 Today we will look at the approximate methods of solution so far we have been looking at rigorous solutions by using similarity methods and solving the resulting similarity equation by the odd by the shooting technique and other numerical methods. Now we will introduce a approximate way of getting solution to different configurations may be flow without pressure gradient like the flat plate case or including the pressure gradient for all of these cases we can get approximate solutions by making suitable approximations to the velocity and temperature profile. So in the last class we had derived the integral form of the momentum and the energy equations okay so the generic form including the pressure gradient and finally we expressed the momentum integral equation in this particular form so where this includes momentum thickness which is nothing but this is the momentum integral sitting inside this and the displacement thickness which is the displacement integral here okay. So if you consider flat plate case where your free stream velocity does not change with the position you can neglect this complete term right here and therefore this is a more familiar form which you have been using in your earlier heat transfer class to solve with approximate methods so you will have only the momentum thickness so this is the rate of change of momentum thickness along the axial direction will balance the shear stress okay so this is wall shear stress basically so this is basically the momentum integral equation very very important equation and why we are using the integral form here is that we will guess some approximations for velocity in terms of some polynomials may be linear or quadratic cubic fourth order whatever we will substitute that into this and we will integrate it out over the entire boundary layer thickness and we will find an expression for calculating the boundary layer thickness ? from here the same way we can also integrate the energy equation across the boundary layer and this is the resulting energy integral equation that you get and once again we can make an approximation guess a value of guess a profile for temperature whichever profile that you want to guess substitute here integrate it out and then you get a resulting equation in terms of the thermal boundary layer thickness ? T so with that we will get an expression for ? ? T therefore with that we can use that to calculate the wall shear stress in friction coefficient heat transfer coefficient and your nusselt number okay so this is how our approach will be okay and yesterday we did this for the flow problem we guessed initially a linear profile okay so linear profile of the form u is equal to a plus by and finally when we plug in the profile comes out very simply as u by u 8 is equal to y by ? okay so we were putting boundary conditions to satisfy this particular profile and determine the constants a and b so that is at y equal to 0 u equal to 0 and at y equal to ? u equal to u 8 okay and the and when we when we use the linear profile so that means we have assumed we assume that now the variation in the boundary layer is completely like this okay which is totally unlike the solution what blushes got okay this is your ? and this varies from 0 to u 8 okay but even in spite of this big assumption when you substitute this into the momentum integral and determine an expression for ? we got the expression for ? to be 3. ? by x is 3.47 by square root of rex okay compare this to the Blasius solution where ? by x is 5 by square root of rex okay so still this is a reasonable way to get started to get an order of magnitude of the boundary layer thickness okay so this is about 31% less than the exact solution and so was your local variation of skin friction coefficient which came out to 0.576 by square root of Reynolds number the actual value was 0.664 okay so this is about 13% less than the exact solution okay so then what we did we improved the profile assumption from a linear profile to a cubic profile okay so we assumed a profile something like a plus by plus C y square plus d y cube so now that it is a higher order polynomial we need to satisfy more boundary conditions to determine these additional constants and apart from these two boundary conditions them see this these are the most important boundary conditions in terms of the order of relevance okay so you need to first satisfy no slip at the wall and the condition at y equal to ? and apart from that we have also used that at y equal to 0 d square u by dy square equal to 0 which comes from the momentum equation at the wall okay also at y equal to ? the condition du by dy equal to 0 has to be satisfied okay so these four conditions have to be satisfied in this particular order and with that you will get a profile which is u by u infinity is equal to y by ? minus y by 3 by 2 y by ? minus 1 by 2 y by ? the whole Q right this was what we saw yesterday and if you plug in this velocity profile now instead of the linear profile into the momentum integral integrate it out across the boundary layer thickness we found that the resulting boundary layer thickness is actually improving a great deal now from 3.47 here we finally reach a value which is 4 point anybody remember 4.64 okay so which is 7% less than the exact but it is a very close very close to the exact solution okay the same way with the skin friction coefficient if you calculate this wall shear stress and therefore the non-dimensional skin friction coefficient this value comes out to be 0.646 okay which is also very close 3% less than the actual okay therefore you can see the higher order polynomial is definitely better than using a linear polynomial and you can also see that it is important to satisfy all these four boundary conditions if you want to get a reasonably accurate solution okay so especially at the wall you have to also satisfy the curvature equal to 0 and at y equal to ? du by dy has to be 0 shear stress has to be 0 at the interface so therefore these are the important things now the question you may ask is whether if you go to fourth order or fifth order polynomial if you can we get a more closer solution it need not be okay because these are the four most important boundary conditions to be satisfied and if you are introducing additional constants and polynomial so you have to introduce higher order derivatives to be 0 okay nothing more than that for example if you introduce an additional constant you have to say d2u by dy2 equal to 0 and further than d3 u by dy3 at y equal to 0 is 0 and so on okay so they are relatively less important and therefore you will find you will not get a significant improvement beyond this in fact using quadratic polynomial will in fact even deteriorate the solution okay that is that is also because of the fact in quadratic case you use this boundary condition at y equal to ? okay rather than using this boundary condition because you have only up to quadratic term you cannot take derivative up to the second order you have to take up to first order apply it so therefore you do this before applying this and this is a very important boundary condition okay so to be satisfied so which is not applied in the second order polynomial case and therefore the second order polynomial will not be as accurate maybe I in my opinion I think if you do the calculations it may come out to be somewhat even worse than the linear polynomial okay and definitely not when you anyway close to the third order polynomial okay so this is what we did yesterday with the flow solution so today we will look at such kind of approximations for temperature for a flat plate case okay so so now we will look at the thermal boundary layer on an isothermal flat plate so apart from the development of the velocity boundary layer now you also maintain a temperature fixed to all temperature which is uniform and this is your coordinate X and Y so your velocity boundary layer grows and your boundary layer thickness keeps increasing with the axial location now when you apply this T the T wall is constant it it need not mean that you are heating all the way from the leading edge okay so you can for a more generic case you can actually not heat the initial portion okay so this could be what is called the unheated starting length okay and you can start the heating after some certain distance so that distance will be X not okay so this X not is your unheated length okay in order to get a generic solution you know for the case where your X not equal to 0 the plate is completely heated right from the beginning okay now in this case the boundary the velocity boundary layer may grow all the way from the leading edge but the thermal boundary layer will not grow unless it sees the wall temperature okay so therefore it will start from here and it will grow okay so this is your ? T function of X okay so depending on the case whether your parental number is greater than one much greater than one or approximately one or much lesser than one once it starts growing so this boundary layer may overtake the momentum boundary layer thickness okay so if your parental number is much greater than one so in that case this will be lower than this okay the momentum boundary layer thickness will be growing at a much faster pace so this will not be able to catch up okay now for the case when your parental number is much lesser than one so then you will find that this will rapidly overtake and finally somewhere here your thermal boundary layer thickness will be greater than exceeding the momentum boundary layer thickness beyond a certain position okay if your parental number is equal to one so they will be both growing at the same rate however this is already shifted here so once again this will never be able to meet up to the growth of the momentum boundary layer okay so these are certain things that you should know very clearly for sure which for which parental number case how the rates of the growth of the either boundary layers appear okay so right now we will consider a case where your parental number is greater than one okay we can also use the integral method for the other case where your parental number is much lesser than one but we will start with this particular case so let us now also assume directly a third degree polynomial for temperature something of this sort now you have to tell me which are the boundary conditions that can be applied to satisfy this okay y equal to 0 t equal to T wall okay this is still dimensional okay I am not going to non-dimension okay and then y equal to ? or T okay so T equal to T because if you look at suppose you are assuming that your T wall is greater than your T infinity so this is your T infinity this is your T wall okay so similarly T infinity T infinity the same way with the elastic boundary layer also so this is your U infinity this is 0 so we need two more boundary conditions okay so in y equal to ? T by ? y equal to 0 so you can see that once this reaches T infinity after that the gradient will be 0 right after the thermal boundary layer so at the interface you have to satisfy this interface continuity condition continuity in temperature and slope should be continuous okay and then we need one more heat flux at the wall but we do not know this is a case where we have constant wall temperature yeah the same way if you look at the momentum the same way that you did for momentum if you look at the energy equation if you do not include the viscous dissipation term so at wall these are 0 therefore d square T by dy square has to be 0 okay so you can you can see that you have four boundary conditions in four most important boundary conditions and if you substitute them into the profile you should be getting a profile something like T minus T wall by T infinity minus T wall will be 3 by 2 y by ? T minus half y by ? T the whole Q now we can define this as my non-dimensional temperature ? which I also defined in the similarity solution okay so this is the resulting non-dimensional temperature profile just like I have a non-dimensional velocity profile exactly the same expression only in term rather than boundary layer thickness here you have the thermal boundary layer thickness ? T okay the same cubic approximation so the next step will be to substitute this into the energy integral and do the same procedure that we did for the momentum integral just a little bit I am going to rewrite the energy integral in terms of non-dimensional temperature ? okay so right now this is my T infinity minus T I can express this in terms of ? as d by dx 1 minus ? also I can multiply and divide by U infinity so U by U infinity and I can transform my variables from Y space to ? space where I am going to define a variable ? subscript T as y by ? T okay so this is not the similarity variable okay it is just some notation that I am giving here okay coincidentally this is also for the similarity solution the similarity variable exactly related as y by ? T so I am just transforming in terms of ? so this will be dy by d ? into d ? so dy by d ? is nothing but your ? T which will come out and the limits of the integral 0 to ? T will now become 0 to 1 so you have d ? T all right on the right hand side of course you multiplied and divided by U infinity so that I will take it to the denominator on the right hand side this will become a by U infinity and dt by dy I will write in terms of ? and ? so this I can substitute in terms of ? so this will be d ? by d ? into d ? by dy okay so d ? by d ? okay so here I have to be yeah I think it is fine so ? equal to 0 this is ? T equal to 0 into d ? by dy which is again 1 by ? T I will keep the ? T here okay so this is the non-dimensional form of the energy integral let me call this is number one okay everybody is clear okay so so now this is ? so 1- ? will be what T- T 8 by T wall by T wall- T 8 is that right so I can I can just substitute for T 8- T okay so that is 1- of 1- ? and on this side dt by dy should be – d ? by dy okay so – – cancels on both sides the T wall is a constant okay so is that right see okay so T 8- T now I can write it as – of 1- ? into T wall – T 8 okay so T wall and T 8 their constants so – of T wall – T 8 D by DX 0 to ? T into 1- ? u dy okay on this side I have dt by dy is nothing but – d ? by dy so this is – a d ? by dy at y equal to 0 to you have T wall – T 8 okay so I think now this cancels all right so now you now the rest of the things is I am multiplying dividing by u 8 and this is u by u 8 here and the other u 8 I am taking to the denominator here okay and I am transforming from y to ? T that is all okay any doubts I hope everything is clear so if you have any doubts please let me know I am going a little bit fast assuming that you could follow all right so this is the final expression because we have in terms of ? the profile and also u by u 8 so therefore we can directly substitute that into this expression so if you do that so now let me erase this so substituting cubic temperature and cubic velocity profiles into one so D by DX ? T 0 to 1 so my my cubic temperature profile is now 1- ? so that is 1-3 by 2 ? T ? T Q okay so I can write it as 1- 3 by 2 ? T and – of – is plus plus half ? T Q okay this is my temperature profile 1- ? and what is my velocity profile 3 by 2 let me also define the same way defined ? T let me define ? where ? equal to y by ? okay so my velocity profile can be written as 3 by 2 ? – 1 by 2 ? Q okay so into D ? T so this is integrated over D ? T okay so this is my left hand side on the right hand side yeah on the right hand side now what is D ? by D ? T okay so this is my ? D D ? by D ? T at ? T equal to 0 3 by 2 that is it okay so 3 by 2 into a by ? T into U 8 okay so this is therefore the final expression after we substitute now we have to simply integrate it out before we integrate it we have to write ? in terms of ? T because ? and ? T are both connected okay so therefore what we are going to do is we are going to introduce a variable another non-dimensional variable I will call this as ? which is nothing but the ratio of the thermal boundary layer thickness to the momentum boundary layer thickness so since this is the ratio of this and this this is this has to be a function of frontal number okay because this is the ratio of the two boundary layer thicknesses so therefore since I define my ? as y by ? and ? T as y by ? T you can also express this as the ratio of ? T and ? okay so this will be nothing but ? by ? T okay so now therefore I can substitute ? as the ? into ? T okay so this can be substituted into this and we can write this as D by DX so I can write this as ? T and this ? can be taken out of the integral because it is fixed for at a particular Y location okay so we need not now integrate it where whereas in this case ? is a function of Y okay so that therefore that has to be integrated across the boundary layer now ? is fixed value at a given X location so therefore it can be taken out outside the integral so this will be ? T ? and inside the integral you have 0 to 1 1 – 3 by 2 ? T plus 1 by 2 ? T Q into 3 by 2 this is again ? T – 1 by 2 ? T Q D ? T okay so I am just substituting ? as Z into ? ? T so okay so this should actually be written as yeah so this is ? here this should be ? Q okay so I just keep one more step okay so on this side you can retain the terms as it is so now you can integrate integrate it from 0 to 1 so integrate it with respect to ? T okay so if you do that I am not now going to do step by step but I will so this should come out to be D by DX and ? T ? if you take common this should become 3 by 20 – 3 by 280 ? square okay so this ? Q term we can write it as ? common out and ? square inside on this side you have 3 by 2 ? by ? T okay so if you integrate it you can you can do that yourself so each of these terms you have to multiply and integrate it okay so you have terms here ? here ? square ? T power 4 again again ? Q ? ? T power 4 ? T power 6 okay so you have to integrate each of them between 0 to 1 so finally the resultant solution will be will be independent of ? T and this is what you will get okay so from here we can make an approximation now already I said this is for a case where Prandtl number is greater than 1 if you make the approximation so so far we have not introduced any approximation on the Prandtl number okay so if you now bring in the approximation that Prandtl number is greater than 1 here if Prandtl number is greater than 1 what should be ? T with respect to ? less therefore your ? should be less than 1 okay so for Prandtl number much greater than 1 your ? should be much lesser than 1 okay so if you bring in that approximation you can neglect the higher order terms in terms of ? okay they are very very small so therefore if you do that if you retain only the most important term most significant term will be the first term so this will be 3 by 20 x ? ? T once again I can write ? T as ? x ? so therefore this has become ? square ? on this side I am taking 3 by 20 and factoring it out this will become 10 okay so this is this cancels this becomes 10 a by u 8 and once again ? T I can write in terms of ? and ? so finally I am reducing the entire equation to an equation only in terms of ? and ? okay so I know now the expression for ? coming from the solution to the momentum integral equation which I can substitute into this and now I will have an ordinary differential equation in terms of the unknown ? okay so I think with that I will be able to calculate also my thermal boundary layer thickness so so this is my equation so if I substitute for the cubic velocity profile ? was 4.64 square root of ? x by u 8 okay so if I substitute this value of ? into this and if I expand this particular ODE so this will be ? Q plus 4 by 3 x d ? Q by DX this should be equal to 0.929 by Prandtl number where Prandtl number is equal to u by a this is what you will finally get us the ordinary differential equation so I am just expanding this particular term right here okay so what I am doing is I am taking this ? here okay I am I am saying this as 1 by 3 x d by DX ? Q okay if you want me to explain this I will say so this can be written as 1 by 3 or I can say this is D by DX 1 by 3 ? Q ? Q ? is equal to 10 a by u 8 x ? okay I think yeah should be right so now I can expand this into two terms so one is ? Q into D by DX 1 by 3 ? so ? you can substitute and you can differentiate it and the other term will be 1 by 3 ? x D ? Q by DX okay and if you substitute for ? so you already have ? you can multiply the entire left hand side by ? for which you have the expression for boundary layer thickness and it will simplify on the right hand side you will have a by ? because you have an expression for new okay here which will come and you will exactly get a factor of ? in the denominator which is nothing but the parental number so couple of steps which you can do yourself you know I am just simplifying that once you simplify you will get this is your final order differential equation okay so this is very straightforward to solve okay this is a non-homogeneous ODE so you can find two solutions one is your complementary function which is the solution to your homogeneous ODE and a particular integral assuming that some particular solution exists you substitute so you will get those two solutions which you can add and that solution will be finally ? Q should be C X power some constant C X power – 3 by 4 plus 0.929 by PR so this is your complementary function and this will be your particular intake this is your solution for ? okay so once you determine the solution now we have to determine this constant so therefore we should give a boundary condition for ? okay so that is we know for the generic case at X is equal to X not your ? should be 0 right so if you put that condition at X is equal to X not my ? equal to 0 okay so what what will be the constant the constant will can you substitute and let me know what the constant comes out to be – 0.929 by PR and this will be X not power – 3 by 4 it will go to the other side this will become X not power 3 by 4 okay so therefore the final solution for ? can be expressed as 0.976 by PR to the power 1 by 3 1 – X by 3 X not – 3 by 4 the whole raise to 1 by 3 okay so I am substituting for C into this into this expression right here and I will get the final expression for ? okay so now this is done therefore we now know the thermal boundary layer thickness expression because ? is nothing but ? T by ? and ? is already known so we can find the expression for ? T and to derive a specific case okay for the case where you do not have the unheated length that is you start maintaining a wall temperature condition right from the beginning at X equal to 0 okay so for for the case where X not equal to 0 okay so this is X not by X the power 3 by 4 so this will go to 0 and therefore your ? will be simply 0.976 divided by Prandtl raise to the power 1 by 3 okay also for the case where there is no unheated length in its heated right from the beginning for a given Prandtl number the ratio of ? T by ? will be constant because Prandtl number is a constant okay so therefore the ratio of ? T by ? will be constant therefore you can see from this expression for a given Prandtl number the ? will be actually a constant okay if you do not have an unheated length if you have an unheated length you can see ? is also function of the position correct makes sense right because if you if you heat it right from the beginning both the boundary layers grow according to the Prandtl number and at any cross section at any section axial location the ratio of the boundary layer thicknesses will be such that it is a constant whereas if you maintain a starting length heated length unheated length then you can see at different locations the ratios will become different okay so that is what is coming out of these expressions so for the case where you do not have unheated length this is your expression so with this we can calculate the heat flux – k dT by y equal to 0 so dT by dy at y equal to 0 can be written as of course you are if you remember your definition of ? T – T wall by T 8 okay so you can you can write this as dT by dy at y equal to 0 and multiplied by T 8 – T wall okay so this will be – of course so – so I am going to write this in terms of T wall – T 8 okay again you can make a transformation so dT by d ? into d ? by dy okay so you can run this is ? T ? T equal to 0 and d ? by dy what is d ? by dy your ? T is equal to y by ? T so 1 by ? T so that will be 1 ? T in the denominator here okay and d ? by d ? ? T at ? T equal to 0 this is nothing but 3 by 2 okay so therefore this can be written as Q wall double prime as 3 by 2 k by ? T into T wall – T 8 okay so from this everything else follows you can define your local heat transfer coefficient as Q wall by T wall – T 8 which will be 3 by 2 k by ? T and therefore you can define your Nusselt number locally as H X by K okay which will be 3 by 2 X by ? T okay so finally you can substitute the expression for ? T and ? so already you know that ? by X is equal to 4.76 by square root of Reynolds number and this is nothing but ? T by ? so finally for ? T if you substitute this should come come out as 0.331 into REX to the power half Prandtl number to the power 1 by 3 okay so I think this should come out and in terms of the unheated starting length this will also add come as a factor 1 – X by X 0 to the power – 3 by 4 the power – 1 by 3 so this will be the expression for the local variation of the Nusselt number okay so far then for the heated length for the heated case right from the beginning X not equal to 0 so any X will become 0.331 REX over half and Prandtl number power 1 by 3 now you compare this with the solution similarity solution that was what 0.332 so it is very very close okay so here I am skipping couple of steps you can just do that yourself okay because you have this is your ? T by ? you can write this ? T as ? into ? and ? you already have the expression you can just substitute that into this and finally you will get a very neat expression in terms of Reynolds number and Prandtl number okay so far the case where you are hitting right from the beginning this is your Paul Haussens case now you remember when we did the similarity solution we cannot make the assumption that you have an unheated length and then get the similarity solution okay the similarity solution was derived based on the assumption that the wall temperature is uniformly applied through the entire length of the plate otherwise we cannot derive the similarity solution so therefore the approximate method the integral method gives us a choice to also impose this condition of unheated length okay see not only it makes our derivation simpler relatively compared to the similarity solution but also it gives some flexibility in terms of applying variable boundary conditions for example we will see later on down the line in a few classes 3 to 4 classes that we can solve use the similar use the approximate solution method to take a case where the temperature is varying along the wall not just something like this you know it could be linearly varying it could be some piecewise constant variation whatever may be that can be handled with the approximate solution okay which is which is quite difficult and which cannot be done in fact with the similarity solution so therefore although this also looks quite tedious but this can be worked out with the hand unlike the similarity solution which needs programming to solve the ordinary differential equation okay so relatively in terms of the total effort this is much better than doing complete similarity procedure and finally you also get a very accurate result in terms of the heat transfer so we will stop here today and I just want to also tell you we can we can also do a similar approximate solution for pranter number less than one which I am not going to do but we can make an approximation there since per pranter number much lesser than one your thermal boundary layer thickness is going to be far greater than your momentum boundary layer thickness so therefore you can make a uniform velocity approximation rather than substituting a profile okay so the velocities boundary layer is so small that most of the thermal boundary layer will be looking at a uniform velocity okay so that is a much simpler case to deal with and I will give you that as a homework assignment so in the next class on Tuesday we will look at flows with pressure gradients where we can apply the approximate solution.