Derivation and Discussion of Bernoulli's equation

Okay, today I'm gonna talk about the Bernoulli's equation. This is one of the most commonly used equations in fluid dynamics. Okay, but I would like you to be very careful about this equation. The reason is that it's a very simple equation, so we tend to overuse it by not realizing the underlying assumptions that we need to check. Okay, be careful about this equation. Alright, so what I did in the conservation of energy equation is I picked a, let's say that I have like a bend, right? And what I did was I picked the control volume to be this entire bend, right? So now I'm gonna take a different angle to it. Okay. So, but before doing that, I need to introduce you to a concept. Okay. And we do discuss this concept in much more detail in the upcoming videos. Okay, and what is this is? It's called streamline. First off the bat, streamline is rarely a line. Okay. It's more like a curve, but that's okay. It's not my definition, right? That's why I say this is a line which is rarely or a curve, okay, tangent to the flow velocity. Okay. So it's actually a curve tangent to the flow velocity. Okay, and one more thing that I need to be careful about is streamlines do not cross each other. The reason for this, streamlines do not cross each other, is the streamlines are obtained being tangent to the velocity. So for that reason, there will be no crossing of the streamlines among the flow. So that's something note. Okay. Okay, so it's a line or a curve tangent to the flow velocity. So in this, let’s say, a bend, the streamline is gonna look like this. Okay. If this is laminar, right, so it's gonna look like this. And there will be infinite number of streamlines. Okay. And if I have, let's say that I have a velocity here, it's like this, the velocity vectors are tangent to this. That's how we obtain the streamline. Okay. So this is a useful terminology, okay, that we use in analysis of the question. Now I'm gonna pick the control volume to be right around the streamline. Okay. Before, I was taking for the entire large bend. Now I'm gonna focus to be this small, let's call it cylinder kind of shape, like a bended cylinder. Okay. Now, and if I look at this case, what I'm gonna obtain is I'm gonna have a one inlet, one outlet. Okay. One inlet being that I'm picking a one streamline. That's why it's a one inlet, one outlet. I don't have an exit from these sides over here or over here. I don't have any exits because streamlines do not cross each other. Okay. So for that reason, this will be a good system and an easy system to analyze. It's a one inlet, one outlet system. Okay. So if I write my conservation of energy for this system, and there's a video for that, so you may want to watch this. P over rho at the exit, we write this way, V square over 2 plus gz plus P over rho at the inlet, will be equal to; we call this l minus w. Right? Now I'm gonna assume a few more things. Okay. To get us going. So I'm going to assume these additional things: assume density is constant. Okay. No work, no loss. So when I do these three assumptions, we need to be careful. Okay. Let's first talk about the, let’s start with this; no loss. Okay. Over here, I have a loss. So when I say no loss, if you think about the real-life applications, I have a tank of gas, I start from school, I'm going to home, and at the end I still have the same amount of gas. Well, that's not quite realistic, right, for this large or grand scale applications. It may be okay for some cases. Okay. But we are assuming, please note that this we are assuming loss to be zero. Okay. So the second thing that we assume is no work. Well, it really depends on the question. In this particular case, for instance, in a bend question, I don't have any of those, so that's actually quite fine. But I get rid of those. So now let's take a look at what happens. Okay. So simply, my V square over 2 plus gz plus P over Rho at the exit minus V square over 2 plus gz plus P over Rho at the inlet turns out to be the same. So if this is the same, what I'm gonna do is I'm gonna move one of them to the other side of the equation, and basically what I'm gonna get is: this V square over 2 plus gz plus P over Rho at the exit is equal to V square over 2 plus gz plus P over Rho at the inlet. Right? So basically what I'm saying is, this is like the energy term. Right? This is the energy term, and I'm saying that the exit and inlet, they have the same magnitude. Okay, hang on a second. Well, let's look at my control volume. I said this is inlet, I said this is exit. What does prevent me from saying that this is the inlet, this is the exit? Nothing. I can, right? It's up to me to choose my control volume. So my point is, I can have my inlet here, here, here, here, here, here, everywhere. And same thing for exit over anywhere. So the point is, what happens is, as this quantity is constant at the inlet and exit, I can simply say that this whole thing is constant this on this entire streamline. This parenthesis is constant on this entire streamline. That's what the Bernoulli's equation says. Okay. It says that this V square over 2 plus gz plus P over Rho is equal to a constant. Because basically I'm saying that this is equal to a constant. Right? Now, it's important to note that this is along any one streamline. It's not across streamlines. I'm on a particular streamline. So this is the Bernoulli's equation, okay. And the C varies from streamline to streamline, okay. So for instance, C is one, two, five, ten, among different streamlines. Okay? That's how we quantify different streamline. Approach that I take was from the originated from the conservation of energy equations. If you look at different resources, you may see that this also can be derived from Newton's Second Law, which is the conservation of momentum, if you remember. So that's quite alright. I want you to show, I want you to understand that. Let's recap the assumptions that we really need to be careful about, okay. The first one is that this is only applicable for any two points on a single streamline. Okay. Two, we said that no work, right? We said that no loss, right? And if you remember the origination of how I derived this conservation of energy formula, we started with the steady flow. Okay. And I also take the density is equal to a constant. Again, this may be a good assumption for liquids and not be so good of an assumption if I have my thermodynamic state changing significantly. So it really depends, okay. And also I have inviscid flow. So viscous effects are neglected. So note that I have a bunch of these assumptions and I really need to be careful about using all these and double-checking that they're applicable to my case before going ahead and using the streamline terminology, okay. One more thing before I close up this video is that the most common version of this Bernoulli's equation is actually not one that I write, okay. The most common version that I personally don't prefer too much, but it's worth noting, is that I can divide both sides of this equation by g. Okay? So can I do that? Yeah, why not, right? So, because C is a constant, if I divide this constant with another constant, I obtain a different constant. So I can say for instance D is equal to C over g, right? Or some other letter, right? And what I get here is, it's gonna be V square over 2g plus z plus P over Rho g. And we have a special name for Rho g, that is the specific weight, okay? So I'll write that down. Divide both sides by g (gravity), okay. And now what I'm gonna obtain is, I'm gonna get V square over 2g plus z plus P over Rho times g, which is a specific weight, will be equal to C over g, and I'm gonna call this a D, which is a constant, right? So, and we have special names for that. So this is called the velocity head, this is called the elevation head, and this is called pressure head. You can imagine, right? Pressure head. So why do we do this? Well, let's look at the units of this, okay? What is the unit of left-hand side? Well, one thing is, the unit of this must be equal to unit of this, must be equal to unit of that. Because I cannot simply go ahead and add Newtons plus Watts plus, I don't know, length. Right? So in this particular case, look at the unit of this, this is length. So this is in meters, this is meters, this is meters. How we see the unit of a constant, but it's still a unit of it is meters as well, right?