 So good morning all of you so today we will look into some basics of Bessel functions because we need to understand the solution for Bessel functions before we do the solution to the Eigen value problem so I am not going to run a class on Bessel functions but just to prime you what should be the form of the Bessel equation and what is the solution to a given Bessel equation and some properties of Bessel function differentiation and integration okay now if you have any second order linear differential equation of this particular form okay the form that I have written here so you have x square d square y by dx square plus x dy by dx plus this coefficient now this coefficient is function of x here okay so times y equal to 0 so this functional form of this equation is referred to as the Bessel equation after the German mathematician Bessel so the solution to the Bessel equation is given as y of x is equal to some constant C1 times Bessel function so you use the Bessel function there are two kinds one is the Bessel function of the first kind Bessel function of the second kind okay the Bessel function of the first kind is represented by the letter j okay so j and the order of the Bessel function is denoted by this value of nu here this is any real number and you use the okay nu here to denote the order of the Bessel function this is a function of M times x where M is the value that you have got here so M times x plus C2 now the Bessel function of the second kind is represented by the letter y the order is represented by the subscript nu Mx okay so this is called the Bessel function of the first kind and of the order nu this is the Bessel function of the second kind and order nu okay so this is so far to say for a Bessel function now you can slightly write this Bessel function in a different way suppose if you are M square here was instead of being a positive value if you have a negative value here so you can replace this with a negative sign and you can write an equation like this okay suppose your M square value has to be a negative value in that case then you write like this then this becomes what is called as a modified Bessel equation it is a same structure as the Bessel equation only the coefficient term here will have a negative sign and correspondingly the solution to this here this is M so now this will become imaginary number okay once this is negative here so this will become M I basically an imaginary number so instead of writing everything in terms of imaginary functions okay so they have introduced what is called as a modified Bessel function okay which takes into account naturally the imaginary part of the coefficient so therefore for that you have a modified Bessel function solution so where you represent the solution as C1 times the modified Bessel function of the first kind is denoted by the letter capital I subscript nu for the order MX plus the modified Bessel function of the second kind is represented by the symbol letter K subscript nu MX okay so these are your modified Bessel functions of the first and second kind and of the order nu okay so as far as the solution to our Eigen value problem is concerned I think this is all sufficient to write how to represent the solution to the Eigen value problem okay so now we will compare this Bessel equation with the Eigen value problem that we have okay so what was what were our solutions basically when we substituted when we use the separation of variables into the energy equation we found out that the PDE can be reduced to two ODEs one is a first order ODE with respect to X so here we have assumed that the the grades problem has originally assumed that the profile velocity profile is a slug flow profile so it is a uniform velocity the value is um and the solution to this along the variation along the X is given by the first order ODE okay and we have also seen the solution to can be written as some constant times e power minus alpha lambda square by um into X right so directly you can integrate it out and the actual Eigen value problem in terms of R was this right and the boundary condition for R what were the boundary conditions to solve this second order ODE in terms of R R at R equal to 0 do you have any boundary condition at R equal to 0 okay finite R I would say d R by d R should be 0 okay or R at R equal to 0 should be finite both are equivalent the other condition R at R equal to 1 should be 0 okay because the temperature there at the wall is constant wall temperature so ? has to be 0 therefore R at R equal to 1 has to be 0 so now you compare this with your Bessel equation okay so the original Bessel equation not the modified one so you multiply throughout by R square so you have R here here you have again R square so now you see these and these coefficients are exactly identical instead of X you have R okay now as far as this coefficient is concerned you have M square X square Y okay instead of that you have ? square R square capital R and there is nothing like a new here the new is equal to 0 so therefore the order of the Bessel functions is 0th order okay so that is one thing and your M square is equal to ? square so therefore the solution to this OD will be in terms of the Bessel functions of the first kind and second kind 0th order to C 1 J 0 into ? R plus C 2 you have Y 0 into ? R okay so this is your solution to the eigenvalue problem so this is your eigenvalue problem so any eigenvalue problem should have two homogeneous boundary conditions okay so this is the solution to the eigenvalue problem so now we apply the two boundary conditions to get the two constants okay so now directly we come to the condition at R equal to 0 R should be finite capital R should be finite so I will also show you how these Bessel functions behave so if you draw the Bessel function J as a function of your X or R okay so it will start from 1 and then so this is your J 0 of R okay and J 1 will have a behavior like this and J 2 will have another behavior and so similarly if you draw the Bessel function of the second kind that is your Y as a function of R so this is negative this is positive here so the first the 0th order Bessel function of the second kind will at R equal to 0 will start from infinity and it will increase to positive value and it will be oscillatory again okay so this is your Y 0 of X and you will find that Y 1 will have another behavior like this Y 1 of R Y 0 of R so this is how your Bessel functions behave okay and your modified Bessel functions will have a different behavior so your I versus R so this is 0 this will always be positive here so this will be your I 0 of R and then this will be your I 1 of R similarly your Bessel modified Bessel function of the second kind that is your K function of R so this is your K 0 of R this is your K 1 of R okay this is to just give you an idea how the Bessel functions you take any Bessel function chart you will find the tabulated values corresponding to different values of X or R you will find the corresponding variation in the first kind and second kind all the different orders starting from the 0th order 1st order 2nd order the 0th order Bessel function here means it is the highest order okay does not mean it is the lower order it is the highest order and 1 2 3 4 there are the lower order Bessel function okay so now coming back to this particular eigenvalue problem so at R equal to 0 if your capital R has to be finite now you can see that at R equal to 0 my Bessel function of the second kind will be going to infinity right therefore in order to make my capital R finite C 2 has to be 0 okay so at R equal to 0 my Y 0 of 0 goes to infinity so this gives that C 2 has to be 0 for the solution to be finite so therefore my R of R directly reduces to C 1 J 0 ? R okay so this this is how my eigen eigenvalue problem reduces okay now again I can find out the constant C 1 by applying the other boundary condition so now the thing is I should also know what is the eigenvalue here because the eigenvalue is also undetermined so the remaining constant whatever is left out that is at R equal to 1 capital R is equal to 0 that will be used to find the eigenvalue okay so one of the constants is determined from the one boundary condition and the eigenvalue is determined from the other boundary condition so this constant is not a problem because final solution for ? I can multiply this and this into a single constant and I can use the remaining boundary condition that is at x is equal to 0 that is one more boundary condition which I have not utilized so totally there are three boundary conditions okay so that I can determine it later but right now I can use the second boundary condition for the eigenvalue problem and determine the value of ? here so therefore the condition that R R at R equal to 1 is equal to 0 so this should give me the fact that J 1 C 1 J 0 ? should be equal to 0 or in other words so the eigenvalues should satisfy this particular equation so that means you look at the Bessel function so wherever it becomes 0 so these are the solution which will give you the corresponding value of ? right so okay so this should be R is equal to sorry not 1 R 0 okay these are all dimensional so I should write in terms of J R 0 here please correct it so this is at R equal to R these are all dimensional radii so I should use that R equal to R 0 so therefore it should satisfy this particular equation so the corresponding value of ? R 0 will be the ones where J become 0 so you can see there are several values right there are several places where J become 0 so that this has now multiple solutions okay so I will just give you the first few roots where a where J become 0 so let me call ? N R 0 as some factor ? N okay so the 0s so these are the 0s of J 0 ? N equal to 0s of J 0 ? N okay so the first 6 0s correspond to ? 1 is equal to 2.4048 then ? 2 5.5207 ? 3 is 8.6537 ? 4 is 11.7915 and then ? 5 14.9309 and the last one ? 6 is 18.071 so these are the first 6 0s so the corresponding value of ? into R 0 okay so now I have determined my eigen values because these eigen values are all nothing but these values so this is ? 1 R 0 ? 2 R 0 ? 3 R 0 and so on and so forth you can have many number of solutions but I have given the first 6 these are the most relevant okay the other ones will be of the lower order which you can neglect all right so I will give you two more properties when you differentiate and integrate the Bessel functions which are important now so I will give you the derivative identities of the Bessel function so I will call this as derivative identities so the identity number 1 if you differentiate the 0th order Bessel function of the first kind that is your J 0 ? n R so that gives you minus ? n J 1 ? n R okay this is your first identity the second identity if you differentiate R into J 1 ? n R so if you multiply R with the Bessel function of the first order okay the first kind Bessel function of the first order you will be getting ? n R into Bessel function of the 0th order ? n R this is your second identity we will use all these identities so number 3 is the fact the Bessel functions have another important property that is the principle of orthogonality okay so if you look at any Eigen function Eigen value problem and you determine any Eigen function so in this case the Eigen functions are Bessel functions of the first kind of the 0th order these are the Eigen functions right so these are the corresponding Eigen function to the Eigen value problem okay so this is your Eigen value problem so the solution to R is in terms of Bessel function of the 0th order of the first kind so these are your Eigen function now any Eigen function any Eigen value problem where you determine your Eigen function should satisfy the principle of orthogonality okay we will see that that is very useful to determine the remaining constant okay now that principle of orthogonality for this case can be written as 0 to R 0 R J 0 square d R is equal to R 0 square by 2 J 1 O square okay so this is another important property when you integrate to multiply R with J 0 square ? n R so you get this particular expression okay so this is another thing which you should remember now the principle of orthogonality states that if the Eigen functions are orthogonal if you integrate them if you multiply an Eigen function J 0 ? n R with the Eigen function ? m R so n being different from m so these are different integers okay so your n may go from 1 2 3 and for a fixed value of m okay now if you multiply these two Eigen functions and you also multiply R okay and if you integrate it this will be equal to 0 if m is not equal to n this is the principle of orthogonality okay so only if m equal to n the Eigen value integer corresponds exactly same so then only you have some finite value non-trivial value so this will be equal to 0 to R 0 R into J 0 ? n R the whole square into d R if m equal to n so this is the principle of orthogonality basically so that means see these Eigen values are like principle directions like your XYZ Cartesian directions so these represent the variation of solution in those principle directions so if you if you multiply the solution corresponding to one principle direction to another so those two are mutually orthogonal directions so therefore the product will be 0 okay so whereas if you multiply in the same direction then this is where you get a non-trivial solution so this is where the principle of orthogonality plays importance and when you are integrating this 0 to R 0 R J 0 square we will use this particular identity here alright so now we will erase this we do not require is the principle of orthogonality clear I think if you take any course on linear algebra I think you will be taught this it is like saying your I ? I is equal to 1 your I ? J equal to 0 so these are mutually orthogonal directions so the Eigen values correspond to principle directions which are mutually orthogonal and the Eigen function should respect that orthogonality alright so now we have all the necessary background therefore you can write your solution for ? X ? R so you have to now tell me what will be the solution this is nothing but according to separation of variable and now we found the solution for both X and R so how can we write it so let us multiply C 1 and C and call this is another constant C okay some constant C or maybe a or whatever you want to call let me have used in my notes as a so I will use a and what is the remaining J 0 ? R C 1 J 0 ? R is your Eigen function okay and what is the remaining part X of X will be e power – ? ? square by Q M X the finally your solution reduces to this so in terms of X it is exponentially decaying function your ? starts at 1 at the inlet where the temperature equal to T I and then somewhere downstream your temperature should approach the wall temperature therefore the difference should become 0 T – T wall should become 0 so that is your exponentially decaying function with respect to X right now with respect to R it is a Bessel function variation now you see that when you found out the Eigen values you had multiple Eigen values right so you had first I had written down the first six Eigen values okay but there are multiple Eigen values and they are all solutions so then how do you include them so you assume that they are all linearly super post and you can denote your final solution as a summation of all those from n equal to 1 to infinity and you have ? n okay so this is your final solution so which is a linear superposition of all your solutions for different values of Eigen value for each value of Eigen value you have a particular Bessel function you have a particular exponential term so like that for each ? 1 ? 2 ? 3 you now get all the value sum them up that will give you the final value of ? of course the first few terms will be the significant term the first say three or four terms after that they will be insignificant and even if you do not include them it is not going to change the solution much okay so therefore this is your final solution and now the thing is how do we calculate the remaining constant a so this a also becomes a n so corresponding to each value of ? n you have particular value of these constant a n so now how do you calculate this constant now we use the remaining boundary condition for ? x is equal to 0 right so ? at x is equal to 0 at any R should be 1 now how do we apply this boundary condition here so therefore we say that summation n equal to 1 to infinity a n j 0 ? n R and this becomes x is equal to 0 so that will be 1 so this this should be equal to but still we have not determined what is a n because this is within the summation so now we make use of the orthogonality principle okay so what we will do is we will multiply both sides with j 0 ? m R where m is different from n okay and integrate them so we so for orthogonality condition in this case of Bessel function you should be having R into j 0 ? m R so we will multiply both sides with that so we have 0 to R 0 so my right hand side I am writing on the left hand side here I multiply with R j 0 ? m R dr that is my RHS right on the LHS I have this 0 to R 0 summation n equal to 1 to infinity a n into R into j 0 ? n R j 0 ? m R into dr okay so now of course it has to satisfy orthogonality condition here therefore for any values of m which is not equal to n this will be 0 and this will be equal to R j 0 ? n R the whole square dr for m equal to n so only there it will be a non-trivial solution so therefore this will become 0 to R 0 so for m equal to n so only then so you can imagine the summation here so now I sum from n equal to 1 to infinity for a given value of m if your n equal is not equal to m so those values are all 0 so therefore this summation will reduce to the fact that it will be equal to j 0 ? n R the whole square only for m equal to n so that should be a n into 0 to R 0 R j 0 square ? n R dr okay so now you understood so this reduces to this particular expression here so the summation is now gone because for m not equal to n so those values are all 0 so the final whatever remains is where your m equal to n so your summation is now removed and from here directly you can calculate the value of a n so this gives a n as 0 to R 0 j R 0 j ? n R dr divided by 0 to R 0 R j 0 square ? n R dr okay so now we have to simplify this further so that is where we make use of the derivative identities so numerator integral R j 0 ? n R so you integrate both sides so that should be equal to R j 1 ? n R right R j 1 ? n R divided by ? n and now you have integral R j 0 square ? n R so that is you use this identity number 3 that should be R 0 square by 2 now of course when you integrate it you have to apply the limits here this is between 0 to R 0 and when you integrate this from 0 to R 0 you directly get this particular expression R 0 square by 2 into j 1 square ? n R 0 okay so at 0 j 1 will be what 0 if you look at the curve so therefore this will be R 0 square R 0 into j 1 ? n R 0 divided by R 0 square by 2 so this j 1 cancels here R 0 cancels so this can be written as 2 here R 0 and j 1 ? n R 0 okay so this is your expression for a n final expression so therefore you can substitute this into the let me call this as 1 number 1 for a n you can substitute and then you can write the final expression for ? as so 2 can be taken out summation n equal to 1 to infinity 1 by ? n R 0 you have j 0 ? n R divided by j 1 ? n R 0 exponential minus ? by u m ? by u m ? n square x so this is your final solution okay what we can do is we can cast this into a completely non-dimensional representation okay so that you do not work in terms of x or ? but something like ? n R we can we have already used the notation ? okay and even will non-dimensional is this term so how I am going to non-dimensional is this the following way so I can write this minus ? by u m ? n square x as I can write this as minus ? by u m into x by R 0 square so I am multiplying and dividing by R 0 square so ? n R 0 the whole square okay and I am also going to rewrite this a little bit as so I can write this as x by D 0 into ? n R 0 the whole square divided by u m D 0 by ? that is a factor of 4 okay so this R 0 square I have replaced this by D 0 square by 4 okay so and I am grouping x by D 0 as one non-dimensional term and I am left with u m D 0 by ? so what is u m D 0 by ? Reynolds number times frontal number okay so therefore I can replace this entire thing as a non-dimensional group which is minus 4 x by D 0 ? n R 0 I have used the notation ? n okay because ? n as the units of what what is the unit of ? 1 by R so therefore ? n R 0 will be a non-dimensional group okay so that therefore I replace that with ? n square divided by RE into PR now I use another non-dimensional group called Peclet number which is the product of Reynolds number times the Prandtl number which is nothing but u m D 0 by ? okay so generally rather than all the time doing RE PR RE PR the people refer to that as a Peclet number so the product of RE times PR therefore in terms of the non-dimensional form you can describe your ? as twice n equal to 1 to infinity 1 by ? n J 0 now how do you write this ? n R so you divide and multiply by R 0 so then that will become ? n times R by R 0 okay divided by J 1 ? n okay into e power this entire thing can be written as minus 4 x by D 0 into ? n R 0 is nothing but ? n square divided by Peclet number okay so this is completely non-dimensional so once you know the particular value of Peclet number that you are solving so that directly has both Reynolds number and Prandtl number for different values of non-dimensional R by R 0 and different values of non-dimensional x by x by D 0 you can directly get the solution for ? so everything is now non-dimensional okay so you can plot the solution for ? as a variation with respect to R by R 0 and x by D 0 for a given value of Peclet number okay so once we got the solution now still we are not done so ultimately we need to find what is the Nusselt number expression for Nusselt number right so therefore for Nusselt number we need to do a little bit further work further quantities are required okay the first quantity now when you define Nusselt number here so your Nusselt number is basically H x D by K now you should be careful here that in the case of thermally developing flow your H is not a constant H is a function of the axial location so this should be strictly speaking H subscript X okay so now that you define this as K DT DR at R equal to R 0 into D by K into T wall minus T mean this is how you express your heat transfer coefficient wall flux divided by temperature difference so now I can substitute in terms of non dimensional ? so ? is defined as T minus T wall by TI minus T wall okay so I can write this numerator as now minus K into D ? by DR R equal to R 0 times TI minus T wall into D divided by K times now T wall minus T mean so I can define ? mean as T mean minus T wall by TI minus T wall so I can write T mean minus T wall as TI minus T wall into ? so into ? M so K into ? M into TI minus T wall so already I put a minus sign because this is T wall minus T M I am putting minus of T M minus T wall okay so this cancels right here and of course your K cancels so this will be so left with Nu X is equal to minus D ? by DR at R equal to R 0 into D by ? M so therefore now to calculate your Nusselt number what are the quantities that you need one is your mean temperature ? M the other is your derivative at the wall so you have your temperature profile now we can calculate these two quantities so first we will calculate your mean temperature so very quickly we will do that so your ? M is nothing but your T M minus T wall by TI minus T wall that will be what so you integrate you take the mass weighted average of ? M of ? ? into R into U so this will be ? into U into R DR of course you have 2 pi divided by integral 0 to R 0 you have 2 pi U of R DR right so this is your this is how you define your bulk temperature or mean temperature mass weighted average okay now in your present case your U is a constant this is a slug plug flow case so therefore the U can be taken out of the integral and cancelled of straight away so integral 0 to R 0 R DR will be R 0 square by 2 okay so this can be written as 2 by R 0 square integral 0 to R 0 ? R DR okay now if you substitute for ? from the final expression that we have you can write your ? mean as 4 times n equal to 1 to 8 1 by ? N R 0 cube 0 to R 0 J 0 ? N R divided by J 1 ? N R 0 into e power minus ? by U M into ? N square x so I am just substituting for ? and multiplying with R here okay so you can just verify it I am going a little bit fast but it straight forward there is no difficulty here so integral 0 to R 0 J 0 ? N into R so already we have that identity so that will be R into J 1 ? N R divided by ? N so if you put that and you manipulate it finally you get your expression for ? M as 4 times you please check this 1 by ? N R 0 the whole square into e power minus ? by U M ? N square x so this is the final expression for ? M that you will be getting okay so if you if you just use this identity and then you substitute it here you will cancel of J 0 ? N R 0 J this J 0 ? N R 0 will be nothing but J 1 ? N R so that and this will get cancelled off okay so it is just a very simple manipulation and from your ? you can also calculate your d ? by d R at R equal to R 0 which comes to twice summation n equal to 1 to 8 1 by ? N so now when you differentiate you have to differentiate J 0 into ? into R right so J 0 ? into R as I said that is nothing but minus ? N J 1 ? N R okay so that will put a minus sign here so this will be ? N J 1 ? N R and at R equal to R 0 this will be at R 0 divided by you already have J 1 ? N R 0 into e power minus 4 ? N square x by d by peclanum once again these two will cancel off and if you substitute your derivative as well as your mean temperature into this expression let me call this as expression number 2 so this will give me muscle number substituting for ? M and d ? by T R at R equal to R 0 into 2 the final expression I think if you cancel off some common terms which is straight forward you can do that you will get minus 4 ? N square into x by d by peclanum divided by summation n equal to 1 to 8 ? N square power minus 4 ? N square into x by d by peclanum okay this is the final expression for muscle number as a function of x okay so all you are doing you have this ? M you have taken the derivative of ? with respect to R at R equal to R 0 just you substitute into this it is a very simple manipulation just one step and then you finally get the expression for any way so as you can see it is a function of only x therefore all the functions of R have cancelled off you have only variation with respect to x and these summations are separate in the numerator and denominator you cannot simply cancel this term here okay so this is a separate summation this is a separate summation okay now this is a general expression for the completely for fully developed hydrodynamically and thermally developing flow okay now we should asymptotically retrieve the case for thermally fully developed flows from this expression what is the asymptotic case what is the limit at which we can retrieve the fully developed case when your x by d goes to infinity for large values okay so for large value or you can say for large values of x by d by peclanumber or you can just simply say for large values of x by d you go asymptotically because you now are in the thermal entry length region so if you keep going down further and further and further okay so somewhere so you have your thermal boundary layers will be meeting somewhere here so now your x is somewhere here so if you go to large values of x okay so that will give you a asymptotically it will reach the limit of thermally fully developed flows so in that limit what happens is see these values of beta n will be smaller and smaller as you go for larger values of x so therefore we retain only the first value of beta that will be the most dominating term okay so only the first term is will be significant when you look at large values of x by d so that will reduce this to n u going to x by d going to infinity this will reduce to just beta 1 square okay for for large values this anyway will cancel off no you have this going to 0 this going to 0 you have only in the denominator 1 by beta n square and only the first term will be the most significant term okay so that if you substitute the value of beta 1 square as 2.4048 the whole square that comes out as 5.783 okay so this is the limiting solution for thermally fully developed flows with this is your limiting solution for thermally fully developed with flux flow whatever we have derived earlier was with a parabolic flow with a parabolic flow constant wall temperature boundary condition what was the nusselt number 3.6 so now this is much higher than that value because now instead of parabolic profile you have a uniform profile everywhere you see wherever you take the profile so now what is the significance of a uniform profile earlier when you had a parabolic profile near the wall the velocities were small so due to that the heat transfer coefficient is smaller now when you replace that with the parabolic with the uniform plug flow profile the velocities are very high near the wall and this will improve the convective heat transfer therefore you see the nusselt number has gone up like anything okay the same thing if you had if you had asked you to do instead of a parabolic profile with a plug flow for the thermally fully developed case you would have reached the same value so that you are reaching as an asymptotic limit to the thermally developing flow okay rather than doing the thermally fully developed flow you could do this and then we can go for large values of x pi d and directly get the limiting solution. So tomorrow so this is the classical grades problem that he did that he assumed a plug flow and then he did it so tomorrow what we will do is we will do an extension of grades problem so we will do the extension first to parabolic velocity profile so then you should reach the value earlier what you got 3.6 okay and the next we will look at the extension of grades problem to other boundary conditions like constant wall flux as well as linear variation so I have already posted a particular solution manual in the website that is a basically an extension of grades problem by cellars the 3 people who did this extension work and it is a original classical work in 1954 and we look at some of that solution as well okay.