 We're now going to take a look at an example problem involving the solution of a manometer. And the problem that we're going to solve is one where we're trying to solve for the pressure at some location with a two-fluid system. And this is a pretty common problem that you'll find in any textbook on fluid mechanics. So what we're going to do, we're going to walk through the procedure. We're dealing with liquids, hydrostatics, and liquids. So I'll draw out the problem with the schematic of it and then we'll work it through. Okay, so there is our problem. It's a schematic. And what we are told to do is to solve for the pressure at A, which is this location right here. And in this problem, we can see that we have two fluids. We have one fluid, fluid number one, and with density one, and that's the red fluid. And then we have a second fluid with density row two. And we can see that we have an interface here. And we also have an interface up here. And for the interface on the upper right, what we're also told is that above the fluid interface, we have atmospheric air. And consequently, given that we're talking, you know, probably millimeters or inches of air above this interface, we can then assume that the pressure right at this surface is P atmosphere. So that's why we write this approximation P2 is approximately equal to pressure atmosphere. So in this problem, what we're after, we want to be able to solve for the pressure at A at this location here. The things that it appears that we know, well, what we'll do will work out the solution. And then we'll see what we need to know in order to solve it. So if you recall, we had equations for hydrostatic pressure distribution and fluids. And I'll write out the equations here. And remember, I said that the order of these we have to be careful with. So on the left hand side, we have pressure, that's PA minus P1. And that's going to be equal to row. Now, which density are we dealing with? Well, if you look here, we're dealing with PA with respect to P1. So we're dealing with fluid one, it's the fluid shown in red. So we write row one, multiplied by G. And then we multiply it by some difference in height. And remember the first height, you know what I should do here, let me put our coordinates. So Z is positive in that direction. So the first variable we have corresponds to PA, so we write ZA minus. And then the second one is one, so we write Z1. And so that's the first equation that we have. The second equation that we have, well, when we look at this, we have one interface here, and we have one interface here. So how can we figure out the difference between pressure at this location at P1 versus the P2 location up here? Let's write that out. And then we'll try to figure out how to solve that. And so what we can write then is P1 minus P2. Let me get rid of those lines there because that might be confusing. And that will be equal to now the density that we're dealing with here between one and two is the blue fluid. And that has density row two. So we write minus row two times G, the gravitational constant. And then with our convention, it would be Z1 minus Z2. We have those two equations. Now, how can we make this reference? Well, we have Z1 and we have Z2 here. But one of the things that we can do in manometry, if you have the same fluid within a YouTube, we can go directly across here. So let's move over to that point. And the pressure at this point, oops, sorry about that. The pressure at this point is identical to the pressure on the left. So we can write at this point as well, that as well as Z1 and P1. If we look at that location, although there's no fluid interface there, at that location within the YouTube, the pressures would be the same. So with that, what we can write is that we have these two equations here. So let's combine them and see what we get. And if you recall, let's look back on our schematic. We said that P2 was approximately atmospheric pressure. So we can write the pressure at A, which is the area that we're trying to measure, is equal to P atmosphere minus the density of fluid 1 times the difference from ZA to Z1, the first interface between the red and the blue liquid, and then minus rho 2G and then Z1 minus Z2. So those are the last interfaces we had with the air. So what this tells us is that in order to measure the pressure at A, in order to calculate PA, we need to measure the following. We need to know ZA, and that would come out of the convention of by which the manometer is designed. Given the geometry of this manometer, we would know ZA exactly. So that comes from there. And then the other things that we need to measure is interface Z1 and Z2. So let's take a look back. We need to be able to measure this interface here, and we need to be able to measure that interface there. So when we do those, and the last thing we need to know, we need to know the atmospheric pressure. And we also need P atmosphere. So with that, in order to get PA, we need to measure the geometry, interface, interface, and pressure atmospheric. And so what this example demonstrates is the ability to measure pressure at a location that is not necessarily exposed directly to the fluid within A. So you might have a case where you have a fluid that you don't want to be able to measure or expose that to atmospheric. You could have a case like this where you then measure that. So you have a two-fluid manometer. So it allows us to measure PA without being directly exposed to fluid A, and that would be one advantage of it. A couple of rules of thumb that we'll write out. Okay, so a couple of rules of thumb we have here. The first one pertains to the fact that we are able to jump across the U2 manometer. But what it says is any points at the same vertical location in a continuous length of the same liquid are at the same pressures. So if we look back at our schematic, that is what enabled us to do this jump across here. We have a continuous liquid going between those two locations, and consequently that was how we were able to do that. The other one is that pressure increases as one goes down a liquid column. So be careful with the way that you have the order. Remember A corresponds to A there and one to one. If you do that and you assign your coordinate system and be consistent with that, you should not have any problems with solving manometer problems. So the other thing is that it's just natural as you go further down the column the pressure gets higher and higher. If you've ever gone swimming in a swimming pool or in the ocean or anything like that, as you dive down the liquid pressure on your ears becomes more and more greater and greater, and you can feel that exerted on your body. And so we know through intuition that as you go down and down the liquid column will exert higher and higher pressure. And that is solving problems with manometers.