 Now, we will start talking about the integral analysis that we employ in fluid dynamics. As I mentioned in the morning session, the integral analysis may not be necessarily required for a CFD background, but it is a useful analysis tool in fluid mechanics. And also as I mentioned, the differential equations of motion can be obtained through the integral equations of motion using some mathematical manipulations, which we will do in the later part of the course. So, it is a useful thing to complete in our opinion. So, that is what we will be doing now in this session. So, many of you will probably recall that we employ these Eulerian and Lagrangian approaches of description of fluid motion in fluid dynamics. And if you talk about Eulerian approach, what we are doing is we are referring to a control volume or an open system sort of an approach. And typically, this is the preferred approach in fluid mechanics. The Lagrangian approach will essentially involve a control mass or equivalently what people call in thermodynamics as a systems approach, which is essentially identical to a solid body dynamics in the sense that we are always talking about the same material or the same mass, which we will follow in the fluid domain. Whereas, in case of an Eulerian approach, what we are doing is we are identifying a control volume, which is a region of space within the domain. And we monitor what changes are happening because of the fluid flow in and out of that control volume. Now, there are two more ways you can classify the fluid dynamics analysis, which is the integral analysis and the differential analysis. When it comes to integral analysis, we are interested really in overall transfer rates, whether they are mass flow in and out or momentum flow in and out or energy flow in and out of a control volume. Whereas, when it goes through a differential analysis, we are actually interested in finding out the detail, which will be essentially point to point information about the flow field. So, in case of a differential analysis, we want to find out the point to point variation of pressure, density, temperature, etc. Whereas, in the integral case, we are interested to know the overall transfer rate of energy or momentum or mass in an integral control volume. In principle, both Eulerian and Lagrangian approaches can be employed on both integral and differential approaches. So, what we are probably going to do right now is what you can say is an Eulerian approach on an integral analysis. So, the backbone or the cornerstone of the integral analysis is essentially a balanced statement that we write for a control volume of a finite size. So, why I am talking about a finite size control volume here is because we are dealing with an integral analysis. Essentially, a similar statement can be written for a control volume of an infinitesimal size when we want to obtain the governing equations of motion on a differential basis, which we will do later in the course. But right now, for an integral or a finite size control volume, we are going to essentially write a balanced statement and this balanced statement is what really that you need to know in order to understand the integral analysis. As I said, this is the cornerstone of the integral analysis. So, the balanced statement is actually fairly intuitive if you look at it. What it simply says is that the rate of accumulation of mass or momentum or energy let us say. So, let us consider mass for now. If you consider a control volume, the rate at which mass will be accumulated in that control volume will be essentially equal to the rate at which the mass is flowing into the control volume minus the rate at which the mass is flowing out of the control volume plus if at all, if there is a rate of increase of mass due to some sort of a source. So, we will talk about the source little later, but this balanced statement essentially is a general balanced statement. So, it does not have to be necessarily restricted to only mass balance and that is why I have added mass or momentum or energy. So, when it comes to a conservation of mass statement, we will introduce this balanced statement for the mass. When it comes to a momentum equation, we will introduce this balanced statement for momentum and when it comes to a conservation of energy principle, we will use this balanced statement for energy. But as far as the integral analysis is concerned, this is it. This is the physics of the integral analysis and we are in a position to say something about the balance of a quantity with respect to a control volume. So, now having written the balanced statement in English or in words, let us say, now our job is to try to figure out if we can come up with corresponding mathematical expressions, which will tell what is the expression for the inflow of mass, outflow of mass, inflow of energy and so on. So, let us begin the discussion with the mass balance statement. If you go back here for a minute, this is entirely written in words whereas the first line here is written in symbols that is all. So, it is exactly the same statement as was written on the previous slide. The term on the left hand side denotes the rate of accumulation of the mass within a control volume and an arbitrary looking control volume has been drawn here, which has the volume of V, let us say and the surface area that is surrounding the control volume is A, let us say. And going back to our balance statement, so rate of accumulation of mass within the control volume is equal to the rate at which mass is coming into the control volume. So, there is a fluid flow in and out of control volume in general because of which mass of fluid is coming in at some rate and similarly mass of fluid is going out at some rate. So, m dot in is the mass flow rate into the control volume minus the mass flow rate out of the control volume and remember that there was the last term here rate of increase of mass due to a source. Now, in this case the mass source that we are referring to is essentially not to be considered for most situations in fluid mechanics unless we are actually talking about a creation of a mass through a nuclear type reaction. Now, this is a very special situation. This can be a mass source or a sink by the way. So, the nuclear reaction can actually create a mass or destroy a mass through the Einstein's relation, but that seems to be a very very special situation and in general we are not talking about this at all when we come to standard fluid mechanics problems. So, for standard fluid mechanics problems this mass source term which will be either a positive term that is increase or decrease which is a sink term which will be because of nuclear type reactions will be assumed to be not present at all and therefore I have really cancelled it out altogether. Now, for each of these remaining three terms the accumulation the inflow and the outflow we will come up with mathematical expressions and then put the lot together. So, if you look at the first term on the left hand side that is only term on the left hand side which is the rate at which mass is getting accumulated in the control volume. So, it is a time rate of accumulation and that is the reason we are denoting this as a time derivative and m suffix c v in here is the mass content within the control volume at any instant of time. So, instead of writing this as m suffix c v I simply use an integral representation to denote that mass content within the c v which is simply integral over the control volume and elemental mass within the c v. So, d m is an elemental mass within the c v if you integrate that over the entire control volume you will get the total mass content within the control volume and the elemental mass within the control volume is further represented as the density times the elemental volume for a mass element. So, we know that mass is equal to rho times volume and therefore the infinitesimal mass that we are talking about say here somewhere within the control volume is simply the density times the elemental volume of that small elemental mass and then you integrate that over the entire control volume and therefore this d d t of integral over the control volume rho times d v is nothing but the integral representation of the mass rate of mass accumulation within the control volume. Now, what I have shown here on the figure is that over the entire control surface A I have identified a suffix o as the outlet area which means that it is only through this area a suffix o that the fluid is going out of the control volume. Similarly, there is a suffix I which is the inlet area this is the only inlet in terms of fluid coming into the control volume. Remaining part of the control volume the surface area that is surrounding the control volume can be considered to be walls if you want. So, that there is no inflow and outflow across the remaining part this is just for the purpose of derivation. Now, if you talk about the mass flow rate out then we are only restricting ourselves to this a suffix o. Again starting from the basic expression m suffix m dot out rather is the mass flow rate that is going out of the control volume will be simply the area integral over only now this a suffix o because that is our only area over which the mass is flowing out of the elemental mass flow rate out of the control volume. In other words what we are doing is that we are actually dividing this a suffix o into several small area elements and for each area elements within that a suffix o we are essentially writing a small mass flow rate that is going out and then we integrate this over the entire a suffix o which is the outlet area and that is precisely what is written in the next part of the statement. So, if you look at any one elemental area within a o I will call that d a o that is because it is an elemental area. What is the mass flow rate crossing an elemental area d a o? It is simply the density at that elemental area d a o which I am calling as rho suffix o times the normal component of the outflow velocity at the elemental area d a o. So, keep in mind that it is the normal component of a velocity with respect to an area that will actually carry the mass flow across the area. So, what I have shown here in the picture is in general the outflow velocity is oriented somehow which is shown by this vector v suffix o and the unit outward normal for the elemental area d a o is your n hat o which is always pointing outward which is positive. So, now if you look at this expression for the elemental mass flow rate out rho suffix o times the normal component of the velocity at the elemental area outlet. You can actually replace this with a dot product appropriately form dot product of the outflow velocity which is in general v0 dotted with the unit outward normal which is n hat o and that is precisely what is written as the third part of or the fourth part of this statement. Everything else is the same. So, we are performing the integral over a o which is our outflow area rho naught is the rho suffix o is the outflow density. So, that is that is fine v0 or v suffix o dotted with n hat o is essentially giving you the normal component of the velocity with respect to this elemental area d a o and then we integrate over the entire outflow area meaning that we add all these elemental d a o's together over the entire outflow area and that gives me the total mass flow rate going out of the control volume. When it comes to the mass flow rate that is flowing into the control volume we essentially repeat the same analysis on the inflow area a suffix i. Now, here if you see the third expression which the pointer is right now pointing at the mass flow rate in then is an integral over the inlet area the density at the inlet. So, in general we are saying that the density can vary from the inlet to outlet which is fine which is a general case multiplied by the normal component of the velocity of the flow that is coming in now at the elemental inlet area d a i and then you integrate that over the entire inlet area. Now, one thing that you need to notice here is we said that the unit outward normal will always be positive when it is pointing away from the surface. So, if you see the orientations of this n hat i which is the inlet unit outward normal and v i which is the inlet velocity the angle between these two is greater than 90 degrees because you will attach these two vectors at their basis. So, you have to point out the velocity here and the unit outward normal here. So, that the angle between those two will be greater than 90 and therefore, if you form a dot product it will be a cosine of 90 plus which will be a minus. However, what we want is we want finally, the dot product to give rho suffix i v n i times d a i if you form the dot product it is going to introduce one negative sign because of the angle being greater than 90. To cancel that you add another minus sign here so that minus from here and the one minus which will come from the dot product will cancel each other and will give you the scalar quantity rho i v n i d a i as the mass flow rate coming in and that is about it. All that you need to do now is we go back to our balance statement and wherever you see our term such as this accumulation term you simply replace that with the mathematical expression that we have obtained in this integral representation. So, if you do that the left hand side is the accumulation term. The first term on the right hand side is m dot in which is minus this should have been a suffix i there is a typo here. So, please correct this this should be a suffix i the integral rho i v i n dotted with n i d a i which is exactly how it is written here. So, this should have been i also and the second term which is the outflow term has a minus anyway and the outflow term does not have any minus in it. So, if you go ahead with it the minus is there and then the outflow term does not have a minus in it and therefore the integral representation can be completed here. So, what you have here on the first line is the left hand side denoting the rate of accumulation of mass within the control volume. The first term on the right hand side along with the minus here is the rate at which mass flow rate is coming into the control volume and again as I said this integral should have been over a suffix i. So, please correct that the second term without the minus is the mass flow rate going out of the control volume and that is correctly represented. Now, what is usually done is that the two surface integrals on the right hand side are usually combined into one area integral and that one area integral will be denoted as the area integral over the entire control surface area. So, here is a minus sign, here is a minus sign. So, then the minus sign is common between the two integrals. So, we maintain that and these two integrals rather than writing those two separately as an integral over the inlet area and an integral over the outlet area. We simply combine without really then prescribing any inlet and outlet subscripts as rho times v dot n dA. So, what this means is that we are carrying out physically now, we are carrying out this integral over the entire surface area A that is surrounding the control volume. If you go back the integral will return a non-zero value only over A naught and A i because that is the only two areas where you have flow going out and flow coming in. We said that the remaining part of the area A will not contribute to anything and it will automatically contribute a value of 0. So, that 0 essentially is getting added in this entire integral representation and therefore, combining these two individual integrals into one integral over the entire area A makes sense because though the contribution is going to come only from the inlet area and the outlet area. So, now, if you see the standard way in which the integral mass balance equation is written for a control volume is by bringing this term on to the left hand side and then writing this as d dt of the integral over C v rho d v plus the integral over A rho times v dot n dA equal to 0. And the way to interpret these terms physically is that the first term on the left hand side then is still our rate of accumulation of mass within the C v and the total area integral which is the second term on the left hand side is now to be interpreted as rate of outflow minus rate of inflow. The reason is because if you go back it is still the same balance statement that we are writing only thing is that we are bringing all terms to the left hand side. So, if you bring both these terms to the left hand side you will have d dt of m C v plus m dot out minus m dot n and that is the way this term has been written and to be interpreted therefore it is the rate of outflow of mass minus the rate of inflow of mass. So, whatever is boxed here is the so called integral mass balance relation which on a purely physical basis simply reads that it is the rate of accumulation of mass within the control volume plus the rate of outflow of mass minus the rate of inflow of mass is equal to 0. So, if you just rearrange these terms it is pretty obvious and intuitive the mass balance statement it is something coming in minus something going out the difference between those two that is getting accumulated in the control volume. So, that way it is physically very very intuitive is just that these mathematical representations have to be slightly carefully looked at especially adding this minus sign because of the inflow area being such that outward normal and the velocities are oriented in a particular manner. Here the angle between that is at the outlet the angle between the velocity and the unit outward normal is less than 90. So, that the cosine of 90 which will come through this dot product will automatically be positive we do not have to do anything there. On the other hand on the inlet area we need to assign a minus sign. So, that minus from here and minus from here the dot product will return a positive value. So, what we have done is that we start with a balance statement written out in words plain English statement and then we have gone ahead to write that statement in words in terms of symbols first and then for those symbols we have come up with a mathematical representation. So, technically nothing has changed even if you see this particular boxed equation it is still the same balance statement that we knew earlier is just that the mathematical expression for each of those terms has been incorporated in there fine. So, if you are dealing with typically what we call a steady flow situation then in a steady flow situation there is no accumulation of mass in the control volume which means that whatever is the mass flow rate coming in will be the mass flow rate that will be going out. So, that there is no accumulation within the control volume. So, this is as simple as that the integral mass balance equation for a control volume. Now, we want to extend the same treatment to a linear momentum balance for a control volume to do this let us begin with the linear momentum balance in one particular direction let it be x direction or y direction or whatever and then after having done this for a particular direction as a scalar quantity we can generalize this as a vector quantity because as you know the linear momentum is a vector quantity. However, if we are looking at one particular direction essentially we are looking at only one component of this vector quantity. So, let us go in that fashion. So, let us begin with something simpler and then try to generalize it as a vector relation. So, now going back let me for the purpose of reminding everyone we still go back to the same balance statement. Now, instead of mass we are talking about linear momentum. So, rate of accumulation of linear momentum in a control volume is equal to the rate of inflow of linear momentum in the control volume minus the rate of outflow of linear momentum from the control volume let us say plus the rate of increase of linear momentum due to a source. So, let us try to see what that means right here. So, if you recall from basic mechanics I am utilizing the symbol P for linear momentum and at the bottom of the slide here the vector representation is shown. So, the linear momentum is simply mass times velocity as we know. So, that is all that is written out here. We write the balance statement for the linear momentum. So, the left hand side here is simply the rate of accumulation of linear momentum in the control volume. The first two terms on the right hand side are those associated with the fluid flow in and out of the control volume much in the same fashion as there was a mass flow rate in and mass flow rate out in the previous discussion of the mass balance statement that we talked about. So, there is in principle nothing different there. What if at all that needs to be discussed here little bit is this source term so to say. So, the way the source term in this case can be interpreted is that if there is a net force acting in whatever direction that we have chosen x, y, z etcetera on the material that is contained within the control volume ok. So, what we are always looking at is that the control volume is isolating a certain amount of material and if on that material in the given direction of interest at the instant of interest if there is a net force acting this will actually give rise to a source of momentum that is the idea. And therefore, what is this force coming from or where is this force coming from? This is essentially the from the Newton's second law of motion that that the force is coming from you can say. And in order for the source to be specified as the force you essentially treat the material inside the control volume as a free body. I hope you remember this terminology from your strength of materials or solid mechanics classes that we isolate in those analysis if you remember we isolate a certain amount of material as a free body. And when we isolate a certain amount of material as a free body we have to mark all sorts of forces that are acting on that isolated material because of its interaction with the surrounding material. So, the idea is exactly the same here what we say is that the control volume is containing and isolating a certain material and that certain material perhaps is getting acted upon by a net force in some specific direction which will give rise to a source term as far as this balance statement is concerned. So, that is about it. So, let us see how we can now put some mathematical expression for each of these equations. I have chosen to do this in the x direction. So, I am talking about the rate of accumulation of x linear momentum within the control volume that is my first term or the term on the left hand side let us say only now that I am adding additional subscript of x which is essentially the x direction. And that is nothing but time rate of change of the momentum linear momentum that is in the x direction contained within the control volume. So, now you see what I am doing here inside the integral over the control volume what I have is dm which is an elemental mass and a u velocity associated with that elemental mass. The product of these two will actually give me an elemental linear momentum associated with that elemental mass dm within the control volume. And this elemental linear momentum in the x direction is then integrated over the entire control volume to obtain my total x direction linear momentum within the control volume that is getting accumulated. And dm within the control volume from our previous discussion of mass balance you will remember that dm was expressed as rho multiplied by the elemental volume. So, I will simply take this same expression that dm is equal to rho times dv and substitute it for this dm suffix cv u as it is. So, u times rho times dv integrated over the entire control volume will give me the x linear momentum contained within the control volume. And then the time rate of change will give me how it is getting accumulated within the control volume and that is our left hand side term. In exactly the same manner as we came up with the mass flow rate in minus mass flow rate out term which let me remind you it is this right hand side of the second equation on the screen right now minus over the area that is surrounding the control volume rho multiplied by the velocity of the fluid dotted with the unit vector on the surface area times the elemental surface area is going to give me m dot in minus m dot out remember in the same manner I am now interested in p dot in the x direction in minus p dot in the x direction out which is rate at which the x linear momentum is coming in minus the rate at which the x linear momentum is going out. If you see this expression on the right hand side it is exactly the same as what we had in the case of the mass flow rates only thing is that u which is the x direction velocity is getting multiplied by the entire same expression that we had for the mass flow rate. So, rho times v dotted with n dot dA is essentially the mass flow rate crossing the control volume and that if you multiplied by the x direction velocity will give me the rate of x linear momentum with a minus sign and the integral over the area the net term is essentially interpreting as the rate at which x linear momentum comes in minus the rate at which the x linear momentum goes out. So, the analogy between this expression here and the mass balance expression here in here it is m dot in minus m dot out should be noted carefully because then you really understand that it is the same balance statement that we are utilizing in essentially repeated manner only thing is that we are making minor changes going from mass flow rates to momentum flow rates that is about it. And therefore, putting it together what we have our balance is rate of accumulation equal to the rate of inflow minus rate of outflow plus the net force acting in the given direction. So, we have talked about first term on the left hand side, first term on the right hand side and second term on the right hand side in here. And the last term on the right hand side is essentially that net force which I was talking about. So, that I have been writing here now with a symbol f suffix x. So, that is nothing but the net force which will include a surface force which is acting on the surrounding surface of the control volume and a body force which is essentially acting on the entire material contained within the control volume on the entire material within the control volume acting of course in the x direction because we are still talking about the x momentum. The first term on the left hand side is our rate of accumulation of x linear momentum in the control volume. The second term on the left hand side which is now a combined effect of both inflow and outflow only thing is that see what has been done it has been brought on to the left hand side. So, that that it carries a plus sign with it and therefore we interpret this as the rate of outflow of x linear momentum minus the rate of inflow of x linear momentum. When you write it with a minus sign here it is to be interpreted as rate of inflow minus rate out of rate of outflow. Now that the minus sign is gone because you bring it to the left hand side you interpret this as the rate of outflow minus rate of inflow and that is it. This is what your integral linear momentum equation in the x direction for the control volume. Now if I want to immediately write the y momentum balance let us say for the control volume all that I will change is instead of this u here I will replace that with the y direction velocity which is v. So, v will come here v will come here and instead of net force in the x direction obviously I am going to talk about net force in the y direction. So, it will be f suffix y and that is precisely what I have written out here. If you see v here v here and f suffix y again the physical interpretations remain exactly the same all I am doing is that instead of x linear momentum now I am talking about y linear momentum. So, now having done this x and y I am immediately in a position to generalize this to a vector form where I am not talking about necessarily x direction or y direction or z direction now. So, instead what I will do is I will talk about the general velocity vector replacing either this v here or u in the previous slide. So, therefore what you see at the bottom whatever is boxed here is the velocity vector velocity vector rho times dv is still our mass contained within the control volume rho times v dotted with n d a is essentially the mass flow rate crossing the control volume and the physical interpretation remains exactly the same. Now, we cannot talk about f suffix x or f suffix y as the net x force or the net y force acting it will be a general force vector that will be acting as the net force on the control volume essentially the material that is getting isolated by the control volume as a free body. So, these are the two basic integral balance equations that we need to know in fluid mechanics. If you see textbooks usually they will talk about something called a Reynolds transport theorem using which they will express these. I have tried to avoid the word Reynolds transport theorem, but in principle whatever has been done through these balance statements is exactly the same. So, if you are getting if you are going to get confused with the words Reynolds transport theorem if you open a book please do not because it is exactly the same material that we have just talked about without utilizing the words Reynolds transport theorem that is about it. So, let me try to summarize these two right now. We start with a balance statement for mass and linear momentum. We have a rate of accumulation term and that is equal to rate of inflow minus rate of outflow plus a rate of increase due to a source for mass flow that source is missing altogether if you do not have nuclear type reactions which we will not in our analysis. So, what is left is simply rate of accumulation is equal to mass flow rate in minus mass flow rate out what we have done is we have expressed each of these terms through appropriate mathematical relations and we come up with what we call here in the box as our standard mass balance equation on an integral basis for a control volume. Exactly the same analysis has been extended for a linear momentum equation. So, the balance statement remains the same the only additional part that needs to be understood is that the source of linear momentum can be interpreted as the net force acting in whatever direction that you are choosing to analyze on the material that is contained within the control volume. So, we treat the material inside the control volume as a free body and as we do in case of solid mechanics or strength of material we mark all sorts of forces that will come on this free body because now you are isolating this material from its surrounding. If you write it for the x direction you need to include the x velocity here if you write it for y direction you need to include y velocity and the general vector form is written out here at the bottom where we are writing it with a velocity vector and the force vector as the net force acting on the material contained within the control volume. So, what I want to do now is this has been some sort of a derivation as to what you start with namely a balance statement you come up with mathematical expressions to replace the symbols in that statement and then you obtain the expressions.