 So good morning all of you so today we will look at in fact continuation of the last class which was extending great solution the great solution was originally proposed by grades for slug flow profile and it looks at the thermally developing region or the thermal entry region wherein you are hydrodynamically fully developed and you are looking at only the entrance region where the thermal boundary layer is developing for such kind of a that is what we call as region 2. So grades assumed a slug flow so where anyway the slug flow does not vary and it is a uniform everywhere and here developed solution for constant wall temperature which we had seen earlier so there the Eigen functions where what were the Eigen functions essential in the original grades problem so the Eigen the Eigen value problem there was actually a Bessel equation right so now the same problem can be extended to a case which is more realistic that is for a parabolic velocity profile okay it was this extension was done by a group of people cellars at all cellars tribe us and I have posted that in the moodle you can just have a look originally 1954 how they did the extension to the grades problem of course nowadays the solution is more by numerical methods they try to do approximate technique where they did something for our close to the wall the R which is somewhere in the middle and R which is far away from the wall and they patched up all the solutions together so what we are doing nowadays is directly go for a numerical solution to the Eigen value problem so the easier way to start is to introduce these non-dimensional variables for temperature as usual we define ? as T – T wall by Ti – T wall since this is a constant wall temperature boundary condition and your inlet temperature is also assumed constant so this is an appropriate definition for ? once you have a constant wall flux boundary condition then you cannot define ? this way okay because your when you take the differential with respect to X your wall temperature will not be a constant and therefore they will not cancel out on both the sides and this is the non-dimensional radial coordinate and your axial non-dimensional axial coordinate can be non-dimensionalized in this manner which is somewhat similar to an inverse of the grades number your grades number 1 by grades number is actually X by D by Peclet number okay so this is somewhat similar exactly now we are not using X by D here but X by R not so when we substitute into the energy equation okay the energy equation for thermally developing flow is this and if it had been fully developed your D ? by DX would have been 0 but in this case your flow is still developing and therefore if you substitute you get a partial differential equation in terms of ? which is a function of both the axial coordinate ? and your non-dimensional radial coordinate ? and the boundary conditions are at the entry region that is at ? equal to 0 your T equal to Ti so therefore your ? will be 1 and at the location R equal to R not that is at the wall okay so your coordinate system that you are looking is something like this this is your R and this is your X okay so this is your R not so at R equal to R not that is your ? equal to 1 that is where your ? equal to 0 where your T is equal to T wall so this is where you apply constant wall temperature and at ? equal to 0 that is at R equal to 0 center line there is a symmetry in the profile so therefore your gradient at the center line should be 0 so now once you assume that you can use separation of variables to solve this we introduce ? and we break up the solution as a product of two independent solutions one is a function of only ? the other is a function of ? so we introduce X as a function of ? and R as a function of ? and then substitute into the PDE and then we get to we reduce the PDE to two ordinary differential equations through the eigenvalue ? square okay and the cell the eigenvalue problem here is basically the one with homogeneous boundary conditions which is basically in the ? direction and if you compare this to the Bessel equation you can find that this last term here is not same so therefore this is not a Bessel equation if you look at these first two terms they appear similar but the third term is not the same as it is a function of you have ? if you multiply by ? square will have ? square into 1- ? square that comes different from the Bessel equation and these kind of general eigenvalue problems are called Sturm-Lüweil eigenvalue problems okay so any kind of an eigenvalue problem whether this is a Bessel equation or a basic ODE can be represented as a Sturm-Lüweil eigenvalue problem so the the Sturm-Lüweil eigenvalue problem can be cast into a differential equation like this of course you can express this also in this particular form and the Sturm-Lüweil problems have a particular property that when you integrate your eigenfunction you multiply your eigenfunction and you multiply it with the weighing function the weighing function is this which appears on the right hand side so this is the particular property this is your chronicle ? so if m equal to n then this will be equal to 1 otherwise it will be 0 okay so now if you compare your eigenvalue problem to the Sturm-Lüweil problem okay so you can write your eigenvalue problem into the Sturm-Lüweil form so that will come out as – D by d ? x ? dr by d ? plus the term corresponding to Q of x is 0 here and this term can go to the right hand side and that can be expressed as ? square x 1 – ? square x ? x r so I am multiplying throughout by ? and the first two terms I can write and combine and write it in this particular format okay now the format is the same as the Sturm-Lüweil format you can compare the coefficients your P of x is nothing but ? Q of x is 0 and your weighing function is 1 – ? square x ? so this is your weighing function as a function of ? okay so this can be written as 0 to 1 so this is strictly speaking a to b okay in your case your ? goes from 0 to 1 that is why I have just used 0 to 1 there so this is 0 to 1 and this will be R n of ? x r m of ? x weighing function here which is ? x 1 – ? d ? is equal to ? m n times 1 by 2 ? because here your ? square is your ? so this is square root of ? square which is ? n x dr n by d ? n x dr n by d ? at ? equal to 1 okay so this is this is how the corresponding property of Sturm-Lüweil comes out to be in this case and if you want to write the final solution for ? which is basically x into R so you know the eigen functions now in this particular case this is an OD and to get the eigen function you have no other option but to solve it numerical okay you can once again go back to your shooting method you can reduce it to do first order OD is and then here guess the value of ? because you do not know the value of ? unless you know the value of ? you cannot find out the eigen functions so guess the value of ? and once again you have to satisfy the other boundary condition and that is iteratively found out and that is the suitable value of ? so like this for the given value of ? you have a particular value of eigen function so for each value of ? you have to find the eigen functions and finally the solution will be a superposition of all these eigen functions okay so that can be written as a constant times the eigen function into the other solution for X the other for solution for X as a function of ? that is a very straight forward OD which can be directly integrated and that will be in terms of constant times e power minus ? square ? okay so this is your final solution for ? having determined your eigen functions and your eigen values you can therefore plug in into this expression and find the variation of the temperature I will just give you a representation of how the eigen functions look if you solve for them and if you plot the first three eigen functions RN 0 it varies between negative minus 1 to 1 0.5 so this is your R0 R1 and R2 this is a this is a representation of so these are not R1 R2 so here you have plotted with respect to ? here okay so this the so this is the eigen function corresponding to n equal to 0 that is for ? 0 okay and this is the eigen function corresponding to ? 1 so for different each eigen eigen value sorry for eigen value ? 0 ? 1 ? 2 you substitute into the OD and you can get the corresponding eigen function variation with respect to ? so these are the first three eigen functions so these are the most important ones there are other higher order eigen functions but their contribution will be relatively smaller so when you sum them you take only the first three or four important eigen functions into account okay so now the thing is this is the solution but still we have to find the constant CN okay so for this we have to apply the initial condition that is at ? equal to 0 and we make use of the property of the Sturm Louis when you integrate with the weighing function so this is basically since this is an orthogonal all Sturm Louis problems a lot orthogonal so this is a property which satisfies the orthogonal condition so this is an orthogonality condition of Sturm Louis system of problems and we will make use of this in calculating the constant so let me call this as equation number one and I am going to multiply both sides by RM and integrate so so for first before doing that I will apply the condition at ? equal to 0 which is equal to 1 therefore 1 equal to summation n equal to 0 to 8 CN you have Rn of ? into e power – ? n square 0 that is 1 so I can now multiply both sides by RM of ? and the weighing function okay so the weighing function is here ? x 1 – ? square d ? and integrate from 0 to 1 so this is also the same I have RM ? x ? x 1 – ? square d okay so multiply by this RM ? x the weighing function and integrate both sides so now I make use of my orthogonality property so therefore only for m equal to n this will be nonzero okay so this will turn out to be 0 to 1 Rn of ? x ? x 1 – ? square d ? on this side you can sum only if m equal to n so therefore this will be 0 to 1 and the constant can come out okay so this will be Rn square of ? because if m equal to n then only this will be 1 so this will be Rn square ? x ? x 1 – ? square d ? okay so that is therefore my constant CN will be 0 to 1 Rn ? ? x 1 – ? square d ? divided by 0 to 1 integral Rn square ? x so now I have to evaluate these integrals how do I do that for example the integral in the denominator how do I calculate integral 0 to 1 Rn square ? x 1 – ? square d ? yes so that will be 1 by 2 ? n x this particular thing right so I already know from the property of storm Louis will system of equations that the denominator 0 to 1 Rn square x ? x ? x 1 – ? square d ? should be equal to 1 by 2 ? n x dr n by d ? n x dr n by d ? at ? equal to 1 okay so this is what I get from integrating the denominator and how about the numerator how about the numerator you have a very nice clue you have basically the eigenvalue problem here from this if you integrate both the sites okay so if you integrate this that is basically you have already R x ? x 1 – ? square d ? and this will be therefore on this side minus 1 by ? square x you have ? x dr by d ? and you are integrating between 0 and 1 so at 0 ? will be 0 so therefore this should be at ? equal to 1 this will be 1 right so this comes from the eigenvalue problem itself so I can just integrate and I can find out the value of this okay so therefore now I can write my constant CN so if I substitute for this integral in this integral what will be the constant – 2 by ? n and so this will be dr by d ? at ? equal to 1 and denominator dr by d ? these two will cancel off so you have only this in the denominator so you have into 1 by dr n by d ? n corresponding to ? equal to 1 okay so this and this cancels off you have – 2 by ? n and this in the denominator so now for calculating this constant see as a function of n this is a function of n so for different values of eigenvalue ? 0 ? 1 ? 2 I need to know what is the derivative of the eigenfunction with respect to ? okay corresponding to ? equal to 1 so therefore now for different values of ? for ? equal to 1 I should know what is the value of the eigenfunction and then I should fit some approximate curve and calculate the slope okay so that basically corresponding to ? equal to 1 from there I can calculate my constant C okay so I can finally substitute for C into my equation number one for the solution so therefore the solution for ? can be written as – 2 this can be taken out and I have summation n equal to 0 to infinity and for CN so the other these are all functions of n so I have to just keep it inside the summation I have e power – ? n square ? into Rn of ? divided by ? n into dr n by d ? n at ? equal to 1 okay so I have simply substituted for C or C of n from what I have obtained here all right so finally one so once I know my eigenfunctions my eigenvalues and this derivative I can now finally find the solution for ? so this is my final solution so I will go further and I will calculate the expression for the nusselt number the local nusselt number so therefore now I will define my local heat transfer coefficient as K dT by dr at R equal to R0 divided by T wall – T mean this is my definition of local H and from substituting the temperatures in terms of ? and everything in terms of non-dimensional radial coordinates ? okay so this will be if you substitute for T in terms of ? you remember that ? is T – Ti by T – T wall by Ti – T wall okay so this will be Ti – T wall into d ? by dr at R equal to R0 divided by T wall – Tm okay now I can define a mean temperature non-dimensional mean temperature which is ? m as Tm – T wall by Ti – T wall so here T is a function of both R and X here T mean will be a function of only X okay so I can define a non-dimensional mean temperature and you see that T wall – Tm by Ti – T wall is nothing but – ? m okay so I can write this as – K d ? by dr at R equal to R0 by ? m what I can also do is replace R in terms of ? so therefore this will be ? into R0 and this will be at ? equal to 1 because your ? is R by R0 so I can replace directly with respect to ? so all I need to know is my mean temperature ? m and also the non-dimensional gradient of temperature at the wall so once these two are calculated I can find the expression for Hx so from the equation for ? let me call this is equation number 2 now I can go on and calculate the expression for first ? m so therefore I can write my ? m as follows so how do I define my bulk mean temperature according to the basic definition I take my non-dimensional temperature multiply it by the velocity and integrate across the cross sectional area so that is 2 ? into R dr okay so this is from 0 to R0 so this I divide it again by the mass flow rate so that is 2 ? integral 0 to R0 into u of R into R dr okay this in this cancels now I define my bulk mean velocity or mean velocity as once again 1 by so this is 0 to R0 u of R into R dr so this will be 1 2 this will be 2 by R0 square correct okay so I can now substitute for this right here in terms of the mean velocity so this is basically integral 0 to R0 ? into u of R into R dr divided by this will be R0 square into um by 2 okay so since um is only a function of X I can take this inside the integral and I can write this as u by um into R dr so I can now convert this completely in terms of non-dimensional coordinates okay so this will be 2 0 to 1 ? into u by um into ? into d ? because I have R by R0 here dr by R0 okay so this will be my expression for ? m as a function of non-dimensional ? and u by um okay so I can now substitute I already have my expression for u by um what is the expression for u by um that is from the fully developed parabolic velocity profile right that is twice 1- ? square in terms of the non-dimensional coordinate okay so I can substitute for ? coming from equation 2 and the velocity profile from this into ? m so so this will become – 2 summation of n equal to 0 to 8 e power – ? n square ? divided by ? n into dr n by d ? n corresponding to ? equal to 1 into 0 to 1 then I have so 2 into 2 4 okay into ? into 1 – ? square into R which is also a function of ? okay rn of ? into d ? just erase okay so I am just substituting for ? and u by um into this and I am grouping all the terms which are function of ? and that I am integrating from 0 to 1 so that is basically 4 times ? into 1 – ? square into R into d ? right so this ? into 1 – ? square rn ? d ? what is this value so we have already seen that this is nothing but 1 by ? n square into dr by by d ? at ? equal to 1 okay so we will just substitute that and my ? m now becomes so 4 into 2 becomes 8 here this is a – sign here okay just take note of it so – and – become plus here so 8 into ? n equal to 0 to 8 into so that is a dr by dr by d ? n at ? equal to 1 no this should be with respect to ? sorry therefore this comes out with respect to ? please correct it okay so this is dr n by d ? here okay so this this is one of the terms multiplied by e power – ? n square into ? and divided by you have ? n square here and there is a ? n here so this becomes ? n cube into already you have a term dr n by d ? n so this becomes dr n by d ? n corresponding to ? equal to 1 okay so this is your expression for ? m so now you have a closed form expression for ? m as a function of the eigen value and the derivative of eigen function both derivative as a with respect to ? as well as with respect to ? okay so now we need to still find d ? by d ? with respect to ? equal to 1 what is this value directly you can differentiate that is – 2 summation n equal to 0 to 8 e power – ? n square ? into this will be your dr n by d ? at ? equal to 1 divided by ? n into dr n by d ? n at ? equal to 1 okay so therefore we will substitute for ? m and d ? by d ? so therefore if I substitute I can so I can also write my so I have my expression for H x as – k by r 0 I can directly get an expression for n u which is defined as H into r 0 by k so it should be actually defined with respect to D therefore I will write this as H into D 0 and so this will be D 0 by 2 therefore this will be – 2 into d ? by d ? ? equal to 1 divided by ? m okay so I can substitute for D ? by d ? and my ? m so that gives my n ux as summation n equal to 0 to 8 a n e power – ? n square ? divided by twice n equal to 0 to 8 a n by ? n square e power – ? n square zeta where a n is nothing but dr n by d ? at ? equal to 1 divided by ? n into dr n by okay so I will give you a couple of minutes you can substitute and check for yourself it is just straightforward I am just grouping this entire term dr by d ? divided by ? into dr by d ? this has a constant which is a function of n okay so I call this as a n okay so this will be nothing but therefore a n into summation of n equal to 0 to 8 power – ? n square that is the numerator divided by the ? m also as the same term so in addition I have ? n cube here so therefore this will be a n by ? n square into this okay the numerator has basically 2 into 24 denominator as 8 therefore there is 1 by 2 there okay I hope all of you are clear I am going pretty slow here so therefore this is your final expression now to calculate my nusselt number here I need to basically know numerically the values of all these slopes of eigen function with respect to ? as well as with respect to ? and also the eigen values so from there I can calculate my nusselt number once I know the location axial location where I need to so my nusselt number is now only a function of my axial location non-dimensional axial location so what cellars did if you look at that particular paper which I posted so he did it numerically and he has predetermined the values of all these constants and tabulated them so I am just going to give you only the final values tabulated by cellars okay so you can also see the for the value of corresponding to n ? n the value of CN and AN okay so CN is required where in your solution for ? okay so that you also everything has been computed numerical so for n equal to 0 the most important eigen value that is your 2.7043 the value of CN is 1.466 0.748 and similarly for 1 2 3 I will also give the 4th value this is 6.6790 10.67 14.67 and 18.66-0.802 0.587 is 0.474 plus 0.404 0.544 so 0.462 0.38 so the most important the first five eigen values the corresponding constant CN and AN they were all numerically calculated and tabulated by cellars at all 1954 so therefore you can just directly use these values you do not have to sum them to so many terms you can just sum them to the first four or five terms okay you can directly substitute the corresponding value of ? A and in the solution for temperature the value of C then you can get an expression directly in terms of ? okay so which is basically your non-dimensional axial coordinates so for different values of ? you can actually plot and see how the Nusselt number varies right from the location where your thermal entry length starts okay now for the limiting case if you look at a very far away distance axially so that is for large values of ? for ? very large you can see that this is an exponentially decaying function okay so except the smallest value of ? if you go for larger values of ? that is corresponding to N equal to 1 2 3 4 so these are large values and it is already an exponentially decaying function so they will be very small okay so only the first term corresponding N equal to 0 will be important then you look at large values of ? so for that this will be reducing to your a 0 into so basically you have e power minus ? 0 square into something which is very large okay divided by two times you have a 0 by ? 0 square into e power minus ? 0 square ? right while I am retaining only the first term and neglecting all the higher term so these cancel out and this will become so your Nusselt number corresponding to large values of ? will be nothing but 2 ? 0 square divided by 2 so if you substitute the value of ? 0 corresponding to N equal to 0 what will be the value of NU so that is basically 2.7 the whole square divided by any of you anyone who is having a calculator can quickly check that 3.6566 anybody remembers the significance of this number this is for the case where we started fully developed flows in the fully developed both thermally and hydro dynamically fully developed region 3 and constant wall temperature boundary condition this was the Nusselt number okay so now that we are getting as a asymptotic solution to the thermal thermally developing case for large values of ? the same same value so this is here fully developed okay that comes out naturally as an asymptotic solution now wrap up this particular case as you can see this term Louisville problems any general term Louisville problems you cannot find direct closed form solution you have to do a numerical solution and therefore sometimes it is difficult to do so so once the numerical values are calculated and tabulated there was a person called housing who actually fitted the empirical curves to this numerical solution for different values of N he took and then he substituted them there and then he fitted set of numerical empirical curves to the numerical solution and then he directly proposed an empirical correlation which is independent of all these constants and the eigenvalues okay so that is a very famous correlation so that goes as a new is equal to 3.66 plus 0.0668 into grades number divided by 1 plus 0.04 grades number to the power two thirds okay where your grades number is defined as 1 over or we can say 1 by grades number is basically X by D by Peclet number okay so this is a very simpler correlation you just substitute the axial location non-dimensional form as a crates number into this expression and you directly get your local Nusselt number okay you do not have to find out eigenvalues the corresponding constant corresponding to n equal to 0 1 2 3 for this okay this is an empirical correlation which fits very well with the exact solution okay so you can see for the limiting case where your X by D goes to very large values your 1 by grades number is basically see here your grades number becomes very small for large values of X by D so then this particular term here disappears and it will lead to the limiting case of 3.66 which is the Nusselt number for fully developed flow okay so if you want to just plot the axial variation of the local Nusselt number with respect to 1 over grades number so you will find that this is for the constant heat flux case anybody remember what is the asymptotic solution 4.4.3 4.4.3 roughly this is for a constant heat flux and this dotted line here is for constant wall temperature and the asymptotic case leads to the value of 3.6 okay so if you plot the local variation you know once you have the first five dominant terms you plot it as a function of 1 over grades number and you will find that exponentially it is a decaying function okay and also you can plot both the constant wall flux boundary condition case constant wall temperature boundary condition case the constant wall flux case as a slightly higher Nusselt number and asymptotically that will reach to a value of 4.3 and constant wall temperature case reaches to 3.6 okay so this is to give an idea about the thermally developing region now as I said how do we get the constant heat flux case okay so that is a slightly different problem okay the problems that we have solved in the class corresponding to constant wall temperature boundary condition okay and parabolic velocity distribution this was an extension which was proposed by sellers extension of the great's problem now the other extension is for a non uniform wall temperature that is for case where you can have an axial variation of wall temperature which is a linear variation or you can also have a constant heat flux boundary condition so these were proposed also by sellers and I have uploaded document on the Moodle which gives the extended solution for the case of a plug flow the velocity profile is plug flow but the boundary condition is a constant wall flux boundary condition okay so there I will probably in the tomorrow's class give you hints how to approach that problem there the boundary condition is non homogenous because you have a defined heat flux so the thing is how do we homogenize that boundary condition for getting an eigenvalue problem so that that is the key once you know how to do that after that the rest of the things are straightforward so to get an idea you can just look into the Moodle where I have posted for the case of parallel plates okay that is a much simpler case to deal with the Cartesian coordinate system and taken case of plug flow and constant wall flux condition okay so I will give you I will just describe the procedure briefly and you can go over the document and you can do the same thing for the case of circular tube balls okay so with that the developing case will be over the thermally developing case or the thermal entry length cases will be done and finally we look at the case of simultaneously developing simultaneous entry length problems so those problems we cannot do analytical solutions because you cannot neglect any terms strictly speaking okay we have to go for a full numerical solution to the Navier-Stokes equations there have been some solutions where some approximations have been made but still they were involving a numerical solution so I will just give you the empirical correlations coming out of those solutions and I think tomorrow we should be able to complete this part and the last two classes on Thursday and on Saturday we look at the approximate solution to the internal flow problems so how do we use the integral method here okay.