Radiation Heat Transfer Prof. J. Srinivasan Centre for...

Radiation Heat Transfer Prof. J. Srinivasan

Centre for Atmospheric and Oceanic Sciences Indian Institute of Science, Bangalore

Lecture - 5

Shape factor

(Refer Slide Time: 00:27)

Today we start a new part of the course, which is radiative transfer between surfaces. So

far, we have looked at emission by blackbodies and their radiative properties of rear

surfaces. Now, we are ready to ask this question, what fraction of radiation leaving a

surface actually reaches another surface?

Now, this is a very important issue in engineering application. You would like to know

of the radiation leaving a flame, in what fraction reaches to the certain wall of a boiler or

you would like know of the radiation leaving a certain wall of an enclosure, in what

fraction it will reach another wall. So, this problem is one of Geometry; three

dimensional geometry, because as you know radiation leaves the surface in all direction,

all the three dimensions. And we would like to know what fraction actually reaches

another surface in order to calculate the radiation impinging on the surface. So, this

problem is fundamentally a geometry problem; purely a theorem in geometry problem.

But if the radiation leaving the surface depends strongly on an angle, then not only

geometry is the issue, you also have to worry about the directional properties of the

radiation emitted or reflected by the surface. But, there are many examples in real life,

where one can assume that these surfaces we are dealing with are the diffuse-isotropic.

emitters and reflectors. In such a case, we will show that the geometry part of the

problem gets separated from the radiation part of the problem. And that makes life

somewhat easy.

So, in today’s class we will define what is known as geometric configuration factor. So,

this factor is a very important quantity. It tells you what is the fraction of radiation that

leaves one surface and reaches another surface. And under certain conditions, this is

purely dependent on geometric configuration of the two surfaces, and nothing else. So, it

makes the analysis of radiation problem somewhat simpler because we can separate the

radiation part of the problem and the geometry part of the problem easily. And you will

see some examples, where this makes our computation somewhat easier.

But, of course if you are unlucky and the surfaces we are dealing with are not diffuse

isotropic, then the geometry and the surface property gets entangled and you have to do

very complicated kind of analysis. So, if you encounter such condition where the

properties of surfaces are very complex function angle, and of course we are going to

discuss in next few lectures, which may not be that useful. You maybe better of using

what is known as Montecarlo. We will discuss this Montecarlo method near the end of

the course. These methods actually follow every photon that is emitted or reflected by a

surface and follow the photon till… If it is reflected, you will follow the photon after

reflection and ultimately until the photon is absorbed in the another surface.

So, today because of the availability of high speed computers, it is possible for you to

follow millions of photons from their birth and their subsequent reflection and

absorption. And you can do the life history of all the photons and compute the total

radiative flux. So, this method is very computer intensive, but has an advantage that it

can deal with any complex geometry, any complex situation. But, you would not resort

to these techniques for simple problem which we will deal with in the next few lectures.

For that, you should be quite satisfied in using the geometric configuration factor. Now,

let us now look at the example here.


So, we want to ask of the radiation leaving the surface d A 1, elemental surface, what

fraction reaches the surface two, d A 2. So, radiation leaving the surface and a certain

fraction is intercepted by surface d A 2, and what fraction of the radiation leaving surface

one is intercepted the surface two depends upon the orientation of the surface two with

respect to surface one; the distance between the two surfaces. So, it depends on the

distances, that is, the distance between the center of the surface one and surface two and

also the angle theta two and theta one. These are the angles between the line joining

these two elemental surfaces and the local normal. These are local normal.

Now, we will derive that expression. It tells you in what fraction radiation leaving the

surface reaches the surface. After that, we will ask under what condition this quantity is a

purely geometric factor. And we will show that if the radiation leaving the surface is

diffuse-isotropic. at independent angle theta one and phi one, then we will show that

indeed this fraction is a geometric factor. And that makes our life very easy. So, it is a

purely geometric problem.

And then in the next few lectures we will discuss techniques of calculating this factor

purely from the geometry of the problem. There are several interesting techniques

available to calculate this factor. And we will go through those techniques.


Now, let us derive this expression for geometry factor. And so, now I will draw a simple

figure to remind you of that situation. So, let us say, call this now surface d A 1 has a

normal; this is surface d A 2 it has (( )) normal and the line joining them is here, present

there. And this is theta one and that is theta two; the angle between the line joining these

two surfaces and the normal to these surfaces. And, we asked what is the solid angle

subtended by surface two and surface one. By definition of the solid angle, d omega is

nothing but area projector of surface two divided by the distance S squared. So, this is

the solid angle subtended by surface two in surface one. Now, let us ask what is the

energy radiation leaving surface one.

This, from the definition of intensity is i prime lambda one surface one cos theta one d

omega d lambda. This is the radiation leaving surface one in the direction towards

surface two and radiation that leaves one, which arrives in surface two is this times the

solid angle subtended with surface two, which is d A 2 cos theta two by S square.

So, this is the solid angle that subtended by two in one. So, remember the fact, now I do

not want to keep this because I have to define what is d omega 1 here. So, so this is solid

angle subtended by. So, this let me rewrite that for convenience. So, we wrote this as d

omega one. This can be further expanded to i prime lambda one cos theta one and d

omega one I take it from there, cos theta two d A 1 d A 2 by S square. So, this is the

radiation that is leaving one and reaching two. This is the radiation leaving one and

reaching two. Now, let us ask this question, What is the total radiation leaving one in all

directions?


If we take all directions and integrate to all directions, then radiation leaving surface one

in all directions is nothing but pi times i prime lambda one d A 1 d lambda. While doing

this integration, we are assuming that radiation leaving one is diffuse-isotropic. Only

then you can calculate this quantity. Without assumption of diffuse-isotropic, this

quantity looks function of data and phi; you can integrate. If it gives me the data in phi,

you can integrate and get the answer. Now, we say what is the fraction of radiation

leaving one and reaching two. That is the definition of F d A 1 d A 2. So, now we will

write down what this expression says.


So, F d A 1 d A 2 equals what is the radiation leaving one and reaching two; that is the

numerator. This is what we saw just now. This is the radiation which leaves one reaches

two divided by radiation leaving one in all direction. This is...

So, you can see that i prime lambda one will cancel out here. And so will d lambda, d

lambda I am sorry. The d lambda also does not play any role here. So, d A 1 also cancels

out. So, there is a simple relation for differential geometric configuration factor; has

nothing but cos theta one cos theta two by pi S square. That is the d A 2. Of course, d A

2 is not gone; d A 2.

So, you should notice that this is a non-dimensional number because it is a fraction. This

is non-dimensional because it represents the fraction and it depends only on the angle

theta one, theta two, area of surface two and the distance. So, there will be some relation.


Now, this relation has some interesting features. Notice that d A 1 F d A 1 d A 2 will be

equal to cos theta one cos theta two d A 1 d A 2 by pi S square. So, this is a symmetric

function. It can interchange theta one, theta two. It has not changed. Therefore by

symmetry, it is also equal to d A 2 F d A 2 d A 1. This is known as reciprocity. The

reciprocity is a very important property telling you that the total radiation leaving one

and arrives at two; now, the same as leaving at two and arriving at one. And so this will

be very useful because if we are able to calculate this quantity, then you also know this

quantity; because this depends only on the area ratio.

Now, this geometric configuration factor between differential areas requires only one

condition; that is, you are dealing with diffuse-isotropic emitters and reflectors. So, once

you have made that assumption, this geometric factor comes out naturally. So, as long as

the radiation emitted and reflected by a surface is diffuse-isotropic, then you are

guaranteed that the relation between radiation leaving surface d A 1 and arriving surface

d A 2 is purely a geometric factor. But as you can realize, this equation although very

useful, is not very practical; because in application in Engineering you need not F d A 1

d A 2, but you need F A 1 A 2. That is the geometric configuration factor between finite

areas.


So, this quantity is a quantity that is really of great interest to us. Now the question is,

we have just shown that the geometric configuration factor between differential areas is

completely a geometric factor, if the surfaces emit and reflect in a diffuse-isotropic

fashion. But is that conditions sufficient for calculating this factor between finite areas?

The answer is, no, you have to apply some more conditions. Why is that because we look

at the definitions.

If you want to calculate within finite area, then you have to integrate over the two

surfaces A 1 and A 2. This intensive leaving surface one cos theta one cos theta two d A

1 d A 2 by S square; this is the quantity in numerator. In the denominator, we are looking

at pi times what is the total radiation leaving surface one. Now, if you look at this general

expression for the geometric configuration factor between finite areas, you will find that

this will be a purely geometric factor, only if i prime lambda not a function of space in A

1 and A 2.

If the intensive leaving surface one is varying with space, then you are in trouble. Then

this integration, you cannot take out this term out of the integration. If you want to take

out this term i prime lambda one out of the integration, then it should not be a function of

space in one and two. Now, when will that happen? It will happen if emission is uniform,

reflection is uniform, does not vary with space and finally incident radiation is uniform.

So, only then you can guarantee that this quantity is independent of angle, independent

of space, and you take it to the integration. Then this also comes out. And then you have

a simple expression. Now, this is not commonly met in real situations. Especially, the

condition that the incident radiation should be reformed is rarely met because most

surfaces you are dealing with, gets radiation from all directions and the radiation

incoming from different surfaces impinging on a given surface will not be uniform.

So, if the incoming radiation coming at surface is not spatially uniform, then there is a

chance that the radiation leaving the surface will be uniform in space. Even though the

properties of surface like reflectivity are uniform, even then because incoming radiation

is non-uniform, the reflecting radiation will be non-uniform.


So, you must remember that this condition, finally we get for F A 1 A 2. Once you

remove the i prime lambda, it is simple; A 1 A 2 cos theta one cos theta two d A 1 d A 2

by pi, sorry S square (( )) in the numerator, by pi A 1. Then you can write it more neatly

as one over A 1 integral A 1 to A 2 cos theta one cos theta two d A 1 d A 2 by pi S

square. So, this is now a truly geometric factor. It depends on the relative orientation of

the theta one, theta two of the two surfaces. And this is a very useful quantity to have.

But, remember this is useful only if incoming radiation is uniform in space.

And secondly, surface properties like emissivity and reflectivity are invariant in space.

So, you want radiation to come uniformly on that surface. And the surface properties like

emissivity and reflectivity should also do not vary in space. So, this condition is not met

commonly in many practical applications. So, you have to keep that in mind that, using,

use of geometric configuration factor between finite areas is not always valid, unless you

have a new condition that the incoming radiation is uniform spatially, and this surface

properties like emissivity and reflectivity do not vary in space. But if you are lucky and

these are satisfied, then you can use this factor.


Now, you can also see from this symmetry of this particular equation. We can also see

the following conditions are satisfied; A 1 F A 1 A 2 equals A 2 F A 2 A 1; this is called

reciprocity. So, this is a very useful property. If you happened to get hold of F A 1 A 2

from some source, we also know F A 2 A 1 immediately. So, it helps you to generate

new shape factors or geometric factors based on available values. So, this is a basic

parameter which you will be now studying in the next few lectures. And try to

understand under what condition these can be applied and what are the real life

examples, where this can be implemented.

Now in today’s computer world, in principle, this integration which we talked about, this

integration can be done on the computer. But, you must recall that although this can be

done on the computer fairly easily, but it is important for you to actually tabulate

correctly the limits of the integration because that the computer will not know. You have

to indicate what are limits of the integration; A 1, A 2. And once you have given those

limits, of course there are many software packages which will do the integration for you.

But for simple geometries, these configurations have already been done and given in

various tables in books. So, we will take a few examples to illustrate the values of

geometric factors. So, very common geometry is which I encountered quite often in the

real world are quantities for which integrations are already been conducted and the

relationship is already available to you.


And, one example that you will now look at is the… Now, let us look at this example

here; between a small area d A 1 and this long element here. So, F d A 1 d A 2 is

parallelly well defined quantity from the definition given there. Now, you need to

integrate over this length to get the configuration between in element d A 1 and a strip A

2. So, that is our interest. And in this case, the basic d A 1 d A 2 we know from the

definition, but we need to integrate over one of the elements. That is the element d A 2 to

get a relation between F d A d A 2 to strip one.

Now, this integration can be carried out fairly easily. And this is a purely a geometric

problem. So, I will now write down a few steps indicating how this integration is done.

So, we will look at this integration. From, this is now area F d A 1 to A 2; A 2 is the strip

that you are looking at. So, you can look at the basic geometry of the problem and

wherein the distance, the normal distance between the two elements. So, one element is

here; that is, d A 1 here. And the other element, if you look at the end view, is here. And

this projected distance between these two is L. It is not the three dimension, this is

distance projected normal to this plane here.

So, you can do all the simple geometry that this cause for and one can easily show that.

And this angle between the normal to the surface and this is beta, this is not theta one;

theta two is the three dimensional angle. So, if you go back to the picture, you can see

theta one is the angle when you draw the vector S joining d A 1 and d A 2, while beta is

the projection of this on this plane. And so that is why you get beta. And once you know

that beta and this distance l, so l cos beta is the vertical distance here.

And, so we will now use the geometry and I will speak to notation in the figure. So, I

will draw this; small l cube cos beta, this is purely problem in geometry, by pi. And then

you integrate in the x direction from minus infinity to plus infinity, then that normal to

the plane of this board. And that will come out as d x by l square plus x square the whole

square.

Now, all of you can do the integration that is fairly straight forward. And you will get

that answer now as cos beta d beta by two. So, this is a very simple result indicating that

the geometric factor between the element d A 1 and a finite area A 2 which is the strip, is

nothing but cost beta d beta by two. It can also be written as half of d sin beta. Now, you

can go further and ask the next question as to what will happen if you now integrate over

both A 1 and A 2 and that result is a (( )).

And, we will examine that issue when we look at the geometric factor between two

surfaces, which are infinite in one direction. That is, they are very long in one direction.

And so, we look at the… such examples to occur in practical situations, where one

dimension may be in furnace or some other situation or a cloud; where one dimension is

very large compared to the other dimension.

Then, the problem is essentially a two dimensional. But, remember that in radiation the

basic phenomena in three dimensional, although one dimensional is long compared to the

other dimension, you still have to take into account the three dimensional nature of the

phenomena, which you cannot ignore. And, so that issue we will now address; in our

example, now we will take up. And issue we are going to addresses is how to calculate

geometric configuration factor in those situations, wherein one dimension is extremely

long compared to the other two dimensions. We want to know whether some

simplification we have arrived at; which will enable us to calculate things without

actually doing the integration. And this method is quite useful and powerful and has a

simple geometry of simple interpretation in terms of energy.


So, we will look at an example now. And to do that, I will take a simple example of a

triangular enclosure triangular system. Imagine, we have a triangular enclosure A 1, A 2

and A 3. So, all dimensions are perpendicular to the board are very long; are infinite. So,

you are looking at the problem where this dimension is extremely long compared to what

is there in this plane. And the question is, such a configuration does simplify your

problem? Now, we will appeal to reciprocity. And see we know F A 1 A 2, F A 1 A 3.

We want to calculate these quantities without actually performing integration. We want

to calculate these quantities without actually doing integration of the equation. Now, it is

possible because we can write down energy balance; radiation leaving surface one and

reaching surface two as well as radiation leaving surface one and reaching surface three

as equal to radiation in the surface one.

So, this shows that F A 1 A 2 plus F A 1 A 3 equals one; because these are plane

surfaces. And a plane surface cannot see itself. So for plane surface, that is, which are

not concave, F A 1 A 1 is 0. So, this is called as self-viewing factor. So, the self-viewing

factor is the fraction of radiation leaving a surface which comes back to itself. Now, this

can happen only when the surface of concave, the surface of plane or they are convex,

they cannot see themselves. So, self-viewing factor is 0. So, we are considering a case,

where all the three surfaces A 1, A 2, A 3 are completely...


So, this is surface A 1, A 2 and A 3. And all three surfaces are non-concave. And

therefore F A 1 A 1 equals F A 2 A 2 equals F A 3 A 3 equals 0. So, they do not see

themselves. Now, let us write down the first law of each case. the radiant leaving surface

one and surface two are leaving one and ending in surface three equals one by first law;

because whatever leaving reaches; one has to reach two or three, no other way. And this

dimension is infinity. Similarly, F A 2 A 1 plus F A 2 A 3 equals one. And of course F A

3 A 1 plus F A 3 A 2 equals one.

So, we have three equations here with six unknowns. right. So, you can solve for all

these values. But we know reciprocity, which is A 1 F A 1 A 2 equals A 2 F A 2 A 1,

similarly A 2 F A 2 A 3 equals A 3 F A 3 A 2 and finally A 1 F A 1 A 3 is equal to A 3 F

A 3 A 1. So, now, we have six equations in six unknowns. You have (( )) six equations in

six unknowns. And hence we can solve this problem completely without doing any

integration. You required no integration to be done. That is only there, which is quite

complicated. So, in such a situation the results are very simple; you can just eliminate

what you do not want. And the key point what I want to remind is that this simplification

implies that this integration is not required. This integration is fairly complicated and

will take a lot of time and effort.


So, if you encounter a situation with three surfaces which forms enclosure, and so we

want to say that. If there are three surface enclosure with non-concave surfaces, then you

can show that F A 1 A 2 equals A 1 plus A, sorry (( )), that is A 1 A 2 plus A 3 minus

two A 1. A 1 plus A 2 (( )). If you want to write F A 1 A 2, it is A 1 plus A 2 minus two

A 2 A 1… A 1 plus A 2 minus A 3 by two A 1.

So, this configuration factor between two finite areas A 1 and A 2, if they are part of an

enclosure containing three surfaces and all surfaces are non-concave, it is sum of the area

of the two surfaces involved in the interaction minus the area of the third surface divided

by twice A 1. Similarly, you can write the other two. F A 1 A 3 is A 1 plus A 3 minus A

2 by two A 1. And this is a very powerful result because we have avoided all integration.

You need to know only the surface area of the three surfaces. So, that is fairly easy

quantity for surfaces of known shape. So, you have managed to avoid the difficult three

dimensional integration that is usually called for. So, this is a very powerful result.

And, so this can be used in situation where one dimension is very large compared to the

other two dimensions; because the problem becomes essentially a two dimensional

problem. But, still the integration that you have to do is three dimensional. But, you can

avoid the integration by appealing to Law of conservation and energy and reciprocity.

So, combining energy conservation and reciprocity, you get this relationship between the

shape factors. And this is a very useful result and it has application which we can adopt

very easily. Now, there could be situations where you actually do not have three

surfaces, but you can always construct a three surface enclosure, even though…


Let me take an example. Suppose there are two parallel plates, may be a furnace for

example; one is surface one, another is surface two. And the distance between the plates

is D. And this dimension, the dimension normal to the board here is infinity. Question is,

Can we calculate F A 1 A 2 without actually doing any integration? That is the issue.

Now, the way we will do this is, we will construct an enclosure; your surface two here,

surface one here. We draw a dotted line; this is an imaginary surface. We draw another

dotted line; that is also imaginary surface. Right. So, this is surface one, let us call this

surface three and let us call this surface four.

Then, we apply the rule we just derived to this triangular enclosure; three, four, one. And

what we know from this? from our earlier formula we know that A 1 F A 1 to A 3 is

equal to area one area three minus area four by two area one. Right. Similarly, you can

construct other triangle here. So, let us call this as five and six. So, we can also say A 1 F

A 1 A 6 is equal to A 1 plus A 6 minus A 5 by two A 1. So, I constructed two triangles

on the same base of surface one and applied the triangle rule for non-concave surfaces

and arrived at two expressions. Now, what I really want is F A 1 A 2. That is what I

want. I have got F A 1 A 3, I got F A 1 A 6. Now, I can apply first law of

Thermodynamics.


So this enclosure, you can say radiation leaving one reaching two plus radiation leaving

one reaching three, radiation leaving one reaching six has to be equal to one. Therefore,

the quantity I what is F A 1 A 2 equals one minus F A 1 A 3 minus F A 1 A 6. So, this is

from the first law. This first law, now we have already derived the expression for these

two. So, we can write F A 1 A 2 because one minus F A 1 A 3, if we recall was A 1 plus

A 3 minus A 4, which was this quantity, by two A 1 minus A 1 plus A 6 minus A 5 by

two A 1.

So, we have been able to get the configuration factor between surface one and two in

terms of just the areas of one, three, four, five and six. And these can be easily

calculated. And since this dimensions is infinite in this direction, all these are essentially

(( )) of various surfaces here. And you can see here there is a minus half here, there is a

minus half here; the one is cancelled out. So, finally you are; let me go to that which is

write down and then we go next page. And you can see that this and this will cancel out

and we can add four and five. So, we will get A 4 plus A 5 minus A 3 plus A 6 whole

thing divided by two A 1. So, it is a very simple result, which has several physical

insights.

So, in the next lecture we will proceed further in using this method. So, what we get in

this lecture is define a quantity called geometric configuration factor, which is a purely

geometric factor if it is applied between finite areas, if the radiation leaving that surface

is uniform in space. And then we showed that for certain situation in which one

dimension is very long compared to the other, there is a simple way to evaluate this

geometric configuration factor without actually doing the integration. So, we will… in

the next lecture we will look at how this simplification helps us to solve some interesting

problem.

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