 Okay, so now I think we are more or less ready to get on to the differential analysis which we will do now. Okay, so as I told you in the morning session what I am planning to do here is utilize those integral forms that we already brought out in the morning and using those integral forms bring about these differential equations. This is just one approach, by no means this is the only approach. What you can do as I said alternatively is that those balance equations were written before we actually brought about these mathematical expressions. The balance equations were written in plain English and then we placed certain mathematics to it to bring about these expressions. What you can do is that you can utilize the same balance equation written in English but you employ those on a differential control volume. So differential control volume would essentially mean that we are talking about the control volume which has dimensions delta x by delta y with the delta x and delta y as we have been talking about. So there you can think about a mass flow rate coming in, a mass flow rate going out. The difference between those two is what you are going to say that we will get accumulated in the control volume and then you bring about the differential form. That is more or less what many of the textbooks will do and if you are going to read a book it is very easy to follow. So I do not want to do that. Slightly differently one can do it with the integral forms by manipulating the mathematics of it. I think it is worthwhile pointing that couple of those mathematics are points which I think everyone probably has seen somewhere or the other just to put them in perspective when it comes to these equations of fluid mechanics. Let us just see how we do it. So let us say we start with the conservation of mass statement where the very first line obviously is the integral form exactly the way it is written earlier. It is just that we are now going to talk about only stationary control volume for one reason that this is the most common utility stationary control volume and with stationary control volumes you can do some manipulations without really getting into too much trouble. So what are those manipulations? These are the two manipulations. There is something called an application of a Leibniz rule and then there is something called an application of a divergence here. So how many of you have seen these two terms? The Leibniz rule and the divergence here. How many of you have not seen at all? How many of you have heard the term differentiation under integral sign in the applied mathematics class? I think that is how it is popularly known. The Leibniz rule is many times popularly known as DUIS, differentiation under integral sign. If you have not, it does not matter. It is a pretty straightforward expression from the calculus and we are not going to get into the proof of it. What it does is that you can just see this first integral. What we have is a time derivative of an integral. As long as we are dealing with a stationary control volume, using the Leibniz rule what you can do is you can bring this time derivative which is standing outside the integral in the integral inside the integral. Only thing is that in that case it becomes a partial derivative with respect to time. That is all it is. If you want to read about the Leibniz rule and how it comes about, you can look at some of the applied mathematics books but for our purpose all that we are saying is that the Leibniz rule for the time derivative of the control volume integral will be utilized such that the time derivative is taken inside the integral and it becomes a partial derivative. If the control volume is not stationary then there are some additional terms that will come in. However, if the control volume is stationary as what we have been assuming that is it. Nothing else is to be done. And the divergence theorem basically relates an area integral as it is written out here to a volume integral over the control volume for which that area is the bounding surface. So if you see the expression for the divergence theorem what we have is a certain vector expression rho is a scalar. So I can always club that with the velocity vector. So rho times V is the vector that I am talking about which is dotted with the area element given through the unit vector, unit normal vector and this is an area integral. So the divergence theorem says that you can convert this into a volume integral if you take the divergence of whatever this function was which was getting dotted with the area element. So in that sense what I write then is instead of writing as area integral I write it as a integral over the control volume del dot this function rho V and then I am integrating it over the control volume and therefore the volume element must stay in here that is it. These are the two rules that we will use all the time when we want to convert these integral expressions to differential expressions and see how. Now you see that both integrals are now volume integrals they are over the control volume. So I actually can combine those okay so if I combine those that partial derivative is really operating only on the density here plus this divergence of rho V the whole thing to be integrated over the control volume and that must be equal to 0. So the only way this is possible for any arbitrary control volume is that whatever we are integrating must identically be equal to 0 otherwise this equality will not hold for any arbitrary control volume okay and that is why we say that if this has to be true namely this so called integrand integrated over the control volume has to be equal to 0 then that integrand itself has to be 0. In other words partial derivative of rho with respect to time plus del dot of this rho V equal to 0 must be satisfied and this is what is popularly called as the conservation of mass on a differential basis. Fundamentally you can see that the physics cannot be lost physics is the same what we did was we did some mathematical manipulations to cast an integral form into something like this. So what was the integral form initially telling us it was basically the rate of accumulation term and a net rate of outflow of mass term from the control volume. Those have been now written as partial derivative of rho with respect to time multiplied by the volume element if you want plus del dot of rho V multiplied by the volume element. So the physical interpretation associated with these two terms is really the same as how it was for those two terms. In other words what I am saying is that you look at whatever is inside this bracket that is integrated over the entire control volume and then it becomes 0 that is the different thing. In other words whatever is this quantity is essentially the mass balance on a per unit volume basis because you essentially look at the entire control volume as it was on an integral basis you are seeing now a large number of these elemental control volumes if you want inside the big control volume and you are now integrating really each of over each of those elemental control volumes this differential equation and you are getting it as 0 that is the idea. So the physics is exactly the same what I can point out is that this is nothing but a conservation of mass on a per unit volume basis. You multiply this by the unit volume and integrate it over the entire control volume and you get exactly what you started with namely the basic integral equation for the mass conservation. But this is now a differential equation and this differential equation as the conservation of mass statement which is by the way called the continuity equation as many of you would know has to be satisfied point by point in the fluid domain. Now that point by point has the same interpretation as how we were talking earlier it is not really a mathematical point that you can think about is that small fluid element that surrounds a given point over which we are really solving or satisfying but from a mathematical point this becomes a differential equation and so be it. So the thing to note here is that the continuity equation is nothing but the conservation of mass statement on a per unit volume basis that is about it. It just comes about in this particular form but so be it. You can get the exact same form by employing that mass balance as I was talking about something coming in something going out the difference between those two getting accumulated for a differential control volume of delta x times delta y. Those are the things that are normally outlined in the book formally you will get exactly the same thing. The reason I wanted to do this is because I want all of you to be familiar with various ways of obtaining these governing equations. It is quite likely that some students read something and they half understand something and they will come to you say hey I read it here this way so can you tell me how it is. So you should be at least reasonably familiar with what is done if the approach like this is for. Okay now let us see little more of the mathematics. We start with this d rho dt plus del dot rho equal to 0 and this del dot rho v is then expanded using the vector identity. So I will do that here it is written in the vector form but I will do that in a Cartesian form which I think will give out the clear picture. So let me write this it is first let me first write it in the Cartesian form. So any help and now each of these will be written as rho times du dx plus u times 0 dx plus same thing for here which is rho times dv dy plus v times d rho dy. Let us say we are dealing with a two dimensional situation for simplicity so that I would not have to bother about. So what I see now is rho dt as it is plus I will collect like terms so I will say that this is rho times du dx plus dv dy plus u f u dx plus e dy operating on rho I hope you see where it is coming from these two guys are coming here these two guys are going there is that fine. So now I will combine this sorry I will combine that in what is this rho times du dx plus dv dy I forget the rho what is inside in here yes correct thank you yeah that is the divergence of the velocity of course if I had taken the third term that there will be a dw dz in here that is perfect so let me immediately write that as and what is all this substantial derivative of then that is dt with a substantial plus this is the alternative form. So you started with the very first form that we obtained when we converted that integral form into differential form through that Leibniz rule and the divergence theorem and now I wrote here in the slide all this manipulation directly in the vector form here though I am writing it explicitly in the Cartesian form so that you can follow it nicely finally what I am getting is this substantial derivative of rho plus rho times the divergence of velocity equal to 0 which is another statement for the continuity equation. Now there is some perhaps you will know many of you will know I am sure if you write it in this form as we obtained initially versus if you write it in this form any idea of what this form so we call this situation as the conservative statement or a conservation form of the continuity equation this is something that the CFD community uses a lot that this is a conservation form and this is a non-conservation form. I am not going to get into the discussion of that but keep that in mind right now that this is the conservative and this is the so-called non-conservative form. Now let us see one other way whether we can bring about this form to see that now we say that let us try to employ a mass conservation statement for a fluid particle. So for a fluid particle the mass of that fluid particle as it keeps moving about remains the same right that is the conservation of mass statement for that fluid particle of course this is in absence of our nuclear and some other special situation fine. So how do I write the mass of a fluid particle it is simply written as rho times volume element for the particle and since I am following the fluid particle I must use the substantial derivative to say that the substantial derivative of this rho times delta v has to be equal to 0 and the substantial derivative if you remember from our previous earlier afternoon discussion is just an operator. So it operates just like a derivative operate on these two guys. So the way it will operate then is that delta v remains a constant and this capital dd operates on rho in one plus rho remains the same and then this substantial derivative operates on delta v just like a derivative product rule from here you if you divide this entire equation by rho times delta v what you will see is that 1 over rho substantial derivative of rho plus 1 over delta v substantial derivative of delta v equal to 0 but what is this where we just finished that we just evaluated that volumetric strain rate and volumetric strain rate is exactly what you have here which is 1 over delta v substantial derivative of delta v so that entire term is simply replaced with the divergence of the velocity. So this is coming from the kinematical consideration and then what you get is exactly what you can have. So this is another way of obtaining the same equation. So my take on this so called conservative or sorry this is non-conservative or conservative form is if you employ a statement from the Lagrangian point of view and obtain the corresponding equation you get a non-conservative statement just the way it was done here. This last part on this slide is basically a Lagrangian formulation of conservation of mass statement because we are following a given particle. So if you employ a certain law such as the conservation of mass statement using a Lagrangian approach you will get a non-conservation form. On the other hand fundamentally if you see how was this obtained? This we started with a integral approach but the integral approach was actually written using the Eulerian approach. It was written for a control volume where we were actually just monitoring what comes out, what goes out, there is some source and then therefore things are happening in a certain balance. So fundamentally what we are doing here is employing the Eulerian approach and if you employ Eulerian approach for these conservation of mass, momentum, energy, etc. the equations that you will obtain will be in the conservative form. So this is something that you can note down. Clearly you can always convert from one form to the other just like what we did here where we started from this conservative form and got the non-conservative form through some manipulation. So that is not a big deal but fundamentally I will say that if you want to attach some physics to where the conservation form is coming from and where the non-conservation form is coming from I will say that you can employ this Eulerian business for conservation form and Lagrangian for non-conservation. Is that fine? I hope it was followable. So now what we say is that another way of looking at incompressible flow is that we are following a fluid particle which means that necessarily it has the same mass that we are following. Now we say that incompressibility means that the volume does not really change which means that the density also has to be the same for the particle which means that substantial derivative of the density must be equal to 0 and if you look at the non-conservation form if you place the substantial derivative equal to 0 automatically you will get the divergence of velocity equal to 0 as before. So we have seen this divergence of velocity equal to 0 as the condition for incompressibility from two different points of view. Earlier it was through that kinematics consideration where we had worked out the expression for the volumetric strain rate and we simply said that that has to be 0 for incompressibility and it brings about that del dot v equal to 0. Here we say that it is the substantial derivative of density has to be 0 because we are following the same fluid particle with the same mass and if its volume does not change substantial derivative of density has to be 0 which again brings about del dot v equal to 0. So sometimes now if you are dealing with a 2D incompressible situation in especially in the Cartesian coordinate we use something called a stream function which I think most of you probably have seen. So what is this stream function? Stream function is simply defined as a mathematical function whose partial derivative with respect to y gives me the x component of velocity and partial derivative with respect to x with a minus sign gives me the v velocity. Why do you think this is of any use doing it this way? Do you think there is any use to this? In particular if you employ these definitions in the incompressibility or equivalently the continuity equation what happens? Exactly. So it automatically satisfies the continuity equation or the mass conservation equation. So many times you can actually get rid of the mass conservation equation altogether and work with a formulation which will involve stream function. That is fine. Now let us see what else happens if you employ that. If you recall our definition of stream line which was dx over u equal to dy over v from kinematic you just cross multiply and you get udy minus min dx equal to 0. If you just substitute these definitions in here what will happen? This will be partial derivative of psi with respect to y times dy plus partial derivative of psi with respect to x dx equal to 0 which is essentially the total derivative of psi equal to 0 or in other words if this needs to be followed then psi has to be a constant for a given stream line. So that is another utility of this stream function psi that if you have a bunch of stream lines then for each of those stream lines there is a certain constant value of this psi which you can assign. So for example if for whatever reason if you are getting a stream line pattern like this there will be a value of psi 1 which will be constant for this stream line number 1. There will be a value of psi 2 which will be constant for this 2 and so on. So this is something that you will utilize when you bring about your CFD result and when you want to graphically show how the flow field looks like. Many times you will actually be plotting essentially these stream functions because that will give you a family of stream lines and then you can see how the flow field is looking. There is also a physical interpretation associated with a stream function which I have not outlined here very briefly. I am sure most of you will know it. If I just look at the difference between this psi 1 and psi 2 for example what does it give me? So essentially the volume flow rate per unit depth and since you know it that is perfectly fine. So that is more or less what we have to say about the differential mass conservation equation. Let me move on because this is exactly the way it will work out for the remaining equations as well. The energy part I will look later because the momentum equation is going to take the most of the time when it comes to forming the differential momentum equation. So let us begin today. So we have about 15 minutes. Let us go as much as we can go and then tomorrow we will pick up from here. Exact same procedure. I started with a integral balance for mass, Leibniz rule, divergence theorem. Everything was converted into a volume integral and since the right hand side was 0, we claimed that the integrand has to be identically 0. So you do the same thing here. Here now we start with the linear momentum equation. So all I am doing is that I am adding as you can see here that velocity vector. Use the Leibniz rule. So dd time gets inside the integral with a partial. Whatever this quantity is, let us not get into the interpretation of it right now. Rho times the vector velocity, times the vector velocity again, you form a del dot of that and integrate over dv using the divergence theorem. On the right hand side what we had for an integral statement is the net force acting on the control volume. So all that I am doing now is slightly trickery you can say that since I want all these integrals to be eventually expressed as volume integrals, I will somehow want to express that force also as a volume integral. So I simply write that as some sort of a force per unit volume integrated over the entire volume. So that small f here is essentially the net force acting on the control volume on a per unit volume basis. Now I am in business because then I can combine all these integrals as one volume integral with this f brought onto the left side equal to 0. Again the same argument if this has to hold for any arbitrary cv whatever integrand we have must be 0 and then I bring about that small f and put it on the right hand side again. That is just trickery. So this is it. This is what I will call a momentum equation which is essentially the linear momentum principle on a per unit volume basis written in again our conservation form. Now let us not bother with the vectors too much because it can become troublesome. So let us see how we can write it for an x direction. So what I will do is here was a vector velocity v. I will simply replace that with the x component of the velocity and accordingly one of these guys will also go as the x component of the velocity. So that is it. Partial derivative of rho times x component of velocity with respect to time plus del dot of rho uv this remains as the vector equal to sorry here is the vector force. Now I am just saying that I will get the x component of it. So that is it. Net x force again on a per unit volume basis acting on the elemental c. So immediately rather than keeping to write this as vector form I am expanding this del dot rho uv as Cartesian expression. So if you expand that you will see that this will expand as d dx of rho uu. Obviously you can write it as rho u squared but I am keeping to write it on purpose in this fashion. The second term will be ddy partial of rho uv d dz of rho uw. This is something that you can just expand this using our expression for the del operator and using velocity vector as ui plus vj plus wk. There is nothing difficult about this. An exact same thing if you write for the y and z directions you will see that this will come out in the forms as your primary. So now the net force again whether it is on a per unit volume or whatever it is still going to be composed of the surface forces and the body forces as was the case in the case of integral situation. Before getting into the formal discussion of how to formulate these force expressions let us just do one more manipulation which is like how we did for the conservative to non-conservative form of the continuity equation. So here what is done is let us look at this x equation which is boxed and that is out written out here. From this you subtract u multiplied by what is this continuity yes correct fine. So from the momentum equation you subtract u multiplied by the continuity equation what will you get is what has been written out here. Basically you can treat terms such as ddx of rho uu as having a product of rho u as one thing and u as another thing and then ddx operates on that that is the idea. So u times ddx of rho u if you subtract what you will get is rho u times dudx that is it is simple straightforward product rule we are subtracting yes no this is fine this is fine this is fine what I am doing is I am taking the entire continuity equation which is is that fine. So I am not changing anything on the right hand side is the idea I am basically adding 0 or subtracting 0 on the right hand side all that I am doing is changing something on the left hand side and then this entire left hand side then you can nicely write as rho common plus the way you want it for the substantial derivative to come in operating on u and therefore what we started with is the so called conservative form is now written as a non conservative form. So here the way to interpret this is you just multiply this box equation here by the volume element of the fluid particle right. So what I am talking about here is rho times let us say delta v for the fluid particle fx times delta v so what is rho times delta v for a fluid particle mass and then I have mass times the acceleration equal to the total force in the x direction obviously we are talking about the x acceleration. So doing this manipulation you are bringing the form in the form of the Newton's second law of motion for the fluid particle. So these two are equivalent the conservative form as was brought about through the utility of employing it on a control volume has been equivalently shown to be the same expression as if you were to take a fluid particle and employ Newton's law of second law of motion f equal to ma you would get the same thing that is the idea here okay. Let me just begin with this but I do not think I can finish this at all today. So far so good we are talking about the forms of these equations in either a so called conservative form or a non-conservative form what we have been talked about in detail is forces the surface forces on the body force. So let us now start getting into that description. Alright so we have talked about this a little bit that if you are talking about a surface force those are essentially expressed in terms of a normal stress and a shear stress on a surface okay. So usually from solid mechanics time we follow this normal stress as sigma and shear stress as tau so I will just continue with that. On the other hand body force is usually expressed as an intensity which is to be interpreted as on a per unit mass basis. So much force per unit mass basis okay. Now let us let us just look at something which is normally perhaps not talked about in detail in the books this is nicely outlined in the book by Gupta and Gupta so that is one good part of that which is how do we specify a so called state of stress at a point in a moving fluid. So to do that we what we look at is a fluid particle which has a specific shape so to say it is not our standard delta x times delta y but it is some sort of a triangular shape. Before we get on with this have you seen such triangular shape fluid elements used for proving some law somewhere exactly absolutely. So if you know that this is exactly the same. So what we are doing now is that there is a delta x delta y and delta s as our three lengths it is say again a two dimensional situation and all that we are showing is whatever surface and body force that can act on this have been shown that is all. So if you look at the vertical surface there is a normal stress and then there is a shear stress same thing for the horizontal surface there is a normal stress and then there is a shear stress same thing for the inclined surface there is a normal stress although perhaps my drawing doesn't look that great in terms of normal but that's fine and then there is a shear stress. There is a body force which essentially will be that by gravity in our case but those have been shown in component form as capital G suffix x and G suffix y in the x and y direction and the expressions for those have been written out here. Remember that the body force we are expressing on a per unit mass basis. So capital Gx is on a per unit mass basis is small gx let's say times the mass of the particle. So the mass is now in this case since it's a triangular element the mass will be density times dx dy or delta x delta y over 2 and that's it. So all I am saying here is that all stress components which are basically the surface forces shown are positive. So is this convention that everyone is familiar with? First of all the double subscripts and second of all the positive or negative stress I think many of you would be those who use solid mechanics as their base line will definitely relate to it. Is there anyone here who is not familiar with double subscript notation and the sign convention for the stress? I am going to talk about it but just to ask. No, fine. So very briefly this is the last slide I think I will discuss and then yeah then we will move on to the tomorrow. So the notation is every stress component will be using a pair of subscript. The first subscript is giving me the direction of the normal of the surface over which it is acting. So going back or I have a few examples here, fine. Second subscript is the direction of the stress component itself whether it is acting in the x or y or whatever. Positive stress component if both the surface normal and the stress component point either the positive coordinate direction or the negative coordinate. So now here I have shown two examples and we will go back to that picture in a minute. Here is an area element which is shown by the full line. The area element is such that the normal unit vector is shown to be in the negative x axis and the normal stress is actually shown to be acting in the positive x axis. But this is a normal sorry negative normal stress component because these two guys are pointing in two different directions, okay. On the other hand if you go and look at this situation the unit outward normal is also in the negative x direction. The stress component, normal stress component is also pointing in the minus x direction. So therefore this sigma xx here is a positive stress component. Let us look at one more. Here I have an area element whose unit outward normal is pointing in the positive x direction and now I am talking about a shear stress which is pointing in the positive y direction. So therefore this is a positive stress component. Same exactly the same thing in here for example. The bottom surface if you see the unit outward normal is in minus x direction sorry minus y direction. The shear stress is minus x direction therefore it is a positive component. The first subscript of this tau yx y is pointing where the unit normal for the area is pointing that is in the y direction. The second subscript is its own direction which is in the x axis that is it. So with that we have essentially come up with the notation and the sign convention. Everything here that has been shown is positive. What we will do tomorrow is we will first claim that if you are able to specify two stress components which are mutually perpendicular on two planes which are mutually perpendicular in a 2D situation you are in a position to define the state of stress completely. So we will do that tomorrow.