 Good morning everyone. Let us continue with what we started yesterday which is differential analysis. We had already started the discussion and more or less completed it also for the differential mass conservation equation which is what we call a continuity equation. So, let me just briefly look at that. So, what we have done here is that we have employed our balance statement for an elemental control volume which is shown on the slide right now as the rectangular control volume with elemental length delta x and delta y. And then we identified the mass flow rates coming in to this control volume, mass flow rates going out from the control volume which were expressed using the first order Taylor series. And then completed the balance statement by realizing that the mass flow rate coming in that is a total mass flow rate coming in which will be m dot x plus m dot y minus the total mass flow rate going out would be then the mass flow rate that will be sorry the mass that will be getting accumulated. So, this is how the balance statement was written and by substituting the expressions for the mass flow rates m dot x and m dot y we simplified it to the form that is at the bottom of slide number 3 here. So, this is what we call a conservative form several questions have been asked already about the conservative form and the non-conservative form. So, let me take a minute to say here that when you employ a balance statement such as what we have done here inherently you will generate a governing equation whether it is a governing equation for mass or whether it is for linear momentum or whether it is for energy in the conservative form. So, if you see what I mean by conservative form you just go to the top line here the top box and this is essentially a conservation law employed for the control volume as you can see the first term on the right hand side is the net mass flow rate going out rather coming in to the control volume. Similarly, the second term on the right hand side with the minus sign that is is the net mass flow rate coming into the control volume and essentially the total net mass flow rate coming into the control volume will accumulate at the rate given by the left hand side. So, this is what we call a conservation statement or a conservation law and if you employ a conservation law through a balance statement like what we have done here you will always generate the governing equation in the so called conservative form. Using a mathematical manipulation here we converted the conservative form which is given by equation number 1 into another form which is dealing with a substantial derivative and this substantial derivative form is actually what is called as a non-conservative form. So, let me go ahead and talk about that in a second. Here again we obtained the same conservative form which is at the bottom of the screen, but the way it was obtained you can imagine again that we started from the integral mass balance equation. So, again fundamentally it is a balance equation written for an integral control volume we chose a stationary control volume and performing certain mathematical manipulations which are listed here using the Leibniz rule and the divergence theorem what we did was we converted the area integral into the volume integral. We brought the differentiation with respect to time inside the integral sign and made it partial differentiation and then essentially what you obtain is the same conservative form on a per unit volume basis because we said that the control volume is arbitrary and if a quantity integrated over an arbitrary control volume is giving you a result of 0 then that quantity itself must be identically equal to 0. So, what we mean by that quantity is the essentially the integrand. So, the integrand is nothing but this partial derivative of rho with respect to time plus divergence of rho v and this has to be equal to 0 was the conclusion and which is nothing but the mass conservation on a per unit volume basis or again the continuity equation expressed in a conservative form. So, again the point to emphasize here is we start from a balance statement and we obtain the expression in a conservative form. Again doing some mathematical manipulations which by the way here in this slide number 6 at the top where my highlighter is the mathematical manipulation is essentially opening up this divergence of rho times v and opening up of this divergence of rho time v has been achieved through the mathematical background that was discussed on day number 1. Essentially it is the same operation that was done here on slide number 4. On slide number 4 it has been done explicitly for Cartesian coordinates whereas, in slide number 6 it has been shown in the general vector form. So, please do not get confused about whether we are doing something different here than what we did earlier here. They are exactly the same here it is explicitly written out in Cartesian form whereas, on slide number 6 what we are doing is we are using the vector form directly. The purpose of doing all this is to make you familiar with different ways of manipulating the equations and primarily because different books will follow different approaches and you should know these approaches so that you do not get stuck while going through any particular book. Anyhow, so when you open this divergence of rho v you get a gradient of rho dotted with v plus rho times divergence of velocity and the first two terms which are which are underlined are essentially combined to obtain the substantial derivative of density and we obtain our non-conservative form. So, in this manner the non-conservative form has been obtained from the conservative form and using some mathematical manipulation. Exactly the same way it was obtained here we started with the conservative form in the Cartesian coordinates and opened up these differentiations and then collected certain terms together to obtain the non-conservative form. One last thing that we discussed here was that the non-conservative form can be obtained directly by realizing that we can employ our Lagrangian approach for mass conservation. When we employ Lagrangian approach what we say is that we will identify a fluid particle a given fluid particle which will essentially imply or mean that we are always following the same mass. A fluid particle will necessarily mean that we are following the fixed amount of matter or mass as it moves from one location to the other and therefore by definition since it is the same mass that we are following and we are following it as a Lagrangian point of view the rate of change of that mass itself has to be identically 0. This is from the definition and then rho times delta v is essentially the mass content of the fluid particle where delta v is the volume element for the fluid particle rho is the density and in general everything is assumed to be variable. So that this substantial derivative operator remember we talked about this substantial derivative as an operator also simply operates on a product of rho multiplied by delta v using the standard product rule of differentiation I have written out the next step which is substantial derivative of rho times delta v plus rho times substantial derivative of delta v equal to 0 then we divide the entire equation by rho multiplied by delta v and realize that this 1 over delta v times the substantial derivative of delta t from the discussion on kinematics is nothing but the volumetric rate of strain which was found in kinematics to be equal to the divergence of velocity where my highlighter is right now and therefore we obtain again the non-conservative form of the mass conservation equation directly. So let me just take a minute again and use the whiteboard to write that the mass conservation equation on a differential basis this form is called as the conservative form and if I write it in the form of a substantial derivative this will be the non-conservative form. Keep in mind that the conservative form is obtained by employing a balance equation on a control volume whereas the non-conservative form can be directly obtained by employing the conservation of mass statement to a fluid particle which means that we are necessarily employing Lagrangian approach. So let me continue now and try to wrap up this mass conservation and go ahead with the remaining part. Remember that if we have incompressible flow from kinematics we had obtained the condition that the divergence of the velocity field is 0. If you go back and look at our non-conservative form of the mass conservation equation if the divergence of velocity is equal to 0 the second term will be identically equal to 0 and therefore an alternative statement for incompressible flow is going to be that the substantial derivative of density is also equal to 0. Now let us look at a special situation where we are dealing with a two dimensional incompressible flow in Cartesian coordinates. So in Cartesian coordinates this divergence of velocity equal to 0 in two dimensions will simply mean that partial derivative of u with respect to x plus partial derivative of v with respect to y will be equal to 0. In two dimensions for incompressible flow we many times use a concept called a stream function which is usually denoted by psi such that the way this stream function is defined as the u velocity which is essentially the x component of the velocity will be given by y derivative of psi. So psi remember then is a function of both x as well as y. So the partial derivative of psi will be giving you the x component of the velocity and negative of the partial derivative of psi with respect to x will give you a v velocity. So what will be the utilization of this definition if you take this definition and substitute it in the continuity equation let us see what happens. So let me go to the whiteboard again. So what we end up doing is that we realize that the continuity equation is satisfied automatically assuming that then the psi function is a well behaved function where whether you calculate the cross derivative as first the x derivative and the y derivative or first the y derivative and the x derivative does not really matter. So automatically the continuity equation is satisfied with this usage of stream function psi. So many times what happens is in some classes of flow many times the continuity equation is replaced in terms of an appropriate equation involving psi. So that is the utility of this stream function. Now let us see what else this stream function signifies physically. So if you recall from our kinematics discussion yesterday that we had talked about stream lines and in stream lines we realize that the differential equation which gives us the equation of the stream line is given by dx over the x component of the velocity equal to dy over v component or sorry the y component which is v. This is in two dimensions of course that we are talking about. So if you cross multiply and rearrange the equation you will see that it turns out to be u times dy minus v times dx equal to 0. Now here if you substitute u equal to partial derivative of psi with y and v equal to minus partial derivative of psi with respect to x what you will get is d psi dy dy that is partial derivative of psi with y times dy plus because this minus and this minus will cancel each other. Partial derivative of psi with respect to x times dx equal to 0 but then what will be left out on the left hand side is essentially the total derivative of psi equal to 0 and therefore this will imply that we are talking about d psi equal to 0 or if d of psi is 0 that is the total differential of psi is 0 then psi must be a constant and that is what the conclusion is that for a given streamline the value of the stream function is a constant. So physically if you want to imagine what is going on let me let me draw a sketch. Let us say that we have a streamline field or a flow field which will be represented with a bunch of streamlines. This could be in general some flow field and I have shown four different streamlines. Each of these streamlines will have a constant value of psi associated with it. So this could be psi 1 which is a constant for this entire streamline number 1 this would be psi 2 which would be constant for the entire streamline number 2. This will be psi 3 which will be constant for entire streamline number 3 and so on. So many times when we plot streamlines as a result of the CFD computations, in effect you can say that we are generating this stream function value and then we are plotting these stream function values as a contour plot which will generate these streamline patterns. So this is something that is special for a two-dimensional incompressible flow. Please keep in mind that the stream function is an entity which is defined only for two-dimensional incompressible flow and it is defined in the following fashion. With the use that the continuity equation gets automatically satisfied therefore you can replace the continuity equation in terms of an equation involving psi. I will not talk about that right now. I will want you to think about it a little bit that what will be the equation of psi that certain types of flows will satisfy. Importantly right now what we wanted to point out was psi equal to a constant is an alternative representation for the equation of a streamline and as I drew those few streamlines what we realize is that each streamline will have a value of psi associated with it which will be a constant for that streamline. I think this is more or less here. This is more or less what we have to discuss as far as the differential mass conservation statement is concerned. So the continuity equation again going back non-conservative form and conservative form. These are the two important expressions that you would like to know. So let us move on and start talking about the linear momentum principle on a differential basis. So the ideas are now going to be repeated and therefore in order to avoid boredom in some sense what I will say is that let us see what we are doing here and let us see if you understand the overall idea as to how things are getting computed. The specific expressions for each of these quantities I will not again go in serious detail because they are very very straightforward to understand once you know what is going on and also anyway in each slide everything has been written out. So if you want you can always go back to a given slide and see how it is getting computed. So let us see what is happening. We go back to our elemental control volume which is shown on the top of the slide. The x dimension is again delta x, y dimension is delta y as usual these are our elemental lengths and for the purpose of later connecting the CFD terminology what I have shown is I am marking the bottom face as the south face, I am marking the top face as the north face, marking the left face as the west face and marking the right face as the east face. So this south north west east terminology is quite common in finite volume methods which will be covered in the next week and therefore, let us start getting used to that terminology alright. So what is happening now is we have already discussed mass flow rates coming in and going out. So I have just shown for reference m dot x coming in from the west face and m dot y coming in from the south face. Obviously again we know that using Taylor series we can calculate the mass flow rate going out of north and mass flow rate going out of east face that is not been shown because we have just completed it. What happens is now because the flow is considered to be two dimensional here. The entire derivation is done in two dimensions in reality everything is three dimensions but the ideas are similar two dimensional derivations are easy to handle from the algebra point of view and that is why we will stick to two dimensions. Generalizing to 3D is not really a big deal at all. So since the flow is two dimensional the mass flow rate that is coming in from the west face in particular let us say will carry with it in two components of velocity x as well as y and therefore this mass flow rate m dot x will bring in a linear momentum both in the x direction where my highlighter is and a linear momentum in the y direction as well. So just for the showing of these two linear momenta that are coming in at the west face I am showing the x momentum entering the control volume by a horizontal line and y momentum entering the control volume from the west face using some sort of a L shaped arrow. I hope the idea is clear mass flow rate is coming in but it carries with it two components of velocity and therefore it brings in both x linear momentum as well as y linear momentum. And the same idea is repeated for the south face mass flow rate m dot y will come in two dimensional flow hence it will bring in y momentum as well and an x momentum as well. So the terminology used here on the picture is p is our linear momentum dot is the rate of linear momentum x is the x direction linear momentum and finally I am also representing which face we are talking about. So either it is south or west or north or east alright. So let us look at one expression and hopefully you will understand what is going on. Let us look at the west face y momentum. So what we are looking at is the second line here in the list of expressions y direction momentum coming in from the west face. So fundamentally you should realize that it is the mass flow rate which is entering from the west face that is bringing in this momentum right. So therefore the mass flow rate is m dot x for the west face and that brings in a y direction momentum which will be given by the v velocity or the y velocity and therefore the product of these two will give me the rate of y momentum coming in from the west face. Now what is the expression for the mass flow rate m dot x? Remember that it is the normal component of the flow with respect to area element that carries the mass flow rate. So what is our area element here? It is delta y multiplied by one which is the depth into the depth of the paper. So that delta y shown here multiply that by the normal component of the velocity which is u or the x component multiply that by the density. So rho multiplied by u multiplied by delta y is the m dot x multiply that by the y component of velocity which is v and what you get is the linear momentum coming in from the west face. So putting it together what we realize is that this p dot y from the west face is rho times u times v times delta y this is simply rearranging the previous expression. Same thing you can do for the remaining momenta coming in. So remember that there are two momenta coming in in x and y direction from the west face. Similarly there will be two momenta coming in from the south face again in the x direction sorry x direction and y direction and those four expressions have been written out here. I hope that with the explanation that was provided just now you will easily get to understand how these four expressions have been obtained. So for example if you look at the last line absolute last line on the slide it talks about the y momentum coming in from the south face which is simply the mass flow rate coming in from the south face multiplied by the y direction velocity itself because that is what we are looking at. Now the mass flow rate coming in from the south face is delta x which is the area element multiplied by 1 of course multiplied by the normal component of the velocity which is v times rho and that gives you the mass flow rate. So rho v delta x is the mass flow rate for the south face and multiply that by the y component of the velocity and you will get the y direction linear momentum coming in. So with this we know how to obtain the four linear momenta coming in what we will do on the next slide and I will not do this in detail because it is very very obvious now what we are going to do. We will express the linear momenta going out of the control volume from the east face and the north face in terms of a first order Taylor series expansion about the linear momenta coming in from the west face and from the south face and this is precisely what has been done here. I am repeating the figure for convenience on the left top here but if you see I will look at one expression let us look at the very first expression. I am interested in knowing the linear momentum in the x direction leaving the east face so p dot x east. I will express that in terms of the p dot x west which is coming in plus the first order Taylor series expansion term. So partial derivative of p dot x west with respect to x times the delta x distance between the west face and the east face. We already have the expressions for p dot x w which was rho times u times u times delta y from the previous slide the very first expression here and you substitute it then in these two places and realize that you will generate the volume element delta x delta y as a constant factor multiplying the second term and that is about it. So what you have is the linear momentum leaving the x in the x direction from the east face is simply rho times u times u times delta y plus partial derivative with respect to x of rho times u times u times delta y delta x which is simply the volume element in the two dimension. In exactly the same fashion the remaining four linear momenta remaining three linear momenta are figured out which three would be those y linear momentum leaving from east face x linear momentum leaving from north face and y linear momentum leaving from north face and those have been listed out here. Then we go back to our balance statement which says that here is our standard balance statement now we will employ it for linear momentum rate of accumulation equal to rate of inflow minus rate of outflow plus the source we have done this before for the integral we will now do this for differential mass content within the control volume multiplied by the x direction velocity will give me the linear momentum content within the control volume and a time rate of change of that will give me the rate of accumulation of x linear momentum within the control volume. Now here what is the mass within the control volume mass content within the control volume so let me just go quickly to the whiteboard it is simply the product of the density and the volume element of the control volume. The volume element is delta x delta y multiply that by the density and therefore m suffix c v which is the mass content of or within the c v is simply rho times delta x times delta y which you substitute in here what you will realize is that delta x multiplied by delta y will then be appearing in all three terms. So, on purpose I did not write it on the first term on the left hand side or the term on the left hand side I want you to substitute that expression for the mass content within the c v in here and you will realize that this delta x times delta y delta x times delta y will come in in all terms the first term on the right hand side and first term second term on the right hand side will be corresponding to rate of inflow minus rate of outflow terms. So, remember we are still talking about x direction linear momentum. So, because each because there are four x direction linear momentum that we are talking about one here, one here, one here and one here you will generate two terms. So, please make sure that you take these expressions and incorporate those in the balance statement to realize that these two terms will come in for the x momentum equation and then finally the source term for the linear momentum equation as we had seen in the case of integral analysis is the net force acting in the x direction on the material contained within the elemental c v in this case. So, for convenience because this delta x delta y term is appearing in all terms we choose to express this net force acting on the material in the c v as some sort of a intensity of the force that is per unit volume. So, that is small f of x x multiplied by delta x delta y. So, this is just for the convenience of formulating the equation in a compact form. If you substitute for capital F x in this form you will realize that then the delta x delta y terms will cancel from all terms and what you are left with is something what is shown in the box. So, this is what you will call a differential x momentum equation. The first term here is the rate of accumulation of linear momentum on a per unit volume basis now. So, the rate of accumulation of x linear momentum within the control volume or a per unit volume basis and the two terms next to the left hand side term the remaining two terms on the left hand side will combine combinedly say that this is the net rate of x momentum leaving the control volume. So, the reason two terms come about is because as we said there are four linear momentum in the x direction two coming in and two going out and that is how we end up generating the momentum equation. Finally, on the right hand side we have this small f suffix x which is from our usage is the net x force acting on the material contained in the elemental CV on a per unit volume basis. So, this is something that you may note down right away that all differential equations of motion that we will talk about whether it was earlier the continuity equation or whether now it is this momentum equation whenever we are talking about these differential equations they will be necessarily on a per unit volume basis. Each term here whether it is the accumulation term or whether it is the net rate of x momentum leaving the control volume term signified by these two terms on the left hand side will be all always on a per unit volume basis. So, this was worked out completely for the x direction linear momentum if you go back I already have listed all y direction momenta as well. So, y direction momentum coming in from the west phase, y direction momentum coming in from the south phase, y direction momentum leaving from east phase, y direction momentum leaving from north phase. So, similar to this x direction momentum you can go ahead and formulate this balance statement for the y momentum and what you will realize is you obtain what will be given in the box at the bottom. So, again let me go to the wide board if this is the form of momentum equation that we are generating this is essentially a y direction linear momentum equation where what we have is on the right hand side small f suffix y is the net y force acting on the material within our elemental control volume on a per unit volume basis. So, each term here there are four terms let us say all four terms are on a per unit volume basis. So, the first term which I will now circle signifies the rate of accumulation of y momentum within our elemental control volume on a per unit volume basis. The next two terms together will signify the net rate of y momentum leaving the control volume on a per unit volume basis and the reason we again get two terms is because we are dealing with four linear momentum in the y direction two coming in from the west and south phase and two leaving from the east and north phase. So, keep this in mind that each term in a differential equation of motion is going to be on a per unit volume basis. Again we obtain these two equations in terms of employing a balance statement for our control volume. So, inherently we are following the Eulerian approach where we are using the control volume and then identifying mass flow rates in and momentum flow rates in and going out and so on. So, what we are generating here is the momentum equations in their conservative forms. So, keep that in mind again just like when we did our mass balance equation using an Eulerian format like what we are doing we generated a conservative form. So, so far we have obtained this on a conservative form. One last point is that we still have not talked about in detail the nature of these forces f suffix x and f suffix y the net x force on a per unit volume basis etc. We are going to talk about that little later. So, this was so far done from a an Eulerian point of view where we utilized our balance equation for the elemental control volume. Now, let us see what we can do if we want to employ our integral equation. So, just as we obtained the differential mass conservation statement in the continuity equation from an integral equation we can do this here as well. So, we start with our integral form of the momentum equation which if you go back two days on the first day we had we had discussed this. So, I am directly writing the vector form which is available in your in your notes from the first day for a stationary CV exactly the same procedure as was employed for the mass balance situation namely use the Leibniz rule for first term to bring in this time derivative inside the integral use the divergence theorem to convert the area integral over the control volume integral or volume integral. And just because we want now all terms as volume integrals we for convenience express the net force acting on the material in this control volume in terms of an integral representation where the small f vector is again simply the net force acting on the material inside the control volume on a per unit volume basis because that is what is getting then integrated over the entire control volume. So, now what we have achieved is that all integrals are on the volume basis. So, that we can combine the integrants together and we combine the integrants in the following form the partial derivative with respect to time of rho times v plus divergence of rho times v times v all vectors v minus this vector f which is on a per unit volume basis integrated over the entire control volume is giving us 0 again using the same argument as before arbitrary control volume will necessarily mean that the integrand itself is identically equal to 0. So, again writing that as writing that with the f small f the per unit volume force onto the right hand side we obtain the momentum equation on a per unit volume basis. Now, remember what we have done here we started from employing our balance statement in a Cartesian frame of reference. So, we explicitly written wrote down linear momentum x direction linear momentum in y direction coming in from both west and south faces etcetera and went ahead and then obtained something like this for the x direction something like this for the y direction. So, in some sense I will say that even though may be the procedure is easy to understand it usually involves more algebra. However, if you look at this integral equation to begin with and then employing our Leibniz ruler and divergence theorem we convert that into a differential form you directly obtain within a few steps the same equation as what we have obtained using the Cartesian framework. Only thing is that here it is a vector equation directly, but that is no problem and we can always generate the corresponding Cartesian components by replacing this vector v with either u or small v or small w. So, again since we have obtained the momentum equation here which is again on a per unit volume basis using a balance statement because this was the balance statement for the integral form we have converted that into a per unit volume balance statement which is essentially the differential momentum equation. This again is in the conservative form alright. So, let us see how we can now obtain the component equation again from our vector equation. Just one point I would like to make here that the term that you see here inside the divergence rho times vector v times vector v. So, this is a special product of two vectors it is what is called as a tensor product of two vectors. Let us not get into the details of this right now usually the properties of these vector products of rather tensor product of two vectors etc. We deal with in an advanced fluid mechanics course. At the moment you do not have to worry about it make sure that this is neither a dot product nor a cross product it is what is called as a tensor product of two vectors. We are not going to be bothered about this much because we will usually deal with the component forms. What happens is when we are obtaining the differential equation from the integral form directly this product will come in without really you doing anything when we employ the divergence theorem. So, that is the way it has appeared but do not worry too much about it because immediately we are going to start writing the component equations. So, what I said was if I want from the vector equation the corresponding x direction equation I will simply replace the vector velocity v here and one of the vector velocities v there with the corresponding directional velocity which is in this case u. So, let us just look at this carefully on the whiteboard. So, this is the form of the equation let me go back and make sure that is what it is. This is right here I have not bothered to write the brackets around this rho times small u times the vector v but that is about it. So, now let us see what we how we can expand this into our standard Cartesian form. So, remember that our grad operator is unit vector i times partial derivative with respect to x plus unit vector j times partial derivative with respect to y and unit vector z k rather times partial derivative with respect to x and the velocity vector v is i times u plus j times v plus k times w as we know. So, all that we do is we will substitute in here for the velocity sorry for the gradient operator and for the vector velocity v here what will end up happening is we will get d dt of rho u as it is the first term on the left hand side plus the very first term on the right hand the first term when we expand this divergence of rho u v will be d dx partial of rho times u times u plus d dy partial of rho times u which is common in here. The second term will be essentially j dotted with j and therefore, you will obtain a v the third term will be d dz partial rho times u which is common at the at the top multiplied by w and this will be equal to f of xx which is the net force acting in the x direction on a per unit volume basis. So, forget about right now the third dimension the z dimension. So, let us get rid of this for the time being because earlier we were discussing two dimensional derivations. So, let us get rid of this let us get rid of this. So, what is left is partial derivative of rho u with respect to t plus partial derivative with respect to x of rho u u plus partial derivative with respect to y of rho u v equal to f suffix x on a per unit volume basis and let us make sure that this was exactly the statement that we had obtained through our initial Cartesian derivation and it is right here ok. So, you do not get anything different is just that the different approaches to obtain the same equation and that is that is what you will obtain right. So, coming back to this then now I have actually written this with all three directions. So, therefore, we have this partial derivative with respect to z also term which is fine in exactly the same fashion you can repeat this for the y direction and the z direction and you will realize that this is how things will turn out. So, my suggestion then is just like what we worked out right now for the x direction you start with the vector form and obtain the corresponding expressions for the y direction and the z direction by replacing this vector v with small v and small w and one of these vectors v with small v and small w and you will essentially generate this set of equations. So, this is to sort of get familiar with the way these equations are to be manipulated. What we will start talking about now is the nature of this net force that is acting on the material within the control volume and we said that the way we have expressed this is on a per unit volume basis. So, in general the vector small f will be considered as the net force on a per unit volume basis acting on the control volume the material contained within the control volume and during the integral analysis we had noted that in general the forces are surface forces and body forces as well. So, both surface forces and body forces will contribute to the net force. So, let us start talking about the nature of these forces. But before that there is one another manipulation which I want you to carefully see. See what we had done let me go back to the mass conservation equation. Remember we employed our balance equation and obtained the mass conservation equation. This was obtained in the conservative form because we started with our Eulerian balance approach for the control volume. Then what we did was we did some manipulation and obtained the non-conservative form from the conservative form. Essentially all that was done was opened these derivatives that is about it. Now here in case of momentum equations so far we have been obtaining the momentum equations on a conservative form. So, the one that you see right now here on the board projected or the vector form that you see here or all these forms written on slide number 12 they are all in conservative form because essentially they have been obtained using balance statements. Just the same way as we did some manipulations on the continuity equation to obtain a non-conservative form from the conservative form we can do that here as well and that is what is written on slide number 13. So, we start with the conservative form of the momentum equation which is at the top and subtract from it u multiplied by the conservative form of the continuity equation. So, please realize that inside here what we have is partial derivative of rho with respect to t plus d dx of rho u plus d dy of rho v plus d dz of rho w equal to 0 is nothing but the continuity equation that we derived few minutes earlier on the conservative basis. So, the entire continuity equation which is equal to 0 by the way is multiplied by u and subtracted from the conservative form of the momentum equation. So, because the entire continuity equation is equal to 0 you multiply anything to 0 and add or subtract to the equation at the top it does not really change the value of the equation so to say. So, this is perfectly fine. So, let us see what happens. I will write this only for two dimensions just to save my writing a little bit. Remember that this entire continuity equation is equal to 0 and therefore adding or subtracting any multiple of the 0 to our top equation is not really going to change anything. So, let us just see what will happen. The first term is just let us just look at the time derivative I have d dt of rho u as the first term on the top equation minus u times partial derivative of rho with respect to time what will this give us. Essentially what you need to do is you open this first derivative using product rule. So, when you open you will get u multiplied by d rho dt which is going to cancel with this term and what will be left is rho multiplied by d u dt on a partial basis. I hope you understand this. Then the second term you see what happens what I have is d dx of rho times u times u from the top equation minus u times d dx of rho u this is from the second equation. Let me explicitly write this one term and you will see what is happening. This will imply that for the purpose of now convenience what I am going to do is I am going to split this as two terms inside the product rho u and u and then rho u will be treated as one term u will be treated as another term and then you expand this derivative using the product rule. So, therefore, u times d dx of rho u plus rho u times d u dx will be the first term. Is that fine? Let me ask some of the audience here if they are able to follow it minus u times d dx of rho u and as you can see the first term and the third term will cancel and what you will end up with is rho times u times d u dx. In exactly the same fashion you can manipulate the third term as well and I hope now you can complete this yourself. So, let me go back to my slide and as you can see the first term here where my highlighter is standing is the first term that we had obtained on our whiteboard rho times partial derivative of u with respect to t. The second term where my highlighter is standing was what we obtained on the whiteboard a minute ago which was rho times u times partial derivative of u with respect to x. Similar two terms you can now obtain on your own by subtracting from the third term here u multiplied by third term here and so on. What you realize is that rho is essentially a constant entirely on the left hand side. So, I bring that out and then I will realize that u is getting operated by a bunch of terms all derivatives. So, I will write this in the form that rho times partial derivative with respect to time plus u times partial derivative with respect to x plus v times partial derivative with respect to y plus w times partial derivative with respect to z operating on the x component of the velocity u and that is equal to the x direction net force acting on the material within the control volume on a per unit volume basis. So, that has been carried the way it is the reason is because we were essentially adding 0. So, nothing was changing in the first equation, but realize what this is whatever is incorporated inside the brackets is nothing but the substantial derivative operator. So, I can replace this entire bracket with capital DDT which is nothing but partial derivative with respect to time plus u dot grad as they say or u DDX plus v DDY and w DDZ. And therefore, an alternative way in which the momentum equation has been now obtained in the x direction is that it is rho times substantial derivative of the x direction velocity is equal to the net force acting on the control volume on a per unit volume basis. So, the way to interpret this is as follows. So, let me go to the white board. So, imagine that a fluid particle which has the same volume as the volume of our elemental CV is occupying that elemental CV at the instant of our interest and let the volume element of that fluid particle be delta v. So, let me multiply this entire equation by that volume element delta v which in our case of Cartesian control volume is essentially delta x multiplied by delta y, but let me not write that explicitly I will simply write. So, now we are suddenly seeing something different in the sense that rho multiplied by delta v can be essentially interpreted as the mass content of the fluid particle which is occupying the same position as the control volume that we have been talking about. So, mass multiplied by an acceleration equal to net force in the x direction is nothing but the statement of Newton's second law of motion. And therefore, what we are saying is that the momentum equation which we started out as a balance statement for a control volume we have been able to do some manipulations and we have been able to interpret this as essentially a statement of the Newton's second law of motion for a fluid particle which is having the same volume element as our elemental control volume and that has been essentially occupying the space where our control volume is. So, in that sense what I have written out is rho multiplied by the substantial derivative of u equal to a net force in the x direction per unit volume basis is essentially a statement of Newton's second law of motion for an elemental fluid particle which is supposed to be coinciding with the elemental CV at the time of our interest where everything is getting analyzed or calculated. And so, this entire thing is expressed as a per unit volume statement. The point here to be noted is that since we are talking about now a fluid particle per se we are now suddenly talking about a Lagrangian point of view. And therefore, the last statement which is boxed at the bottom of the slide can be interpreted as a non-conservative statement or form of the momentum equation because now we are talking about this as a Newton's second law of motion in terms of f equal to ma applied for a given elemental fluid particle just so happens that by the design of our derivation we will essentially say that that elemental fluid particle is occupying the same position where our elemental control volume was when we began our discussion. So, just like we obtained a non-conservative statement of continuity equation earlier this way you are able to obtain a non-conservative statement for the momentum equation as well. This was of course done for an x direction similar treatment can be performed for the y direction as well and one can even generalize the statement in a vector form. In fact, let me do that on the whiteboard. So, at the top we have x direction similarly we can write for y direction and finally to generalize it we can write it as rho times substantial derivative of the velocity vector equal to force net force vector on a per unit volume basis. So, this all these statements that have been written out here are all on per volume. So, keep this in mind that same equation you do some manipulations and you can interpret that from a Lagrangian point of view and then you are in a business of saying that this equation now is in the non-conservation form alright. So, this is what we wanted to point out in this case. So, having done this discussion now we are in a position to start talking about the forces that are acting on the fluid particle. So, now what we will do now is because here on this slide number 13 we established that whether we start with a control volume and do a balance on the control volume to obtain the momentum equation or equivalently we consider a fluid particle which is coinciding with the control volume at the instant of interest and choose to employ Newton's second law of motion to that particular fluid element which is coinciding with the control volume we essentially get the same result ok. So, therefore, for the purpose of further discussions what we will say is that we will now start talking about that particular fluid particle for which this non-conservative form of the momentum equation has been derived, but keep in mind that they are interchangeable these two forms. When we talk about the conservative form it is essentially to deal with the control volume formulation when we talk about the non-conservative form it is essentially to deal with a fluid particle which has been identified as a fixed amount of mass where we are employing our Newton's second law of motion. One way or the other we are essentially saying that the fluid particle will be coincident with the elemental CV at the instant of interest. So, from that point of view the net force which has been shown here can be considered to be acting either on the control volume or equivalently on the fluid particle which is exactly coinciding with the control volume. So, please keep this in mind.