 So, I think this was, so we will continue today working on the problem of incorporating the effect of viscous heating in a viscometer, okay. And one of the things we are going to see is that the estimation of the viscosity if you neglect viscous heating what is it going to be and when you include the effect of viscous heating how is it going to be effected, so that is the whole idea. Because viscosity is a function of temperature okay, now in the last class what we did was we worked on the problem of the momentum equation and we simplified it. So, today we are going to take the energy equation which we had derived last time and we are going to simplify this again and do a perturbation series analysis, alright. So we had assumed the viscosity to be varying as 1 divided by 1 plus alpha theta and k as varying linearly with theta, okay. Now it is perfectly fine, this has been done just to keep the algebra simple in some sense. But you know 1 by 1 plus alpha theta can also be approximated as a linear relationship if you are interested in a small interval of theta for example, okay. So, what we will do is we will proceed with simplifying the equation given as dk by dy multiplied by d theta by dy plus k times d square theta by dy square plus mu times du by dy the whole square equals 0. Now I am going to find dk by dy as k is a function of theta and theta is a function of y. So the first term dk by dy becomes dk by d theta multiplied by d theta by dy times d theta by dy. So this is just d theta by dy the whole square plus k times d square theta by dy square plus mu times du by dy the whole square equals 0. Now I think there was some confusion in the last time because I had not taken into account the 0th order term but I am just going to say that we seek theta as being equal to theta 0 plus Brinkman number times theta 1 plus higher order terms theta 2 u as u0 plus Brinkman number times u1 plus Brinkman number squared times u2 plus higher order terms. These are the corrections which we need to find out. Now basically what this means is when the Brinkman number is 0 and that is the small parameter about which I am doing the expansion I am going to have to the solution to the governing the momentum and the energy equation is going to be theta 0 and u0 okay. Physically what is it when there is no viscous heating the temperature is going to be isothermal temperature is going to be equal to t0 everywhere or theta is going to be 0 okay and the velocity profile is going to be linear and when we since we made a dimensionless we know that theta 0 is 0 and u0 is y. Let us just say from the condition that Brinkman number is 0 where Brinkman number is 0 then basically there is no viscous heating the momentum equation there is no temperature change momentum equation temperature equation become decoupled and you can solve separately okay and physically that is what you expect no temperature change linear velocity profile. I am going to substitute all the things that I know dk by d theta is nothing but beta okay. So, I am going to write this as beta multiplied by d theta by dy whole squared d theta by dy whole squared is nothing but d by dy of theta 0 plus Brinkman number times theta 1 plus etcetera whole squared plus k which is nothing but 1 plus beta theta times d square theta by dy squared of theta plus mu is 1 divided by 1 plus alpha theta times du by dy the whole squared equals 0. I am just doing this possibly in a slightly inefficient way but then I hope to minimize mistakes that I may end up making. So, this is beta and now what I need to do is substitute for theta in terms of this expansion because my job is to find out theta 0 theta 1 theta 2 and if I find out this then I can substitute it back here and find out what theta is right. So, I know theta 0 is 0 and what does this give me d by dy of I am just inserting this. So, it simplifies my life otherwise I just if I put do not put theta 0 equal to 0 here and get a 0th order term and then I have to calculate it I am just using the fact that I have already calculated the 0th order term. So, I am just going to use that to directly get the first order term and you can clearly see that this term is not going to be first order because the lowest power that this is going to contribute to is the Brinkman number squared. So, this is actually going to give me a second order term. So, for the first order thing this is basically 0 this is not a first order term. Let us come to this term here 1 plus beta and theta is written as theta 0 which is 0 plus theta 1 multiplied by Brinkman number plus higher order terms which I do not want to write because that is going to give me Brinkman number squared okay and d square theta by dy squared is going to be when I substitute this back inside here I am going to get theta 0 0 I get Brinkman number times d square theta 1 by dy squared okay plus higher order terms which we do not worry about and this term here represents what plus 1 plus alpha theta inverse times du by dy squared okay du by dy squared is going to be du 0 by dy which is 1 plus Brinkman number times du 1 by dy that is when I substitute this back inside here du 0 by dy is 1 plus Brinkman number times that plus higher order terms whole squared equals 0. I am telling you here that this term is of order Brinkman number squared this term because the leading order term will be Brinkman number squared. So, this is going to drop off okay so I do not worry about this what about this term here this term here gives me a 1 multiplied by this will give me a term of order Brinkman number this multiplied by this is going to give me a term of order Brinkman number squared okay. So, to order Brinkman number what I have is so the second term on the left hand side at order of Brinkman number is equal to 1 multiplied by d squared theta 1 by dy squared okay because the this term is of order Brinkman number squared because Brinkman number multiplied by Brinkman number and what about the term on the right hand side this I am going to do a binomial series expansion it is also on the left hand side. So, the third term on the LHS yeah the third term on the LHS is this but doing a binomial series expansion like 1-alpha theta okay etc multiplied by 1-brinkman number times du1 by dy the whole squared the whole thing squared right the whole thing squared and now you substitute this thing back inside this is equal to theta is theta 0 plus Brinkman number times theta 1 1 times theta 1 plus higher order terms okay. So, the point is yeah what is the point here this term here gives me this term here gives me 1 plus I keep making this mistake here right this Brinkman number times theta 1 plus higher order terms we neglect and this gives me the du by dy which is 1 plus Brinkman number this whole squared yeah let us keep going. So, I get 1-alpha Brinkman number times theta 1 times 1 plus 2 times Br times du1 by dy the whole squared plus okay. So, what am I left with I have actually got a problem. So, you think I have made a mistake here because this should simplify to 1 times Brinkman number. So, these guys will not contribute only this I have to worry about okay let us go ahead I get 1 multiplied by 1 plus 2 times du1 by dy times Brinkman number okay and minus alpha times Brinkman number times theta 1. So, clearly things are not as good as they should have been because I have made a mistake somewhere here I should have taken the the simplification in the algebra I could have taken this in the denominator I would have got 1 plus alpha theta here I would have got 1 plus alpha theta here and then things are fine. So, what is this 1 plus alpha theta yeah. So, that is why do you think I made a mistake because this should actually simplify to 1 or this was simplified to Brinkman number and then I have the second term should be an order Brinkman number this should give me 1 okay. Now, clearly I am not getting 1 here and I am unable to find my mistake let us do it like this yeah. This is order 0 yeah the order 0 term is not a problem because I am not taking that into account but this is let us do like this I am going to redo this thing yeah no problem. So, we come back to this equation and we start here okay 1 plus beta multiplied by I am going to take this 1 plus alpha theta to the other side yeah we should get this thing do not worry we will figure this out hopefully soon. So, all I am doing is taking the 1 plus alpha theta 1 reciprocal here I am neglecting that because that is going to contribute everything of order Br square okay and let us see this helps if it does not help then we really do have a problem plus. So, now we come back to this one and say this is 1 plus alpha theta 1 multiplied by Br and this is 1 plus beta times theta 1 Br plus Br d square theta 1 by dy square. So, this there is already a Brinkman number here and the only term which will be of order Brinkman number which contributes to this is going to be when I take the 1 and 1 here because everything else is going to give me a higher order term okay. So, that definitely makes my life simple then do not seem to have solved my problem because I really have not found my mistake right. So, this guy here gives me d square theta 1 by dy square and if you do it right what you should get is plus 1 here okay and so you guys have to do it right and get plus 1 and so now we move on assuming it is in it plus 1 equals 0 let us not spend time on this the point is d square theta 1 by dy square plus 1 equals 0 okay if you do it right and so now what I can do is I can solve this for theta 1 and that gives me d square theta 1 by dy square equals minus 1 and you can integrate this out you get d theta 1 by dy equals minus y plus c 1 okay theta 1 is again integrate out one more time you get minus y square by 2 plus c 1 y plus c 2 okay. The boundary conditions on theta are the theta has to be 0 at the 2 walls okay. So, every term in my perturbation expansion has to satisfy this. So, boundary conditions are theta 1 equals 0 at y equals 0 and 1 okay I use that at y equal to 0 I should get theta 1 is 0. So, I get c 2 equals 0 okay and I get half equals c 1. So, this implies that theta 1 is so this implies theta 1 is half of y minus y square that is theta 1. Now, you go back to the equation which we have derived yesterday which actually related u 1 to theta 1 okay and you can actually solve this equation for u 1. Yesterday from the momentum balance to the first order we derived an equation which said something like d square u 1 by dy square and you have to tell me what we got because this equals alpha times d theta 1 by dy is it can you check is the plus or minus this is plus okay. So, now I already found out what theta 1 is okay I can substitute this here and I can solve this equation for u 1. So, this way I am getting my collection for u 1 for the velocity and the first order. So, substituting the theta 1 which I have got here and that is what we want to do remember solve a bunch of equations sequentially is alpha multiplied by d theta 1 by dy which is half minus 2y by 2 which is y okay. So, again you can do the same thing integrate this out and find u 1. So, the point is d square theta 1 by dy square plus 1 equals 0 okay if you do it right and so now what I can do is I can solve this for theta 1 and that gives me d square theta 1 by dy square equals minus 1 and you can integrate this out you can d theta 1 by dy equals minus y plus c 1 okay theta 1 is again integrated out one more time you can minus y square by 2 plus c 1 y plus c 2 okay. The boundary conditions on theta are the theta has to be 0 at the 2 walls okay. So, every term in my perturbation expansion has to satisfy this. So, boundary conditions are theta 1 equals 0 at y equals 0 and 1 okay. I use that and y equal to 0 I should get theta 1 is 0. So, I get c 2 equals 0 okay and I get half equals c 1. So, this implies that theta 1 is so this implies theta 1 is half of y minus y square that is theta 1. Now, you go back to the equation which we had derived yesterday which actually related u 1 to theta 1 okay and you can actually solve this equation for u 1. Yesterday from the momentum balance to the first order we derived an equation which said something like d square u 1 by dy square and you have to tell me what we got because this equals alpha times d theta 1 by dy is it. Can you check? Is it a plus or minus? This is plus okay. So, now I already found out what theta 1 is okay. I can substitute this here and I can solve this equation for u 1. So, this way I am getting my collection for u 1 for the velocity and the first order. So, substituting the theta 1 which I have got here and that is what we want to do remember solve a bunch of equations sequentially is alpha multiplied by d theta 1 by dy which is half minus 2y by 2 which is y okay. So, again you can do the same thing integrate this out and find u 1. So, d u 1 by dy equals alpha times y by 2 minus y square by 2 plus a constant c 1 and you can integrate this one more time to get u 1 equals alpha multiplied by integrate this one more time y square by 4 minus y cube by 6 plus c 1 y plus c 2 okay. So, I think the chances of me making mistakes here is very minimal but you never know. The point I want to make here now is using the boundary conditions, what are the boundary conditions at y equals 0 and 1 u 1 is 0 the mostly boundary conditions okay. So, this implies what c 2 is 0 okay and what is c 1 0 equals alpha multiplied by 1 by 4 minus 1 by 6 plus c 1 which gives me c 1 is alpha times 1 by 6 minus 1 by 4 which is alpha times minus alpha by 12 okay. So, I have got u 1 as alpha times y square by 4 minus y cube by 6 minus y by 12 okay. So, this is my first order correction to the velocity profile u 1 and you already know the temperature equation. So, basically what I am trying to tell you here is when you have viscous heating there is going to be a modification in the linear velocity profile without the viscous heating when the thing is isothermal your velocity profile is linear. If you have actually a temperature dependency heat heat being generated viscosity being a function of temperature and let us say it does not vary too much a small temperature rise and you will have a difference in the velocity profile that is being caused by the variation of the viscosity with temperature okay. The variation of the viscosity with temperature makes the velocity deviate from the straight line okay. So, that is the idea. So, what is the deviation that you get what is the actual velocity profile? So, the temperature so what we have done is we have calculated the temperature first and we are substituting it and getting the velocity profile. So, the actual velocity is going to be u 0 which is nothing but y plus Brinkman number times u 1 and u 1 is what we have just calculated which is alpha multiplied by y squared by 4 minus y cube by 6. I should have written this in a slightly better way but then in increasing powers or something like that but let it be as it is. So, that is the correction which I have to the velocity profile. So, remember if Brinkman number is 0 I get the Kuwait flow linear profile. If alpha is 0 which means my viscosity is constant viscosity does not change the temperature then again there is no correction to the profile. So, the basically the modification the velocity profile is occurring only because of the Brinkman number or the alpha if one of them is 0 you will have no change okay. So, you may ask so what of course you are interested you can go ahead and do higher order terms and get better corrections right. Now, the question is the objective of any viscometric study is we are trying to estimate the viscosity of the liquid okay and if you remember suppose we have an isothermal situation that means Brinkman number is 0. No temperature rise and how do you estimate the viscosity mu is tau xy the plane was x right. So, tau xy was it yx yeah tau yx is not this y is in this direction tau yx at y equals 0 divided by the shear rate. The shear rate is du by dy and what you know is you know the bottom plate is at velocity 0 upper plate is the velocity capital U and U divided by D because it is a linear profile you are going to assume that this is just U by D okay. That is this is the shear rate. So, this is the stress at the wall and this is the shear rate okay. So, this would mean the actual viscosity which you are going to measure because viscosity is not changing. Now, supposing it is actually a non-isothermal situation what you are going to do is you are going to you can do 2 things. One is you can assume that there is still a linear velocity profile inside that is you just know what the velocity is at the bottom you know what the velocity is at the top and assuming that the variation is linear. Now, of course that is wrong but what you are going to get is something like an apparent viscosity something like an effective viscosity okay. So, basically what I am saying is if we have a temperature rise then the effective viscosity or the apparent viscosity is going to be given by tau yx at y equals 0 divided by U by D. Why is there an effective viscosity? Because I am assuming that the profile is linear but it is actually not linear. I am not correcting for the non-linear velocity profile here okay. So, this is viscosity since we are not correcting for the deviation from the linear profile. So, in order for you to get the actual velocity actual viscosity what would you have to do? You would have to find out the shear stress at the wall and you have to find the shear rate also at the wall. But to find the shear rate at the wall you have to include the correction term as well okay. So, when you actually use the entire velocity field and get the shear rate then you the viscosity that you are going to be calculating is the one which is the actual viscosity not the effective viscosity okay. So, the reason I am calling this apparent is because I have assumed it is still linear and that is wrong okay because actually it is varying. So, what about the actual shear rate at y equal to 0 that is du by dy at y equal to 0 that is going to be 1 plus sorry and y equal to 0 right only this guy is going to contribute 1 minus alpha Brinkman number by 12 because I am evaluating this at y equal to 0 because that is what I want to find the viscosity right. So, now the actual viscosity is going to be y x and y equal to 0 divided by the actual shear rate. The shear rate remember this is a dimensionless shear rate I want to convert it to a dimensional which means I am going to multiply this by the characteristic velocity and the characteristic length I get my u by d again okay. It is u by d multiplied by 1 minus alpha Br divided by 12. So, this is my actual viscosity that I have I want to basically relate the mu apparent to the actual mu okay and what I am going to do is I am just going to eliminate the tau y x the shear stress between these 2 equations and I am going to find that mu. So, from these 2 expressions I find that mu apparent is going to be mu multiplied by 1 minus alpha Br number by 12 okay. So, that is the correction which you have to do so if you do not want to worry about say as an engineer you do not want to worry about what the actual velocity profile is you know my top plate velocity you know my your bottom plate velocity. So, what you would do is you would just use that to find the shear rate and the instrument is going to give you your shear stress you divide to get the apparent viscosity. Once you know the apparent viscosity and if you know the properties of your fluid you know your alpha and you know your Brinkman number you can use this to get the actual viscosity okay. So, this is the correction which you have to do if your fluid has a viscosity as a thermal conductivity which depends on temperature. So, the point I am trying to make here is that this is the actual viscosity the actual strain rate at the lower wall at y equals 0 is used okay. So, that is the reason this is the actual viscosity which is there and if you do not want to take into account the correction and assume that the velocity profile is still linear then you get the apparent viscosity. So, that is the 2 different things that we are doing here. So, this is the correction or this is the relationship between the 2 viscosities okay. The apparent viscosity is calculated using the overall shear rate u by d I do not remember what dimension I used last time but maybe it was d and mu is calculated using the actual shear rate du by dy at y equals 0. This is basically done in Gary Lille and you can take a look at it but only thing I realized after the last class was I think I switched the alpha and the beta. So, I have the alpha with the viscosity here alpha with the thermal conductivity those are some small adjustments you have to make possibly okay. Okay so, what we have done is we have basically seen 2 problems where we just illustrated the idea of doing this perturbation series analysis okay when we can get analytical solutions for non-linear equations. The first one was this pulsatile flow problem and the second one was this problem with the viscous heating. Now the approach that we have been using is what is called a regular perturbation series is what we have illustrated so far. Now there is another class of problems where this regular perturbation series will not work. So, I am just going to tell you what the name is first and the other class is what is called a singular perturbation series is. So, to give you an idea just again to illustrate this idea in a very simple way we will do what we always do we start with the polynomial and the simplest polynomial is a quadratic right and everybody understands what a quadratic is. So, you already we start off with the quadratic to begin with if you remember and then we start came to a differential equation. So, now we go back to our quadratic again because that is my comfort zone right. So, consider this quadratic epsilon x square plus x minus 1 equals 0 okay. Clearly because this is a quadratic I need to have two roots okay and so this has two roots since it is a quadratic and what you would do is you would say oh this is a small parameter epsilon. So, let us do a perturbation series and let us get these roots right. Of course you know all the roots okay let us get the roots right. So, if we do a perturbation series if we seek x like we did last time x0 plus epsilon x1 plus etcetera and now if you were to put epsilon equal to 0 because that is the solution to that is what gives you x0 okay. If we put epsilon equal to 0 then what do I get to find x0 we have x equals 1 or we have only one root. So, the other root the information is not even there. So, if you remember we need to have the last earlier problem and we did we had two roots and then we got corrections to each of those roots okay. But now what has happened the other root has disappeared the other root has disappeared because this problem the parameter epsilon is now multiplying my leading order term. If my epsilon had been in one of the other order terms my x to the power 1 term of my constant term I could have proceeded even then there are situations when the perturbation series may not work okay. So, the point is let us look at why this problem is arising okay and one way to do that is since I already know the solution to this equation because it is a quadratic you know the formula for the quadratic equation. So, let us use that and see what happens to the two roots okay. So, the exact solution is minus b which is minus 1 plus or minus square root of b squared minus 4ac divided by 2 epsilon correct and we will do for small epsilons we do our famous binomials series expansion and what do we get for small epsilon this is approximately minus 1 divided by 2 epsilon plus or minus 1 minus 2 epsilon plus etcetera okay. You know 1 plus is this plus you are right it is plus yeah it is plus what do I get what are my 2 roots my 2 roots are 2 epsilon divided by 2. So, x the 2 roots are 1 2 epsilon divided by 2 epsilon and then I have minus and minus cancels I have minus 1 wow something is wrong yeah minus 2 minus 2 epsilon divided by 2 epsilon correct and take plus I get 2 epsilon by 2 epsilon yeah wonderful. So, this gives me 1 and minus 1 by epsilon minus 1. So, in the limit of epsilon tending to 0 this root is going to go to infinity okay and that is what it means is normally what we are doing is we are assuming that the x is going to very smoothly small changes and x gives you small changes and small changes in epsilon give you small changes in x but what we say is this guy is pushing off to infinity as epsilon tends to 0. So, x is actually becoming unbounded. So, the usual way in which you can actually solve this problem and these kinds of problems usually arise when you have different scales in the problem okay when you have multiple length scales or multiple time scales in the problem. So, what we have to do is we have to remember that there are multiple scales and in the limit of epsilon tending to 0 since x is actually tending to infinity you actually go about defining a new variable and then you calculate okay. So, this here as epsilon tends to 0 x tends to infinity okay. So, we cannot do a perturbation about this other root is actually infinity okay. So, how do you overcome this problem by exploiting the fact that this implies that we have some multiple scales we define and as a new variable y I want to make sure y does not push off to infinity. I want to make sure y is you know going to be bounded x is going off to infinity that means x is you can see x is going as 1 by epsilon okay. The lower the epsilon is x is pushing off to infinity and is kind of inversely proportional to epsilon. So, I am going to define y as x multiplied by epsilon. So, as epsilon tends to 0 x goes as 1 by epsilon and so this guy is going to be bounded finite in the limit of epsilon tending to 0. So, this is a new variable which I am defining okay and basically what I am doing is I am scaling if I actually had a physical problem that means I am defining my length scale in a different way to actually get a bounded solution in that region okay. Now, of course you are looking at this as a mathematical problem but tomorrow when you are actually solving a physical problem this would correspond to something like a length scale or a dimensionless number which means like we saw earlier your time scale can be different okay. So, now if you were to substitute this in your quadratic equation and you will get a quadratic equation in y. So, let us do that. So, x is y by epsilon. So, now this gives me epsilon multiplied by y by epsilon whole squared plus y by epsilon minus 1 equal to 0. So, when I expand this out I get this implies y squared plus y minus epsilon equal to 0. So, this of course has gotten rid of the epsilon from a second power but it is not the coefficient of my quadratic term my y squared term is not moved to my constant term. So, now I can be bold enough to try and seek a perturbation series solution okay and I can get y as an approximate solution. Once I get y I can scale it back and get x okay. So, this is one way in which you can actually solve problems just singular perturbation problems. So, singular perturbation problems occur in many areas. For example, if you have viscous flow in a channel or along a wall and you used to boundary layer flows right. In a boundary layer flow what do you do? You assume that very close to the boundary layer we have a small region where the effects of viscosity is important. Far away from the boundary layer the effect of the viscosity is not important right. Remember in your Navier-Stokes equation your viscous term occurs as a second order term it is a differential equation now. So, that means far away when the viscous terms are not important if you would actually drop the viscosity effect your second order equation has now become a first order equation because you are dropping off all your del square v terms okay. So, effectively to solve the equation you had to have two boundary conditions but now since it is the first order equation you need to satisfy only one boundary condition. So, the fact that you are unable to satisfy both the boundary conditions tells you that you will not be able to satisfy the boundary condition near the wall where the viscous effects are important. So, what you have to do is just like we did here for low epsilon we did a rescaling what you do in boundary layer theory is in near the boundary layer you rescale your variables to find the solution in the boundary layer then you find the solution away from the boundary layer and then you try to match okay. So, basically boundary layer flows is another example where singular perturbations are important. So, I just wanted to mention this because the perturbation series method has some maybe I should not call it limitations it is not so simple and straightforward to apply all the time it is not that every time you can seek you know a perturbation series solution and expect it to work there are times when it will not work okay. And one example is this singular perturbation. So, I will just say that when you do boundary layer flows okay we assume far from the wall the flow is inviscid near the wall it is viscous. So, if you remember you have this boundary layer over a flat plate you have u0 uniform. So, effectively there is no velocity gradient far away from the wall but here it is varying parabolically or in some shape eventually it matches with this. So, the point is here I would actually have to retain the effect of viscosity here I have to retain the effect of viscosity and here I have a second order equation and here I have a first order equation and I will not worry about trying to satisfy the no slip boundary condition. This profile I will worry about satisfying the no slip boundary condition because this is my second order equation I need to have two conditions one is no slip here and u equal to u0 at the boundary layer here I am just going to not worry about satisfying this. So, it looks like I am allowing it to slip okay. So, these are problems which are actually solved by the method of what is called matched asymptotic expansions help us solve such problems okay and I am just giving you some keywords here for you to you know pick up and read those of you who are interested how this mass asymptotic expansion done. So, once you get the idea you can possibly apply to some problem okay what we want to do tomorrow is talk about what is called the domain perturbation method okay. So far what we have done is we have always talked about the small parameter occurring in the differential equation and then we said how do we go about doing this perturbation series. We need to now worry about the situation when maybe the small parameter is occurring in the boundary condition and the boundary is not going to be coinciding with one of your coordinates y equal to constant z equal to constant and the boundary itself is actually having a small parameter. How do you go about solving that we need this method to be able to actually solve the stability problems we will be worrying about later okay.