 Now, going back we perform a overall energy balance in the sense that we choose what is called as a semi infinitesimal control volume. So, the semi infinitesimal control volume is spanning the entire width of the channel, but it has a length of only delta x. If you perform the energy balance what you will realize is that the energy flow into this control volume it is a steady flow. So, there is no accumulation. What is coming in is given by m dot multiplied by the specific heat multiplied by the mean temperature at let us say the left phase of the control volume. From the walls there is a heat flux which is providing a heat input over a length of delta x and then the addition of these two guys will go out as the energy content that the fluid is carrying with it. That energy content with the fluid is carrying out of the control volume is as usual expressed in terms of what is coming in and the first order Taylor series expansion about that expression or about that formula. So, what we have is the energy content of the fluid leaving from this semi infinitesimal control volume is m dot c t m which was the energy content of then the fluid coming into the control volume plus m dot is a constant it is a constant mass flow rate situation c is a constant. So, I take that out from the Taylor series. So, only differentiation left is with respect to x of the mean temperature multiply that by the length over which we are carrying out this Taylor expansion which is delta x that is our first order Taylor expansion. So, you perform the energy balance in the sense that the wall heat flux providing the energy into the control volume plus the fluid that is bringing certain energy that into the control volume is equal to whatever is going out of it because there is no accumulation, no other form of energy and work are being considered either because there is none. If you carry out this balance you will come to conclusion that the rate at which the mean temperature is changing with respect to x will be given by this 2 times q double prime s over the mass flow rate times c. And from the previous slide we had just figured out that the temperature at any given y location is also changing at the same rate as the mean temperature is changing. So, therefore, I am equating these two again and saying that partial derivative with respect to x of temperature which is some general temperature at any y within the fluid is also equal to this 2 times q double prime s over m dot c. So, this we obtain from the energy balance employed on this so called semi infinitesimal CV why semi infinitesimal because it is delta x, but the entire span of the channel is chosen. One last thing that we assume here is that the y direction conduction heat transfer would be considered to be much more than the x direction heat transfer conduction heat transfer. Convection of course is happening in the y direction or technically I should say advection is happening in the y direction, but as far as conduction heat transfer is concerned we will assume that the y direction conduction heat transfer that is the lateral conduction heat transfer is much higher than the longitudinal conduction heat transfer. And from again conduction heat transfer you may recall that the second derivative of temperature with respect to x provides us a net amount of heat transfer into the control volume due to conduction in the x direction. Similarly, second derivative of T with respect to y will imply that it is the net conduction heat transfer in the y direction within our control volume. So, because we assume that the y direction conduction heat transfer is much larger than the x 1 we will employ this simplification and with all this background now we are in a position to start looking at the energy equation. It is a steady flow situation so that the time derivative which is our local derivative is identically equal to 0 only the convective derivative is left. Within the convective derivative you remember that when we talked about the flow solution we had figured out that the v velocity is identically equal to 0. So, that this v is also the third term on the left hand side is also equal to 0. From the assumption that the y direction conduction heat transfer is much larger than the x direction conduction heat transfer I am immediately throwing out this term I am not putting this equal to 0, but it is some sort of a comparison between these two conduction terms. And I am saying that d 2 T dx squared is much smaller in comparison with d 2 T dy squared and therefore, I am simply neglecting it from the equation. So, the reduced energy equation finally with all this discussion and background turns out to be u multiplied by partial derivative of t with respect to x equal to alpha which is what is called as the thermal diffusivity which is k the conductivity divided by rho multiplied by c which is a constant coming from the left hand side here times d 2 T dy squared that is the only surviving term from the right hand side. So, the equation that we are boxing now is what the reduced energy equation is and this is to be solved subject to the u velocity profile being already determined from the flow solution as given on the bottom right on the right hand side on the slide. And this energy balance giving us 2 times q double prime which is the wall heat flux divided by mass flow rate times the specific heat equal to partial derivative of t with respect to x. So, these two results are put together because we have a d T dx here we have a u here substitute that d T dx equal to 2 q double prime s over m dot c in here substitute this u in here and what is left is a reasonable amount of algebra which I am not going to carry out here what you end up doing is that having substituted these two terms on the left hand side you integrate then this equation with respect to y. So, I have just outlined the procedure this is actually a reasonable amount of algebra and those who want to do it can do it at their leisure, but I do not want to do it right now we integrate the reduced energy equation with respect to y and then we will obviously since it is a integration with respect to y we will do it twice having done that we will have to eventually employ a couple of boundary conditions. The two boundary conditions that we will employ here in this case are that because the problem is symmetric about y equal to h by 2 as we have drawn in the sketch the variation of temperature with y at y equal to h by 2 which is the y gradient of temperature will be essentially equal to 0. So, that is one condition which is d T dy at y equal to h by 2 equal to 0 and then even though we are not really specifying a surface temperature in this particular problem let us assume that there will be some surface temperature at a given actual location and use that as the other boundary condition in particular at the bottom plate it is a symmetric problem. So, it does not matter where you choose it let us choose it at y equal to 0. Then the resulting temperature profile is plugged back into the expression for the bulk mean temperature which is given here. So, this T here that you see is what you first obtain from integration and then plug it in here to carry out the expression for bulk mean temperature this is also reasonably T d s algebra. Finally, if you have the patience to go through all this you will realize that the final expression results to what is called as the formation of a Nusselt number which many of you who are dealing with heat transfer would know and that Nusselt number is defined as the heat transfer coefficient h suffix c is what I am calling that multiplied by a length scale which in this particular case I am treating this as a hydraulic diameter which is given for the present situation as 2 multiplied by the separation distance divided by the conductivity k and this entire collection is what is called as the Nusselt number based on the hydraulic diameter and if you completely work out this algebra you will say that or you will show that the Nusselt number turns out to be equal to 8.24 in this case. So, the solution of the energy equation finally, let us look at what we have obtained the solution of energy equation is fundamentally to obtain the temperature profile one may stop at that that is perfectly fine. So, that is the top bullet here. However, from a heat transfer engineering point of view what is important to know is the heat transfer coefficient value which is non dimensionalized in the form of this Nusselt number. So, therefore, the second part which is taking that expression from the temperature and plugging it into the bulk temperature expression and carrying out the further analysis to bring about this Nusselt number value is required. If some of you are interested to actually trace this entire development starting from right here actually because this is where we began. We first obtained the velocity profile for our plane Poisson flow then we took up the heat transfer problem. If someone is really willing to go through all these details make sure that you understand how each of these simplifications is being done, perform the final integration and the calculation of the bulk mean temperature etcetera. I will be really impressed and happy if you get to this number. Normally these types of problems we are not solving in undergraduate heat transfer or fluid mechanics courses. Typically, we are solving these for post graduate heat transfer courses in particular this problem that we have just talked about is really a classic problem that we solve in a post graduate heat transfer problem. It is a purely convection situation as you can see. So, in that sense it is not a per say fluid mechanics problem. What we are solving is an energy equation using the velocity profile that we obtain from our fluid mechanics analysis and that is the reason I have gone through slightly fast in this case. But I am very sure that many of you if you just go back and go through the steps you will be able to follow how it has been done. If you want to look at some of these details, you must open a convective heat transfer or convection heat transfer book because typically this problem in the plane channel case typically at least I have not come across this solution in the standard heat transfer books in an undergraduate level book. However, if you look at a convection heat transfer book which is slightly at higher level I am sure you will find this analysis in one of the books. Finally, the number that we are going to obtain as a part of the solution here is the Nusselt number defined in this particular fashion is equal to 8.24. So, those who are interested I really suggest go back and do all this. It is worthwhile doing it once partly because it is a test of patience a little bit in some sense. The other part will be that you will appreciate the kind of work required if you want to work out some of these analytical solutions. In fact, to be honest with you this is one of the more straightforward analytical solution. Many of the analytical solutions that people have worked out in the literature are far more complicated in terms of the level of mathematics and the level of analysis required and that is the reason I would like you to work out this to satisfy yourself that it is indeed this 8.24 that comes up when you perform the analysis. So, here what I am doing now is I will take up the problem of circular Poiseuille flow. So, we have just completed the discussion of plane Poiseuille flow which was a channel formed by two parallel plates and the flow was happening between these two plates. Now, what we have is a pipe of circular cross section and in the cross section it is going to look exactly like a pair of parallel plates. So, it is a pipe of radius r and we are essentially solving a pure fluid mechanics problem right now in the sense that we are after determining the velocity field which will turn out to be again a velocity profile in this case. So, we will assume that this is a steady constant density flow, we will neglect body forces. The thing that is going to happen here is that because it is a pipe of circular cross section I am going to utilize those governing equations in the polar coordinates or the r theta z coordinates and that is why what I am showing here is the radial coordinate as a lateral coordinate and the axial coordinate in this case is z. We will assume a few more things here to simplify our situation otherwise we cannot solve this analytically. So, we will assume that as the flow moves there is no tangential component of the velocity. In other words the flow is not swirling as it moves through the pipe. So, that v suffix theta that tangential component of the velocity will be assumed to be 0. We will assume that it says axis symmetric flow meaning that no matter which theta position around the periphery that you go you get the same situation ok. So, this is in some sense equivalent to that two dimensional approximation that we employed for the plane channel. This is in some sense equivalent to that that it is an axis symmetric situation. In other words what will happen is that if you form a derivative of any quantity with respect to theta you will essentially see that there is no change and that is the reason partial derivative of any quantity with respect to theta is equal to 0 due to the axis symmetric nature of the flow. And we will finally assume that the flow is fully developed which again means that the axial derivative of the axial velocity is 0. The axial velocity will not change with axial distance and that mathematically means that it is the partial derivative of vz with respect to z equal to 0 vz being the axial velocity. Immediately we point out that that means that the axial velocity is not a function of z because d dz of vz is equal to 0. So, my point is that you keep these two sets of slides for the plane Poiseuille flow that we talked about a few minutes back and the circular Poiseuille flow that we are now going to do side by side. You will see that these are extremely analogous and the steps that we will follow are also extremely analogous. So, if you have understood what was getting done in the previous case this is actually exactly similar. Only thing is that here we are utilizing these r theta z coordinates and corresponding velocity components, but other than that in principle there is nothing different between what we did for the plane Poiseuille case the channel situation and now what we are doing. So, let us see that. So, I have listed here for the convenience again the entire set of equations which the first one is the continuity equation written for r theta z situation our cylindrical coordinates and the next three are the three momentum equations the Navier-Stokes equations written in r direction theta direction and the axial direction which is the z direction. Remember this is a pure flow situation with a constant density. So, we do not have to bother about dealing with the energy equation at this point. What I am going to do is I will work out the continuity equation and show you what is happening. So, if you go back to the continuity equation which is the first one on slide number 14 and simplify it according to our assumptions the third term is getting to 0 because it is a fully developed flow. So, v dz of v dz equal to 0 fine. Second term is getting equal to 0 because it is a axisymmetric flow and we said that the derivative with respect to theta the tangential direction of any quantity is going to be 0. So, this second term is going to 0 because it is a axisymmetric assumption. So, then what is left is 1 over r you can cancel with 0 obviously and d dr of r multiplied by v r is equal to 0 which then you integrate with respect to r and you will find out that this implies that the radial direction velocity v suffix r can be a function of theta and z as a result of the partial integration with respect to r divided by r. So, let me just show this explicitly. So, what we have is 1 over r d dr of r times v r equal to 0. So, first we get rid of this 1 over r and then what is left is d dr of r v r equal to 0. So, we partially integrate this with respect to r when we partially integrate this with respect to r we have r times v r not as an integration constant on the right hand side, but because we are doing a partial integration with respect to r we are forced to say that this could be a function of theta and z and therefore, we will write that the radial velocity then is a function of theta comma z divided by r and that is precisely what has been written out here. Then let me take the time to go back and establish the equivalence. Here we go back to our first example the plane Poiseuille flow parallel plate equation. We had integrated the continuity equation which was reduced to only partial derivative of v with respect to y. So, we actually integrated this partially with respect to y and therefore, we did not have a integration constant here we had a function of x. So, this is the analogy between the plane situation and what we have just carried out right now I hope you are able to see that ok. So, now we need to figure out just as we did in the case of the plane channel what should be this function in order to do that we employ a boundary condition on the radial velocity in particular what we say is that the radial velocity at r equal to capital R which is essentially equal to the pipe wall is equal to 0. This is same as your v equal to 0 at y equal to 0 or y equal to h in the case of the parallel plate channel. We are still talking about the same situation in the sense that we are employing this boundary condition for this normal velocity at the solid surface and if we are assuming that the solid surface is impermeable we will say that the normal velocity at the solid surface is equal to 0. Remember in this case we are dealing with a radial coordinate. So, let me go to the whiteboard and show this pipe in the cross section in some sense. So, this is my cross sectional area and the radial coordinate is perpendicular to the surface here and this is r equal to capital R and at this point which is the wall we are employing this normal velocity equal to 0 because there is no because the wall pipe wall is considered to be impermeable solid. So, if at the pipe wall the radial velocity has to be 0 the only way you can obtain a 0 value of this radial velocity is by assuming that this function theta comma z is identically 0 that is the only way you can obtain a 0 value in general. So, therefore, just as we had argued and found in the case of the plane Poiseuille flow that the v velocity which is the lateral velocity in that case is identically 0 everywhere. Remember I had remarked that as a result of these fully developed situations you will see that the lateral velocity will always turn out to be 0 which is what is happening again in this case. The lateral velocity in the case of this pipe flow is the radial velocity and which is coming out to be equal to 0. Of course, we have already assumed that there is no swirling and therefore, there is no tangential velocity, but additionally now we are saying that the radial velocity will be always equal to 0 no matter what. So, what we are saying essentially is then the only non-zero velocity is going to be the axial velocity in this case just the way we had only axial velocity left in parallel plate channel here is the same situation. So, therefore, this is the conclusion here only non-zero velocity is the axial velocity and it is a function of R alone. So, why do we say that we already have seen that as part of our assumption the axial velocity is not a function of z nothing is a function of theta anyway because it is an axisymmetric situation. So, what is left is that we simply say that it has to be a function of only the radial coordinate I hope you followed that. Now, this is a very straightforward set of algebra which I will show you how to do it, but I am not showing it here on the slide neither will I show it on the whiteboard. What we do is again look at the analogy in the case of the parallel plate problem after the continuity equation we went to the y momentum equation. So, just like that we are going to visit the tangential and the radial momentum equations in this case. So, let me go back to the set the second equation here is the radial momentum equation the third equation is the tangential momentum equation simply go here plug in the expressions for this del squared which are the Laplacian operators and I had I had written down explicitly the expression for Laplacian in the radial coordinates yesterday. So, go back to those slides plug those expressions here and verify yourself that under the set of assumptions and under the set of results that we have obtained up to this point here you will see that all terms from the radial momentum equation and the tangential momentum equation will drop off leaving you with the only conclusion that there is no tangential pressure gradient and then there is no radial pressure gradient which means that pressure is not going to be a function of r and theta if it is not going to be a function of r and theta the only possibility is that it is going to be a function of z. So, that is what the conclusion will be I hope you understood what I meant go to those two momentum equations simplify terms according to our assumptions and the intermediate results and you will see that p is only a function of z. Finally, we visit the x momentum equation and I am directly now writing the reduced form of the x momentum equation. Since p is at most a function of z we have dp dz in the total form we do not have to talk about the partial derivative you will see that the left hand side can be couched or placed in the form of mu over r d dr of r d v z dr. So, let us see if that turns out to be the case we write this down what was written on the before. So, I have just rewritten the left hand side in the form of this mu over r d dr r d v z dr the reason is because it is easy to integrate this than the way it is written here. All right, but let us see what is happening if you look at the left hand side we had argued that axial velocity can be at most a function of r. If you see the left hand side it involves only r and r derivatives of v z. So, the left hand side can be at most a function of r e is already argued to be a function of z at most therefore, dp dz can be again at most a function of z at most a function of r on the left hand side equal to at most a function of z on the right hand side will mean just like what we had concluded in the case of parallel plate channel each of them has to be equal to the same constant. And therefore, just like what we assume in the case of parallel plate channel here also we will assume that the axial pressure gradient is a known constant. And with respect to this known constant then the left hand side will be integrated with respect to radial distance twice. And we obtain an expression for the axial velocity as a function of r with a known axial pressure gradient with two integration constants. So, this is a pretty straightforward integration and I would like you to do it because it is really straightforward. In order to determine these two constants of integration we realize that there is a problem with this second term which involves this logarithm of r. However, here we invoke this physical boundary condition in some sense that at r equal to 0 which is our pipe center line r equal to capital R is our pipe wall. So, r equal to 0 is our pipe center line and we say that purely based on physical argument in this case we must have a finite value of the axial velocity at r equal to 0 which will mean that if you want to have any finite value this ln r which will be a troublemaker since r is tending to 0 has to be thrown out all together. And the only way you can throw it out all together is if the constant a is equal to 0. So, we conclude that the constant a is equal to 0 the other boundary condition will be at the pipe wall the axial velocity will be 0 exactly because of that so called no slip boundary condition we are essentially assuming that the pipe is a stationary pipe. So, there is no relative tangential velocity at the pipe wall which is solid surface. So, therefore, v is equal to v z equal to 0 the axial velocity 0 at r equal to capital R and if you work out the or if you substitute these in the expression at the top you will realize that we again come up with a parabola sort of a distribution for the velocity as a function of r. I am very sure most of you know this I just and there is a good chance by the way that many of you would have actually gone through this step by step procedure starting from the governing equations and then simplifying it in a step by step to eventually obtain this what we call as the paraboloid of revolution. But in case those of you who have not seen this kind of a procedure to simplify the governing equations one by one and obtaining finally a solution for the velocity in terms of assumed known pressure gradient I think it is a good example finally it turns out to be again that paraboloid of revolution of course it is an axisymmetric situation. So, that is why we call it a paraboloid of revolution. So, this is as I said an analogous situation to that plane Poiseuille flow this is popularly called as the circular Poiseuille flow or also a Hagen Poiseuille flow as many of you would know it in that fashion. I am not going to work out the corresponding energy equation in this case which is again a convective heat transfer problem. This energy equation in the circular Poiseuille flow will follow the exact same procedure as what we did for the energy equation in case of the plane Poiseuille flow however here is a cylindrical coordinate. What you can remember perhaps is that if you have seen this in a heat transfer class you will remember that for the case of a constant heat flux thermally and fully sorry thermally and hydrodynamically fully developed conditions the Nusselt number based on the diameter of the pipe turns out to be 4.36. So, that solution is many times worked out in standard text books and I think if I remember correctly the heat transfer undergraduate text books will actually work out this solution completely in the convection chapters. So, those who are interested to read this and those who already by the way do not know this there is a very good chance that many of you have seen this in the heat transfer course. Let me just write a couple of names of heat transfer text where you can see that solution. So, there is a very famous text written by Professor Sukhathmi of this department I think it is a text book on or off I think of heat transfer and one more text that we routinely follow I think in the heat transfer courses here is by in Kropera and do it this is I think fundamentals of heat and mass transfer. So, if you look at any of these books and of course there are plenty more, but these are the two that we follow quite regularly in classes here for the undergraduate heat transfer. You will see that in the chapter on convection this energy equation solution for the case of a circular pipe cross sectional pipe with a constant heat flux imposed on the wall is worked out to finally show that the Nusselt number is equal to 4.36 in a situation like this. So, remember that in the case of plane channels earlier here we had obtained the result that it is 8.24 of course the definition is based on a hydraulic diameter which is defined as 2 times the separation distance in this case, but it is essentially an analogous result to what we are going to see in a circular Poiseuille flow which is this 4.36. So, since this is a heat transfer convection type problem I did not want to really discuss this too much I have given you a couple of references where if you are really interested you can go and see this. On the other hand that plane channel problem which I worked out earlier is possibly not to be seen in those undergraduate text books, but if you see one of the postgraduate level convection book most likely you will find it there. What I will do is I will take a break now. Thank you.