 In this lecture, we shall take up laminar developing flow heat transfer. As you will recall, in the entrance region of a duct, the velocity profile develops like saw. This is the velocity profile, velocity boundary layer will develop like saw, but the temperature boundary layer development would be governed by the value of Prandtl number. So, if Prandtl number was very, very small, you will get very rapid development. This is for Prandtl number very much less than 1, because the thermal boundary layers will develop much faster than the velocity boundary layer. This is the case of liquid metals. On the other hand, if Prandtl number was very, very large, that is oils, then the thermal boundary layer development will be much, much slower and this would be the temperature boundary layer for Prandtl number much, much greater than 1. So, much so that we can make some very simple approximations for these two cases. For example, for liquid metals, say where Prandtl number is of the order of 0.001, over greater part of the length, u will be simply equal to u bar, whereas Prandtl number very, very greater than 1 over greater part of the thermal development, u will be simply equal to u fully developed, and which will be function of r, which is for a circular tube. It will be 2 times u bar into 1 minus r square by r square, for example. So, one can make very suitable approximations for these two extreme cases and obtain solutions, but when Prandtl number is say between 0.5 and 10, then both velocity and temperature profiles will develop at comparable rates, and that is the case called the simultaneous development of flow and heat transfer. So, that is the importance of Prandtl number in study of laminar developing heat transfer. So, for Prandtl number close to 1 in a small range, let us say from 0.5 to 10, one must consider both flow and heat transfer development simultaneously. When Prandtl number is very, very greater than 1, then flow will be fully developed, and we will be calling it a thermal entry length problem, because the velocity is specified. Likewise, when Prandtl number is very, very less than 1, then u will be specified at the inlet value, and it is almost like a piston or a slug flow thermal entry length problem. So, we will consider these cases separately. So, this is much of what I have already explained to you that for Prandtl numbers much, much greater than that is oils, the flow will be fully developed over greater part of thermal development length, and in liquid metals it will be taken as almost equal to u bar. The real problem comes when there is a Prandtl number is moderate, and the two layers can be expected to develop at the same rate or comparable rates. Simultaneous development of both flow and heat transfer. So, we consider by way of an example, the entry flow between two parallel plates to be apart. Then the governing equations will be, this will be the continuity equation. This you will recall is the momentum equation, and now we will have this as the temperature equation. Of course, if we make boundary layer approximations, and when Reynolds Prandtl as you will recall from lecture number 17, that when Reynolds Prandtl is greater than 100, then temperature gradients in the y direction are much, much greater than the temperature gradients in the x direction. So, this term would be dropped. We will be assuming that the Reynolds multiplied by Prandtl number is greater than 100. The variables are shown here. So, this is the convection term, and there is one diffusion term that is this multiplied by 1 over Reynolds Prandtl. From lecture 14, you will recall we had adopted Langar's solution method, and we have already obtained the velocity solution as given by that C 1, C 2 were given. For different values of beta, we printed out values of friction factor and others, but since u is known, we can also get v from the continuity equation that is here. Knowing u star and v star, we have these two quantities are unknown, and therefore, we now intend to solve this equation only for temperature using the velocity solution from Langar's case. Again, that resulting equation can again be solved by Langar's method of linearization. However, the algebra turns out to be very, very cumbersome as you will see from a paper by Heaton and Reynolds, in case international general of Heaton mass transfer, volume 7, 1964. So, the method is very, very cumbersome, and therefore, what I am going to do is to present only the solution in order that you appreciate what the solutions under signable tennis development look like. I am going to consider the case of flow between two parallel plates. Q top is finite, but Q bottom is 0. This I will call heated side, and this is unheated. The flow and temperature profiles are now simultaneously developing in this particular case, and x plus here is simply x divided by d h Reynolds Prandtl. Theta is defined as temperature minus temperature in inlet divided by Q on the hot side, I mean Q heated d h divided by k. Of course, the bulk temperature will vary simply linearly. As you will recall, the bulk temperature will vary regularly as 2 x plus, and therefore, the Nusselt number would be defined as h d h by k as 1 over delta theta, where delta theta is T wall minus T bulk. So, you will see that plotted for each Prandtl number. There are three Prandtl numbers I have taken, and you have Nuh on the heated side, theta wall on the heated side minus theta bulk, and theta wall on the unheated side. I am showing here the solutions for x plus equal to 0.001, 0.0025, 0.005, 0.01, 0.05 and 0.1, and ultimately of course, at infinity. The value is 5.39 for the Nusselt number, and it is almost reached around let us say 0.09 or you can say at about 0.1. At 0.7 Prandtl number, again Nusselt number starts off from a very high value at close to the entrance and gradually goes on decreasing, and again at about 0.1 you get very close to fully developed value, but at 0.01 even at 0.1 the value is not very close to fully developed value, but it will be tend to close to fully developed let us say a little longer at 0.12 at x plus equal to 0.1. Now, this can be deceptive x plus equal to 0.1 at Prandtl number of 0.01 actually represents a shorter physical length when the Prandtl number is 10. So, in as much as I say x plus equal to 0.1, it does not mean same physical length at low Prandtl number. At low Prandtl number the length is longer, the physical length is shorter than it is for Prandtl number greater than 1 at 10 for example. So, these are very instructive solutions. Notice also that on the unheated side the wall temperature is less than the bulk value as you can see in all cases. It is only on the heated side that the wall temperature exceeds the bulk temperature and the value of bulk temperature as evaluated from this is given right here. So, this is how the heat transfer coefficient would vary on the heated side it will vary like that this is nu h versus x plus and it reaches 0.1. We said when Nusselt number becomes constant with x that is when we say fully developed heat transfer has been achieved. So, similar thing has been now done for circular tube. Now, the cross section is a circular tube and of course, it is uniform all around and also actually q wall is constant actually as well as in circumferential direction and let us see what happens when the flow and heat transfer develops simultaneously. Again as you will see in this case theta bulk would vary as 4x plus u star and v star are again taken from Langar's solution to solve the temperature problem. Nusselt number is given by this and delta theta is theta wall minus theta bulk. So, in this case for pranzler equal to 10 and x plus equal to the same values you get very close to fully developed Nusselt number at x plus equal to 0.1. At 0.7 again very much so. In fact, it indicates that the development lengths are pranzler number being very close to 1 it is almost the same, but at 0.01 again at x plus equal to 0.1 the Nusselt number is still higher and it would require larger x plus get to very close to fully developed value. These are the theta bulk and the wall temperatures are 0.697 to start with at pranzler number 0.10, 0.12, 0.15, 0.22, I beg your pardon this should be 0.0697. I have made an error here and then the temperature difference goes on increasing, but becomes then progressively constant which results into constant Nusselt number in all pranzler numbers. As you will recall in the fully developed state the Nusselt number is constant at 4.36. So, for both parallel plates and circular tubes thermal development length is L H by d H equal to approximately equal to 0.1 times Reynolds Prandtl. So, obviously when the Prandtl number is small L H by d H L H the physical length would be smaller than when the Prandtl number is very large let us say about 10. This is a typical behavior for ducts of nearly all cross sections. You can easily take x plus equal to 0.1 as indicative of thermal development length, but for flow development length you will recall we had taken for parallel plates it was 0.01 times Reynolds and 0.05 times Reynolds for circular tube. So, but for thermal development and in the range of Prandtl number close to 1 as I said from 0.5 to 10 it is very good to approximate that as 0.1 times Reynolds Prandtl. Now, I consider case of both plates of parallel plates here also T w and here also T w. So, we have a case of uniform wall temperature and I am considering again Prandtl numbers in the vicinity of 1. So, you can see here 0.725 and again Langar's method has been applied for constant wall temperature case and you will see, but both plates are heated now unlike the first problem in which only one plate was heated. The plate temperatures are constant x plus equal to 1 raise to minus 4 up to infinity Nusselt number goes from high of 40.9 and close to inlet to 7.54 and this is how the theta bulk varies. Theta bulk decreases to 0, x plus because theta bulk essentially is theta bulk minus T wall and therefore, that goes on reducing. x plus is equal to this here for 2.5 and you can see that nearly fully developed value is reached at about 0.027. You could even say 0.012 it depends on how you define the fully developed at 0.5 Prandtl equal to 0.7. Again, you get very, very good results from 0.012 onwards very close to fully developed value results at this value. You can see that behavior of Nusselt number is very, very similar and thermal development length in such a case could be taken as let us say about 0.03 if you like or 0.02 is good enough to be close to 7.54. Similarly, we consider the case of T w equal to constant. Now, actually the circular tube has actually T w equal to constant and simultaneous development is being considered for 3 Prandtl numbers. Prandtl number 0.7, 2 and 5 and again you will see that at 0.7 you get local Nusselt number as 16.8 and so on so forth to 3.6 you see the mean Nusselt number goes on like that. The mean Nusselt number is defined as 1 over x 0 to x nu x dx. Similarly, at Prandtl 2 you get that again at fully developed state you get 3.66 which is an unknown value. We had already computed that when we considered fully developed heat transfer. Prandtl number 2 again is around at 0.05 you get 4.1 and 3.9. So, in terms of local Nusselt numbers the development length of 0.05 is a very good indicator of thermal development x plus equal to 0.05. It is very good for near unity Prandtl number fluids that for circular tube. Now, we turn our attention to oils and Prandtl number very much greater than 1 and as I said over greater part of the thermal development length the velocity profile can be assumed to be fully developed. Hence, for parallel plates for example, it will be u f d d t by dx equal to alpha d 2 t by dy square and u f d by u bar square here will be 3 by 2 1 minus y by b square where b is the half distance between the 2 plates. And for a circular tube it will be u f d d t by dx will be equal to alpha divided by r d by sorry bigger this is d by dr into r d t by dr and u f d by u bar is equal to 2 into 1 minus r by r whole square. The boundary conditions are at the symmetry plane that is y equal to 0 or r equal to 0 and y equal to b or y equal to r you have the wall condition which must be specified with the inlet condition t equal to t i at x equal to 0. So, let us first of all take the case of flow between parallel plates and t wall is equal to constant flow between parallel plates and t wall is constant on both sides the fluid enters let us say at t i, but the velocity profile is already fully developed the velocity profile and it remains. So, throughout the length how do we obtain solution for this case the governing equation then would be if I substitute for u over u bar equal to 3 by 2 into 1 minus y by b square then defining y star equal to y divided by b x star equal to x by b Reynolds Prandtl and t minus t w t i minus t w equal to theta then the governing equation would simply be this and the boundary condition will be at the wall theta would be 0 because that is how we have defined theta d theta by dy star would be 0 at the symmetric plane and then the inlet plane theta would be equal to 1. This problem with this boundary conditions is called the Gritz problem is very famous Gritz problem in heat transfer and it is solved by the method of separation of variables. Of course, nowadays you can solve this problem by finite difference method quite easily, but I am deliberately presenting here the method of separation of variables and indicating the solutions that are obtained and you again all this is for Prandtl number very greater than 1 or oils. So, we say let theta be product of two functions x of x star and y of y star then if I substitute in this then I would get two od's x dash plus 8 by 3 lambda square x equal to 0 with x 0 equal to 1 which is the inlet condition and y double prime plus delta square 1 minus y star square y equal to 0 with y equal to 1 and y dash equal to 0 is equal to 0 lambda square is the Eigen values. Now, this solution set is called the Sturm-Lewel equation set and the solution to that is simply a product solution. For example, the first equation would have a straight forward solution x would be proportional to exponential of minus 8 by 3 lambda n square x star and multiplied by y n star which is the solution to that equation taking for different values of n you will have different values of constant of proportionality and the functions y n. The C n values the in this expression coefficient is evaluated from 0 to 1 1 minus y star square y n d y star over that equal to minus 2 by lambda n d y by lambda n nu star equal to 1. Now, how do we obtain lambda n's that is simply by solving this equation by shooting method. So, you start let us say at axis of symmetry and this is the wall. So, you start with y dash equal to 0 as the known boundary condition y itself can be anything and you choose different values of lambda and you may arrive at here and then you may arrive at here and then the correct value of lambda will be the one this is lambda correct. So, this will be the first value of lambda likewise you go on changing the values of lambda the next value correct value will come out at in that fashion this is lambda 1 this is lambda 2 the third value will come out to be like so the fourth correct value will come out to be so and so on and so forth. So, this is lambda 3 this is lambda 4 and so on. So, this is how we determine the lambda n till in all cases I mean y star equal to 1 will turn out to be 0 which is the wall condition. So, once you have determined lambda n's in this way you can see Nusselt number itself will be h times hydraulic diameter because the plate width is 2B the distance between the plates is 2B. So, hydraulic diameter is 4B and that would equal minus 4 theta prime 1 divided by theta prime at the wall minus divided by theta bulk and theta bulk itself would be evaluated in this manner which gives you 3 by 2 a n by lambda n square exponential of minus h by 3 lambda n square x plus and theta 1 theta prime it is will be given in this fashion where a n is equal to minus c n y n dash 1 the y n dash 1 are noted down every time you get the solution at 1. So, the slope of the y function at this for all correct solutions is noted down and that is stored into an array called y n 1. So, that is how you get the a n coefficient in this expression and therefore, your Nusselt number expression looks like 8 by 3 into sum of that divided by sum of this and the mean Nusselt number would be evaluated in this manner which is simply is minus l n theta bulk divided by x star. Here are the values of lambda n c n and a n to be used in solving. So, this can be calculated once and for all for parallel plates 0 1 2 3 4 and for n greater than 1 they can be curve fitted in this manner. Incidentally, it so turns out that these coefficients also apply to the circular tube case if I had circular tube with uniform wall temperature then the same coefficients would again apply to the circular tube case and this was shown by Brown in 1960. Here are the computations of T wall equal to constant case. So, here I have got x star divided by 4 going from 0 to infinity theta bulk of course, at inlet would be 1 and it is reducing down to 0. Nusselt number would start with infinity of course, at x equal to 0 but would begin to decline and you can see by about 0.02 the Nusselt number has almost become equal to the fully developed heat transfer and whereas, the mean Nusselt number takes much longer let us say something like 0.3 or 0.4 and you fully developed of course, can be simply evaluated from 8 by 3 into lambda naught square which is lambda naught was given as 1.6816. So, you can see that it will turn out to be 7.5407 and that is what is very well predicted here because only first term is required for fully developed heat transfer. So, you can see Nusselt number does fall even in thermal entrance length problem of this type where velocity has been assumed to be fully developed and the solution is applicable to oils where frontal number is very very large. We now take the case of constant wall heat flux. Now, when q w is constant on both the plates ultimately the solution would become like this under the fully developed heat transfer the solution would be T wall there and there would be T bulk and although T wall and T bulk will increase with x the difference between T wall minus T bulk would remain constant in the fully developed whereas, in the earlier cases T wall and T bulk will increase at different rates and therefore, at individual points also individual values of y T wall and T bulk will change actually at different rates. Once you reach the fully developed then for all values of y the temperatures would increase linearly and we would again therefore, the solution can be split up in this manner as I showed. Let us say in this case the dimensionless temperature is called psi instead of theta and that would be T x y minus T fully developed q wall minus q b by k that is the up to in the developing part length and then onwards it is fully developed. So, I can split it up as but of course, as yet I do not know what T f d is but we will discover that in a minute. So, this is taken as T theta x y plus theta fully developed x y because q wall equal to constant is given temperature has been normalized with respect to q wall b by k as you can see here. In the fully developed part theta f d by theta x star would be 4 as I indicated here all of them would be varying linearly and that would be exactly equal to d T bulk by d x and that is equal to x by b Reynolds Prandtl x star and theta f d by d x star would be 4. Then we have two equations this would be the equation for the fully developed part 3 by 2 1 minus y star having substituted for theta f d by T x you will see this will be I have substituted 4 here. So, 3 by 8 into 4 is 3 by 2 d 2 theta f d by d y star square this is the fully developed part of the solution and 3 by 8 1 minus y star square T theta by d x equal to d 2 theta by d y square is the developing part and this equation is very similar to what we had in the constant wall temperature case. So, for the fully developed part straight forward integration gives theta f d equal to 3 by 4 a function of y plus 4 x star minus 39 by 80 very straight forward integration. For the developing part it will be very similar to what we had for the constant wall temperature is a function of coefficient c n y n y star in a function of y and this would be a function of x and c n for would be determined again in this fashion where theta f d at x star equal to 0 that is putting 4 x star equal to 0 here I will have a function of y into 1 minus y star y n y star d y star into all that and y n distribution with y is already known. So, I can determine c n. So, the complete solution will look like this 3 by 4 y star square minus y 4 by 6 plus 4 x star minus 39 by 280 plus c n y n which is the developing part. If I substitute y equal to 1 I will get psi wall which will vary in this fashion and if I carry out integration to evaluate bulk temperature I will get psi bulk, but q wall is already constant and therefore, psi bulk will be 4 x star, but you can also verify that by integrating this with respect in the usual manner. b n here now is c n y n 1 then the Nusselt number would then become this 4 divided by psi wall minus psi bulk and 1 over Nusselt number would become 1 over 4 17 by 35 all this. So, this is the complete solution to the constant wall heat flux problem. Again here are this Eigen values and Eigen constants for the constant wall heat flux case and these are lambda n values these are minus b n values they can be correlated for n greater than by this expression. The Nusselt number values themselves look like this x stars by 4 local Nusselt number varies from as high as 32 going down to 8.235 and you can see that the under constant wall heat flux case the thermal development length of for very high Prandtl number fluids is about 0.05 x star by 4 equal to 0.05 or x star equal to 0.2 mean Nusselt number varies in this fashion. This value we have already noted before that for parallel plates with constant wall heat flux on both sides it will be 8.235. We now take up the final case of Prandtl number very much less than 1 that is liquid metal then in this case then in the entrance region the velocity profile can be taken as equal to u bar and therefore, it will be u bar d t by dx equal to alpha d 2 t by dy square or 1 over 4 d theta by dx square d 2 theta by dy star square and these are the definition you will recall this is nothing but the heat conduction equation or unsteady heat conduction equation if you replace x star by time. So, solutions to this can be obtained by method of separation of variables. So, I will not develop that solution it is a fairly well known solution and the solution is given for t wall equal to constant the solution will turn out to be this. So, theta bulk is calculated in this fashion d theta by dy star is the wall heat flux is calculated in this fashion and the Nusselt number is calculated in this fashion. So, the evaluation of the series is what is required to obtain the Nusselt number for large x star and you fully developed turns out to be equal to pi square and equal to 9.87 for liquid metals for Prandtl number very much greater than 1 it was 7.545 and the fully developed Nusselt number here is 9.87 for constant wall heat flux case again you can develop a solution which is psi equal to theta plus theta f d in this manner this is the developing part of the solution and this is the fully developed part of the solution and you can evaluate psi wall psi bulk is this and therefore, Nusselt can be evaluated in this fashion. Again for large x star Nusselt now turns up to be 12 which is much greater than 8.235 for Prandtl number very much greater than 1. So, with this I conclude everything on laminar developing heat transfer as I said for Prandtl number close to 1 they say between Prandtl number range 0.5 to 10 one must solve the velocity N and the temperature equation simultaneously and it is now best done by using CFD codes or you can write your own finite difference codes are very easy to write for this particular class of problems. When Prandtl number is very much greater than 1 then you can use fully developed velocity profiles which are available now to you from our previous lectures. For Prandtl number very less than 1 you can easily take u equal to u bar and the problem becomes essentially that of heat conduction. So, it is very straight forward to evaluate for Prandtl number very much less than 1.