 This is a recap of what we did in the last class on Thursday for those of you who could not make it in the morning class we were looking at the case of constant wall temperature boundary condition and when you apply earlier when we did the case of constant wall flux okay so there we came to the conclusion that your DTM by DX is equal to DT wall by DX and these two terms got cancelled and your DT by DX is equal to DT wall by DX so all the three can be equal only if they are equal to a constant and therefore the variation of the temperature at any radial location as well as your wall temperature and mean temperature they have to be parallel to each other and they have to be a linear line okay when it comes to the case of constant wall temperature so this comes to the fact that DT wall by DX is a constant and therefore these two terms will get nullified so you have DT by DX equal to T minus T wall by TM minus T wall into DTM by DX so therefore your DT by DX now related to your DTM by DX now coming to the energy equation so we will substitute this whatever conclusion that we made so we will instead of DT by DX we write that in terms of T minus T wall by T mean minus T wall into DTM by DX okay also since we have already calculated the velocity profile which is a parabolic profile we have substituted the parabolic profile into this expression now what is the problem with this unlike the case of constant wall flux boundary condition there your DTM by DX was a constant so therefore so this side this was not there you had only DTM by DX this side it was T as a function of R so you could directly integrate this this was this was integrable straight away and you can find the temperature profile with respect to R and apply the boundary condition okay so that was a very straightforward way now if you look at it this DTM by DX is not a constant okay so this is now changing and we do not know how exactly it is changing strictly speaking I I we last time derived how the profile should vary if you plot your profile as a function of X the wall temperature is a constant however your mean temperature will vary such that the difference between the mean temperature and wall temperature X exponentially decaying function so at X going to infinity the difference goes to 0 correct so so therefore this is the way that the mean temperature varies okay so having known this you have again on the right hand side T which is a function of both X and R so you look at this equation now it is a partial differential equation and there is a chance that you can convert this into an OD provided you can express your DTM by DX as a function of DT by DR okay so then you can also substitute for T-T wall by Tm-T wall in terms of ? and now since D ? by DX is 0 so ? is only a function of R so then you can convert this PDE into a nice OD so this is how we are doing it so from energy balance my DTM by DR can be related in terms of the wall heat flux as DT by DR at R equal to 0 and when I substitute that into the earlier energy equation so now I have an equation where I can use ? to denote T-T wall by Tm-T wall okay so therefore I can non-dimensionalize my radial coordinate also I can introduce a non-dimensional radial coordinate which is R by R0 which is which is R0 is the radius of the duct so final resulting expression if I substitute that will come out in the form that is presented here this is nothing but an OD which can be solved by shooting method once again iterative so you look at the boundary conditions so ? at R star equal to 1 that is corresponding to R equal to R0 so that should be 0 because there your T will be T wall and at R star equal to 0 you should make use of the fact that the profile is symmetric and therefore the slope has to be 0 so these are the two boundary conditions this is a second-order ordinary differential equation so we can solve the OD with the two boundary conditions so how do we do this once again we can make use of the shooting technique to get the to solve this equation and also to get the value of Nusselt number okay I will just briefly explain how you will be doing this it is very similar to the earlier shooting technique problems only thing see now the coordinate system what you have been working with starts with the center so what I would like to do is when you solve this with shooting technique you can start working from the coordinate system which starts from the wall okay so why we can do that is that at the wall you do not know the gradient okay also you do not the gradient is related directly to your Nusselt number so in this equation you do not know what is the Nusselt number also you do not know what is the gradient at R star equal to 1 correct so both are not known so in order to simplify this problem if you start working directly with a coordinate system starting from the wall and marching to the center you can directly guess the value of Nu and therefore that will give you a guess for directly the slope at the wall right so you understood the problem so right now if you are marching from the center okay you need a boundary condition for Phi at R star equal to 0 so you need to guess a value of Phi at R star equal to 0 and again you need to guess the value of Nusselt number both are unknowns that will involve two guesses and that will become little tedious whereas if you start from a coordinate system from the wall and proceed to the center you can you should need a guess for the slope at the wall and that is directly related to a Nusselt number so both in one shot you get it okay so therefore I am going to shift the coordinate system from the center to the wall okay so I will have a coordinate system right now I have an R star non-dimensional coordinate system ranging from 0 to 1 I will shift that to R prime how do I shift it 1 minus R star so that when R star is equal to 0 R prime equal to 1 when R star equal to 1 R prime will be 0 so then I can substitute that into let me call this as equation number 1 substitute into equation number 1 and I can rewrite in terms of R prime okay that will come as 1 minus R prime D by DR 1 minus R prime D Phi by DR prime of course your DR prime will be minus DR star right so that should be equal to minus twice Nusselt number into Phi into 1 minus 1 minus 1 minus R prime the whole square instead of R star square I have 1 minus R prime the whole square so this is how my ODE will be and the boundary conditions will be Phi at R prime equal to 0 will be what so now I should also make the boundary condition transform to R prime instead of R star this should be 0 and D Phi by DR prime at R prime equal to 1 should be 0 okay so now I can here your Nusselt number is defined as twice D Phi by DR prime at R prime equal to 0 this is how your Nusselt number is defined correct so earlier it was defined as in this case your Nusselt number was defined as minus twice D Phi by DR star at R star equal to 1 now when you substitute for D for DR star as minus DR prime so that will become Nusselt number will be twice D Phi by DR prime at R prime equal to 0 so that is starting from the top from the wall and towards the center so now this equation is easy to be solved by the shooting method so we will call this as 2 so so by shooting method first I should reduce before for applying shooting technique I should reduce the second order ODE into 2 first order ODE so by introducing the fact that my D Phi by DR prime is equal to ? this is one of the this is one of the ODE's the therefore if you substitute into equation number 2 so you should be getting D by DR this is all your DR prime into 1 – R prime into ? should be equal to – NU Phi into 1 – R prime into 1 – 1 – R prime the whole square okay so this is equation number 3 this is equation number 4 so now you have reduced equation 2 to 2 first order ODE is one you have to solve for Phi the other you have to solve for ? so the boundary condition for this so Phi at R prime equal to 0 so that basically 0 so that is sufficient for solving this now for solving this you need ? which is nothing but ? at R prime equal to 0 which is D Phi by DR prime at R prime equal to 0 this is not known this is nothing but nusseld by 2 therefore now you see you do not know the nusseld number anyway so you can guess a value of nusseld number and that is the guess for ? at R prime equal to 0 but what do you know is basically ? at R prime equal to 1 okay so now this becomes an iterative process again so you guess a value of nusseld number therefore you guess a value for ? at R prime equal to 0 you keep marching by the shooting technique you shoot in the March and you should make sure that your ? at R prime equal to 0 is equal to 0 so that satisfies the other boundary condition so you have to do this iteratively you keep guessing the value of nusseld number you change the value of nusseld number until you satisfy this condition so finally you end up directly getting the correct nusseld number that is it okay so once again I will just write down the procedure so solve 3 and 4 by shooting the step 1 is to guess any and therefore hence your ? at R prime equal to 0 is calculated as any by 2 that is the first step number 2 you solve 3 and 4 by say Euler method okay so you can take a large value of so you go you start from R prime equal to 0 up to R prime equal to 1 okay so unlike your external flows where your ? was going till say 10 or 15 now your range is defined so your R prime equal to 0 to 1 and then check if whatever from your solution if your ? at R prime equal to 1 is equal to 0 if that is true then whatever you guessed is correct that directly gives your any if not you have to iterate till step 3 is satisfied now again for iteration we make use of an intelligent guess using the Newton Raphson method so here the Newton Raphson algorithm will be ? you need to basically guess the value of ? at R prime equal to 0 therefore ? at R prime equal to 0 at K plus 2 so you first guess once you guess twice and then for the third guess you use the Newton Raphson method because again it needs at least two guesses to calculate the difference so this will be equal to ? at R prime equal to 0 so K plus 1th guess plus so this will be f of x divided by f prime of x now f of x is a condition that ? at R prime equal to 0 should be equal to 0 so anyway that so this minus 0 should be 0 or this this is equal to 0 right sorry R prime equal to 1 so in this case this comes out as 0 therefore this term should be I think this entire term then should come out as 0 here because your f of x should be anyway 0 right where f of x is anyway 0 so ? at R star equal to 1 should be equal to 0 so therefore so this numerator completely so I think using the Newton's method here may not be that beneficial because anyway so this this does not this term is absent here right is that right because you have your ? at R star equal to 1 this is your function which it has to be satisfied this is equal to 0 your f of x is basically 0 okay so therefore you have to guess your nusselt number so you can guess it iteratively okay trial and error yeah okay but this is actually the function that it has to right so the function it has to satisfy is f of x is equal to 0 so this is basically on the right this this is plus f of x by f prime of x so this we are writing as a difference but your f of x itself is 0 here okay so if you if you had a condition that this is equal to 1 then you can use it this minus 1 equal to 0 is your f of x they are similar to the external boundary layer correct so here we have directly your f of x is 0 so there is no point in using this particular method will not give you lead you anywhere yeah but this is what you have to satisfy so you have to say that your f of when you say that this is the equation it has to satisfy this has to anyway you have to force it as 0 here right you can just try and check but I am myself I am thinking whether this will work out for this kind of a problem this boundary condition because see when you had the condition clearly that say for example ? equal to 1 so you make sure that it has to satisfy ? minus 1 equal to 0 so that is a function there which is to which it has to satisfy whereas here ? is equal to 0 directly okay so anyway you can try your whether you can apply your Newton's method and see but in my opinion this may not lead to the correct solution using this so you can just do a wild guess anyway I will give you the solution you can start with some guess close to the solution and you will find that your nusselt number for this case comes out to be 3.658 so it is a constant value once again as I said if you have a fully developed thermally and hydrodynamically fully developed case that is in region 3 you have a constant value whether it isothermal or isoflux case for isoflux case what do you remember the value 4.36 so for the isothermal case if you do a trial and error you will finally see that at nusselt number 3.658 this exactly satisfies this boundary condition ? at r star at r prime equal to 1 will be 0 okay so therefore if you compare this to the isoflux case you see the isoflux case has a higher nusselt number okay so there has been no concrete reason why people observe this experimentally also that this is higher but some kind of an explanation can be given from the thermal boundary layer now when you talk about the fully developed thermally so you talk about a region where the thermal boundary layers and the velocity boundary layers have met so this is your region 3 okay so what could possibly happen if you apply a constant wall flux is that the wall temperature keeps rising gradually and due to that if you look at the property calculations in the case of external flows all the properties that we calculated were based on T mean which was T wall plus T 8 by 2 it is a linear average between the wall temperature and the free stream temperature in the case of internal flows we defined a T mean based on your mass weighted average for temperature u of r r dr so you have divided by integral u of r r dr this is 0 to r 0 to r so you have to calculate the properties based on this mean temperature and this temperature varies from the wall till the center line okay and if you have a case where your wall temperature keeps varying then therefore and this will consequently result if you look at the property calculation the properties will be keep on changing due to the wall temperature variation whereas if you have a wall temperature which is constant if you take this kind of an average it will not vary the way that it varies for the case of Q all equal to constant so this will result possibly due to due to this dependence of the thermal boundary layer on the Prandtl number when it when this to merge there will be a possible fluctuations in the thermal boundary layer okay so it will not be exactly located and it will not be exactly merging at the center but there will be some small fluctuations due to the property variation okay because these boundary layer thickness are functions of Prandtl number and locally the Prandtl number keeps changing due to difference different temperature values axial so they attribute that these kind of disturbances in the thermal boundary layer will result in possibly a higher diffusion heat diffusion in the case of constant heat flux resulting in a higher nusselt number okay so this is about what 16% higher than the constant wall temperature case so one possible reason this is a possible reason although you cannot add repeat it to a particular reason this is one possible reason what people say is that the variation in the properties due to varying much you know the variation properties are much more in the constant wall flux boundary condition than the wall temperature case and this will result in instabilities in the thermal boundary layer which will possibly drive the heat transfer to be much higher than in the case of constant wall temperature okay so this is one possible explanation although in the textbooks it is not even mentioned they just say generally that this is higher and they stop there there is no concrete reason why they should be higher but you should all remember that a constant heat flux case will generally result in a higher heat transfer coefficient than so this is true even for external boundary layers you observed it in external boundary layers the two boundary layers never merge they keep the boundary layer keeps on growing so there the fluctuations will be quite dominantly seen whereas here the two boundary layers merged and therefore you can only attribute some fluctuations towards the center line okay so this is something that you have to pay attention now the next thing what we will do is we will move from region 3 so in region 3 we looked at constant wall temperature and constant heat flux separately and we will now focus our attention to region 2 that is this region right here so where your hydrodynamically fully developed and the boundary layer still have not merged in terms of you know the thermal boundary layers have not merged or you cannot say that your d ? by dx is equal to 0 okay so you can only say du by dx is 0 the velocity profiles are fully developed but you cannot say the temperature profiles are fully developed so we for this particular region is a more interesting region we will focus the rest of the classes another three or four classes towards the region to so any questions on this so far so I think I will give an assignment where you will be doing the shooting method you can you can try out with an iterative guess you can start with 1 2 3 4 and you will find that progressively this will be satisfying so once you reach you know close to the solution you can take the guess values much closer you know 4.1 4.2 then you should see that it should converge okay so we move on to the thermal entry length region or the region 2 which I denote it where your hydrodynamically fully developed but thermally it is not developed yet so for this case the first solutions were given by grades so he did the first solutions for the region 2 as early as 1835 not 1835 it is 1883 and 1885 okay so he produced the first solutions in fact some of his papers are still available but you have to go to a library and get it if you have the corresponding journal but whatever grades did he did it for only one case for a very simplistic assumption that your velocity is completely uniform that is a slug flow or plug flow case okay so later on people extended the grades flow solution to other generic cases where you can have a parabolic velocity distribution and you can have other boundary conditions such as constant wall flux and so on in fact there is a person called sellers have also uploaded a document on Moodle yesterday so you can just have a look at it where the great solution was extended by this person this group of people sellers is one of them is the first author so there is a document where they have proposed the solutions to parabolic velocity and also to cases where you have linear variation of wall temperature wall flux is equal to constant so there are further extensions of these first original great solution so this is popularly called as the great's problem also and I will just list down what are the four assumptions that he made the first assumption that he made is that your radial velocity is 0 everywhere the second assumption is the ratio of K by OCP into you is a constant now this is a big assumption okay when you say K by rho CP this is your alpha so this you can understand that as a property could be constant but when you say also you is a constant that means is assuming everywhere plug flow okay so this is for a case where you have only plug flow and therefore you can assume that this is your slug flow or a rod like flow he calls it okay he calls it as a rod like flow so he assumes as it is like a solid rod which is passing through a circular tube okay so where you have a velocity which is a constant and properties are also constant and at the starting point where you where you are looking at the thermal boundary layer okay so there the inlet temperature is a constant so the region that you are looking at now is where your velocity boundary layers merged and then you start the thermal problem okay so suppose you start your thermal problem somewhere here okay so strictly speaking you have to start it somewhere after the velocity boundary layers have merged okay so from here you look at the thermal problem where the two thermal boundary layers grow and then merge okay so this is completely your region 2 so now this is region 3 so you can now see that if you solve region 2 where you start from some initial temperature here correct from where the two boundary layers start growing thermally already the velocity boundary layers have grown and they have merged then you start the growth of the thermal boundary layers at the start inception of the growth of the thermal boundary layer your temperature is equal to the inlet temperature so from there they grow so you are developing solution for this region down thermally developing region and asymptotically once the two boundary layers merge you should asymptotically reach the solution that you had already developed for region 3 correct so that is an asymptotic solution of the solution to the thermally developing region so we will see that when we develop the final expression for asymptotic case you will directly find that we will be reaching these two limiting values okay for region 3 okay therefore we could not be strictly speaking we need not look at region 3 separately we could have directly looked at region 2 and said that for the limiting case of x by d going to infinity you directly reach your fully developed solution okay so this is your assumption that at the inlet everywhere inlet where you start the growth of the thermal boundary layer your inlet temperature is uniform it is constant and so this is for the case if at x is equal to 0 so now your coordinate starts from the point where you start the thermal boundary layer growth okay so this is your coordinate system so it does not start from the inlet of the pipe it starts from the point where your thermal boundary layer starts growing and where your fully developed velocity profiles are present okay and T equal to T wall at R equal to R not if your x is greater than 0 so he has looked at a constant wall temperature boundary condition okay so at x greater than 0 the wall temperature is applied less than that you do not have any thermal condition okay and the other assumption he makes is that your thermal conductivity in the axial direction is 0 or you can say you neglect your axial conduction in comparison to the radial conduction okay so then we can write down the energy equation for this case so now if you say the energy equation is U DT by DX is equal to I am expanding originally you have 1 by R D by DR of R DT by DR I can expand it I can say a is common so I can differentiate with respect to R keeping DT by DR constant okay so that will be 1 by R DT by DR plus I can differentiate DT by DR keeping R constant so that will be D square T by DR square so now this is your energy equation so you are neglecting your axial conduction term with respect to radial conduction you have neglected your radial velocities your radial velocity is 0 so only your axial velocity is there so these are all the assumption based on the great's problem so now you can assume a non-dimensional temperature ? like the way that you assume before a T- T wall now rather than using T mean here I would like to use TI okay so why I am going to use here is that in your earlier case you somehow manipulated such that your DT by DX was a constant okay so therefore you do not have to apply any boundary condition corresponding to X is equal to 0 in this case this is a partial differential equation so for this even you need a boundary condition for T at X is equal to X equal to 0 so therefore at X equal to 0 your T equal to TI so to do that you can non-dimensionalize with respect to TI such that at X equal to 0 so ? at X equal to 0 will be 1 okay and ? at R equal to 1 will be 0 that will be T wall which is a constant so you need two boundary conditions with respect to radial direction so what is the other boundary condition should be finite or D ? by DR should be 0 okay so substituting this we can rewrite this as I can say my U equal to U M which is a constant based on the plug flow assumption so this can be written as U M D ? by DX is equal to K by ? CP or I can just I can just leave it as a here D2 ? by DR2 plus 1 by R D ? by DR so this is my non-dimensional temperature and these are the boundary conditions so now in the case of thermally fully developed case depending on whether it is a uniform wall flux or a uniform temperature I can solve this directly I can convert this into an ordinary differential equation and I could have solved it but now in this case my ? is a function of both X and R and I cannot put the condition that D ? by DX equal to 0 so therefore this is a partial differential equation and have to solve the PD as it is so how do I solve the PD I think most of you have done the solution of PD's before what is the simplest technique to solve PD's huh separation of variables okay so how many of you have done separation of variables how many of you have done PD solution by separation of variable so how about the other M tech students I think at least in the heat transfer advanced heat and mass transfer it should be taught to you I think so I am not going to spend too much time explaining the separation of variables but you should understand that you assume now this is a linear equation once you have your velocity so any linear equation now you can assume the solution for ? in this case which is a function of X and R you can split it into two solutions you can assume that this is a product of two solutions one which is only a function of X and the other which is only a function of R this is the starting point of any problem for separation of variables so then this this is the assumed solution you put it into the PD let me call this PD as number 1 so substitute the assumed solution into 1 what do you get so this will be when you differentiate with respect to X R will be held constant so this will be U M R x DX by DX on the right hand side you have when you differentiate with respect to R your X will become constant here okay so you say X D2R by DR2 plus 1 by R DR by DR so now I divide both sides by X into R X into R so this will be U M by ? bring it here 1 by X you have D now now this since X is a function of only X all this partial differentials will get converted into ordinary differentials my X is a function of only X and R is a function of only R so all these partial differentials here should be written in terms of the normal differential so this will be DX by DX on this side it will be 1 by R D2R by DR2 plus 1 by small R DR by DR so on this side I have everything as a function of X on this side I have everything as a function of R so on these two have to be equal so this can be equal only if they are equal to a constant so I assume that this constant is negative minus ?2 so why I put this is that if you look at the solution for X then it will be an exponentially decaying function if I put a positive constant there it will be an exponentially increasing function and what what is this X standing for here that is basically variation of ? with respect to X okay so if you look at the variation of ? you can see that as you start from this point your T will be equal to Ti okay and that that will be one so your ? will be one so if you plot your variation of ? with respect to X so it starts from some value 1 and from there it has to increase or decrease it has to decrease till it reaches the value of T wall where it becomes 0 so that has to happen exponentially right so this can be possible only if your constant is negative here so this has to be an exponentially decaying function only then your ? will behave in that way okay so this ?2 what I am using these are called as eigenvalues okay this is the principle solution to the problem these are called as eigenvalues so therefore now I have two ordinary differential equations so one is DX by DX plus a by UM I have ?2 X equal to 0 this I call as equation number 2 and D2 R by DR2 plus 1 by R DR by plus ?2 R equal to 0 so I can equate this separately to the constant this separately to the constant I have two equations to OD so ultimately what I have done I have converted the partial differential equation into two OD now OD is I can always solve them you know I can at least use a simple numerical technique to solve the OD now what is the solution to the OD number 2 it is an exponential solution okay so it will be X of X should be some constant into E power minus ?2 by ? ?2 by UM so this is my solution with respect to X okay the other solution into X sorry okay the other solution is a solution to this OD so how do I how do I solve this OD now that is the question anybody can recognize what kind of an equation is this I think in heat transfer I have taught so this is called as a Bessel equation we so if for people who do not know what Bessel equations are I will probably give a brief overview tomorrow about Bessel equation and the solution general solution to the Bessel equation okay generally for problems in cylindrical coordinate systems okay when you try to do separation of variables you end up converting that into a Bessel equation and what are the boundary conditions to solve this so how many boundary conditions you need for this how many boundary conditions for this for X 1 this is the first order now this is a second order OD you need two boundary conditions for R so what is the boundary condition for X X at X is equal to 0 should be do we have any concrete boundary condition for X now we do not have we have only boundary condition for ? so we know that ? at X is equal to 0 should be 1 so you can write this as X into R X at X is equal to 0 and R should be 1 but from there we cannot deduce anything for X so we will hold on and apply this boundary condition in the end okay so now if you look at this problem we can apply the boundary conditions directly because they are homogeneous boundary condition okay so already we know that your ? equal to 1 at R equal to 1 ? equal to 0 and it should be finite at R equal to 0 so correspondingly if you write ? as R x X you can say that R at R equal to 1 should be 0 and your DR by DR at R equal to 0 should be 0 or your solution for R should be finite okay so this is a ODE with two homogeneous boundary conditions and this becomes what is called the Eigen function problem so how do you identify which direction is the Eigen function problem you identify the direction where you have two homogeneous boundary conditions okay and that gives you the direction where you find the Eigen function problem why do you need the Eigen function problem because you apply the boundary conditions you find the roots of the Eigen values so that is why this is this equation is the Eigen function equation so once you know the Eigen function Eigen values then the R solution to R is the Eigen function okay so and then you can express your solution as a product of your Eigen function x okay so this we will I will explain give you a brief introduction to Bessel functions and then I will give you the solution and then I will show you how to combine these two solutions okay so any questions on this so far so for people who have not had any course on partial differential I suggest you can you try to learn up at least the separation of variables this is a very fundamental thing the rest of the classes will be working only with the separation of variables.