 0 pressure gradient and we came up with this form of the expression and the whole objective of doing the analysis was to calculate the wall shear stress. Now if you look into this expression, you will see that you may calculate the wall shear stress provided you know what is the velocity profile within the boundary layer because that is what is an unknown in the integral expression which is there. The big question is that what should be the velocity profile? Any velocity profile which you may choose or approximate in general will not be the correct one because the correct one should come out of the detailed solution of the flow field and then what is exactly there within the boundary layer but you may always substitute this velocity profile with some approximate velocity profile which sort of satisfies your boundary conditions. So when it satisfies your boundary conditions and the velocity profile is substituted here, our expectation will be that the velocity profile may be inaccurate but in a somewhat integral form the velocity profile should give not a bad estimate of the wall shear stress and this expectation is from the fact that after all the boundary layer is very thin. So error in the velocity profile may be important when you consider the velocity profile as such but in the integral sense the integral of the velocity profile might not be that erroneous. So that is one of the expectations. So whether that expectation is justified or not the best way in which we may look into it by some example. So let us say that we want to have a polynomial approximation of the velocity profile. So for example you may have different forms of this expression like one may assume a polynomial like this even you may consider a fourth order term and that is what was done by an engineer known as Pohal Hussain. So his particular method was dedicated in the name of his honour but here we are not going for a fourth order polynomial, we will leave that on you as an exercise in your assignment problem but we will consider a third order polynomial. Even one may consider a first order polynomial, one may consider a sinusoidal function, so many different forms of the functions are possible. Within the constraints of the function you have to satisfy the most important boundary conditions. So that these unknown constants a0, a1, a2, a3 whatever these are determined. So let us look into this special case as an example problem and see that how do you calculate the boundary layer thickness, the wall shear stress and the total drag force on the plate on the basis of this. So to do that first of all we have to understand that how do we calculate these coefficients. So if you have 4 coefficients you must satisfy 4 boundary conditions to get these 4 coefficients. Let us see what are the most essential boundary conditions that you need to satisfy. What is the physical problem we are considering? You have a flat plate on which you have a free stream flow coming with a velocity u infinity and the boundary layer is growing like this and you are interested in describing a velocity profile at some section x. So what are the boundary conditions that you want to satisfy? Boundary conditions. What is the most essential boundary condition that you would always like to satisfy? See the boundary conditions should come in order of priority because if you choose lower order polynomials you may not be able to satisfy all the boundary conditions because you have less number of coefficients but the most essential one you should always satisfy. What is the most essential one? No slip boundary condition at the wall. So u equal to 0 at y equal to 0. What happens in the outer stream? So let us say that y equal to delta in an engineering sense is equivalent to y tends to infinity in a mathematical sense because outside delta whatever happens is like first stream. So literally it is the variable y by delta or eta tends to infinity but it is as good as y equal to delta. So at y equal to delta what are the boundary conditions? u equal to u infinity at y equal to delta. Any other boundary condition at y equal to delta? So u does not change further with y at that. That means the gradient of u with respect to y equal to 0 at y equal to delta, a fourth boundary condition. Of course you may go on adding here like in terms of a boundary condition at y equal to delta you may also have the second order derivative of u is equal to 0. But more important boundary condition which has a priority over this is what happens at the wall because at the wall you have one boundary condition but if you really have a liberty to put more the priority should come there because these two are quite sufficient to describe what happens at the boundary layer but this alone is not very sufficient to describe what happens at the wall. I mean if there are scopes of incorporating more boundary conditions at the wall it would be better. So look at the governing equation let us try to satisfy the governing equation at the wall. See this velocity profile does not understand the momentum equation. It is just an approximate velocity profile. So we will try to teach the velocity profile in such a way that it is in a way satisfies the constraints of the governing equation at the wall. So at the wall if you look at the governing equation so if you have the let us write the momentum equation the boundary layer equation basically this is the boundary layer equation. When you consider flow over a flat plate essentially it is a 0 pressure gradient. So this term is not there. Let us try to satisfy this at the wall. At the wall u is equal to 0 by no slip boundary condition. At the wall v is equal to 0 by no penetration boundary condition. That means you must have the second derivative of u with respect to y 0 at the wall right which follows from the governing equation. So this is not a boundary condition that you directly find out intuitively but if you want to make sure that the momentum conservation is satisfied then that is what you get. So the remaining work is easy that you substitute these 4 conditions. So you will get 4 algebraic equations involving a 0, a 1, a 2 and a 3. So from that you find out a 0, a 1, a 2, a 3. So I am not going into the details of the algebra because it is too trivial. So at the end what you will get out of this is u by u infinity is equal to 3 by 2 y by delta minus half y by delta whole cube. So the remaining terms will go away okay. Now at least once you get out of this whether your algebra is correct or not at least you should check that it satisfies most of the essential boundary conditions that at y equal to 0 equal to 0 and at y equal to delta u is u infinity. Now what we will do with this velocity profile? We will use this velocity profile for evaluating the integral which is there in the expression. So to do that we will write tau wall by rho u infinity square is equal to ddx of integral 0 to delta. Now we will write 3 by 2 y by delta minus half y by delta whole cube into 1 minus 3 by 2 y by delta plus half. With a obvious substitution of variable that is eta equal to y by delta you may write this as tau wall by rho u infinity square is equal to ddx of integral 0 to 1 then 3 by 2 eta minus half eta cube into 1 minus 3 by 2 eta plus half eta cube d eta and then multiplied with a parameter delta because dy is equal to delta into d eta okay. So change of variable is there. So important thing is first of all this delta is dependent on x but this integral evaluation is not dependent on delta. So you can take this delta out of the integral. So it boils down to evaluating some sort of a polynomial expression and then coming to the integral again let us not waste any time in like doing this very simple algebra and simple integration. Let us say that out of the integration what you will get? You will get a number say you get a number k which comes out of the integral of this entire expression. It is a definite integral so it will come like a number. Now what is the tau wall? Tau wall is mu into this one at y equal to 0. So that also you may calculate from your assumed velocity profile. So what is your assumed velocity profile? So that is this one and you may calculate the first order derivative at y equal to 0. So that will be 3 by 2 delta because remaining term will have y which will be 0 at y equal to 0. So you will have mu into 3 by 2 delta rho u infinity square is also there in the denominator is equal to k into d delta dx okay. So then you can integrate this and once you integrate this you will get delta as a function of x. So let us write the integration. So you have delta d delta is equal to by rho u infinity square to 3 by 2 k dx. So if you integrate it delta square by 2 is equal to 3 by 2 k mu by rho u infinity square x plus constant of integration. As we know that in the limit as x tends to 0 you have delta tends to 0 that is the edge of the plate where from the boundary layer grows and therefore you must have c equal to 0. So you see the growth of the boundary layer is like an initial value problem where you start with x equal to 0 and whatever condition is at x equal to 0 all the subsequent conditions for delta may be evaluated. So the x coordinate here acts like a time coordinate as if you have started with time equal to 0 instead of time equal to 0 it is x equal to 0 and as you march along x in the positive direction you see that the boundary layer grows. So this is called as a marching problem. So instead of marching with respect to time you are really marching with respect to space or position. So now you can find out what is delta by x. So delta square will be 3 by k. We will always write mu by rho as mu because kinematic viscosity is what that governs the physical situation. So mu by u infinity square x. So you can see clearly that delta varies with square root of x which we have also seen from the previous analysis like the order of magnitude analysis and the blacier solution. Now if you want to find out how delta by x is scaled so basically what you have to do you have to divide by x square. So if you divide it by x square then okay one important thing is we have missed 1 u infinity here because this u should be I mean it is u into u infinity is this one. So somewhere u infinity has to be there. So yes 2 by yes into u infinity that is 1 u infinity we have missed. So this u infinity and this u infinity will cancel. So eventually we will have only 1 u infinity here right. So for evaluation of the tau wall there was 1 u infinity. So this is u infinity that has to be there. Now if you divide it by x so delta by x square is 3 by k into nu by u infinity x that means delta by x is square root of 3 by k into Reynolds number to the power – half okay. So if this is calculated this comes out to be this particular number roughly if I remember correctly roughly 4.64. So see this is somewhat erroneous because if you remember the blacier solution this was 5. So instead of 5 it is coming out to be 4.6 or something like that but the important thing is that the scaling behavior is similar. See the order of magnitude gave delta by x of the order of Reynolds number to the power – half. The exact solution is like 5 into Reynolds number to the power – half. Some approximation it is deviated from the exact solution but we are not really that much bothered about this one. This is somewhat disturbing because the approximate solution is somewhat deviated from the exact one but this is what is not our focus. Our focus is the wall shear stress calculation. We are never expecting that it will give the correct boundary layer growth but our expectation is that at least integral of the velocity profile will nullify the error in the velocity profile to some extent and get a better estimation of the wall shear stress. So let us calculate the wall shear stress. So if you write the wall shear stress tau wall that is mu 3 mu u infinity by 2 delta that is what and delta you may substitute as a function of x. So delta is nothing but 3 mu by 2 3 mu u infinity by in place of delta we will write square root of 3 by k into square root of mu by u infinity x. So what we will get essentially say tau wall, what is our matter of interest is the non-dimensional form of the tau wall. So the non-dimensional form of the tau wall is given by this CF which is the friction coefficient which is tau wall by half rho u infinity square. So you divide this by half rho u infinity square. So one of the u infinities get cancelled then you are left with there is one mu by rho that means it is basically something into which one? This x is no just I am using this expression. So delta is square root of this one whatever into square root of x okay. So that is what is substituted here. So this one something into mu by u infinity into root over mu x by u infinity. So that means CF is coming out to be something into Reynolds number to the power – half. See this becomes square root of u infinity this numerator becomes square root of mu. So square root of mu by u infinity x. So this constant if it is evaluated this comes out to be 0.646 and the exact solution the Blasius solution gives this one as 0.664 and this is really quite close. So the error between these 2 may not be that significant for a calculation for an engineering calculation and therefore you see that remarkably although the velocity profile was erroneous although the boundary layer thickness was quite erroneous this is also erroneous but the error has somewhat gone down and that is because of using this in the integral form. So the whole expectation is that the functions may be inaccurate but area under the function as area under the curve that may be the approximately the same despite the error in the function themselves. So this sometimes is called as skin friction coefficient. So sometimes this is given with a subscript x. So CF with subscript x for CF as a function of x and the name of this is known as skin friction act coefficient. Why it is so? Because we will later on see that there may be other mechanisms of having a drag force on a body. So here this drag force is originated out of the frictional action or the viscous action. So this is given a name of skin friction drag coefficient. The subscript x to indicate that it is local friction coefficient that means as you change x this will vary. Now one is interested at the end to calculate the total drag force on the plate. So how do you calculate the total drag force on the plate? To calculate the total drag force on the plate basically you have to keep in mind that the wall shear stress varies locally with x and it is important to take care of that local variation. So to take care of that local variation what we may consider is let us say at a distance x. We take a small strip on the plate of thickness dx. So the plate is having some width so it is basically there is a strip which has some width. So let us say that the width of the plate is b. So the element that we have considered is something like this and the shaded portion which is there, this is where it is exposed to the fluid and that is where there is a shear stress between the fluid and the solid. So how do you calculate that what is the total shear force on this element, whatever is the shaded element. So let us say that df is the shear force on the element. So what is df? So tau wall is the wall shear stress that into b into dx. If you want to find out the total one of course you have to integrate it. So the total f, so let us give it a subscript d to indicate that it is a drag force. So fd will be integral of tau wall bdx from x equal to 0 to x equal to l where you know that how tau wall varies with x. So that is already known to us. So tau fd let us write integral 0 to l in place of this one we write 3 by 2 mu by square root of 3 by k u infinity divided by nu by u infinity x into b into d. So again if you consider the form like there will be some numbers based on what is the value of k then mu u infinity divided by square root of nu by u infinity b then just integral of x to the power minus half dx. So it will become this one. Now just like the local skin friction coefficient is a non-dimensional representation of the wall shear stress locally. There is also a non-dimensional representation of the drag force and that is given by coefficient of drag. So let us write what is the coefficient of drag. So the coefficient of drag is given by fd divided by something reference. See this is a force. So this should be represented by some equivalent half rho u infinity square that is there times some reference area. So what area you will choose here? So if l is the length of the plate b into l we will see later on that depending on the shape of the body there are engineering conventions of how what reference area should be taken. So this fundamentally is like some reference area. For flow over a flat plate it is like this b into l but for different shaped bodies we will see if it just flow past a circular cylinder maybe flow past a aerofoil section how these reference areas will change from one to the other. It is nothing very fundamental it is an engineering convention that what reference area you choose for designating the coefficient of drag. Now here you can write it in like you will get some number times mu u infinity by square root of nu by u infinity then you get u infinity square into you will get a square root of l because numerator square root of l denominator is l. So again you will get one of the u infinity gets cancelled out. So this is something into Reynolds number to the power – 1.5 but Reynolds number is based on l. So this Reynolds number is not based on any local coordinate but the total length of the plate. So this gives the total effect and this coefficient is just double of this 0.646. So it is whatever like 0.646 into 2 and it is therefore also quite close to the result that one gets from the exact solution or the Blasius solution. So what we get out of this exercise is we have seen that how by using an approximate velocity profile we may estimate how the boundary layer thickness grows with x, how we calculate the wall shear stress and how we calculate the total drag force and their non-dimensional versions. Again because this is originating out of a skin friction force so sometime this is given with a subscript cdf. So f for skin friction because as I have mentioned we will encounter such examples where there will be also other mechanisms of the drag force not just the wall shear stress okay. Next we will try to go through some of the important concepts or important terminologies related to the boundary layer and these terminologies are displacement thickness and momentum thickness. So first terminology that we will try to understand is displacement thickness delta star. What it is? Let us say that you have a flat plate but there is no growth of boundary layer. So then what happens if you have uniform flow, uniform flow will remain uniform and that means the plate is having no effect. So if there was a streamline say there was a streamline like this, this streamline would move parallel to the original direction. So when you have uniform flow the streamlines are actually parallel to the flow direction. So let us say this is one of such undisturbed streamlines. Now this is what that does not happen in reality. In reality what happens? In reality you have a growth of the boundary layer. Let us say that the boundary layer is growing and the boundary layer is growing like this. So because of the growth of the boundary layer what happens? What will happen to the streamlines? The streamlines will originally be parallel to each other but when they come close to the plate will the streamline, so we have decided or we have come to a conclusion that the streamline cannot remain just like this. So it has to get shifted or deflected. Question is it will be deflected upwards or downwards? Upwards. So let us try to see that why it should get deflected upwards. So let us say that because of the effect of the boundary layer the streamline has got deflected upwards. So this same streamline which was undisturbed now it has got deflected upwards. The reasoning is very straightforward. If you consider 2 sections let us say that we consider one section here say section A and another section here say B. If the flow is steady whatever is the flow rate through section A should be the same flow rate through section B because see this black line this is a streamline. So there cannot be any flow across it okay. Now if the same flow rate has to be maintained see this is the region where the flow is slowed down. To compensate for that the streamline should be upward so that this extra flow plus this flow becomes same as this flow right. So the streamline has to move upward not downward. Question is how much does the streamline move upward? Let us say that we have chosen such a reference that this is the delta at the given x and let us say that this shift of the streamline is delta star. We want to find out what is this delta star. Nature is not important for us. This is just a rough sketch. What is important is that whether it is shifted upwards or downwards because like we are not having any attention on what happens across the streamline because by definition of the streamline there is no flow across the streamline okay. So let us write let us try to find out this delta star. What would be the basis by which we may find it out that the mass flow rate at the sections A and B must be the same. So you must have m dot at the section A same as m dot at the section B. So if the uniform velocity was u infinity at the section A it is rho into u infinity into delta into the width. Let us say the width is what? B. Now the width will get cancelled in both the sides. So let us not just write the width. For B you see there is one portion within the boundary layer. So that is integral 0 to delta u dy where u is the velocity as a function of y here. Outside the boundary layer you have u equal to u infinity. So plus rho integral of delta to delta plus delta star into u infinity dy. Basically that into u infinity into delta star. So from here if we simplify one more step what we get? You will get u infinity into delta is equal to integral 0 to delta u dy plus u infinity into delta star. This term you can also write as integral 0 to delta u infinity dy all the same. The whole idea is that we want to club these 2 terms together. So from here you can write delta star is equal to 0 to delta 1-u by u infinity dy by dividing all the terms by u infinity. So this is known as displacement thickness. So physically what it is indicating? Physically it is indicating maybe a displacement or a shift in the streamline because of the existence of the boundary layer. It may also be looked from a different angle. Let us say that we are trying to consider a case when this is a flat plate with boundary layer and retardation of the flow close to the wall and we consider an equivalent case when we do not consider the effect of the wall directly. But what we do is we make a shift of the wall. So the new location of the wall becomes say like this. What should be this shift? So that this problem with a boundary layer is equivalent to a problem with a uniform flow. So this is a problem with a boundary layer. This is a problem where we are trying to have the same flow rate but no boundary layer that means there will be a uniform flow. So let us say that the uniform flow is like this with a velocity u infinity. So here there is a growth of the boundary layer. So you have a delta as a function of x. Here you do not consider the growth of the boundary layer but to avoid that analysis what you do? You shift the plate a bit upwards because the effect of the boundary layer is it has a reduction in the flow rate. So if you still want to use u infinity, you make a shift of the plate somewhat so that your effective area of the flow gets reduced. So that multiplied with u infinity should give the same as this one. So let us say that this shift of the plate is a. What is our constraint? Our constraint is that the flow rate should be the same. This is the hypothetical uniform flow. This is the real boundary layer flow. So here the velocity profile is like this. This is u infinity. So what is the flow rate here? The flow rate here is integral of 0 to delta u dy. What is the flow rate here? u infinity into delta – a. Of course I am just considering the volume flow rate per unit width of the plate. So rho into b that term I am not considering. So if you equate these 2, what you will get from here is what is a? a is nothing but integral 0 to delta 1 – u by u infinity dy. So this is nothing but delta star. So what does it mean? It means that the displacement thickness also may be looked at as a hypothetical displacement of the solid boundary. So that the remaining flow within the length delta may be perceived as a hypothetical uniform flow still predicting the same correct flow rate. So the advantage of this is that sometimes you do not want to analyze the details of the boundary layer but you just want to have a gross estimate of the flow rate. If you do not know what is delta star then what you may say is that within this length this – delta star whatever is there the velocity is uniform. So it is not that it is actually uniform. It is a pseudo situation with which you are matching with the exact situation. What you are not compromising with is the flow rate prediction. So the pseudo situation and the correct situation are giving the same flow rate that is the basis of this. Now although the mass flow rates over the sections a and b are the same but the momentum flow rates are not the same and we will see that what is the difference in the momentum flow rates over the sections a and b. So let us write what is the momentum flux or maybe momentum rate of momentum transfer. So the rate of momentum transfer at a what is this? So you have what is the rate of mass transfer at a? Rho u infinity into delta that multiplied by u infinity is the rate of momentum transfer at a. What is the rate of momentum transfer at b? Again one part within the boundary layer another part outside the boundary layer. So whatever is within the boundary layer you have rho integral of u into u that is u square dy from 0 to delta plus it is basically rho u infinity square into delta star that is the top portion. Top portion outside the boundary layer velocity u infinity. So if you want to find out what is the difference between these 2 momentum flux or momentum rate of momentum transfer at a and b the difference of these 2. So that you have rho into u infinity square delta minus integral 0 to delta u square dy minus rho u infinity square into delta star is what? Delta star is u infinity minus u by u infinity integral of that dy from 0 to delta. So one of the u infinity get cancelled out. So when one of the u infinity get cancelled out the next step we can have a simplification. Let us go for that simplification. So this becomes first of all this becomes minus rho u infinity square delta. So there is one u infinity square delta then there is minus one u infinity square delta. So this term gets cancelled out with the first term. So what remains is rho u infinity minus u square dy from 0 to delta. u infinity from the last term and minus u square from the second term. So now what you see that what this is not 0 right. Therefore there is a difference. Now is this positive or negative? This is positive because u infinity is greater than u. So u into infinity is greater than u square. So that means this is actually a representative of the very important fact that there is a momentum deficit at the section B. So whatever was the rate of momentum transport at the section A, the rate of momentum transport at the section B is somewhat less than that. And this is indicator of how less it is. So this you can write as rho into u infinity square. Let me just taking u infinity square as a reference. So u by u infinity into 1-u by u infinity dy. Just a non-normalized or non-dimensional way of writing is the term which is there in the integral. And see this term in the integral is a term that we have encountered in the momentum integral equation. Tau wall by rho u infinity square was d dx of this quantity. And this we call as theta which is also given as symbol of momentum thickness. So momentum thickness that is given by theta. And what is important is that this momentum thickness what it physically indicates? It physically indicates the deficit in the rate of momentum transport across 2 sections because of the development of the boundary layer. That is the physical implication of this. But mathematically this is expressed just by this integral. Sometimes one also uses third parameter which is known as the shape factor h which is given by delta star by theta. The ratio of these 2. Why it is given a name of a shape factor is somewhat obvious. The obvious is that this ratio depends on the shape of the velocity profile. So only on the shape of the velocity profile because if you see the integrals, these integrals of course they themselves depend on the shape of the velocity profile. But sort of even their ratio also therefore depends on the only on the shape of the velocity profile. Shape of the velocity profile means u by u infinity as a function of y by delta. That is the shape of the velocity profile. So once the shape is fixed this is also automatically fixed. So that is why we call it a shape factor and given a velocity profile it is possible to determine the shape factor. Now the important thing is that we have till now considered the boundary layer over a flat plate and that is one of the very simple cases. But even in this very simple case we have not considered one thing that is the effect of turbulence. We have considered that means implicitly that the boundary layer develops as a laminar boundary layer. But in reality the boundary layer may have a transition towards turbulence and what is the Reynolds number that is important here? The Reynolds number here is dependent on the parameter the axial location x. So you see that as you move along the plate this Reynolds number increases. Beyond the critical Reynolds number you have the inertia forces dominating so much that a slight disturbance may trigger the onset of turbulence and then the laminar boundary layer changes its characteristic to a turbulent boundary layer. May not be abruptly but at least over a given distance and the critical x or the critical Reynolds number at which this transition occurs for flow over a flat plate is roughly 5 x 10 to the power 5. So you see that the critical Reynolds number it depends on different geometries. See for flow through a circular pipe the critical Reynolds number was roughly of the order of 2000. There the reference length was the diameter of the pipe. Here the reference length is the axial coordinate. There the reference velocity was the average velocity. Here the reference velocity is the free stream velocity u infinity. So the critical Reynolds number is therefore not a magical number which is true for all cases like depending on the situation your reference length changes, your reference velocity changes and the convention changes all together. Now what happens to the growth of the boundary layer if you have such a transition. To understand that let us work out a problem by which we will base illustrate it. So just note down this problem that consider the turbulent boundary layer over a flat plate for which u by u infinity is given by y by delta to the power of 1 by 7. So this is the velocity profile for the turbulent boundary layer. So when you say u remember we are not writing it explicitly but this is the average velocity. This is not the instantaneous velocity with fluctuations. So u by u infinity is y by delta to the power of 1 by 7. This is given for the turbulent boundary layer. What you have to determine? Number 1 the Cf and Cd that is the local skin friction drag coefficient and total drag coefficient on the plate for the flow over the plate if the entire boundary layer is turbulent. Number 2 the same things Cf and Cd if the boundary layer undergoes a transition to turbulence Re critical equal to 5 into 10 to the power 5 okay. Then what is given? Given is that at the wall this is what is given at the wall. We will see that why this is given at the wall. This is from experiments. This is actually real data from experiments conducted by Blasius. So this is from experiments and for laminar flow for laminar boundary layer it is given that Cf is equal to 1.328 into Reynolds number that is Cdf right Cd 1.328 into Reynolds number to the power minus half Cdf okay. So with this problem statement let us try to work out this problem and this will give us some idea. First of all this is the velocity profile which is again for turbulent flow it is not possible to determine the velocity profile exactly. So this is just a curve fit of the experimental velocity profiles and therefore it is not a very accurate one. The loss of accuracy is important at some place and let us see what is the place. So let us first write tau all by rho u infinity square is equal to ddx of integral 0 to delta u by u infinity into 1 minus u by u infinity dy. So this part you may substitute u by u infinity as a function of y by delta and then you will get this as something into d delta dx that is fine. Substitute u by u infinity as y by delta to the power this. Now what about tau all? See if you want to calculate mu du dy at the wall you see that you cannot calculate using this velocity profile. So this velocity profile is not exactly valid at the wall for calculating the wall shear stress and that is why the wall shear stress has to be separately determined from experimental conditions and that is what it has been experimentally determined. See fluid mechanics is such a subject where you cannot independently go or grow with theory or experiment you have to somehow make a good combination of these for understanding the physical principles. So tau all by rho u infinity square you will have 0.0 225 into mu by u infinity into delta to the power 1 by 4. So then it is straight forward you integrate this and get delta as a function of x. So if you integrate this you will get let me just tell you the answer quickly you will get delta by x is equal to 0.37 into Reynolds number to the power – 1 5th okay. So once you get delta by x it is easy to calculate the Cf and Cd just as we did for the previous cases. So let me give you the answers at least so that you can verify later on. So Cf will be 0.058 into Reynolds number to the power – 1 5th and Cd is 0.075 into Reynolds number to the power – 1 5th okay. So these are the answers for the first part of the problem. But this is not a very realistic representation why? Because we know that the boundary layer does not become turbulent from the very beginning. The Reynolds number is based on the local x. Local x is initially very small so it is always initially laminar and then it becomes turbulent. So the real picture maybe something like this. So you may have the flat plate in this way. You have first a laminar boundary layer then it undergoes a transition to turbulent boundary layer. Let us say that this is the length x critical over which it is laminar and then beyond that it is turbulent. So the question will be that what is the length of your plate? If the length is greater than x critical how do you find out x critical? x critical is given by u infinity x critical by nu is 5 into 10 to the power 5 that is re critical. So given u infinity you can always find out what is x critical. If your length of the plate is greater than x critical then in reality there is a great chance that the remaining flow is turbulent. So your actual drag force actual is what? If you say calculate the total drag force as an artifact or outcome of that the entire boundary layer is turbulent there is an error because of the presence of the laminar part. So that error we have to nullify. So what we can do? We may consider that it is f drag force that is the drag force with from x equal to 0 to x equal to l considering it as turbulent, fully turbulent. Then subtract the drag force from x equal to 0 to x equal to x critical for turbulent and add the drag force for that part for the laminar okay. So as if the full thing was turbulent then you subtract the turbulent part from the initial and add the laminar part okay. And you have to keep in mind that when you are writing the corresponding the cd's so this will be cd into half rho u infinity square into b into l. Here it will be cd into half rho u infinity square b into x critical and here also it will be the same cd into half rho u infinity square b into x critical. The difference is that in one case the cd is the laminar cd in another case this is the turbulent cd. So this is turbulent cd this is the turbulent cd and this is the laminar cd. In the problem already it is given what is the expression for the laminar cd you have to be careful in place of l you have to now use the x critical as the reference length. And then if you make a simplification of this the net drag force is the f drag force actual by half rho u infinity square into b into l that is the net cd and the answer of this will come out to be 0.0725 Reynolds number to the power –1.5 –1740 by Reynolds number. This is the answer to this. So if you just substitute and make a simplification just check that whether you get this final expression. So what this final expression says that there is a correction because of a part of the boundary layer being laminar and not the entire boundary layer being turbulent and this is the expression of the composite cd that takes care of the fact that a part is laminar and then it is turbulent. So with this we stop our discussion today and in the next class we will start with the discussions on boundary layer where you have the effect of the pressure gradient till now we have considered the boundary layer without the effect of pressure gradient and boundary layer with the effect of the pressure gradient we will start discussing from the next class. Thank you.