 We will come all of you to the question answer session, we are eagerly waiting for your questions especially we are having after a long time. Before that I would like to introduce the TA to this course who has done an excellent job as far as the lab session is concerned, he has single handedly developed all the codes for this course, he is Vishay Shagraval, he has done his master's years and now he is working with us. We are happy to say that all the codes have been developed by him, so let us go to the all. Mofakam ja college. Sir please refer to topic number 1 and slide number 73, see whether the flow is along the axis of the cylinder or it is, or it is in the theta direction of the cylinder. The flow direction is horizontal direction, it is, so the inlet flow is horizontal although when it reaches close to the cylinder it moves up and down then it becomes in velocities in the angular direction, but the inlet velocity is horizontal, is this answers your question? Is it a flow over the cylinder or is it a pipe or cylinder? It is a cylinder, it is a solid cylinder, we are taking a two dimensional flow and the length of the cylinder perpendicular to this plane is infinite, it is an external flow, it is not an internal flow, external flow across a circular solid cylinder. In that case the flow direction, I mean the description is not so easy the way it is written and it becomes more complicated. I think what you are trying to point out is that this is the problem which is not in the Cartesian or a cylindrical coordinate system because the there are two domains outer domain which is Cartesian in nature and the inner domain which is circular in nature. So, we need to use a body fitted curvilinear grid which are not aligned along the any coordinate system. So, we need to have a complex geometric formulation, whether this answers your question I am not sure about it. The cylinder it is written free slip and no slip, so what is the condition that is persisting there? Ok, this question is the boundary conditions which I am showing here is no slip here on the surface of the cylinder and free slip on the top and bottom wall of the boundary. This is a solid surface on a solid surface you always have a no slip boundary condition. So, I hope this is fine to you now whenever we do a CFD competition even if it is an external flow. So, when we take an external flow strictly speaking the size of the domain is infinite. However, we have to limit to region in space to do a CFD simulation. So, we need to close this circular cylinder. So, there is a bottom wall and the top wall of the domain where we have to use a boundary condition and you can also use u is equals to u infinity but I had mentioned that this free slip boundary condition is better as compared to u is equals to infinity on the top and the bottom wall. So, this is a no slip boundary condition this is what is called as a free slip boundary condition. Note that however this is not an internal flow this boundary condition is used for external flow only. Sir, please refer to topic number 2 and slide number 36 and slide number 36 sir could you just give a brief clarification of the calculations. I think the question is on the values which I am showing you here. So, what I am saying is let us take this is just a sample calculation let me show the problem first. In this case, we are taking the length of the domain as 5. So, in this we are doing 5 equal divisions so delta x is equal to 1. So, this is so if I tell the x coordinate this is x equals to 0 this is x is equals to 0.5 this is 1.5 this is 2.5 this is 3.5 this is 4.5 and this is 5. So, with this distribution of the x coordinate at different grade points just to show a sample calculation we are taking a case where let us suppose the surface area as it is a one-dimensional problem the dimension perpendicular in the vertical direction as well as perpendicular to this plane the dimensions we are taking unity. So, surface area you can easily take as unity for simplicity in the calculations in the number distribution we are just taking the thermo physical property density specific it conductivity as unity delta x in this problem the way have we have discretized delta x you know it is equal to 1 time step also for simplicity we are taking unity. So, if you substitute this values into this discretized equation you end up with an equation where temperature at a new time level is equals to temperature of the old time level plus the difference of the heat fluxes on the west face of the control volume and on the east face of the control volume. So, what I to show the sample calculation this is the boundary condition 0 degree centigrade on the left side 100 degree centigrade on the right side this is the initial condition where let us say initially the plate everywhere it is at 50 degree centigrade. So, you have 5 yellow circles corresponding to which you have here you have 50 degree centigrade. Now, then what we are doing is as I said that the if you go by law of conservation of energy if you know this temperature distribution we can calculate the heat fluxes at the different faces. Now, what are the faces? So, if you look into the faces of the control volume if you want to calculate how many control volumes you have this is the first real control volume this is the border control volume this is the first real control volume this is second this is third this is fourth and this is fifth. So, you have 5 interior control volumes. So, and how many faces are there in this domain this is the first face west face of this control volume this is the east face of this control volume, but this becomes the west face of this control volume. So, this is a second face which is a common face this is a third face this is a fourth face this is a fifth face and this is a sixth face. So, you see 6 different values of q x and the way we calculate the q x when you want to calculate q x at a at a face which is which is interior because note that there are 6 faces, but 2 faces are lying at the boundary and there are 4 faces which are showed by the small red lines. So, whenever you want to calculate q x at the red face the calculation is that you take the temperature of this point minus temperature of this point divided by the distance between the these 2 point which is delta x delta x in this problem is taken as unity. So, you have in this case when you start with an initial condition you take the difference it comes out to be 0. So, this is 0 here here also it is 0 here also it is 0 at this face also it will come out to be 0, but when you look into the face which is on the left side left boundary to calculate dT by dx we here what we are doing is we are using a first order method which is called as forward difference on this side we use a forward difference. So, to calculate dT by dx it is 50 minus 0 divided by here the distance is half so it will be 1 delta x is 1 delta x by 2 will be 0.5. So, 50 divided by 0.5 give you 100, but when you want to calculate q x it is minus k. So, k is taken as unity so it becomes minus 100 when you go on the right hand side to calculate dT by dx here again I would like to point out here you have neighbors only on the one side one side here you have you did forward difference here you have to do what you call as a backward difference. So, here we are using finite difference method and by backward difference what the way we calculate is when to calculate dT by dx it will be 100 minus 50 divided by here again distance is 0.5 which is 50 divided by 0.5 will be 100 when you multiply by minus k it becomes minus 100. So, this way you calculate q x once you have calculated q x then you go to a control volume 1 by 1 all this boundary values will not change. So, this 0 and 100 will remain same what will change will be the value of temperature at this 5 yellow circles not to calculate this we use this equation when we use this equation what is tpn tpn is 50 degree centigrade what is qw it is minus 100 what is qe in this equation it is 0. So, if you use that you get minus 50 degree centigrade similarly you go to the next control volume what is the qw here 0 qe 0 tpn 50. So, that way you calculate all this temperature once you have calculated this temperature you have computed the temperature of the first time step delta T which in this case we are taking as unity. So, this completes the first time step competition in two stages first heat flux second temperature heat computation then you go to the next time step. So, from here you get first picture of your let us say animation or a movie from here you get the second picture then you from this temperature distribution again you calculate qx here what I am showing you here is what is basically an explicit method. So, the qx we are calculating using the temperatures of the previous time level note that. So, once you have calculated temperature of the first time step then we do the same thing first we calculate qx using the procedure which I mentioned then we calculate temperature and keep doing this. So, note that we use this equation and obtain this value any other question. Thank you professor now no more questions from this hand easy college this ask the question. Can I have professor Puranik. Yeah sure. Yeah good evening professor in the differential analysis slide number 15 regarding this sub script notation is it only for normal stress or it to apply for shear stress. Yes. So, the question is on differential analysis slide number 15 and what we have shown here is the sign convention for the surface stresses and the question is whether the notation is applicable for shear stresses also or it is applicable only for normal stresses. So, the answer is that yes it is applicable for shear stresses as well. In fact, at the bottom right of the slide what I have shown is an example of what we call a positive shear stress component. So, what is shown here is it is a vertical surface and vertical surface with the unit surface normal for the surface pointing in the positive x direction. And the shear stress which is acting on the vertical surface is shown in the vertical that is y direction and the sense or direction in which the shear stresses acting is positive y axis. So, since both the surface normal is pointing in a positive coordinate axis and the shear stress is also pointing out in a positive coordinate axis, positive x for the surface normal positive y for the shear stress we call this situation a positive shear stress. Let me just so that basically answers your question I hope. In general this sign convention that I have explained in this differential analysis has been taken as I mentioned during the lectures directly from the sign convention that we have in case of solid mechanics also. So, there is absolutely no difference between a solid mechanics stress analysis type sign convention and what we are doing here. So, just to re-emphasize that point thank you. Next question regarding this conservative and non-conservative form I need a small clarification on that like when we have this substantial derivative is there any physical significance what we can explain to our students the conservative or non-conservative means in this physical significance can you please elaborate the physical significance thank you. Yes, so the next question is on conservative versus non-conservative ways of expressing the governing equation and if we can clearly point out to our students any specific physical significance if we have a substantial derivative in the governing equation. Yes, you can we actually discussed this a few times but let me again point that out. If we are employing a Lagrangian way of deriving governing equations we will inherently or automatically end up generating the equation in a non-conservative form and you will see that it will be containing the substantial derivative. Let me quickly try to point this out slide number 6 which is now projected on the screen right at the bottom what we are doing here is that we are deriving the governing equation for conservation of mass by employing the Lagrangian approach specifically what we are doing is we are following a given fluid particle which will be always containing the same amount of mass and therefore as we follow this particle in the fluid what we say is that we are employing our Lagrangian approach and therefore the substantial derivative of the content of the mass within this fluid particle will be equal to 0. So in this way the conservation of mass statement was derived and as you can see the final expression came in the form of a substantial derivative. On the other hand if you employed a balance statement as we had done before where we identify a control volume and therefore we are employing this as a Eulerian approach when we employ the balance statement and the difference in the mass flux coming in minus the mass I should say mass flow rates going out is what we say is what is getting stored etc. Then we automatically generate the conservative form of the same governing equation. So to just summarize then non-conservative will always come about if you employ Lagrangian approach and it will always contain substantial derivative which is necessarily employing that we are employing that governing equation or the conservation law to a particular fluid particle by following that particle that is the way to express it. Thank you. Last question the finite different solution slide number 11 about this tridiagonal matrix algorithm in this on the slide you are mentioning during your lecture that subscripts i, that is a not clearly visible like is it j minus 1 or is it i comma j minus 1 or yes so can you please clarify. Yes, yes sure. So the question is on this TDMA or the Thomas algorithm which is the Gauss elimination algorithm for a tridiagonal system and the subscripts were not clear is what the point is. So yes I will try to point this out. So what we do in case of a tridiagonal matrix algorithm is we simply replace the diagonal elements by changing their values according to the expression given on the on the screen in this forward elimination process. So capital A subscript i comma i which is where my highlighter is standing is simply the diagonal element in the i comma i location. So if i is equal to 2 we are talking about the matrix element at 2 comma 2. If you go back to the matrix representation you will see that all main element sorry main diagonal locations have the value of the element equal to b. So they are all the same really. So in each of these situations along the main diagonal if you look at a 2 comma 2 or a 3 comma 3 or 4 comma 4 or whatever the element value is always b to begin with and then that element value is changed as its old value minus the ratio is a i comma i minus 1 divided by a i minus 1 comma i minus 1 and that ratio is multiplying a i comma sorry a i minus 1 comma i. So there is no j here everything is in terms of i is only. Thank you sir. One small observation towards professor Sharma regarding this animation what he shown it was very effective for the understanding. So can we get those animations so that we can use for our lectures. Yes let me pass the mic on to professor Sharma. Thank you. Yes I will be happy to share the animations to all of you what I am planning I thought about it and I am planning to create a video of those animations and put it on the model and I will be feel I will be happy if you have any questions on the procedure how to create those animations. Thank you. Yeah thank you professor. Amrita Kolam go ahead. So my my question is with regard to slide number 75 topic 1 on integral parameters. In that the expression for integral expressions for drag force and lift force are given. So in that expression d is equal to integral n dot sigma dot i. So from my understanding n is the unit vector normal to the surface. So if the surface is horizontal then n would be j. So j dot i would become 0. So that means drag force is 0 if the horizontal plate is so that is my doubt I should be replaced by j. Let us go to the expression for the drag force. His question is I have taken an example for over a flat plate and I am showing you that the drag force is given by this expression. Now here I am when I am doing calculating the drag force I am let us go to the expression. So the expression is n dot sigma now n dot sigma is the expression for drag force is this a. Now a sigma xx plus b sigma yx. Now what are this a's and b's in case of a flow over a flat plate? So flow over a flat plate a is equal to 0 and b is equal to plus 1 for the top surface and minus 1 for the bottom surface. So in this case a will come out to be 0 and you will get b which will be plus 1 or minus 1 sigma yx. Now the sigma yx is equal to why I am showing you the 2 twice because you have drag force on the top surface as well as on the bottom surface. So this is sigma yx it is mu del u by del y plus del v by del x but on this solid surface at all del v by del x is equal to 0. So you only have del u by del y so it is not equal to 0 because here a is equal to 0 and b is equal to plus 1 on the top surface and we are using this expression. Now my question is as you have elaborated on this implicit and explicit solution methods can you explain this segregated and coupled solver with some application? This question is I discussed the explicit method and implicit method. Can I discuss the segregated solver and coupled solver? This terms comes into picture when we solve the Neuer-Stokes equation. I would suggest you to please hold on till I come to that topic because right now if I discuss I had not told you the background of those so please hold on to that point I will discuss that this question I will try to cover this topic during that lecture. Can I ask to Professor Pyranox? Yes please. So slide number 22 of differential analysis. So it is very easy for us to understand this pressure contribution offering normal stresses but can you explain how this viscous contribution will offer normal stress? Yeah so the question is on slide number 22 differential analysis right at the top of the slide I have shown the normal stress as getting a contribution from pressure as well as from a viscous contribution and the question is it is relatively easy to understand the contribution from pressure but it is not that easy to understand the contribution from the viscous or the viscous contribution to the normal stress and if we can have a little discussion on that. Actually that is a really good question. Let me say that I was trying to avoid this question to be honest with you it is a very important part of the derivation of the fluid mechanics in particular these momentum equations. To be honest with you again if you want to understand the complete development of these derivations we need to go into the details of the Stokes analysis. So let me try to briefly describe what the Stokes analysis was although we have not really gone into the detail. So if you go back to the previous slide I have written here Stokes's assumptions and the fundamental assumption on which the Stokes's relations are worked out is that the stress is proportional to the strain rate which is simply a Newton's law of viscosity. However in general we saw that there are you can imagine that there were let me go back to yeah here yeah here. If you see we remarked that out of these nine components in case of a two-dimensional situation for example only six are independent entries you can say here. So those six independent entries each of those is expressed as a in terms of the strain rate in different directions. So what ends up happening is that you generate a very large system of equations where each stress component is expressed as a linear combination of all sorts of strain rates that you can think about in the flow situation. So this is what the basis of the Stokes's development is and then as you keep on simplifying you realize that many of those constants which in that linear combination result will actually drop out under the assumption of this isotropic behavior and finally what is left out is something that was shown on the top of the slide number 23. However if you want to simply have some sort of a physical feel for a viscous contribution what is what can be pointed out is that you imagine a coordinate transformation into which any surface can be made into a normal surface. So in general any surface has a normal stress and a viscous stress but in the form of a coordinate transformation any of that surface can be considered to be a normal surface in some rotated coordinate system and the state of stress should remain the same in the rotated coordinate system as well so that there happens to be a viscous component to the normal stress. Unfortunately I cannot get into any more detail than this other than pointing out the fact that we have to go to that Stokes's development in its entirety to understand where this viscous contribution to the normal stress is coming from. But roughly speaking what you can imagine is that any surface which will have normal as well as shear stress can be considered to be only a normal surface in some rotated coordinate system. So therefore to keep the state of stress the same you always will have some sort of a viscous contribution to the normal stress when you think about the surface in the rotated coordinate system. That's all really I can point out with the limited background that we have covered. What I will do is I will put up on Moodle a small discussion where I will point out a few references where you can go and read the Stokes's development in completeness where you will actually get to the bottom of this issue in completeness. Thank you. Sir exact solution 22. So we have obtained this du by dy at y equal to h as approximately equal to 0 by comparing the viscosity that mu times of air and the liquid. Can the same be obtained by considering the fact that at the free surface u will be u max and by maximum minimum in calculus du by dy shouldn't it be 0 there since it is maximum u is maximum at the free surface. So the point that is getting raised here is exact solution slide number 22 the boundary condition on the free surface was initially written in terms of the continuity of shear stress across the free surface and then utilizing the much higher value of viscosity for the liquid it was obtained as a simplified situation of du dy equal to 0 at y equal to h. And the question is if we can simply say that at the free surface the velocity is going to be a maximum u equal to u max and therefore by since it is a maximum automatically du dy would be equal to 0 at y equal to h. That is actually quite acceptable that is perfectly fine if you can visualize the fact that the velocity is going to be maximum at the free surface if you can do that there is absolutely nothing wrong it is the one and the same thing that what we have done at least mathematically. What I have tried to point out here is the fundamental physical boundary condition that exists at a liquid air interface if there is no curvature to the interface and the fundamental physical boundary condition is the continuity of shear stress across the interface and that gets appropriately simplified in the form that you see on the board. However what you pointed out is essentially equivalent condition so that is perfectly fine. Thank you. Sir it is a general question sir we usually consider a free vortex to be an irrotational flow even though there is there are velocity gradients in the field still we consider free vortex as irrotational so can you just clarify that. So the question is we consider free vortex as an irrotational flow and even though there are velocity gradients in the flow so if we can provide a clarification to that. Actually if you we haven't covered potential flow at all here but if you go to the potential flow theory in any of the standard text you will see that the condition of irrotationality which is the curl of the velocity field equal to 0 gets identically satisfied in the case of your irrotational vortex or the free vortex situation. So because del cross v is identically equal to 0 in the case of a free vortex we end up calling that as a irrotational vortex so I don't remember exactly the expressions are but even if the velocity gradients exist they should be such that the curl of the velocity will be equal to 0. Now we haven't explicitly written out the expression for the curl of velocity in cylindrical polar coordinates which would be required for the case that we are talking about but if you go back to it it's normally available in any of the standard fluid mechanics books the expressions for the curl of velocity. If you just open it and express the curl of velocity using the potential that describes this free vortex you will actually realize that the velocity gradients if at all they exist they will act in such a manner that finally the vorticity or the angular velocity will come out to be exactly equal to 0 and that's the way we will classify this situation as an irrotational flow situation. We have a question related to slide number 31 topic number 2. Sir in the you have made a statement that for transient problem this method that implicit approach may not be as accurate as an explicit approach can you please elaborate on this point? The question is on the statement which I am making here where I am saying that for transient problem this method may not be as accurate as an explicit approach. Note the word may which I am using here I would like to point out that in a steady state problem you have to do what we call as a grid independent study. Moreover if you want to capture true transient in fluid mechanics you have two class of problems as far as time wise variation is concerned there are problems which finally reaches to a steady state and there are certain class of problems which do not reach to a steady state they reach to an unsteady state which are of different types it could be periodic state it could be a chaotic state. So, there is a route to chaos in a transient flow situation. So, when you are want to simulate or capture a movie or an animation or we want to understand those unsteady flow features like if you want to have capture a movie of let us say flow situation when things are changing very fast and if you have a video camera you not only need a good pixel as far as the spatial resolution is concerned, but you also need a video camera with a very good frame rate also. So, analogously in computational fluid dynamics analogous to pixel we have what I called as a grid size and analogous to frame rate which is the time instant between the two consecutive picture analogously here we have what we call as a time step. So, you can appreciate that we need a video camera which is a very good frame rate to capture the true transient when the things are changing with respect to time very fast. Analogously in computational fluid dynamics when you want to capture true transient you need a very time small time step and in those cases if you use an implicit approach your temporal resolution may not be accurate enough your frame rate not may not be that accurate enough. So, that way you may not capture the transient growth in an accurate fashion because here you know the way the computational method evolves is that the data of one picture is used to calculate the data of the next picture. So, the temporal resolution may not be good enough UVU is a larger time step with an implicit approach. Thank you. Yeah, this is Balchandar Puranik I'd like to add one more comment to this discussion and would like to point out in fact that the finite difference lab session that was carried out yesterday we solved this diffusion equation, unsteady diffusion equation with both explicit as well as implicit approaches. If you want to go back and look at the implicit solution that was worked out one of the things that were tried was using different time steps in that implicit solution and along with the numerical solution the analytical solution obtained through the series was also getting plotted and superimposed on the numerical solution. So, if you want you can just go back and carefully see that if you take larger time steps in the implicit solution it doesn't actually follow the analytical solution closely. So, the accuracy in the implicit solution is lost if you if you take larger time steps which is exactly what was getting pointed out by professor Sharma right now. So, just as another point to re-emphasize you can verify that yourself through the lab session from yesterday. Thank you. So, one more thing I want to ask her like the point you made sir when we talk about implicit method we call it as unconditionally stable then I don't think it is like I don't understand how it is dependent upon the time step which we choose. Yeah, so the question is about the nomenclature used along with implicit method as unconditionally stable. See the way to interpret this stability is that no matter what time step value you choose the implicit method does not explore and it will always work and it will give you a solution whereas the explicit method actually will crash. So, that's the difference between implicit and explicit. The implicit method would always work whatever the time step value that you may provide to it. It may not give you an accurate solution that's a different thing but it is it is always going to work also to add to that very quickly remember what professor Sharma just told you a few minutes back that many times a steady state is true steady state is achieved at the end of the transient situation and if you are interested in actually capturing the true steady state of a transient problem going there quickly is what many times is required and then you can take these larger time steps using the implicit method and reach the steady state fast. Here obviously the transient solution is not going to be accurate but you don't care about that because your ultimate objective is to capture the final steady state. So that way many people utilize the implicit method as a vehicle to reach the steady state faster. Thank you. So one more so one more question sir as we have talked about the validity of continuum model sir can we have some criteria to to choose when the flow is incompressible or compressible do we have any number or such criteria? Yeah so the question is we talked about the continuum approach and a number was pointed out the Knudsen number based on which we decided the roughly when the situation is continuum or not. So similarly if there is any criterion to decide whether a flow is incompressible or compressible and yeah the the most standard way to classify the the flow being compressible or incompressible is using the the Mach number which is the ratio of velocity of fluid to the ratio of sound speed in the fluid so that that ratio that Mach number if it is roughly less than 0.3 this is again a rule of thumb but roughly it if it is less than 0.3 we treat the flow to be incompressible if the Mach number in the flow is greater than 0.3 roughly we have to treat the flow as compressible so that's the the standard rule of thumb classification. Thank you. It is it is regarding differential analysis 21 yes sir in which in the Stokes and assumptions the second assumption was isotropic fluid. So can you please explain briefly what is isotropic fluid and what is non isotropic fluid and give examples for each. So the question is on slide number 21 differential analysis where we talk about the the fluid being isotropic so what does what does that mean the the the standard meaning associated with the words isotropic is that if you look at any direction within the flow at a given point the fluid will have the same behavior in terms of the stress and strain rate relation that is what the assumption is and to some extent the assumption is a simplification no fluid in real life will behave exactly as what is described using this isotropic assumption. However most fluids come very very close the the fluid such as air or water which are our more standard fluids and most other fluids that are used as working fluids in most of the technological applications are actually conforming quite reasonably well to this isotropic nature which is again any direction within the fluid that you can think of at a point the stress and strain rate relation remains the same. To be honest with you I'm not too aware of an or a non isotropic fluid most likely my if I have to guess actually it will be in some sort of a chemical engineering situation where perhaps the the kinds of fluids that end up getting used are perhaps non isotropic. I myself never have dealt with a situation where there is a non isotropic fluid to be to be handled. As I said most of the fluids that mechanical engineers and aerospace engineers in particular will deal with are conforming to this direction independent behavior quite quite reasonably. Remember everything is a model so there is nothing absolutely true about any of it. Finally everything is a model and we say that whatever comes closest to the model is good and and we move on. Same is the case with your Navier-Stokes equations also. Thank you. Sir the second question is regarding differential analysis 33. It is on the simplified form of energy equation. So the equation which is written in the box of Salomon 33 in differential analysis from that you subtracted momentum equation by taking divergence on it and obtained an equation which is given in Salomon 34 of differential analysis. So in the momentum equation you are not considered but viscous terms and body forms. I don't know I could not follow exactly how it was going on. Can you please explain? Yeah so the question is on the energy equation derivation in the differential analysis and in particular the one manipulation that we did on slide number 33 where we obtained the equation for the total specific energy and then subtracted what we called a mechanical energy equation that was formed by taking the dot product of the momentum equation with velocity. And the question is in the momentum equation we are taking only the pressure gradient term and not anything else the viscous terms and the body force term and the question is why. So actually it's simply an assumption or a simplification under which we are carrying out this derivation. Let me go back to the slide number 30 where we have listed the simplifying assumptions at the bottom if you now see where my highlighter is. The first point is what says that we are neglecting the viscous forces altogether in comparison with other terms. This is again a simplification and the simplification is to be interpreted in the fashion that for low speed flows which typically exist in mechanical engineering I just talked about that Mach number being less than 0.3 etc in general. If you want to look at the Mach numbers that the standard mechanical engineering flows are working at it is actually very very close to 0 in fact. So it is something like 0.05 or 0.08 or something of that sort. In situations like this what we call these as low speed flows and the velocity gradients are actually very very small. If the velocity gradients are small the viscous dissipation work which discusses the or which describes I should say the conversion of kinetic energy into thermal energy is actually really really negligible in comparison with other terms in the equation. So that's an implicit assumption under which we are working here. What we are saying is that we will simply assume. So under the assumption of these low speed flows wherein we can reasonably well argue that the viscous effects are altogether neglected we are working this. Similarly the body force have been simply neglected in this simplifying derivation. In the sense that the body forces will be considered to be much smaller in comparison with the pressure forces. That's the implicit built-in assumption in the entire derivation and that is why you don't see those terms in this equation here. That's all it. Thank you. Sir my last question sir. In fact I was confused because you know on the side number 35 you have like you know taken body forces into the account and whereas in the energy equation you have not taken those things in the account that's why I was having that confusion. Anyhow my third question is. Let me let me let me talk to you about that for a minute. Actually it's a good point it's a good point I'm sorry to interrupt but it's a good point that that you are bringing out on slide number 35 I have written the momentum equation including the body forces whereas in energy term I am I'm I have not included it. The idea is that the energy term the energy equation was derived as an independent equation looking at only the energy contents and even though you may have the body forces in the momentum equation when it comes to the energy calculation the idea is that the energy terms related to body forces which are essentially potential energy are negligible and therefore it has been neglected but I really want to point out that it's a very good observation and thanks for that. Thank you sir and my third question is regarding this recent chain relationships and you we like we have derived all the Navier-Stokes equations under the very simplified assumptions so can you please guide or give couple of references where we can see all these detailed analysis or in fact general analysis. Yeah so the the question is on these stress strain rates relationships or the stokes development and without really good getting into each and every of those details we have really looked at only the final expressions more or less and talked a little bit about those and the question is whether we can point out a reference here where the entire development can be seen. So the best way to to look at this is if you if you don't mind noting down as I'm writing right now I'll I'll mention two books which are which are very good to to look into both of them are actually somewhat old fluid mechanics books and one is by Shames S H A M E S Irving Shames let me write it on the board whiteboard so so this is one book and the other one which is also a somewhat old book where a nice treatment on this Stokes's development is provided is by S W Yuan and the title of the book is Foundations of Fluid Mechanics. So these two I will from my point of view recommend for detailed Stokes's development and once you start going through it you will realize why I have decided to omit that here. It's a it's a fairly long drawn out derivation and one has to have sufficient patience to go through the derivation and it's really up to the individual to to work through all the steps as as they are getting worked out so I would suggest that those who are really interested should should go to these two and perhaps they can they can follow it from the background that we have already generated here thank you. Thank you sir thank you so much and there's one more question from one of my colleagues. How the body forces are developed? So the the question is how body forces are developed so that in general the body forces are developed if if if your fluid material or any material for example is operating in some sort of a force field and that force field can be typically the gravitational force field as as what we have been we have been looking at. So the the the body force for example that we can talk about for ourselves that is if you consider the human body as a fluid particle or solid particle whichever way you want to look at. We are in a we are situated in a gravitational force field and the weight which is our mass multiplied by the gravitational acceleration is nothing but the body force that we are going to experience in this gravitational force field. So similarly any material which has a non-zero mass content if it is operating in a gravitational force field will experience a body force which will be given by the mass of that particle times the the gravitational acceleration and by by and large the the most common body force that we need to talk about in fluid mechanics is because of the the gravitational force. Some additional body forces in some very special situations can be incorporated where you have something like a charged particle or charged electrically charged fluid moving in an electrostatic field where again because of the electrical charges introduced in the fluid and the motion of those in an electrostatic field additional electromagnetic forces are generated which are also acting as body forces in the sense that they are acting over the entire mass of the fluid. So this is a couple of standard situations where we have to deal with the body forces. For all practical purposes the only important body force that most mechanical engineering people will ever face is really only the gravitational force. Thank you.