 Okay, this begins the discussion on differential analysis and by far perhaps this is the longest topic that we will cover primarily because the differential analysis is the most important part that one should understand before getting on with the CFD analysis. What we have discussed so far the integral analysis and the kinematics will serve as auxiliary discussions to the differential analysis. Let me just begin this differential analysis with some general comments. We saw that in the case of integral analysis what we were after was calculation of overall rates of transfer such as a mass flow rate from an integral control volume or a force transmitted from one object to the fluid or vice versa, fluid to an object. We did not really care about the detailed flow information in those situations. However, if you are interested in knowing the detailed flow information which usually we will call a point to point information what we need to resort to is the differential analysis we cannot really utilize integral analysis. It so turns out that if you have the results of integral sorry the differential analysis you can always find out by appropriate integrations the overall rates of transfer as we were looking at yesterday. So in that sense perhaps you can say that differential analysis will give you far more information about the flow field than the integral analysis will give and differential analysis always will be a more sophisticated analysis tool than integral analysis perhaps that can be said. So as the name suggests differential analysis will deal with differential equations for the conservation laws. Now we have already seen the conservation laws when we talked about and derived the integral equations. These are the same conservation laws that we will utilize namely a conservation of mass statement, a linear momentum statement and a conservation of energy statement. Only thing is that now here we will derive these in their differential forms. To do these derivations we will actually follow two approaches at least for the mass balance and the momentum equation we will employ balance statements as we have seen in the case of integral analysis to an elemental control volume. So if you employ the same balance statement for an elemental control volume we will generate differential equations of motion and one other way of obtaining the differential equations of motion are from the integral equations directly by employing certain mathematical manipulations such as that Leibniz rule and the divergence theorem that we had seen yesterday in the opening session. So this is how we will proceed in order for you to get a complete idea both approaches are being followed namely a balance equation and conversion of the integral form into a differential form directly using some mathematical manipulations. You can decide which one appeals to you more and you can stick to it. As far as I can understand when the second half of this workshop begins some of these equations will be briefly re derived in the form of a balance statement. So that will reinforce your understanding once again in the second half of the semester. But there are several books which follow the approach of obtaining these differential equations from the integral equations directly using those divergence theorem and Leibniz rule etc. And that is the reason we thought that let us incorporate both approaches so that you should know if you read one of those how to how to interpret that. So let us just revisit our balance statement as we had seen in the case of integral analysis which was again the rate of accumulation of mass momentum or energy within a control volume equal to the rate of inflow minus the rate of outflow plus the rate of increase due to a source. So now what we will do is that we will take this balance statement and we will employ it to an elemental control volume. So this is not an integral control volume. This is an elemental control volume in the sense that the 2D control volume that I have shown on the slide here has lengths delta x in the x direction and delta y in the y direction with the understanding that both delta x and delta y are sufficiently small tending to 0 in the calculus sense. And because these lengths are very small we are in a position to employ the Taylor series expansions without any problem when we want to express the change in the variable from one location to the neighboring location. So let us see how this is done. Many of you are perhaps familiar with this so it will be a good revision but those who are not familiar with this should perhaps pay good amount of attention. So let us start with the conservation of mass statement for the differential or the elemental control volume. So what I have shown here is from the left face and the bottom face mass flow rates are flowing into this control volume and from the top face and from the right face they are leaving the control volume. And because the distances delta x here and delta y are elemental meaning they are tending to 0 we are in a position to express the mass flow rate for the right face in terms of the mass flow rate from the left face and a first order Taylor series expansion. So you see that m dot x is the mass flow rate coming in from the left vertical face here and using a first order Taylor series expansion when we go from the left face to the right face the mass flow rate that is leaving the right face is expressed as m dot x plus the partial derivative with respect to x of m dot x multiplied by the distance delta x as the standard first order Taylor series expansion. Exactly the same thing is done for the y direction what is coming in from the bottom face is m dot y and what is leaving from the top face is expressed as a first order Taylor series expansion m dot y plus partial derivative of m dot y with respect to y times delta y. So in terms of mass flow rates coming in and going out for the control volume m dot x and m dot y are coming in and these two are going out. So if you want to go back to our balance statement we have rate of accumulation of mass within the control volume equal to rate of inflow minus the rate of outflow plus the rate of increase due to a source. Now clearly again as far as the mass conservation is concerned we are not going to include this term due to a source because only way you can include the source term is if you have nuclear type reactions in your flow which we are going to discard. So we will include essentially only these three terms one on the left and two on the right and that is precisely what has been simplified here. So the term on the left is simply the rate at which mass is getting accumulated within the control volume and I have written out the mass content within the control volume as m suffix cv equal to the density rho times the volume of this control volume which is a two dimensional control volume. So we simply have the volume as delta x multiplied by delta y if you want there is unit depth in the z direction. So that is the mass content within the control volume and we differentiate that with respect to time and remember that this is a partial derivative. The reason for using partial derivatives is that these are infinitesimally small control volumes because delta x and delta y are considered to be very small tending to 0. So therefore essentially what we are doing on a physical basis is that we are looking at a very very small region equivalent to some sort of a point in space where we are trying to come up with this conservation of mass statement. So since we are focusing on one particular point location we are going to talk about differentiating this with respect to time in the partial manner since we are focusing on one particular location. If you go back to the balance statement what we have is rate of inflow of mass minus the rate of outflow of mass. So if you see in the x direction inflow is m dot x outflow is m dot x plus this business. So if you subtract the term going from the right face from the term coming in the left face what you will be left out is minus of partial derivative of m dot x with dx of with x times delta x is exactly what is written out as the first term here. Similarly minus partial derivative with respect to y of m dot y is the times delta y is the term that will be left out in the y direction. So now what is required is if we want to complete this derivation we need to come up with expressions for the mass flow rate in the x direction this m dot x and mass flow rate in the y direction that is m dot y coming into the control volume because then you substitute it here and here and you are done. So let us look at the left vertical face. Remember last yesterday's discussion in general the velocity here will have two components an x component and a y component but as far as a vertical face is concerned mass flow rate across a vertical face will be carried only by the horizontal velocity component which in this case will be the x velocity component. So therefore m dot x is simply that x velocity component which is u multiplied by the area of cross section available for the flow to come in which is delta y multiplied by 1 in the depth of the plane multiply by the density of the fluid let us say rho and so m dot x is simply rho times u times delta y which is the mass flow rate coming through the left face which is vertical face into the control volume. Similarly, we need the expression for m dot y here which is the mass flow rate coming in from the bottom horizontal face. Since we are talking about a bottom horizontal face the vertical component of the velocity which will be the y component of the velocity will carry out a mass transfer across this flow or mass flow across this flow and therefore m dot y then is that y component of the velocity multiplied by the element area which is delta x multiplied by 1 in the depth multiplied by the density and that is all you need to know. So if you plug this m dot x and m dot y and m c v which is the mass content within this elemental control volume in our balance statement and simplify because what will happen is when you substitute these terms in each of these expressions you will realize that the volume element delta x multiplied by delta y will appear in each of these terms which then you can cancel from the equation all together and what you will be left out is partial derivative of density with respect to time equal to minus partial derivative with respect to x of the quantity rho times u minus partial derivative with respect to y of the quantity rho times v and the standard way of expressing this conservation of mass statement is bringing everything on to the left hand side and that is what is shown in the boxed equation at the bottom. So this is nothing but the conservation of mass statement on a differential basis written out in a Cartesian form in two dimensional space. There is another alternative form that you can bring about. So let us first look at the form that we obtained right now which is boxed out here and repeated as the top left equation. If you want to immediately write this in the vector form you will realize that this can be written as partial derivative of rho with respect to time plus the divergence of the quantity rho times the velocity vector and that is equal to 0. So the box term which I am boxed equation which I am calling as equation number 1 is the general vector form of the Cartesian equation that we wrote here. Now what we can do is we can expand this left top equation by opening these derivatives with respect to x. Remember so far we have not made any assumption in terms of the density being constant or not. So far we are assuming that the density is indeed not a constant and therefore all these derivatives of density with respect to either time or space will be in general nonzero. So if you open this partial derivative of rho times u with respect to x you will generate two terms. Similarly if you open the derivative here you will generate two terms. What I have done here is that I have combined the three terms that are underlined which are partial derivative of rho with respect to t plus u times partial derivative of rho with respect to x plus v times partial derivative of rho with respect to y and if you go back to our kinematics discussion these three underlined terms will essentially combine into the substantial derivative or the material derivative of density which is capital DDT of rho. The remaining two terms have density as a common factor so that will be rho the entire thing multiplied by du dx plus dv dy. Remember we are dealing with a two dimensional situation so from the just concluded discussion on kinematics we will remember that this du dx plus dv dy was essentially the volumetric rate of strain which was also given by the divergence of the velocity. In fact this is the divergence of the velocity du dx plus dv dy in Cartesian two dimensional Cartesian coordinates and therefore we have an alternative expression for the statement of conservation of mass on a differential basis which is now given in terms of the substantial derivative of density plus rho multiplied by the divergence of velocity equal to 0. So, this I am calling as equation number 2. Essentially equation number 1 and equation number 2 can be formed from each other so if you take this you can form this or if you take this you can form this and therefore equations 1 and 2 are the two alternative forms of the differential conservation of mass and popularly either of these equations is what is called as the continuity equation as some of you may already know. So, this was from the balance statement approach employed for a elemental control volume which had a size delta x multiplied by delta y. So, let us see if we can do the same thing that is obtain the governing equation from the integral equation. So, if you go back to your slides of the integral equation and if you see the integral statement of the mass conservation you will have d dt that is the time derivative of rho dv integrated over the control volume plus an area integral over a rho v dot n dA equal to 0. So, let us employ this for a stationary or a fixed control volume. So, far we are still talking about an integral equation. So, this control volume that we are talking about here which is fixed for stationary is essentially an integral control volume. So, what we do here is that because the control volume is stationary and fixed you can imagine that the limits on this control volume are essentially constants and therefore if you want to utilize that Leibniz rule which was shown only for one dimension now here really we are dealing with three dimensions in general, but that is fine that is not a problem. The Leibniz rule said that you can take this derivative outside the integral inside the integral by making it into a partial derivative, but then there were two more terms that will come because the limits in general would be functions of the parameter. However, in this particular case we are choosing that the control volume is a fixed control volume. Therefore, the limits are fixed and you are not going to have any contribution from those two terms in the Leibniz rule. The only term that will survive is the partial derivative with respect to time of rho times dv when you are bringing this derivative inside. For the second term which is an area integral of rho v dotted with n dA what we do is we employ the divergence theorem as was outlined yesterday in doing so the quantity rho times v is changed to a divergence of that rho times v and then the entire divergence term is integrated over the control volume. So, what we are doing is that we are utilizing the divergence theorem to convert an area integral into a volume integral. The reason being that then both integrals are on the same control volume so that you can combine them. If you combine them you will realize that they can be written as partial derivative of rho with respect to t plus divergence of the quantity rho times v the entire quantity integrated over the control volume equal to 0. Here we never specified anything special about the control volume if some quantity which we are calling this as d rho dt plus del dot rho v which is what we will call an integrand inside the integral. If for any arbitrary control volume if this integrand when integrated over the control volume is giving you a 0 result it necessarily means that the integrand itself is identically equal to 0 because the control volume is arbitrary. And therefore, what we end up concluding is that the integrand which is d rho dt partial plus del dot rho v which is the divergence of rho v equal to 0 and this is nothing but the continuity equation that we had already seen. So, what we have done here is that we started from an integral approach and then we essentially converted the integrand to be the mass conservation principle on a per unit volume basis because if you see this particular step the integrand is getting integrated over the entire control volume. So, the integrand then you can associate with an elemental control volume within the arbitrary integral control volume and that is why we can interpret this entire continuity equation as is boxed here as a conservation of mass on a per unit volume basis because you multiply this entire equation by the elemental volume and then integrate over the entire control volume you will go back to your integral equation for the mass balance. So, in that sense this will be treated as a continuity equation which is a conservation of mass on a per unit volume basis. In fact, if you saw this equation as we derived it earlier, I had remarked that the volume element delta x multiplied by delta y will occur in each of these terms which you can then cancel. So, therefore, this also is giving you the idea that we are talking about a per unit volume basis because each term here is getting multiplied by the volume element delta x delta y when you substitute of course all these in here. So, in that sense the boxed equations one and two here or which is the same as here is essentially a continuity equation on a per unit volume basis. In fact, when we say that we are dealing with a continuity equation it is necessarily a conservation of mass statement on a per unit volume basis. So, again doing some manipulations starting from our continuity equation here and opening this del dot rho v you can actually obtain your alternative form of the continuity equation which was obtained earlier in here as second equation substantial derivative of rho plus rho times del dot v equal to 0. One last point I would like to make here is that we can directly obtain this alternative form which involves the substantial derivative in the following manner. So, what we can do is we can imagine a fluid particle which is essentially an elemental mass which always has the same amount of mass that we are following as it goes from one location to the other. So, because we are following the same mass from one location to the other inherently there is no change in the mass content for that same mass and because we are following it as one fluid particle as an entity we are essentially going to say that in a Lagrangian frame of reference we will realize that its mass content is not changing with time or equivalently the substantial rate of change of the mass content for a fluid particle is 0. So, let delta v be the volume of the fluid particle multiply that by the density of the fluid particle. So, that rho multiplied by delta v is the mass content for a fluid particle by definition it is the same mass that we are following everywhere when we talk about fluid particle. Therefore, the substantial derivative of that is identically equal to 0. The substantial derivative will act just like a standard derivative. So, you can open up this product. So, this derivative using the product rule and what you will see is that the first term will come as substantial derivative of rho times delta v plus rho times substantial derivative of delta v and if you divide this entire equation by rho multiplied by delta v what we will see is that the first term turns out to be 1 over rho substantial derivative of rho plus 1 over delta v substantial derivative of delta v. But I hope you remember from the discussion that we completed earlier in the morning on kinematics that this 1 over delta v substantial rate of change of delta v is nothing but volumetric rate of strain which is nothing but divergence of velocity as we had shown. So, therefore this 1 over delta v times substantial rate of change of delta v is simply replaced with the divergence of velocity and then multiplying through again by rho you obtain substantial derivative of rho plus rho times divergence of velocity equal to 0 as what we had seen here as well as what we had seen here. So, there are various ways in which you can obtain this differential equation for the mass conservation. Essentially we have seen three different ways one was employing the balance equation for an elemental control volume. The other approach was to convert an integral form into a differential form using the Leibniz rule and the divergence theorem and third approach in this case is to identify that the mass content for a given fluid particle is not changing at all. So, therefore the substantial derivative of the mass content would be 0 and then you obtain the equation in the form of the substantial derivative as was obtained here which you can always go back and put it in a form which was obtained earlier. So, this is what this is what we have for the differential mass conservation equation what I will do is I will stop here right now. Thank you very much.