 So basically, we are going to discuss here Bernoulli's principle write down Bernoulli's Principle fine, so now before we get down to the mathematics of the Bernoulli's principle Let me let us discuss few things about it first. Okay, so we have Work energy theorem when we talk about the solids Okay Similar to work energy theorem as in the same concept of conserving energy We can somehow use it in the case of fluid also Okay, but the thing is that right now the way we know work energy theorem that is work done is equal to Change in potential energy plus change in kinetic energy. It is not very comfortable to use the formula of work energy theorem as it is in the case of Fluid when it is flowing. All right, hence, we need to modify the work energy theorem So as to make it suitable to use it for the fluids All right. Now when you write in for the fluids, you need to use terms like pressure terms like Density right so in terms of these I want my work energy theorem for the fluid flow Okay, so this is what we are going to do over here. All right Now when we apply basically work energy theorem for the Bernoulli's principle There are certain assumptions also. Okay, so first we'll list down all the things related to it. So write down Bernoulli's principle is nothing but application of application of work energy theorem Streamline flow. Okay. There are other assumptions also. We'll discuss it So basically we need to modify We need to modify work energy theorem. So as to make it comfortable to use for fluids Okay, and when we talk about fluids Basically, I want I want to basically use work energy theorem in terms of What are the terms I should need? Pressure density you can use velocity of it and the height Okay, so basically instead of force. I want pressure in the equation instead of mass I want density in the equation fine So that is the thing that we have to do here now the assumptions write down these are very important Before you apply Bernoulli's theorem you need to make sure that These things are valid for that flow. So the first assumption is that it is a streamlined flow Okay, it's a streamlined flow Okay, second is that it is a non viscous flow I'm sure you guys have a little bit idea about the viscosity So basically when one laminar or one layer of the fluid move on the other layer of the fluid if they move relative to each other You can say it is like a friction between the two layers So if viscous flow is there So there is a chance that some mechanical energy gets converted into heat So the energy loss has happened. So if energy loss happens, you can't account it You know in a work magic theorem the way we know it fine So you need to include only mechanical energy you can't account for the thermal losses, right? So you need to make sure that it is a non viscous flow or At least no frictional losses are there fine then incompressible flow write down Why incompressible flow as in we have derived it like that. Okay, you'll see the derivation then you will understand. Okay, and then it It should be It should not be rotational flow. Okay, so Irrotational flow What's a rotational flow tornado you might have seen in the movies. Okay, it rotates like that So this kind of flow is not included for the Bernoulli's theorem because there are losses in those kind of flow Also, fine. So these assumptions we Have to keep in our mind and now let us see how we can derive the Bernoulli's principle. Okay So this is a very popular let's say Thing which is asked at least in your school exam a lot of you know Teachers they ask this pay attention for your school also. It is very important So what I'm going to do is I am going to consider a Section of a pipe Fine. So all of you draw with me. It's a part of a bigger pipe Fine. So the pipe goes on like this It's a part of a bigger pipe now The parameters which are known to us right for example area of cross-section a1 is known area of cross-section a2 is known The velocity over here. It is known V1 velocity over there It is known as V2 Okay. Now when you write the mechanical energy, let's talk about the potential energy What is the first thing you do? You take a level a horizontal level and then say that that is my zero potential energy, right? So we need to define that over here also So let us say that this horizontal line represents zero gravitational potential energy okay, and The heights should be given so that we can write the potential energy terms These are the two heights that are given to us fine, this is let's say height h2 and This is height h1 All right, and The pressure over here pressure inside the pipe over here is p2 pressure inside the pipe or there is P2 over here P1 over there it is P2. Okay, all of you understand this Any question in the scenario the scenario which we have taken quickly type in clear fine now Do you remember how we have used work music theorem in case of solids What we do we take two points We take two points and apply a verge theorem between the two points, right? These two points are Two points in time or two points in space what I'm trying to say is that You are using work music theorem between what is happening right now and what is going to happen later Is that the case or you are applying work music theorem? For the same time between two places How you I mean do you understand what I'm trying to ask here all of you Can you answer? Different times right different times. What we do is that we find out What is the what is happening right now and then what will happen later on and Use work music theorem between these two points in time Isn't it for example a ball is moving down the inclined plane. So right now it is at the top Later on it comes at the bottom and you have to find. Let's say velocity over there, right? So you use two points in time Okay, so remember that okay So I'll write here What we do is that we apply work energy theorem between what is Happening right now what will happen later Okay. Yeah, hold on for some time now Tell me when we apply the work energy theorem Okay, when we apply the work energy theorem The first thing what we do We select our mass on which I am applying work energy theorem Right. So where is that mass over here on which you're applying work energy theorem? All of you on which mass I'm applying work energy theorem the fluid entire fluid Or a particle of the fluid what it is Entire fluid as in the big pipe entire pipe or only the pipe which is shown to you Pipe shown to see yeah only the shown part Right. So yeah, you can apply work energy theorem for a particle also No problem with it But the way we have derived the Bernoulli's theorem our system is the fluid inside this pipe Inside this yellow pipe, which is shown here. This is a section of a bigger pipe and my system is this I'm going to study this what happens to it. Right. What is happening right now? The fluid inside it and what will happen later on With the fluid that is inside it, okay So right now, this is what is happening. This is the snapshot right now later on what will happen later on Some fluid will go inside. Let us say from this side Okay, and some fluid will go out From this side Like that Okay, now I'm going to ask you something very very important This this is let's say This part is zone one Z one This part is let's say zone two and this part is Let's say Zone three. Okay. So your mass your mass on which you are applying working with your theorem Initially was what Z 1 plus Z 2 Z 2 plus Z 3 or Z 1 plus Z 3 Where was your entire mass initially at equal to 0 your mass was what? Z 1 plus Z 2 all of you agree Z 1 plus Z 2 right Entire fluid inside it. That was your T equal to 0 right After a small time interval dt at T equal to delta T. Let's say Where your mass went now. What is your system a system become what? Z 1 plus Z 2 Z 2 plus Z 3 or what? Yes, good. So Z 2 plus Z 3. All right. So keep in your mind All right. Keep in your mind that Initially my system was inside Z 1 Z 2 zone 1 zone 2 after delta T my system shifts and It has become now Z 2 plus Z 3 because I have to trace where my mass went right when we apply work as a theorem Okay, great Now the way we know work energy theorem will write it as it is. Okay. So work energy theorem says that w is equal to U 2 minus u 1 Plus K 2 minus k 1 Do you all understand this everyone? Okay, let's say this distance This distance is delta X 1 Okay, and this distance Let's say that one is Delta X 2 Fine, so first I'll find out the work done Tell me how much is the work done when my system which was Z 1 plus Z 2 Became Z 2 plus Z 3. What is the work done by the external force other than gravity on my system? What is the external force? Tell me external forces what apart from gravity there is This pressure over here which will create force and pressure over there that will create the force So how much is the work done by these forces? We force into displacement is the work done Isn't it keep it simple the forces p 1 a 1 Okay, displacement is delta X 1 work done be positive or not this side Left hand side work done is positive or should I add negative here? Everyone force into displacement positive or negative work done Positive what about over here? What is the work done by P2? What's the work done by P2 quickly type in work done by P2 is what I'm not asking positive negative Tell me the value how much? Okay, people are saying Positive work done P2 a 2 delta X 2 is it correct is a work done positive negative It is negative guys Pressure will push it from this side like that Right pressure will try to compress Isn't it so it is a negative work forces this way displacement that way Displacement is away from the force. So that's why I work done is negative. So and this is the total work done Okay, now comes the potential energy. I'll first write down the Initial potential energy Okay, so potential energy initially is z1 plus z2 so potential energy of z1 Plus potential energy of z2 Okay, what is the potential energy of z1? It's a small zone over here. How much it is? Tell me in terms of density you can write Ro, this is my zero gravitation potential energy Ug is zero What is the mass of this? This portions masses how much oh a1 delta x1 correct. So Ro a1 Delta x1. This is mass Mg Into h1 right mg h1 all of you agree Plus potential energy of z2 Why I am not writing potential of z2. Why I am not expanding it. I'm keeping it like that only anyone Because it is going to get cancelled Right you do is what similarly tell me you do you do is potential you have z2 Plus potential you have z3 you can see at equal to delta t z2 plus x3. It has become what it is you do is potential G of z2 plus Ro a1 sorry Ro a2 Delta x2 g h2 This is my you to Getting it now. Tell me. What is my k1? kinetic energy of zone z1 plus kind of g of zone z2 This is my k1. How much it is equal to? Half mv square is kinetic energy use that keep it simple. What should I write? zone Z1's kinetic energy half Ro a1 delta x1. This is mass into v square Plus kinetic energy of zone z2. This is initial kind energy, right? Final kind energy is what? Connecting a g of zone z2 plus kind energy of zone z3 This is equal to K z2 plus half Ro a1 sorry a2 Delta x2 v2 square This was v1 square Everybody understood till now all of you have understood what is w we have found out we have found out u1 u2 k1 k2 Everybody understood all of you type in quick Now tell me is the volume of z1 Volume of z1. Is it equal to volume of z3 or not? Whatever volume of liquid goes in equals to same volume of liquid goes out That is a continuity question. Isn't it? So I can say that I can say a1 Delta x1 is equal to a2 delta x2 It comes from a1 v1 is equal to a2 v2 v1 is delta x1 by delta t So delta t get cancelled away. So a1 delta x1 is equal to a2 delta x2 now I want you to Substitute the value of w u1 u2 k1 k2 over here Use this continuity equation to simplify and tell me what you get. Okay quickly do it your own Quick so once you substitute it you're going to get this P1 plus half row v1 square Plus rho gh1 is equal to p2 plus half row v2 square Plus rho gh2 All right, so this is the modified form of work in a theorem suitable for the flow kind of scenario Okay, I hope everybody Understood over here. So let us you know Understand what are these terms imply? So clearly, you know when you look at half row v square you like oh This looks like kinetic energy rho gh you look like you know potential energy fine So basically you are adding all the energies of point one and equating it to Energies of point two. Do you all see that all the energies of point one in a way you have equated it to energies of point number two so basically You know when we talk about half Row v square It is nothing but kinetic energy per unit volume Row is m by v mass divided by volume, right? So this is kind of energy per in volume or you can say density of kinetic energy density of kinetic energy Okay, then the second term Sorry, the third term row gh This is the gravitational potential energy per unit volume Which is nothing but density of the Potential energy All right, then you have P Okay, P is a new kind of energy, which is pressure energy per unit volume Okay, pressure energy per unit volume. So this is density of the pressure energy All right, so you are nothing but what you are doing is equating the density of The energies on the left hand side to the density of the energy on the right hand side So that's another reason why liquid should not be compressible if liquid compresses density increases So you can't equate the density of the energies, right? So p1 plus half row v1 square plus row gh is nothing but density of the total energy Okay, and one more thing if you can understand properly over here that When we are applying work with the theorem for the flow kind of scenario we are applying Two points in space not two points in time. Look at the expression P1 half row u1 square rho gh1 p2 half row to v2 square plus rho gh2 Everything is happening right now Right now whatever is happening at point number one You're equating it to right now whatever happens at point number two. Okay So what is your theorem for a flow kind of scenario? You're applying it in two points in space at the same time Okay, all right few more things Talking about how you can utilize Bernoulli's theorem to Basically solve questions. First of all, there has to be a flow. Okay, it has to be laminar Non-viscous incompressible Irrotational flow. Okay, if all those things are happening then what you do is that since it is a streamline flow we need to We need to first draw draw a streamline then pick two points on it two points on it to apply Bernoulli's theorem This should be very clear Fine, and one more thing you need to understand over here is that h1 h2 they are heights Okay, they are height Why I'm emphasizing that because you know h comes in this kind of expression also variation of pressure with height or depth P a plus rho gh This edge is measured from the top. This edge is measured from the Depth sorry from the bottom. Okay, so when you use Bernoulli's theorem, you have to use edge Just like potential energy when you're finding the pressure This edge is depth Okay, so Depth and height are two different things. Okay, suppose this is you The height of your head is this from the bottom The depth of the your head is zero. It is right at the top only Okay, depth of your head is zero, but height is maximum. So I Hope you got it All right Fine, so this is the Bernoulli's theorem discussion. All right