Fluids (part 9) Second part of the Bernoulli's Equation proof. Beginning of a problem that uses the equation.
Fluids (part 9)
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- 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.
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