 In the previous lecture, we saw how to account for the pressure gradient variation. In principle, that method can also be extended to other types where V w is present or where viscous dissipation is present, but the algebra simply becomes extremely complex. Therefore, I will not deal with it, but there is a simple method also to account for wall temperature variation which we have not considered in the previous lectures. So, my purpose today is to consider how to account for wall temperature variation in integral analysis and one makes use of what is called as superposition theory. What is superposition theory? You will know very shortly. So, I will develop the theory very briefly and then obtain solutions with arbitrary variation of wall temperature using unheated starting length x naught solution for a flat plate. Just to recall your memory, the solution for the flat plate with unheated starting length x naught read as stenton x equal to 0.331 re x to the minus 0.5, Prandtl to the minus 0.6, 1 minus x naught by x 0.75 raise to minus 0.33. For constant fluid properties, you will recall our energy equation is actually u dT by dx plus v dT by dy equal to alpha dT dy square plus nu by Cp du by dy whole square. If the fluid properties are constant alpha, nu and Cp, if these are constants, then the temperature solution is independent of the velocity solution. So, even u v and du by dy whole square, all these are given in the energy equation and therefore, what remains is a simple homogeneous equation in temperature T. Such an equation is called linear homogeneous equation and in that, if T equal to T 1 is a solution and T equal to T 2 is a solution, then T equal to C 1 T 1 plus C 2 T 2 and so on so forth is also a solution. So, the sum of the solutions is also a solution. This property of a linear equation is exploited to derive Stanton x results for arbitrary variation of T w x knowing the solution for T w equal to constant. So, to do that, we shall define theta x y equal to T w minus T over T w minus T infinity as we have done before. Thus, theta x y x naught will be the unheated starting length solution for T w equal to constant for x greater than x naught, the situation being what we had described earlier. So, the situation is very clear. This is the flat plate. The boundary layer velocity boundary layer grows so, but the temperature boundary layer starts here. This is delta and Prandtl number is greater than 1 and this is the unheated starting length x naught. So, we have defined theta is equal to T w minus T. As I said, this is T w, but here it is T infinity over T w minus T infinity. Therefore, the solution for this particular case characterized by x naught choice of x naught will be theta x y x naught. Then, that solution will read as T minus T infinity equal to 1 minus theta x y x naught into T w minus T infinity T w minus T infinity. Now, we want to capture the effect of variation of T w minus T infinity. In reality, T w can change abruptly like a step change or it can change in a continuous manner. Both are possible, but the case we have here is precisely the case of a step change because if I plot T w, well it was equal to T infinity up to x naught and has been suddenly raised to T w from x equal to x naught. In other words, we have a solution which responds to a step change in wall temperature, a step change in wall temperature, but I may also have after some length x that it may vary linearly or it may go like that. Any variation is possible. These sort of variations occur over gas turbine blade surfaces. When you are cooling electronic boards like printed circuit boards, there are wires carrying currents, there are condensers, there are capacitors, there we generate heat and are at different temperatures and you have a cooling air flowing over it. So, the wall temperature can either vary abruptly or can vary continuously and we want to be able to predict how would the heat flux change or how would the heat transfer coefficient change along this length when the wall temperature varies in an arbitrary fashion. For continuous variation, the response of this function to an infinitesimal change dT w will be simply dT minus T infinity equal to 1 minus theta x y x naught dT w minus T infinity. It is straight forward. I have just differentiated this and this part keeping this constant. However, for a finite change or a discrete change, the response will be delta times T minus T infinity 1 minus theta x y x naught delta times T w minus T infinity. These are response functions to a small change either discrete or continuous. Therefore, supposing I have a temperature variation of T w with x, let us say it is first a step change, then another step change, then some variation continuous variation, then constant, then another continuous variation, then another continuous variation and so on and so forth. So, then I have several situations of… So, there is a discrete change here, there is a change delta here, but there is a continuous change here, then there is a constant value and then there is a continuous change like that and then there is a continuous change here. But from here I have a little continuous discrete change here, let us say another delta somewhere. So, I may have variation in which there are 1, 2, 3, 4 I discrete changes and there are several portions over which the change is continuous. So, T minus T infinity x naught equal to 0 to x equal to 1 minus theta x naught plus D T w plus… So, the total response for such a change would be little bit from the continuous change and a little bit from discrete change is I. So, sum of I equal to 1 to I equal to I into T w minus T infinity. But notice that in the continuous part D T w is nothing but D T w d x naught into d x naught and therefore, I can also write the total response temperature function as x naught equal to 0 to x naught equal to x into 1 minus theta x y x naught d T w by d x naught d x naught into 1 minus theta x y x naught into delta T w minus T infinity I. If I differentiate this equation with respect to y, d T by d y at y equal to 0 with a negative sign then q w, remember if I do q w will be simply minus k d T by d y equal to 0, h x x naught will be simply q wall x over T w minus T infinity and that will become equal to minus k d theta by d y y equal to 0. Because the way we have defined theta, theta by remember was defined as T w minus T over T w minus T infinity. Therefore, you will get this as d theta by d y equal to 0. Here this will turn to when I differentiate this temperature with respect to y, I will get minus d theta by d y x y x naught and here also minus d theta by d y x y x naught, which will be nothing but the heat transfer coefficient x x naught as a continuous variation d T w by d x naught d x naught h x x naught into delta T w minus T infinity I. Therefore, flat plate and Prandtl-Lomond ghetto h x x naught will be evaluated from our relationship here. So, it will be simply h x x naught rho C p u infinity into 0.33 a 1 Reynolds x to the power half Prandtl to the minus 0.6 and so on and so forth. So, I can get now for a given wall temperature variation the response of the heat flux as a function of x. And once I know the response of the heat flux I simply divide by the local value of T w minus T infinity to get the heat transfer coefficient variation over the entire surface. So, how do we do this now? So, I am going to take a very simple problem here. It shows a flat plate boundary layer in which the surface temperature varies as for 0 to x 1 where x 1 is 0.1 meters the temperature is varying linearly as 40 plus 100 x T infinity is 90. So, there is a hot gas flowing over a colder surface. So, remember the temperature here is 40 it will change to 50 at x equal to 0.1. Now, at x 1 the temperature suddenly rises to 80 and remains so up to 0.2. So, there is a constant temperature, but still less than the free stream temperature x 2 to x 3 the temperature is lower now to 0.265 degree centigrade for x greater than x 3 the temperature again rises linearly let us say as 200 x minus x 3 plus 65. So, remember the wall is throughout at a lower temperature than the free stream temperature and therefore, one would expect the heat flux to be continuously pouring into the surface, but you will see as we because of the wall temperature variation in x direction this will not so be so and there will be situations where although T infinity is greater than T wall the heat flux will actually flow out of the surface. This kind of situations we considered even in similarity solution where T w minus T infinity was allowed to vary as C x raise to gamma. Nu infinity is given as 7.5 meters per second the kinetic viscosity is this thermal conductivity is this and Prandtl number is 0.696. Now, remember although I said my our analysis is valid for Prandtl greater than 1 essentially where delta by delta is less than or equal to 1. So, 0.696 is not too bad one can use the same results even for air no harm done and we would still capture the essential features of the solution. So, our task is to determine how the wall heat flux and the Nusrat number would vary with x in response to such a variation of wall temperature and all we know is for a single step change the how the temperature should how the heat transfer coefficient should vary. For 0 x x 1 the temperature varies like this this is x 1 equal to 0.1 meters. So, the temperature is 40 and it goes up to 50 the free stream temperature T infinity is 90. So, I can say that at x equal x not equal to 0 itself there is a step change from 90 to 40 and therefore and then there is a continuous change. So, the continuous change part would be written as 1 minus x not by x 0.75 into minus 0.33 dt wall by dx not in this region is 100. I will show you the previous line see 100 dt w by dx is 100, but there is also a step change 40 minus 90 equal to minus 50. So, delta T w 0 will be minus 50 and this is down a would be 0.331 re x to the half Prandtl to the minus 0.6 would remain as a multiplier and I would integrate this from 0 to x where x not varies into dx not delta T w the temperature has increased from 50 to 80 and in this region of course, the effect of the previous solution would survive a into 0 to x 1 part that that is the first part that would still be the solution at x 1 this is the solution at x 1 plus now the solution sorry there should be a plus sign here there should be plus sign here plus delta T w 1 and plus a 2 this is the solution x 1 to x integral there is an error here which we shall correct plus delta T w 1 and delta T w 1 is 80 minus 50. In fact, 80 minus 50 should be read in correctly as 80 minus 90 minus 50 minus 90 equal to 80 minus 50 which is plus 30. So, remember there is an error here it should be 0 x 1 to x 1 minus x 1 by x 0.75 dx 1 plus delta T w 1 there is no error because the between x 1 and x 2 there is no continuous variation only a step change and therefore this result is correct it is 1 minus x 1 by x 0.75 0.33 into delta T w 1. So, that is correct that is absolutely correct and a would be as before from x 2 to x 3 now there is a delta T w 2 which is 65 minus 80 is minus 50. So, remember here the solution up to x 1 is correct this is the solution up to x 2 now this is x 1 divided by x 2 0.75 0.30 3 delta T w 1 and now you have another step change which is 1 minus x 2 by x 0.75 and it will go up to x 3 x will go from x 2 to x 3 with a with a step change delta T w equal to minus 15 because 65 minus 80 is minus 15 a is again that. So, this is the solution up to x 1 this is the solution up to x 2 and from x 2 onwards, but less than x 3 this is the solution for x greater than x 3 this is the solution up to x 1 this is the solution up to x 2 this is the solution up to x 3 now because x 2 has gone up to x 3 and you will have simply integral x 3 to x 0.75 there is no discrete change here simply from 65 there is no discrete change there is only a continuous change and therefore, in the last part in the last part you will have 200 times d x 3 this is nothing, but d T w by d x 0 and 1 minus x 3 by x 0.75 where a is equal to all that. So, notice that we have slowly built up the solution by superposition this is what is meant by superposition you started with the solution for the first step change which was valid for x greater between 0 and x 1. So, you could integrate this relation then from x 1 to x 2 you had solution at x 1 plus this discrete change solution where x varies from x 1 to x 2. So, the solutions are valid from there. So, x is simply taken from x 1 to x 2 then from x 2 to x 3 you had discrete solution up to x 1 discrete solution up to x 2 and then a third discrete solution from x 2 to x 3 and then you had the continuous variation where this is discrete solution up to x 1 this is up to x 2 x 2 x 3 and then the continuous solution. Now, wherever the integrals are involved like in this case and here the integration 0 to x 1 minus x naught by x raise to 0.75 raise to minus 0.33 d x naught is actually a beta function and it can be shown that it is equal to 4 by 3 beta 2 by 3 4 by 3 x and that is equal to 1.612 x very useful result very very useful. So, one can make use of that here to simply replace these quantities by 1.612 x that is all there is to it. Which will be 1.612 x 1 to 0 x 1 minus 0 multiplied by 100 and this would be 200 into 1.612 x minus x 3 x minus x 3 into 200. So, this is how the solution proceed and I will get q w x variation for different segments of the solution. What do they look like? The squares show the variation of q w x. As you expect in the first part the wall temperature varied from 40 to 50 the free stream was at 90 and therefore heat was simply flowing into the wall and therefore q w x is negative, but rising because as the temperature difference goes on reducing the amount of heat flow is also going on reducing. The moment however the temperature moves of the wall moves from 50 to 80. Remember 80 is less than 90. Then what happens is that the layers of fluid close to the wall are still at a temperature close to 50 and although the wall temperature is 80 now. So, the heat actually flows from the wall to the fluid although the free stream temperature is higher than the wall temperature. So, one would expect heat transfer to be positive from the wall to the fluid because the fluid layers close to the wall are close to 50 degree centigrade. Therefore, you see suddenly the heat flux becomes positive and starts to fall in this manner. Then when you come to from 85 to 65 there is now the wall is at 65, but the fluid close to the wall is close to 80 degree centigrade and is likely to be at a higher temperature than 65 degree centigrade. Therefore, again the heat transfer goes into the fluid and ultimately up to 3 it is negative and then it continues when the wall temperature begins to rise you get positive heat transfer. Now, this is a case corresponding to that will be Nusselt numbers. You will see Nusselt number is more or less very slow decline and then it Nusselt number has turned negative and then again positive. Now, this is a case really that can be handled by numerical methods or the finite difference method and I have done calculations of this type for this case using a computer program for a finite difference method. What I have done is plotted the finite different solutions by a dotted line for q wall x and Nusselt x, this shows you the power of the integral method and the superposition theory to capture effects of arbitrary variations of wall temperature, but the case we have considered is that of a flat plate Bamburaya, very simple case free stream remains constant. So, notice the changes in q wall n u x although over the entire length t infinity is greater than t w. So, you could get situations where actually the heat transfer takes place from the solid surface to the fluid although the fluid bulk mean free stream temperature is higher than the negative q wall x implies heat transfer to the wall and vice versa of course. So, problems of this type are important in electronic cooling such as printed circuit boards. So, much so that many a time the heat generating surface itself would begin to get if it was in the wake of a certain type of temperature variation then heat generating condenser would actually instead of giving out heat would actually receive heat from the gas and either it can go into an adiabatic case or actually positive heat flux from the gas air to the component and that the temperature of that component would suddenly shoot up and a drift would occur as you know in electronic equipment the components should not vary too much in that temperature drifts in order to deliver a certain performance. Very important consideration in printed circuit board design is the thermal management of how to put the current carrying conductors as well as the condensers and other pieces of electronic equipment. Now, as I said we have developed this solution for a flow over a flat plate with arbitrary variation. What about flow over a gas turbine plate where the free stream will vary or a flow over a cylinder where free stream would vary in an arbitrary fashion and so would the wall temperature vary in arbitrary fashion. Well, you have to simply go back to the velocity and temperature equations that we have derived evaluate the delta 2 in the most laborious way and then allow for dT w by dx term in the integral equation and from there on calculate the variation of delta 2 with x and from which you would extract the heat transfer coefficient variation. The procedure is quite laborious there are some simplifications possible and one of them is published by Spaulding dv heat transfer from surfaces of non-uniform temperature in journal of fluid mechanics in 1957. One could look at those equations but as I must say that today no one does integral solutions of this complexity anymore people have turned to numerical solutions and that was my purpose in showing you how good the numerical solutions in fact are the finite different solution and for the simple cases at least the agreement is simply excellent. With this I complete our discussion on integral solutions to laminar temperature boundary equations.