 Let us move on with the second part of this kinematics discussion. One aspect of kinematics is that it describes how a fluid particle is behaving as it is moving in the fluid domain because of the various types of forces that are acting on it. So, in case of kinematics we do not really talk about the forces and what kind of forces, their details and so on. We simply say that let there be some forces acting because of which the fluid particle is behaving in a certain manner and we want to analyze only that behavior of the fluid particle in terms of a purely kinematic quantities. So, in that sense what I have written out here is that usually you will see that in general a fluid particle will simultaneously undergo translation which will mean that it will go from one location to the next. Rotation by rotation I mean that while it is moving from one location to the next it may spin about its own axis and it will also undergo deformation and typically you will see that there are two types of deformations that we will come across. One is what we will call volumetric deformation. So, it will shrink or expand under the action of forces as it is moving and also it will undergo a shear deformation when there are shear forces acting. So, now you remember yesterday that we had taken a very very simple situation where it was a one-dimensional pure shear flow wherein we had u as a function of only y coordinate. Now, we will try to come up with more general expressions where we will relax that assumption and the velocity field will be multi-dimensional meaning that u v components etc. will be functions of x, y, z and in general the time as well. So, what we are going to do now is we will try to obtain relations which will describe these different types of motions that the fluid particle will experience under the action of different types of forces. So, we are still not talking about the forces on acting on the particle. The description of the forces will come only when we go to the dynamics of fluid. This is still a kinematics part which is without regard to the forces we are discussing on this. So, here what I have shown is a two-dimensional fluid particle for the sake of simplicity which is usually shown as a rectangle with sides delta x in the x direction and delta y in the y direction. So, it is a two-dimensional fluid particle as we say. Normally in such cases what we do is we mark such a two-dimensional particle with two mutually perpendicular line segments. The horizontal line segment is AB and the vertical line segment is AC. So, with these two segments AB and AC we are marking our fluid particle. What I have shown on the figure on the left hand side is that at point A let the x velocity be u and y velocity be v. If we travel a distance of delta x to the right we will arrive at point B. Delta x and delta y will be in general assumed to be tending to 0. So, these are infinitesimally small distances. So, therefore, using our standard Taylor series expansion we express the x velocity at point B with respect to the x velocity at point A as u plus partial derivative of u with respect to x multiplied by the distance delta x. Remember that u is in general a function of x, y and z and that is the reason we go with partial derivatives. Similarly, as we go from A to B over the distance delta x, v velocity changes from v here to v plus partial derivative of v with respect to x times delta x. Exactly the same thing we do for the y direction we go from A to C because of which the x velocity changes by an amount du dy which is partial times delta y and the y velocity changes by the amount partial v partial y times delta y. So, this is our setting of the situation. So, what is going to happen is as I wrote in the previous slide this particle will undergo all these things simultaneously. It will undergo translation, rotation, deformation which will be of two types and in doing so it will go from this location one let us say where it is shown in the blue color to some other location in the period of time period of delta t. But because of these velocities being different at different points the particle will get distorted in a shape which will be given by this A prime, B prime, C prime and this could be if you want a D prime. So, the original rectangular shape will get distorted into a shape such as this. So, what I have shown here is that the segment horizontal segment AB has now rotated through an angle of delta alpha and now it stands at A prime, B prime. Similarly, the vertical segment AC has rotated through an angle of delta beta and it stands at A prime, C prime. Remember that this is the final state of the two dimensional fluid particle at the end of the period delta t and going from location one to location two it has experienced not only translation as it is pretty obvious, but also these other three effects which is rotation, volumetric deformation and shear deformation. In general the amount by which the horizontal segment rotates which is delta alpha in the final location is not equal to the amount that the vertical segment rotates which is delta beta. In general this need not be the same and the reason is because in general there is no restriction on the way these velocities u and v are going to behave. So, you can imagine that you know as you go from A to B the y direction velocity will push point number A or point A by certain extent and point B by certain other extent because of which we are getting this delta alpha as the angle of inclination. Similarly, this u velocity will push point A in the x direction by certain amount. At point C u plus some other value will push point C by some other amount and that is why you get an angle of inclination of delta beta for this rotated segment A prime C prime. So, I hope the setting of this situation is clear. What we will do is we will decompose this motion which is shown as a cumulative effect in the present slide going from location 1 to location 2 into these individual rotation deformation etcetera. Translation is pretty obvious we have shown that the particles physically moved from location 1 to location 2. So, translation is something that we need not really talk about much. However, let us try to get expressions for the other effects. So, let us first begin with a calculation of a linear strain and a linear strain rate. So, some of these expressions are worked out in sufficient details on the slide. Some of them are not completely worked out. What I will suggest is that from the expressions that are shown here you should be in a position to easily complete the derivation and I would really suggest that you please do it so that you understand how the algebra is worked out. So, let us begin with the discussion on linear strain and in particular for the horizontal segment AB. So, what is going to happen is linear strain meaning that if you look at point A it is going to be moved by the u velocity here within a time period of delta t by a distance of u multiplied by delta t. Similarly, the point B will be moved by a distance of u plus partial derivative of u with respect to x times delta x the whole thing multiplied by delta t. In general u and u plus du dx times delta x are different and that is why you will see that as the particle moves the segment AB the horizontal segment AB will be stretched. And because of that stretching we are talking about a linear strain for segment AB and that linear strain for segment AB will be defined using our standard expression for this strain using our solid mechanics or strength of materials expressions that it is the change in the length of the segment divided by its original length. So, delta L AB divided by L AB and I am calling this linear strain for the segment AB by the symbol epsilon suffix xx it does not have to be 2 x's but for some reason I am choosing 2 x's it can be single x as well if you want. So, now in order to evaluate this change in the length that is experienced by the segment AB I simply figure out how much the point B is moved that is simply u plus du dx times delta x that is the velocity at point B and it is going to move that point B by a distance of this velocity multiplied by the time interval delta t. So, that is what the first term on the numerator is minus u times delta t is simply the distance moved by point A. If it was a pure translation without stretching there will not be any change in the length of AB it will remain as delta x but now it is going to change the original length is delta x. So, that is in the numerator if you simplify this you will realize that the linear strain will simply come out as du dx that is the partial derivative of u with respect to x times delta t you will realize that this u multiplied by delta t will cancel and this delta x will cancel with this delta x here. Now, we are interested in fluid mechanics in strain rates. So, what we have calculated right now is the linear strain the corresponding strain rate will be calculated by simply dividing this through by delta t the time interval and if you do that the delta t in here will simply go away the corresponding strain rate I am denoting by the symbol e suffix xx with a dot over epsilon rather sorry. So, the dot over a symbol is usually the time derivative and that is what the strain rate turns out to be which is du dx. So, it is simply the change in the x direction velocity with respect to the x coordinate that turns out to be the linear strain rate in the x direction. If I repeat this same exercise for the vertical segment y I will be able to come up with a linear strain rate experienced by the vertical segment y and just by analogy you can do that exercise yourself in this fashion but just by analogy since the x linear strain rate was equal to the x derivative of the x component of the velocity y strain rate will be the y derivative of the y component of the velocity. Similarly, the z strain rate linear strain rate that is will be equal to the z derivative of the z component of the velocity. So, this is happening because the segments marking the fluid particle are getting stretched as the particle moves from one location to the next location. Having determined the linear strain rates we can actually go ahead and find out what should be the volumetric strain rate experienced by this fluid particle. So, volumetric strain rate will be essentially the change in the volume of the fluid element divided by the original volume. So, if you simply look at this ratio it will be the volumetric strain and then divide by the time interval over which this is occurring take the limit as delta d tends to 0 as usual and what you will get is the volumetric strain rate. So, what is happening now is because AB is getting stretched AC is getting stretched you can define a new volume for the particle which will be the original length plus the change in the length of AB whole thing multiplied by the original length AC plus the change in the length AC everything is on a per unit depth basis because we are talking about a two dimensional particle. So, the first part here on the numerator talks about the new volume if you want of the fluid particle subtract from it the old volume which is simply the original length of AB times the original length of AC and divide that by the original volume which is again the length AB times length AC the original ones on a per unit volume basis on a per unit depth basis I am sorry and divide that by delta t and this entire expression will give you the volumetric strain rate. Now here I have not worked out the algebra completely however I would like you to complete this analysis by substituting for this LAB delta LAB LAC delta LAC etc and then evaluating these products while evaluating these products you will realize that you will generate quantities which will be of second order accuracy what I mean by second order accuracy is that some quantities will generate where there will be a product of delta x squared multiplied by delta y or alternatively it will be delta y squared times delta x. So, since delta x and delta y are typically these elemental lengths which are tending to 0 whenever there appears a square of each of these quantities in one term those will be considered as second order accurate and we will discard those just the way we discard the second order term in the Taylor series expansion when you have a delta x squared or a delta y squared and then you simplify the entire expression and what you will see is that the expression simplifies in the present case of a 2D particle to du dx partial plus dv dy partial and the reason I have employed here an approximately equal sign is to signify that those second order terms have been discarded when you evaluate the numerator in this expression. So, keep that in mind please mark it on your notes if you want, but I would really suggest that go ahead and perform this algebra and make sure that you do end up with a expression of du dx plus dv dy as a result of the simplification here and that is it. So, what we were after was the volumetric strain rate experienced by the fluid particle which turns out to be then du dx plus dv dy. Now, I am not bothered with saying that this is approximate everything is a first order accurate expression in fluid mechanics. So, that is what we are sticking to and therefore I am using an equal sign. So, 2D particle experienced a volumetric strain rate which was x derivative of x velocity plus y derivative of y velocity. If I want to generalize this through to a three dimensional situation, I will simply add to this the z derivative of the z velocity and if you see this expression it is nothing, but the divergence of the velocity field del dot v you can evaluate this quickly using the Cartesian coordinates and you will see that it turns out to be du dx plus dv dy plus dw dz. So, yesterday during the question and answer session there was one question that what can be the physical significance of divergence of a vector field. This is one particular situation where we are talking about the divergence of velocity field in particular and the divergence of the velocity field is nothing, but the volumetric strain rate experienced by a fluid particle as it is moving in the fluid under the action of forces. So, that is one clear physical significance of the divergence of certain vector field in this case it is the velocity field. Now, if it turns out that in a special situation if the divergence of the velocity field is exactly equal to 0 then that situation is what we call an incompressible flow situation. Essentially what that means is that a fluid particle as it is moving from one location to the next if the flow is incompressible it will not experience any volumetric rate of strain. It will not experience any volumetric changes that is what is to be understood by incompressible flow. Alternatively we say that the divergence of velocity is equal to 0 when we have incompressible flow. So, du dx plus dv dy plus dw dz is 0 when we have incompressible flow. Let us quickly look at an example where we can use some of these derived results. Here a special velocity field is given where the u component is given as 2x minus 4 the v component is given as some constant multiplied by y and the w component is given as 5 multiplied by t which is the time. First part of the problem is that find the value of this constant c if the flow is to be incompressible and second we have to find out the acceleration that a fluid particle experiences while passing the point 3 comma 2 comma 0. So, we just saw that if we have to deal with an incompressible flow the divergence of the velocity will be 0 that is what I have written here. So, du dx plus dv dy plus dw dz must be 0 if the flow is incompressible. From the given expressions for the velocity fields you can calculate du dx calculate dv dy in terms of c dw dz equate the addition to 0 and you will realize that c has to be equal to minus 2 if the flow is to be incompressible. First part asks the calculation of an acceleration that if fluid particle experiences while passing a certain point. Now since the problem is asking for the acceleration that a fluid particle is experiencing you should realize that what is getting asked is that we need to calculate the substantial rate of change of velocity at this point 3 comma 2 comma 0. And for the purpose of showing this explicitly I have calculated the three different components of the acceleration in the x direction, y direction and z direction as substantial rates of change of u, v and w. So, if you go to the x direction it is partial u partial t plus u times partial u partial x plus v times partial u partial y and w times partial u partial z. This is our standard expression that we had discussed may be happen or back or so. The local rate of change of velocity and the three terms after that are the convective rate of change of velocity. Now given the particular flow field what has been figured out is that there is no local rate of change for x direction similarly du dy and du dz are identically equal to 0. So, only component that is left out in the convective acceleration is u multiplied by du dx to be evaluated at 3 comma 2 comma 0 and similarly for the two other directions. So, what I would like you to do is once make sure that each of these terms has been correctly calculated according to the specification of this velocity field and then in this particular way you can calculate the x direction acceleration separately, y direction acceleration separately, z direction acceleration separately and if you want to report this as a vector acceleration you simply add the unit vector i, unit vector j and unit vector k and add them together and report that this is the acceleration in the vector form that the fluid particle experiences. So, this is an example very simple again, but immediately after having derived the expression for the volumetric strain rates and what is meant by an incompressible flow we thought that let us just look at this quickly and make sure that everyone knows how to utilize these expressions that have been derived. So, that was about acceleration sorry volumetric rate of change which comes from essentially linear rates of strain. In fact, if you see the linear rate of strain was du dx in the x direction, the linear rate of strain was dv dy in the y direction and dw dz in the z direction. If you go to the next page you will see that you are simply adding those three independent or individual linear strain rates in the three directions and that will give you your volumetric strain rate du dx plus dv dy plus dw dz all partial derivatives. So, let me just go back to where we started in the present half session. What we are doing right now is we are trying to develop expressions which will quantify the rotation, the volumetric deformation and the shear deformation experienced by a fluid particle as it is moving under the action of different types of forces. So, we talked about what will happen to the fluid particle in a period of time delta t when it goes from one location to the next and as a result of all of these rotation deformation and translation happening simultaneously the fluid particle will attain a shape which was provided on the right part of this transparency here. So, as we said this is the result of everything happening together and what we are trying to do now is to decompose this motion one by one. The first part that we looked at was a formulation of expression for linear strain rates. So, we first found the expression for the linear strain experienced by these two reference segments A B and A C which we are using to mark the particle and we said that because the x direction velocity in case of segment A B and y direction velocity in case of segment A C will keep changing in the x direction and y direction respectively there will be a stretching of segment A B and segment A C involved which will give rise to a linear strain and then a corresponding linear strain rate for these segments. So, here we have worked out the expression for the linear strain rate experienced by segment A B and it turned out to be the x derivative of the x direction velocity and immediately then analogously we wrote that the corresponding strain rates in the y and z direction are simply the y derivative of y velocity and z derivative of z velocity. Then we formulated the expression for the volumetric strain rate experienced by this fluid particle and it turned out to be summation of the linear strain rates in the three directions x, y and z. You see the expression here for the volumetric strain rate written out in the Cartesian form and we realize that it is nothing but the divergence of the velocity field and therefore a physical interpretation for the divergence of velocity field was provided as that simply signifies volumetric strain rate experienced by a fluid particle and we said that if the volumetric strain rate is equal to 0 or equivalently the divergence of the velocity field is equal to 0, we have what we call an incompressible flow wherein there is no volumetric rate of strain experienced by the fluid particle and therefore that part was complete and then we quickly looked at an example which was a fairly simple example to see how these expressions can be utilized. Now let us get on with the rotation part. So it turns out that the rotation and the shear deformation or the angular deformation you can see sort of go hand in hand in the sense that the final position of the particle that is shown here where the horizontal segment AB is now finally in the position A prime B prime where it is inclined at angle delta alpha with respect to the horizontal and the segment AC which is now in its final position A prime C prime which is inclined at an angle delta beta with respect to the vertical. They are essentially the result of both the shear deformation and the angular movement of the particle or the rotational movement of the particle happening together. One more thing that I would like to point out is that what we are showing here is that the segment AB which now sits as A prime B prime has rotated counter clockwise through an angle delta alpha whereas segment AC which sits now as A prime C prime has rotated or moved I should say through an angular displacement of delta beta which is counter sorry which is clockwise. In general delta alpha and delta beta are not equal as we mentioned but if you want to assign a sign for delta alpha and delta beta because delta alpha is counter clockwise we assign that as a positive sign and because delta beta is clockwise we assign that as a negative sign. So with this in mind let us see how we can analyze this rotational plus shear deformation together. If you look at the angular displacement of segment AB we just saw that the total angular displacement experience was delta alpha negative because it was counter clockwise. So I have split that into two components which are written as one half of delta alpha minus delta beta plus one half of delta alpha plus delta beta. So if you add this together you get delta alpha as is required. In a similar fashion for the angular displacement of segment AC which has experienced a total angular displacement of delta beta in the clockwise sense I split that as one half of delta alpha minus delta beta plus minus of one half of delta alpha plus delta beta. So if you can see the colors for the boxes the quantity one half times delta alpha minus delta beta which appears in both these expressions is identical and is marked in blue whereas the quantity which appears in this purple boxes the second part are one half of delta alpha plus delta beta once with a positive sign and once with a negative sign. So magnitude wise if you want to talk about these are identical one has a negative sign one has a positive sign. So therefore with this background what we can interpret this entire situation is that both AB and AC can be essentially thought to have undergone an anticlockwise rotation shown by the magnitude enclosed in the blue boxes which is one half of delta alpha minus delta beta followed by a shear displacement the magnitude of each of which again is the same namely half of delta alpha plus delta beta. However in case of segment AB it is a counter clockwise or anticlockwise displacement for AB and in case of AC it is clockwise because of this minus sign. So this is the way it is useful to interpret this entire total angular displacements experienced by segments AB and segment AC. So what I have tried to do is I have tried to show this in a graphical form or a sketch form where the left most figure shows the original position of the horizontal segment AB and the vertical segment AC. As we said that we can argue that both segments AB and segment AC have undergone a counter clockwise rotation of the magnitude one half times delta alpha minus delta beta and that is precisely what is shown in the second figure. So the dotted line is obviously the original position AB and AC. The rotated locations of AB and AC by the same magnitude delta alpha minus delta beta divided by 2 and also in the same sense which is anticlockwise is shown by the dashed lines. So this is essentially the rigid body like angular displacement of the fluid particle. The reason is because both segments AB and AC which are marking our fluid particle have undergone the angular displacement in the same sense. So that is why we call this the rigid body like angular displacement component and that is then followed by segment AC experiencing a clockwise angular displacement because of the shear deformation and the segment AB experiencing a further counter clockwise or anticlockwise angular displacement both of them are of the same magnitude one half of delta alpha plus delta beta. And therefore finally the location of AB and AC will be shown in these dark lines continuous lines. So that is the idea into interpreting the way this motion can be decomposed. So if it is understood then then what we can do is we go ahead and we find out the mathematical expressions which will now give us the rotational motion in terms of the angular velocity for the fluid particle. We just noted that the rigid body like angular displacement of the fluid particle was one half of delta alpha minus delta beta. If you want to convert this into an angular velocity you will divide that by delta t which is how it is. And now what is left is if we can figure out from the geometry of the situation expressions for delta alpha and delta beta we are in a position to find out what is the angular velocity that the fluid particle is experiencing. So going back to the sketch let me first see how the expressions for delta alpha are calculated. So delta alpha is written as the length of the segment B prime B double prime divided by the length of the segment A prime B double prime. So going back, sorry I will have to go back here. So this is the length B prime B double prime and this is the length A prime B double prime. So B prime B double prime divided by A prime B double prime will give you delta alpha. And similarly C prime C double prime divided by A prime C double prime will give you the angle delta beta. And all then you need to do is figure out the lengths B prime B double prime. So the way this length B prime B double prime has been obtained is as follows. So let me go again back. See if you look at the segment AB it is going to be pushed by velocity V at point A and velocity V plus dV dx times delta x at point B. So the point A will be moving through a distance of V multiplied by delta t, point B will be moved by a distance of V plus dV dx times delta x whole thing multiplied by delta t. And therefore the difference between these two will give you the required length B prime B double prime. Similarly the length A prime B double prime can be interpreted as the extended length because of the linear strain that AB will change too. With this ideas we go back to our expressions again. The numerator is exactly as I described 30 seconds back and the denominator is the original length AB plus the change in the original length AB that is experienced because of the stretching that we talked about few minutes earlier before the t break where we were talking about the linear strains and the linear strain rates. And with this all you need to do is you substitute for each of these expressions. So length AB is simply the original length delta x delta L that is the change in the length of AB is what we had already figured out when we were discussing the linear strain rates. To substitute in here and again discard the second order terms which I would like you to do as part of completing this algebra and what you will realize is that delta alpha simply then turns out again with an approximately equal sign to signify that we have discarded the second order terms. Delta alpha will turn out to be partial derivative of V with respect to x. So you can see this is a cross derivative now. It is the y direction of velocity differentiated with the x direction multiplied by delta t. So I would request all of you to make sure that you take this expression and make sure that you are able to simplify it by discarding the second order terms into something like this. This will give you delta alpha what we are interested now is delta beta which is obtained in an analogous fashion. Only thing is that here we have to talk about the segment AC now. If you see again it involves a cross derivative. This should have been u. This is a typo. Please make sure that you change this to u. So this is the partial derivative of u with respect to x. That should have been done correctly here. This should have been u. So it is the cross derivative again. Derivative of the x direction velocity with respect to y direction again multiplying by delta t. And finally what we are interested is one half of delta alpha minus delta beta over delta t as the angular velocity of the particle for its rotation about the z axis. And which I am going to call as omega suffix z. So small omega suffix z. And if you simply substitute all these expressions that we are talking about you will realize that omega z simply comes out to be one half of partial derivative of y velocity with respect to x minus partial derivative of x velocity with respect to y. So again this velocity here where my highlighter is standing should be actually u. Here it is correctly written but from here to here I seem to have made a typographical error. So please make sure that this is correctly changed. Similarly make sure that this expression is also correct. Eventually what you should get is partial derivative of u with respect to y. And that is what the angular velocity of the fluid particle for its rotation about the z axis would be which is half the whole thing multiplied by partial derivative of y velocity with respect to x minus partial derivative of x velocity with respect to y. So in general this was so far discussed for a two dimensional fluid particle which is moving in the xy plane and therefore it can experience only an angular velocity about the z axis. If you want to generalize it for a three dimensional particle you can say that in general the fluid particle will experience all three angular velocity components which will be given by omega suffix x, omega suffix y and omega suffix z. So there is a typo here as well. Those expressions for omega suffix x and omega suffix y you can find out analogously using the expression that we have derived for the z direction angular velocity. And I have directly written out those expressions here and I would request you to verify that these are correctly written. If you want to express this entire angular velocity in the form of a vector you write that as a vector omega equal to one half which is coming in each of these. The remaining three Cartesian components you can actually combine and bring them in the form of a curl of the velocity. So here if you want you can you can evaluate this curl of the velocity using Cartesian coordinates. We have you know how to expand this curl using that determinant expression. So if you do that with velocity being ui plus vj plus wk and the gradient operator or the del operator being i times ddx partial plus j times ddy partial etcetera you will realize that you are recovering all these three components through this curl of velocity. So once again there was a question yesterday about the physical interpretation associated with the curl of a vector. So in this particular case what we can point out is that the curl of the velocity field divided by 2 is essentially the angular velocity experienced by a fluid particle. This angular velocity is essentially related to the spinning of the fluid particle as it moves either along the straight streamline or a curved streamline that is in material. As long as we are talking about either the vorticity as it is called del cross v or the curl of velocity or equivalently the angular velocity small omega we are talking about spinning of a fluid particle as it is moving in the fluid. So as I mentioned this curl of velocity field has a special name which is called the vorticity and I am denoting that by capital omega here. So del cross v which is the curl is vorticity. If either the omega small which is the angular velocity or omega large which is the vorticity which are essentially related to each other by only that factor of one half. If either of them are or is 0 we treat the flow to be irrotational meaning that the fluid particles in an irrotational flow as they are moving along either a streamline of straight nature or curved nature or whatever those particles are not spinning about their own axis. So that is what I have tried to show here through these two sketches at the bottom. So if we are talking about a rotational flow where omega or the vorticity is non-zero we will see that a fluid particle will actually spin about its own axis as it moves along a streamline that is shown here. If we are dealing with an irrotational flow situation there will be no spinning involved as the particle moves it will remain in its orientation the way it is as it moves along the streamline. So please keep in mind that the rotationality or irrotationality has nothing to do with whether a particle is moving along the straight streamline or a curved streamline it can be either what we are referring to when we say rotational or irrotational is whether the particle is spinning about its own axis or whether the particle is not spinning about its own axis. So irrotational means that it is not spinning about its own axis. So this was about the angular displacement and the resultant angular velocities that the fluid particle experiences. Keep in mind that the linear strain rates and the corresponding volumetric strain rates were because of derivatives of velocity components in their own direction. So we had x direction derivative of x velocity y direction derivative of y velocity and so on which contributed to the volumetric strain rate. In case of rotation here you note that these are the cross derivatives. So the x direction derivative of y velocity y direction derivative of x velocity here in the first expression. If we are generalizing to 3D then we are talking about y direction derivative of z velocity and z direction derivative of y velocity and so on. Those will contribute to rotational motion and the spinning of the particle that is somewhat important to know. Coming to the last point here remember that we said that the fluid particle is experiencing not only rotation which we just quantified but also a shear deformation. And we had said that each of these segments AB and AC is experiencing a shear deformation of magnitude delta alpha plus delta beta over 2. Segment AC was essentially argued to experience a shear deformation in the clockwise sense. Magnitude is the same and segment AB was considered to experience a shear deformation in the anticlockwise sense both with the magnitude of one half delta alpha plus delta beta. So many times and many books I should say bring out a total shear deformation in the xy plane in terms of its magnitude. So that half of delta alpha plus delta beta is simply multiplied by twice because it is the same experience by both segments AB and AC. So therefore the total shear deformation experienced in the xy plane in terms of its magnitude is simply delta alpha plus delta beta that is given a symbol epsilon suffix xy because it is in the xy plane. And in order for its rate of shear deformation to be found we divide this entirely by delta t the time interval over which all this is happening and that is given a symbol epsilon suffix xy with a dot which signifies time rate of change. So essentially it is delta alpha delta divided by delta t plus delta alpha divided by delta t and going back we have the expressions delta alpha divided by delta t is simply partial derivative of v with respect to x, delta beta divided by delta t is partial derivative of u with respect to y. So putting it together the total shear deformation rate in the xy plane is simply dv dx partial plus du dy partial. And immediately you can extend this to a three dimensional situation. So we will talk about shear deformations in not only just xy plane but also xz plane and yz plane. So correspondingly the expressions in the xz plane and the yz plane are written out here. So these are not independently derived but simply inferred from the derivation that we have carried out in the xy plane. And that more or less wraps up the discussion on kinematics. What was important to note here is the expressions that we have formed for the rotation and the shear deformation. Many times as I mentioned yesterday also in the beginning of our workshop that once a CFD analysis is carried out you generate lot of data as part of your numerical solution. So typically the velocity data that will be generated as part of the CFD simulation will be utilized to generate auxiliary fields such as a vorticity field. So what you will need to remember is that in case the velocity field needs to be converted into a vorticity field as part of your post processing in the CFD simulation. You will be numerically evaluating such derivatives. So dv dx, du dy etc will be evaluated numerically. They will be put together in this fashion and then the vorticity field can be generated from the velocity field data as your simulation gives you the velocity data. So then the utilization of all this is many times to understand how the flow field is behaving where regions of large rotation are perhaps existing and so on. So the kinematics part that we have discussed here is particularly useful in the post processing of the CFD results. Earlier also when we were discussing the streamlines and the pathlines and so on I said that as part of the post processing in the CFD you will generate actually many of these streamline patterns and the pathline patterns if at all if it is in an steady flow from which you get an idea as to how the flow field is behaving. So that is the reason this kinematics is very important not only from the fundamental fluid mechanics but also as something useful for a CFD post processing activity. So I just have one more example which I am not going to complete here. It is very straight forward in the sense that again some sort of a velocity field is provided. So the x velocity here is 20 multiplied by y square the y velocity here is minus 20 times x y and for this particular velocity field you are asked to determine angular velocity, vorticity vector and all rates of strain at the point 1 comma minus 1 comma 2. All that you need to do is you need to go back to the specific slide which has the expressions for the angular velocity and the vorticity and the linear strain rates and the shear strain rates and simply substitute there the appropriate derivatives either the cross derivatives when it comes to vorticity angular velocity and shear deformation or the derivatives of a velocity in a given direction with respect to its own direction that is du dx sort of a situation and then you can obtain all these required quantities at the point 1 comma minus 1 comma 2. So this is an example which I would like you to complete. If you want you can complete it before the afternoon session pretty straight forward all that you need to do is you need to substitute correctly for the expressions for all these quantities. So this is essentially the end of the kinematics discussion and to summarize this again what we have done is we have discussed how to describe a fluid flow without regards to the forces acting on it. We talked about the Eulerian point of view, the Lagrangian point of view we connected the rates of change that you experience in an Eulerian frame and a Lagrangian frame. Then we talked about the graphical descriptors which are our path lines, stream lines and streak lines and then the last part has been quantification of the rotational rates, the shear rates and the volumetric strain rates as the particle moves under the action of various forces. So this is what we will end here the discussion of kinematics with.