# Enthalpy of Moist Air

## The enthalpy of humid air consists of sensible and latent heat

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*and

^{o}F)*100*

^{o}C*(212*the specific heat can be set to

^{o}F)

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}= 1.84 (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 [(1.84 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*and^{o}F)*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*and^{o}C)*x = 0 lb/lb*. The evaporation heat for water at*0*is^{o}F*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) [(1.84 kJ/kg^{o}C) (25^{o}C) + (2501 kJ/kg)]

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