 Then we were discussing about the displacement thickness and the momentum thickness. Just to get some idea of how these concepts may be conveniently used, let us look into one example. Let us say that you have duct, this duct may be of a square shape or whatever shape. Let us assume that the section of this duct is a square just for convenience. And we are focusing our attention on 2 sections, 1 and 2 which are located relative to each other by a distance L along the axial direction. And our interest is to find out that what is the pressure gradient, average pressure gradient acting over this length. Since it is not a very impractical situation, this is a very practical situation actually if you somehow want to assess the flow past different bodies in something called as a wind tunnel. So if you have heard of something called as a wind tunnel, wind tunnel is a artificial place or a location where you have controlled flow of air and you may have or you may study the effect of flow of air past bodies of different shapes. So the key is that if you have a body of a particular shape and the body is moving in air, the fact that dictates that what should be the resultant force is the relative velocity between the solid and the fluid. So you may keep the solid stationary and make the fluid flow past over it. It is as good as having the solid moving through the fluid with the same relative velocity. So if you have a control arrangements or tunnel, in the tunnel you put the solid somewhere and have a control flow of air over the solid and measure various parameters. Some of the parameters of interest may be say pressure drop and let us say that at the sections 1 and 2, we know or by some measurement we have got an idea of the velocity profile within the boundary layer. So that is you know that what is u by u infinity as a function of y by delta. Remember here there is nothing called as u infinity because here it is a confined flow. So it may be u outside the boundary layer, u outside the boundary layer is the u corresponding to the central line maybe. So u by u outside the boundary layer as a function of maybe y by delta. So let us say that that profile is known. So then also it is known say that what is the total rate at which the fluid flow is entering the system. So that is also known because that is what is controllable. So to get a simple estimate of this delta P by L, what one may do is one may get an estimate of the displacement thicknesses at 1 and 2, delta 1 star and delta 2 star. What are these? Of course we have the corresponding expressions like as we have derived in the previous class. So if you either have a measured velocity profile or if you have some approximate velocity profile that is also fine but you should have a fair enough estimation of the boundary layer thickness and that is possible because if you have a velocity probe across the section you will find that beyond a certain distance from the wall the velocity gradients are not there and that distance you may estimate as the boundary layer thickness. And so this is how you get delta star. Once you get delta star for average predictions you may forget about the boundary layer characteristics altogether because then your situation is equivalent to that you have as if displaced the solid boundary by an amount say delta star so that the flow now is taking place within the narrowed domain but almost in an inviscid fashion. So then you may write if say if you consider that v1 is the velocity of the fluid in the core of the inviscid core through section 1 and may be v2 is the velocity of the fluid which is flowing through the inviscid core at section 2 then you may write from the continuity equation q equal to v1 into what is the equivalent a1. Equivalent a1 let us say that the square has a section a by a this is the section of the wind tunnel. So v1 into a-2 delta 1 star square right that is equal to that is how you relate v1 and v2. So this is one of the relationships you get not only that since these are inviscid cores and let us say that you have negligible difference in height you can always use the Bernoulli's equation if it is an incompressible flow. So p1 by rho plus v1 square by 2 is equal to p2 by rho see these p1 p2 are not very accurate or local p1 p2 these are like sort of average p1 p2 over a section but if you get these ones then you will see that you may write it in terms of v1 square and v2 square. You get another expression involving v1 and v2 and depending on whatever is measured and whatever is not measured by using these very 2 simple relationships you can find out the pressure difference or whatever it depends on what you measure and what you estimate from the calculations. But the basic principles is you may use these equations depending on what is known and what is not known. Again the important understanding is that this is not very accurate but for many engineering estimations that kind of local variation and accuracy of local variation is not what is required and what may be required is just an average pressure difference and in such cases this type of simple estimation is fine. Now the other important part is that see what we assume implicitly while looking into this problem may be that we assume that the boundary layer is thin. Now it might so happen that the boundary layer is thick obviously if the boundary layer is very thick the boundary layer theory will not be valid that is one of the things. But despite the validity or invalidity of the boundary layer theory you may have thick boundary layer only the corresponding theory that we have developed might not work but it does not mean that boundary layer may not be there if it is thick. But what is our expectation in these cases whenever we have put or whenever we have almost presumed that the boundary layer is very thin is the Reynolds number is very very large. So in these cases we are looking for experiments of large Reynolds number flows towards the end of such discussions we will come into such cases when the Reynolds numbers are very small and we will try to see that what are the corresponding demarketing features we will not go into the mathematical details of such cases. Sometimes the mathematical details of low Reynolds number flows although apparently they might be easier sometimes they may be more involved we will not go into that details but what we will try to see is that what is the characteristic demarcation between the very low Reynolds number flows and high Reynolds number flows. But one of the important hallmarks of high Reynolds number flows that we have identified that the boundary layer is very thin and that is what we have assumed for this case so that boundary layer if it is thick and it is thicker than the width of the of course it it cannot be thicker than the half of this one but like I mean if it is as good as almost the half of this one then this estimation will be more and more erroneous. Because this entire estimation was based on boundary layer equations which were based on the assumption that delta is much much small as compared to the axial length scale. So all those assumptions we may not be able to justify if we have a low Reynolds number flow. Now the second big question I mean this question always appear to my mind when I first was introduced into the boundary layer theory that if the boundary layer is so thin why not neglect it at all because after all if you have a flow domain say you have a some body and fluid flow past it now if the boundary layer is very very thin and the remaining parts does not understand the effect of the wall to get a gross effect of the flow why not neglect the boundary layer all together because at least mathematically when something is very small in comparison to many other things we have neglected in many times. So why we cannot do it for the analysis of the boundary layers to understand that let us look into a very simple problem. So we are trying to answer a question that why not neglect the boundary layer all together if it is so thin that is the question that we are trying to answer. That question we will try to answer not through a direct use of the boundary layer theory but through a separate example. Let us consider this as an example say you have 2 parallel plates and let us say that these plates have some holes through which fluid may flow vertically that means let us say as an example that some fluid with a velocity enters the gap in the top plate enters the pores in the top plate with a uniform velocity V0 and leaves through some pores in the bottom that is one of the things the other thing is let us say that it is basically a quit flow so that the top plate is pulled towards the right with a velocity relative to the bottom plate let us say that the velocity is ut. So this is and let us also assume that it is a fully developed flow. So basically flow between 2 parallel plates just like the quit flow that we have discussed while we were discussing the exact solution of the Navier-Stokes equation assume that there is 0 pressure gradient that means the pressures at the inlet and the outlet are the same. Only extra thing beyond what we considered in that problem is now you are having a transverse motion because of some suction effect which is sucking some fluid from the top and releasing it through the bottom. So there is a forced vertical component of velocity. We will later on see that this type of suction is not something which is very irrelevant it has some consequence with the boundary layer theory. We will come into that later on but right now just we are treating it as a simple problem which may be treated mathematically in a elegant way. So let us write the governing equations for this. So the governing equations to set that up let us set the coordinate axis say x and y like this and let us say that the gap between the 2 plates is h. So the continuity equation first let us write. So we are assuming 2 dimensional incompressible flow. So you have this as the continuity equation. Also the flow is fully developed. What is the outcome of the fully developed flow? So this will be 0 for fully developed flow. So from here we conclude that v is not a function of y. So from this can we predict that what should be v for this problem? v is equal to-v0 right because at y equal to h v is equal to-v0. Since v is not a function of y therefore it should be same for all values of y. So v is equal to-v0 that is what we get from the continuity equation. Next let us go into the momentum equation say x momentum equation. So let us simplify it first of all we are assuming that no pressure gradient is acting on this pressure at the inlet and exit are the same. It is fully developed flow. So you have the derivatives of u and the higher order derivatives all with respect to x vanish. So u is a function of y only. v becomes equal to-v0. So the governing equation that you get is v0 du dy plus nu d2u dy2 equal to 0. Now sometimes this analysis becomes a bit more insightful if we non-dimensionalize the parameter. So let us just use some non-dimensional parameters. Let us say u non-dimensional is equal to u divided by u of the top plate and let us say y non-dimensional is equal to y by h. So this governing equation will become v0 u top plate by h du dy plus nu u top plate by h square d2u dy2 all non-dimensional is equal to 0. So we may write it in a simplified form as d2u dy2 plus v0h by nu du dy equal to 0. v0h by nu is a sort of Reynolds number. If you recognize this first of all it is a non-dimensional number that you must recognize because all other terms are non-dimensional. So it has to be non-dimensional and it is something as the Reynolds number we can say just as a symbol we are writing this. So our objective will be to solve this equation. Solution of this equation is very easy actually because you just it is a second order ordinary differential equation and you may just use the standard technique like substitute u equal to e to the power my and then get the auxiliary equation and so on get the solution of this equation. We will not go into the solution in that way but we will try to have a different insight by going through a different method of solution. So what we will do first of all for writing convenience let us just omit the bar at the top. So we will just write d2u dy2 plus Reynolds number into du dy equal to 0. Remember these are all actually u bar y bar just for the convenience in writing we are dropping the bar. Now what we will do is we will consider to limiting cases. What are the limiting cases? One is Reynolds number is small. This is the first case that we will consider. Let us say that is equal to epsilon okay. When the Reynolds number is small or it is epsilon see this problem may be viewed upon in this way that you have say undisturbed state when there is no vertical component of velocity. So it is like a regular quet flow. Now you are adding a small disturbance velocity to it and that disturbance velocity may be thought of as a perturbation to the usual quet flow and that perturbation is going to affect the solution of the velocity. So in that case there is a method known as method of perturbation where what you do is you may expand u as u0 plus epsilon u1 plus epsilon square u2 like this. So what you are doing? You are expanding u in the form of a series in powers of epsilon where u1, u2 these are not u0, u1, u2 are not constant. These are functions of y okay. So what you are having? You are having term by term higher and higher order terms may be less and less important because epsilon is small. So it is as good as a power series expansion in terms of small number. This is known as a perturbation expansion. So what you are doing? So u0 is the so-called base state which does not understand the effect of v0. Now because of the effect of v0 the additional perturbations in u0 come into the picture and these perturbations are given by this. Now the way in which you may solve this is very simple. It is just algebra. So what you do is you substitute these expressions of u in the governing equation. So if you substitute that so what you have d2 u0 dy2 plus epsilon d2 u1 dy2 plus epsilon square d2 u2 dy2 plus epsilon du0 dy plus epsilon square du1 dy plus epsilon cube du2 dy that is equal to 0 right. So we have done nothing special just substituted this series expansion of u in the governing equation. Now what we may do? We may yes Reynolds number is epsilon. So epsilon into epsilon has become epsilon square like that. So now what we may do? We may equate the like powers of u. So that means what you have or like powers of epsilon that means you have d2 u0 dy2 that is equal to 0 that is one equation that is of the order of 1. Then what is of the order of epsilon d2 u1 dy2 plus du0 dy that is equal to 0. This is of the order of epsilon. Then of the order of epsilon square d2 u2 dy2 plus du1 dy equal to 0 and so on. So depending on how many terms you take you may go on writing terms of various orders which will give their individual governing equations. Then u0 u1 u2 should also have their own boundary conditions. So how do you assess that what should be their boundary conditions? What are the boundary conditions? Boundary conditions are at y equal to 0 u equal to 0 that means u0 plus epsilon u1 plus epsilon square u2 whatever this equal to 0. So at y equal to 0 you must have individually u0 u1 0 u2 0 like that okay. What is the other boundary condition? Second boundary condition? This is a non-dimensional y remember. So at y equal to 1 what is u? u equal to 1 non-dimensional okay. So that means you have 1 is equal to u0 1 plus epsilon u1 1 plus epsilon square u2 1 like that. These are all at 0 you have to understand this. So then what are the boundary conditions? What are the boundary conditions on u0? So u0 0 equal to 0 and u0 1 equal to 1 by equating the like coefficients from the 2 sides of this series expansion for the boundary condition okay. So then what is the solution of this? You have du0 dy is equal to some constant c1. So u0 equal to c1 y plus c2. At y equal to 0 u0 equal to 0 therefore you have c2 equal to 0 and at y equal to 1 u0 equal to 1 that means c1 equal to 1. So you have u0 equal to y okay. Next you come to the second one order of epsilon. So what it will give you? du1 dy2 plus du0 dy is 1 equal to 0. So if you integrate it du1 dy is equal to minus y say plus c3 and u1 is equal to minus y square by 2 plus c3 y plus c4. What are the boundary conditions on u1? u1 at 0 is 0 that means you have c4 equal to 0 and u1 at 1 is equal to what? 0 from this expansion. So at y equal to 0 at y equal to 1 so this becomes so what is c3? At y equal to 1 this is 0 so c3 becomes 1. In this way you may have expansions of higher and higher order terms. So at the end what is the solution that you are going to get? The solution that you are going to get is u equal to u0 plus epsilon u1 plus like this. So u0 is y plus epsilon u1 is this is 1.5 into y by 2 into 1 minus y right. In this way you have a series of terms. If you compare this with the exact solution let us say that the exact let us find out the exact solution of this. So if you substitute u equal to e to the power my as a trial solution. So you have m square plus epsilon m equal to 0 that is the auxiliary equation. So m equal to 0, minus epsilon. So you will have u equal to some a plus b e to the power minus epsilon y. Boundary conditions at y equal to 0 u equal to 0 so u0 equal to 0 will give you 0 is equal to a plus b and u1 equal to 1 will give you 1 is equal to a plus b e to the power minus epsilon. So from here you can find out a and b. It will be nice and interesting to see that it is an exponential of course you can clearly see it is an exponential variation. Now you may write it in the form of an exponential series e to the power x like 1 plus x by factorial 1 x square by factorial 2 like that and you will see that these terms will be some of the terms of the initial part of the exponential series. So more and more number of terms you take actually you can cover the more and more accurate part of the proper exponential series but this might be the dominating terms. So in this way you can see that if you have a small perturbation you may get a solution in this series expansion and that in a limiting sense of small epsilon may match quite accurately with the limiting value of the exact solution that is fine. So this is for low Reynolds number. Let us consider the other case the Reynolds number as large. So the second limiting case that we are considering is the large Reynolds number. So Reynolds number as large. So this now what we have seen is that one of the tricks in which you may use the perturbation method is by expanding it in the form of a series where you have powers of small quantities like powers of epsilon where epsilon is a small number. Now here if you treat it as epsilon, epsilon is not small because Reynolds number is large but you may introduce a delta equal to 1 by epsilon that will then be small. So then it will be so just replace epsilon with 1 by delta. So you have du dy plus delta into d2u dy2 that is equal to 0. We have just switched the variable from epsilon to delta. Why? The expectation is that now delta is small. So we should be able to write a power series expansion as we did for the previous case. Let us try to do that. So if we try to do that so try for expansion of u. So you write u equal to u0 plus delta u1 plus delta square u2 in this way. Now in place of epsilon your small variable is delta. Okay. Substitute it here. Yes which one? Okay. No this is 1 by delta. So this delta has gone here. So now you try for expansion of u in this way. So you substitute u as a function of y here. So you have du0 dy plus delta du1 dy plus delta square du2 dy plus delta du2 u0 dy2 plus delta square du2 u1 dy2 plus delta cube d2 u2 dy2 that is equal to 0. So if you isolate terms of different orders the leading order term will be what will be the leading order term here du0 dy equal to 0 that is the leading order term. What are the corresponding boundary conditions? Let us again expand the boundary conditions. So you have u0 which is equal to 0. Here we are writing the boundary conditions u0 plus u0 at 0 plus delta u1 at 0 plus delta square u2 at 0 like that in this way. Similarly u at 1 which is equal to 1 equal to u0 1 plus delta u1 1 plus delta square u2 1 like that. So again by equating the like coefficients what are the corresponding boundary conditions for this? What is u0 0? Yes? What is u0 1? 1. Can you satisfy these 2 boundary conditions by solving this? This gives u0 equal to constant. You cannot simultaneously satisfy these 2 constants. These are very simple examples see how interestingly mathematics gives you beautiful physics. Where are we lacking here? See as what I was telling you as a part of the boundary layer theory? See in the boundary layer theory you have the coefficient of the viscous term may be small in comparison to the coefficient of the inertia term. This is like an inertia term. This is like a viscous term. So this coefficient is small because of the smallness of the coefficient you may be tempted to discard this term altogether and that is what has happened. In the process what has happened mathematically? Mathematically your second order differential equation has got converted into a first order differential equation. So it cannot satisfy the boundary 2 boundary conditions with the second order differential equation demands. So in whatever way you make an approximation you cannot reduce the order of your governing equation because then you cannot match with the boundary conditions which must be satisfied or which must be given by the proper higher order governing differential equation. So here forcefully the governing differential equation has got reduced to first order although you have to satisfy the corresponding boundary conditions given by the requirements of the second order equation. That is one of the basic mathematical reasons why you cannot neglect this term altogether although the boundary layer is thin. See this has something to do with the equivalence of the thinness of the boundary layer because this is Reynolds number. This is 1 by Reynolds number. We know that in boundary layer theory Reynolds number is large. So this has some sort of analogy with the boundary layer theory. Full analogy mathematically partial analogy physically but mathematically it is fully analogous because you are having a small coefficient of the viscous term so to say and that is what we have to remember that in no way no matter how small it is you can neglect it outright because then it becomes mathematically an ill posed problem. Physically that means what? That means physically also you cannot justify such an assumption. So when you physically cannot justify that assumption that means that you must give a due importance to the boundary layer. What is the difficulty in giving a due importance to the boundary layer? The difficulty is that the boundary layer is so thin. So maybe you are getting an illusion you are thinking that well I should not capture the boundary layer because it is so thin but if you want to capture the boundary layer properly one way is like you magnify the boundary layer. So as if the boundary layer is very thin but you are sitting with a magnifying glass which zooms up the boundary layer to a large extent so that you can see whatever is happening within it you can resolve whatever is happening within it. So then what you have to do? You have to within the boundary layer say if the coordinate transfers coordinate is y you have to apply a magnification factor to the coordinate. So that it becomes large enough blown in proportion and although it is thin within that boundary layer if you resolve it properly it has to be given due importance that is what we have mathematically recognized. Now we are trying to see that physically how we can give it its due importance only by resolving it properly. To resolve it properly you have to understand that since it is thin you have to apply a magnification factor to it. So if y say y has a range within the boundary layer which is 0 to a very small number which is so small as compared to your flow domain that it is not coming well within your resolution. So what you may do? You may apply a magnification factor and a natural magnification factor is 1 by delta okay. So here if you introduce a new variable eta equal to y by delta then what happens? This is a small number you are dividing it by something small to blow it up or to stretch it up. So that means the boundary layer whatever is this resolution you are trying to stretch up its resolution by applying this transformation. So this is known as a stretching transformation in mathematics. So the whole idea is that since you are having a very small region which you desperately want to resolve physically because if you cannot resolve it you cannot solve the problem but physically if you want to resolve you must use a different coordinate for that. That coordinate should be a blown up coordinate not the original coordinate and that blowing up factor is 1 by delta keeping in mind that delta is small. Now if you see if you now recast the variable so instead of y you use the eta. So then what is du dy? So du dy is du d eta into d eta dy right. So this is 1 by delta then this term delta d2u dy2 is 1 by delta square d2u d eta 2 right. So now you see that from both the terms you may cancel 1 by delta. See 1 by delta is not 0 it may be small but it is not 0. So there is a difference between small and 0 that is what we are trying to highlight here. It may be limitingly small but that does not mean that you may have the liberty of taking it as 0. So then what is the governing equation? The governing equation becomes d2u d eta 2 plus du d eta equal to 0. This we say is the appropriate governing equation for the inner region. So now we are demarcating the flow region into 2 parts. One is the inner region, inner region is what? That is the boundary layer that means a region very close to the wall where you really want to resolve this high gradient. So that is the inner region. Outer region that is outside the boundary layer this equation is fine. It should have no problem because outside the boundary layer the resolution of the boundary layer is not important. So for the outer region you can still use the original differential equation. I mean you are still using the original differential equation but with a rescaled variable. This is the outer region. So you have 2 variables. One is y another is eta. These variables directly do not know each other. One is the inner variable which is eta is the inner variable which sort of has the responsibility of resolving only the boundary layer. This is what? This is like the outer variable which sort of has the responsibility of resolving what is there outside the boundary layer. So what it means is that these 2 variables may be treated independently. So when you want to treat the inner variable you have to keep in mind that anything outside the boundary layer is like eta tends to infinity because inner variable is only confined within the boundary layer. Outside the boundary layer is something which is very large for that guy who is sitting in the boundary layer. So for that it becomes a coordinate like infinity. So if you want to solve this of course it is possible to solve this now with the perturbation. So this is now known now if you want to apply the perturbation method with the inner and the outer regions separately by resolving the rescaling the inner region. This is known as a singular perturbation method. It is different from the regular perturbation method that we have seen. Singular perturbation method is important because you want to resolve what is happening in a very thin region close to the wall and therefore you are having a rescaled variable. We are not going to the details of the singular perturbation method just for giving you the information on the name of the method I am giving it. But we will just try to solve these 2 equations independently and try to match them and see that what solution we get out of that. So if you solve this equation so if you use u equal to e to the power m eta as the trial solution. Our whole objective now is not to just do the full perturbation analysis but to see that how you match the inner and the outer solution. Because they directly are 2 different variables but somehow the outer solution must know what is the inner solution. That is how to get the complete solution. So u equal to e to the power m eta if it is the trial solution then you have m square plus is equal to that means m is equal to 0, – 1. So u equal to some c1 plus c2 e to the power – eta. What is the boundary condition that you can apply on u? See at the wall definitely this is the thing that you are looking for within the boundary layer. So at the wall is u at eta equal to 0 that must be equal to 0 that means you have 0 is equal to c1 plus c2 that means c2 is equal to – c1. So u is equal to c1 into 1 – e to the power – eta. Question is how do you get c1? For that you have to match these with the outer solution. So you have to know what is the outer solution. So for the outer solution you may use the perturbation method. Let us apply the perturbation method for the outer solution at least. So this is the outer solution. This we had just written earlier but where we failed we tried to apply it for both the boundary layer as well as the whatever is outside the boundary layer. Now we know that it fails within the boundary layer we are not trying to use it within the boundary layer but outside the boundary layer anyway the viscous effect is not there. So you may totally disregard the viscous effect outside the boundary layer. So that physics we are trying to use here and if you recall that from here what we got is du0 dy equal to 0 obviously from the highest order term and what were the boundary conditions for this if you recall for u0. See u0 at 1, u0 at 0 we cannot use because this is not valid at 0. This is not this is only in the outer region. So only we can use that u01 equal to 1. That means the solution of u0 is 1 because u0 is a constant. So now you have to make a matching. What is the matching? The matching is y tends to 0 for the outer region is equivalent to eta tends to infinity for the inner region. So as if where the outer region starts is beyond the tractability of the inner region coordinate. So it goes so this is the limiting sense in which so this is called as a matching condition between the outer and the inner region. So when you have y tends to 0 so you do not care what is y tends to 0 because this is same for all y but eta tends to infinity is c1. So u becomes c1 that means from here you will get 1 equal to c1. That means your complete solution in the boundary layer becomes u is equal to 1-e to the power – eta. So what is the trick of this method? Trick is you introduce a separate coordinate for the boundary layer which you call as a inner region where basically you stretch up the local coordinate. You use an outer region coordinate for which you may sacrifice even by lowering the order of the equation because higher order equation you may only preserve within the boundary layer because higher order term is important that is the second order term in the governing differential equation is important only within the boundary layer where viscous effects are only important and then match up the 2 solutions by considering the limiting values of the inner and the outer coordinates okay. So at least from this exercise this is reasonably simple and tractable mathematically but with this we have understood a very important concept. What is that concept? That you cannot neglect the effect of the boundary layer altogether although that may be thin and how to mathematically track that you see that y by delta is something in our similarity variables we have introduced y by delta and see start from the experimental observations of Blasius. Blasius observed that u by u infinity is a single valued function of y by delta. So y by delta did not come up to Blasius as a mathematical entity. It came up to him as a physical entity from the experiments. Then similarity transformations were automatically giving rise to a tractable solution if you rescale the variable as eta is equal to y into some function of x gx where that scales with 1 by delta and therefore we see that y by delta is a sort of a magic coordinate system within the boundary layer which gets confirmed by this analysis also. So what it essentially does physically it zooms up the boundary layer coordinate by applying it as stretching by an amount 1 by delta okay. That we have understood that the boundary layer is important and it cannot be discarded altogether. So we have to understand that what are the importances of the boundary layer. So we have to now see that if there is a body which is immersed in a fluid and if there is some force which is there acting on the body then what are the consequences of these forces under a dynamic condition. That means if a fluid is flowing past a body then there is a boundary layer maybe at high Reynolds number the boundary layer is very thin but if there is a fluid element or maybe a fluid particle located within the boundary layer what are the forces that the fluid particle fills from within the boundary layer. So let us identify the forces experienced by a fluid particle located within the boundary layer. See when there is a fluid particle located within the boundary layer it is being subjected to at least a couple of effects one effect coming from the top of it another effect coming from the bottom of it. The general understanding is whatever is there at the top of it is trying to pull it forward because it is there with a higher velocity that may be the free stream velocity for flow over a flat plate. So you have a forward pull the outer stream how this forward pull of the outer stream is transmitted into the boundary layer by viscous action for laminar flow directly and by the momentum transport between eddies for turbulent flows. So transmitted into the boundary layer by viscous action or momentum transport. Then obviously there is a wall at the bottom which tries to slow it down. So slowing down effect by the wall shear stress that is also one of the forcing parameters okay. This effect is always present because the fluid is viscous and at least within the boundary layer there are velocity gradients that means you have both the viscosity as well as the rate of deformation. So the shear stress will be there. A third point that we have not considered till now but we will consider subsequently is any effect of pressure gradient on the flow force due to pressure gradient on the flow. When we consider flow over a flat plate the first two situations were still working. You had the outer stream trying to pull the thing forward because it is moving with the higher velocity. The wall is trying to slow it down. So these two for the laminar flow these are like artifacts of the shear in the flow. But not only you have a shear in the flow you may also have a pressure gradient in the flow. Pressure gradient in the flow is not important for flow over a flat plate. You have seen that because for that if you have u infinity as constant you have dp infinity dx equal to 0. And what is the important thing that we could understand of the general boundary layer theory that whatever is the pressure gradient imposed from the outer stream the same pressure gradient acts on the fluid in the boundary layer. Therefore if there is no pressure gradient in the outer stream obviously the boundary layer fluid is not subjected to any pressure gradient. But if instead of a flat plate you have a curved boundary because of the curvature effect you will have a pressure gradient. Why because of a curvature effect you will have a pressure gradient you will come into the theoretical aspect of it. But as engineers we must understand the basic physics out of it. Let us say you have flow over a body of a circular shape. So think in this way. Consider a fluid which is far away from the boundary. So it is like moving in a free stream like this. Now when you come close to the wall the wall will have some effect so it cannot move like this. So if you consider extreme streamlines may be one located at a very far distance it is not feeling its effect so it is moving like this. And the shape of a body is itself a streamline because there is no flow across it. So you have as if a confinement like this but this is one end of the confinement this is an outer end of the confinement. So from here to here you see that the area of cross section is gradually decreasing. So it is like a converging cone of a venturimeter so to say. And here you see that the area of cross section that the flow gets is continuously increasing. So it is like a diffuser. And therefore the pressure gradients acting on these sections are different. In one case it is favourable in another case it is adverse. In which side it is favourable left or right? Left side it is favourable. On the other hand on this side it is adverse. So you can clearly see that the effect of the curvature of the body itself is sort of introducing a pressure gradient. And that pressure gradient must exert some resultant force on the body. So that we must understand that what is the force due to pressure gradient that is there on in the flow pressure gradient in the flow. So the important thing is that the pressure gradient in the flow may be induced by curvature of the body itself. Our next objective will be to study that what is the effect of the pressure gradient in terms of the force acting on a body which is immersed in a fluid and fluid is passing across it. It is very very important because almost all engineering objects are not like flat plates. So they have certain curvatures and there will be effects of pressure gradient starting from engineering objects to say sports balls like tennis balls and cricket balls and so on. These are having certain curvatures. So when they move in the flow what types of forces act on that and how these things are deviated from their original trajectory because of that will be a matter of great interest and that we will study in our subsequent classes. Thank you.