 So, as I just said a few minutes back, we can employ some extremely simplifying situations and assumptions and obtain the analytical solutions to the governing equations that we just looked at. So, we will go through a few of these situations, there are plenty more available in the literature. Normally in an undergraduate fluid mechanics course, we discuss only one or two of these. Some others which I have taken here, so there are some five or six that we will be talking about. Some others are really from heat transfer, if you look at them carefully. But since we have chosen to include the energy equation as part of our governing equations, the solution of the energy equation will be also taken up in a couple of situations. Formally those would be considered as heat transfer situations. However, this course is a CFD-HT course and in that sense we are going to deal with little bit of heat transfer as well. We would not spend too much time on the heat transfer part. Some of the people who are already familiar with the heat transfer courses would probably recognize those solutions very fast. But other than that, we are not going to spend too much time on the heat transfer solutions. We will just outline the procedure and I will point out if you want to look up into any of the books later where you can see the complete detail. I have worked out actually a reasonable amount of detail, but the complete detail then I will point out where to look for as far as the heat transfer problems are concerned. So let us just start looking at some of the problems. I have listed which ones that we are going to talk about here, 5 or 6 situations. Normally some of these are taught in our Advanced Fluid Mechanics course which is a postgraduate course, but nonetheless it is worthwhile looking at those right now. So the first situation that we will take up is what is called as a fully developed flow between infinite parallel plates. So the problem setting is shown on the screen right now. What we have is let us say two very large plates. By large I mean that their area of cross sections are very large or the area in their own plane is very large and they are separated by a small distance which is h and this space between the two parallel plates is filled with a viscous fluid. Now we call this infinite for the purpose of formulation of the problem. Essentially infinite is supposed to mean that the spacing between the two plates is much smaller than either the length or the width into the plane of the paper of these plates. The coordinate system is shown. We will choose y equal to 0 at the bottom plate, y equal to h at the top plate. The assumptions that we are going to employ and these are very important to note because under these assumptions only we are in a position to solve these equations. So assumptions are that we will deal with steady constant density flow. We will neglect body forces. What infinitely large plate is supposed to mean is that essentially there is no third component of the velocity in the z direction which is perpendicular to the plane of the board. The reason for this in one way you can say is that there are no boundary conditions to be employed on the z direction and in absence of any boundary conditions you will not generate any variation in the z direction. So, because of the 2D nature of the flow then we will treat third component w equal to 0 and any variable that you can think of will not have any variation in the z direction. So, that the differentiation with respect to z of any variable is equal to 0. This is because there is no boundary condition that can be employed in the z direction, z direction being infinitely large. So, this is how that d dz of an empty bracket is supposed to be implying that the empty bracket can contain any variable and let it be pressure, let it be density whatever it is there will be no variation in the z direction. Finally, we will employ what is called as the fully developed flow situation. What fully developed flow means is that physically speaking the axial velocity will not be changing with axial distance that is what is formally meant by fully developed flow. So, in terms of mathematical representation it is the partial derivative of u which is our axial velocity with respect to x which is our axial distance equal to 0 and immediately that implies that the u velocity is not going to be a function of x. In fact, if you remember this was the setting really that we used right on day 1 when we discussed the expression for dynamic viscosity mu tau is equal to mu times du dy is what we had obtained long time back and that setting really was exactly what you see right now here. So, it is good to revisit that. Now the where is the fully developed condition coming from and there is no analytical basis for it really speaking. It is a purely experimental observation that if you have a sufficiently large length of a channel or a pipe experimentally you find that after a certain distance after the flow enters in the channel or pipe after certain distance it simply seems to follow this expression namely that the axial velocity becomes independent of axial direction. So, this fully developed condition has a purely experimental basis it has no theoretical basis so to say it is found to occur time and again in large pipelines or large channels in terms of the length I am talking about large meaning and therefore, we will employ this as a simplifying assumption in our problem. So, what we are going to do is we will take the continuity equation we will take the momentum equation and we will try to step by step simplify these to see what happens under the given assumptions. So, begin with the continuity equation we have divergence of velocity equal to 0 in our 2D situation it is du dx plus dv dy equal to 0 with the fully developed condition in we already know that du dx equal to 0. So, I have crossed that out which means that what is left out of the continuity equation is only that partial derivative of v with respect to y 0 which immediately implies that v is not a function of y. So, v can be a function of only x because partial derivative of v with respect to y is 0. So, v can be only a function of x. So, let us see what this function of x has to be. Now so far when we discussed our governing equations yesterday and earlier in the morning we never talked about the boundary conditions and in some sense I did that on purpose because as we are going to solve these analytical solutions we will actually talk about the boundary conditions that we normally employ in typical fluid flow situations. So, going back to our setting what we have is 2 plates. So, we will essentially assume that these 2 plates are solid in the sense that they are not porous and therefore there can be no flow flowing in a normal direction across the plate it is a impermeable solid surface which then means that at both y equal to 0 or y equal to h because there is no flow across the plates we have v velocity equal to 0. Remember that if there was supposed to be any mass flow rate across a horizontal surface such as the bottom plate or top plate there has to be a non-zero vertical component of velocity because only a vertical component will carry mass flow rate across a horizontal surface keep that in mind. Therefore the boundary condition that is employed is that v is equal to 0 at y equal to 0 or h and the only way you can satisfy this boundary condition is basically to assume that this f of x has to be identically 0 otherwise you will never satisfy v equal to 0 at y equal to 0 or h. So, the conclusion that we obtain from here is that this so called function x which the velocity v was is actually identically equal to 0 and in fact that implies that the velocity in the y direction in this situation itself is identically equal to 0. So, this is an important conclusion and it actually comes from the fact that we have a fully developed situation. So, we are going to talk about these fully developed situations a couple of times right now we are talking about them in a pair of flat plates later we will talk about it in a pipe of circular cross section. You will see that whenever you employ this fully developed condition which is axial velocity is independent of axial direction you will see that the lateral velocity. So, in this case it is the v velocity turns out to be 0 if you have impermeable bounding surfaces. So, it is something that you should keep in mind. So, the boundary condition in words is written on the slide as well that it is 0 technically it is relative velocity with respect to the solid surface. If our surface is stationary it will be the absolute velocity, but when we want to write it in a more general fashion we write it as 0 relative normal velocity at the solid surface as our v velocity and that is equal to 0 at both y equal to 0 and at y equal to h right. So, the conclusion from continuity equation is that there is no y component of velocity. Then we will go to the y momentum equation which is the lateral momentum equation. Under our governing equations we had seen this of course and under our assumptions that there will be no body force and so on. We write the momentum equation in the y direction in a non conservative form. So, for such analytical solutions it does not matter what form you use. In fact, many times it is convenient to use the non conservative form. So, that is what you will see that I am using most of the times during this set of slides. So, the first line writes the y momentum equation in Cartesian coordinates. Remember we are dealing with 2D situation because we said that w etcetera are 0. So, I am just going to write 2D form and then I am writing this explicitly opening the substantial derivative. Remember that we assumed steady flow. So, when we look at the substantial derivative there is a local derivative and then there is a convective derivative. Local derivative will be identically equal to 0 because we are dealing with a steady flow situation. So, what is written out on the last line on the slide is simply the opened up substantial derivative and that will show only the convective derivative. But let us see what is happening to the convective derivative. We said that from the continuity equation we arrived at the conclusion that v velocity is identically equal to 0 everywhere. So, if you look at v velocity as a mathematical function it is equal to 0 everywhere. So, the derivative of a 0 function everywhere is going to be also 0 and therefore, this dv dx term in the first term of the convective derivative will go away. Similarly, the second term will go away because v itself is 0. Exactly the same thing will happen on the second term on the right hand side which is our viscous term mu multiplied by Laplacian operating on v. So, since the function v is identically equal to 0 everywhere we say that we will see that rather the derivative and the second derivative are also equal to 0. This is important to note because this is an identically 0 function. So, that is what we are trying to point out. So, what is left out then here is only the pressure gradient dp dy and all other terms are found to be 0 and therefore, we are going to conclude that there is no pressure gradient in the y direction or dp dy is equal to 0. So, this is the way these analytical solutions are worked out. One by one you take the continuity equation, simplify it based on our assumptions, see what conclusion comes out, move on to the next equation which is presently we have just looked at the y momentum equation. Finally, going to the x momentum equation again for steady flow I have written it only as convective derivative. Now, with the discussion that we have already had you will see that the entire convective derivative will go away because either du dx is 0 or v is equal to 0. Similarly, du dx is 0 which is our fully developed condition which means that du dx as a function is identically equal to 0 is what that means. So, the first term in this viscous term the Laplacian of u we have d 2 u dx squared which you can write as d dx of du dx, but if du dx is an identically 0 function everywhere the derivative of du dx is also going to be 0 everywhere identical. We are not talking about a local point where du dx is 0 it is 0 everywhere in the domain. So, it is identically 0 function and that is the reason the derivative secondary derivative with respect to u is also set to 0. We have already derived from the y momentum equation the conclusion that pressure is not a function of y which means that it can be a function of only x. Therefore, there is no need to write a partial derivative for the x and that is why I have started writing this as a total derivative of x with for pressure. So, dp dx is what is written with the total derivative and that remains to be equal to mu multiplied by d 2 u dy squared that is only surviving term from our x momentum equation. Now, again if you go here write at the beginning as a result of our fully developed flow assumption we had immediately concluded that u is not a function of x. So, it is only a function of y and therefore, this partial derivative of u with respect to y the second one second derivative is not really required to be written as a partial derivative and that is precisely what I have written. So, now let us look at what has happened this is a very important step again on the left I have dp dx which we are calling as at most a function of x. The reason is because we have already worked out that p is not a function of y from our previous equation. So, therefore, the derivative of p with respect to x can be at most a function of x. Similarly, the second derivative of y second derivative of u with respect to y can be at most a function of y, but the way the equation states is that the left hand side which can be a function of at most x is equal to a function of at most y. So, the only way this can be possible is that both sides are equal to a constant that is the only way this equation will have any physical meaning whatsoever and therefore, we are in a position to conclude here that both the left hand side which is written in terms of the pressure gradient or the right hand side which is written in terms of mu times the second derivative of u with respect to y are independently equal to the same constant. So, what is done now is that in order for us to find the velocity field in this problem we then assume that the pressure gradient in the x direction is a known constant. So, let us assume that it is a known constant and therefore, after assuming this to be the known constant we are in a position to integrate this mu times d 2 u d y squared equal to a known constant with respect to y twice and that is precisely what I have done and written the expression for u as a function of y which will obviously involve two constants of integration because we have we have we have integrated this twice with respect to y. So, what remains in this expression is the pressure gradient term in the axial direction and that pressure gradient term is going to be assumed to be known or specified as part of the problem. Although it was not really given here I wanted to point out that as a result of the simplifications you come to a conclusion where you are forced to essentially say that unless I assume one of the two to be constant I cannot go anywhere. So, the customary procedure in these in these analytical solutions is to assume that the pressure gradient is known and then you are in a position to integrate the reduced x momentum equation as we call it. So, this is the reduced x momentum equation and obtain the velocity field which as we expected right from the slide one number one is going to be only a function of y. So, the only thing that remains to be determined now is the two integration constants a and b for which we will consider it the two separate cases. So, the case one would be that both plates are stationary and this is what is called as a plane Poiseuille flow in which case the boundary conditions are the following. Now, just like we had a boundary condition on the v velocity we assume that there is no relative normal velocity because the plates were considered to be impermeable solid. The other boundary condition for the axial velocity or the x direction velocity is essentially that the relative tangential velocity meaning in the x direction at a solid surface which is either at y equal to 0 or at y equal to h will be equal to 0 and this is what is called as the no slip condition. So, the no slip condition is again a boundary condition that is coming from purely experimental observations. So, people have observed time and again that in continuum flows what appears to happen is that as you go closer and closer to a solid surface which is stationary and which is bounding the flow what is found is that the relative tangential velocity if it is a stationary surface it becomes an absolute velocity. So, the absolute velocity in case of a stationary solid surface tangential direction at the surface is equal to 0 this is what is popularly called as the no slip condition and again there is no theoretical basis for this no slip condition for continuum flows the kinds of flows that we are talking about under continuum assumptions this is found to be always the case from a purely experimental point of view and therefore, we are choosing this as one of the boundary conditions to be employed on the actual velocity in the x direction in this case. So, if you go back and substitute u is equal to 0 at both y equal to 0 and at y equal to h you can easily evaluate these constants a and b and finally, the velocity field which turns out to be a velocity profile because it is only a function of y is written in the middle box on the slide right now. So, again just to summarize how we have gone about we started with the continuity equation, simplified it under the given assumptions got some conclusions then moved on to the y momentum equation simplified it got some conclusion, moved on to the x momentum equation simplified it realized that it was a situation where one of the two terms that was left in the x momentum equation had to be assumed as a constant. So, for the purpose of sticking to how the literature moves we have chosen the axial gradient as actual pressure gradient as a known constant and assuming the actual pressure gradient as a known constant we have integrated this reduced x momentum to find out the velocity field which turns out to be as we say a velocity profile. So, this is case number one the other case which is considered here is that the lower plate is stationary and the upper plate at y equal to h is moving with a constant velocity u and capital U that is and this is what is called as a quet flow situation. So, the boundary conditions then is that there is again 0 relative tangential velocity at a solid surface. So, the lower surface is stationary. So, the relative velocity is essentially equal to 0 for the fluid it will also be having that means a 0 velocity. At the top surface at y equal to h because the top plate itself is moving at a constant speed of capital U and if there has to be no relative tangential velocity between the fluid and the plate at y equal to h we essentially say that the fluid will also have the same velocity as the surface which is equal to capital U and therefore, the boundary condition here at y equal to h which is written out at the bottom of the transparency is that the u velocity is equal to capital U at y equal to h. So, with these two conditions go back to your general form and evaluate the constants a and b and you will realize that this is how the velocity field which is again a velocity profile shows up. So, let us just see how these two situations look like that is what I have tried to try to show here on the on the sketch. So, our case number one was the plane Poiseuille flow and if you go back remember h is a constant mu is a constant and we assumed dp dx also as a constant as part of the solution. So, the entire multiplying factor to this square bracket is a constant what is left is essentially a square variation or y squared variation I should say which is essentially a parabolic variation and that is what is shown as the u velocity profile in the Poiseuille flow plane Poiseuille flow which is a parabolic profile. Obviously, many of you have seen a similar parabolic profile in flow inside a pipe of circular cross section. So, this plane Poiseuille flow is exactly same is just that it is a counter part of that circular Poiseuille flow as it is called fully developed laminar flow in a pipe of circular cross section. So, here we are doing the same problem in two dimensions where we are talking about a channel that that is about it going back to the quit flow part you you see this at the bottom you will see that the u velocity has two parts one is a linear variation with y and then there is a parabolic variation again which comes through as the second part. So, in case of quit flow and there was a question asked on quit flow I think couple of days back as well. So, we can we can note this down right now in case of quit flow even if you assume this axial pressure gradient to be 0 you still get a non-zero solution and the reason is because it is that upper plate which is moving at a constant velocity equal to capital U will transmit its momentum through the action of viscosity to the bottom layers in the fluid subsequently lower layers in the fluid and that motion of the top layer through the action of viscosity will maintain actually create as well as maintain the flow. In addition if you have the axial pressure gradient that is not a problem because you can add that through the second term, but even if the axial pressure gradient is identically equal to 0 in case of quit flow the flow will still be non-zero because of the motion of the top plate and that is what I have tried to show that the typical situations where you have no axial pressure gradient you will get that linear profile as was shown by u multiplied by y over h the second term will be exactly equal to 0 in that case and then there are two possibilities one which we call as a favorable pressure gradient which is given by dp dx less than 0 which simply means that as you go in the direction of the flow the pressure is decreasing. So, it is going from a higher value to a lower value which means that the pressure gradient which is defined as p later minus p earlier divided by the distance is negative this is what is called as a favorable pressure gradient. So, let me just write that on the whiteboard. So, if I want to roughly plotted it could be in general this is our x direction the pressure could be something like this it does not have to be necessarily linear or whatever. So, I have just shown it as a dropping pressure as you move in the direction of the flow. So, this is what is called as a favorable pressure gradient and it will try to accelerate the flow in the direction of positive x axis. So, going back again so, in the case of a favorable pressure gradient what you generate is a velocity profile that is given by this curve B the third option is that the pressure gradient can be positive and that is what is called as an adverse pressure gradient. So, let me go back to my whiteboard. So, again if I want to come up with a sketch it can be in general something like this. So, the flow is going this way along the positive x direction and the pressure as a function of x can look something like this which means that as you are moving in the positive x direction the flow actually encounters increasing pressure. So, in that sense the flow is moving against a higher and higher pressure value in a situation like an adverse pressure gradient. So, the only way it is possible is because you have the motion of the top net driving the flow that is the only reason it is possible that the flow can move against an adverse pressure gradient. So, keep that in mind that if you remove the motion from the top plate if you make the plate stationary just like the plane positively flow and if you have an adverse pressure gradient as what is shown by this C expression it will actually reverse this direction and it will start moving in the negative x direction. So, the only reason why the flow is able to move against an adverse pressure gradient is because the top plate is trying to overcome it through the motion of its own motion of its own. So, in case of this adverse pressure gradient the kind of velocity profile that you generate in this squared flow is given by this C curve. So, what we are going to do is we will actually solve during our lab session tomorrow a transient problem which is a transient quid problem. So, this is a steady state problem meaning that it is a time independent problem. What we are going to do is we will solve a transient quid flow in the lab session. So, what that means is that we will start with a situation when at time equal to 0 because it is a transient or time dependent problem at time equal to 0 both plates will be stationary and then at time equal to 0 suddenly you let the top plate start moving with a velocity equal to capital U. At the same time you impose an axial pressure gradient and then as time evolves you find out how the flow develops eventually the flow will become steady and depending on whether you have 0 axial pressure gradient, favorable axial pressure gradient or adverse axial pressure gradient in the steady state you will generate the solution which will be looking like either A or B or C and we will verify that this indeed happens in the lab session tomorrow afternoon. In the meantime this was the first case which I wanted to discuss as part of our exact solutions. So, moving on the same problem actually this one I do not want to discuss completely because this is not per say a fluid mechanics problem this is actually a convection heat transfer problem. However, the idea was that since we have talked about the energy equation let me try to outline how the energy equations situation will be solved in the same problem that we were discussing namely the plane Poiseuille flow situation. So, I have chosen that we will deal with the plane Poiseuille flow where both plates are essentially now stationary. So, that the velocity profile has already been determined which is our parabolic profile and the specific expression is here on slide number 5. So, that profile is already existing in the flow. In addition what we are saying is that let there be a constant heat flux imposed on both the top plate and the bottom plate. So, that constant heat flux over the entire length of course I am showing by this set of arrows and I have written this q suffix x with 2 primes which is supposed to mean that this is a heat flux and let us assume that this is a constant heat flux. So, what is going to happen is that the heat transfer into the fluid is going to start heating up the fluid as you can imagine and because of that the temperature within the fluid will start increasing as you go from the left end to the right end. So, at any particular axial location if you want to look because the surface is heating the fluid a temperature profile can be imagined where you have a higher temperature at the surface and then as you get into the bulk of the fluid the temperature will gradually reduce will reach some sort of a minimum and since the problem is symmetric about y equal to h by 2 it will again go back and reach a surface temperature which is higher. So, this is what we expect in a situation like this. Again you know those people who are familiar with convective heat transfer who have taught or been through heat transfer course will realize that we deal with a very very similar situation for flow inside a circular pipe meaning pipe of circular cross section where there is a laminar fully developed flow already established and then we start heating the pipe wall from an external source thereby imposing some sort of a constant heat flux situation. So, this is a very similar situation only thing is that we are dealing with a two dimensional parallel channel type situation as we said which is our plane Poiseuille flow. So, in order to find now the solution for the temperature distribution within the flow we need to solve the energy equation. So, here the way it is going is already we have solved the flow problem in the first few slides. Now taking that flow problem we will move on and we will try to solve the energy equation to obtain the temperature profile within the flow. So, therefore what we are going to do is we are going to solve the energy equation for what is called as hydrodynamically fully developed which was our parabolic profile what we obtained earlier and also thermally fully developed flow. So, in addition to assuming that the flow is hydrodynamically fully developed we will also assume that the flow is what is called as thermally fully developed and again the people who are familiar with convective heat transfer those who have already gone through the heat transfer course that happened in November and December of last year will actually remember this terminology of thermally fully developed situation. So, let us see what that means I have also shown on the sketch here is a sort of mean temperature value at the given axial location given by a uniform cross section equal to T suffix m the magnitude of the mean temperature is given as T suffix m. Essentially what you are going to do is you are going to realize that this profile is a function of y. So, you can appropriately find out an average value and we will discuss that in a minute how to how to obtain that average value for the temperature profile at a given axial location we are calling that as either the mean temperature or it is also called as the bulk temperature it is also called as the bulk mean temperature. So, the assumptions are listed at the bottom which is that a constant heat flux is imposed on both the top and bottom walls remember the z dimension is infinite. So, there are no boundary conditions on the z direction and we assume what is called as a thermally fully developed flow and again from heat transfer I am directly utilizing the mathematical expression which determines the nature of this fully developed situations again fully developed condition in a thermal situation such as what we are discussing is also a purely experimental observation. People have determined that this is how the flow behaves if you are if you are heating it either as a constant temperature or a constant heat flux boundary condition and sufficiently downstream from the inlet the flow starts behaving as if it is fully developed in the sense that a relative temperature profile remains unchanged. You can say that either the temperature profile or relative temperature profile it does not matter what it means is that although the fluid is going to get continuously heated thereby it will keep on increasing its mean temperature continuously what we see is that a profile such as this shown for the temperature remains unchanged with respect to the axial direction and the specific mathematical expression that I have simply picked from the heat transfer part is that the axial gradient or axial gradient of a typical ratio defined as the surface temperature minus the temperature within the flow divided by the surface temperature minus the mean bulk temperature. So it is some way in normalized profile you can say so surface temperature is obviously at the surface on the plate at some axial location. The mean temperature is of course as I have shown here as the mean of the temperature profile in some sense we will talk about that mean in a minute and t comma t of y comma x is any general point on the profile. So that is the way to to interpret it. So this fully developed condition for thermal situations is again an experimentally found condition and that is what we are going to assume to be the case in this situation. So if you recall those who are familiar with the bulk mean temperature idea that it is essentially defined on the basis of the same energy content travelling across a cross section. So it is what we call a mass averaged temperature if you want in the sense that if you consider a mass flow rate of m dot with a specific heat of C and a mean temperature of Tm you will see that m dot times C times Tm is the energy content of that flow crossing a given cross section and that is equated to the energy content that a given situation such as the one that the profile shows here is carrying and therefore we equate those energy contents to say that the mean bulk temperature then is defined in that fashion. So if you actually work out the algebra and again I do not want to spend too much time on these heat transfer problems because it is slightly out of the scope but because we are using energy equation as part of our governing equations I am discussing these. Those who really want to know more details at the end of this problem I will suggest which book you can look at if you want to get the exact more detail. So with this definition of the mean temperature what we realize is that we can go and do some simplifications or some manipulations on that condition which is our fully developed condition. So let us go back to our thermally fully developed condition and what we have here is that the derivative with respect to x of some quantity is equal to 0. So because of that we will say that that quantity is essentially not a function of x if the derivative with respect to f with respect to x if it is 0 of a certain quantity that quantity is not going to be a function of x and therefore you can take a y derivative of that quantity and that y derivative is also not going to be a function of x. Having said that we will carry out this further you actually employ this y derivative on each of the terms inside what you will realize is that the surface temperature is only a function of x divided by the denominator which is a completely function of x. So y derivative operating on the first term is not going to get you anything it will basically be 0. What is left is then is that the y derivative will operate on the second term in the numerator which is a function of both y and x. So therefore that partial derivative with respect to y will operate on the t which is the second derivative in the second term in the numerator and divide that of course by a term which is the surface temperature minus the mean temperature which is just a function of x. So as far as derivatives with respect to y are concerned it is a constant. So what will be left is minus dT dy divided by surface temperature minus the mean temperature and this we are saying is not a function of x. You multiply this by the constant thermal conductivity of the fluid and you will still have the same expression which will say that minus k times dT dy over that difference between the surface temperature and the mean temperature is not a function of x. However minus k times dT dy if you evaluate at either y equal to 0 or y equal to h will give you using the Fourier's law of heat conduction the heat flux at the surface and that is what precisely what I have written that the heat flux then divided by the surface temperature minus the bulk mean temperature is not a function of x. But if you remember from heat transfer heat flux divided by surface temperature minus bulk mean temperature is nothing but what is called as a convective heat transfer coefficient. So what we basically obtain through this set of manipulations is that for fully developed condition in this case the heat transfer coefficient is not a function of x it turns out to be a constant. So now furthermore if we are dealing with a constant heat flux as what we are saying it turns out that the heat transfer coefficient is a constant the heat flux is a constant. Therefore the denominator here which is the surface temperature minus the bulk mean temperature will also be a constant and then performing a x derivation on this T s minus T m you realize that the rate at which the surface temperature is changing along the x direction is the same as the rate at which the mean temperature is changing. So that is one key result. Going back to our standard expression for the thermally fully developed situation which is the x derivative of surface temperature minus the fluid temperature at any given point the whole thing divided by surface temperature minus the mean temperature equal to 0. Now what we just saw is that under the situation of a constant heat flux surface temperature minus the mean temperature is essentially a constant. So you can completely remove the denominator from the differentiation process and what is left is that the d by dx will operate on the surface temperature minus it will operate on the general temperature inside the fluid. And therefore from here for the case of constant heat flux we also obtain that the rate at which the surface temperature is changing along the x direction is equal to the rate at which the fluid temperature at any point within the fluid domain is also changing in the x direction. So putting these two together we essentially obtain a very important condition here is that the rate at which the mean temperature or the bulk mean temperature is changing with x is equal to the rate at which the fluid temperature at any particular y location you can say or any y location is also changing with respect to x. So let me just go back to my whiteboard and show this in a slightly more clearer fashion. So let me draw that profile which roughly looks like this for temperature. So at the top we have the surface temperature which is a function of x. So we are talking about a given axial location at which everything is happening. Since the problem is symmetric we have surface temperature on the other side as well at y equal to 0 and at y equal to h. And now so I will just show you a couple of arrows which we usually used to mark the profile that general temperature within the fluid is any general point on the temperature profile. So it can be here or it can be here or it can be here etc. So keep that in mind. On the other hand the mean temperature is some sort of an average value and we have already discussed how we calculate that average by equating the energy content of the uniform flow in terms of the uniform temperature Ptm and the actual energy content of what is happening inside the channel. And so that is what this final expression is supposed to mean that if you look at how the temperature at any y location is changing with respect to x it turns out to be exactly the same way as how the mean temperature is also changing with respect to x. So that is what we have obtained here through these initial discussions of this particular problem.