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Concept #1: How Manometers Work

Transcript

Hey guys. So in this video we're going to talk about the manometer which is a type of pressure gauge, let's check it out. Alright, so monitors are pressure gauges and pressure gauges are just devices that use height differences, for example, here's this gray liquid. Notice that there's a height difference between here and here, you can come across to this side and you can calculate this height difference, you can measure this height difference and if you know this height difference, you got it, you can use that to calculate pressure using this equation right here, where h is my measured height, okay? So, a manometer have some weird shaped tube typically making some sort of u-shape it has a gas of the left, let's call this gas1, has a gas on the right gas2 and it has a liquid in the middle section here. Now, sometimes that gas is simply air or sometimes there's no gas at all, it's got vacuum which is what's going on here, also this top part here, could be closed as it is or it could be open and if it's open it's exposed to whatever air, whatever gas is right outside of it which most of the times if it's open we're going to have the atmospheric pressure pushing down on the open side. Now, this first two examples here are pretty trivial, let me write this here, these two are pretty trivial, pretty silly the pressure is the same therefore the height will be the same. So, here there's no pressure so the pressure 0 and here the pressure is 0 as well. So, because the pressure at the top of, both sides of the liquid are 0 because they're the same actually they're going to, the liquid is going to level itself out, same thing here, it's open. So, you have the atmospheric pressure here which is 1 atm, the standard one. So, if you have 1 atm pushing here and then you have 1 atm push here the liquid level itself out. So, equal pressure means equal heights, okay? Now, before we move into the second one here which is where it actually gets interesting, where it actually gets useful, I want to make a quick point here. Remember that the pressure, remember that the pressure in a gas doesn't change much, let me write this here, doesn't change much with height differences or depth differences, what does that mean? That means that the pressure here is 1 atm, the pressure here is also going to be 1 atm, the pressure is also going to be 1 atm so it's the same pressure everywhere, this is not the case with liquids, with liquids pressure changes pretty rapidly even with subtle differences in heights with gases the pressures, we're going to consider not to change in height differences that are too small, okay? So, let's look here. So, here you have the pressure difference, you can come across here. So, here it's vacuum so the pressure here is 1 and then 0 and then the pressure here is 1, that's why the liquid it looks like this because here you have a 0 so it's not pushing at all on this column but this column is being pushed with 1. So, there's more pressure here so it does this, okay? So, that's why there's this difference, same thing here, this here is closed this with no air here so it's vacuum, this is open. So, you have atmospheric pressure which is 1 atm but on the other side there's a gas that has 2 atm. So, same thing, this is pushing harder it's going to do this, cool? Let's do a quick example, this is very straightforward, here we have to find the density of a known liquid, you pour some of it into a manometer and eventually by the way you're going to calculate the density of that liquid. So, let me just write here, what is the density of the liquid you're putting on a manometer just as shown above and it is close at one of its ends and it has, it's a vacuum there. So, let's draw a manometer similar to what we had before by the way the bulb sometimes could be sideways, it really doesn't matter, and it's closed here and there's no pressure here, pressure is going to be 0 and it has 1 atm air. So, the pressure, it doesn't matter that it's air by the way it just matters that the pressure is 1 atm, at the bulb side, this is called the bulb usually where you have the gas, you measure the height difference to be 60 centimeter. So, the way I like to approach this, I think of the two sides mean the same and then I look at the pressures, this is 0, this is 1 so it's going to do this, right? So, you can imagine that they were, initially the same but then now this rode up a little bit and the other one rode down a little bit. So, that's this, it says here that the pressure, the height difference I can go across to the other side and this height difference here, h is 0.6 meters and we want to know the density of the liquid. So, every time you have a height difference in a liquid problem you're going to write the pressure difference equation. So, p bottom equals p top plus Rho g. h. Remember, this is also referred to as the absolute pressure, this is also referred to as the relative pressure and this is also referred to as the gauge pressure, in this case we're looking for a rho of the liquid. So, pressure at the bottom is always this pressure over here at the bottom of the column of liquid. Now, I can go over to the other side, let's give these names, let's call this pressure1 pressure2, let's call this pressure3 and let's call this pressure4, pressure1 is simply 0 because it's touching, it's touching vacuum, pressure2 is the same thing as pressure3 because you are within the same liquid at the same height. So, I'm going to write this down because it's very important, within the same liquid at the same height you have the same pressure, that means I can go across here and say hey p2 the same as p3 and by the way p3 is touching the air, the the gas here so it's the same as p4. So, really p2 which was the bottom pressure is the same as the gas on the other side here, it's p gas of the second gas. So, when I say, when I have p bottom here, it's simply going to be 1 atm, you go all the way across and p bottom is simply 1 atm, I'm going to rewrite this equation here, 1 atm, the pressure at the top is this over here, at the top of the column so that's going to be 0 atm and then Rho which is what we're looking for and then gravity 9.8 and the height which is 0.6. Notice that I have all the numbers so I can just move some stuff around to find density but before you do that. Remember that you cannot use 1 atm in equations, that's just a shortcut unit, in reality have to write this in pascal and atm, you have to memorize this, is 1.01 times 10 to the fifth pascal. So I'm going to rewrite, this is 1.01 or sometimes it's easier to just write the big number 101,000 Pascal divided by 9.8 meters per second squared and I'm just moving stuff to the other side, 0.6. So, this is my density, okay? And by the way, if you multiply all of this you get 17,177 and the unit's, this is pressure, I'm sorry, this density. So, the units since I'm using standard units everywhere, you're going to get the standard units of density which is kilograms per cubic meter, okay? So, that's it, that's the answer for Part A, all you have to do is plug it into the big equation and we have all the numbers, just have to move some stuff around.

Let's do Part B and I'm now going to get out of the way here. So, for Part B it says, you replace the 1 atm air. So, whatever air, whatever was here air, you replace that with a unknown gas. So, what does that mean? Well, remember the pressure, you knew this was 1 atm, if you replace it with something else you no longer know what that pressure is and that's in fact what the problem is asking, what is the pressure of this new gas on the right side, okay? Now, you replace that and remember that this pressure is the same pressure as the pressure at the bottom so this is going to be p bottom and you have to figure out what p bottom is. Now, keep in mind that these problems aren't going to say find p bottom or find p top sometimes they will but a lot of times they're going to say find a pressure of this gas or find the pressure of air, find the pressure of whatever. So, it's important you can relate that you can translate that the pressure of the gas is p bottom on your equation, okay? It's very important that you're able to do that, let me get out of the way again. So, really what we're looking for is p bottom and we're going to write that equation again, p bottom equals p top plus Rho g, h, p top is the pressure at the top of the column. So, that's 0 and density it says here that you to replace the air with something else, it doesn't say anything about the liquid actually says here it's same liquid, same liquid means same density. So, the density here is going to be the same one we just found so it's going to be 17,177, gravity 9.8 and the height it says that the height is now 80 centimeters, okay? So, if you multiply all of this you get 1.35 times 10 to the fifth Pascal, pressure is in Pascals and this is my friends the answer. Now, I want to talk about this a little bit, if you changed this to atms, atm, you don't have to but I'm doing it to you real quick, this is going to give you 1.33 atm and let's actually talk about this real quick. So, for the first part, for Part A, when you had a pressure, when you had a pressure of gas that was 1 atm the height was, the height was 60 centimeters, okay? Here, we had a different height, the height was 80 centimeters and that's because the pressure was 1.33 atm and this should make sense, the first time I had this pressure here, 1 so it did this, on the second situation this pressure over here is 1.33. So, you should expect that the height will be increasing, okay? So, this is a way that you can kind of validate that your answer makes sense, in fact this difference is proportional to each other. So, notice that 80, 80 is 1.33 times more than 60. So if you do the ratios 80 divided by 60 is 1.33. So, if you're 1.33 times higher that is because the pressure outside or the pressure of that gas is 1.33 stronger, okay? So, that's it for this one, let's keep going.

Practice: A classic manometer (as shown below) has one of its ends open, and a 2 atm gas on the other. When mercury (13,600 kg/m3 ) is added to the manometer, you measure the top of the mercury column on the left to be 40 cm higher than the mercury column on the right. Calculate the atmospheric pressure that the manometer is exposed to.