PhD., University of Maryland
Published author
Jonathan is a published author and recently completed a book on physics and applied mathematics.
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PhD., University of Maryland
Published author
Jonathan is a published author and recently completed a book on physics and applied mathematics.
Liquid pressure is the increase in pressure at increasing depths in a liquid. This pressure increases because the liquid at lower depths has to support all of the water above it. We calculate liquid pressure using the equation liquid pressure = mass x acceleration due to g density x depth in fluid.
Alright let's talk about liquid pressure. We've talked about pressure before which is just force over area but it turns out that when you descend down into a fluid, down into a liquid the pressure gets bigger and bigger and bigger. Well why is that? Well you've got to support the weight of all the fluid that's on top of you. So that's where liquid pressure comes from. So let's just go ahead and look, let's say that we're h meters down below the surface of water, it could be another fluid but let's just pretend it's water for now and we want to look at how much pressure is exerted because of the weight of all this water above us. Well pressure is equal to force over area, well what I'm I supporting? Well I'm supporting the weight of the water. The weight of the water is always the mass of the water above me times the acceleration due to gravity. And then we divide by the area and now we're going to do something a little bit sneaky that you not have thought of doing.
The mass of all this water above us is just the density of water times the volume of this cylinder. But the volume of this cylinder is the cross sectional area times the height, so that means that we've got density times volume times acceleration due to gravity divided by area. But we said, the volume was the cross sectional area times the height, so the cross sectional area cancels and that gives us the following expression for liquid pressure. Density of the fluid in our case water, but it works just as well with other fluids times acceleration due to gravity times how deep you are in the fluid. Alright so let's do a sample problem with this, just to see how it goes these problems really aren't that difficult. So how much pressure is exerted by water that's 4 kilometers deep? And then I've got a second question about this, how much denser is the water at the bottom of this pool, I [guess really this] ocean then on its surface.
Alright so the first question is easy to do, we've got pressure equals rho, g, h. Now the density of water as long as it's not sea water, sea water is a little bit more, but let's just ignore that to make the numbers nice. The density of pure water is 1,000 kilograms per cubic meter, the acceleration due to gravity is 9.8 meters per second squared. But you know what, let's make the numbers nice, I don't care let's just call it 10, it's only 2% off right, so we'll have 10 meters per second squared and then what's the height, well it's 4 kilometers, that's not alright. You can't tell me kilometers, you've got to tell me in SI units or else the numbers will be off. So I need to write 4 times 10 to the 3, and then we just multiply and it's easy to multiply these numbers together because look at all the 10's. So we'll have a 4 and then how many 10's well I got 3, 4, 7.
Alright now what's the unit? Look at all these units floating around, I don't want all that business, but wait a minute we were careful to always work in SI units. And that means that we don't have to think about it. It's a pressure what's the SI unit of pressure? Pascal's, done right okay so I hope that, that one went pretty well. I mean you just kind of take the density, take the exertion due to gravity, take the height. If you've got a difference in height, if you want to know the difference in pressure between 2 different depths in the ocean then you'll just use the difference in height not the whole height okay easy enough. Well what if I want to know how much denser these water is. I mean look at this, this is a pretty big pressure, 40 million Pascal's right?
And so we've seen with both modulus that when you put all that pressure on the water it's going to decrease, it's going to get smaller. It's going to increase its density or decrease its volume. So of course that's going to give us a little bit of an issue. Because that means, if this density that we used here 1,000 kilograms per cubic won't actually be correct for the whole thing, it's going to be a little bit denser at the bottom. So the pressure will actually be a little bit bigger. But how much of an effect is that? Well let's see, this is the pressure that we're applying in order to find out how much that compresses the water, we look at the ratio. What's this divided by the bulk modulus? Well it'll be 4 times 10 to the 7 Pascal's divided by the bulk modulus which we can look up 2.2 times 10 to the 9 Pascal's and if we type that into a calculator then we'll get 1.9 times 10 to the minus 2 okay?
Now remember if this was equal then it would imply 100% decrease in density, sorry increase in density. Here since it's 1.9 times 10 to the minus 2, that's a 2% increase in density. It's not a huge increase so can say that this answer is pretty accurate. So that's it liquid pressure.
Unit
Solids, Liquids and Gases