Heating Humid Air

Calculating enthalpy change and temperature rise when sensible heating moist and humid air

The process of sensible heating of air can be expressed in the Mollier diagram as:

Heating of air moves the air condition from A to B along a constant specific humidity - x - line. The supplied heat - dH - can be read in the diagram as shown.

The heating process of air expressed can also be expressed in the psychrometric chart as:

Note! When heating air the specific moisture remains constant and the relative humidity decreases.

Calculating Enthalpy

The enthalpy of moist air can be expressed 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.^oC, kWs/kg.K)

t = air temperature (^oC)

x = humidity ratio (kg/kg)

c_pw = 1.84 - specific heat capacity of water vapor at constant pressure (kJ/kg.^oC, kWs/kg.K)

h_we = 2,502 - evaporation heat of water at 0^oC (kJ/kg)

(1) can be modified as:

h = 1.01 (kJ/kg.^oC) t + x [1.84 (kJ/kg.^oC) t + 2,502 (kJ/kg)] (1b)

The Enthalpy Difference

The enthalpy difference when heating air without changing moisture content can be expressed as:

dh_A-B = 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 Difference Heating Air

The specific humidity of air at 25^oC and relative moisture 50% is 0.0115 kg/kg - check the Mollier diagram. The change in enthalpy when heating the air to 35^oC can be calculated as:

dh_A-B = (1.01 kJ/kg.^oC)(35^oC - 25^oC) + (0.0115 kg/kg) (1.84 kJ/kg.^oC) (35^oC - 25^oC)

= 10.1 kJ/kg + 0.2 kJ/kg

= 10.3 kJ/kg

Note! The contribution from the water vapor is relatively small and for practical purposes it may often be neglected. (2) can then be modified to:

dh_A-B = c_pa( t_B - t_A) (2b)

Increase in Temperature when Heating Air

If heat is added to humid air the increase air temperature can be calculated by modifying (2b) [(2) to be more exact]:

t_B - t_A = dh_A-B / c_pa (2b)

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.^oC)

= 10^oC

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 = v ρ (h_B - h_A) (3a)

where

v = volume flow (m³/s)

ρ = density of air (kg/m³)

Note! The density vary with temperature. At 0^oC the density is 1.293 kg/m³. At 80^oC the density is 1.0 kg/m³.

It's common to express the sensible heat flow rate as:

q = m c_pa (t_B - t_A) (3b)

or alternatively:

q = v 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 (^oC)

Example - Heating Air

1 m³/s of air at 15^oC and relative humidity 60% (A) is heated to 30^oC (B). The surface temperature of the heating coil is 80^oC. The density of air at 20^oC is 1.205 kg/m³.

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^oC - 15^oC) / (80^oC - 15^oC)

= 0.23

The heat flow can be calculated as:

q = (1 m³/s) (1.205 kg/m³) ((46 kJ/kg) - (31 kJ/kg))

= 18 (kJ/s, kW)

As an alternative, as one of the most common methods:

q = (1 m³/s) (1.205 kg/m³) (1.01 kJ/kg.^oC) (30^oC - 15^oC)

= 18.3 (kJ/s, kW)

Note! Due to inaccuracy when working diagrams there is a small difference between the total heat flow and the sum of the latent and sensible heat. This inaccuracy is in general within acceptable limits.

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