 Good morning to you in the last few lectures about six or seven lectures so far we were focusing on deriving the governing equations describing the flow and heat transfer of course we have done that in a Cartesian coordinate framework as well as a coordinate free framework based on applying the Reynolds transport theorem and finally so we were looking at approximations to two-dimensional incompressible flows to Navier-Stokes equations one is casting that into form of stream function vorticity equations which we have derived and if you apply that to a typical problem which is for a external boundary layer flow okay so far that we can use some scaling arguments do some order of magnitude analysis and then conclusively show that certain terms can be dropped out provided that you are dealing with sufficiently high Reynolds number flows okay so under that approximation we have derived what are called as the boundary layer equations okay so let us summarize the boundary layer equations which we derived in the last class okay so let us write it in a dimensional form not the non-dimensional form so finally the continuity equation stays as it is okay so the boundary layer equations what we are writing here is for 2d incompressible for steady state flows okay so of course and this is also we are assuming they are laminar okay that is another important assumption so something I have to tell you about this when we derive the Navier-Stokes equation we do not really derive one set of equation for laminar flows and another for turbulent flows the Navier-Stokes equations are valid for all kinds of flows okay so the problem is in solving those equations numerically so most of the times if you are applying the same set of equations for turbulent flow and turbulence has a lot of features it you have to resolve all kinds of scales in turbulence length and time scales and therefore numerically it may not be possible to do that kind of resolution so then what we do in order to model the turbulent equations turbulent flows we construct what are called as Reynolds averaged Navier-Stokes equations okay so we average the equation that we have and we can do different kinds of averaging of course for steady state it does not matter we have ensemble average for unsteady problems we can do time averaging and when we do this averaging we are trying to get information only on the mean properties of the flow such as mean velocities mean pressures and mean density we are not interested in all the microscopic turbulent fluctuations and turbulent quantities okay so therefore we are going to when we apply these kind of equations to model turbulent flows I think when professor Kohler comes I think he will discuss how we average the corresponding equations and then we get the Reynolds averaged Navier-Stokes the structurally everything will look similar only apart from the laminar stresses you will also end up with turbulent stress term okay and the problem is how do we close this particular turbulent stress okay so that is all the different turbulence models are about okay so they are there are different there are many books written on you know turbulence modeling many so many the hundreds and hundreds of papers published on that so this is a very very important issue where most of the practical flows are turbulent okay as is the if you solve the Navier-Stokes equations as it is there is no problem but most of the times you cannot resolve all the scales and therefore you have to approximately model what the water called as averaged quantities and that is that is where the Rans equations come so far we have not discussed those things I am just looking at plain Navier-Stokes equations and approximations so you can say safely that this is mostly apply to laminar solutions okay so when we solve this the resulting solutions are all for laminar flow okay for turbulent flow again the characteristics become a little bit different and therefore we have to solve the Rans equations and most of the times you do not have analytical solutions okay you have to solve them numerically but a few of few extreme cases have analytical solutions which we can do okay so this is the continuity equation this is your momentum only your X momentum stays your Y momentum tells you that your DP by DY is approximately 0 and finally your energy equation okay so what should be the discuss dissipation term here okay ? but what is the ? Cp right I am dividing by ? Cp everywhere so this is K by ? Cp which is a ? U by ? Cp times DU DY the whole square okay so this is the only term which is staying there all right okay so these are the boundary layer equations finally all of you agree and you know why only this term stays right okay so if not you please go back and revise so now in order to solve these equations what do we do we need definitely some boundary conditions right these are partial differential equations so can you tell me what are the boundary conditions that we can we can apply say we take the case of our flat plate the same yesterday we have a momentum boundary layer ? thickness thermal boundary layer ? T and the wall is maintained at a constant temperature and you have a flow U 8 T 8 okay so now you have to tell me what are the boundary conditions which we can apply for first the momentum and next the energy okay so at y equal to 0 so how many boundary conditions we need in the x direction for for example the momentum as far as the velocity is concerned along the x direction how many boundary condition 3 why okay so if you are too much bothered about okay I am just asking for velocity okay so 2 2 in along the x y so why do we need to and how about in the y 2 plus 2 what is this combination y 2 in x direction what is the order what is the order of the PDE in x direction first order what is the order in y direction second order okay so any so how many boundary conditions does the PDE need equal to number of what is the order of the PDE right so therefore for along the y direction how many we need to give 2 okay so 1 we are saying at y equal to 0 U is equal to 0 so this is what no slip boundary condition okay so this is the biggest contribution to Brantel okay who has invented the boundary layer theory we discovered that the fluid velocity should be exactly identical to the solid wall velocity okay now also when you complete saying no slip and there is no vertical velocity or there is no transpiration through the wall okay the vertical velocity also has to be 0 so and what is the other boundary condition along why we can give why okay let us say y going to infinity okay so you are you infinity okay and so therefore you have two boundary conditions with respect to you and now what is the remaining boundary condition we need to give one in long x okay so we will say at x equal to 0 you should be you infinity okay so similarly for the temperature the temperature also requires two boundary conditions along why one boundary condition along x okay so what is the boundary condition at y equal to 0 either this or if I apply a uniform heat flux instead of uniform wall temperature if I say my Q all is constant so then it should be minus k dt by dy at y equal to 0 should be equal to 0 why it should what happens if it is 0 so adiabatic boundary condition okay so therefore this is the condition at y equal to 0 how about at y going to infinity t should be equal to T infinity and x is equal to 0 at t should be T infinity okay so this is these are the required boundary conditions to solve the boundary layer equations all right so now we will first see the solution for the flat plate flow so this is also called the Blasio solution I think most of you are familiar with Blasio solution okay so this is anyway a flow problem you must have done in your fluid mechanics as well as in the heat transfer anyway for the convective heat transfer we have to revisit the flow problem before going to the heat transfer so I am going to go through the solution okay in probably in a slightly different way from how it was done before so for the flat plate case what are the approximations we can make okay still you should understand that this is a considerably complicated set of equations even if you write it for a boundary layer and of course we can transform this into a form which we can solve okay and we are going to do that but before we go to a complicated case where we have a pressure gradient okay so now we are going to first apply that to a very simple geometry the flat plate okay so for this case what kind of approximation that you can make can you see some some approximation further to this pressure gradient is 0 okay so therefore since your free stream velocity is constant okay so this term can be knocked out that we have seen yesterday and also right now we could retain the viscous dissipation term and solve okay but that is a slightly complicated way of solving it and we are not going to do now if time permits towards the end of this course I will take up this particular case okay so what it means even if you have an adiabatic boundary suppose you put minus K DT DY Y equal to 0 is 0 and you have this viscous dissipation you will find a temperature profile okay so this is something which we will be taking up when we have if we have time okay otherwise for this particular case when is the viscous dissipation important as we said when you have the ratio of your Eckert number by Reynolds number which is kind of very high okay so when that can be possible when your flow speed is high because a record number is directly function of the flow velocity or a Reynolds number is very small that is extremely viscous flows or very high speed flows okay so in those two cases you have to consider the viscous dissipation okay so in the other cases the moderate velocity regimes okay you can safely neglect this term also okay so the resulting equations now look slightly much more simplified and this is what we are going to solve for the Blasius solution so before going into the solution just to give you a brief brief history how this was originally proposed so first it was Prandtl who pointed out in 1904 okay Ludwig Prandtl he was a very famous German physicist mathematician and aerodynamic who was actually credited with the invention of boundary layer concept of boundary layer and in 1904 he says that the boundary layer equations for flat plate okay can be transformed into a ordinary differential equation okay so in those days they did not have very sophisticated computers and therefore anybody was looking at set of PD is that look like no they could not solve it because numerical methods were not that popular there was no computer at that time okay so they could not find a direct closed form solution to the PD but Prandtl proposed that if we can convert the PD to a ODE okay so then ODE is always solvable even numerically you can do it by hand also okay so therefore he proposed that the boundary layer equations can be transformed to an ordinary differential equation and this transformation is through what is called similarity variable so this is the it is a great concept that he introduces that all the time we do not have to go for a numerical solution if you are clever enough to see some kind of relationship between the flow velocity non-dimensional flow velocity and the property and the for example the coordinate that it is dependent on if you cast them in a correct form that you can transform the PD directly into a ODE okay so this is called as a similarity solution and in fact it was a student blushes okay in 1908 who did it actually who did this similarity solution for the flat plate which we are going to do so what we are going to do next is we are going to identically trace what blushes exactly did it exactly in the way that he solved it okay so this he got the solution for the flat plate velocity profiles okay so he was a student and later on I think it was full house and full house and in the year 1921 who extended the blushes solution for velocity profile to solve the energy equation okay okay so it took almost a decade to solve the energy equation from the original blushes solution and then in 1930 was Faulkner and scan there were two people Faulkner and scan who actually extended the similarity method of solving the PD is to flows with pressure gradient also okay so far we are neglecting this particular term here but they have extended for a generalized set of flows you know flows including pressure gradient without pressure gradient so all of them they constructed a similarity solution which we are also going to see okay so these are general family of problems okay so therefore you can see that it had a considerably long history and in fact in the first in this course when we are talking about external flows we will be doing all of this okay starting from the blushes solution till Faulkner and scan and probably an extension of this to different boundary conditions okay so all these are basically closed form solutions which you can do but of course the resulting OD cannot be solved analytically blushes himself struggled with that we will be using some elegant numerical techniques to do that okay so now we will start with the blushes solution okay now what blushes thought now if you look at the velocity profiles okay so you can draw the velocity profile something like this here if you go a little bit downstream so if I were to calculate the gradient of velocity at the wall du by dy okay so what will happen to this gradient as I go downstream will it increase or will it decrease it will decrease right decrease why what is that it is always 0 always at the wall it is 0 but why the gradient should decrease yeah the boundary layer thickness keeps on increasing and you have more flow which is entraining from outside okay so therefore the profile becomes more gradual and gradual as you go down so initially you have a very sharp gradient at x is equal to 0 the gradient will be infinite okay so you do not have any profile which is developed okay slowly your profile develops and it becomes more gradual and it has to satisfy continuity also right downstream so therefore what he has observed if you can find a variable okay so a non-dimensional form of the velocity which we can say u by u 8 because we can see the order of magnitude of use of the order of u 8 right so if we say u by u 8 this should be some function of some variables now you can see the variables which you are dependent on it is it is a function of two variables one is your x and the other is your y okay now when you look at when you look at ? you can also see ? as a function of x clearly the boundary layer thickness is a function of x so therefore now also we saw that y scales with ? very well y is of y y is of the order of magnitude of ? so therefore if we combine these two variables as a non-dimensional group so he says that if we can write this as some function of y over ? which in turn ? is a function of x okay so this is a non-dimensional group and if we can find this function so this should directly explain the dependence of y on this single variable rather than on variables y and x right so if you say y by ? is some variable ? which we call as similarity variable okay so now your u by u 8 is just a function of ? directly and once you use the similarity variable you know to have do not have to solve this in terms of x and y you can just directly solve in terms of ? okay that will help you in reducing the PDE to an ODE so that was this basic observation okay now how he obtained the basically the relationship between ? y and ? so now we should know how ? is a function of x okay one directly you can use simple scaling arguments like what Beja and does look at the inertia and this terms balance and then you can directly get the functionality of ? okay the other long way that I am going to do is helpful because anyway we do not have to derive the same set of equations again okay so and that is what I am going to do now so what I am going to do is I am starting with the x momentum equations and I am going to integrate it along y from the wall surface all the way till the boundary layer thickness ? okay so if you integrate it out so directly you reach to this particular form and what I am going to do now is a little bit jugglery here to eliminate in terms of V and construct that in terms of you so how do I do that I want to come from the continuity okay so your continuity says that your du by dx this is equal to minus dv by dy okay so I can write for V I can integrate this and I can say this is minus integral 0 to ? du by dx into so integrated 0 to ? over dy and now we can so this is your kinematic viscosity so we can substitute for so I can substitute into that okay so what I am going to do a little bit more I am going to integrate by parts this particular term I can write it as so d by I can write this as d by dy u v dy integral 0 to ? right minus 0 to ? u dv by dy dy right d what is that I think this is right okay which one okay so I am integrating by parts okay if I do that this I can directly write it as u v between the limits 0 and ? okay so this will give me of course at 0 u and v are 0 at ? u is equal to u 8 so I can say that this is u 8 times v at ? okay so minus this particular term here and of course from continuity I know d v by dy can be written in terms of du by dx okay so therefore this can be further written as minus u 8 so v at ? is this right here this is my v at ? right so I have integrated it from 0 to ? and at 0 v is anyway 0 so v at ? will be this so I can substitute for v of ? from here so this will be minus u 8 times 0 to ? du dx dy okay again d v by dy as minus du by dx so this will be plus integral 0 to ? u du by dx dy okay so this is how my second term so now we can see I have eliminated v completely I have written everything in terms of u okay is that clear which part integration by part okay so basically I have integrated by part here okay and then the first part if you integrate it you will end up with between the limits 0 and ? at 0 the velocities are 0 at ? u is nothing but u 8 multiplied by v at ? okay now from the continuity equation when you integrated between 0 and ? your v of ? can be expressed like this so I am just substituting for this directly here and d v by dy is nothing but minus du by dx okay so I am just using continuity to plug in okay so this is the resulting equation so if I substitute for this term and then write the complete equation so you have to tell me how the resulting equation should look like okay so already you have u du by dx there is another u du by dx okay so that will be two times integral 0 to ? u du by dx dy and then you have minus u 8 0 to ? du by dx dy that should be equal to u u by ? y at now this has to be integrated from 0 to so this is the limits I am applying okay this is nothing but okay you can call this put this as equation number one and please store it in your mind somewhere that we have derived this equation because I am not going to do it again when we do the integral method okay so this is nothing but the momentum integral equation so that is slightly cast in another form you can say this is du2 by 2 dx and 2 to cancel so you can write du2 dy- u 8 so that slightly cast in a different form but anyway that is basically coming from this particular equation yes this is your momentum integral your momentum integral is written in only in terms of u your entire equation is in terms of you there is no v coming there okay and of course your you know your momentum integral right what is your momentum integral ? how is it denoted u by u 8 integral 1- u by u 8 dy 0 to this is this is the definition of momentum integral okay or people call it as momentum thickness sometimes okay so we can you can you can do this as a nice exercise you can combine these two terms bring it to this form and then write in terms of d ? by dx okay so that is another way of writing casting this equation okay so anyway so that we are not interested in solving for the momentum integral right now what we are going to do we have already assumed that my u by u 8 is a function of my G which is function of y by ? this is nothing but my similarity variable ? okay so I am going to plug in this particular solution form of solution which blushes as assumed into this particular momentum integral so why I am going to do all this is finally I have to solve for ? okay so first if you calculate your du by dx from here how will you calculate du by dx so that is basically u 8 times okay so you should go back and revise all your differentiation rules okay G ? minus y by ?2 into d ? by dx okay so how you got it you can you can say that this is some function of ? right and so your du by dx is nothing but u 8 times d G by dx okay so this is nothing but d G u 8 times d G by d ? into d ? by dx and now d G by d ? is nothing but your G prime okay and d ? by dx so this it has nothing but y by ? okay so then you get minus u by y by ?2 into d ? by dx okay so you please go back and revise your differentiation rules okay so once you get this now the next term that we need here what is the other term du by dy okay so what what will be du by dy here du by dy is much easier u 8 into G prime G prime by by ? okay so but ? is a function of X you do not have to differentiate it okay so now you substitute for this du dx and du dy okay and you can group all the terms together which I am not going to do step by step I will only show the equation after it is grouped okay so if you substitute into 1 and you group all the terms I am just going to write the final form you get something like a – b into ? d ? by dx is equal to ? by u 8 times C now this a b and C are constants where a is equal to 0 to 1 G prime y by ? into d y by ? so what I am doing is this integral dy wherever I have I am writing this as dy by ? into ? okay so therefore the integral limits become 0 to 1 okay so you can check that it is a nice exercise you can does not take too much of time you know you are just rewriting everything in terms of y by ? grouping the terms together and B is 2 2 times 0 to 1 G G prime y by ? into d y by ? and your C can you now guess and tell me what C should be this you should be able to tell 0 to 1 why it is already so already integrated and the limits are to be applied what is du by dy u 8 G bar by ? okay so actually when you put that all of this u 8 can be cancelled off so you have new times G bar now G prime G prime between the limits ? – ? 0 divided by you have another ? and that ? also will cancel off so finally C will come out to be G prime of 1 – G prime of 0 okay so this because you are transforming all your y with respect to y by ? okay so wherever you are applying limit ? that becomes 1 all right so this is your ODE can you tell me what is the solution for this how do we solve this ODE separation okay so what will be the solution for ? okay so I can write this as d ?2 by 2 by dx is equal to C by a – B into so I can separate it as you said u by u 8 x dx all right so if I integrate it from x is equal to 0 to some position x okay so I can use a dummy variable here and I can say I can integrate from 0 x is equal to 0 to some position where you are interested to calculate the boundary layer thickness okay so this will give my ? is equal to so 2 times square root of 2 C by a – B into square root of ux by u 8 okay so therefore you can see clearly that we have arrived at a relationship between ? and x okay so your ? therefore is a function of x through this particular relationship and therefore if you if you say your y by ? ? is equal to y by ? so that will become y times square root of u 8 by ux therefore this is my similarity variable okay this is my similarity variable which goes like this so this similarity variable now you can see is constructed as a function of both x and y okay so now if I do this way what it means my u by u 8 is going to be a function of only ? anymore so not on x and y respectively all right and this is the similarity variable now the same result you can get by very very just one step process scaling process which you can do it yourself okay and now what I am going to do is introduce a similarity transformation now still I have not converted the PDE into OD and still I do not know whether the similarity variable is correct this is just what Blasius has assumed so what is the check that this variable is correct so when we substitute that variable into the PDE that it should get converted into an OD that is the proof that you have a similarity solution otherwise the similarity variable that you guessed is wrong correct okay so let us go and check that so I am going to see now I have to introduce a function so such that I can make the continuity equation redundant I do not want to solve this so what kind of function should I introduce stream function okay so stream function all naturally satisfies the continuity okay so therefore I will introduce a stream function such that your u is equal to d ? by dy and v is equal to minus d ? by dx okay now if I take this particular and integrate it along y so I get 0 to y u dy and already I know my u by u 8 is nothing but g of ? or g of y by ? so I can just substitute in terms of ? here this is 0 to y okay u d y by ? which is d ? into ? right or I can just say dy by d ? into okay so I can directly transform transform it this way okay so dy by d ? you know so now we have constructed a relationship for ? which is your y square root of u 8 by ? x okay so you can calculate what dy by d ? is what what is dy by d ? here square root of ? x by okay so that I can substitute into this and I can write this in terms of g okay so this will be 0 to ? so why I am transforming to ? and u is equal to u 8 which I can take out into g of ? into d ? into you have square root of ? x by u 8 so this is nothing but your square root of u 8 x times 0 to ? g of ? d ? okay so this is your ? and now I am going to introduce another function okay function which is says that this integral is nothing but which is nothing but a function of ? so you can use dummy variables here and you are integrating from 0 to ? so the resulting function which is a function of ? I am doing it denoting it by f of ? so therefore your ? is equal to square root of u 8 ? u x f of ? okay so this is the relationship between the stream function which I am going to introduce and your similarity variable ? okay I hope all of you got it is just a very simple substitution and writing in terms of ? any questions on this okay so now we have done this so just couple of more minutes and I am going to write down the velocities and derivatives everything in terms of the similarity variable ? okay any questions is it clear okay so all I am doing is I am transforming from y to ? okay so I already know the relationship between ? and y so I substitute for dy by d ? and I already know use a function of ? this way so I substitute and finally get the relationship between ? in ? so now first you are u okay your u by u 8 already we know it is a function of g of ? okay and your g of ? is nothing but if you look at this particular definition is nothing but df by d ? right because integral of g of ? is nothing but f ? okay so therefore g of ? has to be what df by d ? that is right now similarly my v velocity is nothing but – d ? by dx so now you have to tell me so you have ? as a function of ? so this we have to differentiate with respect to x okay so you have to do this and tell me I think will be that will stop today okay so if I group these terms together so finally I should be getting v by u 8 this should be equal to 1 by 2 square root of ? by u 8 x into so this term right here I can write this as ? x df by d ? okay – f okay so you can check for yourself this particular it is a little bit of jugglery so you can cast this in terms of ? and you can check whether this particular form comes out okay so this is you have to play a little bit around there okay so with that we will stop here tomorrow we will plug so why we are doing all this we are first getting an expression for similarity this is your stream function as a function of the similarity variable and then in your boundary layer equations you have your velocities your derivatives all of them have to be now cast in terms of the similarity variable and how we are doing that because we know velocity is a function of ? and ? is a function of similarity variable okay so we have to use that put them in the PDE and finally you will be ending up with a nice ordinary differential equation okay so I suggest all of you whenever we do these kind of derivations especially in the beginning where you are not used to so many differentials you please go back and pay some I know spend some half not 45 minutes checking all these equations and I think slowly you will get accustomed to this and after that you will become faster okay.