Humid Air  Heating
Enthalpy change and temperature rise when heating humid air without adding moisture.
The process of sensible heating of air  heating without adding moisture  can be visualized in the Mollier diagram as:
Sensible heating of air changes the state of the air from A to B along the constant specific humidity  x  line. The supplied heat  dH  can be estimated as indicated in the diagram above.
The heating process can also be visualized in the psychrometric chart
Note!  when sensible heating of air  the specific moisture remains constant  the relative humidity is decreased.
Calculating Enthalpy
The enthalpy of moist air can be calculated as:
h = c _{ pa } t + x [c _{ pw } t + h _{ we } ] (1)
where
h = specific enthalpy of moist air (kJ/kg)
c _{ pa } = 1.01  specific heat capacity of air at constant pressure (kJ/kg ^{o}C, kWs/kgK)
t = air temperature (^{o}C)
x = humidity ratio (kg/kg)
c _{ pw } = 1.84  specific heat capacity of water vapor at constant pressure (kJ/kg. ^{o}C, kWs/kg.K)
h _{ we } = 2502  evaporation heat of water at 0 ^{o}C (kJ/kg)
(1) can be modified to:
h = 1.01 (kJ/kg. ^{o}C) t + x [1.84 (kJ/kg. ^{o}C) t + 2502 (kJ/kg)] (1b)
The Enthalpy Difference
The enthalpy difference when heating air without adding moisture can be calculated as:
dh _{ AB } = c _{ pa } t _{ B } + x [c _{ pw } t _{ B } + h _{ we } ]  c _{ pa } t _{ A } + x [c _{ pw } t _{ A } + h _{ we } ]
= c _{ pa } (t _{ B }  t _{ A } ) + x c _{ pw } (t _{ B }  t _{ A } ) (2)
Example  Enthalpy Change when Heating Air
The specific humidity of air at 25 ^{o}C and relative moisture 50% is 0.0115 kg/kg  check the Mollier diagram . The change in enthalpy when heating the air to 35 ^{o}C can be calculated as:
dh _{ AB } = (1.01 kJ/kg ^{o}C)(35 ^{o}C  25 ^{o}C) + (0.0115 kg/kg) (1.84 kJ/kg ^{o}C) (35 ^{o}C  25 ^{o}C)
= (10.1 kJ/kg) + (0.2 kJ/kg)
= 10.3 (kJ/kg)
Note!  the contribution from the water vapor is relatively small and can for practical purposes often be neglected. (2) can then be modified to:
dh _{ AB } = c _{ pa } ( t _{ B }  t _{ A } ) (2b)
Increase in Temperature when Heating Air
If heat is added to humid air the increase in air temperature can be calculated by modifying (2b) to:
t _{ B }  t _{ A } = dh _{ AB } / c _{ pa } (2c)
Example  Heating Air and Temperature Rise
If 10.1 kJ is added to 1 kg air the temperature rise can be calculated as:
t _{ B }  t _{ A } = (10.1 kJ/kg) / (1.01 kJ/kg ^{o}C)
= 10 (^{o}C)
Heat Flow in a Heating Coil
The total heat flow rate through a heating coil can be calculated as:
q = m (h _{ B }  h _{ A } ) (3)
where
q = heat flow rate (kJ/s, kW)
m = mass flow rate of air (kg/s)
The total heat flow can also be expressed as:
q _{ s } = L ρ (h _{ B }  h _{ A } ) (3a)
where
L = air flow rate (m^{3} /s)
ρ = density of air (kg/m^{3} )
Note! The density of air varies with temperature. At 0 ^{o}C the density is 1.293 kg/m^{3} . At 80 ^{o}C the density is 1.0 kg/m^{3} .
It's common to express sensible heat flow rate as:
q = m c _{ pa } (t _{ B }  t _{ A } ) (3b)
or alternatively:
q = L ρ c _{ pa } (t _{ B }  t _{ A } ) (3c)
Heating Coil Effectiveness
For a limited heating coil surface the average surface temperature will always be higher than the outlet air temperature. The effectiveness of a heating coil can be expressed as:
μ = (t _{ B }  t _{ A } ) / (t _{ HC }  t _{ A } ) (4)
where
μ = heating coil effectiveness
t _{ HC } = mean surface temperature of the heating coil (^{o}C)
Example  Heating Air
1 m^{3} /s of air at 15 ^{o}C and relative humidity 60% (A) is heated to 30 ^{o}C (B). The surface temperature of the heating coil is 80 ^{o}C . The density of air at 20 ^{o}C is 1.205 kg/m^{3} .
From the Mollier diagram the enthalpy in (A) is 31 kJ/kg and in (B) 46 kJ/kg .
The heating coil effectiveness can be calculated as:
μ = (30 ^{o}C  15 ^{o}C) / (80 ^{o}C  15 ^{o}C)
= 0.23
The heat flow can be calculated as:
q = (1 m^{3} /s) (1.205 kg/m^{3} ) ((46 kJ/kg)  (31 kJ/kg))
= 18 (kJ/s, kW)
As an alternative, as one of the most common methods:
q = (1 m^{3} /s) (1.205 kg/m^{3} ) (1.01 kJ/kg. ^{o}C) (30 ^{o}C  15 ^{o}C)
= 18.3 (kJ/s, kW)
Note! Due to the inaccuracy when working with diagrams there is a small difference between the total heat flow and the sum of the latent and sensible heat. In general  this inaccuracy is within acceptable limits.
Related Topics

Air Psychrometrics
Moist and humid air calculations. Psychrometric charts and Mollier diagrams. Aircondition systems temperatures, absolute and relative humidities and moisture content in air.
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