 Good morning all of you and welcome to the last lecture of this course most of you must have been quite happy and relieved also that you know you have to go through a lot of assignments I know and also computer assignments as well as detailed derivations probably putting lot of effort compared to the other three credit course but this is course where you learn by applying you know so I mean the classroom lectures are just kind of helping you to understand in a broader sense but as long as you solve a particular problem yourself you will not be understanding because we have so many different types of problems in convective heat transfer it is not a one particular type of solution and then you just change the boundary conditions and then you find you know different equation solutions and so on. So we have external flows where we have a particular type of solution and then internal flows all together again different regimes you see how so it ends up from starting from a simple you know integral equation all the way to solving a PD anything we can do within the internal flows okay and again when we are talking about natural convection again the solution methodology changes because you have coupling of heat and mass transfer right so then we same method the approximate solution or for example the similarity solution when we apply it to natural convection has to be slightly modified in a kind of unknowns that we are dealing again when you talk about turbulence it is altogether different it looks very different from the kind of analysis we have applied to the laminar boundary layer equations we have not looked at any similarity solutions in turbulent flows so whatever was possible in terms of simple analogies you know we have been trying to direct so you know in that sense this course convective heat transfer has lot of things to be covered you know so it is a it is quite a bit of challenge for me also and to make justice to this course and still in spite of that we have not covered topics like you know porous media convection in porous media then we have face change condensation and boiling so these are also parts of you know really speaking convective heat transfer but I would suggest that you know we since in our department we are also offering specific courses related to those topics you know such as boiling and condensation porous media so anybody who has taken this course now should be able to easily you know get on with get along with those courses as well as the foundation has been very well established so the rigor will be more or less similar it is only you are just changing the approach a little bit when you do it for porous media for example okay you have additional resistances which have to be counted for boiling and condensation is also highly empirical subject so the kind of analytical solutions are very limited to extremely simple cases especially in condensation in case of boiling you cannot find any analytical solution more or less everything has to be empirically you know there is not even a rigorous derivation but the thing is the physics of boiling is very complex so there is a combined heat and mass transfer so far we have not looked at mass transfer in this course again when we say convective heat it is also convective heat and mass transfer there is there are cases where you know mass transfer is an analogy to heat transfer just like we are doing an analogy between heat and momentum through Reynolds Prandtl all these analogies so we are solving the flow field and finding the heat transfer solution same way we can also find an analogy between heat transfer and mass transfer so we introduce another non-dimensional number called the Schmidt number which is the ratio of momentum to mass diffusivity and once we know this then once you know momentum diffusivity you can extract the mass diffusivity and a similar kind of analysis instead of Nusselt number we define a Sherwood number things like the mass transfer coefficient instead of heat transfer coefficient so it becomes very similar instead of Prandtl number you have a Schmidt number which becomes the governing parameter there but there are cases where the heat and mass transfer gets coupled that is a good example is the phase change evaporation for example so in evaporation you also remove certain amount of mass from one phase and you know add it to the other phase so that becomes more complex now when you couple heat and mass transfer you cannot find analytical solutions again so that requires a very rigorous treatment even numerical solutions are kind of you know hand-waving approaches you know whatever we have you know in a commercial computational fluid dynamics packages and so on they are not very rigorously established models and we are still elusive okay so I mean so boiling and condensation in that sense it is a highly empirical subject you have to deal with a lot of correlations and experiments which have been conducted and the understandings of that okay so yeah overall this is a very vast course you know so I think it is really hard to cover all these topics in one semester if at all can be done but it can be done but you will not be able to appreciate the depth of this subject you know so which I think already most of you are aware of the breadth of this course so when you take a heat transfer course I am sure you are exposed to the breadth of heat transfer but unless you probe deep into a particular topic you will not be able to appreciate the fine aspects of finding rigorous solutions especially analytical solutions okay so when it comes to turbulence again people have different approaches to it my personal approach is also I do not look for analytical solutions everywhere but you should understand that for certain problems there are analytical solutions available which you should not overlook okay for example I have seen cases where people solve the flat plate boundary layer numerically and they do not know where to look for the Blasier solution to validate okay so you should know now by now for which cases you can actually compare with exact solutions even when you talk about turbulent flat plate boundary layer so we have correlations derived from approximate methods okay in the last class I was talking about the use of analogy and also knowledge so in that case we have established an empirical relation for the skin friction coefficient right so if you remember what was the relation for CF so this is the this is an empirical one okay this we cannot rigorously derive this okay this is coming from set of experiments okay for the hydrodynamics part do you remember what is correlation is 0.046 into now when we define RE now rather than defining it as RE because RE you might actually misinterpret the length scale used so here it is whatever U M or U 8 depending on whether this is an external or internal flow if you are dealing with external flow you have U 8 and then length scale is now the boundary layer thickness divided by what is the minus point point 2 so this is a kind of most commonly used relationship for internal flows it becomes easier you replace your ? with D right and then the next step is suppose you take the case of internal flows the simplest analogy that we have derived so far is the Reynolds analogy right which considers the entire boundary layer to be purely turbulent okay so for this case Reynolds analogy what does it say your standard number is equal to CF by 2 for this in this case you are actually neglecting the role of molecular diffusivities compared to the turbulent diffusivity which you know that for there is an influence of molecular parental number even in the turbulent muscle number case because this is a wall phenomena and near the wall the molecular diffusivities cannot be neglected right so therefore Colburn extended this Reynolds analogy to also cases where he considers the effect of molecular parental number okay then it becomes the Reynolds Colburn analogy and what does it say your standard number times we are power 2 by 3 is equal to CF by 2 now when we substitute the expression for CF into this therefore we find we can derive an expression for the Nusselt number do you remember what it is 0.023 times R E D power 0.8 PR power 0.3 okay so this is called the detus bolt detus bolt okay so detus bolt der correlation many people think it is a completely you know based on empirical formulation it is semi empirical you can say that this skin friction coefficient is empirical but after that we use the Reynolds analogy to get this and again the textbooks there is a variation of exponent of parental number for heating we use value cooling we use 0.4 but it does not matter overall the structure of this comes from the Reynolds analogy okay so then it has been slightly modified depending on the heating or cooling case because the parental number dependence is not a very exact dependence it has been introduced later on by Colburn but it depends whether you have a heating or cooling there is been shown to be a slight difference but not too much okay some people use an average value of 0.33 and they are still okay with that right so now for the external flows however you have in terms of the boundary layer thickness ? right which has to be first determined because we know that in the external flows boundary layer thickness ? itself is a function of Reynolds number so therefore we cannot simply define Reynolds number based on ? and stop here so we have to therefore how do we do this we have to use the approximate methods okay so the only way to find ? as a function of the local Reynolds number is to use the approximate methods and then substitute this for the wall shear stress and then we will have an expression for ? and then we apply the Reynolds analogy then we find the Nusselt number correlation so there also you will have something similar except that this constant yet slightly modified and the Reynolds number dependence still remains the same only this becomes a local Reynolds number okay and the Nusselt number also becomes a local Nusselt number now because in the external flows your boundary layer thickness keeps varying so therefore the Nusselt number and Reynolds number have to be defined based on the local coordinate okay so from this you can understand even for internal flows in the laminar case your Nusselt number was constant in the fully developed region but what this says is that even if it is fully developed in the turbulent flow this is still a function of Reynolds number and Prandtl number correct because what is fully developed it is only the turbulent boundary layer when you say these two boundary layers these are the turbulent boundary layers so only these are merging but the viscous sublayer will not merge this is very close to the wall it will not merge okay so therefore there is an influence of Reynolds number now because Reynolds number controls the amount of this viscous sublayer thickness okay which in turn affects your Nusselt number so although you are turbulent boundary layers are fully merged your viscous sublayer is still remaining which is now governed by the value of Reynolds number and therefore that comes into the expression and so is the case now since this is a heat transfer case Prandtl number will also be important okay so please remember that in the turbulent fully developed flow you do not have a constant value of Nusselt number like in the laminar case laminar case does not depend on Reynolds and Prandtl number because you have only laminar boundary layer once it is merged there is no effect of what you are putting upstream whatever Reynolds number may be once the two boundary layers are merged you will have the same profiles and only thing that will decide the Nusselt number will be boundary condition whether it is a constant wall temperature or constant heat flux only that alters the nature of the temperature profile whereas in the turbulent case still this is dictated by Reynolds and Prandtl numbers okay so this is a very popular correlation you know so you find in most of the engineering applications people use the Detas-Volter correlation without having to worry what is the accuracy of this how it has been derived and still it gives a reasonably good prediction within something like plus or minus 15 to 20% of the experimental data right so within for simple flows like pipe flows you know flat plate boundary layers these people have been successfully using this right for more complex flows they have to look at numerical solutions there is also we should however you know we should look at modification to the basic analogy okay we know that although this is a very useful analogy widely used but nevertheless since this considers the entire boundary layer as turbulent it is not the most accurate representation of the turbulent boundary layer profile so we have seen that that the entire turbulent boundary layer profile can be divided into three regions so if you plot U plus and Y plus and Y plus on a log scale for example right so for example if you are talking about so you have 5 here you have something like 50 let us say you have 100 let us say this is like 500 and so on so based on this we have identified a profile in the viscous sub layer which is linear plotted on a logarithmic x axis becomes a curve like this okay so and then we have another expression for the log layer which is extending above Y plus of 30 okay so that can actually if you extend it it goes something like this okay and here it is given as a logarithmic profile so do you remember how the profile looks plus a constant and this is your von Karman constant which is 0.41 and this is 5.5 okay these two we have derived from the basics okay now also there is a patching layer which has to be connected between the linear and the logarithmic variation another logarithmic profile so this is called the buffer layer okay so also represented with a logarithmic variation remember what the profile is 5 Y plus minus 3.05 so this is your buffer layer so this is valid for Y plus greater than 5 less than 30 right this is valid for Y plus less than 5 and this is valid for Y plus greater than 30 these are all kind of you can really measure the viscous sub layer and the log layer very clear okay when you have a very fine probe which can resolve the viscous sub layer you can actually measure and show that the data also falls with these so this is been generally well accepted again variations will happen when you have pressure gradients other when you have flow separation okay then when the boundary layer itself is detached so then the this kind of profiles will not be there so according to the Reynolds analogy we all together neglect this and only consider the entire turbulent log law profile okay and then we make use of the calculations of no turbulent wall shear stress and also the heat flux everything to be having only one component which is nothing but the turbulent diffusivities we integrate it all the way from the wall to the edge of the boundary layer okay we ignore completely the molecular diffusion and then we derive the Reynolds analogy so naturally then extension to this this is called Prandtl Taylor so I am not really sure about who really did it but looks like it has been also named after GI Taylor so it should have been I think maybe a joint kind of intuitive discovery of both Prandtl and Taylor so it has been credited to both of them so in this case the point is that we assume a two layer model where the inner layer which is the viscous sub layer is also considered which is the important thing you know the molecular diffusion has to be dominant near the wall it cannot be ignored so therefore so we consider the viscous sub layer and we consider the rest of the boundary layer to be turbulent so when we draw the velocity for example as a function of Y and also the temperature so we will have a variation like this for example so till a certain location so let me let me give it a much so till a certain location this will be dominated by viscous effects from the wall and from this point let us call this ul where it transitions from a laminar sub layer to a fully turbulent layer okay the corresponding thickness of the viscous sub layer let us call this as ? air okay so above this the profile becomes governed by fully turbulent diffusivity right and finally at the edge of the turbulent boundary layer okay so you reach the velocity which is either the mean velocity inside the duct or it is a free stream velocity in the external flows okay so this is at ? right so therefore now you have two regions one is your viscous sub layer and the other is your turbulent boundary layer similarly if you draw the temperature profile something like this so your temperature profile is governed by again let us say till up to the laminar sub layer you have a value which will call TL okay and from that it becomes fully turbulent and then outside it reaches the value Tm so this is your temperature variation with Y and this is your velocity variation with Y so this is the picture of a two layer model which was which was used by Prandtl to derive the two layer analogy so naturally the assumption here is that so we are picturing the entire within the turbulent boundary layer is divided into two layers okay one is the viscous sub layer near the wall and we have the turbulent boundary layer outside okay and again when we derive it we also use condition that turbulent Prandtl number is equal to one so when we say turbulent Prandtl number is equal to one we say that your ? is equal to ? T okay because even though if you assume turbulent Prandtl number is equal to one all the processes at the wall are governed by the molecular diffusion okay so therefore the assumption of ? T is equal to ? outside will not affect what is happening near the wall okay so this same thing was is also used in the Reynolds analogy right so in the Reynolds Colburn analogy also we use turbulent Prandtl number equal to one and then only we derive the analogy okay so the assumption is that inherently this is not bad and especially when we are doing even turbulence modeling most of the time we assume this turbulent Prandtl number to be close to one okay so the starting point of this is now we have the shear stress which has now both the molecular and the turbulent diffusivities so we can write this as ? T into du by dy and similarly the heat flux also consists of minus k plus kT by dy okay so if you are looking at Reynolds analogy it assumes fully turbulent so therefore we neglected ? compared to ? Tk compared to kT and then we divided we integrated from the wall to the boundary layer and then divided one over the other and then we ended up with this so now we have divided this into two regions so we will first apply this for region one which is the laminar viscous sublayer okay so within the laminar sublayer can you integrate let us say equations 1 and 2 within the laminar sublayer where we can neglect the turbulent diffusivities integrate your velocity and temperature profiles from your wall till the edge of the laminar sublayer ? and then take the ratio of 2 by 1 so that is the heat flux divided by shear stress first integrate it and then you divide it so what do you get if you do that what will happen to ? so at wall it becomes ? wall right and this is a ? T is neglected you have ? and then the U gets integrated from 0 all the way to UL okay and why all the way from 0 to ? L the same way here also you have Q wall and this gets integrated from T wall to TL and then this is also 0 to ? L and then we divide 2 by 1 so ? L gets cancelled essentially we will have therefore Q wall divided by ? wall is equal to minus K by ? into TL minus T wall by UL okay so therefore we can write this in terms of molecular Pantel number multiply and divide by CP so ? CP by K so we can therefore write this as CP divided by Pantel number okay so let me call this as equation number 3 so now next we will apply equations 1 and 2 in the turbulent boundary layer okay so for turbulent boundary layer we can of course do the other approximation where the turbulent diffusivities are much larger than the molecular diffusivities okay and therefore we can integrate from ? L all the way till ? right so you do the integration of these profiles from ? L all the way till ? and then divide again 2 by 1 okay let us see what what is the expression that you get so let us say again you have Q double ? by ? will be equal to minus CP into Tm minus TL divided by Pantel number into UM so where do you have Pantel number here because this is turbulent so this is yeah PRT is equal to 1 so therefore we can so therefore now we have this is your equation number 4 so for continuity the same wall shear stress or wall heat flux has to be taken all the way from the wall to the edge of the turbulent boundary layer correct so we are right now assuming a one-dimensional transport of heat so whatever you apply at the wall has to be transported all the way till the edge of the boundary layer so therefore this has to be the same as the wall quantities here correct there is no discontinuity in the heat flux or shear stress will also be felt by the turbulent boundary layer same with the wall shear stress okay so therefore we can just replace Q and T with Q all and T all and now the unknown quantity in this we know for example for a given isothermal condition we know what is the wall temperature T wall okay but what we do not know here is TL the temperature at the edge of the laminar sublayer okay now UL is also obtained from measurements okay we will see how do we get it but when we are solving the heat transfer problem TL is difficult to obtain so therefore we will eliminate TL by using these two equations so therefore eliminating TL between 3 and 4 so how do we do that we just add 3 plus 4 so if you add 3 plus 4 you have minus Cp you have plus Cp TL there you have minus Cp TL so the TL gets eliminated right away okay so therefore you please do 3 plus 4 if you do that so you will be ending up with the following expression which is Q wall double prime by okay so I just take this to the left hand side okay and then this also to the left hand side and then I add 3 and 4 so on the right hand side I have Cp into T wall minus Tm okay so this is let us say equation number 5 in this case the TL is eliminated so what I ask you to do now that we have something like Q wall by T wall okay you can take this T wall to the right hand side and write this in terms of stanton number okay so if you write this in terms that means Q wall by T wall minus Tm is nothing but H H by rho Cp into Um is your stanton number so write this in terms of stanton number on the left hand side take the other extra terms to the right hand side and see what kind of expression you get okay so you should be able to get an expression for stanton number is equal to function of the other terms including parental number so now this entire term here is stanton number so Q wall by T wall minus Tm is H divided by rho Cp Um okay so your stanton number therefore is equal to now what I am going to do is multiply and divide by Um okay so therefore I will have T wall by rho Um square is nothing but your CF by 2 okay so this is your CF by 2 and when you take Um common so you have parental number into UL by Um correct plus I have I will just rewrite it 1 plus UL by Um into parental number minus 1 so finally I end up with a relation between stanton number and CF and you see in this case naturally the parental number is coming so I am not forcing parental number to come in like the case of Reynolds-Colburn analogy since we have not neglected the viscous sublayer we have also considered that so naturally the parental number is built in inside this analogy right so now the only thing that needs to be closed is UL by Um so UL by Um actually is measured for turbulent flows turbulent boundary layers and we have some correlations for that now let us see for the case of a circular duct for example so the UL by Um is expressed as a function of the CF written as 5 into square root of 1 by 2 CF this is the kind of correlation so let us call this as equation number 6 and this is number 7 so if you plug in equation 7 into 6 so you get the expression for a circular duct you get the analogy CF by 2 divided by 1 plus 5 times CF again into parental number minus 1 for example so this is the analogy for the circular duct similarly if you are talking about maybe a rectangular duct you will have a different relation between UL by Um and CF which you will substitute and get the final analogy okay so therefore once you know the CF we can use the previous CF that we have written down okay for the Reynolds analogy that can be used substituted here and we can find an expression for Nusselt number right so this is how the Prandtl Taylor analogy is developed so for the case limiting case where Prandtl number equal to 1 so where we have a fluid which is having a Prandtl number exactly equal to 1 the molecular and a molecular laminar and the thermal diffusivities are identical so what happens to this reduces to the Reynolds analogy so in this case this gives you that Prandtl number is equal to so you see that Reynolds analogy is not bad after all especially when your Prandtl number is equal to 1 that is your molecular Prandtl number equal to 1 Reynolds analogy is the correct analogy because all the complex analogies are finally collapsing to that okay it only makes a difference where your molecular Prandtl number is either law very large or very small that is when you start deviating from the accurate results so most of the gases you do not have to worry so you can actually therefore most of the gases you will find Reynolds analogy it is reasonably accurate whereas if you apply it to very large Prandtl numbers or very small Prandtl of a liquid metals and so on very small Prandtl numbers so then you will find that Prandtl analogy is more accurate than Reynolds analogy is that clear so this is more complex expression naturally so the final case will be to consider all three layers including the buffer layer right so if you also consider the buffer layer then you are tracking the transition more accurate so I will not derive that but I will only give you the final analogy so this three layer analogy is called the Von Karman analogy so finally the most complex analogy that you can derive so this is based on the three layer model which includes the laminar the fully turbulent and the buffer layer and this case the expression for the final expression for standard number for the case of circular duct okay so you have the same term till now till now this is the same as the Prandtl Taylor analogy okay now the additional term comes because of the buffer layer so this additional term is coming in due to the buffer layer okay so this is the difference between the two analogies and again for the limiting case of Prandtl number equal to 1 what happens same this becomes 6 by 6 so on of 1 0 this again becomes 0 so you have CF by 2 so finally it collapses to the Reynolds analogy again right so for more complex cases where none of these analogies will work okay so we cannot find a simple analytical solution again then we have to use numerical solutions for heat runs okay so we will kind of stop here because although turbulent heat transfer involves lot of modeling I do not want to talk about in detail because these are covered in other courses related to turbulence modeling and so on where you can learn how to use different models for different problems and they are also available in many of the numerical solvers okay so we can you can also go through those documentations try to understand which kind of turbulence model work for different kinds of problems now practically when you are working with turbulence you will not be using too many too much of analogies the most of the complex engineering problems cannot be solved with analogies so there you have to find numerical solution you will be using some kind of turbulence model right okay so we will I hope that you know I could cover with the limited number of hours at least some overview of about turbulent heat transfer okay thank you so much.