Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications!

This is an AMP page - Open full page! for all features.

• the most efficient way to navigate the Engineering ToolBox!

# Heating Humid Air

## 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 = cpa t + x [cpw t + hwe]                            (1)

where

h = specific enthalpy of moist air (kJ/kg)

cpa = 1.01 - specific heat capacity of air at constant pressure (kJ/kgoC, kWs/kgK)

t = air temperature (oC)

x = humidity ratio (kg/kg)

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

hwe = 2502 - evaporation heat of water at 0oC (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:

dhA-B = cpa tB + x [cpw tB + hwe] - cpa tA + x [cpw tA + hwe]

= cpa(tB - tA) + x cpw (tB - tA)                                                (2)

#### Example - Enthalpy Change when Heating Air

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

dhA-B = (1.01 kJ/kgoC)(35oC - 25oC) + (0.0115 kg/kg) (1.84 kJ/kgoC) (35oC - 25oC)

= (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:

dhA-B = cpa( tB - tA)                                 (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:

tB - tA = dhA-B / cpa                                  (2c)

#### Example - Heating Air and Temperature Rise

If 10.1 kJ is added to 1 kg air the temperature rise can be calculated as:

tB - tA = (10.1 kJ/kg) / (1.01 kJ/kgoC)

= 10 (oC)

### Heat Flow in a Heating Coil

The total heat flow rate through a heating coil can be calculated as:

q = m (hB - hA)                                     (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:

qs = L ρ (hB - hA)                                     (3a)

where

L = air flow rate (m3/s)

ρ = density of air (kg/m3)

Note! The density of air varies with temperature. At 0oC the density is 1.293 kg/m3. At 80oC the density is 1.0 kg/m3.

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

q = m cpa (tB - tA)                                  (3b)

or alternatively:

q = L ρ cpa (tB - tA)                                   (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 = (tB - tA) / (tHC - tA)                                   (4)

where

μ = heating coil effectiveness

tHC = mean surface temperature of the heating coil (oC)

### Example - Heating Air

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

= 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) (30oC - 15oC)

= 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 - The study of moist and humid air - psychrometric charts, Mollier diagrams, air-condition temperatures and absolute and relative humidity and moisture content

## Search Engineering ToolBox

• the most efficient way to navigate the Engineering ToolBox!

## SketchUp Extension - Online 3D modeling!

Add standard and customized parametric components - like flange beams, lumbers, piping, stairs and more - to your Sketchup model with the Engineering ToolBox - SketchUp Extension - enabled for use with the amazing, fun and free SketchUp Make and SketchUp Pro .Add the Engineering ToolBox extension to your SketchUp from the Sketchup Extension Warehouse!

## Privacy

We don't collect information from our users. Only emails and answers are saved in our archive. Cookies are only used in the browser to improve user experience.

Some of our calculators and applications let you save application data to your local computer. These applications will - due to browser restrictions - send data between your browser and our server. We don't save this data.

Temperature

oC
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

kgf
N
lbf

Velocity

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

Pressure

Pa (N/m2)
bar
mm H2O
kg/cm2
psi
inches H2O

Flow

m3/s
m3/h
US gpm
cfm

1 24