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## Physics library

### Unit 9: Lesson 3

Fluid Dynamics- Volume flow rate and equation of continuity
- What is volume flow rate?
- 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
- What is Bernoulli's equation?
- Viscosity and Poiseuille flow
- Turbulence at high velocities and Reynold's number
- Venturi effect and Pitot tubes
- Surface Tension and Adhesion

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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.

## 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.