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Finding fluid speed exiting hole

Sal finishes the example problem where liquid exits a hole in a container.  Created by Sal Khan.

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  • blobby green style avatar for user Padraigh.Docherty
    What about the atmospheric pressure outside the canister? did we assume that there was no air outside as well as on top of the liquid? if there is air outside, pressure P2 should be unlike 0 but 1 atm, shouldnt it?
    (79 votes)
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    • piceratops ultimate style avatar for user ∫∫ Greg Boyle  dG dB
      Okoye.nnenna is right about Sal changing the problem so the entire canister is at a vacuum both inside and out. However, your intuition is correct about P2 being at one atmosphere. The driving force for the water coming out of the hole would be due to the height of the column of fluid known as static head. P2 in Sal's original setup would be 1 atmosphere. This means that all of the water would blow out of the canister until the forces balanced at the hole. Essentially, the canister would become a barometer with the fluid height being pushed up by the atmospheric pressure.
      (39 votes)
  • aqualine seed style avatar for user Kevin Lin
    I thought flux was volume/time. Why did he do area times velocity to get flux?
    (13 votes)
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  • male robot hal style avatar for user The Last Guy
    Why wouldnt the external pressure of the water leaving the hole be atmospheric pressure?
    (4 votes)
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    • duskpin ultimate style avatar for user Danielle
      Because the cup is in a vacuum.

      Edit: Here's an experiment that could explain it better: Take a drinking straw, start drinking some water, but in mid-sip take the straw out of your mouth and put your finger on the tip. There's a vacuum made when your finger prevents air from entering through one end and no water leaves because the atmosphere is pressing on the water and keeping it in the "pipe", or drinking straw.

      If this cup was under a normal Earth atmosphere, with a hole that small the water would not be leaving the "pipe" because of the air pressing on it. But when the cup is in a vacuum, the water can leave because there is no pressure on either end. Gravity pulls it out.

      I hope that helped explain some of it.
      (16 votes)
  • purple pi purple style avatar for user VIHU2210
    while considering the pressure at the hole, why don't we involve the pressure exerted by the fluid coming out of the hole. won't it have some impact to the external pressyre at the hole.
    (4 votes)
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  • spunky sam blue style avatar for user Manav Mishra
    Okay... I didn't got the last part.... How did rate of flow become equal to area * velocity...?
    Rate = A2*(root 2gh)

    Anyways, Thanks for making such great videos,. U r a boon to the students here.... Thanks
    (2 votes)
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    • leafers ultimate style avatar for user Kim Jensen
      Flux can be said to be the rate at which the fluid passes through a given area. In this case that area is A2 and the flux is the volume of water passing through the hole during time t. That volume is a cylinder with area A2 and length/height L. L is equal to the distance that the water have traveled during t and since its velocity is v2, L = t · v2. Therefore the volume is V = A2 · t · v2. The flux is volume per time which in this case is R = V/t = A2 · v2 = A2 · sqrt(2gh).
      (3 votes)
  • leaf green style avatar for user Dan Brunsink
    Starting at , Sal cancels out two rho (p) on the left side of the equation, but only one rho (p) on the right side. shouldn't there still be a rho (p) on the left side of the equation?
    (1 vote)
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    • blobby green style avatar for user Brina.Goyette
      Sal cancels out one rho from each term on both sides, so that's ok. Another way to think about it is to rearrange the left side first. You start with:

      rho*g*h + rho*(v1^2)/2 = rho*(v2^2)

      There is a rho in each term on the left, so we can write:

      rho*(g*h + (v1^2)/2) = rho*(v2^2)/2

      Now, we divide both sides by rho:

      g*h + (v1^2)/2 = (v2^2)/2

      Does that help?
      (4 votes)
  • blobby green style avatar for user Shabir  Ahmed
    Why sometimes the P2 is referred as the EXTERNAL PRESSURE while in the first part of derivation it was referred as THE PRESSURE WHICH FLUID EXERTS?
    (2 votes)
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    • piceratops ultimate style avatar for user Timothy Morris
      That is not the actual reason that the pressure at the hole is zero. The true reason is that the pressure that otherwise would have existed above the hole is released by the water escaping from the hole. Basically, in terms of depth pressure, the level of the hole becomes the new surface level of the water.
      (1 vote)
  • leaf green style avatar for user pooja gopinathan
    when we talk about applications of Bernoullis theoram say for example during a storm the roof of a house flies of , we talk about increase in air velocity and decrease in pressure .How does Bernoullis equation explain this ?
    (2 votes)
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  • starky sapling style avatar for user Katie
    Even if this entire system is in a vacuum, wouldn't there still be pressure present in the system due to all the fluid behind a single fluid particle at point P2? Like the deeper you go in the sea, the more pressure there is on you because there is more water above you. So at P2, there is more fluid above this point hence more pressure? (I realise that P2=0 as this is outside pressure. I mean is there still pressure INSIDE the cup?)

    Also, correct me if I'm wrong, but if this system wasn't in a vacuum then no fluid would leave the hole because the atmospheric pressure outside exerts a force on the fluid, preventing it from leaving the cup. Is that correct?
    (2 votes)
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    • mr pink red style avatar for user Jean Rambo
      1) There is pressure inside the cup. This is given by rho*g*h, but it was taken into account on the left side of the equation where Sal said h1 = h which is why you don't see it on the right side. He chose the frame of reference as point 2 equals zero height. If he were to choose point 1 equals zero height, then that rho*g*h1 would be zero and you would have a non-zero value on that same term on the right side. So it would give you the same result.

      2)If the system was not in a vacuum, it would come out at the same velocity as in the vacuum because the pressure on the left side P1 and the right side P2 would cancel each other out. There's force acting on Area 2, but that same force is acting on Area 1, so neither one wins.
      (1 vote)
  • piceratops ultimate style avatar for user Mustaghees Butt
    Why doesn't he takes p(ro)gh1 = P1 , so the Bernolli's equation becomes 2P1 + p(ro)v1(sqr)/2 = 2P2 + p(ro)v2(sqr)/2
    (2 votes)
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Video transcript

Where we left off, we had this canister, because it had a closed top and it had a vacuum above the fluid. The fluid on top had an area of A1, and I poked a little hole with a super-small area A2. I said that the area of A2 is so small, it's 1/1,000 of area 1. Then we used the continuity equation. We said the velocity, the rate at which the surface is moving up here, V1, times area 1, the whole surface area of the liquid, has to be equal to the output velocity, which we're trying to figure out as a function of everything else times this output area. I made a mistake. I don't know if I did this in the last video, or I did this is in the mistake video. So we know that the initial top velocity times this top area is equal to the output velocity times-- instead of writing area 2, we could write area 1 over 1,000. You can get rid of the area 1 on both sides, and then you're saying that the velocity up here is equal to the rate at which the top of the surface moves down and is equal to 1/1,000 of the velocity of the liquid spurting out of this little hole. With that, we actually have the three variables for the left-hand side of Bernoulli's equation. What are the variables on the left-hand side? What is the pressure at this point where we have a hole? This is an important thin. When we talk about Bernoulli's-- let me rewrite Bernoulli's equation. It's P1 plus rho gh1 plus rho V1 squared over 2 is equal to P2 plus rho gh2 plus rho V2 squared over 2. We figured out all of these terms. Now let's figure out the things that we have to input here. What is the pressure at point two? This is the important thing. You might want to say, and this was my initial reaction, too, and that why I made a mistake, is that what 's the pressure at this depth in the fluid? That's not what Bernoulli's equation is telling us. Bernoulli's equation is telling us actually what is the external pressure at that hole. When we did the derivation, we were saying how much work-- this was kind of the work term, although we played around with it a little bit. But if we look at the water that's spurting out of the hole, it's not doing any work, because it's not actually exerting force against anything so it's not actually doing work. When we think about the pressure, the output pressure, it's not the pressure at that depth of the fluid. You should think of it as the external pressure at the hole. In this case, there is no external pressure at the hole. Let's say that if we closed the hole, then at that point, sure. The pressure would be the pressure that's being exerted by the outside of the canister to contain the water, in which case, we would end up with no velocity. The water wouldn't spurt anywhere. But now we're seeing the external pressure is zero. That's what the hole essentially creates. We're going to say that P2 is zero, so this pressure was zero, because we're in a vacuum. P2 is also zero, so both of these are zero. Remember, that's the external pressure. P1 is the external pressure to the input to the pipe, and you can view this as a pipe. I could redraw it as a pipe that looks like it has a big hole on the top, and it goes down to some level to a super-small hole like that. This would be a vacuum, and the fluid is just going in and it's spurting out of this end. Anyway, the pressure going into the pipe is zero, and we said since we put a hole, the pressure coming out of the pipe is zero, so we're doing no work. What is this term? This was the potential energy term, and we said that h1 is equal to h. We're saying that this is zero height, so now this simplifies to rho times gravity times h plus rho times V1 squared. V1, we said, is equal to this, so this is rho over 2 times V2 over 1,000 squared. I just substituted V2 over 1,000 for V1. That equals the pressure at the hole, the external pressure at the hole, which is zero, plus h2. This is h2 right here, which we said is zero. We determined that the hole was poked at height zero, so this is also zero. That equals this kinetic energy-like term. It's not exactly kinetic energy. It's rho times V squared divided by 2. One thing that we can immediately see is that we have all these rhos on both sides of the equation, so we can divide both sides by rho and get rid of all of those. Then we can multiply both sides of the equation by 2, and we get 2gh plus V2 squared over-- what's 1,000 squared-- over 1 million. That is equal to V2 squared. We could do the exact thing. We could subtract 1 over 1 million V2 squared from both sides, and we would get 0.999999 V2 squared, but let's just say for the sake of simplicity, or let's say, if this wasn't 1,000, but 1 million, and that this surface was much bigger, we see that this term becomes very, very, very small. If that hole is one millionth of the surface area, then it becomes really insignificant, so we can ignore this term because it just makes things complicated, and we're assuming this is a really, really large number, and that this hole is much smaller than the surface area of the fluid. This is like poking a hole in Hoover Dam. Hoover Dam is backing up this huge lake, and you poke a hole in it, so that hole is going to be this very small fraction of the surface area of the fluid. You can only make this assumption when that output hole is much smaller than the input hole. With that said, what is the output velocity? The velocity-- you just take the square root of both sides-- is the square root of 2gh. That is the output velocity. What is the amount of liquid that flows out per second? We figured that out already. It's a column of fluid that comes out, so per second, the length of the column of fluid will be the velocity times time, and then the cross-section of that column is equal to the output area. If I wanted to know the flow coming out, the flow coming out or the flux coming out would be equal to the hole's area times the hole's output velocity. That would equal the area times the square root of 2gh. We could use that actually solve problems in the future if we had actual numbers. I only have a minute and a half left. I'll see you in the next video.