WEBVTT physics/ap-physics-1-2/fullerton
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Welcome back to Educator.com.
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Today we are going to finish up our study of fluids, by talking about Bernoulli's principle.
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Our objectives are going to be to understand how Bernoulli's principle describes the conservation of energy in fluid flow and apply Bernoulli's principle to problems of fluids in motion.
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Let us start by talking about Bernoulli's principle qualitatively.
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Bernoulli's principle states that fluids moving at higher velocities lead to lower pressure and fluids moving at lower velocities lead to higher pressures.
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This comes into play quite a bit when you talk about the design of an airplane wing.
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If the airplane's moving forward to the right, then as the air goes over the wing, it has a longer path up above the wing.
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And because it has a longer path compared to below the wing, where it has a shorter path above the wing with a longer path, it has a higher relative velocity.
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If you have the higher air velocity over that wing, you get lower pressure and where you have the shorter path, you end up with a lower velocity where the air does not have to travel as far in the same amount of time, so you end up with a higher pressure.
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If you have higher pressure below and lower pressure above that wants to equal out and you get a net force up.
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That force is one of the components of lift.
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Now note that it is not the only component and that there is a lot more to lift and flight than just Bernoulli's principle in the airplane wing shape, but that is one component of it -- one application of Bernoulli's principle.
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Another one is called a venturi pump. The idea here is it that you can make a vacuum pump or another type of pump using fluid flow.
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If you have fluid or an incompressible fluid in a closed system coming through from the left to the right here, and as it goes there, you have a narrow opening, you must have faster flow here -- we know that from our continuity equations for fluids.
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And if we have faster flow, we must have lower pressures here.
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As we have lower pressures there, what we are going to get is a pumping action in which we can start sucking things up this way to join that fluid flow.
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So we force a lot of water or some other fluid through the pipe this way, constricted here and you are going to get a sucking action, you are going to get a pumping action.
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These are used in venturi pumps -- applications of those -- things like carburetors...
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...It is responsible for sailboat propulsion, gas delivery systems, or even for some folks who have sump pumps in case the power ever goes out.
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Often times you will see these as a backup sump pump.
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When the power goes out, if the water level gets too high in the sump, what happens it is lifts a float that turns on a bunch of water pressure and the water pressure comes running through here and this is connected to that cistern and sucks up the water to pull it out with that pumping action.
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This is a way to help keep your basement dry even when you do not have power to keep that sump pump running.
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All right. Bernoulli's equation quantitatively looks a little bit scarier.
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It relates the pressure velocity and height of a liquid in a tube at various points.
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Do not let it scare you, it is all a fairly simple equation, it just looks like a lot all at once.
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Pressure 1 plus 1/2 the fluid density times the square of its velocity plus the fluid density times (g) times its height equals (P2 plus 1/2 ρv2² plus ρgy2².
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Really what this is is a statement of conservation of energy.
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Notice how similar this looks -- 1/2 ρv² to 1/2mv² -- similar to kinetic energy -- ρgy is similar to (mgh), which is similar to gravitational potential energy.
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These are some pretty close parallels here.
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It really is a version of conservation of energy.
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What it says is the pressure at any point in the tube plus 1/2 the density times the square of the velocity added to (ρgy) must be the same anywhere at any point in the tube.
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As we check this out, let us take a look and use it to derive what is known as Torricelli's theorem.
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If we have water sitting in a large jug at a height of 0.2 m above the spigot, what is the pressure on the spigot and at what velocity will the water leave the spigot when the spigot is opened?
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Well, the first thing we need to realize is that (P1) up here and (P2) up here are both open to atmosphere.
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So we are going to say P1 = P2 = Atmosphere, and we are not going to worry about the difference in height compared to the overall atmospheric pressure as that is going to be negligible.
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Now as we start to look at this, we will start by writing down Bernoulli's equation -- P1 + 1/2ρv1² + ρgy1 = P2 + 1/2ρv2² + ρgy2.
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Now as we look at this, up here at the top of our fluid, because we have so much fluid there, we can assume that (v1) is approximately equal to 0.
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Now, (P1) and (P2) are both atmosphere, so we can subtract both of those right out and that will simplify right there.
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We said (v1) is approximately 0, and that term goes away, so what we are left with now is ρgy1 = 1/2ρv2² + ρgy2.
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All right, we are getting closer already.
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I can divide out the density to say that gy1 = 1/2v2² + gy2...
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...or solving for (v2) -- v2² = 2g × y1 - y2 (the quantity) = 2 × 10 and our height difference from y1 to y2 is just 0.2 m...
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...so that implies that v2² is going to be equal to 4 or take the square root of v2 must equal 2 m/s.
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And what we have really done here is we have derived at what is known as Torricelli's theorem -- this part right here v2² = 2gy1 - y2 or more commonly written -- take the square root, v2 = 2g × y1 - y2, square root...
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...calculating the velocity coming out of a container of liquid like this, Torricelli's theorem.
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We can look at an example with gauge pressure here too.
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Water flows through a large diameter pipe at Point (A) before it is constricted into a smaller diameter pipe at Point (B).
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How does the gauge pressure compare at Points A and B?
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Well, if the water is going through this pipe and it is being constricted, it must be going faster right?
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So the velocity at (B) must be greater than the velocity of (A).
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We know that from out continuity equations for fluids.
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If it is going faster at (B) then (B) must have lower pressure than (A), therefore (A) must have a higher pressure than (B).
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Let us take a look at the shower problem.
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A water main of area 0.003 m² at ground level flows at 2 m/s into Kate's house.
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At the second floor shower head, 5 m above ground level, the pipe has an area of 0.001 m².
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Find the velocity of the water in the pipe as well as the gauge pressure just prior to the shower head if the water main's pressure gauge reads 2 atmosphere.
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Well let us start with a diagram here -- water main of area 0.003 -- so we are going to start over here with a pipe over here at ground level and we know if we put a gauge on it, that it is going to read 2 atmosphere's right there.
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It has a cross-sectional area of 0.003 m² and it is flowing at 2 m/s into the house -- that is at section one.
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Now, somewhere up here it has a height difference of about 5 m and before it comes out, it is now down to a cross-sectional area of 0.001 m².
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We need to figure out, here at section two, area two, what the velocity is of the water in the pipe as well as the gauge pressure just prior to that shower head.
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To start off to find the velocity of the pipe, I am going to use the continuity equation for fluids.
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a1v1 = a2v2, therefore the velocity here at Point 2 is going to be equal to a1/a2 × v1 or 0.003/0.001 × 2 m/s (v1) = 6 m/s.
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All right, now let us see if we can find out the gauge pressure, just prior to that shower head.
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We will use Bernoulli's equation, P1 + 1/2ρv1² + ρgy1 = P2 + 1/2ρv2² + ρgy2.
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Now, if we set (y1) over here and call this ground level, y1 = 0, that term becomes 0, and that goes away.
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Now let us start substituting in our values to see if we cannot find what (P2) is going to be.
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Over here at (P1), we already know our pressure is 2 atmospheres or 200,000 Pa + 1/2 × 100,000 kg/m³ (density of freshwater) -- v1 is 2 m/s²...
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...equal to P2 + 1/2(1,000) (density), v2² (6 m/s²) + ρ (1,000) × g(10) and y2 is 5 m higher, so 5.
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So a little bit of math here -- 200,000 + 1,000 × 4 -- 1/2 of that will be 2,000 = P2 + 36,000 × 1/2 (18,000) + 10,000 × 5 = 50,000...
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... so solving for P2 then = 202,000 - 68,000 or 134,000 Pa...
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...or approximately 1.34 atmospheres.
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Great! Let us take a look at a water fountain example.
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Sandy is designing a water fountain for her front yard.
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She would like the fountain to spray to a height of 10 m -- that is a pretty impressive water fountain.
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What gauge pressure must her water pump develop?
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Well, let us start with a diagram again.
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We will start over here at her pump and that is going to go to a point where it is going to release the water up at ground level and once it is there, we want the water to go up to a height of 10 m or so.
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We will start again with Bernoulli's equation -- P1 + 1/2ρv1² + ρgy1 = P2 + 1/2ρv2² + ρgy2.
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If we call this section over here on the left (1) and over here (2), right away we can make some simplifications.
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At (1), we will assume that we have so much water that the velocity there in the pipe is roughly 0 -- that term goes away.
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We are also doing this at ground level, so y1 = 0.
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On the right hand side at its highest point right here, we want the velocity of the water to be 0 and that is what happens when it gets to its highest point, so velocity (2) will go to 0, therefore, P1 = P2 + ρgy2.
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We are looking for P1 and P2 is open to atmosphere, so we know that is going to be 100,000 Pa + ρ(1000) × g (10 m/s²) × y2...
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...we want that 10 m high, so P1 = 100,000 + 1,000 × 100 for a total of 200,000 Pa.
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So that is the pressure that we need total, so 200,000 Pa is equal to P1, which is equal to atmospheric pressure + ρgh...
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...where P0, here, is our atmospheric pressure (100,000 Pa) and ρgh here is our gauge pressure.
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So if 200,000 = 100,000 + gauge pressure, that means our gauge pressure must be 100,000 Pa in order to make the water fountain shoot the water 10 m high.
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Let us take a look at one more -- an elevated cistern problem.
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We have a water cistern that is elevated 15 m above the ground and it feeds a pipe that terminates horizontally 5 m above the ground as shown.
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With what velocity will the water leave the pipe and how far from the end of the pipe, will the water strike the ground?
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The first thing I am going to do is try to come up with a strategy here.
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I think I can use Bernoulli's principle to find the velocity of the water right here at what we will call Point (2) and then it becomes a projectile problem as to where it is going to land.
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If we call this our Point (1), we have a height of 15 m here and a height of 5 meters here, to find the velocity, why do I not just bring this back and if I call Point (2) ground level for the first part of the problem for figuring out the velocity, then the height here will be 10 m because that is the difference.
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So let us apply Bernoulli's equation and see how this is all going to look and work out.
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Bernoulli's equation -- P1 + 1/2ρv1² + ρgy1 = P2 + 1/2ρv2² + ρgy2.
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As we look at that, some simplifications we can make -- P1 is open to atmosphere; P2 is open to atmosphere, so they will have the same pressure and we can subtract those out of both sides -- v1 is going to be roughly 0, so we can make that go away.
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On the right hand side, if we are calling this the 0 height level for the first part of our problem, setting that as our 0 and the height here is 10, so we can make that term go away.
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So we have simplified Bernoulli's equation to say that ρgy1 = 1/2ρv2² or as we substitute in our values -- first off we can get rid of the ρ's.
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We have gy1 = 1/2v2² or v2² = 2gy1 or v2 = the square root of 2gy1.
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Notice how similar that looks -- v = square root of (2gh), the conservation of energy value we found in order to determine how fast something is moving after it has been dropped some distance...
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...V = square root of 2gh by kinematics or by conservation of energy approach.
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It is the same idea, so (v2) will be equal to the square root of 2 × g (10 m/s²) × y1 (10 m) or the square root of 200, that is about 14.1 m/s.
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Our water is going to be leaving the pipe down here with a horizontal velocity of 14.1 m/s.
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Now we have ourselves a projectile problem, where we have a height of 5 m and we need to find the horizontal distance the water travels.
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All right. Well let us first figure out how long that water is going to be in the air.
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That is a vertical kinematics problem, where V-initial vertically is 0, Δy will be 5 meters, acceleration will be 10 m/s² and we need to find the time.
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I would use the equation Δy = V-initial(t) + 1/2 at², but again, V-initial is 0, so that term goes away.
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(T) then becomes 2Δy/a (square root) or 2 ×5/10 (square root) or 1 s.
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Now I can use my horizontal kinematics to figure out how far it goes.
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Horizontally, the velocity is going to be a constant 14.1 m/s and it is going to be in the air for 1 s, so Δx is just going to be velocity × time, 14.1 × 1 or 14.1 m.
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Putting a couple of these concepts together to get a big picture solution.
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All right. Hopefully that gets you a great start with Bernoulli's principles and Bernoulli's equations.
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Thanks for watching. Make it a great day everyone!