 In the previous lecture, we saw how the law of the wall for velocity that is for u plus can be used to evaluate friction coefficient C f x for both external boundary layers as well as for pipe flows irrespective of the pressure gradients and also the effect of surface velocity B w or the surface roughness. In all those cases, we were able to obtain coefficient of friction as a function of the Reynolds number, the local value of Reynolds number. Today, we are going to look at how the temperature law of the wall can be used to predict the Stanton number in external boundary layers and Nusselt number in internal flows. Again, most of these methods are essentially analogy methods that is similarity between heat transfer and momentum transfer is one way or the other assumed. We would be able to predict the variations of Stanton number and Nusselt number as functions of Reynolds number as well as Prandtl number. So, I would first deal with the external boundary layers in four different ways. One is use of the law of the wall for situations in which there is no suction or blowing. I would also use analogy methods. Then, I will show you how to apply integral method that is the solution of the integral energy equation to take care of effects of pressure gradients wall temperature variations. Then, finally, we will extend the method to take care of the effects of roughness and suction and blowing. Likewise, we would look at prediction of Nusselt number in pipe flows, internal flows and again use make use of the law of the wall as well as the analogy methods. So, let us start with the external boundary layer case. From lecture 28, you will recall that the temperature law of the wall is written as T plus equal to Prandtl U plus plus P f which is a function of U plus. If we write that equation in the infinity state, we would get T infinity plus equal to Prandtl T into U infinity plus plus P f infinity which as you recall is a function of Prandtl number only. So, U infinity plus here would be simply U infinity by U tau which is U infinity under root tau wall by rho or it can be written as under root rho U infinity square divided by tau wall which is nothing but under root 2 by C f x. T infinity plus as you will recall is defined as minus T infinity minus T wall divided by Q wall by rho C P U tau. So, if I divide and multiply this by U infinity and note that Q wall divided by T wall minus T infinity is H x, the heat transfer coefficient. Then you can see that this part is the one over Stanton number and U tau or U infinity is nothing but under root C f x by 2 and as a result this equation simply transforms to Stanton x equal to C f x by 2 square root into Prandtl T into under root 2 by C f x plus P f infinity and as you will recall I said the turbulent Prandtl number is approximately 0.9 some people take it 0.85, but that is still one can make Prandtl T also a function of Prandtl number itself as I indicated what the possible correlation could be. For Prandtl number 1 you recall P f infinity is 0. So, that is 0 for Prandtl number equal to 1 and Reynolds actually used Prandtl T equal to 1 and hence then you will see the Stanton x would simply be C f x by 2 which implies perfect analogy between heat transfer and momentum transfer. From experiments for near unity Prandtl numbers Stanton x correlates as C f x by 2 into Prandtl raise to 0.4 minus 0.4 and hence for 0 pressure gradient boundary layer Stanton x is equal to 0.0286 Reynolds x to the power of minus 0.2 Prandtl to the power of minus 0.2 this was the correlation you use routinely in your undergraduate work and we have shown that it can be derived from this equation the temperature law of the wall. Of course, if you had a rough surface then one must evaluate C f x for a rough surface and P f infinity also to be must be used for a rough surface. So, as to get Stanton number for a rough surface now in all this evaluation you can see that C f x must be evaluated from the methods of the previous lecture. So, whatever the pressure gradient or V w or whatever is present you simply use that to evaluate C f x and straight away use evaluate Stanton x from that this is the simplest way to evaluate Stanton number, but then we can also apply somewhat more rigorously analogy method in which we call that the effective Prandtl number is essentially d t plus by d u plus which we can write as d t plus by d y plus multiplied by d u plus by d y plus then sorry this should be raised to d y plus by d u plus it should be not d u plus by d y plus d y plus by d u plus and hence using the relation tau tau divided by tau wall is approximately equal to 1 equal to 1 plus nu t by nu d u plus by d y plus you will recall we had derived this equation gives d t plus by d y plus equal to 1 plus nu t by nu d u plus by d y plus Prandtl raise to minus 1 plus nu t by nu into Prandtl t or simply 1 over Prandtl number into 1 over d u plus by d y plus minus 1 1 over Prandtl t raise to minus 1 integrating from y equal to 0 to infinity and using the three layer law for u plus and hence for d u plus by d y plus it follows that now if I pollute d u plus by d y plus for the laminar sub layer is equal to 1 then of course, that quantity vanishes and I simply get t age of the sub layer plus minus 0 equal to Prandtl y s l plus equal to Prandtl u s l plus and as you know u s l plus and y s l plus are 5. So, therefore, you get t s l plus is equal to 5 Prandtl extend the integration further from sub layer to transitional layer. So, t transitional layer plus minus t s l plus and here d u plus by d y plus would become 1 over kappa y plus where kappa is 0.2 and therefore, that will become 5. So, 5 Prandtl t l n 1 plus 5 Prandtl by Prandtl t and here I have used transitional layer as y plus for transitional layer as 30 then you get that relationship and then from the age of the transitional layer till the age of the boundary layer you will get t infinity plus minus t transitional layer plus again d u plus by d y plus will be 1 over kappa y plus and therefore, where kappa is 0.4 then you will get that relationship that involving delta plus the boundary layer thickness. So, we got essentially layer by layer contributions to t infinity plus and if I add these three equations as I show on the next slide and rearrange if I add these things then you will see t s l plus here gets cancelled with that this gets cancelled with that and I would get a relationship for t infinity plus which as you will recall from the previous slide t infinity plus is nothing but c f x by 2 by Stanton then I get c f x by 2 divided by Stanton x equal to 5 Prandtl plus all this this is the transitional layer contribution this is the laminar sub layer contribution and this is the fully turbulent layer contribution. Now, how do we evaluate delta plus here? Well as you will recall in the outer layers delta power law very well applies. So, instead of logarithmic law if we apply power law then simply delta plus is equal to u infinity plus divided by 8.75 raise to 7 or that is equal to this quantity the c f x is evaluated by integral methods of equation of lecture 29. So, you substitute delta plus here as c f x and c f x that appears here and here are both first evaluated from lecture 29 for a given situation and that gives you the variation of Stanton number as a function of Reynolds number and Prandtl number. Now, when u infinity and t w minus t infinity vary arbitrarily with x then one must invoke the integral energy equation which reads as 1 over u infinity into t w minus t infinity by dx delta 2 u infinity t w minus t infinity equal to Stanton x. So, you recall this integral energy equation when v w is 0. Now, for further analysis let Stanton x be equal to c Reynolds x to the power of minus n. Now, this kind of this method is often called the Ambrokes methods. Ambrokes was a Soviet scientist and he published a paper on this. So, for the moment we will simply assume that the Stanton x will vary as c times Reynolds x to the power of minus n and that is substituted here. Then for u constant u infinity and t w minus t infinity boundary layer that is let us say fat plate it will simply mean d delta 2 by dx equal to Stanton x equal to c u infinity x by nu raise to minus n and if we integrate that I will get delta 2 equal to c over 1 minus n u infinity by nu raise to minus n x raise to 1 minus n and if I use the idea that delta 2 is 0 at x equal to 0 that is right at the leading edge then Stanton x would simply become that. Stanton x would simply be function of delta 2 raise to n over n minus 1. Now, this is a very important relationship because we will be subsequently using this relationship for a variety of situations. So, we assume the validity of the last relationship regardless of the previous history of the boundary layer. Then the integral energy equation becomes d delta 2 u infinity t w minus t infinity of d by dx and then that is equal to u infinity t w minus t infinity c 1 minus n by c u infinity delta 2 by nu n over n minus 1 and if we were to integrate this equation you will get c nu raise to n 1 minus n u infinity t w minus t infinity 0 to x u infinity t w minus t infinity 1 over n. So, in effect then I can calculate delta 2 for any arbitrary variation of u infinity n t w minus t infinity with respect to x. If I use now the previous Stanton x delta 2 relationship of this type then you can see I can get Stanton x equal to c nu n t w minus t infinity raise to n over n minus 1 divided by this integral and this essentially allows you to calculate Stanton x for any arbitrary variation of free stream velocity and wall temperature. Now, assuming the flat plate data for c equal to 0.0284 Prandtl raise to minus 0.4 and n equal to 2 you will recall I showed this to be the case in the first slide here that Stanton x can actually written as 0.0286 and all x to the power of 1. Essentially if I say c is equal to 0.0286 Prandtl raise to minus 0.4 and n equal to 0.2 then that is what I have done here. I would get a relationship for Stanton x equal to this quantity and you will recall that u w is u infinity is actually the pressure gradient. Although these constants are strictly valid only for flat plate, but this Stanton number expression has used the same value of c and n and you get this relationship for Stanton x. Now, Crawford and K's have actually experimented with the constant wall temperature boundary layers in which the free stream varies arbitrarily. That means only free stream varies arbitrarily not T w minus T infinity and their experimental data fit this correlation very well. This is the pressure gradient parameter nu infinity square du infinity by dx is the pressure gradient parameter less than tenders to plus 6 and the Stanton number evaluated from this relation this integration and that evaluated from this agrees extremely well. Therefore, we can say that use of c and n in this manner appears to be quite valid even for situations in which u infinity is not constant or T w is not constant. So, I am with this experience we now move forward and look at situations in which there is suction and blowing. So, now again for a flat plate and T w minus T infinity constant Crawford and K show that for finite v w Stanton x when v w is finite is equal to divided by Stanton x when v w equal to 0 can be written as l n 1 plus b h by b h and b h is nothing, but our blowing parameter, but this time based on Stanton x. We shall derive this relationship later on when we consider mass transfer problem in which the suction and blowing would be viewed as a problem of mass transfer and the b h then is defined in this fashion and Stanton x v w then and substituting for Stanton v w equal to 0 which is 0.0284 Prandtl point power minus 4 and all x to the power become into l n 1 plus b h by b h that is what become. And the energy equation with for a flat plate where u infinity is constant and T w minus T infinity equal constant will energy equation would simply be d delta 2 by dx Stanton x v w plus v w by u infinity, but if I substitute this for Stanton x v w sorry if I substitute for v w by u infinity is equal to b h into Stanton x v w then you will see it simply becomes that 1 plus b h or the total expression then can be written in this fashion Reynolds x to the power of minus. So, there is l n 1 plus b h multiplied by 1 plus b h divided by b h Reynolds x to the power of minus point minus point 2. If b h was constant then the entire term here on the inside the bracket would be constant and it is not very difficult to integrate this equation using delta 2 equal to 0 x equal to 0 gives the integration gives r e x to the power of minus point 2 equal to this relationship multiplied by r e delta 2 raise to minus point 25 or using again the Stanton x Reynolds r e x relationship of the previous slide here. The solution can be written as Stanton x for a finite v w is written as 0.0125 Prandtl raise to minus 0.5 Reynolds delta 2 raise to minus 0.25 1 plus b h raise to 0.25 and then this factor raise to 1.25 of course, this integration and was made possible by assuming b h equal to constant. Now, like in the previous case we shall assume the validity of this relationship between Stanton x and delta 2 even when b h u infinity and T w minus T infinity vary arbitrarily with x. So, that is what I have done next slide. So, for this case then the integral energy equation we will read like this into all this quantity into that and then if we were to integrate as in the previous case you get Stanton x will be equal to 0.0284 Prandtl raise to minus 0.4 then b h now varies with x. So, that is included and you can see I can perform this integration for any arbitrary variation of u infinity T w minus T infinity and b h. So, that I can get variation of Stanton x. Crawford and K's have show remarkably good fit to experimental data and predictions using mixing length. So, they had a situation in which a highly accelerated boundary layer was considered with v w present and it was changing arbitrarily and therefore, the problem was solved by mixing length model and predictions were obtained. Experimental data were available for the same case and then this expression this approximate expression derived from Ambrox procedure was used and very very good agreement was shown between experimental data and the correlation as well as predictions using mixing length. With this I end the methods in which law of the wall is used or integral energy equation is used and now I turn to the differential equation based methods which are and it so happens that you can use similarity type methods for turbulent boundary layers as well. And governing equation for the temperature boundary layer would read as u dT by dx v dT dv dv d y plus nu times sorry equal to plus nu d by d y b Prandtl dT by d y where b Prandtl is alpha by nu and alpha T by nu which is Prandtl raise to minus 1 plus Prandtl T raise to minus 1 into nu T plus and nu T plus as you remember is nothing but nu T by nu and nu T would be given by Prandtl's mixing length as a function of y the distance from the wall. In lecture 29 I introduced the similarity variables to be used for turbulent boundary layers and if we use the same similarity variables then the equation for turbulent heat transfer boundary layer would be given by this d by d eta into b Prandtl into theta prime plus f theta prime plus 2 n over m plus 1 f dash 1 minus theta equal to again a function of x on the right hand side. This was also found in case of momentum equation that you do get things on the right hand side which are functions of x where the things on the left hand side are essentially functions of eta. So, here m is the pressure gradient parameter defined as x over u infinity d u infinity by dx n is the parameter related to wall temperature variation. So, n would be 0 of course, if T w was constant and theta is defined as T w minus T over T w minus T infinity and again like in the velocity boundary layer case you need to of course, f f dash are available already from the velocity boundary layer solutions and therefore, solutions for theta would be obtained by iterative method just in the manner in which the similarity solution for velocity was solved. The boundary conditions are of course, theta equal to 0 at eta equal to 0 and theta equal to infinity would be equal to 1. That completes discussion of the external boundary layers and the similarity method. You need to do is that every x you solve the left hand side by shooting method the right hand side is evaluated from values of available at the one step before it and so that the right hand side can be formulated as a constant for that step and one simply solves the left hand side again by shooting method. Using the intermediate solution for theta the right hand side is evaluated again and till convergence is obtained and we accept the solution at that position x and move to the next step. I now move to internal flows. So, we wish to use the wall law for pipe flow. So, if I write the T plus equal to Prandtl T into u plus plus p f infinity corresponding to central line then T cl plus will be that and u cl plus you will recall is nothing but u bar plus 1.5 by kappa plus p f infinity. So, where T cl plus now is would get defined in this manner T w minus actually it will get defined as T w minus T cl divided by q w, but if I multiply and divide by T w minus T bulk then you will see this gets T w minus T bulk over q wall T w minus T cl over T w minus T bulk multiplied by rho Cp and u tau that would be the definition of T cl plus. Now, I multiply by k and divide by k I multiply by u bar and divide by u bar then you will see this can be written as this is nothing but 1 over h the heat transfer coefficient into k divided by diameter which I have divided by and now again multiplied by. So, I get u bar d by alpha multiplied by u tau over u bar multiplied by T wall minus T cl divided by T wall minus T bulk. Now, k over h d is nothing but 1 over Nusselt number u bar d by alpha is nothing but Peclet number or product of Reynolds number and Prandtl number. So, these two factors are nothing but Reynolds Prandtl during Nusselt number u tau over u bar would be simply under root f by 2 the friction factor for a pipe flow multiplied by T wall minus T cl divided by T wall minus T bulk and hence equating T cl plus from this expression and from this expression that is the law of the wall then you will see I can write Nusselt number equal to Reynolds Prandtl under root f by 2 divided by Prandtl T into under root 2 by f plus 1.5 kappa plus P f infinity and this becomes T w minus T cl over T w minus T bulk. To use this relationship we would of course, need estimate of T w minus T central line divided by T w minus T bulk which we expect to be somewhat greater than 1 in a turbulent pipe flow because remember the temperature profiles are very flat inside the core of the flow and there are sharp gradients of temperature near the wall and therefore, the ratio of T w minus T cl divided by T w minus T bulk would be only slightly greater than 1. So, that is what we shall show now many times quite analogous to what is power law for velocity you assume validity of the power law even for temperature T minus T w divided by T central line minus T w equal to y by r raise to 1 by 7 and which we said would be equal to u over u cl then using definition of T bulk as you know which is simply integral 0 to r ut r dr divided by integral u r dr divided by r dr it is easy to show that T w minus T central line divided by T w minus T bulk would be about 6 by 5 which is close to 1 let us say or 1.2 1 to 1.2 remember this relationship is not absolutely exact, but we can take it to be between 1 and 1 it is a function of Reynolds number you will notice that for higher Reynolds number you need to take this as 1 over 9 whereas, below 50000 it can be taken as 1 over 7. So, as a result T w minus T cl over T w minus T bulk is actually a function of Reynolds number this factor turns out to be function of Reynolds number the higher the Reynolds number closer it gets to the value of 1 and likewise u cl by u bar is 1.22, but that is too high an estimate but nonetheless can be taken to be approximately solved close to 1 then you can substitute these values for u cl by u bar and I mean mainly you want this value, but that evaluation of that value requires this value. So, that is why I have quoted it the most widely used correlation for pipe flow which is considered to be very accurate with experiment like that is the one by Gnian Lenski and it reads like this n u equal to Reynolds minus 1000 Prandtl under root f by 2 into 2 by f plus 12.7 Prandtl risk 2 by 3 minus 1 and it is valid for gases to heavy oils 0.5 to 2000 and 2300 to 5 million Reynolds number. So, notice the similarity between the equation we have derived here. So, instead of Reynolds the Gnian Lenski correlation has Reynolds minus 1000 Prandtl number is still there under root f by 2 is very much there Prandtl T is perhaps taken as 1 and then there is a factor of 2 by f and then there is remember this is 3.66 in our calculation and then the P f infinity which we use from our correlation, but Gnian Lenski takes Prandtl T equal to 1 is also taking the ratio if you like of this to be nearly 1 and then this is the function of the Prandtl number. So, this most widely accepted correlation has a form which is very similar to what we derived from the temperature law. We can also apply analogy method for pipe flow. So, for fully developed pipe flow for example, dp dx is equal to constant hence the axial momentum equation and its consequences are that 1 over r d r tau thought by dr would be equal to minus dp by dx and integration would give tau thought by tau wall equal to r by r which meaning the total stress divided by the wall stress is a linear function of r by r and this we had shown from the experimental data in an earlier side and replacing r equal to r minus y the capital radius minus y as a distance from the wall it will be 1 minus y over r. So, tau thought is equal to rho times nu nu t du dr and that would become minus rho nu plus nu t du by dy and therefore, 1 plus nu t by nu can be shown to be 1 minus y plus by r plus divided by du plus by dy plus. So, that is what I shall use to substitute in our from slide 2 here I would use that to substitute for du plus by dy plus and nu t by nu sorry I would use that to do this 1 over nu t by nu then you will see that dt plus by dy plus now is 1 minus y plus by r plus over 1 over Prandtl number 1 minus y plus by r plus over du plus by dy plus minus 1 over 1 plus 1 over Prandtl t raise to minus 1 and now we will evaluate du plus by dy plus for each of the three layers that is sublayer the transitional layer and the fully turbulent layer and this is what the integration gives you for three layer law. Tsl plus minus 0 equal to that as before this is also everything is as before for an external boundary layer and the Tcl plus minus T transitional layer would give you 2.5 Prandtl t ln r plus by 30 again for Prandtl greater than or equal to 1, but also gases can be included which is very close to 1. If I add these three I would get Tsl plus before adding of course Tcl plus by definition is Reynolds Prandtl over nu over this this we showed on the previous slide if I equate this equal to the summation of the three I would get the expression for the Nusselt number where r plus is now expressed as r e by 2 under root f by 2 and therefore Nusselt number would become r e Prandtl f by 2 T wall minus Tcl T wall minus T bulb divided by the entire quantity here and this will be the second expression we have got now for representing Nusselt number. Now of course you have been using quite routinely the Dittus-Bolter correlation for very simple as nu equal to 0.023 r e raise to 0.8 and Prandtl raise to N where N is equal to 0.4 for heating case and N equal to 3 for cooling case. In chemical engineering literature most often the correlation due to slisher and Rouse is used and it says Nusselt number is equal to 5 plus 0.015 Reynolds raise to A and Prandtl raise to B and this is valid for 0.1 to I mean 10000 and Reynolds number from 10000 to a million and A is equal to B is made a function of Prandtl number and B is made again a function of Prandtl number. So these are some of the experimental correlations which are routinely used. Now as I said all this in our analysis is actually for Prandtl number greater than 1 but for liquid metals Nusselt number is correlated in this fashion A plus B raise to 0.85 Prandtl raise to 0.93 and where A and B take these values. Now remember I said in my lectures when I introduced turbulent flow that Nusselt numbers do not respond to the boundary conditions like in laminar flow. Nusselt numbers in turbulent flow are relatively insensitive to whether it is a constant wall heat flux boundary condition or constant wall temperature but that argument applies only to gases and situations in which Prandtl number is greater than 1. When you come to liquid metals they develop character the thermal boundary layer thickness now enters the transitional layer and also inner part of the turbulent layer many times depends on the value of Prandtl number and in such situations the boundary condition begins to influence events even in turbulent flow and that is what is shown here that the constants A and B should be modified as shown here for q wall equal to constant and t wall equal to constant. So what it shows is again the heat Nusselt number for constant wall heat flux will be greater than the Nusselt number for constant wall temperature a circumstance very similar to that found in laminar flows. By taking temperature ratio as 1.1 and Prandtl t equal to 0.943, 887 and you recall I had given you the relationship that Prandtl t actually can be modeled as Prandtl t equal to 0.85 plus 0.0309 into Prandtl divided by Prandtl plus 1 sorry this should be Prandtl 1 divided by Prandtl. Then you will see that this relationship assumes for Prandtl greater than or equal to 1 will give you for example, at Prandtl equal to 1 this will simply make it about 0.06 and Prandtl t will be about 0.91 for Prandtl equal to 1. For very large Prandtl numbers also it will be around 0.88 because this ratio would be 1. But now imagine for liquid metals where Prandtl number is let us say of the order of 0.001 then you will see this quantity becomes 1.01 divided by 0.001 or nearly 1000 and therefore, this will become almost equal to 31.0. So, Prandtl t can be very large for very small Prandtl number that is liquid metal. But I am presently considering cases in which only say from gases to organic liquids. So, Prandtl 0.5 to 5 and 25 and I have calculated Prandtl t using the relationship that I just showed. So, it is 0.943 here for 0.5 Prandtl t equal to 0.87 for Prandtl equal to 5 and 0.82 for Prandtl equal to 25 and the temperature ratio I have always taken as 1.1 in each of those cases then just see what happens. So, at Reynolds number 3000, Ginalanski correlation predicts 8.13 whereas Ditters-Volter predicts 10.5, Sieger-Rausch-Fletcher and Rausch produced 11.9 whereas the analogy method produces 10.3 and similarly at 10,000 and 50,000 and 1 lakh and 1 million then you will see that Ginalanski and other correlation predict very well are comparable. But Ditters-Volter is way out as you can see for higher Reynolds numbers, Ditters-Volter relationship predicts much higher than predicted by any of the other correlations for gases. What about water? Something very similar you will see that sufficiently higher Reynolds numbers say 10th this correlations are close to each error but again at higher Reynolds number Ditters-Volter under predicts the Nusset number compared to other correlations whereas for gases it was over predicting for water it is under predicting and that trend continues even for organic liquids. So, for Prandtl greater than 1 apparently Ditters-Volter under predicts whereas for Prandtl less than 1 it over predicts compared to the correlations that are well accepted Ginalanski in particular is very well accepted correlation. So, from these relative comparisons we say that the correlations for pipe flow can be applied to of course, a non-circular ducts by evaluating F Reynolds and Nusset number based on hydraulic diameter. This you have routinely you done in your undergraduate work of course, the theory to support this assumption requires solution of Reynolds stress equations so that the secondary flows predicted in the cross section by the Reynolds stress model can actually explain why hydraulic diameter concept works for non-circular ducts. The easy to use Ditters-Volter correlation although it is very easy to use actually over predicts Nusset number for Prandtl greater than 1 and under predicts for Prandtl greater than 1. For complete description of flow and heat transfer involving complex ducts, strong and changing strain rates due to body forces and it is best to use CFD techniques with two equation or stress equation models. This completes our discussion on turbulent flow and heat transfer and henceforth now I will begin with convective mass transfer.