# Physics Siphoning - Torricellis Law

###### Explanation

Siphons are tubes which draw fluid over the rim of a tank to a lower point. After an initial pressure change to initiate the flow, siphons operate continuously due to the pull of gravity. Torricelli's law says if a liquid flows from an opening in a container, its speed is the same as the speed that fluid would have if it dropped in free fall from the top of that container. Torricelli's Law derives directly from Bernoulli's principle.

###### Transcript

Alright let's talk Siphons and Torricelli's law. Now first off what is a siphon? Well a siphon is a tube that I stick into a pool of water or some other fluid and then pull up out of the top and then I have the other end of that tube lower down closer to the earth than the tube in the water. And so what happens is, if I can initiate a flow, if I can get that fluid flowing up through there, so that it comes over the top and then falls down it's going to pull the rest of the fluid with it. And so people will use this to empty pools and sometimes to empty gas tanks but it is a very important physical process that uses Bernoulli's Principle. Now what's Torricelli's law? Torricelli's law is named after the Italian physicist Evangelista Torricelli who derived it in 1643 and what he said was that if you take a container of liquid and you've got a hole in the bottom of it, then the liquid will flow out of that hole at the same speed that it would move at if would just take the liquid from the top and drop it.

Now the direction is going to be different, if I take liquid from the top and I drop it, it's going to be going down. But if I've got a tank and I poke a hole on the side of it the liquid is going to be going horizontally is going to be going over, but the speed is the same and that's Torricelli's law. Alright so let's go ahead and look at a problem involving a siphon and we'll see where these Torricelli stuff comes up and also where Bernoulli comes up it'll come up all the time. Alright so let's look at this siphon problem right here, I've got a pool of water, we'll just take it to be water, it could be any other fluid though and I've got my hose I've put it one end of the hose at point c down at the bottom of the tank of water which is 5 meters below the surface, then I pull it up and out. It's got to go over the edge of the container and that's a point b which is 2 meters above the surface and then I bring down the other end to point d which is 8 meters below the surface. Now the reason that the siphon is going to work is because the fluid is going to come out here at a lower point in the gravitational field that the earth than here where it came in.

Alright and so that what's going to give me the energy that I need to get over this potential barrier. Now if I'm just going to leave that thing sitting here, it's not going to do anything. I got to initiate the flow first, alright and the way that I do that is I have to put a pump over here and pump out some of the air at point d or I'll just use my mouth and I suck on it real quick to pull the liquid up to point b so that it realizes that it can then fall down. And then once I've initiated that flow I just stand back and the whole container will empty by itself. Alright now how are we going to determine the actual numbers associated with this problem? Well it turns out that Bernoulli's principle is exactly what we need. Bernoulli's principle let's, I've written it over here, now I've written it in the case of constant density which is fine with water because water is almost incompressible. Very, very difficult to change the density of water you need a lot of pressure to do that and that's something we're not going to have here. So we've got pressure plus density times acceleration due to gravity times height plus one half density speed squared is a constant.

Now we can interpret this in the exactly the same way as conservation of energy. Pressure represents a sort of potential energy associated with the fluid itself, rho gh represents gravitational potential energy just like mgh except I divided by the volume which is essentially what we always do with fluids. One half rho d squared is just like one half mv squared that's kinetic energy. So this is just conservation of energy, now as with essentially all conservation of energy problems the way we're going to approach this is by finding 2 different points. 1 point where this whole expression is really easy and I can evaluate it without even having to try and I'll know all the numbers I can just write it down and that's what gives me the constant. And the other point which contains information that I want to know, and then I just say okay they got to be the same. And then I'll solve the equation and get everything I want. Alright so the most complicated part of this problem essentially any part of it, is setting it up, trying to decide which parts we want to use and what we know about those parts. So what I want to do is just kind of show you how to set it up and I'll give you the answers but I'm going to go through a full detailed calculation.

Alright, so let's go ahead to part a, part a asks us for the gauge pressure to initiate the flow. Now remember gauge pressure is pressure minus atmospheric pressure. Now we said that in order to initiate flow we're going to have to suck on the end of this hose at point d. So that means that the pressure up here at point b is going to have to be less than atmospheric, so the gauge pressure will be negative. Alright now when we've got flow initiated the water has got to come up to this point, just right at that point. It won't be moving yet but it's going to get right up to that point. And I want to know the pressure right here okay. So I need 2 different points in this fluid flow to apply Bernoulli's principle to. Okay so let's see obviously I got to use point b as one of them because I want the pressure at point b and how else I'm I going to get that? What's the other point that I should use? Well it turns out that point a is the best point why? Well because whether or not I've started flow point a is associated with water that is barely moving right there at 0 height. Notice all the heights are measured relative to the surface of the water. So that means the surface of the water I can take at 0 height, then the height of point b will be 2 meters, the height of point c will be negative 5 and the height of point d will be negative 8.

Right so my 8 just going to be 0 at point a, my speed is going to be 0 at point a and what's my pressure going to be? Well the only thing that's pushing on the water a point a is the atmosphere. So it's going to be atmospheric pressure, so that means that all 3 of these added together the Bernoulli sum will just be p atmosphere okay so that's at point a. At point b what do I have? Well jeez I got the pressure now I've just wanted to initiate flow, the flow has not initiated yet. So my speed is 0 at point b but my height is 2 and so all I need to do is solve for minus p atmospheric and if you go through the numbers, you'll find that p gauge which is p minus p atmospheric is negative 19,600 Pascal's notice it's negative and that means that the pressure is less than p atmospheric. Alright let's go ahead to part b, so part b asks us for the speed of the flow. Once the flow has been started, so now it's flowing alright so this isn't like at part a where I took the speed to be 0.

Okay, so I need to know the speed of the flow, alright well again I want to use point a as one of my points because point a still is essentially not moving because the pool is so much bigger than the hose. So I mean the flow can be flowing through the hose but that surface of the whole pool is going down real slow so I'm just going to ignore that. Alright so I still have p atmospheric at point a but which point should I use to determine the speed of the flow? Alright this is a little tricky, alright now once you've seen a couple of times you get used to it, it makes perfect sense but it might not be something that you thought of or that you would've thought of. We're going to use point d, the reason that we're going to use point d is because point d is open to the atmosphere and that means that its pressure has to equal the atmospheric pressure. Notice that, that's not true at point b or at point c. I don't know the pressure there so I don't want to use those points. But at point d I know the pressure is p atmospheric, I know the height it's negative 8 and I know, well I don't know the speed but that's what I want. One half rho v squared, now notice what happens here and this actually is Torricelli's law right here notice it comes directly from Bernoulli's principle which is one of the nice things about Bernoulli's principle. Once you got that one, you can sweep everybody else off the table because Bernoulli gives you everything.

Alright p atmospheric gone, and look what we got here, the density is going to cancel so this is going to give us rho g8 equals one half rho v squared. The densities cancel, and this is going to give us v equals square root 2g8. Now that 8 is really the height, so this should be familiar from kinematics. Speed equals square root 2 gh and that's exactly what we get when we drop something, so that's Torricelli's law. That's why it's true, it's because these 2 pressures cancel. Alright that's why it's true, it comes straight from Bernoulli very, very nice of course the answer if you just plug in the numbers I think it's 12.54 meters per second. Alright let's go ahead to part c, I want to know the pressure at point b during flow. Alright this is fairly straight forward, alright now that we've kind of gotten through part b which I think is the trickiest one. Part c is pretty straight forward, I want to know the pressure at point b, so what point I'm I going to use? I'm going to use point b, what other point should I use?

Right if I want to know the pressure at point b I'm going to use a and b because a is easy and b has the information I want. So I'll just plug in for a obviously again it'll just be p atmospheric and then for b I'll have pressure which is what I'm looking for, I'll have height which is 2 and I'll have v which is the answer that I got in part b because the flow through the whole hose is at the same speed, so we're good to go with that. Plug in all the numbers and we end up with a pressure for part c of 3,325 Pascal's. Alright one important point to make about this answer for part c is that this pressure has to be positive. Notice that it doesn't ask for the gauge pressure, we're not solving for the gauge pressure here, we're solving for the total pressure. The total pressure must always be positive, now this pressure here is very small I mean 3,325 isn't necessarily a small number but when you compare it to the atmospheric pressure 101,325 this is really, really small. So it's so close to 0 why? Well because point b and point d here the highest part and the part where it's leaving the hose are 10 meters apart.

The highest column of water that the atmosphere on earth can support is 10.34 meters just a little bit bigger than this. So that's why this pressure is so small, if this difference in height will be more than 10.34 meters then the siphoning process wouldn't work in the way that we've assumed. Essentially the speed of the flow through the hose wouldn't be constant all the way through the water would actually speed up as it flow down here and it wouldn't fill up the whole hose it would be a more complicated problem. And that's because it's the atmosphere that's having to drive this whole thing and the atmosphere cannot support more than 10.34 meters. So that's just something to think about, not all problems that you set up this way will actually work. Alright at this height, this difference in heights that needs to be small in enough to be supported by the atmosphere. Alright let's go ahead to part d, part d asks us for the gauge pressure at point c during the flow. Again fairly simple, I want to know some piece of information about point c, I'm going to use point c. Point a is easy like it always is, I'm going to use point a and there we go. So we'll have p atmospheric at point a, at point c we'll have the pressure which we're looking for, we'll have the height negative 5 and we'll have that one half rho v squared. Where that v is the same thing that we found in part b, we'll just plug those things in, solve for the gauge pressure which is pressure minus p atmospheric.

Remember that, that's what gauge pressure is and the answer that we get for part d is negative 29,400 Pascal's notice it's negative that means that the pressure is less than the atmospheric pressure. So what that means is, that if I were to take my finger and stick it right there, the hose will try to suck that in. It's this negative right here which is telling me that the hose is sucking the water in. Alright so that's the way siphon's work and that's how Torricelli's law comes in.