water when t ≥ 0; 611.15 × exp((23.036 − t ÷ 333.7) × (t ÷ (279.82 + t))) over ice when t < 0.
  • Vapour pressure p_w: from relative humidity, RH ÷ 100 × p_sat(t); from a dew point t_d, p_sat(t_d); from a wet bulb t*, see below. p_w is never taken above p_sat(t).
  • Humidity ratio W = 0.62198 × p_w ÷ (p − p_w) kg/kg; shown × 1,000 as g/kg and × 7,000 as grains per lb.
  • Relative humidity = p_w ÷ p_sat(t) × 100%. Degree of saturation = W ÷ W_s × 100%, with W_s = 0.62198 × p_sat(t) ÷ (p − p_sat(t)).
  • Enthalpy h = 1.006 × t + W × (2501 + 1.86 × t) kJ per kg of dry air.
  • Specific volume v = 287.058 × T ÷ (p − p_w) m³ per kg of dry air; ft³/lb = v × 0.45359237 ÷ 0.028316846592. Density = (1 + W) ÷ v.
  • Dew point solves p_sat(t_d) = p_w with the same Buck formula: with L = ln(p_w ÷ 611.21) and B = L − 18.678 over water (p_w ≥ 611.21 Pa), t_d = (−B − √(B² − 4 × 257.14 × L ÷ 234.5)) × 234.5 ÷ 2. Below 611.21 Pa use the ice constants (611.15, 23.036, 333.7, 279.82); a result above 0 there (p_w from 611.15 to 611.21 Pa) is 0 °C.
  • Wet bulb t* is the adiabatic saturation temperature: adding water at t* until saturated keeps the energy balance h(t, W) + (W_s* − W) × 4.186 × t* = 1.006 × t* + W_s* × (2501 + 1.86 × t*), with W_s* = 0.62198 × p_sat(t*) ÷ (p − p_sat(t*)). It is found by bisection between the dew point and the dry bulb.
  • From a wet bulb, the same balance gives W = ((2501 − 2.326 × t*) × W_s* − 1.006 × (t − t*)) ÷ (2501 + 1.86 × t − 4.186 × t*), then p_w = p × W ÷ (0.62198 + W).

Rules

  • Dry bulb, wet bulb and dew point from −40 °C to 60 °C; pressure from 50 kPa to 110 kPa; relative humidity from 0.1% to 100%.
  • No answer when the wet bulb or dew point is above the dry bulb, when the wet bulb is so low that W ≤ 0, or when p_sat(t) is at least the pressure.

Output format

Humidity ratio to 2 decimals (g/kg) and 1 decimal (gr/lb); RH and degree of saturation to 1 decimal; dew point and wet bulb to 1 decimal in °F (US) or °C (Metric); enthalpy to 2 decimals; specific volume to 4 decimals (m³/kg) and 3 (ft³/lb); density to 4 decimals; pressures in kPa to 4 decimals. All arithmetic is in floats.

Worked examples by hand

25 °C and 50% RH at 101,325 Pa. p_sat = 611.21 × exp((18.678 − 0.10661) × 0.088660) = 3,168.53 Pa; p_w = 1,584.27 Pa; W = 0.62198 × 1,584.27 ÷ 99,740.73 = 0.0098794 (9.88 g/kg); h = 25.15 + 0.0098794 × (2501 + 46.5) = 50.318 kJ/kg; v = 287.058 × 298.15 ÷ 99,740.73 = 0.8581 m³/kg. The wet bulb, by bisection, is 17.889 °C.

20 °C with a 10 °C dew point. p_w = p_sat(10) = 1,227.86 Pa; p_sat(20) = 2,338.34 Pa; RH = 52.51%; W = 0.62198 × 1,227.86 ÷ 100,097.14 = 0.0076296 (7.63 g/kg).

Saturated air. Dry and wet bulb both 30 °C: the balance gives W = W_s, so RH = 100% and the dew point is 30 °C.

Other questions people ask

What is a psychrometric calculator?

It does the work of a psychrometric chart: from two properties of moist air (here the dry bulb and one humidity value) and the pressure, it finds all the others, such as humidity ratio, enthalpy and dew point.

How do you calculate the humidity ratio?

W = 0.62198 × p_w ÷ (p − p_w), with p_w the vapour pressure and p the air pressure. At 25 °C and 50% RH at sea level, p_w = 1,584 Pa and W = 0.00988 kg/kg, or 9.88 g of water per kg of dry air.

How do you calculate the enthalpy of moist air?

h = 1.006 × t + W × (2501 + 1.86 × t) kJ per kg of dry air, with t in °C. At 25 °C and W = 0.00988 that is 25.15 + 25.17 = 50.32 kJ/kg.

What is the difference between wet bulb and dew point?

The dew point is the temperature at which the air becomes saturated when cooled with no water added. The wet bulb is the temperature reached by evaporating water into it until it is saturated. The dew point is never above the wet bulb, and both equal the dry bulb at 100% RH.

How does air pressure change the results?

At a lower pressure (higher altitude) the same vapour pressure is a larger share of the air, so the humidity ratio, the enthalpy and the specific volume are higher at the same temperature and RH.

What is the specific volume?

The volume of moist air that holds 1 kg of dry air: v = 287.058 × T ÷ (p − p_w). It is about 0.86 m³/kg (13.7 ft³/lb) at 25 °C and 50% RH.