 So, I think we will get started with our session now. So, what we will do is now we will take standard flow over a flat plate. So, in most of the cases when we take flow over a flat plate what are the let us just write what are the let me just take flow over a flat plate that is this is my flat plate and let this be u infinity comma t infinity. Now, what is the continuity I am taking two dimensional incompressible steady flow two dimensional incompressible steady flow that means velocity is varying only with x and y in x it is going to be u and in y it is going to be v. So, what is continuity equation for incompressible 2D flow del u by del x plus del v by del y equal to 0 is that right. Now, what is the momentum equation x momentum equation there is no z momentum equation because w is u. So, I have u del u by del x plus v del u by del y del u by del t is not there because my flow is steady equal to minus of 1 by rho del p by del x plus mu by rho into del squared u by del x squared plus del squared u by del y squared. If I write y momentum equation what is that u into del v by del x plus v into del v by del y equal to minus of 1 by rho del p by del y plus mu by rho del squared v by del x squared plus del squared v by del y. Professor Arun has already told I will not write energy equation right now I will go ahead and write that also what is there energy equation is u del t by del x plus v del t by del y is equal to 1 by alpha into del squared t by del x squared plus del squared t by del y squared let me check whether I have written it rightly or not I have made a mistake that is k by rho c p actually this is this is alpha not 1 by alpha. So, now what happens to my y momentum equation what happens to my y momentum equation we know very well that v is very small compared to u v is very small compared to u. So, that makes my all this term will be vanishing this term will vanish and this term will vanish then what is what does this mean implies that del p by del y is equal to 0 that means pressure is a function of pressure is a function of x only x only. So, what does this my x momentum equation reduced now here let us go to the next term what happens to flow over a flat plate for del p by del x yes let us d p by d x, but that is not the answer see let us write let us write the x momentum equation continuity equation is del u by del x plus del v by del y equal to 0 u del u by del x plus v del u by del y equal to minus of 1 by rho d p by d x plus mu by rho into del square d u by del x square plus del square d u by del y. Now, first thing is for a flow over a flat plate for flow over a flat plate if you see the equation what I have written here I have not written d p by d x and I have also not written del square d u by del x square, but I need to now show why they can be neglected. So, now for flow over a flat plate for flow over a flat plate what is u infinity all over u infinity is constant it does not vary. So, now if I apply as professor was telling I have two portion whenever a flow takes place over a plate there are two portions one is viscid portion another one is inviscid portion that means in the inviscid portion viscosity effects are not there. If I apply this equation for the inviscid portion in the inviscid portion what is u infinity? u infinity is constant u infinity is constant if you ask for proof for that I would say that if you put a put a tube and measure very much outside the boundary layer u infinity is constant you can check it that way. Now of course, from potential flow solutions also one can show that u infinity is constant because inviscid flows can always be handled by potential flow. So, u infinity is constant means what will be my x momentum equation reducing to u infinity d u infinity by d x plus v into d u infinity by d y equal to minus of one by rho infinity d p infinity by d x u infinity is constant means what will happen to this term u infinity is constant I have not written the next terms why viscosity is 0 no viscosity. So, that is why this term becomes 0 because viscosity is 0 now what happens to this term u infinity is constant. So, this term is 0 again here u infinity is constant this term is 0 that means what what did I end up getting d p infinity by d x is equal to 0 this is only applicable for whom for all those flows where in which u infinity is constant. But now if I take u infinity is constant for flow over a flat plate if you take flow around a cylinder if you take flow around a cylinder u infinity is not constant if you take this is u infinity flow in the potential core that is very much away you will get u by u infinity as some sin theta or something sin theta no I would say some function of theta let me put it like that I do not recollect the function exactly u by u infinity is a function of theta that means u is not a function of u is not constant anymore it is a function of location where I am then in that case d p by d x is not equal to 0 is that ok. So, that is why for flow over a flat plate d p infinity by d x. So, what is this d p infinity by d x this is d p infinity by d x is the pressure with which is the pressure with which the inviscid portion is impressing upon impressing upon whom impressing upon upon boundary layer see because we said that there is inviscid portion this is viscid portion and this is inviscid portion this d p infinity by d x is acting on the boundary that is how the inviscid portion and the viscid portion get connected with each other. So, one solution is required for the inviscid portion to get d p by d x in the boundary layer equation that is for this equation d p by d x in the boundary layer for getting this equation I need to get d p by d p infinity by d x using the inviscid solution using the potential core solution of course we are not taught you how to get the potential core solution it becomes too much I would be ending up teaching only fluid mechanics, but not heat transfer. So, this comes from potential flow solution. So, this means to say that inviscid solutions are also as important as the solution because that is how both the viscid and the inviscid portion get connected with each other usually this point is not told by most of the text books directly this is something which comes out of inside of course, Schlichting tells this and various specialized book tell, but usually undergraduate books do not tell this kind of insights. So, this is what we need to impress upon the students otherwise he will be asking us always why for a flow or a flat plate d p by d x equal to 0 I will not be getting convincing answer. Next let me tell what is this del squared u by I have neglected del squared u by del x squared equal to 0 why now let me do that for doing that I will be using the recourse of scale analysis. If you see here I have neglected del squared u by del x squared and similarly del squared t by del x squared I will just answer this in a little while, but I am going to start off something very interesting what is called as scale analysis or order of magnitude analysis order of magnitude sis or it is also called as scale analysis this analysis what I am teaching you can get it from Professor Bejan's book either heat transfer book or convective heat transfer book John Wiley publications. So, what is my continuity equation del u by del x what am I up to what am I up to I am going to take only the continuity equation and the x momentum equation and I am going to get the velocity boundary layer thickness and skin friction coefficient without solving the equation. Let us see how do we do that it appears like a magic, but yes indeed it is a magic let us see how do we apply that magical idea of scale analysis del u by del x plus del v by del y equal to 0. Now, what am I doing here I am going to take if I take flat plate again and I have u infinity and this is y direction and this is x direction u is of the order of note this I am using the notation tilde u is of the order of u is of the order of what u is of the order of what or x is of the order of what it is of the order of length of the plate means if it is 1 meter I can say that anywhere I take at least it will be 0.1 meter, 0.2 meter, 0.2 meter it is of the order of meter. So, x is of the order of L x is of the order of L u would be of the order of u infinity that means if u infinity is meters per second let us say 10 meters per second throughout the boundary layer u infinity will be of the order of 1, 2, 3, 4, 5 meters per second. So, u is of the order of u infinity what is the order of y we are interested in boundary layer thickness. So, I am interested not outside the boundary layer I am interested only within the boundary layer thickness. So, y is of the order of delta. Now, let us find the order of v from continuity equation what do I get what is the order of del u u infinity minus 0 that is u infinity upon delta x L minus 0 that is L these two of the same order know they are equal to another. So, this is of the order of v upon what is the order of delta y delta del y is of the order of delta. So, v is of the order of what is the v what is the order of v u infinity delta by L remember it is not equal to if you have to make equal to you have to put some constant number that is equal to u infinity delta by L. But remember I am not putting equal to here I am putting tilde that means it says that it is of the order of that is v infinity what does this say what do I understand from this how does delta compare with L delta by L is always less than 1 is not it. So, that means v is always less than u infinity how much times delta by L times is that ok. So, now, I have got the scale of v now let us take x momentum equation that is u del u by del x plus v del u by del y equal to p is not there for flow over a flat plate only I am doing d p by d x we have made it as 0 I have explained you just a little while ago why because it is u infinity is constant. So, 1 upon or mu upon rho into del squared u by del x squared plus del squared u by del y square. Now what is the scale of del squared u by del x squared mu by rho into u infinity upon upon x x squared is L squared. Now, what is the scale of the next term comma mu upon rho into u infinity upon delta. So, which of these two terms are important now which of these two terms are important anything divided by delta squared delta is very small 0.1 meter. So, it will get amplified L is 1 meter or 2 meter. So, this will be of the order of meters this will be at least 10 times larger than this. So, that is why I can neglect what I can neglect del squared u by del x squared after neglecting that let me see what do I get that is u del u by del x plus v del u by del y equal to mu by rho del squared u by del y squared. I have answered the question why del squared u by del x squared is neglected is neglected here and the same arguments you can do this here also, but let us do that little later on. So, now if I do that if I do that what do I let me put the scales of each one what is the scale of u u infinity what is the scale of del u u infinity what is the scale of del x L comma what is the scale of v u infinity delta by L I want all of you to derive with derive with me all over u infinity delta by L what is del u by del y scale u infinity by L sorry delta u infinity by delta by delta square delta someone said delta square delta this is of the order of left hand side is of the order of right hand side is what mu by rho into u infinity del squared u is u infinity no u infinity upon del y squared is delta squared. Now let us see this is important or not important because we get a doubt v del u by del y v is very small why am I not throwing it out why am I not throwing it out now you see what do I get delta delta cancels out what do I get here u infinity squared by L which is same as u infinity squared by L that means what this term is of the same order as the second term that is why although v is small which one is large del u by del y is large that is why I cannot neglect. So, the overall scale of the left hand side is what u infinity squared by L which is equal to mu by rho u infinity by delta square what does this imply what does this imply delta squared by L is of the order of mu upon rho u infinity one u infinity has got cancelled out is that ok if I write now delta squared by L squared that means both sides I am dividing it by L. So, what will I get mu upon rho u infinity L what is this mu upon rho u infinity L this is nothing but one by R e defined on the basis of characteristic length L. So, delta by L whole square delta by L is of the order of R e L to the power of minus half is that ok. So, now this answers the question delta is of the order of L R e L to the power of minus half. So, now I can tell boundary layer increases with the increase of the length you can see here delta by L if you see if I from here if I write directly delta is of the order or delta squared is of the order of what mu L by rho u infinity delta increases with the increase of viscosity it increases with the increase of length it decreases with the increase of velocity and it decreases with the increase of ok. So, yesterday and day before yesterday we had so many questions how does boundary layer thickness depend on velocity thermo physical properties location this is the answer for that remember I have not even solved if I solve this equation I am going to get the constant I am just going to get the constant if I have to show you that constant I am going to get the constant little later in the day that point it is going to be 5 delta by x equal to 5 times that is delta by L equal to 5 times R e x to the power of minus half that 5 is the only thing which is missing, but otherwise I have captured all the parameters now let us see how do I get c of x that is shear stress what is c of x c f x equal to tau wall upon half rho u infinity square is that right what is tau wall what is tau wall tau wall equal to mu del u by del y upon half rho u infinity square ok. So, now c f x is of the order of equal to I cannot write mu is of the order of means I do not have to worry about this half ok mu what is the order of del u by del y u infinity by delta upon rho u infinity square. What will I get mu by rho into 1 upon u infinity into delta, but I have just now found what is delta what is delta delta is c f x equal to mu by rho upon u infinity what is delta L R e L to the power of minus half just now we have found is that right what will this give me u infinity L upon mu what is that 1 by no I think I have made a mistake here somewhere rho u infinity square yeah it is right nothing is wrong nothing is wrong nothing is wrong yeah that is right. So, what do I get 1 upon R e L into R e L to the power of minus half so, what do I get c f x is of the order of R e L to the power of R e L to the power of minus half. So, what happens to my shear stress with the increase of the Reynolds number it decreases shear stress decreases with the increase of the increase of Reynolds number is that ok. Now, let us see you see so nicely we have found all that you are going to do is in the derivation that this constant is going to be 0.332 after you derived without deriving we have got the functional form the beauty is we have got the functional form that is the beauty let us see can we do this for boundary layer thermal boundary layer thickness and also nusselt number. So, there I have to specify this is what I have said delta by L is R e L to the power of minus half and c f is of the order of R e L to the power of minus half is that ok. So, now, let us see what do we do if in case of boundary layer thermal boundary layer I am taking Prandtl number less than 1 if I take Prandtl number less than 1 what will I get Prandtl number less than 1 is always results in thick thermal boundary layer Prandtl number less than 1 gives me thick thermal boundary layer ok. So, what is my energy equation first let me write the thick boundary layer. So, that is I am taking Prandtl number less than 1 where does this occur it occurs in liquid metals like lead mercury. So, these are the indium tin these are the ones which will typically have Prandtl number of 0.02, 0.003 and things like that. So, very less Prandtl number if you have such a low Prandtl number what will happen we have said that already delta by delta t is of the order of Prandtl number to the power of n where n is a positive number right. So, what will happen delta is how does delta we compare with delta t delta is lower than delta because my Prandtl number is less that means thick thermal boundary layer thick thermal boundary thick thermal boundary layer. If I take thick thermal boundary layer how does that boundary layer look like actually boundary layer thickness will be this is delta and this is delta t ok. This is hydrodynamic boundary layer this is thermal boundary layer tbl and hbl or vbl velocity boundary layer. If I take the velocity profile how do they look like you see the velocity profile looks like this I am taking I am going to take linear relationships because my life becomes slightly easier. So, that is this is my velocity boundary layer or the velocity profile within the boundary layer and also outside and this thickness is what is this thickness hydrodynamic boundary layer delta this is u infinity and remember I am maintaining this at constant temperature ok this is my plate. Now how does the thermal boundary layer look like thermal boundary layer we have realized already it is thicker than hydrodynamic boundary layer. So, I will get a temperature gradient this is t infinity and this is what I am going to get as temperature and this boundary layer is delta t thermal boundary layer thickness ok. So, now if you go to the document back and see here what is my energy equation u del t by del x plus v del t by del y is equal to alpha del square t by del y square. I think I can write this directly now on the same arguments the way I wrote for del square u by del x square plus del square u by del y square. If you see del square u by del x square what did I do I neglected that because I end up getting u infinity by l square. So, that is going to be smaller than this. So, similar argument here if I take for if I take for del square t by del x square I will get delta t by l square as opposed to delta t by delta t square. So, delta t square is very much small which is sitting in the denominator which will blow to my complete number ok. So, now what is the scale of u what is the scale of u yes within the thermal boundary layer what am I going to do is that within the thermal boundary layer because my thermal boundary layer thickness is much larger than the hydrodynamic boundary layer. So, within the thermal boundary layer I am going ahead and making the velocity is same and that velocity is equal to u infinity. In fact, this point was told by professor Arun when he had touched upon the energy equation earlier in the yesterday yesterday he had told he said it is little early, but this is what he was meaning when he told when he told that. So, that is u infinity. So, what is the scale of u u infinity what is the scale of del t delta t what is this delta t this delta t is nothing, but t s minus t infinity it is of that order what is the scale of delta x del x l what is the scale of v we have done already u infinity delta by l remember this is hydrodynamic boundary layer then del t is delta t what is the scale of del y only delta we are in the thermal boundary layer the energy equation is for thermal boundary layer y here means for us delta t thermal boundary layer thickness is of that order of alpha delta t upon again delta t. Now, which of these terms are important and which of these terms are not important let us just have a look at it see you see here what is that I am saying. So, you see here what do I get delta by delta t how does this delta by delta t compare is delta by delta t larger small number it is a small number because parental number is less than 1. So, I can neglect this term compare to u infinity delta by l is that right delta by delta t is a small. So, anything which I multiply with that has to be a smaller number. So, what is there left hand side I have u infinity delta t by l is of the order of alpha delta t by delta t square. So, what do I get for delta t square delta t square is of the order of alpha l by u infinity is that right now what is this alpha l k by rho c p into l upon u infinity. Let me multiply and divide by mu multiply and divide by mu there is l here what is mu c p by k given by mu c p by k is what parental number Reynolds number Reynolds number is u infinity l by rho infinite mu infinity by mu. So, let me rewrite this I am going to rewrite this whole thing in the next board that is delta t square equal is of the order of delta t square is of the order of k by rho c p l by rho c p l by rho mu upon u infinity mu. Now, let me divide both sides by l square. So, delta t square by l square is of the order of if I do rearrangement mu c p by k into u infinity l rho by mu this is what I get is that right what is this by the way 1 upon parental number what is this 1 upon r e l. So, delta t by l is of the order of r e l to the power of minus half p r to the power of minus half. Now, what is Nusselt number let us go to heat transfer coefficient quickly I am I am not explaining this here much because I have already explained how you can look at all that I am saying is this constant is missing this constant is missing. So, h equal to minus k del t by del y at y equal to 0 upon t s minus t infinity what is the scale I get for this h k into del t by what is the scale of del y delta t upon delta t t s minus t infinity of the scale of delta t. So, this delta t this delta t cancels out. So, what do I get h is of the order of k into what is 1 upon k upon delta t that is h is of the order of k upon delta t is what delta t is what l r e l to the power of minus half p r to the power of minus half this means what this implies that h l by k h l by k is of the order of r e to the power of half and p r to the power of half we have reached where we are supposed to reach all that is missing is here constant which we are going to derive in couple of minutes from now. This is the beauty of scale analysis we are skipping the Prandtl number greater than 1 and I would like you to ask plenty of questions I would like to take this tomorrow I would like all of you to work on this and derive and show that delta t by l is of the order of p r to the power of minus 1 by 3 r e to the power of minus half for thin thermal boundary layer that is p r greater than 1 and nusselt number is r e to the power of half and p r to the power of 1 by 3. Let us see how many of you can do it. Now I am handing over the charge to professor Arun who will teach us the closed form solution of flow over a flat plate. So, good afternoon we will quickly now that we have understood professor Prabhu has very elaborately explained why pressure gradient over a flat plate necessarily has to be equal to 0. So, d p by d x is equal to 0 for flow over a flat plate and therefore, in case of in case of flow around a cylinder or a sphere d p by d x is not equal to 0, but d p d p infinity by d x pressure with which the inviscid portion is impressing upon the boundary layer I am using whatever he has written that basically is related to the potential flow solution. Now he has done the order of magnitude or scale analysis and this is something which unfortunately we do not tell our students at undergraduate level which is ok, but the students usually have this doubt when to use what why is it r e to the power half p r to the power half r e to the power one third p r to the power one third so on and so forth. And all those questions I think have been answered by the scale analysis which was done so elegantly professor Bejan's book on convective heat transfer has this scale analysis done in detail with this scale analysis we do not have to teach this to our students, but at least those who are interested where people come and ask why how do I remember how do I know that it is r e to the power half p r to the power one third means we say that yeah it is given like that it is an empirical correlation this is what we give an answer, but it has come primarily from scale analysis and this I think if this two or three sets of transparencies which have been flashed these things are the heart of the scaling thing ok. So, you can with confidence hopefully answer questions related to Nusselt number being a function of r e p r that we showed when we showed the non dimensional solution now we are showing that it is r e to the power half p r to the power one third for one of the case and half and half to the other case. Now this dimensionless similarity solution and all these things these are essentially the mathematics related to flow over a flat plate fluid mechanics has basically taught us all this I do not I am not going to spend more than 5 minutes on this it is essentially what is the similarity variable we have done similarity analysis when we did this semi infinite medium essentially eta was defined to take care of the x and the t dependency x and the t dependency was there in a semi infinite medium because our governing equation was d square t by dx square is equal to 1 by alpha dt by dt and then we said let me define a variable eta which was something related to under root alpha t x by under root 4 alpha t or something like that which then took care of the variation in both x and t such that a PDE was converted to a ordinary differential equation and that ordinary differential equation we could get a solution analytically in the form of some error function or conjugate error function we could do that very easily and then obtain a close form solution. Now what we are doing going to do here for this laminar flat plate solution which most fluid mechanics books will have if you take fluid mechanics by Frank White or Fox and McDonald's, Engels, Engels and Simbala any of the standard good books they will have this Blasius solution. So, I am not going to spend time on it we are just going to see the similarity variable let eta be defined in terms of this y under root u infinity by nu x and that u by u infinity is a function f prime of t. It is given as f prime for convenience f prime of eta therefore, you get y over delta is essentially of this form because of the scale analysis delta over x is of the order of Reynolds number to the power minus half you can write this in this form and then the stream function definition you use u is equal to d psi by d y v is equal to minus d psi by d x continuity equation you can get this psi is equal to integral u d y substitute for u from here etcetera. So, this maths has been every step here has been given. So, we are not going to spend time because we are already running behind schedule on this does not matter what we say is our governing equation what this does our partial differential equation d u by d x plus d v by d y equal to 0 this is our continuity equation. Now, my thing is I have u as a function of x comma y v as a function of x comma y this eta will convert this p d e into a o d e such that my d u by d x therefore, gets defined in this form u infinity eta by 2 x f double prime and this functional form is f double prime etcetera we will see what it is. So, therefore, I will get v if upon integration I will get v and this was one of the things which we had asked in the previous tutorial and if you do the maths you will get d v by d y as eta by 2 u infinity by x f double prime d u by d x you will get continuity equation is automatically satisfied therefore, u by u prime is a function of eta is a correct assumption why are we doing this we will just understand a bit. So, with this assuming that this f prime of n is a correct so called similarity variable that is a variable which transposes or which converges p d e to a o d e my momentum equation therefore, on applying this whole thing this becomes d u by d y is equal to d by d y of u infinity times f prime this is the definition of u. So, I want to expand this. So, summation rule I will get u infinity is a constant so this is just going to be u infinity d by d y of f double prime y, but then I do not want x is n y that is the basic idea of the similarity solution all the x is n y's have to go away that means I have to write let me just go back here see eta is a function of y and x. So, this when I want to write something as a derivative with respect to y it is essentially a derivative with respect to eta, but it includes this aspect also so d u infinity f double prime this d eta by d y this d eta by d y comes because of the change in the coordinate system. So, x and y gets transposed to eta so because of that we are going to get d eta by d y I am not consciously spending any time on this you can do all the max every step is outlined here. Therefore, I will get an expression for d u by d y d square u by d y square everything is boxed in red u by u infinity is known. So, d u by d x can be obtained. So, this x momentum equation basically got transformed to a ODE third order ordinary differential equation f triple prime plus f f double prime by 2 equal to 0. So, when I do the transposition my boundary conditions also get transposed get converted. So, this is what is given here. So, 2 f triple prime plus f this one 2 f triple prime plus f f double prime equal to 0 that is what is put in a different form and third order non-linear third order because it is to the power 3 non-linear because f is also occurring here differential equation look at the beauty of it p d e in x and y d u by d x d d u by d y d square u by d y square that has been come converted into a ordinary differential equation. And the boundary condition this is third order. So, I need 3 boundary conditions in eta when eta equal to 0 f is equal to 0 what is this how did we get that when eta equal to 0 f it equal to 0 came from the gradient at the what is f here f of eta is nothing but psi divided by u infinity under root nu x by u infinity. So, when this eta equal to 0 f of eta equal to 0 that is the first boundary condition d f by d eta that is u by u infinity at eta equal to 0 that is also equal to 0 and this is d f by d infinity d f by d eta at eta tending to infinity that is equal to 1 because d f by d eta is nothing but this quantity d f by d eta is this quantity f prime is nothing but d f by d eta. So, eta going to 0 refers to y equal to 0 that is the wall surface u equal to 0 no slip condition eta tending to infinity refers to y tending to infinity which is far away from the boundary layer. So, u is equal to u infinity f prime equal to 1. So, with this idea I will say this ODE with the three boundary condition was first solved by Blashe's in 1908 and this solution all of us know gives us this nice elegant form d f by d eta is equal to u by u infinity is equal to 0.992 and that eta equal to 5 therefore, delta which comes out to be 5 divided by u infinity by mu x and this is how all of us have seen this delta over x is equal to 5 over Reynolds number that we have seen that is this Blashe's solution delta over x is equal to 5 over under root Reynolds number where x is the distance from the leading edge of the plate leading edge means if I have a flat plate going from left to right this is the leading edge the starting. So, that is why there were questions on Moodle which asked why should a boundary layer grow how far does a boundary layer grow with respect to x the answer is given here delta is directly related to x to the power one half x divided by square root of x is x to the power one half. So, delta grows as x increases. So, it will grow as long as there is a solid flat plate associated as long as the viscous effects are going to be there as long as a no slip condition is imposed at a solid surface this as long as there is a flat plate which is there and that boundary layer essentially is seeing the effect of the flat plate in a y direction as long as that is there we are going to have the growth of the boundary layer. I know I have gone fast, but this whole this 4 5 transparencies are essentially algebra there is hardly any physics to be explained. So, start with this and just proceed in this direction what are we checking first is we are defining this and defining this and then we are saying is this going to satisfy as a similarity variable a similarity variable basically we have to see whether it is going to satisfy this equation. So, that once it is satisfied we said it is then we substitute for these d u by d x. This is u v d u by d y squared everything u v nu from here u is u infinity f prime of n that is all we know u is u infinity f prime of n. So, continuity equation will give me v like this. So, I will substitute all those things here u d u by d x v d u by d y d square d u by d x v d u by d y d square u by d y square this all these things I will substitute and then I will cancel off the terms finally, I will get this as the form. So, please I we urge you to go through this I know there is lot of go through this at home kind of things today, but most of these things are algebra. So, what are you what are the three things that you will do at home? This one the scale analysis for case which Prabhu outline where for the Prandtl number is much greater than 1 that you are going to and do then you are going to do the energy equation non-dimensionalization. So, we told in the morning that you will be tested on that thing also. So, energy equation non-dimensionalization scale analysis for P r much greater than 1 and this Blasius solution Blasius solution blindly you can take from any of the good fluid mechanics book there is nothing to teach here. So, this basically gives a shear stress which is mu d u by d y at the wall and that shear stress go grows like this R e to the power minus half thus unlike boundary layer thickness wall shear stress and the skin friction coefficient decrease along the plate to the in the form x to the power half. So, boundary layer thickness increase. So, if I want to draw it like this whiteboard boundary layer thickness for a laminar boundary layer definitely we are going to have this kind of a functional form. How did this come? This came from fundamentals what are the fundamentals solution of the governing equation? Solution of the governing equation how could I solve this? I had a bad looking equation momentum equation which was cast in a convenient form by the definition of a similarity variable basically what it did was the following I will just summarize this I had u d u by d x plus v d u by d y is equal to mu d square u by d y square I had this form this with the associated boundary conditions was there. We what happened to the y momentum equation this is the x momentum equation y momentum equation professor showed us that it gave us only the following d p by d y equal to 0. So, we have already exhausted it. So, only this one remains now this equation I cannot solve directly. So, what I do is I have go to this so called similarity analysis where I say I define f of eta which is nothing, but f prime of eta is u by u infinity and then I define eta is equal to y u infinity mu x under root. So, y is here x under root and then I go through the whole algebra. So, I get u I get v where do I get this from this is from continuity I get d square I mean I get d u by d x I get d v by d y sorry d u by d y then I get d square u by d y square all these things I will get one by one after I get that I will substitute all this back into the x momentum equation. Once I substitute into the x momentum equation I will get a equation of this form f triple prime or 2 f triple prime plus f double prime equal to 0. This with the appropriate transformed boundary conditions will give me a solution what will this give me it will give me f as a function of eta correct because this is my independent variable and this is my dependent variable. So, this is going to give me f as a function of eta. So, this f as a function of eta when why did I start this I forgot to tell this has come from fundamental solution. So, after having obtained this one after having obtained this it is not come from magic it is a solution only thing is that we are not understanding the solution because it looks more different because you have f eta so on and so forth, but it is essentially the same equation. So, once I get this solution I recast this in the form of u d u by d y so on and so forth and see here we have d u by d y in terms of d square f by d eta square at eta equal to 0 d square f by d eta square has been tabulated here this is f prime this is f double prime this is the header of the table at eta equal to 0 is 0.332. So, this 0.332 is the second derivative of f with respect to eta. So, second derivative of f with respect to eta d u by d y first derivative of u with respect to y has become second derivative of f with respect to eta, but this was partial this was ordinary this value is 0.332 which has come from here and that is why I will get tau wall is of this functional form. So, why is this 0.332 why is it not something else that question is answered directly by looking at this table of solution where what is this table of solution this table of solution is a solution for this differential equation with the boundary condition and what is this let us not lose lose site this is the transformed bound governing equation of the x momentum equation along with the boundary condition. So, I have solved for the x momentum equation only closed form laminar flow solution, but I cannot directly see it because it is in terms of f and eta and that is what I am getting here. So, solution of the x momentum equation gave me the velocity distribution which in turn gave me the wall shear stress this is the functional form of the wall shear stress and that is why you get 0.332 and not some other constant and therefore, c of x varies as x to the minus one half. So, I just want to take over and do this part and then we will quickly finish this and go to internal. So, now actually I just want to give a historical perspective of this in fact, what we whatever taught us just little while ago was actually the Blasius solution Blasius did this solution as a part of his PhD with Prandtl. So, in fact, the solution is called as Blasius solution. In fact, the solution does not it did not come the way it has been taught now. In fact, this whatever we gave you the solution directly is actually based on numerical methods that is there is something called shooting method technique, but let us not get into that the point is at that time Blasius did not have that time means what it was in may be around 1920s 1908 1908 there is no question of calculator or computer at the most you have logarithmic table. So, what did it do is actually the solution is obtained by power series solution. So, you just take f as a function of eta that is a polynomial function f dash is a function of eta f double prime as a function of eta. So, let me just write just I will write one equation. So, that how painful it was for Blasius, but he was ingenious I mean there is no no word we can explain the intelligence of Blasius f equal to a naught plus a 1 eta plus a 2 upon 2 factorial eta square plus a 3 upon 3 factorial eta cube plus dot dot dot. So, likewise we can write equation for f prime and f double prime and f prime equal to a 1 plus a 2 eta let me cancel this out f prime equal to a 1 plus a 2 eta plus I am just differentiating this if you just see this I am just differentiating I am doing nothing a 3 upon 2 factorial eta square plus a 4 upon 3 factorial eta cube and so on so like that I can write for f double prime and also f triple prime why I need f f prime f double prime f triple prime my equation looks like 2 d cube f upon d eta cube plus f d f by d eta equal to 0 what is this this is nothing but my f triple prime what is this this nothing but f prime. So, I have this equation just solve now differential equation has got reduced to d squared f by d eta square. So, this becomes f double prime now differential equation has got reduced to just an algebraic equation which we need to solve and remember all the coefficients of this equation for each of 1 eta, eta square, eta cube, eta to the power of 4, eta to the power of 5 all of those coefficients should be such that I should be getting this back to 0. So, if I equate that I will be getting a 1 a 2 a 3. So, that is how he got so intelligent is not it we do not have computer also what we got this solution. So, that is how bless has gave even today that 0.3302 which we have got it from computer he has given it from simple power series solution he has just done what is this whatever I wrote this is what is called as which we study any in mathematics actually what is called as power series solution beautiful it is what beauty also cannot contain what the meaning we mean actually. So, this is how we get this 0.332 and if I integrate if I put that I am going to get 0.332 here 2 is where that gets multiplied and I get 0.664 now another important thing we need to realize is how does one get eta equal to y square root of u infinity by new x I would remind you what did we do for semi infinite medium, semi infinite medium also we took the scale analysis and derived this similarly we have already done the scale analysis what is my scale analysis telling y upon delta what is delta delta is x re x to the power of minus half that is what I am substituting x upon square root of u infinity x of new so I get y into square root of u infinity upon this is essentially if I put that is not like a magic it has come from scale analysis without scale analysis the point is actually this there is a one important point I we want to emphasize here is always a question is whether we should teach to teach scale analysis or not to teach scale analysis I would say for under graduates please do not teach scale analysis it will be little too much for them but if you can teach and spend time and make them understand we teach here in UG in IIT we do teach scale analysis myself and professor Arun do teach scale analysis why because of this simple reason otherwise I cannot explain the student how did I come upon eta equal to y into square root of u infinity upon new it is just otherwise impossible to convince my student how do I know that I have to take u infinity upon square root of u infinity upon new so that is the reason why I say that scale analysis is very important order of magnitude analysis is very important in fact why only here for a post graduate approach I would say that for natural convection for forced convection for all cases one can derive boundary layer thicknesses and nusselt numbers from simple bound scale analysis okay okay so now with this we will move on to energy equation at the same pace what we did it for and definitely we do we have not covered it in we are not solved what we are not done is we are not solved this equation for you we have set up the equation and we have given the boundary condition similarly we are going to do for energy equation so in fact this can be solved by if you are if you are very particular about numerical method I am sure many of you are very good in numerical method this can be solved by rangay kutta method by what is called a shooting technique that is I will assume f prime and then go on iterating so this is combination of rangay kutta method and shooting technique if it is a post graduate class I would ask the student to solve this and submit this as an assignment if it is UG class I think it is little too much for ask them but if someone is interested do not discourage them for them to solve okay so that is about the shear stress it is very logical now bless us solve the temperature profile and now sorry velocity profile by solving the momentum equation now it is very logical for us to get to energy equation so I have energy equation U del t by del x plus V del t by del y equal to alpha del square t by del eta del y square there is nothing difficult here all that I have to do is I have to differentiate or I have to express t in terms of theta this is not new for us we have done this already in transient conduction so theta of x comma y is t of x comma y minus t s upon t infinity minus t s if I do that if I do this express this t in terms of theta and I anyway have eta already eta is y square root of U infinity by nu x if I do the transformation I have already have U V if I do the transformation for del t by del x and del t by del y and del square t by del y square I know I am going very fast what I am not doing is I am not differentiating all that you need to do is del t by del x can be written as del theta by del eta into del eta by del x so differentiate this with respect to eta theta with respect to eta and eta with respect to x you end up with this if you do similarly del t by del y you end up with this again you differentiate the same thing with respect to y you get del square t by del y square if you put that and simplify you are going to get what is as the way it is looking like this that is theta double prime theta double prime plus Prandtl by 2 into f theta prime f theta prime equal to 0 this is my equation and again I have boundary condition I am going to solve only for constant wall temperature that is if I take constant wall temperature case and substitute those boundary conditions I am going to get delta t again I am skipping all the mathematics I am skipping all the mathematics I intentionally I am doing please go back home and do this so delta t I get 5 I want to write this because delta t I get as equal to 5 re x to the power of half we are to the power of minus 1 by 3 does this equation look familiar with us does this equation look familiar by scale analysis we had found that I think I have made a mistake here it is delta t by x that is delta t by x equal to okay so that is we had found already delta t by L is of the order of re l to the power of minus half pr to the power of minus 1 by 3 so now you can write a cell number equal to just we need to flip that is 5 re to the power of re x to the power of half and pr to the power of 1 by 3 that is what we are going to see in a cell number you see here you are going to get 0.332 re x to the power of 0.5 and pr to the power of 1 by 3 why because I need to substitute minus k del t by del y k by delta t yeah k by delta t if I put delta t 1 upon 5 okay 1 upon 5 I should be getting yeah but k by delta I should get so if you substitute that you are going to get let me come back to you how exactly I get that that is I get 0.332 re x to the power of 0.5 and pr to the power of 1 by 3 so here what is missing is yeah yeah yeah yes 0.332 I get here because I am from this this is what so I get directly 0.332 and from that I am inferring delta t I think there is a small problem here no no there is no problem everything is fine 0.332 is what we had got from f prime at eta equal to 0 f prime eta equal to 0 that is what we have substituted and if I do that I am going to get the nusselt number the point is we have demonstrated through this that nusselt number is going to be a function of Reynolds and Prandtl we have shown that it works for it works for flat plate case any other configuration we take as long as the flow is incompressible we can represent the nusselt number as a function of Reynolds and Prandtl number we should be able to come up that is how we see hundreds of correlations put in data handbook after understanding this now you use any correlation left and right absolutely there is no problem I think I will stop at this point.