Fluids (part 8) Beginning of the proof of Bernoulli's Equation.
Fluids (part 8)
- 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.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
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This is great, I finally understand quadratic functions!
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At 2:33, Sal said "single bonds" but meant "covalent bonds."
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