Introduction

In the previous lessons, we considered ‘Thermal Conduction’, and ‘Convection’. The third and final mode of heat transfer we shall consider is ‘Thermal Radiation’. Radiation is the process in which heat is transferred from a solid to its surroundings by electromagnetic wave dispersion.

Both conduction and convection rely on contact with matter such as neighbouring particles to energy transfer from high energy material (hot objects) to low energy material (cooler objects or surroundings). Thermal radiation is a type of electromagnetic energy (energy transferred by light waves) and therefore does not require mass exchange or any medium through which to act.

The rate of heat transfer by thermal radiation depends upon the ‘emissivity’ (ε)’ of the material (a constant between 0 and 1) and is used in the Stefan-Boltzmann Law of Radiation.

via GIPHY

Stefan-Boltzmann Law of Radiation

The Stefan-Boltzmann Law of Radiation is captured by the following equation

$q = ε σ A T^{4}$ eq.(11)

Which states that the rate of heat transfer ‘q’ is equal to the product of

the emissivity of the surface, (0 ≤ ‘ε’ ≤ 1).
the Stefan-Boltzmann constant, ‘σ’, with a value of 5.670 x 10^-8Wm^-2 K^-4.
the Area of the surface, ‘A’.
the fourth power of the Temperature (in Kelvin) of the surface ‘T’.

Note that theoretical materials that are have an emissivity ‘ε’ of 1 are termed ‘black body’ materials. A black body material absorbs all wavelengths of thermal radiation incident upon it and therefore no incident radiation is reflected.

law of radiation

In the real-world, all common materials are termed ‘grey body’ materials and have an emissivity ‘ε’ between 0 and 1. The emissivity of some common materials are listed in the table below:

Material	Emissivity (ε)	Material	Emissivity (ε)
Aluminium (commercial sheet)	0.09	Brick (fireclay)	0.75
Copper (electroplated)	390	Concrete	0.85
Steel (oxidised)	0.79	Glass (smooth)	0.92-0.94

Engineering toolbox

It should also be noted that as the Stefan-Boltzmann constant, ‘σ’ is very small, the rate of heat transfer by thermal radiation is generally small at low temperatures but as it is also dependent on the fourth power of temperature ‘T’ and therefore at high temperatures, the rate of heat transfer by thermal radiation increases drastically. We shall illustrate this in the following worked examples.

Example 6

Heat Radiation from the Surface of the Sun

If we assume that the surface temperature of the sun is 5505°C and that the sun can be regarded as a black body object, what is the rate of heat transfer per unit area of thermal radiation emitted from the sun’s surface?

Image from NASA

View Solution

What is the rate of heat transfer per unit area of thermal radiation emitted from the sun’s surface?

To answer this question, we shall rearrange the Stefan-Boltzmann Law of Thermal Radiation to describe rate of heat transfer per unit area:

$\frac{q}{A} = ε σ T^{4}$

$ε = 1$ and $T = 5505 ° C$ $= 5778 K$

$\frac{q}{A} = (5.67 \times 10^{-8}) \times {(5778)}^{4} = 6.32 \times 10^{7} W / m^{2}$

Net Radiation Loss Rate

A point to note is that Example 6 assumes that the surroundings of the sun are unheated, i.e., the temperature of the surrounding material is zero. If a hot object is radiating heat to cooler surroundings the net rate of heat transfer by thermal radiation is given by a modified Stefan-Boltzmann Law of Radiation equation:

$q = ε σ A_{h} (T_{h}^{4} - T_{c}^{4})$ eq.(11)

where

T_h is the Absolute Temperature (K) of the hot body.
T_c is the Absolute Temperature (K) of the cold body.
A_h is the Area (m²) of the of the hot body.

The following graph illustrates heat loss rates from a heated surface to unheated surrounding with mean radiant temperatures:

radiation heat transfer graph