 A 12-foot swimming pool is filled with a fluid which has a known specific weight of 62.4 pounds of force per cubic foot. Determine the gauge pressure at the bottom of the pool in both imperial units and metric units, and discuss the difference between atmospheric and relative pressure. I'll start by drawing a crude diagram. The swimming pool has a depth of 12 feet, and the fluid inside has a specific weight, which we abbreviate with gamma, of 62.4 pounds of force per cubic foot. I want to know the gauge pressure at the bottom of the pool. Remember that gauge pressure is going to be the pressure difference between the absolute pressure and atmospheric pressure, which means that I'm essentially looking for the pressure increase as a result of the water in the swimming pool. I'm going to be using my PAH equation again to relate the pressure difference across the 12-foot height of the fluid to its density. Gravitational acceleration isn't given, but I'm assuming it is 9.81 meters per second squared, standard earth gravity, and the height difference here is 12 feet. So the pressure difference, delta P here, is going to be the absolute pressure at the bottom minus atmospheric pressure. And when I rearrange absolute pressure as gauge pressure plus atmospheric pressure, then I'm left with just gauge pressure on the left-hand side of this equation. So the gauge pressure, the thing that I'm actually looking for, is the density of the fluid times gravity times 12 feet of height. Now I don't know the density, but I do know the specific weight. Remember that the specific weight is like density, except with weight instead of mass. We can write that as the force of the weight divided by volume, which would be mass times gravity divided by volume, which would just be density times gravity. Therefore, the density of this fluid is going to be at specific weight divided by gravity. So specific weight divided by gravity times gravity times height. Gravitational acceleration cancels and I'm left with just specific weight times height. So I begin with imperial units, let's say. Yep, imperial units in PSI-G. So I'm going to plug in 62.4 pounds of force per cubic foot. And I'm going to multiply by 12 feet. And my goal is to come up with PSI on the other end. Remember that G or A at the end of PSI is just an indication as to whether or not it is a gauge pressure or an absolute pressure. And a PSI is going to be a pound of force per square inch. So in order to get inches squared and feet squared to cancel, I'm going to have to convert 12 inches to one foot and square everything. One squared is boring, inches squared cancels inches squared, feet squared and feet cancel cubic feet giving me PSI. So I will take 62.4 multiplied by 12 divided by 12 squared. That gives me 5.2. So the gauge pressure at the bottom of the swimming pool is 5.2 PSI. We are now halfway there. To get the rest of the way there, let's look at our conversion factor sheet. On our conversion factor sheet, we do have a direct conversion for pressure between PSI, pound of force per square inch and Pascals. So we could use that to convert our answer from imperial units into metric units. Let's start there, I guess. I'm going to say 5.2 PSI from the conversion factor sheet. One PSI, one pound of force per square inch is equivalent to 6,894.8 Pascals. So PSI cancels PSI and I'm left with just a multiplication. I think the problem asked for Pascals, not kilopascals, so I am just going to be multiplying these two numbers together. So calculator, if you would join us again, we need 5.2 multiplied by 6,894.8 and we get 35,853 or 35.85 kilopascals. And that conversion is perfectly valid, but what if we hadn't calculated the answer first in imperial? Just for character building, let's step back to this step. So we start off with what we had earlier, 62.4 pound of force per cubic foot and we are multiplying by 12 feet and we are trying to get to Pascals. Well I know Pascal is a pressure unit so I could convert this into the imperial pressure unit and handle the same unit conversion as earlier, but that's boring and the point of this exercise is to get more practiced at dimensional analysis and the unit conversion factors. So let's go the long way, shall we? Let's start with Pascals because that's our goal and I want to work backwards from our destination. And let's define a Pascals a Newton per square meter, which is on your conversion factor sheet under pressure. And then I recognize that I can convert between pound force and Newtons, so I could make my life pretty easy here, but since we're doing this, let's do it all the way. So let's break the Newton apart into its components. A Newton is a kilogram meter per second squared. Now we are all the way down into our primary dimensions. So to get the rest of the primary dimensions out of the pound force, I'm going to have to go back to my conversion factor sheet. I had to calculate it for a second. 32.174 pound mass times a foot per second squared is a pound of force. So I'm going to write that as 1 pound of force is equal to 32.174 pound mass times feet per second squared. Note that I'm adding the M to try to keep my brain on track for pound mass versus pound force. And at this point pound force cancels pound force and second squared cancels second squared. So to get the rest of the primary dimensions to cancel, I have to convert feet to meters and kilograms to pound mass. So let's start with the mass. 1 kilogram is equal to 2.2046 pound mass. Kilograms cancels kilograms, pound mass cancels pound mass. And then feet, feet, feet. So I need one foot in the numerator. And I'll just write that as 1 meter to kind of give this more convenience. 1 meter is 3.2808 feet. And now I can cancel feet, feet and feet with the cubic feet in the denominator. And then I have meters and meters which cancels square meters. Therefore the only thing left is Pascal's. So calculator if you join us again. Southwest, Southwest, Southwest. I want 62.4 multiplied by 12 multiplied by 32.174 multiplied by 3.2808 divided by 2.2046. And look, we got 35,852.6 Pascal's which you could round to 35,853. So we added in like four unnecessary unit conversions here and we incurred a little bit of rounding error as a result because those numbers on the conversion factor sheet are rounded. But more importantly, we learned how we could go about a unit conversion process that was particularly sticky. One in doubt and break everything down into primary dimensions and you can convert those pretty straightforwardly. The last thing I asked for here was a discussion of the difference between atmospheric and relative pressure. Well, that atmospheric pressure here is indicating the pressure at the top of the pool, at the interface between the pool and what is presumably air around the pool. And the relative pressure at that point is zero because relative pressure is describing the pressure difference between the absolute pressure that is the actual pressure experienced and atmospheric pressure. The reason that that might be useful is that if you had, let's say, a watch, the watch has a face that is experiencing two pressures, a pressure on the inside of the watch and a pressure on the outside of the watch. The inside of the watch is presumably atmospheric pressure or whatever the pressure was when it was built and sealed somehow, or it has a pressure that's pretty close to atmospheric pressure as a result of temperature fluctuations within the watch and the pressure on the outside of the watch, which if you were standing at the top of this pool, would be atmospheric pressure. So at that interface, the pressure on this hypothetical watch face would be zero. There'd be no pressure difference between the inside and the outside. And then as you got deeper, there would be a relative pressure applied to the watch. But the relative pressure experienced is the pressure difference between the actual pressure and the inside pressure, which in this case is assumed to be atmospheric. It's the same as if I'm riding my bicycle down a street at 15 miles an hour and a car comes up behind me going in the same direction at 30 miles an hour and if it were to strike me, the velocity of that collision is not 45, it is 15, because it would be the relative velocity between the two moving entities.