 d square t by d y square plus mu by k d u by d y whole square equal to 0. How did you get this that term from the phi term? You have the phi term with you. In that if you look at the phi, you have the phi term with you or no? So if you look at that all del by del x terms vanish, del by del z terms vanish, v terms vanish, w terms vanish. All you have is from that second del u by del y plus del v by del x whole square. From that particular term del u by del y whole square remains. So now look at the expression now, the energy equation, the entire set. Now all of them actually are ordinary differential equations. This is the point number 1. You should notice observe, appreciate a complete set of non-linear partial differential equations in its entirety by in this particular case of this particular physical model we have taken with certain fairly reasonable acceptable justifiable assumptions is brought down to the simplest. We cannot getting get anything more simpler than this in convection and d v by d y equal to 0 says v equal to 0. The solution is already there with the boundary conditions of course. d square u by d y square equal to 0 remains. The y component is a hydrostatic pressure description w is 0 everywhere and then you are only 2 terms in the energy equation. So now you can change all the dou u by d y dou t by d y to simply dou by d y d t by d y number 1. This actually tells you how before you formulate a during the formulation of the problem, you think what assumptions are justifiable. The first thing you should there is no necessity for directly hitting the complex partial differential equations. Also please understand these are all done when there were no computers. They said how do we simplify the problem for our hand calculations. Computers came in only in the forties early forties during the second world war it was basically developed by the Manhattan project basically the nuclear project. Before that what were they doing? And also that was the time when heat transfer was really picking up. Heat transfer is a subject to be taught in colleges came only in the nineteen sixties actually. If you look at the first book by Jacob Hawkins nineteen fifty five or so that is the very first textbook for heat transfer nobody people did not want to put into that. Later it has now come to the incomparable David and Bajan level action. So, making assumptions simplifications was a necessity at that time. And this apparently simple problem you can make it into a three dimensional thing. If you do not put in that del by del x tends to 0 del by del z tends to 0 it becomes a right solution starting from the leading edge to the trailing edge and the two sides. What you have done is I will not worry about the flow near the edges always you see even when you do the experiments you will see near the edges you will have two dimensionality three dimensionality. In a what do you think is a one dimensional four there will be two dimensionality what do you think is a two dimensional four is a three dimensionality. And boundary layer assumptions are not valid very close to the leading edge as all of you know. There are certain complications both physical and mathematical. So, first take away from as far as I am concerned is first of all the procedure of a convection problem starting from the physical model. The second take away is how do I simplify before I start saga. Now you have computers and codes so you write down the how many times have you really written down the physical model to really hit the mathematical model. You simply so flat plate you start del by del x plus del by del equal to 0 and start putting your boundary conditions. Boundary conditions if you look at this velocity conditions are fixed not much can be done either it is 0 or it takes the velocity conditions over the plate. But look at the thermal we have now taken we did not write it down please write it down isothermal plates that it skip my mind both are isothermal. But if you look at the thermal part of it you can put various kinds of boundary conditions on both the plates with various combinations both isothermal both non-isothermal one isothermal one non-isothermal one with constant heat flux both with constant heat flux one adiabatic how many kinds of boundary conditions you can have combinations each one of those boundary conditions will give you a different solution for the same set of equations any number of times I will be repeating this the importance of formulating the problem rightly formulation of the problem correctly is 50 percent of the solution especially it is true important rather when you hit a numerical techniques when you use certain equations appear to give you the right solution but they are really not the right solution because your numerical equivalents that you have written do not really simulate the mathematical equation that you have written. So, whenever you write numerical you should check for most of the time you should check only for convergence stability criteria you have to develop but third most important thing is consistency whether your finite difference of finite element representation of the partial differential equation or though it is consistent with that equation because of this round off errors and truncation errors sometimes what you think is the right finite difference formulation will simply go off in a different direction but give you a convergent stable solution this has been experience of many people it would not happen in every case but there are situations where this happens you should bear in mind. So, formulation of the problem is extremely important simplification is extremely important I like this problem because in its entirety you can get this idea of what simplification all solutions for neighbor stokes equations base is based on simplification what is Pantel's boundary layer this is simplification here we did not invoke the boundary layer that necessary we have solved the entire neighbor stokes equations in order to add then you can of course add the heat transfer what is it called heat generation term you can add we have kept a very important term in energy equation the viscous dissipation we will see later. Now write down your written the boundary conditions so these four equations with the boundary condition u equal to 0 v equal to 0 w equal to 0 at the wall bottom wall t equal to t0 at the bottom wall u equal to u1 at the top wall v equal to 0 w equal to 0 at the top wall these are the boundary conditions there are no initial conditions because we have taken as a steady state. Now let us proceed I hope some of you have solved it but let us go step by step solve the velocity equation in any heat transfer problem first you have to solve the velocity equation because you require that for the energy equation in the case of force convection these are different in the case of free convection they have to be solved together. So what is the solution v square u by d y square equal to 0 y equal to 0 u equal to 0 y equal to l u equal to u1 so now this also tells you how when you are actually applying a numerical technique to a partial differential equation what are you doing actually after putting in the finite difference for finite difference or finite element then you go ahead with the solution point by point actually you are I want you to you are dash the partial differential equation you are dash doing this to the no no no no no no when you get the solution a differential equation what should you do for you to get the solution no that is a numeric that is your you are directly jumping to new discretization because you want to use a numerical method even after that when you really solve what is a pde do partial differential equation let us take t for example it is a function of x y z and tau 1 what is the difference between a pde and a finite difference equation finite difference equation of that pde what is the difference what do you get when you solve these things in terms of the solution tell me where where in the other one so you are now I will I will to save time you are integrating you are integrating the pde you have to integrate now you do that here that is what I am saying d square u by d y square twice you have to integrate and two constants the two boundary conditions you must have at y equal to 0 y equal to 0 please by looking at the equation you should be able to actually write it is nothing but u by u1 is equal to so now we will go to what is this step now solution now for now we are actually going to the solution so the momentum equation that is the I will say velocity profile is nothing but u by u1 equal to please see and what does this say what kind of equation is this now use the word whatever you used earlier the solution now is linear so plot this please in your physical model what it says is at any given this is let us say x here so here this value here is u1 and this is 0 again I want you to appreciate this such as originally what we thought was extremely nonlinear it is a linear profile basically so one more point v equal to 0 everywhere we have said w equal to 0 everywhere u is this profile now you come out with a name for this what kind of a flow is this it is not a potential flow it is not a boundary layer flow all the layers are can you look at this all the layer stream layers are to each other in what fashion are they oriented to each other it is parallel flow v equal to 0 w equal to 0 it is simply parallel flow number 1 it is also fully developed flow it is not going to change del u by del x equal to 0 so it is a parallel flow it is a parallel plate geometry it is okay but you can have nonlinear velocity profile in parallel plates as you will see later and this happens only for dp by dx equal to 0 please see that so therefore it is called a simple or plain quiet flow and therefore you get a linear velocity profile number 1 it is also called a parallel flow as different from boundary layer flow the stream lines are all parallel to each other this is something again you get from this profile very easy the really exciting thing is go to the temperature profile write down the energy equation d square t by dy square plus mu by k du by du by whole square equal to 0 boundary conditions at y equal to 0 t equal to t0 y equal to l t equal to t1 we have not said what are the relative values of t0 and t1 leave it let us see what happens later now you have to integrate this twice I had yesterday I asked you to do this but some of you have not done go ahead and do it please this actual procedure that you follow this is a classical solution now classical complete exact solution no approximations you get two constants take two minutes give me an expression for t of y equal to what is a function of only y now finished can you give me the terms I will write down tell me plus t1 into y by n t1 into y by n is it minus mu by k minus y by n what is it minus is it no y by n here bracket bracket y by minus y by k square something else is missing here mu u square by 2 k yeah this is okay this is minus r plus check it out or you should redo all of you get this this equation is important because you have to differentiate it later two three times tell me some manipulation of this can be done I am not sure I will write how many of you got this do not copy that do not I am checking whether it is right or wrong I will give you my expression you check it out may be slight rearrangement will be all right please check this out t0 that is okay actually this is also okay here please see whether we would like to correct it please check whether something is missing there and then the sign also check it out please we will again later for something else something is missing there was it okay ty equal to t0 plus mu by 2k u1 square y by l y minus y square by l square plus t1 minus t0 y by l did you miss one term that check it out how is it somebody tell me this second one is right then I will proceed because you gave me the first answer I am asking you everything is that except that that t0 was missing t1 y by l minus t0 just check it out okay now you see a little bit this is this is the equation which please keep it give it a number I do not know what is the number of this instead of what was it can you give me a number okay today I will give you this number now you see you have got a y by l without you are trying for it you have got a non-dimensional distance can you get a non-dimensional temperature difference by reman by manipulating this just look at that expression there is a t here which is what we want there is a t0 there is a t1 minus t0 so I can rewrite this expression as t of y minus t0 bring t1 minus t0 to the left hand side now you have a non-dimensional temperature difference it is not non-dimensional temperature in non-dimensional temperature difference always non-dimensionalization makes things easier for various reasons now so this is y by l is y by l plus this whole thing divided by see now you have got a non-dimensional temperature difference temperature is a non-dimensional form as a function of a non-dimensional distance which is very nice on its own it is coming out slight manipulation and as a function of the fluid the viscosity the velocity the thermal conductivity we will do a small further manipulation and see something else very important will come out and then we will go further can you okay temperature difference is non-dimensionalized this is non-dimensionalized therefore one would think this term is non-dimensional please write down what this term would be mu u1 square by 2k t1 minus t0 I am communicating here I am communicating please that term I am saying t minus t0 by t1 minus t0 you have your y by l why do not you put the other see whether non-dimensionalization possible what is that term mu u square by 2k is there and you have divided by t1 minus t0 how can you non-dimensionalize what is it that you can do here I will give you a hint you will get a very well known non-dimensional number out of this by slight manipulation you are jumping fine fine we will get the cut but a more familiar non-dimensional number you have a mu here k how to get a plant number there multiply divided by cp and now you see what you are getting only now you get the cut number so now you will see the non-dimensional temperature difference can you give me the second term here t minus t0 by t1 minus t0 equal to y by l plus mu cp by k is now plant number then what else you have there so you have mu cp by k half is there of course you have u1 square t1 minus t0 is there you have multiplied and divided it by cp now look at that plant number you know now you are coming out with a new non-dimensional number you did not try for it it is coming on its own with a slight manipulation and what is that number this number is the Eckert number we did not plan to get this we did not say we will get the Eckert number even when they did it first time they did not plan for Eckert number it was that name was given that is a different matter later but you see how from again and again I emphasize this from very simple problem you are coming out with plant number you knew that of course Eckert number is very new and it comes out because of you will look at the expression u1 square by cp delta t what is the physical significance of a plant number mu cp by k what is the physical significance of the Eckert number u1 square by cp delta t cp delta t is a imposed temperature difference so cp is a imposed thermal energy kind of thing and what is u1 square it is a kinetic energy so kinetic energy divided by a imposed thermal energy form there cp into actually it is a cpt not minus cpt1 so there is a certain in a problem in this problem there is a certain energy in terms of cpt that is now being related to the velocity of the plate if the plate was stationary it would not have come up in fact that would be 0 obviously so a new number has come up now called Eckert some books give you the symbol as ec some give you as a symbol e you use as you like but now you just change the final equation as y by l plus half prantle Eckert into y by l minus y square by you do not have to do that but this gives you a new you can visually see the new non-dimensional number and you also realize that it is important only when u1 is significant u1 can be 0 then it is not significant u1 can be very small value it has its own role but you also know it changes with u1 square so from 5 meters per second u1 you go to 10 the effect is squared actually so that means a small change in the velocity finite change will have a very that power law dependent effect on the Eckert number which is nothing but the manifestation of the u1 square the kinetic energy if you go into the previous one here this this actually is much more according to me representative of the situation mu u1 square we are talking about viscous dissipation if you look at only u1 square it is kinetic energy that is also right but the real effect is the viscous dissipation is mu u1 square and that can be in a mu is already here in the prantle number so you can try to give it a physical significance this is extremely important why do we do all this we would like to know the temperature difference by the way suppose you have u1 equal to 0 here what is the temperature profile linear same thing in fact the velocity profile is linear temperature profile is linear when u equal to 0 that means it is simply conduction it is just conduction that is happening the moment you start adding u1 convection is coming into picture the profile is becoming non-linear this is a non-linear term now second term is u1 square is being introduced now this is the temperature profile after getting the temperature profile in a convection problem what is it that we want to do find out what is the heat transfer at the top plate what is the heat transfer at the bottom plate get the expression how do we do this therefore you should first get now I will say this is step whatever now I have last track of this heat transfer heat transfer q I will say flux equal to minus k dt I can easily write dt by dy there is no problem not necessarily del t by del y this is a general expression if I put q0 here this is the wall so I have to put dt by dy at y equal to 0 and flux at l at the top plate is minus k dt by dy at y equal to l can you get two simple expressions first get dt by dy and get two expressions please that will complete this part of the quit problem there are two more important things which I want to refer to get the general expression for dt by dy from the t equation you know how to do this is not it what I will do you do it it takes some time I will I will give the expressions check it out but I want you to do it separately q0 double dot the heat transfer at the wall this is equal to minus mu u1 square by 2 k 2 l minus k by l tl minus t0 t1 minus t0 minus mu u1 square by 2 l minus k by l t1 minus t0 and the heat transfer at the top wall is plus mu 1 square by 2 l minus please check it out so you get the temperature profile then take dt by dy dt by dy at 0 multiplied by k minus k you get the heat transfer at the bottom wall and then heat transfer at the top wall minus k dt by dy at y equal to l this is all fine these are all equations but you have to have numbers fine so what I will do I will give you a very simple problem I want all of you to write a simple program and now study this solution for the effect of various variables here tell me looking at these expressions what are the important variables that you that affect the heat transfer number one we will do the various so all other conditions being the same u1 controls the heat transfer influences the heat transfer then what else suppose I keep u1 constant fluid frontal number okay then temperature right of course you have to say l but we will fix l now I want you to do the following please take a problem I want you to write a small program for this we have these two plates distance small we have never talked about what is l actually let us take this l equal to 3 millimeters and this is u1 please switch off your mobiles please 10 meters per second we will see the frontal number here. This is at 10 degrees T0 10 T1 30 is everything there frontal number now I want you to do this take air here frontal number take water take what is referred to as engine oil different kinds of oil are there fuel oil engine oil that gives the frontal number with these conditions please get the temperature profile but basically the heat transfers at the two walls find out what is the effect of frontal number then in the same program please change this u1 up to 100 meters per second or so in 10 meters per second intervals find out what is the effect of this velocity okay now keep it at 10 meters per second change the bottom temperature T0 take 5 degrees centigrade and 50 degrees centigrade you get the point what I am trying to do here you get the point what I am trying to do here frontal number three frontal numbers air water and oil higher velocities 10 meters per second is very small temperature you see what I have done the first problem I have given is 10 degrees centigrade keeping that at 30 why I have given you 5 and 50 if I take 50 here what is the difference between this problem and that problem this is 30 and 10 so T1 is greater than T0 actually what you could do you could also take 30 degrees T1 equal to T0 and then T0 greater than T1 see the difference in heat transfers but actually there is a very important application of this and therefore another factor is extremely important for us to calculate the moment there is a nonlinearity in the temperature profile can you visualize another factor which comes into may come into picture another kind of a temperature within the fluid by the way please understand all this temperature profile is for the fluid plate temperatures are fixed if there is a nonlinearity what would you like to look at as a means of as a way of understanding the problem linear you know how it is now there is a nonlinearity whenever there is a nonlinearity we would like to find out one particular I am I do not know whether I am able to communicate this question properly you want to look at something else in terms of the temperature profile how does it will it continuously nonlinearity vary in one direction I mean I am trying to give a hint or could there be a in the curve what do you expect you have 10 degrees at the bottom and 30 at the top linear profile you can easily say how it is how would be the nonlinear profile I will come to that you have to plot it then only you may be understand okay maybe I am I am not able to communicate let me assume I am not able to communicate the problem I will state it what is the maximum temperature in the fluid find it out find out if if there were to be maximum temperature because it is a nonlinear problem exact the answer is actually my reason for asking you to do this how do you find out whether there is a maximum or a minimum what do you do to the equation what differentiate what so dt by dy should be what find out if there is a dt by dy equal to 0 at what condition how do you find out whether it is a maximum or a minimum do that see if there is a maximum and if it is there find out if there is a maximum what would you like to find out first of all find out whether there is a maximum if there is a maximum what is the next question that should arise in your mind okay but something else related to this please find this out this is very important and therefore I want you to see what is the t maximum for air for water for oil oils can be anything plant number from 1000 to 50000 if you want you can check it out but take any one or two plant numbers it is okay and please write a small program and plot all these things we go back to the temperature profile how do we plot plot the temperature profile what should be the ordinate and the abscissa what can be look at the physical problem and how do you want the profile to be plotted what should be the abscissa what should be the ordinate that is fine so where do you want theta where do you want to looking at the problem which would be okay for you so you please plot what will be the maximum value of y by of course this is 0 what is the maximum value of y by L what is this what is the linear profile if there is no u1 u1 is 0 static plate how will this be here to where will it go there what I want you to do is really plot when I say u1 equal to 0 look at this expression please theta if this is theta if we have to refer to it theta equal to y by L that profile is for second term being absent the moment I bring in u1 that means Eckhart is coming into picture tell me give me the effect of plant lacker that will give the effect of the u1 more importantly viscous dissipation what numbers can it take you do not know you can assume now at this point in time I am going to suggest to you use this expression now this is a very nice non dimensional expression it came up just like that plot y by L and this for various plant lacker numbers plant lacker numbers and I want to convey to you unless you plot it you will not understand what I want to talk about in the next class please do this not only plot it think about it what is happening I have to still bring out two important aspects of this whole problem we will do it on Monday no problem I am not in hurry I want you to understand this especially one two three concepts will come out of this which you can then relate it to normal convection so please plot this y by L plotting is no problem it is the same thing y by L and theta are the same u by u1 equal to y by L theta equal to y by L so there is no problem but for varying plant lacker number make sure that is when you do that you will know there is a maximum now I am giving you actually the answer but I want you to do that it is a maximum find out where that maximum is why think about it that is one aspect of the solution if a maximum comes into picture what is the effect of this on the heat transfer not the rate heat transfer direction at the two plates we have when I when we said how to plot you said something when I ask what is the maximum temperature you said t1 you will find it is not t1 it is something different it is t1 if there is no u1 because of u1 something is happening that something is there is a maximum which occurs I am giving you that answer find it out at some location but then what is the effect of that maximum coming into first of all y and what is the effect that will affect gentleman the direction of heat transfer itself this is extremely important point I want to make little bit more force in the next class but for that I want you to plot the theta with respect to plant lacker also it is not simply lacker it is plant lacker when you then fix up a plant lacker numbers will change that is different but give a number to plant lacker as 1 2 3 4 5 6 7 8 in particular tell me what happens at plant lacker equal to 2 study that also therefore find out what happens after plant lacker equal to 2 what happens means what happens to the heat transfer that is what I am looking you will find very new information coming out of this is very extremely extremely important I will ask you something else now before I leave you look at the quit problem what does that look at look at that please look at that is there is a bottom plate fixed there is a gap there is a top plate which is moving there is a temperature here that is u equal to 0 u equal to 1 what does this this physical model remind you of which you are more familiar with you will be that is because you are not thought which is no problem we have said there is a plate suppose there is no plate but there is a velocity what would that problem boil down to look at the physical picture look at the boundary conditions Baba flat plate flat plate ok flat plate what is the flow there what is the type of flow on do not simply say flat plate add something to that what type of a flow on the flat plate sometimes you know you have to give the teacher the answer that he wants but in different ways we can discuss about it this point you have not thought of so far I am making you think requesting you to think rather you are right but flat plate boundary layer flow do not simply say I am there that is also correct that is why I want the bright words to come look at this problem quit problem you thought it is all restricted but one of the takeaways from quit flow is this can be related to a boundary layer flow very simple bottom plate fixed y equal to 0 u equal to 0 t equal to tw y equal to L is y equal to delta u equal to infinity whatever that is t equal to t infinity which one small qualification what could that you are right now you can write a sketch of the boundary layer flow and compare these two where will this quit flow situation will be really applicable in a boundary layer flow in what zone or tell me where it is not applicable delta is always varying delta is a function delta is y by x root Reynolds x near the entry it is not valid I say I wanted that word that is actually now the quit flow is nothing but a representation of a fully developed boundary layer flow fully developed I would say on a flat plate still but you can extend it to the pipe flows later where after that nothing is changing it is a constant you can it varies as delta by x is y by x root Reynolds x x power half it varies but as you go along half it the rate reduces so away from the leading edge the flow is very similar to quit flow once you appreciate this the effect of viscous dissipation you can actually super impose upon the boundary layer flow post convection mode and therefore any information that comes out of it is just a modified it has to be modified for a regular flat plate this is one of the major takeaways from looking at a quit flow in some detail we could have done it by same I could have given you the equations and then you say solve it would solve it but I want you to please think about these things so that is another takeaway I will stop here now Monday we will meet I will draw the profiles solve this problem I know you have other subjects but also please do this you have a what is it quiz is a quiz week but do this as much as possible at least for one plant number especially for oil you do I am giving just a hint you know why I am saying this this problem okay tell me in in in machines in mechanical equipment where would you I have given you a hint actually where would you apply this especially therefore look at the size of three millimeters that we have taken this is journal bearing problem actually but where is journal bearing problem how can you and go to infinity I mean a constant you one a plate moving at you one actually that is replaced by the journal so that is you can take it as a flat plate moving continuously and then there is a bearing in between is now tell me why T max is important that simply for you to do some calculations here and says I got T max now related to journal bearing problem why is T max important that is why I ask you to do for oil journal bearing if I put oil what is that why do why do I put the oil huh what is the word used for that lubrication and for you are putting oil why it has a certain kind of what viscosity now how does temperature affect viscosity find out you have to maintain the temperatures properly for the oil film for you to have that lubrication effect lubrication will fail under these conditions so find out what is T max now you see this square problem has a practical application although not a very great it is important actually but it is not earth shattering application but it is a very important application you will know how to formulate the problem how to make simplifications how to make it the simplest of equations in convection but how much information you can get out of it especially the next class after you draw the temperature profiles we will discuss a little bit further and see something very interesting that is happening thank you.