Special Cases

If air contains excess water beyond saturation (fog), the enthalpy is given by:

h = c_p^a t + x_s (c_p^w t + h_{we}) + (x - x_s) c_w t \quad (7)

where $x_s$ is the humidity ratio at saturation (kg/kg), and $c_w = 4.19 \, \text{kJ/(kg·°C)}$ is the specific heat of liquid water.

For air containing water as ice or snow:

h = c_p^a t + x_s (c_p^w t + h_{we}) + (x - x_s) c_i t - (x - x_s) h_{im} \quad (8)

where $c_i = 2.05 \, \text{kJ/(kg·°C)}$ is the specific heat of ice, and $h_{im} = 335 \, \text{kJ/kg}$ is the melting heat of ice.

Examples

Example: Enthalpy of saturated air at 25 °C

Given:

Calculation:

h = (1.006 \, \text{kJ/(kg·°C)}) \cdot 25 + 0.0203 \cdot ((1.86 \, \text{kJ/(kg·°C)}) \cdot 25 + 2501 \, \text{kJ/kg})

h = 25.15 + (0.94 + 50.77) = 76.9 \, \text{kJ/kg}

Note: The latent heat component ( $50.77 \, \text{kJ/kg}$ ) dominates the total enthalpy.

Reference Points: The reference points for metric and imperial enthalpies differ.
- Metric (Eq. 5): Reference is $h = 0 \, \text{kJ/kg}$ at $t = 0 \, \text{°C}$ and $x = 0 \, \text{kg/kg}$ .
- Imperial (Eq. 6): Reference is $h = 0 \, \text{Btu/lb}$ at $t = 0 \, \text{°F}$ and $x = 0 \, \text{lb/lb}$ .
- Consequence: Enthalpy values cannot be directly converted between metric and imperial units.
Assumption for Sensible Heat: The enthalpy of dry air is set to zero at $0 \, \text{°C}$ for practical psychrometric calculations, simplifying the focus to enthalpy differences.