# 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^{ 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_{ 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 ^{ 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_{ A-B }=(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_{ 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^{ 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*. At

^{ 3 }*80*the density is

^{ o }C*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*and relative humidity

^{ o }C*60%*(A) is heated to

*30*(B). The surface temperature of the heating coil is

^{ o }C*80*. The density of air at

^{ o }C*20*is

^{ o }C*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 - psychrometric charts, Mollier diagrams, air-condition temperatures and absolute and relative humidity and moisture content.

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