Pressure Gauge: Manometer - Video Tutorials & Practice Problems

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Pressure Gauges: Manometer

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Hey, guys. So in this video, we're gonna talk about the Manama ter, which is a type of pressure gauge. Let's check it out. Alright. So Manama tres our pressure gauges and pressure gauges are just devices that use height differences. For example, here's this great liquid notice that there's a height difference between here and here. You can come across to this side and you can calculate this high difference. You could measure this high difference. And if you know this high difference, you got it. You can use that to calculate pressure using this equation right here where h is my measured heights. Okay, so, um, a nominator has some weird shaped tube typically making some sort of U shape. It has a gas on the left. Let's call this gas One has a gas on the rights gas, too, 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. Whatever gas is right outside of it, which most of the time, if it's open, we're gonna have the atmospheric pressure pushing down on the open side. Now, this first two examples here are pretty trivial. Let me write this year. These two are pretty trivial. Pretty silly. The pressure is the same. Therefore the heights will be the same. So here there's no pressure for the pressure zero. And here the pressure zero as well. So because the pressure at the top of both sides of the liquid are zero because they're the same, actually, they're gonna The liquid is gonna level itself out. Same thing here, it's open. So you have the atmospheric pressure here, which is one ATM, the standard one. So if you have 1. 80 m pushing here and then you have 1 18 pushing 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 actually gets useful. I wanna 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 heights, differences or depth differences. What does that mean? That means that the pressure here is 1. 80 m. The pressure here is also gonna be won a t m. The pressure here is always gonna also gonna be won t m. 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 gas is the pressures. Um, we're gonna consider not to change, um, in height differences that are the small. Okay, so let's look here. So here you have pressure difference. You can come across here, So here it's vacuum. So the pressure here is one and then zero. And then the pressure here is one. That's why the liquid looks like this, because here you have a zero. So it's not pushing at all in this column, but this column is being pushed with one, So there's more pressure here. So it does this. Okay, so that's why there's this different same thing here this year is closed. This with no air here since vacuum. This is open. So you have atmospheric pressure, which is 1. 80 m. But on the other side. There's a gas that has to a t m. So same thing, this is pushing harder. So it's gonna do this. Cool. Let's go quick. Example. This is very straightforward. Um, here we have to find a density of unknown liquid. You pour some of it into a min ometer, and eventually, by the way, you're gonna want to calculate the density of that liquid. So let me just right here. What is the density of the liquid? You're putting it on a man ometer 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 men ometer similar to what we had before. By the way, the bulbs sometimes could be sideways. It really doesn't matter. And it's closed here and there is no pressure here. Pressure is going to be zero. Um, it has won a t m air. So the pressure, it doesn't matter that it's air, by the way, just matters that the pressure is 1 80 m at the bulb side. This is called the bulb usually where you have the gas. You mentioned the height difference to be 60 centimeters. So the way I like to approach this, I think of the two sides of the same. And then I look at the pressures. This is zero. This is one. So it's gonna do this right? So you can imagine that they were initially the same. But then now this road up a little bit and the other one wrote 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 high difference here h is 0.6 m 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 equations. So Pete bottom equals P top plus row G. H. Remember, this is also referred to 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 row 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 pressure one pressure to Let's call this pressure three. And let's call this pressure four. Pressure one is simply zero because it's touching. It's touching. Vacuum pressure to is the same thing is pressure three because you are within the same liquid at the same height. So I'm gonna write this now because it's very important within the same liquid at the same height. You have the same pressure. Let me taken. Go across here and say, Hey, P two is the same sp three and by the way, p three is touching the air. The gas here. So it's the same S p four. So really p two 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 gonna be won a t m. You go all the way across and P bottom is simply 1 80 m. I'm gonna rewrite this equation here. One a team. The pressure at the top Is this over here at the top of the column s Oh, that's gonna be zero a t m. And then row, which is what we're looking for. And then gravity 9.8 in 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 one Attman equations. That's just a shortcut. Unit in reality. Have to write this in Moscow in 1 80 m you have to memorize. This is 1.1 times 10 to the fifth Moscow. So I'm gonna be right. This is 1.1 Or sometimes it's easier to just write the big number one on 1000 pass cow divided by 9.8 m per second square. And I'm just moving stuff to the other side six. So this is my density, okay? And by the way, if you multiply all of this, um, you get 17. 177 um, 17 177 And the units, This is pressure. I'm sorry. This is density. So the units, since I'm using standard units everywhere, you're gonna 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 we have to do is plug it into the big equation. And we had all the numbers. You just have to move some stuff around. Let's do part B. And I'm now gonna get out of the way. Here s O for part B. It says you replace the 1 80 m air. So whatever air, whatever eyes was here air, you replace that with an unknown gas. So what does that mean? Well, remember the pressure you knew this was one ATM? If you're replacing 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 gonna be P bottom, and you have to figure out what P bottom is. Now, keep in mind that these problems aren't gonna say find P bottom or find Pete stop. Sometimes they will. But a lot of times you're gonna say, find the pressure of this gas or find the pressure of air, find the pressure or whatever. Eso It's important that you can relate that you can translate that the pressure of the gas is p bottom on your equation. Okay, so it's very important that you're able to do that when you go out of the way again. So really, what we're looking for is P bottom and we're gonna write that equation again. P bottom equals p top plus row. G H P Top is the pressure at the top of the column. So that zero and density, it says here that you replace the air with something else. It doesn't say anything about the liquid. Actually, it says here it's same liquid, same liquid means same density. So the density here is gonna be the same one we just found. So it's gonna be 17 177 gravity 9. and the height. It says that the height is now 80 centimeters. Okay, so if you multiply all of this, you get one point 35 times 10 to the fifth. Pascal pressures and pass cows. And this is my friends. The answer Now I want to talk about this a little bit. If you change this into a t m e a T m. You don't have to, but I'm doing in here real quick. Uh, this is gonna give you 1.33 80 m. 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 one a T. M. The height plus the height was centimeters. Okay, um, here we had a different height. The height was 80 centimeters. And that's because the pressure was 1.33. 80 m. And this should make sense the first time I had this pressure here one. So we 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 the way that you can kind of validate that you answer makes sense. Um, in fact, this difference is proportional to each other. So notice that 80 80 is 11.33 times more than 60. So if you do the ratios 80 divided by 60 is 1. 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.

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Problem

Problem

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/m^{3} ) 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, in units of atm. (Use g=9.8 m/s^{2}.)

A

0.477 atm

B

1.0 atm

C

1.47 atm

D

2.53 atm

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