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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
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 moist air.
Specific enthalpy of moist air can be expressed as:
h = ha + x hw (1)
where
h = specific enthalpy of moist air (kJ/kg, Btu/lb)
ha = specific enthalpy of dry air (kJ/kg, Btu/lb)
x = humidity ratio (kg/kg, lb/lb)
hw = specific enthalpy of water vapor (kJ/kg, Btu/lb)
Assuming constant pressure conditions the specific enthalpy of dry air can be expressed as:
ha = cpa t (2)
where
cpa = specific heat capacity of air at constant pressure (kJ/kgoC, kWs/kgK, Btu/lboF)
t = air temperature (oC, oF)
For air temperatures between -100oC (-150oF) and 100oC (212oF) the specific heat capacity can be set to
cpa = 1.006 (kJ/kgoC)
= 0.240 (Btu/lboF)
Note! that the enthalpy is 0 kJ/kg at 0oC. 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.
Assuming constant pressure conditions the specific enthalpy of water vapor can be expressed as:
hw = cpw t + hwe (3)
where
cpw = specific heat capacity of water vapor at constant pressure (kJ/kgoC, kWs/kgK)
t = water vapor temperature (oC)
hwe = evaporation heat of water at 0oC (kJ/kg)
For water vapor the specific heat capacity can be set to
cpw = 1.84 (kJ/kgoC)
= 0.444 (Btu/lboF)
The evaporation heat can be set to
hwe = 2502 (kJ/kg)
= 1061 (Btu/lb)
Using (2) and (3), (1) can be modified to
h = cpa t + x [cpw t + hwe] (1b)
(1b) in metric units
h = (1.006 kJ/kgoC) t + x [(1.84 kJ/kgoC) t + (2,502 kJ/kg)] (1c)
where
h = enthalpy (kJ/kg)
x = mass of water vapor (kg/kg)
t = temperature (oC)
(1b) in Imperial units
h = (0.240 Btu/lboF) t + x [(0.444 Btu/lboF) t + (1061 Btu/lb)] (1d)
where
h = enthalpy (Btu/lb)
x = mass of water vapor (lb/lb)
t = temperature (oF)
The enthalpy of humid air at 25oC with specific moisture content x = 0.0203 kg/kg, can be calculated as:
h = (1.006 kJ/kgoC) (25oC) + (0.0203 kg/kg) [(1.84 kJ/kgoC) (25oC) + (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 be almost neglected.
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 = cpa t + xs [cpw t + hwe] + (x - xs) cw t (2)
where
xs = humidity ratio at saturation (kg/kg)
cw = 4.19 - specific heat capacity of water (kJ/kgoC)
If the air contains water as ice or snow, the enthalpy of air can be expressed as:
h = cpa t + xs [cpw t + hwe] + (x - xs) ci t - (x - xs) him (3)
where
ci = 2.05 - specific heat capacity of ice (kJ/kg.oC)
him= 335 - melting heat of ice (kJ/kg)
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