 In the previous two lectures using phenomenology, we were able to derive near universal laws for both velocity as well as temperature variations as distance from the wall. The task now is to use these laws to predict friction factors and Nusselt numbers. In today's lecture however, I will concentrate only on the friction factor and I will do so to firstly to predict the friction factor or the coefficient of friction as you say in external boundary layers using integral method in purely turbulent boundary layers. But then our interest would turn to using integral method for complete boundary layer development that is starting from laminar through transition to turbulent boundary layers. We shall also see how complete laminar to turbulent and through transition boundary layers can also be predicted by similarity method. And then I will turn my attention to the internal flows like flow in a pipe for example, and again use the law of the wall there to predict the friction factor. So, let us begin with the prediction of CFX for a purely turbulent boundary layer using integral method. We begin of course with the integral momentum equation which as you know is applicable to laminar, transition and turbulent boundary layers. I mean there is no separate integral equation because the differential equations have already been integrated from the wall to the infinity state. And the integral equation reads as d delta 2 by dx plus 1 over u infinity d u infinity by dx which is the pressure gradient parameter, delta 2 is the momentum thickness, delta 1 the displacement thickness, CFX is the coefficient of friction and V w is the wall velocity. And of course definitions of delta 1 and delta 2 you will readily recall that the delta 1 for example is 0 to delta 1 minus u over u infinity d y and delta 2 likewise is 0 to delta u over u infinity into 1 minus u infinity d y for a constant property boundary layer. Now, in order to solve really the complete problem of laminar transition and turbulent boundary layers what is needed is the expressions for u over u infinity in each of these three regimes. Now, of course the laminar one you already know that one you already know it was given by a polynomial. We shall see how transitional one can be handled and the turbulent one also is something I shall elaborate on. So, before going to complete laminar transitional and turbulent boundary layers what I am going to do first of all is to consider only a fully turbulent boundary layer as if it was originating from x equal to 0 that is the leading edge. For example, if the leading edge was rough then there is no possibility of laminar flow developing immediately and the flow becomes turbulent boundary layer right from the origin itself. On the other hand it may originate from the end of transition with a finite thickness of the boundary layer, but that case we will see a little later. So, let us turn our attention on the next slide to the issue of calculating delta 1 and delta 2. Now, first of all you must recognize that delta 1 and delta 2 as you will know require integration of these from 0 to delta, but as you already remember that delta plus would be of the order of 800 to 1000 or 1200. Whereas, laminar sublayer would extend only up to y plus of 5 and transitional layer would extend up to y plus of 30 and the fully turbulent layer begins at 30 and goes up to 800 to 1000 or 1200 or whatever depends on the Reynolds number at that location. Therefore, even if I ignore this 0 to 30 region it would not matter to the evaluation of delta 1 and likewise of delta 2. In other words the low velocity regions are dropped out and therefore, we essentially compute delta 1 and delta 2 using turbulent velocity profiles only. So, now in the fully turbulent profile is a logarithmic law for the inner layer, inner part of the turbulent layer and plus the law of the wake as we saw in the in the previous lecture. However, if we were to use this law there are two difficulties. First of all because of the logarithmic expression the evaluation of integration becomes somewhat difficult and secondly CFX is not very easy to extract from the universal logarithmic law of the wall. So, there are two difficulties simply because we cannot use that law to evaluate shear stress as mu du dy equal to 0 and therefore, shear stress has to be specially evaluated. On both these accounts the use of the universal logarithmic law plus law of the wake is somewhat not less fruitful and therefore, what is done is a shortcut. Now, instead of for a smooth impermeable wall at any rate that is V w equals 0 we assume a power law instead of a logarithmic law plus law of the wake we assume a power law as u plus equal to a times y plus raise to b and a and b can be functions of Reynolds number, but I have given here some values of Reynolds number some values of a and b which are usually found to be quite good in practically encountered turbulent boundary layers. So, a is 8.75 and b is 1 by 7 and this is called the one seventh power law. Now, this law actually fits the experimental data up to y plus of 1500 quite well in fact, better than the log law as you will see in the next slide. So, here is the plot of u plus versus y plus and y plus is on the log scale. Here are the experimental data for 0 favorable and adverse pressure gradients. Of course, our interest is now only in turbulent boundary layers because that is where we are going to integrate things and you can see that the dashed line is the power law expression and the solid line is the logarithmic law plus the law of the wake. Now, you can see that the dashed line in fact predicts the velocity profile at least in 0 pressure gradient quite well right in up to the wake region whereas in the wake region the logarithmic law does not really predict that well the velocities. What we see is that we have some confidence in use of the power law in as much as it will not influence the calculations of delta 1 and delta 2 in spite of these little departures that you see here. Essentially, our law u plus equal to a y plus raise to b is essentially u over u infinity equal to y by delta 1 by 7 and if I substitute this expression in our expressions for delta 1 and delta 2 then you get ready evaluations as shown here delta 1 by delta 2 will be 1 by 8 or 0.125 delta 2 by delta would be 7 by 7 2 or equal to 0.097 and the shape factor which is the ratio of delta 1 by delta 2 is equal to 1.29. Now, this is quite characteristic of turbulent boundary layers that they usually have much smaller shape factor than in laminar boundary layers you will recall that in laminar boundary layers we had h of the order of 2.5 whereas in turbulent boundary layers it is of the order of 1.3 or so on. Now, in order to evaluate the CFX part CFX that you need here essentially that it equals CFX by 2 equals tau wall by rho u infinity square and that can be evaluated now quite easily from this expression u over u tau equal to 8.75 delta u tau by nu. So, you will see if I multiply this both sides by u tau I get u tau raise to 8 by 7 equal to the rest and therefore, I can reform reorganize this equation because u tau as we know is square root of tau wall by rho and therefore, CFX by 2 tau wall by rho u infinity square would simply become 0.0 225 u infinity delta by nu raise to minus 0.225. Now, if I replace delta in terms of delta 2 using this relationship 7 by 72 then it will become 0.0125 u infinity delta 2 by nu raise to minus 0.25. Now, both these expressions where either in terms of delta or delta 2 are very good approximations to mildly adverse pressure gradients through to very highly favorable pressure gradients and up to about Reynolds x of 10 raise to 7 or 10 million. So, we can confidently use CFX expression CFX by 2 in this manner and use h delta 1 by delta 2 equal to 1.29 then you will see that the momentum equation would become like this d delta 2 by dx this is the CFX term in terms of delta 2 and this is really 2 plus delta 1 by delta 2 which is h. So, which was 1.29 so that becomes 3.29 into delta 2 by u infinity du infinity by dx. If I transfer this term on the left hand side then you will see I can reorganize it as d by dx of u infinity raise to 4.11 into delta 2 raise to 1.25 equal to 0.0156 nu 0.25 u infinity raise to 3.86 and if I integrate that then I would get u infinity 4.11 delta 2 raise to 1.25 at any x would be equal to the same quantity at the lower limit of integration plus 0.0156 nu raise to 0.25 0 to x u infinity 3.86 dx. Now, if the turbulent boundary layer originates at the leading edge where so that the x in is 0 the lower limit of integration is 0 then I can readily evaluate delta 2 for any arbitrary variation of u infinity with x because this integration can easily be done either by hand or by numerical integration knowing of course, the variation of u infinity with respect to x. Then once you know delta 2 then you can evaluate recover your CFX from this expression that we have already written on the previous slide. So, as long as the pressure gradient is mildly adverse through to highly favorable pressure gradient then of course, all this derivations are applicable. Now, suppose u infinity was constant that is flat plate boundary layer or pressure gradient being equal to 0 which means that term is 0 then you will see u infinity raise to 3.86 would simply come out of the integration and if you can readily show then that CFX then would be equal to 0.0574 Reynolds x raise to minus 0.2. Now, this is an expression you have routinely used in your undergraduate work for calculating skin fiction coefficient variation with x in a flat plate boundary layer and we have recovered that through integral method. What about cases when the pressure gradient is highly adverse? We said the what we have done so far is valid only from mildly adverse pressure gradient to highly accelerated. So, let us look at the lower limit of the adverse pressure gradient when the adverse pressure gradient is high how do we do that? We again recall that integral momentum equation would be like this and also now we would also like to include the effect of V w should it be present that is suction and blowing are present. Now, in this case what is done is H the shift factor is tuned and the manner in which it is tuned is quite empirical whatever I am going to say now is very very empirical but it has been validated through a series of computations by earlier workers and it says that the shift factor would be made a function of firstly G where G is a function of beta which is a pressure gradient parameter as well as suction and blowing parameter. So, beta is given by that B is given by that and secondly it would be function of the skin friction coefficient C f x itself. Now, how do we evaluate C f x for example, C f x would be related to C f x for d p d x equal to 0 into 1 plus 0.2 beta raise to minus 1 this is a correlation given by Crawford in case to account for the effect of pressure gradient the parameter beta. But there is also another one given by Ludwig and Tillman and in that C f x itself is made a function of H. So, 0.246 into 10 raise to minus 67 at H into Reynolds delta 2 raise to minus 0.268 both these are meant for smooth surfaces, but there is also another recommendation in which C f x here is to be written in this fashion 3.336 into logarithm of it 54.6 delta 2 divided by equivalent surface roughness raise to minus 2. So, depending on the problem and hand one would use either the smooth surface expression or the rough surface expression these expressions are valid for 1.43 minus 1.43 beta plus B varying from minus 1.43 to plus 12. Now, of course, because H is a function of C f x and C f x is a function of delta 2 as you know close form solutions cannot be obtained in this case and therefore, you need to do iterative solution of the integral momentum equation. This equation has to be solved iteratively at each x. So, now, let us turn our attention to complete boundary layer prediction in which we begin with laminar regime and for laminar regime of course, you know all this very well you do not really need computer programming because you already have the solutions, but since we are going to compute through transition and turbulent layers we would include the laminar calculations in a computer program based calculation and that is what I am showing now. So, if you know variation of u infinity and v w x then of course, you would evaluate delta 2 laminar x suffix l represents laminar and therefore, you evaluate kappa the pressure gradient parameter based on delta 2 and evaluate H which is delta 1 by delta 2 and then the shear factor which is delta 2 by delta 4. So, once you got your delta 4 out of this then readily you can evaluate the C f x for the laminar part of the boundary layer. We continue this calculation until the onset of transition and the onset of transition is recognized either by using Sebesi formula or by the Fraser and Milne criterion which we discussed in the previous lecture. Now, both these formulae give you start of transition as well as the criterion for end of transition. So, as soon as we come to the end of laminar boundary layer we already know the total length that would be occupied by the transitional layer x t e minus x t s. So, now, let us go to the turbulent transition. Now, in the transitional part again we need a velocity profile which is taken as this u over u infinity transition is taken as 1 minus intermitten c gamma into u over u infinity sub l which is laminar which as you know was a polynomial plus gamma times u over u infinity of turbulent. Now, how to tackle the turbulent part we will see in a short while gamma as you know is an intermitten c parameter which varies with x and since you know x t s and x t e already you can readily evaluate psi and therefore, intermitten c distribution over the transitional length is already known. If you substitute this transitional distribution here then using u over u infinity transition would give you delta 1 transition equal to 1 minus gamma delta 1 laminar plus gamma times delta 1 turbulent. Delta 2 transition again has a similar looking term multiplied by 1 minus gamma plus gamma times gamma of delta 2 t minus 1 minus gamma of delta 1 t. So, the delta 2 evaluation is somewhat involved plus it has a term which requires integration of product of laminar and turbulent boundaries. As you can see therefore, that closed form solutions are again extremely difficult to obtain and computer solution becomes very very valuable. Then the h transition for the transitional part h transition is simply evaluated as delta 1 transition divided by delta 2 transition and that is what I have shown here. The c f x in the transitional regime would evaluate as 1 minus gamma of c f x due to laminar profile and gamma times c f x due to turbulent profile. Now the turbulent profile itself is taken in this manner u over u infinity t is equal to y over delta t 1 over n which means again a power law profile, but the value of n is taken as 2 over h t minus 1 which is the shape factor of the turbulent part minus 1 which gives you delta t equal to delta 2 t equal to h t plus h t plus 1 divided by h t minus 1. So, if I were to substitute this you can get delta 2 t for the turbulent part as a function of delta t or the other way round. So, these are the evaluating equations for the transitional part. However, a little trick is required to carry on the computations through laminar, through traditional and turbulent layers and that is what we shall see on the next slide. So, first of all in order to be able to compute the transitional the turbulent contribution through transitional layer, what is done is that since we know x t e already we somehow back extrapolate the turbulent development and identify what is called as a virtual origin of the turbulent boundary layer x v o and x v o minus start of transition x t s that is this length is taken as 12.6 percent of the total transitional length. So, that is the first thing that is done is to identify the virtual origin. So, in order to calculate delta 1 t and delta 2 t and so on and so forth we first of all define x dash equal to x minus x v o the virtual origin a new variable x dash and commence the turbulent calculations at x dash equal to 0 that means the laminar boundary layer calculations are continued in fact up to x v o there is no contribution of turbulent at very low intermittency levels. We simply have no contribution of the turbulent boundary layer, but the turbulent boundary layer contributions begin from x v o onwards. Now, this would require some value of delta 1 and delta 2 t at these points x v o and that is taken as delta 2 t at this quite empirically is taken as 20 percent of the delta 2 as computed from laminar velocity profile h t is taken as 1.5 and c f x of the turbulent part in this region at this point rather is taken as 0.99 of c f x of laminar. So, these are all empirical settings, but that have been found to be quite good in predicting flows through the transitional region. So, now both the turbulent and laminar velocity profiles are used as shown in the previous slide and we calculate the c f x of transition delta 1 of transition delta 2 of transition and so on and so forth and therefore, the h of transition by calculating using both laminar and turbulent velocity profiles. So, that is how we continue up to x t e because x t is already known before we started the transitional calculations. So, at x dash t e equal to x t e minus x v o appropriate specifications then are that the start of fully turbulent calculation delta 2 t would be equal to delta 2 t transition then h t would be equal to h transition at this point and c f x t would be equal to c f x transition at this point. And therefore, now laminar flow calculations are completely stopped and we only integrate the turbulent part the turbulent part and in the turbulent part we use again the power law as set in on the previous slide. So, for x dash greater than x dash t e turbulent integral momentum energy equation is solved iteratively as described why the iterations are required particularly when h is a function now of pressure gradient and the suction and blowing parameter as well. So, we just use this iterative method to compute the flow through the turbulent layer. Now, in order that the number of iterations required is small typical advice for the step size is that delta x here should be taken as one quarter of the momentum thickness at the previous step in the boundary layer. So, it is taken as one fourth of the momentum thickness and usually converges obtain in less than four iterations per step. So, very elegance computation and very fast computation can be performed using this method right from laminar through transition two turbulent boundary layers. So, now let us see some example and in what I am going to do is to consider flow over a family of ellipses ellipses is a wonderful family because it has the minor axis 2 b it is a major axis is 2 a and if b by a is greater than 0 then you get an ellipse. If it is exactly equal to 1 you will get a cylinder and if it is equal to 0 that is if b is equal to 0 you simply get the flat plate. So, in this case you specify a b approach velocity u approach the density and viscosity and therefore, the Reynolds number based on u approach and the major axis length 2 a divided by mu would be the Reynolds number in case of a flat plate 2 a would be the length of the flat plate. So, 2 a would be simply equal to l. Now, for family of ellipses u infinity that is the variation of u infinity along the periphery of the ellipse would be u approach into 1 plus b by a cos beta where cos beta is evaluated like this. At each point on the surface you draw a tangent wherever it intersects the x axis you note the angle beta and that is you take the cosine of that to get the free stream velocity u infinity variation with respect to x. S x is the distance along the surface of the ellipse whereas, x itself is the distance along the axis as shown here x is the distance along the axis whereas, S x is measured from the forward stagnation point and along the surface of the ellipse. So, we shall see all the 3 cases 0 1 and b by a which would be less than 1 and greater than 0 which would be the ellipse. So, let us take the first case I have taken the flat plate case. So, u infinity is constant as you can see the Reynolds number based on length is 10 raise to 7 and what do I find? I find that the start of transition is identified at about 0.31 x transition divided by l is identified point transition. End of transition is identified as 0.4342 as you can see here and the virtual origin is bit closer to start of transition is 0.325 x v o by l equal to 0.325 and you can see now that the C f x in fact decreases along the length in the laminar boundary layer up to x v o and then it suddenly rises in the transitional layer because both turbulent and laminar contributions have been begun and it begins to fall again and meets the end of transition here and then the turbulent part of the skin friction factor continues. Now, I have multiplied here C f x by 500 to get everything on proper scaling done. The shape factor is of the order of 2.5 in the laminar boundary layer laminar case then it dips and slowly goes on decreasing and becomes almost equal to 1.29 as we had shown in our calculations earlier. Here are the rate of growth of delta and we always said that all the integral thicknesses vary as x to the power of 0.5 in laminar case whereas, they vary as x to the power of 0.8 and that is borne out very nicely in this figure. So, this is how you can complete the calculation of a flow over a flat plate. Now, let us look at flow over a cylinder. You can see now the free stream velocity variation is like this. This is the forward stagnation point is somewhere here where the velocity would be 0 of course, but on a log scale you will not see 0 and then it accelerates till the top of the cylinder and then decelerates as you go along here. So, in other words 2 a is the diameter then the Reynolds number based on diameter is 10 raised to 7 and again I have plotted values of C f x multiplied by 500. Now, very interestingly at this Reynolds number you see the C f x falls in the laminar regime, but it does not turn into transition at all and instead laminar separation occurs at a somewhat after the midpoint. To show this what I mean is the following that if this is the cylinder and if this is the approach velocity then the laminar boundary layer begins to develop here and pass the midpoint somewhere here it passes through the boundary layer. So, it separates. So, you have separation taking place it is very close to the wall that separation and what we do now is to cheat the flow and say laminar separation equal to start off turbulent boundary layer. So, we simply start compute assuming that the now the boundary layer will turn turbulent after this point of separation and continue the calculations. So, that is what I have shown here that is what I have shown here and you can see there is an abrupt shift from a very low C f x which goes down because of separation to 0 and then very quickly rises up again to the turbulent part of the calculation which the C f x now again decreases even in the decelerating part of the flow, but now what happens is at x by d equal to 0.812 the boundary layer separates in the turbulent part which because of the deceleration it separates right here that is at the back of the somewhere at this point the turbulent boundary layer also separates. Now of course, we cannot continue the calculation and the calculations are stopped of course, this neglect of this much region say from 0.812 to 1 would not affect the integral quantity of C f x that is our belief that 0 to s which is equal to pi times d by 2 1 over pi d by 2 will be equal to C f average and that value will not be severely affected by neglect of this calculation and therefore, we can still recover the drag of a cylinder from integral method. So, laminar separation in this case occurs at 0.597 and it is taken as turbulent reattachment is taken as turbulent reattachment straight away. So, in other words the transitional length is simply absent here and you simply begin the turbulent calculations immediately after laminar calculations have yielded C f x going down to 0 very interesting phenomenon no transition at all. Now, let us look at ellipse as I have done here and I have chosen b by a equal to 0.5, but you could choose any other again the flow accelerates till top of the ellipse and then decelerates, but the region here is much more flatter. So, there is a considerable region of 0 pressure gradient as it were in case of flow over an ellipse again you will see the C f x value here decreases and it does encounter transition at x t s by major axis equal to 0.3588 that is where it encounters transition the end of transition therefore, is predicted at 0.4284 and the virtual origin therefore, is 0.3678 or the 12.6 percent of the x t e minus x t s length and there you will see the C f x now increases again and then starts declining, but then at 0.958 it experiences again turbulent separation and the computations cannot be taken any forward further because the separation has occurred. So, you can see C f x behavior is quite different from that for a cylinder, but in both cases the turbulent separation is encountered. Here laminar transition length finite laminar transition length has been encountered whereas, then the flow over a cylinder we found no transitional length and the laminar separation was straight away taken as turbulent start of turbulent layer. The shape factor again is of the order of 2.5 in the laminar regime and then it decreases to around 1.3, 1.4 so of that order computations of this type are very valuable because one can now take the case of flow over let us say any blade or so. Flow over a blade and one can begin calculations on the pressure on the suction side for example, as well as on the pressure side this is the pressure side, this is the suction side and as long as you know variation of u infinity as a function of x you can compute the development of the boundary layer you can identify where transition occurs and you can identify the length of transition on such blades and so on and so forth and should any separation occur then of course, that is undesirable and therefore, one would like to adjust the shape of the blade in such a way that no separation is found near the trailing edge. One can do the same way on the pressure side, so calculations of this type have been used in the pre CFD days these type of calculations were extensively used to design and shape the compressor and turbine blades. Now, we turn to the differential method for calculation of CFX through all the three layers and the time average RANS equations for the boundary layer would be like this. These are the convection term, this is the pressure gradient term and this is the diffusion term where I have taken the nu out and that which would make it 1 plus nu plus t. So, this is the turbulent part turbulent viscosity divided by nu and the variation of nu t by nu as a function of y is given by Prandtl's mixing length for example, that we have already seen. Now, in order to convert these equations, this differential equation where this is a function of y to an ordinary differential equation, we invoke these parameters in the so called similarity parameters in which we take L and V as some reference velocity scales then eta is defined as y into u infinity to nu L V psi where psi itself is 1 over L V 0 to x u infinity x u infinity dx where u infinity is a function of x. The stream function is defined in that fashion and the pressure gradient parameter beta is defined in this fashion. So, if you make use of this definition of psi and substitute here and calculate the gradients and so on and so forth in the usual manner say about 2 pages of algebra, you can show that this equation would transform to d by d eta of 1 plus nu t plus f double prime plus f f double prime plus beta into 1 minus f dash square equal to not 0, but 2 psi into f dash d f dash by d psi minus f double prime d f by d psi. These are the inconvenient terms because these are functions of psi and psi itself is a function of x. In other words, our equation is not ordinary differential equation. All we have done is separated out the eta dependent variables on the left hand side and the eta n x dependent variables on the right hand side. The boundary conditions of course will be f psi 0 equal to 0 because we are assuming v equal to 0 at the moment f dash psi 0 because u itself is 0 and f dash infinity would be 1. Now u infinity x is a prescribed arbitrary variation in such a formulation because it allows you to handle any arbitrary variation of u infinity where and beta simply takes different values at different actual stations. So, how do we solve such an equation? In other words you can see now the similarity method has been made amenable to arbitrary variation of u infinity. How do we handle such an equation which has all the left hand side is a function of eta only whereas all the right hand side is a function of both x eta as well as psi or x. The equation of the previous slide can be used for flow over an ellipse for example in which u infinity varies arbitrarily with x in this manner and it will have nu t plus equal to 0 in laminar range nu t transition would be 1 minus gamma plus gamma time nu t plus in the transitional region and of course in the turbulent regime gamma would be 1. So, that would be nu t plus would be simply nu t plus. So, we can use the previous slide equation to calculate flow over a family of ellipses but presently what I am going to do is let us consider what I call equilibrium boundary layers in which u infinity has this special form of variation the wedge flow variation u infinity equal to c x to the power of m. In which case if I use this expression in all these definition for u infinity everywhere then you will see the eta variable would become y into under root of u infinity by nu x m plus 1 by 2 psi would become that and the equation itself would be something like this with this as the pressure gradient parameter and now you can see x explicitly appearing f dash d f dash by d x minus f double prime f d f by d x and at x equal to 0 sorry at y eta equal to 0 you have f 0 equal to 0 that is v equal to 0 as well as u equal to 0 and f x infinity equal to 1. How do we solve such a mixed equation in which the left hand side is a function of eta where at the right hand side is a function of both eta and x. So, let us look at that how we solve such an equation it is as follows. So, let us say we have a surface then at x equal to 0 that is your starting point and u infinity can vary anyway with respect to x and let us say you have chosen first delta x you have chosen the first value of delta x then what you do is you say at the first step which is very very small you assume as is shown on the slide here you assume x is very very small and therefore, you assume that this is 0 the right hand side is 0 in which case of course, this equation can be readily solved because the right hand side is 0 and it is a perfect ODE and therefore, you will generate values of f f dash and f double prime at the first step at the end of the first step at the end of the first step now you go to the second step here. So, let us say these are values at the first step now you calculate in order to go to the next step and solve the left hand side you need the right hand side. So, the right hand side which is equal to x times f dash d f dash by d x minus f double prime d f by d x we write that as x at 2 into f dash at 1 into f dash at 2 minus f dash at 1 divided by delta x. So, this is how we write. So, in other words the right hand side will have values of f and f dash 2 at location 2 that is a second location and therefore, since the left hand side is being solved also at location 2 you will have essentially an implicit equation in values of f at location 2 and therefore, you need iterations. So, once you do the iterations you can get the values. So, the procedure as I said here at the first step simply say right hand side is equal to 0 at subsequent steps the right hand side is evaluated from d f by d x equal to f x minus f x minus delta x over delta x and likewise for f dash itself and solve the third order ODE by Ranga-Kutta method and then since this is an implicit in f you will get new values of f f dash and f double dash and evaluate the right hand side again and solve the ODE again and if you find that at this iteration level the values of these quantities were same as the values at the previous iteration level then you say the convergence has been obtained you then move to the next step. Now, there are many refinements in order to reduce these kind of number of iterations and so on and so forth and a recent book by Sibasian acoustics gives you the details of that. Now, we turn to the internal flow that is flow in a pipe and we are going to make use of the log law to assume that it is the velocity profile is governed by the log law then we evaluate first the mean velocity u bar would be equal to 2 over r square 0 to r u r t r and if I change r to r minus y then you will see that u bar plus would become that. Now, I substitute u bar plus equal to log law but that integration becomes difficult and therefore, you simply make the y plus as a function of u plus e raise to minus 1 kappa u plus and simply ignore as before 0 to 30 region that is the up to the transitional layer you simply ignore that and assume that this the integration would not be affected if you only use the turbulent part of the law you will see e would be equal to 9.1.2 kappa equal to 0.41 and therefore, if I were to substitute that convert first of all d y plus to d u plus and then integrate then I will get that expression and which is equal to u plus at the central line minus 3 by 2 kappa which would be 3.66 plus 2 over a large quantity because r plus is of the order of 1000 or so e is of the order of 9.1 pi and even bigger quantity here and therefore, for all practical purposes I could drop both these terms and I would get this equal to approximately equal to central line plus minus 3 by 2 kappa and the C l is a central line. So, then what does that mean you get central line plus equal to u bar plus plus 3.66 and u bar plus as you know it would be equal to 2 by f plus 3.66 and that shows you what the magnitude of u C l by u bar will be in a typical turbulent boundary layer about 1.19 at 50000. Now, if I substitute in this u C l plus equal to kappa raise to minus 1 then e r plus we can organize u bar plus would be equal to that minus 3.66 or that would be equal to root 2 by f which is u bar plus and the same quantity again and that transforms to this implicit formula for prediction of friction factor as a function of Reynolds number. Now, you have used this formula in your undergraduate work, but the origin of that is in the logarithmic law near the wall. So, that is very important to recognize that the logarithmic law has been recovered for the friction factor. If you wanted to derive the explicit form of friction factor relationship like for example, this or this then you use the power law instead of the log law then again if you evaluate u bar plus then it will integrate like this. What I mean is u plus equal to a y plus raise to b then if I take a equal to 8.75 and b by 7 it predicts this expression which we routinely use for Reynolds number less than 50000 and if I use a equal to 10.3 and b equal to 1 by 9 then it will predict this expression which we routinely use for Reynolds number greater than 50000. If it is a rough pipe again we go back to the log law and use this log law which can also be written as kappa raise to minus 129.73 y plus divided by y re plus then the integration will show that f by 2 is equal to 2.5 ln d by y re plus 3 raise to minus 2. Now, this is a very interesting result that in rough pipes the integration of the logarithmic law of this type shows that the friction factor is no longer function of Reynolds number and that is what you had learnt in your undergraduate work when you studied Moody's chart that this is the laminar friction factor f versus ln of Reynolds up to about 2000 and then you have a transitional regime and then you have a smooth pipe this is the smooth pipe regime but when the pipes are rough you get friction factors varying like that where roughness increasing and these lines were almost horizontal meaning friction factor was no longer function of Reynolds number and that is something we have recovered in our derivation. So, everything that you have used in your undergraduate work for internal flows has also been recovered from our universal laws of the wall. In the next lecture I will use universal temperature law to predict the set number.