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Current time:0:00Total duration:10:08

Let's say we have a pipe again--
this is the opening-- and we have fluid going
through it. The fluid is going with a
velocity of v1, the pressure entering the pipe is P1, and
then the area of this opening of the pipe is A1. It could even go up, and
the other end is actually even smaller. The fluid-- the liquid-- is
exiting the pipe with velocity v2, the pressure that it
exerts as it goes out. If there was a membrane on the
outside, how much pressure would it exert on it as it
pushes it out on the adjacent water is P2, and the area of
the smaller opening-- it doesn't have to be
smaller-- is A2. Let's say that this opening is
at a height, on average, of h1, and the water exiting this
opening is on average at a height of h2. We won't worry too much about
the differential between the top of the pipe and the bottom
of the pipe-- we'll assume that these h's are much
bigger relative to the size of the pipe. With that set up-- and remember,
there's fluid going through this thing-- let's go
back to what keeps showing up, which is the law of conservation
of energy, which is in any closed system, the
amount of energy that you put into something is equal to
the amount of energy that you get out. So energy in is equal
to energy out. What's the energy that you put
into a system, or that the system starts off with
at this end? It's the work that you input
plus the potential energy at that point of the system, plus
the kinetic energy at that point of the system. Then we know from the
conservation of energy that that has to equal the output
work plus the output potential energy plus the output
kinetic energy. A lot of times in the past,
we've just said that the potential energy input plus the
kinetic energy input is equal to the potential energy
output plus the kinetic energy output, but the initial energy
in the system can also be done by work. So we just added work to this
equation that says that the energy in is equal to
the energy out. With that information, let's
see if we can do anything interesting with this pipe
that I've drawn. So what's the work that's being
put into this system? Work is force times distance,
so let's just focus on this. It's the force in times the
distance in, and so over a period of time, t, what
has been done? We learned in the last video
that over a period of time, t, the fluid here might have
moved this far. What is this distance? This distance is the input
velocity times whatever amount of time we're dealing with, so
T-- so that's the distance. What's the force? The force is just pressure times
area, and we can figure that out by just dividing
force by, area and then multiply by area, so we get
input force divided by area input, times area input. It's divided and multiplied by
the same number-- that's pressure, that's area. It's equal to the input distance
over that amount of time, and that's velocity times
time, so the work input is equal to the input pressure
times the input area times input velocity times time. What is this area times velocity
times time, times this distance? That's the volume of fluid
that flowed in over that amount of time. So that equals the volume of
fluid over that period of time, so we could call that
volume in, or volume i-- that's the input volume. We know that density is just
mass per volume, or that volume times density is equal
to mass, or we know that volume is equal to mass
divided by density. The work that I'm putting into
the system-- I know I'm doing a lot of crazy things, but it'll
make sense so far-- is equal to the input pressure
times the amount of volume of fluid that moved over
that period of time. That volume of fluid is equal to
the mass of the fluid that went in at that period of time,
and we'll call that the input mass, divided
by the density. Hopefully, that makes a
little bit of sense. As we know, the input volume
is going to be equal to the output volume, so the input
mass-- because the density doesn't change-- is equal to the
output mass, so we don't have to write an input and
output for the mass. The mass is going to be
constant; in any given amount of time, the mass that enters
the system will be equivalent to the mass that exits
the system. There we go: we have an
expression, an interesting expression, for the work being
put into the system. What is the potential energy
of the system on the left-hand side? The potential energy of the
system is going to be equal to that same mass of fluid that I
talked about times gravity times this input height-- the
initial height-- times h1. The initial kinetic energy of
the fluid equals the mass of the fluid-- this mass right
here, of that same cylinder volume that I keep pointing to--
times the velocity of the fluid squared. We remember this from kinetic
energy divided by 2. So what's the total energy at
this point in the system over this period of time? How much energy has gone
into the system? It's going to be the work
done, which is the input pressure-- I'm running out
of space, so let me erase all of this. I'll probably have to run out of
time, too, but that's OK-- it's better than
being confused. Back to what we were doing. So, the total energy going into
the system is the work being done into the system,
and I rewrote it in this format, which is the input
pressure-- we'll call that P1-- times the mass divided by
the density of the liquid, whatever it is. This is work in plus-- and
what's the potential energy? I wrote it right here-- that's
just mgh, where m is the mass of this volume of fluid, h is
its average height, and you could almost think of how high
the center of mass above the surface of the planet. Since we have a g here, we
assume we're on Earth, so this is h1, because the height
actually changes, so this is potential energy input plus
the kinetic energy mv1 squared over 2. That is the kinetic
energy input. We know that this has
to equal the energy coming out of the system. This is going to be equal
to the same thing on the output side. This is going to be equal to
the work out, so that'll be the output pressure times the
mass divided by the density plus the output potential
energy, which will just be mg h2, plus the outbound kinetic
energy, which will be mv2 squared divided by 2. I just realized I'm
out of time. I will continue this
in the next video. See you soon.