 A monometer is used to indicate the pressure inside of a tank. The monometer is filled with two gauge fluids, oil and mercury, and the tank itself contains water. The ambient atmospheric pressure is 85.6 kilopascals. If H1, H2, H3 are 0.2, 0.3, and 0.46 meters respectively, determine the absolute and gauge pressure of the air at the top of the tank in kilopascals. So let's just pause and think about this for a moment. We have multiple gauge fluids in our monometer. Why would we possibly have multiple gauge fluids? Well, in this situation, you must have a rather large pressure change depending on the operation of the tank. Maybe you want an operator to be able to see a lot of different pressures on a relatively small pressure gauge. In order to achieve that, you need a dense gauge fluid. The dense gauge fluid of choice for these applications is mercury. It has a nice meniscus, it's unlikely to evaporate and affect your gauge as much. The downside is you presumably don't want mercury to leak into your water. So you would separate the water from the mercury with oil. Oil isn't going to mix with the mercury, oil isn't going to mix with the water, it would just serve as a barrier between the two. Next, even though this has multiple gauge fluids, the analysis itself isn't any more complicated than it was with a single gauge fluid monometer. We are repeating that approach several times. We're going to use our PAH equation to work across this height difference, which will give us the pressure here, which is the same as the pressure here. And then we are going to recognize that the Pascal's law simplification will also allow us to say the pressure here is the same as the pressure here. Therefore if I work across this height using a PAH equation, I will get this pressure. Then Pascal's law is also going to say that the pressure here is the same as the pressure here, even though the areas are different, just like the hydraulics problem. Or we can use a PAH equation to figure out the pressure difference between here and here, and that will give us our answer. So we are working from two down to here, which I can call a state point. I will call that state point three. Three is the same as this state point. And I can step from this state point up to here, which I can call four. And then four is the same as this state point here. And I can step from that point over to this point directly. So if four is equal to four B and four C, then I can step all the way up to one in a third PAH equation. Does that make sense? We're going from two down to three. Three is the same as three B. We're using a PAH equation to go from three B up to four. That's the same as four B, which is the same as four C. And a third PAH equation gets us up to one. Perfectly clear, I'm sure. OK, I'm going to get rid of all my extra arrows here. The point is we are doing three PAH equations. We are doing the PAH equation that gets us from two to three, doing the PAH equation that gets us from three to four, and then the PAH equation that gets us from four to one. And each of those is the same structure, density times the acceleration experienced, which I'm assuming is gravity, times the height that's relevant. So the first PAH equation is between two and three. Three is going to be the higher pressure because it's lower. So I'll write that as P3 minus P2. And that is going to be the density of Mercury times gravity times the relevant height, which is H3. The next PAH equation is going to be from three up to four. So I can say P3 minus P4 because, again, I'm going from lower to top, from a higher pressure to lower pressure, always in these deltas. I'm saying density of oil this time, times gravity, times the relevant height there, which I believe is H2. It is indeed. And then one more PAH equation to step from four, which remember is the same as over here, up to one. So that's going to be P4 minus P1, density of water this time, times gravity times H1. And P1 is what I'm actually looking for. And P2 is equal to atmospheric pressure because it's open to the atmosphere. So I can rearrange this and write it as a single equation. I can say P1 is equal to P4 minus density of water, times gravity times H1. And then I can substitute in for P4 and write P3 minus density of oil, times gravity times H2. And then I can substitute in for P3 and write that as P2 plus density of oil, excuse me, mercury times gravity times H3. Note that once you get a little bit more proficient at this, you can step your way through one of these monometers and add any time you go down and subtract any time you go up. So in this situation, you could have thought of it like, well, we're starting with P2 and then we are adding density of mercury times gravity times H3 because we're going down and then we are subtracting density of oil times gravity times H2 because we're going up and then we're subtracting again density of water times gravity times H1 because we're going up and that would have yielded the same result. But that's enough talking, it's time, math. P2 I know is 85.6 kPa. And then I'm going to add the density of mercury times gravity times H3. Well, I don't know gravity, but I can probably reasonably assume it. I will say gravity is about 9.81 meters per second squared. But for my density of mercury and my density of oil, we're going to have to look those up. And once again, we're going to use table A19 for that. So I will begin with the density of mercury, density of mercury. We'll do density of oil and then density of water. So jumping into table A19, I can see that at 300 kelvin, I have a density for mercury of 13,529. And note that because we weren't told the temperature, we're just using the values at 300 kelvin. You can write that down as an assumption as well. After mercury comes oil, and I only have one density of oil here, it is the density of unused engine oil. Therefore, we're going to assume that the oil in question here has properties similar to unused engine oil. And it's not even begin to get into the conversation about different properties of different types of engine oil and what qualifies as unused. But a density for that oil at 300 kelvin is 884.1. Then water, because we're grabbing 300 kelvin, we're going to use 996.5. And with those three densities, we can continue our math. So I will say 13,529 kilograms per cubic meter. Multiplying by gravity. Note that you could factor out gravity if you wanted to. I want to leave it in because I want to calculate these quantities separately for reasons. Reasons that I'll make clear in about five minutes here. H3 was given as 0.46 meters. And we want an answer in kilopascals. Yep. So I'm going to convert this to kilopascals. Again, you could also factor out the unit conversion, but I want to do this separately for reasons. So a kilopascal is 1,000 Pascals, a Pascale is a Newton per square meter, and a Newton is a kilogram meter per second squared. Pascale cancels Pascals, Newton cancels Newtons, kilograms cancels kilograms, meters, meters, and square meters cancels cubic meters times meters, and second squared cancels second squared giving me kilopascals. And in fact, I'm going to break this across separate lines. Okay. So sorry, those of you completing this with a pencil or pen, I'm going to abuse the fact that this is digital a little bit and get out of writing that two more times. So next was properties of oil and that's subtracted. And then properties of water, which is also subtracted. And that density was 884.1 and its height was 0.3 and the density of water was 996.5 and our height was 0.2. So like I said, I'm going to calculate these quantities separately because I want to look at them. Calculator, if you would please what is 884.1 multiplied by 9.81 multiplied by 0.3 divided by 1000, it is 2.6 long density. No, that wasn't the wrong density. That's just the wrong line. Uh-huh. And then what is 13,529 and 0.46, I get 61. And then lastly scroll down a little bit 996.5 multiplied by 0.2 and we get 1.955. So the reason I'm writing them separately is to indicate to you how much bigger the effect of mercury is than oil and water. We could arguably get a close enough answer by only considering mercury and neglecting the other two because they are two orders of magnitude smaller densities. The reason I'm pointing that out is because oftentimes when we are considering the effects of a height of a column of fluid, we neglect it when it's densities of gas. That is, we neglect the effect of pressure increases as you get deeper into a gas unless the height is very large, like on the order of several miles if you're considering how pressure changes in the atmosphere of a planet, for example. The reason we do that is because the densities of gases are at least three orders of magnitude smaller than the densities of liquids that we're considering. This same difference in the property change in a column of mercury versus water is even more magnified when you're considering water and air. That is why we neglect the gas effects. With this, we have enough information to actually compute an answer. So I will say 85.6 plus 61.051 minus 2.605, excuse me, 2.6019 minus 1.955. And we get 142. And once again, I'll pose the question, is this a gauge pressure or an absolute pressure? It is an absolute pressure because we are adding in atmospheric pressure. If you had treated the problem as though the atmospheric pressure was zero, what you're doing is determining the pressure effects relative to atmosphere, which would be a gauge pressure. That would be just considering these three quantities instead of all four. And then once again, I will pose the question, how do we convert this to a gauge pressure? You're right. We subtract 85.6 kilopascals. We do not just always subtract one atmosphere. We subtract the actual atmospheric pressure we have in the problem. You could also think of it like neglecting 85.6 in our calculation. If we were to do that, we would end up with the same result, 56.494. So the logic here is the pressure at the interface between air and water, which is P1, is going to be sufficient for an answer because we are neglecting the pressure changes across the height of the column of air between state one and wherever the pressure gauge might be. And if a pressure gauge were to be read here, it would indicate the pressure difference between the air and atmosphere because that's how those pressure gauges work. And we would get a readout of 56.5.