3a6 radiation
TRANSCRIPT
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Cambridge University Engineering Department
Engineering Tripos Part IIA
Module 3A6: Heat and Mass TransferRadiation
Prof. Simone Hochgreb
email: [email protected]
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Schedule
L5 Blackbody and gray body limits, radiation energy balances
L6 Radiation between surfaces and electrical analogy
Recommended reading
Holman, Ch. 8
Incropera Ch. 12,13
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Chapter 2
Radiation
3
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Contents
2 Radiation 3
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Radiation Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.3 Electromagnetic Spectrum . . . . . . . . . . . . . . . . . . . . . . . 7
2.4 Radiation Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.5 Radiation Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.6 Emissive Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.7 Irradiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.8 Radiosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.9 Blackbody Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.10 Plancks Spectral Distribution . . . . . . . . . . . . . . . . . . . . . 14
2.11 Stefan-Boltzmann Law . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.11.1 Stefan-Boltzmann Law and Band Emission . . . . . . . . . . 17
2.12 Radiative Behaviour of Real Surfaces . . . . . . . . . . . . . . . . . 20
2.12.1 Emissivity . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.12.2 Real Surface Emissivity . . . . . . . . . . . . . . . . . . . . 20
2.12.3 Energy Balance at a Surface: Absorptivity, Reflectivity, Trans-
missivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.12.4 Opaque and Diffuse Surface Properties . . . . . . . . . . . . 23
2.12.5 Kirchoff s Law . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.12.6 Gray Surface . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.12.7 IR Thermography . . . . . . . . . . . . . . . . . . . . . . . . 25
2.13 Radiation Exchange Between Surfaces . . . . . . . . . . . . . . . . . 30
2.13.1 View Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.13.2 Calculating View Factors . . . . . . . . . . . . . . . . . . . . 31
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2.14 Radiation Exchange at a Surface . . . . . . . . . . . . . . . . . . . . 34
2.14.1 Two-surface Enclosure: Radiation Shields . . . . . . . . . . . 37
2.15 Gas and particle radiation . . . . . . . . . . . . . . . . . . . . . . . . 37
2.16 Solar Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
2.17 Earth Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
2.18 Overall Summary: Radiation . . . . . . . . . . . . . . . . . . . . . . 43
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2.1 Introduction
In part IB you have seen an introduction to radiation. Here we explore the subject
in greater depth, including the spectral characteristics, the calculation of view factors,
multi-surface networks and gas absorption.
2.2 Radiation Properties
Radiation is energy transferred via electromagnetic waves. It is emitted and absorbed
by all matter at a finite temperature. Radiation is transmitted in discrete packets called
photons, which carry quantized amounts of energy and travel at the speed of light:
E= h= hc/ (2.1)
whereE energy per unit photon transferred, in Jh = 6.625 1034 J s Plancks constant,c = 2.998 108 m/s speed of light, frequency of light in Hz = (s1) wavelength in m.
The very small value of the constant h means that for many practical purposes inthe macro scale, radiation transfer can be considered a continuum of directional waves
propagating in a straight line from a source. The propagation of electromagnetic waves
can is affected by the medium through which it propagates, and the absorption andreflection of energy can depend on the direction of radiation.
Semi-transparent liquids and gases such as air or water are only weakly absorbing,
so we can actually see through them. Solids and liquids absorb very strongly, so
that most frequencies of radiation only penetrate within a few microns of the surface.
Therefore, heat transfer via radiation can usually be treated as a surface phenomenon.
2.3 Electromagnetic Spectrum
The electromagnetic spectrum shown in Fig. 2.1 shows the entire range of energytransmission. At the higher wavelengths, i.e. lower frequencies and energies, we find
radiowaves and the infrared spectrum, which composes much of what we usually call
radiative heat. In the middle of the spectrum, over a narrow range from 400-800
nm, we find the visible range. Radiation reflected in that range composes the range of
colours that we observe in the universe. At higher energies still we find the ultraviolet
range, which can penetrate a variety of materials (including human skin), and at the
highest frequencies and lowest wavelengths, observed, below 1 nm, X-rays, which can
travel across materials.
The range usually responsible for most of the radiation energy transferred via heat at
typical energies finds itself in the visible to far infrared range, as discussed below.
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Figure 2.1: Electromagnetic spectrum
2.4 Radiation Intensity
For all purposes in this paper, radiation can be considered as a unidirectional propaga-
tion of energy at a particular frequency, which can be represented by a vector pointing
in a particular direction from a source. Let us consider an element of source of area
dA1, which can radiate in all directions it can see, or the hemisphere delimited by theplane containing the source.
n
dq
dA
r
1dA cos
dA1
d
Figure 2.2: Radiation intensity emitted by a differential surface
Spectral Intensity I,e(,,) is the energy rate per unit area dA1, solid angle d andwavelength transmitted in a particular direction r, defined by angles and . The
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intensity can be regarded as a vector quantity, carrying energy emitted per unit solid
plane emitting the radiation, over a direction normal to the direction of propagation:
I,e(,,) =dq
dA1 cos dd(2.2)
where d = dA/r2 = cos dd
dA1 is the element emitting the radiation, and dA1 cos is the area dA1 projected inthe direction of the radiation. The solid angle d is defined as:
d = dA/r2
where dA is the projected area normal to the direction of propagation over which theradiation is incident.
The spectral intensity can vary with direction and wavelength.
2.5 Radiation Flux 0000000011111111n r1dA cos
dA1
d
r
d
r d
dr sin
Figure 2.3: Radiation intensity emitted and solid angle
Surface interactions are greatly dependent on the particular wavelength emitted or in-
cident. Therefore, we are often interested in the total energy flux leaving a surface per
unit wavelength, as follows:
Spectral Radiation: Total energy rate per unit wavelength emitted by surface dA1
dq =dq
d= I,e(,,)dA1 cos d (2.3)
Spectral Radiation Flux from dA1: Total energy rate per unit area of surface dA1
dq =dqdA1
= I,e(,,)cos sin dd (2.4)
Hemispherical Spectral Radiation Flux from dA1: total energy per unit area dA1,integrated over a hemisphere.
q() =
20
/20
I,e(,,)cos sin dd (2.5)
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Example: Radiation intensity and emissive power
A2
A1
A3
5 m
4 m
60o
Surface A1 is a diffuse emitter. The total intensity associated with A1 is Ie = 1000W/m2sr. A1 = A2 = A3 = 10
3m2. Determine the rate at which radiation emittedby A1 is intercepted by the other surfaces.
Solution
q=
Ie cos d Ie cos
where is the angle between the emitted ray and the normal to the direction of emis-sion.
2 =A2r22
=1 103
52= 4 105 sr
3 =A3,n
r23=A3 cos 30
r23=
1 103 cos 30
42= 5.41 105 sr
q2 = Ie cos 22 = (1000 W/m2)(cos 0)(4 105 sr) = 4 102 W/m2
q3 = Ie cos 33 = (1000 W/m2)(cos 60)(5.41 105 sr) = 2.7 102 W/m2
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2.7 Irradiation
The converse of emitted radiation is the incident radiation incident on a surface, which
can be the result of the combination of multiple incident radiative sources.
ndA
r
dA1
dI ,i
Figure 2.4: Irradiation of differential surface element
Irradiation is the sum all radiation rate incident on the surface. It can also be rep-
resented by a directional, spectral intensity I,i(,,) per unit wavelength and solidangle about direction and unit intercepting surface normal to this direction. In general,
irradiation intensity depends on direction and wavelength.
Spectral irradiation
G() =20
/20
I,i(,,)cos sin dd (2.12)
Total irradiation
G =
0G()d (2.13)
Diffuse Irradiation: Irradiation can in some cases not depend on direction, so that a
diffuse or hemispherical radiosity can be defined.
G() = I,i (2.14)
G = Ii (2.15)
2.8 Radiosity
Since radiation can originate from direct emission of from reflected radiation, it is
useful to consider the total radiation leaving a surface.
Radiosity is the sum ofemitted and reflected radiation. It is represented by a spectral
intensity I,e(,,) per unit wavelength, solid angle about direction and unit emittingsurface normal to this direction. As any other form of radiation, it can be direction or
wavelength dependent.
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00000000
00000000
11111111
11111111
emissionirradiation radiosityreflected
irradiation
Figure 2.5: Radiosity as sum of emission and reflected radiation
Spectral radiosity
J() =20
/20 I,e+r(,,)cos sin dd (2.16)
Total radiosity
J=
0J()d (2.17)
Diffuse radiosity: Similarly to diffuse emission and incidence, radiosity can in some
cases not depend on direction, so that a hemispherical radiosity can be defined.
J() = I,iJ= Ii (2.18)
2.9 Blackbody Radiation
The concept of blackbody radiation is central to the understanding of radiation, yet is
is an idealization of the surface properties encountered in practice. In order to describe
the idea of radiation emission and absorption, it is useful to have a limiting case. The
idealization of a blackbody includes the following characteristics (Fig. 2.9):
All incident radiation is absorbed, at all directions and wavelengths
For a given temperature, no surface can emit more
Surface is a diffuse emitter.
This idealization allows the calculation of the spectral distribution of blackbody radia-
tion to take place. The idea is to assume that a blackbody is a closed insulated cavity at
a fixed temperature, in which allowable stationary radiation modes exist (waves), with
minimal interaction with the environment via a small opening (which represents the
surface of the black body), through which observations of radiation can be made (Fig.
2.9). In this approximation, a spectrum of standing waves can be derived by counting
all possible wave modes that can exist at a given temperature in a cavity and obtaining
the most likely distribution, by using statistical thermodynamics.
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T
,i
Figure 2.6: Blackbody idealization: ideal absorber as a cavity with a small opening.
All radiation is absorbed, negligible radiation emitted.
Clearly, all incident radiation incident onto the blackbody must be absorbed, as the
probability of emission through the hole is very small. For a given temperature, the
system is at equilibrium, thus the maximum possible radiation is emitted through the
small hole. Finally, blackbody radiation emission is also independent of direction. No
radiation is reflected (all is absorbed), and any radiated energy can come from any
direction inside the cavity, with the same mean intensity (Fig. 2.9).
T I = I,e ,b
Figure 2.7: Diffuse radiation by a blackbody
2.10 Plancks Spectral Distribution
Using statistical mechanics, it is possible to enumerate all the waves that can exist in a
given volume at temperature T, and calculate the mean energy of a radiation beam asa function of wavelength. The result, originally derived by Planck (1901), is known as
Plancks spectral distribution for blackbody radiation.
Blackbody emission intensity
I,b(, T) =2hc2o
5[exp(hco/(kT)) 1](2.19)
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whereh Plancks constant 6.6256 1034 J.sco Speed of light in vacuum 2.998 10
8 m/s
k Boltzmanns constant 1.3805 1023 J/KT Absolute temperature K wavelength m
Blackbody spectral emissive power
E,b(, T) = I,b(, T) =C1
5[exp(C2/(T)) 1](2.20)
whereC1 = 2c
2o = 3.742 10
8 W/m4/m2
C2 = hco/k = 1.439 104 m.K
0 2 4 6 8 10 12 14 16 18 200
1
2
3
4
5
6
7
8x 10
5
I ,b
[(m.
K.sr)
1]
Figure 2.8: Blackbody radiation intensity Ib/T5 as a function of = C2/(T)
Plancks distribution has the following features:
The entire distribution is a function of the parameter T, for a fixed temperature.
The radiation intensity is low both at high and low wavelengths (or frequencies),
emission is smaller at extreme low and high wavelengths, and peaks at some
intermediate value.
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101
100
101
102
100
101
102
103
104
105
106
107
108
( m)
E(
,T)(W/(m
2.
m))
1000 K
5000 K
500 K
max. ( T)= cst
Figure 2.9: Total hemispherical emission Eb,()
At any wavelength, emission increases strongly with temperature.
Increasingly more radiation is emitted at low wavelengths (high frequencies and
energies) as temperature increases.
Peak emission takes place at maxT = C3, where C3 = C2/4.965 = 2897.8m.K (Wiens Law)
Example The sun emits with a spectrum close to T = 5000 K, and the Earth close to
T = 300 K. What is the ratio of wavelengths at peak intensity?
Solution
T = const
1T1 = 2T2
2/1 = T1/T2 = 5000/300 = 16.67
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2.11 Stefan-Boltzmann Law
By integrating Plancks spectral distribution for all wavelengths, we obtain:
Eb =
0E,b(, T) d =
0I,b(, T)d =
0
C15[exp(C2/(T) 1]
d
(2.21)
Eb = T4 (2.22)
where =25
15
k4
c2h3= 5.669 108W/m2.K4 is the Stefan-Boltzmann constant.
This is the famous radiation law, with a T4 dependence. Notice that the integrationthat yields the constant is only valid for integration over the entire spectrum.
2.11.1 Stefan-Boltzmann Law and Band Emission
Many surfaces only emit or radiate over a narrow range of frequencies. Therefore,
it is useful to know the fraction of total blackbody emission contained in a particular
frequency range.
Band Emission
The fraction of total emission over a certain wavelength range F(12) is given by
integrating E,b over the range. This can only be done numerically, and typicallytabulated for 0 as a function ofT:
F(0) =
0 E,b(, T)d
0 E,b(, T)(2.23)
F(12) = F(02) F(01) (2.24)
F(0) is a continuously increasing function ofT, or the non-dimensional = T/C2.
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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.5
1
1.5
2
2.5
3
3.5
f()
Figure 2.10: Normalized blackbody radiation intensity distribution as a function of
= T/C2
Example Using the provided Tables, determine the total and fraction of emission in-
tensity that is emitted in the visible range (400-700 nm) by a blackbody at: (a) 300 K
(b) 5000 KSolution
The total emission intensity between a range of frequencies is given by:
I= Ib(T)(F(2T) F(1T))
We have:
1Ta = (300 K)(0.4m) = 120 K m
2Ta = (300 K)(0.7m) = 210 K m
1Tb = (5000 K)(0.4m) = 2000 K m
2Tb = (5000 K)(0.7m) = 3500 K m
From the Tables in Incropera & DeWitt for the cumulative fractions F(T), we obtain:
F(1Ta) 0
F(2Tb) 0
F(1Tb) = 0.06672
F(2Tb) = 0.382815
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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
F()
Figure 2.11: Cumulative normalized blackbody radiation intensity distribution as a
function of= T/C2
The fraction and total emission emitted in the visible range at 300 K is approximately
zero.
The fraction and total emission emitted in the visible range at 5000 K are
F(2Tb) F(2Tb) = (0.06672 0.382815) = 0.31609
Ib(Tb, 1 2) = (F(2Tb) F(2Tb))T4b =
= (0.31609)(5.669 108 MW/m2K4)(5000 K)4 = 11.2 MW/m2
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2.12 Radiative Behaviour of Real Surfaces
2.12.1 Emissivity
The emissivity of a surface relates the actual emitted intensity at a particular wave-
length and angle to that of a black body.
Emissivity
(,,,T) =I,e(,,,T)
I,b(, T)(2.25)
The emissivity can often be considered primarily a property associated with wave-
length, with a minor dependence on angle. In general, it is possible to average the
emissivity over a hemisphere, to obtain the directionally averaged emissivity.
Emissivity, directionally averaged
(, T) =E(, T)
E,b(, T)(2.26)
In general, there is a strong dependence of emissivity with wavelength and tempera-
ture. Nevertheless, it is possible to average the emissivity over the entire spectrum to
obtain the total hemispherical emissivity, which is integrated over both direction and
wavelength:
Total hemispherical emissivity
(T) = E(T)Eb(T)(2.27)
2.12.2 Real Surface Emissivity
1.0
highly polished
metals
metals
oxides, ceramics
carbon, graphite
vegetation, water, skin
specialty paints, finishes
0.0
Figure 2.12: Typical range of emissivities
The emission intensity from real surfaces is lower than blackbody emission, and de-
pends on wavelength and direction. It is useful therefore to define a surface emissivity
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as an index relative to the maximum possible emission. The emissivity is defined as
the ratio of the directional, spectral intensity to that of a black body.
Real surfaces can have quite complex emissivities, which depend on the type of ma-terial, surface finish, texture, coatings and so on. Many optical materials are actually
tailored to be active over ranges of frequencies, or to completely absorb radiation.
Nevertheless, some broad generalizations can be made about total emissivities.
Metals have the lowest emissivities, and highly polished metals have the lowest emis-
sivities. Oxidized surfaces (silicates such as brick and stone) and carbon black have
the highest emissivities. In general, non-conducting materials (non-metals) have much
higher emissivities than conducting materials.
Figure 2.13: Emissivity of high temperature material SiC
2.12.3 Energy Balance at a Surface: Absorptivity, Reflectivity, Transmis-sivity
Not all irradiation that reaches a surface is absorbed. Part of the energy is absorbed,
part reflected, and in the case of semi-transparent media, also transmitted.
G = G,a + G,r + G,t (2.28)
Different materials behave differently with regards to their ability to absorb, reflect and
transmit radiation. In order to quantify the ratios, it is useful to to define the following
fractions:
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Figure 2.14: Emissivity of thermocouple wire Pt10%Rh [2]
G,i
G ,a
irradiation reflection
G ,r,r
G,t
absorption(semitransparent)
transmission
Figure 2.15: Energy balance for irradiation
Absorptivity
(,,) =I,i,a(,,,T)
I,i(,,)() =
G,a()
G()(2.29)
Reflectivity
(,,) =I,i,r(,,,T)
I,i(,,)() =
Gr()
G()(2.30)
Transmissivity
() =G,t()
G()
(2.31)
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Therefore,
+ + = 1 (2.32)
An opaque medium transmits no radiation, so that = 0 and + = 1. This is thecase with most common (non-optical) materials. Since radiation in the ultraviolet to
infrared can always penetrate over the first fewer microns, thin films are always some-
what transparent. Partially transmitting materials such as optical filters and sunglasses
can be tailored to transmit only certain wavelengths.
Absorptivity and reflectivity are responsible for the colours we see on objects: red
objects absorb all radiation in the visible range but in the red, black objects absorb all
radiation in the visible, and so on. Most materials absorb strongly in the UV range.
Oxygen and nitrogen do not, but ozone does, thus providing a protective layer in the
stratosphere against UV sun rays.
2.12.4 Opaque and Diffuse Surface Properties
00000000
00000000
11111111
11111111
00000000
00000000
11111111
11111111
nc ent
reflected
nc ent
reflected
diffuse reflection specular reflection
Figure 2.16: Diffuse and specular reflection
Surfaces can have diffuse or specular reflection. In the case of diffuse reflection, the
incident radiation is reflected uniformly at all angles around the surface. In the case
of specular materials, the radiation is reflected with the same angle to the surface as
the incident radiation. Most materials are somewhere in between, except for highlypolished materials for which the incoming radiation is specularly reflected.
2.12.5 Kirchoffs Law
The fundamental reason why radiation is absorbed or emitted by a material is the in-
trinsic ability of matter to absorb or emit radiation at discrete energy levels. Quantum
mechanics postulates that there is equal probability of absorption as that of emission
of a photon. Emissivity and absorptivity can be interpreted as the mean probability of
energy emission or absorption, Kirchoffs law expresses the equality of these proba-
bilities, for each wavelength, for a particular direction:
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h h
j
Figure 2.17: Absorption or emission of a photon by a molecule
, = , (2.33)
that is, the fraction of the total radiation that can be emitted by a body is also thatwhich can be absorbed.
2.12.6 Gray Surface
Consider a situation in which the following conditions apply:
The irradiation is diffuse: I,i independent of direction (depends on the situation,but fine for many engineering applications), so that I,i = G/
Surface is diffuse: independent of direction (usually good for non-conductors)
Then, by integrating the intensity over a hemisphere, we have:
= (2.34)
Under what conditions is the mean emissivity equal to the mean absorptivity over a
given wavelength range considered?. In other words, when is the following equality
valid?
=
21
G
G
(2.35)
=
21
E,b(, T)
Eb(T)= ,o = (2.36)
By inspection, we can see that this is the case if either:
If and are independent of.
=
21
G()
G=
= ,o = ,o =
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If the irradiation field corresponds to that of a blackbody at the same temperature
T of the surface, then
=2
1 E,b(, T)Eb(T)
(2.37)
=
21
E,b(, T)
Eb(T)= (2.38)
A Gray Surface, which satisfy either of the conditions above, is such that and are independent of the irradiation and surface emission.
This is often true in some region of the spectrum for some situations, but not always
over the whole spectrum.
2.12.7 IR Thermography
The dependence of the radiosity of a surface on temperature leads to a straightforward
method of temperature measurement based on detecting the wavelength and inten-
sity of radiation by a surface of known emissivity. This method is called pyrometry
or thermography. If the materials have known emissivities (or can be calibrated), a
quantitative measure of the temperature can be obtained. In many cases, however, a
qualitative idea of where hot or cold spots are is sufficient to identify faults or flaws.
Figure 2.18: This is a computer chip under test. Imaging radiometers provide identifi-
cation of a fault (Source: FLIR)
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(a) For the white body:
T4 =0.1
0.9
(700 W/m2)
5.669 108 W/m2K4+ (298 K)4
T = 310 K
(b) For the black body:
T4 =(700 W/m2)
5.669 108 W/m2K4+ (298 K)4
T4 = 377 K
Notice that the temperature depends only on the ratios of absortivities rather
than on their values.
Clearly, the white body remains cooler than the black body, but primarily be-
cause of the ratios of emissivities at different wavelengths. If the emissivity
remained constant throughout the spectrum, albeit low, the temperatures would
be no different.
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Example
Consider a wall of surface temperature Ts = 500 K exposed to black body radiation
from a source at Tc = 2000 K. The spectral emissivity of the surface is equal to 0.1 inthe range 0 to 1 = 1.5 m, 0.5 from 1.5 to 2= 10 m and 0.8 from 10 m on. Allsurfaces are assumed to be diffuse. Determine:
(a) The total hemispherical emissivity of the wall
(b) Total emissive power of the wall
(c) The total absorptivity of the wall to irradiation from the source
(d) Is this surface gray?
Solution
(a) From the definition of emissivity:
=
0
E(, Ts)
Eb(Ts)d
=
10
E(, Ts)
Eb(Ts)d +
21
E(, Ts)
Eb(Ts)d+
+
1
E(, Ts)
Eb(Ts) d
= 1F01(Ts) + 2F12(Ts) + 2F2(Ts)
We calculate the fractions from the values ofT2 and look up on the tables:
1Ts = (1.5m)(500 K) = 750m F01 = 0
2Ts = (10m)(500 K) = 5000m F02 = 0.634
= (0.1)(0) + (0.5)(0.634 0) + (0.8)(1 0.634) = 0.610
(b) The total emissive power is given by:
E(Ts) = T4s = (0.610)(5.669 10
8 W/m2K4)(500 K)4 = 2161 W/m2
(c) Now we must average the absorptivity over the irradiated spectral power rather
than the emitted power:
=
0 G()d
0 G() d
For a diffuse surface, = .
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The surface at 500 K receives radiation from a blackbody source at Tc = 2000 K:
G() = E,b(, Tc)
=
0 ()E,b(, Tc) d
0 E,b(, Tc) d
= 1F01(Tc) + 2F12(Tc) + 3F2(Tc)
1Tc = (1.5m)(2000 K) = 3000m F01 = 0.273
2Tc = (10m)(2000 K) = 20000m F02 = 0.986
= (0.1)(0.273) + (0.5)(0.986 0.273) + (0.8)(1 0.987) = 0.395
(d) Analysis: and are averaged for different black body spectra, so = , eventhough = , and the body is notgray.
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2.13 Radiation Exchange Between Surfaces
2.13.1 View Factor
dij
dAi
ni
nj
dAj
Ii
Ti
i
j
Figure 2.20: Radiation exchange between two differential elements - diffuse surface
The amount of radiation emitted from one surface that reaches another surface depends
on the geometry connecting the two surfaces. Surfaces that cannot view each other
via a straight line cannot directly exchange radiation.
Consider the radiation exchange between two differential surfaces, defined by their
directing normals and angles. The radiation leaving surface i that is intercepted bysurface j can be expressed as:
dqij = Ii cos idAidji (2.39)
dji =cos jdAj
r2(2.40)
dqij = Ii
cos i cos jdAidAj
r2 (2.41)
where dji is the differential solid angle enclosing surfacej from an origin at elementi.
View Factor: fraction of the radiation leaving surface i that is intercepted by surface j
If surface i emits and reflects diffusely, we have Ii = Ji/ and
dqij = Jicos i cos j
r2dAidAj (2.42)
(2.43)
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The total radiation that leaves surface i and is intercepted by surface j is then:
qij = Ji AiAj
cos i cos j
r2dAi dAj (2.44)
(2.45)
The View Factor1 represents the fraction of the radiation leaving surface i that isintercepted by surface j. For diffuse emitters and reflectors with uniform radiosity, theview factor is therefore defined as:
Fij =qijJiAi
=1
Ai
Ai
Aj
cos i cos jr2
dAidAj (2.46)
2.13.2 Calculating View Factors
The direct calculation of view factors using Eq. (2.46) can be quite complex, depend-
ing on the geometry. A few insights can greatly simplify the calculations. View factors
are tabulated for standard shapes, as this can be a tedious calculation.
Reciprocity relation
From the definition of view factor, we have:
Fji =1
AjAiAj
cos i cos j
r2
dAidAj (2.47)
(2.48)
Therefore
FijAi = FjiAj
Figure 2.21: Radiation within an enclosure
1Sometimes called shape factor or geometry factor.
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Enclosures
In an enclosure, all radiation must reach some surface, so that:
Nj=1
Fij = 1
Convexity
Figure 2.22: Flat surface Figure 2.23: Convex surfaces
For a convex surface, which does not see itself,
Fii = 0
But for a concave surface, Fii is in general non-zero.
Figure 2.24: View from a concave surface
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Example: Concentric Cylinders or Spheres
Figure 2.25: Concentric surfaces
Consider the situation where a cylinder (or sphere) is entirely enclosed inside another
cylinder. What fraction of the radiation emitted by the outer sphere or cylinder reaches
itself?
F11 = 0
F12 = 1
F21 = A1
A2F12 = A
1
A2
F22 = 1 F21 = 1A1A2
By analogy, this conclusion also applies for inner convex surfaces of any shape.
Example: Parallel Circular Disks
Consider a disk of differential element area A1, parallel to a disk of areaA2 and diam-
eter D, and separated by a distance L. Both surfaces can be considered diffuse. Usingthe definition of view factor, we have:
F12 =1
A1
A1
A2
cos 1 cos 2R2
dA1dA2
1, 2 and R are independent of position on A1, since it is a small differential element.Therefore, 1 = 2 = and
A1fdA1 fA1
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Figure 2.26: Example: parallel disks
F12 =1
A1[
A2
cos2
R2dA2]A1
F12 =
D/20
L2
r2 + L21
(r2 + L2)22rdr
F12 =D2
D2 + 4L2
For calculating over a non-differential element, a second integral would be taken with
a factor dependent on (r, ).
2.14 Radiation Exchange at a Surface
Consider a diffuse, opaque surface at with emissivity i, which receives irradiation
equal to Gi under steady state conditions (Fig. 2.27). The net radiation exchange qithat that leaves the surface is the sum between the radiosity and the irradiation:
qi = Ai(Ji Gi) (2.49)
Ji = Ei + iGi (2.50)
For a gray surface
Ji = iEbi + (1 i)Gi (2.51)
Gi =Ji iEbi
(1i)
(2.52)
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Figure 2.27: Radiation exchange
Therefore
qi = Ai(Ji Ji iEbi
1 i) (2.53)
qi =Ebi Ji
(1 i)/(iAi)(2.54)
The latter expression is the analog to an electrical circuit with potential Ebi Ji andresistance (1 i)/(iAi).
Figure 2.28: Radiative resistance
Net radiation is from the surface when the black body emission rate exceeds the ra-
diosity, and to the surface when the opposite is true.
Now let us consider a situation in which the irradiation originates by from the radiosity
from N surfaces:
AiGi =Nj=1
FjiAjJj =Nj=1
FijAiJj (2.55)
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Figure 2.29: Network of radiative resistances
But
qi = Ai(Ji Gi) = Ai(Ji Nj=1
FijJj)
= Ai((N
j=1
Fij)Ji N
j=1
FijJj) (2.56)
so that
qi =Nj=1
AiFij(Ji Jj) =Nj=1
(Ji Jj)
(AiFij)1(2.57)
Finally, we have:
qi =Ebi Ji
(1 i)/(iAi)
=N
j=1
AiFij(Ji Jj) =N
j=1
(Ji Jj)
(AiFij)1
=N
j=1
qij (2.58)
qi =Ebi Ji
(1 i)/(iAi)=Nj=1
AiFij(Ji Jj) =Nj=1
(Ji Jj)
(AiFij)1=Nj=1
qij (2.59)
The last equation is equivalent to a network of potentials Ebi and radiative resis-tances (1 i)/(iAi) and geometric resistances (AiFij)
1. If temperatures or heat
fluxes at surfaces i are known, we can solve the system for the radiosities of eachsurface and finally to the temperatures and heat fluxes at all surfaces.
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Figure 2.30: Radiation shields
2.14.1 Two-surface Enclosure: Radiation Shields
In the same manner that insulation can provide a resistance to heat transfer, radiation
shields using low emissivity material can provide high radiative resistance (1 )/.
Consider the exchange between two diffuse gray planar surfaces at known temper-
atures, with areas A1 and A2, with view factor 1. The heat exchange via radiationwithout any intermediate obstruction would be:
q12 =Eb1 Eb2
(1 1)/(1A) + 1/A + (1 2)/(2A)=
A(T41 T42 )
1
1+ 1
2 1
(2.60)
If we add a thin radiation shield (surface 3) with emissivity 3 on both sides, we have:
q12 =A(T41 T
42 )
11
+ 12 1 + 1 + 2133
(2.61)
Therefore, if the shield emissivity is very small, the radiative resistance significantly
increases and the overall radiative heat transfer decreases.
Notes:
In situations involving insulated surfaces, we have a node where the net heat
flux qi =EbiJi
(1i)/(iAi)= 0. Therefore, Ebi = Ji regardless of the emissivity.
In cases involving the exchange of radiation with the (very large) environment,
the latter can be considered as a black body.
2.15 Gas and particle radiation
Gases can absorb and emit radiation, typically within narrow wavelength bands. Non-
polar gases such as N2 and O2 are essentially transparent at most temperatures, while
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CO2 and H2O, hydrocarbons and other heteroatomic gases interact significantly with
electromagnetic radiation. Similarly, particle-laden gases such as clouds, coal furnaces
and dusty atmosphere also absorb and re-radiate according to their temperatures. Coal
and oil-fired furnaces rely on the intense radiation provided by soot particles for ef-
fective heat transfer to boiler tubes, whereas clean gas flames must effect essentially
all heat transfer via convection only. The warm glow of a fireplace fire is provided via
radiation from soot and char particles travelling through the fire.
dx
I,o
I
x
Figure 2.31: Radiation absorption through a gas column
The extinction of radiation through an absorbing medium can be described rather sim-
ply by considering a band of monochromatic radiation with intensity I travellingthrough a uniform column of gas as in Fig.2.31. Absorption is assumed to be constant
and proportional to the volume of gas, so that:
dI = kI dx (2.62)
where the proportionality constant k is the monochromatic absorption coefficient.Integrating this equation with constant a gives:
II,o
dII
= k
x0dx = x (2.63)
II,o
= exp(kx) (2.64)
Equation 2.64 is called Beers law, and is a suitable approximation for most gases and
semi-transparent materials, and represents the transmissivity of the column. If thegas is non-reflecting (true in all pure gas cases), then its absortivity and emissivity are
given by:
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3A6 Heat and Mass Transfer 2010 39
= = 1 = 1 exp(kx) (2.65)
The absorption coefficient is a function of gas temperature and pressure, and no simple
expressions exist for the absorption coefficient. In all cases, however, k is propor-tional to the volumetric concentration of particles or molecules. For gases, it is often
expressed as k = kp(T)pL, where kp(T) is the temperature dependent coefficient,p is the pressure and Le is a semi-empirical measure of the mean path travelled byradiation.
Example: Parallel plates
Consider the situation in which a hot, absorbing gas medium is present between two
parallel plates of area A at temperatures T1 and T2. For plates 1 and 2 the energy gain
per unit area is:
q1A = G1AEb1A (2.66)
q2A = G2AEb2A (2.67)
The irradiations G1 and G2 must be calculated considering the irradiation by the gas:
G1A = AgFg1g(Tg)Ebg + AF21g(T2)Eb2 (2.68)
G2A = AgFg2g(Tg)Ebg + AF12g(T1)Eb1 (2.69)
We can take the view factors for gas and surfaces to be 1.0, and the areas to be equal.
The emissivities for the gases depend on the gas properties, temperature and pressure,
and the transmissivity is given by Eq.2.65.
For more complex geometries, for example for heat transfer from flames to walls, the
individual contribution of radiation and the individual view factor needs to be sepa-
rately calculated, and numerical methods must be used.
2.16 Solar Radiation
The red line in Fig.2.32 shows measurements of solar irradiance at high altitudes ( i.e.
prior to absorption by atmospheric gases), at a tropical latitude. The measurements are
expressed in terms of the wavenumber = 1/, expressed in cm1 (a common unitin spectroscopic measurements in the infrared region) on the bottom scale. The peak
irradiation is clearly around 12,000 cm1, or a wavelength of 500 nm.
The overlaid line shows the spectrum of a a blackbody at 5780 K, which clearly
matches the spectrum very well, with an maximum given by Wiens Law as =2897.8 m K /5780 K = 501 nm.
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Figure 2.32: Solar radiation measured at the upper atmosphere. Source: NASA God-
dard
The integral over all wavelengths represents the total irradiation incident per unit area.
This is the so called solar constant, equal to S=1370 W/m2. The incident radiationdecreases with increasing latitudes according to
Gs = Sc cos (2.70)
where is the zenith angle.
As the solar radiation makes its way through the atmosphere, its magnitude and spec-
tral/directional distributions are significantly changed by scattering and absorption by
gases.
Absorption reduces the intensity of the total solar radiation slightly (TSI). Overall,
about 30 percent of the total solar energy that strikes the Earth is reflected back intospace by clouds, atmospheric aerosols, snow, ice, desert sand, roof, and ocean. The
remaining 70 percent is absorbed by the land, ocean, and atmosphere. The mean frac-
tion of the incident radiation reflected by the earth, i.e. its reflectivity, is called albedo,
a.
About one percent of the TSI, mostly in the form of UV radiation, is absorbed by
the upper atmosphere, mainly by stratospheric ozone. Twenty to 24 percent of the
TSI and a majority of the near infrared radiation is absorbed in the lower atmosphere
(troposphere), mainly by water vapor, trace gases, clouds, and darker aerosols. The
remaining 46 to 50 percent of predominately visible light penetrates the atmosphere
and is taken in by the land and the oceans. Overall, about 230-280 W/m2 of radiation
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3A6 Heat and Mass Transfer 2010 41
reaches the surface of the Earth on the mean, with great variations from the equator to
very low amounts reaching the poles.
Light scattering changes the intensity and spectral characteristics. Rayleigh scatter oflight (scatter by molecules) is proportional to 4. This leads to a shift of the remaininglight to the red (lower frequencies), and the reddish sunsets as sun rays have to make a
long way through the atmosphere.
The final radiation reaching the Earth is a combination of direct and diffuse radiation.
Diffuse irradiation from gases in the atmosphere (reradiation of absorbed radiation)
approximately Gatm = T4sky, where Tsky 230285K. This is the only irradiation
source at night.
2.17 Earth Radiation
The irradiation from the sun is reflected by the earth, and the earth itself radiates as a
finite temperature body. The final result gives a radiosity spectrum from the earth.
An estimate of the effective temperature of the earth in the absence of other clouds or
other effects can be obtained by considering the incident radiation of the sun as being
constant over the solid angle encompassing a circle of the earths radius R, in steadystate with outer space at approximately 0 K. If we assume that the earth has emissivity
of about 0.9 at the relevant wavelength, we have.
G = S(1 a) (2.71)Q = GR2 = S(1 a)R2 = 4R2T4e (2.72)
Te = (S(1 a)/(4))1/4 (2.73)
and for the values given, we obtain a temperature of 255 K, or -18 oC. However, the
presence of the atmosphere leads to both absorption and scattering on the way in and
out of the ground, and trapping of radiation leaving the earth.
The radiosity of the earth corresponds to a blackbody temperature of around 255 to
280 K. The power emitted is of the order of 230-280 W/m 2. The atmosphere is much
more transparent to incoming radiation than leaving radiation (Fig. 2.33). Much of
the radiation at either end of the wavelength distribution is trapped by water vapour inthe atmosphere (with emissions cut off around 300 cm1 and lower, and 1400 cm1
and up. The key absorbing gases are 2.33 CO2 and H2O. Water plays a dual role in
this balance, with direct absorption and trapping of heat, but also with additional cloud
cover leading to reflection of sunlight. There are large uncertainties in the estimates of
the role of clouds in the global energy balance.
Consider the energy balance in Fig. 2.34. Of the 1370 W/m2, 30 percent is reflected by
atmosphere and clouds, and the earths surface. The average radiation rate absorbed at
the surface is around 235 W/m2. Without absorption by atmospheric cases, the Earths
surface would be expected to have an average temperature of -18oC. The shielding ef-
fect of the absorbing atmosphere in the infrared region of the spectrum means that the
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Figure 2.33: Absorption of sun and earth radiation - calculated using spectroscopic
absorption. Source: Global Warming Art Project, R. A. Rohde
earths atmosphere recycles heat coming from the surface and delivers an additional
324 W/m2, which results in an average surface temperature of roughly +14 oC. There-
fore, this greenhouse effect is essential for life on the planet - yet small imbalances cancreate an undesirable rise in mean the temperature of the earth.
Of the surface heat captured by the atmosphere, more than 75% can be attributed tothe action of greenhouse gases that absorb thermal radiation emitted by the Earths
surface. The atmosphere in turn transfers the energy it receives both into space (38%)and back to the Earths surface (62%).
Recent measurements indicate that the earth atmosphere is presently absorbing about
1-2 W/m2 more than it emits into space. This increase is believed to have been caused
by the recent increase in greenhouse gas concentrations.
Some of the recent suggestions regarding mitigation of global warming include using
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Figure 2.34: Earths energy budget (Source: NASA)
geoengineering by for example, changing the mean albedo earth by adding particles
to the stratosphere around the polar caps to increase its reflectivity. These have been
shown to cause severe cooling, as experienced during volcanic eruptions.
2.18 Overall Summary: Radiation
Emission: Function of surface temperature, direction, wavelength
Radiosity: sum of emission and reflected irradiation, depends also on character-istics of incoming radiation
Diffuse radiation and irradiation are independent of emission or incidence angle
Black body radiation intensity is strong function of wavelength and temperature
Total hemispherical black body radiation is proportional to T4
Kirchoffs law: spectral, directional emissivity is equal to the spectral, direc-
tional absorptivity
Gray surfaces: diffuse, independent of wavelength, so that =
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View factor: geometric property that expresses the fraction of radiant energy
emitted by one surface intercepted by the second surface. Reciprocity relations
apply.
Network analogy: radiation exchange between gray surfaces can be calculated
from blackbody radiation, with radiative resistances for emissivity and view fac-
tors.
References
1. F. P. Incropera and D. P. De Witt, Fundamentals of Heat and Mass Transfer,
John Wiley & sons, 2002.
2. C. P. Cagran, L. M. Hanssen, M. Noorma, A. V. Gura and S. N. Mekhontsev,
International Journal of Thermophysics, Vol. 28, No. 2, April 2007 (2007).