Moist Air  Enthalpy
Sensible and latent heat of moist air.
Moist air is a mixture of dry air and water vapor. In atmospheric air water vapor content varies from 0  3% by mass. The enthalpy of moist and humid air includes the
 enthalpy of the dry air  the sensible heat
 enthalpy of the evaporated water in the air  the latent heat
The total enthalpy  sensible and latent  is used when calculating cooling and heating processes.
Specific enthalpy  h  (J/kg, Btu/lb) of moist air is defined as the total enthalpy (J, Btu) of the dry air and the water vapor mixture  per unit mass (kg, lb) of dry 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, Btu/lb)
h_{a} = specific enthalpy of dry air (kJ/kg, Btu/lb)
x = humidity ratio (kg/kg, lb/lb)
h_{w} = specific enthalpy of water vapor (kJ/kg, Btu/lb)
Specific Enthalpy of Dry Air  the 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 of air at constant pressure (kJ/kg^{o}C, kWs/kgK, Btu/lb^{o}F)
t = air temperature (^{o}C, ^{o}F)
For air temperature between 100^{o}C (150^{o}F) and 100^{o}C (212^{o}F) the specific heat can be set to
c_{pa} = 1.006 (kJ/kg^{o}C)
= 0.240 (Btu/lb^{o}F)
Note!  that the enthalpy is 0 kJ/kg at 0^{o}C. This is not correct according the definition of enthalpy in the thermodynamics, but for practical purposes in air psychrometrics this assumption is good enough since our interest is the enthalpy difference.
Specific Enthalpy of Water Vapor  the 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 of water vapor at constant pressure (kJ/kg^{o}C, kWs/kgK)
t = water vapor temperature (^{o}C)
h_{we} = evaporation heat of water at 0^{o}C (kJ/kg)
For water vapor the specific heat can be set to
c_{pw} = 1.86 (kJ/kg^{o}C)
= 0.444 (Btu/lb^{o}F)
The evaporation heat (water at 0^{o}C, 32^{o}F) can be set to
h_{we} = 2501 (kJ/kg)
= 1061 (Btu/lb)
Using (2) and (3), (1) can be modified to
h = c_{pa} t + x [c_{pw} t + h_{we}] (4)
(1b) in metric units
h = (1.006 kJ/kg^{o}C) t + x [(1.86 kJ/kg^{o}C) t + (2501 kJ/kg)] (5)
where
h = enthalpy (kJ/kg)
x = mass of water vapor (kg/kg)
t = temperature (^{o}C)
(1b) in Imperial units
h = (0.240 Btu/lb^{o}F) t + x [(0.444 Btu/lb^{o}F) t + (1061 Btu/lb)] (6)
where
h = enthalpy (Btu/lb)
x = mass of water vapor (lb/lb)
t = temperature (^{o}F)
Note!  the "reference" points for the metric and imperial enthalpies are different.
 for eq. (5) metric units  the "reference" point for enthalpy h = 0 (kJ/kg) is t = 0 ^{o}C (32^{o}F) and x = 0 kg/kg
 for eq. (6) imperial units  the "reference" point for enthalpy h = 0 (Btu/lb) is t = 0 ^{o}F (17.8^{o}C) and x = 0 lb/lb. The evaporation heat for water at 0 ^{o}F is 1061 Btu/lb as used in eq. (6).
You can not convert from metric to imperial enthalpy or vice versa directly.
Example  Enthalpy in Moist Air
The enthalpy of humid air at 25^{o}C with specific moisture content x = 0.0203 kg/kg (saturation), can be calculated as:
h = (1.006 kJ/kg^{o}C) (25^{o}C) + (0.0203 kg/kg) [(1.86 kJ/kg^{o}C) (25^{o}C) + (2501 kJ/kg)]
= (25.15 kJ/kg) + [(0.94 kJ/kg) + (50.77 kJ/kg)]
= 76.9 (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 be almost neglected.
Enthalpy of Moist Air containing Water as 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 (7)
where
x_{s} = humidity ratio at saturation (kg/kg)
c_{w} = 4.19  specific heat water (kJ/kg^{o}C)
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} (8)
where
c_{i} = 2.05  specific heat ice (kJ/kg^{o}C)
h_{im}= 335  melting heat of ice (kJ/kg)
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