 Okay, so let us continue with this kinematic discussion. So what we are talking about here is that there are certain forces acting on a fluid particle which we are still not bothered to discuss. But as a result of those forces what happens to the fluid particle is that it translates from one location to the other, it undergoes some sort of a rotation and it also undergoes a deformation which normally we will split as a volumetric deformation and a shear deformation. So let us try to look at each of these in a more detail. So the standard way of describing this situation is you show a fluid particle at a certain location at a certain time and for the sake of keeping everything manageable I am showing a two dimensional fluid particle which has a dimension delta x by delta y. So again here is a fluid particle in the sense that we are essentially talking about the same fluid material enclosed within this volume under the action of forces sorry, under the action of forces all these things are going to happen simultaneously, the translation rotation and the two types of deformations and a time equal to delta t later what you will typically find is that the fluid particle has basically become something like this okay. So in particular you can say that this horizontal line AB gets not only translated but also rotated through an angle of delta alpha, similarly the vertical line AC not only gets translated but also gets rotated through an angle delta beta. So why is this happening? Because basically there are velocity gradients that exist in the flow. So in order to show those velocity gradients what I am showing here is that point A let us say is our reference point and at this reference point let the velocities in the x direction be u and v and again using the first order Taylor expansion what we are showing here is the u and v velocities at point B and point C. You can see that at point B it is simply u plus u dx that is partial now times delta x because it is a two dimensional situation now and u is going to be a function of x y t etc. Here the v velocity again likewise is going to be v plus partial derivative with respect to x of v times delta x and so on. So under this situation what is going to happen is that this fluid particle then goes to a later location and also gets deformed in the manner that is shown and in general this angle of rotation delta alpha is not equal to delta beta. So what we do for the purpose of derivation is that we essentially assume that in the positive x direction velocities increase positive y direction velocities increase in reality it could be happening whichever way but for the purpose of coming up with a consistent set of results we assume that. So as I wrote here it is a cumulative effect of all these four things or three things happening simultaneously. What is normally done is an exercise called the decomposition of motion in which each of these effects is separately quantified in some sense. So let us look at that out of which the translation is the most obvious one that it is physically going from one location to the next so that is that point. Let us look at the linear strains and correspondingly then there will be linear strain rates. So if you look at the two segments AB and AC which are essentially marking our fluid particle what is going to happen to AC if I can show this is if this is AC you can say from your side this is where the u velocity is acting this is where the u plus du dx delta f is acting obviously these two are different as this fellow moves translation wise it is also going to get stretched differently in the sense that this end will be stretched differently this end will be stretched meaning pushed differently. In that sense it is undergoing eventually a linear strain. So what is causing this linear strain it is basically these x direction velocity those are causing the linear strain similarly if you look at the vertical segment AC the linear strains for that segment are caused by the vertical velocities v at point A and v plus d dy partial of v delta f. So now let us see what happens. So I am using the standard symbol in some sense epsilon suffix two x's and we will come to this double x double y x y notation little later right now we will go with it. That is basically the change in the length of that segment AB divided by its original length AB that is the standard form of strain expression. So now you see as I was saying that point B will be stretched by an amount u plus du dx delta x times whole thing times delta t in the time period delta t this fellow will be actually pushed by an amount u times delta t that is precisely what I have written the difference between those two will give me the differential stretching for the line element and divided by the original length which is delta x and if you simply go ahead and simplify this you will end up with a du dx partial times delta t. And therefore since this is a linear strain I divide by that delta t altogether take the limit as delta t tends to 0 and I will get a linear strain rate which is what we are interested in in fluid mechanics which will be denoted by e suffix two x's and with a dot which is simply telling me the time derivative and that is after you divide by delta t throughout it will be simply partial derivative of u with respect to x. So this is for the horizontal segment AB exactly the same thing you can talk about for the vertical segment AC and in that case it will simply come out as the y derivative of the y component of the velocity. So for the horizontal segment AB it came out as x derivative of the x component for the y it will come out as y derivative of the y component and if you want to simply generalize it without really thinking much it is going to be simply the z derivative of the z velocity. So these are all individual linear strain rates if you want. Now we normally talk about a volumetric strain rate which is what volumetric strain rate is basically forget about the delta 1 over delta d for now the volumetric strain on a per unit time basis what is volumetric strain change in the volume of the particle divided by its original volume and then you take the 1 over delta t with delta t tending to 0 as simple as that. So the way it is written is basically we are following this fluid particle now okay it goes from location 1 to location 2 and does all sorts of things. So inherently we are talking about a Lagrangian rate here the reason is because we are following a fluid particle. So this delta of delta v obviously the volume element of the fluid particle divided by the time interval with the time interval tending to 0 will give you the limit as the substantial rate of change of the volume element itself. So this volume element here that is dividing the change in the volume element is written outside here the time interval delta t has been taken and inside and then the limit has been taken to write this substantial derivative of the volume element itself that is the idea. So now this is as far as the mathematical notation goes what we are talking about is there is a fluid particle which undergoes linear strain rates thereby its volume changes and we are interested in knowing the new volume you subtract the old volume divide by the old volume and find out that that is the volumetric strain and then divide that by the delta t you get the volumetric strain rate fine. So now the new volume is simply the original length will be a b plus the change in that length times original length a c plus the change in that this will give me the new volume because each a b and a c have been stretched minus this l of a b times l of a c is the old volume and divide that again by the old volume which is l a b times l a c and of course this one delta t is missing which I should which I should write what I have written out here is only the expression for the volumetric strain okay is the new volume minus old volume divided by the old volume that is the strain so I need to have a 1 over delta t here with the understanding that the delta t tends to 0 fine. This is a bit of algebra in the sense that we have actually expressions for all these guys so what is l a b it is it is delta x fine delta of l a b was worked out here this numerator here okay which is simply du dx delta x delta t same thing for the length of a c and the change in the length of a c which were delta y and correspondingly you can work out the change in the length a c. If you actually put together all these and carry out the product what we are interested in is after then subtracting this delta x times delta y divided by delta x delta y etc you simplify all this and eventually you will realize that there will be some terms left which will be multiplied by delta x times delta y type situation so as we want to shrink this fluid element to a small size the terms which will have delta x times delta y multiplying will be considered to be small compared to the terms which do not have that. So this is the algebra which I want you to work out and what you should realize at the end of it is by appropriate simplification of throwing out the so called second order terms as we say we essentially obtain the expression that the volumetric strain rate now is simply d dx partial of u plus d dy partial of for this two dimensional element. So this particular algebra where you substitute all these expressions carry out the product subtract all the previous volumes and such and realize that there are some terms which are second order nature this is something that I want really to you to work out and eventually you should get d u dx plus d v dy and if you want to then generalize to a 3d I will simply add to this the z derivative of the z velocity and if you realize the addition here is simply in the vector form given as the divergence of the velocity field. So it is not the v dot del it is actually the del dot v so you knowing that expression for the del operator or the gradient operator there you can make sure that del dot v really comes out to be d u dx plus d v dy plus d w d z. So that is essentially the volumetric strain rate and if it turns out that the volumetric strain rate is equal to 0 we call that situation as incompressible flow. So the condition that we will look for incompressible flow is this del dot v or the divergence of velocity in Cartesian given by d u dx plus d v dy plus d w d z 3 dimensions is equal to 0 which essentially means that the fluid particle is not really undergoing any volumetric strain rate changes as it moves. So please I will request that this algebra you please work out if you have any problems let me know I will try to help out with that alright. So that is all about the linear strain rates and the corresponding volumetric strain rates. Now we said that the particle can be thought of having undergone a rotational motion also. So there are a couple of different ways of explaining this rotation let me just go back to that picture. So what we have here is that this a prime b prime which you can say is the stretched element having translated now is rotating that is the way you can think about it. Same thing for this a prime c prime which is ac translated and stretched and now it is rotating. So ab has essentially undergone a counter clockwise rotation of delta alpha where ac has gone clockwise rotation of delta beta with delta alpha not necessarily equal to delta beta. So one way of looking at this entire situation is the following you just say that let the angular displacement for ab which is positive delta alpha why because it has undergone a counter clockwise rotation be written as one half of delta alpha minus delta beta plus one half of delta alpha plus delta beta. So if you see it is the same thing as delta it just split in a certain manner is that fine same thing you look for the angular displacement of the segment ac. Here because it has undergone a clockwise rotation of delta beta angular displacement that minus delta beta is written as one half of delta alpha minus delta beta plus and again in some roundabout way I am writing this on purpose minus one half of delta alpha plus delta beta that the reason this perhaps is slightly appealing this way of writing is because just look at the two blue boxes they are the same expressions. So what I am arguing here is that this appears as if both ab and ac which are this way they seem to have undergone a counter clockwise rotation of whatever is the amount enclosed in the blue box because right now we are putting signs also along with everything. So both ab and ac can be thought of having undergone an anticlockwise rotation of the magnitude given by one half of delta alpha minus delta beta and further a shear displacement each of magnitude shown by whatever is enclosed in the red box. Now why am I talking about magnitude one is positive and one is negative as you can see. So which one is positive the ab is positive meaning that you can think about ab as having undergone a rotational motion of half delta alpha minus delta beta and further shear displacement of one half delta alpha plus delta beta and thereby it lands in a position of delta alpha exactly the same thing you can think about for the segment ac where the segment ac can be thought of having undergone a counter clockwise rotation of delta alpha minus delta beta over 2 followed by a counter clockwise shear displacement because there is a minus sign but of the same magnitude as what is in here. This perhaps is little physically appealing you know many times you will see that in most of the books will simply say that the rotation is going to be given by the average of the angular velocity of these two guys and so on. In this way why I can say then that there is a some sort of rigid body rotation and then there is a shear deformation involved is if you see the blue boxes. The blue boxes say that both segments have essentially undergone the same angular displacement both of magnitude as well as sense which means that it is as if it is like a rigid body rotation of the amount one half of delta alpha minus delta beta. On the other hand if you see the other two one definitely is a counter clockwise rotation the first one the other one is a same magnitude but a clockwise one. So this is the way I like to present it just because it gives me a sense of decomposing this rotational motion into a pure rigid body like rotation and then followed by a shear rotation. I hope it is somewhat clear. So if that is the case then we simply say that then the rigid body like angular displacement is simply given by the blue box which is one half of delta alpha minus delta beta and now this is the angular displacement if I divide by the delta t I will get the angular velocity of the fluid particle this is a rigid body like angular velocity and since it is a 2D particle that I am talking about in the xy plane I am essentially talking about rotation about the z axis right. Now what is involved is the expressions for delta alpha and delta beta those need to be found out. If you go back to the geometry this is slightly methodical you have to see delta alpha I will basically say is for small time period delta t it is simply this length B prime B double prime divided by A prime B double prime which is hopefully what I have written here B prime B double prime A prime B double prime and each of those you can work out in complete detail again work out the algebra and you will see that some terms will have left some second order term cancel those and you will get delta alpha equal to dv dx times delta t. This is actually more or less worked out here you just said after this term you have to do the simplification which you can do exactly the same manner I want that delta beta which I am saying it is this length C prime C double prime divided by A prime C double prime is what I have written here correspondingly the expressions you get to find it is dv dy times delta t. So I have the expression for delta alpha I have the expression for delta beta I just put it in here and I realize that the angular velocity then of the fluid particle for the rotation about z axis is simply coming out as half dv dx minus dv dy. So this expression is the correct expression whichever way you look at it whether it is simply looked at as a average of the two angular velocities of the two segments but the way this delta alpha and delta beta are split it gives you a clear idea that there is a rigid body rotation involved followed by the shear deformation. So this is something that you normally may not find in the books the only book which talks about it in this fashion in some sense not completely is by Fox and Mcdonald the latest edition the seventh edition is what I found that it talks about like this but there is a typographical mistake in that chapter on this. So I hope you can find that out but anyway this is the idea behind decomposition of the motion. So this is what I have for the angular velocity about z axis I can now generalize because I know the expressions for the z angular velocity in terms of dv dx and du dy so simply looking at this expression I can go ahead and I form what is called as an angular velocity vector which will have the three components omega x omega z this should be y sorry this is omega z which we just worked out. So the omega x and omega y this should be y omega y is something that I have written out here make sure that those are correct and then you can put it together in a compact vector form where you introduce this so called curl of the velocity del cross v so if you remember from your vector analysis the curl is evaluated as the determinant where the first row is ijk unit vector the second one in this case will be the components of this del operator d dx d dy dz partial and then uvw and you just use your standard determinant rule to obtain the cross product if you put it together you will see that the z component is this here x component is this and y component is this and popularly if you see the books they will call this curl of velocity as the vorticity so you will see that many times this relation is quoted often that the angular velocity is one half of the vorticity or the vorticity is twice of the angular velocity either way whether you are talking about vorticity or whether you are talking about angular velocity what we are talking about is the spinning of the fluid particle about its own axis please keep that in mind the fluid particle may be actually traveling along a straight flat stream line but it can still be spinning about it axis not that it is traveling along necessarily a curved stream line and if either of these guys that is either the angular velocity or the vorticity is 0 then we simply have what is called as irrotational flow which is a special case many times if you are dealing with inviscid flows you will automatically invoke this irrotational assumption as well along with the inviscid flow and then you can deal with what is called as a potential flow which we are not going to get into but this is something that you can note down the key idea though is this and you split it in this fashion and in that sense you can say that it is undergoing what you can say from my side this is the case is undergoing a rigid body rotation by the same amount and then it is doing this is that fine so this that is the idea why half and half is the question that one has to one can split it with your wing I can do a 3 4th and a 1 4th 3 4th and a 1 4th that is something for you to think but I hope that the physical idea is clear this is something that you again as I said normally you may not find it in this fashion in the book but if you just recall this picture always in your mind this and then this. Now I do not want to use that average actually that is precisely what I am saying you can always interpret it that way that it is the average of the two angular velocities that is another way of looking at it there is little more it is little more involved if you see that Kunduz book you will get some clue to it actually what happens is that half if you utilize you can show that this the way this is written is essentially independent of how initially those two lines are oriented as long as they are perpendicular to each other whichever way they are oriented they will always bring about that half but that is fine do not worry about it too much. So that is basically your angular velocity component of that decomposition of motion so let us just quickly go back what we had was translation which we never even talked about because it is pretty easy to visualize rotation and shear deformation is what essentially is linked in some sense so rotation is what we have talked about shear deformation we will now look at and the volumetric deformation is coming through those linear strain rates which we first talked about these are the linear strain rate then the volumetric strain rate now the angular velocity and now finally the shear deformation. So for the shear deformation these two magnitudes remember our picture again it goes through like this rigid body and then it is going like this so the final magnitudes of those two shear deformations are same as one half of delta alpha plus delta beta the sign here simply says it is a clockwise rotation but these two are simply then added as a total shear deformation in the x y plane and that is what is called or that is what is denoted by epsilon suffix x y with delta alpha and delta beta added if you want to compute the rate of shear deformation then you simply divide this by delta t take the limit as delta t tending to 0 call that e dot x y we already have the expressions for delta alpha and delta beta this was for delta beta this was for delta alpha we just plug in those and you will see that the rate of shear deformation in the x y plane simply comes in as dv dx partial plus du dy partial so that half is gone and it is simply a flux and likewise then you can easily generalize the two other shear deformation rates in the other two planes e dot x z and e dot y z just by looking at the expression for the rate of shear deformation in the x y plane that is that is pretty straightforward. You can repeat it formally if you want you can repeat the same exercise by drawing the picture in the x z plane if you want and coming out with complete analysis but once you know the expression here for the x y you can just looking at it analogously you can write down the expressions for the remaining two planes that is more or less what is involved as kinematics part that typically we include in our undergraduate class as I said pretty decent discussion of this is available in the Fox's book unfortunately Gupta and Gupta do not have a good detailed discussion on the kinematics part in Potter's book also there is a reasonable discussion but the best part for this is really the Fox's book but the latest edition the seventh edition as is available today in the Indian market that really has very nicely written explanations for how to get these various rates of deformation there is a little bit of difference in the way they work out the analysis but one way or the other is not very different as long as you are following the logic I think it should be fine any question on this this is the only thing that I really want you to work out is this algebra to make sure that by discarding certain higher order terms second order terms you actually bring about this d u d x plus d v d y that is the only thing that is worthwhile as an algebra to work out and I think apart from one error that I had done here which I will correct yeah here there should have been a 1 over delta t because we are talking about the strain rate whereas what I have written is simply the expression for the strain other than that actually all details I have put on these slides so if you just want to work out the algebra and make sure that you are indeed getting these I think that is plenty so do you know the question is do you always do this in the fluid mechanics class or some people do looks like this is worthwhile doing because most of these things you are later on going to use when it comes to the differential analysis of fluid motion where you connect the stresses to the strain rates you need all these expressions so that is the reason we do it in a reasonable detail right I think that is more or less what I have for the kinematics part okay let's just quickly look at I am not even going to work out the examples because once you know these formulas in some sense is just a plug in given some velocity field and so let's just quickly look at some of these things so I have the first problem here where the temperature is given as a function of y and time and certain functionalities provided with a velocity u velocity as a function of y so this is the only velocity that apparently is present in the flow and the temperature is a function of y and time fine so what is asked is determine the rate of change of temperature of a fluid particle located at y equal to 0 at t equal to 20 so just tell me what is asked to be found out that is good enough rate of change of the temperature let me say experience by a fluid particle correct so it's everything is correct so what is asked is the substantial derivative of temperature so let me write that for one time so what is actually asked is capital D T of temperature is asked and this I will simply write as partial derivative with respect to time plus in general. Now in this case what you have is let me go back to the screen right so now you just tell me from the expression partial derivative of t with temperature with respect to time it will exist because the functionality has time in it that's fine u times dT dx was the next term what about that no because t is not a function of x at all correct same thing for v times dT dy because there is no v velocity given only u is there and there is nothing to talk about w so going back then yep exactly only first term is left this is gone this is gone only this fellow is left and all this was required to be evaluated at y equal to 0 and t equal to 20 so you just have to evaluate this derivative with respect to time of that functionality that is given and substitute these two guys there so whenever you see words like acceleration of the fluid particle it necessarily is implying that we are talking about the substantial acceleration or the material acceleration so wherever that fluid particle word show up we have to think that this is a substantial rate of change fine so that's that in the second example some velocity field is given so this 2x minus 4 is basically your u c times y is your v and 5 times t is the w first part says that determine the value of c if the flow is to be incompressible so if the flow is to be incompressible this thing has to be satisfied so that is it you have all u v w just work out that expression and you can find out whatever that c must be in order for the del dot v to be equal to 0 here the acceleration that a fluid particle experiences while passing the point 3 comma 2 so what is getting asked here again substantial acceleration if you want you can directly write it as a vector expression or you actually split it as an x direction acceleration and a y direction acceleration if you want so let me just point out what I mean by that so if you want you can write this as d e capital D t of u equal to same thing here same thing for v and w and then simply work out whatever these expressions are based on the velocity field that is given to be evaluated at whatever point 3 comma 2 that is that is that okay here some velocity field is given again so u is equal to x and v is equal to minus y so here the first part is equation of the path line for the particle situated at x y equal to 1 comma 2 at time equal to 0 let us just look at this and the second part is equation of the stream line through the same point so u is equal to x v is equal to minus y fine equation of the path line if you recall what we had written was you write d x over d t equal to u d y over d t equal to v and you integrate these with the initial condition that in this case x0 equal to 1 is given at time equal to 0 and y0 equal to 2 at same time t equal to 0 so that is it as simple as that so here then d x over d t equal to x so d x over x equal to d t so then x then will be e raise to t plus some constant let us say if you want for x write it as c e raise to the constant what will happen to the y minus y y then minus t what is c you want to put okay put the initial condition at the initial good for the stream line we had d x over u equal to d y over v that was our in 2d that this is how it will be so it will be d x over x equal to d y over minus y so this will be then this if you bring it to the left hand side etc ln x y equal to let us say ln of some constant if I want to write it that fine so x y equal to constant is the equation how do I evaluate the constant just plug in the coordinates of the point through which this is passing so the point of interest was given as 1, 2 so c is simply equal to 2 so x y equal to 2 is the equation of the stream line before we proceed that equation of the path line is there any way to bring the equation of the stream line from this so what will happen if you let us say you put the initial conditions in t equal to 0 here so e raise to d 1 this is c 1 equal to 1 correct x equal to e raise to t what about y y equal to 2 e raise to minus t can you do something here is there a way to eliminate e raise to t you can if you do that yeah x y equal to 2 same thing if you eliminate t from here you basically get x y equal to 2 which is same as your equation of the stream line what is that exactly that is the point of the example that if you are dealing with a steady flow situation you will actually see that the path line and the stream line will actually come out to be exactly the same and so is the streak line however we have not talked about the equation of the streak line from a mathematical point of view it is slightly more involved than these two guys but this is a point which I did not want to explain earlier I just wanted to show it through some sort of an example so here I hope you understand what was going on all that I have done here is that eliminated then the time dependence from these two guys to get the same exact same expression so this is one more point that normally people will point out in the books that if you are dealing with a steady flow situation the stream lines and the path lines will actually be and then the one final example which is simply plugging in the expressions which I do not want to do now it is again some velocity is given so this 20 y squared is the u minus 20 x y is the v determine the angular velocity, odicity vector and all rates of strain at the point such and such all that you have to do really is that you just use whatever these expressions are and plug those with the appropriate derivatives using this u and v so this is very straight forward so the problems if you see in this kinematics are not really that difficult very straight forward most of the times it is utilizing the expression that we have worked out for the rates of strain or the rotation rates and the deformation rates the idea of how to get those rates is little more important which hopefully they have followed I think that is about it for the kinematics part so all that we need to now do for the differential equations is already with us so before we get into that let us just look at where you can find this information as I said already the better of the lot as far as the kinematic discussion is concerned is by Fox and McDonald in chapters 2 and 5 so most of the material that you saw here is in 5 some of it is in chapter 2 not much honestly in the Gupta's book but it is never discussion again in the Potter's book chapter 3.