Calculating moist air enthalpy: formula, tables and h-x diagram

The enthalpy of moist air is the total heat content of 1 kg of dry air including its water vapor. The formula: $h = 1.006 \cdot t + x \cdot (2501 + 1.86 \cdot t)$ [kJ/kg], where $t$ is the temperature [°C] and $x$ is the humidity ratio [kg/kg dry air]. From the enthalpy you compute the capacity of any process directly: $\Phi = \dot{m} \cdot (h_1 - h_2)$.

Why enthalpy, not just temperature

Temperature measures only the sensible heat, that is the kinetic energy of the molecules. But air also carries latent heat, the energy bound up in the water vapor.

Example: air A (20 °C, φ = 90%) and air B (20 °C, φ = 30%) have the same temperature, but air A holds far more moisture and a higher enthalpy. A cooling coil has to remove more energy from air A.

The formula for enthalpy

The exact ASHRAE formula

$$h = c_{pa} \cdot t + x \cdot (r_0 + c_{pv} \cdot t)$$

where:

• $c_{pa} = 1.006\ \text{kJ/(kg·K)}$ — specific heat capacity of dry air
• $c_{pv} = 1.860\ \text{kJ/(kg·K)}$ — specific heat capacity of water vapor
• $r_0 = 2501.0\ \text{kJ/kg}$ — latent heat of vaporization at 0 °C
• $t$ = temperature [°C]
• $x$ = humidity ratio [kg/kg dry air]

Working form (the same formula with the constants substituted in, used in the examples below):

$$h = 1.006 \cdot t + x \cdot (2501 + 1.86 \cdot t) \quad \text{[kJ/kg dry air]}$$

Worked examples

Outdoor winter air: $t = -10\ ^{\circ}\text{C}$, $\varphi = 80\%$ → $x \approx 0.001285\ \text{kg/kg}$

$$\begin{aligned} h &= 1.006 \cdot (-10) + 0.001285 \cdot (2501 + 1.86 \cdot (-10)) \\ &= -10.06 + 0.001285 \cdot 2482.4 \\ &= -10.06 + 3.19 = \mathbf{-6.87}\ \textbf{kJ/kg} \end{aligned}$$

Summer supply air: $t = 26\ ^{\circ}\text{C}$, $\varphi = 55\%$ → $x \approx 0.01162\ \text{kg/kg}$

$$\begin{aligned} h &= 1.006 \cdot 26 + 0.01162 \cdot (2501 + 1.86 \cdot 26) \\ &= 26.16 + 0.01162 \cdot 2549.4 \\ &= 26.16 + 29.62 = \mathbf{55.78}\ \textbf{kJ/kg} \end{aligned}$$

Enthalpy table

t [°C]	φ = 30%	φ = 50%	φ = 70%	φ = 100%
−10	−8.9	−8.1	−7.3	−6.1
0	2.8	4.7	6.6	9.5
10	15.8	19.7	23.5	29.4
20	31.2	38.6	46.2	57.6
26	42.2	53.0	64.1	80.9
30	50.5	64.4	78.5	100.1

Values [kJ/kg dry air], pressure 101,325 Pa; computed by the PsychroView calculation engine (saturation vapor pressure per Hyland–Wexler, ASHRAE Fundamentals; below 0 °C, saturation over ice). Check: 20 °C / 50% = 38.6 kJ/kg, saturated air at 30 °C = 100.1 kJ/kg.

Capacity calculation for AHU components

Heating or cooling coil capacity

$$\Phi = \dot{m} \cdot (h_1 - h_2) \quad \text{[kW]}$$

where $\dot{m}$ is the mass flow rate [kg/s].

Example — winter heating:

• $\dot{V} = 5\,000\ \text{m}^3/\text{h}$, $\rho = 1.23\ \text{kg/m}^3$ → $\dot{m} = \frac{5000}{3600} \cdot 1.23 = 1.708\ \text{kg/s}$
• $h_1 = 4.0\ \text{kJ/kg}$ (air after heat recovery, t = −3 °C, φ = 95%)
• $h_2 = 34.7\ \text{kJ/kg}$ (supply air, t = 22 °C, φ = 30%)
• $\Phi = 1.708 \cdot (34.7 - 4.0) = 1.708 \cdot 30.7 = \mathbf{52.4}\ \textbf{kW}$

Example — summer cooling:

• $\dot{m} = 1.708\ \text{kg/s}$
• $h_1 = 74.6\ \text{kJ/kg}$ (t = 32 °C, φ = 55%)
• $h_2 = 38.1\ \text{kJ/kg}$ (t = 14 °C, φ = 95%)
• $\Phi_{\text{cool}} = 1.708 \cdot (74.6 - 38.1) = 1.708 \cdot 36.5 = \mathbf{62.3}\ \textbf{kW}$

A common mistake in cooling design is to underestimate the latent part of the load and size the coil on the sensible component alone. Removing the moisture then eats capacity that was never allowed for, and the coil comes up short overall.

Another frequent one is a coil designed with an unnecessarily low ADP: the condensation it forces raises the total cooling capacity required, which would have been lower at a properly chosen ADP. Where dehumidification is not actually needed, a higher ADP gives a higher SHR.

ADP (apparatus dew point) is the temperature the air would reach if the coil took it all the way to saturation. In practice it corresponds to the coil surface temperature, and on the diagram it lies on the saturation curve, on the extension of the cooling line.

SHR is the ratio of the sensible capacity of the coil to its total capacity.

Splitting into sensible and latent components

Total cooling capacity = sensible cooling + dehumidification (latent):

$$\Phi_s = \dot{m} \cdot c_{pa} \cdot (t_1 - t_2) \approx 1.708 \cdot 1.006 \cdot 18 = \mathbf{30.9}\ \textbf{kW}$$

(For the sensible capacity we use only $c_{pa} = 1.006$ here as a simplification; more precisely, the moist specific heat is $c_p \approx 1.006 + 1.86 \cdot x$, which raises the result by a few percent.)

$$\Phi_{lat} = \Phi_c - \Phi_s = 62.3 - 30.9 = \mathbf{31.4}\ \textbf{kW}$$ $$\text{SHR} = \frac{\Phi_s}{\Phi_c} = \frac{30.9}{62.3} = \mathbf{0.50}$$

An SHR of 0.50 means half the coil duty goes into dehumidification, leaving little margin for sensible cooling. Covering the sensible load then calls for a larger coil surface or a higher air flow rate.

SHR range	Regime	What it implies
> 0.80	Predominantly sensible cooling	Little or no condensation
0.50 – 0.80	Balanced	The typical design, trading dehumidification against sensible cooling and efficiency
0.30 – 0.50	Heavy dehumidification	Low ADP (5–7 °C or below), larger surface
< 0.30	Extreme dehumidification	Special coils, possibly a two-stage arrangement

Enthalpy on the h-x diagram

The name of the Mollier diagram, “h-x diagram,” refers directly to the axes: h (enthalpy) and x (humidity ratio). The oblique coordinates allow enthalpy to be read directly from any point.

Reading from the diagram:

1. Find the point (the intersection of a temperature line and a relative-humidity line)
2. Draw a line of constant enthalpy (an isenthalp) from the point to the enthalpy scale at the edge
3. The value you read off is the enthalpy in kJ/kg dry air

Frequently asked questions

Why is the enthalpy reference point 0 °C? Enthalpy is always a relative quantity — only differences matter. ASHRAE defines the reference point as dry air at 0 °C and water in the liquid state at 0 °C (h = 0). This convention is used throughout ASHRAE Fundamentals.

How does enthalpy relate to a building’s energy demand? The annual energy use for ventilation and air conditioning depends on the total enthalpy of the supply air. A high-enthalpy climate (South Asia: t = 34 °C, φ = 80%, h = 105 kJ/kg) requires far more energy for cooling than a Central European climate (summer: t = 28 °C, φ = 55%, h = 62 kJ/kg).

Can enthalpy be calculated without knowing x? Yes — from the temperature and relative humidity you first compute the humidity ratio x (via the saturation vapor pressure), then substitute it into the enthalpy formula. PsychroView does this automatically once you enter t and φ.