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Bernoulli's equation derivation part 2

This is the second of two videos where Sal derives Bernoulli's equation. In the second half of the video Sal also begins an example problem where liquid exits a hole in a container. Created by Sal Khan.

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  • orange juice squid orange style avatar for user ashwin
    HI! I'm Ashwin from India. The service you provide for clarifying doubts is great!!
    I have a question on the application of Bernoulli's principle. When air flows over a foil/paper, why is it that the air flowing on the upper surface of the foil(though has a higher velocity) has a lower pressure than the air flowing on the lower surface of the foil?
    Thank you.
    (22 votes)
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    • blobby green style avatar for user Khashon Haselrig
      It is not because they have to cover the same area in the same time. In fact the air on the top of the wing will reach the trailing(back edge) of the wing FIRST. NASA talks about the incorrect theories floating around. http://www.grc.nasa.gov/WWW/k-12/airplane/wrong3.html


      The way I understand it is(you will want to talk to someone who actually does experiments in a wind tunnel, not just a lecturer and definitely not a pilot as they are the worst offenders), the air impacts the front of the wing imparting a velocity to the fluid both upwards and downwards, creating low pressure directly ahead of the wing and perpendicular to the direction the air at the leading edge of the wing was just accelerated. Now the air moves over the wing through the area of lower pressure that was created by the leading edge. The movement of the air over the wing surfaces again causes low pressure over the top AND bottom of the wing. However the bottom of the wing does not allow air to flow past easily if it is inclined and so the air tends to create high pressure under the wing. Above the wing the air wont be stopped and is free to continue accelerating to the trailing edge of the wing. Because of its high speed the air at the top of the wing will flow downward off the trailing edge because the wing is not there to support it and it is traveling with a lower pressure and higher velocity than the air around it. The high pressure air under the wing will try to flow on top of it but can't get over the trailing edge because sucking is different from blowing(try to suck out a candle)and it can't turn the sharp angle. It can get over the tips of the wings though and does cause drag.

      I am not an expert but that is what I understood someone who worked for NASA to be saying. If you want to know definitively ask someone who DOES ACTUAL EXPERIMENTS. And remember air is a fluid acting and reacting continuously with itself to create vortexes.


      Also I thin the magnus effect has to do with rotation and friction slowing a fluid down and creating a difference in pressure that way, and not how an airfoil would generate lift.


      Oh and the more general idea this falls under is "Bernouli's principle".
      (13 votes)
  • blobby green style avatar for user Festavarian
    In fluids Part 2 we learned that Pressure-in equals Pressure-out. Doesn't Bernoulli's equation contradict this, where there is lower pressure in a zone of restricted diameter?
    (8 votes)
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    • leaf green style avatar for user Tamika
      Good question! I think the difference is that Pascal's principle (pressure in = pressure out) applies to "confined fluids," that is, static fluids. Bernoulli's equation applies to fluid dynamics, or fluids in motion.
      (19 votes)
  • starky ultimate style avatar for user fDash
    i still don't understand why pressure is low when velocity is high ? i get the thing about keeping height fixed but how did we come up with that inverse relation b/w vel. and pressure?
    (11 votes)
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    • duskpin ultimate style avatar for user anita.o
      I like to think about it like this: if the velocity is high, this means that the fluid is moving very fast right? If the fluid is moving fast, the fluid particles have less time to collide into the walls of the tube, so they exert less force on the tube. Pressure = (Force/Area), so less force exerted on the walls of the tube means less pressure for fast moving fluids.

      Of course the other factors come into play, but for quick thinking I hope this helps :)
      (11 votes)
  • starky ultimate style avatar for user Alex the Chrononaut
    At , wouldn't the liquid just stay inside because the hole is the only way for air to get in?
    (8 votes)
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  • leaf green style avatar for user Kris Kalavantavanich
    Is it just me, but why is the (rho)*g*h looks vaguely familiar?
    (2 votes)
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  • male robot hal style avatar for user The Last Guy
    So would an object that is moving underwater experience less pressure than the same object when it is still?
    (3 votes)
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    • leaf green style avatar for user Vinicius Taguchi
      I would imagine that moving forward would cause the object to experience force against its front (like how air slows down a ball flying through the air). This force spread over the area of the front of the object would then be pressure. The pressure the object feels from every direction should remain the same (unless for some reason the pressure on its back would decrease because of the forward motion (I don't know if this is the case). Could anyone elaborate or correct my thinking?
      (3 votes)
  • blobby green style avatar for user Kyler
    How does pressure decrease when going into a more constricted area of the pipe? If pressure is Force/Area, wouldn't having a smaller area just create an even larger pressure? Also, does the pressure increase=velocity decrease apply if there's a change in height?
    (3 votes)
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    • blobby green style avatar for user Teacher Mackenzie (UK)
      You are right, P=F/A when a force is applied over an area. In this case we are actually more concerned with conservation of ENERGY.

      OK, you need to take a look at the Bernoulli equation. See that the total energy in the system remains constant. So if there is an increase in kinetic energy (or gravitaitonal potential) then there will be a decrease elesewhere, hence the decrease in pressure P.
      (2 votes)
  • female robot grace style avatar for user Aelin
    the rho v. wade part was really funny
    (3 votes)
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  • leaf green style avatar for user Chris D
    The formula for pressure is P=rho*g*h. I have also seen P+rho*g*h=constant. I don't get this last formula. I don't get the + and the constant. If you bring (rho*g*h) to the left how can it be positive?

    You could write Bernoullli's law as P+roh*g*h+1/2rho*v^2=constant. But why is it constant?
    (2 votes)
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  • male robot hal style avatar for user Aman kumar
    Why you consider work in equation of law of conservation of mass.
    (2 votes)
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Video transcript

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.