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# Finding fluid speed exiting hole

Explore the principles of fluid dynamics with Bernoulli's equation as Sal finishes the example problem where liquid exits a hole in a container. Understand how fluid velocity and pressure interact in systems like a canister with a small hole. Discover why the external pressure at the hole matters and how the output velocity is calculated. Created by Sal Khan.

## Want to join the conversation?

• 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?
• 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.
• I thought flux was volume/time. Why did he do area times velocity to get flux?
• If you have a cross-sectional area A and a fluid with velocity v moving through it, then every second a fluid volume of V/t = A*v will flow through A.
• Why wouldnt the external pressure of the water leaving the hole be atmospheric pressure?
• 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.
• 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.
• you are right if you are talking about a point in the fluid . but the point we are talking about is outside the fluid so it has no thing to do with p(rho)*g*h , the pressure outside is the force over the unit area which in this problem is 0

hope that's clear
• 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
• 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).
• 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)
• 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?
• 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?
• 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)
• 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 ?
• High velocity means low pressure. Low pressure on top of the roof and normal pressure on the bottom equals a lifting force on the roof.
(1 vote)
• 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?
• 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)
• 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