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

Lecture 8

 - Horizontal Transport


This lecture describes how pollutants mix in the atmosphere. Three cases are considered: confined mixing, unconfined mixing, and unconfined mixing with wind. In a confined volume, the concentration of pollutant in the air depends on the volume and the mass of the air present in the volume. Unconfined mixing is also known as diffusion, in which the pollutant disperses through the air from the source over time. When wind is considered, the pollutant disperses from the source in the direction of the wind. The change in temperature with height in the atmosphere is also discussed.

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The Atmosphere, the Ocean, and Environmental Change

GG 140 - Lecture 8 - Horizontal Transport

Chapter 1: Mixing in the Atmosphere: Confined Valley [00:00:00]

Professor Ron Smith: Well, today we’re going to talk about mixing and dilution. When you add something to the atmosphere, gases or particles, what happens to it? How does it mix in? To what degree does it get diluted? This is a subject that has applications to human-induced air pollution, but also to a large number of natural processes in the atmosphere. So it’s really a set of quite fundamental ideas. And I could be thinking about either small particles or gases added to the atmosphere. Basically, something that’s going to move with the air and mix in, in a way. If the particles are large enough, of course, they might gravitationally fall out after awhile. And that effect would not be included in the calculations I’m doing here. So I’m assuming the particles are small or it’s a gas so that it will just move along with the air itself.

So the easiest case–maybe it’s a little bit, unnatural, but not so much–is that you’re in a confined valley and you know the dimensions of the valley–horizontal dimensions of the valley, and the valley is capped, in some way. I’ve written here that it’s capped by an inversion. Now, I’ll give a more careful and physical interpretation of inversion later on, but an inversion is defined as a layer of air in which the temperature increases with height. And for reasons that are not at all clear, from what I’ve just said, but will become clear, it’s very difficult for mixing to occur across an inversion. An inversion tends to act almost like a rigid lid. It holds air and pollutants below it or in any way, prevents it from crossing. In this case, it traps these pollutants below the inversion. So leaving that as only partially defined concept for a moment, we’ll know then the volume of the air into which the pollution from my little power plant here has reached. The volume into which the substance has mixed is then the product of the two horizontal dimensions, L1 and L2, and the depth of the layer, that’s the depth up to the inversion. The mass of the air into which the pollutant has mixed is just the product of the air density and that volume. And that’s going to have units of kilograms. So here comes a very simple example.

Let’s that over some period of time we put in 40 metric tons of a pollutant. Maybe it’s small particles, for example, or a gas. A metric ton, you recall, is a thousand kilograms. So I’ve written that in SI units as 40x103 kilograms. That’s the amount of pollutant we’ve put in. That’s quite a bit, but it depends on the problem we’re working on. Let’s say that the horizontal dimensions of the valley are 10 kilometers in one dimension and 10 kilometers in the other dimension and that the depth to the inversion is one kilometer.

So then the concentration of the pollutant is going to be the mass of the pollutant divided by the mass of the air into which it has mixed. I didn’t put that definition up here, but be sure you get that in your notes. We’re defining concentration in this case as a mass ratio. The mass of pollutant that you’ve added to the atmosphere divided by the mass of the air into which it has mixed.

So here’s the formula then. There’s the mass of pollutant that we added, 40x103 kilograms, and here’s the mass of the air into which it mixed, it’s the air density multiplied times the volume of the air–L1 L2 times D. So it’s 10 kilometers expressed in meters, 10 kilometers again, and one kilometer expressed in meters. You work all that out, you get 0.4x10-6 kilograms per kilogram. You could cancel it out and say it has no units. That’s fine. Or you could leave it as kilograms per kilogram, just to remind you what it is. It’s a kilogram of pollutant mixed into a certain number of kilograms of the air into which you mixed it. So since this is 10-6 and it’s a mass ratio, I could also write this as 0.4 ppmm, parts per million by mass. The per million, the pm there, refers to the 10-6. And the last m there refers to the fact that it’s a mass ratio. Question? Yes.

Student: Why is it kilograms of pollutant?

Professor Ron Smith: Well, because I’ve got the kilograms here, but it’s kilograms of pollutant divided by the kilograms of air into which it’s been mixed.

Student: So they don’t cancel out?

Professor Ron Smith: They could. That’s what I mentioned. You could cancel that out. I wouldn’t mind it at all if you did. But I sometimes like to keep it there just to remind myself of what that definition is. Either way would be fine. It’d be perfectly fine to cancel that out. Normally, concentration is considered to be a quantity that has no units. But sometimes I like to keep it there just as a reminder. Yeah?

Student: Why are the tens raised to the fourth power?  Oh, it’s meters.  OK.

Professor Ron Smith: Where are you looking?

Student: The concentration calculation.

Professor Ron Smith: Yeah. Well, that’s right. So this was 10 kilometers. And a kilometer is a 1,000 meters. So the 10 times the 3 gives 4, 104 meters. Yeah.

Student: What is L1 and L2 again?

Professor Ron Smith: The horizontal dimensions of the valley.

Student: Oh, so the area.

Professor Ron Smith: Well the product is the area of the valley, but the L1 is the dimension in one direction, L2 is the dimension in the other direction. If you want to know the base area of the valley, that’s L1 times L2. Other questions? Yes?

Student: Where did you get 103?

Professor Ron Smith: 103?

Student: It’s the mass of the air below the inversion?

Professor Ron Smith: Well the mass of the air below the inversion is what’s in the denominator here. It’s the density of the air times the volume of the air and the volume was given by that product, which you see written out here. Yeah. It’s good to get this one really clear, because the next one is even tougher. So any other questions on this one? OK.

Chapter 2: Mixing in the Atmosphere: Unconfined Mixing [00:07:22]

So now we move to case two, which is unconfined mixing. We don’t have this predetermined valley in a nice inversion in which we know things are going to be trapped. We know how much pollutant we’re putting in–I’m going to use the same value as I did here–but we don’t know into what volume will it be mixed. But we’re going to assume that there’s a certain amount of turbulence in this atmosphere. The air is kind of mixing and spinning and moving around a little bit. And we’re going to define something called the dispersion coefficient, which will help us decide how fast that added material will diffuse or will disperse into the atmosphere. I’ll call it capital K and it has units of meters squared per second.

And you can estimate a magnitude for that quantity in the following way. You can take a typical turbulent velocity–let’s say there are eddies in this room that are turning over and that air speed has a certain value, let’s say one meter per second. And a typical eddy size, maybe the eddies are filling the space between the floor and the ceiling, that’s about four meters. So if that were the case, if the typical speed were one meter per second, and the eddy size were four meters, that would give a dispersion coefficient of four meters squared per second. It’s simply that product. See, that’ll give you the right units. That has meters per second. This has meters. And so that’s going to give meters squared per second, when you take that product.

So this is sometimes very difficult to determine, what that dispersion coefficient is going to be for a particular atmospheric application. But you can get a rough estimate, if you can estimate the size of the eddies and the speed at which those eddies are turning over in this turbulent atmosphere. Let’s say somehow we know what that is. Then how fast will that mix things into the atmosphere? If I have a point source sitting on the ground, one direction is as good as any other direction. It’s going to mix to the left, to the right, and up equally fast. So it’s going to produce a kind of hemisphere of pollutant, a half sphere that’s growing in time as that material diffuses outward and upward.

And the radius of that hemisphere is given by this formula, the square root of the dispersion coefficient times time. So the longer you wait, the larger that hemisphere will be, and of course, the more dilute will be the pollutant because you’ve put a fixed amount of pollutant in, unless you’re continuing to add it. I’m assuming, the problem I’m going to work out here, you put in just a fixed amount of pollutant and then gradually you watch it dilute itself as it mixes into a larger and larger part of the atmosphere.

So what does that say? It goes as a square root of time. That function, the square root function, increases rapidly at first and then more slowly later on. And that is the nature of turbulent dispersion. Rapidly at first and then more slowly later in time. And you may have had that experience. If you burn something in the kitchen and you’re right there, you smell it very quickly. A couple of meters away, it takes a lot longer to get there and even further away even longer, but not linearly. It actually gets slower and slower and slower. The advance of that burned smell moves slower and slower as time progresses. So rapidly at first and then more slowly later on describes how things disperse away from their source.

So let’s carry out this example then. Let’s say the dispersion coefficient on this day is a 100 meters squared per second and that the time we’re going to wait after the pollutant was suddenly put in is 10 minutes. Well, that’s 600 seconds. So the first thing is to compute the radius of that hemisphere that contains the pollutant. It’s the square root of 100 times 600. And that’s about 245 meters.

By the way, check the units on this. Is this going to give you a length? How can that be? Well, if I multiply this times the time, the seconds are going to cancel out. I’m going to have meters squared inside the square root. Taking the square root, that gives me units of meters. And that’s just what I need to be physically consistent with this quantity, R, which is the radius of the hemisphere. So these units work out nicely. So the radius of that’s going to be 245 meters. We want to know the volume or the mass of the air into which the pollutant has spread.

So I’ve written here the formula for the air mass. It’s the density of the air, ρ, times the volume of half of a sphere. You may recall the formula for the volume of a sphere is 4/3πr3. Well, I’ve taken half of that, because this is a hemisphere. It’s half of a sphere. So it’s 2/3πr3. And I put the density in front, so I get the mass of the air and it’s 3.7x107 kilograms. So what have I done? I’m going to use the same source here. I’ve mixed in 40x103 kilograms of pollutant into 3.7x107 kilograms of air. So the concentration is given by that ratio and it’s 1.1x10-3. Again, you could cancel that out or you could say kilograms per kilogram. Kilogram of pollutant per kilogram of air. Question? Yes.

Student: So you don’t know the radius of the hemisphere unless you know the dispersion coefficient and the time which–?

Professor Ron Smith: That’s correct. It’s right. So unlike the valley case, where we kind of knew from the initial dimensions of the system, how big or how much the dilution will be, here we have to estimate it based on the dispersion coefficient and the time that’s left. Obviously, for longer times, that’s going to continue to grow and this concentration is going to get less and less and less. We haven’t removed any pollutant, but we’ve diluted it into a larger and larger amount of air. Yes?

Student: And so the air density is not affected by the pollutant?

Professor Ron Smith: No. I’m assuming that this is a small amount of pollutant relative to the air. And so no. It doesn’t change the air density. It just mixes in with the air. Yeah. Was there another question? Yes.

Student: So this only tells you overall concentration rather than concentration within the different smaller hemispheres it could go to?

Professor Ron Smith: Yeah, so that’s a very good question. The question is this only tells you what’s going on generally within that sphere. It doesn’t give you details of what’s inside the sphere. And of course, the model that I’ve described here is not precisely true. If I were to plot concentration across this distance, it would be very small outside. It would begin to rise smoothly, probably would be at maximum somewhere in the middle, and then decrease again.

Whereas what I’ve indicated to you, which is a little bit off, is if there would be zero outside and uniform concentration inside and then zero again outside. And that’s not exactly true. So your vision of it is actually more accurate than mine. It’d be more of a continuous distribution, large closer to the source, smaller further away. But this is an easy-to-compute estimate of what those concentrations would be. You would not have a sharp boundary here with this concentration inside and zero outside. It wouldn’t be like that, there’d be more of a gradual transition. So this is kind of an approximation to the real condition. That’s an excellent question. Other questions on this?

So there would be our concentration after time, 600 seconds. You could redo the computation for a later time and you’d get a larger value in the denominator and therefore, a smaller value of concentration. That’s the way dilution works. Anything else on this? OK.

Chapter 3: Mixing in the Atmosphere: Unconfined Mixing with Wind [00:16:59]

Now we’re going to add another element of complexity here. For case three, we’re going to add a wind. So there’s a wind blowing in this system. So it’s unconfined, no valley walls, but now there’s a wind blowing.

So I’ve drawn some little cartoons here. There is the source. Maybe it’s a–what I’m envisioning here is not just an instantaneous release of pollutant, like I was in case one and case two, but here I’m envisioning a steady state source of pollution, like a power plant that’s putting out a certain amount of smoke from the stack per unit time continuously. If I had a side view of that, and the wind was blowing from left to right, I would find that pollutant would be confined in a kind of–we call this a plume, a plume of pollutant, moving downstream, getting deeper the further it goes downstream due to the turbulent mixing.

If I had a top-view of that same situation–here’s the top-view–there’s the power plant, there’s the source, the wind is from left to right. Now we see that plume is spreading in the horizontal dimensions as well, again by turbulent mixing and dispersion. And if I was to slice this here or here and look at it along the direction of the wind, at any particular distance downstream, x, it would look like half a circle. And the radius of that circle will be larger the further downwind that I go, because dispersion is enlarging that plume the further we get from the source.

So how are we going to work this one out? Again, it’s just a matter of computing dilution, but now this is a steady-state pollution and we’re going to have to take into account the effect of wind moving that material downwind. So I wanted to define a transit time, the time it takes the air to go from the pollutant source to whatever x location we’re interested in. Maybe your house is five kilometers downwind of the power plant. Well then, x is going to be five kilometers. And the time it takes the air to get from the power plant to your house is going to be x divided by the wind speed. Wind speed is capital U in this case. Notice that’ll have the right units, distance over a distance per time. That’s going to work out to have units of time.

The radius of this plume is given by the same formula we had earlier, except now it’s this transit time we put in to that formula. It’s a steady-state situation, so it’s not changing in time, but yet there is this transit time that plays the role of time in the original problem. So I’ll put in x over t into this formula and I get  for this quantity, r, that you see in each of these three diagrams. Question, yes.

Student: What is U?

Professor Ron Smith: U is the wind speed, typically in meters per second. Now, by the way, the cross-sectional area of this plume, if I were to slice across like this, it is half a circle–and you know the area of the circle is πr2–and so the plume cross-sectional area is πr2/2, because it’s half a circle.

Now here comes probably the trickiest part of the derivation. I’m going to assume that I know the source strength and that’s how much pollutant is being added to the air per unit time by the power plant. That might have units, for example, of kilograms per second, kilograms of smoke, let’s say, being added to the atmosphere every second. Well, in a steady state, the amount of pollutant being added at the source must be the same amount that’s being carried away by the wind. And so by equating those two, I can derive a formula for the concentration.

So this is the key step, I’m going to equate the rate at which pollutant’s being added to the rate at which it’s being carried away by the wind. So on the left-hand side is a rate which it’s being added, this is the rate at which it’s being carried away by the wind. Let me go through this right-hand side. That’s the air density, that’s wind speed, that’s the cross-sectional area of the plume. And so that product by itself, those three things together, is the amount of air per unit time passing away within the plume–being carried downwind by the wind. And I just have to multiply that times the concentration to convert that from an air mass flux to a pollutant air mass flux.

So then solving this formula for concentration, I get 2 times the source, I’ve plugged this in there as well, 2 times the source, divided by the density of the air, divided by π times k times x. And you’re going to want to leave an inch or two in your notes to check the units on this as well. For example, the units on the concentration, we expect if previous calculations are right, it’ll be either kilograms per kilogram, if you prefer, or it’ll have no units at all. And so all the units over here should cancel out. And leave an inch or two of your notes to deal with that.

One very interesting thing is that the velocity of the wind canceled out. How did that happen? I had it here because the rate at which material is being blown downwind is going to be proportional to the wind speed. But you see the radius here, well the area, depends on r, and r depends on U underneath. So when you square the radius, you get a U underneath, which cancels that out. So the width of the plume will change with the wind speed, but not the concentration within the plume.

And when you do–you’re going to do a couple of examples like this in your next problem set. When you’re doing that, you might spend a few minutes puzzling over what was the role of the wind speed. The wind speed is very important in this problem, but in that particular formula, it canceled out. And so you should try to resolve that little dilemma–how, if the wind speed is important, how did it cancel out in that particular formula. So think a bit about how that would work. Any questions on this? Yes.

Student: What is A?

Professor Ron Smith: A is the–is this thing, the cross-sectional area of the plume, πr2/2. Yes.

Student:  What is the x in the equation over there?

Professor Ron Smith: X is the distance between the source of the pollutant and the location of interest: your house or the school yard, where the kids are playing, or whatever it is that you’re interested in finding out what the concentration is. It’s the distance away from the pollutant source. Now obviously, if the wind is from the west and you’re living down here somewhere, that plume is not going to hit you at all. So this calculation is irrelevant for you. So the wind direction comes into this in a very important way also, I haven’t mentioned that up until now, but it’s probably quite obvious that the wind direction is a key player in this. It’s only if the wind is blowing generally in the direction from the source to your location that this calculation becomes relevant. Yeah.

Student: Does the plume get wider as you go farther away from it? So if you live south of it, far enough away–

Professor Ron Smith: That’s true. That’s what I’ve sketched. This is a top view. It is getting wider, fast at first, and then slower. But it never stops widening, it keeps getting wider and wider and wider the further you are downwind. So that’s true. If you lived a great distance downwind, but not in the exactly in the wind direction, you might still get the pollutant, because the pollution will widen the plume. So keep track of that when you do these calculations. Yes.

Student: What does source represent in that final equation?

Professor Ron Smith: Source would be the rate at which pollutant is being added to the atmosphere at the power plant, let’s say. It could be in units of kilograms per second. But if it’s particles, it could be the number of particles per second that’s being added. And then your concentration would have to be in the appropriate–by the way, that’s the curve ball I’ve given you the problem set. I think I gave you a problem in which the source was quoted as a number of particles per second. And so the concentration then was in units of particles per cubic meter, rather than kilograms per cubic meter. So just be wary, you have to have consistent units when you do this.

Chapter 4: Lapse Rate [00:27:10]

OK, now we’re going to leave that there and move on to another aspect of mixing which is very important also. And that’s the role of the temperature lapse rate. So I called this section lapse rate and buoyancy effects. First of all, I want to define what I mean by the lapse rate. Lapse rate is the rate of change of temperature with height. If you launch a balloon and get that data back, you can measure the values of temperature with height and you can define the lapse rate. It has units typically of degrees Celsius per meter or perhaps you would want to use degrees Celsius per kilometer. It’s basically how fast is the temperature change in the atmosphere as you go up and down.

Typically, in the troposphere, you remember, it gets colder as you go up. And an average value–I’m using lower gamma (γ) for this–typical lapse rate for the troposphere is about 6.5 degrees Celsius per kilometer. On average, every kilometer you go up, it cools about 6.5 degrees. But don’t take that too literally. That’s just an average. Sometimes it’s very different than that, sometimes it’s even positive. For example, in an inversion–remember how I defined an inversion? An inversion was a layer where the temperature increases with height. Well, that would mean a positive lapse rate, instead of a negative one. So for some layers–and then in the stratosphere, remember you have a positive lapse rate, the temperature gets warmer as you go up.

The other really important quantity is a–more of a theoretical one, but it’s equally important. It’s the adiabatic lapse rate. This is the lapse rate that a parcel of air experiences when it rises. Why would the temperature of an air parcel change when it rises? Well, it’s pretty–it’s fairly straightforward. When you take an air parcel of a certain volume and lift it–you know the pressure decreases as you go up, so when you lift a parcel up, it’s going to expand a little bit. You lift it further, it’s going to expand still more. And when air expands, it does work on its environment by pressing out and expanding. It decreases the amount of energy stored in the parcel because it’s done work on the environment. And its temperature drops. It’s called adiabatic expansion or adiabatic cooling.

The word adiabatic here means without adding or subtracting heat. And you may wonder, how can I change the temperature without adding heat? The point is I’m changing the temperature by having that parcel do work on its environment, not by adding heat from some energy source. It’s purely a mechanical process of expanding the air and watching it cool because of that adiabatic expansion. That’s quite a powerful effect, as you’ll see in just a moment. So the adiabatic lapse rate is defined as a rate of cooling as an air parcel rises.

It is a reversible phenomenon, so if a parcel sinks back down in the atmosphere, it will compress and its temperature will increase at the same rate. I’ll use capital gamma (Γ) for this quantity, adiabatic lapse rate. It can be computed as a ratio of the acceleration of gravity to the heat capacity of the air. If this were a course on thermodynamics, I would derive that for you. For the earth, and our atmosphere made of air, the value is about -9.8x10-3 degrees Celsius per meter or expressing that in kilometers, it’s about -9.8 degrees Celsius per kilometer. Question.

Student: What’s the denominator again?

Professor Ron Smith: The heat capacity of constant pressure for air. You could look it up in a table of physical constants. You can Google it. Its value is about a 1,004, I believe, in SI units. So I have to confess, I’m a little bit sloppy, that number is so close to 10 that I very often round it off to 10 when I’m doing quick calculations. And I’ve done that here in my little example.

Let’s say I’ve got air at sea level that’s at 10 degrees Celsius and I move it up in the atmosphere three kilometers, 3,000 meters. What temperature will it be at when it reaches that higher elevation? Well, if I round that off to 10, it’s cooling 10 degrees for every kilometer that I lift it, so it’s going to cool by 30 degrees Celsius, approximately. Please, I would ask you not to do that rounding. In your problem sets, let it be 9.8, not 10. But for the purposes of illustration, it’s going to cool by approximately 30 degrees Celsius. It starts at 10, so it’s going to be -20 Celsius when it gets to 3 kilometers. You can do that in Kelvins as well, it will work out the same way. If it starts at–let’s see if it’s 10 degrees Celsius, that’s going to be 283 Kelvins, and subtract 30 from that, it’s going to be 253 Kelvins when it gets up there. So you might want to put both the Celsius and the Kelvins in your notes to be sure you’re clear on that question.

Student: So that first one there is it’s shown as expanded in that diagram?

Professor Ron Smith: That’s right. I’ve drawn it that way to remind you that the reason it’s cooled is because it has expanded. It moved into a region where the pressure was less and therefore, it naturally expanded because of that lowered pressure. Question.

Student: So it doesn’t matter the size of the air parcel?

Professor Ron Smith: No. So it’s so simple. It doesn’t matter the size of the parcel, it’s just that elevation difference is all that matters. That tells you how much, on a relative basis, it has expanded, and therefore, how many degrees Celsius it has cooled. Now remember I said this is reversible. If I take that parcel and move it back down, it’ll compress back down to its original volume, and it’ll return to its original temperature. So it’s a reversible process.

Student: Are you assuming the parcel was on the ground?

Professor Ron Smith: The question is whether we’re assuming it was on the ground. No, not necessarily. This would happen, I could start at any elevation and move it to any other elevation. So this is independent of where the ground was. I just happened to give you an example with the ground. Just a three kilometer lifting, no matter where you started, would give you an approximate 330 degrees Celsius cooling. Questions on that? OK.

Chapter 5: Buoyancy Effects of Rising and Descending Air Parcels [00:34:52]

Now, we can begin to do some calculations then of what happens to air as you move it up and down in the atmosphere. Using these two concepts, the measured lapse rate and the adiabatic lapse rate. So normally in the textbooks, you’ll see diagrams like this, where the temperatures have been put on the x-axis, altitude’s been put on the y-axis, and some reference lines are drawn on here, where the slope, given by capital gamma (Γ). And then I’ve drawn in a black curve, which represents the actual lapse rate on the day under consideration.

Let’s say we’ve launched a balloon, measured temperature as a function of height, and I’ve plotted that up as that black line. Now, the question is what’s going to happen to an air parcel as it’s moved upwards or downwards in this kind of a situation. If I take an air parcel from this elevation and lift it, it’s going to cool. It’s going to cool moving along the dashed blue curves, because that’s the adiabatic lapse rate. So this parcel, when I take it out of its environment and lift it a little ways, it’s going to move along like this to its new elevation. Its new temperature is going to be there and the temperature of its new environment is going to be there.

Here’s the tricky part of this argument. When you take an air parcel and lift it, it’s cooling at one rate, its environment is changing at a different rate. And we’re interested in knowing, after we lift it a certain amount, what is its temperature relative to the environment? It’s new, the air that it’s now next to. In this case, the parcel went to that temperature, it moved into cooler air above, but that wasn’t such a great effect. And actually, the parcel is now colder than its new environment. Environment cooled, the parcel cooled, the parcel cooled more, so the parcel is now cold, relative to its new environment.

What’s going to happen to that parcel? It’s going to sink because it’s more dense than its environment. So I lifted it up there, it looks around and says, oh my lord, I am denser than everything up here, back down I go. I don’t belong here in this crowd. I’ve got to go back, I’m negatively buoyant, I’m heavy, I’m going to sink back. You can repeat it. You could push that parcel down in the atmosphere, it’ll move to here. There’s a new temperature. There is a new temperature of its environment. It got warmer as it went down. The environment got warmer too, but it got warmer faster. Therefore, it’s warmer than its new environment. It’s now buoyant. It’s going to want to bob back up to its original condition. This is called a stable atmosphere. This term is very important, this is called a stable atmosphere because a parcel that is lifted upwards wants to sink back down, a parcel that’s pushed down wants to bob back up. And the atmosphere is usually in this state, but not always.

I want to look at this second example, here, which is again, temperature plotted versus altitude. The same reference lines have been drawn on here for capital gamma. But I put a different actual lapse rate on that curve. It’s a lapse rate where the temperature increases with height. This is an example of an inversion.

If I take a parcel from here and lift it, it gets colder. I’m going to take it from that elevation up to that elevation. There’s the parcel temperature. It’s actually been moving into warmer air, however. So the new environment is there. So this is an extreme example of stability. This is a very stable environment where the parcel gets colder as you go up, the environment gets warmer. And so you develop these restoring forces, this tendency for the parcel to want to quickly remove, go back to where it was, are amplified in this case, because of the inversion.

That’s the reason why air cannot mix effectively through an inversion. Because the air parcels want to cool as they rise, and yet there’s warmer air aloft, the two things add to each other, making it virtually impossible for air parcels to mix upwards or downwards, for that matter. The air tends to lay in stable layers, stratified stable layers with very little turbulent mixing in a situation like this.

Now the situation that I didn’t do, and you can fill in your notes, do one last case where now–tilt this curve over so it’s flatter than the reference curves. Leave some space in your notes for this. Take that observed temperature, the black line, rotate it over until it’s more horizontal than the blue reference lines. And then repeat the argument. What you’re going to find there is it if you look to parcel, it suddenly becomes buoyant relative to its environment and it’ll continue to rise. That’s called an unstable atmosphere.

Same thing would happen if you pushed it down. It would become negatively buoyant and it would then drop further, another illustration that it’s an unstable atmosphere. So this sort of thing will have a big impact on how materials mix around in the atmosphere. And as we will see next time, it also has a big impact on how clouds develop. Next week, we’re going to get into the issue of water vapor in the atmosphere, how clouds form. And then by Wednesday or so, how precipitation forms in the atmosphere. And I think we’ll call it quits today. But that will be the subject for next week.

[end of transcript]

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