 Okay, so let us continue from where we left off yesterday just to briefly revise that linear momentum idea for the differential analysis. What we had done was we had started with the integral linear momentum equation and then used the Leibniz rule and the divergence theorem to convert everything on integral on the control volume basis from which we obtained this final momentum equation which is essentially nothing but as we remarked linear momentum principle on a per unit volume basis and this was of course the vector form which was specifically written out for x direction and similarly for the y and z directions. Then what we also looked at is how to obtain a so called non-conservative form of the momentum equation starting from our conservative form and there is a little bit of math manipulation all that we did was we subtracted u multiplied by the continuity equation from the conservative form and we obtained this so called non-conservative form which if you remember if you simply multiply both sides by delta v, delta v being the volume of a fluid particle then we can interpret this nicely as f equal to ma applied to a fluid particle which is the Newton's second law of motion. So the point that one wanted to convey here was that these so called conservation form or conservative form and the non-conservative form are essentially identical you can obtain one from the other. In this form when you write it though the way it is written on the board right now namely the non-conservative form you can interpret this as a Newton's second law applied for a fluid particle. Again the volume element has been cancelled from both sides here so that we end up interpreting this as Newton's second law on a per unit volume basis. Then we introduced the ideas of surface forces and the body forces and a special shape of a fluid particle has been identified with all these stress components as positive and we explained what the notation and the sign convention was. So let us just quickly see that again. The first subscript on a stress component gives you the direction of the normal to the surface over which the stress is acting. So essentially if the first subscript here is x which means that the direction of the surface normal is in the x direction and the second subscript gives the direction of the stress component itself. So that is the double subscript notation and then we also came up with a sign convention how we are going to call something positive, how we are going to call something negative. One thing I did not mention yesterday about this positive and negative and that is that with this convention if you go back and see tensile type forces if you can think about come about as positive and compressive forces come about as negative. So that way it is physically also intuitive and it fits into our picture of what is happening. This is exactly like what we have in solid mechanics many of you I am sure are familiar with the stress analysis part of solid mechanics it is exactly the same. So now let us move on. So I am rewriting this again basically starting from this non-conservative form multiplying both sides by the volume element. What I have done here is for the x direction and y direction nothing different and here since the shape of the particle was chosen to be this triangular we are appropriately multiplying or appropriately calculating the volume element which is simply one half of the delta x times delta y. So after having multiplied by delta v on both sides what we have is again f equal to m a type representation for the fluid particle. So now what is to be done? This is a bit of algebra and I wanted you to do it so I have outlined what needs to be done. If you see the right hand side here we have capital F suffix x and capital F suffix y which are essentially the net x and y forces on the fluid particle. So let us go back to this picture and all that we need to do from here is we sum the forces in the x direction likewise we sum the forces in the y direction. So here it is pretty obvious that we have this sigma xx which is acting in the minus x direction then we have this tau yx which is also acting in the minus x direction anything else? There is the body force gx which is assumed to be acting in the positive x direction fine anything else? So now we need to resolve these two guys that sigma nn and tau nt need to be resolved in a direction that is in the x as well as in the y and those also need to be appropriately included in our summation of the x forces and summation of y forces. So that you can easily do the resolution of these into two components in terms of this theta and the geometrical parameters like delta x, delta y and delta x. Remember again these are stress components so if you want to find out the force you have to multiply by an area element so for example sigma xx then will multiply by delta y multiplied by 1 which is per unit depth correct and so on. So that is a bit of algebra which I am simply writing here needs to be done so sum the forces on the particle. Now what is going to be done is what is written out there that is you actually write out the sum of all x forces and plug them in the right hand side here in the left hand side sorry right hand side here for the x and y forces. What when you do that you will realize that all these force the terms will have some multiplying factors delta x times delta y or delta x times delta y squared or something of that sort okay I do not remember exactly what it is but some volume element will be multiplying on everything. Now what we want to do is the objective of what was happening here is to come up with a statement by which we can specify the state of stress in a moving fluid at a point okay. So to that end we are doing this we have marked all the forces we are going to employ the Newton second law of motion as is known to us and then for the stress at a point we essentially take that fluid particle and shrink it to 0. Mathematically speaking what we are going to do is we will take the limit as these dimensions delta x delta y and hence delta s all value tending to 0. If you carry out algebra what you will see is that the left hand side here okay which contains a volume element on the right hand side there will be also a volume element in the summation of the forces and when you take the limit you will see that the left hand side and the body force term which is going to be present on the right hand side actually will go to the 0 limit faster than the surface forces. So what will be left at the end of the limiting process is you will have only an equation actually couple of equations relating the sigma nn which was a normal force on that slanted surface in terms of sigma xx, sigma yy, tau xy and tau yx. Similarly there will be another equation which will be left out which will relate tau nt which is the shear force or shear stress I should say on the slant surface again as a function of these four stress components. So this is something that is worth doing as a algebra exercise with that limiting process and not all books actually have this outline. It is nicely outlined here although I think if I remember correctly their element is in this fashion, the reverse triangle if you want but in principle there is nothing different in what we are doing. So if you want you can look at it but it is very easy to do. I have hopefully outlined what needs to be done. All that you need to do is sum the forces and take the limit as delta x delta y and hence delta s tend to 0 and you will see that the left hand side which is the body force sorry which is the inertial force as we say and the body force which is embedded in the right hand side will actually go to 0 faster in the limiting process than the surface forces and you will be left out with two expressions one relating sigma nn to all these four stress components and second relating tau nt to all these stress components for a given theta obviously that theta will still be there which is that inclination angle of the inclined surface. Fine this is the final expression that you will get in terms of some mathematical form. How do you want to interpret it? The interpretation is what is written here and that is we wanted the state of stress at a point in the moving fluid which is given by that sigma nn and tau nt on an arbitrary plane. See that theta can be whatever this is exactly the way you probably have seen the Pascal's law of derivation. I think many of you immediately commented yesterday when that triangular element was shown that this is the way that Pascal's law is proved. So what you do there is that inclined surface for the Pascal's law fluid element is also at some arbitrary angle and that is what you argue that angle can be changed whichever way we want and in the limiting process what you show is that pressure is essentially the same whichever direction you look at. So in that sense it is the same argument that the state of stress at a point will be specified in terms of that sigma nn and tau nt which is a pair of normal stress and a shear stress acting on an arbitrary plane. We are talking about 2D so let me write it 2D but one can quickly generalize to 3D once we know what is happening. So whatever is in the bracket is for 3D that is an extension. In 2D flow is completely specified in terms of 4 that is for 2D, 9 for 3D stress components acting on 2 for 2D, 3 for 3D mutually orthogonal planes passing through that point. So which are these 4 stress components acting on 2 mutually orthogonal planes passing through that point? It is basically these 4 sigma xx, sigma yy, tau xy and tau y. And why do we say that those 2 mutually orthogonal planes pass through that point is the limiting process that we are talking about so that the idea that this entire fluid element shrinks to a point will bring both these orthogonal planes essentially at the same point that is the idea. As far as I remember this kind of argument is normally put forth in the stress analysis course in solid mechanics. How many of you have seen it perhaps in solid mechanics? It is exactly the same. There is no difference between what is done here for fluids and what is done for solids. So in case you have not seen it perhaps you can open a solid mechanics book also just for making sure that this is how it is done. So in 2D then we need to have 4 stress components acting on 2 mutually orthogonal planes passing through that point to specify the state of stress completely at any point. In 3D we have 9 stress components on 3 mutually perpendicular planes that are passing through the point. And usually if you see then we for a 3D situation we will represent all these 9 stress components using what is called as a stress tensor representation which is usually in this matrix form and normally you will see that the normal stresses are along the diagonal and off diagonal elements are essentially your shear stresses. And if you see along one particular row any row I have outlined just one. What you notice is that for all these stress components the first subscript is the same. So it essentially implies that stress components in a given row are stress components acting on a given plane because the first subscript is telling you the direction of the normal to the given plane. If you look at any particular column likewise the second subscript is common and the second subscript tells us the direction of the stress component itself. So if you look at any particular column it simply gives the stress components in a given direction. So the idea to introduce this is just for completeness. Many times you will not find this sort of an argument in introductory fluid mechanics textbook. For example the two that I have brought here the Potter's book and the Fox and McDonald. They do not really go through this formally here. In the second chapter for example in Fox and McDonald they simply write that a state of stress in a 3D flow is specified in terms of 9 stress components acting on 3 mutually perpendicular planes passing through that point. But where is that coming from is what I wanted to sort of convey to you that you do this formal analysis on a triangular element like this and thereby you bring about that statement. So there they will simply write a tensor here that this is the 9 stress components that are required etc. Same thing for that book but Gupta and Gupta have outlined this very nicely in my opinion. So I think it is worthwhile going through that. After this is just algebra and that is why I really want you to do it once if you have not done it before. Fine. So this is something that was slightly aside in the sense that as I said I wanted to point out the idea behind this state of stress at a point in a moving fluid. So now let us go back to our momentum equation and come up with our standard shape of the fluid element which is the rectangular delta x delta y type. And now that we know that the state of stress at a point is given in terms of those 4 stress components. What I will say is that this so called center point if you want will be considered as the reference point where I will say that the state of stress is given by all these 4 components since we are doing a 2D analysis. The body force in x and y direction are again shown here. Now the body forces of course are for the volume element of delta x times delta y. So the volume here is delta x times delta y as you can see. What I have shown here is only 1, 2, 3, 4 components. So I think I have on purpose shown 2 normal components and 2 shear components. The idea is that if I wrote all then it will become very cluttered the picture and what you can easily do is you can complete this picture yourself. So for example this normal stress here on the right hand side and this shear stress on the top. Similarly this normal stress and this shear stress you can complete. So how are they obtained anyway? I have the state of stress at point O in terms of sigma x x sigma y y tau x y and tau y x. So I simply do a Taylor expansion from O to delta x over 2 to the right, from O to delta x over 2 to the left, from O to delta y over 2 to the top and O to delta y over 2 to the bottom. So then accordingly when I go from this point on the upper side through a distance of plus delta y over 2 I need to have a plus sign here. When I go from O to delta y over 2 on the bottom side negative I have appropriately a minus sign here. And these are all again first order Taylor expansions which essentially are found to be sufficient to describe our fluid flow situations quite reasonably. I hope that you can complete the remaining 4 which is pretty straight forward. Any question on this? Fine. Now what is the use of this? Before that there is one more relation which is very useful and that gives you the characteristic of these shear stresses. So with respect to this figure here if you take moments of all forces about this center point O, summation of all moment. Clearly what is going to happen all normal forces and the body force will not contribute to the moment because they are all passing through the point O. The only contributing factors will be from the shear stress component. And if you can see the shear stress here and the shear stress here will form a couple which will be in the anticlockwise direction. Similarly the shear stress on the top surface and the one on the bottom surface will form a clockwise couple. So you can think about a net anticlockwise couple which you can find very easily and that is if you do that algebra again you will see that the net counter clockwise couple about the point O will simply come in the form of this tau xy minus tau xyx times the volume element delta x delta y. So this is a bit of algebra which you can simply do by forming the moments of these shear stresses which is fine. But what is that equal to? The net moment is equal to just by Newton's second law of motion we have net force equal to m times the acceleration. We have this as the moment of inertia about z axis because we are doing a 2D element times an angular acceleration. So this is the equivalent form let us say for moments of the Newton's second law. It turns out from calculus that the moment of inertia for a rectangular element such as this is given by an expression like this. So I have not done anything here I have just picked a mathematics book a calculus book essentially and picked the expression for the moment of inertia for a rectangular element and if you plug this fellow in here and rearrange you will obtain an expression for the angular acceleration. Now the idea or the argument here is that if we keep shrinking that particle meaning that if we take the limit as delta x and delta y tend to 0 what is going to happen? The denominator will actually go to 0 which is meaningless unless from mathematical point of view if you have to have a finite limit for this limiting process what has to be done? Continuum should be satisfied. Continuum is indeed satisfied but the idea is that if I take if I look at this expression and take the limit as delta x and delta y tend to 0 meaning that we are shrinking the particle to a really small size. We are forced to come up with the conclusion that this alpha suffix z which is the angular acceleration of the particle about the z axis tends to go to infinity for no reason. There is no absolutely no reason to for us to expect or assume that just because the mathematical limiting process is suggesting that the alpha z which is the angular acceleration shoots up to infinity for no specific reason it is physically unacceptable situation. So the only way that physically unacceptable situation will occur is if alpha z is finite. So how do you obtain a finite limit when you have a denominator 0 from calculus side. So the numerator also needs to go to 0 in the limiting case so that we actually get a 0 over 0 situation which from the mathematical point of view usually has a finite limit. So it is a nice combination of a physical interpretation and the calculus idea which helps us out here and therefore what we have to conclude here that to keep alpha z finite the only possibility to expect here is to have the numerator also 0 so that in the limiting process 0 over 0 has some finite limit and therefore the biggest conclusion of this business is that tau x y turns out to be equal to tau y x which is actually very very important simplification. And this is believe me this is actually not coming just by that limiting process there is a physics involved in this as well and the physics is that we cannot for no reason expect that alpha z tending to infinity. There is no special mention of any situation which will make that alpha z the angular acceleration go to infinity to avoid that we must conclude that tau x y must be equal to tau y x and in general you can repeat this exercise in all planes I have done it in the x y plane you can draw the picture in the y z and the x z planes and come up with essentially the same argument no difference. So therefore to generalize it I will simply write that tau j i or tau y j equal to tau j i where i and j can independently take x and y so tau x y equal to tau y x similarly tau y z equal to tau z y and so on. So if you go back to our stress tensor then you see the off diagonal elements tau x y and tau y x tau x z and tau z x tau y z and tau z y so we just proved here that these occur in pairs that is these are equal so therefore how many independent entries then exist in this stress tensor 6. So we are not dealing with 9 independent entries but we are actually dealing with only 6. So this is a useful information to know. So this was again a little bit of aside from the main thing but nonetheless extremely useful result. Now the idea for this picture was that you show all sorts of stresses which are essentially the surface forces show the body forces some all those in the x and y directions and plug them in the Newton's second law of motion which was written out here. So this is precisely what I have done now that is why I am writing going back to the Newton's second law of motion for the fluid particle and immediately I am going to generalize because this is only an algebra. This is what we wanted to plug in the net x force on the particle for which here is the picture available. Stress components are given you have to multiply those by the appropriate area elements to obtain the forces some all forces in the x direction some all forces in the y direction and plug those in here and here. And immediately to generalize what you will see is that the left hand side from rho times substantial derivative of u to up to here there is nothing new here what is that that was already discussed and completed earlier which is our left hand side here or the left hand side here all now presently we are talking about is the forces which is the right hand side here here here and that is what this is. So the right hand side then which is essentially the net x force on the fluid particle net y force on the fluid particle and likewise the net z force in the fluid particle if you work out the algebra will come about as partial derivative with respect to x of sigma xx plus partial derivative with respect to y of tau yx plus partial derivative with respect to z of tau zx plus rho times that intensity which is per unit mass value of the body force in the x direction and so on. So this is something that usually is worked out I mean most of the books will have this discussion that that you will find sometimes some books do not talk about this moments business sometimes they do sometimes they do not I do not remember whether these people are talking about it or not. Nonetheless I hope that the argument was understood here and I think as far as I remember it is the same argument that solid mechanics use they also have the same relation for the shear stresses that they occur in pairs it is exactly the same manner in which they bring about there is no difference between the solid mechanics argument and the fluid mechanics discussion that we are having here okay. So now we have what we call differential momentum equations which are essentially nothing but again Newton's second law of motion written for a particle if you look at the left hand side written in terms of the substantial derivative and then the net force acting on the right hand side in whatever direction that we are looking at. Alternatively you can utilize the so called conservative form on the left hand side which was brought about through the use of essentially that Euler's approach Eulerian approach which we brought out by converting the integral form into differential form. So the substantial derivative and this middle term are essentially equivalent to each other which we have shown and the right hand side simply here tells me the net x force on the fluid particle but in this boxed set of equations everything here is on a per unit volume basis that is something that you must keep in mind okay. So when you simplify these two equations on the top what you will find out that the volume element actually drops out and what remains is what has been written out here and therefore the boxed equations are what are called as differential momentum equations sometimes they are called Cauchy's equations they are all on a per unit volume basis and this terminology of Cauchy's equations actually comes directly from solid mechanics. So if you see this set of equations is more or less the same for solid mechanics as well. The only difference between solid mechanics and fluid mechanics comes is when we want to now relate these stresses to rates of strain in our case they will utilize the so-called Hooke's law which will relate stresses to strains rather than rates of strain. In fact that is how I want to bring about the next part in this discussion but so far barring the algebra which I am assuming that you will sometime do is the idea reasonably clear as to how we are progressing. Some of these issues like the moments part and the state of stress at a point are I think very very important to have included in the discussion because unless those are clarified it does not make sense to really directly write the stress tensor in terms of these nine components. So students usually ask unless you discuss this first the state of stress that why is it that these nine in a 3D situation are going to completely specify my state of stress and it is very useful from that point of view to do this discussion and show that it is really indeed coming in this point. So now what is left to be done to bring about our final forms that we normally are used to seeing is as I said these stress components that you can see here need to be related to the rates of strain and we already know the expression that we are going to use assuming the Newtonian fluid what we are going to say is that the stress component is going to be linearly proportional to the rate of strain. However the very first lecture what we talked about was a simple one dimensional flow that parallel shear flow we looked at and we had seen that their tau then was written as u times d u d y that was a one dimensional situation. So now it is going to be generalized to a two or three dimensional situation and to do that we have to talk about a little bit more on what is called as the constitutive relations for Newtonian fluid which are nothing but the relations between stress components and the strain rates in a generalized form. So this is also called as the generalized Newton's law of viscosity whatever discussion we are going to do. Now here the formal let us say derivation or in some sense the development it is not really I would not say it is a derivation but the development of this relations is very, very tedious and this was done by Stokes apparently in 1845. Normally you will see that this is normally not there in any of the elementary books like what we are talking about here. Normally you will find this in a one level higher book for example that Kundu and Cohen which I wrote on the board yesterday has this discussion but it is from a completely mathematical point which unless you are very comfortable with knowing the Cartesian tensors and their relations and such it is very, very difficult to understand. So in my opinion if you are ready to sit through a very, very long and tedious discussion the best physical let us say work out process is in a book by S, it is a very old book some of you will know it, SWE on it is I think called the Foundations. It is a very old book, how many of you have heard it? Okay good. So it is really outlined completely there like the treatment is there but you have to resort to it, there is no other option but it is extremely unreadable in my opinion. Shames all I have seen it of course but not to a great extent, just please comment whether it is a completely worked out derivation as how it is in Yuan or I think James's book is called Mechanics of Fluids and of course as he is mentioning this is Muralir and Viswas you are right. The problem that I have with Muralir and Viswas is that there are too many errors in the book and in the sense that there are I think too many typographical errors. So it is very, very irritating if you are working out something and if you see that something else is coming in the book and you have no clue whether you are going wrong or whether the book has been incorrectly printed and that is the reason I did not want to point out this book to you because there are too many typos in the book. However if you have read it and if you think that it is fine you can always look at it. Fine. But I mean I have not really spent too much time with it so there is no problem in looking at it. But this one I absolutely know for sure that the entire algebra is worked out as one would want it and I think from this book itself Professor Prabhu had put together his heat transfer slides wherein he had explained this if I am remembering correctly. James's book unfortunately is not that easily available unless it is in the library. But it does not matter correct. The reason is because James is basically a solid mechanism. Yeah I have seen this one and it is the last edition that I saw was an international edition I think it was a 1988 international edition with very thick volume like this. I have edition I mean our library has an edition which is somewhere 90s which is not a very big book it suggests only this size. But I am actually in fact I have looked at brochures of this Magra Hill but it seems there are recent editions has come up sometime in 2008 or 2007. I do not remember exactly but I am not able to get the book in India. But it does not matter because this is this business has been there for ages so even if you look at an older edition it will be there. Fine. So these three you can look at definitely I have myself gone through this completely. So I am confident about that part. If someone is saying that a physical explanation is provided there perhaps you should look at it. And James is something that I have barely seen but I know that it exists there. That is correct. There are many other advanced texts which have this business but it is all completely mathematical and it is very difficult to make any sense of it as to what is going on unless your background in the tensor calculus is very good. So that is fine. I mean you can spend time on tensor calculus background and then go and read it but it becomes a little bit of work. So what I am going to do is rather superficial discussion in the sense that I am just going to point out to you what are the bases on which this development is done. And it is simply that there are three assumptions that have gone into it which is in really speaking there are only two assumptions which is the first one is stress is proportional to directly proportional to the rates of strain which is simply our standard Newton's law viscosity. Now it is in the generalized form that is all. The main assumption is that the fluid will be considered as isotropic which means that the fluid will have a direction independent behavior. Whichever direction you look at the stress strain relation remains the same that is the main assumption. And the third one is not really an assumption but it is a common sense statement that in absence of strain rate whatever stress expression that you have should automatically reduce to describe the static fluid situation correctly. So that is the common sense statement in that sense. And I will directly give you then the final results here which have come about through this. For the shear stresses we have already seen that the shear stresses occur in pairs. And for example the tau xy equal to tau yx is simply the dynamic viscosity times the strain rate total strain rate in the xy plane which we derived yesterday from the kinematics consideration. And likewise for the remaining two. So this I will say is still a fairly intuitive statement to understand that tau equal to mu times the strain rate is just that now we are talking about a two dimensional situation. So accordingly that strain rate earlier in that pure one dimensional shear flow we only had du dy if you remember. Now that there is that dv dx component also added because of the multi dimensionality of the flow and so on for the remaining two. So this is as far as the shear stress relations are concerned. For the normal stress there is a little more involved business. Basically what I want to point out is that the normal stress that is going to be acting. Let me go back to that picture here for example this one or this one or this one that will have two contributions in general. One is what we will call a pressure contribution and there is a viscous contribution to the normal stress as well. So the total normal stress for example sigma xx is written as minus p y minus again it is the compressive nature of pressure which is brought about through our sign convention that we talked about plus a sigma xx which is contributed from the viscous side. So this viscous side is something that needs to be talked about and here what I am doing for rather simple discussion is that I am just pointing out an analogy between a generalized Hooke's law in solid mechanics and I will simply say that we are going to replace from the Hooke's law statement the strains by strain rates for our fluid mechanics situation. So if you look at a generalized Hooke's law from any solid mechanics text the normal stress there let us say the sigma xx again for an elastic solid is written as two times the shear modulus multiplied by the strain in the x direction plus some other factor where that Poisson's ratio also comes in whole thing multiplied by a summation of all normal strains. So now if you want to use this analogy and say that we want to write directly then the viscous contribution in a fluid flow situation I will simply write by analogy that sigma xx then is two times mu times e dot xx where that first coefficient of viscosity or the dynamic viscosity comes in and if you look at this all business I am combining in some sort of a second coefficient of viscosity as it is popularly called and instead of the normal strains I will simply have the summation of all normal strain rates. So this is as let us say pedestrian way of putting it across as it gets without really getting into too much detail of the Stokes's development and if you recall from our kinematics discussion the summation of the three linear strain rates was essentially equal to the volumetric strain rate which was given by divergence of the velocity. So that this e dot xx plus e dot yy plus e dot dz is simply replaced as del dot v and that is what the expression for the normal stress is let us say here I have written it for the x direction for the viscous part. Similarly if I want to write it for y direction it will be simply this two mu times e dot will be yy here this all remains exactly the same because it is a total addition anyway same thing for zz sigma zz viscous will come out as two times mu times e dot zz same thing here okay. So now let us see what happens then we said that normal stress has a pressure contribution and the viscous contribution viscous contribution has been simply argued from the basis of an analogy with generalized Hooke's law in solid mechanics pressure as it is we know minus p. So we put together sigma xx minus p plus all this from here which is again simply written out in Cartesian form just for clarity in some sense. Similarly for yy and zz and this first set which I am calling as one and the second set which is called as two is what together called the constitutive relations in fluid mechanics all that they are doing is we are relating the stress components to the strain rates through viscosity and something else also has creptin now which is this second coefficient of viscosity. So now let us see what we can still do with this relations one and two as I said called the generalized Newton's law of viscosity or constitutive relations for Newtonian fluid. Now let us say we look at this again second set and if you add all these guys if you add all three relations what we have is sigma xx plus sigma yy plus sigma zz on the left hand side there will be minus p minus p minus p so sorry that will go to minus 3p and these two guys will nicely add together to get into all strain rates sum together multiplied by whole thing multiplied by 2 mu plus 3 lambda that is very straight forward algebraic addition nothing different. So customarily what is done in case of fluid mechanics is we define the summation of these three normal stresses as minus 3 times a so called mechanical pressure. So if you see the way the mechanical pressure is defined it is simply minus the average of the three normal stresses because if you take sigma xx plus sigma yy plus sigma zz divide the entire thing by 3 and put a minus sign to it you will get this p bar let us say which is what is called as a mechanical pressure. So fine let us go with it and see what happens if you replace this summation then by minus 3p and then simplify by dividing through that minus 3 then you will obtain a relation of this sort that is minus p bar which is the mechanical pressure equal to p which is right here minus 3 has been divided out minus if you divide the remaining part by 3 you can rearrange it in this fashion. So this is where some discussion perhaps can be done and here is where what I want to point out there is this so called p bar which is simply the by definition average of the three normal stresses whereas this here is our p which is what we normally call a thermodynamic pressure. So it turns out that there seems to be a difference when we come to a fluid that is flowing that there is a difference between the so called thermodynamic pressure and a so called mechanical pressure which is simply the average of the three normal stresses and the difference between these two is given in terms of whatever is you whatever you see on the right hand side here. My question is what is that you understand by thermodynamic pressure it is also called as static pressure in fluid mechanics terminology so what is your understanding of static pressure how would you interpret static or thermodynamic pressure this is an important concept so static pressure is recorded by a device which is static with respect to flow. Exactly, absolutely right. Where should you do static and steady state of the fluid? It is the same thing actually what he is talking about what he said is absolutely right that is the exact definition of static pressure so in case of a moving fluid the way one wants to interpret static fluid static pressure is let us say you take a probe and you start moving with the flow so that there is no relative motion between the probe and the fluid whatever the pressure that the probe will measure is what is called as the static pressure or essentially it is called as a thermodynamic pressure the reason why it is called thermodynamic pressure is because there is a systems approach in built in here we are essentially making the flow stationary with respect to your probe and as if you are dealing with a closed system at that point so that is what the idea of a thermodynamic pressure is which is absolutely right and so it turns out that the thermodynamic pressure which what we are used to knowing or the static pressure seems to be different in general from this so called mechanical pressure if we simply define it as the average of the three normal component. And as I said the average sorry the difference comes in the form of this two thirds mu plus alpha times these total addition of these strain rates. So here is the biggest modeling let us say input that came from Stokes and Stokes simply assumed that let this two thirds mu plus lambda be equal to 0. This is actually a postulate there is absolutely no theoretical basis for it as far as my understanding goes and that is why it is called as Stokes's postulate. Now let me quickly ask you before we get on with this what about if I have an incompressible flow do I need to bother with this postulate at all no why because automatically this is equal to 0 for an incompressible flow del dot v so in that case the mechanical pressure turns out to be equal to the thermodynamic pressure. However if we are now if we are dealing with a non incompressible flow or compressible flow so to say we actually formally should take into account the difference between the thermodynamic pressure and the mechanical pressure. However to make life simple in some sense Stokes simply postulated that let this two thirds mu plus lambda be equal to 0 and accordingly then we generalize that this p bar is simply equal to p in all situations with a auxiliary relation that lambda that second coefficient of viscosity equal to minus two thirds of mu. So this is what is normally called as the Stokes's postulate again there is absolutely no theoretical basis for it in some situations you cannot really neglect this contribution. Any idea what that could be that situation clearly we are talking about compressible flow because incompressible as we said that entire del dot v will become 0 and we are automatically getting p bar equal to p prime. But incompressible this how many of you are familiar with compressible flow I think some of you are saying that you do gas dynamics teaching etc. High Mach number is correct but additionally not just high Mach number that is not really a sufficient condition it is necessary condition for this to be included but there is a not just high Mach we need to have some special situations occurring what happens typically in high Mach number situation. In case you are dealing with a situation where there are shocks in the flow within the shock region see we treat the shock as usually a discontinuity but shock is really not a discontinuity the shock shock spread as it is called is usually over a distance of about 5 to 7 mean free parts. So if you want to really zoom into the shock you will see that the shock region is about 5 to 7 mean free parts. So within that shock region you actually cannot really neglect or you cannot employ this Stokes's postulate but however in general how many situations we actually deal with where you are getting into the shock and seeing what is happening. So on the whole what is done is that without really bothering too much we simply utilize the Stokes's postulate and it turns out that in most situations except like the one that we just talked about namely a shock structure type situation this works perfectly fine. So please again I will repeat my statement from yesterday that people blindly utilize these differential equations as some holy grail in some sense. I do not know those words or not but there is nothing holy about those. They are just about a model that someone has proposed and it seems to work more or less okay in most situations. So there is no reason to say that differential equations form the basis and you know that is something that is absolutely true and such and so on. There is nothing true in general anyway everything is a model and this model seems to be doing its job quite reasonably. So please keep that in mind and I think it is useful to point this out to students all the time because students will otherwise go out thinking that oh this is some absolute godsend thing that you know you just take it. It is not like that everything is a model. Is f equal to ma anything else? Newton's second law of motion which we have been using left right and center from childhood. Is there any theoretical basis for f equal to ma? Zero there is none. It is a phenomenological law based on observation and you should treat it that way that it simply seems to work and that is about it. So when someone says that Newton's second law is some again absolute truth nothing like that you know it is just a model that seems to be working and we simply keep using that model later and it seems to keep doing the job for us that is about it. So please keep this in mind that Stokes's postulate is just a modeling exercise and it seems to work reasonably well for most fluid situations where it does not work is if you have to actually deal with a shock structure type situation in a compressible flow and you cannot really utilize any of this in a shock structure situation. Any of this means what? Any of this continuum business that we are talking about doesn't work. If you want to actually study the shock structure in a gas where there is high Mach number flow and a shock is getting generated you actually have to give up on this continuum analysis and you have to do something else which is the molecular analysis if you are really keen on that. However in most engineering applications such detail is not required ever and so we can live with this. So that's what my take on this. Alright fine so having accepted then this model we simply substitute back in the Cauchy's equations which were right here. So now I have a relation for sigma xx tau yx tau zx etc from these sets 1 and 2. So again there is a bunch of algebra which I am not bothered to do here. I will simply say that substitute relations in 1 and 2 or substitute relations 1 and 2 in the Cauchy's equations and simply go ahead and do the algebra for the x direction then assuming a constant viscosity you can simplify that into something like this and immediately looking at this expression you can generalize using the vector form where whatever has been written out in the box is the so called Navier-Stokes equation or equations if you want for a viscous but with a constant viscosity compressive flow. So this algebra is again usually it's okay to do it once so that we absolutely are sure that this is how it comes about. I hope you understand what is to be done. Take this set and this set and plug it in these equations and go ahead and simplify and you should obtain something like this. Fine. So if you want to write it in the vector form then it shows up as what is shown in the box which is something that looks perhaps familiar except that since we are talking about a compressible business here this additional mu over 3 times the gradient of the divergence is an additional term that shows up because of the compressibility. So if you want to look at a simplified model then what is outlined here we will say that let us deal with a either a constant density or an incompressible flow. The only thing that you can essentially throw out is this del dot v equal to 0 because of incompressibility and then what is left out is something written out here and this is the most common form of the Navier-Stokes equation that mechanical engineers will end up using. If you are dealing with aerospace applications with truly high speed applications where compressibility is really, really present then you must go to the very first one. And that too there is very high chance that the viscosity may not be a constant. Viscosity for very high temperature gas applications becomes a function of temperature which means that this form either is not good which means that you actually have to go back and use this in the absolute basic form along with these relations here in their form so that some sort of a temperature dependence on viscosity etc can be incorporated right at the beginning and accordingly then you can work out the solution. But for our purpose the standard undergraduate class it is okay to simplify this with a constant viscosity at least to bring about the compressible form first and then for the incompressible or a constant density flow you simply say that this del dot v is going to be thrown out and what is left out is the most common form that you will see in mechanical engineering. So again this is nothing but a F equal to MA type situation as you can see if you multiply the left most part with volume element of the fluid delta v it becomes essentially m times acceleration. On the right hand side then what we have is absolute right hand side here all the forces the net forces acting on the net force acting on the fluid particle on a per unit volume basis again this equation is essentially on a per unit volume basis so you multiply that by the volume element and you will get the forces in their standard form fine any question on this since I am writing here this is a very small homework exercise for you if you want to do it. I wrote here constant density or incompressible are they the same or are they different if I say that a constant density flow or if I say an incompressible flow you say it is the same okay I will make a statement and you think about it and see if you agree with it or disagree with it and not now take your time whenever you want to do it. I will say that if I am dealing with a constant density flow it necessarily is an incompressible flow but if I am dealing with an incompressible flow it does not have to be a constant density flow this is my statement and you can note it down and think about it no it does not have to be even a natural convection situation I will make a general statement here that if it is a constant density flow it will necessarily be an incompressible flow whereas if I am dealing with an incompressible flow without resorting to any natural convection or anything it does not necessarily have to be a constant no no it has nothing to do with it. Is it about density variations yes it has to do with density variations absolutely but you make sure that you understand what I mean by this so my point is if I have a constant density flow I will necessarily have an incompressible flow which means that that del dot v here is going to be 0 okay on the other hand incompressibility condition itself from kinematics is del dot v equal to 0 so in either case I will be able to get rid of this del dot v and thereby simplify it to a situation like this. Let us not discuss this further but please those who want to really think about it note it down and let me know even by the time the wind works up from if you agree with me okay let us move on and look at another simplified form which is what we will call an inviscid flow so quite simply inviscid flow is where we will simply neglect the presence of viscosity and the only term that takes into account that is this mu times del squared v so if you get rid of that what is left out is what is popularly called as the Euler equation which is written out in the box here. Note here that the density need not be a constant in this equation it can be a compressible inviscid flow as well the way it is written okay now quick question see usually people will students will ask this is this just a absolute simplification in the sense that I want to somehow get rid of one term which is this mu times del squared v here and therefore I am saying that let the situation be inviscid meaning that mu itself we are putting equal to 0 and thereby getting rid of it is that how it is or is there any justification for treating a flow to be inviscid how do you yeah I think the flow can be inviscid in sense that it is not necessary mu is equal to 0 the velocity gains can be there especially when we are dealing with flows away from the boundary correct so that is precisely what he was pointing out here that if you are away from boundary you can normally neglect the action of viscosity but is it so in a particular situation more so second operator of velocity is 0 for irrotational flows oh no I am not talking about irrotational yet let us say that without getting into the irrotationality so far I agree with you though whatever you have pointed out for irrotational flow is absolutely right but without resorting to irrotationality someone already said from here and which is what I think the high Reynolds number so the necessary condition is that the flow has to be a higher Reynolds number situation in which case you can argue that the viscous effects can be neglected and then additionally you have to be away from walls fine so if that is the case then you can use this inviscid flow situation to a good extent and the equation that is written out here which is the governing equation for inviscid flow is what we will call the Euler equation irrotationality comes usually as a next step along with the inviscid nature of the flow you will normally see that irrotational and inviscid go hand in hand and it makes sense to me that is fine.