 Sir, this is a question related to dynamic viscosity. I want to know that physical significance of N that will comes in the dynamic viscosity relations, sir. Yes, okay. So, the question is about dynamic viscosity relation that we talked about in the morning. Specifically, the power N that the strain rate is raised to, that is the question about. Actually, some of the fluids that we deal with which are the non-Newtonian fluids behave such that they have a non-linear relation between the shear stress and the rate of strain within them. So, that is all that we were trying to point out to you that some fluids will behave such that tau will be proportional to the rate of strain raised to some power which will be not equal to 1. So, physical significance is essentially that some fluids are found to behave in that fashion and some fluids are found to behave in the fashion that tau is equal to mu times du dy which is tau is directly proportional to du dy or the rate of strain which are the Newtonian fluids. So, in nature you are or in some technological applications as well, you find that some fluids behave like the Newtonian fluids, some fluid behave in the non-Newtonian fashion that is about it really. I mean that N is simply to signify that it is a non-Newtonian fluid if it is not equal to 1 that is about it. Amrita Colm, you have a question please go ahead. Sir, can we apply this momentum conservation in Caute flow? Yes. So, the question is whether you can employ momentum conservation in Couette flow. The answer is yes, momentum conservation and mass conservations are universal principles and they will be employed in any flow that you are talking about. Couette flow is just a specific example of a flow. In fact, what we are going to do is two days from now when we talk about the exact solutions, Couette flow will be one of the exact solutions that we will completely work out in terms of the velocity profile determination. So, at that time you will realize how we will employ the mass conservation and the momentum conservation for Couette flow. So, the answer is yes and we will do that in a few days of time. But sir, in that case there is no pressure force that is causing a flow to take place. Yes. So, in the case, well I will really request you to wait for a few days, but let me answer it quickly. So, in case of Couette flow when there is no pressure gradient in the axial direction, it is actually the if you remember the Couette flow setting, there are two plates parallel plates. One is stationary and the other one is set in motion. It is actually the plate that is set in motion that transmits its momentum through the action of viscosity to the subsequent fluid layers and that is how you generate and maintain the flow. So, you do not really necessarily need a pressure gradient to maintain a flow in that situation. We will look at this in detail when we solve the Couette flow problem in a couple of days from now. Thank you. PVEPS Vijay Vada, you have a question? Please go ahead. I have two questions, Professor. Yes. Here is the integral analysis. Hello. Yes. Yes, please go ahead. This is regarding the selection of control volume which is very important for the analysis. What I observed here is in the example two and three, the flow of flat plate and the flow of body, you have selected control volume in a different way that in the example two, flow of flat plate, there is no mass flow across the control volume. There is your set to the streamline. And in the third example, you have selected the control volume in such a way that there is a mass flow across the control volume on the top surface. Yes, so the question is about selection of control volumes. In particular, the two examples, example number two and example number three, where in example two, the control volume is chosen in one manner and the example number three, the control volume is chosen in another manner. So, actually the example number two, you can do the same example with a different choice of a control volume altogether. You can actually re-solve example number two in exactly the same manner as the choice of CV is concerned using the kind of control volume that we chose for example number three. So, for example in example two here, the control volume could be essentially having a horizontal top surface as well, in which case there will be a mass flow rate that will cross the top surface and that needs to be properly accounted for. So, in fact, I have chosen the two different control volumes for these two different examples differently on purpose. So, that in one example, I could show it that you could do it in this manner. In other example, I could show that you can do it with some other choice. But your question is exactly right and what I suggest is that if you are up to it, you can try to redo this example number two with a control volume such that the top surface of the control volume is a horizontal line at the level h. But then accordingly you have to take into account the mass flow rate that will be crossing that top surface. The answer will be the same whether you are choosing the control volume in one fashion or the other. This is regarding jet pump example, example five. Yes. Example five. At the inlet section, there is a jet pressure as well as a surrounding fluid pressure. Yes. But the given condition assumption at the section one pressure is taken as uniform. Is it practically possible? No, no, I mean these are simplified examples in the sense that these are supposed to mimic the real life situation to some extent. Obviously, what you are pointing out is correct that we cannot really have a completely uniform pressure in the jet section as well as the surrounding fluid section. But to simplify the problem to a reasonable extent and yet obtain some reasonable numbers as an estimate in this kind of an example, the problem statement says that you assume as simply or for simplicity the pressure to be uniform in section one. That is essentially the idea. Thank you, sir. Okay, thank you. Yes, Perrier, many am I. You have a question, please go ahead. Yes sir, that example three, the integral analysis 14, the slide integral analysis 15. Yes. So, how do we replace actually the small u with the 2 into capital U y divided by h? Yes. So, yes. So, the question is in example number three of the integral analysis, the profile of the actual velocity at the outlet has been found out. And basically if you realize that let me actually switch to the whiteboard for a second because I can draw it there. So, this is the profile that we are basically looking at. So, y equal to 0 is the center line where the actual velocity is 0. And then it linearly increases to capital U by the time the y value comes to h by 2. So, essentially it is a linear profile. So, you assume that u is equal to some constant times y. And then this constant you can evaluate based on the value of u equal to capital U at y equal to h by 2. So, then u is equal to simply c multiplied by h by 2, which implies that the constant is 2 times capital U over h. And therefore, small u, which is the actual velocity as a function of y, then is 2 times capital U over h multiplied by y, which has been then used in the in the expression. Thank you, sir. Yes, V i t well or please go ahead if you have a question. The question is about the outlet velocity profile in example number 3. And the specific question is how realistic the velocity profile is. And the answer is it is not very realistic, it is actually some sort of a devised profile for the purpose of designing a problem such as this, where you can actually then implement the mass balance and the momentum balance in an appropriate manner. Otherwise, it is not really a very realistic profile that you can actually experimentally observe. Okay. K I T Kolhapur, you have a question please go ahead. Sir, one question for our today morning session sir. Sir, as I am a chemical engineer, what is the difference approach wise, difficulty level wise, application wise in safety for mechanical engineering and safety for chemical engineering. And second question in say up to one session on interior analysis slide number 12, you have used mass rate is equal to U over h. But dimension leap check, dimension comes kg per meter per second. Fine, yeah. So, there are two questions. One is if I read you correctly, how is the CFD analysis to be employed for a chemical engineering type problem. So, I am assuming that you are referring to the fluid being non-Newtonian, if you are dealing with a chemical engineering application situation. In principle, there is no difference between how you employ a CFD analysis for a mechanical engineering application versus how you will employ it for a chemical engineering application. All that you need to do is you need to remember when we said that a CFD analysis always begins with an appropriate set of governing equations. So, the governing equations that we are now going to derive in the next couple of days are for Newtonian fluids. However, similar type of governing equations can be derived from first principles of conservation of mass, momentum and energy for fluids that are non-Newtonian as well. So, once the governing equations have been derived in that fashion, you can go ahead and employ one kind of discretization that we will talk about here and carry out the CFD analysis. So, really speaking, there is no difference in principle between how a mechanical engineering analysis goes and how a chemical engineering analysis goes. It always starts with an appropriate set of governing equations and as long as you have the correct governing equation with you, you can always employ the CFD analysis. That should not be any problem. On the second question, what I have talked about in this example number 3 is that the entire analysis is done on a per unit depth basis so that if you for example, right now I am projecting slide number 14 on the board, you will see that the mass flow rate as you correctly point out will be missing one length dimension. However, that one length dimension is essentially that unit length into the plane of the board or the paper that we are talking about. So, if you want you can always multiply by 1 and that 1 will have the dimension of the length so that the mass flow rate will correctly come out as kg per second. So, there is one more question. Yes, please go ahead. Sir, in I will define viscosity tau is equal to tau is proportional to du by dy raise to n. So, how one should experimentally check whether n is 1 or not so that we can carry out experiments? Well, actually the way you can do it is perhaps you can measure the viscosity in a viscometer and see if it is really following a Newtonian law or not. To be honest with you, I have not really dealt with non-Newtonian fluids so that way I cannot perhaps directly answer your question, but in order for you to realize whether a fluid is Newtonian or not, you will necessarily need to find out its rheological behavior meaning that for various different types of or different values of shear stresses, you measure the strain rates and then you plot a curve of the shear stress versus strain rate and find out whether it is coming linear as a Newtonian fluid should or a non-linear in case of a non-Newtonian fluid. That is the only way to figure out whether the fluid is Newtonian or non-Newtonian and then you can proceed from there. Generally, we have in our laboratory to find, we have viscometers. We can find mu like Reynolds viscometer, Sabert and all. But we never, we do not have any such set up to find strain rate and all those things. So, that is what I got this question exactly. Yes, yes. I actually understand your question. I am not really in a position to answer it immediately. Let me think about it and let me get back to you maybe if not today but by tomorrow. Let me note this question down because I do not think we have the answer right now other than generating a rheological curve which is the curve of tau versus du dy. That is the only way you can figure out whether the fluid is behaving in a Newtonian or non-Newtonian fashion. But let me, let me get back to you on that. Thank you. Yeah. Sviya Naiti Sura, you have a question. Please go ahead. My question here is in the integral analysis 14 seat, how mass moves out of the control volume that is m dot out top? Yes. So, that is the. Second question. Yeah, please, sorry, go ahead. Suppose, here's values are given in these problems, then whether we need to include them in any of the equations? Let me answer the second part first. The question is about in these types of problems, the integral problems if shear stress values are given, if we need to include those. Actually, if you see the drag that will come on an object such as this will be due to both the pressure drag as well as a shear drag because of the shear stress. Now, in this particular situation, the pressure is given to be uniform everywhere. So, therefore, there will be no drag due to pressure. So, in fact, if shear stress values are given on the surface of this particular object, your problem is already solved because then from the shear stress values, you can directly calculate the drag that will come on the object. All you need to do is that at every point on the surface, you will need to calculate the shear force and the axial component of that force, you will need to add in order to get the drag on the object. So, that way, if the shear stress values are given on the surface, your problem is actually already solved. You do not have to really do anything. On the first question, the question was about how do we calculate the mass flow rate that is going across the top surface in this particular example. So, I have already projected the slide on the board. So, if you see here, this is a symmetric problem as we saw. So, let me go back. This is symmetric problem about the central horizontal line and therefore, we have chosen only the top half. We could have chosen the bottom half as well. The bottom surface of this control volume is such that because the problem is symmetric, there is no mass exchange happening across the bottom surface. I hope that is clear. Now, having done that or having said that, if you look at the mass flow rate that is entering this control volume from the left hand side, from a geometric interpretation point of view, if you want to get an idea of the mass flow rate, it is basically the shaded area of the control, of the rectangle that I am highlighting. It is the rectangle which has a length or one dimension of capital U and the other dimension of h by 2. The area under this is what the mass flow rate that is coming into the control volume. That is the geometric way of looking at it. If you look at the outlet, what you see is that, if you calculate then the area of this triangle which has a base of h by 2, let us say, and a height of U, you will see that the area of this triangle is less than the area of the rectangle at the beginning which will represent the mass flow rate coming in. Therefore, there is a mass deficit. There is more mass coming in than what is going out. So, the difference between those two has to be going out from the top surface. There is no other way it can go anywhere is the point that we would like to make here. I hope it is understood. Sir, but whether it is moving out perpendicularly to the control volume or it is coming out at particular direction, that is the more specific question. It is immaterial. It is immaterial to know whether it is moving perpendicular or otherwise. In general, it need not move perpendicular to the control volume surface. In general, you can assume that it is going at some sort of an inclined manner as I have shown here, but that is really immaterial as far as the problem solution is concerned. All you need to realize is that there is a certain non-zero mass flow rate that is leaking past this top surface of the control volume. That is the most essential thing to understand here. Whether it is going in a perpendicular manner or not is really not an important thing. In fact, if it is going in a perpendicular manner, then it actually will mean that in the X momentum, you need not account for it. If it is going in a inclined manner, so to say, it will have a velocity component in the X direction which will cause an X momentum going out from the top surface as what we have done in the problem. Thank you, sir. Okay, thank you. Ruba Indore, you have a question. Please go ahead. Sir, how will establish momentum conservation in fluids which are exceptionally high at viscosity? The question is how will we ensure momentum conservation in fluids which have very high viscosity? And actually the answer is that you do not have to worry about whatever value of the viscosity, whether it is less or more. The basic momentum principle, whether it is on integral basis or later, whether it is on differential basis as we will see in a couple of days, it will always employ or apply. So the viscosity, whether it is low or high, it simply does not matter. It will automatically be taken care of. Sir, because one experiment I read was pitch drop experiment. In that one fluid regarding tar was, one derivative of tar was used and it was, I think it travels inches in years. So I was a bit curious about it. Yes, so it is a very special kind of a fluid and looks like from the description that you have given, it appears to be a non-linear fluid, meaning that it is a non-Newtonian fluid as well. Even if that is the case, the fundamental principles of mass conservation and momentum conservation or momentum equation as we said, they are not going to change whether the fluid is highly viscous or otherwise. So they can still employ or apply without any problem. I do not see any trouble with that. It is just that the time scales over which the flow is happening is very, very large as you say, but that does not violate the principles of mass conservation and momentum equations. Sir, thank you. Okay, thank you. P.C.O.E. Baramati, please go ahead if you have a question. My question is, sir, can change in inlet and outlet portion of control volume affects to all the three mass, energy and momentum balance. Okay, I think I understood your question. So the question is, if the control volume is chosen differently especially with respect to the position of the inlet and outlets, if the balance statements will change. See, as long as the velocity profiles that you are dealing with are not going to be different at the inlet and outlet, the position at which you take the inlet as well as outlet is not really going to change the analysis. Only thing that will change is if the extent, the lateral extent which will permit lesser or larger mass and momentum flow rates coming in, then obviously things will change, but then automatically the balances will make sure that your final answer is correct. So in that sense, there is nothing to worry about. Okay, sir, thank you. Thank you. Yeah, Mufakum Ja, please go ahead if you have a question. To integrate analysis, 14 PPT, 14, in that mass conservation equation has been written, that is correct. That is in that box equation is given. Okay, and when we come to 15, block number 15, there this velocity u is given for the mass that is going out, but that is not correct because the mass is going across the area, across the top area. So we should take that cross-sectional area, not this cross-sectional area so that velocity will be different. Okay, so the question is about whether the expression which I am highlighting right now is correctly worked out or not, especially the value of the velocity chosen is correct or not. So let me just go back to the control volume. So if you look at this slide number 14 which is projected right now, you mentioned that the mass flow rate is correctly calculated, which is good. Now the mass flow rate is actually carried out by a component of the velocity which is perpendicular to the flow, to the surface, control surface. Okay, however in this particular case what we are doing is we are employing the x direction momentum equation. So in the x direction momentum equation what we are interested to know is that the x momentum on a rate basis flowing out and in from the control volume. So if you look at the mass flow rate that is flowing from the top surface of the control volume here, as far as the x momentum that it carries, it has to be multiplied by the velocity in the x direction which exists at this location which is at y equal to h by 2 and if you see the velocity plots that have been shown at the inlet and outlet, you will see that at y equal to h by 2 the x direction velocity is equal to indeed capital U. Therefore the x momentum that the mass flow rate is carrying out of the control volume is equal to the mass flow rate from the top multiplied by the capital U velocity. That is the only thing that needs to be understood here that at y equal to h by 2 which is the top surface of the control volume, the x velocity is actually equal to capital U and therefore the mass flow rate will carry an x momentum which will be the mass flow rate multiplied by that capital U and that is the reason it has been calculated in this fashion. Okay, thank you and the other one is instead of integral analysis, whether differential analysis also could be used for safety? So the question is whether differential analysis could be used in place of integral analysis. See in a problem like this, when the overall objective was to find out the drag, in a situation that is shown on the board right now where you have an inlet velocity profile given, an exit velocity profile given at certain upstream and downstream locations etc. Without resorting to the differential analysis, you are in a position to calculate the drag directly with an integral analysis like this. Now obviously the differential analysis is a detailed analysis, much more detailed analysis. So there what will actually be happening is that you will find out the entire velocity field as it will exist within the domain defined by the control volume in a point wise fashion. So that will be far more information that you will have available with you. Using that information, you can always find out the drag force on this object. So that way differential analysis will always give what you want. It is just that the amount of work required for differential analysis will be enormous whereas in this particular case, using this simple integral analysis, you are able to find out the drag force on this object without really using the differential analysis at all. So that is the advantage of such problems in integral analysis. But the point is that differential analysis will always give what you want. Amritha, Bengaluru, please go ahead if you have a question. Sir, I have two questions. So one is connected to Taylor series of expansion. Is there any mathematical and physical constraints to consider Taylor series? Why? Because most of the cases we consider Taylor series of expansion. Okay, let me try to answer this right away. The Taylor series expansion is used for situations where functions are smooth and nicely behaving as they say when there are no discontinuities and issues such as that. What happens in case of fluid mechanic situations is that as long as you are dealing with a continuum situation in an incompressible situation in particular, what you have is a nice smooth distribution of properties from one point in the domain to the next. So that there is no restriction on the use of Taylor series expansion as far as these types of problems are concerned. So that way there is absolutely no restriction. The second question is we come across conservation form and non-conservation form. In principally though equations looks like simple through mathematical manipulations, but in physically what different meanings it sort about? Yeah, so the question is about conservation form and non-conservation form of governing equations. We are actually going to derive these in the next two days, the governing differential equations both in conservation form as well as in non-conservation form. So hopefully the question will be answered in due course of time, but let me say that at this point what you should perhaps realize that if you employ the inherent Eulerian approach of balancing mass flow rates and momentum flow rates etc. for a control volume, you will automatically get the governing equations in conservative form whereas if you employ the Lagrangian form of derivations, you will always get the equations in non-conservative form. So that way what we will do is we will do actually both and in principle there is no difference between both. In some CFD situations especially when you are using a finite volume technique, it seems that the conservation form of the equations is more useful to be utilized than the non-conservation form. So when it comes to implementation detail using a technique such as finite volume technique, the conservation form works perfectly well, non-conservation form has a problem. However, non-conservation form people have used with finite differencing and that also is an option if you want to pursue. But in a couple of days, we will derive these equations in both forms so that hopefully this question will be clear. Thank you, over and out. Okay, thank you. Yes, Manipal please go ahead if you have a question. And the streamline is diverging from the flat plate at section 2. So naturally when uniform pressure is considered in the x momentum equation for reaction due to the plate that p into a in minus a out should be out should have been considered. But there has been neglected that due to the pressure two terms will be coming p2a2 minus p2a p1a1. Actually p is same and a1 minus a2 should have appeared. Why the areas have been neglected? The streamline is diverging from the flat plate. Inlet flow area is different, outlet flow area is different. Yes, the question is about example number 2 in the integral analysis. The question is whether the pressure forces have been neglected correctly or incorrectly because what is getting pointed out is that the streamline is getting wider so to say from the inlet to the outlet. Therefore the outlet area is larger than the inlet area and therefore the pressure forces should have been considered as higher or the pressure difference between the inlet and outlet is the point. So the answer to that is that the pressure is actually uniform everywhere. It is given as part of the problem and furthermore the control volume is a closed control volume completely. So what ends up happening in a situation such as this where you are dealing with a closed control volume with a uniform pressure everywhere that the pressure contribution cancels out completely when it is acting on a completely closed control volume from all sides. So the one example that you can think about is rather than a continuously widening area of cross section, if you want you can think about a step change in the area that may help you. Let me in fact try to draw it on a whiteboard and try to explain what I am talking about. So let us say that if we were dealing with a closed control volume in this form where you have a step change in the cross section rather than a continuous change and what is given is that the pressure is completely uniform surrounding this control volume. In this particular case you will appreciate the fact that there is no net force due to the pressure on to this control volume. The situation as I have drawn on this whiteboard is no different than what was shown earlier. You can say that the continuously increasing or decreasing cross sectional area is the most general case of a closed control volume whereas what I have drawn here is a special case where you are seeing a step change. See what happens in case of a continuously increasing or decreasing area as what we had in the example is that you need to only look at the projected areas and then the projected areas will make sure that the uniform pressure contribution will cancel completely and there will be no net pressure force because of that on the control volume. VNIT Nagpur if you have a question please go ahead. I have two questions one is related to the morning session on dynamic viscosity. Why does the dynamic viscosity does not change along x direction in this flow and does it remain same for any type of flow? And another second question is on integral analysis 9 P1 is taken as plus while P2 is taken as negative why it is soon? Let me let me try to answer your second question first the integral analysis. The question was on integral analysis slide number 9 P1 was taken as positive whereas P2 was taken as negative and what was the reason for that? See we talked about pressure being a compressive force so as far as the control volume is concerned and the material that is enclosed within the control volume is concerned. Pressure 1 if it has to act in a compressive manner it has to act on the control volume in the positive x direction. So when it comes to calculating the or adding the forces in the x direction it will be basically acting in the plus x direction whereas the force due to pressure at section number 2 here at the outlet will be such that if it has to act as a compressive force it has to act on the control volume in the negative x direction. So when it comes to summing the forces on the control volume in the x direction it will come as a negative force that is acting in the negative x direction. So that is the reason P1 a1 was taken as acting in the x direction whereas P2 a2 was taken as acting in the minus x direction. I hope that answers that yes please go ahead. Why does the dynamic viscosity does not change along x direction in this case? Yeah I am not understanding your question that the dynamic viscosity is not changing in the x direction etc. So can you please point out to which slide or which part you are referring to? Introduction 5 PPT number introduction 5 in that one dynamic viscosity as per the figure the value of dynamic viscosity along x direction is given as mu while along y direction it is given as mu plus dou u by dou y into delta y. Okay now let me try to explain this figure again. So what we are doing here is that this u here is sorry yeah the u here which the highlighter is now pointing out is the axial velocity in the x direction and u plus du dy times delta y is the axial velocity at the point Q. So these are the axial or the x direction velocities and these have nothing to do with the dynamic viscosity. Using these velocities what we are finding out is first we are finding out the shear strain experienced by this vertical segment PQ and then dividing that shear strain by the time interval over which this happens we are finding the rate of shear strain which then turns out to be this du dy. So up to here there is no mention of dynamic viscosity at all here all that we are doing is we are calculating the rate of shear strain in this particular simplified situation of a 1D shear flow. Now if you are asking me why u is only a function of y and u is not a function of x. I will answer this in the fashion that this is what we will call a fully developed situation and in a couple of days of time we will actually talk about the nature of fully developed flows little bit in more detail. At this point you just assume that in a fully developed situation the axial velocity is independent of the axial direction or axial distance. It can be only a function of the lateral distance and that is why u has been chosen to be a function of only y coordinate but u is not a function of x coordinate. I wonder if that answers your question. Sir what is the utility of devising the term kinematic viscosity when as we have defined here dynamic viscosity and understood the significance we have understood the physical significance of dynamic viscosity. As viscosity is broken into dynamic and kinematics so what is the significance of the or why we have devised the term kinematic viscosity and where it is used and like what is the utility of it. Yes, so the question is about the use of term called kinematic viscosity and what is the significance of kinematic viscosity. The kinematic viscosity is defined as the dynamic viscosity divided by the density of the fluid. So, it is mu divided by rho and in principle physically speaking there is no difference in terms of what dynamic viscosity does and what kinematic viscosity does. Both of them are related to momentum transfer within a fluid. So, in some cases what happens in especially in incompressible flows or constant density flows you will see that the governing equations are such that you can divide the governing equations completely through the density and then you end up generating this kinematic viscosity in one of the terms. We will see that in a few days time. However when it comes to the physical interpretation both of them more or less say the same thing namely they tell how momentum is transferred within a fluid and that way there is no difference between those two. Okay, thank you sir. Thank you. Nitte Meenakshi please go ahead if you have a question. Whereas in the example number four why haven't we considered the frictional losses through pipes because in order to conserve we need to consider all the losses also. So, as we come across frictional losses through the pipes flow through pipes why haven't we considered the frictional losses here? So, the question is about whether frictional losses have been considered or not considered in one of the examples. Can you please repeat which example it is? Example number four. Yes, so the see in this case the frictional losses have not been mentioned at all because we are dealing only with mass and momentum conservation. We have not talked about the energy equation at all so far. So, here it was only to deal with what sort of forces the fluid is exerting on the pipe and vice versa. When it comes to the frictional losses as you are mentioning we will we can utilize the energy equation and in the energy equation we incorporate the frictional loss terms. In any case the frictional loss terms is actually inbuilt in the numbers that have been shown in this example in the sense that the frictional loss will cause a pressure loss okay from the inlet to the outlet. So, the pressures that have been given in this case are 400 kPa at the inlet and 300 kPa at the outlet. So, in that sense you can say that the frictional losses through the pipe have already been incorporated by mentioning a pressure loss over the length that has been shown. Uniform pressure there won't be any frictional losses or we neglect the time frictional losses if they are given uniform pressure across the pipe. Uniform pressure. Otherwise it will be rated as an ideal flow wherein there will be no frictional losses. No, no, no this is not an ideal flow. This is indeed a real flow. Only thing is that the mention of frictional losses is not explicitly done here. What has been done is that the problem design is such that there is a pressure drop from the inlet to the outlet as you can see from here 400 kPa to 300 kPa and this pressure loss is supposed to be occurring because of the action of viscosity which will result in frictional losses. So, even though we have not specifically mentioned the frictional losses here because there is a delta P pressure loss from the inlet to the outlet those are supposed to be acting anyway. So, they are built into the problem is what I am trying to say. Kolhapur. Yeah, can I say Kolhapur go ahead please if you have a question. Sir, liquid and gases. What if the solid comes into the picture? Will it, will all the equations follow the same or remains same or there will be any changes? Okay. So, the question is about multi-component flow in some sense where you have solid particles also embedded within the flow and the question is whether anything will change in terms of the equations or not. So, the answer is that yes actually you need to modify the governing equations because there will be terms which will have to be taken into account which take into action or which take into consideration the momentum exchange between the fluid and the solid particles that are suspended in the fluid. So, those have to be incorporated in the governing equations itself. So, that the models that you start with as far as the description of the problem is concerned are going to be different than what we are considering here. In this particular course we are focusing only on a single phase fluid whether it is a gas or whether it is a liquid. We are not really talking about multi-component situations, but the answer is yes there will be changes and the changes are first reflected in the governing equations themselves. The governing equations have to be modified to take into account the momentum exchange and perhaps the energy exchange that occurs between the solid phase and the liquid phase or the fluid phase. DK, T, A, C, S, L, Karanjee please go on. In plastic injection flow analysis, in plastic injection molding process? The question is how CFD is useful in plastic injection molding process? Unfortunately, I am not at all familiar with the manufacturing processes such as what you mentioned and in principle it appears that there will be some sort of a fluid flow involved and therefore, the kinds of equations that we are dealing with should be applicable assuming that the liquid or the fluid that is involved in there follows the Newtonian law reasonably accurately. If the Newtonian law is not obeyed by the fluid, then the entire set of governing equations has to be rewritten for the kind of fluid that we are dealing with in that particular situation. Once you have the correct set of governing equations, then the methods of CFD can work on those governing equations as well without any change. So, that is the best answer that I can provide right now because I am really not aware of any details in a manufacturing process such as that. Sorry about that. It is moving ahead in the mold then its temperature is goes on decreasing. So, its viscosity goes on increasing. So, its viscosity is not constant. So, how we can apply that equation? Yes. So, the question is actually related to temperature dependent viscosity in a fluid and that is not necessarily a problem at all. In general, if you know how the viscosity of the fluid behaves with temperature, somehow if you have characterized that separately or if the data is available in some other literature, you can incorporate that temperature dependence on viscosity in your CFD analysis and you can perform the CFD analysis. So, having a temperature dependent viscosity is not necessarily a problem at all because that is routinely taken care of by the CFD people. Gajanan Maharaj, College of Engineering, please go ahead if you have a question. Publication flow like gas and liquid, which approach we should use for solving it, either Eulerian or Lagrelyan? The question is about a multi-component flow again when we are dealing with a gas flow with solid particles embedded in it and the question is whether we can employ Eulerian or Lagrangian approach to it. So, actually the question is out of the scope as far as this course is concerned, but let me answer it in the best manner that I can. When it comes to such flows, when you have gas flow and then solid particles embedded in it, there is typically an approach called a Lagrangian-Eulerian combined approach where the fluid flow is solved using the Eulerian approach and the solid particles are tracked with the Lagrangian approach. So, it is what is called as a many times it is what is called as an ALE which is the arbitrary Lagrangian Eulerian approach which is the most common approach used for solving such problems and to the best of my knowledge some of the commercial codes do have this kind of a feature available. However, for our course it is completely out of scope and we will deal with only a single phase fluid flow in this case. I hope that answers your question to some extent. I have one question sir, 13, 13 in that case we have gone through the drag force. Suppose we have to go for the lift force, then how the equations will be different for that case. So, the question is about example number 3 in integral analysis and the question is that we have calculated drag force in this situation. If a lift force was to be obtained, how we can proceed with that? So, it turns out that if you look at this particular example, because it is symmetric about a horizontal central line, the lift in this case is exactly equal to 0. So, in this particular case there will not be any lift force on the object. If it was asymmetric in terms of the flow about the central line, then we will have to think about a lift type force and in which case what we have to do is we have to employ a y direction momentum equation. So, the y direction momentum equation appropriately incorporated should give you a lift force, but in this particular case, because the problem is symmetric about the horizontal line here, horizontal center line, there will be no lift force whatsoever. So, the lift here will be 0. Yes, Amruta School of Engineering column, please go ahead. First of all thank you very much for kindly delivering us very good lecture on food mechanics. I have a question. Of course it is not connected to none of the examples here worked out today. That is, suppose if we take a flow over a cylinder, a circular cylinder, a flow is taking place, of course we know there will be some separation, there will be some shedding of vortices in the downstream. So, applying to this situation, how will you consider the conservative equations? That is, on the upstream side there is only plain flow that is potential flow. Downstream, the flow takes place in three modes, one as vortices, one as a liquid, third one as potential flow. So, there are three modes of mass flows. So, when you consider the conservation equations, how will you treat the conservative equations? Would you please? Yes. So, the question is about flow across a cylinder and what is getting pointed out here is that it is a fairly complex situation in the sense that upstream of the cylinder it will be a plain flow approaching the cylinder, whereas due to the separation over the surface of the cylinder, there will be a vortex shedding at the back of it and then there will be a wake flow and then away from the wake flow there will be again a potential flow sort of a situation. So, how to handle a situation such as this? So, clearly if you want to analyze a problem like this, you have to resort to a CFD analysis which will involve the complete Navier-Stokes equations and if you solve the complete Navier-Stokes equations for a problem like that, all these flow domains that you are referring to will automatically get captured in the appropriate places. In fact, Professor Sharma who is going to deliver the second half of this workshop is actually, he has actually done several such computations for bluff bodies including flow across cylinders. So, perhaps in due course of time what he will show is some of the results from the CFD simulations that he is doing, but in particular what he does is that it is the complete Navier-Stokes system that needs to be solved subject to the appropriate boundary conditions and automatically all these flow features that you are referring to will get captured through a detailed flow simulation. Thank you very much. Yes, VIT Valor please go ahead. Can I go to the example 4? Sir, this axial force, where is acting on the control volume? Yes, the question is where is the axial force acting on the control volume? See, the way this control volume is chosen, we are actually choosing it such that it is cutting across the pipe over the length of interest. So, in principle what is happening is that because of this cutting of the pipe, there will be a reaction from the pipe onto the control volume and that is what is getting calculated as we have worked out the solution in this case. So, if you see the last part on this slide, the reaction which is getting calculated in this particular fashion is the x direction reaction on the control volume due to its interaction with the pipe because the way it has been chosen it is cutting the pipe and the idea is that using the Newton's third law of motion, this force will be exactly equal to the negative of the axial force that the pipe is experiencing because of the fluid that is flowing through it. See, one other way of solving this problem is you choose this control volume such that the top and the bottom surface of the control volume are essentially coinciding with the pipe walls, but they are just inside the pipe walls. In which case what you can argue is that the force that is coming onto the control volume is essentially the frictional force or the reaction of the frictional force let us say onto the control volume and then the negative of that will be the axial force on the pipe. That is one other way of doing it as well. I hope that answers your question. Yes, Nirmah, please go ahead. Sir, I want to know that through porous media, will the flow will be always non-uniform or to you sir? So, the question is about flow through porous medium and if the flow will be always non-uniform, I am not really in a position to answer that because I have never done flow through a porous medium kind of a simulation. However, it appears to me that in general there should be non-uniformity involved if you are dealing with flow through a porous medium. Unfortunately, flow through transport through porous media falls completely out of the scope of this course. So, I may not be able to answer it completely at all, but I think based on what generally I know about this there will be non-uniformity involved in the flow, if it is through a porous medium. That is why I was in doubt that it is a steady non-uniform flow or it will be steady uniform flow. Yes, so the question is coming because the example number 4 dealt with a porous wall pipe. See actually this is a very special situation in the sense that the flow that we are talking about really is happening inside the pipe and it is a completely homogeneous situation, there is no porous medium there. So, the only porous medium that the flow is seeing is the pipe walls and it is not really coming into the analysis as such. What I was trying to answer is if the entire domain within the pipe let us say in this case was a porous medium and then if you were talking about flow through that then whether we will have a non-uniformity or not. But in this particular case the pipe really is completely open and there is no porous medium inside the pipe, the only porous medium is on the walls and that is really not affecting the uniformity of the flow within the pipe. But sir if it is instead of constant cross section, if it is varying cross section then it will affect. Yes, see typically even in this case to assume that the flow is uniform is a fairly big assumption. In general when you are dealing with flows in pipes whether they are porous walled or not, you will not really get a uniform flow type of situation. However for the purpose of simplicity in an example such as this we assume that the flow is uniform across the cross section. In general it will not be the case at all whether the pipe is porous walled or not. Sir what is the main point wise most distribution of the property? Yes, so the question is what is meant by point wise smooth distribution of property. So this refers to our continuum discussion right at the beginning when we said that we will assume that our fluid medium is a continuum medium so that the fluid matter is continuously distributed in the domain of interest. So here if you look at two points within the fluid domain which are very close to each other and if you want to monitor the property such as say density or pressure at these two points which are very close to each other, you will see that you will typically not get a huge jump between the pressure value or the density value from one point to the neighboring point and to the neighboring point and so on. So if you want to represent a pressure as a variable or density as a variable through a mathematical function that mathematical function of pressure as a function of x y z will be a smoothly varying function within the domain. So the value of p or rho or a temperature will vary smoothly from one point to the neighboring point and not drastically is what I was trying to point out. So what is getting pointed out immediately is that the continuum assumption cannot be utilized across a normal shock which is a phenomenon that occurs in a compressible flow type situation where all properties such as density, pressure etc. will change by very very large amount in a very short distance and that is absolutely correct. So within the shock region you cannot employ a continuum assumption and you have to resort to a molecular simulation if you want to actually capture the structure of a shock. In example number three it is written that pressure at those two stations 1 and 2 same so then my question is does flow is possible if the pressure between the two stations remains the same. Yes, so the question is related to example number 3 where in the problem statement it was given that pressure set sections 1 and 2 are the same. The question is whether this is a reasonable realistic assumption or not and the answer is actually it is not. It is absolutely incorrect to assume that the pressure at section 1 and section 2 are going to be more or less the same. In fact, typically you will see that the pressure at section 1 will be higher than how it is at section 2. However, for the purpose of simplicity in an example such as this where the objective was just to sort of show how to implement the integral mass and momentum balances to simplify the situation it has been given that just go ahead and assume that the pressure is the same at section 1 and section 2 that is about it. Thank you. Yes, PVPS Vijay Vada please go ahead. Question where I want to see the role of CFP the application is I want to find out the time to calculate time to for a cooling of a hot coffee in which I am using a sugar and stirring it with a spoon. So, what are the different mechanisms involved with that is it apart from heat transfer mass transfer does CFP has a role in that that is my question. Yes, so the question is pertaining to a specific situation of cooling of a coffee mass contained in a cup when someone is stirring it whether a CFD type simulation can be done of that and the answer is yes it is actually even though we do it on a regular basis day in and day out the fluid mechanics of stirring a coffee contained in a cup is actually very very complex. The reason is one of the things that you may have to consider here is that the flow may be highly turbulent in addition to some other things that are happening. So, for example, as you mentioned if there is some sugar particles in the coffee cup there is perhaps a mass transfer happening as the sugar is getting dissolved in the coffee. So, how much of that affects the flow is another question whether the dissolving of the sugar is actually affecting the flow significantly or not and if it is not perhaps we do not have to bother about this. So, these are what we will call modeling inputs into first deciding what sort of a set of governing equations that we will need to have in order for a problem such as this to be simulated and then because it is a fairly complex problem as far as the present course is concerned we cannot really talk about it much at all, but in principle a CFD simulation can be done to figure out how the coffee will then cool down as you keep stirring about it. The fact is that most likely you will have to resort to a commercial solver to try and do this kind of simulation because the commercial solvers will be fairly sophisticated to try and do this kind of simulation, but that is all that I can say at this point really. Yes, NIT Varangal please go ahead. Sir, one R and D related and small doubts are ahead. So, from the morning itself we are discussing about generally about the convincing systems in CFD generally. Can you apply this CFD analysis to non-conventional energy sources just like solar energy applications or thermal systems, solar heaters like solar cookers like this. Is this concept useful in this area sir? Yes, so the question is about whether we can employ a CFD analysis for the design and let us say analysis of non-conventional energy related equipment such as solar heaters or something of that sort and the answer is yes absolutely there is no question about that. In fact, if you refer to recent literature in many of these energy related journals you will see that people are routinely utilizing CFD analysis for flow inside let us say a parabolic crop collector or any other types of situation including radiation heat transfer. So, people have employed all these kinds of CFD applications for the analysis of non-conventional energy sources as well. So, there is no problem in that. Next question from Michael. Yes, if possible how to model the governing equation considering the flow conditions and the material properties whenever flows are going to be considered in pipes? Yeah, can you please repeat the last part of your question? Yeah, so for example if you are going to be considered the fourth example in the fourth example I would like to generate the governing equation for the consideration of flow conditions as well as flow conditions as well as material properties. If possible can you suggest how I am going to be considered the governing equation? I am sorry I am still unable to understand the last part of the question. So, you are referring to flow through a pipe such as what is shown in example number 4, but then the specific question you are asking is how to incorporate something which I am not getting I am sorry can you please once repeat that again? Next to that fluid structure interaction governing equation for the fourth example. So, the question is about fluid structure interaction type problem whether we can use CFD for fluid structure interaction type problem what sort of governing equations to use is that correct? Yes, so people are using CFD for fluid structure interaction type problems. So, one example I can give you where what we are doing is if there is a thickness to the pipe wall here and there is a high pressure liquid that is flowing through the pipe what we do is we solve the fluid flow problem and we obtain the pressure distribution that is going to be existing on the pipe wall from the inside and then use that pressure distribution as a boundary condition and then solve the structural analysis problem within the pipe wall using the corresponding governing equations from solid mechanics. So, that is one way of doing it which is what we will call a decoupled way meaning that we are first solving the fluid flow problem completely without regard to the solids problem and then utilizing the results from the fluids problem as a boundary condition to solve the governing equations for the solid problem. So, this is one standard way of doing fluid structure interaction problems where you are essentially decoupling the fluid problem from the solid problem and using only the boundary conditions that are generated as a result of the fluid solution for the solid problem, but yes these kinds of applications people are solving with CFD type applications. So, with this we have come to the end of today the first day. Thank you and we will see you tomorrow.