 Now, I want to take a few questions. Yes, Mr. Mufakumjha, you have a question please go ahead. We are considering a control value fixed in space, we are deriving equations, we are getting conservation form of equation. And when the control value is knowing its space, we are getting non-convolution form of equation. So, I want to know the difference between the conservation form and non-convolution form. Yes. So, the question is in these derivations, in one manner when we are carrying out the derivation, we are obtaining what is called as a conservation form and in other manner we are obtaining what is called as a non-convolution form. So, let me first try to go back and point out what is meant by a conservation form. Let me go back here. So, when we employed our mass balance equation for a control volume in the slide that is projected on the screen, we obtained an inherently conservative form of the equation. So, the form that is now shown as equation 1 as partial derivative of rho with respect to time plus the divergence of rho v equal to 0 is what we call a conservative form. So, the conservative form actually comes about automatically when you obtain the differential equation from a balance statement point of view. Or in other words, if you look at the other approach that we employed, here also we started with an integral form which was essentially a balance statement if you recall on an integral basis. And we have simply manipulated that balance statement into a balance statement on a per unit volume basis and therefore you again obtain the form in the conservative form. So, whenever you are employing an Eulerian approach of deriving the expression which will necessarily mean that you are doing a balance statement on a control volume type approach, you will necessarily get the equation in a conservative form. On the other hand, this last part which I described if you go to slide number 6 on differential analysis, what we are doing here is we are utilizing the inherent Lagrangian form because we have decided that we will follow a one given particle and we will simply make sure that the mass content for this one fluid particle is going to remain unchanged which is what we write in terms of the substantial derivative of the mass content being 0. So, my point is that if you employ a Lagrangian approach of deriving these equations, you will always get the equations in a non conservative form which is shown out here. As far as fluid mechanics is concerned it does not matter whether you are obtaining the equations in a non conservative form or you are obtaining them in a conservative form here. On the other hand, when it comes to CFD utilization, especially for a finite volume technique as we will see in the later half of this course, the conservative form is something that is found to be more useful. But the point and the answer that I want to give here is that if you employ Eulerian approach for derivations, you will obtain conservative form. If you employ Lagrangian way of doing things in terms of derivations, you will obtain a non conservative form. SVNIT Suresh, you have a question please go ahead. Sir, the way in which you explain the solid boundary rotation and the shear deformation during the coordinate workshop. Sir, can you repeat the same thing so that the participant can have a better understanding using the two perpendicular ball pens? Yes, I will do that in a second. Actually, I had tried to, so the question is related to the way the rotation and the shear strain rates were described and obtained. What I had done actually was let me project that slide and then also I will show what I mean by that. So, here what I have projected right now is really what I had shown with those two pens in the coordinator workshop. So, let me try to show that again through the video camera. So, if you people are looking at this, so what I have is the two perpendicular segments. One is AB which is the horizontal segment and the one which is black here is the vertical segment AC when you are looking at it. Now, what I mean by the decomposition of this rotational motion and the shear deformation is that initially you can think about both of these guys experiencing a anticlockwise solid body like rotation in this fashion. So, both of them are getting turned by the same amount through an anticlockwise sense and then follow this by one segment experiencing anticlockwise shear rotation and the other segment experiencing a clockwise shear rotation. So, finally what ends up happening is that the segments AB and AC will end up in a situation such as what I am holding these two pens. So, let me repeat this. I start with a mutually perpendicular situation then both of these guys experience a anticlockwise rotation of the same amount and then one segment keeps on going anticlockwise due to the shear deformation. The other segment keeps on going clockwise because of the shear deformation. So, this is the final location for segments AB and AC as I had drawn it. Hopefully, this is carrying the point across and this is what I tried to actually show through the sketch that I have prepared here in the slide number 22 on kinematics. Institute of Road and Transport erode, you have a question. This is Dr. Dharmish Moghiddin from IRT to Remote Centre Tamil Nadu. I have one question in connection with various types of lines. While explaining the kinematics of fluid flow, you have explained something about path lines, streamline and streak lines. Just I want to know what are the practical importance of these lines in your flow field and how they are helpful in CFD analysis. This is question number one and I have one more question. Question number two, you have explained about Lagrangian method and the Eulerian approach to solve the flow field problems and how to select the particular type of solution method for a particular application and what are the limitations of each in your two methods. These are the two questions. Thank you. Yes. So, there are two questions. One is related to the different lines that I talked about, the path line, streamline and streak line. In particular, how are they used in practice as well as from the CFD point of view. I tried to explain that during the lecture that if you are performing an experiment for example, if you are actually performing a fluid dynamics experiment in a lab, many times what you will do is you will employ a flow visualization type technique and one of the very popular flow visualization techniques perhaps slightly older is creating hydrogen bubbles in a liquid flow in particular in water flow and then letting those hydrogen bubbles flow along with the flow and then taking a video camera and then recording their motion. So, each of these hydrogen bubbles that are created in that particular technique are basically tracing a path line which gets recorded when a video recording of such a flow field is carried out. So, that is one practical application where people have employed these ideas in a lab setting. When it comes to CFD application, what I said was that usually as a result of the CFD simulation you will generate a velocity field in the flow domain. So, what I mean by velocity field is that you will generate the velocity numbers at the various nodes or the computational points where you are calculating all these. So, using this velocity field, you can actually generate the streamlines and in fact, most of the times the graphics plotting software that nowadays people use will automatically generate the path lines and streamlines once you provide the velocity field as the input. So, that way you can easily get the streamlines from your computational fluid dynamics analysis. So, that is what I would like to say on question number one. Yes, question number two was about Lagrangian versus Eulerian point of view and where they are utilized. So, typically if you see fluid flow analysis, pure fluid flow motions, you will always see that an Eulerian point of view is employed. The reason is because the Lagrangian point of view is very cumbersome to utilize. Where you will see a Lagrangian point of view employed is the kinds of situation that I was talking about yesterday, especially these rarefied gas dynamic situations where the density is very very low and correspondingly the number of gas molecules in the domain of interest are very low. In which case you cannot really employ continuum methods and you actually have to go to a Lagrangian method where you actually select a large number of particles which are your gas molecules and then track them as they move about in the domain to find out the flow field information. So, that is where you will see the utility of Lagrangian approach. Also in situations as were being pointed out yesterday in the discussion, there are some situations when we have solid particles embedded in fluid flow. So, to analyze such situations, people utilize Lagrangian approach to track the solid particles and Eulerian equations of motion for the solution of the fluid problem. So, that is another example where Lagrangian is mixed with Eulerian in some sense. Which method will be more accurate whether it is Eulerian or Eulerian method for a particular application? The question is whether Lagrangian or Eulerian method will be more accurate in a given application. Accuracy wise both are probably the same. It is just that to obtain the same accuracy the amount of computational time for example, in case of Lagrangian approach is very, very large unless you are dealing with rarefied gas dynamics type situations. But accuracy wise I do not see if any one is going to be more accurate than the other. It is just that practically speaking using Lagrangian approach is essentially impossible for standard continuum flow situations. Thank you professor. Thank you. I got answers for the questions. Thank you very much. Thank you. M.E.S. Pillai Panvel please go ahead if you have a question. My question is regarding slide number 20. Acceleration of the fluid particle where the local rate of change plus convective part that is spatial part has been described. Whether this application is applicable I mean true for other conservation flow variables also. So the question is about what was worked out on slide number 20 which was the acceleration of fluid particle and the specifically the question is about the local rates and the convective rates and whether they are applicable to other flow variables also if I understand the question correctly. And yes the local rate of change and the convective rate of change will in general be applicable for any flow variable. So in this particular instance when I worked out this problem the flow variable was taken as the velocity and in particular the three velocity components were separately considered as u, v and w. But if you are desirable you can always try to find out the substantial rate of change of pressure for example in which case there will be a local rate of change of pressure and a convective rate of change of pressure. So in which case wherever you see this u inside the derivative you will replace that inside the derivative u with p as the value or the quantity of interest. Similarly you can obtain substantial rate of change of temperature as well. So there is no problem with that any flow variable can have a substantial rate of change and it will be always composed of the local rate of change and the convective rate of change. Now in situations special situations it is possible that the local rate of change is 0 for example if you are dealing with steady flow. It is possible that in some special situations may be the convective rate of change is 0. But in general both of these are present and both of these are applicable for any flow variable. KJ Sumaiya Mumbai you have a question please go ahead. In kinematics 18 slide hello kinematics 18 slide you have written that second step 1 upon elemental volume times substantial derivative of elemental volume is equal to del u by del x plus del v by del y. How the left hand side has arrived? This is one question and the second question is in the differential analysis six slide from the top the second step del rho by del t plus grad rho dot v bar plus rho grad dot v bar is equal to 0 in that how the third term has arrived? Yeah so there are two questions one is related to the slide that is projected right now on the screen which talks about the volumetric strain rate for the fluid particle. So the volumetric strain rate in its basic sense has been defined as the change in the volume of the fluid particle divided by its original volume and the whole thing then divided by the time interval over which this is happening. So what has been done is first of all you should note that we are following a particular fluid particle as it goes from one location to the next and while doing it it is experiencing the volumetric rate of change. So since we are following a one particular fluid particle we will necessarily be talking about the substantial rate of change. Now if you see the expression at the top what we have is 1 over delta t times delta of the volume element divided by delta v which has been rewritten in the sense that what I have done is I have taken this 1 over delta t inside and the delta v on the denominator outside. So you can read the entire left hand side as 1 over delta v change in the volume element over the time interval which is over which it is happening. And as delta t which is a time interval tending to 0 change in the volume element over that delta t is essentially going to be a rate of change of delta v with respect to time and that is how this substantial derivative has showed up here and the delta v has gone out. So this delta v here has gone out delta t has come in as delta t tends to 0 change in the volume element divided by the delta t the time interval essentially becomes a time derivative of the volume and that is why we have we have written it as substantial derivative because we are following the fluid particle in its own sense. So we are we are necessarily employing a Lagrangian method right now because we are following a given single fluid particle. So that is the that is the first question. Let me let me load the other yeah. So there was there was a question on slide number 6 of differential analysis and in particular how this step is getting achieved if I understand the question correctly. So the way it is been done is that if I if I go back one slide here we have obtained the continuity equation from the integral form by doing these mathematical manipulations and the form of the continuity equation comes in the form of divergence of rho times v. Now keep in mind that divergence of rho times v essentially implies that rho here is not assumed to be a constant it is a variable. So therefore I am in a position to expand this del dot rho v by expanding the del dot whatever is inside and if you want to look at the specific form of the expansion what you should do is you go back to the the lectures yesterday and when I had passed out the slides for the mathematical background you will see that I have given an expression for del dot a multiplied by a vector b and the expansion of that. So using that expansion formula which was provided yesterday in the mathematical background this del dot rho v has been expanded and that is about it really. If you want you can actually redo this entirely in Cartesian coordinates. So if you employ Cartesian coordinates you will see it far more easily. Right now what I am trying to do here is to save space I was using the the expansions directly in the vector form however you can do this entirely in Cartesian coordinate as well. But having said that I have given you the relevant expansion formulas yesterday in the mathematical background. So if you do not mind going back to it you will realize that there is one formula which says del dot a multiplied by vector b and how to expand that using that this particular expansion has been obtained that is about it. So slide number 15 you have written under the action of forces a fluid particle simultaneously undergoes translation rotation deformation and after that you have derived independent equations for each particle undergoing translation rotation deformation. What about the net composite effect of all these transformation? So what will be the net transformation due to translation rotation and deformation on a fluid particle? So the question is about what I what I said on slide number 15 in kinematics that we said that under the action of forces a fluid particle simultaneously undergoes all this and then the question is that we have independently analyzed each of these motions but the question is what happens as a cumulative action of this. In fact what I did was when I wrote this I immediately showed the sketch here in slide number 16 which actually shows the cumulative effect and then what we did was we decompose this cumulative effect into independent actions one was rotation one was shear deformation and one was volumetric deformation. So we went in the reverse manner in some sense what we showed was the cumulative effect was shown on slide number 16. So when the fluid particle goes from the location number 1 where it is rectangular to the location number 2 when it becomes this Rombus type situation this is actually the cumulative effect. So the fluid particle will translate it will change its volume it will also rotate and undergo shear deformation and finally over a time of delta t it will look like this. Having said this we have gone in the next few slides for example slide number 17, 18, etc and then try to analyze each of these components separately but the cumulative effect is already shown in slide number 16. It is the diagram that you are showing but can we have a composed transform equation of it? Yes you can yes yes yes the equation or analyze each one separately but it is about the net equation. You can indeed have an equation which can show all these effects in one shot where you can put together all these rotation translation etc together. I have not actually put together that equation in this set of slides the reason is because I wanted to show these effects separately. In fact the issue is that if you put it together as one equation most of the times it is not really easy to see what sort of detailed issues are happening as a result of each of these different motions. So indeed you can show this as one equation by putting it together. In fact some of the advanced textbooks in fluid mechanics if you read you will see that they will show this combined equation. However in our experience what happens is that if that combined equation is shown it is somewhat difficult to follow what has been happening in each of these individual effects and that is what I have tried to show separately rather than putting it together. I hope that answers your question. Okay thank you sir. Thank you. So right now it is one o'clock so what we will do is we will stop for lunch right now. Thank you very much.