Along with Professor Puranik, welcome all of you to the question and answer session. So, we will start the question and answer session with Mukhopkam Jha, Hyderabad. Sir, regarding fluid mechanics and heat transfer, what our equations are derived as they are all available in minute textbooks, but when it comes to the CFD part, there is a grid generation and all those things which have been covering for the part two days, could you kindly name a couple of books which could be referred to, because as I have told during the period that you are teaching, of course that absorption rate could be low or even if it is high, there is a likelihood that we may not retain for longer time. So, for subsequent reference, could you kindly name a couple of references so that we can go through and subsequent we can interact with you. Thank you for your question. I am happy to see your interest in taking whatever we are teaching into the practical application that you are ready to teach this course in future. So, thank you for your interest. So, the question is that the material which is being covered, especially computational fluid dynamics part, the reference book, I can give you a good reference book like computational fluid flow and heat transfer by Patankar and computational fluid dynamics by Parik and Fersigar. However, I would like to point out that the details which I am giving in this course or the finite volume procedure which I had shown is slightly different from what is given in those books. So, what I would suggest is that after you have gone through this course, you can refer this lecture notes, you can refer those books and moreover, we have given a detailed lecture slide and even all this lecture, right now it is a video recording is being done and it will be put on the web so and we will be available for modal interaction at least for the next few months. So, I hope with all this you should be in a better position to understand. I would like to point out that right now you, it is situation is such that we are just delivering the materials and we are not giving you time to assimilate and CFD is such that you need to, it is not only that you can listen and understand everything, you need to work it out slide by slide each and everything, you will have definitely have questions, you can ask us it was and we will be more than happy to help you out. So, it will take time to assimilate the material, but I think we are giving you sufficient material in slides format, code format, lab problems and this video recording of this lecture will be available to you. You can go through the reference material, we will give a detailed list of the reference book after the end of this course, both for the fluid mechanics parts as well as CFD part. So, thank you. Thank you. Sir, you have shown in a finite different solution slide number 19, convergence of iteration, it depends upon one parameter relaxation parameter. If the value of relaxation parameter is less than one, then it makes the process lower and if it is greater than one, then it makes the process faster. However, the relaxation parameter is in between 1 and 2. So, what are the parameters which affects the value of relaxation parameter? So, the question is on the use of this relaxation parameter that we had implemented in the iterative solution process for the del squared t equal to 0, the two dimensional steady state conduction that we saw. And the question is on what parameters this relaxation parameter depends. Actually, there is no hard and fast answer for that. This relaxation parameter is a quantity which is absolutely problem dependent and from problem to problem, the optimum value so to say of the relaxation parameter will change. The specific problem that we have looked at, the two dimensional steady state conduction problem, I think if you want to achieve the solution in the least number of iterations, there will be one value of omega, somewhere around 1.7 I think and that can be obtained through a numerical analysis for this simple situation. But if you want to look at more complicated situations like the fluid flow problems which Professor Sharma will discuss in the next couple of days, when you go to solve Navier-Stokes equations, if an iterative solution of the Navier-Stokes equation is involved, you will realize that it is really a situation where there is no hard and fast answer. In fact, most of the times non-linear equations like the Navier-Stokes equations require a under relaxation of the solution process and usually you will be required to implement a relaxation parameter which will be much less than 1 in order for those non-linear problems to be solved. So the answer is that there is no unique way to find out what is the correct or the good value of relaxation parameter. It is totally problem dependent and most of the times as the literature says, you have to actually find out the optimum value through what we call numerical experiments which simply mean that you have to try out a few values and find out how it is going. And rule of thumb for sure is that non-linear problems like the fluid flow situations will require necessarily an under relaxation. It will never work with an over relaxation. Simple problem like what we have done here can work with over relaxation also. Thank you. Yeah, so the question is on the convergence criterion for those iterations or iterative solution that we talked about in finite difference. So what is this epsilon? So epsilon is actually a user input. Epsilon is something typically that we are using as a convergence criterion number. Usually we are using this as 10 to the power minus 4 or 10 to the power minus 5, 10 to the power minus 6, etc. It depends on how accurate the solution requirement is and the way it is implemented at least one way it is shown here is that what is done is you have the values for temperature available at the end of one iteration. What you do is you carry out the next iteration and during this present iteration, let us say k to k plus 1, you find out at each node point what is the absolute difference between the value of temperature that was available at the end of previous iteration and what is the value of temperature that you are obtaining after the present iteration is over. To find this difference in the absolute sense at each grid point, there will be some grid point where this temperature difference will be maximum and if you say that the maximum temperature difference at any grid point is less than say 10 to the power minus 5, we will claim that our solution is converged. It simply means that the temperature values are not really changing at all from one iteration to the next iteration which essentially implies from a numerical point of view that you have achieved a converged solution. So that epsilon is a user input and depending on what accuracy you want, you utilize that as 10 to the power minus 4 or minus 5. Thank you. Last question from differential solution, have you tried sir principle of resolution for inclined stress, shear stress? Yes, yes, of course I have, so the question is on the positive shear stress on that inclined surface which we talked about when we described the sign convention. Yes, actually I have discussed that with a few solid mechanics people in my department as well and unfortunately or fortunately whichever way you want to look at it, they are not able to solve it for me. So the answer from one of the solid mechanics people is that it is really immaterial which way you call that as a positive direction. So it can be the way I have shown or it can be in the opposite direction as I have shown. Finally what happens is you work out the algebra and you take the limit as the element size tends to zero. The final conclusion which is that the state of stress is represented uniquely in terms of the four stress components acting on two mutually perpendicular planes passing through the same point remains the same whether you are actually looking at the shear stress in one direction or the other. So that is the answer that I got from the solid mechanics people and that is what I am passing on to you right now but we will still keep on looking at it from a more convincing point of view. Sir, in this case if we resolve shear stress into directions x and y then we have to have the resolutions of normal vector in y and x and y because the direction of or sign convention of shear stress it depends upon normal vector also. That is correct you need to resolve the normal vector for the surface also along with the shear stress but if you look at the resolution carefully it brings about some sort of an inconsistency in the way the positive shear stress is defined. So let me not take too much time on this. At the moment what you can note is that it appears that it is immaterial whichever way you define the positive direction for the shear stress on that inclined plane because the conclusion remains the same as far as the state of stress is concerned. Technically there was one professor who had pointed out in the discussion last time from VIT Vellore if you actually work out the entire algebra using the tensor notations which are slightly higher level to follow it automatically fits into the picture. What I was trying to do was that I was trying to work this out in a Cartesian type situation in the two dimensional x y plane and that's where the problems are coming. Nonetheless we will keep discussing this on Moodle I suppose because that's the only option that we have right now. Thank you. My question is that when we calculate the diffusive flux for the boundary node value then in that book the author has taken the value as k by k into 2 by 3 upon delta x into 9 times pi of a minus 8 times pi of e minus 8 times pi of p minus pi of e. So the author has said that he derived this from the expression used for the same expression used for deriving the quick scheme. So I tried to derive that equation but I wasn't able to derive the expression. So I was confused whether for quick scheme we always use that expression or it is okay to use the normal difference between the two values for calculating the boundary node value at the boundary node diffuser flux at the boundary node. First of all you are saying it is a diffusion flux but as you have put this let me first say that what the question is asking there is a book on computational fluid dynamic by Wursteck and Malisekara and in that book on chapter number 5 in problem number 5.4 he has shown what happens when you use this advection scheme near to the boundary is some special treatment is needed. So what he has done he has created a mirror grid points near to the boundary and he said that this was first done by Leonhard he has given a reference also and for that mirror grid point he is doing a linear extrapolation and from that linear extrapolation he has obtained an expression I will open a slide and then discuss this. What happens is that if you use this advection scheme if you are using inside for the grid point which are inside let us say if you are using for let us say if you want to calculate use advection scheme and this phase you get one downstream neighbor to upstream neighbor but near to the boundary I am showing you the enlarge view of what happens near to the boundary here near to the boundary what happens is that if the mass flux is invert from the boundary then this is your downstream neighbor this is your upstream neighbor and let us suppose this is your boundary grid point. So the distance between the upstream and downstream neighbor is let us say delta x now the distance between upstream and boundary grid point is delta x by 2 but the advections but the quick scheme which we have derived where we have 37.5 percent weightage to this downstream node 75 percent weightage to this and minus 12.5 percent weightage to this that we have done when the distance between this two point is delta x then this two point is also delta x. So near to the boundary some special treatment is required so for that in problem number 5.4 what is given in the book of Malasekara, Wastik and Malasekara is that using these two points they have done what they are calling as a linear extrapolation the distance between this two point is delta x by 2. So they take a mirror point which is away on this side at a distance delta x by 2. So the idea is you use phi is equals to A x plus B the way I had shown you you put phi is equals to 0 sorry x equals to 0 phi is equals to phi B at x equals to delta x by 2 phi is equals to phi U then you get an expression for phi here. Now using the phi which you get from at this point this point and this point you use the advection scheme which has been given. He has given this procedure however tomorrow I will discuss this separately where the procedure I will propose is slightly different from what they have proposed but I think you are struggling with the derivation which you have done I would like to point out just use an expression phi is equals to A x plus B and substitute x equals to 0 phi is equals to phi B at x equals to delta x by 2 phi is equals to phi U here and then you will get the mirror node phi value which is substituting in the usual expression for quick scheme. I hope I have answered your question. This question is about partial differentiation. Suppose phi that is stream function is a function of x, y and t it is quite possible that y is not equally dependent on x, y and t. The strength wise impact of x, y and t may be different on I mean phi stream function. My question is that how to how these weights of independent variables are taken care of while performing the partial differences. So, the question is on the stream function where in general the stream function for a two dimensional situation can be a function of x, y and time and the specific question is when we perform the partial differentiation how do we take into account which variable amongst x, y and t affects the velocities more. I think that is the way at least I understood it. In fact once you have a closed form relation let us say for psi as a function of x, y and t you do not have to bother about anything. Actually all you need to do is perform the partial integration partial differentiation I am sorry with respect to y to find the x velocity component and the partial differentiation with respect to x with a minus sign to obtain the y velocity component. The function psi which has all the dependence on x, y and time built into it will automatically take care of which of the independent variables decides or determines or influences the velocities more. So, this is from a purely mathematical point of view. In a real analysis of a real life system complex fluid mechanics type problem we do not ever have really a closed form solution available for a stream function. What we do is actually we solve the problem in the so called primitive variables formulation meaning that we do not really talk about the stream function at all when we solve the problem. We solve it using our Navier-Stokes equations which are written in terms of the pressure and the velocity components. Having solved the equations then you can actually find out numerically the stream function if you want at all and that is the way to look at it. If some sort of a artificial problem like what we have been discussing in the fluid mechanics part is given where there is a function specified for psi then all the dependence of x, y, etcetera is built into that function. So, it is already there and you do not have to worry about it at all. All you need to do is differentiate it partially with y and x to find the velocity components. Thank you. Sir, my question is regarding this flow along stream line. Suppose that is curve surface is there then energy equation for this one this case what will be that means what changes will be there. Energy equation for flow along stream line and suppose that surface is curve surface and for that centripetal acceleration term may be there in energy equation. So, the question is if we are talking about a fluid particle travelling along a curved surface if we have to modify the energy equation and actually the modification if at all will come in probably the momentum equation if there are any additional centrifugal type forces. But as far as the energy equation is concerned I do not think there will be any change. You just have to implement it along the stream line which is your curved surface. But perhaps if the centrifugal forces are large enough then you may have to include that in the momentum equation. Regarding that heat transfer taking place generally what we are considering for heat conduction and heat advection in that basically we are considering only flow means that we are not considering any relevant number or neselt number or Weber number. That means will that laminar turbulent flow consideration will be there. The question is we are talked about right now pure conduction and pure advection. So the question is we have not considered the governing parameter like Reynolds number and Weber number. The answer to this question is right now this parameter does not come into the equations or the physics which we have tried to formulate. In conduction you cannot expect Reynolds number. In pure advection also you do not have any diffusion term. So you cannot have Reynolds number. Reynolds number is the ratio of the inertia and the viscous forces. This terms will definitely come when we talk about the full Niemann stroke equation. Just hold on for a couple of days. We will have this terms in the when we talk about the Niemann stroke equation. However I would like to add that right now I am doing everything in a dimensional form so that you get feel of the terms because many times if you use a non-dimensional form your diffusion coefficient becomes 1 by Re. Then you do not get feel of here it is a thermal diffusivity. So right now everything I am doing it in a dimensional form but the answer to your question is the Reynolds number terms are not relevant to the problems which we have discussed till now. Thank you. One more question sir to Puranik sir. Sir my question is something mathematical more of mathematical. The exact solution 27 we have seen Eigen value problem and in finite different solutions we have seen boundary value problem. So I want to know how Eigen value problems are different from boundary value problems application wise maybe. So the question is about once we saw in the exact solution an Eigen value problem and in the finite difference solution we saw what is a boundary value problem. In fact let's just take that example the del square t equal to 0 the 2 d steady state conduction equation that we solved. So physically speaking when you want to classify that as a problem it is actually a boundary value problem. It is the absolute classification for that two dimensional steady state conduction problem is that it is a boundary value problem. What that means is that it is related to a steady state process in certain domain we are talking about a steady state conduction in a two dimensional domain and because it is a steady state problem the temperature distribution within that domain is governed only by the boundary conditions and that is what we mean by the boundary value problem. So boundary value problem is a physical classification if you want for a given problem. On the other hand the first problem that we saw which was a diffusion equation it is actually what is called as an initial and boundary value problem. So mathematically people will call it in that fashion. Physically speaking you can interpret that as a problem where the variable value will be influenced not only by the boundary conditions but also by the initial value of that variable at time equal to 0. So it turns out that when you want to solve these either the boundary value problems or the initial and boundary value problems which I will still call as physical classifications. The mathematical techniques that we end up employing will generate these so called Eigen values of the problem which essentially means that only at certain values of those what we call Eigen values the solution will be non-zero and the general solution if you recall from your differential equations the general solution is essentially a summation of all possible solutions. So in case of the solution technique mathematically speaking each Eigen value provides one solution to the problem and since there are infinitely many Eigen values in the situation we actually end up adding all these individual solutions to form the most general possible solution for our problem. So I hope the distinction between an Eigen value problem and the boundary value problem is clear. Eigen value problem will come in automatically if you want to solve either the boundary value problem or the initial and boundary value problem using any standard mathematical technique. So Eigen value problem simply shows up as part of the solution procedure if you want to employ a mathematical technique. Hopefully I have answered to some extent at least. Yeah, come in to college Pune please go ahead if you have a question. I have a question to professor Puranee on exact solutions 25. Yeah as you said this is initial and boundary value problem. The boundary conditions are u at 0, t equal to 0 and u h comma t equal to capital U and the initial condition is u at y comma 0 is 0. So I have question like at t equal to 0 I saw in all the problems in the lab that boundary condition is always winning over initial condition at t equal to 0. So is that so for every problem for exterior nodes? So the question relates to the way we actually interpret initial conditions and boundary conditions. I think the way at least I have been interpreting this as at time equal to 0 when we set the initial condition we look at all possible nodes or all possible locations. In this case it is a one-dimensional domain. So all possible y locations including even the boundary locations can be considered to be part of the solution as far as time equal to 0 is concerned. And at time equal to 0 you set the initial condition which is in this case equal to 0 to all y locations including the boundary locations also. Whereas the boundary conditions are supposed to come into play and picture only when you initiate the problem. So in some sense you can imagine at time equal to 0 you can think about time equal to 0 minus which is one side of the 0 and time equal to 0 plus which is the other side of 0 let's say. So that 0 plus is when the boundary conditions come into picture and then they start employing whereas up to time equal to 0 minus the initial conditions are prevailing. That's the way to interpret it at least from my point of view. Hopefully that is giving some insight into it. Yes sir but like the other question similar question is to Professor Shama where we solved this problem similar problem with FVM finite volume method. It's topic number two slide number 51. He has shown at T equal to 0 the graph where the boundary conditions are applied like at x equal to 0 the temperature is 0 and x equal to 1 temperature is 100. So the question is on the example problem which had taken in the computational heat conduction lecture number two slide number 51. So in that had taken lot of problems where I had given the boundary condition. So the way I take it is that you had shown you if you remember let me go to the slide. Here what I the way I had discussed this topic is that the boundary condition is applied at all times for this blue circles which are the grid points at the boundary and the initial condition is applied at all the points which are inside this domain. Remember that I had shown you an animation where if you see the animation at time T equals to 0 there is a step change from 0 to 30 degree centigrade. Then for what is all these nodes it is constant at 30 degree centigrade. So if you remember the first picture was something like this so on the first node it is 0 and the second node it goes to 30. Let me use the pointer on the first node it is 0 on the second node is 30 same 30 values continues for all the interior nodes which are yellow circles. Then there is a sharp there is a change because there is a discrete representation from 30 to 100. So this is the way we apply the initial condition and boundary condition. Yeah so this is yeah this is Balchandar Puranic actually let me add to that the it is again consistent with what I was what I was talking about. So you can imagine that the CFD simulation initiates itself at the time 0 plus so to say when the boundary conditions are coming into picture and they get employed. So we will always talk about the time interval for the CFD simulation to be going from T equal to 0 plus to whatever maximum time that we are setting so that when we start at 0 plus we are essentially instantaneously employing the boundary conditions and they will remain there for the remainder of the time. That is the way that is the way to interpret this I would think. Sir I have another question about that the same slide that at T equal to 0 as you said 0 plus what type of distribution is going from x equal to 0 to 0.1 like what type of curve is that on topic number 2 slide number 51. So the question is that when we start the competition how is the variation of temperature between x equal to 0 and x equal to 0.1 just to tell you if you take the length as 1 this is 0 this is 0.1 this is 0.3 0.5 0.7 0.9 and 1. So here we are using a discrete solution so there is there we do not have any information that what is the temperature distribution between 2 nodes to be honest. So we have a just discrete representation however when we plot the plotting software the way it works it just when we give 2 values let us suppose if I give 0 here and if I give 30 here when a plotting you take it to a plotting software it just joins by a line. So physically if you want to ask that what is the temperature distribution between these 2 point I would say that we do not have any idea we are not calculating and our solutions are discrete so we cannot say what is the temperature between these 2. Let me add one small point to that in general on a purely physical basis you can expect some sort of a non-linear distribution from the boundary point to the first point but as has already been pointed out we are looking at only discrete nodes as far as the solution is concerned. So many times some CFD analysts prefer plotting solutions only as discrete points. So they do not use any lines joining the discrete points and this is to emphasize the fact that we are actually finding solution only at discrete points. So some people actually like to follow that procedure as well. So this is just for information thank you. I have just last question on topic number 2 and slide number 4 in grid generation to Professor Sharma. So what happens to the corner nodes? It is not shown in this figure but do we take average of the neighboring nodes or there should be some corner nodes to maintain it 7 by 7? It is one of the most question asked in this slide that what happens to the corner. Now in the way we have discretized the corner even the corner node does not come into our expressions. If you remember our computational stencil is such that value at a particular point is a function of its north-south east-west neighbor. It is not a function of its corner neighbor. Like the value at this node is not a function of this corner values. So when you go to the corner control volumes it wants value from this north neighbor, this east neighbor, this south and this west. So the corner point does not come into the expression. I would like to add that I use the boundary condition like 100 on the left wall let us say 400 on the top wall. So you may again ask a question what about corner? Corner is common to this too whether it is 100 or 400 but I would say that it does not come into the calculation. So let us not bother about it. Thank you. Yes, Panvel please ask your question. I have a question for Professor Puranek. This is a general question not related to a specific topic. Can you elaborate the difference between material rate of change and convective rate of change? And whether material rate of change is the same as the temporal rate of change? Yes, so the question is on our kinematics discussion where we had defined the material rate of change, convective rate of change and the temporal rate of change. Specifically if material rate of change is equal to convective rate of change or if it is equal to temporal rate of change. Actually the most general answer is that the material rate of change is equal to the temporal rate of change plus the convective rate of change that is the relation. The way to interpret material rate of change is if you are following a given fluid particle along with it. So imagine that you are sitting on a fluid particle and moving along with the fluid particle with a sensor in your hand. The sensor will keep on noting the values of whichever quantity that you are trying to monitor. And as you look at the sensor the change in the quantity shown by the sensor will give rise to what is called as a material rate of change. So by definition it is the rate of change as observed by an observer who is moving along with the fluid. Now when it comes to convective rate of change it relates to the difference in the quantity let us say temperature or density or whatever between two points which are separated by certain distance in a space at a given time. So in general at a given time instant if you look at two different locations in the fluid domain they will have in general different values for any quantity say temperature or pressure. So this difference which is with respect to the space which is called the spatial rate of change let us say gives rise to this convective rate of change. And the third thing is the local rate of change which is if you focus only on one given location in the space and simply monitor what is happening at that particular location for any quantity such as temperature or pressure or whatever. So the rate of change as observed by a sensor mounted at a given location fixed in space is what is called as the temporal rate of change or the local rate of change. So in general the substantial rate or the material rate is equal to the temporal or the local rate plus the convective rate of change. If you actually go back to the relevant slides on kinematics in the beginning itself this was explained reasonably and I think hopefully now this discussion can help you a little bit more. Nirma University if you have a question please go ahead. Sir my question is related to the differential analysis 25. Sir in this slide we have so we have del by del t of rho u plus del by del x of rho u u so that term and in generalizing on this form we have written in the box that del by del t of rho v plus del dot rho v v which is a tensor form. So would you please re-explain the physical how to physically interpret this terms. So the question is on differential analysis 25 especially with respect to the conservative form of the expression on the left hand side. So we have written the Navier-Stokes equations here now. The left most part talks about the non-conservative approach and the middle part which we can replace the left hand side with is the conservative approach. So if you look at any particular term here this is actually written for the x direction the equation that my highlighter is standing on. So each of these terms really added together so partial derivative with respect to x of rho u u partial derivative of rho u v with respect to y and that of rho u w with respect to z added together is to be expressed or interpreted as the net rate of x momentum leaving the elemental control volume on a per unit volume basis. So that is how we had brought out these expressions. So it's nothing different than let me just go back yeah right here. So if you just look at the slides number 8 and 9 what we have done is we have we have found out our balances momentum balances and if you look at in the boxed equation in the center of the slide second term and third term it is exactly the same as what we are talking about on slide number 25. So from there we interpret this three terms together which are appearing in this x momentum equation as the net rate of x momentum leaving our elemental control volume on a per unit volume basis. Thank you. Sir my another question is the stream function is cannot be defined for the three dimensional whereas we have for the streamline x, y and z component. So would you please point out the how to differentiate with this streamline and stream function yeah so that's a good question I think. The question is we have used the equation of streamline in three dimensions but we pointed that the stream function is defined only for two dimensions. See these are two different quantities physically you can say. Streamline on its own we have defined as a line the tangent to which is pointing in the local velocity vector and this is a general definition it can be employed for a three dimensional or two dimensional flow. On the other hand by virtue of the definition of the stream function it is restricted to two dimensions because only in the two dimensions it satisfies the incompressible continuity equation identically if you utilize the stream function definition. So that's the only difference it's restrictive from the definition itself to two dimensions the stream function idea. To be honest with you again let me re-emphasize here that the idea of a stream function is somewhat old idea if you look at the modern fluid dynamics especially with respect to the CFD that we do nowadays. Stream function is something of an outdated concept it's not really that useful nowadays it's just for historical purpose I think for two dimensional flows I think people will still want to look at it but the definition is restrictive for stream function because it satisfies the continuity equation only in two dimensions and that's the only difference between 3D and 2D that's all. And sir differential analysis is 19 I have asked one question so what is a physical interpret how to physically interpret that tau ij is equal to tau ji. So the question is on differential analysis 19 where we obtained this result tau ij equal to tau ji and how to interpret this. By the way I didn't mention this in the lectures but later on in one of the discussions I think I mentioned it. The manipulations that are getting done on slide number 19 are what are popularly called as the angular momentum equation for differential analysis. So what we are doing is that we look at the stresses acting on our given fluid particle and we simply take the moments of this and then we realize that there is only a net counter clockwise moment on the fluid particle given by the top expression. The top expression is just an equivalent expression to F equal to MA the Newton second law of motion this is for moments. So we have the net moment equal to the moment of inertia times the angular acceleration experienced by the fluid particle. And then from calculus we simply borrowed the expression for the moment of inertia for this rectangular fluid particle and that's what my highlighter is showing right now. Having put together we obtain a relation for the angular acceleration and the argument is that in order for us to think about is very small infinitesimal fluid particle we tend to shrink the size of the fluid particle to smaller and smaller sizes thereby making that delta x and delta y very very small. That doesn't mean that we are actually geometrically going to zero but we simply realize that as delta x and delta y become smaller and smaller it appears that for no specific reason the angular acceleration of the fluid particle seems to shoot to infinity because the numerator tends to a very very small value and physically speaking such a situation is considered unacceptable because this is general discussion there is absolutely no specific constraint that we are putting on the fluid particle and in that sense there is no reason for the particle to experience any extremely high value of angular acceleration. So in order to resolve this paradox the only conclusion that seems to be occurring if the mathematics also has to go in hand in hand with the physics is that there has to be a zero over zero limit which which has at least a possibility of a finite limit in order for alpha z to be finite and therefore the conclusion that seems to come about is that the numerator which is in terms of tau xy minus tau yx must also be tending to a very very small value in order for the expression to have a zero over zero limit so that mathematically also it satisfies the the limiting condition and physically also we will realize that angular acceleration is not unnecessarily and for no reason shooting to infinity. In my understanding that's the only the way to to interpret this analysis usually you won't see this in in a standard undergraduate fluid mechanics textbook usually you will see this in a slightly advanced fluid mechanics textbook and normally this kind of an explanation is also not really provided so it's up to us to interpret it whichever way we want and what I have given you is at least my understanding of how to how to interpret this. Thank you. I have question in topic number 2 slide number 30 in this the nodes are taken at the corners so this method is applicable of this implicit and implicit are applicable to finite difference only or because we are using it for finite volume also. I agree with you this question is in this animations which I am showing you actually I created this animation when I was teaching finite difference method because we are here we have a two course one is numerical method where we teach finite difference method. So I made lot of effort in creating this animation and the same animation I am using in a finite volume method. So the position of the node is you are right it seems to be a finite difference method but the animation which I am showing is indeed applicable in finite volume method. So it's just that it needs to be staggered slightly so I agree with what you are saying. Thank you. I have one more question in topic number 2 slide number 4 in this in the interior nodes we have considered border cv's also in interior cv's we have considered border cv's also both are overlapping. This question is that border control volume we are considered as as interior control volume we are defining two types of control volumes you can say boundary control volume and interior control volume and boundary control sorry interior control volume constitute two types of control volume border control volume as well as the control volume which is away from the border okay. So this is just a classification I think it should be acceptable. Discretization of the domain or or discretization there are we are generating a grid and there are two types of the grid as far I know that is a triangular grid and rectangular grid. So for a particular problem or particular geometry which type of discretization is much more gives gives a result much more accurate. So how to decide how to discretize the domain? The question is on grid generation and his that there are the background is that there are there are triangular grids there are rectangular grids triangular grids typically is an unstructured grid formulation the rectangular grids are the Cartesian grid formulation. Now the question is which one is more accurate the unstructured grid formulation if you take the same number of let us say Cartesian grid and do a Cartesian grid formulation for the same problem the unstructured grid formulation when you have unstructured grid is slightly less accurate as compared to the Cartesian grid formulation because the approximations which are used tends to make it more it is less accurate. However there are problems where you cannot have a Cartesian grid and you have no option other than to use the triangular grid or unstructured grid. So there of course unstructured grid is is the only option and it is a it gives the good results accurate thank you yes I have three doubts first one is for Puranik sir it is in differential analysis 25 momentum conservation equation in differential vector form the term del dot rho V rho into vector V into vector V is there can we write del dot rho into vector V into vector V is equal to vector V into can we write the term vector V into del dot rho into vector V if not how can we multiply vector V with vector V. Yeah so the the question is on slide number 25 differential analysis when I have written the momentum equation in the in the vector form and specifically the question is related to this tensor product of two vectors which is appearing inside this divergence so divergence of rho times V times V is what what is written here. Let me let me actually not spend the the time on this right now because what we are interested right now is really the corresponding x y and z momentum equations so if you look at the x momentum equation which is which is available right at the top all we do is we replace one V vector with the x component of the velocity which is u and thereby you obtain the top equation. It is much easier to interpret and write the equation in the Cartesian form however if you are if you are interested to know this tensor product it is actually a it's actually a it's actually what is called as a dyadic or a tensor product which is a second-order tensor and it turns out to be having nine components of the velocity so if you want let's say the the first vector velocity has component u1 u2 u3 let's say and the second velocity has components v1 v2 v3 in the three directions then the tensor product V times vector V is going to have nine components written in a matrix form so the the the matrix form will read the first row will read u1 u1 v1 u1 v2 u1 v3 the second row reads u2 v1 u2 v2 u2 v3 and the third row will read u3 v1 u3 v2 u3 v3 so that's the way to interpret the the tensor product of two vectors that it will it will provide you a second-order tensor so that there will be nine components corresponding to this tensor product of two vectors but my suggestion at the moment right now is I have written it just for the purpose of information that rather than writing always the Cartesian components we can write it in a compact form there is no difference in the physical interpretation associated with either the Cartesian form or the vector form so divergence of row times V times V can be interpreted still as the net rate of momentum leaving a control volume now this is not in any particular direction it is the vector momentum that we are talking about on a per unit volume basis but my request right now is to to actually concentrate more on the the Cartesian components and make sure that the the derivation of the Cartesian components is is understood with with time then you can easily get on to the vector forms and start manipulating those with the mathematical background that was provided in on day number one so this is this is what I would like to request though at this point thank you sir transient heat conduction problems we are taking convergence criteria as 10 power minus 6 how can we use this parameter to transient problems I think it should be represented with some other parameter this is one doubt and second out is topic to slide number 4 border CV's border control volumes and inter interior control volumes are shown border control volumes are subset of interior control volumes is there any significance of using separately this border control volume so that is second I'll answer the second question first so the question is what is the significance of using border control volume there is a significance of using border control volume because if you remember there was one question from College of Engineering Pune that when we use advection schemes especially the higher order advection scheme for this border cell you if we remind if I want to remind you when you use a quick scheme there is a 37.5 percent weight to the downstream neighbor 75 percent weight to the upstream neighbor and minus 12.5 percent weight to upstream of upstream neighbor so that weight we have obtained when the nodes are equispaced let us suppose if you want to calculate use that advection scheme here and this are two upstream neighbor then you can see this circle this circle and this circle are equispaced in the x direction however if you want to use here then if you consider this three circle the distance between this three circle is delta x by 2 the distance between this two circle is delta x so many times when you are using even in diffusion also the distance between this two node is delta x by 2 for the border control volumes when you want to implement when you want to develop a code you have to do some special treatment so that was the reason that I had given it a separate name so that we when we want to discuss the implementation details we can we have to be very specific on which control volume we are talking of thank you and regarding your first question the question is that we are using a convergence criteria and we said it 10 to power minus 3 and the question is some other value should be used or some other parameter should be used let me tell you convergence criteria which we are defining epsilon in we are we want that this is based on whether if you are following unsteady state formulation then we want to check for steady state steady state when we want to check we have to stop that is a used as a stopping criteria and computer does not stand understand what 0 is so user has to define what he means by 0 you can define as 10 to power minus 3 another person can define it as a 10 to power minus 4 but it is up to you because the user has to make a decision that what he wants to call as a practically 0 and at the same time he has to check whether if he has used 10 to power minus 3 is practically 0 for a particular problem and if I if he uses 10 to power minus 4 if the change in the result is negligible then he can come to a conclusion I hope I had answered your question actually my question is transient problem we are using convergence criteria means the convergence means it is the value nearer to the value what we got by got by analytical analysis for example if a 2D conduction problem is there steady state so there we use the convergence criteria as some old value minus new value by new value should be less than some value like that we will put but actually in transient problems we will get a particular value at particular node at different time time intervals time steps so this convergence criteria encode what we are using is the value new what we got at like a next time step minus previous time step like that we are using so this is the difference between the value of previous times difference between present time step to the previous time step so for this can we use this name convergence criteria I think it should be named with some other parameter or notation I agree with you I think you have a right point I should not call it as a convergence criteria probably convergence criteria word is good if there is a steady state formulation I think the better word will be the steady state criteria study nest criteria so I agree with you you are right it should not be a convergence criteria it should be a study nest criteria or a steady state criteria thank you for your suggestion my question is related to lab session 2 that is one one-dimensional conduction with heat generation we have solved this problem with positive heat generation so while executing the code in the lab I have given negative value of heat generation right the temperature distribution was upward dome with positive and with negative heat generation the dome was downward my concern is that when I have given a sufficiently large volumetric heat generation that I have reached below I have gone below absolute zero which is a violation of basic law of the question is that for the one-dimensional lab session we had a problem one-dimensional conduction with volumetric heat generation so you had used a positive value you got a maxima and when you use a negative value you have got a minima that's what you are saying for the dip in the curve and then you are saying that for very large value you got that you found the temperature distribution is going less than zero now my question is I would like to ask that when you say very large value you have used a very large negative value or very large positive value when you got less than zero use a very large negative value then it is expected that it should although because although negative value of volumetric heat generation is unphysical but if you use that it should go it will be unbounded as as I said I had mentioned that if use a volumetric heat generation you solution will become unbounded unbounded means it will be outside the bounds of the boundary values here the boundary condition has 0 and 100 degree centigrade for large positive value it will go more than 100 and for large negative value it can give and go less than zero degree although the problem is unphysical but if it is there the question the result is correct for the lab session giving boundary conditions we are getting a temperature counter plot so how to read that counter plot exactly that we are not getting if possible explain with one case so how to interpret it or how to write that results the question is how to interpret the isotherm contours isotherm lines I will take the isotherm contours and discuss how to interpret let me take yeah this is a good picture so let us suppose you got a steady state temperature contour where this which problem is this this problem corresponds to left wall 100 degree centigrade bottom ball 200 degree centigrade right wall 300 degree centigrade and top wall 400 degree centigrade so we got a temperature contour like this now the message you can take qualitatively by looking into this picture is you can see that near to this left wall left wall is at 100 this is 150 this is a problem with no volumetric heat generation so that temperature distribution will be monotonic all the lines in between 100 and 150 let me use a pointer all the line this is 150 this is 100 so this is 110 this is 120 this is 130 140 and 150 so all the lines are having temperatures which are greater than 100 so at least you can understand if the temperature near inside this plate near to the boundary is greater than 100 then the heat transfer will be outward this is the way you can interpret when you go to the bottom wall it is more interesting what you see is that there are two parts of the bottom wall left part where the temperature above is less than 200 like here it is 190 180 170 160 and 150 and on the right half you find that temperature is 210 220 230 so what you are seeing is on the left part of the bottom wall heat is going in because the temperature above is less on the right part heat is coming out similar is the case when you go to this wall if you go to isotherms for the on a second problem which I consider where left wall is 100 bottom wall insulated right wall constant heat flux and top wall convective boundary condition in this case you can have a different interpretation in this case you should this is a good case where you can check whether the plot which you have got is correct because bottom wall you are having an insulated boundary condition so when you want to interpret this result you first need to check whether the line intersecting the bottom wall is vertical or at wall you can see du by dy sorry dt by dy here is 0 here also it is 0 so you can check that this is you have implemented insulated boundary condition and the steady state temperature distribution is indeed giving dt by dy is equals to 0 at bottom wall on the right wall it is a constant heat flux boundary condition so dt by dx is minus qw by k so dt by dx here should be so the slope here should be constant which is indeed being followed at the this wall so that way this is a good test as far as this picture is concerned and as far as heat transfer is concerned you can draw similar conclusion here that there on the left wall there is a top part where the temperature nearby is less than 100 so the heat transfer is inward and bottom type bottom side it is outward this is the way you can interpret you when you draw the engineering when you look into the plot of let us say variation of local heat fluxes on this wall so with respect to why how heat flux is varying if you plot that you indeed get a results where there is change in the sign of the heat flux so in the engineering parameter this is reflected but the reason why that change is happening you can confirm through this isotherm contour if you go to the fluid flow problem if you let us say if you are a car designer you design different shapes of the car you create a movie of the flow structures and you calculate the engineering parameter and let us suppose a particular design gives you the least drag force but the reason why that is giving a least drag force will come if you try to make a story out of the movie which you have created for that best design as far as the flow structure is concerned so this flow structures are very important in scientific understanding of why something is happening thank you. We will take the last question from J N T U Hyderabad. Hello sir this is regarding topic 2 slide number 66 it is about two-dimensional heat conduction problem where the boundary conditions have been shown particularly for the right boundary where the constant heat flux is applied instead of going with the boundary nodes shown by blue circles can we just balance the heat fluxes for the boundary control volumes on the right boundary okay yeah the question is interesting when we are doing an energy balance for the boundary control volume that is a border control volume in this case let us say this control volume the question is rather than using this boundary condition insulated boundary condition can we take this is the east phase this is the west phase the heat flux on the east phase and the west phase are 0 so basically if you do this way what you are saying is that del by del x of q x is equals to 0 which makes del square t by del x square is equals to 0 and it indeed follows this insulated boundary condition so yes you are right you can do that way also thank you we will stop here and we will have the questions are shared tomorrow at the same time 4 to 5 30


Dear students, in this module, we shall define synapses and discuss their types and discuss one type that is electrical synapses in detail. The synapses are the junctions between exon terminals of one exon and the dendrites of an other exon. The transmission of signals takes place from one neuron to the other through these junctions. There are two types of synapses. One are the electrical synapses and the second type is of chemical synapses. We shall discuss electrical synapses in detail. At the electrical synapses, the plasma membranes of pre- and post-synaptic cells are in close opposition. Pre- or post-synaptic cells are very close to each other and they are electrically coupled through the gap junction proteins. Dear students, the electric current can flow directly from one cell into the other through the gap junction proteins purely by electrical means without involving any chemical transmitter. It is just like a signal transmission along a single exon involving passive spread of local circuit current that depolarizes and excites the neuron ahead. So, the transfer of information through these synapses is much rapid. Dear students, the electrical synapses provide a mean of very rapid transmission without involving any chemical in between. However, it has limited efficacy because the signal that is transferred from one neuron to the other through the gap junction proteins is somewhat attenuated. So, a single pre-synaptic action potential that arrives in the post-synaptic potential cannot generate an action potential in the post-synaptic cell. When a weak signal reaches, the post-synaptic cell membrane will not be so depolarized that it generates action potential and the propagation of action potential. This is the reason that most electrical synapses require more than one action potential to propagate. This limitation reduces the efficacy of electrical synapses. But in spite of their limited efficacy, they have certain advantages. They are particularly effective in synchronization of electrical activity within a group of cells. They can transmit information across a series of cell-cell junction. This requires a series of cells to convey an electrical signal in the entire series and immediately play a role in the electrical synapses. For example, in the myocardium of the heart, there are electrical signals and in the muscle cells of the heart, they are very fast transmissions. Dear students, the electrical synapses are relatively rare and most of the signaling between neurons takes place through the chemical synapses. However, electrical transmission has been discovered in many cells of the central nervous system. Moreover, in many receptors like the retina of the eye, electrical synapses are present. Similarly, many smooth muscles have electrical synapses and cardiac muscle fibers and many other receptor cells and some exones of peripheral nervous system also contain electrical synapses through which transmission of signals takes place rapidly.


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