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Current time:0:00Total duration:10:09

Bernoulli's equation derivation part 2

Video transcript

so just a quick review of what we're doing in the last video we had this oddly shaped pipe and you know I have this the fluid coming in at an input velocity v1 the pressure on the the left-hand side pushing to the right is p1 and it's velocity or I already said the boss in the area this hole is a 1 and what we and and of everything as you know the same variables with the 2 on it is coming out of the pipe and what we just set up in that last video as we've said a little by the law of conservation of energy the essentially the joules 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 and so we use 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 divided by the density of whatever type of liquid we we had the potential energy this is just typically MGH where the mass is the mass of this column of fluid right we're kind of saying how much work was done over some period of time T essentially that that's the way I would think about it how much energy was there over some period of time T and the kinetic energy over that period of time would have been the mass of this volume of a fluid times its velocity squared divided by 2 that's difficult kinetic energy right and of course that has to be equal to essentially the output energy and so this is the output work the output work or essentially how much work a column of water could do on the output side so and it's an equivalent volume of water remember that so in some period of time T whatever volume of water this was in in some period of time T an equivalent volume of water maybe it'll be a longer cylinder now and it's going to be going faster so on the output side is 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 output 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 it's 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 hasn't changed so and the density hasn't changed so they're the same mass although now this column has more potential energy right it's it's up at h2 which I'm assuming is higher than h1 and then as kinetic energy is just the mass of this of this cylinder of fluid times its velocity squared which is the output velocity divided by two so this is this is potential energy out and this is kinetic energy out and these equal each other so we this set up this is Bernoulli's equation but let's 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 so one thing that we see there's an M in every term so let's get rid of them divide both sides this equation by M dividing both sides of this equation by M we get that and let's I don't know 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 we're multiplying everything by this Rho this you know this density so we have input pressure plus Rho G H 1 the input height the initial height plus Rho V squared over 2 you'll probably expect me to say roe v wade but now this is Rho on bad joke this is Rho V squared over 2 and that equals we multiplied both sides by Rho so we get D on this is the input velocity the row the so that equals the pressure out plus the density times gravity times the output height I don't know we can well let's make everything consistent I wrote twos here so let's just miss its pressure - this is height - plus Rho times the velocity squared the output velocity squared this is Bernoulli's equation and it has all sorts of what I would say is fairly neat repercussions for example it tells us that let's assume that the height stays constant so we could ignore these middle terms if the height is constant if I have a higher velocity if I have a higher velocity and this whole term is constant then my pressure is going to be lower right think about it if let's ignore if height is constant this doesn't change but if this increases if 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 and all sorts of neat things happen but we'll get more into that into a second but let's see if we can if we can use Bernoulli's equation to do something useful and you should memorize this and it shouldn't be too hard to memorize you know it's pressure and then you have this kind of potential energy term but instead of mass you have density and you have this kind of kinetic energy term it's not getting energy anymore because we manipulated it some but instead of mass you have density so with that said let's let's do a problem I 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 but let me see if I can that's how I wanted to erase it I wanted a reason like that without getting rid of anything useful okay that's good enough and then let me clean up clean up all this stuff you know this stuff so let's say that I have a a cup a cup let me well I'll just draw a cup easier to draw sometimes in two straight lines and all of that that's too dark do purple that I'm using the super wide tool have to switch to the link okay so that's my cup it has some fluid actually let's say it has a top to it it has a top to it and I have some some fluid in it maybe it's happens to be red we haven't been dealing with the red fluids as yet let me oh I didn't want to do that so let's just you know you know there's a fluid there and let's say that there's no air here so this is a vacuum and let's say that H we don't know what units are but let's say H meters below the surface of the fluid so this is all fluid here right I poked a hole I poked a hole right there and and fluid starts spurting out fluid starts burning out and my question to you is what is this output velocity of the fluid as a function of this height and let's say let me tell you something else let's say that this hole is so small let's call the area of that hole a two and let's say that the surface area of the water is a 1 and let's say that hole is so small that the surface area of the water let's say that a two is equal to one one thousandth of a one so this is a small hole relative to the surface area of this cup right so with that said let's see what we can do about figuring out the velocity coming out well Bernoulli's equation tells us that the input pressure plus the input potential energy plus the input kinetic energy is equal to the output and etc etc etc 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 in pressure is zero what is the input height well let me let's just assume that this is that this that the hole is done at height zero H equals zero so the input height h1 is just H right if this is zero then this height right here is H and what is the input velocity well we know from the continuity equation or whatever that thing was called we know that the input velocity times the input area is equal to the output velocity times the output area output area and we also know that we also know that the output area is equal to one one thousandth of the input area right so this is area two so we know that the input velocity times area one is equal to the output velocity times one one thousandth of area one so we could say area one over a thousand divide both sides by area one so we know that the input velocity is equal to v2 over a thousand right so that's good to know so these are the three input three inputs into the left-hand side of Bernoulli's equation and what's on the right-hand side of Bernoulli's equation well what's p2 what's p2 what's the pressure at this point well oh I just ran out of time I'll continue this into the next video