 So today we will take up a slightly different problem until yesterday we were doing looking at the fundamental grades problem the fundamental grades problem simply looks at only a slug velocity profile and then you look at the thermally developing region and calculate the solution for the temperature as well as for the Nusselt number and finally for the asymptotic case where your X goes to large values you can recover your Nusselt number for the thermally fully developed region okay. So the grades profile originally which was which was done the great solution done by grades himself was a simplistic case and it was actually 1954 when three people you know cellars and tribus I think I have uploaded the solution from report original report from 1954 on the Moodle you can just have a look at it so these three people cellars et al have extended the original grades problem to other cases so the most obvious case will be to look at a parabolic profile okay rather than a plug flow which was a simplistic case and do the same solution with a constant wall temperature boundary condition okay so that is the first extension they did and also they extended it to other boundary conditions like uniform wall flux and also case where the wall temperature is linearly varying okay so these are some extensions of grades problem so we will try to do a couple of these extensions the first extension that we will do today is for the parabolic velocity profile so I will call this as great solution extended for parabolic velocity profile now obviously when the great solution was extended the original great solution for a plug flow was having Bessel functions as the Eigen functions and once the extension was done it was observed that the solution was not as simple as just getting a Bessel function okay so we will look at the nature of the Eigen function how it looks and these Eigen functions have to be estimated in some way or the other that's when sellers did it they did not have access to computers 1954 so they try to do a kind of no mix and match approach so they split the problem solution into three parts one for small values of R one for middle wall values of R one for large values of R they got three different asymptotic cases and they had patched up the solution okay but nowadays you can solve the ordinary differential equation numerically to get the Eigen functions okay so anyway so the thing is this the solution to the Eigen functions are not as straight forward as what we saw in the original greats problem so now we know the energy equation so we will write down the energy equation still remains the same that is no change 1 by R D by DR R DT DR okay so this is the form of the energy equation will be still working will neglect the axial conduction with respect to the radial conduction and still you have the convection term due to the axial velocity so we assume that there are no radial velocity and no variation in the ? direction now the question is earlier we just assumed you to be a constant and just substituted as some U M now we have to put the actual parabolic velocity profile coming from the fully developed hydrodynamically fully developed case to do that first we you can also do this in a dimensional form like what we did for the greats problem you can substitute for you in terms of U M and leave everything in a dimensional form and then you can proceed to solve solution but it will also be a little bit simpler to do it in non-dimensional form so we are going to introduce some non-dimensional variables in the beginning and then substitute those non-dimensional variables so I will introduce for temperature ? like the way I introduced before it is T – T wall by T inlet – T wall so that at X is equal to 0 you have ? equal to 1 and it and R equal to R 0 your ? will be 0 okay and at R equal to 0 your ? should be finite so these are the boundary conditions for ? define this way and we will also introduce a non-dimensional radial coordinate we will use the symbol ? so that will be R by R 0 that is the radius of the duct okay and we will introduce another non-dimensional coordinate for the axial location or axial coordinate so we will use the symbol ? for that and we will define this as X by R 0 by Peclet number so this is the non-dimensional axial coordinate this is the non-dimensional radial coordinate ? okay the way that we have defined this X by R 0 by Peclet number okay this is this is convenient because once you get the solution for variation in the X you saw that in the greats problem you had a exponential minus you had something like X by D 0 by Peclet number right so in order to accommodate this variation and group this together in one non-dimensional form it is convenient to define this entire group as so this group appears to gather okay you do not find X separately D separately and Peclet number they all come as one single group for variation along X so therefore it is wise to define a non-dimensional variable ? grouping this together okay now there is another non-dimensional number named after grades okay it is called the grades number and actually one over grades number is defined as basically X by D by Peclet number so this is basically how the grades number is defined so this this particular non-dimensional grouping of X by D by Peclet number is referred to as grades number inverse okay so it basically tells you about the non-dimensional axial variation is represented by this non-dimensional number called the grades number inverse and therefore you can see that if I define X by R not by Pe as one non-dimensional group so this is nothing but a function of grades number okay so this comes out as grades number by D by 2 exactly okay so this is a this is a one form of convenient form of grouping okay so we will stick to this particular non-dimensional formulation for ? ? and ? and therefore we will substitute this and we will see how the governing equation reduces to we will also use the parabolic velocity profile so basically that is U by U M is equal to twice 1 – R by R not the whole square okay so in terms of non-dimensional radial coordinate that is 1 – ? square okay so X from here this will be already you know Peclet number is what this is ? U M or U M D by ? okay so D or D not you can use this is the diameter of the duct so therefore we can express X in terms of ? so this is U M D not by ? into ? okay so you have into R not which I will write it as D not by 2 so you have a term something like this okay or I can express everything in terms of R so I can write this as 4 R not square so this becomes 2 R not square okay so this is the transformation I have to do from I have to substitute for X from this expression in so it becomes in terms of ? and for R I can use this transformation so that I can substitute in terms of ? okay so if you do that so you will be ending up with for you you have already 2 U M into 1 – ? square if you substitute the non-dimensional variables let me call this as number equation number 1 the energy equation so divided by ? into DT by DX so T in terms of ? that is basically TI – T wall D ? by now once again X we can write in terms of ? from here okay so this will be D ? and you have the other terms you have to U M R not square and there is an A coming in the numerator okay so that should be equal to so when you substitute for R in terms of ? so the R not R not cancels between these two and then you have an R not square coming here okay so you have 1 by R not square okay D by D ? into ? D so I substitute once again in terms of ? so this becomes D ? by D ? and then I have TI – T Y okay the reason why I use this ? here because my inlet temperature is a constant wall temperature is a constant okay so it is not a function of either X or ? so for the case of constant wall temperature boundary condition if you non-dimensional is this way so my T TI – T will come out of all these derivatives and get cancelled on both sides okay so therefore this cancels of you have R not square which cancels of here you have a which cancels of and then to U M cancels of right away so finally you are left with a expression 1 – ? square into D ? by D ? which is equal to whatever is there on the right hand side okay 1 by okay I have left 1 by R so that should be 1 by ? here 1 by ? D by D ? ? D ? by okay let me call this as equation number 2 so therefore this is the non-dimensional form of the energy equation after I substitute for the fully developed velocity profile so this has become a little more compact than when I work with the dimensional form right I have so many terms in terms of you I have U M I have a all those things appearing now all those things are eliminated okay so this is a much so I always recommend that you can if you like to work with a more compact form you can define non-dimensional variables in the beginning and then you can use it for the constant wall flux boundary condition however you cannot define a non-dimensional ? like this because there your wall temperature will be varying and therefore you cannot define so you have to just say T – Ti or something like that okay so you have to be careful for the case of constant wall temperature condition you can define a non-dimensional ? for a case of constant wall flux you cannot do that okay so but still you can for that case also you can define your non-dimensional ? and ? okay that will save some put some save some effort in removing all this U M and variation ? and all those terms okay so now the boundary conditions for this in terms of ? to solve this how many boundary conditions we need 3 so in direction in ? direction how many in ? and how many one in ? and 2 in ? direction right so therefore ? now at any value of ? corresponding to ? equal to 0 that is basically saying that T at x is equal to 0 so the non-dimensional ? will be what will happen at if you look at the one so if you look at the problem configuration so this is your region which you are looking now so this is your ? T this is your ? okay and in fact what you are doing right here it is not something exactly like this but you are starting from a point where it is fully developed that is your coordinate so you can consider that your thermally fully developed region meet somewhere further down okay and this is your coordinate right here this is your X this is your R so once the velocity profile is fully developed then you start looking at the thermal entry length problem okay so you start basically heating from here or maintaining a isothermal condition so in this particular case at X is equal to 0 this is where you are calling this as T I the temperature which is entering the thermally fully developed region okay so at X is equal to 0 ? should be 1 because T you should be T I and the other boundary conditions with respect to ? so at ? equal to 0 your d ? by d ? should be 0 or ? should be finite both are equivalent boundary conditions and the remaining boundary condition at R equal to R 0 so that corresponds to ? equal to what any value of ? at ? equal to 1 what is the boundary condition 0 okay so therefore now you have a partial differential equation like before for the grades problem only thing you can see earlier you did not have this 1- ? square that was a constant plug flow case now you have this extra term and the other boundary conditions are all same okay there is no difference so now we have to solve this PD so the same method that we use separation of variables we will do it here okay so how do we start we say ? ? ? and ? we will break up into two parts 1 X which is a function of ? and R which is a function of ? so the actual solution is product of two solutions one which is so these are two independent solution okay one function of only X the other function of R or ? all right so therefore if you substitute this into your equation number 2 okay assume that this is your solution and substitute so you end up getting 1- ? square into DX by D ? and your R is constant on the other side you have 1 by ? you have X constant D by D by D ? ? into DR by D ? okay so I divide both sides by X into R so this becomes 1- ? square by X into I will also take 1- ? square on the right hand side so this becomes 1 by X DX by D ? this is equal to 1 by R into 1- ? square into I have already a ? here into D by D ? into ? DR by D okay so now I can see that this is a function of ? this is a function of ? so these both have to be equal and therefore they have to be equal to a constant which is always negative okay so in order to have an exponentially decaying function along ? this has to be a negative constant okay and this is the eigen value so now I can write this into form of 2 OD is one where I can say DX by D ? plus ? square X equal to 0 I will call this as number 3 and the second OD I will expand this term right here if I expand it I can write this as D square R by D ? square plus 1 by ? DR by D ? plus you have ? square and this entire term I am taking it to the other side so this is 1- ? square into R this is equal to 0 I will call this as my so equation for so this is the first OD with respect to X second OD with respect to R so I just expanded this term right here so this is ? into D square R by D ? square plus DR by D ? into 1 okay so that is what I did and I already have ? at the bottom so I divide everywhere by ? and so you have this term ? square into R into 1- ? square okay so therefore I use the same boundary conditions there and with respect to R now you can see that you have two homogeneous boundary conditions because with respect to ? you can substitute as R into X okay since X has to be finite you know it cannot be trivial therefore the boundary condition should have should apply to R right so whatever you say here so your D ? by DX so this will be D by D ? into you have X of ? into R of ? so this should be equal to 0 at anyway so I can write this as ? equal to 0 okay so this can be 0 only if I can say D DX by D ? if you X is equal to 0 also this can become trivial but the problem is if X equal to 0 there can be no solution okay therefore this implies that your D by D ? at R D by D ? R at ? equal to 0 this can only be 0 okay right because this can be 0 only if either of this is 0 if X is 0 that means my entire solution becomes trivial okay whereas R at ? equal to 0 can be 0 that is the only other option okay so therefore this is the boundary condition which falls on R the other boundary condition that is ? at ? ? ? equal to 1 this is equal to 0 the same way so once again I can say my X of ? into R of ? equal to 1 is equal to 0 so once again X cannot be 0 so therefore this indicates my R ? equal to 1 should be 0 so therefore these are the two boundary conditions with respect to so this will be anyway normal derivative not the partial one because R is a function of only ? so to solve the eigenvalue problem now you can see that eigenvalue problem has two homogeneous boundary conditions with respect to ? okay so therefore equation number 4 is the eigenvalue problem which I have to solve and apply the boundary condition now if you look at equation number 4 what kind of an equation is that this is where you have to be a little bit careful you look at the structure of the Bessel equation I have given it is very deceptive but if you look at it a little bit more carefully if you multiply everywhere with ? square okay the first to the second order and the first order derivatives appear to be similar to the Bessel equation but the term the third term here will have ? square ? square into 1 minus ? square into R so that structure is not the same as the Bessel equation okay so the Bessel equation has minus some constant term minus some constant number minus ? square okay it does not have minus a function of X right so therefore this is not the actual Bessel equation so what we can call this as a general form of any eigenvalue problem which is called a Sturm-Lewel equation anybody has heard about this equation Sturm-Lewel problem Sturm anybody has taken partial differential equation course anything to do with differential equation how many have done only one so you must be taught Sturm-Lewel equation because that is the most fundamental part when you talk about separation of variables any general eigenvalue problem okay to for which you can find the eigenfunctions it is generally called as a Sturm-Lewel equation that includes even the Bessel equation okay so this is a generic name given to any eigenvalue problem that includes all kinds of equations all kinds of second order ODEs all kinds of Bessel equations whichever that you have encountered okay so that is a generic category of Sturm-Lewel problem so therefore we cannot pinpoint this to any particular familiar equation okay this is not a Bessel equation but it can be generally called as some Sturm-Lewel equation okay so and I will just give you the generic form of the Sturm-Lewel equation and yourself can see that it is generally of this particular form okay where these are the this is the weight function so any ordinary differential equation which can be cast into this particular form okay forms a Sturm-Lewel system of equation and this includes most of the all the eigenvalue problems okay even this equation right here we can just cast it in this particular form okay so for this case you can put it as – D by D ? ? D R by ? plus 0 equal to ? square into 1 – ? square into ? into R so you just expand and see so this is ? into D square R by D ? square okay plus you have D R by D ? okay so I am multiplying everywhere by ? okay plus you have ? square 1 – ? square into ? right so I can write that equation in this particular form so this comes to a Sturm-Lewel kind of an equation so now if you compare the coefficients okay the coefficient here P of X is nothing but ? and Q of X is nothing but 0 and of course ? here is ? square okay and if you look at this weighing function this is a function of X here so that this has to be a function of ? so this entire thing is the weighing function or weight function so the property of any eigen function as I said yesterday it has to be orthogonal okay that includes even Bessel functions and any general eigen functions which are of the Sturm-Lewel type have the orthogonal property and for the Sturm-Lewel kind of equation set the orthogonal property will be you have integral a to b y of n into x y of m into x into w of x dx okay is equal to ? m so this is the orthogonal orthogonality okay so what it says when m is not equal to n this Kronecker ? this will be 0 right then I mean m is not equal to n this will be 0 therefore they have to be orthogonal means if m equal to n then only this will be equal to 1 so in that case this will be y n square into the weight function dx integral should be equal to 1 if m is not equal to n this integral will be equal to 0 just like what we saw yesterday if you multiply Bessel function j n x into j m y okay if m is not equal to n so there will be 0 the integral product will be 0 okay so this is the principle of orthogonality for any Sturm-Lewel problem so here you have identified what is the weight function correct so in this particular case what should be the orthogonality condition so if I integrate what should be the limits of the integral 0 to 1 good and then instead of y I have my r n of ? into r m of ? and what is the weighing function here ? into 1- ? square into d ? this is nothing but your Kronecker ? this is a ? function I hope all of you know what a ? function is so it is equal to 1 at only one value everywhere else it is 0 so for the case where your m equal to n then this is equal to 1 otherwise it is 0 okay so therefore this is my orthogonality condition which I will use where should I use the orthogonality condition when I solve the ODE when I want to calculate one of the constants right so this I keep it now so therefore you would you see that the solution to the Sturm-Lewel problems can be represented as so you can say in this case that you can write this as summation of some CN times r n of ? where n can go from 0 to 8 so once you know the eigen function okay so you can write this solution for ? or whatever you call you can write this as r prime or whatever so your r of ? can be written as the eigen function times the constant summed over all the values of eigen value okay now the question is how to basically find the eigen values so for that we have to it is not as simple as solving the Bessel equation because now this is a general Sturm-Lewel problem and you have to solve the ordinary differential equation that is equation number 4 we have to solve this numerically okay you cannot this is not a generic this is not a particular form of any equation where you can get a ready made solution okay so any general Sturm-Lewel problem if it is familiar then you can get a solution straight away like the Bessel equation if it is unfamiliar then you have to apply these boundary conditions to boundary conditions and you have to solve for the eigen functions which are in terms of r okay so therefore this is a little bit rigorous procedure you know I am not asking you to do it over this sellers they have done it in a slightly different way they use some approximate methods to patch it up I do not think that is a very good technique right now because we have access to computers you can solve this equation straight away numerical now the question is for solving this you need the eigen value okay so that becomes like a constraint and the eigen value can be related to the derivative of r okay I am not going into the detail but there is something called as a Rayleigh coefficient okay for Sturm-Lewel problems so which relates your eigen value to the derivatives of r and therefore you can guess the value of lambda right just like your shooting method you guess your value of lambda and then you apply the boundary conditions okay and then keep marching until the other boundary condition is satisfied and then again you use that derivative and then check if your lambda is correct okay so this has to be done iteratively till you guess the right value of lambda okay so overall what I would like to say is that you have to solve equation for numerically to obtain the eigen values and the eigen function the eigen function is basically r variation of r with respect to ? so once you do that so now your final solution so can be expressed now directly as ? which is a function of ? and ? now what is the solution to this equation number 3 this is a straight forward ODE x equal to some constant C e power – lambda square x okay so that is a straight forward solution so therefore your final solution ? will be product of x and r okay so as I said you can have multiple values of eigen values and for each of this you will have a solution okay eigen function so therefore you have to linearly superpose all these multiple solutions so therefore you use a summation from n equal to 0 to infinity and the constants coming from this C and here you just club to club them together as one C into the eigen function coming out of that that is Rn of ? and the solution which is coming out of this this is basically ? here that is e power – ? square ? this is your form of your solution right the product of this and this okay so now the thing is finally so now you have to do it numerically to solve for the eigen value and the eigen function the remaining constant C can be solved by applying what yeah we you make use of the orthogonality property but we have to use the remaining condition for ? okay that is ? at ? equal to 0 which is 1 okay so that is nothing but summation n equal to 0 to infinity Cn Rn of ? and for ? equal to 0 that becomes 1 okay so now to get Cn out of this I make use of the orthogonality condition so I multiply both sides by the way weighing function times some Rm of ? okay both sides and integrate so from 0 to 1 so the weighing function is ? into 1 – ? square into some Rm of ? into d ? okay I multiply this from both sides so I have integral 0 to 1 summation ? equal to n equal to 0 to infinity ? into 1 – ? square I have Rm of ? Rn of ? into Cn so Cn will be here into d ? right so now you can see that based on the orthogonality property if m is not equal to n then this entire thing will be 0 okay only for m equal to n this will be 1 okay so therefore we can say this is integral 0 to 1 some there would not be any summation because for any value of n which is not equal to m this summation will be anyway leading to 0 so this will be Cn into ? into 1 – ? square into Rn square d ? so if I say 0 to 1 ? into 1 – ? square into Rn square of ? d ? will be actually – 1 by this is the actual value if you know the eigen function you have to take the derivative with respect to ? and this will be at ? equal to 1 and this is the value of this complete product and integrate okay so this is the property so this is also property of this term Louisville system of equations so therefore you can calculate your C so your Cn comes out as integral 0 to 1 ? into 1 – ? square into Rn of ? d ? divided by so the summation is gone you have integral 0 to 1 into ? into 1 – ? square into Rn square of ? d ? right for m equal to n so this becomes Rn square otherwise this is 0 so basically we will stop with the fact that you can calculate your Cn here I will give you the actual calculation of these integrals two of them I have to check whether this is with respect to ? or whether this respect to ? so let me just hold on for that in the next class so I will give you the expressions for evaluating these two integrals and then from there it is mere substitution okay so once you get your constant Cn therefore your solution so for ? is known once you know your eigenfunctions and your eigenvalues okay so these people sellers at all they have with some approximate method they have evaluated the values of eigenvalues and the corresponding constants I will give those tables also in the next class when we do the solution and from there you can asymptotically reach the case of large values of X and you can recover your earlier result for nusselt number that was some 3.6 for constant wall temperature with a parabolic profile so you can exactly reach that asymptotic value so in the next class on Tuesday we will complete this solution and so we will look at very briefly the extension of the grades problem to uniform wall flux boundary condition there I have posted already in the moodle a solution for channel flow where I assume a plug flow velocity distribution and constant wall flux boundary condition and I have posted the form of solution I will explain very briefly how to do it and you can look at that solution and you can very easily understand it for the case of constant wall flux boundary condition you have to tweak in such a way that the eigenvalue problem has homogeneous boundary conditions whereas if you have a constant wall flux boundary condition that is not homogeneous okay so therefore we play around with the solution in such a way you reach a homogeneous boundary condition and that is explained in the derivation which I posted on moodle and you can do a similar kind of a derivation for the duct flow so what I have done is for a channel very similar procedure for duct flow can be done so that will be a extension which will be useful to you and that we will go towards the integral solution the approximate solutions to internal flow problems okay it is very similar to your external flow integral method okay that will be the last topic and with that we will wrap the internal flows laminar internal flow okay.