 So, good morning and in today's lecture we will be looking at flow and curve channels. So the whole problem of a fluid flowing through a curve channel is one that is very interesting. Firstly for theoretical reasons which we will be covering today and secondly because of the wide range of applications that the flow has in industry and even in biology. So for example if we start with the natural sciences first blood flow through our body happens through a lot of curve channels especially in the arteries and in many cases it is creeping type flow and that is one motivation initially by people started looking at flow and curve channels. Then practically when you look at industries you have so many curve channels and heat exchangers and mass transfer equipment and just so that you can enhance mass transfer. And why that happens also we will see you know in today's class the other aspect is that very often you just do not have space to have state channels all the time. So then you need to have curves again. So to understand what goes on in curve channels what are the new features of the flow once again we can apply perturbation techniques to see how the curvature comes in how the centrifugal force comes in and what effect it has on the flow. And naturally once again we can look at stability issues. So in this class I will give an introduction about flow and curve channels and then we will work on one of the problems. So right off there are two cases we can consider. One is the simple case of flow inside a annulus. So you have two cylinders of radius let us say R1 and R2 and the fluid is flowing into the plane. So this is one situation where basically the walls of the channel are infinite in the z direction and fluid is going around. So that is also a type of curve channel. In fact it is one of the first curve channels that was studied because of its simplicity. Something like the stratified flow problem you have this for curve channels. So that is one problem. The finite analog of this would be actually flow in a square duct. Equivalently for a slightly different cross section we could have I mean a curve pipe that should not be too very different. So in most of our analysis in the past we have seen that when you look at infinite cases like the stratified flow problem the results we got for a semi infinite case and the finite case were very different. We just have to account for wall effects. So for example in the stratified flow problem you have two flat plates and basically this was trying to mimic this situation. So when this dimension went to infinity we had the stratified plate problem and if it is finite which is a finite box then we would have to solve it in a square duct. But I showed you in the earlier presentation that the solutions that we got with the infinite plates and that with the square duct were quite similar especially if you were to cut through the centre. So that is the same reason why people thought of studying this and that problem. But as you would know now if we solve this problem and most of us have already done this in earlier courses in fluid mechanics you would get a solution where there is nothing very different. You will have a velocity in the theta direction but there will be no other components of velocity in the base solution if you see and the pressure will vary as a function of r. That is the classic Taylor-Couette base state where you have purely azimuthal flow and that is it. So that is the only directional azimuthal flow and you can solve that almost straight forward implicitly directly. So there is nothing very fancy about that. On the other hand if you come here to the finite wall case you will find that and that is what we look at today that there is no base state that is unidirectional when the duct is finite. You see here there is a unidirectional base state with flow only in theta but in this case you will immediately get circulations just because the duct is finite. So there is a big qualitative change when we go from the infinite walls case to the case where we have a finite height duct and that is not something we have come across so far. So the story of these two channels just tells us that we need to be very careful whenever we go to a limit. So actually what it is is you can think of a very tall duct not infinite but very tall then this would be a kind of limiting perturbation of a very tall duct or a duct with very high aspect ratio. So what happens is when we take that aspect ratio to infinity you have a singular perturbation type solution a type problem where the qualitative behavior changes. So here there is absolutely no circulations in the base state but as here there will be finite circulation simply because of the finite height. Now why does that happen before we look at the math we can look at the physics and that will help us in what we are going to do today. So in this the infinite duct case and the finite duct case so what happens is in both situations when the fluid is flowing around the curve it experiences a centrifugal force. So there is a centrifugal force pushing the fluid outside right so centrifugal force. I have the same thing happening here there is a centrifugal force but now heuristically speaking of the centrifugal force is v theta square over r of course you will have density in some. So if we have uniform v theta by some reason then the centrifugal force the fluid experiences is of course as v theta square by r. Now what happens is that in this situation you can conceive of a v theta that is uniform in the vertical direction. So let me call that in this case z I will do the same thing here okay. So just take this as the x direction for now. So now in the z direction in this case you can because it is infinite it is either going to be periodic in z or it is going to be independent of z. So the base state we say v theta is independent of z not a function of z. So in that case the centrifugal force which goes as v theta square by r is going to be independent of the z direction which means that the fluid gets pushed uniformly throughout z. So if I look at how does this force get balanced it gets balanced by the pressure and that is what you will see if you solve the problem that the centrifugal force pushes the fluid out because there is a wall here it gets balanced by pressure. So pressure becomes a function of r so p is a function of r but it is not a function of z but in this case v theta will have a value at the centre of the channel but at the wall it has to go to 0. So here because you have bounding walls v theta cannot be independent of z v theta will be a function of z. So my centrifugal force is going to become less at the wall while it is greater at the centre. Because of that the pressure that is trying to balance it will be a function of r and it will be a function of z also. So because it is a function of z immediately it will set up a vertical flow and I will have circulation but here there will be no vertical flow set up because p is not a function of z. So qualitatively thinking that is the basic reason and mathematically you will see it worked out that if I take this problem and try to obtain the simpler steady state and I put these assumptions in I will find that I have a unidirectional flow in the theta direction whereas that set of assumptions will not give me a solution here. And immediately you will know that there is no unidirectional solution the base solution itself has all three components of velocity. And so today we will be focusing on this problem where for the first time a base state itself is quite complicated. In this case the base state is unidirectional but then as you can imagine that as you keep increasing the velocity the centrifugal force increases ultimately the flow becomes unstable and you will have circulations even in this case. And that is the famous Taylor-Cohet problem where as you keep increasing the centrifugal force you will get different fluid patterns and different vortices. So that is a stability problem here whereas this is not a stability problem because the flow is always circulated. So we will focus on that. So in this class we look at the circulations that are set up naturally inside a square duct at arbitrarily small centrifugal forces. And in the future class we look at the flow to the infinite duct and see how it I mean how you get the base state which has no circulations and how it becomes unstable. So there are two parts of the same coin and at later stage we can come back and see what we learn from it. So let us focus on, for simplicity we will look at the circular problem. So first things first we need to write down the governing equations, the full non-linear set. So before we can do that also we need an appropriate coordinate system. So what we can do is if this is the origin and you have the x, y, z Cartesian axis here then we can set up a new system which has an translating origin o prime and this distance from zeta is actually the radius of curvature. So it is a radius from this origin to the centre line of the curved channel. So within this thing we can mount a polar coordinate system. So you have r here and I will call the angle from the vertical eta. So r, eta gives me any point, let us look at this point p. So the position of p inside the cross section is given by polar coordinates r, eta and then the position of the cross section itself comes from theta. So if I take x I measure theta from the x axis, a y does not have to be here also, this dash line is just the line vector joining the centre of the cross section of interest to the origin and y could be here also. So theta measures the location of the cross section so as I move through the curve channel my theta keeps changing and inside any particular cross sectional slice the point p is given by r and eta. So totally I have a coordinate system which has r, eta, eta and this coordinate system can be related back to the x, y, z coordinate system by mapping. So with this coordinate system we can now look at writing the Navier stokes in r, eta, theta. So how do you go about doing this? We have seen in earlier lecture how we can write down the Navier stokes equations from a general vectorial form in any coordinate system using the vector identities and vector methods. So we will do just that not in the class, right now I will directly begin with the equations but you can go back and see how it is done like you had done before for symmetrical and spherical. Once you find the mapping from x, y, z to r, eta, eta you can go apply those methods and find the gradient, the divergence and so on and get the Navier stokes equations. So the velocity components are u, v and w in the r, eta, theta directions and the equations as you see will bear some similarity to the cylindrical coordinate equations naturally because the coordinate system is similar. So that is the equation in the u direction. The one thing to point out immediately is it is very similar to cylindrical coordinates. So when you look at the any component or any derivative in the eta direction I will get the 1 by r factor that is the length factor. When I look at derivatives in theta or quantities to do with w, you will find the scale factor zeta plus r sin eta. The reason for that is that zeta plus r sin eta is actually the, is this radius scale factor. This is zeta plus r sin eta will give me the distance along the vertical, r cos eta gives me the vertical, r sin eta gives me the horizontal in the plane. So zeta plus r sin eta is this total projected distance to any point that I am looking at. So that is the denominator scale factor that will come throughout the equations. So this part is like exactly like cylindrical. This comes from like v theta square by r sort of thing but here v theta is just w. So it is like w square by r but then the r is zeta plus r sin eta. So we will go to the next. So those are the 3 momentum equations. What else do I have to write? The continuity equation. Those are the equations in the toroidal coordinate system and all their glory. Of course these are the, okay. So one thing I have to say is that today we look at only fully developed flow. So that means that I have not considered any derivatives in the theta direction except for pressure. These are equations for steady fully developed flow. If you do not look at fully developed you will have additional terms that take care of the derivatives of velocity with theta, alright. So if you will have written them down we will proceed, okay. So the one thing I want to point out here is that this term is the analog of that v theta square by r term I was talking about. So that is one of the main centrifugal force terms that sort of drives momentum from the theta direction velocity. It creates the force that will drive my v and u components of velocity because this is the u component of velocity is momentum equation in the r direction, the eta direction. So both in both these directions the force acceleration, centrifugal acceleration is coming from here, centripetal acceleration. So that is one thing. Second thing is that you see this guy gets multiplied by sin eta whereas here you have cos eta. So it is almost like the sin eta gives me the component of that centrifugal, centripetal acceleration along the radial direction. So it makes sense that sin eta will come you know in the outward r direction and the other component comes in the cos direction. So that is that, right. So the first thing we should do is scale the equations and when we do that we are again looking at, alright. So these are the equations and the first thing we should do now before we proceed is to scale them. So for that we need to select our characteristic scales. Do you have any questions? So I should be having a question right. Okay, maybe you all do not know what the equations look like that cannot be expected but you should be, I would expect you all to know the general form. So the right hand side seems fine. You have u dou by dou r v dou by dou eta the similar sort of thing that is these are the centrifugal, Coriolis terms and so on. Continuity equation looks fine. What about the viscous terms? Are they of the form that you are used to seeing or they are not, you would have thought at least a dou square by dou r square will be there somewhere because that is just rectilinear r, right, so especially in this equation. So what I have actually done here, this is not del square v that much is obvious. So what the thing is that normally you have mu del square v in the, v is the vector, del square v in the Navier stokes, the stokes I mean the viscous terms. So it turns out that of course del square v can be written as gradient of the divergence of the velocity minus the curl of the curl of the velocity, alright. So what happens here is because it is incompressible, this gets knocked off. So directly del square v will be negative of the curl of the velocity and that is this whole term here is just that, alright. The reason for doing that is that it is much easier to write the equations here. If you actually go compute del square v, it will be much longer. You can do the same thing in cylindrical coordinates and see that it will give you a more compact representation and for the purposes of this study it works out easier, like especially when you use steam function formulations, this curl of curl of velocity is a better way of looking at it. But if you had proceeded with del square v, you would not have gone wrong. It is just another alternative representation, alright. So that should be clear, fine. So now let us scale the equations with the right scales, okay. So what would be my scale and r, length scale? So I have not said, yeah. So it should be the, let us say this is the radius, I will call the radius A. So the radius of this channel is small a and I have another length scale which is zeta, correct. So which scale should I choose for the length scale, right? Should I choose zeta or should I choose A? A. A, why A? But even this is a proper length scale in the problem, right. So what he has said basically is that the reason which we are interested in the length scale at which we want to see changes in the velocity and so on are happening here. See there is nothing to tell us that variations in the velocity do not happen in the zeta scale. Of course here I am saying the flow is fully developed but if it is not, the flow could vary in the axial direction that you know with the scale of zeta. So the real reason is that we want to focus on the circulations that are happening in the cross section of the channel, that is one thing. The second thing is of course in this case we have already looked at fully developed flow. So we are not going to be interested in variations in zeta and that is where really the zeta thing would come in. The third thing is if it will allow us to go to the limit of a state channel because what we are trying to do here is study the influences of curvature of a curved channel and we are going to use perturbation techniques that we always have. So if we want to perturb a curved channel about a state one, we need to send zeta to infinity, right because if zeta is infinite then this curvature becomes very small and it ultimately reaches a state channel or zeta for a state channel would be infinity. So if we are going to send this to infinity we cannot very well be scaling things with zeta. It also tells us that what we are really interested in is happening at the scale of the radius. So lc will take as a, so that is our length scale throughout. Velocity scales will come from the equation. So once again the pressure is, I mean the flow is a pressure driven flow and at fully developed flow there will be a constant pressure which I will call g, pressure drop. So if you look at it that way, this is the term that is driving the flow, this guy and it is not immediately obvious I will admit but if you do the usual scaling that we go through you will see that this g which is the pressure gradient along the center line is what ultimately gives me the scale for the velocity and it will look very similar to that for what we do when we did a state pipe. So you will actually get, suppose g is the pressure gradient here then you will get g times a square because of two length scales from here by mu. So the velocity scale will take as g a square by mu and this 4 is a factor that simplifies calculations later on. So dimensionally it should be consistent. This is the same scale we use in Hagen-Poisel flow of that. So that is it. So with these scales let us scale the problem. So you should find here now that if I, of course pressure will be, should be g times a because g is the scale of the pressure gradient, the pressure drop. So it will be g times the length scale. Now the important thing is because I have, because I have taken Fourier developed flow I cannot actually retain this term in a general sense because I have to realize that dopey by dote eta is a constant otherwise it would not be consistent. So how can we see that, yeah miss something, yeah w would not be there in the continuity because it is, why, why won't it be there, w won't be there, think about it. So the w will only come if w is changing with eta so it is not, so Fourier developed flow you will lose that derivative. All right where was I, yeah I was talking about this term. So because you are saying it is Fourier developed flow you will always be able to show that the pressure drop in that direction is a constant. So how can you see that? If you differentiate this equation with eta okay you will find that everything here will drop off because all the velocities are theta invariant and the same thing will happen there. So I will get dopey by dote eta it is second derivative will be equal to 0. That you will find immediately just by differentiating with eta all because all the velocities will drop off, dopey by dote eta, dopey by dote eta, dopey by dote eta, dopey by dote eta all 0. So I can come here and do the same thing okay. Again I will get right because once again all the velocities are theta invariant. So what will happen is now of course I can interchange the derivatives in these 2 cases. So what I will get is that dopey by dote eta taken as a function alright is independent of theta alright it is also independent of eta and of r. So this dopey by dote eta is constant throughout the channel not only with length also with cross section. So at any point in the channel you go and you check dopey by dote eta is the same. So that is what is actually related to my g. So to be very specific g is 1 by zeta where this dopey by dote eta is a constant which we have now shown holes for any fully developed flow regardless of whether there were circulations or no circulations or base state whatever it may be if the flow is fully developed this is a constant and that is what my g is. It is the pressure gradient along the center line right but see although dopey by dote eta is a constant the gradient is given by 1 by zeta plus r sin eta times dopey by dote eta. So the driving force depends on this length scale and that will vary in the channel because parts closer to the channel have a smaller you know arc length parts further away have a larger one larger radial distance. So the gradient the driving force will vary but dopey by dote eta is the same see if I scale it in that way but see what I am doing is I am fixing the value not of dopey by dote eta of the of the driving force. So if I take zeta larger and larger this guy will just change its value but I am putting the total thing to a constant if I just did this then what he said will be true. So these are all subtle points that you have to be careful about especially when you read papers some guys will give this value and then they will have this 1 by zeta coming everywhere in the base state and it will look funny like you put that to 0 something will happen and so you can always you should always go back and you know understand what he has done at the very least. So that is settled so now dopey by dote is a constant so from this I can say that p from these 3 cases I can write down p as z plus p hat of r, theta so that is the general functional form for p I should use theta times g zeta theta minus because dopey by dote is a constant. So this is important to realize in any fully developed flow so in that because of this I can take my pressure scale as g times a if I did not have that because the driving forces coming from here I would actually there would be some problem taking g times small a because I will be scaling the w direction also with small a and that lead to some issues in the scaling but it is only because it is fully developed flow I can do that otherwise I should look at the zeta scale for this part of the equations so that means I can replace this I do not need this guy is nothing but minus g zeta so that is important to realize if we did not do that and we treated dopey by dote as a general variable you will not get the scaling rate so this is actually a constant term and this is the driving force so now that we know this is the driving force we can go and scale our equations and do it properly so let us do that now so I will just work on these equations because they are too big to keep writing again and again so if you look at this term first what I can do is I want I know r by zeta is my scale sorry a is my scale for all the length scales so if I do that here and I replace r by r star times a where r star is the dimension is variable a square will come out from the bottom right I have one w on top so that will give me the w characteristic so I have got my usual group but what will happen here is if I pull out a I should take out this zeta guy and I will get 1 plus a r by zeta what I did was I removed the zeta so I will get 1 plus r by zeta sin eta then I replace r by a r star so I am just not putting the r star in so I will get a by zeta here okay and zeta comes out but then yeah and then because I have pulled out the a square here right so I will actually have an a in the numerator so I will do this something similar 1 a came from here 1 a came from there so if I am removing an a I just multiplied and divided by that a so if now this a by zeta is the is what I am going to take to be my epsilon parameter and if we want to look at a state channel as epsilon goes to 0 I will get a state channel right and as yeah and the larger the epsilon the more curved the channel is so this is called the curvature ratio fine so that is how that comes out naturally so wherever you have zeta plus r sin eta I am going to get right epsilon upon 1 plus epsilon sin eta the same thing will happen here and that makes sense because all those zeta plus r sin eta remember what the scale in that direction so if it goes to infinity that has to become order epsilon so not only you have epsilon here the entire term gets multiplied by an epsilon so all these terms that represent the curvature of the axial direction are going to be of order epsilon all right that is clear that is the first thing so then we get this guy out of course here I should have been I have been a little careful so I will get zeta here wait so then the zeta and the zeta will get knocked off so I will just get so this term is the driving delta p and it is order 0 the largest part of it is not order epsilon even though there is epsilon here I put that off I will just get plus g and that is my driving force so this 0th order term in the whole thing has to be the driving force of the problem and that is retained here right the other pressure if the all the other derivatives in theta order epsilon but that cannot be order epsilon because that is what is driving the flow then on this side if I again I can do the same thing so I will get Wc squared over here right by a and finally what I can do is take this stuff to that side or rather bring that here because I want to retain the viscous terms okay so now we have a choice we always have this choice when we reach this point either I make these terms order 1 and take this to that side that means I will retain all the inertial terms or I can retain all my viscous terms I have no choice about the pressure drop because the pressure drop is the driving force so because I want to look at I mean I have decided to look at viscous forces in this class and that has applications to micro channels and stuff like that so if you want the viscous forces to be there and also we will see later on that this whole thing I am doing has a lot to do with the Reynolds number so far we have not spoken about Reynolds number at all but since we are discussing centrifugal forces they will be proportional to the density as you can see so somewhere the Reynolds number has to come in so far we have not looked at it we were just talking about small epsilon but you will see that I need to scale this in the right way so that we get the right result later so what I will do is I will take this to the other side because I want my viscous terms to be there so I will get a square here Wc mu and this guy becomes 1 right here I will get a square by Wc fine and now this whole group is just the Reynolds number right I will get rho a Wc by mu so that is what you will usually get if you scale by viscosity you will get the viscous terms you will get Reynolds number multiplying the inertial terms right and here we see now that you can find out what the velocity scale is so I have made this order 1 this guy gives me the Reynolds number so they are as important as Re and this I said I do not know Wc but I know that this is the driving force so I will make this whole group equal to 1 so that will happen if Wc is Ga square by mu so that is what I written here and I threw in a factor of 1 by 4 that is purely algebraic if you do not but what you will see I think later on is this 1 by 4 is also closer to the average velocity but since that is what I am following Gary Lee so I will just follow that so this whole guy will go off and I will get just 4 alright so these are the scaled equations we will stop for today tomorrow I will start from the scaled equations and then we will solve the problem by perturbation methods so one important thing to realize is that these are the equations in the same equations we have not done anything to them they are the fully nonlinear coupled equations for flow in a curved channel but we have scaled it in such a way that these epsilon terms are now very clear so we know that all the terms that are coming in because of curvature are going to get multiplied by epsilon so directly on inspection you should see that if you put epsilon to 0 you should get back the equations of the cylindrical coordinate system because this is basically the cylindrical coordinate system if zeta, theta were equal to z zeta goes to infinity so if you put epsilon to 0 here you directly get back the cylindrical coordinates that is from the continuity equation you can see that immediately and this gets knocked off so when you looked at the 0th order problem it will be cylindrical coordinate Hagen-Poisol flow because it just flow on a straight pipe with viscous terms and then when we look at the first order we will start seeing the centrifugal force coming into play.