 We were looking into some examples to illustrate the use of the Reynolds transport theorem for working out problems related to that. We will continue with some more examples. Let us take an example where let us say you have a solid body say a solid body of whatever shape maybe circular shape if you wanted to be so and fluid is flowing it is coming from a free stream with a velocity uniform velocity say u infinity and because of the presence of the solid the velocity is disturbed and if you go a little bit away from the solid and if you draw the velocity profile say the velocity profile is obtained something like let us make a sketch of how the velocity profile is there. Let us say that the velocity profile varies in this way. In one of our later chapters we will see that what are the factors that will determine that what should be this velocity profile or what should be the nature of variation of this velocity profile but for the time being let us say that this is the this is a qualitative sketch of how the velocity profile varies. Assume that it is totally symmetric with respect to the centre line and the velocity profile is such that say like at the middle like if you draw it totally in a symmetric manner at the middle it is like a minimum and then it increases in both sides comes to almost u infinity at a given height. Let us say that this height at which it comes to almost u infinity is a and let us say this velocity profile is given in terms of the x and y coordinates. Let us say that x is the axial direction and y is the transverse direction and the velocity profile say is given by u by u infinity say is equal to sin this is given okay not that it has to be like this this is just an example. We are trying to satisfy the condition that when y equal to a u equal to u infinity that is how this velocity profile is there. So the question is that what is the total drag force on the solid body exerted by the water or the fluid. Let us assume that the density of the fluid is rho and we have to find out based on these dimensions. So how do we go about this? We worked out a very similar problem when we were considering flow over a flat plate. This is not flow over a flat plate this is flow past some body of arbitrary contour but the policy or the philosophy remains the same. So we have to basically find out we have to identify a control volume and see what is the net force on the control volume. So to identify a control volume see what part of the control volume will have some inlet and some outlet. So one inlet is this one which is straight forward that the flow is entering outlet this is straight forward and you can see that outlet is interesting only up to y equal to a or y equal to minus a because beyond that the velocity is uniform. So if we take a control volume say something like this where we consider one inflow boundary one outflow boundary and across other boundaries we do not want any flow. So what should be the edges of the other boundaries? There should be streamlines so that there is no flow across those. So let us say that we consider a streamline like this and these are not horizontal lines these are inclined ones. Let us just magnify those a little bit to represent that. Let us say that this is one extreme streamline this is another extreme streamline. So these are streamlines. Keep in mind that these it is not necessary to choose a control volume which has which contains streamlines but only elegance it gives to us is that we do not have to bother about the cross flows. But if we just take say some horizontal lines at the top and the bottom and constitute say a rectangular piece as the control volume then there will be flow across that and one has to make calculations related to that. So it is just a matter of convenience for choosing the control volume maybe you consider only the water in the control volume so you exclude the solid part. Now let us write the expressions for the mass and the linear momentum conservation for the control volume. So one thing that remains unknown is that what is this height h because we have constructed a streamline from the right from the edge of this layer where u becomes u infinity this is not a boundary layer. We will discuss later that what is the difference between this and the boundary layer as such. So when we have the streamline starting from the edge the streamline will end up here at some arbitrary point which is not known to us so we have to find that out. So let us say that we mark the edges as a b c d and try to identify that what are the expressions for the conservation of mass. So if you write the conservation of mass you have the dm dt for the system I am just writing it straight away without going for the explanation of different terms because we have already encountered that. So dm dt for the system in the left hand side is 0 then we assume that it is a steady flow not changing with time. So the right hand side you have first term 0 and then the term integral of rho vr.n da so that term we have to basically write. So what will be the corresponding expression here? So you have one outflow and one inflow so there is no flow across these 2 that is the advantage of taking the streamline. So what will be this term for the outflow? Say rho what is da, da you can take as dy into the width. Let us say w is the width perpendicular to the plane of the figure that is the width of this body. So rho into u dy into w that is the total when integrated from say y equal to minus a to plus a that means 2 into y equal to 0 to a that is the outflow and then the inflow that is there that is with a minus sign because velocity is along the positive x outward normally is along the negative x. So what will be that minus again 2 rho u infinity into h into w right. So from here you can get what is h because that is the unknown that you have to find out mass conservation tells us how to find out that. So h equal to integral u dy 0 to a divided by 2 infinity. Next conservation of linear momentum what it will tell us the resultant force which is acting on the control volume. So what is the resultant force that acts on the control volume? There is a pressure distribution but because it is open to the ambient the pressure is same all around. So the net effect of pressure distribution may be 0 but in reality it is it may be so that here the pressure is not same as the atmospheric pressure but something which is different from that. But let us assume that there is a uniform pressure distribution just for simplicity. Then the only force that remains along x so let us say we are bothered about the linear momentum transport along x. The only force that remains important is the drag force is equal to again the unsteady term is not there in the right hand side first and then basically these things will be multiplied with another u. So it will be plus 2 rho integral of u square dy w then minus so you can substitute the value of h here sorry this is u infinity square. 2 rho we have already taken as common so that will be the corresponding expression right. So what we have done is we have replaced h with this expression okay. Now what does this force represent? This is the force exerted by what on what? This is the force exerted by the solid body on the fluid control volume and you can easily see it say you do not know it but the mathematical sign will tell you see u square is less than u infinity. So this integral when it is evaluated it will be negative. So that means you have a negative force that means force which is along the negative x direction. So what force is there along the negative x direction? It is definitely force exerted by the solid on the fluid because it is trying to resist the motion of the fluid. On the other hand the force exerted by the fluid on the solid is equal in magnitude to this but opposite in sense. So that is along the positive x direction that we also call as drag force on the solid object. There are certain interesting things that we can observe from this problem. One interesting thing is it appears as if this force does not depend on the shape of this object but that is an illusion. Why? Where the effect of the shape of the object comes into the picture? The velocity profile. So the velocity profile will very much depend on what is the shape of the object. So we have assumed the velocity profile but this is like it does it is not that it is it comes just arbitrarily. This whatever is the velocity profile that velocity profile should come from the shape of the body and therefore that is where the shape of the body becomes critical. Now the other thing is that it will also appear as if the force does not depend on the viscosity of the fluid. It appears so. So it is a it is a it is a kind of like not an intuitive thing that you expect the force due to viscosity because if there is no viscous effect perhaps there would be no drag but there is no there is no viscosity here. There is no there is no there is no presence of the parameter viscosity here. So what what is the viscosity doing here or the question may be posed in this way that is it always necessary that viscosity will directly come into the picture for the drag force calculation. See viscous effect is there. There are 2 important effects which are important which are prevalent here. One is the viscous effect another is the contour of the body as the fluid is flowing over the contour of the body there is a change in pressure. So one is the geometrical effect another is the viscous effect and this velocity profile is a combined consequence of what has taken place. So once you are given a velocity profile you may be abstracted of anything else but where from the velocity profile has originated for that viscosity may be important but once you get a velocity profile see that is what the integral balance is giving integral balance is the net effect. It is not microscopic looking into what has happened at individual points but it has got a gross consequence. What what is the consequence some velocity profile at the outlet and this kind of gross consequence is important because you can measure it experimentally. Experimentally point to point measurement is difficult it is not impossible but it is it is always more expensive to do that but experimentally you can at least find out velocity profile on a given section. So you can have different probes also may not be as simple as a pitot tube probe but you you may have any velocity measurement along this section that is not difficult and uniform velocity which which is the free stream velocity that you know. So from the experimental understanding of what is the velocity profile at the inlet and at the outlet you may be in a position to experimentally calculate or rather to calculate from the experimental data what is the drag force and the limitation of that is it does not pinpoint that how how the flow feel varied from one point to another point to give rise to the drag force but it gives the total effect in an integral sense. Now how it varies from one point to another point for that we have to look into the corresponding differential equation for viscous flows that we will do in our next chapter. Now let us look into another problem let us say that there is a pipe like this it has to be fitted with another pipe which is like this I am trying to give you an industrial perspective of the problem rather than just stating the problem as it is. So the problem is you have to connect these pipelines it is very common that no matter whatever plant you visit you will see that there are lots of pipelines and pipelines are not always straight because they have to connect different systems and there are space constraints and so on. So the pipeline has to be bent many times. So there must be some fitting which connects this pipe to this pipe there are 2 things which have happened one is the direction has changed may be the axis of these are oriented at an angle 90 degree it is not required or necessary that it has to be 90 degree but let us say that the angle between these 2 is 90 degree plus there is a reduction in cross sectional area that also is not a mass sometimes cross sectional area may remain the same or may even may increase but one has to just have a fitting to fit that and that fitting in industry is known as an elbow. So what an elbow does it basically tries to have a fitting like this to fit or match with pipelines of different orientations and different sections. If the angle between these 2 axis is 90 degree it is called as a 90 degree elbow like that. So if we assume that this angle is 90 degree we call it 90 degree elbow. Now this elbow it cannot be free in air because we will see that there is a lot of force that is being exerted on this elbow because of the change of linear momentum of the water that is entering and leaving and because of that there is a force on the elbow and it has to be supported. So it must be supported with a support that provides some necessary reaction force which balances those forces exerted by the water on the elbow so that it is in equilibrium otherwise it might have a tendency to move or to get deflected from its equilibrium configuration and that will disturb the entire stability of the structure. So when you design a structure you have to be careful of what are the support forces that need to be sustained by that structure. For that the support force has to basically balance the force exerted by the fluid on the structure. So we have to know what is the force exerted by the water on the pipe bend. So that is our objective of solving this problem that is we want to know that what is the force exerted on the elbow. What are the data given? Let us try to list that force on the elbow. Let us say that you have points 1 and 2 or let us call these as sections 1 and 2. So at the section 1 say there is a equivalent pressure P1 which is given, area of cross section is given. At the section 2 you have P2 and A2 these are given. Let us say that the velocity profile is uniform. If it is not uniform at least you know what is the average velocity. So you know the average velocity at section 1, you know the average velocity at section 2. Where from you know the average velocity is experimentally what you can always find out it what is the flow rate. And if you know the area of cross section say it is a circular one you know the diameter. So you know the area of cross section. So A into the average velocity is the flow rate. From that you can find out the average velocity. If you neglect the viscous effects then the average velocity is same as the velocity at any point. So you may neglect that let us say that neglect non-uniformity in velocity profile. What else is given that what is the weight of the elbow say it is equal to w. Weight of the elbow is a solid. So it has its own weight. So we are considering that weight and we are given the density of the water which is there inside. Rho is the density and what else we require. Let us say the angle between this inlet and the outlet. So the water is entering like this it is leaving like this. The angle is 90 degree. If it is not 90 degree then also like if it is inclined it will have its horizontal and vertical components of the flux velocity and so on. Now we are interested to write the expression for the force component along x and force component along y. Sometimes you see that the pressure at 1 is given but pressure at 2 may not be given. But if you assume an inviscid flow and you connect a streamline say from the center of the section 1 to center of the section 2 you can use the Bernoulli's equation to find out what is the pressure and if you assume that the pressure is uniformly distributed over the section then that will be the pressure throughout the section 2. You have to keep in mind that what can create a non-uniformity in pressure. So if you have pressure at the center line say at p1 what can make it deviate if you go to different location in the cross section curvature of the stream lines. So if stream lines are almost parallel to each other then the change in pressure is very very small or negligible. So then we are assuming that stream lines here and here are almost parallel to each other. See if you take that on the bend that is not valid. So we are considering the section 2 which has actually crossed the curvature part of the elbow. This is an assumption. In reality the piece may be short so that may not be a very good assumption but this is what we are assuming otherwise you have to also consider a non-uniformity in pressure across the section which itself adds to the complexity. We are not going into the complexities but I am trying to highlight the complexity because these are important, these may be important in some realistic conditions. So 2 important complexities may be non-uniformity of velocity over each section and non-uniformity of pressure over each section and when non-uniformity of velocity over each section is occurring then that means that and it is always there until unless it is a highly turbulent flow where the velocity profile due to high mixing is almost uniform. Otherwise if there is a velocity profile it gives an important understanding that yes viscous effects are important and when viscous effects are important you cannot apply Bernoulli's equation along a streamline between 1 and 2. Still you can use a1 v1 average equal to a2 v2 average that is a conservation of mass that does not depend on how viscous forces are occurring or not but you cannot really relate the pressure at 1 with pressure at 2 using the Bernoulli's equation. One has to solve the viscous flow equations to find out that. Now when you write the resultant force along x, so let us try to write the resultant force in a vector form. So we are using the Reynolds transport theorem, the right hand side the first term due to unsteadiness that is 0, next term is integral of rho v. When you are writing this force f, what is this force f? Let us now write what are the constituents of this? One force is the force exerted by the elbow on the water, so that is the f reaction. Then what other force is there? Force due to pressure is there plus force due to 2 weights. One is the weight of the elbow itself, another is the weight of the water which is instantaneously there within the elbow. So f due to water weight and plus f due to elbow weight, okay. So let us try to write these expressions of these forces, of course this is what you are interested to find out every action, so this is an unknown. Force due to pressure, how do you find out what is the force due to pressure? What is the force due to pressure on the control volume? P1A1 for section 1 along x, P2A2 for section 2 along y like that, it is not like that. I have mentioned it earlier y, when you have a pressure distribution on a surface you have to consider the force due to gauge pressure only because atmospheric pressure is there from all sides and that is nullifying the total force when it is integrated over a closed contour. So when you are writing the force due to pressure it should be the net force because of the pressure over and above the atmospheric pressure. So to calculate the force P1 has to be converted into the gauge pressure at 1, so P1 gauge has to be evaluated that is P1-P atmosphere. Similarly this has to be converted into gauge pressure. These are subtle but very important things. These are places where like in most of the cases students will make mistakes. Of course if you practice enough problems you will never make such a mistake but general tendency is like before the exam you just look into worked out examples. So when then these things are not highlighted, you just look into the gross formula but these are very important things that you have to keep in mind. Do not just take it as a formula, keep in mind that why it should be so that why you have to take the gauge pressure for evaluation of the force. So force due to the pressure what should be the corresponding expression P1 gauge into A1 that is the net force due to pressure at section 1. So in a vector notation we call it this i cap then plus P2 gauge into A2 j cap. You have to keep in mind that pressure is always into the surface. In whatever phase you are considering pressure is always acting towards that okay. Then force due to the water weight what is that? It is not impossible to calculate what is the volume of this given this contour. Let us say the volume of the water is given that is the volume of the elbow basically. So if the volume of the water is given then it is the what is the mass rho into volume of the water that into g is the weight, g is acting along negative y then this is minus of this j and the elbow weight what is the elbow weight minus w j and the right hand side the integral of when you are considering this integral first is what are the surfaces across which fluid is flowing 1 and 2. What is the control volume that we have taken? Since we have represented the elbow weight we have considered the elbow also as a part of the control volume. So the control volume let us draw the control volume. So till you express explicitly that what is the force, what is the resultant force that you are having? It may not be so straight forward to say that what is the resultant like what is the control volume that you have taken? So if you say if you take this as the control volume that means this excludes the elbow and then f elbow weight is not there but once f elbow weight is there you have basically taken including the elbow. So it is elbow plus the water that you have taken as the control volume. So the question is then when you have taken this as the control volume the outer one as the control volume say which includes both the elbow and the water the question is then what is this for f reaction this is provided by it is now not provided by elbow by the water. So what is this? So it consider there is a support which is there outside which exerts the force on this elbow plus water system. So there is some support which is there which is not drawn in the figure but it is highlighting that support. So now for that particular control volume we are having how many inlets and how many outlets we have one inlet and one outlet and let us write that. So the right hand side first let us write for the section 1 for the section 1 if you assume a uniform velocity profile then like this entire integral will be based on the velocity say v1 which is over uniform over the section 1. So rho then for v1 it is v1 i cap and then into v1 a1 so with what sign plus or minus right. You can clearly see that if there is a velocity variation along y then this expression is not valid then you have to integrate the velocity profile and that would give a net momentum flux this is this like a momentum flux. So the net momentum flux will be different and if one assumes a uniform velocity profile and it is really not so then that is an error and one has to adjust that error with some momentum correction factor maybe but here because of uniform profile assumption that such a correction is not necessary otherwise if the velocity profile is given to you you can integrate it to get this expression then there is no correction factor necessary. Then for the surface 2 plus rho what is the velocity at the section 2 let us say v2 it is directed along which direction minus y so minus v2j into this is plus. So the left hand side is equal to the right hand side and that will give you what are the components of the reaction force. This is the force exerted by the support on this system. So it should provide an equal and opposite force on the support and the support must be good enough to sustain that force okay. So if it is unsupported then because of some resultant force see it with the elbow may start moving okay. Let us look into one more example let us say there is a cart like this and a water jet is striking on the cart and it is changing its direction let us say this angle is theta. Let us say that the velocity of the water jet is v and the corresponding area over which the jet is moving here is a cross section area and let us assume that this is smooth. So if this is smooth that means there is no friction that the fluid is encountering as it is moving along the cart only its direction is getting changed. The first question that we would like to answer is that will it be possible to keep the cart stationary if such a water jet falls on it and changes its direction. For simplicity let us assume that this is a frictionless surface and may be assume that the cart is having a particular weight but that is not of great concern for us because we are interested to consider the motion along x whether there will be any motion along x or not. First of all let us say that what do you expect to be the velocity at which the water leaves the cart. Say it enters the cart at 1 and leaves the cart at 2 what is the velocity that we expect if the velocity here is say v1 which is equal to v what is the velocity at 2 there what are the first of all if you consider a streamline that connects some point at the inlet with some point at the outlet then could we apply the Bernoulli's equation along that streamline. If it is the first question is if it is inviscid flow then like that that is the first question that you would like to ask. So assume that it is an inviscid flow if it is an inviscid flow yes provided other conditions are satisfied what are those you have density as constant and steady flow obviously although unsteady version of Bernoulli's equation is also there what let us assume that it is a steady flow. So if this is a smooth one we and the water this is quite thin and this because of the smoothness there is no such wall roughness effect that is propagated into the fluid. So it is as if like a frictionless flow although you this is a great idealization in reality the effect of the solid boundary will always be propagated into the fluid and in all cases it is likely to give viscous resistance but here we are just idealizing it by too much and assuming that that effect is not there if that effect is not there if the velocity here is v1 the velocity v2 should be equal to v1 provided that the difference in height between 1 and 2 is neglected. So we are neglecting the z2-z1 that is neglected it is really a very small height and the corresponding potential energy chain is insignificant as compared to the kinetic energies of the jets. See in engineering when we say that we are neglecting something there is a very important thing that we should keep in mind we are not actually neglecting potential energy we are neglecting the change in potential energy and that change itself may not be negligible in an absolute sense. What we are banking on is that the jet is falling on with a very high kinetic energy with respect to those kinetic energies the potential energy effect is negligible not that it is always in an absolute sense negligible and regarding the pressure both are exposed to atmospheric pressure. So the pressure is like p1 and p2 are same so if the z2-z1 is neglected and if we may apply the Bernoulli's equation with all the assumptions satisfied then you have v1 equal to v2. In general if there is a friction here v2 will be somewhat less than v1 but because the friction less nature v2 is v1 and then you can apply the continuity equation then a2 also must be same as a1. Now let us say that we are interested to find out what is the resultant force along x because that is what is going to make it move maybe. So the resultant force on x resultant force along x what is that so you have 2 sections basically so you have one section like this where the fluid is entering and you have another section at 2 where the fluid is leaving these are only the 2 flow sections. Where do you choose your control volume see since there is no friction on the ground it will not be any difference any different if you include that like all the structural part of the cart and exclude the structural part of the cart for obtaining the force along x definitely for force along y it will be mattering but not force along x. So for force along x the right hand side first the unsteady term is 0 and then integral of rho v vr.nda so this is like fx i plus fyj because this is in a vector form now let us write try to write it in terms of its scalar components at the section 2 what is v v has a magnitude v what is the direction cos theta i plus sin theta j right. So that is the v in the vector form and then the remaining is a into v that is integral of v.nda what for the section 1 for the section 1 what is the velocity vi and-av so what is the force along x you can find out only the x component of that that is rho v rho av square into cos theta-1 this is the force exerted by the solid structure on the fluid right. So if you consider say a control volume like this we just n compasses the fluid z so this is the force exerted by the cart on the fluid the fluid exerts an equal and opposite force on the cart so this force is positive or negative this negative this is along-x so force exerted by water on the cart is along-x so you have a plus fx that is there on the cart and that is quite obvious even if you do not go to go through the mathematics if there is a z striking like this it should exert a force along the x direction. So there is a fx on the cart so the cart under this force may try to move and if this is a frictionless surface it will move always if there is a friction the static friction may just balance it and keep it in equilibrium but if it is not then it has to move. If it moves the question is that then is this consideration valid that is like here we are having to use the relative velocity but we do not know what is the velocity of the cart so how we should go about it that is the first question. Second is whether this velocity then we have to use the absolute velocity or the relative velocity. So these are the questions that we will like to address in a subsequent theoretical development where we consider also the moving reference frames still now we have considered only the stationary reference frames but in the jurisdiction of stationary reference frame if you have to consider it you have to consider somehow that this is stationary. Now how can you design a system such that this remains stationary there could be many ways. Let us say let me give one alternative when you say that whether it is a good alternative or not. Let us say we have a pulley like this and let us say there is a weight mg which is there this pulley is he supported like this is it acceptable will it work no or yes it depends on what is this weight and that you can design exactly because you know what is the exact magnitude of the force. So if you draw the free body diagram of the cart what are the forces that you will see you will see a tension in the string and you will see a force fx exerted by the water on cart okay. So when you have these two of course the other y component is there so y component you have the weight of the cart then you have a normal reaction like that but for us interesting is the x component and if you want to keep it in equilibrium you have to balance t with fx on the cart and if you consider it to be all those idealistic situations that it is a frictionless pulley and then what you get is that you get this tension same as the mg. So this in turn from the mass pulley system is equal to the mg. So you know that what has to balance what so you can put the correct mass here to keep it in equilibrium okay but you can clearly see that this is this is a forceful arrangement to keep it in equilibrium but in general because of these forces it will not be in equilibrium and when it is not in equilibrium with these forces it might have a velocity the that velocity itself might change with time. So it might have a situation when the reference frame which may be attached to the cart itself is moving and moving with arbitrary velocity or arbitrary acceleration. So we have to also be equipped with an analytical ability by which we can encounter such situations that is situations where you can encounter accelerating reference frames in general. When we say an accelerating reference frame we mean accelerating frame reference frame in all respects that means it could be linearly accelerating it might have an angular velocity because of which it has its original acceleration. So we have to next go for an analysis for accelerating reference frames. So we will use certain nomenclature we will consider axis say capital X capital Y capital Z for an inertial reference frame small x small y small z reference frame as an arbitrary it may be it may be inertial may be non-inertial but it is a moving reference frame. If it is moving with an acceleration then it is non-inertial if it is moving with a uniform velocity it is still inertial. What we are interested to find out is that if we have a vector say A here in this reference frame and let us say that this reference frame is moving with an arbitrary angular velocity omega then what is the derivative of this vector with respect to the inertial reference frame. Can you show it if I ask you to show it how do you show it say you have a vector A which is there in a reference frame that is rotating with an angular velocity omega. Let us say that the angular velocity is such that the rotation is taking place in the plane of the board. The rotation will take place in some plane what is that plane the plane is perpendicular to the axis of rotation. It might not be xy plane or yz plane like that but it is some plane. So in that plane this vector A is rotating. So when it is rotating it comes to what state it comes to a new location say this is at time t it is at time t plus delta t. What we are keeping in mind we are keeping in mind this is a fixed vector in a moving reference frame. So this is a fixed vector in xyz that we have to keep in mind. It is not any arbitrary vector that means if you are sitting on this you do not see any change in the at least in the length and if you are outside although it is same in length but because the reference frame is rotating this also rotates. Let us say that it traverses an angle delta theta over this time delta t. So what is the change in the vector the change in the vector is this delta A. So what is this delta A for small delta t the delta theta is small. So this is just like an arc of a circle. So delta A in terms of magnitude is what A delta theta in terms of a vector you have to give it a proper direction and sense. So if let us say that this is A is in a direction of E1 then it should be a direction which is normal to E1 say E2. If you want to find out what is da dt if you are not mentioning any subscript capital xyz that means we are talking about inertial then it is basically we are dividing this by delta t and taking the limit as delta t tends to 0. So A E2 into limit as delta t tends to 0 delta theta by delta t which is nothing but the magnitude of the angular velocity. And what is omega cross A omega is what is the omega vector it is omega scalar times a unit vector E3 which is perpendicular to the plane of the board. So you may take E1 like x E2 like y and E3 like z just like that this cross A is A E1. So it is omega A E2 E1 E2 E3 from orthogonal basis just like x ijk. So you can write that da dt capital xyz is equal to omega E2 E3. Omega cross A but this is only for a vector A which is fixed in the small xyz reference frame. If it is moving in a small xyz reference frame then that velocity also has to be added with this. So in general for an arbitrary vector A in capital x in small xyz you have da dt capital xyz is equal to da dt small xyz plus omega cross A. So this change is felt even if A is fixed but if A is moving relative to small xyz this is an additional change. So that is the total change and this you know from your earlier studies that this is known as Chaisel's theorem. So we will take up from this and try to write an equation of linear momentum conservation for a control volume which is having arbitrary motion. It may have angular motion, it may have linear motion, it may be a non-accelerating reference. It may be a accelerating reference frame in general. So we will take that up in the next class. Thank you.