 In this lecture, I will take up solution of the integral momentum equations or the integral equation of the velocity boundary layer. My task would be to introduce you to the general solution procedure and then using this procedure study the effects of the pressure gradient and suction and blowing. I recall again that integral solutions can be obtained for arbitrary variations of the free stream velocity u infinity at the edge of the boundary layer and the wall velocity at the at the surface of at the solid surface we call suction or blowing. To refresh our memory the integral momentum equation reads as d delta 2 by dx plus 1 over u infinity d u infinity by dx plus 2 delta 2 by delta 1 equal to c f x by 2 plus v w by 2. How do we solve this equation? Just recall it is the procedure here is quite the reverse of what we adopted for similarity method. In the similarity method a third order similarity equation was solved with appropriate boundary conditions to obtain the velocity profile. The velocity profile came out of the solution of the ordinary differential equation and from which the integral parameters such as delta 1, delta 2 and c f x were recovered by integrating and differentiating the profiles. In contrast in integral method the velocity profile is assumed usually a polynomial in y by delta such that it satisfies the boundary conditions. Having assumed this profile we evaluate the integral parameters delta 1, delta 2 and c f x and substitute them in the integral momentum equation. The IME is then solved to obtain delta 2 as a function of x and hence all other parameters as functions of x. So, the procedure is quite the reverse of the similarity method and to the extent that we have assumed a velocity profile that which satisfies boundary conditions but nonetheless is an approximation to what could be the real velocity profile at a given x. So, we say let u over u infinity which is the dimensionless velocity and it would be a function of x and y be a polynomial in eta variable where eta is y by delta a plus b eta plus c eta square plus d eta cube and e eta 4. So, there are five constants to be determined a, b, c, d and e and this we do by invoking five boundary conditions and the boundary conditions are as follows. At the wall y equal to 0 of course, u is equal to 0 which will render u equal to 0 but the second condition is that if you look at the momentum equation. Now, if I write this equation at y equal to 0 then clearly that term will be 0 because u itself is 0. v however will be v w into d u by d y at y equal to 0. This term of course will survive u infinity d u infinity by d x plus nu d 2 u by d y square at y equal to 0. So, I use this as a boundary condition to say that nu d 2 u by d y square at y equal to 0 is equal to v w d u by d y at y equal to 0 minus u infinity d u infinity by d x and that is what I have written here as the boundary condition 4 I mean the equation 4 which is the second boundary condition at the wall. At the edge of the boundary layer y equal to delta u will equal u infinity and therefore the left hand side will be equal to 1 and so will each of these eta's will be 1. So, you will have a condition a plus b plus c plus d plus e equal to 1 also d u by d y is equal to 0 at the edge of the boundary layer d u d y is 0 and to the extent that the velocity approaches u infinity asymptotically the continuity of this first derivative survives d u by d y equal to 0 survives as we increase y and therefore the last boundary condition is that d 2 u by d y square will also be equal to 0. So, I have 1 2 3 4 and 5 5 boundary conditions which will enable me to determine a b c d and e. Let us see what those coefficients look like each of these coefficients can be represented in terms of the coefficient e a equal to 0 of course because of u is equal to 0 at y equal to 0 b will be equal to 3 minus e c will be equal to 3 into e minus 1 d 1 minus 3 e and e will equal 3 times v w star minus lambda plus 6 divided by 6 plus v w star. So, with this 5 boundary conditions if I were to write out the equation the velocity profile will look like u over u infinity 6 by 6 plus v w star a function f 1 a second function f 2 multiplied by v w star and a third function f 3 multiplied by lambda f 1 is simply 2 eta minus 2 eta cube plus eta raise to 4 f 2 is this function f 3 is this function v w star is a dimensionless v w delta by nu. It is a suction and blowing parameter and lambda is delta square by nu d u infinity by dx is a pressure gradient parameter associated with variation of free stream velocity u infinity with respect to x. Of course, I did say that we have to choose the values of v w star and lambda very carefully because in as much as we can allow for any variations of v w and u infinity, we must ensure that the assumed velocity profile is such that u over u infinity is always less than 1 inside the boundary layer for eta less than 1. So, values of v w star and lambda minimum or maximum must be such that this condition holds. Remember v w star can be both positive or negative depends on suction or blowing and lambda depending on whether it is an adverse pressure gradient or a favorable pressure gradient can also be positive or negative and we have to choose these parameters of within a certain restricted range only so that the boundary layer approximations are well held. I have plotted the three graphs. The middle graph is for no suction or blowing that is v w star equal to 0. When lambda is equal to 0, you will have the flat plate because d u infinity by dx is equal to 0. As you can see here, this will be the profile. Lambda equal to positive values means accelerating boundary layer because d u infinity by dx then is positive. Lambda negative would be d u infinity by dx because d u infinity by dx is negative and therefore a decelerating boundary layer. So, if I choose arbitrarily values of lambda, you will see up to lambda equal to 12, the u over u infinity values are well within 1 and up to minus 12 they are very much within 0 to 1. But if I take lambda equal to 30 which is a very highly accelerating flow then of course, u over u infinity exceeds 1. In fact, that is what would happen for all values of lambda greater than 12 and therefore such values of lambda are not admissible. Likewise, values less than lambda equal to minus 12 are not admissible because velocity itself will turn negative. This is the importance of this common which I made here. Remember, if u over u infinity exceeds 1 then delta 1 or delta 2 can become negative. Likewise, if u over u infinity is negative then delta 2 can be negative and that is not admissible. This is similar profiles for v w star equal to minus 2 which means the suction case and again I have plotted several values. Before I do look at that, let me go back again to v w star equal to 0 value. Obviously, when lambda equal to 0, I said this is a flat plate solution, but lambda equal to minus 12 gives a 0 gradient at the wall which means separation must occur. It is a peculiar value lambda equal to 7.052 and the importance of that you will recognize a short while from now. Let me go back again to the suction profile. Here, you will see for lambda equal to 12 plus 12 u over u infinity exceeds 1 and of course, for 28 exceeds very much more. Likewise, here for lambda equal to minus 12, there is a velocity which goes less than 0. On the blowing side however, up to lambda equal to 20 or little more, there is a u over u infinity well within 1. So, blowing permits much higher pressure gradients favorable pressure gradients. On the adverse pressure gradient side, again lambda equal to minus 12 is the limit. Integral method, we evaluate three thicknesses. The first one is very well known to you delta 1 which can be where the integration limit from 0 to infinity is replaced by 0 to delta. Simply because from delta to infinity, u will remain equal to infinity and therefore, both these integrals will be 0 between delta and infinity. Therefore, the limit infinity can be replaced by delta in both of them. In addition, we define a shear thickness mu u infinity divided by shear stress. You will notice that it has a length dimension we call delta 4 and it is called the shear thickness mu u infinity by tau all x. The reason for this you will understand very shortly. So, we have three thicknesses delta 1, delta 2 and delta 4. We could have recovered value of delta 4 even in the similarity method because we know how the tau all x varies with x. So, if I substitute our velocity profile u over u infinity equal to all this as a function of delta, as a function of eta into our definitions, then you will see that delta 1 by delta would become 1 over 4, 1 plus e by 5. If I represent e as 3 v w star minus lambda plus 6 over 6 plus v w star, then it will simply read like that. Delta 2 by delta, this integration is somewhat involved because it involves product of u over u infinity square and that would result into 3 by 28 plus e by 70 minus e square by 252 or what I have given here by simply replacing e. Most importantly, delta 4 by delta would simply result in 1 minus 3 e 6 plus v w star divided by lambda plus 12. This is quite easy to see. If we see our equation, then remember delta 4 is mu over u infinity divided by tau all x and this is equal to mu times u infinity divided by mu times du by dy at y equal to 0. If we look at our velocity equation, then you will see du by dy at y equal to 0 will become equal to u infinity into 6 over 6 plus v w star into 2 by delta plus v w star into 0 plus lambda by 6 into 1 over delta. You will see then if I take delta common, I will get this as u infinity by delta into 6 over 6 plus v w star into 2 plus lambda by 6. Therefore, delta 4 will become equal to mu mu gets cancelled here and u infinity divided by u infinity by delta 6 over 6 plus v w star into 2 plus lambda by 6. Therefore, this and this gets cancelled and delta 4 by delta will read as what I have shown there 6 plus v w star over lambda plus 12. This shows very clearly that if lambda is equal to minus 12, delta 4 will become infinity or if delta 4 becomes infinity, then tau all x must be 0. Therefore, lambda equal to minus 12 represents separation. Values of u over infinity were not applicable in our profiles for lambda less than minus 12. We reorganize the integral momentum equation now by multiplying each term by u delta 2 by mu. So, you will see that our equation would now read as u infinity delta 2 by mu d delta 2 by dx plus u infinity delta 2 by mu divided by 1 over u infinity d u infinity by dx plus delta into delta 2 into 2 plus delta 1 by delta 2 equal to c f x, which is by 2, which is tau all over rho u infinity square into u infinity delta 2 by mu plus v w by u infinity into u infinity delta 2 by mu. Then you will see this term simply becomes u infinity by mu into d delta 2 square by dx divided by 2. In this case, u infinity gets cancelled with this and I get delta 2 square by mu d u infinity by dx equal to now rho into mu becomes mu and u infinity becomes u infinity. So, mu u infinity by tau all is delta 2 by delta 4 plus u infinity gets cancelled with that and I get v w delta 2 by mu. Therefore, multiplying throughout by 2, I get u infinity by mu d delta 2 square by dx equal to 2 times v w delta 2 by mu plus delta 2 by delta 4 minus delta 2 square by mu sorry this should be multiplied by 2 plus delta 1 by delta 2 I forgot d u infinity by dx into 2 plus delta 1 by delta 2. Again, the equation maintains its dimensionless form. Now, all I have done is in our derivation delta 2 by delta 4 is replaced by s and we call it shear factor. Delta 1 by delta 2 is replaced by h, which we call shape factor. Delta 2 square divided by mu d u infinity by dx and that would simply be equal to lambda times delta 2 by delta whole square as a pressure gradient parameter. Remember what was lambda? Lambda was delta square by mu d u infinity by dx. Therefore, this parameter which is kappa will be simply lambda times delta 2 by delta whole square and that is a pressure gradient parameter. v w plus will be v w delta 2 by mu is equal to v w star delta 2 by delta because v w star was simply v w delta by mu. Each term is dimensionless. You can see this is velocity, a length dimension because this is squared and this has a length dimension and this is nu. So, it is a kind of Reynolds number which represents rate of growth of momentum thickness delta 2. Kappa is a pressure gradient parameter. This is really a universal relationship. What we mean therefore, that the relationship applies no matter what is the variation of u infinity or the wall velocity. It is a universal relationship between integral parameters delta 2, delta 4 and delta 1. Let me go to the next slide. By universal, we mean that it is applicable applicable to all types of variation of u infinity including the ones we used in similarity method u infinity equal to c x raise to m. Kappa which is proportional to du infinity by dx implies when it is 0 implies a flat plate solution because u infinity equals constant. On the other hand, the right hand side f k or u infinity over nu d delta 2 square by dx equal to 0 implies that delta 2 is constant with x. As you will recall from our similarity method, it represents stagnation point solution because for m equal to 1, all thicknesses delta 1, delta 2, enthalpy thickness, heat transfer coefficient all are constants with x. That is the characteristic of a stagnation point solution. Shear parameter equal to 0, delta 2 over delta 4 equal to 0 implies that delta 4 is infinite which in turn implies that tau wall x is equal to 0 and it represents separation. All values of V w plus n lambda for which delta 1, delta 2, delta 4 are less than 0 must be discarded because for these values u over u infinity is either greater than 1 or u over u it is less than 0 which means those are inadmissible values of V w plus n delta lambda. Now, for u infinity equal to c x m, it is easy to derive that kappa will be equal to delta 2 star square divided by 2. I will show you how this is the case. If u infinity is equal to c x raise to m and kappa is equal to delta 2 star by nu du infinity by dx, then you will see that this becomes delta 2 square by nu c m x raise to m minus 1 or I can write this as c x raise to m into delta 2 square by x into m. I can write like that and this is nothing but c x raise to m is nothing but u infinity into delta 2 square by x into m which I can write as u infinity sorry there should be a nu here. So, divided by nu I can write this as u infinity x by nu into delta 2 by x whole square into m or that is equal to delta 2 by x square into re x into m. Now, if you recall in our similarity method we had defined delta 2 star as delta 2 by x re x to the half. So, this essentially becomes delta 2 star square into m and that is what I shown here. Similarly, you can show that f kappa which is equal to u infinity by nu d delta 2 square by dx can be shown to be equal to 1 minus m delta 2 square star square for that value of m. Similarly, here delta 2 square is corresponding to the value of m you are concerned with. You can also show S for example, what is S? S is equal to delta 2 by delta 4 and that is equal to delta 2 divided by mu u infinity by tau all x or that is equal to delta 2 tau all x divided by mu u infinity. This term can be shown to be equal to f double prime 0 into delta 2 star. It is simply a question of manipulation here and you will see the level at tau all x. For example, this will be delta 2 into mu times du by dy at 0 divided by u infinity into mu. So, mu and mu gets cancelled and I can construct here delta 2 by x re by x by half and you will get that f double prime 0. v w plus can also be shown to be b f the similarity parameter b f delta 2 star and delta 2 star m are available from similarity methods. These deductions are very important as we shall see shortly. Let me plot the results of remember I have calculated S v w plus kappa I have assumed values of kappa and calculated H also from the velocity profile that we assume that is delta 1 by delta delta 2 by delta 1 delta 4 by delta. For a boundary layer without suction or blowing the acceleration and deceleration parameters. So, acceleration parameter I have gone up to 12 and deceleration I have gone up to minus 12 because I cannot go less than minus 12. Then the kappa values take these values this is 0.095 and this goes on to minus 0.157. Delta 1 by delta 2 as you can see compared to 0 where it was about 0.3 with acceleration the displacement thickness reduces and with deceleration it increases. Same thing holds for a momentum thickness divided by delta it reduces and increases. Inverse is true for shear thickness the shear thickness goes on increasing whereas, on the negative side it when there is suction it goes on reducing. So, much so that at minus 12 it reduces exactly to 0. The shape factor H delta 1 by delta 2 is remarkably constant for highly accelerating on an acceleration side, but on decelerating side it goes on increasing quite significantly. This is the right hand side with some values negative and then positive. Now, what is f k to remember again f k is nothing but that value and when that is equal to 0 it means delta 2 is constant. Therefore, it represents stagnation point solution. If you look at here the stagnation point solution will be somewhere here and its value will be lambda equal to 7.052 and kappa equal to minus 7.824. The similarity solution for delta 2 star m equal to 0 was 0.63 and delta 2 star m equal to 1 was 0.292. Therefore, kappa for m equal to 1 will be 0.0841 and f for k m equal to kappa m equal to 0 would be 0.44. We will make use of these numbers very shortly, but remember this is the flat plate solution. 7.052 lambda is a stagnation point solution and minus 12 is a separation solution. This is what I have plotted here. On the x axis you have kappa, the pressure gradient parameter and f kappa, which is the rate of growth of momentum thickness parameter on the y axis. When kappa is positive, we have accelerating flow or favorable pressure gradient. When kappa is negative, we have the decelerating flow or adverse pressure gradient. The results are plotted when there is no suction and blowing and therefore, V w star is equal to 0. You will notice that when kappa is equal to 0, the value of f k must represent the flat plate solution. And when f k is equal to 0, as we just said, it must represent the stagnation point solution. So the intercept on the y axis represents the flat plate solution, whereas the intercept on the x axis represents the stagnation point solution. And f k values turn negative when you have very highly accelerated flow, whereas when the flow is decelerating, f k values are positive. Now, using the relationship between kappa and f k with integral parameters, I have also plotted the parameters here, the similarity solutions here. You will notice that this is the m equal to 1 solution, this is m equal to 0.33 solution minus 4, m equal to minus 4, minus 0.04, minus 0.065, minus 0.085 and minus 0.09 is the separation. The separation is seen at about kappa equal to minus, let us say about minus 0.07 in similarity solutions, whereas in integral solution, the separation occurs at minus 0.1567. So on the integral solution deviates from similarity solution for kappa less than 0, because we are allowed for arbitrary variation of u infinity, but for specific variation u infinity equal to c x m, the results go along there. Now, in order to develop close form solution, we can see that at least for very moderate decelerating flows and accelerating flows, a near straight line approximation can be made. This was done by a scientist called Thwaites. It is simply f kappa equal to a minus b kappa and the a will be the value of kappa equal to 0 and therefore, represents flat plate solution a delta 2 star square m equal to 0, whereas b will be simply when f k is equal to 0 or the stagnation point solution and therefore, b will equal a divided by delta 2 star square of m equal to 1. If you look at our previous slide, I have said delta 2 star m equal to 0 is 0.663. So delta 2 star square would be square of 0.663 which is 0.44 and therefore, delta 2 star m equal to 1 is 0.292. So 0.44 divided by 0.292 will give me this value of 5.17. So a will become equal to 0.44, whereas b will equal minus 0.4. So f k being equal to u infinity, so rate of growth of momentum thickness is equal to a constant minus another constant times delta 2 square by nu du infinity by dx. This is a Thwaites curve fit, a universal curve fit for the case in which suction and blowing are absent. We will make use of this relationship as you will see on the next slide. So just see, this was the relationship. I can manipulate this two terms, this term and this term as d by dx of du delta 2 square u infinity 5.7 equal to 0.44 nu to convince you. Let me open up again the equation. Then you will see d delta 2 square u infinity equal to 5.17 by dx will equal delta 2 square into 5.17 into u infinity raise to 4.17 du infinity by dx plus u infinity raise to 5.17 into d delta 2 square by dx. This is what it would mean. Therefore, if I divide this 1 over nu u infinity by raise to 4.17, I will get 5.17 into delta 2 square by nu du infinity by dx plus u infinity square by nu d delta 2 square by dx. That is equal to 0.44. Therefore, you will see that I can write this equation in this form. If I were to integrate this equation from 0 to x, then delta 2 square u infinity raise to 5.17 at x will equal delta 2 square u infinity raise to 5 by x equal to 0 equal to 0.44 nu 0 to x u infinity raise to 4.17 dx. The solution is applicable to any arbitrary variation of u infinity and restriction imposed by similarity method is now removed. We use this relationship to calculate delta 2 at any x because u infinity at that x will be known. Of course, you must know delta 2 square at x equal to 0. If you start from x equal to 0 itself where delta 2 is 0, then of course, that term will be 0. Evaluate kappa from du infinity by dx. Now that you know delta 2 square, you can now evaluate our kappa which is… Once you have evaluated delta 2, you can now evaluate delta 2 square nu du infinity by dx because you already know what u infinity x is. For this value of kappa, you can evaluate s value. If you are knowing the kappa value, you can interpolate to get s value. You get a shear stress value from which you can evaluate s and delta 4. From delta 4, you can evaluate the skin friction coefficient to nu over delta 4 u infinity which is what we wish to evaluate anyway. That is the purpose. In other words, knowing u infinity as a function of x, we get delta 2 as a function of x from which we get kappa from which we get s as a function of x. In fact, we get all other parameters delta 1 as a function of x and so on and so forth because we know the shape factor variation with x and we can get C f x also as a function of x. In the previous table, I had held v w equal to the no suction and blowing. Now I am going to set lambda equal to 0 which is the case of a flat plate, but allow for suction and blowing and that is what I have done here. I allow for blowing parameter to go up to 5. On the suction side, I go up to minus 4.2. Delta 1 by delta with blowing compared to v w star equal to 0, the delta 1 by delta increases as we expect. Delta 1 by delta decrease as we increase suction. This also increases. It is not seen here because I have plotted results only up to second decimal place and on this side it reduces, but notice that at minus 4.4 delta 2 has already turned negative and that is not permitted. So, I cannot go below lambda less than minus 4.2. So, feasible solutions are possible only for lambda greater than minus 4.2 as you can see here. The shear stress also has almost vanished here which means this is where separation is about to take place. The shape factor is increased enormously to 47.25 from its average values around 2.7 on the positive side, on the blowing side and on this moderate suction rates, it is about 2.4, but it increases very rapidly to 47.5. This is almost the separation polar profile and these are the values of f k. Likewise, I have now included effects of both lambda and for a certain v w star which is minus 0.2. That means it is a suction case with v w star equal to minus 0.2 and here I have gone up to 15, but notice that beyond 14.8 or so, s has become negative. Therefore, this is not acceptable solution. So, lambda equal to 15 is not acceptable. On the adverse pressure gradient side you will see I have gone up to minus 13, but at minus 12 it is 0 already. So, this is the separation occurs and minus 13 is minus 0.03. So, this is not advisable. So, effect of pressure gradient on at a certain suction state is valid between minus 12 and 14.5 only. The remarkable feature of this solution is that for very mild acceleration to all the adverse pressure gradients, the value of v w plus is almost constant. In this v w plus is almost constant on the suction side. For lambda equal to 0, f k is about 21. You can see that 0.21 and that was equal to delta 2 star square and v w plus equal to minus 0.21. This amounts to v f equal to minus 0.21 divided by root 0.21, 0.458. This is the blowing side and again you will see for adverse pressure gradients less than minus 12 you have negatives. So, you cannot go below that. Again on v w plus is remarkably constant and this will correspond to v f equal to about 0.2619. For simultaneous variations of v w star and lambda, close form solutions can again be developed. In the regions in which v w plus is constant, you can curve it f k equal to a minus b k or a relationship of this type can be established where a and b are functions of v w plus. So, v w star equal to minus 0.2, you get a equal to 0.21, b equal to 4.2, v w star equal to plus 2, you will get 0.84 and 7.4. Manipulation would give again d delta 2 d by dx of this equal to that and therefore, you will get a solution. The procedure remains exactly the same as before. Only the values of a and b change with value of v w star. So, I am taking now a case of flow over a cylinder. It is an impervious cylinder. So, there is no suction or blowing. There is an approach velocity v a. Potential theory will show that the free stream velocity u infinity would vary as u infinity by v a equal to 2 times sin 2 x star where x star is x divided by diameter. So, 2 x star is nothing but x divided by radius and I will call this f x star. Then for this variation of free stream velocity, I can show that delta 2 by d red will be simply from equation here using this relationship. I can determine delta 2. Remember delta 2 at x equal to 0 at the stagnation point will be 0 and from there I integrate. So, I can show that delta 2 by d red will be that and kappa will be this. Our objective is to determine the location of the separation point corresponding to kappa equal to minus 1567. Red is equal to v a d by nu which is the Reynolds number defined for flow over a cylinder. Here are the results. u infinity by v a is a sin function which goes and beyond 90 deceleration sets in whereas, on the below 90 degrees there is a flow acceleration and this is the variation of kappa and you can see it has reached minus 0.1567 at about 108.3 degrees and therefore, it is associated with separation. So, we have located the separation point from stagnation point from the known velocity distribution arbitrary velocity distribution. Similarly, now consider a cylinder in which there is a blowing taking place from the cylinder surface. Then I can curve it as I said f kappa in this manner for different values of v w star and theta separation what v w star equal to 0 was 108 and that goes on reducing. You will expect that a flow over a cylinder with blowing would the separation would occur at an earlier location and that is what you see up to v w star at point 2. The separation point has advanced. Average skin friction up to the separation point defined in this manner of course, is highest at v w star equal to 0, but with blowing skin friction was on reducing. So, as expected separation angle is advanced with increase in blowing rate with reduction in average skin friction due to thickening of the boundary layer. This shows you the power of the integral method. Of course, what it cannot do is to go beyond the point of separation and complete the analysis of flow over a cylinder, but nonetheless it is a very useful tool to determine flow through conversion or diversion nozzles. For example, where the boundary layer where the free stream velocity would either accelerate or decelerate with x and you want to determine the thickness of the boundary layers developing on the wall, because it is this thickness which determines the discharge coefficients of such nozzles.