 So, welcome to the lecture 16 of multi phase flows. We were looking at this particular equation last class where this equation arises in the context of a pulsatile flow in a circular channel okay. And what we did was we concentrated on the limit of rw much lower than 1 was the limit when the frequency is very low okay. And so now what I want to talk about a little bit before I go on to the next problem is talk about the other limit of rw being much greater than 1 because you know you have 2 timescales and we looked at the one limit wherein we said when rw is very much lower than 1 then the inertial terms are not significant. What you have is a balance between the viscous forces and the pressure term and we found that things were in phase, the velocity was going to be in phase with the pressure. So, now the question is what happens when the frequency is very large, when rw is much greater than 1 the other extreme and is it possible for us to take a look at this system. So, clearly when rw is very large the one thing which we need to include we need to include the pressure term because the pressure is the one which is going to be driving the flow and remember the pressure term is coming from here okay. The other term this comes from the inertial term the derivative the time derivative term and this is your viscous term. So now in this limit we need to retain the pressure term the inertial term is also going to be important because the inertial term is negligible in the limit of rw being very much lower than 1. So, when rw is much greater than 1 this has to be important. So, essentially what we want is that we want to be able to simplify this equation so that these 2 terms become dominant and this becomes negligible okay. But the way this equation is written it looks like when rw is much greater than 1 you do not have a balance between the inertial forces and the pressure forces okay. So, how can you possibly get a balance between these 2 terms? One way in which you can get a balance is if we write h as let us say h bar divided by rw that is as rw increases h becomes smaller then h rw becomes h bar okay and this would have a magnitude of which is comparable with this whereas that is h as I increase rw h has to decrease and if it decreases inversely as I have written then this will have a magnitude which is going to be comparable with this term and that I get the balance between the inertial term and the pressure term and what is going to happen when I substitute this here I get h bar divided by rw okay. So, then I have a 1 by rw here which makes this term very negligible okay. So, when I substitute this here what I get is multiplied by 1 by rw I need to put this bar here minus ih bar equals minus 1. So, now I have a situation what I have done is I am basically hypothesizing that in the limit of rw being much greater than 1 h is going to be very small h per se and the way I am going to capture that is by seeking h as h bar by rw okay. Now whether it is inversely with rw or to the power 2 or something is going to be decided by the form of the equation. So, here this gives me this equation and what this means is in the limit of rw being much greater than 1 this guy is now 0 I can knock off this term the viscous term and now I have a balance between the inertial term and the pressure term okay. So, here in this limit the viscous term is negligible we have a balance between the inertial pressure forces okay. So, basically what I will do is I will knock off that and I have minus ih bar equals minus 1 or h bar equals minus i okay. So, that is my h bar and once I know h bar at least to some approximation of rw turning to infinity this is h bar I can go back and calculate h I can go back in substitute and get back my velocity by taking the imaginary part okay. The thing which I want to mention here is that we have neglected the viscous terms remember this is therefore going to be valid as long as you are away from the walls of the channel. Near the wall of the channel this business of neglecting the viscous term is not a good idea physically because you have the no slip boundary condition the liquid is going to be at rest at the wall and immediately next to it you have the liquid moving. So, you are going to have velocity gradients viscosity is going to be an effect. So, point I am trying to make here is that this is valid away from the walls. Near the walls the viscous effect is dominant and need a different approach. So, basically this particular problem of this pulsatile flow in a circular channel is discussed in Gary Lee's book on advanced transport phenomena. I pretty much try to stick to the notation which he has used okay. So, that when you guys go and read the book you will be easy for you to follow okay. So, I am suggesting that you go and look at the reference to this problem is Gary Lee advanced transport phenomena okay and I believe you need to look at chapters 3 towards the end and chapter 4 towards the beginning. So, chapter 3 basically talks about the complete solution which we spoke about and chapter 4 talks about the asymptotic solution okay which we have discussed. So, that is a place and here he will also talk about how to take into account this viscous effect and where this is. So, if time permits towards the end we may come and revisit this problem of how to do mass asymptotic expansions. Right now I just want to talk about this perturbation theory and then move on okay. So, this is one problem which we have solved. The next problem that I am going to talk about is also a solved problem is a workload problem in Gary Lee okay. So, again that is from chapter 4 and this problem has to do with the taking into account the effect of viscous heating in a shear flow which is let us say one dimensional. So, our objective is just to illustrate the idea. So, we are going to keep the math as simple as possible okay. So, we just illustrating our analysis focus to a one dimensional problem and the applications of this are in basically in viscometry okay. So, for example how do you measure viscosity of a fluid? You put it in a viscometer then you have a wall at the bottom which is stationary and usually you have a cylindrical surface of the top which is actually rotating and then you know the shear rate you measure the shear stress and you find the viscosity of the fluid right. So, there are many applications where you have one wall which is stationary another wall which is moving and there is a liquid below. So, here the motion is going to be induced by the motion of the one of the walls. So, one of the walls moves and that induces the motion in the liquid. So, rather than worry right now about the circular geometry what I am going to do is worry about the simplified things at least geometrically which is just talk about two flat parallel plates which are infinitely long okay. So, keep life simple which is take the x direction going to infinity keep the y direction going to infinity and have the z direction also going to infinity okay. So, y and z directions go to infinity oops y and z do not go to infinity x and z go to infinity and y is bounded between 0 and h and y is bounded between 0 and h. In applications like viscometry where we are focusing on determining viscosity of fluids okay one wall is moving relative to the other wall. So, the idea is the same. So, I am just using the same idea to try and explain a concept to you tomorrow if you really want to go and do a cylindrical geometry you should be able to go and do it okay. We move one wall relative to the other and find the viscosity that is the idea okay. The objective is to find viscosity and then we do this by doing the simple experiment of moving one wall relative to the other. Now, so what I am going to do is I am going to keep this lower wall stationary u is 0 here and I am moving it in the x direction and I am going to move this with the velocity capital U okay and just to make sure that there is a difference between the small u and the capital U I am going to put a subscript 0 there. Now, you know this particular classical problem what is this called? This is the Kuwait flow right and you know that if you have a liquid which is between 2 plates and if the upper plate is moving you have a flow profile velocity which is actually linear varying from 0 to u0 okay and what is this profile going to be given by? This profile is going to be given by u equals u0 multiplied by y divided by capital H okay and y equal to 0 it is 0 and y equals capital S it becomes u0 and it satisfies your momentum equation that is no externally imposed pressure gradient. What we want to do now is supposing your fluid happens to be very viscous you are trying to measure the viscosity of oil let us say and very viscous viscosity being something like friction is going to end up generating a lot of heat okay. So, when you are doing this movement there is going to be heat which is generated and this heat is going to be generated in the form of a viscous dissipation term. Normally what we have done in the past is we would end up neglecting this viscous dissipation term as a result of this viscous dissipation term heat is generated and although the wall of the lower the temperature of the lower wall and the temperature of the upper wall would be constant let us say at room temperature because of the heat generation there is going to be a temperature profile which is going to be induced inside okay. Now if your properties your viscosity and your thermal conductivity were actually dependent on temperature okay then what you need to do is you need to include this effect if you say I am not worried about the dependence of the viscosity on temperature because the actual temperature prevailing inside your viscometer is going to be different from what you think it is because and therefore what you would be measuring is possibly an inaccurate value of viscosity okay. So, what is the point I am trying to make here is there can be liquids when you can have a significant amount of heat generation if there is a significant amount of heat generation temperature is going to rise if the temperature rises you would actually be measuring some kind of apparent viscosity or an average viscosity over the temperature range okay because viscosity is a function of temperature thermal conductivity is a function of temperature okay. So, what we want to do is see how we can get an estimate of the velocity field and the temperature field by in this problem where viscous dissipation is the cause of the heat generation okay. So, that is the idea so and clearly you can already see what I am going to do I am going to in the limit of the viscous dissipation not existing I know the solution I know this linear profile temperature is going to be uniform and I know the velocity is linear. So, what I am going to do is include viscous dissipation and treat the dimensionless parameter which is going to be associated with it as my small parameter and do a perturbation series to find how the actual velocity profile and temperature profile is. So, that is the strategy that is the plan okay. So, now let us go back and write some equations and proceed. So, the momentum equation look I am going to keep things one dimensional remember the actual momentum equation will have d by dy of mu du by dy equal to 0 okay. I am keeping the viscosity inside here because normally what you are used to is saying viscosity is a constant and you take it outside the derivative but in general remember what I am interested in is viscosity can be a function of y implicitly because viscosity is a function of temperature and temperature can be a function of y and therefore viscosity comes inside. So, for my energy balance my differential equation is going to be d by dy of k dt by dy okay plus mu times du by dy the whole square equals 0. So, this is my double conductivity and this is my viscous dissipation term. Remember the viscous dissipation term is a source of heat therefore the square there is a positive it is always positive always going to be generating heat okay. Whenever you are going to have flow there is going to be only thing is depending upon the value of the viscosity this can be small or high. Now, I need to have some boundary conditions for both velocity and this and you already know the boundary conditions at y equal to 0 u is 0 and temperature equals t0 and at y equals capital H u equals u0 and t equals t0. So, since we want to talk in terms of doing a perturbation analysis what I want to do is rather in talking terms of the actual magnitudes of k and mu we need to make things dimensionless and then talk in terms of the relative role of time scales etc. So, when you make things dimensionless you get dimensionless numbers which tells you the magnitude of inertial over viscous or this time scale over the other time scale. So, it is a relative magnitude is what we are interested in. So, next thing to do is to make this dimensionless right. So, I am going to make this dimensionless by choosing my new characteristic as u0 because that is going to decide my velocity y characteristic is clearly H okay and rather than talk in terms of t0 I am going to define temperature scale will be t0 but I want to do it in terms of getting my boundary conditions homogeneous. So, what I am going to do is I am going to define theta as t-t0 divided by t0 clearly t0 is my reference temperature whatever is the temperature of the walls okay but rather than talk about the absolute temperature I want to talk about the thing in a relative way. So, I am just going to define this as t-t0 by t0 this will basically give me an idea of how much is the temperature rise inside my viscometer this indicates the temperature rise inside okay. So, let us make this equation dimensionless. What do I get? Let tilde denote dimensionless variables and clearly you would get okay then I am also going to define mu tilde or mu average as being mu divided by mu0 and k as being equal to divided by k0. What is mu0? Mu0 is nothing but the viscosity evaluated at t0 k0 is the thermal conductivity evaluated at t0 okay. So, mu0 is mu of t0 k0 is k of t0 okay. Now, I am going to substitute all this here and I am going to get d by dy tilde mu is written as mu bar times du tilde by dy tilde equal to 0 what I am going to get out here is mu0 by h squared times mu0 right mu is mu0 times mu bar this is mu0 times that h squared. So, this basically this is a constant and this cannot be 0 therefore this has to be 0 okay. This implies d by dy tilde of mu bar du tilde by dy equals 0. We need to make the energy balance dimensionless okay and let us do that I am going to get k0 by h squared times t I am going to replace in terms of theta I am going to get t0 here d by dy tilde of k bar. So, that is basically what I have and this clearly must equal 0. I am going to make the coefficient of this one move these terms to the next to the second term and basically what I will get is a dimensionless number. This dimensionless number is called the Brinkman number okay. So, I will just write that final equation maybe here I am going to write that equation here and I am going to drop the tilde as but for convenience okay. But remember my equations are dimensionless just to keep life simple. So, what do I do I am going to write this as d by dy of k d theta by dy plus mu du by dy the whole square this is going to be multiplied by this Brinkman number equals 0 and let me define what this Brinkman number is it is going to be mu 0 times u0 squared the h squared does not show up and I have a k0 and I have a t0 okay. So, all I have done is move though that coefficient that to the second term and this is my dimensionless group okay. So, I have a one dimensional problem because I just want to keep my life simple and I have this equation which relates the temperature to the viscous dissipation of the velocity gradient. I also have the momentum equation which is this one which is d by dy of mu du by dy equal to 0. I want to tell you a couple of things here that these 2 equations are actually coupled to each other okay. There is how is the coupling? The coupling is occurring because clearly the theta depends upon u and this equation is non-linear mu depends upon theta the temperature. So, you need to actually solve both these equations simultaneously. It is not that I can solve this equation for velocity and I can substitute it here and solve for temperature. The 2 equations have to be solved together if you actually. So, one way to do this is to do a numerical solution do a finite different scheme and you try to get a solution. But the other thing is what we are going to do which is exploit the fact that for small values of Brinkman number we can possibly get an analytical solution. What motivates me to do that is when Brinkman number is 0 when there is no viscous dissipation I know the solution. I know that the temperature is uniform. There is no temperature gradient. The velocity is linear. So, when the Brinkman number is going to be small I expect only a small deviation from these 2 solutions some small correction and what is that correction that is what perturbation theory helps me get to okay. So, that is the idea. Remember these 2 equations are coupled and non-linear. We can do a numerical solution or we can do an approximate analytical solution okay. So, the solution is obtained numerically or approximately by using some perturbation method or we can get a perturbation series solution okay. Why? Because for Brinkman number equal to 0 we know u will be equal to y. When I have scaled everything u will be y is going to be going from 0 to 1 okay and theta will be 0 because temperature will be equal to T0 everywhere. In terms of the dimensionless numbers theta will be 0 because t is equal to T0. So, for Brinkman number equal to 0 this is the solution. So, for small amounts of Brinkman number it is going to be different from this and so we need to find out what is going to happen right okay. A couple of things but before I proceed I need to make an assumption of how does the thermal conductivity and how does the viscosity actually change with temperature. So, clearly the easiest thing to do is to assume a linear variation depending upon how crazy you are you want to make a higher order approximation maybe for second order, third order but we will keep things simple not be too crazy today. Then we will just keep k as k0 times 1 plus just trying to think beta theta. So, basically the idea is or maybe I do not want k0 here because it is already normalized right. So, it is just 1 plus because this is remember k by k0 which is my k tilde. So, this is my k tilde although not killed in the dimensionless equation now but this is the linear form okay. When theta is 0 when T is equal to T0 k tilde is 1 but k tilde is k divided by k0 remember that. Similarly, I am going to do something here which is slightly different rather than say the viscosity is a linear function I am going to say the reciprocal of the viscosity is a linear function because it helps me with the algebra okay. So, I am just going to assume that the reciprocal of the viscosity is 1 plus alpha times theta. No, I mean you can always say mu tilde as 1 by 1 plus alpha theta which is 1 minus alpha theta in a binomial expansion. So, I mean I think it does not really matter because alpha the only thing which is going to decide is whether it is increasing or decreasing function that is going to be decided by the sine of alpha okay. So, the sine of alpha will tell you if it is increasing or decreasing you can choose mu tilde as 1 plus alpha theta as well but my point is the alpha star let us say that alpha star will be negative of this alpha. So, that is the only thing. So, this is only to make my life simple when I substituting this back inside the differential equation okay. I mean you could have chosen mu tilde as 1 plus alpha theta only thing is you have to do some more differentiation and some more algebra okay and possibly a bigger blackboard. So, so we are all set. Now, clearly u is a function of y and Brinkman number okay and what I am going to do is I am going to seek this as u0 plus Brinkman number times u1 that is my correction. So, I mean you can go further as a power series plus higher order terms but we are going to stop only at Brinkman number to the power 1 today okay and as far as theta is concerned that is also a function of y and Brinkman number and that is going to be equal to theta 0 plus Brinkman number theta 1 plus Brinkman number squared theta 2 okay. Restrict v focus up to order of Brinkman number to the power 1 that is I am interested in getting only the corrections u1 and theta 1. You already know what is u0 and theta 0 because u0 and theta 0 correspond to Brinkman number equal to 0 okay. So, I mean and that should come out when you substitute this inside your differential equation okay. So, u0 is clearly equal to y and theta 0 is clearly equal to 0 okay. I am going to since this is the solution for Brinkman number equal to 0 okay. So, now we are all set we will look at the momentum equation. The momentum equation tells me that d by dy of mu du by dy equals 0 and I am going to say this is therefore d mu by dy which is nothing but d mu by d theta sorry. Let me write it this way let me do this step one step at a time d mu by dy times du by dy plus mu times d square u by dy square equals 0 that is just expanding this out okay. All I have done is just expanded this out and now I am going to write substitute for u in terms of u0 and Brinkman number times u1 and I am going to write d mu by dy as d mu by d theta times d theta by dy because mu is a function of theta okay and this I am going to leave it as it is and I am going to use the fact that there is a linear dependency of mu and things like that. So, let us do that. So, what is d mu by dy and it is d mu by d theta times d theta by dy times du by dy plus mu times d square u by dy square equals 0. What is d mu by d theta? Now, we of course remember this is all in dimensional terms. So, somebody has to tell me what is d mu by d theta. So, this is the only one place where I have to do this is 1 plus alpha theta. So, d mu by d theta turns out to be minus 1 by 1 plus alpha theta squared minus alpha okay. Is this right? So, d mu by d theta is that theta squared. Maybe I will do something beyond this. Since I am only interested in powers of Brinkman number to the power 1, what I want to do is I am going to write this using a binomial form minus alpha times 1 plus alpha theta to the power minus 2 okay and I am going to substitute this as minus alpha times 1 plus alpha times theta remember is theta 0 plus Brinkman number times theta 1. Theta 0 is 0. So, I am going to write this as 1 plus alpha times Brinkman number to the power theta 1 to the power minus 2 okay. So, what I am going to do is I am going to, if I did a binomial series expansion I am going to, this is going to be minus alpha times 1 minus 2 alpha Brinkman number theta 1 plus higher order terms which I am not going to worry about because that is going to involve Brinkman number squared and I am not interested in Brinkman number squared. So, what I am going to do is I am going to write d mu by d theta as minus alpha times 1 minus 2 alpha times Brinkman number multiplied by theta 1 okay. What is d theta by dy? d theta by dy is, when I substitute in terms of theta 0, it is going to be Brinkman number times d theta 1 by dy okay. That is Brinkman number times d theta 1 by dy and what is d u by dy? d u by dy will have 2 terms which is the u0 term and the Brinkman number term. d u0 by dy is 1, we know that and plus Brinkman number times du1 by dy. That is, this is this term here plus the viscosity term which is now 1 by 1 plus alpha theta times d square u0 by dy square is 0 because we already know u0 is y and what I am left with is Brinkman number times d square u1 by dy square equals 0 okay. What I am interested in is taking the terms which have Brinkman number to the power 1 right, that is the thing I am interested in. So, what are the terms which are Brinkman number to the power 1? I have this term here and here this, I have Brinkman number here. So, when I multiply this with this and this, I get Brinkman number squared. This with this is going to give me Brinkman number squared. So, the only thing I need to worry about is the contribution of this term and this term because there is already a Brinkman number outside okay. So, this is at order Brinkman number to the power 1. What do I have? This gives me minus alpha d theta 1 by dy plus 1 plus 1 plus alpha theta and I should actually substitute this here again in terms of which is important. I should do okay. Give me a minute. Let me write this later. I am going to write 1 plus 1 by 1 plus alpha theta as what? 1 minus alpha theta plus alpha squared theta squared. By now I am in a series. Remember theta is theta 0 plus Brinkman number theta 1. So, this becomes 1 minus alpha times theta 0 is 0. So, I am just going to directly write Brinkman number theta 1 okay. So, theta I am just going to substitute in terms of this. Clearly, the only term which is going to contribute now when I substitute this here, the only term which is going to contribute is going to be the 1 multiplied by this term because this already has a Brinkman number. So, this multiplied by this is Brinkman number squared. So, I am not interested in that okay. So, what we do is get this equation minus alpha times d theta 1 by dy times plus again there is a minus alpha d squared u 1 by dy squared equal to 0. Do you guys think you agree with me on this? Yeah, the alpha is going to cancel off in both the terms. Yeah, it is 1 multiplied by I think you are right. You are right. There is no alpha here. Yeah okay. So, I believe this is what I am getting d y square or plus this equals 0. I need to go back to my energy equation. Yeah, both are plus right. Yeah, that should be plus. No, no, this guy is minus minus and the second time is plus you are right. So, now this is minus okay. So, that is what we get okay. So, now I need to go back to my energy equation and work on that. Yeah. In both the terms, yeah, you could have possibly done it. After taking the Br out. I should do, I mean if we just remove the Br, then we are left with order 0 and order 1 also. Okay, the reason the order, see the order 0 has disappeared because I am directly substituting the solution which I have. See, I am not trying to find out u 0 and theta 0. If I had left the u 0 and theta 0 as it is, then the Brinkman number, to the power 0 would have contributed, I would have gotten u 0 and theta 0 okay. That would have corresponded to Brinkman number to the power 0, that is order 1. Order Brinkman number will be the first order term which is what I have got. So, what I have done is basically use the find that I already know the solution for the isothermal system okay. So, what I am getting is the first order. This is not the 0th order. You are suspecting this is 0th order. Yeah, if Br is taken up. No, this is not 0th order. They have already substituted the solution okay. Yeah, because it is just going to minimize my algebra. So, that is the reason I wanted to get rid of the theta 0. Otherwise, I keep it and then at the end I have to throw it out. So, since we already knew the solution, I just wanted to use that directly. Let us look at the, you know, I thought we would be done today but I guess we would not be done today. So, let us look at the energy balance and this is possibly a good time to stop because it gives me an opportunity to ask you people to do the energy balance and I will do it in the class next time and then we can verify if indeed it is right okay. So, the energy balance is d by dy of k d theta by dy plus mu du by dy the whole square equals 0 okay. And remember this is actually mu bar mu by mu 0 and this is k by k 0 and what I want to do now is substitute this as I had alpha there. So, 1 plus beta theta multiplied by d theta by dy plus 1 plus alpha theta multiplied by du by dy whole square equals 0 okay. So, all I am doing is substituting the assumed dependency on temperature and now I am going to substitute things in terms of my Brinkman number the expansion that I had and you should be able to equate terms of order Brinkman number to the power 1 that gives you your first order term okay. Brinkman number to the power 0 is implicitly been already built in because I assumed u0 is y and theta 0 is 0. So, when you do that you should be able to so find terms of order Brinkman number to the power 1. I think there is a small problem here what is the problem mu is not 1 plus alpha theta it is 1 plus alpha theta to the power minus 1 in the way I assumed it and when I bring the mu down here I will get 1 plus alpha theta and that simplifies my calculation that is the advantage of choosing the reciprocal to be 1 plus alpha theta. So, you guys were asking me you know saying why is this guy choosing the reciprocal I am choosing the reciprocal because now mu is the reciprocal of that when I bring it to this side it becomes the numerator it simplifies my algebra a little bit okay of course I would differentiate over there a little bit. So, anyway you cannot have everything your way. So, anyway I want you to find the solution to this equation Brinkman number to the power 1 in fact you would find that you will get something for theta 1 you solve for that and then you can substitute this and then get the velocity. So, you should be able to actually decouple. So, what you will find out is that the 2 equations are getting decoupled okay and then you will be able to find the solution.