Fluids (part 10) Second part of the example that uses Bernoulli's equation.
Fluids (part 10)
- 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.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
Have something that's not a question about this content?
This discussion area is not meant for answering homework questions.
Share a tip
When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
Thank the author
This is great, I finally understand quadratic functions!
Have something that's not a tip or thanks about this content?
This discussion area is not meant for answering homework questions.
At 2:33, Sal said "single bonds" but meant "covalent bonds."
For general discussions about Khan Academy, visit our Reddit discussion page.
Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.
- disrespectful or offensive
- an advertisement
- low quality
- not about the video topic
- soliciting votes or seeking badges
- a homework question
- a duplicate answer
- repeatedly making the same post
- a tip or thanks in Questions
- a question in Tips & Thanks
- an answer that should be its own question