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Buoyancy and Density · Acedra
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Buoyancy and Density
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0:00
All right lets finish up the last topic in chapter 2, well be starting on chapter 3. Buoyancy, you know buoyancy, float on your back in your swimming pool or the ocean, float on your back. You’re buoyant. So lets talk about buoyancy a little bit, everybody has good intuition about it. But if… cup of water, I’m not drinking, I’m not going to do this today, today’s not that bad of a day, Monday, I’m not that desperate. I put them in two cups so you could check. Golf ball, it’s down at the bottom.
1:20
Ping-Pong ball, it floats on top. They are roughly the same size, why does one sink? Students: Buoyancy, density. Density and the weight, that’s right it’s the same volume, so density or weight. What makes the Ping-Pong ball, its not half way down even, it’s not half way down, it’s mostly floating. It’s into the water way down here. So there’s something else happening there too, there’s something else happening. What 2 forces do you think acts on it? What do you think acts on that?
2:12
What 2 forces? Student: Buoyancy is one of them. Ok and? Student: Gravity. Which one is heavier? Well gravity yeah, this one is heavier, so this one is lighter. So that’s right, the buoyant force and the weight. If it sinks to the bottom guess which one wins? The weight, the weight. Float on top, what do you have? Buoyant force up, weight down, it’s floating on top. Let this thing go in the ocean, it will sink a long way.
2:43
So yeah it’s something about the weight, well you know what causes the buoyant force? You can say buoyant force, but so what? I mean I can say it’s the bazooga force. What’s that? A bazooga force! No, why is there a buoyant force? Well that’s what this picture is going to show you. Here’s an object, a golf ball, a Ping-Pong ball, whatever. The top, take a slice, a little circular sliver of it, a cheese stick thing. The pressure up here, not too big. The pressure down here, uh-oh, bigger.
3:28
Were deeper, bigger pressure. Bigger pressure, smaller pressure, which one wins? Bigger pressure. What’s the buoyant force want to do? Raise this thing up. What’s holding it back? Obviously the weight going down. So it’s like a balance between the weight, if the weights too big the thing sinks. When you fill your lungs with air and you float on your back; yeah that’s right your density goes down, you have air in your lungs. So yeah so here’s our little circular area DA for instance.
4:04
Do it vertically, here’s the free surface, pressure up here is Po. The equation, pressure goes up as we go down in the fluid. Po is the surface pressure at the very top. And this is in your book, I’m just repeating what’s in your book, I just want to go through it so you can see it more clearly. The sum of the forces in the z direction due to pressure, due to pressure. At the bottom big pressure, which way is that pressure pointing? You know the rule.
4:36
Pressure is a compressive stress it always points on the surface, into the surface. The pressure at the top, minus, pointing down, normal stress. Subtract the two of them, this little piece off here, subtract the two. The Po’s cancel out, rho g times h2 minus h1 times DA, there it is. But notice what the differential volume DV is, DV the differential volume, it’s the area times the height. What’s the height? h2 minus h1.
5:21
So this is like a differential volume, so replace that with the differential volume. So dFz, the z direction, the z direction is vertically in this picture, up and down. The net force in the z direction equals this guy here. Integrate both sides. Integrate over what? Integrate over the volume, total volume V. Bring out the constants. We assume incompressible, rho is a constant. We assume g is a constant, integrate dV over V. V, and we end up with rho g V. So F buoyancy equal rho g V.
6:39
Whats rho g? gamma. What’s gamma? Newton’s per cubic meter. What’s V? cubic meters. Newton’s per cubic meters times cubic meters, Newton’s. Yeah that’s the force, Newton’s. Ok, what’s the volume? It’s the volume occupied by the part of the object submerged into the liquid. Acts through the centroid of the displaced volume, to the centroid of the displaced volume. What does displaced volume mean? It means the volume of the liquid pushed out of the way by the object.
7:49
That’s the displaced volume, it’s this volume inside that line right there. That’s the displaced volume. Oh you can go through the proof just like we did previously, the textbook doesn’t do it, they just say it goes through the centroid of the displaced volume. So, but you could go through all the gory details. Archimedes principle of course, Archimedes principle from science. So which way does the buoyant force act? Upward. What does it act through? The centroid of that volume.
8:27
Ok that’s the concept of buoyancy; you got a couple problems for homework on buoyancy. Ill do an example here. Ok example, a log, a log floats in water. Were told that one-fourth of its volume protruding above the water surface. One-fourth volume above water surface, what is the specific weight of the log? Ok so here’s our log, one-fourth of it is protruding above the water surface. Draw a free body diagram, buoyant force, weight.
10:00
The weight acts through the centroid of the log, the buoyant force acts through the centroid of the displaced volume. Where is the displaced volume? Here. That’s how much water was pushed out of the way when the log went down in the water. That’s called the displaced volume. Where’s the centroid of the displaced volume? Right there somewhere, I don’t know up or down, but I do know it lies along that line right there. That’s why I put FB there acting through, not through that, there’s two things.
11:00
The center of gravity and the centroid of the displaced volume. The weight acts through the center of gravity. FB acts through a different place, it acts through the centroid of the displaced volume. Sometimes they are on the same line, sometimes they are not; it depends on the geometry. Anyway here they are, so sum force in the y-direction equal zero. The weight equal FB. Ok the weight of what? The log, the whole log? Well of course the whole log.
11:38
Ok gamma of the log, times the volume of the log. FB here it is, gamma of water, what did you push out of the way? Water. The volume of what? The volume pushed out of the way. Displaced volume. Gamma of water times the volume displaced. Ok, gamma, we want to find gamma. Gamma equal gamma water times the ratio volume displaced over the volume of the log. 62.4 pounds per cubic foot, now you got to be careful of course. Where’s the displaced volume?
12:44
That’s the line volume here, the line volume, where’s the volume of the log? That’s the circle; that’s 3/4ths . 1/4th is above water, 3/4ths is below water, 3/4ths is below water. Displaced volume ratio, 3/4ths . Evaluate them; it comes out to be 46.8. Ok so you should be able to work the homework, it’s pretty straightforward in that topic. You can also get into what we call stability, when something might tip, or not tip.
13:30
Fairy boat, too many people on one side on the top deck or something, and it’s a wave that goes over, well I want it to come back and right itself. Well then it’s part of the buoyant force. But there’s no homework problems on stability, so they are just on the buoyant force. Ok so that pretty much wraps up chapter 2, it’s pretty long, we’ve been in this chapter since the second-class meeting about. So yeah it’s pretty long, that’s why the first midterm includes Chapters 1 and Chapter
14:00
2, because 2 is such a big long chapter. Ok now lets, were going to move on to Chapter 3, but were not going to have any of Chapter 3 on the first midterm. Some people cover the material in Chapter 3 later, I’m going to follow the textbook on this one, so we’ll follow what’s in your textbook; and were going to look at the Bernoulli equation. All right I’m going to start off, and maybe we will define streamline first.
15:01
Ok so here is some cylindrical object, and we have flow around the object, so it goes around like this. I won’t draw the top and the bottom, ill just draw the top, the bottom is symmetrical. This is like a piece of pipe in an air stream. When people draw pictures like this these lines they draw are called streamlines. Definition, the official definition. Ok that’s the definition of the streamlines. So here’s the velocity vector at this point. Here’s the velocity vector at this point.
17:06
Here’s the velocity vector at this point. Well you can see what were doing. You ok yeah? Ok well the streamlines are lines; I mean do you see them in the real world? Of course not no. Sometimes you do, you’ve seen the ads for cars in the wind tunnel and smoke, and you see the streamlines go around the car and follow the back contour, and they break off in the back. Those like smoke lines; those are streamlines in steady state.
17:49
So we can see them by adding smoke or hydrogen bubbles to a liquid flow field; or smoke to a gaseous flow field. But that’s what they look like, that’s their official definition. We’ll talk more about streamlines, path lines, and streak lines later on, but for right now were just concerned with streamlines. And were going to define N and S. N means normal, N direction. S means along the streamline direction, so we attach our coordinate system to the streamline.
18:47
So S is along a streamline, N is normal to a streamline. So we have 2 components of acceleration. Along a streamline As, dv dt, partial v with respect to s, partial s with respect to t, product rule. Dv ds, and of course ds dt is just the velocity. And the normal acceleration v squared over R where R is called the radius of curvature. Were going back to our dynamics background for those 2 guys. So you can have both, stream wise acceleration, and a normal acceleration.
20:14
So were going to take a streamline out of here, I’ll take one over here maybe. So here’s a streamline, and on that streamline were going to show a very small differential volume of what’s called a liquid. So picture right, I’ll draw a picture like you see in the textbook so you can follow it. Ok so we have our little differential element and lets see were defining delta s this way and delta y, delta n that way. So here’s delta n, here’s delta s, and we have the weight of that little differential
21:05
element acting down. And there is a pressure force on the left. So this is the pressure in the middle is P, the pressure on the left, the pressure on the left, P minus dP multiplied by the area dn dy. The pressure on the right side is P plus delta P dn dy, dn dy is the area over which the pressure acts. And I think that’s all we’ve got in there, in that direction yeah let me make sure. Oh lets get the angle theta, this should be angle theta, here’s the angle theta. We need that.
22:46
So were neglecting viscous effects, no viscosity, otherwise viscosity would act on these surfaces here, we’re neglecting viscosity so we’ll put that down. We call that kind of fluid where we neglect viscosity we call that inviscid fluid, inviscid. So if you see some words on an exam you know what they mean, inviscid, neglect viscosity. Incompressible, density is a constant; you can pull it outside the integral sign.
23:46
Ok inviscid fluid, all right so lets write the sum of the forces in the s direction. You can see we have obviously 2 pressure forces left and right. But we know that when we do the summation of forces that it’ s going to be mass times acceleration, so lets put that down first. Mass times acceleration in the s direction is equal to the acceleration in the s direction up there. So we have delta m V dv ds. Ok that’s got that, ok and let’s see, lets replace delta m with rho delta V, don’t
24:51
forget mass is density times volume. Mass is density times volume. Here’s where I told you there is a V and a V slash, V means velocity, V slash means volume. Differential volume, multiply by V dV ds. Ok and for the weight, delta W the weight in the s direction minus delta W sine theta. That little guy right there is the component of the weight in the s direction, notice which way it points, minus the sign.
25:53
What’s the weight? Gamma times the volume, so minus gamma delta volume sine theta, component of weight in the s direction. The net pressure force, the net means subtract something, net always means subtract something. So if we subtract this guy and this guy, 2 pressure forces, 2 pressure forces, subtract them and we end up with delta F pressure force in the s direction. Ok this guy minus that guy, these cancel out, plus P minus P, minus delta P, minus sign, minus delta P minus 2 delta Ps.
27:27
That delta Ps again just means in the s direction, no need to keep that subscript on there. So this then becomes, replacing the delta Ps 2 this guy right here, that gives us that. Using this, all right summation of the forces equal mass times acceleration, what are the forces? The weight, the force due to weight in the s direction. The force due to pressure, net pressure in the s direction, so here’s our components ok, so next step put all these guys together.
28:59
There’s the weight, here’s net pressure force. That’s this left hand side equals this right hand side, here it is. You can replace V dV ds with differential one half d V squared dS, d V squared dS is 2 V dV ds, 2 V dV ds times the half is V dV ds, so they are the same thing. Finally we get an expression dP, let dP come out of there. Multiply through by ds, multiply through by ds, cancels out, cancels out, cancels out. Your left with dP, change, move these two guys right hand side plus dP.
31:48
He’s over there, one half rho dV squared put him over on the right hand side, change the sign, gamma dz, there it is. So this is only valid though, along a streamline. Go back, what we did, we took a little fluid element along a streamline, like this. That little differential element had forces on it Number one, gravity creates a weight in the s direction. So the weight, the component in the s direction, sine of theta, that’s the weight. Acts down, minus sign of course.
33:00
The pressure acting on the left and right hand faces of the element, number 1 call the pressure in the middle P. If you go in a negative s direction, that’s towards me, of course the pressure is P minus dP. If you go in the positive s direction, that way to my right, then it’s P plus dP. Multiply a pressure by an area that gives you a force. If you take the difference in these two pressure forces, left and right hand sides, it’s going to give you a net pressure force. Net pressure force.
33:40
So take this guy minus that guy and you end up with this one, use this guy in here. Ok delta Ps right here, put that in there. Ok so net pressure force in the s direction, weight in the s direction, put the equation, summation of forces in the s direction. Equal weight plus pressure, equal mass times the acceleration, this term is mass times the acceleration redone in a little better format. Rho dV times velocity dV ds. Put all that stuff in an equation, here it is down here.
34:33
Play with it some more, get in a form that’s useful, get it in a useful form; that’s our object. It’s not in useful form yet, we’re getting there, make some changes here, this guy right here, multiply through by ds, put all the stuff on one side of the equation, there it is. Ok were almost, almost there with Bernoulli. Ok yeah? Student: What happened to the sine theta? Ok let’s see, where did the sine theta go? He’s, yeah, he’s yeah, he’s up here, gamma, what did we do?
35:11
Did I get that one right; let’s see if I got that one right. Sine theta dP ds, over there we got sine theta dP ds, gamma dP ds. Oh yeah this is, let me put down, I didn’t put this, it’s in your textbook, but I’ll put it on here. This is theta, that’s the wrong way; it’s like this, like this. This is ds, this is dz, this is theta. Ok so sine of theta is what? Dz over ds? Yeah. I’m going to draw one more ok? Here, here, here, dn, dz, theta.
36:11
Cosine theta dz over dn, you can play those games to get rid of the sine theta, ok just go back into the geometry again. Thanks for mentioning it, because it looked like it disappeared, it didn’t disappear, I replaced it with a differential. Ok now obviously we don’t want dP d V squared and dz. Next step obviously is to integrate, ok so were going to integrate. Here’s the terms we divide through by rho, divide through by rho, dP over rho, one half V squared, ok.
37:33
Gz, g is a constant, pull it outside, call the constant C. When you integrate we end up with a constant C. What’s gamma? Rho times g. What do we divide by? Rho. Rho g divided by rho is what? G. Times z, yeah g times z. Integrate again, pull the half outside, pull the rho outside, divide by rho, V squared, that’s where it came from. So C is a constant, notice that this is written for a streamline, C has the same value anywhere along a streamline.
38:37
Ok so now we assume incompressible fluid, rho equal constant. Pull rho outside the integral sign, ok end up with P over rho, V squared over 2, gz, equal a constant C. Every term has units meters squared over seconds squared. The famous Bernoulli equation. It’s a pretty straightforward equation because it’s inviscid. Later on we’re going to repeat this, but were going to allow for viscous stress terms, where? On the two top and bottom sides of this, that’s going to be a lot rougher derivation.
40:20
But this one, it is good a lot of times, you have to remember though, remember its assumptions. We pulled rho outside, incompressible. We said early on inviscid. Along a streamline. We didn’t emphasize it, but of course it’s for steady state. Ok only apply Bernoulli’s if those 4 assumptions are satisfied. Oh there’s an exception later on, but for right now that’s the 4 that we need. Ok what’s a streamline? Ok right here, lines that are tangent to the velocity vector at every point in the steady
41:46
flow, oh there’s the word steady right there, steady flow. Ok so it applies to along a streamline, that equation applies if you want to call this point 1, then you can call this point 2, and it apply along a streamline. Here’s what you don’t want to do, if that’s point 1 you say I don’t like point 2 there, I’m going to put point 2 up there. Oh no no you didn’t read the board very well did you? It says it applies along a streamline, not across a streamline, not across a streamline;
42:19
it applies along a streamline. So no you can’t do that. Point 2 has to be somewhere along that same streamline to apply the Bernoulli equation. Ok now if you if you, I’m going to put this over here, because I’m going to save this middle panel for something else, so I’m going to erase this, and save this space here. You can also apply similar kind of derivation normal to the streamline, normal to the streamline. If you go through all those details, this is what you end up getting.
43:49
This guy here, the radius of curvature. It’s more complicated, because to integrate that second term, velocity squared divided by radius of curvature dn, you need to know how both V and R vary with n. So that’s normal to the streamline, and that’s along the streamline. Horizontal flow, what’s our R? Infinity, what’s 1 over infinity? Zero. If it’s horizontal lines that term drops out, that’s the easy case. That’s the easy case. Ok, another form of Bernoulli’s, multiply through by rho.
45:14
Ok rho times P divided by rho, o rhos cancel. One half rho V squared, ok rho times g, gamma, rho times g, gamma. Units pressure ok. Ok. All right, P by itself, it’s called static pressure. Rho V squared over 2, it’s called the dynamic pressure. Of course it depends on velocity, so it’s called dynamic pressure. Gamma times z is called the hydrostatic pressure. It depends on the height you are; we did this back in chapter 2. P plus rho v squared over 2 is called the stagnation pressure.
47:44
The stagnation point in a flow field is where the velocity is zero. Stagnate, stagnate, zero velocity, stagnation point. Let’s see which equation I need here. * Mumbles * Ok, now I need this board here. Now multiply by rho. Take the same equation, take the same equation, now divide by g. Ok there it is, divide by g. P divided by rho g, oh rho g is gamma, it’s P over gamma. Divide by g, V squared divided by 2 g. Divide by g, plain ol’ z. Constant. Ok, box it in.
50:13
Units, ill take this guy, third guy, meters or feet. This is called the head form of Bernoulli. P over gamma, pressure head. V squared over 2g, velocity head. Z, elevation head. Ok you got 3 different forms of Bernoulli equation, they are all the same equation, but they are just packaged in different ways, it depends the problem you got which one you want to use. If somebody asks you for the stagnation pressure you use the middle one.
51:42
But it’s to you which one you use; sometimes some are easier than others. If I know gamma of oil is 8,200 I’ll use this guy, I wont try and get rho, don’t have to. So again it depends which one is more convenient to you, but they are different. If you want to talk about the equivalent pressures, this guy talks about the equivalent pressure. Static dynamic hydrostatic stagnation. If you want to talk about the head form of the equation, the pressure head, the velocity head, the elevation head.
52:20
And this was the one we derived, so this has rho in it, this has rho in it, this has gamma in it. So it depends again which one you want to use, but no matter which one you want to use for right now these 4 things have to be satisfied. Incompressible fluid, inviscid fluid, steady state situation, and only applied along a streamline. Ok so let’s take a look at a problem, let me see if this edition and our earlier edition 3 37, let me find 3 37.
53:12
Yeah it’s in our book 3 37, so ill do 3 37, put it over here. All right here we have a tank, which is pressurized. Lets see what that is, that comes out of there, ok. Ok an inviscid, this is what it says now, an inviscid, incompressible liquid flows steadily. First 6 words, major giveaway. an inviscid, incompressible, flows steadily, ok Yes, yes, yes, not #3 yet, not #3 yet, we’ll get there. In this tank is a liquid, gamma. Pressurized tank, air, pressure gauge, 10 psi.
54:47
Velocity at the exit, 40 feet per second. 16 feet, let me make sure, got 15 written down here, let me check. Oh I’m sorry 10 feet, 10 feet, 5 feet. So 10 feet from here to here, and 5 feet from here to here, got it. Ok what is the specific gravity of liquid at 10. So find the specific gravity of the liquid. Ok, so I can use any one of those 3, I like to use that one, so ill use that one, that’s just a reason not to.
56:16
So lets get rid of this one, and lets get rid of the other one, and we’ll just leave this one on the board. The head form of the Bernoulli equation. Ok, so when you want to apply the Bernoulli equation, you’re going to apply it between 2 points in the flow field, 2 points in the flow field. You want to choose those 2 points so that you know the most stuff about the problem. You say ok if I choose a point can I find P, V, or z there?
57:13
And of course I’m going to give you the obvious point first, I’m going to choose point 1 here. Why would I do that? Because I know what the pressure is, it’s 10. I know what the velocity is; it’s a free surface, zero. I know how far above the x it is, it’s 15 feet. That’s how you decide where to put point 1, where you know the most information. So I’m going to write down that equation put the subscript 1. P1 over gamma plus V1 squared 2g, plus z1.
58:07
Ok now I go along a streamline until I find a point where I know some more stuff. Oh I know this point down here is 15 feet below the free surface. Oh I know the velocity here, 40 feet per second. Oh I know the pressure here it’s zero, voila that’s point 2. I know 3 things there. Now there’s the Bernoulli equation applied between points 1 and 2. All right lets put the zero stuff in first. P2 exhaust atmosphere, why do I know that? It shows in the picture.
59:07
It shows the pipe ending and a velocity vector coming out of that pipe. That means that you are to assume in that situation that it’s exiting to atmospheric pressure. P2 is zero, I’m going to call z2 zero, that’s my choice, that’s called a datum, d-a-t-u-m, d-a-t-u-m. Datum, where you measure things from. If you’re in a plant, a factory or a plant, the datum is typically the floor of the plant. How high is the water tank above the floor? 30 feet etc etc.
59:40
So the datum you’re free to choose, you could take the top I don’t care, take point 1, I’m taking point 2. How far above is point 1? 15 feet. What’s the velocity at 1? Zero. The pressure at 1? 10 psi, 144 divided by what? Gamma, of what? The fluid in the tank, gamma, I don’t know it. Equal, these guys are gone, zero plus V2 squared. 2 32.2 plus zero, we don’t know g always assume 32.2 or 9.81, that’s a default value. Solve for gamma, 146.3.
1:00:48
Remember what specific gravity is, it’s your choice now, here comes your choice. You can define specific gravity in terms of density ratio, or specific weight ratio; your choice, take the easy way out. I just found gamma, I don’t want to find rho, so I’m going to use gamma. 2.34. Student: Once you know the specific gravity, you can usually determine the fluid that it is from the table? You can go to the inside front cover, or back in the tables and look for SG and see what
1:01:52
fluid it is, yeah exactly. Oil is like 0.8, mercury is 13.6, water is 1.0, and in between you know a ton of different fluids, but yeah you can do a search. Pretty much forgot what’s in that tank, good point, that’s a good point. Ok I’m not going to start problem 2 right now, I’ve only got 8 minutes, I don’t want to get halfway through and stop then pick up next time, so well do it next time. Ok I passed homework back, if you didn’t get it, it’s up here in front over there.
1:02:23
Homework is due right now. On blackboard homework all the way through next Monday. Test in 9 days from today.
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