 Today, I will take up the last lecture on laminar flow heat transfer in ducts and what I am going to consider is the effect of actually varying thermal boundary condition, which is particularly of interest in oil flows, where the Prandtl number is very large and the thermal development length is considerably bigger than the velocity development length. In these sort of situations, the actually varying wall temperature as well as actually varying heat flux have profound effect on the manner in which the heat transfer coefficient changes or varies with the actual distance. So, for example, let us consider the solution that we had obtained for Prandtl number very very greater than 1. Now, that solution was obtained for T equal to T w from x equal to 0, but today I am going to consider the case of heat transfer begin from x equal to x naught before that the wall temperature is same as the inlet fluid temperature. So, in fact, we can view our earlier solution of lecture 18 as a step function in T w minus T i, the T i is inlet temperature at x equal to x naught. Therefore, for x greater than x naught, I may write T minus T i equal to 1 minus theta x star minus x naught star y star T w minus T i and theta you will recall was defined as T minus T w over T i minus T w. So, just as we considered the external boundary layer situation, for the internal boundary layer situation it would again look like this. Therefore, for arbitrary variation of T w, we will have T minus T i equal to 0 to x star 1 minus theta x star minus x naught star y star dT w by d x naught star d x naught star. This would be for the continuous part of the variation and this would be 1 minus theta x star minus x naught k y star dT w minus T i sub k. This would be for the discontinuous part of the solution. This is exactly following what we did in the case of external boundary layers. So, therefore, the heat flux response would be q w star and I am now considering the case of flow between parallel plates separated by distance 2 b, then that would be equal to q wall x star would be equal to k dT dy at the wall y equal to b equal to k by b dT by dy star y star equal to 1 and that would read as theta dash here is d theta by dy x star x naught star 1 where y star is equal to 1 this will be for the continuous part and this would be for the discontinuous part. This is simply a differentiation of this equation with respect to y, but you will recall that from lecture 19 we had found that theta dash 1 is actually equal to minus n sum n equal to 0 to infinity a n exponential of minus 8 by 3 lambda n square x star equal to a n equal to minus c n y n dash 1 where c n and y n dash 1 were tabulated in lecture 19. So, we are going to use the same coefficients a n in this series lambda values 2 were tabulated in lecture 19. So, if I substitute that in here for that I would get the heat flux response to arbitrary variation of T w along the wall. So, q w x star would become k by b 0 to x star n equal to 0 to infinity and this function which is written for x star minus x naught star dT w by dx naught star and this would be for the discontinuous part the summation goes from n equal to 0 to infinity over n and for the number of steps a k equal to 1 to n k. Now, the T bulk minus T i would obviously be equal to 4 b by k integral 0 to x star q wall x star dx star and this is simply from the heat balance and therefore, the Nusselt number would be given by q wall x star divided by T wall minus T bulk x star 4 b by k where T wall minus T bulk is written as T wall minus T i plus or minus T bulk minus T i and this is given and this is what we have evaluated just now. So, T wall minus T bulk is written in the fashion I have shown here. So, that is how one gets T wall minus T bulk at any x star. So, let us move on to a problem in which I am going to consider let us say there are two parallel plates separated by distance 2 b and right at the fluid enters at T i at x equal to 0 itself there is a jump and then the temperature is linear. So, therefore, T w minus T i this is T w equal to a plus b x star and a is the jump at x equal to or x naught this is the jump at x naught star equal to 0 in this case. So, let us see how the equations develop. So, if I were to substitute this variation of T w minus T i in the previous expression here. So, dT w by dx naught star would be simply capital B. So, that is what this would be and the integration of that with b here. So, b simply is a constant and it would come out and the integration of this equation with x naught equal to 0 would be given by a n by lambda n square 1 minus exponential of minus 8 by 3 lambda n square x square plus of course, we must add the step jump right at x naught equal to 0 which is the entrance. So, that will be plus a times n minus 0 equal to infinity a n exponential of minus n lambda n square x star. Now, of course, there are no other step jumps in this and therefore, this one there is a continuous right along and there are no further step jumps it is a simple case. So, once you obtain q wall in this manner T wall minus T bulk would simply be given by 9 b by 16 a n by lambda n 4 you can see that here T b minus T i which is an integral of q wall x star. So, I have used that and added T wall minus T i then they will get 9 b by 16 n equal to 0 to infinity a n by lambda n 4 and this term plus the step jump part which is that. So, as a result you will get a new x equal to q wall 4 b divided by k into T wall minus T bulk, but notice one thing that as x tends to infinity you will see that term will go to 0. So, will that term go to 0. So, therefore, T wall minus T bulk would be simply 9 b by 16 n equal to 0 to infinity a n by lambda n 4 and likewise q wall x star would be simply the a n by lambda n square this term going to 0 and this term equal to 0. Now, of course, when what it shows therefore is that the bulk temperature is also going to vary linearly after a long time after a long time although initially it may vary in an arbitrary manner, but after a long time T wall minus T bulk maintains a constant difference and therefore, this would at this point onwards it would be a case of constant heat flux. It would be case of constant heat flux at the wall and that is what we find it by 3 n equal to 0 to infinity 0 to infinity and that would be equal to 8.235. This would be the general case if we were to go to infinity we have verified that the solution obtained appears at least correct in the infinity state. Now, I am going to assign some values to a and b and see what happens. On this slide I consider a case of T wall minus T bulk equal to 1 minus 5 x star. Therefore, the wall temperature declines as x increases. So, you can see that here this is the variation of wall temperature on the graph that is shown and the values are assigned are these. I have gone up to when T wall minus T i that is the wall temperature reduces back to the inlet temperature. The initial jump of course is a equal to 1 and it reduces to minus. So, at x star equal to 0.2 the temperature of the wall is same as that of the inlet fluid you can see T wall decreases. What happens to T bulk? T bulk as you can see on this graph first increases T bulk increases as you can see here and then its rate of increase is somewhat slowed down to this is the one that is shown by dotted line. So, T bulk increases to 0.32, 0.39, 0.4, 0.4 at about 0.12 it is same as 0.4 and then begins to decrease when the wall temperature becomes equal to inlet fluid temperature the bulk temperature that point is 0.27. Notice what happens to Q wall. Q wall is of course because T wall is greater than inlet fluid temperature initially Q wall is positive 8.7 reduces to 0.73 at x star equal to 0.05 reduces to 0.1 and in fact there is one point at 0.11 where there is no heat transfer that means this situation becomes adiabatic and then Q wall changes sign that means the bulk temperature now is greater than the wall temperature and therefore you get negative heat fluxes from x star equal to 0.12 onwards and what happens to Nusselt number? Well the Nusselt number as you can see here the Nusselt number is plotted along there. So, initially the Nusselt number falls as it should for any heating case, but then it suddenly drops to a very low value almost minus 66 at x star equal to minus 12 and then suddenly jumps back at 0.15 to plus 12.5 and then again reduces down to 9.1 and since I have not computed long enough I do not see 8.325, but I would if I were to continue in that fashion but with a negative heat transfer. So, even that is possible. So, indeed strange things happen the wall temperature although is monotonically decreasing linearly decreasing the bulk temperature has a hump at about 0.4 x star equal to 0.11 to 0.12 and then there is a decline and turning sign and likewise Nusselt number is declines, but it goes negative and then becomes positive. So, this is a very interesting case of how T bulk overtakes T wall and all the Q wall is negative you still get Nusselt number positive. So, this positive Nusselt number is for negative heat transfer this positive Nusselt number is for positive heat transfer into the fluid and at this point you get negative heat transfer with negative Nusselt number. So, indeed very very strange things can happen. One can work out solutions for variety of A and B values to see what happens. As I said one could generate these solutions also for a circular tube because as you will recall the A n and lambda n values remain the same for circular tube also and therefore, for such a case one can again carry out the integrations that I have shown here and evaluate Nusselt number for circular tube which I shall not show here, but it can be done and you do indeed get very strange variations of Nusselt number when the wall temperature varies actually. Now let us consider the case of axial variation of heat flux this is of a considerable interest in nuclear reactors as I will show a little later. So, from lecture 19 we know that the temperature response for step jump in Q wall at x star equal to x naught star is given by T minus T i Q wall b by k this is the fully developed solution plus this one is the developing part of the solution. Remember recall that we had said psi equal to psi f d plus psi developing and this part is the fully developed part and this part is the developing part. So, writing this equation at y star equal to 1 would give me psi wall equal to 17 by 35 plus 4 by 4 x square plus b n exponential of 8 minus 8 by 3 lambda n x star where b n is C n multiplied by y n 1 which I have taken from 1 I mean y star equal to 1. If I were to take d psi by dx star it will be 4 minus 8 by 3 into all this into lambda n square and so on so forth. So, here we consider only continuous variation of Q w x star I am not going to consider situations in which step jumps in a Q wall occur because they are rare. Then the response of the bulk and the wall temperature will be T w minus T i b by k d psi by dx naught star Q wall x naught star dx naught star and that would be essentially equal to b by k equal to all that that we derived on the previous slide into Q wall x naught star dx star and T bulk minus T i would of course, again be Q 4 b by k 0 to x star Q wall x naught star dx naught star and therefore, the Nusselt number would be Q wall x star divided by T wall minus T bulk simply take a difference of these two quantities and multiplied by 4 b by k of flow between parallel plates. So, in nuclear reactors is quite common. So, consider flow between parallel plates and these are the nuclear fuel elements and let us say this is 2 b. In fact, the earliest nuclear fuel elements were in fact, flat sheets and not the circular rods that we find today and the coolant would be CO 2 in the earliest nuclear reactor built. Let us say we have this 2 b nuclear fuel elements generate generally heat in this fashion sinusoidally. So, Q wall would be Q wall max let us say this is Q wall max equal to sin pi x by l where l is the length of the channel through which the coolant is flowing. So, I am going to consider this particular case because of its relevance to nuclear reactors. So, as I said Q wall over Q wall max is let say sin pi x by l where l is the channel length then you will see T b minus T i Q wall b by k would be simply 0 to x star 4 sin pi x naught star by l star dx naught star. So, that gives us the value of T bulk at any x star would be simply given by that and T wall minus T i would be given in this fashion this is the 4 x sin x pi x naught star dx naught star minus 8 by 3 all this and that would equal 4 l star by pi cos pi x star by l star. This integration requires little effort because it is a product of exponential term and a sin term. So, the result is you get a very big bracket here with b n 1 plus 3 pi by 8 lambda n square l star whole square into sin pi x star by l star exponential of that term plus this term cos pi x star by l star exponential of minus 8 by 3 lambda n square x star minus 1. So, we have now obtained the variation of bulk temperature along the duct and we have obtained a variation of wall temperature along the duct and therefore, we can calculate the Nusselt number as it is shown here. So, T wall minus T bulk would simply be different from the previous two slides and that would read in this fashion and we then apply Nu x is equal to q wall x which is been specified sin pi x 4 times because hydraulic diameter is 4 b and T wall minus T bulk into all this q wall max b by k. The values of lambda n and b n for constant heat flux cases where of course, given in lecture number 19. So, see now what happens? So, you can see the first column here and I have taken q wall max equal to 1 for convenience and x by l values I have taken are these from 0 to 1. The heat flux of course, according to the sin function would vary in this fashion it reaches maximum of 0.5 at I mean of 1 at x by l equal to 0.5 and then again drops down to 0 value. So, that is what has been shown here that the T wall is given in this fashion. Notice how T bulk varies T bulk starts off with 0 at the inlet because that is when where the inlet fluid temperature it enters in inlet fluid temperature and then it begins to rise it is not a linear rise although in the central part it is linear where q wall is nearly constant and therefore, T bulk varies more or less linearly in the linear part, but then earlier it is not non-linear and so is towards the end it is non-linear. Most importantly see how the wall temperature varies the wall temperature increases from there it changes the slope a little bit and then rises again to that value. This is the wall temperature variation and therefore, Nusselt number divided by 4 for convenience to get a proper scaling here you will see Nusselt number peaks at 1 at x by l equal to 0.5 at 1.87 as you can see, but then decreases down to 0 down to 0. So, the maximum Nusselt number is 7.48 and it occurs at 0.5 where q wall max occurs, but T wall max and T bulk max continue to increase till x y l equal to 0.95 and this problem is of relevance to nuclear reactors again we can use the constants given in lecture 19 for flow in a circular tube. So, you can see that if you were to assume simply that wall temperature was constant we would get a very wrong picture of what happens, but because we have assumed a sine function we get T bulk and T wall variation which are different from what it would be if we had a linear constant distribution of constant wall heat flux. So, in summary I would say we have considered fully developed heat transfer in circular tube and annular and parallel plates. We have also presented general method for flow and heat transfer in singly connected ducts of arbitrary cross section and arbitrary variation of T w, q w and h w. We have also presented developing flow heat transfer solutions for circular tube and parallel plates for q wall equal to constant and T w equal to constant for the entire range of Prandtl numbers and then finally, we extrapolated these solutions to situations involving arbitrary axial variations of heat flux and wall temperature. Of course, I will write for flow between parallel plates and or circular tube. However, for complex ducts it is best to adopt the computational fluid dynamic technique and obtain the solutions. So, with this I complete my discussion on laminar flow heat transfer. We have considered a typical case of what happens in the developing flow region and considered the fully developed flow situations and learn to obtain friction factor and also number of products for fully developed flows in ducts of arbitrary cross section and then we extended that to the case situation of fully developed heat transfer in ducts of arbitrary cross section with arbitrary variations of thermal boundary conditions. We found that for fully developed heat transfer as well as fully developed flow situation, you could use Fourier series methods or Cantorovic variational methods, but we found that the general method based on conversion of the Poisson equations to Laplace equations turns out to be very general and can be applied to variety of ducts including regularly shaped ducts as well as arbitrary variation of boundary conditions along the circumference. And finally, in today's lecture I indicated how solutions of constant wall heat flux and constant wall temperature that were obtained could be extended the wall temperature and heat flux varies actually. So, with this I complete my discussion on laminar duct flow heat transfer and from next lecture onwards, we will be moving on to turbulent flow and heat transfer. Thank you.