PHYS 200: Fundamentals of Physics I
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Fundamentals of Physics I
PHYS 200 - Lecture 20 - Fluid Dynamics and Statics and Bernoulli's Equation
Chapter 1. Introduction to Fluid Dynamics and Statics — The Notion of Pressure [00:00:00]
Professor Ramamurti Shankar: This is a relatively simple topic. In fact, any of you who took any kind of high school physics, you would have done this thing with fluids. So, this is the subject for today. Fluid dynamics and statics. So, I’ll start with the simple static problem. We are going to take–Whenever I say fluid, you are free to imagine water or oil. That’s the–that’s a good enough example of a fluid. One important property of the fluid, the density, denoted by ρ; the density of water I would probably note by a subscript w. And you know that’s mass per unit volume. For water that happens to be 1,000 kilograms per cubic meter. So, let’s not linger there. I think that’s a fairly simple concept.
The more subtle concept is the one of pressure. So, you have the notion of a pressure if you go in a swimming pool and you dive down to the bottom, you know, the pressure is going up. So, what’s the formal definition of pressure? Is it a vector? Is it a scale? Or does it have magnitude? Does it have a direction? That’s what I want to explain to you. So, you can pick a point on the fluid, there, and say the pressure there is such and such. But what do we mean by that? We mean by that the following. If you get into that fluid and you want to carve out a little space for yourself, you know, make a little cube, maybe a glass cube, and you want to live inside that cube. So, I’m going to blow up the cube like this. The water is trying to push you in from all sides and compress this cube. You therefore have to push out on the two, on all the walls. If the force you exert on this wall is some F and the area of that wall is A, that ratio is called a pressure. So, pressure is an intensive measure of how hard the water is trying to push in. If you don’t put the cube, that pressure is still there, but one way to measure the pressure is to try to go in there and push the fluid out and ask how hard does it push you. And the unit for pressure Newton per meters square–and we use another name for that called a Pascal. One Pascal is one Newton per meter squared.
Here’s another example of pressure. You have a gas. We’ll do a lot of this after the break. There is a gas inside a cylinder. There’s a piston. Now, if the pressure of the gas and the pressure of the outside world are the same, there’s nothing you have to do, assuming the piston is massless. But if you want to increase the pressure in the gas, you put some extra weight. That mg will push down, and mg divided by the area of the piston will be extra pressure you apply. That’s also the pressure of the gas. That pressure is the atmospheric pressure, plus the extra weight you apply, divided by the area of this piston. The atmospheric pressure is everywhere. So, when you push down on the piston here to compress the gas further, you’re adding to the atmospheric pressure this extra force divided by area. Sometimes–I mean, this is called the absolute pressure. And that’s called the atmospheric pressure, and that’s called the gauge pressure. So, the gauge pressure is the pressure on top of atmospheric pressure. For example, when your car has a flat, the bright side of it is the pressure inside the tire is in fact equal to atmospheric pressure. It doesn’t help you because there’s a pressure inside and there’s a pressure outside, and if you want to keep your car moving, you really have to increase the pressure in the tire. And when you stick this gauge in and you measure something, 32 pounds per square inch, that’s the gauge pressure. That’s in excess of the atmospheric pressure.
Chapter 2. Fluid Pressure as a Function of Height [00:04:14]
So, you should understand that the pressure is a condition in a fluid, and one of the important properties of pressure, is that if you took a fluid and you went to a certain height, all points at that height have the same pressure. And we understand that as follows. So, I go to this fluid. I imagine in my mind a little cylindrical section of the same fluid. Just draw a dotted line around that region and focus on that little chunk of fluid and say, “Can the pressure on the two sides – this face and that face – be different?” The answer is “no.” Because if the pressure on the left was bigger than the pressure on the right, I take a cylinder of some area A, pressure times area on the left will exceed pressure time area on the right. And therefore, the fluid should move to the right. But it’s not doing anything. It’s in equilibrium, and the only way that can happen is if it’s pushed equally from both sides. So, notice that the pressure on this chunk from the left points this way, and then that chunk points that way. They’re trying to push it in. So, pressure cannot change at a given depth.
But let’s take a cylinder that looks like this [draws on board]. Remember this is not a real cylinder. This is the same water, and mentally, I have isolated a part of the water that looks like a cylinder; maybe you want to draw it in dotted lines. But it’s just a fluid of cylindrical shape, base area A and height h. I can ask the following question. Let’s say h1 is the measure of that one from the surface, and h2 is the depth from the surface to the lower face of the cylinder. We can now argue the pressure in the top and bottom really should not be equal. You should think about it, by the same argument I gave earlier will tell me it cannot be equal. If they were equal, since these two areas are equal, the forces pushing up and pushing down will cancel. With no net force on the cylinder, you can ask then, “What’s keeping the cylinder of water from falling down?” Well, there has to be a force to equal the weight of that cylinder. Therefore, there has to be a net upward force. That means the pressure down here, pushing up, better be higher than the pressure on the top.
We are now going to calculate what the pressure difference is. So, let’s call the pressure downstairs P2 and upstairs P1. So, the upward force is P1 [correction: should have said P2] times A. The downward force is P2 [correction: should have said P1] times A. That’s the net upward force. That’s got to be equal to the weight of that amount of water. The weight is found by first the mass, which is area times h2 - h1. That’s the volume of the cylinder, times the density of water or whatever fluid you have. That’s the mass of the cylinder. That’s the weight of that cylinder of liquid. All I have done is balance the gravitational force on this cylinder with the net upward force due to the different pressures. I think I made one mistake here. Upward force, in magnitude–I want to write this P2 times A, and downward force I want to write as P1. How we keep track of signs is a little subtle. I’m balancing two magnitudes. I’m balancing net upward force, namely P2 is considered positive upward, and this is the force of gravity down. I’m balancing the magnitudes. So, the area cancels out. Because this area – see the pressure between two points – should not depend on the area of this fictitious cylinder I took. In fact, I find it is equal to ρg times h2 - h1, which is the difference in the depth of these two points. Yes?
Student: It’s actually like a [inaudible]
Professor Ramamurti Shankar: Which one?
Professor Ramamurti Shankar: Yeah. In fact, it really happens to be normal, or perpendicular to area because the pressure will push straight up and straight down. How about forces on the sides of cylinder? They cancel at every height. Because at every height the push from the left and right are equal; I already showed you that. So, for the mechanical stability of this chunk, I need all forces to add up to zero. Horizontally, there is no gravity and the pressure at every height is equal. So, that cancels out. Vertically, there is the force of gravity, and it’s cancelled by the difference in pressure.
So, we can see that P2 = P1 plus ρg times the height difference. And people write this formula as follows. It’s very standard to take P1 to be the point right at the surface, and P2 to be any point inside, and to call the depth of that simply as h. Then the pressure at the point P in the fluid is the pressure at the top, which is usually atmospheric pressure, plus ρgh. So, I don’t have an h1 and an h2 because h1 I’ve chosen to be zero. And h2 I’m simply calling h. This says something very simple. If you go to a lake, at the surface of the lake the pressure is due to the atmosphere. You take a dive, you go down some depth h, the pressure goes up by this amount. Now, if you go to the bottom of the ocean, it’s going to be an incredible amount of pressure. That’s why you and I cannot survive in the bottom of the ocean, because the outside pressure – inside is usual atmospheric pressure – you’re breathing the air into your lungs, you go down with that. Outside is atmospheric plus this, and that can kill you. That’s why when you build a submarine, you’ve got to make sure it can withstand the pressure. But fish don’t have the problem, because fish are breathing the water. The water is going into their system and outside their system. So, that’s one way for you to live. If you’re 20,000 feet under the sea, start drinking the water. But it’s not a long-term solution. It’ll work in the short time because you’ll be equalizing the pressure.
Okay. Now, how about the atmospheric pressure? What’s the origin of that? And the origin of the atmospheric pressure is that we are ourselves living in the bottom of a pool, but it’s filled with air. The air above our heads goes on for maybe 100 miles, but the density decreases, eventually vanishes. So, I cannot tell you precisely where the atmosphere ends, but I can say the following. If I go far above where the absolute pressure is zero, in free space there is just vacuum. The pressure here is the atmospheric pressure. Atmospheric pressure at the bottom of Earth is equal to zero plus ρgh, where ρ is the density of air, g is g, and h is the height of the atmosphere. And that happens to be 105 Pascals. So, we are living in the bottom of a pool where the pressure is 105 Pascals, relative to empty space–interstellar space. But as I said, that pressure doesn’t kills us because the pressure can be felt both from the outside pushing in and inside going through your nostrils and everything else, pushing out. But you have all seen the dramatic experiment where you take a can of something and you heat it up so the air goes out; then you seal it. The air that’s been driven out reduces the pressure. So, when you seal it and you cool it, then the pressure drops. And even the drop in pressure–it doesn’t even drop down to zero, but it’s big enough for the while can to implode.
Now, let’s see what you get here. Let’s do the following. This is ρgh, and you can ask yourself the following question. If this was a swimming pool, how high would the water be? In other words, what height of water above the Earth would produce the same pressure as our atmosphere does? Then I say, if I want 105 Pascals, that’s the density of water at 103, g, let’s pretend it’s ten. And h is what I’m looking for. Cancel all the powers of 10; h is like 10 meters. And it turns out it’s – to the best of my knowledge – it’s 32 feet. These are the units we don’t use in the book, but it’s a very common way of thinking, 32 feet. So, we are at the bottom of a pool. If it was filled with water, it would be 32 feet of water.
Alright. So, now we are going to take this formula P = P0 + ρgh and put it to work. Get some mileage out of that. So, what are the things we can do with that formula? First thing you can do is to build yourself a barometer. Barometer you know is a way to tell what the pressure is today. Now, the atmospheric pressure, when I said it’s 105, that’s the typical pressure. It doesn’t really stay locked into the value. Each day there are fluctuations. That’s why the weather person tells you pressure is going up, pressure is going down. So, let’s find a way to measure the pressure, and here is one way to do that. You take a can of something, fill it with some liquid, take a test tube. Evacuate it completely, suck all the air out of it and stick it into this. When you do, there is a complete vacuum here and the atmosphere is pushing down, so the fluid will rise up to some height h.
And you can ask, “How high will it go? What’s going to be the height?” Well, it’ll go to a height so that zero pressure here, plus this ρgh, which is the pressure there, will be the same as the pressure here, because they are two points at the same height. Pressure here is the atmospheric pressure. So pressure here, atmospheric pressure, is zero at the top of the tube, plus ρgh. So, if you build this gadget, this barometer out of water, the water column will rise to the height of 32 feet. But now, nobody wants a gadget 32-feet high, so what you use instead is mercury, because it’s very dense. So, you want to get the same atmospheric pressure, but you want to have a bigger ρ and get a smaller h. If you go look up a book and find the density of mercury, you’re going to find the height is something like, I don’t know, 750, 780 millimeters. That’s why the weather guy says the pressure today is so many millimeters, and the mercury is dropping. Now, I’m not sure why they bother to give the numbers, because for most of us, including me, those numbers don’t mean anything. Here’s a number, 746 millimeters. Does it speak to you? Not to me. So it–it speaks to one of you guys?
Professor Ramamurti Shankar: Yeah. Okay. Some number. It’s not like saying today is 67 degree Fahrenheit. I guess I know what that means, but when I hear the mercury, I think it’s just a waste of time. Anyway, they’re telling you how much the mercury is. Okay? You can use any fluid you like, but you’ve got to agree on using mercury because if it’s 760 millimeters, and it’s water the person is talking about, of course you are in serious trouble. So understood, we’re talking about mercury. Also, mercury is used in thermometers. That’s the subject for after the break. So there, the mercury falling could stand for falling temperatures and rising temperatures, but here it’s for falling pressure and rising pressure. So, this is the first gadget you can build with what I’ve taught you.
The next gadget you can build is–you can imagine this–you trying to drink. This is your ears and nose and what not. You’ve got a straw and you’re trying to drink something. Now, remember the fluid is water. Okay? It’s not mercury now, because you’re doing different experiment. You’re going to drink from a straw. So, how do you do that? You know when you drink from a straw you create a partial vacuum in your mouth. So, the pressure here–;So, in the case of the fluid that you want to drink, this is your head. The pressure here less than the atmosphere. If it’s less than the atmosphere by some amount, then the fluid can start climbing up to some height, and climb to such a height so that the lower pressure, plus that ρgh will equal atmospheric pressure. So, you got to reduce the pressure more and more in your mouth until this fluid can climb up. If you just–if you want to just make it to your mouth, it depends on the length of the straw, the height of the straw. If you want to suck it up a certain height, then that height, ρgh, plus the pressure in your mouth, is the pressure here, and that’s the atmospheric pressure. So, if you want to drink water from a well that is more than 32 feet deep, you are out of luck. Even if your face and your head are a complete vacuum, you cannot get the water to climb more than 32 feet. So, that’s another illustration of this.
So, one more example of this P = P0 + ρgh, is if I give you another fluid, it doesn’t mix with water and I tell you find the density, there are many ways. One is to just find mass and volume of that fluid and divide. But here is another thing people use. You take what’s called a U-tube. Yeah, that’s not–I know, that’s where they post all the embarrassing videos, but this was physics contribution to pop culture long before any of this happened. This is the U-tube. And the U-tube, let’s say you fill it up with one fluid and this is the other fluid. So, this is oil, and this is water. If the two heights were equal, then you know we’re talking about the same fluid. But this is supposed to tell you that oil is less dense than water, and we can check that by comparing two points in this fluid at the same height, and saying the pressure must be the same at those points. You understand how I get that? That pressure and that pressure are equal because you can draw a cylinder there horizontally, which cannot be pushed sideways. From here, to this pressure add that ρgh and that ρgh, and you come to that point; you conclude that pressure is also equal. You cannot jump into the new fluid here because the ρ for here and the ρ for this are different. But these two points have the same pressure and they are in the same fluid. So, let me write that statement that this pressure and this pressure are equal by saying, atmospheric pressure is on the top for both of them. That, plus ρ1gh1 is equal to atmospheric pressure plus ρ2,gh2, where this is the second fluid , height goes to h2, first fluid, the height is equal to h1, the densities are ρ1 and ρ2. Atmospheric pressure is cancelled; then you find h1/h2 = ρ2/ρ1. So, by comparing the heights you can find the relative density. If one of them is water, then, well, whatever it is. If you know the density of one fluid, you can find the density of the other.
Chapter 3. The Hydraulic Press [00:20:49]
Okay, yet another application of this law that the pressure is equal at a given height is the famous hydraulic press. So, here is a–Here are two pistons of different radii. This has got cross section area A1; this has got cross section area A2. And here I want to put some incompressible fluid. Incompressible fluid is something whose volume cannot be changed no matter how much you press it. Now, water is pretty close to incompressible. It does have a compressibility, but it’s not going to change very much for our purposes. So now here, I have a piston, and I have a piston here. And I push down here with a force F1, and I ask, “What will I get at the other side?” You might think F1 = F2, but since these fluids are at the same height, we only know P1 = P2. That means F1/A1 = F2/A2. So, this fluid here will push up here with a force F2, which is equal to F1 times A2 over A1. Did I get it right? Ah.
Professor Ramamurti Shankar: Pardon me.
Professor Ramamurti Shankar: Did I make a mistake here?
Professor Ramamurti Shankar: But I want to amplify the force. So, what do we really want to do? Well, it’s a matter of who you want to emphasize. So, what you want to emphasize here–Maybe if I drew a picture I will know exactly what I’m trying to do. This picture is fine. In practice, this is not what you do. You want to push down here and raise something there. Because what you really imagine is some elephant. There’s an elephant standing here, and you want to lift the elephant by applying a force here. So, let’s still call your force F2, and the force on the other side is the force you apply times A1/A2. So A1/A2 could be a hundred. What that means is, if you apply one Newton here, you’ll get a hundred Newtons on the other side. So, that’s the way to max–to convert a small force into a large force.
But this is the oldest trick in the book. An even older one invented by cave people, is that if you have a support like this you can put a weight here and, uh, let me see, that’s right. So, a tiny weight here can lift a big weight here because that mg times that distance and this mg times this distance will be equal. But you must know even from that example that you don’t get something for nothing. In other words, if you lift the elephant here by pushing down here, the fact that the forces don’t match is perfectly okay. But the work you do here must be the work delivered to the other side. The Law of Conservation of Energy. The work you do on this side is the force multiplied by the distance. Now, force multiplied by distance I’m going to write as the pressure times the cross section area, times distance. But what is A times–A2 times dX2? A2 times dX2–if you push the liquid here–if you moved it a distance ΔX2, area times ΔX2 is the volume of fluid you push down here. That’s the volume that’ll come up on the other side. So, you’re free to write that as P2 times A1 ΔX1.
Okay, but P2A1–I’m sorry, yes, now, let’s write P2 is the same as P1. P1 times A1 is F1. So, it just says that the force times distance on one side is the force times distance on the other side. That means the work you actually do is not amplified by the process. You cannot get more joules out of one side by any device. What you put in is what you’ll get out. But still it’s useful, because in practice you may have to move a whole meter here to lift the elephant by one centimeter. But the point is, you can lift elephants this way. That’s the important thing. Okay, that’s why it’s worth doing. And this is how you–the brake in your car works. You know, you pump the brake pedal. There’s a little cylinder there and there’s a fluid there, and the fluid is pushed by your feet. And you push it quite a bit, several centimeters. At the other end, there’s another cylinder whose piston is right next to the drum that’s rotating, and pushes on the drum. And it exerts an enormous amount of force, but it moves a very tiny amount. The shoes that grab your rotating drum move a very tiny amount, whereas, your feet move a large amount. But that’s the ratio of the force that’s transmitted. The fluid has the same pressure. The brake fluid has the same pressure, but the force you apply with your feet is much, much smaller than the force that the drum will exert on the rotating–that the disc brakes will exert on the rotating drum. So, a lot of hydraulics is based on this simple amplification of force.
Chapter 4. Archimedes’s Principle [00:26:32]
Alright. Now, I move to the next topic in this field, which is the Archimedes’ principle. So, we all know the conditions under which this was discovered. So, I will not go into that, other than to say that Mr. Archimedes noticed that if you immerse something in a fluid it seems to weigh less. What I mean by that is that if you attached it to some kind of a spring balance, and you weighed it so that the kx of the spring was the mg of the object, and if you did the same thing now you will find it seems to weigh less. And the question was, “How much less?” Archimedes’ answer is very simple. The amount by which you lose the weight, or the weight loss, equals the weight of liquid displaced.
Now, how do you show that? Because we don’t, right now, so many years after Archimedes, we will not accept this on faith. We want to be able to show this is the case. There are several ways to prove this. One way, which I like, is to say if you, if the thing you are hanging here, was itself a chunk of water shaped like that, you don’t have to do anything. Because that chunk of water can float at that height for free. But now, if you took the chunk of water and put a stone here of the same shape, the rest of the fluid doesn’t know what you’re doing. It applies the same force that it would to its own colleagues. Namely, if this is water, the rest of the water is in a configuration ready to support that amount of water. So, if you took that water out and put something in, the water will apply the same amount of force and the rest of it is your problem. You apply the remaining force. So, the water is ready to support its own kind of any volume. That’s the meaning of being in equilibrium. Any chunk of water is in equilibrium; therefore, it’s getting an upward force equal to the weight.
What we are saying is, if without disturbing the environment, you take the water out and put something in its place, the rest of the guys will apply the same force. Now, one formal way to prove that–you can take various geometries, but I prefer a cylindrical geometry. So this is, in this example, not a piece of water, but a new material you’ve introduced inside the water. And we want to find the buoyancy force. What is the net force of buoyancy? Well, it’s the pressure at the bottom times the area, minus pressure at the top times the area. We have already seen the difference in the pressure is ρgh A, but h is now this height. Well, h times A is the volume of the water, ρ times that is the weight of the water, the mass of the water. That times g is the weight of the liquid displaced. You can prove this for a cylinder; you can prove this for all oddball shapes, by thinking a little harder. But this is good enough. So, basically the body weighs less in water because the lower part of the body is being pushed up harder than the upper part of the body is being pushed down, because the pressure increases with depth. That’s why you get this result from Archimedes.
Now, you have to be a little careful on writing the equation, because if this was made of a material like iron, then the density of iron is less than–is more than the density of water. So, the weight of that chunk of iron will be more than the weight of the water displaced. So, you will have to provide a net force, and you can support it with the cable. But suppose this was not made of iron, but made of cork? If it’s cork, it won’t want to be there. Right? Because then, the applied force by the water is more than the weight it takes to support it. So, the cork will then bob up to the surface. It will look like this. If you want to keep it down, it’s like a rubber ducky. You want to keep the rubber ducky inside the water level, you’ve got to pull it down, or you want to tie it to the–I think one of the problems I gave you–has somebody tied this piece of whatever to the floor. Then it’ll stay. But things will bob up to the surface. And the question is, “How far up will it go?” We know part of it’s going to be inside and part of it’s going to be outside. And you can ask how much will be outside and how much will be inside?
You can already guess the answer, but let’s prove it. Let f be the fraction immersed, fractional volume immersed. Then, we can say the weight of the liquid displaced is equal to the fractional volume times ρ, times g, times the total volume, is the ρ of water. You understand? This is the weight of water equal to the full volume; this is the fraction of water displaced. So, this is simply the weight of this shaded region here. And that is going to be equal to the weight of the thing that’s floating; that is the ρg times full volume. If you cancel the g, and you cancel the volume, you find that the fraction that is immersed is the density of the material divided by density of water. In other words, if this material is 90 percent the density of water, it will be immersed by 90 percent. That’s exactly what happens with ice. As you all know, ice has a smaller density than water. One of the great mysteries. Normally, when you cool something it condenses and reduces in volume and the density will go up. But ice actually expands when you cool it. That’s why the density of ice is less than the density of water. That’s why ice floats on water. That’s the reason icebergs look like this, because the big part of the iceberg–;that’s the very peculiar property of ice. That’s why you have those movies like the Titanic where you’ve got a huge ice thing and it’s floating. First of all, if the density of ice was more than the density of water, these two actors would still be alive. Okay? But what happens is it is floating, and not only that; what you see is maybe a small fraction of the whole thing. That’s why those big ships went down. Okay, but it’s all thanks to this accident of nature that, here is one substance which, when cold, increases its–decreases in density.
Alright. Now, Archimedes’ principle has got hundreds of applications. If you want to build a boat, here’s how we build a boat. Here is the steel boat. Now, you cannot come to me and say, “How do you make a steel boat float in water?” Okay. It’s not a solid steel boat. Okay? If you are thinking about a solid steel boat, you should get in another line of work. This is a thing made out of steel, but it’s completely hollow. So, what we claim is, this amount of water weighs the same amount as that amount of–see, the shaded region. So, you can easily calculate how deep this one should sink to balance its weight. Right? That, I assume you know how to do that. Take this to be a rectangular boat. The cross section is rectangular; it’s gone down to a depth ρ, and to a depth h. Therefore, the volume of water displaced, is h times the area of the boat, cross sectional of the area of the floor of the boat. That’s the volume of water, that’s the mass of water, that’s the weight of water. That’s the weight of the boat. If you tell me how many tons the boat weighs and you give me the area of the base and g, and density of water, I’ll tell you what height it’ll sink. Then, of course, you can load more and more cargo in this, you know, if you have a top here, start putting more and more cargo; this will go down until the boat looks like this. And that’s as far as you can push it. That’s the kind of simple calculation you will be asked. How much cargo can the boat take? Well, that’s very simple. The weight of the boat plus weight of the cargo is maximal in this critical situation, when it’s just about to go under. Therefore, the total volume of water displaced, times the density, times g, will be the weight of your boat plus cargo.
And these are all elementary applications, and they came from two principles. One is the pressure increases with depth; the second is that the weight of the liquid displaced is the weight of the–is the reduction in weight. So, you can imagine three cases, one where the density of the material is more than the density of water, in which case it’ll start going down and you will have to hold it up with a cable, but you won’t have to apply the same force as you would outside. The second example is when the density of the object is less than the density of water, in which case it will float, with a certain fraction of it immersed, and a certain fraction of it outside. The fraction immersed is simply the same fraction as what you get by dividing its density by the density of water. Also, the density of water changes. In the Dead Sea, because of the salt concentration, density is a lot higher. So, it’s a lot easier for people to float.
Chapter 5. Bernoulli’s Equation [00:36:36]
Okay. Now, for the last and final topic that’s called Bernoulli’s Equation. This is the first time I’m going to consider fluids in motion. So far, my fluids were at rest. And it’s really very simple. But notice one more time, all I ever invoked was Newton’s Law of Motion. You realize that? I just took this fluid and took that fluid, balanced the forces, and said there should be no acceleration. If I can convince you of the one thing, I have accomplished something. To realize that all the mechanics we have done does not appeal to any other law than F = ma. In all these problems of equilibrium, a is zero, F is zero. Just from that fact, and clever applications of F = ma. For example, it’s very clever to think of a piece of the water and demand that it be in equilibrium. That’s how we find how the pressure varies with depth. But there’s no new principle so far. In fact, there’s going to be no new principles at all this term. And relativity was different. The old space-time was modified, but non-relativistic mechanics is all coming from Newton’s law, as is this problem.
So, we’re going to display for the first time, fluid motion. We’re going to take the most general case of fluid motion, where there’s water in some pipe. This is the most famous picture in all the textbooks. We have not thought of a better picture now. We’re all working on it, but this is all we can come up with after 300 years. Water flowing in a pipe. This point is going to be called 1; this point is going to be called 2. Everything here will have a subscript 1. That means, measured from some ground, that’s at a height h1, that’s at a height h2. The velocity of the fluid here is some V1, the velocity of the fluid there is V2. And ρ is the density of the fluid, and that’s not variable. So, imagine now a steady flow of some fluid through a pipe, whose area of cross section is changing, and the overall altitude is also changing. You can have a huge, you know, pipe in the basement where the whole supply to the house comes, branches out into little pipes maybe; this could be the pipe in the attic. Small cross section possibly at a bigger height, it doesn’t matter. But we’re not considering pipes that break up into four or five pipes. This is just a single pipe.
Chapter 6. The Equation of Continuity [00:39:12]
So, here is the first law that you have to satisfy. If the fluid is incompressible, there’s a certain law. And we are going to talk about that law. That’s called the “equation of continuity.” It relates the area here and the velocity here, to the area here and the velocity here. And the relationship should not surprise you. The basic premise is going to be you’re shoving in water from the left, and the water cannot pile up between here and here because it’s incompressible. That means in that volume, you can only pack in so much water. So, what comes in has to go out. It follows that it’s got to go out much faster here because the area is smaller. And what’s the relation? You can almost guess it, but let’s prove that. How much water do you think comes in through this phase, if you wait one second? Can you visualize in your mind that in one second a certain amount of fluid comes in, maybe until that point, and the cross section of that fluid is A, the distance it travels is V. So, the volume of the fluid coming in from the left is really A1V1. That’s called the flow rate coming in. And that’s got to flow out, and that’s the flow rate outside. So, in an incompressible fluid, if I tell you the flow rate at one point, I’ve told you the flow rate everywhere. Think of cars going down a freeway, and the freeway’s getting narrow, but unlike in real life, we don’t allow the cars to pile up. We want the density of cars to be the same. That means if there’s a narrow road, they’ve got to go faster to maintain the traffic. That follows–Then, it follows, that if I go to one checkpoint and see how many cars cross me per second here, the same number will cross anywhere else. But the speeds will be in inverse proportion to the area so that the product remains the same.
Now, let me show this to you in another way. It’s going to be helpful. Let me wait a small time Δt. In a small time Δt, this front that was here would advance there. It’ll go a distance V1 times the Δt. On the other end, this front will advance the distance, V2 times Δt. Now, the volume that got pushed in the time Δt is A1V1 Δt, and that’s the volume that came out on the other side. And you can cancel the Δts and come up with the result I gave you. So wait a short time, see what comes into the left phase, and see what goes out of the right phase and equate them.
Okay. Now, we are going to find a constraint between the state of the fluid here and the state of the fluid there. What I’m going to do–Look, think about what’s going to happen before you derive any formula. Suppose you’re going uphill. Does it make sense to you that when the fluid climbs uphill it’s going to slow down? It will slow down on the way to the top, because it’s got to work against gravity. Even if you threw a rock up there, it’s going to go up and slow down. Therefore, there’s going to be some connection between the gain in height and the velocity of the fluid. Just from the Law of Conservation of Energy. That’s all I’m going to do. Here is what I’m going to do. You remember if there are no external forces on a system, kinetic plus potential is kinetic plus potential. If there are external forces on a system, kinetic plus potential after minus kinetic plus potential, before, is the work done by external forces. That’s what I’m going to use here.
What are the external forces, and what’s the energy is what I’m going to think about. So, here’s what I’m going to focus on. Take this region of fluid trapped between these two cross sections at t = 0. Wait a short time Δt. What happens to the body of fluid? It does what I told you. In the front, it advances to the new region there. Maybe I should draw better pictures here. This is the advance of the rear guard, and in the front, the fluid that used to terminate here has gone up to there. In other words, if I colored this fluid, just this part of the fluid between here and here, a different color from the rest of the fluid, a short time later that colored liquid will be occupying this new volume. It’s the same numerical volume, but it’s a slightly different volume. I’m going to compare the energy of this chunk of fluid before and after. You realize that in that comparison this region in between the shaded regions is common to both the chunks of fluid. Pointwise, the fluid in the same location is going at the same velocity. So, I don’t have to worry about that. That’s common to before and after. The only difference between final minus initial is the energy contained in that region minus the energy contained in this region. The rest of it, like here, it was part of the old fluid; it’s part of the new fluid, it’s at the same height going at the same speed. So, you don’t have to worry about it.
So, what’s the volume? What’s the energy of this region? The energy is going to be ρ times A2 times Δx2. Δx2 is the distance it advances here. That’s the volume of the fluid; that’s the mass of the fluid. Sorry, that times g is the mass of the fluid. And the kinetic energy will be mass times V2 over 2. Potential energy will be–I’m sorry, am I doing something right? I think I made a mistake. Sorry about that. There is no g here. Let me do it more slowly. What’s the mass of the shaded region? It’s got a base A2; it’s got a height Δx2. That’s a volume, times density is the mass, mv2 over 2 is the kinetic energy. Potential energy would be the same mass times gh, but I should call it gh2. Yep?
Professor Ramamurti Shankar: Oh, you’re worried about the height varying over the tube. Yeah, we are neglecting that aspect. That’s an important point. You can ask, when you say h2 are you talking about the middle or the end and so on. But we are going to be dealing with problems where the height differences are much bigger than the cross section of the pipe. So, we don’t worry about that.
Okay. Now, what is the same quantity here? It’s going to be minus ρA1 ΔX1, times V12 over 2 plus ρ times gh1. That’s the final energy minus initial energy. That’s got to be equal to the work done. Now, who is doing the work on this? Can you guys understand? Where is the work coming from? Yep?
Professor Ramamurti Shankar: Ah. No. Gravity you don’t count as a force when you have a potential energy for gravity. Gravity is included. The minute you write ρgh or mgh, you don’t count gravity again.
Professor Ramamurti Shankar: Right. But if I’m looking at the fluid in question, who is acting on that fluid? Not what it’s doing to others. Who are the others doing something to this fluid? Yep.
Professor Ramamurti Shankar: Now, the walls of the pipe will apply a normal force, and that will not do any work. Yes?
Professor Ramamurti Shankar: Yeah, but that person is somewhere here. But remember, in mechanics you don’t have to go to people in remote places. I’ve told you, when you apply Newton’s law, the only kinds of forces are what? Remember from day one? Contact forces. Except for gravity, which reaches out and grabs everything. We’ve included gravity. Who is this body of fluid in contact with? Yes?
Professor Ramamurti Shankar: This fluid? In between? No, but I’m talking about this fluid. I’m talking about all of this fluid. My taking the difference of that shaded region and that shaded region was a convenience, but I’m looking at the energy change in all of this fluid.
Professor Ramamurti Shankar: Yeah, but the point is the fluid to the left here is pushing it, and it is pushing the fluid to the right. Okay? That’s the only way. It is true there’s a pump somewhere pushing the fluid in the beginning, but you don’t have to go all the way to the pump. In the end you only ask, who is in contact with me? Well, it’s the guys to the left. So, they will exert a force P1, times A1, times Δx1 because that’s the force, that’s the distance, that’s the work done on the fluid. And the work done by the fluid is P2, A2, Δx2, because if it moves to the right it is doing work on the left because it has work done on it. There you go. Now you can see that A1 Δx1 and A2 Δx2 are all equal because that is just the volume of the fluid displaced in a short time. That is just the continuity equation. If you divide everything by Δt, if you like, then you will find that Δx2 over Δt is V2. I’m just using the fact A1V1 is A2V2. So, if you cancelled those factors, what do I get? I’m going to just write it for you. Now, you guys can go home and check it. P1 - P2 = ½ ρV22 + ρgh2, minus ½ ρV12 - ρgh1.
Now, this is a derivation that you can go home and study at leisure in any book you like. They’re all the same; the story is the same. I can only tell you, if you like to know, what the trick behind the derivation was. I don’t mind going over that because maybe there are different volumes in question and you may have gotten confused. But first, let me write down the result. By writing everything with 1 on the one side, and 2 on the other side I get this result: P1 + ρV12 over 2 + ρgh1 = P2 + ρgh2 + ½ ρV22. This is nothing other than the Law of Conservation of Energy applied to unit volume. If you take one meter cubed of the fluid, its mass is ρ times 1. So, this is really a mass of one cubic meter of the fluid. This is its kinetic energy; that’s its potential energy. That’s the kinetic and that’s the potential. You might say, “Why aren’t we just equating kinetic plus potential to kinetic plus potential?” It’s because this chunk of fluid that I focused on is not an isolated system. It’s getting pushed from people behind it, and it’s pushing people ahead of it. So, you’ve got to take the difference between the work done on it, and the work done by it. Then, you will find they don’t quite cancel because the pressures are not equal. If the pressures are unequal, you get an extra contribution. But this is simply the Law of Conservation of Energy. I got it by writing the Law of Conservation of Energy.
But it takes a while for you guys to get used to applying the Law of Conservation of Energy. Maybe you are always used to saying kinetic plus potential is kinetic plus potential. You got used to that, or you also remember kinetic plus potential may be less at the end than the beginning, because there is friction. This is yet another thing. Where there’s no friction, but there are external forces acting on a body, there is gravity, which is included the minute you write a potential energy mgh. There’s the walls, too. I think some of you brought up a very good point. The walls are exerting a force, but that’s perpendicular to the fluid. In a real fluid, the walls will in fact exert a force parallel to the fluid, and it’s called viscosity. So, in a real fluid there will be a drag on the fluid because the fluid really doesn’t like to move right up against the walls. It doesn’t mind moving in the middle of the tube. So, as you go near the edges, the speed of the fluid will relax to zero. There’s a region where there’s a lot of dissipation. So, we’re ignoring what’s called viscosity. We’re ignoring all other losses. Then, this is simply the Law of Conservation of Energy.
And as to what I did with those two volumes, maybe I’ll repeat this so you can all follow this. I’m saying, take all that fluid now, find the total energy in it, and ask what happened to that fluid a little later. Well, fortunately, here the pipe also had some problems [laughter]. Okay. Where is this fluid a short time later? I want you to think about it. This end has moved here, this end has moved there. So, the fluid at the end of the day is sitting there. I want the energy of all of these guys minus the energy of all of this. Then, you can see in the comparison, it’s only this part that has gained, and this part that has lost. This part is common. When I say common, not only is the mass common–the location–at every location the height and velocity are the same. So, in the subtraction this will cancel that. That’ll cancel that. This will have nothing to cancel and this will have nothing to cancel. So, just focus on the relative increments, and that difference is what gives you the net change in energy, and you equate that to the work done.
Chapter 7. Applications of Bernoulli’s Equation [00:53:41]
Okay. So now, we are going to learn how to use this Bernoulli. The main thing to notice about Bernoulli is, for a minute, if you focus on fluid at the same height. Just think about the same height. He tells you whenever the fluid picks up speed, it’s going to lose pressure. Because P + V2 or something, is P + V2 afterwards. If you increase V, you’re going to decrease P. And that’s the thrust of Bernoulli on a qualitative level.
I’ll give you some quantitative examples. So, one example is a baseball. Here’s as baseball coursing through air. If you like, you can sit and ride with the baseball, and say the air going backwards and the baseball is going that way. Now, suppose you spin the baseball like this. If you spin the baseball, it carries some of its own air due to friction between the leather and the air. So, that velocity is counter to the velocity of the drift here on the top and additive on the bottom. So, the actual air velocity on the bottom will be more on the bottom and less on the top because you’re subtracting this vector from the top and adding this vector on the bottom. So, if the velocity is different, this is the higher velocity region, this is the lower velocity region. So, pressure here will be less than the pressure there. So, the ball will sink down. If you spun it the opposite way, the ball will rise up. If you spun the ball side to side, the ball will curve from left to right. So, that’s the spin on the ball. If I spin the ball and release it, then it produces extra forces. This rising and falling is on top of the falling due to gravity. Even in a planet without gravity you will have this extra force. It’s coming simply because the velocity of the top and bottom have been modified.
Here’s another example. Here’s an airplane wing. The plane is going like this. You’ll go right with the plane, in which case the air seems to be doing this. Far from the plane, everything looks the same as if the wing were not there. But near the wing, this is the flow of air past the wing. Notice that these guys above the wing have to travel further than the particles below the wing so they can catch up here, where everything is the same. So, the velocity on top of the airfoil is faster; therefore, pressure is lower. Pressure is lower than on the bottom, then the difference in pressure times the area of the wing will push the wing up. That’s the lift you get when you start a plane. When you’re on the runway, this is why the plane goes up. Once you’re airborne, you can tilt your wing. When you tilt your wing to this angle, then of course it’s very clear that you can get a component of lift. But I’m talking about when the wing is not tilted but horizontal. That’s what gives you the lift. Now, what if you made a wing that looks like this? Okay. So, you’re all laughing. But this has a use too. Have you seen it anywhere?
Professor Ramamurti Shankar: Yes? NASCAR. On the racetrack you want the opposite effect, because you want to push down on the car, and therefore you manufacture wings. You go to the factory where by mistake they built a wing that looks like this, and you can bring it to your race car and attach it, then it will keep the race car down. Of course, you also have the other option of going to the wing and tipping it over and applying it to the airplane. But if it did not occur to you, if you thought once the wing is made like I’m stuck with this wing, well, go take it to your race car, attach it. That’ll push it down. Because you want to get traction. So, you can get the upward lift or the downward lift. Now, it turns out this lift theory is actually somewhat naïve. And I believed it for a long time. Now, I know the story is more complicated. There is a lot of truth to this, but it’s not the whole story. If you go into aeronautic engineering, you will find out that it’s a little more complicated. The reason, the way to calculate it precisely–But this general notion that when the air moves fast it loses pressure is true.
So, here’s another example. If you have an atomizer, you know, you have a perfume here and you’ve got a pump. And then, you have a tube here. When you squeeze the pump, the instant you squeeze the pump, you’re driving a lot of air here at high velocity, whereas the air here, is at rest. So, high velocity air has a lower pressure than low velocity air. Therefore, it will suck the perfume and spray it right on your face [laughter]. Yes. Okay. Very good. You know, I’ve taught you so many things, planet and supernovae, and galaxies, but now I got a rise out of this class. They said, now we are telling you something useful [laughter].
Alright. So, now I’m going to do two problems where I really start putting numbers in. This is very qualitative. I’m going to do quantitative problems. And there are only two kinds. So, when we are done, we are done.
Here is a tank of water. And I poke a hole in this somewhere here, at a depth, let’s call it h. The question is, “How fast will the water come out of here?” Turns out you can use this Bernoulli’s, whether it is called Bernoulli’s equation. You can use Bernoulli to derive this. We just have to pick the points one and two cleverly. So, this is going to be point 1; this is going to be point 2. Point 2, in fact, is just outside that hole. So, what does Bernoulli say? P1, which is atmospheric pressure, plus ½ ρV12 + ρgh1 = P2 + ½ ρV22 + ρgh2. Let’s modify this as follows: let’s pick h2 to be zero, then h1 is just this. I mean, you can pick the zero any way you like. So, where the hole is I’m calling zero. So, I don’t have to worry about this one. V2 is what I’m after. What about P1 and P2? P1 is atmospheric pressure there, and right outside the hole it’s also the atmospheric pressure. So, P1 and P2 you cancel because they’re both equal to atmospheric pressure. How about the velocity here? You know that if you punch a hole in the tank and you drain it, it’s going to start moving down. But we’re going to imagine that’s a huge tank, ten meters diameter, and this is a tiny pinprick. Then, this velocity is negligible. You can put that back in if you like. I’m just going to ignore that for now. That tells me ρgh1 = ½ ρV22; cancel the ρ, and you find V2 = 2gh. V22 = 2gh.
Now, you remember this formula from day one. This is the velocity a droplet would have if it, say, spilled over the top and fell straight down. Because that is just the mgh being converted to ½ mv2. Because by the Law of Conversation of Energy, the minute the fluid starts coming out of here and draining out of the top, I have traded droplets of water at the top for droplets of water at the bottom. Therefore, since they had potential energy at the top, they most likely have kinetic energy at the bottom. But be clear about one thing: the drop coming here is not the same drop moving at the top. If I push it over the top, that would be the same drop, and that’s very clear to all of us. The beauty of this is this water is pushing down and something’s coming out here. It’s coming out with the right speed so that drop by drop, the kinetic energy of what comes out here, is the potential energy of what’s draining at the top. So, you can punch the hole at various places. You know, once you punch a hole here it’s going to land there. And you can imagine the fun you can have with this kind of problem. Where do I punch the hole so that if my dog is standing here it’s going to feed the dog? Or don’t feed the dog, go up in front of the dog, go past the dog. Get the biggest range, smallest range, a whole bunch of problems. They will combine this Bernoulli who will tell you how fast it’s coming out with chapter one of how to do trajectories once the fluid is coming out this way.
Okay. Last topic. It’s again–it’s not even a new topic. The last application of Bernoulli has to do with the–what you call the Venturi meter. By the way, you can imagine this is just a cross section of possible applications of fluid dynamics. You can take a whole course on fluid dynamics.
So, let’s take the following problem. So, you’re going in an airplane, and you want to find out how fast the plane is going. How do you think you find out? I will show you know one device people use to find the flow rate, to find the speed of the plane through the atmosphere. So, you go to the plane and you attach the following device to the underside of the plane. Well, this, I’m imagining in my mind a symmetric thing. This is a pipe with a constriction. Here the air is coming in at the speed of the plane itself. See, in real life the plane is going through the air, but go sit with the plane, because the air is going backwards at the speed of the plane. So, that’s a cross section A1 here. V1 is what we are trying to find. I hope you understand. That’s our goal. Then it comes to this region where it has to speed up, because A1V1 is A2V2. If it speeds up, the pressure of the air is going to be lower. Let’s first calculate the pressure difference between this point and this point. Now they’re both at the same altitude.
Here’s another example. You’re 5,000 feet above the ground; don’t worry about altitude to here or there or there. That’s, 5,000 is the big height, and that ρgh cancels on both sides. We don’t worry about the variation in height over this gadget. It’s a very, very tiny thing you attach to the underside of the plane. So then, I can say P1 + ½ ρV12 + ρgh1, I’m not going to write because h1 and h2 are going be equal. Then, it’s P2 ½ ρV22. Therefore, P2 - P1 = ½ ρV12 - V22. Now, V1 is the speed of the plane. Now, V2 is not the speed of the plane, but we know what V2 is because V2 times A2 is equal to V1 times A1. So, we can write here ½ ρV12 - V22 = V12 times A1 over A22. So, pull out the V1, and you write it as ½ ρV12 times one minus (A1/A2)2.
So, let’s digest this formula for a second. In this problem, A1 is bigger than A2. So, you may worry that 1 - A1/A22 is negative, but so is P2 - P1 because P2 here is certainly going to be lower than P1 here. So, if you’re happier, you can flip it backwards and write P1 - P2 is something. You see that? You can flip it over, provided you flip it over here. But ask yourself, “What do I need to know to find the speed of the plane?” The speed of the plane is here. The ρ is the density of what? Can you tell me, ρ is density of what? Air; ρ is the density of air. A1 and A2 are known, because you designed the tube. So, if you can find the pressure difference, somehow, you can read off the speed of the plane. So, what people do to find the pressure difference is they go and they take–they punch a hole there and they put here a fluid, like oil. If the plane is not moving, you can tell these two heights must be equal because that’s a condition for the hydrostatic equilibrium. Because these two pressures are equal. But if the plane starts moving and the pressure here is lower than the pressure here.
Imagine this is a high pressure; that’s low pressure. It’ll push the fluid up and it’ll start looking like this, with a little extra on the other side. And how much extra is it? Well, let’s think about it. The pressure at these two points is the same; the pressure here is what I called P1; the pressure here is P2 + ρg times the height of this fluid. This ρ is not the density of air. This ρ is the density of, oil let’s say. So, the minute the plane begins to move, the pressure here will be higher than pressure here; the fluid will be pushed up. And the difference in these two heights directly is a measure of the pressure difference. That, in turn, is a measure of the velocity. So, what you will try to do is then, forget all about the height difference, and if you’re clever enough you can put marking so that by counting the difference in the markings you can translate to the speed of the plane.
You can also use this to find the rate at which oil is flowing. Suppose oil is flowing in a pipe and you want to know the rate at which oil is flowing. Again, create a constriction in the flow, then put some tubes. This cannot be oil; this has got to be some other fluid. Once again, there’ll be pressure difference and there’ll be height difference. And from the height difference you can find the rate at which the fluid is flowing here. So, what’s the trick we use? Yes?
Professor Ramamurti Shankar: Oh, for this one?
Professor Ramamurti Shankar: Well, I don’t–I think all I’m–As far as I can tell, there’s going to be a pressure difference and that’s going to translate into height difference. And if the two do not mix, I don’t know the danger. I think one danger you may have is if the fluid comes in here, and if it can penetrate this, it will start mixing. Right? So, for that purpose it may be better if this one is a higher density than the one on top.
Now in practice, I am not really sure what kind of fluids people use in any engineering thing. My–I get very shaky when you start going to the real world. It’s been my practice to avoid it as much as possible, which is why I chose this career. But people who really want to build something have to worry about what fluid to use, you know, what starts mixing, what doesn’t mix, what’s the accuracy. So, I don’t. If I thought hard about it, I would try to reduce everything to Newton’s laws or some laws of thermodynamics. But it would take years and years to go from there to something practical. So, I don’t know in practice what fluids people use. So, this is another example of Bernoulli’s principle.
Final one, which is the kind of a problem you sometimes do get. I’m not going to write the equations or do the numbers, but I’ll tell you how to think about it. Suppose you have a tank of water. And I make another tube here and I cap it. And the fluid here comes to some height. First question you can ask is, “What’s the height to which fluid will rise here?” Now, you know from your high school days it will rise to the same height, but what’s the argument that you will give today for why that height has to be the same? You will use Bernoulli’s principle, and you will take two points here, for example. Luckily, there’s no velocity to worry about. There is no height to worry about. So, P1 = P2, which is what we did use by other considerations. Those pressures are equal. If that is atmospheric pressure, that’s atmospheric pressure; then, these columns have to be equal. They’re starting with atmosphere plus ρgh; you want to hit the same number here as there. But now, if you open this pipe and let the fluid really start flowing, then the story is different. The minute fluid stops flowing here–starts flowing–remember from the Bernoulli, if you’ve got a velocity, you lose pressure. Pressure here will be lower than the atmosphere, and then this will drop. It will drop so that atmosphere, plus that height gives a pressure here, that is atmosphere plus the bigger height gives the pressure here. And the velocity here is assumed to be negligible, and the velocity here is some velocity with which the fluid is coming out. Okay. So guys, have a good holiday in spite of what I’ve done to you most reluctantly, and I’ll see you all after the break.
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