 Good morning all of you so any clarifications on integral methods I think I have been going a little bit slow on the last two three classes so takes enough time for you to understand what I had been doing so last class we had looked into the approximate solution for flow with pressure gradient the Von Karman Paul Hausson method and applied that to heat transfer problem and finally we also looked into an approximation to the Von Karman Paul Hausson solution okay the ordinary differential equation can be simplified and directly integrated by what is called as the walls approximation okay so there he has plotted H of K as a function of K and found that is a very linear curve and he has given the approximate profile for the linear curve and that is what if you use it becomes much straightforward to integrate the equation to get the momentum thickness okay so once you get the momentum thickness from there you can get your other thicknesses like displacement thickness and boundary layer thickness which are required to calculate the flow average flow properties like skin friction coefficient and the same way you can solve the heat transfer problem and in the heat transfer problem you will get an ordinary differential equation for ? which is the ratio of your thermal boundary layer thickness to momentum boundary layer thickness once you have the expression for momentum boundary layer thickness so therefore you can calculate your average integral heat transfer quantities like heat transfer coefficient and therefore you can get the expressions for nusselt number okay so this is the standard procedure for all the solutions whether it is similarity solution or integral solution this is the standard procedure in the integral solution you guess a profile you know you approximate the velocity profile and temperature profile and from there you calculate your boundary layer thickness and thermal momentum and the thermal boundary layer thickness okay whereas in the case of similarity solution you have to solve the ordinary differential equation numerically by some shooting technique or whatever and get the curvature at the wall and from there you can get the other properties such as skin friction coefficient and also you can get your boundary layer thickness and other thicknesses and for heat transfer you can also get the slope of the temperature profile at the wall and from there you can calculate your expression for heat transfer coefficient and nusselt number okay so what we will do today so we had so far looked at application of integral techniques to for flows without pressure gradient that is the similar to the Blasius similarity solution we approximated that with the integral method and also we extended that to flows including the pressure gradient terms that is like the Falkner scan similarity solution we can apply the integral methods for a similar problem for a wedge with different wedge angles and we can derive approximate solutions for these okay we had also looked at the case of circular cylinder okay which is which is a limiting case this is a stagnation flow and also the heat transfer for flow past this circular cylinder now we will look at a case where you can consider a simpler case or maybe even the wedge but we will take a simpler case that is a flow past a flat plate however the boundary condition here need not be a uniform wall temperature or uniform heat flux okay so most of the real cases you will find that the temperatures are actually varying along the surface okay so if you want to consider non-uniform temperature or non-uniform heat flux so how do we solve such kind of problems okay we do not have a straight forward similarity solution to such kind of problems but it is quite likely that we can develop a similarity solution for a particular case where your variation is predetermined that is if you have a flat plate and your variation of the wall temperature is something like a power law X power M variation okay where M is some real number and so this is a power law kind of a profile so for this you can develop a similarity solution also okay we can also very easily solve this by integral methods which we will see now and it need not be a variation like this it can be any kind of variation okay that is the advantage of the approximate methods okay the approximate method give you a lot of flexibility in approaching problems with different kinds of boundary conditions and also things like where the boundary condition there is an unheated starting length okay so there is no similarity solution if you have an unheated starting link so these kind of problems with the variation the boundary temperature and the heat flux can be easily solved using the approximate solution okay so today we will look at extending the simpler approximate solutions to a case where you have more complicated boundary conditions so the first case we are going to do is non-uniform surface temperatures okay so here you can consider flat plate where your temperature is varying in an arbitrary manner okay and of course you have your free stream velocity and free stream temperature and we will look at a particular technique called the Duhamel superposition method for solving this problem okay so the specified condition is something like this you can you can maintain locations where you want to specify a particular oil temperature for example let us call this as ? at the location ? equal to 0 you start with some specified temperature okay that could be something like T you T wall one so let me draw how the wall temperature profile will look so this is my plot of T wall of course as a function of my position so at this particular location ? equal to 0 I will have for example a piecewise constant value of all temperature which is T wall one okay so this is existing till a value of ? which is equal to ? 1 okay so like this I will have multiple piecewise constant values of surface temperature okay so this is a simpler case to begin with of course your actual variation need not be piecewise constants it can be a gradual variation smooth continuous variation okay so like this we can look at temperatures which are success successively increasing the wall temperatures which are successfully increasing in a piecewise constant manner okay so like that you can go up to some value of finally ? n all right so this is how your surface temperature is now plotted as a increasingly piecewise constant which is increasing okay so this is like an approximation to a profile which is like this suppose your wall temperature profile was this then the basic approximation for this is to assume piecewise constant you break this continuous curve into piecewise constants okay and you have an increasing trend in the wall temperature okay so for this how do we approach the problem okay thankfully the equation that we are solving the energy equation is linear once you know the corresponding velocity profiles so the velocity profiles are not going to get affected as long as your properties are not affected by the temperature okay so in that case your velocity is decoupled from the temperature and you can plug in the particular value of velocity at a location into the energy equation and your energy equation becomes quasi linear and therefore for any linear partial differential equation if you have varying boundary conditions okay you can break the solution into multiple solutions and superpose the solutions for each boundary conditions linearly okay so the resulting solution is a linear combination or linear superposition of multiple solutions each corresponding to a different boundary condition okay so if you solve by method of separation of variables you will know that for example the heat conduction problem okay the method of separation of variables will work if you have for example homogeneous boundary conditions in three directions and one non-homogeneous boundary condition okay for example in conduction case you will maintain this at some high temperature and the remaining three sides or maybe you can assume that or you can non-dimensionalize the temperature in such a way that the non-dimensional temperature is 0 okay for this you can solve the conduction equation 2D conduction equation by means of separation of variables okay so there will be an eigenvalue problem in this direction basically where you have two homogeneous boundary conditions in the other direction you apply the remaining homogeneous boundary condition and finally whatever non-homogeneous boundary condition is there to get the final constants okay but if you have non-homogeneous boundary conditions on all the four sides so how do you solve this problem you can still solve it by separation of variables but you need eigenvalue problems okay for eigenvalue problems you need homogenous boundary condition in a particular direction so to do that if suppose these were non-zero okay this was something like TC1 TC2 and TC3 which were not zero okay so you can break this into problem where you have TH here and you can put TC1 here TC3 and zero okay or what you can if you want to make it completely homogeneous you can break this as three zeros here plus you can make the other three as zeros and you can make this as TC1 plus of course you can make these three as zeros and this is your TC3 plus this is your TC2 and the other three zeros so you can you can apply no eigenvalue problems in different directions you know in this case you have y direction y direction you have x direction here in this case you have the x direction okay so you create four equivalent eigenvalue problems and you get solutions for each of these case okay with one non-homogeneous boundary condition and finally you superpose all these four solutions and that will give you the solution for this problem okay this can happen only if the partial differential equation is linear and conduction equation is linear right so the same way if you look at the convective heat transfer the energy equation governing convective heat transfer that is also linear so therefore if you have a combination of multiple boundary conditions like this you can superpose solutions where you have for example you have multiple wall temperatures you can solve one problem where you have T wall 1 throughout okay and the solution to that is known plus you have another problem where from zeta equal to zeta 1 to the end you have T wall 1 minus T wall 2 or T wall 2 minus T wall 1 okay that is the delta T wall so that that is that is applied to the entire plate and again you know the solution for that and from zeta equal to zeta 2 till the end so that is you have T wall 3 minus T wall 2 so like that you keep on applying successive delta and you apply that as your boundary condition and you solve the problem now if you do that you have individual solutions where you have an unheated starting length and then the rest of the plate where you have a uniform temperature okay so like that you break up the problem into multiple boundary conditions and then you get the solutions you already have the solution so you add them you superpose all the solutions together and that will give you the solution for this problem okay that is why it is called as a superposition method okay so this is also called the Duhamel method so let me so let me indicate this as T wall 2 this is T wall 3 T wall 4 and so on so this is your T wall and N minus 1 okay or okay let me call this is T wall okay so the solution for these is anyway you have to solve the energy equation boundary conditions are now at y equal to 0 you do not have a constant value of temperature but T equal to T wall which is a function of x okay and the other boundary conditions are the same y going to infinity T is equal to T infinity and at x is equal to 0 okay now following some analogy similar to the conduction problem where also the partial differential equation is linear and you can convert the problem into equivalent four equivalent problems with convenient boundary conditions so you can do the same way here also and you can superpose superpose the solution so therefore the solution for this problem will be something like you can take one problem like these where you have your free stream temperature T infinity okay and this is your starting from your zeta equal to 0 entire plate you maintain at temperature T wall 1 right so this is your first problem and already you have the solution uniformly heated plate right from without any unheated starting length okay so the if you non-dimensionalize the temperature your solution to be found out will be in the form Phi for the case zeta equal to 0 and it is a function of x and y will be T minus T infinity by T wall 1 minus T infinity this is the way I am going to non-dimensionalize okay so and how does the corresponding wall temperature the non-dimensional temperature profile look at the wall if I if I plot this Phi at zeta equal to 0 x y equal to 0 so I am I am plotting the temperature at the wall so this will become T wall 1 okay and throughout it is T wall 1 okay so therefore it will be just one everywhere all right so this is one case now I can break it up into multiple problems okay my second problem will be what so now I have solved one problem where everywhere it is T wall 1 but now you can see from zeta equal to zeta 1 to zeta 2 it is T wall 2 so now what is the next problem that I have to solve okay so to do that what should I what should I do now I should not solve for any heat transfer problem till zeta 1 because I already have solved it here okay so I should maintain what is what is something called as the unheated starting length till zeta 1 right and what should be the free stream velocity that I should take so that there will not be any heat transfer here now this is already at T wall 1 okay so I have to choose a free stream velocity such that there will not be any heat transfer so that this remains unheated so what is that free stream velocity that I have to take I cannot take T infinity if I take T infinity the T infinity is different from T wall 1 T wall 2 minus T wall 1 why it should be T wall 1 okay so if I take a free stream which is at T wall 1 so this is also a T wall 1 so therefore there will not be any heat transfer till here now at this point onwards this will be at T wall 2 till zeta 2 okay so now I can introduce another non-dimensional phi corresponding to zeta equal to zeta 1 onwards okay now how do I non-dimensionalize this T minus T wall 1 by T wall 2 minus T wall 1 exactly okay so if you draw the non-dimensional profile at the wall that is a sex how does it look now initially till zeta 1 there will not be any heat transfer so from here it will start and this will be one throughout so this will be at T wall 2 throughout okay so that means here this portion is now T wall 1 this entire portion is T wall 2 okay so now you are solving for T wall 2 minus T wall 1 that is the difference that you are solving okay so that means now you have solved for T wall 1 here now here you have already solved for T wall 1 plus T wall 2 minus T wall 1 so that is basically as T wall 2 now same way you have to break this into problems depending on the number of piecewise constants that you have all right so now if you extend this to I will just give you a representation for the third one so this plus this plus okay so till zeta equal from zeta equal to 0 to zeta equal to zeta 1 you have T wall 1 okay now from zeta equal to up to zeta 2 you have T wall 2 now you have to maintain T wall 3 throughout okay so to do this again we take assume a free stream velocity where you are maintaining everywhere as now T wall 2 and your free stream not everywhere I am sorry from zeta equal to 0 to zeta equal to zeta 2 as T wall 2 and your free stream velocity is now T wall 2 from zeta 2 onwards it will be T wall 3 so you maintain everywhere as T wall 3 and you define your fee as zeta equal to zeta 2, x, y which is equal to T minus T wall 2 by T wall 3 minus T wall 2 so if you plot again fee at y equal to 0 as a function of x okay so till your zeta equal to zeta 2 there is no heat transfer from here it becomes 1 okay so this is how you get different solutions okay now we have actually solved here for the reminder T wall 3 minus T wall 2 for this particular region okay so already you had solved for T wall 2 minus T wall 1 and also for T wall 1 minus T infinity so finally that will be T wall 3 minus T infinity okay so then you can superpose all these solutions together yes but there will be heat transfer in zeta equal to 0 to zeta equal to 1 of course temperature is different from the world yeah so you have to maintain this at T wall 2 throughout because you have already solved for this till here you are not interested in this region okay so you maintain this entire region at T wall 2 and then this will be from T wall 3 so here also you have already solved here okay so you do not need any solution the solution is already there that is this solution till this region you already have solution from this so you do not have to solve for anything here so you maintain a unheated starting length forcibly by maintaining the free stream temperature and this the same the same way here okay so you already have solved from here to here so you maintain a same temperature equal to the free stream temperature so that there is no heated length okay so therefore you define your phi in generic terms as T minus T wall zeta n minus 1 by T wall zeta n minus T wall zeta n minus 1 this is your generic formula for defining your non-dimensional phi okay so where for example if n equal to 1 so that is basically that is your zeta 0 that is this first case okay n equal to 2 that is your second case n equal to 3 will be this particular case okay so this is a generic formula how you are defining your non-dimensional temperature okay and the non-dimensional form of the energy equation will be U d phi by dx plus at y equal to 0 now what will be the value of phi now till your x is greater than 0 or maybe you can say if you want to write it in generic terms zeta n minus 1 to zeta n okay so this should be what 0 and for x greater than yeah so for x greater than or equal to zeta n it should be 1 okay can you just check that it has to be zeta eta eta n minus 2 minus 1 is that right okay so when you start with of course something negative it means it is you can say free stream okay free stream so there it will be 0 and then when you start with n equal to 1 so up to for this case so up to eta 0 basically you do not have any heat transfer from here onwards you have greater than that you have your phi equal to 1 okay so this is your generic solution okay and this is your generic way of non-dimensionalizing phi now therefore the solution to the problem for temperature T minus T infinity so this is what we finally want to find out so now we have broken this into sub problems and for each already we have the solution okay we have the unheated starting length problem okay so we can just combine all these for the case where your zeta equal to 0 how do we express T minus T infinity T minus T infinity is phi zeta equal to 0 times T wall 1 minus T infinity okay so this is T wall 1 minus T infinity into phi zeta equal to 0 x, y okay plus the next problem will be T1 T minus T wall 1 which is T wall 2 minus T wall 1 into phi zeta equal to zeta 1 x, y plus all the way till T wall eta n minus T wall eta n minus 1 phi eta n x, y so that is your final this thing okay so T wall eta n minus T wall eta n minus 1 that is the last this thing okay so you keep adding up all these solutions such a way that finally if you add up you get the solution to the problem where you have variation like this in the boundary condition okay so already we have the solution to phi, phi is what T minus you can you can look at this T minus T infinity by T wall minus T infinity so what is suppose if you assume a cubic temperature profile so what is the how can we write phi suppose you assume cubic temperature profile you non dimensionalize it why 1 minus correct correct you are what you have said is correct but why 1 minus you are what you are saying is correct but it should be 1 minus why it should be 1 minus signs of change no how did we define the non dimensional temperature ? when we were fitting a cubic profile T minus T wall by T infinity minus T wall so my phi is what so that should be 1 minus ? right so therefore what what it should be 1 minus 3 by 2 y by ? T plus 1 by 2 y by ? T the whole cube correct so this this is the profile satisfying a condition x greater than or equal to ? n minus 1 right so where so this is the corresponding wall temperature so uniform wall temperature correct corresponding to that this is the profile in the boundary layer thermal boundary layer now I already know the solution so only thing I have to now linearly combine all the solutions that is all and how do you check that how do you know that this is the correct solution correct for way of super posing the solutions how do you verify that the simplest verification is getting the boundary condition itself so you apply this at y equal to 0 what it will be now until so this profile will be one everywhere okay so this will be T wall 1 minus T infinity okay suppose you want to check that at between ? equal to ? 1 2 ? 2 that this gives me T wall 2 minus T infinity does it give that is a check right so this is T wall 2 minus T wall 1 and this is 0 before and it is 1 between ? equal to 1 ? 1 ? 2 and the other things are all 0 okay so T wall 1 T wall 1 cancels so this is T wall 2 minus T infinity so this retrieves my boundary condition okay this is a good check that therefore this is the correct solution because your boundary condition also is a part of the solution right so this is the therefore how you have to write it and if you check if the wall boundary condition are retrieved then let me call this as equation number 1 okay then 1 is correct okay or 1 is represented correctly okay so now that is it once we write the solution like this this is valid if you have a piecewise constant variation in the actual case the variation should be continuous so instead of having a summation of piecewise constants we can replace that by an integral which represents a continuous summation okay so that is what we are going to do now so I would like to rather than writing T wall ? minus T wall ? n minus 1 I am going to introduce ? T wall n which is equal to T wall ? n minus T wall ? n minus 1 so this is my representation so therefore equation 1 can be written as T x, y minus T infinity is equal to T wall 1 minus T infinity into ? 0, x, y so this first notation here corresponds to ? okay so ? equal to 0 plus I can write the rest of summation n equal to 1 to totally n ? T wall n ? ? n, x, y so the rest of them I am just summing over those n number of discrete intervals okay so that I can put under a big summation like this is that right so if the temperature variation is not a piecewise constant but actually is something like this which you have approximated as a piecewise constant okay so now I am going to convert this discrete summation into a continuous integral okay that is all I need to do so for wall temperature variation which is not piecewise constant but continuous so that is either linearly increasing or not linearly it is continuously increasing or decreasing you can rewrite this discrete summation as T wall ? equal to 0 minus T infinity into ? 0, x, y plus integral 0 to x ? x ? x y now how do I write this in an integral ? T so I am going to write this as D T wall by D ? which is the dummy variable D ? okay so all I am doing is I am converting this discrete summation into a continuous integral so if I have a finite slope and the slope is continuously varying with the location okay so all I need to do is get the slope of the wall temperature variation with respect to ? ? and that will be used here okay so this will this is also called as the Duhamel's superposition integral okay let us call this as number 2 this is referred to as the Duhamel's superposition integral okay so the general super superposition which we discussed above is a general method and usually it is applied where you have this discrete piecewise constant kind of approximation if you do not make that but directly you put the slope of the wall temperature variation as a continuous variation if you do that then the resulting expression is called the Duhamel's integral method okay so now once you know the local variation of the temperature now we can calculate all the other things like the wall heat flux for example so now for example in this case we had varied the wall temperature and now for the varying wall temperature case we want to calculate what is the corresponding variation in the wall flux the wall flux also varies right so it is not a constant anymore the wall flux also keeps varying with the X location so we can calculate an X write an expression for so the local wall flux can be estimated as now this is a function of X of course and that will be written as T wall 0- T 8 x H 0, X plus integral 0 to X H z x into DT wall by D z x D z where your H of z, X is nothing but – K d5 by dy z x, Y equal to 0 right my this is my definition of heat transfer coefficient – K DT by dy by T wall – T 8 okay so P is nothing but your T – T 8 by T wall – T 8 so this is your heat transfer coefficient so if you differentiate this with respect to Y at Y equal to 0 you can replace your P as directly now that – K D P by dy as directly H okay now so how do you calculate the local heat transfer coefficient so already if we know for cubic profile my D P by dy at Y equal to 0 what is the value – 3 by 2 ? T okay so also I know the ? T okay so how do I know that because ? T is equal to ? into ? okay and I have got an expression earlier which we derived for the flat plate flat plate case with unheated starting length okay so that comes out as 0.976 by Prandtl number to the power 1 3 1- so the unheated starting length there was assumed as X 0 in any generic case where you know the location of the unheated starting length you can replace that X 0 by the Z here the whole power 3 by 4 the entire thing raised to the power 1 3 right so this was what we derived for flat plate with unheated starting length so this has to be substituted into this expression to calculate the heat transfer coefficient and so do you remember how that comes out do you remember the expression for heat transfer coefficient that we derived earlier after we substitute this and also for ? ? we can assume a cubic velocity profile and we got an expression in terms of Reynolds number right so that in this we substitute into this expression for H we get a final expression do you recollect that 0.331 K by X into Prandtl number to the power what 1 by 3 Re power 1 by 2 into 1- ? by X the whole power 3 by 4 the entire thing power – 1 by 3 this is what we had derived all right so therefore you substitute this value of H into this expression now H is a function of ? okay so that ? will tell you what is the unheated starting length for that particular problem so when you are breaking this problem into multiple boundary conditions for each configuration you have an unheated starting length so it starts initially at ? equal to 0 then ? 1 then ? 2 like that you have location of the unheated starting length keep shifting so that value of ? has to be used in calculating the local heat transfer coefficient and that goes into this expression right here okay so that will give us the value of local wall heat flux which is 0.331 K by X so this is a constant term which will come out and the rest of the terms that is T wall 0- T infinity this is your term and H the other terms we have taken out as constant now for the very first case H of 0 there is no unheated starting length therefore this will be 1 plus you have integral 0 to X in this case you have the unheated start starting length so there you put 1- ? by X- 1 by 3 DT wall by D ? easy all right so this is your final expression for calculating the local variation in the heat flux all right so so what we will do is tomorrow we will apply this to a problem I will just do for a simple case where you have a linear variation what happens if you have a linear variation then we will calculate the heat transfer coefficient and the local heat flux wall heat flux okay now what happens now if you if you have a continuous variation like this and apart from that if you also have local jumps something like this so you have a continuous variation and suddenly you have a jump here and again you have a continuous variation so it is what piecewise continuous okay so in such a case you know the location of these jumps may be at ? 1 ? 2 or so the same expression plus you have to account for these jumps local jumps local discontinuities okay so then your expression for Q all double prime will be H of 0, X plus whatever you had before plus what should you do so this is the same as this correct now additionally you have this local discontinuity so what you should do to account for that so once again that becomes locally discreet so then you have to make this again discreet this becomes H into ? T for continuous variation we replace the discreet representation with the continuous thing now once you have a discreet jump once again you will have to replace this integral with the summation okay so summation suppose you have say K number of jumps okay I am going to go from I equal to 1 to K ? T wall into H that is it okay so this will take care of both the continuous variation wherever and wherever you have discreet jumps at these locations you have you have to write as ? T H and some them okay so this will give you the local variation of heat flux so with this will stop tomorrow we will work out solve an example for assuming a linear variation in the wall temperature and we will see how to calculate your wall flux okay.