Fluids (part 10) Second part of the example that uses Bernoulli's equation.
Fluids (part 10)
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- Where we left off, we had this canister, because it had a
- closed top and it had a vacuum above the fluid.
- The fluid on top had an area of A1, and I poked a little
- hole with a super-small area A2.
- I said that the area of A2 is so small, it's
- 1/1,000 of area 1.
- Then we used the continuity equation.
- We said the velocity, the rate at which the surface is moving
- up here, V1, times area 1, the whole surface area of the
- liquid, has to be equal to the output velocity, which we're
- trying to figure out as a function of everything else
- times this output area.
- I made a mistake.
- I don't know if I did this in the last video, or I did this
- is in the mistake video.
- So we know that the initial top velocity times this top
- area is equal to the output velocity times-- instead of
- writing area 2, we could write area 1 over 1,000.
- You can get rid of the area 1 on both sides, and then you're
- saying that the velocity up here is equal to the rate at
- which the top of the surface moves down and is equal to
- 1/1,000 of the velocity of the liquid spurting out of this
- little hole.
- With that, we actually have the three variables for the
- left-hand side of Bernoulli's equation.
- What are the variables on the left-hand side?
- What is the pressure at this point where we have a hole?
- This is an important thin.
- When we talk about Bernoulli's-- let me rewrite
- Bernoulli's equation.
- It's P1 plus rho gh1 plus rho V1 squared over 2 is equal to
- P2 plus rho gh2 plus rho V2 squared over 2.
- We figured out all of these terms. Now let's figure out
- the things that we have to input here.
- What is the pressure at point two?
- This is the important thing.
- You might want to say, and this was my initial reaction,
- too, and that why I made a mistake, is that what 's the
- pressure at this depth in the fluid?
- That's not what Bernoulli's equation is telling us.
- Bernoulli's equation is telling us actually what is
- the external pressure at that hole.
- When we did the derivation, we were saying how much work--
- this was kind of the work term, although we played
- around with it a little bit.
- But if we look at the water that's spurting out of the
- hole, it's not doing any work, because it's not actually
- exerting force against anything so it's not actually
- doing work.
- When we think about the pressure, the output pressure,
- it's not the pressure at that depth of the fluid.
- You should think of it as the external pressure at the hole.
- In this case, there is no external pressure at the hole.
- Let's say that if we closed the hole, then
- at that point, sure.
- The pressure would be the pressure that's being exerted
- by the outside of the canister to contain the water, in which
- case, we would end up with no velocity.
- The water wouldn't spurt anywhere.
- But now we're seeing the external pressure is zero.
- That's what the hole essentially creates.
- We're going to say that P2 is zero, so this pressure was
- zero, because we're in a vacuum.
- P2 is also zero, so both of these are zero.
- Remember, that's the external pressure.
- P1 is the external pressure to the input to the pipe, and you
- can view this as a pipe.
- I could redraw it as a pipe that looks like it has a big
- hole on the top, and it goes down to some level to a
- super-small hole like that.
- This would be a vacuum, and the fluid is just going in and
- it's spurting out of this end.
- Anyway, the pressure going into the pipe is zero, and we
- said since we put a hole, the pressure coming out of the
- pipe is zero, so we're doing no work.
- What is this term?
- This was the potential energy term, and we said that h1 is
- equal to h.
- We're saying that this is zero height, so now this simplifies
- to rho times gravity times h plus rho times V1 squared.
- V1, we said, is equal to this, so this is rho over 2 times V2
- over 1,000 squared.
- I just substituted V2 over 1,000 for V1.
- That equals the pressure at the hole, the external
- pressure at the hole, which is zero, plus h2.
- This is h2 right here, which we said is zero.
- We determined that the hole was poked at height zero, so
- this is also zero.
- That equals this kinetic energy-like term.
- It's not exactly kinetic energy.
- It's rho times V squared divided by 2.
- One thing that we can immediately see is that we
- have all these rhos on both sides of the equation, so we
- can divide both sides by rho and get rid of all of those.
- Then we can multiply both sides of the equation by 2,
- and we get 2gh plus V2 squared over-- what's 1,000 squared--
- over 1 million.
- That is equal to V2 squared.
- We could do the exact thing.
- We could subtract 1 over 1 million V2 squared from both
- sides, and we would get 0.999999 V2 squared, but let's
- just say for the sake of simplicity, or let's say, if
- this wasn't 1,000, but 1 million, and that this surface
- was much bigger, we see that this term becomes very, very,
- very small.
- If that hole is one millionth of the surface area, then it
- becomes really insignificant, so we can ignore this term
- because it just makes things complicated, and we're
- assuming this is a really, really large number, and that
- this hole is much smaller than the surface area of the fluid.
- This is like poking a hole in Hoover Dam.
- Hoover Dam is backing up this huge lake, and you poke a hole
- in it, so that hole is going to be this very small fraction
- of the surface area of the fluid.
- You can only make this assumption when that output
- hole is much smaller than the input hole.
- With that said, what is the output velocity?
- The velocity-- you just take the square root of both
- sides-- is the square root of 2gh.
- That is the output velocity.
- What is the amount of liquid that flows out per second?
- We figured that out already.
- It's a column of fluid that comes out, so per second, the
- length of the column of fluid will be the velocity times
- time, and then the cross-section of that column
- is equal to the output area.
- If I wanted to know the flow coming out, the flow coming
- out or the flux coming out would be equal to the hole's
- area times the hole's output velocity.
- That would equal the area times the square root of 2gh.
- We could use that actually solve problems in the future
- if we had actual numbers.
- I only have a minute and a half left.
- I'll see you in the next video.
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