 Good afternoon sir, please open the topic number 5 slide 19 actually the Peclet number is coming is rho u delta x by gamma and where these two terms coming because if you resolve this it is 1 minus Peclet number it is coming I am not getting exactly how this two value comes there p by 2 what you are saying is that 1 minus Peclet number in curly brackets divided by 2 that is what you are getting whether you mean that this should be 1 minus Peclet number yes sir yeah yeah because that is what I could answer 1 minus Peclet number yeah yeah so that whole should be divided by 2 not Peclet number divided by 2 correct that no no no 2 should not be there no it is 1 minus Peclet number actually I can understand that 2 gamma I had taken as a coefficient ai 2 gamma I had already taken here so I can understand what you are saying is that when you take gamma outside it becomes 1 minus rho u delta x divided by gamma so there should be no two term here that is what you are saying correct yes yes yeah you are right there is a mistake here yes yes yes you are right thank you any other question actually according to the conditions will be changed no because in slide topic number 5 slide 21 the condition should be if the Peclet number is greater than 1 it should be like that oh no no no no let me put it in the whiteboard and then show you let me do one thing I will come back to this in tomorrow's lecture I think there is some mistake in this slide I will work on it and get back to you tomorrow morning just give me some time right now I could see that there is some issue but I will definitely get back to you are right that this should be Peclet number divided by 2 because when only the condition is correct but there is something wrong in this expression this condition is correct there is nothing wrong in this condition it is just that in this expression there is some issue I will resolve it and come back to you thank you okay sir and what is gamma in this equation it is only k k or k by Cp? gamma on the left hand side yeah yeah you are right gamma it will be if it is a heat transfer problem it will be k divided by Cp and for momentum it should be a mu yes yes thank you yes can I do that yes sir good evening my question is about topic number 5 slide number 34 that is in pseudo code second line is that for j equal to 2 j max minus 1 and for i equal to 1 i max minus 1 in heat flux in x direction my question is why we are not calculating heat flux at i max we are doing until i max minus 1 only but why we are not calculating at i max the question is in the pseudo code this maximum value in the loop is always j max minus 1 i max minus 1 for the heat flux in the x direction as well as in the y direction question is why we are not having the i max j max value for that let us go to the previous slide now the convention which we are following here the running indices for this corresponds to this running indices the running indices for this corresponds to the running index of this so the last green square in the horizontal direction is this one and the running indices of this green square corresponds to the running indices of this yellow circle and the running indices of this yellow circle is i max minus 1 the running indices of this yellow circle is i max okay so the last green square is referred by this running indices which is i max minus 1 Now, let me tell you here the loop starts with let us say this is i is equals to 1, 2, 3, 4, 5, 6 whereas for yellow circle the running indices is 1, 2, 3, 4, 5, 6 and 7. So, in the x direction there is difference of 1, there are 6 green square, 7 yellow circles. So, in the next slide you can see for the green square it is 1 to i max minus 1, 1 to 7 minus 1 which is 6 green squares. Similarly, when you go to the red inverted triangles where we calculate heat flux in the y direction, you can see that this is first one, second, third, fourth, fifth, sixth is just that the number of fish centers are 1 less than the number of total nodes. We have 5 interior node, 2 boundary nodes, so total 7 nodes. So, for this last cell at the top, the last red inverted triangle this is running indices for this corresponds to this which is j max minus 1. So, that is the reason that it goes from 1 to j max minus 1. Thank you. I have a question in topic number 6, slide number 37. Here in first level and second level approximation it is mentioned that they are second order approximation. How we can decide again based on Taylor series expansion or? Yeah, so the question first let me repeat the question. The surface averaging, volume averaging or linear interpolations which we do, I had already I had always mentioned that they are second order accurate, but if you want to show the order of accuracy that comes in the numerical analysis and for that you can show it through a Taylor series expansion. Maybe I will put in the Moodle this analysis part where it shows that this is indeed a second order accurate. To practical application, suppose we carry out simulation of slurry flow, say in pump and you want to study the effect of slurry on the impeller blades, it may lead to erosion of the blade. So, it is related to flow analysis as well as some words say force or stress analysis also. So, how we can solve such problem? The question is different from what has been taught here. The question is on a practical application that is a slurry pump where you have fluid as well as solid particles. So, this is what you can call as a granular flow and it is the number of particles are much less as compared to the fluid. So, this is called as a dilute flows. So, the question is how to solve this problem in using CFD? As far as I know the solid particles in this case are solved by Langranian approach and the fluid flow is solved by an Eulerian approach and there is a coupling between the two and that much I know about this problem. Although I can tell you that as far as the accuracy of the CFD solutions are concerned, as I say that there are lot of problem challenging problems where CFD can give you solution, but the level of accuracy may be not that high. So, because in this case there are lot of modeling issues and the solutions which you get from CFD may not be that much accurate as compared to the other let us say single phase flow. This is a multi phase flow problem. So, sir, Puranik will add to it. Yeah, I think you are referring to the effect of slurry on the impeller blade and how the blade will be affected. So, that actually will become what will be called as a fluid structure interaction problem and in this particular case is going to be actually very complicated because the fluid is as you mentioned it is a slurry. So, it is not a single phase fluid, but it is a multi component situation where you have a fluid flow and also the solid particles. This will require a fluid structure interaction type analysis, but we are unfortunately not really exposed to such kinds of applications and I do not think we can really provide any definitive answer on that. I think you may have to refer to some specialized fluid structure interaction related papers for specifically that problem that you are referring to, which is effect of the particles on the impeller blades and how erosion of the impeller blades may occur in this situation. So, yeah, I think sorry, we do not have the expertise to comment on that. It is about a computational stencil that is when we are going to next step and professor Thurman did explain it reasonably well as I want to listen to that again in the explicit scheme he has explained. So, the question is on the computational stencil explanation of the computational stencil again in a one dimensional situation let me start with the computational stencil in a one dimensional situation which is much easier to follow. So, in this case the computational stencil is such. So, let us suppose this is a one dimensional conduction problem. So, the computational stencil is such that when you want to calculate the let me go back to the algebraic equation I think that is a better this slide is better to explain the computational stencil. Actually, what happens is that when you convert from the conservation law or the governing equation to algebraic equation finally, you end up with a linear algebraic equation. So, this is the linear algebraic equations which we get from an explicit method. This is an algebraic equation which we get from explicit method. This is an algebraic equation which we get from implicit method. Now, if you look into the nature of the algebraic equation this equation is for a particular grade point and actually this equation set of equation is applicable for the number of circles. Let us say in this figures number of yellow circles which are interior grade points. So, there are nine yellow circles. So, this equations there are nine such equations. So, this is right now expressed in a generic form where p represents a representative control volume, capital E represent the east neighbor of that representative control volume, capital W represent west neighbor, capital N represent north neighbor and capital S represents the south neighbor. So, these are the neighboring temperatures which are in an explicit method of previous time level. However, on the right hand side although you will not see temperature at point node p because this constant b which has shown as a constant right now actually this consists of temperature at p or at a previous time level. So, what is mean by a computational stencil in an explicit method is by looking into this algebraic equation. Let us suppose if as I said that this equation is applicable to nine yellow circles shown here. So, right now this stencil is shown for this yellow circle where this is the representative node p at time N plus 1 and the way we have this algebraic equation. This is a function of temperature at same point of previous time level which is comes in this constant b and the four neighboring values for of the previous time level which is east neighbor, west neighbor, north neighbor and south neighbor. So, this is a computational stencil for this point. Similarly, when point moves that moves here then the neighbors change. So, that we had shown you an animation where this computational stencil moves point by point nine times because in a computational we have to apply this equation nine times. We get nine equations where in each equation there is only one unknown. So, the you can see that thing in the computational stencil only that in a new time level there is only one grid point value. Whereas, there are five grid point values from the previous time step this is. So, the pictorial representation or let us say an animation of the way this algebraic equation one by one scrolls through all the nine points is what is shown through this animation. When you go to the implicit method you get an algebraic equation where temperature at a particular point of the next time level is a function of its four neighbors and this b constant consists of temperature of the same grid point from previous time level. So, that is why now the this figure stencil almost reverses in this case and in the new time level you have five neighboring values four neighbors and then node value and one value from the previous time step. So, this is what we call as computational stencil which is a pictorial representative of a representative algebraic equations. Thank you for attention.