# Radiation Heat Transfer

## Heat transfer due to emission of electromagnetic waves is known as thermal radiation

Heat transfer through radiation takes place in form of electromagnetic waves mainly in the infrared region. Radiation emitted by a body is a consequence of thermal agitation of its composing molecules. Radiation heat transfer can be described by reference to the **'black body'**.

### The Black Body

The black body is defined as a body that absorbs all radiation that falls on its surface. Actual black bodies don't exist in nature - though its characteristics are approximated by a hole in a box filled with highly absorptive material. The emission spectrum of such a black body was first fully described by Max Planck.

A black body is a hypothetical body that completely absorbs all wavelengths of thermal radiation incident on it. Such bodies do not reflect light, and therefore appear black if their temperatures are low enough so as not to be self-luminous. All black bodies heated to a given temperature emit thermal radiation.

The radiation energy per unit time from a **black body** is proportional to the fourth power of the absolute temperature and can be expressed with **Stefan-Boltzmann Law ** as

q = σ T^{4}A(1)

where

q= heat transfer per unit time (W)

σ= 5.6703 10^{-8}(W/m^{2}K^{4}) -TheStefan-Boltzmann Constant

T= absolute temperature in kelvins (K)

A= area of the emitting body (m^{2})

#### The Stefan-Boltzmann Constant in Imperial Units

σ= 5.6703 10^{-8}(W/m^{2}K^{4})

= 1.714 10^{-9}( Btu/(h ft^{2}^{o}R^{4}) )

= 1.19 10^{-11}( Btu/(h in^{2 o}R^{4}) )

**Example - Heat Radiation from the surface of the Sun**

If the surface temperature of the sun is* 5800 K* and if we assume that the sun can be regarded as a black body the radiation energy per unit area can be expressed by modifying* (1)* to

q / A = σ T^{4}= (

5.6703 10^{-8}W/m^{2}K^{4}) (5800 K)^{4}

= 6.42 10^{7}(W/m^{2})

### Gray Bodies and Emissivity Coefficients

For objects other than ideal black bodies ('gray bodies') the **Stefan-Boltzmann Law** can be expressed as

q = ε σ T^{4}A(2)

where

ε= emissivity coefficient of the object (one - 1 - for a black body)

For the gray body the incident radiation (also called irradiation) is partly reflected, absorbed or transmitted.

The emissivity coefficient is in the range *0 < **ε** < 1,* depending on the type of material and the temperature of the surface.

- oxidized Iron at
*390*^{o}F (199^{o}C) >*ε**= 0.64* - polished Copper at
*100*^{o}F (38^{o}C) >*ε**= 0.03* - emissivity coefficients for some common materials

### Net Radiation Loss Rate

If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as

q = ε σ (T_{h}^{4}- T_{c}^{4}) A_{h}(3)

where

T_{h}= hot body absolute temperature (K)

T_{c}= cold surroundings absolute temperature (K)

A_{h}= area of the hot object (m^{2})

Heat loss from a heated surface to unheated surroundings with mean radiant temperatures are indicated in the chart below.

Download and print Heat Transfer by Radiation chart

### Radiation Heat Transfer Calculator

This calculator is based on equation *(3)* and can be used to calculate the heat radiation from a warm object to colder surroundings.

Note that the input temperatures are in degrees Celsius.

* t _{h} - object hot temperature (^{o}C)*

* t _{c} - surroundings cold temperature (^{o}C)*

* A _{c} - object area (m^{2})*

### Lambert's cosine law

Heat emission from a surface in an angle *β* can be expressed with Lambert's cosine law as

*q*_{β}* = q cos β (4)*

*where *

*q*_{β}* = heat emission in angle β *

*q = heat emission from the surface*

*β = angle *

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