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.