Calculating Moist Air Enthalpy: Formula, Tables, Psychrometric Chart

The enthalpy of moist air is the total heat content of 1 lb of dry air including its water vapor. The IP formula: $h = 0.240 \cdot t + W \cdot (1061 + 0.444 \cdot t)$ [Btu/lb], where $t$ is the temperature [°F] and $W$ is the humidity ratio [lb/lb dry air]. From the enthalpy you compute the capacity of any process directly: $Q = 4.5 \cdot \dot{V} \cdot \Delta h$.

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 (68°F, φ = 90%) and air B (68°F, 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

$$h = c_{pa} \cdot t + W \cdot (h_{g0} + c_{pv} \cdot t)$$

where:

• $c_{pa} = 0.240\ \text{Btu/(lb·°F)}$ — specific heat capacity of dry air
• $c_{pv} = 0.444\ \text{Btu/(lb·°F)}$ — specific heat capacity of water vapor
• $h_{g0} = 1061\ \text{Btu/lb}$ — latent heat of vaporization at the 0°F datum
• $t$ = temperature [°F]
• $W$ = humidity ratio [lb/lb dry air]

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

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

The SI counterpart is $h = 1.006\,t + x\,(2501 + 1.86\,t)$ with $t$ in °C. The two are not a straight unit conversion of one another — they start from different reference points, so only enthalpy differences carry over between the systems unchanged.

Worked examples

Outdoor winter air: $t = 14\ ^{\circ}\text{F}$, $\varphi = 80\%$ → $W \approx 0.00128\ \text{lb/lb}$

$$\begin{aligned} h &= 0.240 \cdot 14 + 0.00128 \cdot (1061 + 0.444 \cdot 14) \\ &= 3.36 + 0.00128 \cdot 1067.2 \\ &= 3.36 + 1.37 = \mathbf{4.73}\ \textbf{Btu/lb} \end{aligned}$$

Summer supply air: $t = 79\ ^{\circ}\text{F}$, $\varphi = 55\%$ → $W \approx 0.01170\ \text{lb/lb}$

$$\begin{aligned} h &= 0.240 \cdot 79 + 0.01170 \cdot (1061 + 0.444 \cdot 79) \\ &= 18.96 + 0.01170 \cdot 1096.1 \\ &= 18.96 + 12.82 = \mathbf{31.78}\ \textbf{Btu/lb} \end{aligned}$$

Enthalpy table

t [°F]	φ = 30%	φ = 50%	φ = 70%	φ = 100%
14	3.9	4.2	4.6	5.1
32	8.9	9.7	10.5	11.8
50	14.5	16.1	17.8	20.3
68	21.1	24.3	27.5	32.4
79	25.9	30.6	35.4	42.6
86	29.4	35.3	41.4	50.7

Values [Btu/lb dry air], standard atmospheric pressure; humidity ratios from the PsychroView calculation engine (saturation vapor pressure per Hyland–Wexler, ASHRAE Fundamentals; below 32°F, saturation over ice), enthalpy per the ASHRAE IP form above. Check: 68°F / 50% = 24.3 Btu/lb, saturated air at 86°F = 50.7 Btu/lb.

Capacity calculation for AHU components

In IP practice the mass flow rate is usually folded into a constant, giving the two equations every design engineer knows by heart:

$$Q_{total} = 4.5 \cdot \dot{V} \cdot \Delta h \qquad Q_{sensible} = 1.08 \cdot \dot{V} \cdot \Delta t$$

with $\dot{V}$ in CFM, $\Delta h$ in Btu/lb and $\Delta t$ in °F, both giving Btu/h. The 4.5 is 60 min/h × 0.075 lb/ft³ (standard air density); the 1.08 is the same product times $c_{pa} = 0.240$. At altitude or at markedly different temperatures, recompute the density rather than trusting 0.075.

Example — winter heating at 3,000 CFM:

• $h_1 = 9.54\ \text{Btu/lb}$ (air after heat recovery, t = 27°F, φ = 95%)
• $h_2 = 22.74\ \text{Btu/lb}$ (supply air, t = 72°F, φ = 30%)
• $Q = 4.5 \cdot 3000 \cdot (22.74 - 9.54) = 4.5 \cdot 3000 \cdot 13.20 = \mathbf{178{,}200}\ \textbf{Btu/h}$ (14.9 tons)

Example — summer cooling at 3,000 CFM:

• $h_1 = 40.08\ \text{Btu/lb}$ (t = 90°F, φ = 55%)
• $h_2 = 23.93\ \text{Btu/lb}$ (t = 57°F, φ = 95%)
• $Q_{total} = 4.5 \cdot 3000 \cdot 16.15 = \mathbf{218{,}000}\ \textbf{Btu/h}$ (18.2 tons)

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 chart 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):

$$Q_s = 1.08 \cdot 3000 \cdot (90 - 57) = \mathbf{106{,}900}\ \textbf{Btu/h}$$

(The 1.08 factor carries only $c_{pa} = 0.240$; more precisely the moist specific heat is $c_p \approx 0.240 + 0.444 \cdot W$, which raises the result by a few percent.)

$$Q_{lat} = Q_{total} - Q_s = 218{,}000 - 106{,}900 = \mathbf{111{,}100}\ \textbf{Btu/h}$$ $$\text{SHR} = \frac{Q_s}{Q_{total}} = \frac{106{,}900}{218{,}000} = \mathbf{0.49}$$

An SHR near 0.5 means half the coil duty goes into wringing water out of the air, leaving little margin for sensible cooling. Covering the sensible load then calls for a larger coil surface or a higher air flow rate, which is worth remembering when someone sizes a coil from the temperature drop alone.

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 (41–45°F or below), larger surface
< 0.30	Extreme dehumidification	Special coils, possibly a two-stage arrangement

Enthalpy on the psychrometric chart

On the Carrier chart, enthalpy is carried by oblique lines running parallel to the wet-bulb lines, with the scale printed along the upper-left edge, outside the saturation curve. The European Mollier chart takes its name “h-x diagram” from putting those same two quantities — h and x — directly on its axes.

Reading from the chart:

1. Find the point (the intersection of a dry-bulb line and a relative-humidity line)
2. Follow the constant-enthalpy line from the point up and to the left, out to the enthalpy scale
3. The value you read off is the enthalpy in Btu/lb dry air

Frequently asked questions

What is the enthalpy reference point in IP units? Enthalpy is always a relative quantity — only differences matter. In IP, ASHRAE sets the enthalpy of dry air to zero at 0°F and the enthalpy of saturated liquid water to zero at 32°F. The SI convention uses 32°F (0 °C) for both, which is why the same air state carries different absolute numbers in the two systems.

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 = 93°F, φ = 80%, h = 52.4 Btu/lb) requires far more energy for cooling than a temperate one (summer: t = 82°F, φ = 55%, h = 33.9 Btu/lb).

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