 afternoon anyway good that at least a few have turned up otherwise I would have to reschedule the entire class anyway. So, as a one more small correction to the previous class that we that I just want to bring forward I think one is one person who observed it and he was correct the mean temperature for the high speed flows I had told that it should be 0.5 T wall minus T infinity plus C into I think it should just reverse the sign it should be T adiabatic wall minus T wall where C equal to 0.22 so I think I had given this as C into T wall minus T adiabatic someone has correctly pointed out that since the adiabatic wall temperature is greater than the wall temperature. So, this has to be positive so that the mean temperature now is increasing rather than it should not decrease because now with the use of adiabatic profiles for example, if you had plotted Y versus T which we had done this is your T infinity from here the profile should start something like this and so this is your T wall. If you had applied isothermal boundary condition without discuss dissipation it would have come something like this therefore you see the average temperature has to now go up earlier the average temperature was between T wall and T infinity now the actual temperature will be like this something like it will go up inside. So therefore the average temperature has to account for this increase therefore this has to be positive here if this was T wall minus so generally your adiabatic wall temperature will be greater than your wall temperature okay. So, this will be positive and therefore this will increase your T mean temperature otherwise it will decrease the T mean temperature which is not physically correct okay. So, this please incorporate this particular correction and now we will start a new topic it is the same solution to external boundary layer flows but we will use what is called as an approximate method of solution okay. So far we have been doing exact solution so called exact solution where we had derived the similarity equations for the different configurations and we had solved it but we cannot solve it exactly we have to solve it using some numerical method but nevertheless the solution is the solution is for an equation which is actually an exact equation okay. Now we will look at techniques where we will approximate the solution by means of some profiles rather than trying to solve by a numerical technique that the way that we were doing till now so we do not know the profiles how they look so but we got that as a solution of the similarity equation. Now we will approximate the profiles as something depending on the order of accuracy you know you can use a linear profile or quadratic or cubic or fourth order whatever polynomial so we will substitute that into the governing equation once you know the profiles we can determine the expression for boundary layer thickness the thermal boundary layer thickness and therefore the heat flux nusselt number and so on. So this kind of technique is also called the integral method many people also refer this as the approximate okay. So the starting point of this is to start with essentially an integral form of why this is called an integral method is you operate with integral form of the governing equation form of boundary layer momentum and okay so why do we use the integral form is of course integral form is more helpful when you are approximating some solution putting it inside and integrating it out directly okay. So therefore this is more convenient for the approximate solution to work with an integral form and I am not going to now derive the integral form because if I remember correctly I had already derived a particular version of the integral form if you may recollect your earlier notes for the flat plate Blasius solution we try to find the expression for boundary layer thickness delta so as something like square root of ux by uinfinity right. So this was derived by integrating the momentum equation from 0 to delta and from there we had solved an ODE to get the expression for delta okay so in fact at the time I told that I am not going to derive it again because we will be using that as a starting point. So the point where we left off so we integrated the x momentum equations I am just going to write it just to show you how it looked du by dy dy is equal to nu du by dy between the limits equal to 0 and equal to delta okay. So we just integrate the boundary layer momentum equation for a flat plate okay we could also include your pressure gradient term so that will be additionally plus uinfinity du infinity by dx and integrated between 0 to y dy it is nothing but delta okay so how we got this because originally this term was nothing but minus 1 by rho dp by dx and we saw in the boundary layer that the pressure gradient along y is negligible therefore we can apply this equation to the inviscid potential flow outside and we can directly find that this is equal to uinfinity du infinity by dx okay and when you integrate it across the boundary layer so that will be this will be constant okay for a particular y location so you can directly say that is multiplied by delta okay. So if you have a pressure gradient that term also enters into this final expression and after we broke this term we integrated by parts and then we use the continuity equation to make some approximations and finally we arrived at this particular form 0 to delta u du by dx especially the left hand side minus uinfinity 0 to delta du by dx dy this is equal to nu du dy 0 to delta plus uinfinity du infinity by dx okay so I think we arrived at this particular form for the entire left hand side term after we use the continuity and we integrated it out and from here we are going to start now okay so this is nothing but the momentum integral equation but still not the final form okay so I am just going to rearrange some of the terms in fact what I am going to do here is I can once again integrate by parts okay so I can do this as d of uinfinity u by dx minus u into du infinity by dx I can I can just re-express re-express that way and I am going to multiply throughout by a negative sign and if I just rewrite this a little bit so this will be d by dx 0 to infinity 0 to delta u infinity minus u into u dy so u infinity into u is basically this term minus this can be neatly written as directly 0 to delta d into uu by dx dy right so du square is nothing but 2u du dx okay so I can just write this as d by dx so I can take d by dx out so 0 to delta so I am multiplying throughout by negative sign okay so I can write this as so this is already negative negative of negative is become positive so u infinity into u minus uu okay so that is this particular term okay plus the other term which I which I have written this is du infinity by dx which I can take common from here as well as from the right hand side term okay so this can be written as du infinity by dx integral 0 to delta okay so here I have a negative sign here and I multiply it with a negative sign so this should become positive okay so this should be a positive again I multiply with a negative so that should become negative and this one is already negative goes to the left hand side term that becomes positive so this becomes u infinity minus u dy okay on the right hand side now if I integrate this from 0 to delta so at this at y equal to delta du by dy should vanish okay so this becomes minus u du by dy with a minus sign so this becomes positive this becomes new du by dy at y equal to 0 okay so I have probably skipped a couple of steps but I think you can understand straight away how that comes out you want me to explain again or is it okay so till here you know how it has come out okay from the previous derivation so from here I am just rewriting this first time as d du u by dx and this I am integrating by parts again okay so I can write this as u infinity u d by dx minus u du infinity by dx so therefore so this term and this term can be combined I can take d by dx out of the integral so 0 to delta u infinity minus u u okay so this is u infinity into u this is u u so I take small u common u infinity minus u dy and the other two terms so here du infinity by dx is common to this and this okay so and I can I can write this as instead of writing this as delta I can write this as 0 to delta dy okay so that is u infinity that is basically this u infinity into dy minus there is again this u dy okay so which is what I am writing here together on the right hand side term this is 0 to delta so at delta du by dy is 0 right so therefore this is minus du by dy at y equal to 0 already I am multiplying throughout by minus so this becomes du by dy at y equal to 0 okay so this is my final expression in fact I can also rewrite in a more familiar form so I can write this as d by dx of I can take multiply and divide by u infinity square for this particular term so this will be u infinity square into I will just denote this as something like delta 2 plus in this case I can multiply by multiply and divide by u infinity okay so I will call that as some delta 1 into du infinity by dx on the right hand side I have this so where my delta 1 will be so multiply and divide by u infinity so this will be u infinity 0 to delta so this is u infinity by u infinity which is 1 1 minus u by u infinity into okay so I am not going to write exactly like this I am just rewriting slightly in different manner just a minute so I am multiplying and dividing it so that delta 1 into u infinity is here and delta 1 is defined in this particular fact okay and delta 2 so I am multiplying and dividing by u infinity square and divide by u infinity square this will be 0 to delta 1 minus u by u infinity into u by u infinity dy okay so this is also another nice form that you find in many of the textbooks nevertheless all the three whether it is this form or this form this form they are all the momentum integral equations or integral momentum equations okay I personally prefer looking at this form this is almost the final form and this gives you a lot of information because here this delta 1 which I have defined here there is a particular name to it this is also called the displacement thickness and delta 2 here is called the momentum thickness okay so and this is your okay I hope you have a good understanding of what is displacement thickness and momentum thickness okay from your incompressible flows I am not going to go into an explanation for that now it is only so this are not really measurable okay this is what you have to understand unlike the boundary layer thickness which you can measure that as a location these are conceptual values which give you a sense if you suppose replace the boundary layer with a potential flow profile so how much the wall has to be pushed up okay so are displaced in order to satisfy continuity or satisfy conservation of momentum okay so that is this location so these are all not measurable okay these are something if you are conceptualizing and usually these are much smaller than your actual boundary layer thickness okay so this is your momentum integral equation so you are writing this in terms of the momentum integral your displacement integral and so on for the case of flat plate without the pressure gradient the second term will be 0 so you do not have anything in terms of the displacement thickness so for flat plate your du infinity by dx is equal to 0 and this will lead to the form you can write u infinity square d by dx 0 to ? okay so this is the integral momentum equation for flat plate let me call this is number one okay this is the one form that we will be using to substitute all our approximate profile now same way we will derive the integral form of boundary layer energy equation okay we start from the conventional energy boundary layer equation and then we integrate it out across the boundary layer thickness and we arrived at a similar integral energy equation okay so we start with our standard energy equation we do not assume any viscous dissipation term here for flat plate case without viscous dissipation this is your equation at y equal to 0 it could be either wall temperature fixed or your prescribed heat flux which is constant and at y going to infinity t is equal to t infinity at x equal to 0 also the same thing follows so what we are simply doing is integrating again now when we are integrating now we cannot integrate it till the boundary layer thickness because now we have to integrate it till the characteristic thickness for the energy which is the thermal boundary layer thickness okay so we integrate it till ? t of course you know we are not making any assumption whether ? t is greater than ? or less than ? whatever extent it may be we are just integrating till the edge of the thermal boundary layer and so we will try to eliminate v so that we write everything in terms of u so we will see that the second term can be integrated by parts so this can be written as 0 to ? t okay I can write this as d by dy of vt integrate between 0 and ? t that is nothing but v into t between the limit 0 and ? t okay minus integral 0 to ? t into t dv by dy okay on the other side you have a ? t by ? y between the limit 0 and ? t and a ? t dt by dy is 0 okay it has to satisfy continuity in slope at the edge of the boundary layer so therefore so this will become minus a dt by dy at y equal to 0 okay so this is coming because at x is equal to y going to infinity t equal to t infinity therefore dt by dy has to be 0 okay so now this particular term again at y equal to 0 v is 0 so this will be valid number only when you look at y equal to ? t so where once again your t will become t infinity there okay so for ? t this will become t infinity okay so this is that ? t is that fine so now what we can do is once again we can integrate this by parts you can write this as 0 to ? t into d by dx of ut dy minus 0 to ? t du by dx dy okay so therefore you can combine this and this term right here so you can take t common this is du by dx plus dv by dy which is nothing but the continuity equation alright therefore I can write this as 0 to ? t d by dx of ut dy plus vt infinity minus I can take 0 to ? t t common du by dx plus dv by dy this is equal to minus a dt by dy at y equal to 0 so and this also satisfies continuity so this goes to okay and also from continuity you know that my v equal to minus 0 to if I integrate from 0 to ? t the continuity equation I can write my v velocity in terms of u velocity so just the continuity I am integrating okay so I can substitute and now I can eliminate v from this equation so if I write this in terms of u velocities this will be 0 to ? t I can take d by dx out d by dx ut dy minus I have t infinity into integral 0 to ? t du by dx minus a dt dy y equal to 0 okay so I just want to combine these two terms I can write this as now d by dx okay 0 to ? t I multiply by minus sign throughout this will be t infinity minus t so u is common for both I can write this as u dy this is equal to a dt by dy at y equal to 0 okay so this is my final form this is the energy integral equation so I can combine these two terms because t infinity is anyway constant I can take this I can write this is d by dx of t infinity u so u is common in both the cases in combine okay so now we have both of them I am going to define my ? in fact I can write in terms of ? but let me do it later so just let me write down the energy equation energy integral 0 to ? t infinity minus t u dy should be equal to a okay so this is your equation number 2 which is your energy integral okay so the first step is that we derived the integral form of momentum and energy and we have put it in this particular pattern and next what we are going to do in the approximate solution we are going to make an approximate assumption for the velocity and temperature as some polynomial okay so it can be any order polynomial and we have to see which boundary conditions that the profiles have to satisfy to calculate the coefficient of these polynomials so this is the next step so now let us take the case where we have a linear velocity profile you know that is the most basic profile that you can start with so for the laminar flow let us assume a linear velocity profile so I am just saying that my u would be something like a plus by okay this is the linear form of the velocity profile that I am assuming so in order to get the constants a and b I have to satisfy boundary conditions okay so the most you have to start with the most basic boundary conditions before we go to satisfying the more higher order boundary conditions okay the the most important boundary conditions start from the wall at y equal to 0 u equal to 0 okay and after this what should be the second most important boundary condition so once you have given at y equal to 0 you have to give something at y equal to ? okay u equal to u these are the most basic boundary conditions which are which are which have to be satisfied I cannot directly say du by dy at y equal to ? 0 has to be satisfied then it does not tell what value it should reach at y equal to ? okay it just only says that a slope is 0 that is it so if you substitute can you tell me what will be the profile that I get so if you directly substitute this y equal to 0 u equal to 0 a will be 0 at y equal to ? u will become u infinity therefore b will be u infinity by ? okay so therefore your u will be equal to u infinity y by ? so u by u infinity will be equal to y by ? so this is my linear velocity profile so if I if I plot this profile for a flat plate at any location it will show only something like this a very linear variation from 0 to u infinity at y equal to 0 u will be 0 and varies linearly at y equal to ? u will become u infinity so this will be ? something like a quad flow profile okay so this is not actually the real profile right your real profile is satisfying in a different relationship which we have seen from the Blasch's solution and definitely the shape does not look like this okay there is a slope there is a curvature terms are all there which we are not accounting for in this particular okay the slope is constant and there is no curvature so therefore this may not be a very good profile but still this is an approximation okay so why we are approximating it is once we may give this kind of an approximate profile we can substitute for velocity in this momentum integral and we can directly calculate the boundary layer thickness the expression for ? can be determined so what I am going to do now is substitute this profile into 1 so if I substitute into 1 so that I will get u infinity square d by dx 0 to ? into 1 minus so u by infinity now will be y by ? into y by ? into dy on the right hand side I have du by dy at y equal to 0 so du by dy now will be what u infinity by ? okay from the linear profile so therefore so anyway that is a constant slope it does not matter whether it is at y equal to 0 or so this is just linear profile so the slope is constant so this will be ? into u infinity by okay so very simple profile and the equation also looks simple now I am going to introduce a non-dimensional variable ? which is equal to y by ? okay do not confuse this with the similarity variable of course I am using the same variable here because coincidentally in the similarity solution also ? was the similarity variable which was related to y in ? exactly like this but here I am just using ? to represent any non-dimensional variable so if I do that I can transform y in terms of the non-dimensional ? this will become u infinity square d by dx now integral from 0 to ? will become 0 to 1 in terms of ? okay so 1- ? into ? into this will be dy by d ? into d ? okay dy by d ? will be ? so basically I can I can put this ? here into d ? okay which will be equal to on the right hand side u infinity by ? okay now this ? has to be with inside d by dx because ? is a function of your position along the flat plate so if you do the integral maybe you can take a couple of minutes and do it but since we do not have time I am just going to give you the final solution so this will come out to be 1 by 6 okay so the integral 0 to 1 1- ? into ? d ? should come out as 1 by 6 okay so in that case this will become I can cancel u infinity on both sides and this will be ? will come to this side so ? d ? by dx will be equal to 6 ? by u infinity okay so now this is a very simple OD which I can solve by separation of variables straight away okay so this will be ? square will be equal to 12 ? by u infinity x plus some constant okay this is the solution for ? of course we can find the constant by applying the boundary condition that at x is equal to 0 ? equal to 0 right the boundary layer thickness is 0 at the leading edge of the plate and therefore your constant will be 0 okay so your expression for ? now reduces to the form ? is equal to square root of 12 u x by u infinity which is nothing but 3.47 square root of u x by infinity or ? by x can be written as 3.47 I can divide by x on both sides and this will be u infinity x by ? which will be nothing but square root of Reynolds number now you can compare this with the similarity solution from which we calculated the expression for ? do you recollect what is the expression that that was 5 okay so how did we do that we calculated for different values of ? f of 0 f prime of 0 and also at all other values of ? we have tabulated and we found out the corresponding value of ? where your u by u infinity or your f prime was 0.99 corresponding to f prime 0.99 we found that ? was equal to exactly 5 and that is nothing but the value of ? okay so this is how we have got it approximately the actual solution is ? by x is equal to 5 by square root of rex so since we are using an approximate profile this is the variation you know so this is about 31% less than the exact solution okay so there is an under estimation that we get by using the linear profile and we can go ahead and calculate the wall shear stress local variation which is ? du by dy at y equal to 0 okay so in this case du by dy at y equal to 0 is nothing but for the linear profile u infinity by ? okay so this is ? u infinity by ? so we have already got the expression for ? which will substitute here and if you do that you will get 0.288 u u infinity square root of ? u x okay so now if you define non-dimensional local skin friction coefficient as ? by ? infinity square so this will come out as 0.576 by square root of Reynolds number okay so and if you remember the exact solution was giving 0.664 okay so this is about 13% less than exact solution okay so if you also go ahead and integrate the local skin friction coefficient over the plate and calculate the average skin friction coefficient for the entire plate that is your CFL is equal to 1 by L 0 to L CFX dx okay so we will end up with expression 1.152 by square root of REL okay so finally the bottom line is this you can you can get an approximate estimate of all these quantities like boundary layer thickness your local shear stress in friction coefficient everything using some approximate profiles and this is much much easier as you can see than doing a rigorous solution to the similarity equations okay and in fact if you use a better profile you will be amazed to find that the agreement will be even better okay so just to give an example if you go from a linear profile to a cubic profile okay which I will just show you and stop there okay so if you assume a cubic velocity profile u is equal to a plus BY plus CY square plus DYQ okay so now you have 1, 2, 3, 4 coefficients and we have to satisfy therefore 4 boundary conditions so what are the possible boundary conditions y equal to 0 u equal to 0 and at y equal to ? u infinity so these are the important boundary conditions and what is the next important boundary condition at y equal to 0 ? u by ? y so if you if you look at the momentum equation okay so u du by dx plus v du by dy is equal to nu d square u by dy square at the wall okay both are 0 so therefore d square u by dy square has to be this is the next important say in the order of importance I am writing it okay and then finally what is the last boundary condition that you have to give cubic in terms of the order of importance okay so these are most important and then at y equal to 0 this is to be given and then y equal to what ? okay so these are the 4 important boundary conditions so if I substitute these boundary conditions into the profile and you calculate all the coefficients finally I will be able to arrive at an expression which is like this u by u infinity will be 3 by 2 y by ? minus 1 by 2 y by ? the whole cube okay so this is the cubic velocity profile that you will get I am sure most of you have taken the heat transfer course have done this also in advanced heat and mass transfer in incompressible fluid flows okay so cubic profile is like the accurate profile that you can do okay so once you get this and if you substitute the same way that we did the linear profile into the momentum equation so let us see what happens to the final expressions the final expression for ? that comes out with the cubic profile will be 4.64 that is the correct ? by x will be 4.64 by ? root of rex so you can see a remarkable improvement okay so this is like 7% less than the exact solution and when you look at the skin friction coefficient so this value comes out as 0.646 by ? root of rex this is like almost there 3% less okay so this is 0.664 this is 0.646 almost there so you can see that with the cubic profile you get a very good approximation very accurate solutions okay rather than going for a very rigorous similarity solution so nowadays I think many of the people they are not so interested in the similarity solution because you have numerical techniques where you can directly solve the governing equations okay and or you can also resort to solutions like approximate methods with the integral equation where you can make use of some approximate polynomials and also get the solution so that is why these are more popular methods okay which are also workable in a very short time but the you have to be cautious that does not mean that if you keep increasing the polynomial now this accuracy will become better and better okay so for example if you go from linear to quadratic and from quadratic to cubic you will find that it does not progressively increase in terms of accuracy okay in fact quadratic in fact spoils the solution a little bit that is mainly because in the case of quadratic boundary conditions we give this boundary condition this boundary condition and finally this boundary condition we do not give this particular boundary condition okay so because we have only up to here we do not give up to this boundary condition because of that you will find there is a D generation in the accuracy the quadratic case again with the cubic it satisfies all the fundamental boundary conditions and the accuracy goes up and if you use a fourth order polynomial again you may have to give something like y equal to delta d square u by dy square equal to 0 you have to introduce higher derivative basically and they may not be very important so therefore you may not get no the error may not come down all the time so that is why maximum you go to cubic and we can stop there okay so in the next class tomorrow we will make an approximation for the temperature profile the same way that we did and we will substitute into the energy equation and calculate the thermal boundary layer thickness and therefore the heat transfer coefficient okay so we will see these aspects in the next class so quickly we will move from flat plate so flat plate I am sure many of you have familiar with the heat transfer course so we are not going to spend much time we will do this analysis for flows with pressure gradient similarity flows is that dealt with in the other courses or no where you have done similarity solution for adverse pressure gradient flows or pressure gradient flows I am sure something must have been done in incompressible flows the Thwaites method Thwaites method and Carman-Pollhausen method okay so maybe we will cover those aspects in the future next four or five