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## MCAT

### Course: MCAT > Unit 8

Lesson 7: Fluids in motion- Fluids in motion questions
- Volume flow rate and equation of continuity
- Bernoulli's equation derivation part 1
- Bernoulli's equation derivation part 2
- Finding fluid speed exiting hole
- More on finding fluid speed from hole
- Finding flow rate from Bernoulli's equation
- Viscosity and Poiseuille flow
- Turbulence at high velocities and Reynold's number
- Surface Tension and Adhesion
- Venturi effect and Pitot tubes
- Two circulations in the body
- Arteries vs. veins - what's the difference?
- Resistance in a tube
- Putting it all together: Pressure, flow, and resistance

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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?(79 votes)
- 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)

- I thought flux was volume/time. Why did he do area times velocity to get flux?(15 votes)
- 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.(4 votes)

- Why wouldnt the external pressure of the water leaving the hole be atmospheric pressure?(5 votes)
- 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.(19 votes)

- 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)
- 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(4 votes)

- 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)- 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)

- Starting at5:50, 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?(4 votes)

- 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)
- 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 ?(2 votes)
- 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?(2 votes)- 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(2 votes)

## 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.