Bernoulli's equation derivation part 2

This is just a quick review of what we were doing in the last video. We had this oddly shaped pipe and the fluid coming in had input velocity V1. The pressure on the left-hand side pushing to the right is P1, and the area of this hole is A1. Everything that the same variables with the 2 on it is coming out of the pipe. What we just set up in that last video is we said by the law of conservation of energy, essentially the joules, the energy at this point in the system, or that we're putting into the system, has to be equal to the energy coming out of the system. We used that information to set up this big equation, but it's not too complicated. We figured out that the work going into the system was the input pressure times the mass of volume over some period of time divided by the density of whatever type of liquid we had, which was the potential energy. This is just typically mgh, where the mass is the mass of this column of fluid. We're saying, how much work was done over some period of time T? That's the way I would think about it. How much energy was there over some period of time T? The kinetic energy over that period of time would have been the mass of this volume of fluid times its velocity squared divided by 2. That's typical kinetic energy. Of course, that has to be equal to essentially the output energy, and so this is the output work, or how much work a column of water could do on the output side. It's an equivalent volume of water, remember that. In some period of time T, whatever volume of water this was, an equivalent volume of water-- maybe it'll be a longer cylinder now, because it's going to be going faster. So on the output side, it's this longer cylinder that we're talking about, but it's going to be the same volume and the same mass. So what we say is that the work that this column can do in that same amount of time would be the output pressure times the mass of this column divided by the density of the column-- which is the same because the density of the liquid is the same throughout-- times the mass of this column, which is the same as the mass of this column because the volume and the density hasn't changed, so they're the same mass. Now, this column has more potential energy. It's up at h2, which I'm assuming is higher than h1. This kinetic energy is just the mass of this cylinder of fluid times its velocity squared, which is the output velocity divided by 2. This is potential energy out, and this is kinetic energy out. These equal each other. This setup is Bernoulli's equation, but let's see if we can clean it up so that we can get rid of variables that we don't have to know about. One thing that we see is that there's an m in every term, so let's get rid of them. Divide both sides of this equation by m. We get that. I don't like this density in the denominator here, so let's multiply both sides of this equation by density, and what we're left with is-- let me write this in a vibrant color. P, the input pressure, plus-- and we're multiplying everything by this rho, this density. So we have input pressure plus rho g h1, the input height, the initial height, plus rho v squared over 2. This is rho v squared over 2, and that equals-- we multiplied both sides by rho, so we get the input velocity, so that equals the pressure out plus the density times gravity times the output height. Let's make everything consistent. I wrote 2's here, so let's just say this is pressure 2, this is height 2, plus rho times the velocity squared. This is Bernoulli's equation, and it has all sorts of what I would say is fairly neat repercussions. For example, let's assume that the height stays constant, so we can ignore these middle terms. If the height is constant, if I have a higher velocity and this whole term is constant, then my pressure is going to be lower. Think about it: If height is constant, this doesn't change, but if this velocity increases, but this whole thing is constant, pressure has to decrease. Similarly, if pressure increases, then velocity is going to decrease. That might be a little unintuitive, but the other way, it makes a lot of sense. When velocity increases, this pressure is going to decrease, and that's actually what makes planes fly and all sorts of neat things happen, but we'll get more into that in a second. Let's see if we can use Bernoulli's equation to do something useful. You should memorize this, and it shouldn't be too hard to memorize. It's pressure, and then you have this potential energy term, but instead of mass, you have density. You have this kinetic energy term. It's not kinetic energy anymore, because we manipulated it some, but instead of mass, you have density. With that said, let's do a problem. I'll keep this down here, since you probably haven't memorized it as yet. Let me erase everything else. That's not how I wanted to erase it. That's how I wanted to erase it. I wanted to erase it like that without getting rid of anything useful. OK, that's good enough. And then let me clean up. Clean up all this stuff. Let's say that I have a cup. I'll just draw a cup. It's easier to draw sometimes then to draw straight lines and all of that. No, that's too dark. Do purple. I'm using a super-wide tool. I have to switch the length. OK, so that's my cup. It has some fluid. Actually, let's say it has a top to it, and I have some fluid in it. Maybe it happens to be red. We haven't been dealing with red fluids as yet,. Let me-- oh, I didn't want to do that. So you know there's a fluid there. And let's say that there's no air here, so this is a vacuum. Let's say that h-- we don't know what units are, but let's say h meters below the surface of the fluid. This is all fluid here. I poke a hole right there, and fluid starts spurting out. My question to you is, what is this output velocity of the fluid as a function of this height? Let me tell you something else. Let's say that this hole is so small, let's call the area of that hole A2, and let's say that the surface area of the water is A1. Let's say that hole is so small that the surface area the water-- let's say that A2 if equal to 1/1,000 of A1. This is a small hole relative to the surface area of this cup. With that said, let's see what we can do about figuring out the velocity coming out. Bernoulli's equation tells us that the input pressure plus the input potential energy plus the input kinetic energy is equal to the output, et cetera. So what is the input pressure? Well, the input pressure, the pressure at this point, there's no air or no fluid above it, so the pressure at that point is zero. What is the input height? Let's just assume that the hole is done at height 0, h equals 0, so the input height h1 is just h. If this is 0, then this height right here is h. What is the input velocity? We know from the continuity equation, or whatever that thing was called, that the input velocity times the input area is equal to the output velocity times the output area. We also know that the output area is equal to 1/1,000 of the input area-- and this is area 2-- so we know that the input velocity times area 1 is equal to the output velocity times 1/1,000 of area 1. We could say area 1 over 1,000 and divide both sides by area 1. We know that the input velocity is equal to V2 over 1,000, so that's good to know. These are the three inputs into the left-hand side of Bernoulli's equation. What's on the right-hand side of Bernoulli's equation? What's P2? What's the pressure at this point? Oh, I just ran out of time. I'll continue this into the next video.