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. Thirumaleshwarformerly:Professor, Dept. of Mechanical Engineering,St. Joseph Engg. College, Vamanjoor,MangaloreIndia

Preface:

• This file contains slides on THERMALRADIATION-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.

• ��


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.

r2. ......(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_dA2cos φ 1 cos φ 2

. dA 2.

π r2.....(!3.31)

Note that the View factor involves geometricalquantities only.


quantities only.

Eqn. (13.31) gives the view factor between twoinfinetisimal areas.Such a situation is encountered even when finite areasare involved, when the distance between these twoareas, ‘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:

22

2111 )cos())(cos(r

dAdAIb φφ�


11

22111

21

)cos())(cos(

dAIr

dAIF

b

b

AdA π

φφ�=−

Since both Ib1 and dA1 are independent of integration, we can write:

)32.13()cos()cos(

22

22121 ��=−

AAdA r

dAF

πφφ

• 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 22

2111 )cos())(cos(� �=

A A b rdA

dAIF

φφ


11 1

1 2 2

21

�

� �=−

A b

A A

AAdAI

rFπ

For constant Ib1, above eqn. becomes:

� �=− 1 2 21221

121 )33.13(

)cos()cos(1A AAA dAdA

rAF ��

φφπ

• In general, we write eqn. (13.33) compactly as:

F 121

π A 1.

0

A 1

A 2

0

A 2

A 1cos φ 1 cos φ 2

.

r2d 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

122 1 212

221 ��dAdA

rAF

A A� �= φφπ

Multiplying eqn. (13.34) by A1, and eqn. (13.35) by A2, andequating 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 1cos φ 1 cos φ 2

.

π r2.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 24. A 2 F 21

. σ. T 14 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 = 1F21 = A1/A2

F11 = 0

2

1F11 = 0 F11� 0


Fig. 13.14 View factors for a few surfaces

(a) Surface 1 completelyenclosed 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)

Receivingsurface, 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 issubdivided into A3 and A4

A1 = A3 + A4

A4 A1

(b) Receiving surface A2 issubdivided 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 forradiation 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 24. A 1 F 13

. σ. T 14 T 3

4. A 1 F 14. σ. T 1

4 T 44.

A 1 F 1n. σ. T 1

4 T n4.+

...

....(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

iL

j

(b) Coaxial parallel disks

YX 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

1View factor for parallel rectangles

Fij


0.1 1 10 1000.01

Y/L = 0.1Y/L = 0.2Y/L = 0.4Y/L = 0.6Y/L = 1.0Y/L = 2.0Y/L = 4.0Y/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

1View factor for coaxial parallel disks

Fij


0.1 1 10

rj/L = 0.3rj/L = 0.6rj/L = 1.0rj/L = 1.5rj/L = 2.0rj/L = 4.0rj/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

0.5View factor for perpendicular rectangles

Fij


0.1 1 100

Y/X = 0.02Y/X = 0.05Y/X = 0.1Y/X = 0.2Y/X = 0.6Y/X = 1.0Y/X = 2.0Y/X = 4.0Y/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 mT = 1000 K

T2 = 600 K


L = 1 m

Fig. Ex.13.10 Coaxial parallel disks

ri = 0.5 mT1 = 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.



A 1 π r i2. i.e. A 1 0.785= m2....area of disk 1

A 2 π r j2. 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 24. A 2 F 21

. σ. T 14 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 j2


R ir i

LR j

r j

LS R i R j, 1

1 R j

R i2

F 12 R i R j, 1

2S 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 24. W...from eqn. (13.40)

i.e. Q net 8.992 103= W....Ans.

0.2

0.4

0.6

0.8

1View factor for coaxial parallel disks

Fij


0.1 1 10

0.2

rj/L = 0.3rj/L = 0.6rj/L = 1.0rj/L = 1.5rj/L = 2.0rj/L = 4.0rj/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 24. A 2 F 21

. σ. T 14 T 2

4. W....(13.40)

A1 (700K)

3 m

2 mA3

AA6

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 61A 6

A 1

. F 41A 4

A 1

. ...since by reciprocity relation, A 1.F16 = A6.F61, and A1.F14 = A4.F41.

i.e. F 12A 6

A 1F 65 F 63

.A 4

A 1F 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:


HZ

XW

Y

XA W( )

1

π W.B W( ) W atan

1

W.

C H( ) H atan1

H. D H W,( ) H2 W2

1

2atan

1

H2 W2

1

2

.

0.1

0.2

0.3

0.4

0.5View factor for perpendicular rectangles

Fij


0.1 1 100

Y/X = 0.02Y/X = 0.05Y/X = 0.1Y/X = 0.2Y/X = 0.6Y/X = 1.0Y/X = 2.0Y/X = 4.0Y/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 W2 1 H2.

1 W2 H2

W2 1 W2 H2.

1 W2 W2 H2.

W2

. H2 1 H2 W2.

1 H2 H2 W2.

H2

.

F ij H W,( ) A W( ) B W( ) C H( ) D H W,( )1

4ln 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

HZ

Xi.e. H 1=

WY

Xi.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


HZ

Xi.e. H 0.6=

WY

Xi.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:


HZ

Xi.e. H 1=

WY

Xi.e. W 0.6=

Therefore,

F ij 1 0.6,( ) 0.269= ....substituting in eqn.(B)


Note: This value also can be verified from Fig. 13.21.


To find F43:


HZ

Xi.e. H 0.6=

WY

Xi.e. W 0.6=

Therefore,

F ij 0.6 0.6,( ) 0.231= ....substituting in eqn.(B)


Note: This value also can be verified from Fig. 13.21.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 12A 6

A 1F 65 F 63

.A 4

A 1F 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 24. 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



Therefore,

Q net A 1 F 12. σ. T 1

4 T 24.

i.e. Q net 4.926 103= 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 35A 3

A 5

. F 45A 4

A 5

. ...since by reciprocity relation, A 5.F53 = A3.F35, and A5.F54 = A4.F45.

i.e. F 56A 3

A 5F 31 F 32

.A 4

A 5F 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 141

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 completelyenclosed by another black body

1

2 3

1

2

12

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

12

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 21A 1

A 2F 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 12A 2

A 1F 21

.

Now, F 11 1 F 12


Now, F 11 1 F 12

i.e. F 11 1A 2

A 1F 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 21A 1

A 2F 12

.

i.e. F 212 L.

L0.5. 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 2F 12

.


21 A 212

i.e. F 21π d2.

6 d2.

π6

...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 1F 21

.


i.e. F 12π r2.

2 π. r2.1. 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 ir i

LR j

r j

LS R i R j, 1

1 R j2

R i2

R 2

1

2


F ij R i R j, 1

2S 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 2F 12

.

i.e. F 21

π d2.

4

π d. L.0.828.

i.e. F 21

π d2.

4

π d2.0.828. ...since L = d

i.e. F0.828

0.207


i.e. F 210.828

40.207

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 12A 2

A 1

Now, F 11 1 F 12 ...from eqn.(a)

i.e. F 11 1A 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 1A 2

A 1

i.e. F 11 1

π d2.

4

π d2.

4π d. h.

...Note that A 1 consists of the area of bottom circular surface and the cylindrical side surfaces

4 h.


i.e. F 114 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 1A 2

A 1

i.e. F 11 1

π d2.

4

π d. L.

2

...where L is the slant height of the cone

i.e. F 11 1d

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 1d

2 L.

i.e. F 11 1d

2 h2 d2

4.

i.e. F 11 1d

4 h2. d2

(c) F11 for a hemispherical bowl of diameter 'd' : See Fig. Ex.(13.15,d)

We have: F 11 1A 2

A 1

i.e. F 11 1

π d2.

4

π d2.

2


2

i.e. F 11 11

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 L4L5 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 viewfactor 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 12L 5 L 6 L 3 L 4

2 L 1.

i.e. F 12Σ 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.