 Amruta Kolam, if you have a question, please go ahead. Yes sir, this is an external flow problem where a blunt body is towed through a channel. It will generate waves. How can we apply our conservation laws to get the height of the waves that is created by this blunt body? Yes, so the question is related to a peculiar fluid mechanic situation. What is being pointed is that there is a situation when there is a blunt body that is getting towed in a channel and as the blunt body moves it is going to create waves and how can the conservation equations be utilized in this case to find out the height of the waves, etc. So the only question I have is whether this is a completely submerged body or it is actually partially submerged body that is getting dragged on the free surface of the liquid. It is a floating body. Yeah, it is a floating body so it is actually a body that is on the free surface of a liquid and that is getting towed. It is actually a very complicated problem obviously and the way to analyze would be, you cannot do this analytically obviously but if you want to employ a CFD technique the standard Navier-Stokes equations can be utilized but some sort of a interface free surface tracking algorithm has also to be incorporated along with the Navier-Stokes solver and some of the standard free surface simulation methods are what are called as volume of fluid method or a level set type method. So utilizing a volume of fluid method along with the Navier-Stokes equations it is probably possible to do this simulation in a fairly sophisticated CFD program. Obviously some of the commercial solvers which have the capability of incorporating this volume of fluid technique can be employed to simulate this flow situation. This is actually pretty complicated problem that you have pointed and it is way out of scope of the present workshop but in general from the knowledge that I have you will need some sort of a free surface tracking algorithm like the volume of fluid method which can be utilized for this purpose. Sir if we assume it is potency flow will the problem be simplified and can it be solved? Even if so if the comment is if we assume this to be a potential flow if the solution will be simplified. Even if you assume it to be potential flow it will be still a multi-dimensional potential flow and perhaps you can get the waves generated in a potential flow situation but since this is a free surface situation where the body is floating on the free surface you will still need some sort of a volume of fluid algorithm that needs to be incorporated along with the potential flow equations. So even if you go ahead with a subset of the Navier-Stokes which is for potential flow you will still need a volume of fluid algorithm for the handling of the free surface. College of Engineering Pune if you have a question please go ahead. Sir my question is firstly how do you in a physical sense differentiate between a path line and a stream line because both of them seem to be an instantaneous picture of the flow field. This is the first question and the second question is in your slide kinematics 17 the linear strain is ratio of delta LAB upon LAB. So is delta LAB the difference in the length of segment AB and A dash B dash? Okay there are two questions one relates to path lines and stream lines and physically what they signify. First of all let me correct you in one sense that the path line is essentially the trajectory of one single given fluid particle as it moves from one location to the next. So as I mentioned earlier in the lecture the path line actually gives you over a time period how the given fluid particle has moved in the flow so that is meant by a path line. So in fact path line does not really give a instantaneous picture of the flow field it gives you a flow field information over a certain period of time. Stream line on the other hand by definition is that if you draw the line with a tangent on to the line at each point on to the line the tangent will point out the or point in the local velocity direction that is how we define a stream line. So at any point on the stream line if you draw a tangent that tangent will point in the direction of velocity at that point. So essentially this is the difference what happens in the case of a stream line is if the flow is unsteady the velocity at each point within the flow field will keep on changing typically it can change direction also or it can change the magnitude also. So because of that the stream lines will keep on changing from instant to instant as the velocity field keeps on changing in an unsteady flow. So therefore stream lines necessarily provide an instantaneous picture of the flow field if the flow field is steady on the other hand both path lines and stream lines turn out to be identical. However for the general case of unsteady flow field the path lines and stream lines are different. So I hope that answers your first question. The second question was related to linear strain rate. So let me go to that. Yes so the slide number 17 is projected the question was related to linear strain and the linear strain rate and the expressions that have been used to calculate the linear strain rates. So let me go back to this picture first slide number 16. If you look at the horizontal segment AB what we pointed out is that the x direction velocity at A is different and x direction velocity at B is different. So what is going to happen is that as the fluid particle moves from one location to the other point A will be pushed by a different amount over the time delta t whereas point B will be pulled by a different amount. So what is going to happen is by the time the particle moves this line AB the segment AB is going to be changed in its length. So that change in the length experienced by AB is what we will call delta AB. So ideally what I would say is that if you want to find delta AB you can look at A prime B double prime in the figure to your right and that would be your changed length AB. From the change length AB you subtract the original length AB which is delta x and that should give you the change in the length AB which will eventually give you the linear strain experienced by that segment. Hopefully this answers your question. Yes so the way to interpret this situation is that see everything is happening simultaneously. What is happening here is that the fluid particle is getting stretched it is getting rotated everything is happening simultaneously. So for the purpose of ease of understanding what we say is that we decompose this motion one by one and we say that let us say that the particle is first stretched and then rotated. So that is why I said that A prime B double prime will be considered as the stretched length and then you say that it has been rotated through the angle delta alpha. It is just for the convenience of explanation and figuring out the relations that this is done. In reality everything is happening simultaneously but what we do is we do this so called decomposition of motion so that each of these motion can be independently analyzed. This differential analysis that slide number 4 in this that continuity equation is there in that continuity equation suppose this flow is unsteady or change in cross section will be there and a non uniform flow will be there what changes will be happen in this equation particularly. So the question is related to the differential mass conservation or the continuity equation that we have derived here and the question asks whether anything will change if the flow is unsteady and also if there is change in the cross section of the area that the flow is passing through. Hopefully I have understood the question correctly. The unsteady part of the flow field has already been built in into the equation and the unsteady part is basically this time derivative of the density. If the flow field is unsteady this is a non-zero part. If the flow field is steady there will not be any time derivative of density at all in the equation. So a special case of the continuity equation for steady flow would be that only divergence of rho v equal to 0 will be left. The other part of the question is whether this can be employed in any manner for a channel type flow where the area is changing. The answer to that is that remember that this equation as we have derived here in fact what I wrote on the next slide here is on a per unit volume basis. So when we actually solve this in a CFD type simulation we will solve this for the entire cross section point wise. So it does not matter whether the cross section of the flow is changing or not. What we will do is we will divide the domain into very small sub domains and over each of these sub domains the continuity equation will be solved and eventually the answer will be clubbed together to find out what is happening in the entire flow field. So keep in mind that the equation has been derived on a per unit volume basis meaning that it is actually applicable from point to point within the domain of interest. So always any cross section can be considered to be composed of several points and that several points at each of those several points this equation will be solved in principle when we go to a CFD type analysis. So it does not matter whether your situation has unsteady flow or whether the cross sections are changing already everything is built in in some sense into the equation. Thank you sir. Divaji University, Kulapur you have a question please go ahead. Sir my question is on slide number 27 for a particular flow how velocity field is decided in terms of x and y coordinates. So the question is on slide number 27 of kinematics and the question is in a particular flow how is the velocity field decided. So let me first answer this question in the following form that the velocity field that has been given in this particular example number 4 on slide number 27 is really an artificial or a cooked up velocity field. Such a simple velocity field will typically not be available in any real life situation. So it is a simple generated artificially generated velocity field for the purpose of the problem so that using our derived expressions for the angular velocity and the vorticity etcetera you can perform the necessary differentiations and find out the rates of strain and so on. So that is as far as this particular velocity field that you see on the screen is concerned. In general in any real life application the velocity field if you want to determine let us say using a CFD type simulation will actually depend on what sort of boundary conditions are imposed in that particular flow situation and depending on those boundary conditions the velocity field will be determined. Typically no velocity field in a real life situation can be represented using a simple expression such as this. I hope that answers your question. Okay, thank you sir, over. Saint Joseph College Kerala, please go ahead. Magnitude of the total shear deformation, while calculating the magnitude of total shear deformation why we are not considering the sign of the shear deformation in a slide 26, kinematics 26. Yes, so the question is when we are calculating the total shear deformation in the XY plane why are we not taking into account the sign. See actually the very fact that we are calculating the magnitude says that we do not have to bother about the sign. Shear deformation is something that is not necessarily a vector type quantity it is a scalar quantity and therefore all we need to know is how that fluid particle is getting deformed due to shear in the XY plane and that is the reason typically we are bothered about only its magnitude. See what is happening is in the case of a two-dimensional flow the let me go back to the figure. In the case of a two-dimensional flow you see that an original rectangular shape of the particle gets deformed on both sides because of changes in the x direction velocities in the y direction and y direction velocities in the x direction. So therefore this essentially results into a scalar quantity of the total shear deformation we do not really have to bother about the sign when we calculate the shear deformation. On the other hand the rotational velocities are inherently vector quantities and that is the reason we need to assign a sign for them. VNIT Nakur please go ahead. My question is regarding slide number 15 of kinematics in which you said under the action of forces a fluid particle simultaneously undergoes translation, rotation and deformation and you have explained regarding translation rotation. So can you please briefly explain deformation which is other than this translation of rotation. The question is about discussion on the deformation part which has been listed as one of the effects of action of forces. In fact the entire deformation both the volumetric deformation and the shear deformation has been discussed in slides number 17 and 18 if you look at that is the volumetric deformation part. So what we mean by volumetric deformation is that under the action of forces as the fluid particle moves it will experience a change in the volume. So essentially as the particle moves it is going to either contract or expand if it is a compressible type flow and that expression has been evaluated as the divergence of velocity. So the volumetric strain rate and volumetric deformation essentially mean the same thing. Perhaps that is what was unclear that when we wrote here volumetric deformation essentially volumetric strain rate which has been calculated on slide number 18 and the linear strain rate on the slide number 17 as first part of it are the same thing as far as the volumetric deformation is concerned. Similarly if you look at the shear deformation it is explained on a later slide the last slide rather 26 slide number 26. So again shear deformation is also explained or expressed in terms of the rates of shear deformation or the rates of shear strain. So they are the same thing really. So all the information that you need about volumetric strain rates which is same as volumetric deformation and the shear strain rate which is the shear deformation the rate of it is already available in slides number 26 and earlier slide number 18. Thank you sir. Nirmah University go ahead if you have a question. My question is related to previous yesterday's session. My question is related to the integral analysis of momentum equation integral analysis number 6. In the first slide we have d by dt into bracket x component of linear momentum for the control volume is equal to d by dt of integral of the control volume u dm then cv. Sir my question is we have considered u as means throughout the control volume the x component of velocity keeps on changing. So we have taken dm but we have not considered the variation in the x component of velocity that is for you. So I have little bit doubt about that. Yes. So the question is on integral analysis which we talked about yesterday specifically slide number 6 and the accumulation term which has been worked out right at the top. So actually what you are pointing out is correct in the sense that if there is a variation of the u velocity within the control volume as a function of time what will happen is that u multiplied by dm is essentially the linear momentum content of an elemental mass which is residing inside the control volume. Now there will be several such elemental masses inside the control volume each of those elemental masses will have associated with them a u velocity component. So that u multiplied by the elemental mass will then give me the elemental x direction momentum associated with that elemental mass within the control volume which then you integrate over the entire control volume to get the total x momentum contained within the control volume. So if it turns out that u is a function of time inside the control volume for different elemental masses then we will actually have a non-zero accumulation term but otherwise as long as u is not really going to be a function of time inside the control volume we are never going to have an accumulation term. I hope that answers your question. So is there any relation or we can say the effect of viscosity or we can say the relation between viscosity and the types of the flow that is rotational flow and irrotational flow. Yes so the question is if there is any relation between viscosity and the types of flow namely rotational and irrotational. Actually that is a very good question that is one of the nicer questions I must say. What I will say is that typically whenever there is viscosity present you will see that the flow is going to be rotational and the primary reason is that viscosity is going to introduce velocity gradients within the flow especially if there is some sort of a solid surface that is adjoining the flow domain. So what will happen is that because of the viscosity the flow will be decelerated to a zero value at the solid wall and then from that zero value it will increase in viscosity when you go into the middle of the domain let us say. So viscosity is going to introduce velocity gradients and non-zero velocity gradients when you look at our expressions for the angular velocities and the vorticity will make sure that the angular velocity and the vorticity is also non-zero therefore it is quite safe to assume always that viscosity will necessarily mean a rotational flow situation. So for irrotational flow we consider viscosity as a zero? Yes, so continuing with the same answer if you are dealing with an inviscid flow typically it is safe to assume that the flow is going to be irrotational. So let me leave that answer there as typically because there are some special situations in which even in case of inviscid flows you may get rotational motion but those are fairly special types of flows and in general you do not really see that kind of a situation happening. Mufakamja please go ahead if you have a question. My question is the difference between a steam line and a stick line actually we are calculating the velocity with the velocity field in each and every point in the domain then afterwards we are drawing an imaginary line that line is we are drawing in such a way that the velocity is tangent to that line. So a streak line is that line it is a locus of all different fluid particles. In steady state we are saying that both these lines are identical. So in unsteady state whether this velocity is tangent to the streak line. So the question is about the concept of stream lines and streak lines and especially when the flow field is unsteady. So yes in case of unsteady flow field the streak lines and stream lines are in general different that you have already pointed out. When it comes to stream lines in an unsteady flow situation the stream lines will keep on changing from instant to instant. However at each instant the stream line will be still such that if you look at the point on the stream line and draw a tangent at that point that will be in the direction of the local velocity vector at that instant. So from instant to instant the stream line field will change but at any instant it will still be tangential to the velocity field. Whether it is a proper streak line the velocity how the velocity will be the straight line. See streak line is a concept which is such that you have to identify one location which is the reference location within the domain and then from this location you actually have to monitor different fluid particles emerging and after certain number of time steps let us say you try to join all these different particles and then that locus is what will be called as a stream line. So if the flow is truly unsteady at one instant of time the velocity at that location will be something else. At a later instant of time the velocity at the same location will be something else. So every instant the particle which is going to pass through that same location will actually keep experiencing different velocities and as such it will go in different directions. So eventually what will happen is that the different sets of particles which have gone through the same point after a period of time will be all over the place but the streak line will be then joining all these different particles by a continuous line so to say. So that is the way it will be in case of a streak line. Thank you sir. J N T U Hyderabad please go ahead if you have a question. In kinematics one one and two in kinematics one is given r bar is equal to function of r suffix not to the bar. So my question is comparing these kinematics one and kinematics two how we can identify the r not bar and r that is my first question and the second question is in kinematics eighteen the volumetric strain rate change in elemental volume by virginal volume. So all are expressed in terms of length. So here it should be on the basis of volume so please explain these two sir. So let me try to first explain the second question which was related to the volumetric strain rate how the volumetric strain rate has been calculated in particular. So if you see the expression on the top of slide number eighteen what we start with is the change in the volume of the fluid element divided by the original volume and then divide that by the time period over which these changes are occurring. So if you see what is happening is the change in the volume as is required on the numerator is getting calculated as the new volume minus the old or the original volume. So the new volume is how it is getting calculated it is the length AB which is changed by an amount delta length AB. So that is the new length AB LAB plus delta LAB new length AB multiplied by new length AC is going to give me the new volume of the particle. Remember that we are talking about a two dimensional particle so that the depth is assumed to be unity or one. So therefore we are talking about only dimensions in the XY plane as you are seeing on the screen right now. Going back to the expression so I have on the numerator new length AB multiplying new length AC that will give me the new volume of the particle. Subtract the old volume of the particle which is given by the original length AB and the original length AC and then divide that by the original volume again which is LAB times LAB. So that is the way the change in the volume divided by the original volume has been calculated as the volumetric strain and then you divide by delta t to obtain the volumetric strain rate. Hopefully that answers the question what I had asked in the lecture is if you can actually plug in the values for LAB delta LAB etc. and work out this algebra thereby simplifying it finally to this divergence of velocity equal to the volumetric strain rate that will be really nice so that you can convince yourself that it is indeed the correct expression for the volumetric strain rate. For the other question it is right at the beginning of kinematics and it relates to Lagrangian approach and Eulerian approach. So what I will do is I will try to draw something on the whiteboard and hopefully with the help of that we will be able to explain this. So in the case of Eulerian approach what I will say is that let us assume that this is our domain of interest. Within this domain of interest we identify several locations and so on. At each of these locations for all times we monitor the properties of interest such as velocity, pressure etc. So all these are simultaneously monitored all these locations for all times. So this is Eulerian. On the other hand when we go to a Lagrangian description what we do is at time equal to 0 let us say which is our reference time we identify a whole bunch of fluid particles. So let me draw a few of them. So I am showing only a few particles obviously if you want to describe a fluid continuum you have to think about a very very large number of fluid particles. So each of these fluid particles will have its own reference coordinate. So let me call this the reference coordinate for particle number 1. This will be a reference coordinate for particle number 2, this will be the reference coordinate for particle number 3 and so on. And what then we do is we that this set of particles as time progresses will keep on moving and visiting different locations at different time. So that location is going to be denoted by r and will be a function of time but it will always be that we are referring to one known particle which was called as particle number 1 which we are following separately. Similarly particle number 2 will be followed separately, particle number 3 will be followed separately, particle number 4 will be followed separately and so on. So in order for the fluid continuum to be described completely you have to actually follow a really really large number of particles and the only way you can do this in any practical situation is through some computer simulations which are very special types of simulations and there what you do is you keep on monitoring each of these fluid particles how they are going to different locations at what time. So that is the way what I mean by a Lagrangian point of view. One more question, we are neglecting second order terms in the volumetric strain rate and angular velocity derivation is there any technical reason? The question is I will actually generalize the question a little bit you will see that in most of these derivations we are neglecting second order terms we are using only a first order Taylor expansion and typically we are neglecting second order terms and the question is whether there is any fundamental technical reason for this. This question is actually very important question and I would like to point out here that it is only our choice as the decision makers how many terms that you want to keep in such analysis. So, it has been found over a period of time that if you are dealing with continuum fluid mechanic situations using only first terms in these Taylor series expansions for describing property variations from one point to the neighboring point and so on is found to be sufficient. So, this is part of what we will call a modeling exercise that we decide how much complexity we want to maintain in our mathematical model which is describing this fluid flow situations. You carry out the model calculations and find out through laboratory experiments if your model is predicting reasonably good results, if it is reasonably good you say that the model is reasonably good and move on. So, it is actually a process like this which has been followed over the last so many years let us say hundreds of years and people have now absolutely convinced that doing a modeling with only first order Taylor series expansion is typically sufficient for most of these continuum fluid flow situations that is all really I can say about this. If you go for a higher order then is there any effects in results is there any effects in the parameter so that the results will be changes that the results will be altered. Yes, so the follow up question is that if we include the higher order terms will the effects be different in terms of results if the results will change and in fact precisely what people found that the second order terms are not really going to do any difference the effects that they introduce if you include them in your analysis are really negligible and that is the reason the first order terms are found to be more than sufficient for describing all these fluid flow situations correctly. Thank you sir. VH, Raishwani, Nakpur if you have a question please go ahead. Sir my question is that we are talking about the Eulerian and Lagrangian method in which we are considering the motion of a point or a particle but in this method we are not defined any physical property of this point or a particle which is considered as a shape and size. Let us taking an example when moving to the seashore the waves comes and when it is going back we are getting that the sand particles are going back with the wave but there is not a proper uniform motion each and every particle we have. So in this method if you are going to or I am going to ask that is there is not any need to define this physical property of the particle or a point in the method. Okay so the question is about the Lagrangian approach of fluid mechanics and if we need to assign a property to each particle as it is moving. So the answer to that is actually that when we set at the reference time t equal to 0 let us say a collection of particles which we later track in time. At that time reference equal to 0 for each of the fluid particles a corresponding property is also assigned and that property could be density that could be temperature that could be pressure etc. And then as we track each of those particles in time correspondingly we also track how the property associated with that particular particle is also changing with time. That is the real Lagrangian way of doing things and as you can see that if we want to really simulate a fluid continuum using a Lagrangian approach it becomes a really really cumbersome and very time consuming affair and that is the reason we actually do not follow this. However in principle a property can be assigned at the reference time to each particle and as we track each of these particles in time we actually are supposed to track the changes in the property associated with that particle as it goes from one location to the next as a function of time. That is the only way you can actually know exactly how the fluid continuum is behaving as a function of time as well. I want to ask that if you are talking that the property like temperature can affect it. So here in the particular equations we are solved we are not getting any particular property like temperature being actually involved in the process. So what we will do is right now we are talking about general terms however the complete set of governing equations on a differential basis will be derived tomorrow and one of the governing equations will be the energy equation which will involve property temperature as well. So once we have completed the governing equations of course we are going to derive everything in Lagrangian sorry in Eulerian form but that is fine as long as you can monitor at different points within the domain of interest properties such as temperature or pressure or whatever as a function of time you have essentially achieved your objective. So the temperature as a property in particular will be taken up tomorrow when we talk about governing equations in particular the energy equation. So far the energy equation has not been really dealt with but we will do that tomorrow and we will deal with temperature at that time. Thank you very much. Jabalpur college please go ahead. Sir I want to know what is Reynolds-Tall-Port program? Is it same as you are saying the balance statement for an elemental control volume and more thing I want to know how the governing equations behave differently when we assume control volume is moving instead of stationary. So there are two questions one is related to Reynolds transport theorem so what is the Reynolds transport theorem and then the second question is if rather than having a fixed control volume if the control volume is moving how will be the changes in the governing equations. So I will assume that this is related to the integral analysis that we discussed yesterday and I mentioned this actually yesterday that without really specifically talking about the Reynolds transport theorem we have discussed everything. In fact the balance statement that right now I have projected on the screen is really nothing but the Reynolds transport theorem. So in the Reynolds transport theorem what we do is we connect the Lagrangian form to the Eulerian form on an integral basis and in the Eulerian form what we are dealing with then is the accumulation term the rates of inflow and outflow terms as far as the control volume is concerned. The last term which I have chosen here in the balance statements is a term that relates to rate of increase of quantity such as mass, momentum or energy due to a source and this is the remaining term in the Reynolds transport theorem as many of you would perhaps know it. So whatever I have written as a balance statement for a control volume the last term which is on the right hand side is actually the term that you will normally see in the Reynolds transport theorem. So as such whatever I have written here as the balance statement is exactly same as the Reynolds transport theorem without explicitly calling this as the Reynolds transport theorem that is the only difference. As far as the second question is concerned you said that if the control volume is moving rather than fixed how things will change and in that case those are special situations which usually fluid mechanics courses will deal with I agree. We are not dealing with that here in this course where we are taking some fluid mechanics part. So in case of control volumes which are moving what ends up happening is that the control volume can be moving in a straight line with uniform velocity or the control volume can be moving in a straight line with an acceleration. If it is a situation where the control volume is moving in a straight line with a uniform velocity or constant velocity all you need to do is you replace all velocities that we have with relative velocities meaning that we will be using the relative velocity of the fluid with respect to that of the control volume. On the other hand if the control volume is accelerating in a straight line we actually have to include more terms there is something called an inertial body force term which will get added to these net force terms that we have been using. So those are special situations of accelerating control volumes and the moving control volumes which we have not taken because this is considered to be a subset of the entire fluid mechanics course. However you will find those cases in the textbooks that I had listed yesterday where an accelerating control volume and a control volume that is moving with uniform velocity are considered. As I mentioned accelerating control volume will require addition of an inertial body force term as they say in this net force acting on the control volume. So it is little more involved than what we have done but what we have done is sufficient for the purpose of going ahead with our CFD analysis that is the only reason we have chosen to go with fixed control volumes. Question is related to the slide number 15 of kinematics whether there are any cases or example where one can neglect out of three either translation rotation or deformation. Yes, so the question is about the effects that I have listed on slide number 15 in kinematics that in general a fluid particle will simultaneously undergo translation rotation and deformation. The question is whether in any special situations some of one or more of these effects can be neglected. That is actually a good question. If you want to neglect translation you are basically saying that the fluid is not moving at all. So it does not really then become a fluid dynamics problem. If you are dealing with a fluid statics problem the translation part can be completely neglected because there is no movement in the fluid. If you want to neglect rotation there are situations where as we were discussing a little bit earlier in the absence of viscosity you can treat a flow to be inviscid and as a result of the inviscid nature of the flow many times what you will see is that the flow becomes irrotational so that there is no rotation at all in the motion. So that is also quite likely in cases. In case of volumetric deformation if I can go a few slides ahead if you really see the rate of volumetric deformation has finally been calculated as the divergence of velocity and I have written out here that if the divergence of velocity is 0 which is what we call incompressible flow we have no volumetric rate of strain or volumetric deformation. So an incompressible flow or a constant density flow is a very common occurrence where the rate of volumetric deformation is actually exactly equal to 0. So there are situations like these when one or more of the effects which I listed here can be actually identically 0 or can be neglected to a good extent. The reason we wanted to discuss these completely is because in general all of these can be present and therefore we want to know how to quantify each of these and that is how they have been taken together. It is given that under the action of forces a fluid particle simultaneously undergoes rotation deformation that is volumetric and shear deformation. So my question is what will be the effect of viscosity on the deformation whether it will be increases decreases or remains the same. So the question is about the effect of viscosity on deformation. The viscosity will actually carry the shear deformation. It is because of the presence of viscosity that a fluid flow will see gradients in the velocity and those gradients in the velocity will cause shear deformation. Also because of the presence of viscosity you will see that the flow becomes rotational and that is another effect of viscosity that you will see on rotation. So the higher the value of viscosity typically you may see larger velocity gradients. It is not necessarily always true but if the velocity is sorry viscosity is fairly high and also the flow speeds are fairly high you may actually see severe shear deformation and severe rotation. So in that sense the viscosity does affect shear deformation and rotation. The poor engineering college please go ahead if you have a question. My question is in a slide kinematic 16 in this case it is delta alpha and delta beta is equal then what will be happen in this case if both are equal. So the question is if by any chance if delta alpha and delta beta that have been shown as the angles of rotation through which segment A B and segment A C have been rotated what would be the effect. So actually you just have to go to the corresponding expressions and try to equate delta alpha and delta beta and see what happens. So if delta alpha is equal to delta beta you will see that the angular velocity is actually not going to be there ok. So because delta alpha and delta beta will cancel out each other in this case and that is about it really. So the flow will behave as if it is irrotational. Thank you sir. Institute of Road and Transport Technology Road please go ahead. You have mentioned that the substantial rate of change of velocity. You told that substantial signifies that the observer has to move with the fluid particles. When you move with the fluid particles how can we differentiate the relative velocity will be 0. Then how can we tell that substantial velocity is. Ok. So the question is actually about the meaning of the substantial derivative. If we are moving with the particle what are we actually sensing. So in fact we are if we are moving with the particle we are not really sensing your velocity with respect to the fluid velocity. You are actually talking about your own velocity as the fluid particle velocity as an absolute velocity in that sense. So there is no notion of the relative velocity of the particle which you yourself are with respect to the fluid. MA, NIT, Bhopal please go ahead. Actually sir I want to to question of first question graphical description. To path line I know a graphical description of path line. I know the path line I have described the trajectory of the fluid particle. But we have a given equation of the path line du upon dt is equal to u and dy upon dt is equal to v and I think slide number ten. So to get the x is equal to function x naught and t y is equal to f to y naught and t then how will we drawn the path line. So the question is about how to generate the path line. What has been shown here is that the equation of the path line has been written for the x coordinate and the y coordinate separately. So because we are following a given fluid particle what we are saying is that dx over dt is simply the change in the x coordinate of the fluid particle in an interval delta t with the delta t interval tending to 0. So that is what we mean by dx by dt and since we are following the fluid particle the change in the x coordinate with respect to time is necessarily the x component of the velocity of the fluid particle and that is the reason dx by dt has been equated to u. In a similar manner the change in the y coordinate over a period of time is the y velocity of the particle and that is why it has been taken as v. Now if you want to integrate these two first order differential equations you need an initial condition which will simply say that at the time equal to let us say 0 which was our reference time to begin with the fluid particle that we are trying to track was placed at the location x naught comma y naught. So the x naught comma y naught coordinates will simply provide the initial condition for the particular particle that we are trying to trace. With respect to that x naught and y naught then as a function of time we will be able to find out the x coordinate and the y coordinate of this fluid particle as it moves from one location to the next. And another question to determine the rate of change temperature suppose we have a two equations a temperature equation and a velocity equation. Then particle with y is equal to 0 and t is equal to I think any we can take t is equal to 20 a slide number 9. So how will be to 0 of this equation u dt upon dx plus b dt upon dy plus w dt upon dz. How do you do 0 of this value? We can take all of value is 0. So the next question is on example number 1 in kinematics slide number 9 and specifically the question is why these three terms as are getting projected right now on the screen have been put to 0. So if you see the example the temperature field has been provided as a function of only y coordinate and time and therefore it is not a function of x and it is not a function of z either. So therefore the differential of t with respect to x and differential of t with respect to z can be immediately put to 0. In addition you can see that the only non-zero velocity component that is provided in this particular situation is the u velocity component. Essentially the remaining two velocity components v and w are 0. So that is why the second and third term have been also put to 0. Since t is not a function of x at all the second term has also been put to 0 even though u is non-zero in here. Thank you. Amrita Coimture please go ahead if you have a question. Between the substantial derivative and the partial derivative I can clearly visualize the substantial derivative even physically but somehow I could not visualize the partial derivative. So could you please explain. So the question is about the difference or the interpretation of the substantial derivative or the material derivative and the partial derivative which is the local derivative. In fact let me go back to my slide number 8 on kinematics using this example I will try to hopefully explain the difference between the local derivative and the material derivative. So if you see in this particular case the fluid particle let us say when it comes in section number 1 it will experience a velocity of v1 and when it comes to the section 2 it will experience a velocity of v2 which is different and greater than v1. So therefore we will say that going from v1 to v2 the particle has experienced an acceleration. So if you are the particle that is if you are sitting on the particle and moving with the particle you will actually experience that you have increased your speed from v1 to v2 as you go from inlet to outlet and if v1 always remains equal to v1 with respect to time and v2 always remains equal to v2 with respect to time you as a particle travelling from v1 to v2 will still experience an acceleration which will be because of only the convective part which now in this case because the flow is steady the material acceleration is equal to the convective acceleration. So that is the interesting part about this problem that when the flow is steady v1 is not changing with time, v2 is not changing with time. However as a particle when you go from section 1 to section 2 you will experience an acceleration which is actually technically the material acceleration since you are following the particle sitting on the particle and experiencing this change but because the flow is steady you are experiencing this material acceleration as a convective acceleration. Now additionally if you say that somehow I start opening or closing this valve so that the velocity in this particular section 1 will also start changing with time. So then v1 will also become a function of time in which case what will happen is that as a fluid particle which let us say the highlighter is now you are moving from somewhere here and you come to the section what is happening is at the instant when you come to this section number 1 or the inlet section the valve is still operating in terms of getting closed or open because of which there is going to be a local change in the velocity meaning that there will be a change in the velocity at the given location at the inlet which will be experienced and sensed by the fluid particle in addition to that convective acceleration that was present. So whenever there is a change in the property occurring at a given location, a fixed location with respect to time you will realize that there is a local rate of change. Another example if you want to talk about is let us say you are in a situation where there is a flow and there is some sort of a chemical reaction that starts happening from time equal to 0 and you take a thermocouple and you place the thermocouple inside the flow at one particular location and as a function of time you monitor the temperature. Now as the reaction proceeds what will happen is let us say it starts releasing heat because of the chemical reaction and because of which the temperature at that particular location will keep on increasing let us assume that. So what will happen is that the thermocouple probe which is mentioned or which is marked at one place will keep on monitoring this change in the temperature at that particular location as a function of time and therefore this change of temperature with respect to time at the given location is what you will call a local rate of change. Yes, so the question is about the nature of the flow being irrotational under what conditions the flow is irrotational and if inviscid flow can be always treated to be irrotational. So actually let me answer it in the following fashion, in reality no flow really can be truly irrotational. However in many cases what happens is there are regions of flow which are very very low velocity gradient regions. In those regions the effect of viscosity is essentially absent and typically these regions are the regions that are sufficiently away from solid surfaces that are bounding the fluid flow. So in such situations when we are sufficiently away from the solid surfaces where the effect of viscosity is negligible what you realize is that velocity gradients are very very low close to 0 you can say and therefore the flow in such regions can be treated as if it is irrotational. If you are talking about inviscid flow you are saying that the viscosity is altogether set to 0 which is an ideal situation clearly. The answer to your question is that most inviscid flows can be safely treated as irrotational flows. There are some very special cases where within an inviscid flow rotationality can be present. So for all practical purposes inviscid flow can be more or less treated as irrotational flow. VIT Pune please go ahead. As we are applying the mass momentum energy equation for the most of the cases my question is related to the vapor bubble. Inside the bubble we are going to be applying the mass conservation. So how can we apply the mass conservation as well as energy and momentum inside the bubble. So the question is about a special situation in a two phase flow kind of flow let us say where we have a vapor bubble moving in a liquid flow if I can assume that and if we can still employ our momentum and mass conservation etcetera. The answer is yes you can still do it there is no question about it. Only thing is that if there is a heat transfer within the flow because of which if the liquid is getting continuously converted into vapor and if the vapor bubble is growing in size you need to include that effect as well through a mass transfer type term. But otherwise there is no problem in employing the standard conservation equations in this situation also. Unfortunately we are not going to be treating two phase flows in this situation neither are we really any expert in a two phase flow kind of situation. So I cannot really answer this more specifically. However conservation equations are universal and they will not have any problem whether there is a vapor bubble type situation in a liquid flow or not. The only careful consideration needs to be given if the liquid is getting continuously converted into vapor. So that some sort of a mass transfer term needs to be added in the balances. But that is about it. Another question is continuation is that only the kinematics 16 number slide whatever you are given. So in that the two dimensionally are shown as a square. So in that if suppose we are keeping a bubble if we keep a bubble so how come the equations are going to be applied because del x and del y is not the length at time. So actually what you need to do in a two phase flow situation. So the question is if there was a vapor mass inside this I am assuming a liquid particle now. How will the analysis change? Actually what you need to do is that you will perhaps have to employ the analysis separately for the liquid mass and separately for the vapor bubble. So it will be a multi component or a multi phase situation where same sort of equations will be employed but separately for the liquid mass and the bubble and appropriate boundary conditions will have to be implemented to take care of the interface between the bubble and the liquid in which this bubble is situated. Then to be honest with you I am not really an expert in two phase flow type situations. So beyond this level unfortunately I cannot answer but I am very certain that what you will need to do there is employ these equations separately for the liquid and for the bubble and then employ appropriate boundary conditions at the interface between these two. What is the difference between substantial derivative and total derivative? Whether these two concepts are same or different? So the question is is there any difference between substantial derivative and total derivative? The answer is no they are actually the same. So some authors choose to use the term substantial derivative, some will use the term total derivative and some will use the words material derivative. They are all supposed to be essentially the same. One more question related to the conservation and non-conservation form of equations as we apply for the Lagrangian as well as for Euler equations, conservation and non-conservation forms. Let me make it more clear about that. So the question is about the conservation form and the non-conservation form of the governing equations. This question has been asked a few times so far. So let me point out that this conservation form and non-conservation form is a terminology that is more used in the CFD literature, more than the fluid mechanics literature. The fundamental difference if it all I can point out is that if you derive a governing equation from an Eulerian point of view you will inherently generate a conservative form of the equation. Whereas if you derive that same equation from a Lagrangian point of view then inherently you will generate that same equation in the non-conservative form. So the conservative form really what it means is that you can interpret the equation in the form of let us say some sort of a mass flux and the difference in the mass flux getting accumulated in the control volume as the accumulation term. So the differences in the mass fluxes on the boundaries getting accumulated as the accumulation term in the control volume. If you can appreciate this form from the governing equation then that governing equation is essentially in the conservative form and as I said if you employ Eulerian method of derivation you will always generate inherently these conservative forms of equations. Thank you sir, over and out from V.I. Tipune. K.K. Wagg please go ahead. Sir, you have shown the schematic of a path line and a streak line. Can you show the schematic of the stream lines on the white board? And one more question that the stream lines I have read that it is the stream lines are fished in space. Can you elaborate on this? Thank you. So the question is if we can show some sort of a sketch of a stream line pattern. So what I will do is I will take a very classic example of flow across a cylindrical object which is a flow past a bluff body sort of a situation. So let me draw that object first and under a steady flow situation what you can obtain is typically a stream line pattern like this if the flow Reynolds number is very very low. So this is a cylindrical object and these are what we will call a stream line pattern for a steady flow situation when the Reynolds number is very low actually in this case. So if you take any point on the stream line here and draw a tangent it will be in the local velocity direction. Similarly if you take a point here and if you draw a tangent it will be in the local velocity direction. So this is a very typical example of stream line pattern. The next part of the question was if these stream lines will be fixed in space I think and the answer to that is if the flow is steady they are fixed in space but if the flow is unsteady the velocity field will keep changing with time and according to that at every instant the stream line pattern will also be kept changing. So in that case it will not be a fixed stream line pattern in time. If you can put a particle under the electric field can we apply the same equation or we have to go for some different approach because due to the electric field the drop gate elongation and the deformation fail. So we have to apply some different equation for that. So the question is if you have a fluid flow with an electrical potential active and if an electrically charged particle is actually moving through a situation like this what can be done in terms of the governing equations. So you are right you cannot really use the same set of governing equations that we are using for fluid flow. What you need to do is you have to also add the electromagnetic set of governing equations which are typically coming from the Maxwell's equation perhaps you are already aware of those Maxwell's equations. Those need to be incorporated along with the fluid flow equations in order for the electromagnetic forces to be calculated correctly. So only the fluid flow equations is not sufficient in that case you need to include the electromagnetic governing equations which are coming from the Maxwell's equation and then you can proceed with the analysis. So if you look at examples such as plasma flows you will have to actually do that. Those are of course very special kinds of situations where you cannot just get by using the standard fluid mechanics governing sets of equations. So with this I will break for T right now. Thank you.