 So, welcome to today's lecture on multiphase where what we are going to do is wrap up the problem in some sense not completely slightly wrap up because the rest of the thing you guys will wrap up as a homework and what I want to do is just take you through the process of finding the base solution as well as the first order solution okay and then we will make some observations and then we will keep moving. So, let us proceed with finding the solution here the base solution remember corresponds to the problem when epsilon is 0. So, if our method is right you expect that the base solution falls back collapses to the flow between 2 infinite parallel plates okay the flat and that is exactly what is happening this is the upper half is what we are looking at and at y equal to 0 you have the no slip boundary condition and y equal to 0 you have the symmetric condition and that is the differential equation which has to be satisfied. And this is actually a total derivative it is not a partial derivative and you can therefore integrate it out directly right. So, before somebody points that out I will just make this remark. So, now dw0 by dy we integrate the differential equation differential equation twice and what do we get w0 equals minus y square by 2 plus c1 y plus c2 okay and when I impose these boundary conditions I have w0 prime at y equal to 0 equals 0 and this implies that c1 is 0 and I differentiate this substitute y equal to 0 I get c1 and if that has to be 0 c1 has to be 0 and from the other boundary condition and y equals half I have w not equal to 0. So, I have w0 of y equals half equals 0 implies 0 equals half I have minus 1 by 8 plus c2 or c2 is 1 by 8 and what this means is that w0 is actually half of 1 4th minus y square okay. So, that is my parabolic profile that is my base solution and I think that is perfect. So, what we want to do now is we want to go ahead and construct the solution for w1 okay and remember I need this information to find out w1 and that is how it is whenever you are doing a perturbation series what you are doing is you are converting a problem into a hierarchy of problems you use this information to construct the next fellow right and where does this information come in this information comes in here. So, I need to calculate dw0 by dy at y equals half and this is evaluated at y equals half. So, basically what I am going to do is calculate dw0 by dy at y equals half and what is dw0 by dy is minus 2y by 2 is nothing but minus y is nothing but minus half okay. So, I substitute minus half here because this entire thing is evaluated at y equals half this is therefore this evaluated at y is equal to half and dw0 by dy I have just found out this minus half okay I know w0 from my earlier solution and so this boundary condition basically reduces to w1 equals sin 2 pi x divided by 4 at y equals half the other fellow remains as it is this guy w1 dash equals 0 at y equals 0. So, what I want you to observe is that now we have a homogeneous differential equation okay this differential equation is homogeneous the right hand side is 0 but the non-homogeneity is in the boundary condition here this boundary condition of course homogeneous. If this also had been 0 if this also had been 0 what would have happened the solution to the system of equation this differential equation and this boundary condition would be the trivial solution w1 equal to 0 the fact that this guy is non-zero gives me a non-zero value for w1 okay. So, I just want to point out that this is non-homogeneous and ensures w1 is not equal to 0 okay this is homogeneous and this is also homogeneous the other nice thing of course is that this problem is linear okay there is no non-linearity in it so now the question arises I have this guy evaluated y equal to half which is nice because I can now substitute this what I have to do now is remember w1 depends upon both x and y okay. So, if you remember the physical problem the periodicity was in the x direction the flow is in the z direction okay. So, one of the things that we can visualize is that this particular velocity at the boundary is periodic is possibly going to persist throughout the domain throughout y that is whatever is the periodicity of the velocity in the z direction in the x direction that is this periodicity is going to be present everywhere. So, what I am saying is but w1 is a function of y as well so we seek we can look for a solution of the form where w1 of x, y is of the form some f of y multiplied by sin 2 pi x okay that means the periodicity in the boundary condition prevails in the domain see I have fully developed one I think it is changing in the z direction the dependence in x and y I am in the x direction if it is varying periodically at the boundary I am just going to say that the same thing happens everywhere. Now first of all what does this mean what I can do is whether this is correct or not or whether such a solution is possible or not I can find by substituting this form there and see if I can actually get an f of y supposing I am not able to proceed further that means this assumption is wrong okay and I got to come back and do something else. So, that is one way to actually find out if this is indeed a possible. So, you know you make an assumption of this kind you proceed further you get stuck you come back and then you make a change. So, clearly what we are going to do is we are going to substitute this here and what does this mean in terms of the differential equation let us substitute this form of the solution in this equation and I get d by l whole square and d square w1 by dx square when I do it do a second derivative I will I get 2 pi cosine I get minus 2 pi again cosine so I get what do I get minus 4 pi square multiplied by f of y times sin 2 pi x plus d square f by dy square times sin 2 pi x equals 0 all I have done is substituted this form in my equation for w okay. So, substitute w1 so that is perfectly fine okay. Now remember so what is happening is sin 2 pi x is present in both the terms that are present and sin 2 pi x is non-zero we can actually knock it off supposing this is not been second derivative this is not been in your equation actually with the first derivative then such a solution would not have existed because on differentiating you are going to cosine and cosine and sin you could not have factored out okay. So, the fact that I have a diffusive process viscosity is actually I have a second derivative process I actually can get this. So, that is something which I want to point out to you okay if it is a first order if it is a convection then this would not have happened okay. So, this sin 2 pi x is not equal to 0 and therefore what I get is f double prime which is that term minus 2 pi minus comes out 2 pi d by L whole square f equals 0 okay that is my equation for f and what are the boundary conditions on f I need to use these boundary conditions w1 dash equals 0 at y equal to 0 that is the derivative with respect to y. So, f dash of 0 equals 0 okay and I have w1 is sin 2 pi x and y equals half so if w1 is sin 2 pi x by 4 so this must be equal to sin 2 pi x by 4 I have f of y equals 1 fourth at y equals half okay. So, at y equals half w1 is sin 2 pi x by 4 so f of y has to be that so f has to be 1 fourth. So, basically what I have done and this is something which I wanted to see because later on when we are doing other problems we will be following the same strategy we have started off with a partial differential equation and we are going to reduce it to a ordinary differential equation okay and that is one way to help find an analytical solution okay. We will see this in more detail later but basically there is only one periodic mode which is present in the system. Since my equation is linear okay I am looking for a solution which has only that mode okay. If my equation was non-linear then the different modes could have interacted and I could have got in different waveforms not just sin 2 pi x I may have got in sin 4 pi x I may have got in sin 8 pi x okay but because my equation is linear and I am giving this kind of disturbance which is having a shape of sin 2 pi x I am expecting my response to also have sin 2 pi x but if my equation would be non-linear I would have got in other terms okay. So one of the things which allows me to do this is actually the fact that the equation is linear okay. Now this equation is something which everybody can solve with the eyes closed and accept me alright. So how do you solve that equation? I mean you just seek a solution of the form e power mx 2 pi d by L remember is a constant. So seek f of y as e power my equation what would you get? You get the characteristic equation just as m squared minus 2 pi d by L the whole squared equals 0. So m is plus or minus 2 pi d by L okay and therefore f of y is of the form a e power 2 pi d by L y plus d e power minus 2 pi d by L y okay that is your thing from calculus whatever you learnt. Now I need to find a and b and if I know a and b then I can go back I know f I know w1 and that is what I wanted to get. So how do you get a and b you have to use those boundary conditions. Let us use the boundary condition which is homogeneous which is f dash of 0 equals 0 okay. So f dash of y is a multiplied by 2 pi d by L e power 2 pi d by L y minus b 2 pi d by L e power minus 2 pi d by L multiplied by y that is f dash of y okay I want to evaluate this as 0. So f dash of 0 is nothing but a minus b times 2 pi d by L and this has to be 0 which means a has to be equal to b. This equals 0 implies a equals b. So that is good I got it out one constant and now I need to put the other boundary condition and get the constant okay. So let us do that. So from the other boundary condition so f of y is now therefore a equals b right. So a multiplied by e power 2 pi d y by L plus e power minus 2 pi d y by L okay. So what is the funda now? f of half is 1 fourth. I have to evaluate this as half and so 1 fourth f of half equals 1 fourth equals a multiplied by e power y is half right pi d by L plus e power minus pi d by L. I am sticking to the exponentials because I think people are comfortable. You are going to work with hyperbolic functions as well. So that tells me what is a. I can use this to get a. I can go back and substitute it back here and get this thing. So this implies that a is 1 by fourth of e power pi d by L minus pi d by L and f of y is therefore a. So you can just observe that this is cos hyperbolic okay and you will get cos hyperbolic and this also I can carry out into cos hyperbolic so that the factor of 2 which comes they get cancelled off in the definition of cos hyperbolic. So this is 1 fourth of cos hyperbolic 2 pi d y by L divided by cos hyperbolic pi d by L okay. That is what we have. So this is w1 for you and now you have found the correction to the velocity to the first order because now w is w0 which is your parabolic profile plus epsilon w1. You should similarly go ahead and find w2 okay. Again you have homogenous equation. One of the boundary conditions is homogenous. The boundary condition, one boundary condition will be non-homogeneous but for this boundary condition you need the information from w0 and w1 and this information you would use to go find the solution okay and just one last observation before we proceed further. I want you to realize that w1 is going to be of the form 1 by 4 cos hyperbolic 2 pi d by L y divided by cos hyperbolic pi d by L times sin 2 pi x. Now like I had mentioned earlier one of the things we are interested in finding out not only is the actual velocity profile but to find out if there is by having these kind of corrugations is there any change in the flow rate which is passing through the channel okay. You can look at flow rate per unit wavelength because you have an infinite channel which is extending to infinity in the x direction of this periodicity. It makes sense to concentrate on one wavelength here and through the gap of d and see what the flow rate is. So how do you find the flow rate? You would find the flow rate by taking this velocity and integrating this out in the x and y direction okay. Now what you are going to observe is that when you have since you have the dependency and the solution is of variable separable form is some function of x multiplied by a function of y. So when you are going to integrate this out in the x and y directions you would be able to integrate this out in the y direction you would be able to integrate that out in the x direction separately okay. When you do this integration out in the x direction you are going to see that this is periodic and therefore sin 2 pi x when you are integrating from 0 to half or minus half to plus half because you are looking at the whole channel remember okay. In the x direction from 0 to L sorry in the x direction from 0 to L in the y direction from minus half to plus half. Here it is from 0 to L you will find that this guy is not going to make any contribution. So what this means is that because this is going to go to 0 this integral will go to 0. So what this means is as a first order effect there is going to be no change in the flow rate that you are going to see. If you want to go to the second order effect which you have to go because that is a homework problem you will find that it is going to make a difference to the flow rate. So the flow rate is going to be effected only at the second order okay. So point I am going to make here is at order epsilon the flow rate is not affected since integral 2 pi x oh sorry sin 2 pi x and if I am working in dimensionless coordinates is from 0 to 1 okay equals 0 order epsilon squared the flow rate will be affected and that is something I am telling you I want you to prove to yourself and that this is indeed true. So all you need to do is find what w2 is what you will find is that the x dependency is in the form of a square okay maybe sin square or cos square. So it is when you integrate out you are not going to get a 0 value okay and you just need to solve again the same procedure same process. So that is the path so this is the path which has been wrapped up and this portion you guys have to wrap up. I think I just want to make one more remark that the domain perturbation method is extremely important when it comes to solving stability problems. I think it is kind of the basis for solving stability problems because you know remember now I had a flat surface and I gave a small perturbation which was periodic and then I was able to actually construct solutions okay. So when we are talking about multi phase flow problems like the problem of the jet breaking up into drops okay what do you have? You have a cylindrical surface which is your base solution without any epsilon in it which is defined as r equals r0 which is constant and now you give a small perturbation which could be periodic and then you are going to ask the question is this perturbation going to make the jet break up or not. So you can see now we are actually interested whenever you are talking about multi phase flow problems where there is an interface and the base state is normally going to correspond to boundary which is like y equals 1 half but when you give a perturbation it is going to be y equal to 1 half plus some perturbation okay. So now when you want to find out whether something is stable or unstable you need to therefore use the domain perturbation method that we just spoke about to actually do the calculation. So that is where all this thing is fitting in I am just trying to tell you this to show you that whatever we are doing here is actually fitting in. So the first part of the course we just did a small revision of some of the concepts you have learnt and we extended it to the boundary conditions when you have actually an interface which is not necessarily corresponding to a coordinate axis okay and I told you how to find the normal stress and the tangential stress when you have some surface of the form y is equal to f of x that is because when I am going to actually solve a problem I am going to be using normal stresses are balancing tangential stresses are balancing okay not stresses in x direction not stresses in the y direction it is the normal stress and the tangential stress. So that is what we use then we came to a perturbation because this is going to be basically the starting point for stability. So let me discuss a little bit about stability and then what we will do is we will start solving some problems in stability okay. So with this perturbation methods is just to give you some insight about how you can convert a non-linear problem into a bunch of linear problems. So you remember that viscous heating problem was actually non-linear because you had a du by dy whole squared term. So but when what we did was we converted it to a bunch of linear problems and then we got an analytical solution okay. So that is one of the things we are going to do when you talk about stability. So let us quickly jump into stability. So first of all I want to clarify that when we talk about stability we are talking about stability is with respect to a state the system. What does this mean? People normally talk about stability of a system this system is stable the system is unstable okay but what I am saying is you should talk about stability of a particular state may be a steady state. So if you have a steady state you have a system where which let us say has a steady state like you have laminar flow in a pipe that might be the easiest example okay. So if you have a laminar flow in a pipe and let us because we are doing a multiphase flow we solve a single phase then we go to multiphase. So laminar flows is a steady state flow in a pipe okay it is always a steady state. What does this mean? You can calculate the like earlier on today we found all this parabolic velocity profile okay. This parabolic velocity profile was dimensionless so it did not have this g and all that the pressure drop but otherwise it would have had a pressure drop. So no matter what the pressure drop is that is a solution to the equation if you have a flat plate you understand. So what I am saying is the parabolic profile is always a solution to the governing equations. Do you all agree with me? The parabolic profile earlier did not have anything it did not have the pressure drop because it was scaled inside my velocity but if I actually write it in terms of an actual velocity the dimensional velocity I would have had a pressure drop just like your Heggen Poisson equation. Heggen Poisson equation has some dp by dz. So if I give a very small dp by dz I have a very low velocity if I give a large dp by dz I have a large velocity right. So basically what this means is always a solution however you all know that for Reynolds number less than 2100 only you will have the laminar flow which is going to be experimentally observed when Reynolds number is more than 2100 the flow becomes turbulent okay and so what is this critical thing about 2100. So that means that what I am doing is just visualize an experiment where you are actually increasing the pressure drop. So there is a parameter in your problem as an experimentalist which you can actually change okay. So you are doing an experiment where you keep increasing the pressure drop. So let us say that you have a pressure drop which corresponds to Reynolds number of 100 which gives you a flow which corresponds to a Reynolds number of 100 okay. So we have a delta p which gives a flow corresponding to Reynolds number equal to 100. What does this mean? I can given a delta p I can find my velocity profile I can find my average velocity right. I can go back I know my properties of the liquid I know the diameter of the channel I can get rho vd by mu and you do this calculation you find Reynolds number is 100. So what does this mean? You can actually when you are doing this experiment you will actually see a laminar profile. So whenever you are doing an experiment remember your experiments are always going to have some disturbances okay. I mean when you do an experiment it is not possible for you to perfectly control your pressure drop it is not possible for you to come perfectly control your flow rate okay. There always be some small disturbances. If you are able to observe this laminar velocity profile then it means in spite of these disturbances which are existing in some kind of a time average sense you are able to get your laminar velocity profile okay. So what I am trying to tell you is that any system there will always be some small disturbances present okay. But it turns out that these disturbances are decaying to 0 they do not get amplified as long as the Reynolds number is less than 100. But suppose you know you slowly increase the pressure drop and let us say Reynolds number is 1500. Again you will see laminar flow. Again there will be disturbances are present okay and again your system is stable but supposing you change the pressure drop a little bit more and you cross the threshold of 2100 your flow becomes turbulent. What is the difference? Your laminar profile is still theoretically a steady state okay. But there are some disturbances which are going to be present but now the disturbances are getting amplified. The disturbances do not decay to 0 and because the disturbances do not decay to 0 you would get a turbulent state you understand. So what we are talking about so what I say is the laminar flow the laminar profile is stable that is the steady state solution the laminar profile is stable for Reynolds number less than 2100. The laminar profile is unstable for Reynolds number greater than 2100 okay. So I am talking about stability in the context of steady state. So that is how you should always talk. You cannot say the system is unstable for more than 2100 or less than 2100 it is stable. You are talking about stability in the context of the steady state okay. So what I have done is I have introduced the concept of stability in the form of small disturbances which are always going to be present in an experiment okay. So now let us say we have a delta P which gives let me write on a few things. We have a delta P which gives a flow which corresponds to a Reynolds number of 100. In any experiment we will not be able to have perfect control okay. There will be some small deviations okay. I am talking order epsilon. So now you know so very small deviations let us say of order epsilon where epsilon is very much less than 1 and the question is how does the system respond to these disturbances. So see if you have a pump which is pumping a liquid okay there is going to be some small fluctuation coming in at the pump. So it is not that pressure and outlet is always going to be whatever 5 bar or 10 bar although it has a possibility of you know it has the rating of 5 bar at the outlet pressure maybe it is actually going to be slightly varying 5.001 to 4.999 okay. So that is my perturbation. The question is so say for example the pump outlet pressure is say rated at 5 bar but it could vary from 5 plus or minus epsilon. Clearly it is not going to be exactly 5 bar and if somebody is saying you are going to be exactly 5 bar and you believe them you are crazy right. So there is going to be some fluctuation and depending upon how much money you are paying how accurate the pump is epsilon is going to be smaller and smaller okay. So I am saying that this epsilon is my disturbance the inlet pressure but for Reynolds number less than 2100 these disturbances are deviations in the inlet pressure decay to 0 in the fluid okay that is how the dynamics is when you actually solve the partial differential equations and you are actually trying to find out whether this kind of a deviation of pressure from 5 to 5 plus epsilon okay whether it is going to make any difference or not decay to 0 in the fluid if you know Reynolds number is less than 2100. But if the Reynolds number is greater than 2100 the laminar flow becomes unstable and we have turbulence and there is something you all know. So I just wanted I mean you guys have possibly mugged something up and said oh Reynolds number less than 2100 laminar but actually what is happening is when you are actually crossing you are having a stability problem okay what was stable has become unstable and the way I want you to visualize the stability problem is that there are these small fluctuations always going to be present and what happens is for less than 2100 these deviations decay more than 2100 the deviations get amplified and you have turbulence. Remember the laminar flow is a possible solution for Reynolds number greater than 2100 okay the parabolic velocity profile is possible. So what I am saying the point I am trying to make here is laminar flow is always a possible solution is a possible state possible steady state. In fact there are people who have done experiments very very carefully and they have been able to observe laminar flow up till Reynolds number of 10 power 5 okay. So if you really were interested in this you could do very very careful experiments and see that when you have Reynolds number of as high as you know 100,000 you would get laminar flow but then normally if you are not very careful you just have a regular pump you are doing an undergraduate lab I mean people are possibly trying to cut down on costs so you just will get this transition at 2100 okay. So I am just trying to tell you how do I know this careful experiments reveal that laminar flow can be seen at Reynolds number up to 10 to the 5 okay. So all this means is that we are talking about a steady state now and my steady state is my laminar flow profile. So this laminar flow profile is stable I am going to call it stable for Reynolds number less than 2100 it is unstable for Reynolds number greater than 2100 okay. So the laminar flow is stable for Reynolds number less than 2100 since the disturbances decay to 0 okay and the laminar flow is unstable for Reynolds number greater than 2100 as the disturbances now amplified. So this is something we are always going to do when we talk about a stability problem. When we talk about a stability problem what we are going to do is we are going to worry about changing some parameter which you can control as an experimentalist and you are going to find out how does the system behavior change how does the response to disturbances change when you have one parameter which is being varied. So just like you have done now when Reynolds number is low when disturbance are present they are going to decay everything is fine and stable when Reynolds number is higher and that is the parameter I am changing okay the parameter I am changing is the flow rate or the pressure drop. So density the pipe is of course the same pipe so diameter is constant fluid is the same density and viscosity are the same the velocity has to change for changing the Reynolds number I have changed the velocity by changing the pressure drop okay and as you keep changing the parameter which is the inlet pressure you will find that above a critical threshold value which in this case happens to be 2100 for Reynolds number you have a change in the behavior. So what was stable has become unstable so that is what I have written down here. So the stability is always talked in terms of responses to disturbances okay so what you do is you have a particular steady state and the steady state would normally correspond to what in my perturbation method I had as the 0th order solution okay we did perturbation series now usually I have a 0th order solution for which I know the solution like I have the parabolic profile for the laminar. Then what I do is I start giving disturbances and whatever I am talking about experimentally I have to do it using my model to find out what this threshold is. So what we are going to do is we determine stability in terms of the response or disturbance okay. And experimentally if you are actually able to observe a particular state experimentally you can conclude that that particular state is stable I mean without doing anything special you are actually able to observe because disturbances are always going to be present when you are doing an experiment and if you still see that particular steady state that means the system is ensuring that the disturbances are decaying to 0 okay. So any experimentally observed state as long as you have not done something to control it okay to make an unstable system stable if you have not done anything special you are just letting a system run okay and if you observe it experimentally then that particular state is stable. So any experimentally observed state is stable this is seen in the context of in the presence of disturbances. So what we are going to do in this course is try and transform this simple experiment which you are comfortable with into a mathematical language so that we can make a prediction of these threshold values okay. So what we want to do is we want to convert this to a mathematical framework to be able to predict onset of instability in different contexts specially in the context of multiphase flows okay that is the whole idea.