This is just a quick review of what we were
doing in the last video.
We had this oddly shaped pipe and the fluid coming in had
input velocity V1.
The pressure on the left-hand side pushing to the right is
P1, and the area of this hole is A1.
Everything that the same variables with the 2 on it is
coming out of the pipe.
What we just set up in that last video is we said by the
law of conservation of energy, essentially the joules, the
energy at this point in the system, or that we're putting
into the system, has to be equal to the energy coming out
of the system.
We used that information to set up this big equation, but
it's not too complicated.
We figured out that the work going into the system was the
input pressure times the mass of volume over some period of
time divided by the density of whatever type of liquid we
had, which was the potential energy.
This is just typically mgh, where the mass is the mass of
this column of fluid.
We're saying, how much work was done over some
period of time T?
That's the way I would think about it.
How much energy was there over some period of time T?
The kinetic energy over that period of time would have been
the mass of this volume of fluid times its velocity
squared divided by 2.
That's typical kinetic energy.
Of course, that has to be equal to essentially the
output energy, and so this is the output work, or how much
work a column of water could do on the output side.
It's an equivalent volume of water, remember that.
In some period of time T, whatever volume of water this
was, an equivalent volume of water-- maybe it'll be a
longer cylinder now, because it's going to be going faster.
So on the output side, it's this longer cylinder that
we're talking about, but it's going to be the same volume
and the same mass.
So what we say is that the work that this column can do
in that same amount of time would be the output pressure
times the mass of this column divided by the density of the
column-- which is the same because the density of the
liquid is the same throughout-- times the mass of
this column, which is the same as the mass of this column
because the volume and the density hasn't changed, so
they're the same mass.
Now, this column has more potential energy.
It's up at h2, which I'm assuming is higher than h1.
This kinetic energy is just the mass of this cylinder of
fluid times its velocity squared, which is the output
velocity divided by 2.
This is potential energy out, and this is
kinetic energy out.
These equal each other.
This setup is Bernoulli's equation, but let's see if we
can clean it up so that we can get rid of variables that we
don't have to know about.
One thing that we see is that there's an m in every term, so
let's get rid of them.
Divide both sides of this equation by m.
We get that.
I don't like this density in the denominator here, so let's
multiply both sides of this equation by density, and what
we're left with is-- let me write this in a vibrant color.
P, the input pressure, plus-- and we're multiplying
everything by this rho, this density.
So we have input pressure plus rho g h1, the input height,
the initial height, plus rho v squared over 2.
This is rho v squared over 2, and that equals-- we
multiplied both sides by rho, so we get the input velocity,
so that equals the pressure out plus the density times
gravity times the output height.
Let's make everything consistent.
I wrote 2's here, so let's just say this is pressure 2,
this is height 2, plus rho times the velocity squared.
This is Bernoulli's equation, and it has all sorts of what I
would say is fairly neat repercussions.
For example, let's assume that the height stays constant, so
we can ignore these middle terms. If the height is
constant, if I have a higher velocity and this whole term
is constant, then my pressure is going to be lower.
Think about it: If height is constant, this doesn't change,
but if this velocity increases, but this whole
thing is constant, pressure has to decrease.
Similarly, if pressure increases, then velocity is
going to decrease.
That might be a little unintuitive, but the other
way, it makes a lot of sense.
When velocity increases, this pressure is going to decrease,
and that's actually what makes planes fly and all sorts of
neat things happen, but we'll get more
into that in a second.
Let's see if we can use Bernoulli's equation to do
something useful.
You should memorize this, and it shouldn't
be too hard to memorize.
It's pressure, and then you have this potential energy
term, but instead of mass, you have density.
You have this kinetic energy term.
It's not kinetic energy anymore, because we
manipulated it some, but instead of
mass, you have density.
With that said, let's do a problem.
I'll keep this down here, since you probably haven't
memorized it as yet.
Let me erase everything else.
That's not how I wanted to erase it.
That's how I wanted to erase it.
I wanted to erase it like that without getting rid of
anything useful.
OK, that's good enough.
And then let me clean up.
Clean up all this stuff.
Let's say that I have a cup.
I'll just draw a cup.
It's easier to draw sometimes then to draw straight lines
and all of that.
No, that's too dark.
Do purple.
I'm using a super-wide tool.
I have to switch the length.
OK, so that's my cup.
It has some fluid.
Actually, let's say it has a top to it, and I have some
fluid in it.
Maybe it happens to be red.
We haven't been dealing with red fluids as yet,.
Let me-- oh, I didn't want to do that.
So you know there's a fluid there.
And let's say that there's no air here, so this is a vacuum.
Let's say that h-- we don't know what units are, but let's
say h meters below the surface of the fluid.
This is all fluid here.
I poke a hole right there, and fluid starts spurting out.
My question to you is, what is this output velocity of the
fluid as a function of this height?
Let me tell you something else.
Let's say that this hole is so small, let's call the area of
that hole A2, and let's say that the surface area of the
water is A1.
Let's say that hole is so small that the surface area
the water-- let's say that A2 if equal to 1/1,000 of A1.
This is a small hole relative to the surface
area of this cup.
With that said, let's see what we can do about figuring out
the velocity coming out.
Bernoulli's equation tells us that the input pressure plus
the input potential energy plus the input kinetic energy
is equal to the output, et cetera.
So what is the input pressure?
Well, the input pressure, the pressure at this point,
there's no air or no fluid above it, so the pressure at
that point is zero.
What is the input height?
Let's just assume that the hole is done at height 0, h
equals 0, so the input height h1 is just h.
If this is 0, then this height right here is h.
What is the input velocity?
We know from the continuity equation, or whatever that
thing was called, that the input velocity times the input
area is equal to the output velocity
times the output area.
We also know that the output area is equal to 1/1,000 of
the input area-- and this is area 2-- so we know that the
input velocity times area 1 is equal to the output velocity
times 1/1,000 of area 1.
We could say area 1 over 1,000 and divide both
sides by area 1.
We know that the input velocity is equal to V2 over
1,000, so that's good to know.
These are the three inputs into the left-hand side of
Bernoulli's equation.
What's on the right-hand side of Bernoulli's equation?
What's P2?
What's the pressure at this point?
Oh, I just ran out of time.
I'll continue this into the next video.