### Mod-03 Lec-15 Plane parallel model

In the last lecture we look at a Plain Parallel Model for radiation. This is essentially a one-dimensional model, but remember that in radiation even in one-dimensional model it has account of all the rays traveling in all the different directions. So, recall we started with the basic equation for radiated transfer, which was for non scattering medium which says that the changing intensity with distance is proportional to the emission by the gas and reduce by the absorption by the gas; second term absorption first term is emission. This is absorption coefficient which has units of beta minus 1. This can be integrated, so that along a given direction in get the intensity, but we understood in fluxes not intensity So, we took a parallel plane parallel model, and we depended normal and we look to the angle theta, we use in everything is symmetry with this for 5 the assume with angle. And hence we from here related the intensity to flux by the following relation. So, this equation has to be integrated and integration done in two parts. So, in this part upper going race that is theta between 0 and pi by two or all race which are moving up words And so we solve this equation for theta between 0 and pi by 2 and that one we called as high lambda plus race moving upwards and for the race going downwards from the top, that is theta greater than pi by 2 and up to pi, we call it i lambda minus. We have to distinguish between the race going up and race going down, because the lower surface 1 here will only emit race going upwards and the upper surface 2 will only emit race going downwards So, when integrating this equation for upward going race we start from the surface and go up, and when we look at the downward going the race we start from the top look at the emission of the upper surface And we saw that the 4 terms in the equation Two terms involved emission by the surface and attenuation the gas and other two term involved emission by the gas upwards and downwards And so will just remind you, I will write the flux expression. So, the flux expression had 4 terms. So, the q r lambda the net radioactive flux between lambda and lambda plus de lambda consist of Radiation living surface one and it is attenuated in all direction, minus radiation

living surface two, minus because it is downwards, going the race and the difference between the total optical depth. So, this is 0 and this is kappa lambda 0 and any location here is kappa lambda. So, the distance travel here for the second one is kappa lambda, 0 minus kappa lambda while for the race going up; this is kappa lambda. So, this are the two terms is impression the boundary contribution, contribution from a mission and deflection by the boundary which than most to the gas Then there are two more terms involving gas a mission first is from upward 1 it is, it is equal to 0 to from here to here all the upward dimension. And it involves the black body a massive power of a gas, times a function which of science angular integration. And the last term is again minus the downward emission by the gas between this point of the top and that involves So, this is the contribution from the surface 1. This contribution from all gas elements between this surface and kappa lambda, this is our where we calculated flux. This is the downward emission surface 2; this is downward emission by all gas elements between kappa lambda and kappa lambda 0. Now you should look at the argument. Here you can see that you are return this, such that is always positive This is by convention that the argument here has to always to positive. So, in the reason between 0 and kappa lambda, kappa lambda theta has to be less than kappa lambda so, you write it this way. In the second term kappa lambda where is between kappa at kappa lambda 0 that for it has the reverse kappa lambda remain kappa lambda. So, this is a next question which we are going to use a lot In this expression, there are two terms call the extensional integral functions and the general definition is E n affects is equal to 0 to 1, mu to the power of n, n minus 2 e to the power of minus x by mu. So, this accounts for all the angle integration that one has already done and these orgel tabulated functions available in fox ordination transfer usually in the appendix. So, you can look it up in any time and in today’s world this can be computed, filly easily by any soft So, what we are done essentially is in problems read it your transfer, there are three levels of integration world. You have to integrate over a angle; you have to integrate wave length and you have to integrate over a space. So, we have since they separated this three actions the angular integration is built into this function E n of x. Special integration is involved here in this, in this equation here and we need to do somewhat later the wave in that duration. The major challenge we face, in not the angular integration. The solid state forward as set can be done by any completed very easily. A major challenge we solve in to the equation is a fact that, if you want to know the temperature distribution of a gas in an enclosure, then the unknown temperature

variation inside the interval here So, this is a peculiarity in radiated transfer; in both conduction and conversion heat transfer You encounter different equation. And there is a large laboratory of techniques available to solve different equations of mini kinds So, that hold methodology I am use very effectively, in solve in the large number of problems in conduction an converted transfer. In radiated heat has in gases. The problem has been more difficult, because unknown temperature does not appear of the part of different equation, but is a unknown is not integral, this is an integral equation So, are we do not has many techniques for solving integral equation analytically as we have for different equations. So, most of solutions that will be normally done will be numerical, but in subsequent lecture will do a few solutions which are analytical, because analytical solution give you a great physical insight as regards what is happening and in a given problem. So, all though will be dealing with fairly simple and somewhat not realistic real world situation, ideal situations. Still we believe that the simple cases are very useful to you help you understand the phenomenal operating heat transfer. And the key point here number is that because in this equation we have integral equation Essentially what you see here in radiation is something very different from is a conduction or convection. Conduction and convection are governed built different equations. So, that the temperature of given reason is influence only by the temperature of the surrounding reasons; immediate surrounding reasons. And that is why we are able to ride everything in terms of derivatives in different equation On the other hand in the case of radiation the temperature, in the middle of a gas in enclosure made depend on what is happening at the wall. Because some of the protons are is imitate at the wall can go right through the medium without being observed and directly interact with the reason you have concern with. So, this could be thought of as action at a distance So, so radiation problems are essentially are non local in character; that is, temperature was given reason this is influence not only by the temperature. Surrounding reason is also a influence by temperatures of surfaces or gases far away from the reason ventral And you will see that this equation which is more general kinds, can be shown to ultimately become as a different equation in certain limits. So, this formulation be general and when we takes of limiting case limiting condition and we will show that this equation can be become like and equation in conditionally transfer, but the real world situations which we encounter in radiation, this is the furnishers of that atmosphere or a in metallurgy We do not always have situation where we can convert this two a different equation. So we must know how to solve integral equation And so, we will illustrate a few examples in the next few lectures. So, that your familiar with the techniques of solving integral equations But you must realize these are only to as teaching tools to enable your understand phenomena when you solve any real world problem involving radiation, you will solve it numerically So, the enter power of high speed computing and brought in to solve fairly complex radiation transfer problems. Those of solved by lots of to a packages, but those will not give you in a real insight into what is going on in this radiation phenomena. So, we need simple analytical solution; just show that you appreciate what is going on in this mode away transfer,

in contrast to either conduction or convection So, I want to always what do you to compare the wane which radiant transfers heats with that you already know what occurs in the case of conduction or convection Now, before we go for there to solve this equation we not only need the flux you also need the divergence of flux, because we already we recognize that the divergence of the flux is what is important in solving for temperature variation in a medium. Because ultimately you are applying the first law of thermodynamics and in the first of thermodynamics on the right hand side you have the divergence of all heat fluxes. So, that includes conduction convection and radiation. So, ultimately we must know how does divergence of the radio de flux looks like and so, we derived expression, but differentiating this expression; we derived an expression for that diverges of radio de flux And we convene is write as minus. We write a minus because minus divergence represents the energy added to the system. So, each term are right hand side, that we want to write now, can be thought of as the term representing an has he deposition or an has he removal from this system. The first term is the radiation living surface one and observed by the gas So this is heat addition to the gas, heat absorption The second term is energy living the top surface and reaching the level kappa lambda and being observed. See this are two terms represent the gain by the medium on account of absorption of sole radiation. Then you have turned presenting the gain from all other gas elements below kappa lambda. So, these are emissions by layers below the level kappa lambda which are observed at that level and similarly we have terms which represent all the mission by gas elements, above these layers graded in downwards an observed by the layer

So, all the I am not showing the detail that derivation. You must able to do this from the previous equation by q r lambda, but differentiating by applying the leave needs rule, for differentiation of integral which was discussed in the last class. So, we have all do it and look at the first are representing the observe radiation on the top surface, second term observe radiation from the foreign surface, third term observe emitted by all gas space below this layer and the last term radiation observe by the layer of all the emission downwards so they will above kappa lambda. And finally, you have the emission term So, this is a important term this obeisance in the four five directions, how the radiation is emitted in all directions by the gas. So, this is a expression for radiative heat flux Now this is in the spectral domain we want the total radiation So, all of you know that the total radiation flux is 0 to infinity q r lambda d lambda and therefore, one can write as this quantity and declared to over all may be. So, ultimately quantity of grey data trace use this term minus recurs that is present the energies

total energy ideal system by radiation Now, this is a step which is most difficult step in this whole procedure, because a lambda is a very complex function of lambda. So, this is what makes live very hard to integrate this, because the difficult absorption ban of a gas. That it is a millions of lines and your account for absorption each line and the lake absorption between less all those said, accounted for and this is no mean task and people have been attend for the last almost 100 years. This problem was first looked at by straw members in astronomy absorption caption is not a strong question lambda in astronomy; they are willing to adopt a grey gas model that is assuming a lambda is not a function of lambda. This dash remove one complication that is the intercede wave length, but still have to deal with anglo integration and space integration Now a task here is, now to understand. So, we go back to the divergence radiant flux and recognize the 4 absorbent term the absorption from the bottom surface, emission of the top surface, emission of absorption emission by all gas remains below a level kappa lambda and absorption of radiation imputed downwards by all gas elements towards kappa lambda and my is for e lambda if you emission to Now, this equation quit complicated and they primary complicate that arises that highly pointed out I will want to, repeat that so that understood the main problem is unknown is inside the equation and then mix very hard Now before we set out to give example the how to solve this equation, we want to look at too limiting cases Limiting cases are always to look at because they provide some insight regard the behavior of the system. Because you want to understand they behavior system in certain limits you

better appreciate what the seemed us when is not in the limit. The first thing we look at what is known as optically thin limit that is the total optical depth is Now, that is occur in gal world because we take the atmosphere in which below. It most contains Nitrogen Oxygen and Organ. All these three gases hardly observed any radiation And so, all the radiations observed in the atmosphere by manic gases like water paper Methane and Carbon dioxide. So, these are in such small amounts that atmosphere that you can treat them to be in the optically thin domain. So, there are examples in real world where the medium with optically thin domain and you can almost avoid the integration in this case Now we take of the three limit following conditions are satisfying. One can show that for small value is of kappa lambda which is going to happen because this total value is less than 1, now that will imply that E 2 of approximately 1. So, this useful approximation in the thin limit essentially saying that there is no observation by the gas. Gas absorb is so weak and the dimensional of the enclosure are so small, because you remember kappa lambda 0 by definition is 0 to l A lambda d l

And therefore, kappa lambda 0 can be 0 l is very small a l lambda is very small. So, in this situation, in most places in the enclosure you can assume a 2 is equal to 1 and if you do all that you can show that q r lambda becomes and minus So, in the after the thin limit, both the expression for radiative flux and the expression for the divergence of radiative flux are very simple. In the case radiative flux you say now impact of the gas. So, the fluxes are being exchanged between without measurable deviation. The second expression shows that, even the update settlement that is some absorption of. So, this expression is the expression for absorption by the gas of the radiation living surface 1. The second one is absorption of the gas of radiation living surface 2 And what it shows in the both cases the gas is so little absorption there is in the gas, that the intensity of radiation that comes to the layer kappa lambda is not affected by the passes to the medium, because medium is very weekly observing So, you have the observing term the absorption term and the emission term. Now, optical thin limit is not very common which is understandable, because I pointed earlier, one as the pictures are absorption cooption is that it varies very strongly as function wave length. So,

will one wave length it will very strongly observing and just the adjacent wave length we go to it does absorbed of So, in dealing with this minus gases, that are there both in found as and in the atmosphere, we are recognize the fact that till lament is rather rare. Still it is a very useful approximation because you all dealing with essentially algebraic equation. We do not have either analytical equation. So, this is something which we are solve very easily Now delist rate how this can be solved a take the thin limit and ask a myself what is d q R d x. We have to integrate the previous expression over lambda look at this. So, this expression becomes very simple. So, we are integrating or with now. So, this is the first term. This is the second term and this is the third term. So, all have done here is in integration wave length and ultimately this repo found on the computer Suppose let us assume that, we are access to computer and detail all this calculations for gas typical of fondness containing some carbon dioxide, some water paper and knowing the property is effected all this. Even if you do that it will be useful to write this equation in a more understandable form. We general practice is written as. So, this simple expression shows there are three observe caution here call the plank mean observation cooption I will write down the expression for them for a benefit The first expression is for a p 1. Guide but similarly a p two. So, each of this functions is waited by the wave length variation of the radio city or the corresponding surface

and finally, for the gas. So, the purpose of defining this three absorption cooption one for radiation cooption surface 1, one for the relate communities surface two and the third for the gas. Is that in some situations, these can be calculated a prior before you I say solve the problem and. So, once you have tablet values of a p 1 a p 2 and a p g, as function to pressure you can solve most problem that one encounters in this optical thin limit Now, this limit is not that useful as a pointed out, because there are very few gases which choice by the condition they are optically thin at all wave length is interests. That is very rare. So, there will be situations and the certain wave length, certain gases will observe radiation strongly. So, you may be in position to define the plank been cooption than use you. And I have mention one more thing, last quantity a p g is a plank absorption co efficient waited by the black body function at gas temperature, which should be really will called plank emission coefficient This because this coefficient is waited by the black body function at the gas temperature, so it has all the required condition satisfied we call it and emission coefficient, but historically this has been call at absorption cooption, so that name sticks per said he speaking this two are absorption cooption, because they represent the integration over wave length of the incomplete radiation at one, there are two, they could be very different wave lengths. So, you need an separately, but the plank absorption or emission coefficient, the last term, that depends both on the specific gas and it is a lambda is reapportion as well as the waiting function brought in by in this case as a lambda b So, the thin limit is practice till for certain applications, because this expression is a different equation. And this expression is plug into the general another equation and there will be other terms and this will be going along with them. So, some of those cases, where people solving this equation, they will be extremely happy that the expression for the divergence ready reflux is fairly simple If you know the temperature, a problem becomes even simpler, but normally temperature is an unknown, it needs your valuator. So, it does take some time to get that number done Now, we can ask a simple example you can take illustrate this. Suppose the system is in radiative equilibrium. But this we mean system for which that divergence of the ready reflux

is equal to 0. This is a useful static point, because if you look at the n a g equation an assume study state and neglect all other phenomena, re conduction convection other phenomena and focus on the radiation. Then you do expect that in study state, that is from the n a g equation, in a study state everything will be in equilibrium. So, that is the case, than one can see from first row that the divergence of ready de flux has to go to 0 So, after the case of the after thin limit, this employs that the temperature of the gas then enclosure with, in which you in walk the optically thin limit, will come out as, a very well simple expression for the temperature of the gas, because the equations are algebraic So, all you have to do is to from tabular information available in most text book in radiation. It calculated plank be mean of the absorb cooption with a limit surface and surface two, but it by the corresponding radio city values, calculate numerator and dividing the denominator by the accepted monthly mean radiation value. So, in principle in after a thin medium, you can estimate temperature as function x y and z if, he have information available on the right hand side and if not, you should be doing other can a computation which are much more demanding Now, the next step is after having define the thin limit. So, notice that in the thin limit, the equation for radiant transfer becomes algebraic equation. So, it is easy to solve the equation either analytical or numerically So, one would, people would like to in work the thin approximation. So, that is a dealing with full integral equation. You deal with a simple equation solve algebraic Now, if you look at added equilibrium condition, you can get the temperature So, computing this is felly straight forward and values of the plank thin absorption coefficient are tabulated, in has to be, this has to be, I am sorry here, this has to be four a p g So, once you have an estimate of a p 1 a p 2 and a p g, then based on the information of radio city surface one and surface two, you get T temperature in the gas Now, you will notice that in this limit T g is not a function of x. And this is the peculiarity of the thin limit and one can easily show that is, this surface one and this surface two and surface two has that one temperature likes a T 2 and surface one other temperature T one. You will find the gas will have only one temperature and there will be two jumps. This jumps are cause slips for it says is that in the optical thin limit the gas has a certain temperature in variant with distance x, but at both of boundaries there is the change the temperature that is, are slip And this is the very special future of radiation transfer, which is not encounter that often in the case of conduction and conversion transfer In both of which is normal practice, it assume that the temperature of the gas next to the wall, is same as the wall temperature But this approximation is not valid for now in this thin limit the problem is becomes there are trivial, because the temperature

not varying in the gas is uniform and there are two jumps which can be evaluated And there are not many situations in the engineering practice where you might encounter this kind of condition. This may be partly true in small metallurgical furnaces, where because furnaces small it ability to observe radiation from the walls in somewhat limiter. But what I wanted notices is that, from the radio at equilibrium conditions we are able to get temperature on the gas. But when he calculated fluxes, this is what he did, first in the case of fluxes; there is no influence of radiation So as for as ready to fluxes in this enclosure concern, they are unaffected by the presents of the observing gas, but it come to diverges this is flux they are promptly there and they have accounted for Now, this might posses some people, as to how in the flux calculation you can neglect the role played by the gas, but in the divergence of the flux calculation case properties come into play in a prominent way. So, all you must remember is that ultimately whether you really taken account variation lot, depends upon the importance of variation visa base, other modes of n a g transfer; that is conduction and convection So, that when we took the thin limit and look at the derivative, those may be in small if the gas of the thin, but we must compare this radiative flux, with that due to conduction and convection. Only then we cannot say these quantity of small and is we can neglected, we have to compare this quantities how about small with the completing quantities coming from conditionally transfer and conventionally transfer. And only then we can optionally show that these terms are really small Now, the next limit we considered is the optically thick limit. And the major feature of this limit is this is the total path length is much much greater than one. So, this is the excite opposite of the thin limit thin limit this quantity is very small, so we could make very simple approximation, but now we are dealing with a case where this quantity is very large. Now, in this limit if you want to look at the radiative transfer equation you do lot of work, because in this limit the expression we get a derivable. But not in a way that you have done for thin limit, where the form can very trivial, because you are able to essentially neglect gas absorption So now we are what you can do propos is limit to the factor our kappa lambda we do not look at, that defense in the wall are sufficiently will get a simple equation, but remember that somehow you have get the information about the value of kappa lambda, without that we can proceed further Now, once you assume this it automatically it follows that this is also true. So, you have kappa lambda 0 much greater than 1, kappa lambda much greater than 1 and also kappa lambda 0 minus kappa lambda also we much greater than 1. Essentially what is saying, in that in the two parallel plates I am converting compared to the interial and not going to in this wall or this wall calls again go near a wall. Then I am going to violate the condition mean free path is very small, L that is minimum plates. So, photons are getting observing So, in that limit the gases in the interial will only see adjacent layers from than the emissions are occurring. They will not see the two walls, because they are very far and the radiation between wall is absorb right next to the wall and so the elements which are in the interial do not normally see the two walls So, that be in the case we can expand the black body function, that we encounter in the integral equation it is most difficult one, in terms of local value and a derivative So, this expand want to do of the emission by a gas element kappa lambda delta, in terms of the emission at kappa lambda and the derivative of this quantity revaluated at kappa lambda So, this information plugged into the equation In addition because kappa lambda 0 is very large, we can say that E 3 3 of kappa lambda for kappa lambda 0 is approximately equal to half. That is E three 0 half in you know that. So, we are assume it is approximately going to half. Similarly we talk about, I am sorry this is 0 I am sorry. We because it is very large, sorry kappa lambda 0 is very large, this quantity and so 0. Can be use true of E 3 of kappa lambda 0 minus kappa lambda. It also can do 0. So that means, you are for away at the two walls and so the radiation of the two walls are accommodation so much that you do not see it. And the approximations are mention in the previous slide. Now will do it is following expression by q r lambda will be equal to Now, I wanted notice that many of this, we can this will go to 0 will go to 0, because that is what assumed and this two will adopt Sorry this will two also get cancelled out; half and minus half. So, this two expression will not be there both are gone to them because they are smart is to because they cancelling So, you left only with two term involving gas radiation, which is q r lambda is approximately equal to, the two expression representing gas emission. Now in the limit which is optically thick, this goes we 0 to infinity. This can also be re written and terms of kappa lambda 0 minus kappa lambda to make 0 to infinity and using the fact that 0 to infinity, x E 2 of x is equal to two-third can simplify the expression to write the following very simple and elegant result which is negative 8 flux is equal to minus 4 by 3, d e lambda b, d kappa lambda for thick limit This is a very useful and powerful result, because we converted to the basic complicated continue at this kind of next lecture and we will show how this result is a very useful a nature of radiation heat transfer Thank you

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