radiation heat transfer prof. j. srinivasan center for...

Radiation Heat Transfer Prof. J. Srinivasan

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

Lecture - 2

Blackbody radiation

In the last lecture, we had looked at the Planck blackbody radiation.

(Refer Slide Time: 00:25)

We saw that the Planck blackbody formula for emissive power was 2 pi h nu cube by C

square e 2 the power h nu by k t minus 1 where h is the Planck’s constant, k is the

Boltzmann’s constant and C is the speed of light in the medium. Now, this formula for

the emissive power can be derived using principles in statistical mechanics and quantum

mechanics. But we went through another approach, which was proposed by Einstein and

which is much simpler to understand and how last clearly, how these two terms in these

equations, how this term and this term appear in this equation? To do that, we went

through Einstein’s derivation.


We saw that your upper level i and lower lever level j and we wrote down three

expressions, one is d n i d t equal to minus A i j n i for spontaneous emission. Then, we

had d n i change in number of molecules in time due to induced emission, which is the

new idea that Einstein had. He said that there can also be emission of photons due to

photons coming into the system. So, there is induced emission or spontaneous or

stimulated emission. Finally, there is absorption, which has to be positive and which is

from lower level j to upper level i.

So, it depends on number of molecules in the lower level and the number of photons is

having at the system. Now, Einstein assumed that the incoming radiation as from

blackbody and isotropic in all direction and so he wrote down integral of i nu d nu as

nothing but integral i nu B into 4 pi. So, this is because the radiation is coming, is

becoming (( )) in all direction. Then in equilibrium the sum of these three terms has to be

equal to 0.


So, we are going to assure that d n i d t spontaneous emission plus d n i d t in induced

emission plus d n i d t absorption is equal to 0, because in steady state all the number of

molecule in the level does not change. So, we do this calculation, we will find

simplification of that formula n i into A i j plus i prime nu B into 4 pi into B i j is equal

to n j B j i 4 prime i prime nu p. By the arguments related to symmetry and reciprocity

Einstein assumed that B i j must be equal to B j i. The process of emission on plot have

to be symmetric and reciprocal. Then, we assume that the number of molecules at the

upper level to that in the lower level has to go as e to the power of minus delta E by k T.

This is equal to e to for minus h nu by k T. So, we are assuming that the molecules have

a distribution following the well known distribution that comes from surgical mechanics

that the distribution molecule has to decrease exponentially with height. Now, if you put

all this together you will get i prime nu B as A i j by B i j divided by 4 pi e to the power

of h nu by k T minus 1. So, the beauty of this result now if you want to get a result which

is same as what we derived for the Planck’s formula, we know that e nu B is equal to pi

times r prime nu B.


Therefore, you can write e nu b from Einstein derivation as A i j by B i j divided by 4 e

to the power of h nu by k T minus 1. We do not know right? Now, what this ratio is like,

but if you compare this result with the derivation done by Raline Jeans in the classical

limit, one can show that A i j by B i j is nothing but 8 pi h nu cube by c square. So, we

get back the result that is obtained by Planck, by the advantage of this derivation is that,

it really shows that this 1 comes from induced emission, which was not accorded for the

either Raline Jeans or by v in any derivation.

So, this is the term, which is introduced by Planck, but Planck introduced it somewhat

adhoc fashion, while in Einstein derivation this comes out very naturally as a

contribution of induced emission to the result. So, this derivation write inside regarding

that terms. This shows that, this term is coming from the distribution molecules showing

a decline with the energy levels. So, both these are clearly accounted for in this

derivation. Now, this derivation done both by Einstein and Planck use the frequency on a

system that is we calculate emission power of per unit frequency, but we also would like

to know what are the value per unit wavelength and that as we recall can be easily

obtained from this equation. So, once you know nu B from here and you know what is d

nu here because you know that nu into lambda is C, we know that d nu will be lambda s

minus C by lambda square. So, if you count this and convert all this frequency to

wavelength, we get an expression for lambda B, which is very well known.


So, e lambda b becomes 2 pi h C square lambda to power of 5 e to the power of h C by

lambda k T minus 1. So, this is an expression that is used very often. This expression has

interesting properties, one is that if you divide e lambda b by T to the power of 5, in the

right hand side is only a function of the product lambda T. This is very useful and later

we will see that we will exploit this particular feature of this function to advantage.

The next important issue to know is that where is the maxima of the Planck blackbody

function and that you can obtain by differentiating this equation. You will get a

transidential equation, which you have to solve numerically. If you do that you will get

the following result. The wavelength, at which it is the maximum, will come out as 2898

by T, where T is in degrees Kelvin, this answer is in microns. So, this says that the peak

of the Planck body emission curve, when you look at the emission power per unit

wavelength occurs at this wavelength.

On the other hand, if you look at the previous derivation with recommend by it

frequency and find out what is the maximum frequency. Then from there get the

wavelength. We will get that this will maximize at a wavelength given by 5077 by T

micron. So, what this means is that the maxima of the e lambda b function occurs or a

different wavelength at a maxima of the e nu b function. This is not surprising because

you recall that the e lambda b function goes as 1 over lambda 2 to the power of 5 while

the…


So, e lambda b goes as 1 over lambda to the power of 5, while e nu b goes as nu cube.

So, these are different kinds of a functional dependence, so the maxima of these two

functions will not occur at the same wavelength. So, when you coat the maxima of any

blackbody function, it is important to mention which coordinate system you are working

at, but most books coat the value that is given by e lambda b function. This is called the

Wien’s displacement law and this is very useful information as to where exactly the peak

of the blackbody function lies.

Now, another quantity of gate interest was in addition to the peak of the blackbody

spectrum is also the total amount of radiation line. Now, before we go the let us look a

typical blackbody at the temperature of 298.8 Kelvin, then you find its maximum

radiation a at the in the lambda B coordinate will occur at 10 micron. So, what it means

that, if you look at bodies at room temperature and if they behave like blackbody’s, then

their peak radiation occurs around 10 micron and this is true of most objects at room

temperature. So, the maximation occurs in the region around 10 microns.

On the other hand, if you look at the Sun’s emission, assume sun is a blackbody at 5796

Kelvin for convenience, then you find that the maximum radiation occurs at 0.5 micron.

So, you can say that there is a huge difference where the maximum radiation lies. For

bodies around room temperature the peak is around 10 micron while for bodies as a sun,

which is at high temperature the maximize around 0.5 micron. This information will be

very useful when you want to determine the amount of radiation absorbed from the sun.

For example, then it means that you need to have a surface, which absorbs radiation very

effectively in this region around the visible. So, this is sure we will take up again a little

later. In addition to the where the maxima lies, you also would like to know, where does

how much energy lies in a certain wavelength range?


So, we will like to know the following quantity, what fraction of radiation from

blackbody lies in the… Let me just write down this expression, so you look at the total

radiation emitted a blackbody and then compare with how much it lies in a given

wavelength range. So, this tells you what fraction of the radiation emitted blackbody lies

in certain wavelength range. This is a very important information and to do that this kind

analysis we first. To know what this quantity is 0 to infinity e lambda b lambda or it can

also be 0 to infinity e nu d nu. What is this quantity or this quantity leads us to the well

known Stefan’s Boltzmann formula. Let me do a quick derivation of this. So, e b which

is the total radiation lying in all frequency of wavelength, so we use the Planck’s

expression.


Now, this derivation is quite easily done if you assume n on dimensional parameter eta,

which is h nu by k T and then you write here e b as 2 pi will come out, h will come out

and c square will come out here 0 to infinity. You will have now multiple up and down

by k cube t cube by h cube to get eta cube e to the power eta minus 1. Then you, what

you do d eta there? So, multiply by another k T h, then you get the expression e b is

equal to 2 pi 1 h will go out here h cube c square k to the power of 4 T to the power of 4

0 to infinity eta cube e to the power of e eta minus 1 d eta. Now, this integral that is

given here is just a number.

So, right now you, we may not worry too much about what the number is about? We are

more interested in knowing what is the dependence of the emissive power of temperature

that comes s T to the power of 4? So, if you evaluate this integral this comes out as pie to

the power of 4 by 15, this is this kind of integral is studied in when we study complex

numbers. You find the four’s and zero’s of this function. Then evaluate the integral or

today you can just use that is MATLAB and mathematic and get this number. So, we do

that you will have a simple expression e b equal to 2 pi to the power of 5 by 15 k to the

power of 4 by h cube c square into T to the power of 4 and this quantity is called the

Stefan Boltzmann constant.

The numerical value of this quantity have been derived even before Planck had made his

derivation for the spectral quantity. Because in the laboratory, you can measure e b as

well as T and after that you obtain this quantity empirical. But today, we know this

quantity called Stefan Boltzmann constant with a symbol sigma this quantity. You can

evaluate it from three of the universal constants in physics, they are Planck’s constant,

the Boltzmann’s constant and the speed of light in vacuum or this black body radiation is

for calculation for vacuum.


So, preferring like e b equal sigma T to the power of 4 and this quantity can be

remembered very easily. So, this quantity is 5.67 10 to the power of minus 8 watt’s per

meter square kilometer to the power of 4 and this has a easy number 5678, that you

should never forget it in your life time because it is worth remembering very easily.

Now, having done this now we go back to our expression, how much radiation lies

between two wavelength?

You can write this as lambda 1 to lambda 2 e lambda b d lambda. In then denominator

was 0 to infinity, that comes out now as sigma t to the power 4. Now, this can be

simplified little bit if by writing this as lambda 1 T to lambda 2 T e to the power of

lambda by sigma T to the power of 5 d lambda T. Now, it is useful to write this

expression this way, because if you look at the expression e lambda b.


You are already pointed out the e lambda b by T to the power of 5 is a function only of

lambda T. Therefore, our expression for it is lying between two wavelengths as usually

becomes lambda 1 T to lambda to T F lambda T by sigma d lambda T. So, we need to

know lambda and t together in order to calculate this value. We can also write, based on

the simpler wave like this F 0 to lambda T can also be written as 0 to lambda T F lambda

T by sigma d lambda t. So, this expression is useful to evaluate a numerically today

using math law or any other standard integration procedure. I will give you some simple

values which are useful to know.


Suppose, you are looking at lambda T value and what fraction radiation lies between 0

and that lambda T? Then the important one’s are at 1448 micron Kelvin you will have

0.01 at 2898 it will be 0.25. At 4108 it is 0.5, at 6149 it is 0.75 and at 23,220 and 20 it is

0.99. So, this is a useful table to have, what it means is that most the radiation emitted by

blackbody lies between it is 98 percent of the radiation lies between 1448 micron Kelvin

and 23,220 micron Kelvin. So, in evaluating proper surfaces this information is useful

because you need to know in which wavelength range the radiation primary lies. So, let

us take a few examples to illustrate that.


Let us take the surface at 300 K Kelvin. So, 98 percent of the radiation lies between 1

erase between 1 1448 by 300, which is nothing but approximately 4.81 to 23,220 by 300,

which is approximately 77 micron. So, one can say that most of the radiation as emitter

by blackbody at 300 Kelvin, which is somewhere between 4, 4 micron to 75 micron. On

the other hand, if the temperature of the object is closer to 6000 Kelvin is it closer to

room temperature, it is closer to Sun’s temperature.

So, 98 percent of the radiation lies between, which is around 0.24 micron to 23,220 by

6000, which is approximately 4 micron. It is 4 macron, so we can say with some

confidence that most of the radiation emit by blackbody, 6000 lies between 0.25 and 4.

While most of the radiation from a blackbody around room temperature lies between 4.8

micron and 77 micron. So, what is interesting is that, what this shows is that the two are

almost non overlapping. So, the Sun’s radiation lies between 0.24 and 0.4 while this is

somewhere between 4.8 and 77.

So, the radiation emitted by the Sun and a body at room temperature do not have much

overlap, there are completely non overlapping wavelength domains. This becomes very

useful to know when you want to evaluate properties of surfaces, the properties are

relevant to calculating the absorption of Sun’s radiation will be very different from the

properties that you will have to use to calculate the emission of the surface. Because the

two value domains are completely non overlapping in their in their character. So, this

information is interesting.


The second information is interesting is we noticed we saw that e lambda b by d lambda

was 0, where lambda T was 2898 micron Kelvin. You also see that 0 to 2898 e lambda b

d lambda by sigma T to the power of 4 is 0.25. So, while this means is that, if we plot e

lambda b with lambda, then 25 percent of the radiation, so 25 percent lies before maxima

and 75 percent lies above the maxima. What this implies is that the blackbody functions

is a long tail, it is not symmetric like our Gaussian distribution, it is asymmetric.

So, to left over maxima it is only 25 percent of the radiation while to the right of the

maxima 75 percent of the radiation is there. So, this feature is to be remembered and you

are doing calculation that the functions are not symmetric, as you would have assumed.

So, this is the properties of a blackbody, but remember that we are dealing with real

surfaces not blackbodies. So, we need to now look at the properties of real surfaces.

Then compare this real surface with the blackbody, so one important property we should

remember is that blackbody is the best emitter and the best absorber. So, when you

compare a real surface with an idol surface like the blackbody, both at the same

temperature the emission by real surface will always be lower than that of the blackbody

at the same temperature.


So, this gives us a chance to define what is called emissivity. Now, we talk about

properties of real surfaces. So, real surfaces are different from idol surfaces. So, the first

important property that we will talk about is emissivity. So, emissivity compares the

emission of a real surface to that of a black body at the same temperature. For example,

the directional spectral emissivity of the surface is the emission in a given direction

given wavelength by the actual surface to that of the blackbody at the same temperature.

So, these are very fundamental definition of what is directional spectral emissivity.

Similarly, we can define directional spectral absorptivity, but in this case we define the

directional spectral absorptivity as the amount of radiation absorbed in a given direction,

given wavelength to the radiation incident in the same direction and same wavelength.

So, I want you to notice the difference between the emissivity and absorptivity.

Emissivity is defined as a ratio of the actual emission to blackbody emission, while

apsorptivity is defined a ratio of radiation absorb to the incident.

So, the fundamental difference between these two definitions, but remember at the

microscopic level emission absorption are what is called reversible. That is if a certain

object has a certain property of absorption. Then the same object will have similar

property to emit radiation, because when there is emission molecules fall from the upper

state or lower state and emit a photon. If you send the photon back, then there is an equal

property that the molecule will absorb the radiation and take the molecule from the lower

state to the upper state.


So, Kirchoff’s proposed that under most conditions these two quantities are equal. The

only requirement that he imposed on the equality of directional spectral absorptivity and

directional spectral emissivity is that, the object which must be in local thermodynamic

equilibrium. So, in local thermodynamic equilibrium, the ability of an object to absorb

radiation at one wavelength and one angle will be same as a ability of that object to emit

radiation at this same angle and same wavelength.

Now, I want to warn you that Kirchoff’s law only relates the directional spectral

quantities although this is always true. Just remember that the hemispherical spectral

absorptivity is not always equal to hemispherical spectral. Similarly, directional total

absorptivity is not always equal to directional total emissivity or the hemispherical total

absorptivity is not equal to…

So, these three things you must remember because many books clam that this Kirchoff’s

law or this Kirchoff’s law is not true. Kirchoff’s law only connects the most basic

quantity that is directional spectral quantity between absorption emission. It does not say

anything about quantities which are obtained by averaging. Once you average absorption

emission, then things change substantially. So, I want to illustrate as to why? For

example, alpha prime will not be equal to epsilon prime, so let us now go through the

derivation understand, why it is true? Why epsilon prime?


By definition, it is the directional total emission by the object at a given temperature by

the directional total emission by blackbody at the same temperature. This can be defined

as follows. So, what do you see that we obtain the directional total emissivity from the

directional spectrum emissivity, by waiting the directional spectral emissivity by the

black body function at the temperature of the surface, which we study the radiation. On

the other hand, we define the directional total absorptivity as the total radiation absorbed

at that angle with the total radiation in symmetrical time.

Notice, that the directional total absorptivity does not make any reference to any

blackbody function. So, we write this in terms of intensity, you can write this as 0 to

infinity while the denominator is this from the definition of intensity. Now, you see there

is a very basic difference between this integration and this integration, when I am

integrating the directional spectral emissivity my waiting function is that of the

blackbody as a temperature of the surface while when I am integrating, the directional

spectral absorptivity. I am integrating with the amount of radiation incident on that

object from some source, which may not be a blackbody.

So, this study shows that absorptivity that you obtain for any substance depends very

much upon what is incident on that surface. So, if Sun light is in on a surface its

absorptivity will be one value, while if radiation from some other source is incident on

the surface, it will have a different value. So, alpha prime will differ substantially from

one situation to another because an incident radiation is different. Now, compare these

two and see when you can apply Kirchoff’s law for this kind of situation.


I will first write epsilon prime 0 to infinity epsilon prime lambda i prime lambda b b

lambda by 0 to infinity i prime lambda b d lambda and alpha prime as 0 to infinity than

using Kirchoff’s law epsilon prime lambda i prime lambda i b lambda. Now, these to

look somewhat similar, but not same because in the case of emissivity you are varying it

with the blackbody intensity at the temperature of the surface while in the case of

absorptivity.

We are waiting to the incoming radiation intensity and its distribution with wavelength,

but we can see that if by chance the incoming radiation does it happens to be

proportional to the blackbody intensity at the surface temperature? Then we can say

these two are equal, but in real world, this is very rare will be rare occasion when the

radiation is send on a surface has a wavelength distribution proportional to that of the

black body of surface temperature.

So, this is going satisfied very rarely, other case is by chance. If this quantity is not a

function of lambda is where you can take it out for integration. Then both these are equal

and so this is what do you call as gray surface. So, you see that we can assume the

directional total absorptive rate equal to directional total emissivity. If so you have to

write down clearly, the next page.


So, I will say that this is equal if a surface is gray or b incoming radiation is proportional

to blackbody radiation at the temperature of the surface both these occur very rarely.

Most surfaces that we encountered have strong wavelength dependence of absorptivity or

emissivity enhance on assumption over gray surfaces not very accurate. Secondly it is a

rare occasion, when the radiation coming on object happens to be having a wavelength

dependence, which is following that of the blackbody at the temperature surface. So, I

would say that this is not very useful information to have, but we are highlighting it

because there are many books.

We call this equality as Kirchoff’s law, that is not true. Kirchoff’s law only depend

depicts the relation between absorptivity and emissivity for directional spectral quantities

and not for directional total quantity. So, the validity of this is very specialized at the

gray surface or the incoming radiation proportional blackbody intensity at the

temperature surface, which is rarely satisfied with the real world. So, neither condition

are met in the real world and hence these two are is not equal. So, it is a rare situation

that this is rarely satisfied. Now, let us look at the other question, suppose instead of

integrating to our wavelength you integrate the angle.


Let us ask when is this equal to this? When is the hemispherical spectral emissivity equal

to hemispherical spectral absorptivity? To do that once more we go for basic definitions,

we define this as e lambda of T s by e lambda b of T s in by the 0 to lower omega of

epsilon prime lambda e prime lambda b d omega divided by e prime lambda by divided

by omega e prime lambda b d omega. So, we are obtaining a emissive quantity by

averaging over angle. So, just remind you we start with the basic quantity directional

spectral last time we did the wavelength integration, we got this quantity.

Now, we are doing angular integration and your getting this quantity you can reverse it

later if you want, ultimately we will get epsilon, which is the total hemispherical

emissivity. Now, remembering the fact that E prime lambda b is nothing but i prime

lambda b cos theta on basic definition of intensive emissivity and the fact that i prime

lambda b is not a function of theta and phi both these we know.


So, with this integration we can write down epsilon lambda as integral over omega

epsilon prime lambda i prime lambda b d omega at the bottom integral i prime lambda b

d omega. Now, remember that the definition of intensity is such that we know that the

directional spectral emissivity of blackbody is not a function of angle that, was built into

the definition. So, both these integrals come out and I am sorry, I have forgot a cos theta

there and a cos theta here.

So, if we do the integration this will come out as phi, so final expression for epsilon

lambda we are getting is 1 over phi integral over omega E prime lambda cos theta d

omega. The main lesson you have to learn from this expression is that, we do not obtain

hemispherical value by merely taking the directional value and averaging it over

arithmetically. We have to wait it with cos theta, so cos theta waiting is important, which

means that where weightage is given to emission closer to the normal.

So, this region will have higher weightage than this region. So, this is worth

remembering that the directional quantity when it is integrated. This is a cos theta termed

there, which ensures that more weightage is given to radiations appearing at emitter at

lower values of data.


So, now we move on to absorptivity the hemispherical spectral absorptivity is defined as

the omega, alpha prime lambda, i prime lambda i, d omega divided by… I have forgot a

cos theta there. Now, this comes from the basic definition of absorptivity. This is the

radiation incident in the solid angle and what fraction is absorbed in the radiation the

same solid angle? Now, if you assume Kirchoff’s law, which is always valid, then we

have alpha lambda is equal to…


Now, let us compare the two expressions called emissivity absorptivity. You saw epsilon

lambda is 1 over pi integral omega epsilon prime lambda cos theta d omega and alpha

lambda is somewhat different. So, you can see that in general these two are equal, if 1

epsilon prime lambda is not a function of angle that is the emissivity. Surface emits

diffusely and is (( )) and other possibility is or incoming radiation is not a function of

theta and phi. So, the any incoming radiation is emitted equally in all direction. Then

also this integration becomes say, in one case we take this out going to be handle and

take this out or we take this quantity out.

So, to conclude in this lecture we have focused on properties of blackbody radiation and

the properties of real surfaces. We define emissivity and absorptivity at its most basic

level. Then we involved Kirchoff’s law, which says that the directional spectral

emissivity equals directional spectral absorptivity. Then we went on to show under what

condition you can assume equality between hemispherical spectral emissivity and

hemispectral absorptivity. We saw that there are only equal under very special

conditions, we will continue this discussion further because there is a lot of

misunderstanding about, how to invoke Kirchoff’s law? So, we will spend some more

time highlighting the misuse of Kirchoff’s law.

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