Air humidity calculation: all the psychrometric formulas explained

All the quantities of moist air — relative humidity, humidity ratio, enthalpy, dew point — are linked by a single system of thermodynamic equations. Know any two independent quantities (for example the temperature and the relative humidity) and you can compute all the rest. This article lays out the complete set of psychrometric equations with numerical examples, so that you can verify every calculation or implement it in a spreadsheet or a program.

Step 1: saturation vapor pressure $p_{ws}(t)$

Everything starts with the saturation vapor pressure $p_{ws}(t)$ [Pa] — the maximum partial pressure of water vapor at temperature $t$. It rises exponentially with temperature and is the physical foundation of all psychrometrics.

Buck equation (recommended for practice, over liquid water):

$$p_{ws}(t) = 611.21 \cdot \exp\!\left[\left(18.678 - \frac{t}{234.5}\right) \cdot \frac{t}{257.14 + t}\right] \quad \text{[Pa]}$$

For $t = 20\ ^{\circ}\text{C}$: $p_{ws} = 611.21 \cdot \exp(1.342) \approx 2\,338\ \text{Pa}$. The error against the exact IAPWS-95 tables is < 0.02% over the range 0–50 °C.

Magnus approximation (a quick estimate):

$$p_{ws}(t) \approx 610.78 \cdot \exp\!\left(\frac{17.269 \cdot t}{237.29 + t}\right) \quad \text{[Pa]}$$

For $t = 20\ ^{\circ}\text{C}$: $p_{ws} \approx 2\,338\ \text{Pa}$. The error against the Buck equation is < 0.1% for 0–40 °C — the two approximations practically coincide over this range.

Step 2: partial pressure of water vapor $p_w$

$$p_w = \varphi \cdot p_{ws}(t) \quad \text{[Pa]}$$

where $\varphi$ is the relative humidity as a decimal. For $t = 20\ ^{\circ}\text{C}$ and $\varphi = 0.60$: $p_w = 0.60 \cdot 2\,338 = 1\,403\ \text{Pa}$.

Step 3: humidity ratio x

$$x = 0.62198 \cdot \frac{p_w}{p_{\text{atm}} - p_w} \quad \text{[kg/kg]}$$

where $p_{\text{atm}} = 101\,325\ \text{Pa}$. For $p_w = 1\,403\ \text{Pa}$:

$$x = 0.62198 \cdot \frac{1\,403}{101\,325 - 1\,403} \approx 0.00873\ \text{kg/kg} = 8.73\ \text{g/kg}$$

The constant 0.62198 is the ratio of the molar masses of water and dry air (18.015 / 28.966). At a high-altitude location, substitute the actual atmospheric pressure; how far it moves every psychrometric result is easy to underestimate.

Step 4: relative humidity φ from x and temperature

$$\varphi = \frac{x \cdot p_{\text{atm}}}{(0.62198 + x) \cdot p_{ws}(t)}$$

For $x = 0.00873\ \text{kg/kg}$ and $t = 20\ ^{\circ}\text{C}$: $\varphi = \dfrac{0.00873 \cdot 101\,325}{0.63071 \cdot 2\,338} \approx 0.600 = 60.0\%$ — matching the input value.

Step 5: enthalpy of moist air h

$$h = c_{pa} \cdot t + x \cdot (r_0 + c_{pv} \cdot t) \quad \text{[kJ/kg dry air]}$$

with $c_{pa} = 1.006\ \text{kJ/(kg·K)}$, $c_{pv} = 1.860\ \text{kJ/(kg·K)}$ and $r_0 = 2\,501.0\ \text{kJ/kg}$:

$$h = 1.006 \cdot 20 + 0.00873 \cdot (2\,501 + 1.860 \cdot 20) = 20.12 + 22.16 = 42.28\ \text{kJ/kg}$$

The value matches the ASHRAE Fundamentals tables for air at 20 °C / 60% RH. For the details, see Enthalpy of moist air.

Step 6: dew point $t_d$

The inverse Magnus equation, for $p_w > 611\ \text{Pa}$:

$$t_d = \frac{243.04 \cdot \gamma}{17.625 - \gamma}, \qquad \gamma = \ln\!\left(\frac{p_w}{610.94}\right) \quad \text{[°C]}$$

For $p_w = 1\,403\ \text{Pa}$: $\gamma = \ln(1\,403 / 610.94) = 0.8314$, so $t_d = \dfrac{243.04 \cdot 0.8314}{17.625 - 0.8314} \approx 12.0\ ^{\circ}\text{C}$. See dew point.

The Buck equation cannot be inverted in this form — the Magnus approximation (17.625 / 243.04) is what you use to recover temperature from vapour pressure, as in the dew-point article.

Step 7: density of moist air ρ

$$\rho = \frac{(1 + x) \cdot p_{\text{atm}}}{R_a \cdot T \cdot (1 + 1.6078 \cdot x)} \quad \text{[kg/m}^3\text{]}$$

with $R_a = 287.058\ \text{J/(kg·K)}$ and the absolute temperature $T$ [K]. For 20 °C / 60% RH: $\rho \approx 1.198\ \text{kg/m}^3$. The density of moist air is always lower than that of dry air (1.205 kg/m³ at 20 °C) — moist air is lighter, which is the physical basis of convective motion in the atmosphere.

The $(1 + x)$ term in the numerator matters: without it you get the reciprocal of specific volume per kilogram of dry air (1/v ≈ 1.188 kg/m³), not the density of moist air. The PsychroView calculation core uses this same relation.

Reference summary of equations

Quantity	Relation
Saturation pressure	$p_{ws}(t)$ — Buck or Magnus equation
Partial pressure	$p_w = \varphi \cdot p_{ws}(t)$
Humidity ratio	$x = 0.62198\,p_w / (p_{\text{atm}} - p_w)$
Relative humidity	$\varphi = x\,p_{\text{atm}} / [(0.62198 + x)\,p_{ws}(t)]$
Enthalpy	$h = 1.006\,t + x\,(2\,501 + 1.860\,t)$
Dew point	inverse Magnus from $p_w$
Density	$\rho = (1 + x)\,p_{\text{atm}} / [287.058\,T\,(1 + 1.6078\,x)]$

The interactive h-x diagram in PsychroView performs these calculations instantly for any air state you enter.

Frequently asked questions

Which saturation-pressure equation should I choose? For technical documentation, the Buck equation (error < 0.02%). The Magnus equation is simpler and sufficient for most engineering calculations (error < 0.5%).

Why is the humidity ratio referenced to dry air? Because the mass of dry air does not change when moisture is added or removed, which simplifies balance calculations. The unit is g/kg of dry air (g/kg dry air).

Do I have to recompute the formulas for a different altitude? Yes — in the equations for humidity ratio, relative humidity and density, substitute the actual barometric pressure of the location for 101,325 Pa.