 So, the next part, next topic you can say is what we will call integral analysis of fluid motion. Those who know now this Fox etcetera, this is in chapter number 4 of Fox. I am assuming that many of these things are somewhat familiar. So, I am not going to spend too much time on it in the sense that I hope you know what is a Eulerian versus Lagrangian method. Many of you know, anyone has absolutely not heard it. So, in Eulerian what we do is we decide that this is some region of space that we want to separate out which we will call either a control volume or from those who have been teaching and doing thermodynamics is also called as an open system correct. So, a control volume or an open system approach is what the Eulerian approach is and it is preferred in the influence mechanics and I will try to point out why. On the other hand, you can do it with a Lagrangian approach as well which if you are using the fluid mechanics terminology, it is called the control mass approach. If you are using the thermodynamics terminology, it is called the system approach. Fundamentally there is no difference between Lagrangian approach and the approach that you use for a solid body dynamics. So, example of a solid body dynamics is motion of a projectile. So, motion of a projectile when you describe in your dynamics course whatever method you use there is really nothing but the Lagrangian method. So, we will actually talk more much more about this Eulerian and Lagrangian in the next topic which is kinematics. Right now this is just to sort of introduce what we are getting into. Then there are two other ways of doing it, integral versus differential. Presently we are going to do integral. Integral is basically if you are interested only in some sort of an overall picture of the flow. For example, if you want to know how much force is transmitted, if some jet of a fluid comes and hits some plate and then gets deflected this way that way or whatever. So, how much force is transmitted by the jet which is impinging on this plate? This is an overall picture. I am not interested to know the entire details of the flow field and this and that. All I am interested is what is the overall force transfer? Fine. Similarly, let us say that some device is there. Then there are some three inputs and two outputs, pipes. Three inlet pipes are bringing fluid in, two outlets are taking it out. And I want to know if things are working under steady state operation. I know that there is nothing that is getting accumulated in my system here or that device. Let us say if I know the three inlets mass flow rate, one outlet mass flow rate, I am only interested in knowing what is the second outlet mass flow rate. I am not interested in knowing the detailed flow field information inside the device at all. This is the only overall information that I want. And in such situations, you can utilize the integral analysis without getting into any of these too many details about the flow field. On the other hand, if you are interested to know in a fluid domain, point-by-point information about how the velocity is varying, let us say, how the pressure is varying, how density is varying in general, how temperature is varying. Then you get into what is called as a differential. Now, keep in mind that both, so you can form any combination out of these. So you can use Eulerian and using Eulerian you can do integral. You can use Eulerian and you can do differential. You can use Lagrangian and you can do integral. You can do Lagrangian and you can do differential. So any such combination is fine. What we will be doing mostly and for some part later we will use the Lagrangian description just to make sure that people are understanding what is this and what is that. Mostly right now we are talking about the Eulerian side which means that some sort of a control volume is going to be identified and we are simply monitoring what is happening inside this control volume as fluid is brought in and fluid is taken out and so on. Okay, now I think those who are using that pox, etc., will perhaps immediately say let us mention the Reynolds transport theorem. I am not going to do that. How many of you have heard this? Obviously, you do not have. I know that. You have. Anyone who is using any of those books will know it. My take is you do not need to know it. All I am saying is that let us try to go from a purely physical argument which is intuitively understandable. So we begin our discussion through a physical statement written in English words and then we will try to put the mathematics to it. That is the idea. And after doing this class this way, that way, slightly this way, slightly that way, I think finally I have decided that this is the way I want to do it. It is really up to you how you want to do it but at least try to get the idea that what we are doing here is that I do not want to put mathematics in unless I have some sense of what is happening from a physical point of view and then we will do the math anyway. So all I am saying is that think about a balance statement which is pretty intuitive. I have a control volume which is essentially a region of space that has been identified and I am simply talking about a balance of certain quantity. The only three quantities that we are going to be interested are the mass, momentum and energy. So let us talk about the mass first. Rate of accumulation of mass in your control volume which will be called from now on CV. Quite obviously how much is coming in minus how much is going out plus if there is some sort of a generation happening if at all. That is the third part and that is what I will call as source. So if you are talking about only a mass balance equation like this, do you think that there is any sense of talking of a mass source? We are talking about generation of mass, creation of mass. No unless there is something happening. Some very special situation if it is happening in your control volume it is okay to include that source. What can be that special situation? Not chemical be careful. Not chemical. Some other type. We are either creating mass or destroying mass. Yes. So unless we are dealing with nuclear reactions happening in my control volume we are not going to destroy or create mass. Chemical reactions only rearrange the mass. So it goes from one species to the other but the overall mass is not changed. If you are actually doing a nuclear reaction then through that great Einstein's E equal to mc squared you actually can destroy or create mass. So clearly we are not really doing situations or dealing with situations where we are using some sort of a nuclear reaction happening in the fluid flow. But just for the purpose of completeness that source term as we say is added. Keep these terms in your mind though. The source term because when Professor Charmas takes over on Wednesday talking about CFD such terminology will be constantly coming in there source term. So physically the source term should be pretty intuitive to you. What is the source term? It is simply some sort of a generation or a destruction type situation which I have. Fine. Let us talk about energy then. Momentum we will talk about later. Momentum is slightly troublesome. Energy is again a scalar situation. Scalers such as mass and energy are easier to understand. Just like that mass balance statement which we just talked about I think is very intuitive. I have some system sorry control volume I should say. Something is coming in something is going out. If there is a difference between those two then the fellow has to accumulate. That is pretty obvious. Same thing I can talk about energy. So if I am talking about a control volume some inflow of energy could be there, some outflow of energy could be there. There could be a generation of energy which could be one of your chemical reaction because they can give out some heat. Mass is not changed. It is just that one species is converted into another species. But the overall mass in the system remains exactly the same. If there was 10 kilograms of whatever material it will be 10 kilograms unless of course it is an unsteady process where there is an accumulation because some more mass is taken out and less is coming in. That is a different thing. But let us say that I have a situation where it is a steady flow situation in the sense that same mass flow rate coming in as is going out in which case there is no accumulation absolutely correct. Here now if there is no nuclear type reaction you cannot really have a mass source term unless you have a source term like a nuclear reaction some of that mass will be actually destroyed and be converted into some energy. But for those obviously you have to get into situations which are very very special and those we are not. Chemical reactions are just going to rearrange the distribution of mass into different species but the total amount will not go anywhere. Energy same thing. Let us say a fluid stream is bringing some energy with it. There is a outflow of fluid stream which is taking away some energy and there could be a source term which is actually generating that energy which will be of the type of that chemical reaction type. So where have you seen source terms for the energy equation in any of the classes before? Who has done heat transfer course before here? So we are talking about that kind of a source which is in the conduction equation for example you take curing of concrete slab. So there is a endothermic sorry exothermic reaction there which is happening which you use as a source term. So those are the kinds that we are talking about but not the nuclear reactions where you are actually destroying the mass or creating mass through the use of Einstein type equation. That is not what we are talking about. So mass and energy it is relatively easy to intuitively understand this balance that here is a region something is coming in something is going out perhaps there is a source these will all get into either a rate of accumulation or rate of decrease depending on which one is larger and which one is lower that is fine. What about momentum? I have written it there. I will say that we will try to have the momentum also interpreted in the same manner in the sense that we will have a control volume there will be a fluid stream coming in it will bring with it certain momentum. Let us say we specify which direction momentum we are talking about either the x direction or the y or z whatever similarly there will be a fluid stream that will be taking out certain momentum with it again in x, y, z direction. So the first two terms on the right are basically fine to me there is no difference between what we just talked about for mass and energy versus this is the third term that perhaps can be interpreted in a slightly different manner. So what can you think of a source of a momentum? Which entity or quantity from mechanics can be considered to be a source of momentum? Yes who said that? What? Force, force, force. Who said force? That was the correct answer. You can think of the entity force if you apply a force it will essentially carry out a change in the momentum and in that respect I would like to introduce that force as some sort of a source for momentum as the third term when I write the balance for momentum. We will come back to this in more detail later when we talk about the actual momentum equation and how to use it and so on. But right now this is some sort of a generic or a general balance statement that I want to point out to you without really resorting to the use of terms such as Reynolds transport theorem. Obviously they are related it is not like this balance statement is something totally different from that transport theorem. So those who know it I will say that you please try to establish the relationship between the transport theorem and whatever has been written as the balance statement. Those who do not know the balance statement sorry the transport theorem do not even care about it as long as you are ok with the balance statement or of course later you can read the book and try to establish the equivalence between those two. So here is a situation which I want you to carefully think and make sure that this balance statement on a purely physical basis makes sense to you. Something comes in something goes out there could be a source or a sink all it there could be an endothermic reaction there could be an exothermic reaction as would happen in case of a chemical situation all these guys will balance into either an accumulation of quantity or a reduction of quantity in that space which we are calling control volume. So that is the basic physical statement written in English is what I want you to note down carefully. Now comes the math. So let us begin with the mass balance whatever I have written as the first equation on the top is nothing but the same balance statement now written in symbols let us say. So d dt of m cv is simply the rate of accumulation of mass within the cv m dot in can you read this by the way the subscripts and everything whatever is coming in as the mass flow rate I am calling it as m dot in whatever is going out is m dot out and I have written that source term but immediately said that that is going to be zero if we are not going to deal with any nuclear type reaction. So for our purpose from now on word there will never be this mass source term. We will always deal with term on the left hand side and the two terms on the right hand side. Okay technically this is all that you need but there is always utility in putting everything in the mathematical form so let us do that. So I have drawn some sketch here what I have drawn is a control volume as you can see here this fellow this is the bounding surface which I am calling A and just for the purpose of the derivation so to say I am saying that let there be some sort of an inlet area and let there be some sort of an outlet area A and A. Again the remaining part of A other than the inlet and outlet you can assume that it is essentially a wall where nothing goes in or out. Let us look at this first term on the left accumulation of mass within the control volume d dt as it is no change mass within the control volume I am simply now representing using an integral representation which is what I will simply identify a certain small region here the mass content for that small region will be dm let us say okay and then all I have to do is integrate that dm over the entire control volume to get my mass content in the control volume that is all it is. So this integral here represents whatever is inside the bracket is that fine I mean this is as simple as it gets. Now the mass for that elemental region let us say mass content of that elemental region is simply expressed as the density times the volume of that elemental ring mass equal to rho times the volume that is all. So instead of writing dm then I am simply saying that rho times dv is my mass content of the elemental volume inside the cv and then I integrated over the entire cv to get the total mass content and the time rate of age will give me the rate of accumulation it could be positive it could be negative that is fine. So that is that same thing so you start with an English symbolic statement and then keep on adding complexity to make it the way you want in terms of the final mathematical equation. Let us see where is it leaving from this AO. Imagine that that AO the outlet area you divide it into several elemental areas each of which will be called as DAO. What is the expression for the mass flow rate across a certain area element that you know rho av now that v has to be in a certain manner though which manner normal to the that is all I have written rho at the outlet v normal at the outlet I am talking about dm so therefore I am talking about the elemental area DAO and then I will integrate it over the appropriate outlet area to get the total outflow mass or outflow mass rate if you want. How do you get this normal component in general the velocity of the fluid leaving that DAO is not going to be normal to it it can be whichever way in general so you have to get the normal component of that velocity for the area element. So I have shown here that the unit normal outward for the area O technically it is for the DAO so I can imagine that there is a small DAO here. The unit outward normal is written as n hat O and in order to get this normal component for the DAO I simply form the dot product of the velocity vector with the unit outward normal all I am doing is that I am resolving that v O along the normal. So just as a curiosity question student ask something so what happens to the tangential it is not carrying correct that is fine actually that is good answer it is not really carrying out the mass flow across that area the only component that will carry mass flow across the area is going to be the normal component as you say and therefore we are interested only the normal component which is formed by the dot product of this velocity with the unit outward normal and that is it you are done exactly the same thing you do for m in or m dot in I should say. So here I have again integrated over the inlet area I have not written the AI here but that is how it should be here I am writing rho at the inlet the normal velocity component at the inlet times DAI integrated over the area inlet will give me the m dot in. Now here this is the scalar expression okay rho i times v n i DAI. If you look at the inlet situation it is slightly different than how it is at the outlet what is different here is just that you have to take this dot product properly. So here at the inlet the included angle between the velocity at the inlet and the unit outward normal will always be greater than 90 degrees. So therefore in order to get the positive scalar as the mass flow rate you have to include a negative sign along with the dot product. So this dot v i dot n i is going to give you 1 minus this minus and that minus will cancel each other and you will get this is that fine then you are done that is it. So then we have the expression for left hand side and the two right hand side we simply the first line there is written by simply providing those expressions wherever they belong. Now what I am doing is the second line from the top I am basically combining those two integrals into one single integral over the entire area. So why can I do that? All I am doing is the two separately written integrals over okay the first integral should have been a i that is my mistake. So the first integral should be a i. I am combining both those into one overall integral over the entire area a why can I do that is the question. Yes but. So they will adjust themselves is what you are fine that is correct actually what I wanted to say is that the entire area as you saw in picture before for the purpose of derivation I had just pointed out an outlet area and the inlet area. The remainder was anyway a wall is what we say. So there is no velocity crossing in or out over the remainder of the area. So the integral if you were to define an integral over the remaining area is going to identically equal to 0 there is nothing. So therefore I can always combine it and further to say what he is saying the dot products will automatically take care of the inlet and outlet is that fine. So once I combine it then I will simply write it in standard so to say form where I will bring that area integral on to the left hand side and I say that d dt over the integral c v rho d v plus integral over the entire a of rho times v dot n d a. Where inlet and outlet will be appropriately identified and calculated is equal to 0 and that is what is boxed because that is what we call a mass conservation statement on an integral basis. So again we cannot lose the basic physics with which we began and that is why I am pointing out what those integral terms really represent from a physical point of view. First integral on the left is simply the rate of accumulation of mass in the c v plus the second integral now if it was written on the right hand side it was what? It was m dot in minus m dot out. Now those everything has been brought to the left hand side so it better be written as m dot out minus m dot in and that is how it has been written. So this entire integral over the area simply represents the rate of outflow of mass minus the rate of inflow of mass and that is your mass balance statement on an integral basis. And you may have heard this terminology which I have been on and off referring to steady flow. So if you are dealing with steady flow what it means is that m dot in has to be equal to m dot out which means that there cannot be any accumulation in the control volume. So if you are dealing with steady flow situations this first integral will simply vanish because there is no accumulation in which case what the mass balance statement reduces to is simply that integral over a rho v dot n d a that is all which is m dot in equal to m dot out. So just make sure that if you are writing it that first integral here has to be a i I will make that correction and put it on the. So this is the way that one wants to go really I mean what you should do is always you try to put something in a physical term which everyone is comfortable in understanding and then you try to come up with generalized expressions starting from the physical statement this is how we have gone about it as you can see. Let me just once more the reason is if this is understood then the entire remaining part is a repetition for the energy as well as for the moment that is why I want to spend one more minute on this. What we said again was statement based on a physical argument was written and then the expressions in the most generalized form for each of those terms were brought about through step by step introduction of what is involved in that that is all I can say. Any question on this? Yes. For the first term d by dt we used a substantial derivative. It is not it is not it is actually the okay let me let me answer it in a slightly different manner. You will see that if you look at any text such as Fox or Gupta and Gupta they will actually write partial derivative here with respect to time. I do not agree with it. The reason is because if I am dealing with a differential situation where I am looking at an infinitesimally small control volume for that small control volume if I want to write the mass balance equation then I am okay in writing the partial derivative because you are actually focusing on one spatial location more or less. Here what we are talking about is an integral area has been taken or integral control volume. So in principle all those individual small control volumes have been summed over space. So once you sum over the space the the spatial dependence actually goes away. The integration over a big area is actually taking or getting rid of the the spatial dependence. So the only time dependence remains and that is why to me it is more correct to write d dt with total ds but this is not the substantial perhaps to avoid this confusion they are writing the partial derivatives. I do not know. You will see that both Fox and Gupta they are writing partial derivative. On the other hand Potter is writing exactly the way I have written. So there is some discrepancy Robert and Crow yeah they use this okay I have not seen that much. But the idea is that if I am integrating with respect to space I am actually getting rid of the spatial dependence. So whatever is remaining in other words if I just put a bracket here that integral over the cv rho dv essentially implies that we are doing the integration over the space and therefore the spatial dependence is actually going away. The only dependence that then is left is the time dependence and that is why I do not want to use the partial derivative there. If I am doing it on a differential basis though I will use the partial differential. This is a very subtle point honestly I did not want to point it out but it is good that you have asked. This is my take on it yeah. I think I have maybe a explanation for why we actually most of the textbooks use the partial derivative. Actually most of the textbooks start from Reynolds transport theorem where you start with the system approach where you say total derivative and then go for this one. You have to avoid the confusion of the total derivative in the system. That is precisely what I was mentioning that is quite likely that they are using it as a partial derivative. However if you look at it from a mathematician's point of view it is actually incorrect. Yes it is correct. So what I am trying to point out is that if some mathematician were to look at it he or she will say that this is more correct to write in that sense. It really does not matter from an engineer's point of view to be honest with you because finally in our head what we are talking about is this. The rate of accumulation of mass as a physical quantity is what we are talking about. It does not really matter how you want to represent it. It is just that I have a certain idea for this and that is the reason I am doing it. But I think your point is well taken. Actually a under guy student who comes after doing engineering mechanics or solid mechanics when we say the total derivative of mass he immediately says it is equal to zero. So that is where we need to explain how this actually is different from the total derivative that he is talking about. I think it is a very well taken point. Chances are that that is the reason they are doing it. However I will still say that technically it is more correct to write it this way having done the integration over the space which is an integral space contained by the control volume. It does not matter really. At the end of it it really does not matter because we are always going to write things on a physical term which is then the rate of accumulation of mass within the series. However it is a good point. I mean see such things are good to discuss because these are the kinds of questions that students will poll and then you have to have something reasonable to say and these things will help. So is that fine then? Overall the idea for a mass balance statement and then putting those physical terms written in symbols then in corresponding mathematical terms and simply putting it as one standard equation which many of you know. Let me be very honest even if you have not done this even if you find this slightly troublesome it does not matter. It is just that you have to go back and sit on it. That is all I can say. If you just sit on it and think about it it is very very straight forward. It is just that I think suddenly if people start throwing some integrals and this and that many times it is a little troublesome. I mean it is not like you know when we were students I still have trouble in explaining many of these things to students because some things are so subtle that unless you really have spent years and years thinking about that one small issue you cannot really convincingly explain someone that this is the way it is. Someone like Professor Gaithonday who has years and years of experience of thermodynamics teaching can probably say that very nicely but not everyone can do that and that is fine. I mean you have to accept that things are perhaps slightly gray. You have to make them either black or white. Any questions on that mass balance? So since I said that the scalar ideas are more easy in some sense to comprehend I thought of incorporating the energy balance immediately. Formally speaking you will see that this business is already done before people come to fluid mechanics. Where is it done? Thermodynamics. Yes. So I am not going to spend enormous amount of time on this. I will simply outline how the energy equation is brought about and how we simplify it to that so called 1D steady flow energy equation in case of thermodynamics. That open system analysis if you remember from thermodynamics those who are teaching it. So what I want to do is we want to start with a English statement like this. Follow the same procedure as what we did for the mass balance. We put mathematical forms for each of these, get a general form and then using certain simplifying assumptions we want to then reduce that simplifying sorry the general form to that one dimensional steady flow energy equation. So that the idea is that that is something that you have seen before and just to connect that you can bring it about in this fashion as well is the idea. So then we write again the same thing rate of accumulation of total energy in the CB equals something coming in something going out. The first two terms then the in and out are essentially to do with the fluid flow part of it. So there is again just like the previous sketch for the mass balance fluid will bring certain energy in. It will also take out certain energy with it. This is what many times the heat transfer people will call advocated in and out or sometimes they will use convicted in or out. This is the same idea. In addition to that we have the source as usual and in case of energy because there is a possibility of heat transfer and work done, the mechanical work done. We must include that also as either a source term or some sort of a transfer term. So if you want to look at that in and out term you can split it as the first part which is because of the fluid bringing in and taking it out and the second part through the heat transfer and the work done. That is how I want to at least portray that this in and out of energy is split into this fashion and then there can be a source because of some chemical type reactions that is that are happening in the fluid. Now I have written it in a specific form. I have written it as heat transferred to the CB as plus q dot and then I am writing it as plus w dot as the work done on the CB. Usually I think the thermodynamics people will write it as minus w dot with the w dot then work done by the CB. The reason I want to do this is because again intuitively to me this makes more sense because why? I am writing a balance for the energy content within my control volume. So I am always thinking in my head how is that fellow going to increase let us say just for my aid of thinking. So unless you put in some work that thing is not going to increase. So that is the reason why I am writing it as work done on the CB. Those who are more comfortable with the other way you can work out the exact same derivation with that there is absolutely no problem with it. Now there are two points to be, so is that fine the way it is written. Again heat transferred to the CB makes more sense that if I dump certain amount of heat into it the energy content will go up that is what I am calling the rate of accumulation. So let me come to this in a minute. Total energy is what we are talking about. Usually we will talk about that as the mass content times the specific total energy which I am using symbol small e and that specific total energy will be addition of the standard 3 which is specific internal energy which is the thermodynamic energy through the motion of molecules and atoms and what not. Specific kinetic energy and specific potential energy. Why I am pointing out this here for the work done part is because what is work done? First of all we are always talking about rate of work done. Everything is on a rate basis. From mechanics, basic mechanics what is the expression for the rate of work done that you know? Yes, correct? Force. Force times velocity I will say because it is rate of work done that is absolutely right. So here we are interested in the work done on the CV meaning we will actually identify the forces that are acting on the CV and then multiply those forces by the corresponding velocity component to get the work done on the CV the rate of it. Since we are trying to identify the forces that are acting on the CV what is done usually is that we essentially imagine that the control volume and the material inside it is to be treated as a free body. I hope you know this term from mechanics free body, free body diagram. How many of you have absolutely not heard it free body diagram? I am sure everyone has heard it. All we do is that we essentially isolate that body and show all sorts of forces that are acting on it which is precisely what we are going to do here. Since we are interested in the work done or the rate of work done on the CV. There is absolutely no difference between the first two statements the mathematical statements here and what we wrote for the mass balance. So let us go back to the mass balance and make sure that that is how it is. Here was the set of equation which then is written as minus integral over a rho v dot NDA. Now that was only pure mass flow. What we are now going to talk about here is that that mass flow is bringing certain energy with it and taking certain energy. So all I am doing is that I am multiplying that mass flow rate by the specific energy total energy that I need to get in and that is why that E dot in minus E dot out is simply the integral over a minus sign as it was E times rho v dot NDA. Is that fine? So hopefully the progression is clear to you that you know we start from the simplest thing which is the mass balance and then keep on adding these energy and momentum as we will do later. For the accumulation part d dt again total derivative I am writing not the substantial total derivative with respect to time over the control volume if you see the integral is for E times that is small E times dm which means again that I have identified a certain small element corresponding to which the mass is dm corresponding to which the energy content is E times dm and then I simply integrate it over the entire control volume and that is what is written. And that is it for the standard general form of energy balance equation then using those two expressions I keep everything else for the moment the way it is E dot source plus q dot plus w dot on the right. Whatever is boxed is your called general energy equation if you want. Now I want to show to you how we bring about that 1 d steady flow energy equation of thermodynamics which many of you are familiar starting from this general form. So let us see how we do it. So first of all we assume that it is a steady flow as the name suggests steady flow uniform flow over inlet and outlet with v i parallel to a i v i v o parallel to a o what does that mean? This is extremely loosely written if I want to criticize myself what is missing? Mathematically what is missing from here if at all something is missing yes you are right exactly that n is missing n hat is missing or if I do not want to use the n hat that is fine but I need to put a vector on the area because area is essentially a vector quantity which is given by that unit outward normal. So either as you are correctly saying that I write that n i here or I simply put a vector on the all that I am saying is that let us assume that the control volume has been chosen such that wherever there is an outlet area and wherever there is an inlet area those particular areas have been chosen such that the outlet velocity and the inlet velocity is physically normal to the inlet and outlet area physically normal but mathematically parallel no no you do not like it now you see it physically normal meaning what how should this and this be with respect to this area normal this guy has to be along this which means which means what fine so that is precisely what is written there is that fine so what I am saying is that the outlet inlet areas have been chosen such that the unit outward normal for both the inlet and outlet areas are essentially parallel to the velocities that is all. In doing so we get rid of the dot product then we have to do only a scalar multiplication of the magnitude of the velocity and the area that is all. So this is an inbuilt assumption in getting that 1 d steady flow energy equation fine so then let us see what else is written out there so that is what is meant by this statement that it is your choice how to choose that control volume you have chosen it such that inlet and outlet areas are physically let us say normal to the inlet and outlet velocity thereby essentially getting rid of the dot product also there is something called uniform flow over the inlet and outlet areas what does that mean across that area if I want to show it in a simplified fashion I will show it as a uniform velocity profile like this non-uniform velocity profile would have been something like this over the area so I am not doing this I am doing all that again that happens because of this is one that dot product is simply gone and then there is no integral to talk about if I have a situation like this then I have to formally carry out that integral over the areas now there is no integral to carry out is just a simple scalar multiplication with the total inlet area and the total outlet area with the inlet and outlet velocity that is it so left hand side then with these two assumptions no accumulation because it is a steady situation so this guy is 0 the area integral is simply plus 1 for the outlet area minus 1 for the inlet area because physically speaking you can see that the outlet will be this is n dot o and then the velocity will also be like this so that the angle is 0 actually whereas for the inlet this is the inlet outward normal but the velocity is this way so that the angle is 180 so that dot product involves cosine of the angle between the two vectors so cosine of 0 is 1 cosine of 180 is minus 1 and that is why this is plus 1 and this is minus 1 with simply a scalar multiplication of the areas with velocity so then that is it this will simply become a scalar multiplication over the outlet area scalar multiplication over the inlet area with a minus sign so rho times v a times e on the outlet minus that on the inlet yes see what yeah yeah yeah yeah yeah good point fundamentally when you are wanting to know that the mass flow rate is carrying something out okay whether it is energy or momentum which how do you calculate that mass flow rate itself so the very first calculation that the only the normal component of the velocity will actually carry that mass flow rate in and out is taking care of all that issue so it is only the mass flow rate that comes through the normal component of the velocity for the area then suitably gets multiplied by the total energy or whatever so in that sense you are right I am interested in knowing how much energy is coming in or going out that part is only handled by the normal component so I do not care about what is happening as far as going in and coming in going out and coming in of the energy for the remaining component as far as my control volume is defined for that particular control volume I want to know what is coming in and what is going out that is done by the normal component so that is why I said that the first balance statement for the mass is the most useful one to understand because that is where everything is crystallized really everything then is just an add on this energy as well as the momentum that we will do the mass balance itself when we wrote we said that only the normal component of the velocity will actually carry out the mass or bring in the mass with respect to the control volume that is it rest all is simply multiplying by the specific total energy or the specific momentum as we will do later alright so then now we assume that there is no source because in the standard 1d steady flow energy equation there is no source now comes the bigger issue of evaluating the work done now here we have usually three different components to the rate of work done on the control volume first part is usually called a shaft work which simply means that I have a control volume which houses a machine such as either a pump or a compressor which I am calling is a power absorbing machine or there could be a turbine or an engine which is a power producing machine we are interested in finding the work that is going into the control volume and therefore since that is termed as positive this shaft work is going to be positive if inside the CV we have either a pump or a compressor which is a power absorbing machine ok fine so that is that is one part the remaining two are essentially coming from the forces that are acting on the control volume so there are two types of forces that we will deal with in fluid mechanics one is a surface force and one is a body force so now here is that idea of free body diagram comes in we want to basically identify the forces that are acting on the material which is instantaneously inside the control volume fine let us see what that is let us look at just the surface forces we want surface forces acting on the control volume because we are interested in knowing the rate of work done on the control volume that is first part second thing if we have already chosen that the outlet and inlet velocities are going to be normal to the inlet and outlet areas which component of the force will produce non zero work which component of the surface force in general if you look at a surface there will be a normal force and then there will be a shear force will shear force contribute to any work let us first make sure that the shear force does not contribute to any work why not how is the shear force acting on the surface tangentially how is the velocity that we have chosen to be normal so what is the angle between these two ninety if you are going to use our f dot v as the expression for the rate of work done angle being ninety between the shear force and the velocity vector shear force actually will not contribute to any work done on the control volume is that fine so that is what is written out here that with the choice that we have made only the normal surface force which for the present moment we will approximate to be only the pressure which is acting on the surface now the pressure is how how does it act it is a compressive force so we express it as minus p times let us say the area over which it is acting right and that is it so minus p multiplied by the appropriate area over which it acts will produce non zero work with this choice remember this this choice is important if this choice is not done this way you have to actually compute the work done by the shear force so then let us see what the expression will come out this is over the entire area fine forget the integral for a minute minus p times da what does that give you force due to pressure it has to be a vector right so that vector comes from the vector representation of the area which is a unit outward normal the velocity in general is written as velocity times its own unit vector we already know that these two guys are in the same direction but for the purpose of writing it as a general expression I have written it that way so I just separate it out as minus p v da over a o plus p v da over ai why is that for the outlet this fellow will produce plus one and then there is a minus with it so it is here for the inlet this dot product will produce a minus one and that is why that minus and minus will become so is that fine good so then that the two integrals are first written separately but further we had assumed that it is a uniform velocity over the inlet and outlet so then there is no integral to talk about is just a scalar multiplication of the entire area with the velocity and that is it so this integral simply goes away what is left is p v a over the outlet with a minus sign as it is plus p v a over the inlet is that fine so I am trying to hope that you go in steps one by one you identify terms and then try to argue that this is how it should be where should that the minus sign and plus sign and this and that and carry on fine that is about surface forces I hope there is reasonable understanding of what is happening here yes that is that is a good question we are doing an integral balance ok so the viscous dissipation actually is going to be happening inside the control volume but we are not explicitly calculating it we are actually interested only in the overall behavior so whatever is going to happen will actually be inbuilt in the equation and I will try to point out where the viscous dissipation is acting although we are not calculating it separately in the integral formulation viscous dissipation is not explicitly included because that is actually happening in the entire fluid that is inside the control volume we are only interested in the surface forces over the material that is in the entire control volume and the way we have chosen our surfaces there is no viscous work I will try to point out where the viscous dissipation is hidden though let me complete the derivation and then I will point out where the viscous dissipation is hidden that is a really good question and I think the best answer you want to get is in Gupta and Gupta there is a chapter on energy equations where they have nicely tried to explain this that where is this viscous dissipation I will try to point out where the viscous dissipation is right now you just imagine that I am just least bothered about what is happening inside the control volume as a detailed flow as it is happening all I am interested in is this is my overall thing something is coming in something is going out I am only interested in the overall behavior so that I am only interested in the forces on the surface and let's say the forces in the bulk which will be coming through the body forces but the way we have chosen our inlet and outlet area that viscous work is actually vanishing but it's vanishing only on the surface it's actually in inside there but it's not getting calculated explicitly but just bear with me for five minutes and I'll point out where that is alright so then here is the the body force term what we have done is we have included the specific potential energy in our total specific energy so if you do that and if the gravity force is the only body force that is acting on your control volume then we actually do not calculate the work done by the body force at all because what is then the rate of work done by the body force is exactly equal to the change in the potential energy which already is taken care of in the energy energy term total energy term so to avoid duplication if this is the situation keep this in mind if you include the potential energy in the total specific energy and if the only body force is due to gravity then this is okay if there is some other type of body force what could that be then you have to actually compute that work separately however we are not getting into such complicated situations at all we will say that our fluid mechanics is restricted to having only gravity as the body force and since we have included potential energy in the specific total energy to avoid duplication of the rate of work done by the gravity force we will not include it at all in the work done term it's already taken care of on the other hand if you had not included it in the energy total energy then you have to compute it here the choice is yours right now I have done that okay so then what all I am doing is this was the left hand side sorry this was the the surface force term which has been placed on the right hand side here q dot plus w dot shaft as it is and you just realize that that rho times v times a is simply in this 1d uniform situation is equal to the steady mass flow rate okay so you divide that entire equation by m dot what will happen from the very first term on the left it will be simply e at outlet minus e at inlet q dot divided by m dot and simply calling it as small q so it's some sort of a rate of heat transferred into the control volume per unit mass flow rate that is flowing through the control volume same thing for the w shaft work pva divided by rho va what is remaining p over rho p over rho at the inlet minus p over rho at the outlet so you combine that e on the outlet plus p over rho on the outlet bring it to the left equal to same thing on the right inlet plus q and w and now you include that specifically expression for total specific energy which is what which was v squared over 2 is a specific ke gz is the specific p I am using small i for the specific internal thermodynamic internal energy plus p over rho at the outlet equal to all this business plus q plus w shaft and this is exactly your 1d steady flow energy equation from thermodynamics I hope you agree with that so the way it has been brought about is that we start with a fairly general situation make certain choices make certain simplifications and show that it simply reduces to the 1d steady flow energy equation as we know from okay now coming back to your yes okay no let me let me then let me count you right there let us say that I I deal with an adiabatic system and I deal with a zero shaft work situation is there still the viscous dissipation inbuilt no are you sure to answer your question specifically let us consider that there is no heat transfer meaning we are dealing with an adiabatic system no control no heat transfer in or out of control volume no shaft work these are gone let's say what is left is then exactly that is where it is hidden what is then to be understood is that this I fellow at the outlet is in general different from the I fellow in the inlet the internal energy which you normally represent using some sort of a specific heat times the temperature okay let us say that the specific heat doesn't really change much it's a constant in which case you realize that in an adiabatic situation like what we are talking about with no shaft work the temperature at the outlet is going to be different from the temperature at the inlet that is where you are viscous dissipation is built that is how the temperature has actually increased from and you have to show that that is actually increased from inlet to outlet so if you are dealing with a differential energy equation you actually calculate the viscous dissipation explicitly you start from a differential equation you actually integrate it over the control volume and then what you are getting is what this is so that the the viscous dissipation is really built into the equation through as he correctly pointed out the difference between in general I and O for the inlet energy sorry for the internal energy that is where it is built that is absolutely right so let me ask you one more thing let us say I am dealing with this again adiabatic and no shaft work this equation that is left out look suspiciously like something Bernoulli equation is it a Bernoulli equation yeah what about internal energy yeah if there is no temperature change then it seems that it reduces to Bernoulli equation which means what whatever is left out let us say I get out of this this p over rho plus gz plus v squared over 2 is popularly called as what total of that total which energy which energy it is total energy but which total energy it is total mechanical energy and if you have the same temperatures at the inlet and outlet then the total mechanical energy is conserved that is what you can interpret the Bernoulli equation as okay so in order for this equation to really reduce to a Bernoulli equation you have to say that there is no shaft work then there is no heat transfer and somehow you have to say that the I at the inlet has to be equal to the I at the outlet how do you ensure that your your question correct ideal flow then what happens with ideal flow exactly inviscid flow if there is inviscid flow there will not be viscous dissipation inbuilt which essentially will then mean that I at the outlet is going to be equal to I at the inlet and then the mechanical energy is conserved which is what some people will say is the expression for the Bernoulli equation this is one way of saying what the Bernoulli equation is and this is precisely what professor Gaithonday was pointing out in the morning session if you remember he said that in the thermodynamics class they point out that this is how the Bernoulli equation comes about I am going to derive Bernoulli equation slightly differently later using exactly what you are pointing out which is steady flow inviscid flow and technically flow around a or along a streamline so we will look at that later but that was a really good question and I hope you got the answer to whatever extent I really want you to think about it so thanks for pointing it out that's that's the correct answer actually that if you have the the viscous part even though we never talked about it it's inbuilt there through the difference between I at the outlet and I at the good let me begin with let me begin with the next part which is the momentum and then post lunch will have to be any any other question though this was something that I knew that you have done in thermodynamics I just wanted to sort of point out where is the connection between that in thermodynamics and how we can bring about the same form using the balance statement as we write it for fluid mechanics so note all those assumptions are those are important all these assumptions have gone into bringing that in I don't know when you do it in thermodynamics how you really argue that and I don't teach thermodynamics I don't know but those who teach thermodynamics perhaps will connect it better in the sense that make sure that the same set of assumptions has been also used or have been also used in the derivation in thermodynamics to make sure that it is the same form that comes about all right let's begin with this and then let we'll most likely have to do this discussion later all right so fine let's look at the linear momentum balance written exactly in the same way as the previous two balance statements were I'm talking about P as the linear momentum which is simply from mechanics given as the mass times velocity nothing different here and I write it exactly the same way namely the rate of accumulation of linear momentum in the CV equal to whatever is coming in minus whatever is going out because of the fluid flow in and out plus a source term which as I pointed out earlier to you that the source term can be considered or thought of in terms of in net force which is acting at the instant of your interest on the control volume which essentially means that I'm talking about the net force which is acting on the material that is contained in the control volume at the instant of whatever interest I have and where do you get that that that force really comes from Newton's second law of motion f equal to ma however as far as our control volume balance statement is concerned what we will say is that that mass which I want to use in the Newton's second law of motion is instantaneous mass that exists inside the control volume and therefore you need to treat that material inside the CV as a free body for the instant of interest where for the in whatever instant of interest there exists in net force acting on it this is how I want to sort of explain the situation now let me ask you one question let us say that there is absolutely no mass flow rate in or out of this there's no mass flow rate whatsoever in and out of the control volume so that p dot in and p dot out are both identical equal to 0 then what does it what does it mean then not same velocity what he said is what I will go with right now but you tell me what how should I interpret this it's a close system then first of all it's a close system which means that we are talking about the same set of mass right which is getting acted by certain force let us say so what is going to happen to that mass has to accelerate it has to accelerate so in which case that rate of accumulation term really is not to be interpreted as the rate of accumulation it is actually the rate of change of momentum which is which is essentially the acceleration of that material because of the net force that is acting on it so this is I am I'm really happy that you pointed it out but this is the way you can try to interpret the situation where there is no inflow and outflow of the of the control volume it's a purely control mass type situation just like the solid that you are pointed out it's the same mass that we are talking about certain net force is acting on it it better accelerate there is no other way if Newton's law has to hold which has to hold correct then let us say that there is no accumulation meaning that we are dealing with a steady flow situation what does that mean there is no flow in and out but it's a steady flow situation and then because of that the rate of accumulation is zero so therefore what I have to interpret is that p dot out which is the momentum out linear momentum out minus the linear momentum in has to be equal to whatever force is acting on the control volume now you tell me if further p dot in and p dot out are the same there is no force acting on it there there cannot be a force acting on it and we actually will see hope well I won't probably have the time to see that situation but anyone who can think of a situation which you have dealt in fluid mechanics which is just what I described namely steady flow imagine a control volume p dot in is exactly equal to p dot out and hence there is no net force on the control volume where have you seen this you are right you are right actually there is a net force equal to zero which is gravity and the pressure that is that's correct I'm actually it's but it's like a solid body emotion fluid in the relative motion I'm actually talking about a fluid flow situation where it's a steady flow situation with p dot in equal to p dot out that is linear momentum in equal to linear momentum out the rate of it from the control exactly fluid flowing through a constant diameter pipe fully developed flow it has to be fully developed please go back note it down that if you really want to understand this you make sure that what he just pointed out with a fully developed situation steady flow fully developed flow through a pipe it is exactly what I just described namely it's a steady flow there is same momentum coming in as going out and therefore there is no net force on the fluid so what it's a dynamic equilibrium situation as we say where the dynamic equilibrium is between which which and which force well yeah I'm assuming that it's a horizontal pipe so gravity is really not so it's actually a dynamic balance between pressure and the viscous forces we'll see that example later when we solve some of these exact solutions so that's that's a very good observation I think we may have to stop