 Good morning all of you yesterday we were looking into the Von Karman polehausen solution for heat transfer problem just want to make a small correction probably when we looked into the solution for ? so we had the ? value for the stagnation region as 7.052 and the resulting expression for ? so we have to so we had to solve this ODE and the solution as a combination of the homogeneous and the non-homogeneous parts came out 0.6699 by Prandtl number okay so you can look at the non-homogeneous part is actually based basically a constant so that is why we know the particular integral has to be a constant okay if it is a function of x in any form then we have to assume that particular form and you have to calculate the particular integral and coming to this particular equation now this is for the case where we have flow past a circular cylinder you have the heated circular cylinder okay and this is the stagnation value of ? which corresponds to 7.052 so now at x is equal to 0 so this is where you are starting your x okay so you should be careful that the boundary layer thickness is actually not 0 okay because the stagnation flow comes ahead and then it basically bifurcates like this so you have a certain value of boundary layer thickness and the same thing holds true for your thermal boundary layer thickness also so you have to be careful when we stated that at x is equal to 0 we came to the conclusion that the constant is 0 essentially because of the fact as x to attend towards 0 the solution will go to infinity okay because this is negative power so in order to make this value of ? t that is the thermal boundary layer thickness finite at x is equal to 0 this constant has to go to 0 okay so this is why we put this constant as a probably I did not explain it very carefully so you have to be careful that at the stagnation region both the thermal boundary layer thickness and the momentum boundary layer thickness are both nonzero okay and therefore in order to make give a finite value of ? t so your constant has to be 0 that is why the final solution was Q is equal to 0.6699 by and from this we have calculated the expression for Nusselton okay so this is I just want to clarify before we proceed further today we will introduce a smaller approximation to the entire approximate method and this is also called as walls approximation some people also referred this to as Thwaites approximation I am not sure why this is attributed to two people I do not know the historical reason but some books some of the more recent books they refer this to walls approximation in some of the earlier books they refer this to as Thwaites approximation and what essentially happened is this person walls he pointed out that we had defined this function H of K right so a complex function which was function of F1 F2 K and so on so he said and we were thinking that we can construct a lookup table where we will change the values of ? for different values of ? will plot F1 F2 and therefore we can plot H so he said when he did this plotting we found this H of K was actually a linear function of K okay so and he got a curve fit of a straight line which was point 47-6 K so he just simply plotted H of K as a function of K at K equal to 0 this was like point 47 and then it was decreasing like this okay so this was a very good news because now since this is a linear relationship we can simply directly use this to calculate Z okay so we know the expression for the momentum reduced momentum integral equation which was DZ by DX is equal to H of K by U 8 so this was the expression that we derived after we substituted those velocity approximate velocity profiles finally where what is your Z ? 2 square by ? so this is nothing but a parameter which involves the momentum thickness okay so so this has to be solved in order to get new values of Z and then we found from there you iteratively solve for new values of ? and keep doing this as you keep marching from one region to the other region till it separates okay now this was because we did not know a very good functional relationship for H of K so now when walls plotted H of K as a function of K we found this is nothing but you can fit it very nicely with a linear line and therefore a linear approximation was derived now we will substitute this into this expression so this will be 0.47-6 K by U 8 okay so this is called the walls approximation the rest of the thing is it is just how do we now simplify this equation now directly we have DC by DX is equal to this we can integrate it out so one more step we can write this as U 8 DZ by DX is equal to 0.47 anybody remember what K is Z x DU 8 by DX okay this is coming essentially from the pressure gradient parameter okay Z x DU 8 by DX right so therefore this is a function of K is nothing but basically a function of Z okay so it is YZ to put this in terms of Z DU 8 by DX now you see we have an equation directly for Z okay which we can integrate it out we can also rearrange this as U 8 6 ? 22 by ? so I am going to substitute my Z as ? 22 by ? so this can be rearranged in this manner 0.47 U 8 5 okay so now you can expand and check so this will be the same so you have ? is a constant which can be taken out so you have D by DX 1 by ? so we can take ? 22 and U 8 so this will be 6 into U 8 power 5 DU 8 by DX plus I have U 8 power 6 okay to ? 2 into ? 2 by DX is equal to 0.47 U 8 5 okay so I have 6 and now my Z is nothing but ? 2 2 by ? I have the this so I can divide throughout by U 8 5 so this will give me ? 2 2 by ? DU 8 this is 6 D 8 by DX plus U 8 2 ? 2 D ? 2 by DX is equal to 0.47 so this is U 8 so this is 6 Z D 8 by DX and I have U 8 yeah so this DZ by DX is nothing but 2 ? 2 into D ? 2 by DX that is equal to 1 so therefore you can you know all this can be combined into one neat way of clubbing together this particular term and this will be nothing but the expanded version of this okay so now with this we can we have an expression directly in terms of ? 2 which we can integrate it out okay so if you if you integrate it so integrating so we can say ? 22 is equal to 0.47 nu by U 8 power 6 of X okay and so I am going to integrate the right hand side which will be 0 to X U 8 to the power 5 of some Z DZ okay so this is my final expression that I now get so I simply integrate both sides okay and just I write this in terms of ? 2 square so therefore I will call this as number one so your earlier expression for solving for Z in fact that is solving for ? 2 was doing numerically you have to solve this OD numerically now once you introduce this linear slope approximation okay constant slope approximation into this and you directly integrate it out you get a simpler integral okay if your function is known U 8 you substitute here you directly integrate it out and you can get an expression for ? 2 right away so there is no numerical work involved in this okay although this is an approximation you know nevertheless it is a very good approximation in fact I can give one homework where you can check this whether this linear approximation is valid okay you can calculate H for different values of K plot and check for yourself all right so this is the thing and here we have used the fact of course when you integrate it out we have a constant here constant of integration okay so now what basically X is equal to 0 when you are integrating this with respect to X is equal to 0 your U 8 that you are using so this this entire term has to be 0 because at the stagnation point your free stream velocity is 0 so therefore this entire term will be 0 the 0 therefore the constant will be 0 okay so we have used that that at X is equal to 0 your U 8 power 6 ? 2 square by ? equal to 0 okay so it is not that your boundary layer thickness is 0 but your U 8 is 0 which leads to this particular conclusion that the constant is 0 therefore this is the final expression okay now we will apply this again for the cylinder problem earlier we had I had given you the algorithm to solve for the cylinder case in a more rigorous way solving the ODE numerically okay so first we will look at the cylinder problem near the stagnation region so if you substitute the profile of I for U 8 your U 8 is basically to V 0 V 8 R 0 by sorry X by R not so if you substitute this and you integrate it out you get ? 2 square is equal to 0.235 into ? into R not divided by V 8 X power 6 okay so this is basically 0.47 divided by 2 and you have V 8 X power 6 okay that is U 8 power 6 actually everything is power 6 but inside the integral also when you substitute for U 8 to the power 5 the extra terms cancel out so you have Z power 5 DZ is what is left out inside okay so you are basically substituting for U 8 here U 8 power 6 here U 8 power 5 so you have 2 U in 2 V 8 by R not which is the same okay so then you will be left with the factor that this is V 8 into V 8 divided by R not here you will have X power 6 here you will have C power 5 so this you can integrate out directly so this will be point so if you integrate this this will be nothing but X power 6 by 6 okay so 0.235 divided by 6 will be 0.0391 and so X power 6 X power 6 cancels so you have ? R not by V 8 okay so therefore there you go you directly have gotten expression for the momentum thickness without solving the ODE rigorously you know near the stagnation point directly you find it okay and you can also calculate the value of ? for doing that we need to estimate the boundary layer thickness which is estimated so how do you estimate the boundary layer thickness ?2 there is an expression because ? is a function of ? ? okay what is the original expression how did we define ? ? ? U by D U 8 by DX right so we defined this parameter ? like this okay so ?2 D U 8 by DX by ? is that right okay so therefore for this particular case we can substitute D U 8 by DX which is nothing but a constant to V 8 by R not and that comes out to be ? by 2 into ? R not by V 8 okay so now we know ? ? as a function of ? okay we have an expression for ? 2 so we can calculate the ratio of ? 2 by ? the whole square so you let me know what this expression will be so if you take the ratio of these two we are not by V 8 will be cancelled out okay this will be 0.0391 into 2 by ? which is 0.0782 what we are trying to do is we are trying to solve for ? okay so ? is a function of ? and we have we have an expression for now ? 2 by ? the whole square which is now a function of ? we also have derived ? 2 by ? square we have expressed this as another function of ? do you remember that originally that is 1 by 63 square 37 by 5 minus ? by 15 or you can say that this is equal to square root of this and then you can you can write like this ? square by 144 okay all right so now we have an expression algebraic equation in terms of ? since ? 2 by ? is originally this in terms of ? okay now we have another expression for this particular problem which is like this so we can equate both and you can find out ? if you do that in fact this has to be also solved iteratively so ? will come out as 7.25 okay now you can compare this with the ? that we got earlier that was 7.05 right so it is slightly off but nevertheless it is very close okay so I think this is now much easier to solve than the earlier case what we will do is now go ahead and complete the heat transfer solution so for the heat transfer problem whatever we have derived we have to assume the cubic polynomial and then we have substituted into the energy integral and we have finally calculated the expression for ? which is the ratio of ? T by ? as point this is a this is ? Q this is point 6699 by right so the same expression is valid because as far as the energy integral is concerned all you are doing is you are substituting the velocity profile approximate velocity profile approximate temperature profile and finally you have expression in terms of ? which you are getting the solution for okay now when you are getting the solution you have substituted a value of ? there okay the original expression was something like this so your expression was ? D by DX X ? square of ? 12 plus is equal to 90 up till here everything is the same okay so this is coming from substituting into the energy integral right for the velocity profile and the temperature profile and simplifying a little bit where we are neglecting all the higher order terms of ? and only ? square comes finally so now from here we should use the particular value of ? which we got from the walls approximation okay so that is basically 7.25 instead of 7.05 which we substituted earlier if you do that earlier you got your ? as point 75875 by PR to the power 1 by 3 now with this you will be getting ? as 0.864 by PR power 1 by 3 so only the constant will change a little bit because now your ? is slightly different if you substitute it your solution for ? will be just slightly differing by constant it will be 0.864 by PR power 1 by 3 so that is it so once your ? is determined you can calculate your heat transfer coefficient and your Nusselt number okay so even we know that the heat transfer coefficient is 3 by 2 a by ? ? okay so the expression for ? is the same for both the walls approximation and the original upon Carmen pull house and solution so only the value of ? is differing slightly if you substitute this you will be getting an expression for H which is like point 645 K by R0 into Prandtl number or one third RE power half okay now if you define your Nusselt number as H into D0 by K okay so this will be what 1.29 Prandtl number power one third RE power half okay where your RE is nothing but V infinity D0 by ? so compare this with the expression that we obtained yesterday you remember that expression when we used ? 7.05 what was the expression that we got 1.1 no that that is the exact solution 1.291 correct okay so this is 1.29 so finally in terms of Nusselt number there is any hardly any difference okay whether you use the walls approximation or not so so that is it I see this is to just give a good idea that walls approximation is pretty good especially in terms of the heat transfer calculation you get a very good agreement with your pole house and solution as well as your exact solution is what 1.145 and that has Prandtl number 0.4 dependent so there is some difference between the exact similarity solution and approximate solution but using quite approximate walls approximation is much more judicious you know you get faster solutions than solving the OD now the same problem where you are looking at circular cylinder can be done by using walls approximation right okay so the same way you take the profile the complete profile u 8 of x is equal to 2 v 8 sin x by r 0 where your x is equal to r 0 ? right you take your full profile which is coming from the potential flow flow for describing the free stream velocity for circular cylinder and that can be substituted into this okay expression number 1 right here so that u 8 can be substituted now you can integrate it now you have to be careful when you integrate it you are integrating sin 5 x by r 0 there okay so in the stagnation region you have made the approximation that for small values of x by r 0 sin x by r 0 is equal to x by r 0 but if you are looking at integrating for the entire flow from the stagnation point still the separation point you have to use this profile as it is okay and when you put this profile you have sin 5 term here sin power 6 term so if you are finding it difficult to integrate manual you can use some numerical technique such as you know trapezoidal rule for integration simple trapezoidal rule or even simple rectangular rule will work so you have to start from some point right here so at this point you know the value of ? 2 that that is what you have determined okay you also know the value of ? and everything so you can start from that point 0 to some point next okay so you can keep on integrating that way okay till any x that you want till you go to separation okay for each point that you go so now you calculate the new value of ? 2 square and therefore you can calculate the new value of ? like this because you know the ratio of ? 2 by ? correct for each value of x you know the new value of ? 2 by ? you also know the original form original relationship you can equate those two and solve for ? so that will give you a solution for ? basically okay so this you keep doing till you hit separation point where your ? becomes equal to – 12 so now this is a little bit more easier you can do this on an excel sheet you do not have to program it so you can simply do numerical integration and you can solve this equation also you know you can find roots roots of this in an excel sheet itself so that will give you again the boundary layer thickness and the momentum thickness and also the separation point so everything can be obtained from the walls approximation okay then this will be nearly as good as doing it by the solution numerical solution to the OD okay so in fact I will give the next assignment which will be assignment three which will include all the integral solution problems I will also give ask you to do this for sector cylinder and see for yourself how you can basically find the separation point and get expression for boundary layer thickness momentum thing okay so same thing can be done for any wedge problem you know the Falkner's can solutions where you had assumed some kind of a wedge flow and a velocity profile of this particular sort now you can apply the walls approximation you can substitute for you infinity here you can integrate it out okay and finally you can calculate the value of ? and from there of course your boundary layer thickness and again you can use that into the heat transfer problem where you substitute the expression for ? and the particular value of ? you calculated you can derive the expression for Nusselt number so you can do this for any value of m okay like the way you did your similarity solution you can do this and you can compare the result with the similarity solution you will find the agreement once again is pretty good okay so any problem with pressure gradient can be handled with the walls approximation okay so probably we will I will stop here but I will probably will have some kind of a small discussion if you have any doubts because I will start I have one more last topic left under integral method which is non-uniform temperature boundary condition so far we have assumed that the entire plate is heated with a uniform temperature or uniform heat flux in most of the cases that is not true okay the profile will be varying so what happens if you have a variable temperature profile okay so so we cannot do with the similarity solution this kind of a problem of course you can use similarity solution if it varies only linearly but if it is any other variation you have to do integral solution so we will see for a flat plate case what is the technique to how to introduce a non-uniform temperature boundary condition how do we solve this okay so that will be the last topic I will take that about a couple of hours next week before your quiz one okay so quiz one will have all the topics till your integral solutions so pretty much any doubts that you have in this I hope you understood the walls approximation because you will be now using this to solve this flow past circular cylinder okay so if you have any questions you please ask me is it is it clear you have understood this is just a dummy variable yeah this is just a dummy variable you can if you if you are having say C x power m this will be C ? power m so that will be just a dummy variable and it indicates that whatever upper limit till where you are interested to integrate so that will be used into the dummy variable okay so earlier x indicated some arbitrary location now when you are integrating it you have to use the dummy variable because you should not confuse this with the limit of integration okay so many a times we use the same thing no but we know that we should use the upper limit but to differentiate it clearly strictly speaking you always have to use a dummy variable okay so you are clear with all of all of this like how to calculate your ? to numerically and all that you have to numerically integrated by represental rule you can start from say the stagnation point okay assume this is a trapezoid between these two so you can if you are integrating to the first point it is just only one step if you are integrating somewhere till somewhere here you have to assume many trapezoids and some them all together so the area under each trapezoid some them all the trapezoids together will you will give you the area under the integral basically so that is that is the approximation that we are so at each location so you calculate the value of ? to and so like this you will have an expression for ? to right here right now we have completely there is no dependence on X here okay so there when you are putting a sign profile there it will have a dependence on particular value of X so for that particular value of X you will have a particular constant okay for ? to okay so and then you now this is so this is the expression for ? to this expression for ? will be there this is again a function of the position because you are now differentiating sign profile okay so there will be a X dependence so for each position the value of ? to and ? will be different and that you take the ratio so you will be getting an expression in terms of ? okay only this constant will keep changing for each location and then you can solve this find roots of this okay get the new value of ? and you know that where the region is you know as and when the ? keeps going negative and increasingly negative you know that you are approaching close to the separation point and and then you keep doing this integral till where at a particular point where ? becomes minus 12 and then you know that you have hit the separation okay so so all this can be done in a excel sheet you know you do not have to write any program for that separating.