 In this lecture, we shall now consider integral solutions to the temperature boundary layer equation. The equation you will recall is written here, d delta 2 by dx, where delta 2 is the enthalpy thickness plus delta 2 times the wall temperature variation term, the pressure gradient term that is the variation of u infinity equals the stanton x, which is a dimensionless heat transfer coefficient at the surface, the wall velocity term and the viscous dissipation term. A purpose here is to understand how the solution procedure is conducted. Then, we will look at solutions with effects of pressure gradient and suction and blowing with an application to flow over a cylinder. Recently, the effect of wall temperature variation will not be taken, nor will the effect of dissipation be considered. So, like in velocity boundary layer solution, we first begin by saying, let the temperature be a function of distance from the wall y divided now by the thermal boundary layer thickness delta. I called eta t equal to y divided by delta and the assumed expression is as follows. It is the fourth order polynomial with five coefficients. These five coefficients are to be determined as within in the case of velocity boundary layer equation with five boundary conditions. The first of these is at y equal to 0, course t is equal to t w and all eta t is will be 0 and therefore, a will be equal to t w. The second condition at the wall derives from the differential equation of the boundary layer and it would be alpha times d 2 t by dy square, which is the diffusion term equal to the viscous dissipation term and the wall velocity term vw dt by dy. At the edge of the boundary layer, thermal boundary layer, t will equal to t infinity of course. The temperature derivative will be 0 and the fact that temperature approaches t infinity asymptotically would entail that d 2 t dy square should also be 0. So, we have five boundary conditions and five constants a, b, c, d and e to determine. Here is the temperature profile as it looks. In dimensionless form, the temperature profile looks, reads as t minus t infinity divided by t w minus t infinity equals 1 minus 2 eta delta t eta t 2 times eta cube t minus eta 4 t plus a times again a function of eta t, where a captures the effects of wall velocity viscous dissipation through the cut number and the pressure gradient effect through lambda. Lambda as you will recall is nothing but delta square by nu d u infinity by the x v w star is simply v w delta by nu and the corresponding velocity profile is again reproduced to refresh your memory. f 1 is equal to that function, f 2 multiplied by v w star is that function and f 3 multiplied by lambda is that function, f 3 function is that. So, we now have a temperature profile and a velocity profile. As you can see, the expressions are quite complex, but nonetheless can be worked on a piece of paper. Now, to make further progress in our equation here, we need to evaluate delta 2 the enthalpy thickness and the definition of enthalpy thickness is 0 to l u over u infinity into t minus t infinity divided by t w minus t infinity d y. Essentially therefore, it is an integration of the two profiles, the velocity profile and the temperature profile to 0 to l. So, remember we recall that if this was the boundary layer development delta, then if Prandtl was greater than 1, then thermal boundary layer thickness would be smaller than delta, but if Prandtl number was less than 1, then thermal boundary layer thickness would be greater than delta and we now choose l to be bigger than any of them. This is l, l therefore is equal to delta or delta. You can see that l will be equal to capital delta for Prandtl less than 1 and it would be Prandtl greater than 1. I mean greater than or equal to would be appropriate way of writing it, this is delta. Now, you can imagine because the two profiles are highly complex, the evaluation becomes extremely laborious of this quantity delta 2 and hence usually simplifications are made. For example, for liquid metals where Prandtl number is much less than 1, for all practical purposes u over u infinity is 1 that simplifies u over u infinity equal to 1 that simplifies the integration to only the temperature profile and also delta is very much greater than delta and then you will see therefore, that since delta is very much greater than delta and it appears in the numerator as well as denominator 3 divided by Prandtl number. Prandtl number is of the order of 0.001 and therefore, this number is also very large. In other words the both the numerator and the denominator assume very high values and therefore, a would tend to 2 for liquids V w star equal to 0 is not of interest and in which case a if I drop V w star terms here would simply be that term divided by this term. It would read as Prandtl E c by 3 lambda plus 12 by 6 square into delta by delta whole square and if I were to consider oils in particular for which Prandtl number is very much greater than 1 then delta by delta would be much much smaller than 1 and therefore, a would practically trend to 0. So, one can make such assumptions to simplify this evaluation of the integral. So, in order to explain how the procedure works I am going to consider a very simple case of a flat plate boundary layer. I will ignore wall velocity V w equal to 0 because the flat plate boundary layer U infinity will be a constant and E c would be equal to 0. In other words if you see this equation there is no axial variation of temperature there is no axial variation of velocity. So, that term goes to 0 there is no viscous dissipation consider nor is this consider. So, therefore, d delta 2 by delta d x would equal stanton x is the most simple form of the energy equation and the corresponding velocity boundary layer equation will be d delta 2 by d x c f x by 2 equal to tau infinity tau wall over rho infinity square. I am also going to consider the case of Prandtl greater than 1. So, delta will be bigger than thermal boundary layer thickness delta and I am going to postulate that the constant wall temperature boundary condition T equal to T w starts at x greater than x naught and T w is greater than T infinity let us say. So, the temperature between 0 and x naught of the wall will be simply T infinity, but is suddenly raised to T w from x equal to x naught. X naught is called the unheated starting length. There is a purpose for doing this. This analysis would lead us to considering the effects of wall temperature variation which I shall consider in the next lecture, but here simply appreciate the procedure of how these two equations are solved simultaneously. Again to simplify matters instead of taking the longest fifth order profile I am going to take very simple profiles u over u infinity equal to 3 by 2 eta minus 1 by 2 eta cube. You will see that this profile satisfies the boundary condition u equal to 0 at eta equal to 0 which is at the wall. It also satisfies the condition that u equals u infinity equal to 1 at eta equal to 1 which is correct. It also will satisfy du by dy equal to 0 at y equal to delta. So, you can see that that condition is also satisfied and therefore, the equation is quite ok. Likewise, the temperature profile is T minus T w over T infinity minus T w would be 3 by 2 eta T minus 1 over 2 eta T cubed. So, with these two simple very simple temperature profile and velocity profile, the delta 2 would now read as 0 to delta 3 by 2 eta minus 1 by 2 eta cube into as you can see here we want T minus T infinity over T w minus T infinity and therefore, this will be a T minus T infinity over T w minus T infinity would be equal to 1 minus 1 by minus T w minus T over T w minus T infinity and therefore, this will be 1 minus 3 by 2 eta T plus 1 by 2 eta T cube into dy. This is how delta 2 will be evaluated and the evaluation leads to first of all the momentum thickness delta 2 as you recall is 0 to delta will be u over u infinity into 1 minus u over u infinity whole dy and therefore, that will be equal to 0 to delta 3 by 2 eta minus 1 by 2 eta cube into 1 minus 3 by 2 eta plus 1 by 2 eta cube dy. That evaluation gives you that delta 2 by delta will be 39 by 280 delta 2 by delta is equal to 39 by 280. This is one result from the velocity profile tau wall which is equal to mu times du by dy at 0. You will see this will become mu into u infinity into 3 by 2 1 over delta tau wall over rho u infinity square would be equal to 3 by 2 delta into mu divided by u infinity delta. That is what C f x would be and that is what I have shown here. Now, if I substitute this solution delta by 2 equal to delta 39 by 280 into the momentum equation which is d delta 2 by dx equal to tau wall over rho u infinity square which we showed just now is equal to 3 by 2 mu times u infinity delta and delta 2 we said is 39 by 280 into d delta by dx. Then you will see that I get essentially delta into d delta by dx equal to 140 by 13 mu over u infinity mu over u infinity which is nothing but d delta 2 square d delta square by dx equal to 280 by 13 mu by u infinity. If I integrate that then I get delta square minus 0 equal to 280 by 13 mu x by u infinity and that is what I have shown here as the solution delta is 0 at x equal to 0 which is the start of the boundary layer. So, you get delta equal to under root 280 by 13 mu x by u infinity or which is nothing but 4.64 under root mu x by u infinity. If I now substitute this delta in the definition of C f x 3 by 2 mu over u infinity then it can be shown that C f x is 0.646 Reynolds x to the minus half. Now, you will recall from our exact similarity solution we had obtained C f x equal to 0.664 Reynolds x to the minus half. So, even though we have chosen a very simple velocity profile here 3 by 2 eta minus 1 over 2 eta cube we have obtained a result which is very close to the exact solution. It is for this reason that the integral method is called the approximate method because it depends on the approximation to the velocity profile that we have used. It is not that the equations are in exact it is the method of solution that is approximate. So, this is how one solves the momentum equation you now turn to the energy equation. Then as I said this is the definition of delta 2 and if I integrate that equation if I integrate this equation I would get delta 2 by delta equal to 3 by 20 r minus 3 by 280 r cube. Stanton x like we evaluated this Stanton x is nothing but you will recall H x by rho C p u infinity and that is equal to q wall over rho C p u infinity into T w minus T infinity and that is equal to minus k d T dy at y equal to 0 divided by rho C p u infinity T w minus T infinity which will give me minus alpha theta theta times d T by dy equal to 0 over u infinity T w minus T infinity and therefore, you will see that this will reduce to very simply 3 by 2 because we have the temperature profile and therefore, we can evaluate that as 3 by 2 alpha times u infinity into delta. This is the evaluation of Stanton x the right hand side of the energy equation and r here is delta by delta and since we are considering the case of Prandtl greater than 1 delta by delta will always be less than 1. If I substitute these results essentially d delta 2 by dx what is the energy equation equal to Stanton x. So, if I substitute these results I will get d delta 2 by dx equal to 3 delta by 10 r minus r cube by 7 d r by dx plus 3 by 20 r square minus r raise to 4 by 14 d delta by dx it is a fairly straight forward algebra. But notice each of these brackets because r is less than 1 r cube by 7 will be much much smaller than r. So, this can be ignored likewise r 4 by 14 will be much much smaller than r square. So, even that term can be ignored and then I will get d delta 2 by dx is very nearly equal to 3 delta r by 10 into d r by dx plus 3 r square by 20 d delta by dx 3 by 2 u infinity. So, I get essentially d delta 2 by dx is equal to 3 by 10 delta r d r by dx plus 3 by 20 r square d delta by dx equal to 3 by 2 alpha times delta over u infinity. Where do we get delta from? I need to get delta and d delta by dx and that is what we evaluated on the previous slide. You will see that delta is just 3 by 10 into under root 2 80 by 13 into nu x by u infinity r d r by dx plus 3 by 20 r square d delta by dx will be simply 140 by 13 nu by u infinity divided by. So, in other words this will be multiplied by 13 by 280 into u infinity by nu x equal to 3 by 2 alpha times delta u infinity. This part of the equation simplifies to r cubed plus 4 r square x d r by dx equal to 13 by 14 1 over Prandtl number. Now, this left hand side can be manipulated to read like that. This is nothing but 4 by 3 x raised to 0.25 d by dx of x raised to 0.75 r cubed equal to 13 by 14 Prandtl number. So, I can now integrate this equation integrating and noting that r is remember our we are starting to heat from x equal to x naught and r is delta by delta. So, this is delta whereas, this is delta. So, r is equal to 0 at x equal to x naught. So, if I make use of that then I get the solution that r cubed will equal the solution to that equation is r cubed equal to delta by delta whole cube equal to 13 by 14 Prandtl number 1 minus x naught by x raised to 0.75. Therefore, now if I substitute this value, then you will see that stanton x which was already shown to be equal to stanton x which are shown to be equal to 3 by 2 alpha u infinity delta. This definition delta would be r times small delta and r is given by that expression and delta already is known as under root 280 by 13 as you will see here delta is under root 280 by 13 new x by u infinity will give you this relationship. If you want to see how I have done this, you will see that stanton x therefore, will be 3 by 2 into alpha divided by r which is 13 by 14 Prandtl number into 1 minus x naught by x raised to 75 raised to 1 by 3 into delta which is 4.67 under root new x by u infinity into u infinity which is a constant. This is what stanton x is. So, 1 minus x naught by x 0.75 minus 0.33. So, this then is the solution to the case of unheated starting length x naught. When obtaining similarity solutions, we had set, we had boundary thermal boundary layers and velocity boundary layers growing at the same point x equal to 0. Therefore, this is the solution to the case of unheated starting length x naught. When obtaining similarity solutions, we had set, we had boundary thermal boundary layers and velocity boundary layers growing at the same point x equal to 0. Therefore, in effect this was x naught equal to 0 and the similarity solution was 0.33 Reynolds x to the minus 0.5 Prandtl to the power minus 0.6. You can see that in spite of a very simple temperature profile and velocity profile, we have obtained an extremely accurate solution using integral method. But you can see that whereas, similarity solution requires solution of three ODE's. The integral solution is simply pencil and paper method and produces many a times extremely accurate results. Now, of course, this special relationship x naught for this special solution for heating started from x equal to x naught will be used later on to generate solutions when the wall temperature varies arbitrarily with x. I will be using what is called the super position theory and which is something I will discuss in the next lecture. At the moment, let us turn our attention to the case of pressure gradient. Presently in the flat plate case, the pressure gradient was 0 and therefore, the momentum equation was very simple and so was the energy equation. But now you will see our momentum and energy equations would be more complex. If I turn to the first slide, you will see I am not considering wall temperature variation, I am not considering V w, I am not considering, but I am going to include this particular term which contains the pressure gradient term. So, then you will see that the equation governing this situation is simply this. V w, E c all are 0, T w is constant and therefore, the equation would read as delta 2 by dx delta 2 by u infinity d u infinity by d x. So, the equation is d delta 2 by dx plus delta 2 by u infinity d u infinity by dx equal to stanton x. The corresponding momentum equation was d delta 2 by dx plus delta 2 by u infinity d u infinity by dx. Into 2 plus h equal to c f x, where h was equal to delta 1 by delta 2, you can see the two equations are very similar. Now, here to make further progress, I am going to define what is called as conduction thickness delta 4 equal to k by h x delta 4 by h x delta will be defined as k by h x. You recall that the units of k are watts per meter kelvin, whereas units of h x will be watts per meter square kelvin and this has the units of meters therefore, and therefore, it is called the conduction thickness. It will be proportional to all other thermal boundary layer thicknesses such as delta, delta 2 and so on and so forth. Incidentally, stanton x would readily follow. It will be simply alpha divided by rho u infinity because you know that stanton x is h x over rho c p u infinity, but h x is nothing but k by delta 4 u infinity and therefore, this is equal to thermal diffusivity delta 4 u infinity. Now, because of the similarity of the momentum and thermal boundary layer thickness thermal energy equations, we shall postulate as done by Eckert. The reference is given here the reference is given here. It is a German paper of very old paper of 1942 and he said we will postulate that the rate of growth of conduction thickness d delta 4 by d x will be first of all governed by the rate of growth of momentum thickness delta 2, which in turn was governed by u infinity d u infinity by d x and nu. So, there will be the first three parameters here first three parameters here and then of course, it will also be governed by delta 4 and it would be governed by Prandtl number. So, the rate of growth of the conduction thickness would be done. We are following a strategy very analogous to the results we had derived for integral momentum equation and then now, we shall say we will carry out dimensional analysis. So, let us say if this was the boundary layer growing on a flat plate or any other plate any with a pressure gradient then in the stream wise direction let the this is x, this is the transverse direction y and this is the lateral direction z. Then if I say the characteristic dimension in direction x is capital X, the characteristic dimension in direction y is capital Y and characteristic dimension z in direction z is z. So, in other words this is z, this is x and here y then you will see delta 4 being a transverse thickness will have dimension y u infinity being a stream wise velocity would be would have dimension x by t d u infinity by d x. Of course, would be 1 over t simply divide this by x, nu would be y square by t because this is kinematic viscosity mu by rho and the shear stress this is nothing but tau wall by rho equal to nu times d u by d y at y equal to 0 and therefore, you can see that the shear stress which acts along the surface on an area y z would have units and therefore, the nu will have units of y square by t and then finally, d delta 4 by d x would be simply y divided by x. So, if I essentially then if I were to carry out a dimensional analysis you will see I will get y by x which is d delta 4 by d x equal to first of all delta 4 which is y raise to a then u infinity x by t raise to b this is delta 4, this is u infinity then into y square by t raise to c which is nothing but nu and 1 over t raise to d which is d u infinity by d x. So, I have captured all for a fixed Prandtl number. If I equate the like powers, so power of y on the left hand side is 1, so that will equal a plus 2 c power of x is minus 1 will be equal to b and the power of time is 0 equal to minus b minus c minus d. So, with this I can determine a b c and d everything in terms of d let say and you will get u infinity by nu d delta 4 by d x as a function of d delta 4 by d u d u infinity by d x. Remember this quantity delta 4 square by nu d u infinity by d x is very similar to delta 2 square by nu d u infinity by d x which we had called kappa and also delta square by nu d u infinity by d x which was called lambda. So, it is very analogous and therefore, I call it kappa t to indicate that this is thermal parameter associated with conduction thickness delta 4. So, we have a relationship very similar to what we had in integral momentum equation and therefore, I can evaluate both these quantities from similarity solution u infinity equal to c x m which is nothing but a special case of arbitrary variation of u infinity and the functional must admit similarity wedge flow solutions. For a fixed Prandtl number conduction thickness square divided by nu into d u infinity by d x is a function of u infinity divided by nu d delta 4 by d x is a function of u delta 4 square by d x for fixed Prandtl number. Now of course, this functional must hold for wedge flows as well for which u infinity is equal to c x raise to m and nu x r e x to the power of minus 0.5 is equal to minus c 1 m Prandtl sorry plus c 1 n Prandtl or equal to minus theta 0 m Prandtl. These are the similarity solutions, these are known to us. So, let us see what nu x r e x raise to minus half represents. It is nothing but h x and h x is equal to minus x multiplied by x divided by k into u infinity into x divided by nu raise to minus 0.5 which I can also write as remember k by h x is delta 4. So, this can be written as x divided by delta 4 into c x raise to m into x divided by nu raise to minus 0.5 which I can also write as x by delta 4 c by nu raise to minus 0.5 into x raise to minus m plus 1 by 2 which will equal 1 over delta 4 c by nu raise to minus 0.5 into x raise to minus m plus 1 by 2 which will equal 1 over delta 4 c by nu raise to minus 0.5 into x raise to 1 minus m by 2 and all this is equal to c 1 and therefore delta 4 therefore delta 4 will be c by nu raise to minus 0.5 x raise to 1 minus m by 2 divided by c 1 and delta 4 square therefore delta 4 x will be nu by c into x raise to 1 minus m divided by c 1 square and therefore the left hand side which was delta 4 square divided by nu du infinity by dx will be simply nu by c x raise to 1 minus m by c 1 square into 1 over nu into du infinity by dx will be c m x raise to 1 minus m by c 1 square into 1 over nu into du infinity by dx will be c m x raise to m minus 1. So, that is what I write here c m x raise to m minus 1. You can see c and c will get cancelled nu and nu will get cancelled and x raise to m minus 1 and x raise to 1 minus 1. So, both these get cancelled and I will have essentially m by c 1 square. So, you can see the left hand side delta 4 square by nu du infinity by dx would be m by c 1 square. If I follow the same logic then the right hand side that is kappa t which is function of kappa t then kappa t equal to nu infinity by nu d delta 4 square by dx can be shown to be equal to 1 minus m divided by c 1 square. So, our relationship delta 4 square divided by nu du infinity by dx is a function of u infinity nu d delta 4 square by dx for fixed Prandtl number will simply be equal to m by c 1 square equal to f of 1 minus m by c 1 square where c 1 is a function of m and Prandtl number. So, you can see that by knowing the values of m and the corresponding values of c 1 for the value of Prandtl number I can always construct this functional relationship and that is what I have shown on the slide. Basically, that is equal to kappa t and likewise we can show that u infinity over d delta 4 square by dx nu is 1 minus m c 1 square. So, basically the relationship that we had here is simply equal to 1 minus m c 1 square f m c 1 square f kappa t where c 1 is that belonging to the m chosen. Therefore, by selecting different values of m and correspondingly c 1 I can calculate plot this variation and this is what has done on this slide here. You can see this is kappa t, this is function of kappa t the right hand side and this is the kappa t equal to 0 line and this is f kappa equal to 0 line. So, this is where the axis 0 is at this point. What I have plotted here are the similarity solutions, but you can see that considerable variation of kappa t the variations are actually very close to a linear variation. So, the graph has been plotted while looking at results for Prandtl equal to 0.7. You can create similar graphs for other values of Prandtl number for which we have we have the similarity solutions. So, from known similarity solutions for Prandtl equal to 0.7 the relationship is linear and you can see therefore, the x intercept here will give you the stagnation point solution. The y intercept will give you the du infinity by dx or equal to f kappa t equal to 0 solution and that would be flat plate. I am sorry here there is a little error x intercept would read stagnation y intercept should read flat plate. Then using the value of Prandtl equal to 7, c 1 m equal to 0 is 0.293 and c 1 m equal to 1.493. Basically, what I have done as we did in the wedge flow solutions of plates f kappa t is related to a minus b kappa t or essentially this is a minus b delta 4 square by nu du infinity by dx and this is a u infinity by nu d delta 4 square by dx. So, I have to determine a and b. So, the first thing I do is to look at the flat plate solution for with du infinity by dx and therefore, a will be simply equal to as you can see 1 minus m which is equal to 0 divided by c 1 square. So, that is equal to 1 over 0.293 whole square which is 11.67. Now, b would be determined where this is 0 and you will recall for stagnation point solution all thicknesses are constant and therefore, this is equal to 0. Then b will be simply a times this quantity which is m by c 1 square and m is equal to 1 m by m which is 1 over c 1 square that will be equal to 11.67 into 0.493 whole square which gives you 2.87. So, that is how I generate the linear curve fit to the integral energy equation and this can be manipulated to give delta 4 square equal to 11.67 nu u infinity raise to 2.87 0 to x u infinity raise to 1.87 dx exactly as we did in case of momentum. So, then the closed form solution having obtained delta 4 in this manner. So, delta 4 square will be square root of this. I get the solution for stanton x alpha u infinity by delta 4 equal to all this for Prandtl equation. All these constants this one, this one and this one result from the curve fit which we obtained for Prandtl equal to 0.7. If I had chosen any other Prandtl number, I would get both these constants to be somewhat different and therefore, I can generalize and say that stanton x is k 1 nu 0.5 u infinity k 2 0 to x u infinity k 3 where all these are functions of Prandtl number. This is something you can do with yourself because you already have the similarity solutions available to you for different Prandtl numbers for m equal to 0 and m equal to 1 and you can generate the solution. I will now consider the flow over a cylinder that we had considered earlier for which u infinity by V a is 2 sin 2 x star where x star is x by d and I had called this as f x. Then for Prandtl number 0.7 and T w equal to 0, it is possible to integrate substituting for u infinity our function. It is possible to show that delta 4 by d R e d raised to 0.5 would be read like that and stanton x would read in this fashion. If I were to integrate this function from 0 to separation 1 over x 0 to x separation, I get the result 2.686. To see exactly the values of local values of stanton x and how delta 4 varies, here are the angle along the cylinder starting with the stagnation point at 0. So, you have delta 4 by d conduction thickness is 2.42, it reduces to 1.7, 1.05 and then again starts rising to and this is as you will recall is the point of separation. Stanton x is very high, the Nusselt number is very high at the stand or the heat transfer coefficient is high at the stagnation point, but then gradually reduces as you move towards the vertical axis of the cylinder and becomes 0.388. C f x on the other hand starts with a very low value and rises and again when the flow begins to decelerate, the C f x begins to fall again here whereas, during the acceleration part the C f x goes on increasing with the angle. With this I conclude the effects of pressure gradient on heat transfer in integral solutions.