 Hello. Good afternoon sir. Yeah please go ahead. My question is regarding to the differential analysis 19. In the last line we have that tau ij is equal to tau ji. So how to physically interpret this relation? So I mean is what is actually the meaning of this relation? Yeah so the question is on differential analysis slide number 19, where we showed that the shear stresses occur in pairs. In other words tau ji equal to tau ij, where i is not equal to j of course. So tau xy equal to tau y etcetera. And the question is what is the meaning involved here or what is the physical meaning? To be honest with you, I am not too sure if there is any specific physical meaning. At least I am not aware of any specific physical meaning associated with this. This is a result that we obtain by essentially implementing what is called as the differential momentum angular momentum equation, which is what we are doing in this case that we are taking moments about the point O. And doing so we are just bringing about this result that tau yx is equal to tau xy. And the physical argument involved as you perhaps remember was that as we shrink this particle, we do not really expect for no reason the angular acceleration experience by the particle to start tending toward unnecessarily large values. So to be honest with you, I am not too sure if there is any more physical meaning associated with this analysis than what we have just talked about. Just give me a day or so then I let me go back to some of the texts and see if there is anything else. But as far as my understanding goes, it was essentially to generate this particular property of the shear stresses that they occur in pairs that is all it was. So as per I think we can interpret this relation as like that. If the particle is rotating with constant angular velocity, so it keep rotating with constant angular velocity and if it is irrotational then it means it is not rotating then the particle remains in the same state. Can we interpret this relation like that? I unfortunately do not think so I do not think you can interpret it this way because as of now there was no mention here of irrotational or irrotational motion. The only discussion here is related to the various stresses that are acting on the particle and we have identified those as the normal and viscous stresses. It so turns out that if you take this moment about the center point and so on, you can show that the property of the shear stresses is such that they occur in pairs. So that tau yx equal to tau xy etcetera. I really do not think as far as my understanding goes that it has anything to do with rotationality or irrotationality of the flow. It is just the angular acceleration just used as a vehicle to show that the shear stresses occur in pairs. At least that is what my understanding is. As I said just give me a day or so in case I come across something more informing I will inform you. My next question is regarding to the exact solution number 19. In the last line we have taken a fully developed flow so as a result of that we have del u by del x is equal to 0 and u is not a function of x. If the actually the plate is inclined and according to me that the flow in the x direction is sorry flow velocity in x direction is keep on increasing. Means if you consider two point on a plate say u 1 and u 2 and take u 2 minus u 1 upon x 2 minus x 1 that may be positive. So I think this assumption is valid or what? This is I think a good question. The question is about this problem of fully developed flow inclined down in inclined plane as shown on slide number 19. The question is basically whether this assumption of fully developed flow as we are implementing here is realistic or not. Physically speaking it is not really realistic in the sense that if you talk about a finite length inclined plane and somehow you generate this film it is very likely that it will accelerate. There is no question about that. However, the implicit assumption built into this situation is that the extent of the plate that we are talking about is very very large in the x direction. So that there is no boundary condition so to say in the x direction in which case what ends up happening is if there is no boundary condition in the x direction which was not explicitly mentioned I understand that. But let us assume that if there is no boundary condition that we can think of because the x direction is very very long then this fully developed situation in the sense that you do not have an x variation of the x velocity is actually not too bad. It is at the end of it all these situations are artificial in some sense. So we have to treat all these situations with a pinch of salt as they say none of these is really mimicking any reality that correctly is just that to point out some scoping calculations some of these assumptions are built in and we have to take the assumptions in the right spirit in the sense that we cannot really take them too seriously at the same time these are reasonably ok assumptions in the sense that they give something useful eventually which can help you understand the physics of the problem that that is all really it is in all these cases. See let me let me continue that with with a little more information if you go on to slide number 21 on this what you end up seeing here is that the entire left hand side becomes 0 because of our assumptions and what is left then is only the mu times d 2 u d y squared and the rho g sin theta term physically speaking the way to interpret this situation is that the flow is established such that it is it is flowing with a uniform velocity under the dynamic equilibrium of the body force and the viscous force. So in that sense it is to be interpreted that let us assume that you know you you have a sufficiently long length along the inclined plane. So that some sort of a dynamic equilibrium between at least two types of forces see here in particular the gravity force is important because it is actually causing the flow that is the implicit physical understanding that is built into this problem as I said none of these problems that we have talked about is really mimicking mimicking any real life situation that accurately. So in that sense your answer is also fine but we have to go with these assumptions anyway. Sir in differential analysis slide 14 you have taken a triangular section for analysis is there any special significance attached to it? Yeah so the question is back to the triangular element on differential analysis and the question is is there any special significance associated with why we have chosen a triangular element. The significance is that we wanted to point out the state of stress what is meant by state of stress and the state of stress at a point when you want to call a point in a moving fluid. I do not want to necessarily specify that point in any particular direction it has to be any random direction in that sense. So that is the reason this inclined plane is chosen at some random angle theta. So that eventually when we take the fluid element and shrink it to a very very small size we are talking about some randomly chosen direction in the fluid at a given point. And the objective was to show them that the state of stress at random in a moving fluid can be expressed completely in terms of four stress components which are acting on these two mutually perpendicular planes which are passing through essentially the same point. So the significance is to bring about the fact that we are talking about state of stress at a random point in the sense that a random orientation point and that always can be expressed equivalently in terms of the four stress components acting on the two mutually perpendicular planes passing through the same point. So that is all the physical or the significance in the in general associated with this situation. See normally you do not see this kind of analysis in standard undergraduate fluid mechanics books. If you open the book by Gupta and Gupta which I have listed as one of the one of the reference books for fluid mechanics, you will actually see this analysis there and it is a useful analysis to do to realize that how we connect a state of stress at a point in fluid which is oriented randomly with stress components on two or three mutually perpendicular planes that is all it is to it. Again in the differential analysis slide number 35 in the energy equation the work done related term P del V is dropped. So any reasons for that? The question is if the work done term P times the divergence of velocity is dropped is there any reason for it? If you look at slide number 34 as it is projected right now actually from the equation at the top here to the equation at the bottom it is not dropped. It just get converted this P times divergence velocity term gets converted into the substantial derivative of pressure when we incorporate the definition of enthalpy. So instead of internal energy so that way it is not dropped in this pair of equation. Where it is dropped is when we go to the next page and when we summarize the equations for constant density flows for constant density flows you will remember that the divergence of velocity is identically equal to 0. So only for the case of constant density flows this P dot sorry P times del dot V term is neglected or dropped. But in general that is not the case. In page 35 sorry slide 35 it is not there. That is that is precisely what I just said in slide 35 the equations are derived or rather equations are shown for constant density situation. So for constant density situation del dot V is identically 0 from our continuity equation and that is why it is dropped. The analysis up to slide number 34 does not assume constant density only when slide number 35 comes I am summarizing those for constant density. So that is the difference between material up to slide number 34 and then the result shown on slide 35. Sir can you give any specific reference for the exact solution which you discussed today for fully developed flow between infinite parallel plates where the Nusselt number is 8.24. Can you give a specific reference for getting all the steps in this? So the question is if I can suggest a reference where in the solution for that Nusselt number equal to 8.24 solution in that pair of parallel plates is worked out. Let me write it on the whiteboard. I think these two books which I am mentioning in I mean there are many more convection heat transfer books. But definitely something very very close to what we worked out is I think available in the Pusthausen and Naylor's book on convective heat transfer analysis. K's and Crawford is a very standard convection I do not remember the exact title it is either the convection heat transfer or convective heat transfer. But it is again a one of the most standard books followed for convection heat transfer. To be honest with you I do not remember if the exact solution that we worked out is available in either of those. But I know for a fact that Pusthausen's book has something very very similar maybe it is exactly the same thing that we have worked out. So these are a couple of references which which you can which you can use. Sir I have two questions. My question is we have considered the thermal conductivity in this particular case that is minus k multiplied by delta by del y divided by yes that is that is where we have considered the thermal conductivity. My basic two questions are number one should we consider this thermal conductivity as of the flow is having the relative velocity with respect to the two consecutive layers while flowing from y is equal to 0 to y is equal to h. Because it is not like like same velocity between the two layers. It is a relative velocity between the two layers. And second question is if we multiply in the numerator the conductivity should not be multiplied in the denominator also. Yeah. So the the question is on exact solution 8 where what we are doing is we are trying to show something specific to this problem in particular that minus k times dt dy is obtained at the in the denominator sorry in the numerator here. So the question is where is this k coming from. Let me try to answer this in the sense that if you look at the left most part here this is basically coming from the previous step which is which is here. What we have what we have argued is that the quantity inside the square brackets is essentially not a function of x. So therefore the derivative of that quantity with respect to y is also not a function of x. Now what is done is this derivative is actually operated on the terms inside. So let me try to let me try to go to the whiteboard and I will try to explain what I meant by this. So I split this minus sorry plus minus t of x y divided by t s of x minus t n of x. So this first bracket is completely a function of x. So the y derivative of something that is completely a function of x will go away and what will be left is this d dy operate see again denominator here is only a function of x. So as far as the differentiation with respect to y is concerned it is going to be a constant and therefore this partial differential with respect to y will only operate on the numerator and that is why we obtain first minus del t del y if you want whole thing divided by and of course this is not equal to a function of x on the right hand side t s of x minus t m of x not a function of x. So then I have simply multiplied this by k and actually then I should be multiplying here by k as well. So that step I have not shown that I have really multiplied the right hand side and left hand side let us say but k if it is a constant it is not going to change the nature of the right hand side and it will still remain as not a function of x. On the other hand minus k times d t dy if especially evaluated as a function evaluated at y equal to 0 will be identically the heat flux that is provided at the wall. This is from Fourier's law of conduction because at y equal to 0 what we have is essentially a stationary fluid if you go back at y equal to 0 which is the bottom plate what we have is very close to the wall because of the no slip condition the flow is essentially 0 value and therefore right at the wall when the heat transfer occurs from the wall into the fluid right at the wall that is it is purely by conduction and the heat flux then is given by minus k times d t dy evaluated at y equal to 0. So the idea here was or is in general to show that once you replace this minus k times d t dy as the heat flux at the wall we realize that heat flux divided by the surface temperature minus the mean temperature is actually independent of x but by definition in an internal flow situation like this the heat flux divided by the surface temperature minus the mean temperature is nothing but the convective heat transfer coefficient. So there is nothing incorrect here is just that when I multiplied this minus sorry by the conductivity k on the left hand side it is understood that it was multiplying the right hand side also. So there was some right hand side which is not a function of x let us say and because the conductivity k is assumed to be constant that nature of the right hand side is still remaining the same namely the right hand side is remaining as not a function of x. So that is what it was supposed to supposed to mean I hope that is clear little bit more now. Yes sir. Thank you sir yeah R K college yeah. So am I am I on the right slide that that you had asked something about positive nature of the shear stress on the inclined surface shear stress yeah. So I am not really convinced about my own answer that I gave you in fact I wanted to have a little bit of more discussion on that in the sense that you pointed out that whether we can look at the two components here and with respect to those two components for the shear stress if we can if we can come up with the conclusion that this can be considered as a positive shear stress. If you resolve it so let me go to my white board and draw the picture I suppose this is what you are referring to if we resolve this into these two components and then if we can if we can consider each of these two components to be positive then we can come up with the conclusion that the inclined surface shear stress is positive is that correct what you are trying to say. Yes sir yes sir these two components if we combine with normal yeah I understood so I actually understood it it is a really good thought let me let me say that and also I will say that I am not convinced about my own answer. So I am going to retract that answer and I am going to come back to this may be in a day or so with little more clarity the problem with what you are suggesting right now as I see it right now other than the fact that it is a really really good thought is that we are essentially saying then that the horizontal component can be considered to be acting on a fictitious vertical surface pointing in this fashion and the vertical component can be considered to be acting on a fictitious surface which is also pointing downwards and that is the way we combine the two to say that both of them are positive and therefore the inclined surface is positive. So I think I understood your logic and I really appreciate the thought that you have given to this is just that the only question I want to clarify to myself before I come back to you is that we can indeed combine these two fictitious surfaces to represent the inclined surface. So I have not been able to convince myself so far that the two fictitious surfaces the vertical surface and the horizontal surface can be essentially combined in some sense to represent our inclined surface. So let me give it a thought tonight and I will remember this and I will come back to you tomorrow if you do not mind. Okay no problem sir thank you very much. Yeah so but yeah I really appreciate the thought process though it is very well thought let me just make sure that if this is the convincing answer I will try to explain it to you tomorrow. So thanks a lot again. Yeah come in college Pune if you have a question please go ahead. My question is regarding application of Navier-Stokes equation not for a particular slide rather suppose we are having one fluid rather it is a gas it is moving and by chance by any mean second type of fluid also enters in the first fluid but quantity of second gas is very smaller and I want to predict motion of this second fluid inside the first fluid. Second fluid may be in very smaller quantity as compared to first fluid rather maybe it reduces to molecular level bunch of molecules so can it be possible to predict motion of this second fluid or molecules of second fluid by means of Navier-Stokes equations or shall we go for statistical thermodynamics process which method will gives us accurate result or result possibly very close to accuracy. This can be a real life problem. Yes so the question is about a situation where there is a gas flow and in this gas flow another different gas is entrained as an additional flow let us say. What is getting pointed out is that the second gas flow can be really really small in amount compared to the primary gas flow and if Navier-Stokes equations can be employed to simulate such a situation. So, this is actually a very very tough question to answer to be honest with you but let me try my best if the if as you are saying if the amount of the second gas is really in traces in the sense that it is really a few molecules as you are saying then I will have to say that you must have you will have to resort to some sort of a statistical method rather than using Navier-Stokes equation. If both of those gases are in plenty of quantity with each of them really satisfying the continuum assumptions cleanly then in Navier-Stokes equations enhanced by some sort of a mass diffusion equation can be utilized to simulate this problem but if you are saying that the other gas is in really trace amount my gut feeling is that it is probably going to require a statistical method to obtain accurate solution. Actually if we use any kind of CFD software this gives us probably somewhat good result but if we want to solve them by mathematical process I think if we are using CFD software definitely they are based on Navier-Stokes equations but if we are solving it by mathematical method as per your view I think statistical thermodynamics processes will give us accurate result. Yes that is what my opinion is yes that would be the correct thing to do. Thank you. Can you please solve the equation by putting the value of H in the first equation because I have tried that actually the second equation we are not getting actually. So the question is about converting the top equation into the second equation. The equation that I have written in terms of enthalpy and they tried it but they are not getting it. So let me see if I can do it myself. Yeah let me try that. I hope you have followed so far and that is it. Actually the next step is what you want. Thank you sir. Okay thank you. SGSITS Indore if you have a question please go ahead. Sir my question is from differential analysis slide number 223. I want to ask whether these two coefficients of viscosity the lambda coefficient of viscosity is it kinematic coefficient or if it is not then what is the physical or practical significance of the second coefficient of viscosity. Yeah so the question is on slide number 23 differential analysis where this dynamic coefficient sorry the first coefficient of viscosity or the dynamic viscosity and the second coefficient of viscosity are mentioned and the question is whether the second coefficient is called kinematic coefficient. So the answer to that is no. There is no such nomenclature associated with the second coefficient. It is simply a second coefficient. Actually I need to perhaps apologise for putting up this slide in a slightly half hazard manner in the sense that we have not really discussed the entire Stokes's development which is actually a fairly tedious business. What I have simply showed is that it turns out at the end of the analysis that the viscous component to a normal stress in a fluid appears to be having two coefficients eventually in the expression that is obtained after the Stokes's analysis is over and one of them can be shown to be the first coefficient of viscosity as it is called equal to the dynamic viscosity and there exists the second coefficient of viscosity which is simply called as the second coefficient of viscosity. So what ends up happening is that only if there is a compressible flow where the summation of the strain rates as given by E dot xx plus E dot yy plus E dot zz here which is the divergence of velocity non-zero then we need to bother about the second coefficient of viscosity. If the flow is incompressible we do not have to. One way or the other it is simply called as the second coefficient of viscosity and it will appear only if the flow is extremely compressible in nature. Typically the occurrence of second coefficient of viscosity is required to be considered only if you are looking at structures of shock waves. What I can do is so I apologise for that but right now you simply note down that there are two coefficients that show up after the Stokes's analysis is done. As far as our overall objective is concerned for the CFD we do not need to know these kinds of detail. This is a fairly tedious derivation and it needs to be studied carefully. Thank you. Thank you so much sir. Yes, Mufagamja please go ahead. Sir, in case of flow over surfaces how do we define the detachment of flow that is my first question and my second question is regarding conservation equations, continuity, momentum and energy. How do we solve them? Is it sequentially or yes, so there are two questions. One is a general fluid mechanics question and the question is for flow over surfaces how do we define a detachment of the flow? Actually we have not talked about those kinds of fluid mechanics details. The quick answer to that is this is related to essentially a boundary layer separation kind of situation. If you see when the flow in the boundary layer separates the condition that seems to be satisfied is that the shear stress at the location where there is detachment occurring essentially becomes equal to 0. So, that is the condition that will signify the beginning of detachment of the flow. So, that is as far as your first question is concerned. On the second question I will simply say that why do not you just wait for a few days and by the time we complete our workshop next week all your answers as to how we solve the Navier-Stokes equation, the momentum continuity equation and the energy equation.