Fluids (part 2) Pressure and Pascal's Principal.
Fluids (part 2)
- Welcome back.
- To just review what I was doing on the last video before
- I ran out of time, I said that conservation of energy tells
- us that the work I've put into the system or the energy that
- I've put into the system-- because they're really the
- same thing-- is equal to the work that I get out of the
- system, or the energy that I get out of the system.
- That means that the input work is equal to the output work,
- or that the input force times the input distance is equal to
- the output force times the output distance-- that's just
- the definition of work.
- Let me just rewrite this equation here.
- If I could just rewrite this exact equation, I could say--
- the input force, and let me just divide it by this area.
- The input here-- I'm pressing down this piston that's
- pressing down on this area of water.
- So this input force-- times the input area.
- Let's call the input 1, and call the output 2 for
- Let's say I have a piston on the top here.
- Let me do this in a good color-- brown is good color.
- I have another piston here, and there's going to be some
- outward force F2.
- The general notion is that I'm pushing on this water, the
- water can't be compressed, so the water's going to push up
- on this end.
- The input force times the input distance is going to be
- equal to the output force times the output distance
- right-- this is just the law of conservation of energy and
- everything we did with work, et cetera.
- I'm rewriting this equation, so if I take the input force
- and divide by the input area-- let me switch back to green--
- then I multiply by the area, and then I just
- multiply times D1.
- You see what I did here-- I just multiplied and divided by
- A1, which you can do.
- You can multiply and divide by any number, and these two
- cancel out.
- It's equal to the same thing on the other side, which is
- F2-- I'm not good at managing my space on my whiteboard--
- over A2 times A2 times D2.
- Hopefully that makes sense.
- What's this quantity right here, this F1 divided by A1?
- Force divided by area, if you haven't been familiar with it
- already, and if you're just watching my videos there's no
- reason for you to be, is defined as pressure.
- Pressure is force in a given area, so this is pressure--
- we'll call this the pressure that I'm
- inputting into the system.
- What's area 1 times distance 1?
- That's the area of the tube at this point, the
- cross-sectional area, times this distance.
- That's equal to this volume that I calculated in the
- previous video-- we could say that's the
- input volume, or V1.
- Pressure times V1 is equal to the output pressure-- force 2
- divided by area 2 is the output pressure that the water
- is exerting on this piston.
- So that's the output pressure, P2.
- And what's area 2 times D2?
- The cross sectional area, times the height at which how
- much the water's being displaced upward, that is
- equal to volume 2.
- But what do we know about these two volumes?
- I went over it probably redundantly in the previous
- video-- those two volumes are equal, V1 is equal to V2, so
- we could just divide both sides by that equation.
- You get the pressure input is equal to the pressure output,
- so P1 is equal to P2.
- I did all of that just to show you that this isn't a new
- concept: this is just the conservation of energy.
- The only new thing I did is I divided-- we have this notion
- of the cross-sectional area, and we have this notion of
- pressure-- so where does that help us?
- This actually tells us-- and you can do this example in
- multiple situations, but I like to think of if we didn't
- have gravity first, because gravity tends to confuse
- things, but we'll introduce gravity in a video or two-- is
- that when you have any external pressure onto a
- liquid, onto an incompressible fluid, that pressure is
- distributed evenly throughout the fluid.
- That's what we essentially just proved just using the law
- of conservation of energy, and everything we know about work.
- What I just said is called Pascal's principle: if any
- external pressure is applied to a fluid, that pressure is
- distributed throughout the fluid equally.
- Another way to think about it-- we proved it with this
- little drawing here-- is, let's say that I have a tube,
- and at the end of the tube is a balloon.
- Let's say I'm doing this on the Space Shuttle.
- It's saying that if I increase-- say I have some
- piston here.
- This is stable, and I have water
- throughout this whole thing.
- Let me see if I can use that field function again-- oh no,
- there must have been a hole in my drawing.
- Let me just draw the water.
- I have water throughout this whole thing, and all Pascal's
- principle is telling us that if I were to apply some
- pressure here, that that net pressure, that extra pressure
- I'm applying, is going to compress this little bit.
- That extra compression is going to be distributed
- through the whole balloon.
- Let's say that this right here is rigid-- it's some kind of
- metal structure.
- The rest of the balloon is going to expand uniformly, so
- that increased pressure I'm doing is going through the
- whole thing.
- It's not like the balloon will get longer, or that the
- pressure is just translated down here, or that just up
- here the balloon's going to get wider and it's just going
- to stay the same length there.
- Hopefully, that gives you a little bit of intuition.
- Going back to what I had drawn before, that's actually
- interesting, because that's actually another simple or
- maybe not so simple machine that we've constructed.
- I almost defined it as a simple machine when I
- initially drew it.
- Let's draw that weird thing again, where it looks like
- this, where I have water in it.
- Let's make sure I fill it, so that when I do the fill, it
- will completely fill, and doesn't fill other things.
- This is cool, because this is now another simple machine.
- We know that the pressure in is equal to the pressure out.
- And pressure is force divided by area, so the force in,
- divided by the area in, is equal to the force out divided
- by the area out.
- Let me give you an example: let's say that I were to apply
- with a pressure in equal to 10 pascals.
- That's a new word, and it's named after Pascal's
- principle, for Blaise Pascal.
- What is a pascal?
- That is just equal to 10 newtons per meter squared.
- That's all a pascal is-- it's a newton per meter squared,
- it's a very natural unit.
- Let's say my pressure in is 10 pascals, and let's say that my
- input area is 2 square meters.
- If I looked the surface of the water there it would be 2
- square meters, and let's say that my output area is equal
- to 4 meters squared.
- What I'm saying is that I can push on a piston here, and
- that the water's going to push up with some piston here.
- First of all, I told you what my input pressure is-- what's
- my input force?
- Input pressure is equal to input force divided by input
- area, so 10 pascals is equal to my input force divided by
- my area, so I multiply both sides by 2.
- I get input force is equal to 20 newtons.
- My question to you is what is the output force?
- How much force is the system going to push
- upwards at this end?
- We know that must if my input pressure was 10 pascals, my
- output pressure would also be 10 pascals.
- So I also have 10 pascals is equal to my out force over my
- out cross-sectional area.
- So I'll have a piston here, and it goes up like that.
- That's 4 meters, so I do 4 times 10, and so I get 40
- newtons is equal to my output force.
- So what just happened here?
- I inputted-- so my input force is equal to 20 newtons, and my
- output force is equal to 40 newtons, so I just doubled my
- force, or essentially I had a mechanical advantage of 2.
- This is an example of a simple machine, and
- it's a hydraulic machine.
- Anyway, I've just run out of time.
- I'll see you in the next video.
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