 So today we will look at a different kind of a solution to the extended grids problem. So the original grids problem was done as I said for the case of a plug flow with a constant wall temperature and that was extended later by sellers to parabolic velocity profile and we also saw the solution to that so now we will quickly visit the last of the extensions that is with the constant wall flux boundary condition okay so we will look at the extended so this is of course the grids problem for the thermally developing region for the constant wall flux boundary condition either it can be with a plug flow velocity profile or a parabolic flow velocity profile if you take the case of for example a plug flow kind of a velocity profile you start with your energy equation U DT by DX is equal to alpha into 1 by R into D by DR of R DT by DR okay so this is your basic energy equation let us what we will do here is we will not try to non-dimensionalize the temperature because we have a constant wall flux condition and therefore and any attempt to kind of define a non-dimensional temperature will be futile because if you define your ? something like T – T wall by Ti – T wall so there it was useful when you had a constant wall temperature case so you could take this out of the differential and it will cancel off but in this case your wall temperature is a function of X so therefore you cannot define your non-dimensional temperature this way so it is better to keep it in the dimensional form okay and then try to solve so what we will do however is to non-dimensionalize the coordinates for X and R okay so we will write that in terms of ? and so we will define ? as X by R 0 divided by Peclet number and my non-dimensional R is ? which is R by R 0 so if you substitute this into the given equation energy equation so you get U DT by D ? so from here you can write your X as ? ? into R 0 into Peclet number now Peclet number is nothing but ? into U M into D by or you can say U M D by ? okay so this I can substitute in place of X so this becomes R 0 U M D and the ? goes to the numerator here so this is equal to ? 1 by ? into D by D ? into ? DT by D ? of course the R 0 here and here cancels so I have a R 0 square which I have to multiply it outside so my a cancels here and I can also cancel off my R 0 so this I can write write as to two times R 0 so this becomes R 0 square which also cancels off and therefore I get U by U M into 1 by 2 DT by D ? which is equal to 1 by ? D by D ? into ? into DT by ? okay now come now comes the velocity profile which you want to use if you want to use a fully developed parabolic velocity profile then you substitute for the appropriate relation so you know that U by what is the fully developed profile U by U M is equal to twice 1- ? square in terms of the non-dimensional coordinate so directly you can substitute for U by twice U M directly as 1- ? square here if you want to go for a parabolic flow okay however if you want to go with the classical how great started with a plug flow then you can say that your U is equal to U M so this is for a parabolic profile and this is your plug flow or slug flow whatever you want to so therefore let us assume for the time being that yours is a plug flow so substituting this you get DT by D ? is equal to twice of 1 by ? into D by D ? ? DT by D ? okay so this is how your energy equation can be written and now we have to state the boundary conditions okay so now if you substitute the parabolic flow you have 1- ? square here okay 2 2 cancels off you have 1- ? square so till here it is fine so now we have to state the boundary conditions for the constant wall flux case okay so T corresponding to ? equal to 0 and any value of ? T I and coming to the boundary conditions with respect to ? here you have a constant wall flux at R equal to R0 so therefore you should rather write DT by DR or DT by D ? at ? equal to 1 is equal to Q of wall by K this is your condition okay so this is your wall flux which is K DT by DR at ? R equal to R0 this is basically DT by D ? T equal to 1 so this is a constant and what is the other boundary condition at ? equal to 0 T should be finite or DT by D ? at ? equal to 0 should be 0 so this implies symmetry in the profile okay or this also is equivalent to saying T at ? equal to 0 should be finite okay so now you see the problem so you want to define an eigenvalue problem here however the direction of the eigenvalue problem you have a non-homogeneous boundary condition so therefore the question is how will you convert this into a eigenvalue problem okay so here is where we introduce a particular technique to do that I have posted the solution for a constant wall flux but in a Cartesian coordinate that is for a channel flow on the moodle I have worked out the solution and I have posted it you can just go through it and a very similar treatment has to be done for the pipe flow case also just I will give you the overview I think after that you can go through the document and very straightforward process okay so what I am going to do is I am going to now introduce ? here which is of course not non-dimensional but I will say this as T minus Ti okay so therefore I can replace this since Ti is a constant okay that is the inlet temperature so I can replace this T with ? and the condition here becomes at ? equal to 1 a ? equal to 0 ? will be 1 0 T will become Ti okay so that is the advantage so what I am trying to do is a wherever possible I can introduce 0s I am doing it and then this will be in terms of ? this will also be in terms of ? anyway Ti is a constant so if you differentiate it does not matter now I am going to assume that ? of x, y can be written as independent solutions or sorry here it is x, or ?, ? can be written as some X which is a function of ? plus ? which is a function of only ? plus a perturbation which is a function of both ? and ? so see now the this assumption works well because this is a linear operator the equation is linear and therefore if you assume a linear combination of solutions that should also be a solution okay so you are assuming that your actual solution consists of two independent summation of two independent solutions one which is only a function of ? the other which is only a function of ? and of course you know this is just an assumption on top of it the actual solution can be obtained if you have a perturbation to this and that perturbation is you call it as ? which is of course the perturbation has to be a function of ? and ? so that brings out the interaction between the ? and ? terms okay whereas this purely talks about an ordinary differential equation separately and ordinary differential equation separately and this is a mixed solution okay so this is this is basically independent solutions in terms of ? and ? and this is the perturbation to the independent solutions and since the partial differential equation is linear and the solution is a linear superposition of all these solutions still this will be a solution to the governing equation so now you see the advantage so once you substitute this into the governing equation okay so you can write these as you can separate this into two problems one where you first look at only the independent solution and the other where you look at the perturbation solution to the perturbation and then finally add everything together so first first part will be the solution to independent variables okay so first what I say is that I have a problem now I can substitute X of ? and ? of ? separately and it will satisfy this such a way that you should have you have DX by D ? equal to twice 1 by ? into D by D ? into ? ? D ? by D ? okay so you first assume that your independent variables satisfy this governing equation so you write in terms of the independent variables so you get this as terms of X this in terms of ? and this two can be equal if only they are equal to some constant okay so then you can find this is a ordinary differential equal this is a ordinary differential equation this is a first order second order OD directly you can integrate and write the solutions for ? and X okay so therefore you have two independent solutions and the boundary conditions for this particular problem in terms of X you assume that X corresponding to ? equal to 0 is equal to 0 okay and corresponding to ? you assume that your D ? by D ? at ? equal to 1 is equal to Q all by K and the other boundary condition is your D ? by D ? at ? equal to 0 is equal to 0 okay so you apply these two boundary conditions to this now you get the point why I have written it that way so once the non-homogeneous boundary condition goes to this the remaining part which is the perturbation will have a homogeneous boundary condition okay because once I have found the solutions to the independent variables the solution with respect to ? will take the non-homogeneous boundary condition now therefore the second part will be the solution to the perturbation the second part will be solution to the perturbation here so if I write the equation for the perturbation I can just give the boundary conditions so the boundary conditions for say can be written as ? – X – ? okay so now corresponding to this fact of course when I when I look at D ? by D ? at ? equal to 1 okay so this will be D ? by D ? at ? equal to 1 – D ? by D ? at ? equal to 1 so both will cancel off so therefore this will be 0 because your D ? by D ? is what Q all by K and also D ? by D ? equal to 1 you have forced it to be Q all by K so therefore the boundary condition one of them becomes 0 directly and the other boundary condition D ? by D ? at ? equal to 0 anyway both are 0 so that will also be 0 okay so now now you see that you have reduced your non-homogeneous boundary conditions to the two you are now solving actually for the perturbation which has now homogeneous boundary conditions and now this perturbation ? can again be solved by separation of variables okay so now you can say that I can assume that my ? ? ? ? is actually some X of ? and some Y of ? and then I can proceed with my I can I can now proceed with my regular solution okay so what is the remaining boundary condition that is at X at ? equal to 0 so therefore at ? equal to 0 your ? at ? equal to 0 ? will be ? at ? equal to 0 which is 0 minus X of 0 which is again 0 minus you have P of ? so therefore the boundary condition at ? equal to 0 becomes minus P of ? okay so these are the boundary conditions so you have the boundary condition at ? equal to 0 then with respect to ? you have two homogeneous boundary conditions so therefore if you substitute into this you separate the variables so in terms of why you will get the eigenvalue problem okay so and that will have two homogeneous boundary conditions any guess what will what the eigenvalue problem will be here will that be a Bessel equation it will be a Bessel equation and you already know the solution to Bessel equations you have to just combine the Bessel solutions and then you now the only difference between this and the constant wall temperature case there the temperature was 0 at r equal to r not here you have both gradient 0 boundary conditions okay so this is this is the difference and after that the solution is straight forward so this is an independent solution that you will get from separation of variables and you already have the solution for X and ? which is a straight forward PD to integrate and then finally you superpose all the three solutions and you get the final solution for ? okay so this procedure I have very clearly illustrated in that example for Cartesian coordinate system you please go through that and in the assignment I will ask you to do the parallel thing for the duct flow also okay so only thing there you have cosine and sin here you will be getting in terms of Bessel functions so the final solution that comes out it was done by sellers and sellers did both the case of the case where you have a plug flow as well as the case where you have parabolic flow in the case of parabolic flow you have 1- ? square which is coming here okay when you substitute for the velocity profile you have 1- ? square and there is no 2 here and therefore the eigenvalue problem in this case will be what ? Louis okay this term Louis will kind of a problem so sellers did this and the final solution for Nusselt number 11 by 48 plus okay so this R prime here is basically the gradient of the eigenfunction which is basically the eigenfunction comes from the solution of this term Louis will problem okay and these are coefficients for different values of N okay so this is ? and now he tabulated the values of ? N and R prime ? N so this is the case for parabolic velocity profile so this is the solution by seller set up okay so for different values of N ? N square – R prime – ? N square okay so N 1 2 3 the values go as 25.639 84.624 176.4 8.854-3.062 okay so therefore this is what he has done now you can check for large values of ? okay so any large values from this table you substitute and calculate what should be the N you for large values of ? so if you assume large values of ? you can say that this is an exponentially decaying function goes to 0 the entire term will be very small it will be simply 48 by 11 which is exactly 4.364 just check that that come to 4.364 anyone has a calculator can okay so what was so so do you remember this number this was the case there where we proved very first for a fully developed both hydrodynamically and thermally and for a constant wall flux condition this was the Nusselt number so now this is coming as an asymptotic solution to the sellers for a solution okay okay so so these are the things and of course the sellers also did a case with where he did constant linear temperature wall temperature variation okay so he did all the three so he looked at the parabolic velocity profile and he looked at constant wall flux condition and also linear variation in the wall temperature and he has also given the solution I am not going to give that otherwise it becomes too many correlations so he has done all the three so I am just going to plot the variation of the local Nusselt number with all the three different boundary conditions as a function of X by R0 by Peclet number or X by can plot it as D0 which is 1 by grades number okay so okay so he has done for three different boundary conditions constant wall flux and a linear wall temperature variation which was something like T wall is Ti plus some zeta constant time zeta this was the linear wall temperature variation and of course your constant wall temperature so now I have marked here 1 2 3 so you have to tell me which one corresponds to 1 okay let us start with 3 out of these three which one do you think will be corresponding to number 3 which boundary condition T wall is constant so that is number 3 so between 1 and 2 what could be number 1 Q all is constant how about linear variation of all temperature okay so finally whether it is where linear variation of all temperature or Q all is a constant one it be once it becomes both thermally and hydrodynamically fully developed they reach the same value okay that is why these two merge and you see the corresponding value is 4.3 and for the wall temperature you have 3.6 can you explain why why the two cases give the same value that would be same as T correct exact so when you say Q all is constant in the fully developed case we have shown that the wall temperature varies linearly okay so therefore it should approach the first case where your wall temperature is varying linearly throughout okay so therefore the two Nusselt numbers will have to be the same in the completely fully developed region okay so I think this brings to conclusion all our analytical solutions as far as the internal flow is concerned and the last part today which I would like to with which I would like to conclude the analytical solutions will be the region one that is simultaneous developing entry length so this is your region in the duct where your hydrodynamic boundary layer is developing your thermal boundary layer is also developing okay so this is your region one okay where both are developing so that is why it is called simultaneously developing or simultaneous entry length now this is a really tough problem okay so you cannot make any assumption to the velocity profiles so the only solution to this is to solve the momentum equations and get the velocity profile simultaneously and the velocity profile will not be a constant it will also be changing with respect to x okay this is a very difficult problem and therefore if you want to solve the complete equations so you have to solve the complete Navier stokes equations now there are some approximations made to the solution if you say for a circular tube with axi-symmetric assumption okay so that you neglect your variation with respect to the ? direction so that is your axi-symmetric assumption you can write your momentum equation in the axial direction that is ? u du by dx plus ? v r into du by dr will be equal to minus dp by dx plus 1 by r d by dr into r u into du by dr so this equation which we have written right here okay so this assumes that there is no variation of velocity with respect to ? direction and also the radial momentum is also negligible okay so therefore you write only the axial momentum equation neglecting the variation with respect to the ? direction and this equation is very similar to your boundary layer equation in fact this is your boundary layer equation right so this is your boundary layer assumption remember in the flat plate case we had the same thing u du by dx plus v du by dy is equal to if you have a pressure gradient you have 1 by ? dp by dx plus ? du d square u by du dy square the same way we have constructed the boundary layer equation in the radial coordinate system okay so this is an assumption to the actual problem because what you are doing here you are neglecting d square u by dx square okay so that could be important okay because if you look at the acceleration acceleration in the x direction is important and also the higher order derivative d square u by dx square also becomes important which you are neglecting here and also it sometimes since it is three dimensional you could also have three dimensional effects if you have a non circular cross section okay so these are also not taken into account so in fact but you can still get some kind of an approximate solution if you solve this equation for the velocity profile and but this boundary layer assumption is valid only when you have close to the entry length if you move far away once again once the two boundary layers start to merge then the boundary layer assumption will not be valid anymore okay so this is what comes out of it but there was a person called Langhar by the name Langhar what he did was he even neglected the effect of radial velocity that means he neglected this term straight away okay the big assumption and then he took only the axial velocity variation into account with respect to x and r he solved this equation numerically and so the resulting equation which is all was rho u du by dx equal to minus dp by dx 1 by r into d by dr into mu du by dr so the boundary conditions that he solved at x is equal to 0 u is equal to ui so you had some inlet velocity which is a constant okay so when it approaches the entrance of the tube you have a inlet velocity which is constant and corresponding to r equal to r0 you have u equal to 0 no slip boundary condition and at r equal to 0 the profile has to be symmetric du by dr at r equal to 0 has to be 0 okay so he solved this numerically of course you cannot find close from an analytical solution because of all these terms right here and he got a velocity distribution which was in terms of Bessel functions okay so these are the Bessel function of the 0th order and this is the second order Bessel function the first kind Bessel function of the first kind 0th order first kind second order okay so here ? is basically some ? into it is a function of x by d by Reynolds number okay and your ? was r by r0 so this was his velocity profile this is also called as Langer velocity profile so you see now the velocity profile is a function of x by d through the ? as well as it is a function of ? okay so this is the approximate profile that Langer got by the numerical solution to this equation and with the following boundary conditions of course he neglected lot of things here so he neglected the radial velocity and things like that so this is not a very good assumption when you go too close to x is equal to 0 because here you have strong radial components of velocity which are entraining okay so this is slightly this is valid then somewhere intermediate somewhere may be from here to here okay but nevertheless it is a reasonable assumption so using this profile now you can substitute this into the energy equation so energy equation can also be solved numerically so of course you have your energy equation as u by um into d ? by dx equal to d square ? by d ? square plus 1 by ? into d ? by d ? yeah there should be a 2 here if I define my x as x by r0 by Peclet number okay so now this u by um you can substitute from the Langer velocity profile okay and once again now your velocity profile is a function of both your x and y or is your ? and ? okay therefore okay so I can use the conventional variables which I used before I will call this is ? just to avoid confusion okay so this cannot be again analytically solved because your velocity profile again is not only a function of ? but it is also function of ? so therefore this again has to be solved numerically with boundary conditions whether it is constant wall temperature or constant wall flux and you get the solution for the temperature and of course the final no cell number okay so the solution to this equation was numerically done by was numerically integrated by case okay for different boundary conditions and we will find the solution to all of that one by one okay he is the same case who wrote the book case in Crawford that Stanford University okay so there are lot of solutions to basic fundamental heat transfer problems 1950s and 60s which were done by 50 60 70s which were done by case and what happened was he found out of course solutions but this the person house and okay so he comes to a rescue as we saw already that wherever there is a very complicated solution in terms of Bessel functions or Sturm Lüweil Eigen functions this fellow house and was kind enough to do an empirical curve fitting and then give a more easier solutions in terms of only grades number okay so house and came and he took the profiles which were obtained by case and then he finally cast them into a simpler form for different boundary conditions okay so the first case was the constant wall temperature so all these were from using the longer velocity profiles okay so still they are not the most accurate but reasonable so the Nusselt number variation as expressed as 3.66 plus 0.104 RE PR by X by D which is nothing but grades number okay so you can write this as directly grades number divided by 1 plus 0.016 into grades number to the power 0.8 and constant heat flux case so you see the limiting case where your grades number goes to 0 for large values of zeta so it goes to the fully developed okay so 4.36 plus 0.036 into grades number divided by 1 plus 0.0011 grades number or point 8 and finally for the you also did for constant temperature difference okay the constant temperature difference is basically defined as T wall minus Tm is constant okay so for this he got 4.36 plus 0.036 grades number sorry 0.1 0.1 into grades number divided by 1 plus 0.016 grades number to the power 0.8 okay so constant temperature difference the simplest profile that you can think of where you can have a constant temperature difference is if you have a linear wall temperature variation okay so that is one case which where you can think about this and finally both the Nusselt numbers go to the same asymptotic limits so these were the simplified solutions by Hausen and of course these are correlations which are much easier to work than the exact solutions are the numerical solutions which were obtained using the langur's velocity profile so just to summarize I do not want to now talk too much about how these solutions came because they are all numerical solutions and nowadays the more prudent way of doing this is to solve the complete partial differential equation using computational fluid dynamics and directly get the most accurate solution okay rather than putting so much of effort into finding the numerical solution to these approximate equations okay so as far as thermally developing profile has with fully hydrodynamically developed profile is concerned we can straight get reasonably straightforward solutions analytically but simultaneously developing profiles are much difficult and therefore we need to go for numerical solutions so on a concluding note I will just summarize whatever solutions we developed analytically okay if you look at the analytical solution to the problems first is the geometry or the configuration velocity profile and the corresponding eigenvalue problem or I will say eigenfunctions in fact I can also introduce boundary condition so this is for region 2 so hydrodynamically fully developed and thermally developing so now look at the constant wall temperature and if you look the case of channel flow that is in Cartesian coordinate system and if you assume a slug flow or a plug flow velocity profile the eigenfunctions will be in terms of sines and cosines okay when it comes to constant wall flux the same thing and slug flow huh what is that what will be the eigenfunction still it will be the same only thing you have to go by this approach whatever I described break up into two independent solutions and a perturbation the perturbation solution will still be homogeneous homogeneous with a slug flow will be still sines and cosines now T wall is constant or Q wall is constant channel flow but a parabolic or fully developed velocity profile what will be the eigenfunction huh what is that what what function Bessel function in a Cartesian coordinate system you do not get a Bessel equation okay and when it comes to T wall is constant or Q wall is constant but for pipe flow that the circular cross section okay I will say circular cross section if you have slug flow what will be the eigenfunction Bessel okay and the same boundary conditions circular duct and if you have parabolic fully developed then it will be stumbling so please take note of this so when you solve any kind of problem whether it is a channel flow or pipe flow depending on the boundary conditions depending on the velocity profile this you may alternate between either of these kind of equations okay so one more last table and with that so therefore to summarize all the solutions for fully developed case that is region 3 thermally fully developed so this is all for laminar flow till now okay and when Professor Kohler comes you will start looking at solutions to turbulent flow so geometry velocity profile wall condition and the corresponding fully developed a cell number okay we will start with a parallel plate with a parabolic profile and Q wall is constant the value is 8.23 parallel plate parabolic and T wall is constant 7.60 next circular tube slug flow and Q wall that is 8.0 circular tube and then you have slug flow T wall that is 5.75 circular tube but parabolic and then you have Q wall you have 4.36 circular tube parabolic T wall 3.66 similarly for triangular cross section triangular duct if you have a parabolic velocity profile you have Q wall then you have 3.00 triangular parabolic constant wall temperature you have 2.35 okay so this is the summary of the region 3 results we have already done a circular tube complete more or less okay and parallel plate is much easier than this because you will not have any vessel equation you will have straightforward ordinary differential equation for which you can find sign and cosine functions then you can see the order in which the nusselt number decreases so it is the highest for the parallel plate case and compared to the constant wall temperature constant wall flux has always a higher value of nusselt number then comes your circular cross section okay your slug flow always has the highest nusselt number rather than parabolic flow and your triangular cross section or non-circular cross sections will have lower value of nusselt numbers okay so this is to summarize all the results so we will stop here and tomorrow and on Saturday the last two classes we look at the approximate approximate methods there is integral method to solving the thermally developing region.