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This is just a quick review
of what we were

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doing in the last video.

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We had this oddly shaped pipe
and the fluid coming in had

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input velocity V1.

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The pressure on the left-hand
side pushing to the right is

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P1, and the area of
this hole is A1.

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Everything that the same
variables with the 2 on it is

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coming out of the pipe.

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What we just set up in that last
video is we said by the

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law of conservation of energy,
essentially the joules, the

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energy at this point in the
system, or that we're putting

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into the system, has to be equal
to the energy coming out

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of the system.

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We used that information to set
up this big equation, but

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it's not too complicated.

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We figured out that the work
going into the system was the

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input pressure times the mass of
volume over some period of

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time divided by the density of
whatever type of liquid we

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had, which was the
potential energy.

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This is just typically mgh,
where the mass is the mass of

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this column of fluid.

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We're saying, how much work
was done over some

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period of time T?

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That's the way I would
think about it.

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How much energy was there over
some period of time T?

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The kinetic energy over that
period of time would have been

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the mass of this volume of
fluid times its velocity

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squared divided by 2.

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That's typical kinetic energy.

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Of course, that has to be
equal to essentially the

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output energy, and so this is
the output work, or how much

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work a column of water could
do on the output side.

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It's an equivalent volume
of water, remember that.

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In some period of time T,
whatever volume of water this

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was, an equivalent volume of
water-- maybe it'll be a

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longer cylinder now, because
it's going to be going faster.

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So on the output side, it's
this longer cylinder that

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we're talking about, but it's
going to be the same volume

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and the same mass.

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So what we say is that the work
that this column can do

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in that same amount of time
would be the output pressure

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times the mass of this column
divided by the density of the

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column-- which is the same
because the density of the

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liquid is the same throughout--
times the mass of

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this column, which is the same
as the mass of this column

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because the volume and the
density hasn't changed, so

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they're the same mass.

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Now, this column has more
potential energy.

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It's up at h2, which I'm
assuming is higher than h1.

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This kinetic energy is just the
mass of this cylinder of

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fluid times its velocity
squared, which is the output

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velocity divided by 2.

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This is potential energy
out, and this is

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kinetic energy out.

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These equal each other.

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This setup is Bernoulli's
equation, but let's see if we

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can clean it up so that we can
get rid of variables that we

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don't have to know about.

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One thing that we see is that
there's an m in every term, so

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let's get rid of them.

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Divide both sides of
this equation by m.

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We get that.

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I don't like this density in the
denominator here, so let's

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multiply both sides of this
equation by density, and what

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we're left with is-- let me
write this in a vibrant color.

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P, the input pressure, plus--
and we're multiplying

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everything by this rho,
this density.

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So we have input pressure plus
rho g h1, the input height,

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the initial height, plus
rho v squared over 2.

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This is rho v squared over
2, and that equals-- we

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multiplied both sides by rho, so
we get the input velocity,

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so that equals the pressure
out plus the density times

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gravity times the
output height.

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Let's make everything
consistent.

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I wrote 2's here, so let's just
say this is pressure 2,

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this is height 2, plus rho times
the velocity squared.

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This is Bernoulli's equation,
and it has all sorts of what I

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would say is fairly neat
repercussions.

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For example, let's assume that
the height stays constant, so

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we can ignore these middle
terms. If the height is

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constant, if I have a higher
velocity and this whole term

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is constant, then my pressure
is going to be lower.

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Think about it: If height is
constant, this doesn't change,

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but if this velocity increases,
but this whole

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thing is constant, pressure
has to decrease.

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Similarly, if pressure
increases, then velocity is

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going to decrease.

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That might be a little
unintuitive, but the other

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way, it makes a lot of sense.

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When velocity increases, this
pressure is going to decrease,

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and that's actually what makes
planes fly and all sorts of

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neat things happen,
but we'll get more

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into that in a second.

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Let's see if we can use
Bernoulli's equation to do

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something useful.

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You should memorize this,
and it shouldn't

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be too hard to memorize.

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It's pressure, and then you
have this potential energy

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term, but instead of mass,
you have density.

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You have this kinetic
energy term.

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It's not kinetic energy
anymore, because we

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manipulated it some,
but instead of

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mass, you have density.

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With that said, let's
do a problem.

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I'll keep this down here, since
you probably haven't

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memorized it as yet.

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Let me erase everything else.

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That's not how I wanted
to erase it.

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That's how I wanted
to erase it.

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I wanted to erase it like that
without getting rid of

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anything useful.

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OK, that's good enough.

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And then let me clean up.

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Clean up all this stuff.

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Let's say that I have a cup.

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I'll just draw a cup.

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It's easier to draw sometimes
then to draw straight lines

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and all of that.

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No, that's too dark.

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Do purple.

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I'm using a super-wide tool.

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I have to switch the length.

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OK, so that's my cup.

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It has some fluid.

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Actually, let's say it has a
top to it, and I have some

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fluid in it.

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Maybe it happens to be red.

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We haven't been dealing with
red fluids as yet,.

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Let me-- oh, I didn't
want to do that.

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So you know there's
a fluid there.

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And let's say that there's no
air here, so this is a vacuum.

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Let's say that h-- we don't know
what units are, but let's

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say h meters below the
surface of the fluid.

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This is all fluid here.

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I poke a hole right there, and
fluid starts spurting out.

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My question to you is, what is
this output velocity of the

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fluid as a function
of this height?

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Let me tell you something
else.

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Let's say that this hole is so
small, let's call the area of

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that hole A2, and let's say that
the surface area of the

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water is A1.

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Let's say that hole is so small
that the surface area

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the water-- let's say that A2
if equal to 1/1,000 of A1.

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This is a small hole relative
to the surface

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area of this cup.

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With that said, let's see what
we can do about figuring out

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the velocity coming out.

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Bernoulli's equation tells us
that the input pressure plus

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the input potential energy plus
the input kinetic energy

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is equal to the output,
et cetera.

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So what is the input pressure?

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Well, the input pressure, the
pressure at this point,

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there's no air or no fluid above
it, so the pressure at

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that point is zero.

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What is the input height?

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Let's just assume that the hole
is done at height 0, h

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equals 0, so the input
height h1 is just h.

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If this is 0, then this height
right here is h.

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What is the input velocity?

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We know from the continuity
equation, or whatever that

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thing was called, that the input
velocity times the input

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area is equal to the
output velocity

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times the output area.

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We also know that the output
area is equal to 1/1,000 of

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the input area-- and this is
area 2-- so we know that the

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input velocity times area 1 is
equal to the output velocity

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times 1/1,000 of area 1.

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We could say area 1 over
1,000 and divide both

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sides by area 1.

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We know that the input velocity
is equal to V2 over

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1,000, so that's good to know.

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These are the three inputs into
the left-hand side of

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Bernoulli's equation.

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What's on the right-hand side
of Bernoulli's equation?

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What's P2?

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What's the pressure
at this point?

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Oh, I just ran out of time.

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I'll continue this into
the next video.

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