Unit 42: Heat Transfer and Combustion Lesson 4: Radiation

Aim LO1: Understanding Heat Transfer Rates for Composite Systems.

Radiation Thermal radiation is the energy emitted by matter that is at some finite temperature Although we are primarily interested in radiation from solid substances, radiation can also be emitted by liquids and gases. Thermal radiation is attributed to changes in electron energy within atoms and molecules.

Radiation As electron energy levels change, energy is released which travels in the form of electromagnetic waves of varying wavelength. When striking a body, the emitted radiation is either absorbed by, reflected by or transmitted through the body. If we assume that the quantity of radiation striking a body is unity, then we can state that… α + Υ + τ = 1

Radiation α + γ + τ = 1 α is the fraction of incident radiation energy absorbed called the absorptivity γ is the fraction of energy reflected called reflectivity τ is the fraction of energy transmitted called transmissivity

Black and Grey Body Radiation It is useful to define an ideal body which absorbs all the incident radiation that falls on it. This body in known as a black body Thus for a black body α = 1. Thus a black body is a perfect absorber because it absorbs the entire radiation incident upon it irrespective of wavelength.

Black and Grey Body Radiation We would also expect a black body to be the best possible emitter at any given temperature otherwise its temperature would rise above that of its surroundings which is not the case In practice an almost perfect black body consists of an enclosure such as a cylinder that a dull black interior and a small hole

Black and Grey Body Radiation Small hole Incident radiation Dull black inner surface

Black and Grey Body Radiation Radiation entering the hole has little chance of escaping because any energy not absorbed when the wall of the cylinder is first struck will continue to strike other surfaces, gradually depleting its energy until none is left. When the cylinder is heated by an external source, radiation appears from the hole (this radiation may be any colour according to its wavelength)

Black and Grey Body Radiation Thus a whole spectrum of energy may be produced when a black body is heated with the form of the curve shown below.. Relative energy of emitted radiation Radiated Energy Spectrum Wavelength of radiation in μm 1 2 3 4 5 6 7

Black and Grey Body Radiation As the temperature rises the energy in each wave band increases, so the body will become brighter At some temperature, T, the energy radiated is a maximum for a certain wavelength, which then decreases with rising temperature (around 1000K) only a small fraction of the radiation appears as visible light. (Temperatures of around 4000K are required to produce a maximum in the visible region!).

Black and Grey Body Radiation By modelling black bodies as shown in the previous diagram and noting the criteria for the spectrum of radiated energy emitted, we know that for all practical purposes a black body is both a perfect emitter and absorber. Thus we can use this model to compare the radiation emitted by other bodies without these properties

Black and Grey Body Radiation The emissivity (ε) of a body is the ratio of the energy emitted by the body to that emitted by a black body at the same temperature and wavelength Such matter is known as a grey body!

Relative energy of emitted radiation Grey Body Radiation A comparison of the spectral emissive energy of a tungsten filament lamp (non-black body i,.e. a grey body) is shown below.. Emissivity of non-black body ε = EN / EB Relative energy of emitted radiation Black Body EB Non-black body (eg a tungsten filament lamp) EN Wavelength of radiation in μm 1 2 3 4 5 6 7

Stefan-Boltzmann Law It was found by Stefan and subsequently proved theoretically by Boltzmann that, the total energy radiated by all wavelengths per unit area per unit time by a black body is directly proportional to the fourth power of the absolute temperature of the body E = σT4 Where σ = 5.672 x 10-8 W/m2.K4 is the Stefan-Boltzmann constant of proportionality found bt experimentation

Stefan-Boltzmann Law Note the emissivity ε, the energy emitted by a non-black body (i.e. a grey body) is… E = εσT4 When a body is very small compared to its large surroundings, a negligible amount of radiation is reflected from the surroundings onto the body so in effect the surroundings are black.

Stefan-Boltzmann Law Under these circumstances it can be shown that the rate of heat transfer from the body to its surroundings is… Q = εσ(Ts4 – Tsur4) Where TS is the temperature of the solid body and Tsur is the temperature of the immediate surroundings .

Stefan-Boltzmann Law Apart from the radiation there are many instances where the surface of a body may transfer heat by convection from its surface to the surrounding gas Then the total rate of heat transfer from the surface is the sum of the heat rates due to these two modes. Thus… Q = hA(Ts – Tf) + εσ(Ts4 – Tsur4) .

Stefan-Boltzmann Law There are many occasions where it is necessary to express the net radiation exchange in the form of… Q = hrA(Ts – Tsur) With manipulation of formula it can be shown that… hr = εσ(Ts + Tsur)(Ts2 + Tsur2) This relationship enable the radiation heat transfer coefficient to be determined .

Stefan-Boltzmann Law A steam pipe without insulation passes through a large area of a factory in which the air of the factory walls are at 20oC. The external diameter of the pipe is 80mm, its surface temperature is 220oC and it has an emissivity of 0.75. If the coefficient associated with free convection heat transfer from the surface to the air is 20W/m2K, what is the rate of heat loss from the surface per unit length of pipe?

Stefan-Boltzmann Law Making the assumption that the radiation exchange between the pipe and the factory is between a small surface and a very much larger surface then… Q = εσ(Ts4 – Tsur4) Also heat loss from the pipe to the factory air is by convection and by radiation exchange with the factory walls. .

Stefan-Boltzmann Law . Using Q = hA(Ts – Tf) + εσ(Ts4 – Tsur4) we have… Q = h(πdl)(Ts – Tair) + εσ(πdl)(Ts4 – Twall4) The area A being πdl; note also Tair = Twall Thus the heat loss from the pipe per unit length is… Q/L = Q/1 = (20)(π x 0.08 x 1)(493 – 293) + (0.75)(5.672 x 10-8)(π x 0.08 x1)(4934 – 2934) = 1588 W . . .