 So today we will take up a problem on the variable wall temperature case and we will look at how to get the expression for the wall heat flux variation as well as the heat transfer coefficient so let us take the example of a linear surface temperature variation that is of the form T wall of X is equal to A plus B X okay so if you plot the surface temperature as a function of X so at X is equal to 0 that will be equal to constant A and from there it varies linearly okay so this is the value T wall at X is equal to 0 which is equal to A so if you go back to what we derived yesterday with the Duhamel superposition integral if you have a continuous variation of the wall temperature so we have derived the expression for the wall heat flux this was 0.331 into K by X we are 1 by 3 and you have T wall at ? equal to 0- T 8 plus 0 to X 1- ? by X the whole power 3 by 4- 1 by 3 DT wall D ? into okay so this is our expression you know now if for the limiting case where you do not have any slope the slope is 0 so this is T wall- T 8 so this becomes equal to your flat plate with a uniform temperature expression okay so of course if you have continuous variation but also intermediate temperature jumps okay so there you have to introduce an additional ? T into H you know for those temperature jumps in between okay so apart from that this is the expression now we will substitute for the given temperature profile whatever we require so for for example in order to evaluate this Duhamel's integral we need to first get the slope of the wall temperature profile and since this is a linear profile it is very simple so in this case DT wall by D ? will be for this particular profile B right so we will substitute this into equation let us call this as equation 1 so this is T wall at ? equal to 0 is nothing but a okay so right at X is equal to 0 this is nothing but a so I am just substituting for T wall at ? equal to 0 as a- T 8 plus integral 0 to X 1- ? by X the power 3 by 4 the whole power minus 1 by 3 DT wall by D ? is nothing but B so this will be B into D ? okay so now all we need to do is evaluate this integral because everything else is known T 8 is given for this problem whatever value it is known a is a constant B is a constant so we need to find this particular integral and this particular integral whatever slope that you get and put it here is of the following form which we will reduce it to so we can assume a variable Z which is equal to ? by X to the power 3 by 4 okay so I am just going to transform the variables again so I am saying that ? by X to the power 3 by 4 is equal to some other variable Z so therefore DZ by D ? should be equal to 3 by 4 ? by X to the power 3 by 4- 1 which is minus 1 by 4 into with respect to D ? if I so X is a constant so it will be 1 by X right so this is my derivative okay so therefore now I can substitute for D ? in this integral in terms of DZ okay so this will give me my DZ is equal to 3 by 4 into X power this is X power 1 by 4- 1 so this will be X power minus 3 by 4 okay into Z to the power minus 1 by 4 into DZ all right it is just I am just doing some algebraic manipulation there so therefore now I can substitute for DZ I can transform these variables from DZ to DZ okay so I can substitute for Z by X to the power 3 by 4 or Z and DZ I can substitute in terms of DZ in fact I can also write this in terms of Z okay now Z is Z by X to the power 3 by 4 so Z to the power of 1 by 3 I can put it because this is already a so my DZ will be DZ 4 by 3 here X to the power 3 by 4 Z to the power 1 by 4 okay so ? by X so I can write this as 1 by 4 X power minus 3 by 4 I can take 1 by 4 X yeah so I can write this as Z by X the whole power 1 by 4 this is 1- yeah right yeah right I can write it like that because this is anyway 3 power 3 by 4 I can write it as X divided by X power 1 by 4 okay so now this is nothing but Z by X to the power 1 by 4 is nothing but Z to the power 1 by 3 okay so this I can rewrite as Z to the power this entire thing as Z to the power 1 third okay so what I am doing is I am transforming all my variables from Z plane to Z plane okay so wherever I have Z that also have to include that so therefore if I substitute into this expression 0.331 8 by X PR to the power 1 by 3 REX to the power so you have a- T 8 plus so I have substituting for DZ so 4 by 3 is constant 4 by 3 also B is a constant I can take out and also inside the integral this is integrated with respect to now DZ therefore X also can be taken out of the integral now 0 to this is originally 0 to X here so I can transform this to 0 to 1 okay because at ? equal to X this becomes Z equal to 1 so therefore this will be the upper integral will be 1 upper limit of integration so this will be 1- ? 1- Z sorry 1- Z- 1 by 3 and you have this Z power 1 by 3 here so that will also come outside 1 by 3 into DZ okay just check 1- Z to the power – 1 by 3 into Z to the power 1 by 3 DZ okay so now I have any way transform that to this integral right here now how should I integrate so I will just give you the formula now this is of the form of what is called as a ? function okay the ? function can be expressed so generally the problems with the Duhamel integral will be of the form of a ? function once you transform the variables from ? to DZ okay so you will be ending up with a function something like this and you can express this ? function in terms of constants P and Q 0 to 1 Z to the power P-1 1- Z Q-1 DZ okay this is valid for positive values of P and Q less than 8 okay so this is how your ? function is defined okay now if you compare this with this expression you can see 1- Z to the power Q therefore Q-1 is equal to – 1 by 3 and P-1 is equal to 1 by 3 right therefore Q is equal to 1- 1 by 3 2 by 3 and P equal to 1- 1 by 3 4 by 3 okay so therefore this P and Q will be nothing but 4 by 3, 2 by 3 okay so therefore this entire term can be written in terms of ? function okay so everything here as it is A- T 8 plus 4 by 3 BX into ? function okay this entire integral is ? function the parameters are 4 by 3, 2 by 3 so the integral is replaced by the ? and why do we now need to write this in terms of ? function because we can the ? functions are tabulated okay in fact very specifically the ? function itself can be expressed as a function of another function called the ? function okay so on ? function tables are quite common you know they are tabulated for different values of the function so therefore we will express the ? function in terms of ? function as follows this is your ? function where generally your ? function ? of say some variable S is represented as e power – X X to the power S-1 DX so this is your ? function basically okay and this has been tabulated you know you can do this integral numerically also for different values of S but these have been tabulated ? function charts are there so you can look up for the values that we have here so therefore ? of 4 by 3, 2 by 3 will be ? of 4 by 3 into ? of 2 by 3 by ? of what 6 by 3 that is 2 okay so if you plug in from the ? function tables which are available online you can Google ? function charts and you will find those nice charts for different values of this you know for P and Q so if you plug in you will get this ? of 4 by 3, 2 by 3 as 1.2087 okay so this is the resulting resulting expression for ? 4 by 3, 2 by 3 so if you substitute that value into this so this is 1.2087 into 4 by 3 which comes out as 1.612 okay so now we have a complete expression which gives you the variation of heat flux with respect to X okay provided you know your constant A and B and you know the free stream temperature T infinity alright so now therefore from this the heat transfer coefficient can be defined okay so H of X can be defined as Q of all double prime X by T wall of X minus T infinity okay now T wall of X minus T infinity can be written as A plus B X minus T infinity so therefore you have this entire 0.331 A by X we are to the power one third 612 B X the entire thing divided by T wall minus T infinity which is nothing but A plus B X minus T infinity okay so therefore for the given constants you can now determine the local variation in the heat transfer coefficient and for the limiting case where your B equal to 0 B equal to 0 gives me a uniform wall temperature okay so for that case how does it reduce this becomes A minus T infinity here the denominator this cancels with A minus T infinity here B is anyway 0 okay so then you will what will be the expression a constant wall temperature that is 0.331 K by X PR power 1 by 3 REX power half okay so this is your constant wall temperature boundary condition so for the limiting case where B equal to 0 you retrieve your constant wall temperature heat transfer coefficient okay so it is a very straightforward method as such you know so so what it finally means that so you can also solve this by dividing this into piecewise constants like we did yesterday you know you can assume that you can represent this by piecewise constants like this right you can divide this into piecewise constants and instead of the integration that we did here we will replace this by discrete summation okay so if you do that you can also get somewhat similar expression but that will be in a slightly discrete form so in that case you will have something like 0.331 into K by X REX power half PR power one third so this will be T wall 0 minus T infinity so when you differentiate it the original profile you have a fee so that will be minus K D fee by dy at y equal to 0 it will be nothing but the heat transfer coefficient H okay so that H you have already substituted as 0.331 K by X REX power half PR power one by third into one minus zeta by X the whole power minus 3 by 2 and you go back and revise the expression for H of X 0.331 K by X REX power half PR power one third one minus zeta by X to the power what 3 by 4 the whole power minus minus 1 by 3 okay so this we had to substitute into the expression for Q wall of X which was so you had the original expression for T wall minus T infinity that is your team minus T infinity was T wall 0 minus T infinity into fee of 0, X, Y plus summation of n equal to 1 to N ? T wall and fee of X, Y so this was what we saw yesterday this was for the local variation in temperature we have super post the solutions where you have a uniform temperature okay and one by one so then the incremental temperatures so all of that when we super post so first you have this is the basic solution that you have plus the incremental solutions which is basically this okay so now when we want to calculate the heat flux so we had to say minus K into d 5 by dy at y equal to 0 which was nothing but the heat transfer coefficient H so this we had to substitute into this expression so therefore this is taken out as common you have T wall not minus T infinity as one of the terms and the second term will be the rest of the terms will be an n equal to 1 to N number of discrete intervals you have ? T wall N okay into 1 minus ? by X the whole power 3 by 4 whole power minus 1 by 3 okay so this is what you will have if you have a discrete variation if you had a continuous variation you will replace this by an integral okay integral over d ? okay now you have a discrete variation therefore you just substitute for ? T as it is and of course the heat transfer coefficient for the first location where ? equal to 0 there is no unheated starting line so therefore you do not have this term for that for the subsequent boundaries conditions you maintain at ? equal to ? 1 ? equal to ? 2 so there you have unheated starting length so there you have to substitute the values of so this will be corresponding value of ? ? N okay so by doing this also you can calculate your local wall flux variation instead of using the Duhamel's integral method you can just linearly superpose this is the superposition method right so you can divide this continuous curve into small discrete intervals where you have you know variation of temperature like this you can substitute that into this discreetly and you can also estimate the wall heat flux okay so both will be more or less the same the continuous is the more accurate because you are taking into account the slope accurately okay so this is to just give an example okay how you take a problem where suppose you have a wall temperature variation like this and you can use either the Duhamel's integral or a simple superposition technique and you can calculate your local heat transfer coefficient and your wall heat flux okay so any questions on this okay now for more complex profiles you know it is more likely that most of the wall temperature variation cannot be approximated just by a straight line it will be more complicated profile so for a more complicated wall temperature variation what is common practice is that we can approximate the wall temperature variation as something like power law series so instead of using an a plus bx relationship we can write this as a plus summation n equal to 1 to capital N B and X power n okay so this is a power series expansion so which is which is which can be used to approximate more complicated nature of profiles you know if you have a profile something like this so you can you can use power series expansion you can fit the coefficients to a no you can do a regression fit the coefficient such that you can approximate this curve with the power series expansion okay so this is a better way to represent this than using a straight line right so you can substitute this now into this expression here so you had a term here dt wall by d zeta so now you have to calculate dt wall by d zeta for this so what will be the expression for dt wall by d zeta summation n equal to 1 to n you have n bn x power n-1 or zeta power n-1 okay so this has to be now substituted into the expression where we had d dt by dt wall by d zeta okay and if you do that the resulting expression for the heat flux comes out to be everything is the same only you have the summation term 331 K by X so everything up to here is the same except you have the summation term and everything inside the summation term goes there so n equal to 1 to n you have n bn so now so this will be x power n-1 into there will be an x which will come out of the transformation okay so that will be giving you x power n okay into the other the beta function will be there as it is so you will have beta now this beta function also will become a function of n okay so where this beta function we can be written as ? function of 4 by 3 originally it was 4 by 3 now it becomes 4 by 3 n into ? function of 2 by 3 divided by ? function of 4 by 3 n plus 2 by 3 okay so now depending on the value of n that you use okay so the value of beta will change and then you have to sum them over all the values of n so suppose you are using five terms you have to sum them for each value of n and for all the five terms you have to sum them together so so and that that will give you the variation if you have a more complicated profile now you can approximate that with the power series expansion and you can use this expression to calculate the local heat flux variation okay now the question is given local variation in the temperature profile we can use this to calculate the local heat flux but what about the other way suppose your wall boundary condition is a locally varying wall heat flux okay so how do you calculate the local wall temperature as well as the heat transfer coefficient okay so that is also a little bit more complicated derivation I am not going to do that I will just only give you the final solution for the for the wall temperature variation okay for a non-uniform wall flux so for this case you can calculate the wall temperature variation can express it as follows 0.623 by K we are power minus 1 by 3 rex power minus half you have 0 to x 1 minus zeta by x to the power 3 by 4 the whole raise to the power minus 2 third into Q okay so this is the local variation in the wall flux whatever you have so that can be substituted into this and you can get the corresponding wall temperature variation okay so this is this given in your textbook case and crock for okay he has also not derived it but I think there is a reference to some literature where they have done it and they have shown that shown this kind of an expression anyway so this is beyond the scope of your this thing but you should understand that you can do either of these when using the approximate solutions given non-uniform wall temperature how do you calculate the variation in the wall heat flux or given a non-uniform wall heat flux how can you calculate the variation in the wall temperature so both are possible by using the approximate methods so I think with that we will kind of wrap up the external laminar external flows external boundary layer flows so we have covered quite a bit you know we have got almost covered whatever possible similarity solutions under external boundary layers and also the approximate methods we know whatever external laminar similarity solutions they have a complementary integral methods also integral solutions also like we have seen if you have the Falkner scan solutions for wedge problem you have similar fond-karman-polhausen solution when you use the integral methods okay of course like the Blasher solution there you have you can approximate some velocity profile and very easily find out the expression for say Nusselt local Nusselt number okay so whatever is possible in fact you can also use the approximate solution for for example a flow with transpiration you have boundary where you have porous boundary with suction or blowing so we have derived the similarity solution for that in fact we have we have identified the condition for the variation of the suction profile velocity profile so that you can get a similarity solution okay the same way we can derive an approximate by using approximate method you can we can derive expressions for the local skin friction coefficient as well as the Nusselt number variation with transpiration so every similarity solution has a counterpart in the approximate methods and more than that you can also derive some special cases such as the unheated starting link which you cannot derive by similarity solutions and also cases such as these where you have non-uniform wall temperature variation of any of any given profile which you can approximate as a power series expansion or a non-uniform wall heat flux okay so for these kind of boundary conditions you know it will the similarity solutions are not possible or becomes very rigorous so their approximate solutions are much easier okay so this is in a nutshell giving you an idea what we covered so from the next week onwards we will look at internal laminar internal boundary layer flows okay so they are actually strictly speaking the boundary layer concept does not have a meaning that the way that laminar external flows has okay the strict definition of boundary layer flow does not hold for internal flows okay because once you have a fully developed flow both the boundary layers merge and everywhere you have viscous effects there is no place where you can use potential flow and approximate that with the potential theory and somewhere you can solve with the solving the full navier stokes okay so therefore we have to resort to a complete solution of the navier stokes equations in some cases in some cases we can make approximation to the velocity gradients or the temperature gradients okay so there we can obtain exact solutions of reduced form of the navier stokes equations okay so that is also very important and interesting because most of your practical problems in heat transfer okay although there are many external flow problems you will find most of the heat exchangers they are encounter you will be encountering internal flows there and in those cases you should understand the approximations that you can make and how you can get the expressions for local variation in the Nusselt number and in say fully developed case how the Nusselt number variation there is no variation in the Nusselt number and so on okay so that we will cover in the next 9 to 10 classes starting from next week so in about three weeks I think we should be able to cover the laminar internal flows and then from the following we converts that is about the fourth week of March Professor Kohler will start turbulent flows so that will be for about seven lectures or so and or seven or eight lectures and maybe natural convection for another seven or eight lectures.