 dynamics of inviscid flows. In the last chapter we were discussing about the kinematics. So we were not discussing about the forcing parameters which are involved to influence the flow. So we have discussed about the motion. Now we are going to see that what are the forcing parameters which influence the motion and how they are related to the motion. When we talk about inviscid flows what we essentially mean is initially we will discuss about cases where viscous forces are not present. So it is a simplified situation of the reality but at the same time it will provide us with a lot of important insight which we will use later on when we will be discussing about the dynamics of viscous flows. So when we will be considering or focusing our attention in this particular chapter we will be considering cases when viscous forces are not there or negligible. To start with the discussion on this what we will try to do we will try to write the equation of motion for a fluid element where viscous forces are not present. So when viscous forces are not present the kinds of forces which are there are the surface forces in terms of the normal components which are manifested through pressure and some body forces which may be like the gravity forces. Writing that in mind let us say that we want to write the equation of motion for a fluid element. Let us say that it is a 2 dimensional fluid element. It did not always be 2 dimensional but if we are writing the equation of motion along a particular direction then like for simplicity we can take it as a 2 dimensional one for illustration. So let us say that we take a 2 dimensional fluid element as an example. Fundamentally it is always 3 dimensional. So the third dimension you may consider as one or some uniform third dimension. Let us say that these dimensions are delta x and delta y. We will quickly identify what are the forces which are acting on the fluid element only along x. So we will identify forces along x because we are interested to write the equation of motion along x. So other forces we will not show. So it is not a complete free body diagram only the x component of forces will be shown. So here you have force due to pressure. So if p is the pressure here then p times delta y maybe times 1 where 1 is the width is the force that acts on the left face due to pressure. Force that acts on the right face due to pressure is what? We have encountered such situation earlier. So p plus this into dx times delta y into 1. Along x these are the only surface forces because other faces will have surface forces along y. Body force may be there. Let us say that bx is the body force per unit mass. So rho into bx into delta x into delta y is the body force component along x because rho into delta x into delta y is the mass of the fluid element. So we can write the Newton second law of motion for the fluid element. That is we can write resultant force along x is equal to mass of the fluid element times acceleration along x. Maybe you can write delta m it is a small mass to acknowledge that. So we will try to simplify this expression. Resultant force along x is p into delta y minus p plus this 1 with respect to into delta y. This thing then plus rho bx delta x delta y is equal to the mass of the fluid element is what? Rho delta x delta y times acceleration along x. What is acceleration along x? This we have discussed in the kinematics. So what is that? This is the acceleration along x. This we have derived in the kinematics. See when we were discussing about the rigid body type of motion of fluid elements then we did not use this expression. We were using as an expression as if the entire fluid is having a particular acceleration disregarding the deformation within it. So now the different gradients of velocity will become important which was not there or which we kept ourselves abstracted of when we just wrote some acceleration when it was moving like a rigid body. Now we are more detailing it. So we are looking into the detailed expression that reflects that acceleration. So this is acceleration along x. Now you can cancel various terms. So first this term will go away and then like you have let us just correct it a little bit. It was delta x right. We did not consider it dx delta x. So just change this dx to delta x because we took our element as delta x and then we can cancel delta x into delta y from all the terms because these are small tending to 0 but not actually equal to 0. So what we are left with? We are left with a simplified expression plus rho bx is equal to rho. This very simple expression is also known as Euler's equation of motion along x. Similar expressions we can write for the motion along y and z. We are not repeating it because it is very trivial. Now what does this equation of motion contain? If you look into it it is fundamentally like Newton's second law of motion where viscous forces are not considered. So this right hand side is something like the mass into acceleration. Left hand side is the force effect of the force which is acting. So one force is because of the pressure gradient and another force is because of the body force. These 2 forces are considered. So it is just a different way of writing Newton's second law of motion for a fluid where viscous effects are not present and any other force other than this body force of this particular form we are not considering. Let us take an example to illustrate that how we can make use of this. The example is like this. Let us say that you have a velocity field v given by say axi-ayj where a is a constant number and a dimension number to adjust the dimensions in the 2 sides. We are interested to find out what is the difference between pressure at 2 points given by x1, y1 and x2, y2. It is given that g that is the acceleration due to gravity acts along negative z direction. So the question is what is the difference in pressure between these 2 points, okay. The problem is very simple but it will at least give us some idea of how to make use of this expression. A and B are not functions of time. So it is a steady flow field. Let us write this equation say along x for this one. So if we want to write this Euler's equation along x, so we have minus of partial derivative of pressure with respect to x plus what is Bx? There is no x component of body force. Body force only acts along negative z. So this plus 0 is equal to rho. Velocity field is not a function of time here. A is a time independent constant that is given. So the time derivative will be 0. Then so what is u and v here? This is u and this is v, okay with the minus sign of course, includes the minus sign. So rho, so u is Ax into A that is this term. The other terms are not there because u does not have any dependence on y and z. So the other terms are not there. So this is the equation of motion along x. What will be the equation of motion along y? Just it will be similar to this. There is no body force along y and rho right hand side. What is going to happen? This u is only going to be replaced with v. So the term that will remain relevant is only v into partial derivative of v with respect to y, okay. So it is minus Ay into minus A. Let us consider the z component. Now you have a body force along z. What is that? So bz is equal to minus g. Minus rho g is equal to the right hand side. u will be replaced by w and there is no w component of velocity. It is a 2 dimensional flow field. So it is 0. So it is possible to integrate these expressions to find out how p varies with x, y and z. So let us integrate that. Let us say we integrate this one with respect to x. So we get p here as what? Minus rho A square x dx will become x square by 2 plus function of y and z, right. For this it will become p equal to very similar minus rho A square now y square by 2 plus a function of x and z. And what will this give? p equal to minus rho gz plus a function of x and y. All these 3 expressions are representing the same pressure field. So we can compare these to get these 3 functions. So let us compare these and get the 3 functions. If you compare what functions you get? What is f1? f1 is a function of y and z. So minus rho A square y square by 2 then minus rho gz. This is f1. f2 function of x and z. So minus rho A square x square by 2 minus rho gz and f3 minus rho A square x square by 2 minus rho A square y square by 2, okay. So the expression for pressure becomes minus rho A square x square by 2 minus rho A square y square by 2 minus rho gz, right. So this you can write minus half rho A square x square plus A square y square is u square plus v square. So that is the square of the resultant velocity. So let us say capital V square minus rho gz that means p plus half rho v square plus rho gz equal to 0. Now when we write this there is a lack of generality here. What is the lack of generality? When we consider this f1, f2, f3 we did not consider a constant. So fundamentally we should also have a constant there where that constant may be eliminated depending on a choice of a reference frame but that is not done a priori. That is after you get the general expression then only that is possible. So a very important thing here is that each of these should be augmented with a constant. So what it means is that this plus c will be there. So this will become in place of 0, it will become some constant. This looks very familiar to you. It is like a Bernoulli's equation. Now do not get a wrong impression here that the Bernoulli's equation is always valid and that is why you can write it in this form. There is a specialty of this problem because of which the Bernoulli's equation gets valid between any 2 points 1 and 2. So this is a point 1 and this is a point 2. Solving the problem is trivial you can find out p1-p2 by substituting the velocities respectively at x1, y1 and x2, y2. That is a straight forward exercise. The direct comparison of the equation is possible because they represent the same pressure. This is by observation. See I mean when you write when you say that these 2 are equal or these 3 are equal it should be such that it does not contain any function of x, y or z which falls beyond the functions written here. I mean there are certain things which you can do just by common sense and this is one of the big things in mathematics which you can do by a little common sense and that is what is expected when you solve such problems. Now when you come to this conclusion that it is like a Bernoulli's equation. In fact it is of the same form and we therefore can apply it between any 2 points 1 and 2. It is not a general conclusion that we must remember and we will do it rigorously to show that when it is valid and when it is not valid. This is very very important because all of you are very very habituated in using Bernoulli's equation anywhere and everywhere you like. So we will try to see, we will try to restrict you so that you do not apply it anywhere and everywhere and we will see that when it is applicable and when it is not. But before that this problem at least tells us that this is a very easy problem and it demonstrates that in this case it is possible to apply it between 2 points 1 and 2. So what is the specialty of this problem? Let us look into it. See for every problem there is one aspect that is how to solve a problem that is fine but there is a greater aspect how to develop a more detailed insight on what the problem is about. So we are now trying to do that. Problem is solved but it is not enough. Let us see that what insight it gives us. Try to find out what is the rate of deformation and angular velocity of this flow. So if you recall that if you want to find out the say rate of deformation epsilon dot x is equal to 0. What is that? So in this case what is the value of this identically equal to 0? Angular velocity in xy plane half of this one because each of these terms are 0 you have this also identically equal to 0. So what does it show? It shows that if there is a fluid element located in this flow field it does not have any shear deformation, it does not have any rotation. That means if its edges were originally parallel to x and y those will remain always parallel to x and y. If it is incompressible what will happen? It might stretch along say x. So it should reduce its length along y so that the volume is preserved. So it is like if the fluid element was originally like this maybe it will become once like that and it will change its configuration in such a way that angularly there is no change. Only there are changes in linear dimension but ensuring incompressibility because it also represents an incompressible flow field that you can check by checking that divergence of the velocity vector is 0. So it is an incompressible flow. So that is one important observation. So the important observation is it does not have any shear effect. Next is it does not have any angular velocity so it is like an irrotational flow because irrotational flow has no angular velocity or no vorticity so to say. Now let us try to see that what will be the equations of the stream line in this case. So we are interested to find out the equations of the stream line. It will give us even a deeper insight and we will relate it to one of the movies that we saw in our previous lecture. So if you write the equation of the stream line it is dx by u is equal to dy by v. This is the equation of the stream line. So you have dx by Ax is equal to dy by – Ay. A is not equal to 0. You can cancel that and if you integrate it you will get ln x is equal to – ln y plus a constant let us say ln k. So this gives an equation of the stream line of the form xy equal to k which is like a rectangular hyperbola. That means if you have say if you consider this as x axis and maybe this as y axis it is possible that you have your stream lines in this way. So if you have a fluid element originally like this maybe the fluid element is coming down along the stream line. I am just trying to make you recall one of the movies that we showed to you that the fluid element is coming down like this with no angular change, no rotation, no shear deformation but only the lengths of the respective edges are getting altered. So it is a case of pure linear deformation, no angular deformation. And then in such a case we have 2 things satisfied. One is there is no effective viscous effect because the viscous effect comes through what? Viscosity into the rate of shear deformation. So if the rate of shear deformation is 0 it does not matter whether viscosity is 0 or not. So in viscid effect is not always through viscosity equal to 0. It may be the rate of shear deformation equal to 0 because eventually we are interested about whether the shear stress is 0 or not. If the shear stress is 0 it does not matter whether it is 0 because of mu equal to 0 or because of rate of shear deformation equal to 0. Here it is 0 because the rate of shear deformation is 0. So it does not have any effect of viscous shear, it does not have any effect of rotationality. So it is effectively like an inviscid and irrotational flow and for such a flow we will show later on that you can apply Bernoulli's equation between any 2 points in the flow field disregarding where they are and we will now go into a more decorous way of establishing this very important consideration. So to do that what we will do? We will leave this example and go back to the Euler's equations of motion along the different directions. So we have written the Euler's equation of motion along x which is there in the board. Similar equations are there along y and z. Now what we are interested to? We are interested to write or to find what is the difference in pressure between 2 points. So let us say that we have 2 points 1 and 2 which are quite close so that they are connected by a position vector dL. dL is given by dxi plus dyj plus dzk. So this is the position vector that we are looking for. What is our interest to find out? What is the difference in pressure between points 1 to 2? So whatever we did in the previous problem of bit more informally we will now try to generalize it for a very general case that what happens in that case. To do that we will note that if you want to find out the difference in pressure between the 2 points. Here pressure is a function of what? x, y and z. So you can write this as the sum of these 3 partial derivative terms. Each of these terms we can substitute from each component of the equation of motion. So the first term you can substitute from the x component of the equation of motion which is written below. So let us write that so this will be – of so this you are writing now the plus of this one. So that means this term will become – the right hand side. So that will become – of rho then plus a body force. So plus rho dx dx with dx multiplication will come separately. We are just isolating the dp dx term. So we are not writing the dx together with this. So if you also consider the dx term together with this then it will be the entire thing multiplied by dx. Now we will try to write it in a compact form because like it is possible to utilize some of the very well known identities of vector calculus to simplify it. So what we will do is we will write this particular term in a vector calculus notation. So we can write this as v v dot with gradient of u right. So then you will get these terms. Keeping that in mind that other terms will also give similar expressions like what will change for the second term in place of this u it will be v in place of this u it will be v in place of bx it will be by like that. So it is very very analogous and we can write the general expression for dp as – rho. Now we will collect all the terms. We will keep all the terms of similar type together. This is one term. Then next we will write that acceleration term that is the convective component of the acceleration term. This is the temporal component. Then – rho and the body force term okay. So these 3 types of terms are there. We will just for the writing convenience we will call it term 1 and there is a logic behind that these terms are containing expressions of similar nature. So we can simplify them in groups. Let us write or let us try to simplify terms 1, 2 and 3 separately. We will do that keeping in mind that the term 1 is the transient term and when we are simplifying we will be keeping in mind that we will be utilizing the vector dl which connects the 2 points which are close to each other. So the term 1 is – rho. Now can you write it in terms of the 2 vectors v and dl. Remember v has components uvw, dl has components dx, dy, dz. So if you write for example like this dot with dl then that expression and this is the same. It is a dot product of this with this. Of course you have a partial derivative of that one. So it is just writing the same expression in a vector form. Let us write the term 2. How do you write the term 2? – rho. Again let us try to write using the vectors v dot. Let us check whether this is alright or not. See at the end we have to keep in mind that this is a scalar term. So first of all whatever is a vector operator it should give back a scalar. So you have one dot product and the dot product and the product of that is expected to give back a scalar. You can just check. Let us check that. So you can write this as v dot del then edy as wdz. So it becomes of the same form as that given by the term 2. Now it is possible to simplify the v dot del v using a vector identity. What is that? So v dot del v is equal to one important thing we will see that whether the bracket is to be put here or after the v. I am still this is not complete. So let us tentatively write it. This is a very well known vector identity. Now you see that what is this v dot v is a scalar. The gradient operator operating on it makes it a vector and this is very clear this is a vector. So this should be a vector. So when you have v dot del this is a scalar but this being a vector it keeps it a vector. So whenever you write an identity these are certain common sense things that you should check because depending on what you operate the same thing may look may become a scalar and vector very easily depending on how you put your cross products and the dot products. Now why we are putting in this particular form is because here you get the vorticity vector and we were finding out that the condition of rotationality or irrotationality has some influence on the pressure difference between the points and this vector solely is responsible for whether it is rotational or irrotational. So we will put that simplification here. We will put this as –rho in place of the curl of the velocity vector we will write the vorticity then dot del. For the term 3 what we will assume it is again a very general term but we will assume that the gravity is the only body force which acts along the negative z direction as we considered in the problem that we discussed just before this. So what we will assume that Bx is 0, By is 0 and Bz is –g because that is the common thing that we encountered in many problems but if there are other components of body force you know that how to simplify like you can just put the corresponding components here. So then that will become term 3 will just become –rho g dz since it has just only one scalar component it is not useful to write it in a general vector form it will not give us back many things. So dp is the sum of term 1, term 2 and term 3. We can simplify the term 2 and term 3 further. Let us try to simplify the term 2 one more step. So –rho let us now consider the dot product of this with dl. So half what is v dot v, v dot v is v square where this capital V is the resultant velocity that we are writing this dot with this one sorry that is the first term and you also have a term plus rho v cross vorticity vector dot dl. You can recognize that it is like a scalar triple product of 3 vectors like a dot b cross c okay. So we will keep that simplification for a moment and just consider the first term. So what does the first term look like that is the first term I mean first term of the term 2 and then rho. You can clearly see that the first term of the term 2 will become what? It is like it will become d dx of v square it is the sum of the 3 partial derivatives we will give the total one. So this will become at the end the simplified form – half rho d of v square. So this is like not d dx just the total d. So this is partial derivative with respect to x into dx. This is partial derivative of y into dy and that with respect to z into dz. So that is given the total d plus the whatever term that is remaining. Now let us put back all the terms together in the equation. So what is our equation? Our equation is term 1 plus term 2 plus term 3 is equal to 0 that means – rho that is the term 1. So let us sorry dp not 0 it was dp. Then term 2 in place of term 2 we will write as half rho dv square plus rho maybe let us write dl dot v cross zeta that is the term 2 and term 3 is – rho g dz is equal to dp. This is a compact form and it is possible to simplify it even further based on certain special cases. So what special cases we will be interested in let us see. So what special cases maybe let us take all the terms in the same side. So you have dp plus half rho dv square plus rho g dz up to this you can find some similarity with the Bernoulli's type of equation that you have encountered earlier. But you are getting also some 2 extra terms. Let us write those extra terms. So plus then – rho this equal to 0 because these 2 terms are like beyond what you have encountered many times we will try to put more attention to the last 2 terms. We will the first important attention on the last term because this particular term in a case when it is a steady flow this trivially goes away. So there is no big controversy or there is no big uncertainty in that that is quite understandable. But the last term there are many possibilities when the last term can become 0. What are the cases? So if you just write it in a determinant form when you are having such a scalar triple product you can write it in terms of determinants where each rho of the determinant will represent the components of the vectors taken in the particular order. So you have like dx for dl you have dx dy dz for v you have uvw for the vorticity you have okay. Now let us consider a case when this say rho1 just look it into mathematically say rho1 is a scalar multiple of the rho2 when it is possible then the direction of dl and the direction of v are the same. Then one will just be the scalar multiple of the other because direction wise they are representing vectors oriented identically. So when that is possible what is the special dl for which that is true if it is located along a streamline. So if we consider this as like term a so we will identify certain cases when a becomes 0. So a equal to 0 when certain cases one is dl is along streamline. Let us call the streamline direction as ds s for stream wise coordinate. When that is the case we do not care whether it is an irrotational flow or not it does not matter whether it has components non-zero components of the vorticity vector. Yes there is nothing called as streamline flow first you have to understand there is a streamline in a flow there is nothing called streamline flow okay next this is what this is the length this is the line element that you are considering this is the component these are the components of the velocity vector. What is the definition of a streamline such that tangent to the streamline at any point represents the direction of the velocity vector. So tangent is this direction dl a small elemental direction and this is the velocity vector direction. So if they are located in the same direction that means they are parallel vectors that is the definition of the streamline it is nothing extra. So dl is if dl is located along a streamline then we do not care whether it is a rotational or irrotational flow but if it is not then if the vorticity vector is identically equal to 0 then a will become 0 no matter whatever is like no matter whether dl is located along the streamline or not. So vorticity vector is a null vector this is irrotational flow. So you can clearly see that if it is a steady and irrotational flow these 2 terms go away and then sum of these 3 is 0 that means if you integrate that the integration will give a constant of integration and that is what we actually saw in the example the problem that we discussed before going through this derivation. There is a third case there could be many such cases but a third case say you have the v cross this vorticity vector is perpendicular to dl these 2 cases are more common cases that you encounter this is not a very common case you encounter but this mathematically you cannot rule out. You have a vorticity vector you have a velocity vector you can find the cross product and take an element in a direction which is oriented along that cross product and then if you take such an element then for such an element also for steady flow it will appear that the Bernoulli type of equation is valid. So this is not a Bernoulli type of equation this is in fact one step before that where we do not make any explicit assumption on how the rho or the density varies. So this is still the Euler equation of motion. So this is more general way of writing the Euler equation of motion where you are considering all the individual components and trying to write that in a vector form but at least we can understand that this term becomes 0 under what cases. So let us say that we are considering one such case let us say that we take an example we are considering along a streamline that is what we mean by along a streamline that means we are interested to find out these changes. So see this relates what? This relates change in pressure, change in velocity, change in elevation with respect to a change in position vector from 0.1 to 0.2. So when we are considering along a streamline that means we are interested to evaluate that change by moving along a streamline. So never consider something like a streamline flow again I am repeating there is nothing called a streamline flow. In flow there are streamlines but it is not a streamline flow. So when you have a streamline we are looking for the difference in like these variables along a streamline. So when you have along a streamline and let us say steady flow as the first example this example 1. So then what you have then a term will become 0 this term because of steadiness will become 0. So you have dp plus half rho dv square plus rho g dg equal to 0. This is known as Euler equation of motion along a streamline. Now this is valid both for compressible as well as incompressible flows you are not yet committed of how the density changes. So now we are interested to see that how the density changes. To do that we will let us say we will write it in this form dp by rho plus half dv square plus rho g dg and try to integrate it. So we will try to integrate it the rho will not be there because rho we have already divided by rho. So when we try to integrate it what are the points over which we are integrating? We are integrating with respect to 2.1 and 2 which are located on the same streamline because we have considered along a streamline that is we are considering this particular case which has made the term a equal to 0. So when we do that this equal to 0 that is still valid for any type of flow compressible or incompressible. Now you make an assumption that rho is a constant assume that is a special case of an incompressible flow. So then what you can write? You can take the rho out of the derivatives. So you can write p2 minus p1 by rho plus half v2 square minus v1 square plus g into z2 minus z1 is equal to 0. This is nothing but the Bernoulli's equation that is p1 by rho plus v1 square by 2 plus g z1 is equal to p2 by rho plus v2 square by 2 plus g z2. So it is like the it is in fact the Bernoulli's equation. Now you tell that what are the assumptions that we followed in deriving this. So this is the Bernoulli's equation. We will come into the physical significance of this Bernoulli's equation in the next lecture. But let us at least try to identify that what are the assumptions that we utilize to derive this. So what are the assumptions? So first start with the most basic assumption. When we wrote the equation of motion what we assume? Okay only gravity is the only body force it is okay but it is like it is very very explicit. What is not so explicit is inviscid flow. So inviscid flow is very important. Steady flow density is constant. It is a special case of incompressible flow not irrotational flow. We have not taken this condition 3. When we take irrotational flow we get a more freedom then we need not be restricted along a streamline. But when we are considering along a streamline then it need not be irrotational. If it is irrotational fine if it is not irrotational still okay. So along a streamline that is what we considered in this example. And so these are the 4 ones that these are the 4 assumptions that we have considered in deriving this. Now these are the assumptions that we commonly use because commonly we utilize the Bernoulli's equation along a streamline. At the same time we must understand that these are not always the cases. Inviscid flow is the most important thing. Now can you tell that if you are thinking about Euler's equation along a streamline out of these which assumption is not necessary. Say the Euler's equation of motion along a streamline density equal to constant is not necessary. So density equal to constant is the additional assumption beyond the Euler's equation. So after you make that assumption you have to also keep in mind that I would say the most important assumption is inviscid flow because many times we tend to apply the Bernoulli's equation in cases when viscous effects are very much present. Maybe many times you have solved such problems in your earlier high school exercise problems to solve like to get the velocity pressure and so on. We will see that that is not fundamentally correct. In some cases you can get rid of that and still get some qualitative picture. We will see that when and when not but fundamentally it has to be inviscid flow. Steady flow is for this version of the Bernoulli's equation but you can also have an unsteady version of the Bernoulli's equation that we will see later on maybe in the next class that where we retain this term and we can write at Bernoulli's equation by considering even the steady flow along a streamline, unsteady flow along a streamline. So only for this version it is steady flow and that is the standard Bernoulli's equation but we also have unsteady Bernoulli's equation. So for unsteady Bernoulli's equation the steady flow assumption is not required. Rho equal to constant is always required because you are taking rho equal to constant and taking out of the integral and along a streamline is required for this special case when you are not bothered about whether it is irrotational or not. If it is irrotational then this need not be the case. So maybe relaxed for irrotational flow. So what is the summary? The summary is if it is an irrotational flow and other conditions are satisfied that inviscid steady and rho equal to constant you can write p by rho plus v square by 2 plus gz is constant need not always be along a streamline. So this is constant no matter whether you are considering the points 1 and 2 anywhere in the flow field that is very important. So points 1 and 2 may be located anywhere in the flow field still this equation is satisfied if it is an irrotational flow. If it is not an irrotational flow then 1 and 2 have to be located along the same streamline. So these are very very important fundamental assumptions that go behind the Bernoulli's equation. We will stop here today. We will continue again in the next class. Thank you.