 Good morning all of you so yesterday we were looking at application of the approximate method to flows with pressure gradient and so this method is also called as the von Karman Paul Haas and technique which was originally conceived by von Karman and he suggested that we can use higher order polynomials and we can substitute that into the momentum integral equation and try to find solution for flows including pressure gradient. So this is what the procedure that we did and finally the equation that we obtained from the momentum integral after making all those substitutions for quartic polynomial so this is the final expression and so this cannot be solved just like that analytically because it is a non-linear ODE because here U infinity is a function of X okay so therefore we have to use some numerical method like the Euler method that that we had seen that for shooting technique also. So we can just discretize this particular ordinary differential equation for many number of points starting from the stagnation point so suppose you want to apply this to the flow past a circular cylinder okay so for this we know the variation of the free stream velocity so suppose you are the free stream which is approaching the cylinder is U infinity V infinity so we can express the velocity variation of the free stream velocity variation past the circular cylinder as you to U infinity sign X by R0 or maybe I can use some capital R to indicate the radius okay so this is the velocity profile obtained from the inviscid flow analysis and so now we directly integrate this particular ODE by the Euler method and so we start from the stagnation point so that is where the I equal to 1 corresponds to I equal to 1 2 3 4 5 6 so if you want to look at the coordinate system in terms of X okay so that corresponds to X equal to 0 which is nothing but the stagnation point okay so we have also determined the value of lambda lambda is a function of X as I told you and lambda at X is equal to 0 comes out to be 7.052 because we have seen that U infinity of X at X is equal to 0 becomes 0 so in order for this particular ODE to be finite we have to force H of K to be 0 at X is equal to 0 from which we arrive that the root for lambda which is acceptable root is basically 7.052 okay so now we know the value of lambda therefore we have the expression for Z in terms of lambda for a given flow conditions like your V infinity nu and R0 R0 or capital R so in fact if I use capital R here let me use this here so this is my cylinder radius okay so this this this is fixed for the particular geometry and the flow condition okay so I can calculate my Z at the stagnation point based on the value of lambda so that from there I can start integrating it so now I can find the value of Z at the second point okay that is I equal to 2 and once I do that again I come back to this expression okay to this algebraic expression for which I can iteratively find the value of lambda corresponding to the new value of Z okay so like that I can do keep doing this procedure multiple times okay so for all values of Z I have to keep finding the corresponding value of lambda and then using that to calculate the other parameters like H okay because all of them are functions of lambda okay so if I keep doing this integration till I reach a particular point where I will find that the lambda comes to minus 12 okay that indicates that the flow is separating at that particular point and that is where I stop the integration okay so with this method I can actually track how the boundary layer thickness because this is nothing but ? 2 is what the momentum thickness okay so I can track how the momentum thickness is progressively growing with X and from there I can calculate all the other thicknesses because I know ? 1 by ? I have ? 2 by ? I can calculate my displacement thickness I can calculate my boundary layer thickness everything is a function of lambda okay and consequently I can also determine the point where the flow is separating okay so so the exact angle where the flow detaches about 75 degrees okay so this is a rigorous way of doing it but nevertheless it is a very good useful tool now otherwise when you do experimental studies and flow visualization that is one thing but if you do not have access to those experiments theoretically you can use this technique approximate technique to determine at least to the extent where the flow separation takes place okay or otherwise we have to do an expensive numerical simulation of the entire Navier Stokes equation which is too rigorous okay this is a less rigorous procedure but nevertheless it is useful till the separation point beyond the separation point the boundary layer theory is not valid okay so this was what we have seen yesterday know very useful technique which was proposed by Von Karman and demonstrated by Paul Austin now this is for flow let us extend this technique to heat transfer problem okay you can consider the same adverse pressure gradient or favorable pressure gradient flows let us introduce the approximate solution for temperature okay so heat transfer now what we are going to do when we are looking at the Nusselt number correlation the same way for the similarity solution we derived only for the stagnation region okay the same way we are going to focus as far as the heat transfer is concerned yes I think please correct this you are right okay so this delta x should be multiplying only this and then this is z i plus 1 minus z i so it becomes positive on the other side yeah so okay so if you are looking at heat transfer we will focus our heat transfer problem to region which is close to the stagnation region the same way we did the similarity solution the Falkner scan solution for the case of m equal to 1 was nothing but the stagnation point solution and that we had converted to cylindrical system and that gives us the Nusselt number for the stagnation region for a circular cylinder the same way with the approximate method we are going to do the heat transfer solution for the stagnation region so for the same problem let us assume that the free stream or the ambient temperature is T infinity and the walls of the cylinder is maintained as at an isothermal condition where your T wall is constant okay now this rise gives rise to the heat transfer problem for as far as the temperature profile is concerned okay Fankarman or Polozen did not specify anything about how rigorous the profile should be as far as the flow is concerned you have to take a quartic polynomial but with respect to temperature profile we can still use the third order polynomial that we were using before for Blasch's solution so let us define our ? T- T wall by T- T wall so that it scales between 0 and 1 0 at the wall and 1 in the free stream so this can also be assumed to be a function of some ? T which is a non-dimensional coordinate corresponding to the thermal boundary layer okay so let us assume a cubic polynomial we have four coefficients to be determined and here ? T is nothing but y by ? T okay this so here this is not a similarity variable as I said as far as the approximate solution is concerned your ? T does not denote any similar it is just your non-dimensional you when you substitute in terms of y and you get the profile you will have terms like y by ? in order to make it compact you denote your y by ? as a non-dimensional ? but it does not mean that this is a similarity variable okay so of course also coincidentally for the similarity solution this turns out to be this similarity variable okay but here it does not have anything to do with similarity variable so now we have to propose the boundary conditions to satisfy this particular temperature profile so what are the corresponding boundary conditions at y equal to 0 okay this is basically ? T equal to 0 right what should be the value of ? it should be 0 and at y equal to ? T corresponding to ? T equal to 1 okay ? T should be 1 okay and even at y equal to 0 now if you go go on look at the higher order boundary condition from the energy equation again be equal to a D square T by d y square of course we have neglected the viscous dissipation term so near the wall we can neglect the inertial terms so therefore we can also say that d square T by d y square is equal to 0 or d square ? by d y square is equal to 0 okay in terms of ? also the same thing will apply in terms of ? T same condition applies okay now we need one more boundary condition to close this problem so what will be the fourth condition 0 so at y equal to ? ? ? T so your d ? by d ? T should be equal to 0 so now you have the required boundary conditions you can get the profile for ? and what will be the profile we have already done the same thing for Blasius solution we have assumed a cubic profile same boundary conditions okay so what was the profile 3 by 2 ? T minus 1 by 2 ? T cube okay so as far as the temperature is concerned so it does not matter you know as I mean it only depends on the flow problem the flow problem is different for the pressure gradient and without pressure gradient and as far as the temperature profile is concerned so this is that temperature profile and the energy integral equation also remains the same okay so you have to substitute this into the energy integral do you remember the energy integral equation for without the viscous dissipation term so let me erase this part so I hope you can recollect the energy integral d by dx okay so what should be the term inside integral 0 to ? T now I am going to transform the variables from y to ? to 0 to 1 1- ? u by u ? T into d ? T okay so this will be d ? by so you have basically dy there okay so dy by d ? into d ? by dy okay so basically that will be where ? T that will come out outside and on the right hand side you have your a dt by or d ? by d ? T at ? T equal to 0 and you have also from the transformation of variables you get u ? T into ? T okay so this is your energy integral expression let me call this as now equation let this call let us call the profile as equation number 2 and the energy integral expression as number 1 okay so all of you are familiar with this okay so you have your 1 by so if you want me to expand again from the basic equation so d by dx of 0 to ? T 1- ? u by u ? dy okay so this I am writing in terms of ? so 0 to 1 1- ? u by u ? T into dy by d ? T into d ? T and dy by d ? T is nothing but ? T okay so that is how this factor ? T comes okay so this is my left hand side and on the right hand side I have is equal to a dt or d ? by dy at y equal to 0 okay this I can I can write as a dt by d ? T at ? T equal to 0 into d ? T by dy okay d ? T by dy is 1 by ? T so that is why I have the factor ? T here okay and there should be so yeah so I am multiplying and dividing by u ? T here so that u ? T should be taken to the right hand side okay this is originally 1- ? into u dy so I am multiplying and dividing by u ? T and this u ? T is going to the so this is this is the final form in terms of non-dimensional ? and non-dimensional coordinate ? T okay so I can now substitute my temperature profile and the velocity profile that I obtained from the quartic polynomial into this particular equation 1 so substituting for the temperature profile to and velocity profile from a quartic polynomial into one so I have d by dx okay I am taking my u ? T to the left hand side I have ? T 0 to 1 1- 3 by 2 ? T- plus ? T cube and my velocity profile u by u ? T will be so what will be the velocity profile u by what was that that we derived yesterday 1 by 2 where is this coming from yeah so yeah so if you substitute everything in a compact notation it was f f of ? plus ? G of so ? right so where you are f of ? was what 2 ?- so this was what probably you are 2 ?- 2 ? cube plus ? 4 your G of ? G of ? sorry should be 1 by 6 ? into 1- ? the whole cube okay and your ? is nothing but non-dimensional pressure gradient parameter which can be written as ? square by ? du ? by dx in terms of dp by dx your du ? by dx can be written as – ? square by u ? ? into dp by dx so for flows with adverse pressure gradient your ? will be negative with favorable pressure gradient ? should be positive so this this was the time velocity profile that we derived yesterday assuming a quartic polynomial and satisfying those five boundary conditions okay so so your ? here is the non-dimensional parameter which is giving us something like a ratio of your pressure gradient to our pressure forces to the viscous forces okay so this is one measure of how adverse pressure gradient how adverse the pressure gradient you maintained it okay and what is the criteria for separation is given by ? so ? equal to – 12 indicates the point of separation okay now from this we can substitute for the velocity profile so this will be f of we will put it in terms of f of ? plus ? G of ? okay d ? T d ? T is the variable with which you are integrating and that is it okay so this is your left hand side basically on the right hand side this will be equal to a by you in infinity I have taken to the left hand side you have d ? by d ? T so from this profile what is d ? by d ? T at ? T equal to 0 3 by 2 and you have already a by ? T okay so this will be your equation now we can introduce my Z ? as which is nothing but ? T by ? and from the way that we have introduced y by ? as my ? and ? T is equal to y by ? T so this is nothing but the ratio of ? by ? T okay so I am just introducing a parameter called ? now if I for a given Prandtl number you know that this ? T by ? is a function of Prandtl number so this becomes fixed for a given Prandtl number and also if you have starting if you do not have any unheated static length that means you maintain an isothermal temperature condition right from the beginning okay so this ? is going to be a constant because both the thermal boundary layer and the momentum boundary layer will grow simultaneously and for a given Prandtl number that ratio of the boundary layer thicknesses has to be a constant so therefore this ? will not be a function of X anymore it will be constant if you heat the play heat the cylinder right from the beginning okay so right now we are considering a case which is isothermal throughout okay we do not have any unheated starting length okay so in that case you can take your ? as a constant you can write your ? T in terms of ? and ? and also you can substitute for ? in terms of ? and ? T okay so now you can integrate this now since this is a function of ? T and you are integrating across ? T so this will result final expression will be independent of ? T it will contain only terms with respect to ? and finally ? okay so that comes out to be ? x ? so I am writing I am taking my ? T on the right hand side here so and I can I can write this as ? x ? x D by DX x U 8 x ? T again I can write it as ? x ? and if I integrate this entire expression out in terms of ? T so I will be getting a expression like this 3 by 2 a where my M now M should be a function of what okay now I am substituting here for ? whatever function I have all this in terms of ? T and Z ? okay now if I integrate it over ? T now what should be the resulting expression should be a function of only ? okay so therefore M will be coming out as 1 by 5 ? minus 3 by 70 ? Q plus 1 by 80 ? 4 and my N will be 1 by 6 okay so it is not that difficult you know it is just one step integration which I am not doing it here you can just substitute for all this for f of ? in terms of ? T and ? multiply these two and then you integrate it out so you will get one bunch of expressions in terms of ? the other which is contained within the parameter ? okay so I am separating these two and writing it out okay it is clear till here now if you consider a particular case where your Prandtl number is greater than 1 so now what kind of approximation it does it mean on ? okay so your thermal boundary layer thickness is much smaller than your velocity boundary layer okay so therefore this corresponds to the fact that my ? is less than 1 so therefore all the higher order terms of ? like the second power and higher can will be very small and therefore can be neglected okay so all these terms can be neglected only you can retain the the first power terms okay with that we can considerably simplify this expression okay so so therefore neglect higher order terms of ? and also we are interested in the region which is close to the stagnation region as I said we want to get an expression for nusselt number in the stagnation region so this is for the case where my x by R is much lesser than 1 okay square much lesser than 1 okay so for this case we have also derived the expression for ? square now you have to tell me what will be the expression for ? square here if I make the approximation that is close to the stagnation region okay so my ? square comes from the expression of definition of ? if you remember okay this is ? and you in divided by du infinity by dx and for small values of x by R my u infinity of x will be 2 v 8 x by R okay so this will be ? ? ? by 2 v 8 R okay so this I can substitute in first so I have a ? here so I can substitute my expression for ? from this into this expression and also the value of ? for stagnation point okay we have already determined the value of ? which is what 7.052 so we can substitute the value of ? here as well as the ? from here so and therefore the resulting expression will be so let me call this as number 3 okay so substituting all these approximations now into 3 now you should be getting ? d by dx okay I have already substituted for ? okay ? in terms of so this is a I know the value of ? also so I have substituted directly into ? and this will be x times ? square square root of ? x 12 ? which will be equal to 90 by Prandtl number square root of ? okay so I have substituted for ? from here so everything will be in terms of ? and of course you are ? you have neglected all the higher order terms okay all these terms have been neglected so this will be ? here this will be ? ? ? square only the ? square term will be there okay so this will be 1 by ? plus 1 by 60 x ? x ? okay so if you take the ? term out so that will be ? square here and you substitute for ? in terms of whatever expression that we have so everything will be reduced to this particular form it is called let us call this is 4 now for ? equal to 7.052 this will be getting even more simplified and you can expand this differential equation as x d ? cube by dx plus 3 by 2 ? cube 1.0048 by Prandtl number okay so this is my final expression the ODE in terms of ? okay so all I am substituting is for ? so this entire thing is a constant and it comes out and of course this is also in terms of ? so now I can separate this d by dx I can take this inside so this will be ? cube into if I if I write this in terms of ? cube okay suppose I ignore all the constants I have ? x ? square okay so I can write this as okay so ? d by dx I can take ? cube so this is 2 ? x ? square ? x okay plus I can write this as ? cube x dx d by dx of x will be 1 okay so this is how I am expanding it now once again this term right here ? cube ? can be written as 2 x d by dx of ? cube okay into 1 by 3 okay so I have plus ? cube so basically I am dividing throughout by 2 by 3 and therefore I have this final expression okay so just comes through a couple of steps of algebra which is not that difficult okay so therefore this is the final ODE and you know now how to solve this we can this is a non-homogeneous ODE but it is linear so it is not a problem we can directly give a get an analytical solution for this we have to the solution will be one will be for a homogeneous ODE which is nothing but the complementary function the other including the non-homogeneous part separately so that will be the particular integral okay so we can combine both the solutions together let us do that now you have to tell me what is the solution to the homogeneous part of this ODE all of you take couple of minutes you can assume some ? cube is equal to something like P or Z something and you can say that your ? cube solution will be ? cube complementary function as I Q particular integral okay so this is the solution to the homogeneous part of the ODE and you can do it and tell me what will be the homogeneous part solution okay so that is basically it has to satisfy x d ? cube Cf by dx plus 3 by 2 Cf cube equal to 0 so this is the solution to the homogeneous part yes so what should be the complementary function 0 why solve this by separation of variables you have forgotten earlier ODE basics you can you know if you are confused looking at this CQ we can take this as some Z or something like that no ? cube so now then it becomes the standard form of now ODE which you can separate by variables okay x power minus 3 by 2 okay so some constant that is some constant x power minus 3 by 2 right so now how do you find the particular integral so you say that this is some Z complementary function plus Z particular integral okay and you substitute that into the original equation which is this okay so from there you already know the solution for the complementary function okay so then you sub you calculate what is your particular integral solution from there okay if you do that you will get that your QPI will be a constant basically which will be 0.6699 by PR okay so all you have to do is substitute for this plus some PI particular integral into this ODE 5 okay and you get directly an expression for the particular integral which will be constant finally it will be just 0.66 so you please go back and revise if you forgot how to solve a non-homogeneous ODE okay non-homogeneous linear ordinary differential this is the most basic you know so from this therefore your final solution will be zeta cube will be Cx power minus 3 by 2 plus 0.6699 by PR so this is your final solution now if you had a unheated starting length then you could have substituted that x is at x is equal to x 0 my zeta equal to 0 but in your case for the present case we will assume that it is heated right from the beginning therefore at x is equal to 0 zeta equal to 0 okay therefore this constant has to be 0 okay so in our case directly the solution will be 0.6699 by PR okay or this will be 0.875 by PR to the power one third so once you found out zeta now the entire problem is almost nearly solved because to get the heat transfer coefficient h is minus k dt by dy at y equal to 0 by T wall minus T infinity okay now for this particular temperature profile can you tell me how this will reduce I know my temperature profile for ? which is nothing but 3 by 2 ? T minus 1 by 2 ? T cube so this can be written as minus k in terms of ? this will be defined as T minus T wall by T infinity minus T wall okay so dt by dy will be d ? by dy into T infinity minus T wall okay so minus of that will be T wall minus T infinity so this will be k into d ? by dy so once again I can transform the variable as d ? by d ? T at ? T equal to 0 into d ? T by dy okay so my expression here will be basically d ? by d ? T at ? T equal to 0 will be nothing but 3 by 2 so this will be 3 by 2 into k and what is d ? T by dy 1 by ? T and since my ? is equal to ? T by ? I can write my ? T as ? into ? okay so finally since we have found the expression for ? it is wiser to put everything in terms of ? right so and therefore my Nusselt number can be defined as H into X by K or in the case of cylinder my characteristic length will be the diameter or the radius so I will choose the diameter as my characteristic length so this will be H into d ? this is the diameter of the cylinder okay so this will be nothing but so 3 by 2 into d ? divided by ? into ? so I can substitute my expression for ? into this okay and I can directly simplify to get my Nusselt number expression so my expression for H would turn out to be 0.645 into K by R into PR 1 by 3rd RE power half okay so this is already in terms of PR and my ? I know the expression for ? okay so which is nothing but square root of ? R by 2 V 8 correct so I am substituting everything into this and finally I can write in terms of Prandtl number and Reynolds number where my Reynolds number here is defined as V 8 into d ? by ? so d ? is 2 times R okay so then you can finally get the expression for Nusselt number which is HT 0 by K so which comes out to be 1.291 into Prandtl number power 1 3rd RE power half okay so this is my expression for Nusselt number in the stagnation region okay in terms of Reynolds number which is defined this way and the Prandtl number now compare this with the similarity solution for the similarity solution you also you derived the expression for Nusselt number that was in terms of the radius okay if you remember that was any R was 0.81 times RE based on the radius PR 0.4 please go and go back and recollect from the Falkner's can solution we have determined the expression from the similarity solution for the cylinder case this was defined based on the radius so if you convert this in terms of diameter okay that is you know it is a straight forward conversion so this can be written as 0.81 V 8 into d 0 by 2 times ? over half Prandtl number to the power 0.4 and you can define your Nusselt number based on diameter as HD 0 by K which is nothing but twice of N you are okay so if you put this if you multiply by 2 times so this will be nothing but your Nud 0 and on this side you will have a factor of 2 x 0.81 divided by square root of 2 okay so this will give you my Nud 0 as 1.145 and the number 0.4 RE based on your diameter power half so this is your exact solution right from the similarity from the Falkner's can similarity solution so you can compare that with the approximate expression so this is 1.29 so almost 1.3 this is 1.14 so very close and the dependence on the Reynolds number is the same the Prandtl number dependence is slightly different this is giving 0.33 whereas that is to the power 0.4 okay so so from the similarity solution you can see the agreement is pretty close not that exactly the same but very close to the similarity solution so with that we will stop here today so tomorrow let us make another assumption so that we can simplify the Carman-Pollhausen solution itself okay and with that we will get the same expression for Nusselt number and we will see how close it is so any questions okay so so these approximate solutions are therefore you know very useful you do not have to solve the OD in numerically okay so only guess some profile and substitute and finally you find the resulting expressions correlations for Nusselt number especially the integral quantities like Nusselt number are pretty much similar to the similarity solution okay although the profiles themselves may be different for example if you assume a linear temperature profile it is completely off from the actual temperature profile okay but the integral quantities like skin friction coefficient and the Nusselt number you get a reasonably good agreement with the approximate solution okay.