Bernoulli's equation derivation part 1

Let's say we have a pipe again-- this is the opening-- and we have fluid going through it. The fluid is going with a velocity of v1, the pressure entering the pipe is P1, and then the area of this opening of the pipe is A1. It could even go up, and the other end is actually even smaller. The fluid-- the liquid-- is exiting the pipe with velocity v2, the pressure that it exerts as it goes out. If there was a membrane on the outside, how much pressure would it exert on it as it pushes it out on the adjacent water is P2, and the area of the smaller opening-- it doesn't have to be smaller-- is A2. Let's say that this opening is at a height, on average, of h1, and the water exiting this opening is on average at a height of h2. We won't worry too much about the differential between the top of the pipe and the bottom of the pipe-- we'll assume that these h's are much bigger relative to the size of the pipe. With that set up-- and remember, there's fluid going through this thing-- let's go back to what keeps showing up, which is the law of conservation of energy, which is in any closed system, the amount of energy that you put into something is equal to the amount of energy that you get out. So energy in is equal to energy out. What's the energy that you put into a system, or that the system starts off with at this end? It's the work that you input plus the potential energy at that point of the system, plus the kinetic energy at that point of the system. Then we know from the conservation of energy that that has to equal the output work plus the output potential energy plus the output kinetic energy. A lot of times in the past, we've just said that the potential energy input plus the kinetic energy input is equal to the potential energy output plus the kinetic energy output, but the initial energy in the system can also be done by work. So we just added work to this equation that says that the energy in is equal to the energy out. With that information, let's see if we can do anything interesting with this pipe that I've drawn. So what's the work that's being put into this system? Work is force times distance, so let's just focus on this. It's the force in times the distance in, and so over a period of time, t, what has been done? We learned in the last video that over a period of time, t, the fluid here might have moved this far. What is this distance? This distance is the input velocity times whatever amount of time we're dealing with, so T-- so that's the distance. What's the force? The force is just pressure times area, and we can figure that out by just dividing force by, area and then multiply by area, so we get input force divided by area input, times area input. It's divided and multiplied by the same number-- that's pressure, that's area. It's equal to the input distance over that amount of time, and that's velocity times time, so the work input is equal to the input pressure times the input area times input velocity times time. What is this area times velocity times time, times this distance? That's the volume of fluid that flowed in over that amount of time. So that equals the volume of fluid over that period of time, so we could call that volume in, or volume i-- that's the input volume. We know that density is just mass per volume, or that volume times density is equal to mass, or we know that volume is equal to mass divided by density. The work that I'm putting into the system-- I know I'm doing a lot of crazy things, but it'll make sense so far-- is equal to the input pressure times the amount of volume of fluid that moved over that period of time. That volume of fluid is equal to the mass of the fluid that went in at that period of time, and we'll call that the input mass, divided by the density. Hopefully, that makes a little bit of sense. As we know, the input volume is going to be equal to the output volume, so the input mass-- because the density doesn't change-- is equal to the output mass, so we don't have to write an input and output for the mass. The mass is going to be constant; in any given amount of time, the mass that enters the system will be equivalent to the mass that exits the system. There we go: we have an expression, an interesting expression, for the work being put into the system. What is the potential energy of the system on the left-hand side? The potential energy of the system is going to be equal to that same mass of fluid that I talked about times gravity times this input height-- the initial height-- times h1. The initial kinetic energy of the fluid equals the mass of the fluid-- this mass right here, of that same cylinder volume that I keep pointing to-- times the velocity of the fluid squared. We remember this from kinetic energy divided by 2. So what's the total energy at this point in the system over this period of time? How much energy has gone into the system? It's going to be the work done, which is the input pressure-- I'm running out of space, so let me erase all of this. I'll probably have to run out of time, too, but that's OK-- it's better than being confused. Back to what we were doing. So, the total energy going into the system is the work being done into the system, and I rewrote it in this format, which is the input pressure-- we'll call that P1-- times the mass divided by the density of the liquid, whatever it is. This is work in plus-- and what's the potential energy? I wrote it right here-- that's just mgh, where m is the mass of this volume of fluid, h is its average height, and you could almost think of how high the center of mass above the surface of the planet. Since we have a g here, we assume we're on Earth, so this is h1, because the height actually changes, so this is potential energy input plus the kinetic energy mv1 squared over 2. That is the kinetic energy input. We know that this has to equal the energy coming out of the system. This is going to be equal to the same thing on the output side. This is going to be equal to the work out, so that'll be the output pressure times the mass divided by the density plus the output potential energy, which will just be mg h2, plus the outbound kinetic energy, which will be mv2 squared divided by 2. I just realized I'm out of time. I will continue this in the next video. See you soon.