 A U-tube monometer, not to be confused with a U-tube monometer, is connected to a closed tank as shown in the figure to the right. The air pressure in the tank is 0.5 PSI G. Note that that G denotes a gauge pressure, indicating that the gauge pressure, the pressure read by the gauge, and also the absolute pressure of the air minus the atmospheric pressure, is 0.5 PSI. And the liquid in the tank is an oil with a density of 54 pounds of mass per cubic foot. The pressure at point A is 2 PSI, again that's a gauge pressure, which would be the absolute pressure at point A minus atmospheric pressure. I want to determine the depth of the oil, as indicated as Z on the figure, in feet, and the height of the monometer used to read the pressure of the oil, again in feet. So we could approach this in two ways, and that's actually why we have a part A and a part B. You can think of it as working from atmospheric pressure in to point A, through the air gauge and the height Z, indicating the height difference of the oil between A and the top of the oil, which is the interface between oil and air, just for convenience, I'm going to call this alpha, or we could approach it by working from atmospheric pressure in through the monometer. In that case, the thing that gets us from atmospheric pressure down is H, and I can relate the pressure difference between this point and this point by using the density of the gauge fluid times gravity times the height H, and then I know from Pascal's law that the pressure here is going to be the same as the pressure here, because they are at the same height, and then I can work from here, which I will call beta, up to A by using the density of oil times gravity times two feet of height. So I can approach it in two directions. Let's start by working from the gauge into the air and down through the oil. So I was told the pressure of the air is 0.5 PSI-G. That means the absolute pressure of the air minus atmospheric pressure is 0.5 PSI. Furthermore, because the density of gases is so low relative to liquids, I'm going to assume that the pressure difference across the height of the column of air, like the pressure difference between here and here and here, is negligibly small. Therefore, the pressure of the air is also the pressure at point alpha. So I can say the absolute pressure at point alpha is equal to 0.5 PSI plus the atmospheric pressure. Furthermore, I can use my PAH equation to relate the pressure difference between A and alpha, because A is going to be higher because it has all the pressure of alpha, plus the height of the column of oil pushing down on it, to the density of the oil times gravity. I mean, actually, it's the acceleration experienced by the oil, but because I don't know enough about the acceleration to assume anything different, I'm just assuming standard gravitational acceleration on Earth at about sea level, 9.81 meters per second squared, and the height difference here, which is Z. So I'm using that PAH equation, recognizing that I know A in terms of P atmosphere, and I know the density of the oil and gravity to determine the height Z. So part A is going to be PAH minus what I'm calling P alpha. The pressure at the interface between the oil and the air is equal to the density of oil times gravity times Z. And I know P alpha is equal to the atmospheric pressure plus 0.5 PSI. And I know the absolute pressure at A is atmospheric pressure plus 2. That means when I actually start combining these pressures into one equation, I'm going to have atmospheric pressure plus 2 PSI minus atmospheric pressure plus 0.5 PSI, which means my atmospheric pressure terms are going to cancel. I just have 2 PSI minus 0.5 PSI. This is going to happen frequently. So frequently, in fact, that we have a name for it, when we are working the problem in gauge pressure, what we're doing is assuming that the atmospheric pressure cancels. You can also think of it as assuming the atmospheric pressure is zero, and all the pressures described are relative to that quantity. So when we work a problem in gauge pressure, we are describing all pressures relative to atmospheric pressure. The only time that matters is if we start involving things like proportions of pressure or the ideal gas law, in which case we have to keep track of the fact that our gauge pressures are not the quantity we want to use in that calculation. We have to convert back to an absolute pressure. Anyway, this says equal to the density of the oil times what we are assuming is gravity times Z. So Z then is just going to be 2 PSI minus 0.5 PSI, divided by the density of oil times gravity. So 2 minus 0.5 is 1.5 PSI. I was told the density is 54 pounds of mass per cubic foot. Gravitational acceleration is going to be 9.81 meters per second squared, or I could use the imperial form, which would be 32.2 feet per second squared. Since everything else in this calculation is imperial, it probably makes more sense to use 32.2. My goal is going to be to get to an answer in feet. So I'm going to want to break my PSI into pounds of force per square inch and pound of force into pound of mass. So I'll write this as a PSI is a pound of force per square inch and my pound of force from my conversion factor sheet is 32.174 pounds of mass times feet per second squared. 32.174 pound of mass per feet, excuse me, times feet per second squared. So pound of mass is going to cancel pound of mass, pound of force is going to cancel pound of force, PSI cancels PSI, second squared cancels second squared. I have feet to the fourth power in the numerator and then I have feet times square inches in the denominator. So I need to get rid of those inches, 12 inches in one foot and then I square everything. One squared is moreing, square inches cancels square inches. Square feet is going to get rid of two of the feet in the feet cubed. The feet in the denominator already get rid of the other one, leaving me with an answer in just feet. So if my calculator would agree to help us, we can take 1.5 multiplied by 32.174 times 12 squared divided by 54 times 32.2 and I get 3.99677. So I'm just going to call that four, four feet. Z is about four feet. Then for part B, I'm using the same tools but I'm working through a manometer. Note that the manometer in this sort of application is used to read pressure. I mean, it's an analog pressure readout. You can look at the tube and you can read demarcations that you've created on the tube to see what the height is and if you know the pressure at those different points, you've presumably wrote it on those different points so you would know what the pressure is. The advantage of a manometer like this is that it is analog, it's not going to be affected by power or any sort of battery life. That being said, the downside is it's much more subject to minor changes in fluid levels so you don't see them that often anymore. The gauge fluid here is a denser fluid than the oil. It has a specific gravity of 3.05 which is going to be 3.05 times that of water. If the density of water is 998 or so kilograms per cubic meter then that's going to be just under 3,000 kilograms per cubic meter which is going to be just under 200 pounds of mass per cubic foot. So I'm talking about something that is almost four between three and four times denser than the oil. So the reason you would use that kind of gauge fluid is to allow a smaller area on the tube to read out a bigger pressure difference than if you used oil directly. Also in some cases you might use a gauge fluid that is unlikely to evaporate so that you don't have to worry about your fluid levels changing and moving your meniscus around for a given pressure. It also might be a situation where you don't want the oil or whatever it is inside your tank to actually interface with air. Maybe you're trying to keep a smell out or you're trying to keep air from infiltrating it any sort of gasification process. That's the logic behind your selection of your gauge fluid. Anyway, the density of the gauge fluid can be determined by looking at the density of the water which we'll probably have to do. And because the pressure at beta over here on the left is the same as the pressure here because of Pascal's law, I can write out the delta P across this height H as P beta minus atmospheric pressure. And that is equal to the density of the gauge fluid times gravity times H, literally H because it is H. That would allow me to determine the pressure at beta if I knew the atmospheric pressure and H and then I can relate the pressure difference between A and beta by using this two foot height and the density of oil. So the pressure difference there is P beta minus P A and that is equal to the density of oil times gravity times two feet. So I know the density of oil already, gravity is known, two feet is known, P A is known. So I can substitute for P beta. I will write P beta over here in terms of atmospheric pressure density of GF times gravity times height and substitute that in here along with my P A from earlier. And what I get out is atmospheric pressure plus density of gauge fluid times gravity times height H minus atmospheric pressure plus, I think it was two. Yeah, two is equal to the density of oil times gravity times two feet. Okay, so let's bring this down to the part of the paper that actually makes sense. If someone was trying to read this later on, close my parentheses. Now, atmospheric pressure cancels again. So I have the density of the gauge fluid times gravity times height minus two PSI is equal to the density of the oil times gravity times two feet. I'm going to solve for H. So I will write H is equal to two PSI plus the density of oil times gravity times two feet all divided by the density of the gauge fluid times gravity. And then just for convenience of dimensional analysis, I'm going to split my denominator and then I will cancel the gravity on the right term. So I have two PSI divided by the density of GF times gravity plus the proportion of the density of oil to the density of gauge fluid times two feet. Two feet is known, density of oil is known. TPSI is known, gravity is known. All I need is the density of the gauge fluid. For that, I'm going to step back to my definition of specific gravity from thermal one. Specific gravity is the density of a fluid divided by the density of water at standard temperature and pressure. So the density of our gauge fluid is going to be the specific gravity of our gauge fluid multiplied by the density of water at standard temperature and pressure, 300 Kelvin and 100 kilopascals in most cases. So for the density of water, I will go back into my textbook and for that, let's get rid of the conversion factors and the calculator while we're at it. So water at 300 degrees Celsius in one atmosphere or so has a density of 998 kilograms per cubic meter. So we are going to take a 998 and we are going to multiply it by 3.05. 998 times 3.05 and we get 3,043.9. You guys remember earlier, when I took 998 times 3.05 in my head and I said it was just under 3,000? Good math, John. This is why we bring a calculator. 3,043.9 kilograms per cubic meter. And with that, I have everything I need. So I will take two PSI, divided by our shiny new density, 3,043.9 kilograms per cubic meter times gravity. Let's go with, oh really? Okay, so we just calculated the density in kilograms per cubic meter and I can leave it that way, I could. And then handle the unit conversion between imperial and metric, I could do that. But in hindsight, if I could rewind like 45 seconds, it wouldn't be way easier to just look up the density of water in imperial units and then multiply that by 3.03, 3.05. So let's just pretend that I did that. Let's turn back time. Do-do-do-do. And back to our textbook. I'm not gonna use table A3 this time. Table A3 shows the density of a variety of liquids in metric units, which is what we normally want 99% of the time. Unfortunately, if we are solving this problem in the United States, we need to be able to handle imperial units, even imperial masses. Dare I say the slug? So let's go instead to table A1, which is just properties of water. And we are going to be grabbing the density of water in imperial units, which is gonna be the right-hand side of this table. And 20 degrees Celsius is the same row as 68 degrees Fahrenheit, which means my density is 1.937 slugs per cubic foot. Hooray for slugs. So calculator, if you would please. That would be 1.937 times 3.05, which is 5.90785. Okay, 5.90785. Slugs, hooray, per cubic foot. Okay, just a side note here, remember that one pound of force? Is equal to one slug times one feet per second squared. Anyway, now back over here, I can plug in my shiny new density that we totally calculated only one time, 5.90785, just call it 9.9. 7.9 slugs per cubic foot. And then 32.2 feet per second squared. And we want an answer here in feet. So we'll do the same breakout for PSI. PSI is a pound of force per square inch. And from our little definition over here, I know that one pound of force, is one slug times one foot per second squared. So pound of force, cancels pound of force, PSI cancels PSI, and slugs cancel slugs. I need inches and feet to cancel. So I will write 12 inches is one foot and square everything. One square is boring, that's not a square. Let's try that. One squared, inches squared, cancels inches squared. Then I have feet times cubic feet in the numerator and feet times square feet in the denominator. So feet and square feet cancel cubic feet, leaving me with feet in the numerator. And then for the second part, I'm going to take the density of oil divided by the density of the gauge fluid. Gosh, I'm gonna have a density in slugs and density in pound mass. Hate the imperial unit system. So, density in oil is 54 pounds of mass per cubic foot. And then my density of my gauge fluid is 5.90785, 9., 5.9079 slugs per cubic foot. That's gonna be a unitless proportion for the second unit would be two feet, okay? So slugs and pound mass will eventually cancel once I involve the unit conversion and cubic feet will cancel unit cubic feet. So multiplying by two feet will yield an answer in feet. All I have to do is handle the mass conversion. So what I'm gonna do here is write out one pound force is equal to one slug times one feet per second squared. And then one pound of force is also equal to 32.174. That came from my conversion factor sheet right here. So that was a feet, I guess pound mass feet, pound mass feet per second squared. Therefore, I can write a slug times feet per second squared is equal to 32.174 pound mass feet per second squared feet per second squared cancel. Therefore, one slug is equal to 32.174 pound mass. We did it. So one slug is 32.174 pound mass. Slugs cancel slugs, pound mass, cancels pound mass, leaving me with feet. Now on to the math. So two times 12 squared divided by 5.90785 times 32.2. And then I will add some additional parentheses around that just for good measure. Because I don't trust my calculator. I didn't say that, I trust your calculator. Thank you. And then plus 54 times two divided by 5.907, 90785 times 32.174. And I get a syntax error. Just write that down on my sheet. See, calculator, I thought we had an agreement. I thought we were past this. I would type things, you would compute them. We would have a good relationship. 2.08212.