 In the last lecture, we derived the boundary layer equations of a two-dimensional boundary layer and I ended by saying that there are three methods of solving these equations. The first one is the similarity method, the second the integral method and the third is the finite difference or the finite element method. Today, I am going to develop the similarity method. The topics then are to introduce the notion of similarity of profiles. I will establish what are the conditions for existence of similarity solutions and then I will develop the similarity equation with appropriate boundary conditions. So, let us recall the equations. The equation is the this is the so called mass conservation or the continuity equation for a constant property steady flow d u by d x plus d v by d y equal to 0. These are the convection terms in the momentum equations, the pressure gradient and the viscosity affected term mu d 2 u d y square. We said that if this equation was written in the infinity state, then the pressure gradient could actually be written as rho u infinity d u infinity by d x. I shall be making this substitution here in all subsequent developments. Now, the boundary conditions are obvious at y equal to 0 that is at the wall the stream wise velocity u will be 0, but there may be suction or blowing at the wall. So, we shall say v equal to v w which may be function of x at y tends to infinity u of course, will tend to u infinity at x which is the free stream velocity. So, we have two variables here v w as a function of x and u infinity as a function of x and when u infinity as a function of x of course, represents the pressure gradient according to equation 3 here. So, what is the notion of similarity profiles? The term similarity is associated with the possibility that under certain conditions the velocity profiles at different stream wise locations x 1, x 2, x 3 say in the boundary layer will be similar in shape. So, here is a surface over which the flow is flowing and the boundary layer development is taking place. The velocity profile at x 1 is as shown here at x 2 it as shown here and at x 3 it is as shown here. Now, of course, in terms of boundary conditions they are u equal to 0, u equal to u infinity here, u equal to 0, u equal to infinity and u equal to 0 and u equal to infinity. The question we ask is they look similar yes, but can they be collapsed onto a single curve? Now, this idea of a collapse is very important and that is really what underlies the similarity. The profiles at different positions x 1, x 2, x 3 will collapse on each other if their shapes at each position were equal. What does that mean? It means that the actual magnitudes of u at some y at different locations would differ simply by a stretching factor s that is a function of stream wise distance x. In other words, if the value of y here, the physical value of y here and the physical value of y here were stretched by s as a function of x, then it is quite possible that the velocity profiles would collapse on each other. Basically, then what we are saying is u which was function of x and y can now be written as u as a function of eta bar which is a new variable. It is called the similarity variable v which was a function of x and y is a function of v eta bar and eta bar is equal to y the transverse distance multiplied by some function of x. What we are aiming at is equations which are functions of two variables x and y are going to be made functions of one variable. Therefore, a partial differential equation would now be converted to an ordinary differential equation in eta bar. Let us see how this is done on the next slide. The relation suggests that if similarity exists, we want to search for conditions for similarity to exist. The relations that I showed on the last slide suggest that if similarity exists, then the velocity profiles u and v at any stream wise location can be collapsed on a single curve. This is because u and v that were functions of two independent variables x and y are now functions of a single variable eta bar only and the partial differential equations therefore, will be reduced to ordinary differential equations. But such a reduction would be possible only when u infinity x v w as a function of x and the stretching factor s as a function of x assumes certain restricted forms known as similarity conditions and it is these special forms that we wish to discover and they are called the similarity conditions. So, I have written the momentum equation again u rho u du by dx v du by dy rho u infinity du by dx plus mu d 2 u dy square. We have already introduced a variable called eta bar as y into s x. We introduce another variable psi x y such that d psi by dy is defined as u and d psi by dx is defined as minus v. Then you will notice that if I substitute for u in the continuity equation and for v also in the continuity equation, I will get d 2 psi d x dy minus d 2 psi d x dy equal to 0. Therefore, these definitions have not fallen from the skies. It is simply they satisfy the continuity equation psi is called the stream function. If I now were to substitute for all these quantities, then you will see the momentum equation will become and I am already going to divide by rho. Then the momentum equation will become d psi by dy into du by dx would be d 2 psi by dx dy minus d psi by dx which is v d 2 psi by dy square equal to u infinity du infinity by dx plus mu divided by rho which is nu times d 3 psi by dy cube. This is the psi equation we would get. This is the psi equation we would get where psi already satisfies the continuity equation du by dx plus d v by dy equal to 0. I am now going to make some further definition. Now, it is obvious since u is a function of x and y and v is a function of x and y, psi would also be a function of x and y and that function I am introducing as psi x y equal to some function n x into z times eta bar z as a function of eta bar. I am also going to define u over u infinity equal to d z by d eta bar and since z is a function only of eta bar, I can also write this as z dash. Now, we wish to establish the relationship between eta bar u infinity and so on the n. So, we have introduced another function n x. We have already got one function u s x and we have the third function u infinity x. We want to establish a relationship between these three variables. We will recall that u over u infinity as I define is d z by d eta bar and that is equal to 1 over u infinity into d psi by dy. But notice that I can write this as 1 over u infinity d psi by d eta bar into d eta bar by dy which I can write as now what is d psi by d eta bar? You will see d psi by d eta bar would be d psi by d eta bar would be simply n times d z by d eta bar simply because n is a function of x and z is a function of eta bar. Therefore, you will see I can replace this as n over u infinity into d z by d eta bar into d eta bar by dy. So, I can cancel d z by d eta bar on both sides and I will have that as 1 or d eta bar by dy will be equal to u infinity by n both are functions of x and that would equal s x as per our definition because remember eta bar is y into s x. So, d eta bar by dy would be simply s x very important relation this because it establishes relationship between three functions of x that we have introduced. Now, of course, it follows quite obviously d psi by dy would be equal to u infinity into d z by d eta bar from this definition itself u over u infinity is d z by d eta bar d 2 psi by dy square would then be u infinity square by n into d 2 z by d eta bar square d 3 psi by dy cube will be u infinity cube divided by n square d 3 z by d eta cube d psi by d x that requires a little remember psi is n times z and x times z eta bar. Therefore, d psi by d x would be z times d n by d x plus n x d z by d eta bar d eta bar by d x because remember eta bar is function both of y and x d psi by d x equal to z d n by d x plus n z by d eta bar d eta bar by d x. We also need d 2 psi by d x dy and that would be d by d x of d psi by dy and d psi by dy is simply u infinity into d z by d eta bar and multiplication of the differentiation of a product would give d z by d eta bar d u infinity by d x plus u infinity d 2 z by d eta bar d eta bar by d x. So, if I were to introduce all these quantities in the last equation on the previous slide then you will notice that I would get I am repeating the equation d 2 psi by d x dy minus d psi by d x d 2 psi by dy square equal to u infinity d u infinity by d x plus nu d 3 psi by dy cube and if I go to the next slide again this will d psi by dy will be u infinity into d z by d eta bar into d 2 psi by d x dy is d z by d eta bar into d u infinity by d x plus u infinity into d 2 z by d eta bar square into d eta bar by d x this is the first term minus d psi by d x is z times d n by d x plus n times d z by d eta bar d eta bar by d x into d 2 psi by dy square which is simply u infinity square by n into d 2 z by d eta bar square and that would equal u infinity d u infinity by d x plus nu times u infinity cube divided by n square into d 3 z by d eta cube. Now, observe that u infinity d z by d eta which is really the d psi by dy multiplied by u infinity d 2 z by d eta square d eta bar d x in this first term cancels because of the negative sign here with the second term here n d z by d eta bar d eta bar by d x u infinity square by n d z by d 2 z by d eta bar square because n n n cancels here. So, this term and this term vanish and as a result then I would have the equation would look like this u infinity u infinity into d z by d eta bar whole square into d u infinity by d x would be the first term that is what I have written here. The second term will be minus z d n by d x divided multiplied by u infinity square by n d 2 z by d eta bar square and that is the term I have written here would equal u infinity d u infinity by d x plus nu u infinity cube by n square d t 3 by d z d eta cube and I simply replace these by prime quantities z prime square z z prime z into z double prime and this become the z triple prime here. What are the boundary conditions? First of all at y equal to 0 eta bar equal to 0 I have u equal to 0 or that will be equal to z prime 0 is 0 that is the no slip condition which I mentioned here. In the infinity state u equals u infinity and therefore z prime u infinity will equal 1, but at y equal to 0 there is suction of blowing velocity v w and v w will be d psi by d x n eta bar equal to 0 and since psi is equal to n x z eta bar I will have n x d z by d eta bar plus z d n by d x all that eta bar equal to 0 and if I replace d z by d eta by z prime 0 this will be 0, but then as you know this is due to nu slip condition that is 0 and therefore I get v w is equal to minus z 0 d n by d x or in other words z 0 equal to minus v w divided by d n by d x as shown. Remember this is a third order equation and therefore I need three conditions I have provided two conditions at eta bar equal to 0 and one condition at eta bar equal to infinity and therefore the problem statement now is complete. When would this equation be a perfect ordinary differential equation? It would be a perfect ordinary differential equation only if beta 1 and beta 2 are absolute constants and not functions of x. Remember beta 1 is entirely a function of x, beta 2 likewise is entirely function of x and z 0 which is v w divided by d n by d x is also a function of x, but unless these three quantities are absolute constants beta 1, beta 2 and z 0 are absolute constants we do not have an ordinary differential equation with appropriate boundary conditions. That tells us in a way what the similarity conditions are going to be. The equation z triple prime plus beta 1 z z prime plus beta 2 into 1 minus z prime square 0 will be an ODE if beta 1, beta 2 and z 0 are absolute constant. Consider now 2 beta 1 minus beta 2 we have already written the definitions of beta 1 and beta 2 on the previous slide. I am writing now 2 beta 1 minus beta 2 equal to 2 n divided by nu u infinity d n by d x minus n square by nu u infinity square d u infinity by d x. If I were to write the right hand side you will see it can be written as d by d x of n square divided by nu u infinity where nu is the kinematic viscosity and it is a constant. If I were to integrate this equation from x equal to 0 to x equal to n then I will get n square by nu u infinity equal to 2 beta 1 minus beta 2 times x because beta 1 and beta 2 are now taken as constants with respect to x. I now multiply both sides by 1 over u infinity d u infinity by d x then you will see n square nu u infinity square d u infinity by d x will equal 2 beta 1 minus beta 2 x divided by u infinity d u infinity by d x. But if you see the left hand side now it is nothing but beta 2 and therefore I have a relationship that d u infinity by d x d u infinity by u infinity will be equal to beta 2 over 2 beta 1 minus beta 2 into d x by x. So, if I integrate this it gives me the first similarity condition that u infinity must vary as c x raise to 2 beta 2 over 2 beta 1 minus beta 2. If I integrate this from 0 to x you will see l n u infinity will be constant times l n x and therefore the relationship follows. So, l n u infinity equal to same constant l n x or l n u infinity will be constant times c x raise to beta 2 over 2 beta 1 minus beta 2. What about n x? So, you will see n square over nu infinity is equal to 2 beta 1 minus beta 2 x. So, n will be simply all if I multiply true by nu infinity and take a square root then n will be under root nu infinity 2 beta 1 minus beta 2 x eta you will recall was defined as y s x which was shown as y u infinity by n and if I substitute for n here then you will see y under root u infinity. Then similarly psi which was z eta bar into n x would read like that and z 0 v w over x d n by d x equal to constant. So, now we have found the variable eta bar in terms of beta 1 and beta 2 u infinity in terms of beta 1 and beta 2 and psi again in terms of beta 1 and beta 2 and z 0 again likewise and that must be a constant. Now, there is something very useful we can deduce. Remember the equation is an ordinary differential equation as long as beta 1 beta 2 and z 0 are constants. So, I can arbitrarily choose these values I can set for example, beta 1 equal to 1 and beta 2 equal to beta that is what I have done without loss of generality I can set beta 1 equal to 1 and beta 2 equal to beta. Then the equation would read at z triple prime z double prime plus beta 1 minus z square 0 u infinity will simply read as c x beta over 2 minus beta. Now, this type of variation of u infinity as a function of c x raise to beta u infinity equal to c times x time over beta over 2 minus beta has a very special significance in fluid dynamics and that is what I have shown here. The potential flow theory shows that u infinity c x into a c x raise to beta over 2 minus a presence flow over wedges of included angle pi beta as you can see this is a wedge of included angle pi beta and the flow is this way. So, the free stream velocity variation would follow u infinity into c x beta over 2 minus beta this is what you get from the potential flow theory. If beta is equal to 0 that is u infinity equals constant that would straight away give you the flow over a flat plate which is a wedge with included angle 0, but I can also open up the wedge fully and make it into pi beta equal to 1 or beta equal to 1 and I will have what is called as a stagnation point flow, because the flow would hit a plate perpendicular to it and would tend to go this way as well as this way and in the positive x direction the velocity profiles would be like as shown here, but I can also have negative values of beta and that is what is shown here. Now, you can see quite intuitively you will see that when a flow jumps over a hump it is quite possible that downstream of the hump beginning of the hump the velocity profile will go through a point of zero shear and then the flow actually will recirculate with the negative velocity close to the wall. Now, that actually happens for beta less than minus 0.2 as we shall show. Now, when recirculation occurs of course, our boundary layer theory falls flat in the sense that it is no longer applicable, because we entered the regime of elliptic flows. So, beta equal to minus 0.2 has a very special significance, because that is the point at which the flow at best would separate that is have a zero shear velocity. If you increase beta beyond that value then there will be recirculation. So, this generalization beta 1 equal to 1 and beta 2 equal to beta renders this equation with some physical rigor and of course, z 0 v w d n by d x must be constant. So, the set of equations that you see here it tells you now how u infinity should vary, how n should vary with x, how eta bar be defined, how psi be defined and how z 0 or in other words v w should vary with x, so that the equation is truly an ordinary differential equation. I am now for further discussion, I am going to change the definition. I am going to say u infinity c x raise to beta over 2 minus beta will be written as c x raise to m, where m is equal to beta over 2 minus beta or beta is equal to 2 m over m plus 1 eta. Therefore, I am changing eta bar to eta and eta I am defining as y under root u infinity by nu x which will be equal to eta bar times under root 2 minus beta. Remember f dash eta equal to z dash eta bar equal to u over u infinity. Now, the reason I am doing this is to make further analytical development more elegant, there is no other purpose other than making the analysis more elegant. Having to write beta over 2 minus beta every time makes life somewhat difficult, so I have simply defined m equal to beta over 2 minus beta or beta is equal to 2 m divided by m plus 1. I have simplified eta by removing b 2 minus beta into it, so now the new variable is eta. I am replacing z by f and saying f dash eta will be equal to z dash eta bar equal to u over u infinity as before. With these substitutions then, you will notice that our equation z triple prime plus z double prime plus beta into 1 minus z prime square equal to 0 and I am changing this. Remember I have said the f dash eta equal to z dash eta bar equal to u over u infinity and I have also defined. So, you can see now what would be z double prime z double prime will be d z prime by d eta bar r square z z prime by d eta bar, but z prime is f prime eta, which I can also write as d f prime by d eta into d eta by d eta bar, but remember eta is equal to eta bar under root 2 minus beta and therefore, z double prime will be simply under root 2 minus beta into f double prime that is equal to z double. Likewise, you can show z triple prime will be 2 minus beta f triple prime and you can also show that z will be equal to f under root 2 minus beta. So, with these three substitutions in this equation, so I will have 2 minus beta f double prime I mean triple prime equal to f divided by under root 2 minus beta plus into under root 2 minus beta f double prime plus beta into 1 minus f double square equal to 0 and you will see this and this gets cancelled. So, you get essentially 2 minus beta f triple prime plus f f double prime plus beta into 1 minus f prime square equal to 0. Now, as you know we have already defined beta equal to 2 m over m plus 1, if you were to substitute for 2 minus beta and beta you will see this equation reduces to f triple prime plus m plus 1 by 2 f f double prime plus m times 1 minus f prime square equal to 0. So, in the new similarity variable f and eta appropriate z equation will transform to this equation in f psi is equal to f eta under root nu infinity by x and therefore, v will be equal to d psi by d x. If you differentiate this equation, you will see psi f eta under root nu u infinity x which is nearly the n x. So, n x is equal to under root nu u infinity x, but u infinity is c x raise to m. Therefore, nu c x raise to m plus 1 or that will be equal to nu c x raise to m plus 1 by 2 d n by d x will be under root nu c into m plus 1 by 2 into x raise to m minus 1 by 2. Similarly, eta bar I mean eta which is y times under root u infinity by nu x will become equal to y times under root c by nu x raise to m minus 1 by 2 x raise to m minus 1. This is very important d n by d x and therefore, d eta by d x will be simply y times under root c by nu m minus 1 by 2 into x raise to m minus 3 by 2. This is another equation of great value v which is equal to minus d psi by d x which is minus d by d x of f eta n. Then you will see this will be simply equal to minus into f d n by d x plus n into d f by d eta into d eta by d x. If you substitute the last two derivations of d n by d x this quantity and for d eta by d x this quantity and replace this by f dash then this is nothing but f d n by d x plus n f dash d eta by d x then simply substitute these two quantities. You can show after some algebra that v by u infinity under root u infinity by x by nu which is nothing but v over u infinity r e x to the half is equal to minus m plus 1 by 2 into f plus m minus 1 over m plus 1 into f dash eta. That is what I have shown here. You now get a solution to v in terms of f and f dash and eta. So, v is a function of eta is known in terms of the pressure gradient parameter m. At f 0 you have v w by d n by d x. So, that it will transform into v w x by u infinity r u x by x because all I do is f at f equal to 0. I have simply put f 0 and f dash 0. f dash 0 is 0 already as you now become the no slip condition. So, essentially f 0 then becomes v w divided by u infinity by x r e x to the half is the b f which is called the blowing parameter r e x to the half and r e x is u infinity x by nu. We have an equation f triple prime m plus 1 by 2 f double prime m 1 minus f dash square equal to 0 with the boundary conditions f 0 f dash 0 and f dash infinity. So, let me summarize. So, we have transformed the two rebound layer partial differential equation to a third order ordinary differential equation f triple prime plus m plus 1 half f f double prime m 1 minus f double prime square 0. The ODE is valid for u infinity c x raise to m and v w u infinity r e x equal to b f equal to constant only. So, the solution for any arbitrary variations of u infinity and v w the equation is not an ODE. So, these are called the similarity conditions. So, the solutions can only be obtained for these restricted variations of u infinity and v w. The independent similarity variable is eta y under root u infinity divided by nu x for m greater than 0 because u x c x raise to m for m greater than 0 u infinity will increase with x or we say we have a accelerating flow or a negative pressure gradient. Likewise, if m is equal to 0, we say we have a decelerating flow or a positive pressure gradient or an adverse pressure gradient. For m equal to 0, we have already noticed that u infinity is a constant and therefore, the pressure gradient is 0. It is called the flat plate flow. For m equal to 1, we have a stagnation point flow or an accelerating flow and overall, we say m is a pressure gradient parameter. In the next lecture, I will show you how to solve these ordinary differential equations.