Humid Air - Heating

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^oC, kWs/kgK)

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 = 2502 - evaporation heat of water at 0^oC (kJ/kg)

(1) can be modified to:

h = 1.01 (kJ/kg.^oC) t + x [1.84 (kJ/kg.^oC) t + 2502 (kJ/kg)]                                 (1b)

The Enthalpy Difference

The enthalpy difference when heating air without adding moisture can be calculated 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 Change when 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 can for practical purposes 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 in air temperature can be calculated by modifying (2b) to:

t_B - t_A = dh_A-B / 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^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 = L ρ (h_B - h_A)                                     (3a)

where

L = air flow rate (m³/s)

ρ = density of air (kg/m³)

Note! The density of air varies 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 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 (^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 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.

Temperature ^oC
K
^oF

Length m
km
in
ft
yards
miles
naut miles

Area m²
km²
in²
ft²
miles²
acres

Volume m³
liters
in³
ft³
us gal

Weight kg_f
N
lbf

Velocity m/s
km/h
ft/min
ft/s
mph
knots

Pressure Pa
bar
mm H₂O
kg/cm²
psi
inches H₂O

Flow m³/s
m³/h
US gpm
cfm