 What I want to do now is I want to work out 4 or 5 problems where hopefully things will be more clear as to what we are going to do with this. So, let me start the discussion of those problems. So, here is the first example. What I have is a tank which is on wheels and it is free to roll on a horizontal surface. It has an inlet pipe and an outlet pipe and the inlet pipe has an area of cross section A1, outlet pipe has an area of cross section A2. Flow is coming in with a velocity V1, flow is going out with a velocity V2. What is given is that this is a steady constant density flow. So, you can say this is some sort of a liquid flow if you want steady. What is given is uniform flow at inlet and exit. So, what uniform flow essentially means that across the entire inlet and across the entire outlet the magnitude of the velocity will be the same. So, it will be all V1 across the entire outlet sorry inlet A1 and it will be V2 across the entire outlet A2. So, that is meant by uniform flow at inlet and exit and let us say as a part of the problem statement numbers are given although I am not giving the numbers here. It is given how much is V1, how much is A1, how much is A2 and the pressure is acting in the inlet section and the outlet section. What you have to find out is first the outlet velocity and the second is what should be the force required to keep the tank stationary that is the second part of the problem. So, what we do in these cases is that we identify a control volume and the control volume that has been identified for this particular case is shown using dotted lines. So, the dotted rectangle that you see on your screen is essentially the control volume that has been chosen for the analysis of this problem. So, the idea is that the control volume is cutting across the tank and the wheels and so on. So, that way you can imagine that whatever is inside the boundaries of the control volume is the material that is getting isolated by the control volume. So, as such we need to mark when we come to the momentum equation all forces that are acting on the control volume which will be essentially on the material that is getting isolated by this control volume. So, let us see how it is done. So, on the left here I have shown the isolated control volume and I am not showing any other details. The idea is that there will be whatever material inside this control volume which is isolated by it. I am still showing the inlet and the outlet and the pressures that are acting in the inlet and the outlet as P1 and P2. Now remember that pressure we talked in the morning is a compressive stress. So, therefore, whatever pressure is acting in the inlet section 1 is actually acting in the plus x direction because it is trying to compress the material in that sense. Similarly, the pressure that is acting in the outlet section is actually acting in the negative x direction because as far as the material within the control volume is concerned P1 and P2 have to be acting in the compressive manner. So, P1 has to be then acting in the positive x direction and P2 has to be acting in the negative x direction. One more force has been shown acting on the control volume in the positive x direction. So, what is this force? If you go back again what we said is that the control volume is supposed to be isolating a certain amount of material and we need to show the force that is getting acted on this certain amount of material because of its interaction with the remaining material that is the surrounding material. Now if you see the control volume it is cutting across this tank. So, there is some sort of an interaction that you can imagine between the material inside the control volume and the tank. So, when you cut it you have to appropriately show the reaction that will be coming on to the control volume material because of its interaction with the outside material and that is precisely what is shown by this force R suffix x. Now for the purpose of analysis we show it as a force in the x direction it does not have to be in the plus x direction that is. But the analysis should automatically point out whether it will come in the positive x direction or not. So, when we start with it we will always assume that it is in the positive x direction analysis will automatically tell us where it is coming. The first step in this analysis is the implementation of the mass conservation principle. Remember that it is given as a steady flow. So, steady means that there is no accumulation of mass within the control volume which means that whatever is the mass flow rate in has to be the same as mass flow rate out which is exactly written out here. Now because the flow at the inlet and exit is given to be uniform and also because v1 is perpendicular to a1 and v2 is perpendicular to a2 I am essentially in a position to calculate these mass flow rates as simply the scalar algebraic products of rho multiplied by v1 multiplied by a1 which will be giving me the mass flow rate in and rho multiplied by a2 multiplied by v2 which will give me the mass flow rate out. Also eventually the density will cancel out because it is a constant density flow. So, rho1 is equal to rho2. So, let us see that. So, m dot in is equal to m dot out because of the steady flow situation and m dot in is written as rho times v1 times a1. Remember it is the normal velocity v1 across a1 and the normal velocity v2 across a2 and uniform as well. Rho is the same. So, that gets cancelled and what you are after v2 is immediately to be found as v1 times a1 over a2 as part of problem statement v1 a1 and a2 are given. So, you find out v2. The next part is essentially the use of x momentum equation. Again this is a steady flow. So, there is no accumulation term. If there is no accumulation term what is left out will go back to it here. If there is no accumulation term let me go to the x balance this one here. This is the accumulation term. So, it is not there it is gone altogether. What is simply left out is then the net force in the x direction is equal to the rate of outflow of x linear momentum minus the rate of inflow of x linear momentum which is what has been written out here. Net force acting on the control volume which is f suffix x let us say we do not know what that is yet. We will calculate that in a minute is equal to the rate at which x momentum is leaving the control volume that is out minus the rate at which x momentum is coming in. So, now we will calculate the net x direction force on the control volume. There are three forces essentially acting on the control volume. One is because of the pressure p1, other is because of the pressure p2 and the third one is the so called reaction on the control volume because of its interaction with the tank and whatever else outside the control volume. So, I basically add these three forces we had assumed are in the positive x direction. So, this is positive p1 is acting in the positive direction because it is compressive remember we said that. So, p1 multiplied by a1 is the force in the positive x direction acting in the inlet p2 times a2 is the force that is acting in the minus x direction at the outlet. So, that ends our net force calculation and the x momentum on a rate basis that is going out is simply the mass flow rate out times the velocity out minus the x momentum rate on a rate basis that is coming in is the mass flow rate coming in times the velocity at the inlet. So, what is m dot out and what is m dot in they are actually the same because it is a steady flow. So, the way I have written this is I have written it in terms of the reaction that is coming on to the control volume is equal to take all these terms to the right hand side p2 a2 minus p1 a1 plus m dot out is equal to m dot in which I am calling as m dot and written out here this rho1 and rho2 are the same actually they are equal to rho and multiplied by m dot multiplied by the outlet velocity is v1 minus inlet velocity is v2 and that is it. So, everything on the right hand side here will be known. So, you are in a position to calculate that therefore, what you have done is you have been able to obtain the reaction that is coming on to the CV because of its interaction with the tank. Remember that our CV cut across the tank and therefore, this reaction is coming because of the interaction of the CV with the tank. So, through Newton's third law of motion whatever is the reaction on the CV and equal and opposite action let us say should be on the tank and therefore, the force on the tank will be simply equal to minus of this. Now, if that force is permitted to be acting on the tank the tank will keep on accelerating. So, if you want to know prevent the tank from accelerating which is what the problem said that you find out the force which is required to keep the tank stationary what is the answer it is exactly same as this Rx. The way it has been argued is that R suffix x is the reaction coming on to the CV which is nothing but minus of the force on the tank due to the CV through Newton's third law and therefore, if you employ externally a force equal to Rx on the tank it will not move anywhere. So, that is the way the integral analysis has been employed and although I have not given any numbers here numbers can be given and you can just plug those numbers in and you can calculate everything that you need. The one point that one needs to know here carefully is the P1 and P2 that is getting used in the calculation is actually the gauge pressure. Why is it the gauge pressure that will come only in the calculation is because look at the CV. So, the CV is a closed CV. So, in general what the pressure the pressure that will be acting on the CV will be the atmospheric pressure plus any gauge component that is coming on to it. So, the atmospheric pressure is essentially going to be assumed to be the same everywhere around this control volume and because it is a closed control volume the contribution from the constant atmospheric pressure will actually cancel completely and what is left is only the gauge pressure at the inlet and the gauge pressure at the outlet and that is why this P2 and P1 will finally remain as the gauge pressures. The contribution from the atmospheric pressure which is acting uniformly over this entire closed control volume will simply go away and you do not have to bother about it. So, this is the way a typical integral analysis goes about. At the end of it it is simply using the balance equation for mass and balance equation for the momentum. Here we had to look at only the x momentum. There is a possibility that in general you may have to look at both x momentum as well as y momentum perhaps even z if it is required, but the procedure is no different it is exactly the same as we will see in a few more examples. Before going ahead with the examples though let me point out a few useful tips which we normally ask students to keep in mind while performing this integral analysis. First and foremost is that choose a control volume such that the inlet and outlet velocities are normal to the corresponding areas. So, what this will enable is that when you are calculating the mass flow rates in and out you will simply need to calculate an algebraic product of the velocity with the area. Otherwise if the inlet and outlet velocity is not normal to the inlet and outlet area remember that we will actually have to formally evaluate a dot product like this. In general if velocity and areas are not aligned with each other that is if v0 and n hat here are not in the same direction you will actually have to calculate the dot product. So, it is better if you choose the control volume such that the inlet and outlet velocities are normal to the corresponding area. When I mean by normal to the corresponding areas it is normal to the physical area but remember that the outlet normal to the physical area so that you can say that the inlet and outlet velocity should be aligned with the inlet and outlet normals unit vectors. So, that way you can interpret. The forces which are the surface forces and the body forces acting on the control volume are in general the forces acting on the material isolated by the control volume taken as a free body this we have already discussed sufficiently. The way that normally we end up choosing these control volumes are such that the required force or an equivalent opposite reaction to it acts directly on the control volume. So, let me go back to our example here the interest was actually to find out the force required to keep the tank stationary ok. So, the way this control volume was chosen was that an equivalent opposite reaction to this in some sense was acting on the control volume and that is precisely what is meant by this third statement. Finally, you should realize that the linear momentum in and out of the control volume can be positive or negative depending upon whether the inlet and outlet velocity is positive or negative in the chosen coordinate system. Now, in this particular example the both the inlet and outlet velocities are actually positive because they are coming from negative x to positive x and going from negative x to positive x. So, it turns out that in this example both inlet and outlet velocities are positive, but in some examples it is possible that within the chosen frame of reference that you have or coordinate system one of the two can be negative. So, you have to appropriately account for that mass flow rate is going to be always positive it is a scalar quantity. So, you do not have to bother about that, but when it comes to linear momentum in and out because the velocity at the inlet or the outlet can be positive or negative you have to pay careful attention to that and the term linear momentum can then be either a positive or negative. So, now let us look at a few other examples with these points in mind and of course, the previous example. So, here what I have is a flat plate and a fluid flow going past the flat plate at the leading edge of the flat plate what is given is that it is a uniform flow with a magnitude of capital U and at the trailing edge of the flat plate it is given that there is a profile of the axial velocity, meaning in the x direction given by some function U of y and what is already shown here is a control volume which is shown by the dotted line. The control volume is such that the top surface of the control volume is essentially a streamline. Now we have not talked about streamlines yet, but there is a good chance that many of you know that a streamline as a property that no mass flow rate can cross the streamline. So, therefore, the idea here is that whatever the mass flow rate is coming through this height h on the left is actually getting passed through between the plate and the streamline on the top and the same mass flow rate is passing through the right hand side height of y naught eventually. The bottom surface of the control volume is actually coincident with the bottom plate here, so it is not getting seen, but that is where it is. What is given is that it is a steady constant density flow and also given is that there is uniform pressure everywhere and you are asked to show that the drag force on the plate is given by this particular expression which is integral from 0 to y naught rho multiplied by small u, the whole thing multiplied by capital u minus u dy. So, let us see how this can be done. So for the control volume which was shown by these dotted lines, we first employ the mass conservation. It was given that it is a steady constant density flow. So, again what we have is a mass flow rate coming in is equal to mass flow rate going out. If you go back to the sketch, you will see that this is a two dimensional picture meaning that we can work out everything on a per unit depth into the plane of the paper. So, if you see that then mass flow rate in is simply rho multiplied by that uniform velocity at the inlet multiplied by the height of the control volume h. In principle, this whole thing is multiplied by the unit depth into the plane of the paper which I am not writing. So, this is h multiplied by u multiplied by rho multiplied by 1 into the plane of the board or the paper that is the mass flow rate in. On the other hand, at the outlet because we have a profile of the axial velocity as a function of y, we cannot really do a scalar multiplication without an integration. So, what we do is then the m dot out then is simply rho multiplied by u multiplied by dy which is essentially an elemental mass flow rate going out and then you integrate that between 0 and y naught. If you simplify this expression, I obtain a relation between h, height h on the left hand side and whatever else is left in the equation. So, this intermediate result will be used later. Now, let us look at the x momentum equation. It is a steady flow as we said so that there is no accumulation of x momentum in the control volume. What we have is as before the net force acting on the control volume which is shown by f of xx is equal to the rate of x momentum leaving the control volume minus the rate of x momentum coming into the control volume. So, as far as the rate of x momentum leaving the control volume is, we need to again resort to an integral representation and the way I have written this integral representation for the total x momentum leaving the control volume from the right hand phase here is that rho times u times dy which is the elemental mass flow rate, multiply that by the axial velocity in the x direction which will give me the rate of x momentum leaving through an elemental area of dy multiplied by unit depth. So, dy somewhere here multiplied by unit depth is what the area that I am looking at and then I integrate this entire expression from 0 to y0 to get my total x momentum on a rate basis going out. Fortunately, on the inlet side we have uniform velocity. So, the mass flow rate at the inlet multiplied by multiplied by the uniform velocity at the inlet will give me immediately the x momentum on a rate basis coming into the control volume. So, it is simply rho times capital U times h multiplied by capital U. So, you simplify this what I have inside the integral is rho times u squared dy and the inlet x momentum is rho times capital U squared times h going on to the next step what I will do is this h here where my pointer is standing I will replace this h using the intermediate result that we had obtained using the mass balance. So, that h is replaced as 1 over U times the integral over sorry integral from 0 to y0 to U dy and if you rearrange this entirely what you will get is that the reaction on the control volume because of its interaction with the plate keep that in mind. So, the reaction on the control volume because of its reaction or interaction with the plate is equal to whatever the simplified expression will come from this which is simply integral 0 to y0 rho times small u whole thing multiplied by small u minus capital U integrated over the y direction. One thing that we need to talk still carefully is the net force acting on the control volume. So, let us go back to the problem statement here because pressure was given to be uniform and because we are dealing with a control volume that is completely closed the uniform pressure acting around this completely enclosed control volume will essentially cancel out completely. So, then there is no contribution from the pressure as far as the net force acting on the control volume is concerned in the x direction. The only other force that remains is because of the interaction of this control volume with the plate and that is precisely what has been included as the reaction in the x direction on the control volume. So, that net force here fx is simply only the reaction force that is left out there is no contribution from pressure in the x direction there is no other force contributing to the control volume in the x direction and therefore, what I have now at the end of this analysis is a relation for the x direction reaction on the control volume due to its interaction with the plate which is exactly equal to the negative of the drag force on the plate using our previous argument of using the Newton's third law of motion. And therefore, what we wanted to find was that the drag force on the plate is equal to such and such if you go to this slide what we have is the exact negative of the expression that we wanted to find out. And the argument is that the reaction on the control volume is essentially negative of the drag force on the plate and therefore, you simply multiply this entire expression by negative 1 if you want and you will get the expression that you wanted to find out which is this one. So, again to summarize this problem what we have done is we wanted to find out the drag force on the plate. So, we chose the control volume such that at the inlet and that at the outlet the inlet velocity and the outlet velocity are normal to the inlet area and outlet area remember that there is no flow across the streamline. So, nothing to talk about there we also chose the control volume such that the reaction on the control volume because of its interaction with the plate turned out to be or turns out to be the negative of the drag and therefore, once we find the reaction we can immediately know what the drag is. So, what we have done is we have tried to follow these pointers one by one choosing the control volume such that the velocities are normal at the inlet and outlet we are taking the material as isolated by the control volume marking all forces on it the force the control volume has been chosen such that the required force or an equal and opposite reaction to it acts directly on the force on the control volume and so on. So, we are trying to essentially follow these pointers to come up with what we want in these integral analysis. Let me take one or two more examples. So, this is a very classic example that we end up using in the integral analysis. So, what we have here is some sort of a solid object sufficiently upstream or ahead of the solid object I have a uniform stream of fluid approaching it at the speed of u capital U and I am calling this section as section number 1 here sufficiently downstream from the solid object what has been given is that the axial velocity has a very peculiar form. So, if you look at the center line here from the center line it linearly increases to capital U over a distance of h over 2 both ways so up as well as down. What is given then is that this is a steady constant density flow what is also given is that the pressures at the two sections 1 here and 2 here are the same this is given as part of the problem statement. And what is asked is that you have to determine the drag force on the 2D object which is the object shown here per unit depth into the plane of the paper. Since this is a two dimensional picture we are anyway going to talk about per unit depth as we did for the previous problem. So, let us see how this is done. So, here we choose only half of the domain for analysis and the reason is because as you can see about the center line which is horizontal here the problem is symmetric. So, that we do not have to take into account the entire domain what we will do is we will analyze only the half part and then multiply by 2 when it comes to finally getting the answer. So, I am showing here a control volume which has been chosen again according to our guiding principles inlet is such that the inlet velocity is normal to the inlet area, outlet is such that outlet velocity is normal to the outlet area. We are interested in finding the drag force on this object in this case the control volume is actually cutting through the object. So, therefore what we will say is that the control volume will have an interaction going with the object because of which there will be a reaction force coming onto the control volume and for the purpose of analysis we have shown this reaction force to be acting in the positive x direction. The analysis will show whether it is coming positive x or negative x. So, now the peculiar thing about this problem is that if you see the control volume there is an inflow from the left, there is an outflow from the right very obviously the outflow from the right is less than the inflow from the left because the problem is symmetric about the bottom horizontal line here bottom horizontal boundary you can say nothing is going on the bottom side of the bottom boundary. So, therefore what we are left to deduce is that the difference between the mass flow rate that is coming in and the mass flow rate that is going out must be going across the top border of this control volume which has been shown as the m dot out top. So, the first part of this problem is to find out what is this m dot out at the top and to do that we go to our standard mass conservation statement because the flow is given to be steady we have whatever mass flow rate coming in is equal to whatever mass flow rate that goes out and therefore for this particular control volume at the inlet the mass flow rate coming in is simply rho multiplied by the uniform velocity coming in multiplied by the h by 2 which is the lateral extent of this control volume obviously everything is happening on a per unit depth basis. So, multiply that by 1 if you want. So, on the left hand side here is your mass flow rate coming in mass flow rate going out from the right boundary since this is now a profile of axial velocity with respect to the y coordinate we have to employ our integral representation. So, rho which is a constant that is why I have taken it out from the integral times u times dy which is the elemental length in the y direction integrated from y equal to 0. So, obviously I have chosen y equal to 0 at the center line to y equal to h over 2 will give me the mass flow rate going from the right face of the control volume plus whatever is going out from the mass flow rate at the top which we need to find out. So, if you rearrange this mass flow rate going from the top of the control volume will simply come out as rho times capital U times h divided by 4. Remember that when you evaluated this integral u dy in here you actually had to figure out the function for the axial velocity as a function of y. So, the axial velocity now goes linearly from 0 to capital U over a distance of h over 2 and that is what has been incorporated inside the integral here. Simplify it and you obtain first the expression for the mass flow rate that is crossing this top horizontal surface of the control volume. So, that is first part of the problem. Then going to the x momentum equation because again it is a steady flow we get rid of the accumulation term and what is left out is only the x momentum leaving minus x momentum coming in obviously on a rate basis. Now remember that x momentum leaving here is slightly tricky. Obviously, the x momentum is leaving from the right hand face. However, if you look at the top surface what is the velocity in the axial direction at the top surface it is equal to capital U. So, when this mass flow rate crosses the top surface in here it actually carries with it an x momentum equal to the mass flow rate multiplied by the magnitude of the axial velocity at the top surface here which is exactly equal to U. So, therefore, as far as this control volume is concerned the x momentum leaving the control volume has two contributions one from the right hand face and one from the top face. So, please make sure that you understand this idea and that is precisely what is written out here inside the two square brackets. The integral is rho times U times dy which is the elemental mass flow rate going across the right surface here multiply that by the velocity itself and integrate it from 0 to h by 2 to give me the total x momentum on a rate basis leaving from the right hand face. Plus the other component of the out flowing momentum comes from what we just discussed a minute back the mass flow rate that is escaping from the top of this control volume multiply that by whatever is the axial or x direction velocity at this location which is nothing but capital U. So, m dot out from the top multiplied by U is added to the x momentum that is leaving from the right hand face and these two together will constitute the total x momentum leaving the control volume x momentum coming in is fairly straight forward what we have is rho times U times h by 2 is the mass flow rate coming in multiply that by the uniform velocity at the inlet capital U. Here also remember that all momenta have positive sign because all velocities are positive x velocity here is positive coming in x velocity is positive going out similarly for the top surface x velocity is positive going out. So, all momenta are positive and if you now simply rearrange this you will realize that first of all you get the reaction on the control volume that is the only force that is acting on the control volume because as part of the problem statement sorry it was shown or it was given that the pressures are exactly the same. So, there is no contribution from the pressure in the net force acting on the control volume only component that remains is the reaction force which is nothing but minus one half of the drag on the object through our standard Newton's third law of motion argument and if you go ahead and simplify this what I have worked out is that the drag will actually come out in the following form rho multiplied by capital U squared multiplied by h divided by 6. What I will suggest is that you please work out this algebra from here and here and see whether you are actually obtaining the expression for the drag as what I have obtained. So, this was the third example in this integral analysis let us just discuss quickly a couple more. So, what we have here in example number 4 is a porous walled pipe. So, flow rate is coming into the pipe on the left and the conditions at the inlet are given velocity is given it is a gas flow it is a steady gas flow. So, it is a very high speed gas flow with a density of 6 kgs per meter cube and certain pressure is given at the inlet what is what is given is that over this length whatever it is there is a total leakage of mass flow of 20 kgs per second because the pipe is a porous wall pipe and what is given is that the leakage mass flow through the porous wall leaves in a direction normal to the pipe axis it is part of the problem statement. The area of cross section for the pipe is given as 0.15 meter squared and some conditions at the outlet are given which are pressure and density. So, steady gas flow leakage over this entire pipe is occurring normal to the pipe axis and some conditions are shown and what we have to find out is the axial force of the fluid on the pipe. So, in this case the control volume is shown again using a standard dotted rectangle and note that it is cutting across the pipe. First step as always is our mass conservation. So, mass conservation in this case is going to be written for a steady flow situation that whatever is coming in minus the leakage over the length of the pipe will be the mass flow rate that is going out from the right hand end and that is precisely what is written out here m dot in is essentially equal to m dot out plus m dot leakage. m dot leakage is directly given and m dot in and m dot out assuming that the inlet and outlet velocities are uniform over the corresponding areas inlet and outlet. I am writing this as rho 1 times v 1 time the area of cross section of the pipe equal to rho 2 times v 2 times area of cross section of the pipe plus m dot leakage. Rearrange this and the first intermediate result that you obtain is the exit velocity v 2 from the pipe which is given by the following expression. I have not worked out the numbers which you can do by substituting the values in here. Now, we go to the x momentum equation. Again this is a steady flow situation. So, the net x force acting on the control volume is equal to the rate of x momentum leaving the control volume minus the rate of x momentum coming into the control volume. Now, remember that the leakage mass flow rate is actually going normal to the axis and we are interested in the x direction momentum equation. So, because it is going normal to the x axis we do not have to bother about including the leakage mass flow rate and the corresponding momentum that it may carry because it is all in the y direction. So, the net force then is as usual the reaction coming on to the control volume because of its interaction with the pipe. Now, here there is a pressure that is acting in the inlet section which is P 1 here. P 2 here remember that as far as the control volume is concerned P 1 will act in the positive x direction on the control volume and P 2 will act in the negative direction with respect to the control volume. So, P 1 times the area of cross section of the pipe minus P 2 times area of the cross section of the pipe added to this reaction will give me the total net x direction force acting on the control volume and the two momentum terms leaving and entering the control volume this time are straight forward because uniform velocities are going to be assumed. So, m dot out times v out is the rate at which x momentum is leaving the control volume m dot in times velocity in is the rate at which the control volume is coming into the control volume and if you rearrange this what you will get is the as usual x direction reaction on the control volume because of its interaction with the pipe which is exactly equal to negative of the axial force on the pipe due to the flow. So, my suggestion is the following that between today and tomorrow wherever you had these numbers for example, in this case some numbers are given in the previous problem also some number well not exactly number, but you could work out the final expression. But in this case at least what you can do is you actually find out the values by substituting the numbers and verify that the R x which is the x direction reaction on the control volume comes in the minus x direction and therefore the axial force of the fluid on the pipe must be in the positive x direction. So, please verify that that is coming by substituting the numbers that have been provided and perhaps tomorrow morning we can verify that most of you have actually obtained that. But again to summarize this problem what we did was the first part was again mass balance, steady flow so there was no accumulation in the control volume whatever was coming in was either leaking through the pipe or going through the outlet and therefore we came up with the mass conservation statement in this fashion. When it came to the momentum equation we identified all the forces acting in the x direction on the control volume added those on the left hand side and the x momentum out and in on a rate basis in this case were fairly straight forward because of the assumption of uniform velocity over the inlet and outlet area. So, this is something that we call a jet pump. So, what we have is there is a bigger pipe of cross section 0.07 cm square I am sorry in the centre of that pipe a high speed jet of the same fluid at 30.5 meter per second is introduced and on the annular part so to say if you want there is a slower moving fluid what is given is the jet cross sectional area which is 0.009 meter square velocity of jet as I mentioned velocity of the flow that is surrounding the jet flow it is the same fluid the jet fluid and the surrounding fluid are the same given is the steady flow fluid involved is water in fact pressure at section 1 which is at the entrance here is uniform that is given velocity at section 2 is uniform we are going to also assume that velocities for the surrounding fluid and the jet fluid are also uniform over their respective cross sections and what is told is that you neglect the frictional force between the fluid and the walls altogether you have to determine the velocity V2 which is leaving this so called jet pump and the pressure rise which is delta P or P2 minus P1 is what you need to find out. So, the control volume again is shown here the left phase is at section 1 the right phase is at section 2 and the top and the bottom phase are supposed to be aligned with the top and the bottom wall of this pump just inside you can say imagine first thing is the mass conservation. So, if you see there are two streams of inflow mass if you want inflow mass flow rate and one stream of outflow mass flow rate. So, the two streams are added together making sure that we are using the correct cross sectional areas. So, for the jet it is the area of the jet for the surrounding fluid it is the total area minus the jet area which is taken out here and that is equal to whatever is the mass flow rate that is leaving from the right hand side right here all velocities are assumed to be uniform across their cross sections everywhere the fluid is water. So, we cancel the density and if you rearrange this equation you will obtain the velocity V2 that is going from the right hand phase and the x momentum equation again it is a steady flow. So, no accumulation of x momentum within the control volume the net force acting on the control volume is P1 times A1 remember that what is given is that pressure at the entire section 1 is uniform. So, you simply multiply P1 times this entire A1 is essentially equal to A2 in this case which is 0.07 acting in the positive x direction minus P2 times A2. So, actually I have already written out correctly here that is the total cross section area of the pipe A2 P2 times A2 is acting in the negative x direction. So, that is minus remember that there is in the problem statement a direction that neglect the frictional force between the fluid and the walls. So, because of that there is no additional force to be added on the left hand side. If we were told that there is certain frictional force that is acting on the walls you actually have to include that in the calculation of the net force on the left hand side. But since it is given that neglect it we are neglecting and the momentum flow rates are fairly straightforward in this case only thing is that there are two momenta as far as inlet x momenta are concerned. One because of the jet here and one because of the surrounding fluid and those have been appropriately calculated in here is only one outlet stream corresponding to which there is only one outlet momentum in the x direction. If you rearrange this entire expression you will obtain what you are looking for namely the pressure rise between section 1 and section 2 that is P2 minus P1. So, what again I would like to suggest here is that given these numbers please substitute those numbers in these expressions and find out whether indeed P2 minus P1 is coming as a positive number which means that the pressure is actually rising from section 1 to section 2. Remember that that is the purpose of a pump so we are talking about a jet pump and therefore this P2 minus P1 better be positive. See whether you actually obtain this by substituting the numbers that have been given in this particular expression and we can verify that tomorrow. So, this was a set of examples that I wanted to discuss as the standard implementation of integral analysis for control volume problems. Again from the CFD point of view this is not necessarily important however as an independent tool it is really really useful for many fluid mechanic situations that you know this integral analysis and go in a step by step fashion to employ it for the control volume that you are choosing. Please go through these 5 examples that we have discussed this afternoon between today and tomorrow because something very very similar one of these examples will come in the test tomorrow. So, you should be reasonably familiar with doing the mass balance and the momentum balance on some of these kinds of examples. So, with this it is 3.30 pm so we will break for the for the T. Thank you.