 So, welcome to today's lecture and what we are going to do today is basically talk about the problem of natural convection and in particular the onset of natural convection. Now this is a very classical problem and it was investigated almost a century ago by Rayleigh and Bernard and so this problem also goes by the name of Rayleigh-Bernard convection. What we are going to talk today is how we can determine the onset of natural convection by posing it as a stability problem and this particular problem we are going to discuss over maybe the next set of 2 or 3 lectures which is basically to illustrate the concepts behind how to carry out a linearized stability analysis. So, this is the problem of Rayleigh-Bernard convection or the natural convection. The problem of natural convection is the one where you have a density gradient such that the less dense fluid is at the bottom and the more dense fluid is at the top. So, because the density of the fluid lying below is lower, this has a tendency to rise up and this is essentially due to the buoyancy force. When this packet of fluid which is lighter rises up, you have the heavier fluid from the top which is going to sink in and replace this lighter fluid. Now if you have a mechanism by which you can actually sustain, let us say a temperature difference so that the temperature at the bottom is higher, the temperature at the top is lower, this heavy fluid which was colder is again going to get heated up and it is going to rise. So, what this essentially means is you could have a situation wherein the liquid or the fluid is going to exhibit convection. Now this situation will not arise when you have a density gradient such that you have a heavier liquid at the bottom and a lighter liquid at the top. So, that is a stable stratification and the heavier fluid likes to sit at the bottom, the lighter fluid likes to sit at the top and you are not going to see any convection. So, natural convection is one that is induced by density gradients where you have a heavier liquid at the top sitting on a lighter liquid at the bottom okay. So, basically what I am trying to say here is this is induced by density gradients here a heavier liquid sits on top of a lighter liquid. The lighter liquid has a tendency to rise up okay due to buoyancy effects. So, the heavier liquid now sinks down, it is going to possibly absorb heat and become lighter and rise. So, the convection sustains itself. Now to get on to something very specific and in order to do the analysis we would be looking at fluid which is confined between 2 plates and the coordinate system we are going to use is this is the x axis, this is the y axis and this is the z axis. So, the z axis is actually coming out of the plane of the board, the x axis is pointing rightwards and the y axis is pointing vertically up. What I have drawn here is basically 2 plates and these are solid plates which are impermeable and these are located at y equals 0 and at y equals h. Now to keep life simple we are going to assume that the plates extend to infinity in the x direction as well as in the z direction okay. So, the plates extend to infinity in the x and z directions. You will see later on what the implications of this are but at least in the z direction when I am assuming the extent to infinity I am going to end up neglecting changes in the z direction as well as the velocity in the z direction. Basically this z direction extending to infinity allows me to solve a 2 dimensional problem in the x z plane okay. This is something which helps me illustrate the idea to you today in a very clear manner and it helps me simplify things mathematically and remember that is one of the objectives of the course we want to simplify things mathematically but capture the important physics and the x direction extending to infinity essentially allows me to neglect boundary effects in the x direction which essentially means that I would be looking at periodic solutions possibly in the x direction okay. If I had the x direction extending to a finite extent I would have to worry about the boundary conditions which I would have to impose and how that is going to affect my system behavior okay. Now since we are interested in a problem of natural convection and therefore we need to have a variation of density and one way I am going to impose this density variation is by imposing a temperature variation. So what this means is I am going to keep the plate which is at the bottom at a temperature T is 0 and the plate which is at the top at y equals h at the temperature T h okay. So there are these 2 plates which are basically surrounding my liquid or fluid and they are maintained at temperatures T is 0 and T h. So if T is 0 is greater than T h we have a situation where the density increases as we go up. So T is 0 is higher the fluid here is going to be lighter the fluid here is heavier so the density increases as we go up and remember this can possibly give me natural convection. I call this natural convection because I do not have a pump which is actually making my liquid flow and this is actually being caused by density gradients. So what is going to happen and this is one way for you to visualize this supposing we do have natural convection then this fluid which is lighter will have a tendency to rise up okay and when this fluid packet rises up the heavier fluid from the top is going to come down once this heavier fluid from the top comes down once it reaches the bottom it is going to get heated up it is going to become lighter this fluid which was lighter is going to go up okay it is going to become heavier because this is at a lower temperature and once it becomes heavier at the top this is going to have a tendency to come down. And so basically what I am trying to show here is that you have a sustained convection okay. So this is the lighter fluid rising and this is the heavy liquid heavy fluid descending and this sustains itself because of my boundary conditions I am maintaining the temperature here higher than the temperature here okay. Now this is the kind of situation which is actually prevalent in the atmosphere where the atmospheric air is going to get heated by the sun's radiation and you would have a situation where you can have a hotter liquid sitting at the bottom on the earth's surface and you have you know heavier liquid fluid heavier air at the top okay. So basically this convection is going to be induced even in a natural condition like the atmosphere without any pumping effect. So you have you know because the sun's radiation the surface of the earth being hot it is going to cool down as you go up and so you have a heavy fluid sitting on top of a lighter fluid okay and you can have convection natural convection. So rather than talk about the problem of natural convection in the earth's atmosphere we are going to idealize the situation try to capture the important effect and try to analyze the problem okay. For this idealization I am going to use this particular geometry of a fluid element being fluid being present between these 2 plates okay and so this is going to allow me to capture the important physics and do the analysis. So this is the idealization of the natural convection problem which is prevalent in the earth's atmosphere okay. Now I have used the word that this can possibly give me natural convection. Why do I say possibly? Because clearly if the temperature difference T0-TH if T0-TH is less than a critical value or rather is low the density variation is going to be less as well and what is going to happen is the liquid is going to be or the fluid is going to be stationary. Why is that? Because the viscosity of the fluid prevents it from rising up. So you have 2 effects in this problem you have the buoyancy force which is trying to push the liquid up okay and you have the viscosity which is essentially a frictional force which is going to prevent this liquid from moving. So you have viscous forces and you have buoyancy forces and very low values of T0-TH the viscous forces dominate the buoyancy forces and you have no convection. However if T0-TH is sufficiently high then the buoyancy forces are going to dominate the viscous forces and you will have convection okay. So the point I am trying to make here is if T0-TH is low the viscous forces dominate the buoyancy forces we have no convection. The buoyancy forces are too small and so the liquid cannot rise up. However T0-TH is large the convection will occur and why is that? Because now the buoyancy force dominates the viscous force okay. So clearly what this means is that there is a particular value of this temperature difference and this is going to be decided by the properties of the fluid. This is going to be decided by the spacing between the plates above which you are going to have convection, natural convection and below which you will have no convection and the liquid is going to be stationary. So now the question is can we find out this critical value of T0-TH this temperature difference above which we have natural convection okay. So can we find out the critical value of T0-TH above which we will have convection and the answer is yes we will be able to determine this value by posing the question as a stability problem okay. And so what we will do is we will look at the state where the liquid is not moving the liquid is at rest and ask the question whether this particular state is stable or unstable. So if the state were to be stable that means you would be able to experimentally observe it and if the state is unstable you would not be able to experimentally observe it and you would possibly have a different state which you would observe okay. So now the question is if I want to pose this as a stability problem I need to be able to find out the steady state whose stability I am interested in okay. In this particular problem the steady state which naturally arises is the one where the liquid does not move where the velocity components are 0 and this happens for low values of the temperature difference okay. So we will treat this particular steady state as the base state and we will try to find out the stability of this base state and the way we find out the stability of the base state is by imposing small perturbations infinitesimal perturbations and see how these infinitesimal perturbations evolve in space and in time okay. So just by way of trying to explain to you as to the what the complexity level of this problem is this problem is complex in the sense that it is going to be governed by partial differential equations which are essentially the equations of momentum the Navier-Stokes equations and the equation of continuity and the energy equation which basically tells you how the temperature varies in the bulk liquid. It is not so complicated because we are focusing our attention only on a single liquid single phase okay. It would be more complicated if I had let us say a layer of 2 liquids between these plates in which case it would be a real multi phase flow system. So we would be looking at those problems later but the point I am trying to make here is that we are right now interested in only a single phase system. So we do not really have to worry about the boundary conditions because I have a problem where I have solid walls okay that is my idealization and so the boundary conditions which naturally arise are the ones where I have no slip boundary condition at the walls and the fact that I do not have a perpendicular velocity component okay because the wall is impermeable okay. So that is basically where we are. So what we are going to do now is basically try to solve this idealized problem okay and find out the conditions under which natural convection occurs. But remember we are going to follow the procedure of having to write down the governing equations trying to simplify things mathematically so that we capture only the important physical effects. Then doing a linearization about the steady state find out what the steady state stability is and try to answer the question of what is the critical value of this temperature difference above which we are going to see natural convection okay. So let us look at the idealized problem and let us write the governing equations. First we have the equation of continuity which says the derivative of I am going to write this in an expanded form equals d rho by dt. Rho is the density, u is the x component of the velocity and v is the y component of the velocity okay and what I am going to do is I am going to keep in mind the fact that the density in this problem is actually going to be varying spatially and so I am writing the equation of continuity in the most general form where the density is inside my derivative operator. The next equation which I really have to worry about is the momentum equation and that in the x direction tells me rho du by dt plus u du by dx plus v du by dy equals minus dp by dx plus mu times d square u by dx square plus d square u by dy square plus rho d subscript x that is the body force. So I am just writing it in the most general form right now okay and in fact I have written this for the 2 dimensional case where I have not taken into account the velocity in the z direction and I have not taken into account the variation of the velocities in the z direction okay. Similarly the equation in the y direction is going to give me. Now in addition to these equations I need to write down the energy balance equation because remember the temperature difference is what is actually driving the flow because that is the one which is inducing my density variation. So when I want to write the t by dy equals k times d square t by dx square plus d square t by dy square. So this basically is the accumulation term of temperature. This is the convective flux for heat okay and this is my conductive flux for heat. What I want to tell you is that we actually have these 4 equations which are coupled to each other. The coupling is going to occur because the temperature is something on which all the properties the density, the specific heat, the viscosity, the thermal conductivity depend upon okay. So clearly as you can see we have a bunch of partial differential equations and they are non-linear because of these inertial terms here okay. And what we want to do is we want to ask the question rather than solve this by brute force using some kind of a numerical scheme. Is it possible for us to actually simplify these equations and get some insight okay. So our philosophy now is to simplify these equations, retain the important physics and not worry about you know unnecessary effects. So for example like I mentioned the physical properties viscosity k, Cp all of them depend on temperature. But what we are interested in is the phenomena of natural convection wherein only the effect of density on temperature has to be retained and is important okay. So what we will do is we will assume all other physical properties are constant as far as temperature is concerned and only the density is a function of temperature. So we will assume all physical properties are constant except the density okay. So density is going to be a function of temperature and I need to retain that to be able to see this natural convection. Now the question is can I simplify it further. So for example density occurs in several places in the governing equation. It occurs in the continuity equation at these places at 3 places. It occurs in the momentum equations both in the inertial terms as well as in the body force term. So in all these terms is it necessary for you to actually retain the effect of the variation or the dependency of density on temperature. Is it possible for you to say look it is important to retain the effect only in one term and for all practical purposes treat density as a constant in all the other terms. So can I further simplify my equations that is the question and the answer is yes we can possibly further simplify my equations. I am going to assume that the one term in which the effect of density on temperature has to be retained is the one which is going to give me my buoyancy force and because the buoyancy force is required for my natural convection and so I am going to retain the effect of density on temperature only in my body force terms. At all other places I am going to assume the density is a constant some mean average value okay. So this is a further assumption a further approximation I am making and this is what the cracks of modeling is to try and retain the important physical effect. So what I am saying here is we assume density is a function of temperature only in the gravitational term rho as a constant everywhere else okay. This particular approximation is called the buisiness approximation. So the buisiness approximation tells you you can treat density as a function of temperature only in the gravitational term assume it to be constant everywhere else okay. Now the question is how valid is this and we can find out the validity of this approach that is the accuracy with which you are going to get results under these assumptions only by validating it with actual experiments. So if it turns out that the experiments results are going to be in close agreement with the prediction under these assumptions then you are justified to make this assumption. If it turns out that there is a big mismatch between the predictions using this assumption and the experimental value is observed then clearly this is a bad assumption to make and you need to go back and relax this assumption and work on the more detailed problem. So the only way I can actually find out if this assumption is actually valid is by comparing the predictions of this model with experiments okay and this is an extremely important aspect but the idea is I am hoping that making this assumption is going to be valid because it helps me simplify my model and it helps me get a deep physical insight into the behavior of the system as to what the role of different parameters are. Remember if I do not make the simplifications I have a problem with several parameters and I will not be in a position to understand what the effect of each parameter is in determining the behavior of the system. For example what is the effect of the distance between the plates, the viscosity, the thermal conductivity etc. So now what I am going to do is that the density varies linearly with temperature. This is a linear variation with temperature. What I want is the density to actually decrease with temperature and since density decreases with temperature beta has to be negative okay and I am going to assume rho is equal to rho not in all the other terms. So this basically allows me to simplify my governing equations. For example now my equation of continuity is going to be just du by dx plus dv by dy equals 0 because now density is constant in that equation. This is my equation of continuity. I have to simplify my x component equation. I am going to write it as rho 0 times du by dt plus u du by dx plus v du by dy equals minus dp by dx plus mu del square du where I have just written the del square as a 2-dimensional Laplacian which has the second derivatives in the x and y direction okay plus the body force term in the x direction is 0 because the x direction is horizontal and the gravitational forces in the vertical direction okay and in the y direction I similarly have dp by dt plus u dv by dx plus v dv by dy equals minus dp by dy plus mu del square v and now I am going to have to worry about the body force term because it acts in the negative y direction. Remember my y direction is pointing in the upward direction. So gy is going to be minus g and density now in this gravitational term I am going to retain the dependency on temperature and I am going to write this as minus rho 0 times 1 plus beta times t minus t0 times g. So this is the important point here. The negative sign comes because gravity is acting in the negative y direction. I am retaining the linear dependency of density on temperature in this term okay and that is the assumption we made and then you have the rho 0 times cp times the time derivative of temperature plus u dt by dx plus v dt by dy equals k del square t. Again I repeat that the del square operator is a 2 dimensional Laplacian with variations considered only in the x and the y directions okay. I like for you to now see that these equations are all coupled. For example the temperature equation has velocity in it. The velocity equation has temperature in it and the u velocity and the v velocity are all coupled. There is a pressure term and so all these equations are actually interlinked and it is therefore necessary for you to solve them all together simultaneously okay. But what we will do is we will exploit the fact that we are engineers. We will use physics and try to find a steady state to this problem. This steady state that we are interested in is going to be the one which is stationary, the one where the liquid does not move. So that makes our life simple. What it means is that the 2 velocity components are actually going to be 0, u and v are going to be 0 okay and we are going to use this to find out how the temperature is going to vary. We are going to use this steady state to find out how the pressure is going to vary. So that is going to be the steady state which we are going to determine. Once we find what the steady state is, we would then ask the question what is the stability of that steady state. We have stationary steady state and so u ss equals v ss equals 0 because the liquid is not moving and so clearly this steady state is going to satisfy your continuity equation. What I like to do is see when you have such a steady state, what is going to be the temperature profile, how does the temperature change? And so for this I am going to substitute this in my energy equation and when I substitute it in my energy equation and since I am looking at the steady state, the time derivative goes to 0 okay. This is my time derivative term that goes to 0 because we are looking for a steady state. Remember u and v are 0 at the steady state of interest to me and so these terms are going to vanish and what I am left with therefore is that the steady state temperature is going to be governed by the Laplacian of temperature being 0. So the steady state temperature is going to be governed by this and if I want to write it in an expanded form, this is what I am going to get equal to 0, the steady state. Clearly if you now look at the boundary conditions of this problem, the temperature of the lower wall is a uniform T0, the temperature of the upper wall also is uniform, there is no variation in the x direction. So there is nothing in the problem which is going to induce a variation in the x direction for the steady state. So what this means is the steady state temperature is not going to be varying with x okay and so we have Tss is a function of y alone. This is therefore going to be governed by d square Tss by dy square equals 0 and clearly you can find Tss as being just a linear profile and we can calculate a and b by looking at the boundary conditions. The boundary conditions are that Tss equals T0 at y equals 0 is equal to Th at y equals capital H. So these boundary conditions help me evaluate my constants a and b and you can do this exercise and you will find that Tss of y is going to be given by, so at y equal to 0 Tss is T0, so I have T0 and then I have y divided by H. So you can see that when y is 0 I have T0 and when y is H I have Th. So clearly the temperature is varying linearly in the y direction and that should be expected because the liquid is not moving. So for all practical purposes it is behaving only as a solid and you only have conduction as the mechanism for heat transport and the temperature is varying linearly okay. Now what information am I going to get from my momentum balance equation in the x and y direction? So if I were to now use the momentum balance equation in the x direction in the x direction my momentum balance what does it yield? It tells you that the velocities are 0 and so I get something like a trivial relationship dpss by dx equals 0. This means that there is no variation of pressure in the x direction and this is something you expect because you actually have a hydrostatic phenomena. You have hydrostatics liquid is not moving and you expect that there is no variation of pressure in the x direction okay. What about the equation in the y direction? In the y direction we have are going to get dpss by dy equals –rho0 times 1 plus beta times tss – t0 times g. So here the pressure variation in the y direction is going to be present because that is the vertical direction but since the density is not a constant I am going to incorporate that. I see that the temperature that I am actually going to be using is that of the steady state temperature which I have already determined as my linear profile. So I know how the temperature varies in the y direction. I can substitute that inside here and I can calculate how the pressure varies in the y direction okay. So that is the basic steady state that we are going to find the stability of. So let me just write down what the steady state is whose stability we are interested in. Steady state is uss equals vss equals 0, tss is h plus t0 and dpss by dx is 0, ss by dy equals –rho0 times 1 plus beta times tss – t0 times g and that is the steady state whose stability we will be finding out in the next class. Thank you.