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.