GG 140: The Atmosphere, the Ocean, and Environmental Change

Lecture 7

 - Hydrostatic Balance


The hydrostatic law describes the weight of a fluid overlying a given area, or the pressure at a particular point. It can be used to calculate the approximate atmospheric mass over a particular area, or to calculate the change in pressure over a given change in altitude. A calculation of the pressure difference from the ground to the twelfth floor of Klein Biology Tower is found to agree well with measurements taken at both locations. The hydrostatic law also applies to pressure changes with depth in the ocean.

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The Atmosphere, the Ocean, and Environmental Change

GG 140 - Lecture 7 - Hydrostatic Balance

Chapter 1: Recap of Planet Temperature [00:00:00]

Professor Ron Smith: So last time, we derived a formula for the equilibrium temperature of a planet. And I’m not going to go back over that much. The first column of information here is a continuation of that. But the way we did it was to consider the Earth as a kind of reservoir of heat that can store heat. It can receive heat from the Sun, and it can radiate heat to space. And then we tried to consider how that system might come into a steady-state balance.

So we were thinking of this very much like the water tank. If I plotted temperature versus time–oh, I have to excuse myself for the slow writing today. I think I’ve dislocated my shoulder, so I’m going to be doing it two-handed. The lesson of the day is wear your helmet when you’re biking. I did, so my head is OK, but my shoulder’s not.

So if I started the Earth at some temperature but I was–let’s say I started it at zero temperature, zero Kelvin. This is unrealistic. It probably had some temperature when it was first formed. But let’s imagine I started at zero Kelvin. Now, the Stefan-Boltzmann Law says that an object at zero temperature does not radiate. So the Earth would be receiving heat from the Sun but not radiating it.

So the amount of heat stored in the Earth would be increasing. That is to say, its temperature would be increasing with time, getting hotter and hotter each day that went by, receiving more heat from the Sun, not radiating it to space. Well, of course, as soon as it develops some temperature, it’s going to start to radiate. Remember, the Stefan-Boltzmann Law says the power per unit area is the sigma times T to the fourth. So as soon as T begins to rise, you get a little bit of radiation to space.

So this curve will begin to arc over a little bit, but eventually, it’ll reach a constant temperature. And that’ll be the case when the rate of heat received from the Sun equals the rate at which energy is radiating to space. That’s the steady state. So we made the assumption that we were in this state. We equated the incoming radiation and the outgoing radiation and found the temperature that would allow that balance to be reached.

Remember, if the temperature’s too cold, it won’t radiate enough to keep up with what the Sun is delivering. If the temperature is too hot, it’s radiating more than the Sun is delivering, and the temperature of the Earth would cool with time. But in steady state, that balance is reached. And what we did was solve for that temperature, that single temperature that would allow the balance to be reached. And we derived this nice little formula, and we computed some values from it.

So we can summarize that calculation by looking at this little list of the things that control Earth’s temperature. The Sun diameter and the Sun’s temperature, because that controls how much radiation is emitted from the Sun. You can use the Stefan-Boltzmann Law to figure out how much each square meter of the Sun is emitting and then multiply times the surface area of the Sun to get the total amount of Sun’s radiant emission.

It also depends on the Earth-Sun distance, because as that radiation moves out away from the Sun, it diverges, it spreads out. And so the local intensity–that is the watts per square meter–decreases. So if you take a square meter and put it right at the Sun’s surface, you’ll get one value. You take that same square meter and move it further way, it’ll have a lesser value, because some of that radiation has spread out and is missing your little one square meter. So the Earth-Sun distance is important.

These two things together gave us the solar constant, the thing we called capital S, which you’ll recall is about 1,380 watts per square meter. It depends on the Sun diameter, the Sun temperature, and the Earth-Sun distance.

The albedo also came in. The albedo was the fraction of the radiation that hits the Earth that is reflected back to space. Now, you know what are the bright objects and the dark objects. Clouds are very bright. They reflect light back to space. Snow fields are very bright. Bright desert sands and even bright sand beaches are very bright. They have a high albedo. Forests have a very low albedo. Ocean has a very low albedo. The value we used here–we used 33% for the Earth–that’s an average of all of those. The bright objects, the dark objects, sum them all up on average, for Earth, it’s about 33%. For other planets, it’s different. That was important.

And from these considerations, we came up with that formula predicting the planet’s temperature, and we found that it was a little bit low. And we argued that the reason it was low is because we had neglected the greenhouse effect. We had assumed that the long-wave radiation emitted from the Earth’s surface could just escape to space. Whereas, in fact, the greenhouse gases in our atmosphere, especially water vapor, CO2, N2O, NO, some ozone as well, absorbed some of that outgoing long-wave radiation in the atmosphere, heats up the atmosphere, and some of it gets re-radiated back to the surface of the Earth. That’s an extra warming, if you like.

So our estimates for planetary temperatures were universally too cold, because we had neglected that. Now, for Earth, it was a pretty serious discrepancy. You’ve got the values in your notes, but I think it was something like 30 degrees–we were 30 degrees too cold in our simple estimates using these items. And that makes a difference between a habitable planet and an uninhabitable planet, at least for those living creatures that depend on having water in the liquid form.

So we’ve learned a lot of lessons in last time’s lecture, but one of them is that, well, the greenhouse effect is our friend. In other words, it makes this planet habitable. Were it not for that, the temperature would be universally below the freezing point for water, and we wouldn’t have the kind of life that we have. Now, later in the course when we’re talking about global warming, we’ll come back to this subject and see maybe the greenhouse effect is increasing, which may not be a good thing. But on a broad-brush analysis, looking at the habitability of the planet, the greenhouse effect is great. It warms us up to a place where we can live.

Any questions on that? So that’s in the way of review.

Chapter 2: Hydrostatic Balance [00:11:20]

Now, I want to go on and do a different subject today. You may have noticed, or maybe you haven’t, but I’m marching through these first couple of weeks of the course some of the most fundamental ideas in atmospheric science. They’re not in any particular order. I’ve tried to organize them a bit so they make sense, but they’re kind of my picks for the half-dozen or so most important concepts in atmospheric science. And this basic heat budget thing was one of them. We did that last time. Today I want to do another one called hydrostatic balance.

The basic idea behind hydrostatic balance is that air has weight. Air is heavy. And the man who first made this so obvious was the Frenchman Pascal, who has the pressure unit named after him. He took a mercury barometer like the one that you had in your lab the other day and carried that from the base up to the top of a mountain in southern France. Puy-de-Dome is the name of the mountain. I’ve been up there. I drove up there as a historical gesture to Pascal. I wanted to see the place.

And he noted that the pressure decreases as you go up. That was a great thing to notice. But he also explained it. He said, listen, the pressure at each level pushing up has to support the weight of the air above it. And so as you climb up a mountain, you’re putting some air–every time you step up, you’re putting some air below you. So the amount of air above you decreases the higher you go, and therefore the pressure will decrease as well. Because the less air you have above you, the less weight you have to support with that pressure pushing up. This was remarkable. It seems so simple today, but yet it really did revolutionize the way we thought about the atmosphere and so on.

So here’s the idea. We’re going to repeat his calculation, hydrostatic balance. “Static” here meaning there’s no acceleration–it’s just a static balance–“hydro” referring to fluid. Now, we’re talking about air, but this applies to liquids as well.

So here is a column of air extending from sea level up until we’re outside the atmosphere. So of course, the atmosphere doesn’t have a definite top, but we’ll go up 10, 20 scale heights, and so we’re above virtually all the atmosphere. So there’s no pressure pushing down at the top of this, because there’s no air up there. But there is a pressure pushing up. It’s the ground pushing up on the air or the air pushing down on the ground, either one. They have to be in balance. The pressure force pushing up on this column of fluid is the value of the pressure times the base area of that cylinder. Think of it as a cylinder going up to space.

So P times A has units of force. So remember, pressure is a force per unit area. So when I multiply it times an area, I’ll get something with units of force–that is to say, Newtons in the SI system. Now, this air column has a certain amount of mass, and I’ll call that capital M. And any time you’ve got mass and you put it in the gravitational field of a planet, that object will have weight. And you compute the weight by multiplying the mass times the acceleration of gravity.

So now, here’s where the word “balance” comes in. I’m going to equate those two things. I’m going to say that this is a static system, so the weight of that air is held up by the pressure force pushing up at the bottom of the column. So there I’ve done it. The pressure times the area–that’s the vector up–is equal to M times g–that’s the weight vector pulling down. I’ve just rearranged that. I’ve divided through by A and divided through by g to get M over A. That’s the mass per unit area is equal to the surface pressure divided by the acceleration of gravity (M/A=P/g).

This is the fundamental form in which we use it. This has units of kilograms per square meter. And now leave an inch in your notes and work out these units. This should have the same units as that. This one is obvious, kilograms per square meter. This one’s not so obvious. You’re going to have to remember what the units for pressure are and for g and work that out to see if the units check on that.

So that’s the basic assumption, and that’s the basic equation that results from it. The mass per unit area in a column reaching from sea level to outside the atmosphere is just given by the pressure at sea level divided by the acceleration of gravity.

Now, we could do a quick calculation of this. For surface pressure, I would put in about 1,000–let’s see–101,300 Pascals, and I would divide that by 9.81. Let’s round that off to 10. So I’ll just knock a decimal place off here, and that’ll be one 3 with one 0 (10130kg/m2). That’s going to be kilograms per square meter. And that’s about 10 metric tons, 10 tons per square meter. In other words, if you walk outside and imagine this one-square-meter base area and that column going up, there’s about 10 tons of air in that column.

Any questions on that? A pretty simple calculation. And it works. It’s very accurate, actually.

Now, that’s a big step, and it was pretty easy. I’ve got another big step that’s awful easy too. Now that I know the amount of mass per unit area, I can quickly compute the total amount of mass in the Earth’s atmosphere. I just have to know the area of a sphere. We used that the other day for another purpose. That’s 4 pi r squared. So if each square meter of Earth’s surface has this much mass standing above it, then I just multiply that by 4 pi r squared, which is the total number of square meters on the surface of the Earth, and I get the atmospheric mass.

So the atmospheric mass is given by P over g, which is the same as M over A, times 4 pi r squared. Now, this r is the radius of the planet. So I’ve done the calculation here again. I’ve put in 1,013, 9.81, like I did there. 4 pi. There is the radius of the planet. Now remember, I had to get that in meters. Everything has to be in consistent units. You square that out, and you get 54 times 10 to the 17 kilograms. So just like that, we got the mass of the Earth’s atmosphere.

Chapter 3: Calculation of CO2 Mass in the Atmosphere [00:19:15]

Questions on that? I’m going to extend that calculation a little bit further now. If we’ve measured with some device, the mixing ratio of CO2 in the atmosphere–how much CO2 is mixed in–we measure it locally and assume that the ratio of CO2 to air is the same everywhere. It’s not a bad approximation. CO2 is pretty well mixed in the atmosphere. It varies by a few percent, but not much more than that, around the whole Earth’s atmosphere. Then I could compute the amount of mass of CO2, basically.

Remember, though, there’s a couple of ways to represent mixing ratio. For example, you might see the mixing ratio of CO2 in square brackets sometimes written at about 390 parts per million by volume (390ppmv). That’s a molecular count. That’s how many molecules of CO2 per molecule of air. Or you can forget the pp–that’s parts per million–and write it 390x10-6 by volume.

But you might want it by mass instead, in which case you’d have to multiply up front by the ratio of the molecular weights. CO2 is 44. Air is 29. You put the same number in there, you get 550. That’s the concentration of carbon dioxide by mass, parts per million by mass (ppmm). It ends with an m instead of a v.

Well then, if you want to know the amount of carbon dioxide in the atmosphere, you’ve got the total amount here, just multiply that number times that number. Don’t forget, there’s a 10-6 in this parts per million thing here. That’s 10-6. And bingo, you’ve got the amount of CO2 in the Earth’s atmosphere. So from a few simple ideas, our quantitative understanding of the atmosphere is advancing by leaps and bounds with these simple calculations.


Student: I’m sorry. What was the 29, again?

Professor Ron Smith: The 29 is the molecular weight of air. You’ll recall that the molecular weight of nitrogen is 28, and the molecular weight of oxygen is 32. There’s more nitrogen than oxygen, so the average is a little closer to 28 than to 32. It comes out to be almost exactly 29 for the average molecular weight of air, being a mixture of N2 and O2. That’s how that number comes from. Good question. Other questions or comments?

Chapter 4: Derivation of the Differential Form of the Hydrostatic Balance Equation [00:22:08]

Now, there’s another form of the hydrostatic relation. This is for the total column. There’s another form which involves just looking at a little segment of the atmosphere. So imagine a little cylinder. You might want to still consider it a cylinder, perhaps, but you slice it here, and you slice it there, and you just consider this little stub of an atmosphere. And you want to know how the pressure is different at the bottom than the top.

And while I do this derivation, I have a volunteer who is going to take the Kestrel that you’re familiar with and run up to the top of KBT and measure the pressure at the ground floor and the top floor. And when she comes back in class, we will have finished a prediction of what she will measure for that pressure difference, based on the derivation I’m about to do here.

Thanks very much.

Now, in many ways–in some ways, the calculation is similar to this one. But instead of looking at the whole atmospheric column, we’re just going to consider this little stub of a column. So there’s a pressure at the bottom, which I’ll call PB. That’s the pressure at the bottom of that little cylinder. The pressure at the top is PT. The horizontal area of that little chunk is capital A. Its height is Δz. In other words, that’s the difference between the altitude at the bottom and the altitude at the top. I usually use z in this course for altitude.

Student: What is A?

Professor Ron Smith: Sorry?

Student: What is A?

Professor Ron Smith: A is the top area of the cylinder. So now there are three forces acting on this thing. There’s a pressure at the bottom acting on area A pushing up, and that’s the little vector I’ve drawn in there. There’s a pressure at the top acting over the same area A, so it’s PT times A. That’s the vector pointing down. And there’s another down vector, and that’s the weight of the mass in that chunk.

And if you want to know the mass of a fluid, you multiply the density, if you have it, times the volume. So I’ve written it here as Greek letter rho (ρ) for density, and then the volume is the product of the area and the height of the cylinder. Do you remember that from trigonometry? The volume of the cylinder is just the area times the height of it. So that’s the volume. So those three things together–ρ, A, Δz–that’s the mass of the air. It has units of kilograms, like the M over here. And then I multiply it times g to get the weight.

Now, I want to simplify that formula a little bit. Notice A appears in each of the three terms. So I cancel that out, I bring the PT over, and I keep the ρ g Δz on the right-hand side. And this is the formula that I’ve then derived for the hydrostatic law extending over some small range of altitudes. It says that the pressure at the bottom is greater than the pressure at the top by an amount given by the product of the air density, the acceleration of gravity, and the height of the object, the height of the air chunk, the air parcel.

Questions on that?

Chapter 5: Hydrostatic Law Experiment [00:26:05]

Let’s do a calculation then. She’s gone up to Kline Biology Tower. She’ll take a measurement here, take the elevator up to the 12th floor, take a measurement there, and then come back to class. So we’ve got to hurry up and get a prediction of what that is.

So we’re interested in the difference between P at the bottom and P at the top. So here’s what we need to do. We’re going to use the standard value for air density at sea level. It varies a little bit, but for most purposes we can assume that it’s 1.2 kilograms per cubic meter. g is 9.81. Units on that are meters per second squared. And we have to get the height, the Δz.

Now, this is 12 floors, and I’m assuming that there’s about four meters from floor to floor. For example, here, I think that’s about nine or 10 feet, which is about three meters. But then there’s a certain thickness before you get up to the next floor. So this is a pretty crude estimate. I’ve never measured it, but I’m going to assume that there’s four meters per floor, so that’s 48 meters for the height of Kline Biology Tower. We could measure this, and I probably would find that I have some error here. But to a rough approximation, that’s going to be 48 meters.

Now, check the units on this, because when I multiply these things together, I’ve got to get something with units of Pascals. And let me see what I come up with when I do that calculation. Yes, this gives me 565. If anyone has a calculator, please check me on this, because I’m not sure I can read my own handwriting here. 565 Pascals. Would someone check me on that, please, with a calculator? 1.2 times 9.81 times 48. Is that about right? It’s 565? OK.

Now, the instrument she’s reading, the Kestrel reads out in hectoPascals, which is a hundredth of a Pascal. So we are predicting that she will find a pressure difference of 5.65 hectoPascals, which is also millibars, by the way. And so we’ll see when she comes back.

Now, this is a pretty useful idea, and you’ve used it already in your lab. For example, when you were using the–a question. Yes.

Student: Sorry. Is a hectoPascal 100 Pascals or one hundredth of a Pascal?

Professor Ron Smith: A hectoPascal is 100 Pascals, yes. So 565 Pascal is 5.65 hectoPascals. I get that switched all the time.

We’ve used the hydrostatic law in your lab a couple of times. For example, when you’re using the mercury barometer, you are assuming that hydrostatic balance held within the mercury fluid. The atmospheric pressure was pushing down on the top of the reservoir. There’s a vacuum above, so there’s nothing pushing down on the mercury. But the weight of that mercury column is balanced by atmospheric pressure pushing on the reservoir.

Let me just sketch that out. I know you did this in class also, but the mercury lies in here and up to there. So atmospheric pressure acting down here supports that column of mercury. And this calculation would work perfectly for that. Just remember, there’s a vacuum up here, so there’s no pressure pushing down, but the pressure pushing up is exactly balancing the weight of the column of mercury standing above the base level.

I believe you also used it when you were correcting your pressure measurement to go down to sea level. You measured the pressure here in the laboratory, but that’s a few meters above sea level, and so you needed to compute a correction to add to that to get the pressure at sea level.

Are there any questions about the hydrostatic law before I change gears? I’m waiting for her to come back, but I’ve got a few more things I want to say about this. But I’d prefer to take your questions now. Here we go. Could you just call out the pressures for this?

Student: For the ground level, it was 1,007.9.

Professor Ron Smith: 1,007.9.

Student: And then the 12th floor was 1,002.2.

Professor Ron Smith: 1,002.2. So that difference is 5.7. Wow. Did we do good? Look at that. 5.65, 5.7. We did good on that. It doesn’t usually come out that well, actually. Congratulations. You’re good. You’re good. You jiggied those numbers just right, so we got the right answer. You get the point.

Now, you didn’t scale up the outside of the building. You went in and took the elevator up, I presume, right?

Student: Yes.

Professor Ron Smith: So the question is how was she able to get good numbers like this inside the building? Isn’t the inside of the building different than outside, where things are free to the rest of the atmosphere? Not really, because buildings leak around the windows and everywhere. So when you walk into a building, the pressure changes very, very little. The outside pressure, inside pressure is about the same, with a couple of exceptions.

If you play tennis at one of these inflatable tennis courts–you know what I’m talking about, they blow these things up?–you walk into that building, you have to kind of pull the door–high pressure inside, so you kind of have to push to get in there. If you feel when you’re entering that building a pressure resistance when you go through the doors–and most buildings don’t have that–but if you feel it, there’s some kind of active air conditioning that’s keeping the pressure inside higher or lower than outside. And you might want to worry then about whether your measurements are the same inside and outside.

But for the most part, you can measure the pressure inside a building, and it’s the same as it is outside. So in the laboratory, when I had you measure the pressure using the mercury barometer in Room 120, that’s the same as the atmospheric pressure outside. There’s no pressure differential in that building. You couldn’t do that with temperature. Temperature’s different inside, humidity’s different inside, but the pressure likes to equalize. And so this is a very robust calculation, as the agreement shows there.

Other questions about hydrostatic? Let’s spend a few minutes on this. Questions on hydrostatic law? Yeah.

Student: Where do you get the 1.2?

Professor Ron Smith: That is a standard value, sea-level density. You can compute it from the perfect gas law if you know the pressure and the temperature. I think we did that. If you measure the temperature in this room and the pressure in this room, you can use the perfect gas law to solve for that. And I think we did that earlier. Now, as you go up in the atmosphere, of course, that value changes.

By the way, this simple formula has a value for density in it. That means you couldn’t use it to predict pressure difference over a very great height difference, because the density changes over a great height difference. And so you wouldn’t know what value to put in for ρ. So this formula is best used over distances of a couple of hundred meters or less. If you’re going to compute pressure difference over a much greater distance, I would use that exponential formula that we had in class a week ago. So this is for short distances Δz, because it’s assumed a constant value for density.

Chapter 6: Application of Hydrostatic Law [00:35:14]

Now, by the way, this same law operates in the oceans. As you go down in the oceans, the pressure increases at a rate given quite precisely by the hydrostatic law. It’s accurate to within at least one part in 1,000, probably one part in 10,000, so a very accurate calculation. And for that, it’s a bit easier, because the density of water doesn’t change very much as you go down in the ocean. Water is nearly an incompressible liquid. It doesn’t change its volume when you put it under a high pressure. But remember, the density for water is much, much greater than that for air.

What is the density of water? Anybody know? I bet you know from your problem set. One–

Student: 1,000 kilograms–

Professor Ron Smith: 1,000 kilograms per cubic meter, about 1,000 times that value. So water is about 1,000 times denser than air. So I’ve been trying to impress on you today that air has mass, but of course, water has much more mass. Water is much denser liquid than air is.

Other comments or questions about the hydrostatic law? Yes.

Student: I understand the practical applications of this for the ocean, but what are some ways this is used in meteorology?

Professor Ron Smith: Let me give you an example of that. Notice that the rate at which the pressure increases as you go down depends on the density of the air. So let’s say in the center of a hurricane, you had a lot of warm air. If you look at the perfect gas law, you can find that warm air at the same pressure is less dense. So you have a value of ρ here that is relatively small compared to the ρ out here, which is larger. Small ρ in the center of the hurricane, large ρ outside the hurricane.

Now, let’s say you’ve got the same pressure everywhere here, but as you come down in the atmosphere here, through the low-density region, pressure increases. But it increases slowly, because rho is smaller. Here, pressure increases, but it increases more rapidly, because rho is larger. Remember, the story here in this box is that the pressure increases as you go down at a rate given by the air density. So if the pressure is the same at all locations up here, by the time I work my way down under here, I’m going to have low pressure and relatively higher pressure outside the hurricane.

So this hydrostatic law is the reason why there is low pressure in the center of a hurricane. We’ll talk about this later on in the course. We’ll talk about hurricanes. But we see it now, this basic law is responsible for this very important property of hurricanes having low pressure in the center. Because there’s warm air aloft, and the warm air is less dense, has a smaller value of density. Then just use that formula, and you’ll get that very nice and very practical, very important result.

In fact, we typically–very often, we measure the strength of a hurricane by how low that pressure reaches in the center of the hurricane. What you’re really measuring when you do that is the amount of warm air aloft. And that’s a good thing, because that’s an important aspect of the structure of a hurricane. So there’s a good example for that question.

Other issues, other questions? Let’s quit a bit early today. I’ll see you on Friday. 

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