 Let us begin, so we were talking about the linear momentum balance and we have discussed enough on it. So let us just move on and see what this means when we want to apply it in a specific direction. So here now using the same ideas as the mass balance and the energy balance, I am writing the accumulation term and the momentum in minus momentum out term, exactly the same fashion and specifically I am writing it for the x direction. Therefore, this term here you can see that it is essentially the x linear momentum that is getting accumulated inside the CV which is simply written as d dt as before. Now it is for a element of mass dm multiplied by the u velocity. So here now earlier we had this total energy for example, specific total energy, small e. All that I am doing now is I am replacing that with u for the x momentum and integrating it over the control volume and that dm which is essentially the elemental mass of a small volume element which is simply again written as rho times dv as before. So the consistency in representing the balance equations is maintained. The way it was done for mass balance and then later it was done for energy balance is exactly the same. If you see then the momentum in minus momentum out term is exactly the same way minus over the, this should have been actually area here, the integral actually over area, so please make that note and make sure that earlier that was written correctly. Here it was area that has to be, this happens when you do cut and paste. This has to be area and this u is the x component of velocity and then you put together the x momentum balance equation using the accumulation term, when you bring this term on the left hand side it will become out minus in with a plus sign here and this source term as we were discussing before the lunch break, it is to be interpreted as the force acting, the net force acting. Now here we are talking about x direction, so it will be in the x direction. On whatever material you have instantaneously inside the control volume which will be treated essentially as a free body and that is what is written out here as fx which now in words I am writing net force which includes both surface as well as body force just the way we had discussed earlier acting in the x direction on the material inside the ceiling. So it is easy to see this in a scalar form just the way it was easy to look at the mass balance equation and the energy balance equation which were anyway purely scalar. The momentum equation is not a scalar but we are actually looking at it from a scalar first which was written for the x direction using the x component of the velocity u here. Having written this we now simply want to generalize it to general velocity vector v. To do that all I am doing is wherever you see u I am simply replacing that with the velocity vector v. So this u here is replaced with velocity vector v, this u here is replaced with velocity vector v and there is no x force now, this is now vector equation. So we simply write it as the net force as a vector which is again the surface force plus the net body force that is all it is. So if you go back to the mass balance equation quickly once more we started with something like written in English then we wrote the corresponding mathematical expressions you see that the accumulation term here is expressed as an area integral plus then you add on whatever you do to get the overall integral balance. Same thing was done for energy equation, this is the accumulation term, this is the area integral which is the e dot out minus e dot in now that it has been brought on the left hand side. And then for a simplified situation of 1D steady flow energy equation with certain assumptions we could show it to be a known expression from the thermodynamics and now what we are doing is we are simply extending the same analysis to the momentum balance. To do that we first look at a certain direction so that some sort of a scalar component of the momentum equation is written and then we simply generalize it and typically what you will see is in the books for example Fox 4 chapter or whatever you will see that the general expressions are written first and then the component equations are brought about later. I actually prefer doing exactly the opposite many times the reason is because it is easy to visualize what is happening in a scalar component wise situation and then generalize it to a vector form which is what I have been doing here for the momentum balance. So that is more or less what we really want to talk about in integral analysis of fluid motion. There is little more than what we have done that you will find in the books. In particular there is a angular momentum balance also that you will find normally available in the books. You remember Professor Gaithundey in the morning was mentioning that Lawn sprinkler problem where the jet of liquid is coming out with two nozzles and then it simply rotates because of the reaction. So those kinds of problems you can solve if you do what we call an integral angular momentum balance. So the angular momentum really is simply if you want to just modify this very quickly instead of just v which is giving you a linear momentum you just have to do r cross v where r is the position vector of the point where you are looking at the angular momentum. However we are not doing that in this class this is not really a fluid mechanics class whatever if you remember in the morning whatever is required for the CFD plus a little bit more is what we are doing. So I will want to limit the discussion here to only the linear momentum part and now that we have hopefully some idea of how to utilize these integral balance equations. What I thought was let us just look at a few quick problems where we use these integral balance equations. In particular the mass balance and this momentum balance so why not energy balance you think you can use it it is not like it cannot be done for some problems. Usually energy balance is handled as a thermodynamic application if you are looking at a pure fluid flow situation normally you will be dealing with at least for the mechanical energy engineering type applications will normally dealing with the momentum and mass balance. So that is why the discussion on the energy equation was basically up to here I do not want to discuss anymore this was just an introduction to how it is to be done and to bring about that SFE from the general form. So let us just look at a few examples again I think earlier in the morning I said that many of you probably know exactly how to deal with this so I will go through this relatively quickly just to point out how the equations that we have brought about need to be employed for some specific situations where some specific output is required. So here we have what I have drawn here is a cart there is an inlet pipe and then there is an outlet pipe what is given that this is a steady constant density flow it is also given that the flow is uniform at the inlet and exit. So we do not have to bother about those formal integrations with the area and even though I am not giving any numbers here what is given is the inlet velocity is given the inlet area is given outlet area is given the pressure is acting on the inlet and the outlet are given and we are asked to find out the exit velocity V2 and the force required to keep the tank stationary. So to solve this problem first of all the tank is on wheels so unless let us say according to the problem statement we provide a restraining force this fellow is going to start accelerating some which means that it will have an imbalance force coming on so we will see how that works out. So what I will do is I have actually written down the solution so let me just switch this to the other screen so the dotted rectangle is essentially the control volume and the weight is drawn you can see that it is cutting across the tank here, here, here etc. So as far as showing this as a free body is concerned what we will say is that since the control volume and the material that is included in the control volume is actually to be isolated from the remaining part there has to be a reaction force which will come on the control volume which is due to its interaction with whatever is the remaining part. Now what is the remaining part? Remaining part is essentially this say pipe section or the tank where it is cutting and so on. So we are not really completely isolating and showing it as a free body as you would normally do in static type problems but we are simply showing that let there be a reaction force on the control volume because of its interaction with the tank. That is what is written down I will show that. So now we first employ the mass conservation and since this is a steady flow there is no accumulation which essentially implies that whatever m dot is coming in has to be the one that is going out. It is also given from the problem statement if you see that it is a uniform flow over the inlet and outlet. So all I am doing is I am simply calculating the scalar products of rho v1 and a1, rho is also constant given so there is no need for rho1 and rho2 is just the same rho and that is it rho cancels out where you get v2 as v1 a1 over a2 as simple as that. This is something that everyone knows really. Now we will get to the momentum equation. If you see the momentum equation we had written it as the rate of accumulation plus net rate of outflow if you want that is out minus in equal to the net force acting. Now here there is no accumulation steady flow so that accumulation term is gone. So what is left is only the rate of x momentum leaving minus the rate of x momentum entering that must be equal to the net x force on the ceiling. So I am writing this out specifically as mass flow rate times a corresponding velocity which will give me the linear momentum. So going out is from here which is rho v2 a2 is the mass flow rate. The velocity that is going out is in this direction which is positive x and therefore it is multiplied by v2 minus the momentum that is coming in which is simply rho a1 v1 times the corresponding velocity which is v1. So that simply is your left hand side that is the momentum out minus momentum in rate of it in the x direction. Now here on the right hand side we have to basically sum all the x direction forces that are coming on the ceiling which also includes this reaction force which we have marked so that is what now will be written down. So now can you see why I am writing p1 a1 minus p2 a2 free body diagram correct. So you see that p1 here is sorry p1 here is acting in the x direction p2 is acting on the minus x direction p1 acts over a1 so that the total force is p1 times a1 and it is in the plus x direction so that is why it is plus p1 a1 here p2 a2 likewise is minus plus I am saying that let simply be assuming that the reaction force is coming in the x direction the analysis should automatically tell you whether it is coming in the positive x or not that is the idea okay. So this reaction is essentially the reaction force on the control volume as I am writing here as it cuts through the tank so this is the idea of the free body also note that I am writing here p1 and p2 as the gauge pressures why am I writing that basically you can say that over this entire control volume there will be pressure acting that pressure will be summation of the atmospheric pressure plus whatever gauge pressure it is. Now what happens in this case is that atmospheric pressure is acting on this entire control volume which is a closed control volume so any closed control volume with the same p atmosphere acting on it will actually produce zero net effect okay. So that p atm even though it may be acting it is not really contributing to any net force in any direction so we do not have to bother with it only the net force that will come will be because of the gauge pressure. So therefore these p1 and p2 are essentially the gauge pressure and that is it you have calculated v2 from the mass balance you simply go ahead and simplify this and what will come about by solving this part here is the R suffix x which is again the reaction force on the control volume as it cuts through the tank. Now what we want really is the force on the tank let us say okay. So by Newton's third law action and reaction better be equal and opposite so therefore the force on the tank will be simply equal to the negative of this Rx okay so whatever these numbers work out if I provide some numbers they will come out and you simply have to do a negative of that Rx to realize that that is the force on the tank. Now with this minus of Rx acting on the tank unless you restrain it the tank is going to accelerate in either positive x or negative x direction so in order for the tank to be stationary you basically have to provide negative of negative Rx externally yourself let us say on the tank and that is what you want. So even though it turns out to be the same value as this Rx here in the equation I want you to think through the problem as step by step argument okay remember that when we write these momentum equations for the control volume the force here the net force will be always on the control volume so depending on how you have chosen the control volume you need to treat that as a free body and accordingly transfer the forces to whichever other bodies that you want that is all it is to it okay. So since I have not given any numbers here hopefully the idea is reasonably understood let us just see a couple more okay this is another problem where there are no numbers provided but let us see what this is okay. Any one here recognizes what this could be just from the schematic yeah yeah yeah correct but where do you utilize an analysis like this which boundary layer exactly that is right this is essentially a boundary layer situation on the on a plate which can be assumed to be here on the lower right so what we have here is even though I have not shown the plate I have written it there that along the bottom surface of this control volume there could be a plate okay so just below the CV bottom surface what is given is that there is a steady constant density flow and pressure is given to be uniform everywhere alright at the inlet to this control volume what is given is that there is a uniform velocity profile with a magnitude of capital U over let us say a certain height H on the left hand side and we are just tracking one streamline although we have not formally talked about streamline here I think more or less everyone knows what we are mentioning here. So if I am basically tracking a streamline here from the inlet to the exit so to say where now the total height is y0 which is different from H what has to be the case because it is a streamline exactly there will not be any mass that will cross the streamline by definition velocity is essentially always tangential to streamline so whatever mass you have contained in this height H mass flow rate that is will be the same mass flow rate that will be transferred throughout this entire section eventually it shows up at the exit as a velocity profile looks like over the distance y0 where the velocity profile has some functionality which I have not provided but it can be whatever u as a function of y it so happens that it is been given to you as problem statement that at the end of this y0 distance you are reaching essentially again that capital U will not care this is all given to you and with all this you have to show that the drag force on the plate is given by this integral if I specify certain functionality you can actually evaluate that integral but right now I am leaving it as just a general statement okay I hope the problem statement is understood so what will you do here it was given as a steady constant density flow again the same steps as the previous problem we go with the mass conservation first for steady flow it is m dot n equal to m dot out no accumulation in the control volume so now here we are talking about a 2D situation 2D in the sense that we are showing the picture in the xy plane let us say and the assumption is that whatever you see as the picture is repeating exactly the same way along the depth or along the z direction so that on a per unit depth is what we can talk about is that fine so therefore m dot in then is going to be since at the inlet we have uniform velocity profile it will be simply the density times whatever the magnitude of the uniform velocity profile times the area across which this is happening so the area now is going to be height h multiplied by the depth 1 so that is why I am writing that 1 specifically on the m dot outside we have a slightly different situation now it is not a uniform profile but it is a profile where the u velocity which is the actual velocity is given as a function of y it is good that we are talking about only the u velocity because that is the normal velocity to your outlet area so that is already done is just that because it is a function of y coordinate in order to write the total m dot out in principle what I have to think about is I think about a small length dy through that dy how much mass flow rate is flowing which will be simply rho times u times dy times again that depth 1 and then I simply integrate that from y equal to 0 to y equal to y not so what is this giving me it is giving me if you simplify this a relation between that inlet height h in terms of whatever this is something that we will use later coming to the x momentum equation again it is a steady flow situation so we do not have to bother about the accumulation part the x momentum equation simply again reads out the same thing as it did in the previous example which is the x momentum out minus x momentum in on a rate basis has to be equal to the net x force acting on the control volume so again now I am not writing that now here in the previous example we noted that whatever this suffix x is is the net force acting in the x direction on the control volume that you have to keep in mind fine for the linear momentum leaving since it is a non uniform velocity profile we need to resolve to that integral so you simply write it again for an elemental length dy what is the momentum leaving it will be simply mass flow rate leaving times the velocity so rho u dy times 1 from our previous expression is the mass flow rate leaving you can call it dm dot if you want multiplied by u which is again the x velocity and you simply integrate it from 0 to y not that is the p dot x leaving the control volume it is easier is basically that mass flow rate rho times capital U times h times 1 and it has a uniform velocity with a magnitude u so it simply multiplies that and again this is equal to what is the net force acting on the control volume I have written it there but you agree with it look the way the control volume has been placed it is basically you can say that this is the bottom plate and the control volume's bottom surface just about hugs the top surface of the plate so in principle what I can imagine is that there is some interaction going between the plate and the control volume which is essentially coming in the form of that reaction force r suffix x on the control volume again we do not know anything so we assume for the purpose of analysis that it is coming in the positive x direction the analysis will tell us where it is coming fine so then therefore that is the only force that is left out why because there is given in the problem statement that the pressure is uniform because it is a uniform pressure acting on the entire closed control volume there cannot be any net component of that pressure force is that fine remember we are writing this for the x direction okay so there is no other x force that we can think of here fine so then this simply turns out to be rho times u squared dy that one I am not writing anymore this here is simply rho times capital u square h this is rx you simplify this in the sense that this rho times u square times h where this h was obtained previously through the mass conservation so I am simply substituting for that h expression and you simplify it to realize that that r suffix x which is the reaction force on the control volume comes out as integral 0 to y0 rho times small u whole thing multiplied by small u minus capital U dy is that the force that we want to what do we want as the problem the problem statement asked us what is the drag force on the plate so it is actually the reverse of this or the minus of this is what we are looking for and that is it if you just make that minus of rx substitution you will see that this capital U minus small u will show up inside the integral which was exactly what was asked as the problem statement so again here the problem is relatively straight forward but the idea is that you should go in steps identify a control volume identify the mass flow rate going in and coming out identify all forces most of the times the momentum equation is what kills the problem because one way or the other we tend to forget appropriate forces that is all let us see if there is something more from the algebraic point of view this is fine this is slightly more in terms of algebra which I have not actually worked out completely but I will tell you what to do okay here is a situation which is a pipe flow but not a fully developed pipe flow what is given is diameter is something 25 millimeter the length of the pipe is given 2.25 meter at the inlet which is this section 1 there is a uniform velocity coming in at 0.87 meters per second by the time you go to the outlet section 2 is been given that we have a parabolic profile so you can assume that from your previous knowledge that flow has developed now here that is fine and is already been given to you that the pressure drop over this entire length is 1.92 Newton per meter square given to you this is steady constant density air flow and what you have to determine is the total force of friction exerted by the tube on the air and second the average wall shear stress over the length of the pipe fine so the way I have drawn the control volume is basically just inside the pipe you can say that it is hugging the inner surface of the pipe now let us see what needs to be done when it comes to the solution conservation of mass steady flow again you know m dot n equal to m dot out that way it is repeating from as it was before the inlet mass flow rate because it is a uniform flow you can write it as pi r squared u1 pi r square being the area of cross section over which this entire uniform velocity u1 is acting on the m dot outside because we have a profile you have to use the integral representation so how is that integral good thing is that the axial velocity is given because the axial velocity is normal again to the outlet area one headache is already resolved however that axial velocity is a function of the radial distance so what do you identify here again is a small element of area now because this is a circular cross section pipe that element will be in what form annular form and that annular form is written out here which is 2 pi r dr that is it so that 2 pi r dr is basically your dA the elemental area multiply that by the axial velocity multiply that by the density integrate this from 0 to r and you get the total mass flow rate and if you substitute this u sub xe1 minus r over r square etc you have to perform this integral which is straight forward if you perform that you will realize and perhaps you would know this anyway from before that this center line velocity which is uc is coming out as 2 times the uniform velocity which is entering so many times I have seen that students know this relay but they do not know how it has come about I do not know if I am right or wrong who wants to agree with me on this they know that it is twice of the inlet uniform velocity but they do not know how it is whatever it is but they do not know that there is a mass balance to be employed between the inlet section and this fully developed section and you simply get it through the mass balance they do not know that so I think it is worthwhile pointing out that you carry out this formal analysis and then you will clearly see that it is coming out to be twice of the uniform inlet velocity anyway for the x momentum part what we have is again because it is a steady flow I am sorry actually I have chosen all steady flow problems perhaps I will give you something for homework as unsteady x momentum out minus x momentum in equal to the net x force acting on the control volume here there is a little bit of more algebra because what you need to do is you have to first of all write the expression for the x momentum leaving the control volume which you again use that elemental area which is in the form of that annular ring 2 pi r dr multiplied by u multiplied by rho so this first part simply gives you the mass flow rate multiply again by the actual velocity which again is a function of r and that dr is here so that is it integrate from 0 to r the inlet is actually fine so you simply have the mass flow rate times the velocity so it simply shows up as a square of that u1 having already talked about that p1 minus p2 in the previous example the x force due to the pressure difference is simply p1 acting in this direction and p2 acting in this direction on the control volume so it is again p1 minus p2 in the x positive x direction and as usual we take this r suffix x as the reaction force on the cv due to again its interaction with the pipe this integral will be slightly troublesome in the sense that and because of that I am not doing it here there is little more algebra here but it is if you just work out the steps you should get it there is nothing other than just a little bit of few steps of algebra when you solve this everything is available p1 minus p2 is given as problem statement that rx which is the reaction force on the control volume due to its interaction with the pipe will be obtained but solved it I am asking you can you tell me what the sign would come about why can you take fx equal to 0 but it is not right it is a it is a flow fine so then you get the direct answer that you want that is fine that is perfectly fine however since I have written out this I am just asking you or people what could be the in this way if you work out what should come out the sign for that rx is the reaction force on the control volume due to its interaction with the pipe correct so what what what should it be negative so you better work out this the analysis should already tell you that this rx will come out as minus something which is what you want because what was asked as the problem statement is the total force of friction exerted by the tube on the air so on the air mean whatever is the material inside the control volume which is exactly what we want and now the second part of the problem asked you what was the average wall shear stress over the length so you simply express the magnitude of this rx as some sort of an average wall shear stress times the area over which this is acting which is simply pi d times the l and and you should get that so please if you are really enthusiastic you will tell me tomorrow that this really came out negative at least I am hoping one person should tell me is that fine yeah let me just tell you what the other one was so we looked at this one the other one was essentially a porous walled pipe where some inlet conditions were given pressure density velocity this is uniform velocity area of cross section of the pipe was given and it is been told that there is a leakage through this porous wall of the pipe that is happening and over the entire length of this pipe the total mass flow rate that is leaked out is 20 kilograms per second that is the rate of the leakage it is also told that the leakage mass flow through the porous wall leaves in a direction normal to the pipe access this is important to note and I will ask you why some exit conditions are also given p equal to some absolute pressure of 300 kPa rho is given and you are asked to find out the axial force on the off the fluid on the pipe now this is exactly the opposite of what the previous one was why do you think this this is important to mention that the leakage mass flow rate through the porous wall leaves in a direction normal to the pipe access correct so since we are essentially interested in the axial force this leakage mass flow rate is not going to participate for contribute to the axial force if the leakage mass flow rate was leaving with some angle velocity let us say this way then we will have to take into account properly but in this case the leakage mass flow rate will come only in the mass balance equation okay it is not going to come in the momentum equation that is all that is to be noted if you know that then there is nothing to be so I will let us go to the next one which I think is something that most of you have seen this is jet pump the popular device what is done is in this case there is a central jet which has a higher velocity in this case 30.5 meters per second and essentially drags a lower velocity fluid from the annular region to eventually deliver it the total of it that is at a higher pressure what is given as the section 2 here it is given that whatever total fluid that is coming out is coming in the form of a uniform velocity v2 total area of cross section is given as 0.07 the secondary flow velocity is given as 3 the area of the jet is given as 0.009 so the area for the secondary flow can be immediately calculated 0.07 minus 0.09 is given that it is a steady flow and the pressure at section 1 is uniform is given to you because you can see that it is essentially mixing two fluid streams but to simplify the problem is given is been given that the pressure is uniform on the section 1 velocity at section 2 is uniform you are asked to find out what is the velocity v2 and the pressure rise p2 minus p1 so here I have drawn the control volume here again just inside the pump inner walls in some sense and if you do the standard mass balance and momentum balance because it is a steady flow you can write that m.in equal to m.out where m.in now has two contributions to the jet everything is given as a uniform velocity so you do not have to bother much about it the jet contribution and the secondary flow contribution what is going out is a uniform flow with a velocity v2 it is all this you directly get the velocity v2 for the x momentum part exactly the same way as we have been writing before only thing now here you realize that the momentum has two contributions again coming in from the jet as well as from the secondary flow p1 minus p2 times a what will be acting on the control volume in the x direction and this rx which is essentially some sort of a viscous force acting on the control volume because of its interaction with the pipe I am neglecting this here without having specifically said that in the problem statement but I think someone was already mentioning to me in the break that this is a reasonable assumption to do so any comment Professor Gowdy since you were saying it so why can I go ahead and neglect this without any trouble so that you solve this is that your when we do fluid mechanics probably mostly for power consumption they are the total power consumption values or force values we are getting through power input are relatively of higher magnitude than the pressure drop values so in most of the analysis we are neglecting these forces what I felt it thank you the the primary purpose what is the primary purpose of a pump pressure rise isn't it so you actually if you if you were to do a order of magnitude analysis of the pressure forces versus any other kinds of forces that are acting on it usually you will end up with the conclusion that the viscous force would be much much smaller than the the net pressure force that is acting here and that is why you can easily neglect this even though the problem statement never really said this so sometimes we expect that students actually comment in this fashion that you don't really write that neglect the viscous force you ask them that you comment on it whether you want to include or not if you want to include it obviously you cannot solve the problem that's one thing but that's fine at least you can you can ask them to say such comments so that you can test out a little bit of understanding on that all right so this was more or less a few quick examples which set of examples which I wanted to point out normally in the fluid mechanics undergraduate course that we teach we spend a lot of time on solving such problems which I am very sure many of you are also doing if you are not I will suggest that this is really something that you must think about the integral analysis is very very useful as as we said in overall force balance overall mass balance and the this is the absolute first tool that the student should really know quite well so please if you are not including this material in the in the fluid mechanics course that you are teaching I strongly request you that you think about including it let's just see where you can find this integral analysis material if you if you look at Potter's book it's in chapter 4 it's all in one chapter all mass momentum energy everything is in one chapter so is the case with Fox it's in chapter 4 on other hand if you look at Gupta and Gupta they have actually split it over a few chapters so they talk about mass conservation in one momentum balance in something else energy in something else so it's I think split over the four chapters that I have listed 3 4 5 and 7 and by we believe me some of the problems are not at all easy I think I have chosen here extremely easy set just to demonstrate the the utility of the equations how to use it but what I can do is if if you people feel that this is worthwhile mentioning it in reasonable detail when it comes to the main workshop what I'll do is by then I will come up with more complicated problems so I think on Friday of this week there will be a feedback session where you can suggest what we can do additionally or what we can even get rid of from the material that we are talking about here when it comes to the final workshop so I'll request that please just make a note in your book right now that on this integral analysis part let me know on on Friday whether you would like to look at little more problems there is nothing more on the derivations part it's just the utility of those equations to solve problems and I'll try to dig up a few more complicated looking problems and discuss those directly now then in the main workshop okay so as I said I'm sorry actually this this was all set of steady problem actually the thing is that most of the situations that you deal with are indeed steady very very few situations are unsteady for example in this case very easy to to make this unsteady here how will you make this unsteady so then that the level will keep going up or down whatever so that's all that you can think other classic example of making things unsteady is you can think about no inlet basically very large tank with an outlet and that there is a constant discharge so that the level is going down and using Bernoulli which we haven't talked about so far or some ways we have you can relate the change in the height to the velocity of the exit of the jet which will then be a function of time then you can come up with a problem where you say that as this tank is discharging I want to keep this tank stationary so what should be the requirement of a force to be applied to keep it stationary so that force will be actually a function of time as the tank is discharging because the level in the tank is function of time the velocity of the discharge will be a function of time and so will be the force which will be required to keep the tank so such extensions you can easily yeah what is that yeah you are right so it's it's not exactly an unsteady I will call it a pseudo steady situation I agree with your point it will be really considered as a pseudo steady situation the idea there is that really the area of cross section of the tank would be considered to be much larger than the discharge area of the pipe so that it can be roughly taken as a pseudo steady problem you are right though I fully agree with it but there are some very interesting unsteady problems if you are really keen on that let me know by Friday and then on the main workshop I will definitely include those for more discussion but this integral analysis is very important and and I find it actually far more interesting than the differential analysis which is just a bunch of mathematics so think about it I think if you are not including it in the coursework it is nice to do it I think I realize that some of you are indeed doing this in the classes who is not doing it can you please okay I actually think that oh yeah but please use the mic I think that is essential the issue is that I mean students feel overburdened with such kind of mathematical these things and undergraduate level students especially okay my take is that this is actually fairly elementary mathematics the way you look at it I can tell you that most people here who already knew that this was done for their undergraduate will actually say that this needs to be done so I think the people at the back are probably saying that this can be done easily yeah in fact I find it's more easy if we start with this integral analysis the students get interested and then go to differential analysis instead of going to differential analysis first time going to integral analysis I will add on to that in the sense that the differential analysis at the end of it is a bunch of modeling exercise in the sense that you may think that oh Navier-Stokes equation whatever is nothing but a model so how much emphasize or emphasis you want to give to a model even though it may be working quite okay for the fluid flows and so on that's fine the integral balance actually is the most fundamental thought process that one needs to know as a student when it comes to analyzing fluid problems that is at least my take on it and think about it I mean I am not saying that you should do it necessarily or not but if you do it will be a really good value addition to the undergraduate curriculum now having said this I don't know how much you have the control over your syllabus I know but how many other people have the control over their course syllabus I don't know for example usually the university system has certain syllabus you cannot really change it so that's fine that there you don't have the control and I don't expect them to do it but autonomous places like what where you are I think think about it it's worthwhile doing it because by and large I can tell you that students are really more interested in doing this because they get some feel for it otherwise it's just some mathematics that is getting worked out and they'll say okay so what I think more over actually when we solve this kind of problems only integral analysis will help the student to solve the problems better than going for differential analysis actually I think it's easy for them to understand the problem and then understand how to solve the problems I think integral analysis is differently very very useful for the students at the undergraduate level so for example the overall fourth requirement here like let's say here if one were to do this from a differential point of view one would actually need to solve the entire flow field here okay then at each of these actual locations you will have to find the shear stress and then integrate it over the length. But since you are probably aware of the Von Karman integral method that parabolic or maybe even a cubic gives a really really good approximation so at least like he said students will immediately get something useful out of this and that is why I think it's worthwhile doing it I agree with you that it's an approximation which will which will not give you the exactly the same answer as what you want or the real answer but still you know some feel for it yes. Consider the area at the inlet and area at the outlet they are different so even if the pressure is uniform the force force has to end that's what you think so any comment on this okay yeah go ahead if you have a projected area it will be equal here the point your question is that the area of cross section is essentially changing and therefore even though the pressure is uniform given why is it not giving you a net force I'll point out though that there it was the gauge pressure yeah here what is given is that the pressure is uniform which essentially means that I can treat this as a atmospheric pressure which is acting on this entire control volume yeah what is the nature of the control volume it's a closed completely closed control volume you can actually very easily see by looking at the components see for example here this is a slanted area so if you want to show a pressure force acting on it you will want to show this using a normal here so that normal will have a x component and a y component if you do a component wise analysis you will essentially realize that the projected area as as he was pointing out take care of the the component so that the net contribution come out to be so the easiest way to really look at this is if the pressure is uniform okay what can you say about I mean I don't want to use the math mathematics unnecessarily too much but what can you say about this quantity yeah here is my control volume closed one and the area surrounding it is a and it's the same pressure now this is actually the same pressure that is acting on everywhere and I integrate it over this closed that is what is happening there there is nothing more than in fact that's a rule of thumb you can you don't have to even think why if you are dealing with a closed control volume where the pressure is uniform everywhere throw it out there is nothing that you can think about the reason I am saying this requesting again the undergraduates to being exposed to this is when we see the m tech student coming into IIT from various colleges unfortunately they don't have this and in some sense this is selfish motive that we have to do all this before getting on to the other part yeah so it's a selfish motive to some extent but it's a very interesting business as well so I think one way or the other you can think about