Thermal Radiation-II- View factors and Radiation energy exchange between black bodies

Lectures on Heat Transfer –
THERMAL RADIATION-II:
View factors and Radiation energy
exchange between black bodies
by
Dr. M. ThirumaleshwarDr. M. Thirumaleshwar
formerly:
Professor, Dept. of Mechanical Engineering,
St. Joseph Engg. College, Vamanjoor,
Mangalore
India

Preface:
• This file contains slides on THERMAL
RADIATION-II: View factors and Radiation
energy exchange between black bodies.
• The slides were prepared while teaching
Heat Transfer course to the M.Tech.Heat Transfer course to the M.Tech.
students in Mechanical Engineering Dept.
of St. Joseph Engineering College,
Vamanjoor, Mangalore, India, during Sept.
– Dec. 2010.
Aug. 2016 2MT/SJEC/M.Tech.

• It is hoped that these Slides will be useful
to teachers, students, researchers and
professionals working in this field.
• For students, it should be particularly
useful to study, quickly review the subject,useful to study, quickly review the subject,
and to prepare for the examinations.
•
Aug. 2016 3MT/SJEC/M.Tech.

References:
• 1. M. Thirumaleshwar: Fundamentals of Heat &
Mass Transfer, Pearson Edu., 2006
• https://books.google.co.in/books?id=b2238B-
AsqcC&printsec=frontcover&source=gbs_atb#v=onepage&q&f=false
• 2. Cengel Y. A. Heat Transfer: A Practical
Approach, 2nd Ed. McGraw Hill Co., 2003
Aug. 2016 MT/SJEC/M.Tech. 4
Approach, 2nd Ed. McGraw Hill Co., 2003
• 3. Cengel, Y. A. and Ghajar, A. J., Heat and
Mass Transfer - Fundamentals and Applications,
5th Ed., McGraw-Hill, New York, NY, 2014.

References… contd.
• 4. Incropera , Dewitt, Bergman, Lavine:
Fundamentals of Heat and Mass Transfer, 6th
Ed., Wiley Intl.
• 5. M. Thirumaleshwar: Software Solutions to• 5. M. Thirumaleshwar: Software Solutions to
Problems on Heat Transfer – Radiation-Part-I,
Bookboon, 2013
• http://bookboon.com/en/software-solutions-heat-transfer-
radiation-i-ebook

Thermal Radiation – II:
View factors and Radiation energy exchange
between black bodies:
Outline…
• View factor – general relations – radiation
energy exchange between black bodies –
properties of view factor and view factor
algebra – view factor formulas and graphs
– Problems

The ‘View factor’ and radiation energy
exchange between black bodies:
• Radiative heat exchange depends on:
• Absolute temperatures of surfaces
• Radiative properties of surfaces, and
• Geometry and relative orientation of the
• Geometry and relative orientation of the
surfaces involved – this is quantified by
what is known as ‘View factor’.

View factor:
• View factor is defined as the fraction of
radiant energy leaving one surface which
strikes a second surface directly.
• Here, ‘directly’ means that reflection or re-
radiated energy is not considered.
radiated energy is not considered.
• View factor is denoted by F12 where the
first subscript, 1 stands for the emitting
surface, and the second subscript, 2
stands for the receiving surface.

• We have:
• F12 = (Direct radiation from surface1
incident on surface2) divided by (Total
radiation from emitting surface1).
• General relation for view factor
between two surfaces:
between two surfaces:
• Infinetisimal areas:
• Consider differential areas dA1 and dA2 on
two black surfaces A1 and A2 exchanging
heat by radiation only. See Fig. 13.13.

• dA1 and dA2 are at a distance ‘r’ apart and the
normals to these areas make angles with the
line connecting them, as shown. Then, using the
definition of Intensity, we can write:

• Energy leaving dA1 and falling on dA2 = dQ12 =
Intensity of black body dA1 x projected area of dA1 on a
plane perpendicular to line joining dA1 and dA2 x solid
angle subtended by dA2 at dA1.
i.e. dQ 12 I b1 dA 1 cos φ 1
..
dA 2 cos φ 2
.
r
2
. ......(13.29)
Now, total energy radiated from dA1 is given by:
dQ 1 E b1 dA 1
.
i.e. dQ 1 π I b1
. dA 1
. .....(13.30)

• Then, by definition, the view factor FdA1-dA2 is the ratio of
dQ12 to dQ1:
F dA1_dA2
cos φ 1 cos φ 2
. dA 2
.
π r
2.
....(!3.31)
Note that the View factor involves geometrical
quantities only.
quantities only.
Eqn. (13.31) gives the view factor between two
infinetisimal areas.
Such a situation is encountered even when finite areas
are involved, when the distance between these two
areas, ‘r’, is very large.

• Infinitesimal to finite area:
• i.e. the emitter is very small and the receiving surface is
of finite size. Here, integration over the entire surface A2
has to be considered.
• Again, remembering the definition of view factor, and
forming the ratio of dQ12 to dQ1:
2
2
2111 )cos())(cos(
r
dA
dAIb φφ
11
22111
21
)cos())(cos(
dAI
r
dAI
F
b
b
AdA
π
φφ
=−
Since both Ib1 and dA1 are independent of integration,
we can write:
)32.13(
)cos()cos(
2
2
221
21 =−
A
AdA
r
dA
F
π
φφ

• Above case is for a small thermocouple bead located
inside a pipe or a small, spherical point source radiator
located by the side of a wall etc.
• Finite to finite area:
• Once again, from the definition of view factor:
1 2 2
2
2111 )cos())(cos(
=
A A
b
r
dA
dAI
F
φφ
11
1
1 2 2
21 =−
A
b
A A
AA
dAI
rF
π
For constant Ib1, above eqn. becomes:
=−
1 2
212
21
1
21 )33.13(
)cos()cos(1
A A
AA dAdA
rA
F
φφ
π

• In general, we write eqn. (13.33) compactly as:
F 12
1
π A 1
.
0
A 1
A 2
0
A 2
A 1
cos φ 1 cos φ 2
.
r
2
d d. ....(13.34)
Here, F12 means ‘the view factor from surface 1 to surface 2’.
Similarly, the view factor from surface 2 to surface 1:
)35.13(
)cos()cos(1 12
dAdAF =
φφ
)35.13(
)cos()cos(1
12
2 1 2
12
2
21 dAdA
rA
F
A A
=
φφ
π
Multiplying eqn. (13.34) by A1, and eqn. (13.35) by A2, and
equating the double integrals, we get:
A 1 F 12
. A 2 F 21
. .....(13.36)
Eqn. (13.36) is known as ‘reciprocity theorem’ and is a very useful
and important relation.

• Note: It is easier to remember the view factor relation
given in eqn. (13.34) as:
A 1 F 12
.
0
A 1
A 2
0
A 2
A 1
cos φ 1 cos φ 2
.
π r
2.
d d ....(13.37)

Radiation energy exchange
between black bodies:
• Consider two black surfaces A1 and A2 exchanging
radiation energy with each other.
• Then, rate of energy emitted by surface 1, which directly
strikes surface 2 is given by:
4.
Q 12 A 1 F 12
. E b1
. A 1 F 12
. σ. T 1
4. .....(13.38)
This energy is completely absorbed by surface 2, since
surface 2 is black.
Similarly, of energy emitted by surface 2, which directly
strikes surface 1 is given by:
Q 21 A 2 F 21
. E b2
. A 2 F 21
. σ. T 2
4. .....(13.39)

• And, net radiation exchange between the two surfaces
is:
Q net A 1 F 12
. σ. T 1
4. A 2 F 21
. σ. T 2
4.
But, A 1 F 12
. A 2 F 21
. by reciprocity theorem
Therefore,
Therefore,
Q net A 1 F 12
. σ. T 1
4
T 2
4. A 2 F 21
. σ. T 1
4
T 2
4. W....(13.40)

Properties of View factor and
view factor algebra:
• View factors of some geometries are easily
calculated;
• However, more often, calculation of view factors
for more complex shapes is quite difficult.
• In many cases, the complex shapes could be
• In many cases, the complex shapes could be
broken down into simpler shapes for whom the
view factors are already known or could easily
be calculated.
• Then, view factor for the desired complex shape
is calculated using ‘view factor algebra’.

Properties of view factor:
• The view factor depends only on the geometrics of
bodies involved and not on their temperatures or
surface properties.
• Between two surfaces that exchange energy by
radiation, the mutual shape factors are governed by
the ‘reciprocity relation’, viz. A1.F12 = A2.F21.
the ‘reciprocity relation’, viz. A1.F12 = A2.F21.
• When a convex surface 1 is completely enclosed by
another surface 2, it is clear from Fig. 13.14(a) that all
of the radiant energy emitted by surface 1 is
intercepted by the enclosing surface 2. Therefore,
view factor of surface 1 w.r.t. surface 2 is equal to
unity. i.e. F12 = 1. And, the view factor of surface 2
w.r.t. surface 1 is then easily calculated by applying
the reciprocity relation, i.e. A1.1 = A2.F21, or, F21 =
A1/A2.

F12 = 1
F21 = A1/A2
F11 = 0
2
1
F11 = 0 F11 0
Fig. 13.14View factors for a few surfaces
(a) Surface 1 completely
enclosed by surface 2
(b) Flat surface (c) Convex surface (d) Concave surface

• Radiation emitted from a flat surface
never falls directly on that surface (see
Fig. 13. 14 (b)), i.e. View factor of a flat
surface w.r.t. itself is equal to zero, i.e.
F11 = 0. This is valid for a convex surface
too, as shown in Fig. 13.14(c).
• For a concave surface, it is clear from
Fig. 13.14(d) that F11 is not equal to zero
since some fraction of radiation emitted
by a concave surface does fall on that
surface directly.

• If two, plane surfaces A1 and A2 are
parallel to each other and separated by a
short distance between them, practically
all the radiation issuing from surface 1
falls directly on surface 2, and vice –
versa. Therefore, F12 = F21 = 1
• When the radiating surface 1 is divided
• When the radiating surface 1 is divided
into, say, two sub-areas A3 and A4, as
shown in Fig. 13.15(a), we have:
•
• Obviously, F12 ≠ F32 + F42.
A 1 F 12
. A 3 F 32
. A 4 F 42
. .....(13.41)

Receiving
surface, A2
F13A2
A3
A4
A1.F12 = A3.F32+ A4.F42
A1
A3
A4
F14
A1.F12 = A1.F13+ A1.F14
Fig. 13.15 View factors for subdivided surfaces
(a) Radiating surface A1 is
subdivided into A3 and A4
A1 = A3 + A4
A4 A1
(b) Receiving surface A2 is
subdivided into A3 and A4
12 13 14

• Instead, if the receiving surface A2 is sub-
divided into parts A3 and A4 as shown in Fig.
13.15 (b), we have:
A 1 F 12
. A 1 F 13
. A 1 F 14
. .....(13.42)
i.e. F 12 F 13 F 14
• i.e. view factor from the emitting surface 1 to a
sub-divided receiving surface is simply equal
to the sum of the individual shape factors from
the surface 1 to the respective parts of the
receiving surface. This is known as
‘Superposition rule’.

• Symmetry rule: If two (or more) surfaces are
symmetrically located w.r.t. the radiating
surface 1, then the view factors from surface 1
to these symmetrically located surfaces are
identical. A close inspection of the geometry
will reveal if there is any symmetry in a given
problem.
• Summation rule: Since radiation energy is
emitted from a surface in all directions. Then,
emitted from a surface in all directions. Then,
the conservation of energy principle requires
that sum of all the view factors from the
surface 1 to all other surfaces forming the
enclosure, must be equal to 1.
• See Fig. 13.16, where the interior surface of a
completely enclosed space is sub-divided into
n parts, each of area A1, A2, A3…An.

2
1
3
n-1
n
n
Fig. 13.16 View factor--summation rule for
radiation in an enclosure
1
n
j
F ij
=
1 i = 1, 2, 3,.....n

• Then,
• i.e.
F 11 F 12 ..... F 1n 1
F 21 F 22 ..... F 2n 1
..............................................
F n1 F n2 ..... F nn 1
• i.e.
1
n
j
F ij
=
1 i = 1, 2, 3,.....n......(13.44)

• In an enclosure of ‘n’ black surfaces,
maintained at temperatures T1, T2,….Tn,
net radiation from any surface, say
surface 1, is given by summing up the
net radiation heat transfers from surface
1 to each of the other surfaces of the
enclosure:
Q 1net
A 1 F 12
. σ. T 1
4
T 2
4. A 1 F 13
. σ. T 1
4
T 3
4. A 1 F 14
. σ. T 1
4
T 4
4.
A 1 F 1n
. σ. T 1
4
T n
4.+
...
....(13.45)

• Note: Often, while solving radiation
problems, determination of the view factor
is the most difficult part.
• It will be useful to keep in mind the
definition of view factor, summation rule,
reciprocity relation, superposition rule and
reciprocity relation, superposition rule and
symmetry rule while attempting to find out
the view factors.

Methods of determining view
factors:
• 1. By performing the necessary integrations in
equations (13.31), (13.32) or (13.33). However, except
in very simple cases, most of the time, the direct
integration procedure is quite difficult.
• 2. Use of readily available analytical formulas or
graphs prepared by researchers for the specific
graphs prepared by researchers for the specific
geometry in question.
• 3. Use of view factor algebra in conjunction with
definition of view factor, summation rule, reciprocity
relation, superposition rule and symmetry rule.
• 4. Experimental and graphical techniques.

2. View Factors by analytical
formulae and graphs:
• Figs. (13.17) and (13.18) show a few two-dimensional
and three-dimensional geometries and Tables (13.4) and
(13.5) give corresponding view factor relations.

L
(a) Aligned parallel rectangles
j
i
L
j
(b) Coaxial parallel disks
Y
X i
rj
ri
Fig. 13.18 Few three-dimensional geometries
j
(c) Perpendicular rectangles with a common edge
i
XY
Z

• Table 13.5 gives view factor relations for three important
three-dimensional geometries, often required in practice:
• For example, view factors between aligned parallel
rectangles will be useful to calculate heat transfer
between the floor and ceiling of a room or a furnace.
• view factors between coaxial parallel disks will be
required to calculate the heat transfer between the top
and bottom of a cylindrical furnace, and
• view factors between perpendicular rectangles is
• view factors between perpendicular rectangles is
necessary to calculate the fraction of energy entering a
floor through a window on the adjacent wall, or to
determine the fraction of energy radiated from the door
of a furnace to the floor outside etc.

• View factors for a few three-dimensional geometries (contd.)

• View factor relation
for aligned, parallel
rectangles of Fig.
(13.18,a), is shown in
graphical form in Fig.
13.19. This graph is
drawn with Mathcad.
0.1
1
View factor for parallel rectangles
Fij
0.1 1 10 100
0.01
Y/L = 0.1
Y/L = 0.2
Y/L = 0.4
Y/L = 0.6
Y/L = 1.0
Y/L = 2.0
Y/L = 4.0
Y/L = 10.0
X/L
Fig. 13.19 View factor for aligned, parallel rectangles (See Fig. 13.18,a)

• Graph of View
factor for coaxial
parallel disks (of
Fig. 13.18,b) is
drawn using
Mathcad and is
shown in Fig.
13.20.
0.2
0.4
0.6
0.8
1
View factor for coaxial parallel disks
Fij
0.1 1 10
rj/L = 0.3
rj/L = 0.6
rj/L = 1.0
rj/L = 1.5
rj/L = 2.0
rj/L = 4.0
rj/L = 8.0
L/ri
Fig. 13.20 View factor for coaxial, parallel disks (See Fig. 13.18,b)

• And, graph of View
factors for
perpendicular
rectangles with a
common edge (of
Fig. 13.18,c),
drawn using
Mathcad, is shown
in Fig. 13.21.
0.1
0.2
0.3
0.4
0.5
View factor for perpendicular rectangles
Fij
0.1 1 10
0
Y/X = 0.02
Y/X = 0.05
Y/X = 0.1
Y/X = 0.2
Y/X = 0.6
Y/X = 1.0
Y/X = 2.0
Y/X = 4.0
Y/X = 10.0
Z/X
Fig. 13.21 View factors for coaxial, perpendicular rectangles with a common
edge (See Fig. 13.18,c)

• Another practically important geometry is that of two concentric cylinders
of finite length. View factors associated with this geometry are shown in
Fig. 13.22. below [Ref: Cengel]:

• Example 13.10: Find out the net heat
transferred between two circular disks
1 and 2, oriented one above the other,
parallel to each other on the same
centre line, as shown in Fig. Ex. 13.10.
Disk 1 has a radius of 0.5 m and is
maintained at 1000 K, and disk 2 has a
maintained at 1000 K, and disk 2 has a
radius of 0.6 m and is maintained at 600
K . Assume both the disks to be black
surfaces.

L = 1 m
rj = 0.6 m
ri = 0.5 m
T = 1000 K
T2 = 600 K
L = 1 m
Fig. Ex.13.10 Coaxial parallel disks
ri = 0.5 m
T1 = 1000 K

Data:
r i 0.5 m....radius of disk 1
r j 0.6 m....radius of disk 2
L 1 m....distance between disk 1 and disk 2
T 1 1000 K....temp. of disk 1
T 2 600 K...temp. of disk 2
σ 5.67 10
8. W/m2.K....Stefan-Boltzmann const.
σ 5.67 10
A 1 π r i
2. i.e. A 1 0.785= m2....area of disk 1
A 2 π r j
2. i.e. A 2 1.131= m2....area of disk 2
This is the case of heat transfer between two black surfaces. So, we use eqn. (13.40 ), viz.
Q net A 1 F 12
. σ. T 1
4
T 2
4. A 2 F 21
. σ. T 1
4
T 2
4. W....(13.40)

So, the problem reduces to calculating the view factor F12 or F21. We can easily find out F 12
using Fig. 13.20. However, we can determine F12 analytially more accurately with Mathcad
using the view factor relation given in Table 13.5 for coaxial parallel disks.
We re-write the view factor relation given in Table 13.5 as follows, for ease of calculation with
Mathcad:
r i r j 1 R j
2
R i
r i
L
R j
r j
L
S R i R j, 1
1 R j
R i
2
F 12 R i R j,
1
2
S R i R j, S R i R j,
2
4
R j
R i
2
.
1
2
. ...View factor for coaxial parallel disks

Here, first, S is written as a function of Ri and Rj where Ri = ri/L and Rj = rj/L. Then, F12 is
expressed as a function of Ri and Rj. Now, F12 is easily obtained for any values of Ri and Rj,
by simply writing F 12(Ri,Rj) = .
Therefore, R i 0.5=
and, R j 0.6=
We get: F 12 0.5 0.6,( ) 0.232=
Verify: This result may be verified from Fig. 13.20 where F12 is plotted against L/r i for various
values of rj/L. Now, for our problem, L/ri = 1/0.5 = 2, and r j/L = 0.6/1 = 0.6. Then, from Fig. 13.20,
values of rj/L. Now, for our problem, L/ri = 1/0.5 = 2, and r j/L = 0.6/1 = 0.6. Then, from Fig. 13.20,
we read: F12 = 0.232, approx.
i.e. F 12 0.232
Therefore, net heat transfer between disks 1 and 2:
Q net A 1 F 12
. σ. T 1
4
T 2
4. W...from eqn. (13.40)
i.e. Q net 8.992 10
3
= W....Ans.

0.2
0.4
0.6
0.8
1
View factor for coaxial parallel disks
Fij
0.1 1 10
0.2
rj/L = 0.3
rj/L = 0.6
rj/L = 1.0
rj/L = 1.5
rj/L = 2.0
rj/L = 4.0
rj/L = 8.0
L/ri
Fig. 13.20 View factor for coaxial, parallel disks (See Fig. 13.18,b)

3. View Factors by use of View
factor algebra:
• Often, we have to find out view factors for geometries for
which readily no analytical relations or graphs are
available.
• In such cases, some times, it may be possible to get the
required view factor in terms of view factors of already
known geometries, by suitable manipulation using view
factor algebra.
known geometries, by suitable manipulation using view
factor algebra.
• For this purpose, we remember the definition of view
factor (as the fraction of energy emitted by surface 1 and
directly falling on surface 2), and invoke the summation
rule, reciprocity relation, and inspection of geometry.
• We shall illustrate this procedure with some important
examples:

• Example 13.12: Find out the net heat
transferred between the areas A1 and A2
shown in Fig. Ex. 13.12. Area 1 is
maintained at 700 K, and area 2 is
maintained at 400 K . Assume both the
surfaces to be black.
• Solution:
• Solution:
This is the case of heat transfer between two black surfaces. So, we use eqn. (13.40 ), viz.
Q net A 1 F 12
. σ. T 1
4
T 2
4. A 2 F 21
. σ. T 1
4
T 2
4. W....(13.40)

A1 (700K)
3 m
2 m
A3
A
A6
A5
Fig. Ex.13.12 Perpendicular rectangles with a common edge
2 m
3 m
5 m
A4
A2
(400K)
6

So, the problem reduces to calculating the view factor F12 or F21. We see that to calculate
F12 for areas A1 and A2 as oriented in the Fig., we do not readily have an analytical relation
or a graph.Let us denote the combined areas (A1 + A3) by A5 and (A2 + A4) by A6. Then, we
see that A5 and A6 are perpendicular rectangles which have a common edge, and we have
graphs or analytical relation for the view factor for such an orientation. Then, we resort to
view factor algebra, as follows:
Remember that by definition, view factor F12 is the fraction of radiant energy emitted by
surface 1 which falls directly on surface 2. Looking at the fig., we can say that fraction of
energy leaving A1 and falling on A2 is equal to the fraction falling on A6 minus the fraction
falling on A4.
i.e. F 12 F 16 F 14 ...by definition of view factor
i.e. F 12 F 61
A 6
A 1
. F 41
A 4
A 1
. ...since by reciprocity relation, A 1.F16 = A6.F61, and
A1.F14 = A4.F41.

i.e. F 12
A 6
A 1
F 65 F 63
.
A 4
A 1
F 45 F 43
. eqn.(A)....using the definition of view
factor, as done in first step above
Now, observe that view factors F65, F63, F45 and F43 refer to perpendicular rectangles with a
comon edge, and can be readily obtained from Fig. 13.21, or by analytical relation given in
Table 13.5.
We re-write the view factor relation for perpendicular rectangles with a comon edge, given in
Table 13.5 as follows, for ease of calculation with Mathcad:
H
Z
X
W
Y
X
A W( )
1
π W.
B W( ) W atan
1
W
.
C H( ) H atan
1
H
. D H W,( ) H
2
W
2
1
2
atan
1
H
2
W
2
1
2
.

0.1
0.2
0.3
0.4
0.5
View factor for perpendicular rectangles
Fij
0.1 1 10
0
Y/X = 0.02
Y/X = 0.05
Y/X = 0.1
Y/X = 0.2
Y/X = 0.6
Y/X = 1.0
Y/X = 2.0
Y/X = 4.0
Y/X = 10.0
Z/X
Fig. 13.21 View factors for coaxial, perpendicular rectangles with a common
edge (See Fig. 13.18,c)

E H W,( )
1 W
2
1 H
2.
1 W
2
H
2
W
2
1 W
2
H
2.
1 W
2
W
2
H
2.
W
2
. H
2
1 H
2
W
2.
1 H
2
H
2
W
2.
H
2
.
F ij H W,( ) A W( ) B W( ) C H( ) D H W,( )
1
4
ln E H W,( )( ).. .....eqn.(B)
...View factor for coaxial perpendicular rectangles
with a comon edge
To find F65:
X 5 Y 5 Z 5 ....w.r.t. Fig. 13.18(c) and Fig.Ex.13.12
H
Z
X
i.e. H 1=
W
Y
X
i.e. W 1=

Therefore,
F ij 1 1,( ) 0.2= ....substituting in eqn.(B)
i.e. F 65 0.2 ...view factor from area A6 to A5
Note: This value can be verified from Fig. 13.21 also.
To find F63:
63
H
Z
X
i.e. H 0.6=
W
Y
X
i.e. W 1=

Therefore,
F ij 0.6 1,( ) 0.161= ....substituting in eqn.(B)
Note: This value also can be verified from Fig. 13.21.
To find F45:
To find F45:
H
Z
X
i.e. H 1=
W
Y
X
i.e. W 0.6=

Therefore,
F ij 1 0.6,( ) 0.269= ....substituting in eqn.(B)
To find F43:
H
Z
X
i.e. H 0.6=
W
Y
X
i.e. W 0.6=

Therefore,
F ij 0.6 0.6,( ) 0.231= ....substituting in eqn.(B)
Areas:
From Fig.Ex.13.12, we have:
A 1 10 A 2 10 A 3 15 A 4 15
A 1 10 A 2 10 A 3 15 A 4 15
A 5 25 A 6 25 m2
Then, from eqn.(A):
F 12
A 6
A 1
F 65 F 63
.
A 4
A 1
F 45 F 43
.
i.e. F 12 0.041= ....view factor from A 1 to A2.....Ans.
Note: F21 can be calculated,if required, by reciprocity relation, i.e. A 1.F12 = A2.F21

Therefore, net heat transfer between surfaces 1 and 2:
Q net A 1 F 12
. σ. T 1
4
T 2
4. W...from eqn. (13.40)
Here, we have:
T 1 700 K....temp. of surface A 1
T 2 400 K...temp. of surface A 2
σ 5.67 10
Therefore,
Q net A 1 F 12
. σ. T 1
4
T 2
4.
i.e. Q net 4.926 10
3
= W....Ans.

• Example 13.13: Find out the view factor F14 between
the areas A1 and A4 shown in Fig. Ex. 13.13.
A3
A4
A6
Fig. Ex.13.13 Surface arrangement for Ex. 13.13
A1
A2
A5

We observe that to calculate F14 between areas A1 and A4 as oriented in the Fig., we do not
readily have an analytical relation or a graph. Let us denote the combined areas (A1 + A2) by
A5 and (A3 + A4) by A6. Then, we see that A5 and A6 are perpendicular rectangles which have
a common edge, and we have graphs or analytical relation for the view factor for such an
orientation. Then, we resort to view factor algebra, as follows:
Remember the general definition of view factor: F12 is the fraction of radiant energy emitted
by surface 1 which falls directly on surface 2. Looking at the fig., we can say that fraction of
energy leaving A5 and falling on A6 is equal to the fraction falling on A3 plus the fraction falling
on A4.
i.e. ...by definition of view factor
i.e. F 56 F 53 F 54 ...by definition of view factor
i.e. F 56 F 35
A 3
A 5
. F 45
A 4
A 5
. ...since by reciprocity relation, A 5.F53 = A3.F35, and
A5.F54 = A4.F45.
i.e. F 56
A 3
A 5
F 31 F 32
.
A 4
A 5
F 41 F 42
. ...using the rule for subdivision of
receiving surface

i.e. A 5 F 56
. A 3 F 31
. A 3 F 32
. A 4 F 41
. A 4 F 42
. .....(A)
Further, it can be proved that A1.F14 = A2.F23
Also, by reciprociy theorem, we can write:
A 1 F 14
. A 4 F 41
.
A 3 F 31
. A 1 F 13
.
A 4 F 42
. A 2 F 24
.
and, A 2 F 23
. A 3 F 32
.

Using these relations in eqn. (A):
A 5 F 56
. A 1 F 13
. A 1 F 14
. A 1 F 14
. A 2 F 24
.
i.e. 2 A 1
. F 14
. A 5 F 56
. A 1 F 13
. A 2 F 24
.
i.e. F 14
1
2 A 1
.
A 5 F 56
. A 1 F 13
. A 2 F 24
.. .....(B)
• From eqn. (B), F14 can easily be calculated
since, all the three view factors appearing on the
RHS are perpendicular rectangles with a
common edge, which may be obtained readily
from the graphs or analytical relations.

• Example 13.14: Find out the relevant view
factors for the geometries shown in Fig. Ex.
13.14:
• (a) a long tube with cross-section of an
equilateral triangle
• (b) a blackbody completely enclosed by another
black body
• (c) diagonal partition inside a long square duct
• (d) sphere of diameter d inside a cubical box of
sides, L = d
• (e) hemispherical surface closed by a plane
surface, and
• (f) the end and surface of a circular cylinder
whose length is equal to diameter.

(a) Equilateral triangle (b) A blackbody completely
enclosed by another black body
1
2 3
1
2
1
2
3
L
(c) Diagonal partition in
a long square duct
Fig. Ex.13.14 Different geometries for Ex. 13.14
(d) Sphere inside a
cubical box
L = d r
1
2
1
2
3
L = d
(e) Hemispherical bowl (f) Cylinder with length
= diameter

Solution:
General principle in solving these problems is to invoke: Summation rule, reciprocity theorem,
inspection of geometry for symmetry, and of course, remembering the definition of view factor:
(a) Long tube wih cross-section of equilateral triangle: See Fig. (13.14,a)
For surface 1: F 11 F 12 F 13 1 ....Summation rule
But, F 11 0 ...since surface 1 is flat and can not see itself.
Therefore, F 12 F 13 1
Now, by inspection of geometry, we find that surfaces 2 and 3 are located symmetrically w.r.t.
surface 1, since it is an equilateral triangle. Therefore, radiation from surface 1 is divided equally
between surfaces 2 and 3.

i.e. F 12 F 13 0.5
Similarly, for surface 2, we write:
F 21 F 23 1
i.e. F 23 1 F 21
A 1
But, F 21
A 1
A 2
F 12
. F 12 ...since A 1 = A2
Therefore, F 23 1 F 12 0.5
Similarly for surface 3.

(b) Black body enclosed inside a black enlosure: See Fig. (13.14,b)
For surface 1: F 11 F 12 1 ...by Summation rule
and, A 1 F 12
. A 2 F 21
. ...by reciprocity
i.e. F 12
A 2
A 1
F 21
.
Now, F 11 1 F 12
Now, F 11 1 F 12
i.e. F 11 1
A 2
A 1
F 21
.
But, F 21 1 since all the energy radiated by surface 2 is directly intercepted by
surface 1.

Therefore, F 12 F 13 1
By symmetry: F 3 F 12 0.5 ..since radiation emitted by surface 1 is divided equally
between surfaces 2 and 3
between surfaces 2 and 3
By reciprocity: F 21
A 1
A 2
F 12
.
i.e. F 21
2 L.
L
0.5. 0.71

(d) Sphere inside a cubical box: See Fig. (13.14,d)
For surface 1: F 11 0 ...since surface of the sphere is convex and can not 'see'
itself.
and, F 11 F 12 1 ...by Summation rule
Therefore, F 12 1
A 1
A 2
F 12
.
21
A 2
12
i.e. F 21
π d
2.
6 d
2.
π
6
...since L = d
i.e. F 21 0.524

(e) Hemispherical surface closed by a flat surface: See Fig. (13.14,e)
For surface 1: F 11 F 12 1 ....Summation rule
Also, F 21 1 ...since surface 2 is flat and can not 'see'itself, and all
radiation emitted by surface 2 falls directly on the
hemispherical surface 1.
A 2
A 1
F 21
.
i.e. F 12
π r
2.
2 π. r
2.
1. 0.5
Therefore, F 11 1 F 12 0.5

(f) End and sides of a circular cylinder (L = d): See Fig. (13.14,f)
From the fig., note that the two end surfaces are denoted by 1 and 3 and the side surface
is denoted by 2.
View factor F13: Surfaces 1 and 3 can be considered as two concentric parallel disks.
Therefore, F13 can be found out from Fig. 13.20 or by analytical relation given in Table
13.5. Let us use the analytical relation:
We have: R i
r i
L
R j
r j
L
S R i R j, 1
1 R j
2
R i
2
R
2
1
2
F ij R i R j,
1
2
S R i R j, S R i R j,
2
4
R j
R i
2
.
2
.
...View factor for coaxial parallel disks
Now, for a cylinder with L = d:
R i 0.5 R j 0.5
Therefore, F ij R i R j, 0.172=
i.e. F 13 0.172 ...view factor from surface 1 to surface 3

For surface 1: F 11 F 12 F 13 1 ...by Summation rule
But, F 11 0 ...since surface 1 is flat and can not 'see'itself.
Therefore, F 12 1 F 13
i.e. F 12 0.828
12
A 1
A 2
F 12
.

i.e. F 21
π d
2.
4
π d. L.
0.828.
i.e. F 21
π d
2.
4
π d
2.
0.828. ...since L = d
i.e. F
0.828
0.207
i.e. F 21
0.828
4
0.207
Also, by symmetry: F 32 F 12 0.828
and, F 23 F 21 0.207

• Example 13.15: Find out the view factor
(F11) of a cavity with respect to itself.
Hence, find out the view factor F11 for the
following:
• (a) a cylindrical cavity of diameter 'd'and
depth 'h'with vertex angle 2
• (b) a conical cavity of diameter 'd'and
depth 'h'
• (c) a hemispherical bowl of diameter 'd'

View factor of a general cavity w.r.t. itself:
See Fig. Ex. (13.15,a).
We desire to find F11. It is obvious from the fig. that part of the radiation emitted by the cavity
surface 1, falls on itself and therefore, F11 exists.
Close the opening (or mouth) of the cavity by a hypothetical flat surface 2. Then, surfaces 1
and 2 together form an enclosure. We can write:
For surface 1: F 11 F 12 1 ...(a)....by Summation rule
For surface 2: F 21 F 22 1 ....by Summation rule
But, F 22 0 ..since surface 2 is flat and can not 'see'itself.
Therefore, F 21 1

Further, A 1 F 12
. A 2 F 21
. ....by reciprocity
i.e. F 12
A 2
A 1
Now, F 11 1 F 12 ...from eqn.(a)
i.e. F 11 1
A 2
A 1
.....eqn.(b)
A 1
Eqn.(b) is an important result, since it gives the shape factor of any general cavity w.r.t.
itself.
Now, this result will be applied to following specific cavities:

(a) F11 for a cylindrical cavity of diameter 'd'and depth 'h': See Fig. Ex.(13.15,b)
We have: F 11 1
A 2
A 1
i.e. F 11 1
π d
2.
4
π d
2.
4
π d. h.
...Note that A1 consists of the area of bottom
circular surface and the cylindrical side surfaces
4 h.
i.e. F 11
4 h.
4 h. d
....Ans.
(b) F11 for a conical cavity of diameter 'd'and depth 'h': See Fig. Ex.(13.15,c)
We have: F 11 1
A 2
A 1

i.e. F 11 1
π d
2.
4
π d. L.
2
...where L is the slant height of the cone
i.e. F 11 1
d
2 L.
i.e. F 11 1 sin α( ) ...where α is the half-vertex angle of the cone.
Alternatively: To get F11 in terms of depth 'h', we can write:
We have: F 11 1
d
2 L.
i.e. F 11 1
d
2 h
2 d
2
4
.
i.e. F 11 1
d
4 h
2. d
2

(c) F11 for a hemispherical bowl of diameter 'd': See Fig. Ex.(13.15,d)
We have: F 11 1
A 2
A 1
i.e. F 11 1
π d
2.
4
π d
2.
2
2
i.e. F 11 1
1
2
i.e. F 11 0.5 ...Ans.
This result means that for any hemispherical cavity, half of the radiation emitted by the
surface 1 falls on itself; it also means that the remaining half falls on the closing surface 2.

Hottel’s crossed strings method:
• This is an extremely simple method to find
out the view factors between infinitely long
surfaces;
• Generally, channels and ducts which have
• Generally, channels and ducts which have
a constant cross-section and are very long
can be modeled as two dimensional and
infinitely long.
• Consider Fig. 13.24 shown below:

Fig. 13.24 Crossed strings method to determine view
C L2
BA
D
L1
L3 L4
L5 L6
1
2
• A,B and C,D are the end points of two surfaces
1 and 2. These are connected by tightly
stretched strings as shown. Then, the view
factor F12 between surfaces 1 and 2 is given by:
Fig. 13.24 Crossed strings method to determine view
factor between two infinitely long surfaces

• Note that this method can be applied even when
the two surfaces 1 and 2 have a common edge
F 12
L 5 L 6 L 3 L 4
2 L 1
.
i.e. F 12
Σ Crossed_strings( ) Σ Uncrossed_strings( )
2 string_on_surface1( ).
....(13.46)
the two surfaces 1 and 2 have a common edge
(as in the case of a triangle); then, the common
edge is treated as an imaginary string of zero
length.
• Also, note that surfaces 1 and 2 may be curved
surfaces, but L1 and L2 are the straight lengths
connecting the edges of the respective surfaces.

Thermal Radiation-II- View factors and Radiation energy exchange between black bodies

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