 So, good afternoon. Today we will be carrying on with the second half of our work on curved channel flow. So, just to recap we are looking at single phase flow through a curved circular pipe and we have adopted a coordinate system where we have a polar coordinate system in the cross section and the position of the cross section is given by the angle theta from the x axis. So, we quantified the curvature of the channel with the curvature ratio epsilon and that was given by the radius of the channel A divided by the radius of curvature zeta and so we realize that if the quantity zeta tends to infinity then we will approach the case of a straight channel which means that when epsilon is 0 curvature ratio is 0 I will have a straight channel and when epsilon greater than 0 but of course less than 1 I will have a curved channel. So, this one bound is physical characteristic of the geometry the this radius cannot be more than this radius. So, that is the general idea. So, realizing this fact we wanted to apply what we have studied in the course basically perturbation theory and try and understand flow in a curved channel as a small approximation of perturbation about the case of epsilon equal to 0 which means about a straight channel. So, yesterday we had derived the Navistokes equations in this toroidal coordinate system scale the equations and got the governing equations in terms of epsilon. So, today what we are going to do is carry or forward the usual procedure whereby we take an asymptotic expansion in epsilon substituted back in and derive the equations for the zeroth order problem first order problem and so on. So, before I go to that I want to make a point here that I have said the straight channel case we approach when zeta tends to infinity but even for a finite zeta for a reasonable zeta you could still approach the case of a straight channel if the radius became very small. So, why is that important you imagine you have a micro channel a micro channel which has a bend like this now the curvature ratio of this bend zeta is definitely not infinity it is a very finite value it is quite a sharp bend in fact but you could still approach the limit of epsilon tending to 0 provided the a is small. So, even though one way of looking at the problem is a gently curved channel one way of looking at small epsilon another way is a finite curvature but a very small channel or a narrow gap approximation. So, this is where scaling comes into picture. So, even though epsilon is not absolutely tending to infinity if it is relatively large compared to a that is good enough because the length scale in the problem is a. So, with respect to a zeta should be large. So, this is once again where the question is what is the physical scale that is important that is what we do when we scale the problem essentially. So, that is why by scaling with a throughout the problem we got epsilon. So, we put epsilon to 0 it could mean either that a is very small or that zeta is large and that does not really matter as far as the equations are concerned. So, I could basically have a thin channel with sharp curvature or a large channel with a very gentle curvature and both these guys will have the same epsilon kind of geometric similarity. So, with this in mind what I will do now is directly move on with the perturbation calculation. So, let us consider order of epsilon 0 and in the previous class you had written down the equations the full governing non-linear equations. So, the first step of course is to consider the solution for v u v w p. So, that is our standard power series expansion where we expanded the vector field as the zeroth order field plus epsilon times first order variables. So, we take the asymptotic expansion and substitute back in those governing equations that we had yesterday and we will arrive at the equations for various orders. So, I will start with the zeroth order problem and that we can when you do the calculations to the equations and even by inspection you will get simply the equations for the straight channel and that simply Hagen-Poiswell flow. So, what we will do is I will not write down the full set of equations that you can derive, but the main idea is in those equations you will have all the inertial terms and the viscous terms, but that is when we will say that we are interested in looking at unidirectional flow through the channel as the base case. So, in that case all the inertial terms will get knocked off and we will be left with the normal Stokes equations that we get when we solve Hagen-Poiswell flow. So, the equation on epsilon will simply be that is because you remember we scaled the equations with G. So, we would have got a plus 4 over there and only the radial variation of W naught persists. So, U and V have gone to 0 because we are saying that the flow is unidirectional. So, this is a standard Hagen-Poiswell approximation. So, finally we will get the solution at zeroth order to be 1-R square. Actually this is – 4 because delta P and del square V are on the same side take it to the other side it is – 4. So, that is the standard parabolic velocity profile if we plot it that we get in Hagen-Poiswell flow through a state pipe. Most importantly U naught V naught above 0 which means that there is no cross velocity in the pipe at all. There is no circulation nothing else is happening it is just flowing straight. So, that is a state pipe solution. Now what we are going to do is when we go to order epsilon 1 this axial flow W naught is actually going to move around the bend. So, W naught will give rise to centrifugal forces that will come into the problem at order epsilon 1. So, this is the typical stepwise procedure in perturbation calculations the zeroth order effect is just flow and at first order we will start feeling the effects of centrifugal forces. So, once again if we take the asymptotic expansion and go to order epsilon 1 we can derive the equations. So, I will directly write those down and I leave it as an exercise for you to derive them. It is straight forward the usual procedure that we have done so far. So, at order epsilon 1. So, while I am writing this down you should look at the equations yesterday and see if you can identify where each term is coming from the original non-linear equations. So, for example this term is coming from the centrifugal force term in the governing equations and these are the viscous terms which are order epsilon. So, just look and see if you can identify where it is coming from. So, these are the 4 equations that we get at order epsilon 1. So, you have the momentum equations in U in the radial direction in the eta direction in the axial direction and the continuity equation which is equation number 4. So, there are some interesting things that we need to see in these equations. The first point is that centrifugal force is an inertial term it is basically a non-linear term in the Navier-Stokes equations. So, when you go back you will see that you had W square you know in the governing equations. So, that term actually is non-linear and if you want to solve it you would have to solve the problem numerically. But what we have done here is by doing the perturbation approximation we find that that term which was non-linear before has come into the problem now as a linear term because although we have W naught squared W naught is already known from the base problem as 1-R square. So, if you look at this problem which is in terms of the variables U 1, V 1, W 1 and P 1 you will find that it is completely linear set of equations firstly. Secondly it is non-homogeneous with the non-homogeneity being the centrifugal force term which has W naught squared. So, this thing is typical of perturbation calculations in general and you can think of it as a weakly non-linear calculation. So, the purely linear calculation would have never given us circulations because that would be equivalent to putting Reynolds number equal to 0. That is the creeping solutions or stoke solutions just puts Re equal to 0. So, you neglect all the inertia. If we did that even at first order we would have got 0 and 0 here and the whole set of equations would be completely homogeneous which means that even U 1, V 1, W 1 would have been 0. So, I would have just persisted with you know flow going around a bend but without any circulations. If you take Re actually to be 0 but if you allow for non-zero Re but say that wait even though there is some Reynolds numbers effects that those effects are very weak. We said that by saying small epsilon. So, when we did that you do retain Re and you retain its contribution but you are able to do that in a way that allows you to calculate step by step analytically. So, in that sense you are counting for the first effects of the non-linear centrifugal force terms. So, these are those two terms that are basically in homogeneity is now in this set. This is the equation for the correction to the velocity to the axial velocity. So, W naught is what it is in the state channel when it curves it will be corrected by W 1. So, we actually find this is an equation for W 1 and that of course is the continuity equation. So, the third important thing to note is that if you consider equation 1, 2 and 4 you will see that nowhere do they contain W 1 right. So, these three equations the momentum equations for U and V U 1, V 1 and the continuity equation are a self consistent set for you know the two the three variables U 1, V 1, P 1 and you have three governing equations. So, we can actually solve 1, 2 and 4 first and then come here and solve equation 3 for W 1 yes sorry yeah good question yeah I will come to this term actually you will actually end up retaining yeah that is it. But I will come to that later. So, this is the change in the pressure gradient in the theta direction alright. So, what I was saying previously is that we have three equations for three unknowns that we can solve first for U 1, V 1, P 1 and then we can come and solve for W 1 is that clear. So, this will always happen when you have fully developed flow because in fully developed flow you want have changes in W 1 changes in W with theta in the continuity equation. So, the continuity equation will only involve U and V. So, that along with these two equations allows me to solve for U V first and then I can solve for W 1 which is the correction to the axial velocity. So, once again we come back to the I mean the whole methodology of perturbation series. So, what we have done here if you look at it physically is to say that I have a flow the primarily flow is axial just going through the channel. So, that you first calculate that is W naught. Then using W naught you get a centrifugal force that is W naught square and you use that to calculate the circulations which are U 1 and V 1 that is coming from the centrifugal force W naught square. So, you will get U 1 V 1 which will be some circulations and some pressure gradients P 1 and that you use to calculate now the change on W itself which is W 1 and if we keep going forward we will keep doing this. So, first calculating the circulation then coming back and seeing its effect on axial velocity then again going back in calculating the circulations and so on. So, this iterative procedure I mean makes physical sense because you get the base flow find the centrifugal force find the circulations come back to the base flow and by doing that we will increasingly improve our approximation. So, today we will just look at order epsilon which means we will just solve these equations up to W 1 alright. So, now that we have got the equation and understood what we are about we will get down to the mathematical solution. So, as I said we can solve 1 2 and 4 for U 1 V 1 and P 1. So, how do we go about this you can see that the country equation actually has the form of a 2 dimensional system because there is no W 1. So, we can use the stream function formulation that we have done previously for 2D flows. So, stream function formulation works not only for 2D flows but also when you have free developed flow because the form of the continuity equation allows me to do that. So, if I define I will call it psi 1 because it is at order 1 or sorry right r eta and this will be identical to the stream function formulation we have in cylindrical coordinates. So, in fact so the ideas we have replaced 2 of the variables in terms of 1 variable psi 1. So, now that we have this what is the next step that we do when we are looking at stream functions right. We have to eliminate pressure from the equations which amounts to taking the curl of the vectorial form. So, what we need to do is differentiate we want to eliminate these 2 terms right. So, we differentiate this with respect to r and differentiate this with respect to eta multiplied by 1 over r and subtract right. So, that will knock off these 2 terms the other important thing is remember this equation will be will operate on the whole equation right and on this guy we will do derivative with r. The important thing to realize is that when we take the derivative with eta we will get cos eta here whereas derivative with r will leave cos eta free. So, when we differentiate and add you will get cos eta together. So, you will not have sin and cos both the forms will be cos only. So, that will tell us that ultimately your stream function psi 1 will be of this form you can split it up into its r dependence and the cos eta dependence because both the inhomogeneous terms will end up having cos eta right and cos is a even function about eta. So, that means that my stream lines will be symmetric about the left half and the right half right. See eta equal to 0 is here. So, if you have cos eta minus eta and plus eta will be equal for cos. So, the stream functions or whatever the circulations we have will be symmetric about the vertical plane. So, that is something to keep in mind, but just proceed with the calculation differentiate this with eta this with respect to r and then subtract the two equations. Now, I am differentiating this with 1 over here we will get some minus 1 by r square or something second equation. Yeah, okay fine yeah yeah. So, first multiply by r and differentiate. Yeah, that will be better. Yeah, that will be good. So, you first multiply by r and then differentiate with r. All right. So, after you eliminate those two equations if you have not been able to get it so far you all can go back and work it out, but it will work itself out. You will land up with an equation for psi 1. So, most important thing to check is since these equations were second order naturally the equation for psi should be fourth order because we have combined two equations into one. So, you can see that it is in fact a fourth order problem. So, you have del square operating on del square psi where del square is given by this operator it is just the del square and cylindrical coordinates and the right hand side contains your inhomogeneity which will have Reynolds number because that gives me the amount of centrifugal force and you have this cos eta dependency like I pointed out earlier because a sin eta gets differentiated to cos eta. So, now whenever you have these types of PDs you need to see whether you can separate the r dependency and the eta dependency. So, in this case if the right hand side the inhomogeneous term was 0 then psi would be simply 0 because you just have no slip homogenous conditions on the wall but it is not 0. So, that is why you will have some circulation but the eta part is everything in cos eta. So, it makes sense to propose a solution that you have psi 1 equal to let me say some function of r which we do not know times cos eta because if we consider this form here the cos eta dependency would get cancelled out. So, we will just check that now whether if we can put this solution back here whether the cos eta parts go off then we will get an ODE for r. So, we can just either. So, if easy way is to consider what happens to del 1 square because whatever happens here will get repeated again right in the other operator. So, if you put this back here in del 1 square what will you get you operate this on psi 1 right you are going to operate this on psi 1 that is this inner operation. So, f of r will be retained here and cos eta will just pop out right plus 1 by r dou by dou r operated on this term will retain the f inside the operator and again cos eta will come out. And the final term however you are differentiating with respect to eta twice right. So, first derivative with cos eta will give me sin eta with a minus sign second derivative will give me back cos. So, I will simply get minus right. So, this whole thing del 1 square of psi 1 can be written as del 1 square bar which is a different operator acting on f into cos eta ok where del 1 bar operator is dou square by dou r square plus 1 by r dou by dou r minus 1 by r square. So, I hope that is clear you have any doubts about this. So, the cos eta just pops out. So, if we substitute this solution here what will end up happening is you will get del 1 bar square of it because the cos terms will get cancelled off from either side alright. So, what we ended up with is a fourth order ODE for r. So, at this point this ODE still look a little intimidating but actually is very easy to solve. So, that will come by simply realizing that we can rewrite the operator in a more compact way that allows us to integrate it 4 times and get the solution. So, how can we do that if you look at this expression it can be written as dou by dou r of something. So, what is that of this will give us the first term right plus u by r I think right. So, this will give me the first term and if I differentiate taking r out I will get sorry f I differentiate taking r out I will get this term. Then if I differentiate r I will get minus 1 by r square times f now this itself can be written as another nested derivative I guess that is right. So, now we have written del 1 square in a nested form I am saying del f this one this one this one this term no this one correct. So, finally the operator is written in the nested form. So, now you can see that if you want to solve del 1 square equal to an inhomogeneous term all we need to do is integrate with respect to r right multiply by r integrate again and then divide by r. So, now it is really easy to solve this problem is it clear now how will you I mean once you got this you have this here. So, to remove the derivatives you just integrate both sides integrate both sides is like to r this guy will go off then multiply with r then integrate again then divide by r that will leave you with del 1 square f equal to all those operations on this side then you again repeat that 4 more times you will get f 1. So, f 1 will be obtained finally after 4 integrals with respect to r and then multiplying dividing by r as appropriate. So, now the problem has been simplified I mean it is such an easy solution that you can get immediately. So, you can work that out. So, finally after integrating and then applying the boundary conditions we will get the following solution. So, this is what f 1 looks like you should go back and do the calculations for practice alright. So, now let us pay attention to what f 1 is. So, f 1 if you remember sorry this is not f 1 this is psi 1 the stream function. So, now if you remember fluid mechanics theory and stream functions the contours of the stream function will give us the stream lines in the steady state situation. So, like I said again the eta dependency is cos eta that means it will be symmetric about the centre line. So, if you look at just the cross section forgetting about w we just have the cross section with the radial coordinate r and eta. So, if you remember eta goes like this this is eta equal to 0 right and the radial distance is r and if you plot these contours in math lab or mathematical you will get something that looks like this and they will repeat at the bottom. So, because of cos eta it is symmetric about the mid plane while the value of the function is not symmetric on this side it is however the contours have the same shape and from here you can get the velocity field by applying the stream function formulation. So, u is 1 bar – 1 bar dou psi 1 by dou eta. So, what does this tell us now if you look at the u you will find that the stream functions give us the circulatory flow. So, what we have seen now to order epsilon 1 is that the flow through this curve channel the axial flow creates a centrifugal force that results in secondary a secondary flow imposed on that which has these circulations. So, this is the this is how the channel curves. So, this is the inside of the channel and this is the outside. So, fluid is pushed along the centre right and once it hits this wall it recirculates back and the whole reason why we have these circulations at all is remember that the centrifugal force here fc and this centrifugal force is much larger because here w not square was much higher here w not square is small because it is near the wall. So, the push in this direction is much stronger than the push here. So, it is basically like as the fluid gets pushed here and gets pulled back. So, you get circulations but that is because it is finite here if you had the infinite case you would not get the circulations at all because the force everywhere would be the same. So, that we look at a later class but this is essentially the problem and if you combine this u 1 v 1 and add it with w 1 w not you will get fluid a fluid particle moving through the channel and also drifting along these streamlines. So, you will get helical vortices. So, as it moves to the channel it will do a helix and these 2 vortices are called dean vortices and is because of these guys that you have enhanced mixing in curve channels. So, his first paper was in 1927 when he applied the same procedure for same perturbation calculation for flow and he calculated these symmetric streamlines and that was the first theoretical work that was done on the problem. So, I will not calculate to w 1 you can do that as an exercise and the entire thing is given and gary lead. So, you all can work it out and maybe hunt down his paper he has got some interesting results that he has and this started off a huge analysis into single phase flow and curve channels. Some of you all must have come for my seminar. So, I did that is when I did the same calculation for the 2 phase flow where instead of just having a single phase in the whole domain you have a core fluid which is phase 1 and surrounded by an annulus of phase 2 and then you do the perturbation calculation. So, now in essential to the governing equations and in addition to the governing equations you have the normal stress condition, tangential stress condition and all of those being put up. So, this similar idea can be used in so many different problems and so many variations of this other. In fact on a slightly funny in fact how far you can take perturbation. So, there is a paper by Peterson recent one I think in the maybe 2007 or 2010 what he does is he looks at again single phase flow in a curve channel, but now he makes these walls wavy which will remind you of your quiz too. So, in the second quiz we gave state flow through a channel which had a waviness here he took the waviness and the curve. So, he has 2 perturbation parameters one in the curvature and one in the wavy wall and that is the physics of fluids Peterson can check it out. So, then he does it is a double perturbation. So, first you perturb to order epsilon square he went in epsilon which in the curvature and then each of those guys are perturbed you know in a 2 dimensional perturbation series. So, then he has a solution. So, these vortices then become if you look at the stream lines the stream lines also become wavy and it is quite interesting. So, in this sense you can keep going on you know adding different effects alright. So, I think with that we will wrap up our class on Dean vortices and in the next class we will look at the instability problem which is also studied by Dean in 1928 where there he looks at you know 2 infinite walls where circulations do not come immediately, but circulations will set in as an instability. So, there is a certain difference there and there we can apply a linear stability analysis.