Air Humidity Calculation: All the Psychrometric Formulas

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.

A word on units. The empirical equations below are published in SI — degrees Celsius and pascals — and that is how they are used here and inside PsychroView. The worked example starts from an IP input, converts once at the top, runs the chain, and reports each result in both systems. Only the enthalpy has a genuine IP form of its own, and it is given alongside.

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):

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

Our example air is 68°F, which is 20 °C. For $t = 20\ ^{\circ}\text{C}$: $p_{ws} = 611.21 \cdot \exp(1.342) \approx 2\,338\ \text{Pa}$ (0.339 psi). The error against the exact IAPWS-95 tables is < 0.02% over the range 32–122°F.

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 32–104°F — 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 $t = 20\ ^{\circ}\text{C}$ (68°F) and $\varphi = 0.60$: $p_w = 0.60 \cdot 2\,338 = 1\,403\ \text{Pa}$ (0.204 psi).

Step 3: humidity ratio W

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

where $p_{\text{atm}} = 101\,325\ \text{Pa}$ (14.696 psi). For $p_w = 1\,403\ \text{Pa}$:

$$W = 0.62198 \cdot \frac{1\,403}{101\,325 - 1\,403} \approx 0.00873\ \text{lb/lb} = 61.1\ \text{gr/lb}$$

The ratio is dimensionless, so it reads the same in both systems — only the customary display unit changes, from g/kg to grains per pound. 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 — Denver runs about 12.1 psi, and that shifts every humidity ratio on the job.

Step 4: relative humidity φ from W and temperature

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

For $W = 0.00873\ \text{lb/lb}$ 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

This is the one relation with an established IP form of its own:

$$h = 0.240 \cdot t + W \cdot (1061 + 0.444 \cdot t) \quad \text{[Btu/lb dry air]}$$

with $t$ in °F. For 68°F and $W = 0.00873$:

$$h = 0.240 \cdot 68 + 0.00873 \cdot (1061 + 0.444 \cdot 68) = 16.32 + 9.53 = 25.85\ \text{Btu/lb}$$

The SI form $h = 1.006\,t + x\,(2\,501 + 1.86\,t)$ gives 42.28 kJ/kg for the same air. The two numbers are not a unit conversion of one another — the IP form counts from a 0°F datum for dry air, the SI form from 0 °C. 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 $t_d = \dfrac{243.04 \cdot 0.8314}{17.625 - 0.8314} \approx 12.0\ ^{\circ}\text{C} = 53.6\ ^{\circ}\text{F}$. 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 vapor pressure, as in the dew-point article.

Step 7: density of moist air ρ

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

with $R_a = 287.058\ \text{J/(kg·K)}$ and the absolute temperature $T$ [K]. For 68°F / 60% RH: $\rho \approx 1.198\ \text{kg/m}^3 = 0.0748\ \text{lb/ft}^3$. The density of moist air is always lower than that of dry air (0.0752 lb/ft³ at 68°F) — moist air is lighter, which is the physical basis of convective motion in the atmosphere. Note how close this sits to the 0.075 lb/ft³ baked into the 1.08 and 4.5 shortcut constants.

The $(1 + W)$ term in the numerator matters: without it you get the reciprocal of specific volume per pound of dry air (1/v ≈ 0.0742 lb/ft³), 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, $t$ in °C
Partial pressure	$p_w = \varphi \cdot p_{ws}(t)$
Humidity ratio	$W = 0.62198\,p_w / (p_{\text{atm}} - p_w)$
Relative humidity	$\varphi = W\,p_{\text{atm}} / [(0.62198 + W)\,p_{ws}(t)]$
Enthalpy (IP)	$h = 0.240\,t + W\,(1061 + 0.444\,t)$, $t$ in °F
Dew point	inverse Magnus from $p_w$
Density	$\rho = (1 + W)\,p_{\text{atm}} / [287.058\,T\,(1 + 1.6078\,W)]$

The interactive psychrometric chart in PsychroView performs these calculations instantly for any air state you enter.

Frequently asked questions

Which saturation-pressure equation should I choose? For submittal 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 customary unit is grains of water per pound of dry air (gr/lb), with 7,000 grains to the pound.

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. This is not a rounding matter in the Mountain West: at Denver’s elevation the humidity ratio for the same temperature and RH comes out roughly a fifth higher.