 So, welcome to this lecture on multiphase flows. We saw some introductory concepts about stability analysis in the last class. Jason spoke about a 2 dimensional system and it took an example and he just worked out how you go about calculating the stability of a system. And especially if there is a parameter working you know in the system, then that parameter could be something which you can control as an experimentalist and what you want to do is you want to find out how the stability of a system changes as you vary that parameter. So, that parameter could be flow rate, it could be temperature, it could be concentration okay. So, like was mentioned in the last class, what we really have to do as far as finding stability of a system, I mean it is decided by the response to perturbations, response to disturbances right. This is defined on the basis of the response to disturbances, if you give a disturbance and if the system deviates from the steady state you say it is unstable, if it comes back to the steady state you say it is stable okay. So, if the system comes back to the steady state, the state is stable else it is unstable. So, I just wanted to emphasize again that when I am talking about stability, I am talking about stability of a state okay. Now this is what you understand in English and what we want to do is we want to develop a mathematical framework and what you got is some idea about this mathematical framework with the in the context of a system of 2 variables which are dependent only on time but not on space. So, just to keep the mathematics simple, so the example that was illustrated in the last class was that of 2 ordinary differential equations. So, you need the time dependency because whenever you are talking about stability, you are talking about response with respect to time. So, you always are talking about an unstable or an unsteady state system okay and therefore you need the time dependency to keep the math simple we made it well stirred and then you neglected the spatial variations. So, the basic idea, so whenever you want to find out the stability of system, what are the steps you need? The steps to find stability is first write the governing equations. This could be continuity equation of momentum, conservation of mass whatever it is which are going to describe the system okay. Then once you have the governing equations what is the next thing you have to do? The next thing you have to do is find the steady state okay and this will be dependent upon the problem you are considering the first step. The second thing is find the steady state okay because it is the stability of the steady state which you are interested in. So, you need to find the steady state whose stability you are talking about and then what do we do? We spoke about disturbances being given. So, what we want to do is we want to give disturbances to the steady state okay and what we do is we impose small and I say small I mean infinite symbol disturbances and here is where the order of epsilon comes in. So, these are basically of order epsilon very small disturbances and that is the connection which I want you to make with the perturbation series thing that you have seen earlier when you are trying to actually do things at order epsilon you will be neglecting higher order terms okay. So, what we want to do is we want to impose small disturbances and what this basically allows me to do is this order epsilon disturbance linearization of the governing equations. So, the original equations will typically be non-linear okay and since you are talking only about small disturbances what you will be doing is linearization that is something which you also saw you have a begin with a non-linear system and you linearize it because you are giving only small disturbances that is what you are doing you do is you do a Taylor series expansion and you need only the first order term you do not go beyond the first order term you do not take higher order terms because higher order terms will be non-linear and then we look at the growth rate of the disturbances. Now, the growth rate the disturbances growth rate growth or decay rate is given is exponential in time whether it is of the form e power lambda t and if lambda and since lambda will be some eigenvalue of some linear operator which you will see lambda it will turn out to be something like an eigenvalue if the real part of lambda is negative then it is going to decay and you have a stable system if the real part of lambda is positive it is going to grow okay. So, basically what this means is the lambdas will be say eigenvalues of a linear operator if the real part of lambda is negative the steady state is stable and if the real part of lambda is positive it is unstable there is just one subtle thing which I want to mention here that when you are doing this calculation if it turns out that the real part of lambda is positive that means even when you give a small disturbance it is unstable. So, when you give any disturbance a large disturbance it will be unstable. So, if the real part of lambda is greater than 0 you can definitely conclude that the system is unstable okay for all disturbances but if the real part of lambda is less than 0 you only know that it is stable for small disturbances but suppose you give a big disturbance it could be unstable okay. So you have to be kind of careful here. So, this is whatever I am doing here is in context of small disturbances okay. So, real part of lambda being greater than 0 is sufficient to guarantee the instability while real part of lambda is less than 0 is necessary to have stability. Let us just say that you can have the condition can be a necessary condition or a sufficient condition for something to happen. So, I am just trying to tell you the real part of lambda is greater than 0 is sufficient if it is satisfied your guarantee it is unstable that is enough you do not have to do anything else in life. If this is less than 0 you only know that the necessary condition has been satisfied it is not assuring you and you are unstable because of this thing of small disturbances you know. So, whatever stability we are going to be doing is actually going to be based on this linearized analysis. So, whatever we are going to be doing in this course is going to be based on this linearized analysis what this means is these are what are called local stability conditions okay that is only restricted to small disturbances okay. So, the stability analysis we will see in this class is based on linearization and hence local in nature as opposed to what is the other one global okay. So, whatever stability we are talking about is the local stability condition or a linearized stability condition okay. So, the analysis is called linearized stability analysis. So, it turns out these eigenvalues which are going to basically decide stability they are going to be dependent upon your operating conditions your flow rate or your temperature or your this thing concentration and as you vary the parameter these eigenvalues can actually change from having a negative real part to a positive real part the growth rate I want to use a more general term the growth rate which basically tells you how disturbances at a particular point change with time okay. They can change from decay to growth and that is the thing that we are interested in we are trying to find out this critical value when that is going to be a change in the behavior of the system okay. So, that is basically this is the mathematical framework. So, we spoke something in English now we are trying to put things in mathematical framework. So, I have given any problem you can actually work and proceed and address the question of finding the stability okay because that is the advantage of doing mathematics you can apply to any system. So, it helps you generalize things okay. So, what we will do today is work on a particular problem which is basically going to take you to a partial differential equation but it is not a flow problem okay and then we will do a flow problem later on. Because normally when you do a flow problem in a stability there is more than one equation. So, we are just going to slowly add up the complexity. So, I am going to talk about stability in a reaction diffusion system okay. So, since I belong to the chemical engineering department I need to have a chemical engineering example and one of the reasons why I am choosing this example is because I only have to deal with one variable either a concentration or temperature. So, in this case there is a concentration. So, we can see there are catalyst particles which sustain chemical reactions and most of you know that catalysts exist because you have done this in our course in chemistry what it does is it basically provides an alternative path for the reaction to take place and makes the reaction go faster okay. So, but the thing is the catalyst is usually deposited on a surface and it provides the site for the reaction to take place okay. Now, in order for you to maximize the surface area available for the reaction to occur what is normally done is that these catalysts are usually very porous okay. So, basically what I am saying is the catalysts are porous in nature and this is to maximize the surface area available for reaction okay. So, we will keep life simple and rather than talk about a spherical particle we will just say that the particle is a rectangular slab okay. So, imagine just for the sake of working with Cartesian geometry you can work with spherical coordinates. We assume one dimensional problem because I just want to illustrate some ideas dimensional problem. You see my one dimensional system which means take a rectangular slab of thickness l in the x direction extending to infinity in the y and z directions okay. So, I am just extending to infinity in the other direction. So, what this helps me do is keep my life simple that is I am just going to worry about variations in the x direction and since I am talking about a stability problem I want to worry about the time dependency as well okay. So, now I am just going to draw this lab here. Imagine so this is porous okay all kinds of pores all over the place some random network of pores will be there and this is a porous solid. So, the transport of gas assuming this is the gas reaction which takes place on the solid surface transport of gas inside the solid is going to be only by diffusion okay. There is not going to be any great velocity inside this pore. So, what I am trying to talk to you say here is that only flux we are going to transport the species inside this slab is the diffusive process okay. We are also going to keep life simple and say that there is no exothermicity and so we are talking about the temperature being constant okay. So, point is isothermal reaction okay. If we have a negligible heat of reaction diffusion is the only mechanism for transport of mass. Now you can write down the governing equations in many different ways but we will just do a shortcut in the sense you know how a species balance looks okay. You have the convective term, you have the diffusive term, you have the accumulation term and you have the generation term. So, we will just simplify things here. We just say that there is no convection it is only accumulation term. The species balance written as the partial derivative u equals okay. So, this is in some sense this is Reynolds transport theorem for you but I am doing a species balance here. I am doing it for a small infinitesimal control volume. What does this term represent? This term represents the what is this? The accumulation term okay. This is the flux okay because of only diffusion convection I am saying is not present. When we did the Navier-Stokes equation you had the convective term on the left hand side. You only had the viscous transport here. So, this is your diffusion flux. So, basically if you had your overall species balance which you must have done in earlier courses if you write it down and you drop off terms which do not exist then you will get this equation. That is the idea. I am dropping off all the velocity terms that goes off. I have diffusion only in the x direction and the y and z direction have taken it to infinity. So, I do not worry about it and this here I am just saying is my reaction kinetics. So, this kind of reaction kinetics would actually arise when you have an autocatalytic reaction okay. So, this is you are used to things like first order reaction, second order reaction. If it had been a first order reaction you would have had just u or something like that but since it is an autocatalytic reaction it is u times 1-u. So, u is basically representing something like a concentration okay. So, here u is concentration of a species. Clearly this is going to be a subject to some boundary conditions. The boundary conditions are taken as u equals 0 at x equals 0 and l. So, the thickness of the slab is l okay. The thickness of the slab is l and both the ends I am just saying u is 0. So, that is my governing equation. So, what I did is I just did my first step which is write out the governing equation for the system. So, I have a catalytic reaction which is isothermal taking place in the slab. I do not have any temperature. So, isothermal no energy balance. I only need to worry about how the concentration is changing okay. And I need to forget about velocity because there is no convection inside the porous catalyst okay. It is very negligible. So, now the next thing is we need to find a steady state right. So, for the steady state what do you do? Steady state means you need to put us as implies d by dt of us is 0 which means okay subject to. So, since I am looking at a steady state the time derivative goes off and my steady state has to satisfy this equation. So, one of the reasons why we actually chose this problem. So, the one steady state which immediately pops out is one which can be spatially uniform. If supposing you do not have any variation in the x direction okay. Supposing you do not have any variation x direction then this second derivative is going to be 0. So, that is a spatially uniform state or homogeneous state where the concentration everywhere inside my pallet is equal. So, what kind of a spatially homogeneous solution can this system have? Clearly if this is 0 then USS could be 0 or it could be 1 okay. So, if we look for a spatially homogeneous solution okay then d square USS by dx square equals 0 which implies USS times 1-USS equals 0 or USS equals 0 or 1. But then your solution should also satisfy the boundary conditions okay. So, this guy USS equal to 0 satisfies the boundary conditions whereas USS equal to 1 does not satisfy the boundary conditions. So, USS equal to 1 is not a solution. The only thing that is possible is USS equal to 0 is the only solution possible you understand. So, USS equal to 0 is the only spatially homogeneous solution okay. So, my question now is when would this the concentration inside my catalyst be spatially uniform okay. I mean this is a possible solution. So, now what are the different processes which are actually taking place inside your system? One is you have the diffusive process and one is you have the reaction process okay. So, if the diffusion is very very fast diffusion remember works to make things spatially uniform. So, basically if the diffusion is very very fast your concentration is going to become uniform whereas if the diffusion is slow you are going to have you can see a concentration gradient. But this is just English diffusion being fast and diffusion being small. What you want to do is so in relation to the rate of reaction I expect that there is if the diffusion is larger than a critical value I expect that U equal to 0 is going to be USS equal to 0 is going to be something which I can observe. Because if even if there was a disturbance diffusion is going to make it even whereas if there is a disturbance and if diffusion is very slow there will be a non uniform solution okay. So, there will be a variation of concentration inside the catalyst pellet. So, that is the question which you are trying to answer we are trying to find out whether that is some kind of what is the critical value quantitatively can I quantify this just like you were able to quantify terms of the Reynolds number 2100 okay. Can I quantify the value of this diffusion coefficient in some kind of a dimensionless group which will tell me when is this stable when is this unstable. So, that is the question okay and the reason why I talked about diffusion is because say at the end of the day whatever mathematics tells us has to be in line with whatever our physics tells us it is not that they are two completely different things as engineers we had to put mathematics and physics together okay. So, now so what I have given you is a possible solution which is spatially homogeneous which USS is equal to 0 it satisfies the differential equation satisfies the boundary condition okay and I am telling you that look this is a spatially homogeneous solution I expect the solution to be stable when diffusion is very fast as compared to reaction okay. But these are the critical threshold value which decides what is fast and what is slow and that is what mathematics will tell us okay and that is what we are going to find out by doing the linear stability analysis. So, now what we are going to do is we are going to assume okay. So, let me write this down. So, if D is large I am going to put this in you know invert commas because I want to compare that with something else which has the same units of meter square per second okay. D is large then disturbances in the concentration get smeared out we can expect spatially uniform solution to be stable okay. But D is large means what D is larger so what is the critical threshold of D above which we have stability that is the question okay and of course one should not talk in terms of D because the diffusion coefficient will depend upon your slab and all that it is good to work in dimensionless coordinates okay. It is preferable to work with dimensionless groups for having greater validity that is what I want otherwise we will say diffusion that is specific to a particular system rather than in a dimensionless group then it becomes more general okay. So, let us go to the next step. So, this is the first step write out the governing equation I wrote the I did not derive it but I mean it is okay. I think we just wrote out the steady state now we have to do the linearization right. So, I just want to mention that this corresponds to x equal to 0 here and x equals L here where that was my thickness of my slab. So, what we are going to do is look at imposing small disturbances and what do these disturbances measure they measure the deviation from the steady state okay. The disturbances basically tells you how far the system is away from the steady state. Your steady state is USS equals 0. So, now I am going to this is my disturbance variable I am just putting the tilde on top to signify as a disturbance and this tells me how far is u from the steady state. u is the actual concentration and USS is the steady state concentration in this case of course USS stands out to be 0 but you could have a problem where USS is non-zero okay. So, I mean just to keep things general I am writing it like this. So, I have u tilde written as u- USS and what I am going to ask you to remember is that this is of order epsilon this is very small okay. So, if it is of order epsilon I can I may want to specify that is of order epsilon by writing it as to explicitly show this I can write this as epsilon times u tilde equals u- USS but now u tilde will be of order 1 because epsilon u tilde is of order epsilon okay. So, this is to explicitly show that this is of order epsilon. I just want to put things in perspective with what you already done in perturbation analysis. So, now it is like saying I am seeking us USS plus epsilon u1 sorry u tilde okay. This is just to make you relate what you did in your regular perturbation series you sought us USS plus epsilon u tilde etcetera okay. The absence of disturbances it is equal to USS now epsilon tells you the magnitude of the disturbance that epsilon told you something about the parameter value okay. So, we write it like this and now I just have to substitute this in my differential equation okay. I need to substitute this particular form in my differential equation and since I am considering only small deviations I am going to linearize the term the only term which is nonlinear is my reaction term which has the quadratic dependency u multiplied by 1-u okay. So, let me just go back and write this equation here equals this is my equation x square maybe I should just write it here itself okay. Wait a second u and I am going to substitute for u this expression USS plus epsilon u tilde okay substitute for u in terms of u tilde what do you get d by dt of USS plus u tilde times epsilon equals d times d square USS plus epsilon u tilde by dx square plus a times multiplied by 1-okay this is multiplying that. What I want to do is group all the terms together of order epsilon to the power 0 and of order epsilon order epsilon to the power 0 will be my base state my steady state solution order epsilon to the power 1 will be my linearized solution because I am going to basically take only the linear terms okay order epsilon to the power 0 gives the steady state and should give the steady state order epsilon to the power 1 should give the linearized equations okay. So, let us just do that. So, I have d at order epsilon to the power 0 d by dt of USS equals d d square USS by dx square plus a times USS times 1- USS that is what I get because this multiplied by this will give me epsilon term this multiplied by this gives me the 0th order term okay this not x to the okay and this is multiplied by this will give me order epsilon. So, this is my steady state solution about which I am doing the linearization okay because steady state that goes out to 0 and the solution that we are looking at is USS equal to 0 which satisfies the boundary conditions. What about the order epsilon to the power 1 term? I get d by dt of u tilde equals d by d square by dx square of u tilde plus now I have USS times u tilde plus a u tilde times 1- USS okay. When I have this multiplying this I have one term and this multiplying this I have the other term the other term is order epsilon squared. So, I neglect it clearly now what I am going to do is use the information which I already have about USS just like you did your perturbation series analysis you have to use this information about the steady state being 0 and I put steady state equal to 0 and then I am fine is that okay minus yeah minus what is this? This is a minus here Sino second term should be negative a u tilde multiplied by 1- USS it looks okay to me this one yeah yeah second term should be negative you are right. This should be negative yeah you are right that is negative yeah now put USS equal to 0 okay and your linearized equation is d by dt of u tilde equals d times d square u tilde by dx square plus a u tilde that is your equation at order epsilon to the power 1 and what are the boundary conditions? The boundary conditions are going to be u is going to satisfy 0 0 you steady state satisfies 0 0 so u tilde also has to satisfy 0 0 at the boundary okay and u tilde equals 0 at x equals 0 comma l okay. So, this comes from the boundary condition you have to put this perturbation again in the boundary condition and get this what I want to emphasize here is whenever that is a linear equation for you because that is what we have done we have essentially done a linearization you have done this linearization using Jacobian which is what was discussed last time you can you will get the same result okay you should do that and verify for yourself. So, at order epsilon to the power 1 the equation is linear and what do I get a u tilde and u tilde equals 0 at x equals 0 comma l the other not only is it linear it is also homogeneous and I think if you do a linearization whenever you are working out a problem after doing the linearization and after looking at things at order epsilon to the power 1 if you find that your equation is not linear then something is wrong what you do god is wrong you find that there is a non-homogeneity that means something is wrong. The point I am trying to make here is that the linearized equation should always admit 0 as a solution okay. So, this linearized equation must admit 0 as a solution always. So, that is a simple check you can make now just because it admits 0 the solution does not mean you have it right but if it does not admit it is wrong okay. So, I think that is the only thing you can do. So, what I am trying to tell you is that this is a linear and homogeneous and now you know once I started off with a non-linear problem and I got a steady state I have done a linearization and now that is a linear equation I am kind of comfortably I mean my comfort zone because I know how to solve this equation using some of the things that you people have learned in your class like separation of variables okay. So, now I have a linear equation subject to homogeneous my boundary conditions also have to be homogeneous everything has to be homogeneous only then I can have this. So, this is something like an eigenvalue problem if you remember x equals lambda x x equal to 0 is always a solution. There are some values of lambda for which we have non-zero solution that is the kind of thing we are looking at here. So, 0 as a solution must always be satisfied this no matter what the parameters are and what we are trying to find out is because it is linear since it is linear we can solve this analytically. So, for example of course a linear equation is a partial differential equation. So, this is one level of complexity more than what you did yesterday. Yesterday you do not have this x square term you only have the time derivative term after linearization but you have 2 equations. Today I have only one equation but it is partial differential. When we start doing fluid flow problems it is going to be partial differential and more than one variable okay and by the way this is also very relevant to multi phase flow because I am talking about a gas in a solid. So, there are 2 phases right there is a solid and there is a gas. So, if you do not ask me how is this relevant to fluid flow it is definitely there is a fluid flowing and there is a solid particle okay. So, this is very relevant to fluid I was beginning to justify to myself that this is indeed relevant for this course and it is okay. So, now how do you go about solving this yeah who remembers his calculus course on partial differential equations. We will solve right now using separation of variables. So, this u is a function of x and t and what I am going to do is I am going to write seek u is a function of x and t as x of x multiply t of t okay the product of 2 functions one which depends only on x and the other which depends only on time. What we will do is we will what you normally do is you substitute this in that partial differential equation get the x dependency get the time dependency and then you will be in a position to understand by looking at the growth rate the exponential term which I spoke about earlier whether the thing is stable or unstable right. So, that is what we will do substitute this in the actually this will be u tilde in the u tilde equation what do you get x of x multiplied by t dash equals d multiplied by x double dash of x times t plus a times x of x times t of t okay. What we do now is we divide throughout by x multiplied by t okay divide by x t and what do you get t dash by t equals d times x double prime plus a x divided by x okay we just divide by x t and that is what we get. And so the classical argument that this left hand side is a function only of time the right hand side is a function only of x. So, the only way these guys can be equal is this is equal to a constant and what this basically tells me is I can solve for this ordinary differential equation which is first order in time I can solve for this ordinary differential equation which is second order in space and then hope to proceed to get my solution. So, we will do this tomorrow and we will wrap up this problem okay.