CHAPTER 6: RADIATION HEAT TRANSFER PROCESS
Course outline
6.1 Introduction
6.2 Objectives
6.3 Electromagnetic spectrum
6.3.1 Thermal radiation
6.4 Radiative properties
6.4.1 Absorptivity, reflectivity, transmissivity
6.4.2 Specular and diffuse surfaces
6.5 Emissive power
6.6 Blackbody radiation
6.7 Intensity of Radiation
6.7.1 Lamberts cosine law
6.8 Spectral distribution
6.8.1 Planck’s law
6.8.2 Wein’s displacement law
6.8.3 Stefan Boltzman law
6.8.4 The emmissivity
6.8.5 Kirchhoff’s law
6.9 Heat exchange between black surfaces
6.10 The view factor
6.10.1 Properties of view factors
6.11 Summary
References.
6.1 Introduction:
Recall that in the unit FME 422, conduction heat transfer was discussed. And in
preceding chapters, convection heat transfer mode has been considered. Both modes
require a medium for propagation of the heat energy. In this chapter we consider
radiation – the third mode of heat transfer which requires no medium.
Radiation is a term applied to many processes, which involve energy transfer, by
electromagnetic wave phenomena. The waves are propagated in straight lines at a
velocity of light. c   v  3 108 m / s , where  is the wavelength and v the frequency
of the radiation. Radiant energy being electro-magnetic radiation requires no medium
for its propagation and will pass through a vacuum. At temperatures higher than
absolute zero, all matter emits electro-magnetic radiation. The higher the temperature
then the greater is the amount of energy radiated. If two bodies are so placed that the
radiation from each body is intercepted by the other, then the body at the lower
temperature will receive more energy than it is radiating and its internal energy will
increase. Thus there will be a net transfer of energy from the high-temperature body
to the low-temperature body by virtue of the temperature difference between the
bodies. The rate of radiative energy transfer is proportional to the fourth power of the
temperatures of the bodies involved.
We begin this chapter by examining the electromagnetic spectrum where we identify
the portion of radiation that is thermal radiation. We next highlight the various
radiation properties exhibited by surfaces. Finally, we describe the concept of view
factor, important in estimating the radiant heat exchange between surfaces.
1
6.2 Objectives:
The objectives of this chapter are to,
1) Understand means by which thermal radiation is generated and the specific
nature of radiation.
2) Explain the various properties of thermal radiation
3) Explain black body radiation concept
4) Determine heat exchange between two black bodies.
6.3 Electromagnetic spectrum:
Shown in Fig. 6.1 is a portion of the electromagnetic spectrum. The short wavelength
gamma rays, X rays and ultraviolet (UV) radiation are primarily of interest to the
high-energy physicist and the nuclear engineer.
Cosmic rays
Ultra violet
Infrared
X- rays
10-12
10-8
Visible
10-6
10-4
10-2
Hertzian
Microwave
Thermal
102
1
104
Radio
Figure 6.1 Electromagnetic spectrum
The long wavelength microwaves and radio waves are of concern to the electrical
engineer.
6.3.1 Thermal Radiation:
The intermediate portion of the spectrum, which extends from approximately 0.1 to
100μm (1m = 10-6m) and includes a portion of UV and all of the visible and infrared
(IR), is termed thermal radiation and is pertinent to heat transfer. The visible light
portion of the spectrum is very narrow and ranges between 0.35 and 0.75m.
Take Note
Another common unit of wavelength is the Angstrom Ao (1Ao = 10-10m).
Thermal energy is associated with energy emissions from all bodies due to thermally
exited conditions within the matter. The mechanism of emission is related to energy
released as a result of oscillations or transitions of the many electrons that constitute
matter. All forms of matter emit radiation at temperature above absolute zero.
2
6.4 Radiative properties:
In this section we discuss the radiative properties of thermal radiation.
6.4.1 Absorptivity, Reflectivity and Transmissivity:
Part of the radiant energy incident on any surface may be absorbed, part may be
reflected and part may be transmitted through the body. Figure 6.2 shows the
components of incident radiation on a body.
Incident
radiation
I
Reflected
I
Absorbed
I
Transmitted
I
Figure 6.2 Components of incident radiation on an object.
 - Absorptivity is the fraction of incident radiation absorbed.
 - Reflectivity is the fraction of incident radiation reflected.
 - Transmissivity is the fraction of incident radiation transmitted.
     1
(6.1)
Except for those visibly transparent or translucent, most solids do not transmit
radiation. An opaque body is one that does not transmit any radiation. The incident
radiation is either reflected or absorbed. For such a body;
   1
(6.2)
Gases in general reflect very little radiant thermal energy and hence,
   1
(6.3)
Some gases, especially water vapour and carbondioxide absorbs a significant
radiation while most of the rest of gases found in atmosphere such as nitrogen,
oxygen and air transmit most radiation incident on them.
For a transparent body,     0 and   1 . A typical example is air.
6.4.2 Specular and diffuse surfaces:
In reflection of incident radiation, two phenomena occur. Reflection of radiant
thermal energy from a surface can be specular or diffuse. A perfect specular reflector
is where the angle of incidence i is equal to the angle made by the reflected ray r .
A diffuse reflector is such that the magnitude of the reflected energy is distributed
uniformly in all directions. If a surface roughness (height) for a real surface is much
smaller than the wavelength of incident radiation, surfaces behave as specular one. If
the roughness is large with respect to wavelength, the surface reflects diffusely. Many
surfaces have diffuse radiation.
3
Source
Source
i
r
Specular
Diffuse
Figure 7.3 Specular and Diffuse surfaces.
6.5 Emissive power:
The emissive power is the emitted thermal radiation leaving a surface per unit time
per unit area of surface. The total hemispherical emissive power of a surface is all the
emitted energy summed over all directions and all wavelengths, and is denoted by the
symbol E. The total emissive power is a function of temperature of the emitting
surface, the material of which the surface is composed and the nature of the surface
structure (roughness).
The monochromatic emissive power E, is the rate per unit area, at which a surface
emits thermal radiation at a particular wavelength, . Thus, the total and
monochromatic hemispherical emissive powers are related by,

E   E d 
(6.4)
0
and the functional dependence of E on  must be known to evaluate E.
The emissive power does not include any energy leaving the surface due to reflection
of the incident radiation but only of original emission leaving a surface.
Intext Question:
1. What is the nature of radiation? What two important features
characterize radiation?
2. What is the physical origin of radiation emission from a surface?
How does emission affect the thermal energy of a material?
3. In what region of the electromagnetic spectrum is thermal
radiation concentrated?
6.6 Black body radiation:
A black body is an ideal surface that absorbs all incident radiation. I.e.  = 1,  =  =
0. It absorbs all incident thermal radiation regardless of spectral or directional
characteristics. It is termed a black body because a body that absorbs all radiation
appears black to the eye. However, it may be noted that the ‘blackness’ of a surface to
thermal radiation may not relate to the visual definitions of the surface. A surface
coated with lampblack appears black to the eye and turns out to be black for the
thermal radiation spectrum. On the other hand, snow and ice, which appear bright to
the eye, are essentially ‘black’ for long-wavelength thermal radiation. Many white
paints are also essentially black for long wavelength radiation.
4
No actual body is perfectly black; the concept of a black body is an idealization with
which the radiation characteristics of real bodies can be conveniently compared. The
properties of a black body are:
i)
It absorbs all incident radiation falling on it and does not transmit or reflect
regardless of wavelength and direction
ii)
It emits maximum amount of thermal radiation at all wavelengths at any
specified temperature
iii)
It is a diffuse emitter i.e. the radiation is independent of direction.
A black body may be approximated by a cavity (hollow space) with a very small hole
opening (Fig. 6.4).
Ray of
radiation in
Figure 6.4 A black body approximation
The walls of the chamber successively absorb rays of thermal radiation entering the
hole such that a negligible radiation is emitted from the hole. When a body is placed
in large surroundings the later are approximately black to thermal radiation.
6.7 Intensity of radiation I:
Most surfaces do not emit radiation strongly in all directions; the greater part of the
energy emitted is in a direction normal to the surface. The intensity of radiation
describes the directional distribution of radiant energy leaving a surface (either by
emission or reflection) or the directional distribution of the energy incident upon a
surface. For ease of understanding, the definition of intensity may be made in terms of
the energy leaving a ‘point source’ on a surface and how it distributes directionally in
hemisphere above the surface. The rate of energy emission from unit surface area
through unit solid angle, along a normal to the surface is called the intensity of normal
radiation, In. The intensity of radiation in any other direction at any angle θ to the
normal, is denoted by Iθ.
6.7.1 Lambert’s cosine law:
The variation in the intensity of radiation is given by Lambert’s cosine law (Fig. 6.5),
I   I n cos  .
(6.5)
This law is known as Lambert’s law of diffuse radiation.
The emissive power E from a radiating plane surface in any direction is directly
proportional to the cosine of the angle of emission.
E  En cos
This relation is true only for diffuse radiation surface.
5
(6.6)
In
I=Incos

Figure 6.5 Illustration of Lamberts cosine law
The definition of intensity may apply for a given wavelength, resulting in a
monochromatic intensity, or as a quantity summed over all wavelengths, resulting in a
total intensity.
A solid angle is defined as a portion of the space inside a sphere enclosed by a conical
surface with the vertex of the cone at the center of the sphere. It is the ratio of the
spherical surface enclosed by the cone to the square of the radius of the sphere, Fig
6.6. The units of solid angle are steradian (sr). The solid angle subtended by the
2 r 2
complete hemisphere is given by
 2
r2
Solid
angle w
Emitted
radiation
Normal

Black
surface
emitter
Radiation
collector
dA

Figure 6.6 Directional emission from a black surface emitter.
Thermal radiation emitted by a body per unit surface area per unit time, in all
directions is the emissive power. The emissive power is therefore E   I  dw , where
dw is the solid angle subtended by the radiation intensity Iθ.
Consider a small black surface of area dA1 in Fig. 6.7, emitting radiation in different
directions. A black body collector is located at an angular position at zenith angle 
towards the surface normal and at an angle  to the horizontal. The collector subtends
a solid angle dw when viewed from the area dA1.
6
rd
rsin
dA2=r2sindd

r
d
rsind
dA1

d
Figure 6.7 Spatial distribution of energy emitted from surface dA1.
The area, dA2 is given by
dA2  r sin  d  rd  r 2 sin  d d
(6.7)
Let the radiation from dA1 is absorbed by the elemental area dA2, a portion of the
hemispherical surface.
dA2
. The energy emitted from element
r2
dA1 and intercepted by element dA2 is given by dQ12  I n cos dwdA1 .
The solid angle subtended by dA2 , is dw 
dA2
.
r2
 I n dA1 sin  cos dd .
Substituting dw ,
dQ12  I n cos dA1 
Substituting dA2,
dQ12
The total radiation through the hemisphere will be given by,
  2   / 2
Q  IdA1
 
 0
sin  cos  d d ,
0
7
 2 IdA1
  / 2

sin  cos  d
 0
  IdA1
  / 2

2sin  cos  d
(6.8)
 0
  IdA1
  / 2

sin 2 d   IdA1
 0
Or, Q  I n dA1 . Also, Q  E  dA1
That is, EdA1  I n dA1
Or, E  I n
The total emissive power of a diffuse surface is equal to  times its normal intensity
of radiation.
6.8 Spectral distribution:
A surface emits different amounts of energy at different wavelengths. This leads to
variation of radiation intensity and emissive power with wavelength.
6.8.1 Plancks’ Law:
The radiation intensity from a body is not constant for all wavelengths. To calculate
the heat radiated by a black body, Max Planck formulated (using quantum theory) an
expression for the normal radiation intensity at a wavelength  is given by equation
6.9,
2C1'
I b n 
(6.9)
 CT2

  e  1


where, T is the temperature in Kelvin, C1, and C2 are constants having the numerical
values, C1’ = 0.52525  103 W m4/m2
C2 = 1.4387  104 m.K
The monochromatic emissive power of a black body is given by
5
Eb   I b n

2 C1'
 5  e


C1
  e

5
C2
T
 1

C2
T
 1

,
(6.10)
where, C1 = 2C1’= 3.743  105 W. m4/m2, and C2 = 1.4387  104 m.K
The monochromatic emissive power Eb is the energy emitted by the black surface in
all directions at a given wavelength  per unit wavelength interval around . The
energy emitted in the frequency band between wavelength  and +d is equal to
Eb d  .
8

For a black body therefore, Eb   Eb d  . The quantity E d  is the energy emitted in
0
the frequency band between  and  + d.
Intext Question:
4. What is spectral intensity of radiation emitted by a surface? On
what variable does it depend? How may knowledge of this
dependence be used to determine the rate at which matter loses
thermal energy due to emission from its surface?
5. What is steradian? How many steradians are associated with a
hemisphere?
6. What is the distinction between spectral and total radiation?
Between directional and hemispherical?
7. What is total emissive power? What role does it play in a surface
energy balance.
6.8.2 Wien’s displacement law: When the monochromatic emissive power of a black
body is plotted against wavelength at different temperatures, it is noted (Fig. 6.8) that
the peak of the radiation is shifted to shorter wavelength for higher temperatures.
Eb
T1 > T2 >T3
T1
T2
T3

Figure 6.8 Emissive power variation with wavelength.
Wein’s law describes the displacement of the maximum value of the radiation to
2897.6  m .K
shorter wavelengths at higher temperature. Viz., max 
, where max, is
T
the wavelength for maximum radiation intensity at temperature T. As shown in Fig.
6.8, as the temperature increases more and more of the radiation emitted falls in the
low wavelength range.
For example the sun temperature is approximately 6000K. Therefore,
2897.6
max 
 0.483 m . Hence most of the thermal radiation from the sun is in
6000
visible waveband.
9
The temperature of a light bulb filament is approx. 2800K, and therefore only about
10% of the energy is emitted in the visible waveband.
Example 6.1:
Determine (a) the wavelength at which the monochromatic emissive power of a
tungsten filament at 1400K is maximum, (b) the monochromatic power at that
wavelength and (c) the monochromatic power at 5 μm.
Soln:
From Wein’s law, max 
2896.7
2896.7
m . Hence,  max 
 2.07 m .
T
1400
From equation 7.11, the monochromatic power at λmax is
C1
Eb max 
 6.92  1010 W/m 3 .
C2


5max  e maxT  1


At λ of 5 μm, λT = 5 × 1400 × 10-6 = 7.0 × 10-3 m K.
Hence, Eb  1.758  1010 W/m 3 .which is 25.4% of the maximum.
Activity
Substitute C1 and C2 and confirm the values of Ebλmax and Ebλ, in example
7.1.
6.8.3 Stefan Boltzman law of black body radiation:
The emissive power of a black body over all wavelengths can be found by integrating
the monochromatic emissive power.


C1
Eb   Eb d   
 CT2

  e  1


which reduces to:
0
0
d
5
Eb   T 4 ,
(6.11)
where  is called the Stefan-Boltzman constant. Its value was found experimentally
by Stefan and theoretically by Boltzman and is  = 5.672  10-8 Wm-2K4.
For a black body, equation 7.8, Eb  I n . Therefore from equation 6.11, I n  T 4
And, I n 
T 4
.

(6.12)
10
Take Note
Note that the foregoing laws are concerned with blackbody radiation
because their derivation is based on the assumption that the body is
capable of absorbing completely any radiation incident upon it.
6.8.4 Emissivity: Radiation from real surfaces is different from black body radiation.
The emissive power of a real surface is always less than that of a black body at the
same temperature. If the ratio of the monochromatic emissive power of a body to the
emissive power of a black body at the same wavelength is constant over the entire
wavelength spectrum, the body is said to be gray.
The ratio of the emissive power of a real body, E to that of a black body, Eb at the
same temperature is called emissivity. That is,
E

(6.13)
Eb
The emissive power of a real body is E   Eb   T 4 . For a grey body,  is constant.
For purpose of heat transfer calculations, most real surfaces are regarded as grey.
Example 6.2:
The temperature of a black surface, 0.2 m2 in area, is 540°C. Calculate the total rate of
energy emission, the intensity of normal radiation, and the wavelength of maximum
monochromatic emmisive power.
Soln:
4
 813 
From the Stefan-Boltzman law, Q  Eb  T A . = 56.7
  0.2  4.95kW
 1000 
The intensity of normal black radiation is
4
T 4 56.7  813 
2


  7.88kW / m , steradian.

  1000 
4
In 
From Wien’s displacement law, max 
0.0029 0.0029

 3.57  10 6 m
T
813
6.8.5 Kirchhoff’s law: This law relates the emissivity of a grey surface to its
absorptivity in thermal equilibrium.
Consider a real body of surface area A enclosed in a black enclosure at temperature T
shown in Fig. 6.9.
EA
sample
Black
enclosure
A
Figure 6.9 Heat exchange between a sample body and a black enclosure
11
The energy emitted by the sample body is EA. Let qi represent the radiant flux
arriving from the enclosure on to unit surface of the body. If  is the absorptivity of
the surface, the energy absorbed by the body is qi  A  . For the body to be in
thermal equilibrium, EA  qi A
If the sample body is replaced with a black body of same shape and size and allow it
to come to equilibrium, then, Eb A  qi  A 1  qi A since the absorptivity of the black
sample would be unity.
E
  .
Therefore, EA  Eb A and
Eb
This is the Kirchhoff’s law which states that the emissivity of a grey body is equal to
its absorptivity at thermal equilibrium with the surroundings. Since  < 1, then the
emissive power E of a real surface is always less than that of a black body at the same
temperature.
6.9 Heat exchange between black surfaces:
i)
Black enclosure: Consider a body at a temperature T1, completely
surrounded by black surroundings at a lower temperature T2.
Black
enclosure
T1
T2
Figure 6.10 Black body in a black enclosure.
The energy leaving the body is completely absorbed by the black enclosure.
This energy is given by Ebb = T14. The energy emitted by the black surroundings
is Ebs = T24 per unit area. Since the body is black it absorbs the entire energy
incident on it. Therefore energy absorbed by the body is T24. Energy transferred
from the body to the low temperature surroundings is,
q  Ebb  Ebs   T14  T24  , per unit area. For the body surface area A,
Q  A T14  T24 
(6.14)
Example 7.3
A body at 1100°C in black surrounding at 550°C has an emissivity of 0.4 at
1100°C and an emissivity of 0.7 at 550°C. Calculate the rate of heat loss by
radiation per m2,
a) when the body is assumed to be grey with ε = 0.4;
b) when the body is not grey.
Soln:
12
Energy emitted by black surrounding is Eb  T24 . The fraction of this energy
absorbed by the body is  T24 . The energy emitted by the body is E  T14 .
Heat transferred from the body to its surrounding per m2 of the body is,
Q12  T14  T24 .
a) For grey surface, ε = α. Therefore, Q12   T14  T24 .
T1 = 1100 + 273 = 1373K
T2 = 550 + 273 = 823K
Hence, Q12  0.4  5.67  10 8 13734  8234  70.22kW .
b) When the body is not grey.
Absorptivity when the source is at 550°C = emissivity when the body is at
550°C = 0.7.
Energy emitted = T14  0.4  5.67  10 8  13734  80630W
Energy absorbed = T24  0.7  5.67  10 8  8234  18210W .
Heat lost per m2 by radiation is 80630 – 18210 = 62.42 kW.
The grey body assumption over – estimates the heat loss by 12.49%.


ii)


Black surfaces: Let us now consider heat exchange between two black
surfaces. Shown Fig. 7.11 are two black surfaces A1 and A2 separated by a
non-absorbing medium.
dA2
A2
T2
Normal

r

A1
T1
Normal
dA1
Figure 6.11 Orientations of two radiating surfaces.
Consider heat exchange between two elemental areas dA1 and dA2 on the two
surfaces at temperature T1 and T2 respectively. The distance between the two surfaces
is r. The angles the normals to the two area elements make with the line joining them
are  and  respectively.
The projected area of dA1 normal to the line joining the centers is dA1 cos 

dA1cos
13
dA1
Figure 6.12: Projection of area dA1 at an angle θ from the normal.
Energy leaving dA1 in the direction of  is = Ib1dA1 cos  , where Ib1 is the black body
E
intensity of surface A1 and equal to b1 . Radiation arriving at any area normal to r

will depend on the solid angle subtended by it. Let surface dA2 subtend a solid angle
dA cos 
dw1 at the center of surface dA1. The solid angle is given by dw1  2 2
, where
r
dA2cos is the projection of dA2 normal to r.
Radiation arriving at dA2 from dA1 is = Ib1dA1 cos   dw1 = I b1dA1 cos 
dA2 cos 
r2
dA1dA2
 r2
dA dA
Similarly, energy leaving dA2 and arriving at dA1 is = dQ21  Eb 2 cos  cos  2 2 1
r
Hence energy leaving dA1 and arriving at dA2 is = dQ12  Eb1 cos  cos 
The net heat exchange by radiation between dA1 and dA2 is = dQ12  dQ21 .
= dQ1 2   Eb1  Eb 2  cos  cos 
dA1dA2
,
 r2
and between areas A1 and A2,
= Q12   Eb1  Eb 2  
A1
 cos cos 
A2
dA1dA2
 r2
(6.15)
To evaluate the integrals, specific geometry of the two surfaces must be known.
Example 6.4:
A flat black surface of area A1 = 10 cm2 emits 1000 W/m2 sr in the normal direction.
A small surface A2 having the same area as A1 is placed relative to A1 as shown in
Fig. ex 7.4 at a distance of 0.5m. Determine the solid angle subtended by A2 and the
rate at which A2 is irradiated by A1.
Soln.:
Since both areas are quite small, they can be approximated as differential surface
areas and the solid angle can be calculated as,
14
A2 = 10cm2
n2
θ2 = 30°
Normal, n1,
to A1.
0.5m
θ1 = 60°
A1 = 10cm2
Fig. ex 7.4
dAn 2
. dAn2 is the projection of dA2 in the direction normal to the
r2
incident radiation from dA1. Therefore, dAn 2  dA2 cos  2 .
dw21 
A2 cos  2 10 3 cos 30 0

 0.0034 sr
r2
(0.5) 2
The irradiation of A2 by A1 is Q12  I n A1 cos 1dw21 . Substituting,
Thus, dw21 
Q12  1000  10 3  cos 60 0  0.0034  0.00173W .
Activity
A small surface of area A1 = 10-3 m2 is known to emit diffusely, and from
measurements the total intensity associated with emission in the normal
direction is I n  7000 W/m 2 . sr.
Radiation emitted from the surface is intercepted by three other surfaces
of area A2 = A3 = A4 = 10-3 m2, which are 0.5m from A1 and are oriented
as shown.
i) What are the solid angles subtended by the three surfaces when viewed
from A1?
ii) What is the intensity associated with emission in each of the three
directions?
iii) What is the rate at which radiation emitted by A1 is intercepted by the
three surfaces?
15
Intext Question
8. What is a diffuse emitter? For such an emitter how is the intensity
related to the total emissive power?
9. What are the characteristics of a black body? Does such a body
actually exist in nature? What is the principal role of a black body
behaviour in radiation analysis?
10. What is planck distribution?
11. What is Wein’s displacement law?
12. From memory, sketch the spectral distribution of radiation
emission from a blackbody at three temperatures T1 < T2 < T3.
Identify salient features of the distribution.
13. In what region of the electromagnetic spectrum is radiation from a
surface at room temperature concentrated? What is the spectral
region of concentration for a surface at 1000°C. For the surface of
the sun?
6.10 The view factor: With reference to figure 6.11, the fraction of energy received by
A2 to the energy emitted by A1 is called the view factor or sometimes called the
geometric factor or shape factor.
Total energy emitted by A1 is Q1  Eb1 A1 .
dA dA
Energy leaving A1 and arriving at A2 is =  dQ12  Eb1   cos  cos  1 2 2
A1
A2
r
Q
dA dA
1
Hence view factor is given by F1 2  12    cos  cos  1 2 2 . (6.16)
A
A
Q1
A1 1 2
r
Q21 1
dA dA

cos  cos  2 2 1 .
(6.17)


Q2
A2 A2 A1
r
The view factors, F1-2 and F2-1 may be evaluated when the specific geometries are
known. However, it can be seen that,
Similarly, the view factor F21 
A1 F1 2  A2 F21  
A1

A2
cos  cos 
Q1 2   Eb1  Eb 2  A1 F1 2
  Eb1  Eb 2  A2 F21
dA1dA2
 r2
, and therefore,
. But since Eb1   T14 , and Eb 2   T24 , then
Q1 2  A1 F1 2 T14  T24 
(6.18)
the net energy exchange between black bodies A1 and A2.
6.10.1 Properties of view factors:
1) Reciprocal property – the most important property. For two surfaces, A1 and
A2, F1-2  F2-1. But in general A1F12  A2 F21
16
2) Additive property – If one of the surfaces, say Ai is subdivided into sub areas,
Ai1, Ai2, ………….Ain, so that  Ain  Ai , Then Ai Fi  j   Ain Fin  j , and
n
Also, Fj i   Fj in
n
That is, the shape factor of a subdivided transmitting surface with respect to a
receiving surface is not just the sum of the individual shape factors. While the
shape factor from a surface to a subdivided surface is simply the sum of the
individual shape factors.
Receiving
surface
A2
A3
Radiating
surface
A1 = A3 +A4
A4
F12  F32  F42 , But A1F12  A3 F32  A4 F42
F1-3
A1
F1-4
A3
A4
A1F12  A1 F13  A1 F14
and therefore, F12  F13  F14
Figure 6.12 View factor relations of simple geometry
3) Convex surface: When all the radiation emanating from a convex surface 1 is
intercepted by the enclosing surface 2, the shape factor of the convex surface
with respect to the enclosure F1-2 is unity.
A1F12  A2 F21 . If F1-2 = 1, i.e. surface 1 completely sees surface 2, then
A
F21  1 , and F11  0
A2
4) Concave surface: A concave surface has a shape factor with itself because the
radiant energy coming from one part of the surface is intercepted by the other
part of the same surface. Hence F1-1  0.
5) Parallel surfaces: If two surfaces A1 and A2 are parallel and large, radiation
occurs across the gap between them so that A1 = A2 and all radiation emitted
by one falls on the other. That is F1-2 = F2-1 = 1.
17
Evaluation of shape factors requires solution to double integral functions. For
complex geometry the procedure is rather cumbersome. However, the values have
been computed and presented in charts and graphs for shapes commonly encountered
in engineering practice.
Intext Question
14. What is a view factor? What assumptions are typically associated
with computing the view factor between surfaces?
15. What is reciprocity relation for view factors? What is the
summation rule?
16. Can the view factor of a surface with respect to itself be nonzero?
If so what kind of surface exhibits such behavoiur?
6.11 Summary:
In this chapter we have considered various characteristics of thermal
radiation. We first showed the position of thermal radiation within the
electromagnetic spectrum. The wavelength varies within the range, 0.1 100μm.We then considered the radiative properties which included the
absorptivity α, reflectivity ρ and transmissivity τ expressed as a fraction
of the total incident. i.e.       1 . The radiation also depicts diffuse,
and specular characteristics that relates to the surface finish. Specular
reflection is associated with smooth surfaces while rough or matt surfaces
exhibit diffuse reflection. Most surfaces are diffuse. We introduced the
concept of black body radiation. A black body is an ideal surface that
absorbs all radiation incident upon it. It is also the best emitter. The total
energy emitted per unit time by unit area of the black surface is
proportional to the fourth power of the absolute temperature T. This
relation is expressed by the Stefan – Boltzman law, T 4 . The
composition of radiant energy is non-uniform. A surface element does
not radiate energy with equal intensity in all directions and the radiation
emitted consists of electromagnetic waves of various wavelengths and
the energy is not distributed uniformly over the whole range of
wavelengths. The rate of energy leaving a surface in a given direction per
unit solid angle is called the intensity of radiation. The spatial distribution
of intensity is expressed by Lambert’s cosine law, I   I n cos  . The
intensity In normal to the surface of a black body is given as, I n 
T 4

The radiation energy emitted by a black surface is not emitted at one
frequency, but over a wide and continuous range of frequencies. The
spectral distribution i.e. its variation over all wavelengths is given by
C1
Planck’s law, the monochromatic emissive power qb 
C2
 5  e T  1


The higher the temperature T of a surface, the larger the proportion of
energy emitted at the shorter wavelengths, and also the shorter is the
wavelength λmax at which the emissive power is maximum. The position
18
of the maximum emissive power in the waveband is expressed by Wein’s
displacement law, max T  0.00290 m K .
The total hemispherical emissivity or simply emissivity, ε is the ratio of
the total energy emitted by a surface to the total energy emitted by a
E
black surface at the same temperature. i.e.  
.
Eb
The relation between emissivity of a surface and its absorptivity is given
by the kirchoff’s law, ‘the monochromatic emissivity of a surface at T1 is
equal to its monochromatic absorptivity for radiation received from
another surface at the same temperature. i.e.      . A surface, to
which this is true, is known as a grey body.
The net heat exchange between two black surfaces is given as,
Q1 2  A1 F1 2 T14  T24  . F1-2 is the view factor expressed as
F1 2 
1
A1
  cos cos 
A1
A2
dA1 dA2
. The view factor is a function of the
r 2
geometric configuration of the two surfaces. It is a measure of how much
of the field of view of A1 is occupied by A2. Some of the properties
associated with view factors include Reciprocal, Additive, convex and
concave surface considerations and parallel surfaces.
1. Which of the following statements is true? Heat transfer by radiation.
 Only occurs in space
 Is negligible in space in free convection
 Is a fluid phenomenon and travels at the speed of the fluid
 Is an acoustic phenomenon and travels at the speed of sound
 Is an electromagnetic phenomenon and travels at the speed of light
2. The true value of Stefan – Boltzmann constant σ is,
 56.7 × 10-9 w/k
 56.7 × 10-9 w/m2
 56.7 × 10-9 w/mk
 56.7 × 10-9 w/m2k
 56.7 × 10-9 w/m2 k4
3. The range of wavelength over which thermal radiation takes place is,
 0.1 ≤ λ ≤ 100 nm
 0.1 ≤ λ ≤ 100 μm
 0.1 ≤ λ ≤ 100 mm
 0.1 ≤ λ ≤ 100 m
 0.1 ≤ λ ≤ 100 km
4. Which of the following statements applies to black body radiation?
 As the temperature increases the wavelength at which peak – emission
occurs decreases
 As the temperature increases, the wavelength at which peak – emission
occurs remains the same
 As the temperature increases the peak – emission shifts towards the
infrared
19

As the temperature increases the frequency at which the peak –
emission occurs decreases
5. The ratio; thermal radiation emitted by a surface to that emitted by a black
body at the same temperature is known as,
 Reflectivity
 Radiosity
 Emissivity
 Solar irradiation
 Transvissivity
6. τ + α + ρ = ?
a) √-1,
b) π, c) 0, d) 1, e) -1
7. The view factor Fij is defined as,
 The ratio of the emissivity of surface i to surface j
 The ratio of the absolute temperature of surface i to j
 The ratio of the area of surface i to surface j
 The fraction of radiation emitted by surface j which is received by
surface i
 The fraction of radiation emitted by surface i which is received by
surface j
8. The reciprocity rule of view factor algebra states that,
n
a)
 Ai Fij  0 ,
n
b)
j 1
 Ai Fij  1,
n
c)
j 1
F
j 1
ij
1
d) Fij  F ji
1
d) Fij  F ji
e) Ai Fij  A j F ji .
9. The summation rule of view factor states that,
n
a)
AF
j 1
i
ij
0,
n
b)
AF
j 1
i
ij
 1,
n
c)
F
j 1
e) Ai Fij  A j F ji .
10. For concave surface;
 Fii = 0
 Fii ≤ 0
 Fii ≥ 0
 Fii = 1
 Fii = ∞
11. For convex surface;
 Fii = 0
 Fii ≤ 0
 Fii ≥ 0
 Fii = 1
 Fii = ∞
12. A1 = 1m2, F12 = 0.2, A2 = 2m2, so F21 = ?
 1
 0.1
 0.2
 2
20
ij

∞
13. Kirchoff’s law of radiation states that;
 Algebraic sum of the currents at a node is zero
 J1 = E 1
 ε1 = α1 (whatever the temperature)
 ε1 = α1 (providing these processes take at comparable temperature.
 Ei = Ebi.
14. Which of the following is the correct formulation of Wein’s displacement law;
 λmaxT = 2.8978 mm K
 λmaxT2 = 2.8978 mm K2
 λmax /T = 2.8978 mm/K
 λmax /T2 = 2.8978 mm/K2
 λmax /T4 = 2.8978 mm/K4
15. The emissivity of a polished aluminium surface is (approximately)
a) 0.002
b) 0.9
c) 0.7,
d) 0.5,
e) 0.1
16 The emissivity of a polished aluminium surface which is painted matt black is
(approximately)
a) 0.002
b) 0.9
c) 0.7,
d) 0.5,
e) 0.1
17 In a four surface enclosure, how many view factors are there?
 8
 4
 12
 15
 16
References:
1. Introduction to Heat Transfer, Frank P. Incropera, David P.
DeWitt, Third Edition, John Wiley & Sons, 1996
2. Heat and Mass Transfer, C. P. ARoRa, Third Edition, Khanna
Publishers, 1986.
3. Introduction to Thermodynamics and Heat Transfer, Yunus A.
Cengel, Second Edition, McGraw-Hill Primis, 2006.
4. Heat Transfer, J.P Holman, Tenth Edition, McGraw-Hill, 2010.
5. Heat Transfer, Chris long, Nasser Sayma, Ventus Publishing Aps
2009.
6. Engineering Thermodynamics, Work & Heat Transfer, G.F.C
Rogers & Y.R. Mayhew, Fourth Edition, Pearson Prentice Hall,
1992.
7. Applied Thermodynamics, for Engineering Technologists, T.D.
Eastop & A. McConkey, Fourth Edition, Longman Scientific &
Technical, 1986.
8. Principles of Heat Transfer, Frank Kreith, Raj M. Manglik, Mark
S. Bohn. Seventh Edition, Global Engineering, Cengage
Learning, 2011.
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