 In the previous lecture we were discussing about like the how to write the basic principle of angular momentum conservation for a control volume and we will look into some examples related to that but just to physically appreciate that what kind of examples are there maybe let us look into 1 or 2 cases. We will see that what are the differences in like practical cases when you have a linear momentum and angular momentum conservation. So first we will look into an example which we discussed in some way that basically you have a flat object on which there is water which is falling and you can see that the reality is so different from what we considered. The water jet may not be thin, the water jet may be having different thicknesses at different sections. So it is a gross simplification of what we could see. Now think about a look into a situation when the water jet is falling on something which is mounted on with respect to one axis and when it is doing like that it creates a moment of the linear momentum or the angular momentum and that mounted wheel starts rotating and that is the basic principle of a turbine that we will see later on when we will be discussing about the hydraulic machines. But now before going into such complicated things of fluid machines see this example this also gives a very important example of angular momentum conservation this is a lawn sprinkler. So this is commonly used for watering the gardens and if you see that water is coming out of the limbs of the sprinkler and when water is coming out that water comes out with a velocity because water is supplied at a given velocity and when water is coming out with that velocity what is happening basically it is having a linear momentum that linear momentum has a moment with respect to the axis of the sprinkler. So there is an angular momentum and there is a rate of change of angular momentum that makes it rotate even if it is like if there is no frictional resistance that will make it rotate very fast because of some frictional resistance against rotation in the pipe what which is there it will not rotate as easily as we might think but still you have seen in this example that it might rotate quite easily. So if you see that see initially there is no rotation but when the water is coming out see we are trying to develop an understanding that it is actually an unsteady problem. So that is the first thing to appreciate it is not a steady problem because you see that the entire physical behavior what we are observing here is a function of time and our important objective will be to see that if it develops an angular velocity from say time equal to 0 to some time equal to t then how that angular velocity evolves with time. So this lawn sprinkler example is a very simple example but it can demonstrate a good use of the angular momentum conservation principle and this principle in a more elaborate way is used for analyzing those fluid machines which are basically of rotating nature like there are 2 very common devices which we will be discussing later on one is a centrifugal pump and another is a rotating turbine. So we will be discussing about these things in more details that will probably be our last chapter in this course of fluid mechanics that is the fundamentals of fluid machines. Right now we will not go into that application but we will concentrate more on a simple example. So we will use this for a case so let us take that example when we consider that example we will simplify whatever thing which we saw in the practicality we will consider that there is only one limb like this of the sprinkler and this entire thing is in a horizontal plane rotating in a horizontal plane just think that it is one arm there could be many such arms but we are just concentrating on one arm. So the water is moving at a uniform velocity relative to the rotating arm so the arm is rotating if you had looked into the examples very carefully no matter wherever the arms are there the water related to the arm is coming was coming out with a fixed velocity because that is the rate at which it is being provided. So the water it comes to the center and then gets distributed along the arm moves along the arm with a uniform or a constant velocity relative to the arm and eventually it goes out like what shown in the figure and let us say that it starts rotating with a angular velocity omega which itself may be a function of time and our objective is to find out what is this omega as a function of time. Let us say that we know that rho is the density of the water and q is the total rate of supply of water volume flow rate so this is volume flow rate like meter cube per second. So we will try to analyze this problem in 2 ways one is with an inertial reference frame another is with a moving reference frame and we will try to see whether we converge to the same conclusion or not. We start with the moving reference frame that is we as if we attach the reference frame with the sprinkler arm which is rotating with a given angular velocity with not a given with some angular velocity which is a function of time and therefore it is subjected also to some angular acceleration but it is not translating. So now we are having a control volume which is a rotating control volume but not a translating one. See this problem is a symmetric problem with respect to the centre of the arm. So we may take say a control volume say half of the entire thing when you have a problem which has symmetry in certain ways and the physics is also symmetrical geometry is also symmetrical then you can you can reduce your task or burden by taking a symmetrical part of the problem. So we have taken one symmetrical half of the problem in this problem we have neglected one thing we have assumed that the vertical part of the limb is very very small. So it is mainly a horizontal one and the vertical is very very small remember that this is actually in a horizontal plane. So thus because we have to draw it in the plane of the board it appears to be like it is upwards but it is like in a in a plane it is it is a bit bend like that and that bend part is quite small that we are assuming. It is just to give it a sufficient change in direction but not enough length that is what is the assumption and let us say that the radius is capital R which is the radius of each limb and that is symmetry. So we will write first the equation for the moving reference frame. So when we write the equation for the moving reference frame we will use this form whatever we discussed in the previous class. One important correction which one of your friends has mentioned and it is quite correct that this term is not going to be there in the a relative expression because this like when we wrote a relative that was basically a capital XYz-a small xyz and that is how this actually got nullified in that expression. So we need not duplicate it. So this term is not there in the a relative one just keep in mind that was a mistake and you please correct that. Now next is so let us try to apply this one with respect to this expression. So the resultant again we are considering that it is a frictionless pi what? So what it tells is that there is no frictional resistance moment when this is rotating. So this is rotating like freely so to say. So when this is rotating freely you have the resultant moment of all external forces because of what? One may be because of the frictional torque which is present so frictional torque we are assuming to be 0 but there could be torque because of the weight distribution and because it is symmetrically distributed the resultant torque due to weight it is there or not there. So let us consider a small element in that so you have a symmetrically located thing a small element in this side will always be counter balance by an equivalent small element in the other side in terms of the rotational thing. So there is actually no resultant moment of external forces because only 2 external forces could be that rotational resistance plus the moment of the distributed weight. So that is not there so that is a null vector then let us if you consider the moving reference frame so in the moving reference frame we have this correction term so a relative. So in the a relative first term is acceleration of the control volume that is the linear acceleration of the control volume here that is 0. Next are the 3 rotational terms that we need to consider so what are the rotational terms? 2 omega cross v small xyz so v small xyz is what? It is the velocity of the fluid relative to the control volume we are assuming that it is moving at a constant velocity relative to the control volume. So v small xyz is like a constant but if you want to write this term properly let us write minus integral of r cross a relative into dm for the control volume. So what we will do we will let us say we will split this a relative into different parts instead of just writing it in a cumbersome single expression. So we will find out the contribution of the 3 different terms which are there. So 2 so before doing that what is this dm that you have to keep in mind dm is some small element at a distance at a radial distance of small r you can take a strip of width dr and let us say that capital A is the area of cross section of the sprinkler this area of cross section. So dm is what? If dm is the elemental mass of the shaded volume then what is that dm is rho A dr. So basically when you are writing the expression this will be integral of whatever you write as a relative with this entire thing that multiplied by rho r dr integrated from small r equal to 0 to small r equal to capital R that will be the that correction term in the left hand side. So before applying that correction term let us write these things in a vector form. So what is omega in a vector form? Let us say this is xy plane yes omega k what is v small x y z no this is by positive sign convention if it comes out to be minus something then that will be like that. So then what is v small x y z? So it is moving no no no v small x y z is what? Velocity of the water relative to the sprinkler. So if say it is moving at a velocity v r relative to the sprinkler in a magnitude sense vector sense yes with this 2 options i or j yes only 2 options are there it cannot be k. So only i and j yes i or j you are not very sure it should be i see when you are considering this will be a volume integral this part is negligible in comparison to this part. So it should be ideally for this part i and for this part j but because I have already mentioned that this is like of negligible length in the volume integral only this horizontal part is mattering and there it is moving radially. So it is like v r i okay then omega dot that is the only other thing that is remaining let us say that is omega dot k let us say omega dot also has a positive sense like that. So a relative first 2 omega cross v small x y z. So these 2 cross products so 2 omega v r j then omega cross omega cross r what is omega cross r what is r in a vector form r i. So omega cross r is omega r j and omega cross omega cross r is minus omega square r i. So it is just like centripetal acceleration so minus omega square r i then next is omega dot r j. Now if you make r cross a relative so you will clearly see that this centripetal term will not be there because it is directed along r so its moment will be 0 but other terms will be there. So 2 omega v r k plus omega dot r k with r multiplied right so let us write r r square. So when you write this particular term now we are in a position to write it in a proper integral form so that will be minus of 2 omega v r r plus omega dot r square into rho a dr that is dv 0 to r. So that will become minus rho a omega v r r square plus omega dot r cube by 3 okay. Then the right hand side let us write the right hand side now you tell the left hand side we have completed right hand side what will be the first term you have to see what are the things which are functions of time v small x y z is not a function of time because we are assuming that relative to the sprinkler arm the water is moving at a constant velocity. So no matter how the arm is moving but relative to that and we are writing it only relative to that it is not changing with time neither the volume of the control volume is changing with time so that this term will be 0. So in the right hand side the first term will be 0 and what will be the second term? Second term is the flow boundary it represents what happens of the flow boundary. So for the second term when you substitute that so let us assume that these velocities are again uniformly distributed over the flow boundary. So v dot eta da this term if you just keep it inside now r cross v small x y z can you take it to be independent of the area and bring it out of the integral yes or no. What is this r now? This r where this is some r located at the flow boundary that means at this location somewhere because this is very thin it does not really matter exactly that where is the variation but it is somewhere here. So that is basically capital R i plus something j plus some little something j that is the radius outside. So let us say that it is capital R let us let us give it a name say a keeping in mind v small. So r i plus aj that is r cross v small x y z is a constant so that also we can take it out of the integral provided with also consider it to be constant over the area uniform over the area. So if you assume it to be uniform over the area we just take it out and that is v relative and then rho is there of course rho let us write the remaining thing which is there in the integral is the volume flow rate over that section. So what is the volume flow rate over that section? Yes it is q by 2 see we have only taken the half of this that half of the q flows through this half of the q flows through this there is no exception because it is perfectly symmetrical. So this into q by 2 that is the right hand side and the question is what will you put this in a vector form. So v r what i or j this one v r j right. So you will see that this little part a will not matter because of the cross product. So it will become let us write it completely it is equal to rho v r r q by 2 k since the left hand side is equal to the right hand side. So the left hand side equal to the right hand side which implies what minus rho a omega v r r square plus omega dot r cube by 3 is equal to rho v r r q by 2 this we means that k cap the vector sign that was there. So this omega is equal to what d omega d t v r you know because what is v r v r is the flow rate divided by the area. So q by 2 divided by a. So given the flow rate you should be in a position to substitute v r which is a constant. So only the variables are d omega d t and omega. So it is a differential equation of the form something into d omega d t plus something into omega is some constant. So it is you may separate the variables easily to solve it because these are constant coefficients here. Now let us try to look into the problem from an inertial reference view point and see that whether we can solve the same problem in that way. So now we will get rid of the non-inertial corrections and we will just consider that we are looking for control volume which is maybe schematically you draw it like this but that is not moving with the that is not moving with the arm. So even if the arm goes to a different place the control volume is fixed. So at that instant we are trying to we are considering one particular instant when the control volume and the arm they are coinciding and at that instant we are trying to figure out what is happening. Now if you consider the inertial problem that is the inertial reference frame way of looking into it the left hand side is 0 right. The left hand side is 0 or null vector the right hand side now you see that with respect to an inertial reference frame it will be unsteady right. If it was moving at a constant velocity relative to the reference frame then sitting on the reference frame you sit it you look it as a constant but standing from outside which is inertial you see that that velocity is changing with time because the frame velocity itself is changing with time. So this time derivative term with respect to an inertial frame will not go to 0. So the right hand side therefore you have to write this time derivative term properly. So when you write it then let us write R again R you write as R i cap because the volume which is involved for the control volume integral this is the control volume integral. So their major part is the horizontal part then what is this v what is this v this is velocity of the solid plus velocity of the fluid relative to the solid. So what is the velocity of the solid at the coincident point at a radial location R it is omega cross R. So this is the velocity of the solid plus the velocity of the fluid or v relative or vR. So this is omega rj plus vR i right. So we substitute that but before substituting maybe it is useful to find out how it is R cross v. So that is omega R square k. So you write this as omega R square then in place of dv a dr this thing k right then v plus again you assume uniform velocity distribution. So when you write the next term then what should you substitute for this v this one yes remember this v has to be inertial reference frame v. So what is that? So omega k cross with capital R i plus aj plus vR j right. So what will be that omega rj plus omega a minus i plus vR j. So if you make another R cross with that then what is that R that R is again this capital R i plus aj. So what is that R cross with that one omega rj minus omega a i plus vR j right. So here if you neglect small a then like that is what is given in the statement of the problem. So it is omega R plus vR into capital R k. So the next term we will write yes this is omega cross R is what? So if you consider a point here the point first how do you consider a coincident point in the solid plus velocity of the fluid relative to that coincident point okay that is the absolute velocity of the fluid vR is v relative yes. So the same vR that we used in the even the previous method. So the next term is omega R square then integral of rho vR dot eta over the control surface with a k cap. So now what we do? We simplify the different terms just like what we did in the previous case or the in the previous method. Rho is a constant that we are considering for this problem. So we are taking that rho here out of the left hand side and right hand side. So left hand side equal to right hand side when we write. So 0 equal to now you see first of all we integrate with respect to position. So when you integrate with respect to position it is from 0 to capital R. So you have basically first let us complete the integral. So integral will be rho A omega capital R cube by 3. Now a partial derivative of that with respect to time and then the remaining 1 plus omega R square into rho in the remaining term is q by 2. Because remember no matter whether it is inertial or non-inertial this velocity is always v relative for flow rate part calculation. Now rho being constant, A being constant, R being constant but omega is not a constant. So you have rho A omega maybe by divided by 3 also sorry rho A R cube by 3 then you have d omega dt plus now there has been a point where I have omitted one term where is that term this is your work to identify. You may get a clue by comparing with the previous method and this method they should at the end give the same result no matter what is the control volume. So some term is omitted somewhere which term v r into where is that term r cross v so what is there plus rho you tell I will just write r into v r into q by d. So one term because of multiplication of this with r because of this with r that was omitted and not omitted just somehow not written here. So that is equal to 0 and this should give the same differential equation as what we obtain with a non-inertial reference frame it should not be fundamentally different okay. So what we have understood is that you may use different reference frames but when you go to your final analysis the final analysis should converge to the same conclusion. Now what you may do a little bit of modification of this that when you take the part small a not very small that is it is it is of substantial length then neglecting it and not neglecting it will matter more for approach 1 or approach 2 that is will matter more for the inertial frame based analysis or non-inertial frame based analysis that is a bit of more thing that you can add with this one but again the principle will be very similar then you do not neglect the terms with small a and just do the algebra and then you will easily find it out. Now what we have seen here so in this particular chapter what we have discussed in this particular chapter we have found out how or we have learnt how to write the integral forms of conservation equations for mass conservation linear momentum conservation and angular momentum conservation. In one case we have at least shown that you may convert the integral form to a differential form and that is the mass conservation example which gave back the continuity equation. Let us try to see that whether we may do it for other cases or not. So now our objective will be let us say we take an example of linear momentum conservation can we write a differential equation for linear momentum conservation starting from the integral form so that we will do. So the objective of the subsequent analysis is to figure out that is there a root from integral form of momentum conservation to differential momentum conservation. The differential equations are important because in many cases you want to get a point by point variation of the velocity field or the pressure field like that. So then you have to solve the differential form not the integral form. So let us say let us write let us make certain assumptions let us make first assumption is that we have a stationary reference frame number 2 that non deformable control volume based on these let us write the linear momentum conservation. So if you write the linear momentum conservation integral form resultant force on the control volume is equal to so we have assumed straight away that it is a stationary reference frame. So that vr is equal to v that is the first assumption that we had made and regarding the second assumption it will be possible for us to take this inside the integral. Now we will concentrate more on the left hand side and see this is a vector equation so it has its different components. So it may be convenient to express it in terms of a component say in the direction i. We will again start using the index notation somewhat so i equal to 1 will mean x i equal to 2 will mean y and i equal to 3 will mean z. So let us write this for the direction i. So we are interested to find out what is the resultant force acting on the control volume in the direction i. Let us make a sketch and try to see. If you recall our very basic discussions that we classified the force in continuum mechanics in 2 categories. One is a body force another is the surface force. So what is the body force? The body force is the force that is distributed over the extent of the volume of the body. So let us call this as f surface for surface force plus f body. So when you have the surface force that is distributed over the area of the surface. So when you have area of the body so when the body is like a control volume it is a force distributed on the surface of the control volume. So now if you want to find out that what is the body force or what is the surface force first. So if you consider a small chunk of an element like this with the direction normal of eta. So when you have the direction normal as eta and how do you express the surface force at a particular location? You represent it in terms of the traction vector that is the force per unit area but it is dependent on the choice of orientation of the area. So let us say that ti let us say that t with superscript eta is the traction vector at this point based on the chosen area say da. So this traction vector we may write in terms of what? We may write in terms of the stress tensor components that we have discussed and provided we know the direction normal of the area that is under consideration and which theorem gives this? Cauchy's theorem that is you have the traction vector the ith component of the traction vector is given by tau ij into this. Remember that invisible summation sign for j equal to 1, 2, 3 this is actually not usually put in the sign in the representation convention. Fundamentally it was tau ji into nj but because tau ij equal to tau ji how we have got tau ij equal to tau ji from angular momentum conservation. So usually you see that for fluids we do not separately write many times angular momentum conservation in a differential form because it is already inbuilt with some of the considerations in the linear momentum conservation whenever we write tau ij equal to tau ji. So that is already inbuilt here and that is how it becomes like this. So if you just expand it so what is this? This is like if you have the unit vector it has its components x, y, z as say n1, n2, n3 are the direction cosines. Now tau therefore this one we can write see i is a index which is there in the left hand side that should remain also in the right hand side. So tau i1, n1 plus tau i2, n2 plus tau i3, n3. Can you express this as a dot product of some vector with the n vector? What is our objective? The objective is we will replace the area integral in terms of the volume integral as we did for the continuity equation and because we know the divergence theorem that relates these two. For that we have to know we have to get something of a form of like some vector function dot with n dA. So that is why we are looking for a vector which dot with n gives the same thing. So the objective should be clear whenever we are doing an analysis it is not that out of nothing we are doing this manipulation that is the objective of the manipulation that we should keep in mind. So whenever you convert from an integral to a differential form the main way in which you always do is by converting some of the area integrals into the volume integrals and that by using the divergence theorem. So you should make the all the terms compatible for use of the divergence theorem all the area integral terms at least. So when you are trying to represent the surface force what is the surface force f surface is f surface i. So let us just write the i component x component. So when we call the components we will not put the vector sign because the components are just the scalar components. So f surface i will be what? Traction vector is the force per unit area. So it is the traction vector ith component times dA integral over the control surface the surface of the control volume. So that will represent what? See the control volume surface the control volume is like a free body in mechanics. So if you have the internal action reaction forces within the control volume they get cancelled out. Only the surface forces at the outer boundary remains. So it is just consider it equivalent to a free body. So this now our objective is to write it as some vector dot with n. So clearly if you see that if you like write a vector say tau i just give a name of a vector like that tau i1 i cap plus tau i2 j cap plus tau i3 k cap. Then this term is what? This term is nothing but this tau i dot with n that you can clearly see because when you make it a dot with n you see tau i1 is coming with n1 tau i2 is coming with n2 and tau i3 is coming with n3. So we have found an artificial vector just to make use of the divergence theorem. So we may write this as integral of this new vector say tau i dot n dA. What about the body force? Let us say that bi is the body force per unit mass. So bi is body force per unit mass. So how do you express the f body? So basically you take a small volume inside of dv. The mass of that is rho into dv and the corresponding body force along i is bi into rho into dv. If you integrate it over the control volume it will give the total body force along i. So we have been successful in writing the left hand side in a certain way for force components along i. So the left hand side will become integral of over the control surface tau i dot with eta dA plus control volume rho bi. What about the right hand side? We will take the time derivative term inside the integral because of non-deformable control volume. Now see we are writing the scalar components. So in place of v we will write vi or ui. Usually the notations for velocities we call ui. So u1 is like u what we have learnt, u2 is like v and u3 is like w. The usual notation for the velocity components. So this in general when you consider the i-th direction it is ui. i equal to 1 means x, i equal to 2 means y, i equal to 3 means z. So this of rho ui. This v is replaced by its component along i and then the remaining i is replaced by integral of rho. This v is replaced by ui. The next steps are quite simple. The next steps are converting the area integrals into the volume integral. So that we may convert all the area integrals into the corresponding volume integrals. So we should keep in mind that if you have a vector function say a. So a dot n dA over the control surface is the divergence of a over the control volume. So this surface has to entirely be bounding the control volume. So if you consider that and if you consider the left hand side is equal to the right hand side then what you get? So the first term this. So here this tau i is like the vector function that we are talking about. So then what will be the left hand side? First term if you convert into a control volume term divergence of tau i dv then integral of rho bi dv equal to the right hand side integral of rho bi dv equal to the right hand side. First term is already control volume term. Next term divergence of rho uiv. So we now collect all the terms and if we collect all the terms you will get therefore some integral of some function dv equal to 0 because all terms are of integral dv nature and we have seen that when we had discussed about the continuity equation same thing applies that this volume that we have selected is arbitrary. So for an arbitrary elemental volume if this has to be satisfied the integrand has to be 0. So that means you have this integrand let us say this as i. So this i has to be 0. So if i is equal to 0 then you are left with this particular form. Now it is possible to express this is partially written in index form partially written in vector form but it is possible to write it in a proper fully index form. So if you just write divergence of rho uiv this in a proper index notation is written in this way. See divergence is what? Divergence is what you will have a term with partial derivative with respect to x then partial derivative with respect to y and partial derivative with respect to z. So just like x1, x2, x3 some of these 3. So in this so you can see that when you have a partial derivative so when you have the divergence here the partial derivative with respect to x1 should be involving x1 component of this v. When it is with respect to x2 it should involve x2 component of v. So in general when it is with respect to xj it should involve the j component of the velocity. So actually it is a invisible summation from j equal to 1 to 3. So you can clearly understand the notation it is very important and similarly how do you write this one? Now you tell remember that what is this one? What is this tau i? Tau i is it is so you have the partial derivative with respect to that component. So for j equal to 1 it is one component for j equal to 2 that is 2, j equal to 3 that is 3. So in general tau ij again there is a invisible summation j equal to 1 to 3. So you can see from this that is how their tau was defined from that. So in that index notation you can write it like this. So this is an equation of motion, equation of dynamic equilibrium in a differential form and this is known as Navier's equation of equilibrium or simply Navier's equation. So we have been successful in writing an equation of motion along the direction i. If you put i equal to 1 it is x, if you put i equal to 2 it is y, if you put i equal to 3 it is z. Now you see what is the complication of this equation? The complication of the equation is as follows. You do not know this tau ij that is you do not know the stress tensor components. So you must know to make it is a close system of equations you must know the stress tensor components as a function of the velocities or their gradients or pressure on all those quantities. So it should be expressed in terms of this primary variables which are velocity and pressure or their gradients and that varies from one fluid to the other till now we have not assumed that it is a Newtonian fluid or whatever. So we will see in the next class that now we will make some special assumptions of the type of fluid because based on the constitutive behavior of the fluid this tau ij will be dependent on the velocities, pressures or may be the gradients of velocities and we will assume a special case of a fluid known as Newtonian and Stokesian fluid and in that special case this equation may be simplified to a form known as Navier-Stokes equation. So that we will do from the next class. Thank you.