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# More on finding fluid speed from hole

Clarification of, and more thoughts on, the Bernoulli's equation example problem where liquid exits a hole in a container. Created by Sal Khan.

## Want to join the conversation?

• Does this hold true for vessels NOT in a vacuum? I'm guessing on the surface of earth, P_in would equal 1 atmosphere. And P_out would be almost the same. So both pressures would cancel each other out.
(23 votes)
• I think it is important to specify here that if both ends are open or in a vacuum Pi = Po however if one end is in a vacuum and the other is open to the atmosphere that atmospheric pressure is acting on that (hole or opening) so Po becomes 1atm .

just to be extra clear
(29 votes)
• If there is vacuum everywhere around the canister, shouldn't the liquid be boiling?
(18 votes)
• yes it would but we just pretend it doesn't. The reason Sal made it a vacuum is to simplify the calculations.
(8 votes)
• Sal says that the external pressure P2 = 0 , since it's vacuum all around. I don't get it.. Shouldn't P2 be the pressure with which the fluid comes out and not the external pressure acting at that point? Please help.
(9 votes)
• Remember, we are interested in finding the flow velocity (v at point 2), which is just out side of the hole. Therefore the pressure just outside of the hole is what we are interested in. P2 = Patm (in this case it is 0, because it is a vacuum)
(3 votes)
• So is the pressure output always the pressure exerted on the outgoing fluid and not the pressure by the outgoing fluid?.. In the previous video explaining Bernoulli equation it was given the direction of P2 pressure to be along the direction of the flow of the fluid.. I am confused.. Thanks for replying in advance..
(8 votes)
• if P1=P2=0 then why is the fluid moving out of the hole?
(1 vote)
• P1 and P2 are the pressures outside the canister. And P1=P2.

But inside the canister, there'll be some pressure and it'll be (Rho)gh. And since fluids move from high pressure to low pressure, the fluid will move outside to P2.
(6 votes)
• Why isn't the pressure from outside = 1atm??
(1 vote)
• It depends on the frame of reference. If I'm out in space looking at the earth. I see all the gasses of the atmosphere being pulled toward earth. The combined weight of the gasses at sea level translates to 1 atm of pressure. This is sometimes referred to as "absolute pressure."

Now, If I'm sitting at sea level, and measuring the ambient pressure with a bicycle pump pressure gauge, I see zero pressure. That's because when we measure pressure at sea level, we are really comparing it to the 1 atm that is already around us. We are seeing how much greater than 1 atm the pressure in the bicycle tire (or any other object that can hold compressed air) is. This is sometimes referred to as the "gauge pressure."
(4 votes)
• hi, I am not sure this is the right place to ask the following qsn...sorry in advance,
Here goes,
' Does surface tension of a fluid depend on g(9.8ms^-2) and is it affected by gravity ? '
So if i take a spoonful of fruit juice and try to form a bubble on eart it eill not happen,but would the same be possible on a zero gravity situation like moon ? ---thanks !
(2 votes)
• At the end of the video, the equation can be rearranged to solve for v=sqrt(2*g*h), which is the standard potential to kinetic energy result. Is this affected by the diameter of the hole/ pipe attached to the reservoir?
(2 votes)
• as long as area_top/area_hole is large enough (=1000 in this case), v_hole is independent from the size and shape of the hole alone

cause on the right side, we get 1000/1000 - 1/1000 = 999/1000 as a coefficient at the end of the day, which is close to 1

thus the larger the ratio of areas, the closer the coefficient would be to 1, so the smaller effect on the final equation
(1 vote)
• Why is there no pressure at P2 given there is some type of velocity and flux across the P2 hole? Could height #2 be a negative number since it is below the surface given the mass or weight of the water above? It seems that P2 should have more pressure than P1 due to the divergence field within the fluid. Just a thought...
(2 votes)
• My physics textbook uses d instead of rho. Why? What's the more recognised one?
(1 vote)
• rho is more recognised one cause d is mostly used for differentiation
(2 votes)

## Video transcript

Before we move on, I just wanted to make sure that you understood that last point that I made at the end of that last video. We said that the pressure inputting into this, that we could view this cup with a hole in it as essentially a pipe, where the opening on the top of the cup is the input to the pipe, and this little mini-hole is the output to the pipe, and we said that this is a vacuum. Let's say this is vacuum all around. I know when I drew it last time, I closed it, but we have a vacuum everywhere. Since there's a vacuum everywhere, the pressure at this point P1 is equal to zero. The point I wanted to make is because we have a hole here, the pressure at that point at P2 is also equal to zero. You can almost view it as maybe the atmospheric pressure at that point, but since we're in a vacuum, that pressure is zero. That might have been a little confusing to you, because you said, well, wait, I thought at depth, if I had a point at that same height, that I would actually have a pressure at that point of rho gh. That's true. That's completely true. You do have an innate pressure in the liquid at that point of rho gh, and actually, that's what's causing the liquid to come out. But that's actually taken care of in the potential energy part of the equation. Let me rewrite Bernoulli's equation. The input pressure plus rho g h1 plus rho V1 squared over 2 is equal to the output pressure plus rho g h2 plus PV2 squared over 2. I think you understand that this term is pretty close to zero if the rate at which the surface moves is very slow if this surface area is much bigger than this hole. It's like if you poked a hole in Hoover Dam, that whole lake is going to move down very, very slowly, like 1 trillionth of the speed at which the water's coming out at the other end, so you could ignore this term. We also defined that the hole was at zero, so the height of h2 is zero. It simplified down to the input pressure, the pressure at the top of the pipe, or at the left side of the pipe, plus rho gh1. This isn't potential energy, but this was kind of the potential energy term when we derived Bernoulli's equation, and that equals the output pressure, or the pressure at the output of the hole, at the right side of the hole, plus the kinetic energy PV2. It's the kinetic energy term, because it doesn't actually doesn't add up completely to kinetic energy, because we manipulated it. I just wanted to really make the point that is definitely zero. I think that is clear to you, because we have a vacuum up here. The pressure at that point is zero, so we can ignore that. The question is what is the pressure here? This pressure is zero, because we have a vacuum here. If I were to say that the pressure over here at this hole is equal to pgh, then I would have the situation where pgh is equal to pgh plus PV squared over 2. What does that mean? When I say that that pressure at the output of the pipe is pgh, that means that I'm applying some pressure into that hole. Essentially, that pressure I'm applying into the hole is exactly just enough offset to offset the pressure at this depth. Because of that, none of the water will move. You could imagine that if this is the hole, let's say that's the opening of the hole, and I have some water particles, or some fluid particles, let's say that these are the atoms, we're saying innately at any point that there is a pressure at this point that's equal to rho gh, but this is P2. How much pressure am I exerting on this end of the hole? If I exert rho gh at this end, then these molecules that were just about to exit the hole aren't going to exit, because they're going to get the same pressure from every direction. What we said in the last video, and I really want to-- because this is a subtle point-- is that the outside pressure, being on the outside part of the hole, is zero, and because of that, we end up-- this term is zero, and we essentially end up with that the change in the potential energy all becomes kinetic energy, which is something we're familiar with from just our kinematics and our energy equations. With that out of the way, let me do another problem. Actually, I will do that next problem in the next video, just so we have a clean cut between videos. See you soon.