Enthalpy of Moist and Humid Air

Engineering ToolBox > Enthalpy of Moist and Humid Air

The enthalpy of moist and humid air consist of sensible heat and latent heat. The enthalpy is important for calculating cooling and heating processes

Moist air is a mixture of dry air and water vapor. In atmospheric air, water vapor content varies from 0 to 3% by mass. The enthalpy of moist and humid air includes the

enthalpy of the dry air - the sensible heat - and
the enthalpy of the evaporated water - the latent heat

Specific enthalpy - h - (kJ/kg) of moist air is defined as the total enthalpy of the dry air and the water vapor mixture per kg of moist air.

Specific Enthalpy of Moist Air

Specific enthalpy of moist air can be expressed as:

h = h_a + x h_w (1)

where

h = specific enthalpy of moist air (kJ/kg)

h_a = specific enthalpy of dry air (kJ/kg)

x = humidity ratio (kg/kg)

h_w = specific enthalpy of water vapor (kJ/kg)

Specific Enthalpy of Dry Air - Sensible Heat

Assuming constant pressure conditions the specific enthalpy of dry air can be expressed as:

h_a = c_pa t (2)

where

c_pa = specific heat capacity of air at constant pressure (kJ/kg.^oC, kWs/kg.K)

t = air temperature (^oC)

For air temperatures between -100^oC and 100^oC the specific heat capacity can be set to

c_pa = 1.006 (kJ/kg^oC)

Note! that the enthalpy is 0 at 0^oC. This is of course not correct according the enthalpy definition in the thermodynamics, but for practical purposes in air psychrometrics it do because our interest is the enthalpy difference.

Specific Enthalpy of Water Vapor - Latent Heat

Assuming constant pressure conditions the specific enthalpy of water vapor can be expressed as:

h_w = c_pw t + h_we (3)

where

c_pw = specific heat capacity of water vapor at constant pressure (kJ/kg.^oC, kWs/kg.K)

t = water vapor temperature (^oC)

h_we = evaporation heat of water at 0^oC (kJ/kg)

For water vapor the specific heat capacity can be set to

c_pw = 1.84 (kJ/kg.^oC)

The evaporation heat can be set to

h_we = 2,502 (kJ/kg)

Using (2) and (3), (1) can be modified to

h = c_pa t + x [c_pw t + h_we] (1b)

h = 1.006 (kJ/kg.^oC) t + x [1.84 (kJ/kg.^oC) t + 2,502 (kJ/kg)] (1c)

Example - Enthalpy in Moist Air

The enthalpy of humid air at 25^oC with specific moisture content x = 0.0203 kg/kg, can be calculated as:

h = 1.006 (kJ/kg.^oC) 25^oC + 0.0203 (kg/kg) [1.84 (kJ/kg.^oC) 25^oC + 2,502 (kJ/kg)]

= 25.15 (kJ/kg) + 0.93 (kJ/kg) + 50.79 (kJ/kg)

= 76.87 (kJ/kg)

Note! The latent heat due to evaporation of water is the major part of the enthalpy. The sensible heat due to heating evaporated water vapor can almost be neglected.

Enthalpy of Moist Air containing Water - Fog

If the air contains more water than limited by saturation, some of the water exists as droplets - as fog. The enthalpy of moist air with fog can be expressed as:

h = c_pa t + x_s [c_pw t + h_we] + (x - x_s) c_w t (2)

where

x_s = humidity ratio at saturation (kg/kg)

c_w = 4.19 - specific heat capacity of water (kJ/kg.^oC)

Enthalpy of Moist Air containing Ice or Snow

If the air contains water as ice or snow, the enthalpy of air can be expressed as:

h = c_pa t + x_s [c_pw t + h_we] + (x - x_s) c_i t - (x - x_s) h_im (3)

where

c_i = 2.05 - specific heat capacity of ice (kJ/kg.^oC)

h_im= 335 - melting heat of ice (kJ/kg)