 Good morning all of you so now we will look into the one of the last solutions that we can do in this course on similarity methods that is on thermal boundary layer with high speed flows so far in the earlier similarity solutions we had looked at either momentum equations without any pressure gradient term that is the simplest flat plate case and the energy equation without the viscous dissipation term also we had looked at some of the similarity solutions for flows with pressure gradient okay but still the energy equation was the same throughout in all of these configurations you know so now we want to consider a case where you can for example apply either falconess can flows for wedge or you can take a simple flat plate boundary layer and you can see the effect of adding a viscous dissipation term so when is this term is going to be important primarily if you are looking at the non-dimensional variables it turns out to be the ratio of a cut number by Reynolds number so for high values of this so the viscous dissipation term is going to be relatively important so we are looking at typically what are called as high speed flows it could be incompressible and somewhat compressible but not supersonic so it is relatively high high speed subsonic flows basically that is what we have that is what we are interested in we are also making an important assumption that the properties are constant even though they are high speed flows where relatively the temperature fluctuations could be strong and the properties also could be compressible like density could be compressible we are ignoring that we are assuming that density is constant and also all the other properties are constant and therefore we are taking all this out of the derivatives okay so that is another constant property assumption is another assumption that we are making okay so these are the boundary conditions so we are writing this for the flat plate case right now we can also do the same thing for falconess can solution okay the only the similarity solution for the momentum will change right now the similarity solution for momentum is still the blaster solution okay so if you are doing this for which flows then you have to rewrite that with the similarity solution for falconess can problem okay so now the modification will be in terms of the energy equation so now you have included the viscous dissipation therefore the energy is coupled to the velocity velocity gradients earlier in the falconess can or blashes the energy was completely decoupled okay in the sense only the one once you solve this you substitute the velocities and that is it here you also need the velocity gradients okay so little bit more coupling coming out and these are the corresponding boundary conditions for the flow it is very straightforward as far as the temperature is concerned you can either have two conditions one you can have an isothermal wall okay and in fact this is what we are going to see in this particular case we are going to consider flat plate where the temperature at the wall is isothermal it is constant and you have a very high speed flow you have your boundary layer development ? and ? T now you are interested in this case when when you include the viscous dissipation and with a isothermal wall what will be the effect of including the viscous dissipation okay so you can consider this problem into two sub problems since the energy equation is quasi linear you can actually split this into two sub problems with two different boundary conditions one problem is where you know don't take the viscous dissipation term into account and then you assume purely isothermal wall in the other you take the viscous dissipation into account and you maintain an adiabatic wall okay so in this case even with an adiabatic boundary condition you have heat transfer possible because of the viscous dissipation okay so if you do that of course the similarity solution for the flow is still the blaster solution now if you substitute for the similarity variables for u and v and du by dy into the energy equation and you don't write a non-dimensional form of temperature you just retain the dimensional form so this is the similarity ordinary differential equation for energy so you now see compared to the other case the falconous can or the flat plate case you have now this additional term coming from the viscous dissipation okay if this term was not there it will reduce to your earlier a more simpler form and now this can be subjected to either of these boundary conditions depending on the case so what I am saying is rather than solving this directly so there is one more way of doing it you can directly solve it I mean you can define a ? which is T – T wall by T ? T – T wall okay plug it in you can divide throughout by T ? T ? T wall and you can define this as your occurred number u ? T square by CP T wall – T ? is your occurred number so in that case you can you will find a non-dimensional form of this okay so this will become your non-dimensional form in terms of ? defined this way of course you can solve this by shooting technique just like any other ODE that you had seen before and you can get the solution for ? but we are not simply interested only in getting a temperature profile or the slope of the temperature in this case at the wall but we are more interested also in knowing what if what happens if you put an adiabatic boundary condition at the wall and what will be the value of this adiabatic temperature okay so this gives us an opportunity to analyze the case where your DT by DOI is 0 and you have viscous dissipation and what will be the value of this adiabatic temperature okay so in order to do that we want to separate the problems into two sub problems as I said do them separately also know get information about the adiabatic wall temperature and combine these two and get the final solution for the global problem you know which involves viscous dissipation with an isothermal boundary condition okay so this is what we are going to see today so the first the sub problem we can write as the isothermal flat plate with south viscous dissipation that is your case a that I have drawn on the figure okay so if you say have a simple profile like this now that means you do not include the viscous dissipation this is nothing but the Paul Haussens similarity solution okay so there you define your ? the same way that I have defined here T – T wall by T 8 – T wall and you substitute in your energy equation without the viscous dissipation you give you write the Paul Haussens equation which is nothing but the left hand side term of this so this is D2 ? by D ?2 plus half PRF into D ? by D ? equal to 0 subject to boundary condition at ? equal to 0 what ? equal to 0 and ? going to 8 ? equal to 1 okay so this is your solution which have already done you have solved by shooting technique or whatever and now the second part of the solution so this is one problem which you have considered what will be the second problem so one we have split this into two solutions one so this is now without viscous dissipation with isothermal wall the other should be including viscous dissipation but adiabatic boundary condition okay so this is adiabatic so this is what I call as case B and if you look at the temperature profile typically this is how it looks you know so this has to satisfy the zero flux at the wall therefore it becomes normal okay at the wall is you are okay so what we are trying to say I you can solve this equation as it is okay so this is for the case where you have a isothermal boundary condition with the viscous dissipation okay this is what we need the solution but we are also interested to know the solution for the adiabatic wall case we want to actually measure what is the adiabatic wall temperature so in order to do that I want to take a case where I have an adiabatic boundary condition with the viscous dissipation and also the fact that for this particular problem you can linearly superpose two solutions it works out and mean I mean that is what we are going to check you can have one problem where it is isothermal the other where you do where you include this and you make it adiabatic and the entire solution to this problem is a linear combination of those two solutions okay so we are going by that method now okay although we could have directly solve this and got the temperature profile which is nothing but this which I have drawn here we are also interested in calculating the adiabatic boundary the adiabatic temperature okay so in order to do that we have to solve case B now it is also possible that you can solve case A case B separately and keep it and you can linearly combine and get the solution for this generic problem so this is what this is what why I am doing this way okay so assume that you have two solutions one for isothermal case the other for adiabatic case okay you can combine them directly we will say how we are going to combine it and that will give your case which is your generic case isothermal wall with risk is dissipation okay otherwise you could have directly solved this equation by shooting method alright so coming to the second problem how are we going to define ? here because we do not have anything like fixed wall temperature okay so now the first question is how do so I can say this as something like T minus T 8 I can define but the denominator is a problem I do not have any T wall so what I would like to do generally in high speed flows there is a difference between the static temperature and the stagnation temperature okay so if you are looking at compressible flows in fact what is the relationship between the stagnation and static temperature okay so of course that low speed flows the velocities are so small that you know you can neglect this and therefore you can approximately say your static and stagnation temperatures are same but high speed flows you cannot do that okay so I am going to call something like my ? T which is your say T 0 minus T and you can say dimensionless quantity will be something like U 8 square by 2 CP so this is the ? T that I am going to use to scale this okay so this is what I am going to how I am going to scale my ? and of course what I am going to do is I am going to define ? which is different from this ? because this is the isothermal ? so I am going to define the notation a for adiabatic boundary condition okay so I am going to substitute now for T a into this and write in terms of ? so that will give me d2 ? a so you have U 8 square by CP and when you substitute there is a U 8 square by 2 CP so finally you will be ending up with what minus 2 times Prandtl number into F double prime the whole square okay so the boundary condition now at ? equal to 0 what will be the boundary condition ? equal to 0 why why it should be 0 your gradient of yeah so d ? a by d ? should be equal to 0 and your ? going to ? is equal to 0 okay so now you have two definitions of ? one for the isothermal boundary condition the other for adiabatic boundary condition and finally we have to mix these two solutions okay so now this again can be solved by shooting method right so this is another OD which you can write into two first order OD is you have boundary condition for of course you do not have the boundary condition here for ? at ? equal to 0 so you can guess that you have boundary condition for ? prime at ? equal to 0 so you can guess the value of ? at ? equal to 0 such that the ? at some ? equal to 10 become 0 okay so this is the other way of doing the guess guess work so that you match this condition so you keep guessing the value of ? at ? equal to 0 okay so anyway so you can solve this problem by shooting method this problem already you have the solution by shooting method right so once you have these two now I have two solutions one for the adiabatic case one for the isothermal case I want to now find how do I combine this how do I propose a solution which is a linear superposition of these two solutions okay ? is still the same ? is because the ? here is coming from the flow part okay the flow similarity variable is still the same Y by ? which is Y square root of U 8 by ? X that we can wait only because in that case we consider the viscous dissipation to be we neglect the viscous dissipation now this is coming purely from momentum that has nothing to do with viscous dissipation viscous dissipation is affecting only the energy okay as I said the momentum part is still your blashes equation okay so the same thing applies the same similarity variable is put into the energy yes so no okay only possible if viscous dissipation is neglected no then you would not have been able to derive a similarity variable similar similarity solution here this see the fact that you are able to reach a similarity equation shows that this is the right similarity variable okay I what I meant by when that was the way when Paul Hausson guessed that he could derive a similarity equation for energy okay because he found for Prandtl number 1 U by U 8 is exactly equal to ? and they both the equations are similar but does not matter that does not mean that Prandtl number greater than 1 okay the form of ? will be different no use the same ? only and he found that it still gives a similarity equation so here also when you put all your ? finally it comes to this which is still a similarity equation okay so therefore therefore the energy equation does not care much about the ? so the same ? what you use for the momentum is still retained here this is your boundary layer similarity variable that is it okay once again if you go for different kinds of flows like say free shear flows there will be different similarity variable okay so there you will be defining something like Y by Y M by 2 in the case of jets okay so that is that is your jet half width so this is your maximum velocity somewhere you get half of the maximum velocity and corresponding to that is your Y M by 2 so you use that to non-dimensionalize your Y okay so in different kinds of flows as far as boundary layer flow is concerned this is the similarity variable that is it okay so now I am going to just give you a general solution and before that let us try to understand what how this definition of ? what it means at ? equal to 0 so at equal to 0 okay so I can calculate the value of TA at the wall okay that is nothing but what the adiabatic wall temperature right so that is my TA wall minus T infinity will be equal to ? a at ? equal to 0 times U infinity square by 2 CP okay so this is how I can calculate my dimensional value of adiabatic wall temperature once I know value of ? a at ? equal to 0 and how how this this value come by guess work okay so you keep iteratively guessing the value of ? and still it matches this and that final value is your ? at ? equal to 0 so this factor is also called as a recovery factor although this is a non-dimensional temperature okay this relates basically your dynamic head to the adiabatic wall temperature so this is also called as the recovery factor okay so what you have defined as a stagnation temperature is what something like say T infinity plus U infinity square by 2 CP now T adiabatic wall is T infinity plus some factor times this okay so it is not equal to 1 okay but less than 1 so that factor is called recovery factor okay so therefore you can you can imagine that this is something similar to your stagnation temperature that you have defined but maybe of a different factor okay so that factor comes through this recovery factor here okay now I am going to propose a general solution which is a combination of the two cases so I am going to define T as a function of ? minus T infinity okay so T minus T infinity now already for the case B T minus T infinity is defined as ? a into U infinity square by 2 CP that is one part of the solution okay now I want to blend these two solutions there is a linear combination I want to propose so this is one solution now here I have defined ? as T minus T wall by T infinity minus T wall so if I want to get to the form of T minus T infinity then I will say this is 1 minus ? okay 1 minus ? into T infinity minus T wall or if I say 1 minus ? it becomes T minus T 1 minus ? will be what T minus T infinity by T wall minus T infinity is that right the science will flip okay so therefore I can just write this T minus T infinity as 1 minus ? into T wall minus T infinity okay so this can be written as T wall minus T infinity into 1 minus ? okay still this is not a linear combination because I have just simply added it okay so I have to propose some factor some constant which gives a weightage for this part of the solution and this part of the solution so this is a relative weightage I am giving a weightage of 1 for this corresponding to that what is the weightage to this okay it can be less than 1 it can be greater than 1 so this is your final form of the general solution okay so what you are saying is your general solution is a linear combination of these two solutions and how do you know that this is the correct one so you can simply substitute this back into this equation number one and you will check that it satisfies that equation that means this is the right combination okay so I am not going to do that you can do that yourself and check simply substitute this and you will get you can group into two forms okay so one form will be of this which is already satisfied by itself okay the other form will be this which is also satisfied okay so those two will be perfectly satisfying the ODE so this solution satisfies equation one okay I want you to just check that so therefore this is the right solution now one more thing already when we propose this solution we have satisfied the boundary condition that it ? going to infinity T equal to T infinity okay because the way that we have defined my ? so at large values T will become T infinity and this will become 1 here and here the value of ? becomes 0 so this naturally satisfies the boundary condition that this is this okay but still I have not come clearly specified the wall boundary condition here okay so this constant can be determined by satisfying the boundary condition at the wall okay so to calculate C so the BC should be at ? equal to 0 for the solution which involves ? T wall should be T infinity and the solution which inverse ? a so that DT by D ? should be equal to 0 okay so if you can can you substitute this boundary condition and tell me what will be the expression for C so T at ? equal to 0 minus T infinity okay so this is nothing but what T at ? equal to 0 is what T wall okay that is a fixed boundary condition so T wall minus T infinity will be equal to C into T wall minus T infinity into what will be ? at y equal to 0 0 right therefore this will be just 1 plus U infinity square by 2 CP into ? a ? equal to 0 okay now from this definition ? a ? equal to 0 into this is nothing but PA wall minus T infinity okay okay so this will give me C equal to what 1 minus so I can just take this common so this will be 1 minus C ? equal to this so C will be 1 minus TA wall minus T infinity divided by T wall minus T infinity okay or I can rewrite this as C is equal to T wall minus T adiabatic wall by T wall minus T infinity okay so now that I have determined C now the solution is now complete okay so once I find two separate solutions one for ? one for ? a I found out the constant factor which combines this linearly okay so therefore I can write my final solution ? minus T infinity equal to T wall minus so minus T wall minus T infinity cancels when I substitute this it will be T wall minus T adiabatic wall into 1 minus ? plus okay so I can still retain this as U infinity square by 2 CP into ? as a function of ? and so this is my final solution okay I can also go one more step I can non-dimensionalize this with T minus T infinity by T wall minus T infinity okay so I can divide throughout by T wall minus T infinity so that I get us final expression for non-dimensional temperature and this will be still 1 minus ? this is a function of ? and rental number now this will be I can write this as if I divided by T wall minus T infinity this will be U infinity square by CP T wall minus T infinity is a cut number so a cut number by 2 into ? a now this part coming to T wall minus T adiabatic wall by T wall minus T infinity okay so that can be written as you can just check 1 minus ? a at ? equal to 0 into a cut number by 2 okay because ? a and into a cut number that is that is basically this term a cut number is U infinity square CP ? T into this okay so that will be T adiabatic wall minus T infinity so you can you can say 1 minus this divided by T wall minus T infinity will give you this final you can just check that okay so this is your final non-dimensional solution okay so all that requires is your solution for ? okay which is coming from case A and for ? a coming from case B and you combine this like this and you get your final solution also this this is nothing but your ? right so this was this could have been directly obtained by solving the similarity solution that I had written before okay directly okay the same thing can also be done by linear combination of these two fundamental solutions so let us see how if you plot the solution for temperature how this profiles look like okay any doubts on this so far so if you plot the temperature profile first let us look at case B solution because case A we already know how the temperature profile looks case B if you plot it as non-dimensionalize the adiabatic temperature is T a minus T infinity by T a wall minus T infinity such that it scales between 0 and 1 okay so at y equal to 0 this becomes T adiabatic wall it becomes 1 at y going to infinity this becomes 0 okay so on the x axis you have ? so for large values of ? so this becomes 0 okay small values at y equal to 0 this becomes 1 okay it starts from 1 and you will see the solution so this is with increasing Prandtl number your slope at the wall keeps increasing this Prandtl number 0.613 this is like 300 this is like 1000 okay so this is your case B solution that is only for the case with adiabatic boundary condition at the wall this is how the non-dimensional temperature profile behaves once you once you solve this if you want to call this as number 2 and number 3 so if you solve 3 by shooting method this is what you get finally okay so this is coming from solution of 3 by now for the general solution for this the combination of these two solutions if you can plot for a given value of Prandtl number that is my T of ? minus by T wall minus T infinity okay as a function of ? this will also be between 0 and 1 and this will show a behavior something like this okay my 0 actually somewhere here this is for different values of a cut number into ? a ? equal to 0 this corresponds to a heated plate okay so in the case of heated plate your wall temperature is greater than your T infinity in the case of cool plate wall temperature is less than T infinity so you can get negative values of this when you plot it and these are for different values of a cut number into ? a ? equal to 0 so you can write this as something like a cut number into ? a ? equal to 0 so if you look at the solution for this problem 3 and if you look at the generic solution here this depends on ? ? a and a cut number so you have to plot this for a particular value of Prandtl number for a particular value of ? a ? equal to 0 a cut number okay so this is could be for say Prandtl number around 1 okay so you have to fix the Prandtl number here because everything there are too many parameters you have Prandtl number we have the a cut number here so you fix the a cut number to some value fix the Prandtl number and then plot the variation of this non-dimensional temperature as a function of ? and this is how it looks now there could be different values of a cut number into ? a for a heated plate you have profiles which are like this for the cool plate you have profiles which are like this okay so all this can be verified once you solve this equation by shooting technique okay so for a particular value of a cut number you can combine these two solutions and you can get this kind of a temperature profile okay so finally we are interested in the calculation of E-transfer coefficients we will see how that can be obtained so if you are confused about the definition of this your a cut number into ? a at ? equal to 0 is nothing but your T adiabatic wall minus T infinity by T wall minus T infinity okay so because this is your TA wall minus T infinity so this should be a cut number is again U infinity square by CP into T wall minus T infinity so this should lead to some non-dimensional form of temperature like this okay so for different values of this you are plotting plotting this if you are not clear about how the a cut number comes into picture okay so that is how the adiabatic wall temperature enters here through the a cut number times this now let me also write down the form of heat flux okay so now when we write down the heat flux we look at the temperature profile which is your general profile here so therefore I have to differentiate it with respect to the general profile so this will be a minus K first you can write this as D ? minus D ? by D ? so minus and minus will get cancelled here so this will be K into T wall minus T adiabatic wall into D ? by D ? into D ? by dy this is at ? equal to 0 okay plus you have U infinity square plus K into it will be a minus here U infinity square by 2 CP into D ? a by D ? at ? equal to 0 into D ? by dy right so this is coming from the general solution now what is the value of D ? a by D ? D ? equal to 0 so this entire term and we knocked off and this can be written as Q wall into D ? by dy is square root of U infinity by Ux okay into ? prime at ? equal to 0 okay now once you determine the value of ? prime at ? equal to 0 that is for the Paul Hausens condition this is coming from the Paul Hausens case so when you solve this you naturally determine ? prime at ? equal to 0 so once you do that you substitute it and this will give you the wall heat flux all right so we have already done that okay if you can perhaps recollect if you can recollect for Prandtl number range in fact Paul Hausen did this for small Prandtl numbers for intermediate Prandtl numbers for large Prandtl numbers okay the correlation for small Prandtl do you remember what it was minus half not for small Prandtl for intermediate Prandtl number of mass okay so that was 0.332 Prandtl number raise to the power 1 by 3 okay so this can be substituted to calculate Q all double prime and from there you can calculate heat transfer coefficient you can now define heat transfer coefficient how do you now define it now this can be T wall minus whether you want to use T infinity here or T adiabatic wall T infinity why because if you look at this form right here if you had defined based on T wall minus T adiabatic wall this will get cancelled straight away okay so therefore we would like to define my heat transfer coefficient based on T wall minus T adiabatic wall so in that case and therefore my so this will come out to be 0.332 K Prandtl number 1 by 3 where sort of new U infinity new x this will exactly cancel the temperature differences and therefore my local Nusselt number I can define as Hx by K will be 0.332 into Reynolds number power half Prandtl number 1 by 3 okay so now you see the Nusselt number relation is exactly identical to the flat plate case which is isothermal there is no difference okay but the definition of heat transfer coefficient is different okay now you are defining this as T wall minus T adiabatic wall not T wall minus T infinity okay now this is the difference between the high speed flows and if you neglect the viscous dissipation okay apart from this the final expression for Nusselt number is still the same all right so the important conclusion here is that for non-dimensionalizing okay you are there are many temperatures you have now T infinity now you have T wall you have T adiabatic wall so all the three has to be considered some way okay and to define heat transfer coefficient you have defined based on T wall minus T infinity okay so everything is fine so therefore since you are arriving at the same correlation your Reynolds analogy is still valid okay so the Reynolds analogy which says that Stanton number is equal to CF by 2 for Prandtl number of one okay for otherwise the Stanton number 2 by 3 is equal to CF by 2 so this is still valid for high speed flows also because for Prandtl number equal to 1 you are still retain the same expression for between Stanton number and CF so therefore the Reynolds analogy is still valid now one thing is for calculation of properties so far for low speed flows how did we define the property how did we calculate the properties we calculated them at some mean temperature which we also call as film temperature okay so this is called the film temperature which is the average between the wall temperature and the free stream temperature okay so if you if you can also extend this and write like T infinity plus okay so you can write this in this form okay now the same way in the case of high speed flows we have to define some mean temperature to calculate the properties okay so now we have as I said three different temperatures to non-dimensional is okay now so already you have 0.5 into T wall minus T infinity now you also want to account the adiabatic wall temperature somehow so therefore we define plus another constant into T wall minus T adiabatic wall okay so this constant generally is empirical it is taken somewhat something like 0.22 okay so we calculate mean temperature or film temperature this way in the case of high speed flows and then we calculate all the properties based on at this value of property at this temperature okay now for the limiting case where you do not have viscous dissipation the T wall will become equal to exactly T adiabatic wall okay so it will reduce to the form of your earlier case that is your film temperature okay so you have to be careful that the definition of heat transfer coefficient is different but finally the form of nusselt number is exactly the same and the properties where you calculate the property corresponding to the temperature the temperature has to be calculated in this particular fashion okay so this is a very very important thing that you so with this we have completed all the similarity solutions for external boundary layer flows if you want I can just quickly summarize the similarity solutions that that we did before we stop okay so summary of similarity solutions just two more minutes so for velocity I am just writing the global similarity solution which is nothing but the falconous can from there all the other special cases can be derived so for the falconous can you have F triple prime okay the boundary conditions are for the similarity problem your f of 0 f prime of 0 should be 0 f prime at infinity should be equal to 1 now what is the condition most general condition for f of 0 do you remember from our transpiration problem this is minus blowing ratio times 2 by m plus 1 okay the blowing ratio is a constant value okay when you blow blowing ratio equal to 0 that becomes the case without transpiration where f of 0 equal to 0 okay now we in all these cases we are we have made the assumption that your u infinity is of the form Cx power m okay and your blowing ratio is essentially a constant and the similarity variable is right so these are some of the assumptions under which you get a similarity solution once you get it you can calculate your skin friction coefficient as 2 times f double prime of 0 divided by square root of local Reynolds number so same way for temperature the most general similarity solution is ? this is your most general form okay including high-speed flows okay if you are a cut number is small you can neglect the right hand side terms and it reduces to your pull out sense equation okay the boundary conditions ? 0 equal to 1 ? at infinity equal to how did we define ? in the earlier case it was you remember was it T minus T infinity by T wall minus T infinity and you just check how we defined it then you may have to flip the for the general problem T minus T infinity by T wall minus T infinity that is how I think I defined it so in that case the boundary conditions flip okay so ? at the wall now will become equal to 1 okay ? of free stream will become 0 if I define as T minus T wall by T infinity minus T wall it will be 0 and 1 okay so here my definition of ? equal to T minus T infinity by T wall minus T infinity so only then I can write in terms of a cut number here that is why I did like this okay so for this the Nusselt number relation finally will be my minus ? at 0 into rex square root of rex or rex per half okay so these are the most general similarity solutions of course for particular cases we have derived and also we have derived for the general case from which we can get those specific cases so we will stop here today and in the next class we will look at the integral method solution.