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

Lecture 08.?Horizontal Transport

https://oyc.yale.edu/geology-and-geophysics/gg-140/lecture-8

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

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,?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?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.?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. 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?and?prevents it from crossing. In this case, it traps these pollutants below the inversion.?

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 (Volumn=L1*L1*D). The mass of the air into which the pollutant has mixed is just the product of the air density and that volume?(Mass of Air, M=ρ*V).?And that's going to have units of kilograms.

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?is a thousand kilograms. So I've written that in SI units as 40x10^3?kilograms. That's the amount of pollutant we've put in.?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.?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. Concentration=40*10^3 kg/1.2*10^4*10^4*10^3.?There's the mass of pollutant that we added, 40x10^3?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.?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.?

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

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?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. That has meters per second. This has meters. And so that's going to give meters squared per second, when you take that product. K=disperse coefficient=turbulent velocity*eddy size.

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. 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. 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.

Let's say the dispersion coefficient on this day is a 100 ?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. The radius R?=?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 3.7x10^7?kilograms. I'm going to use the same source here. I've mixed in 40x10^3?kilograms of pollutant into 3.7x10^7?kilograms of air. So the concentration is given by that ratio and it's 1.1x10^-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.

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 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, 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.

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?(t=x/U).?Wind speed is capital U in this case. Notice that'll have the right units, distance over a distance per 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.?().

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 A=πr^2/2, because it's half a circle.?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?of kilograms per second. 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.

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.?Source=ρ*U*A*[concentration],?ρ*U*A=airmass flux.?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 [concentration]=2*Source/ρπKx. One very interesting thing is that the velocity of the wind canceled out.?So the width of the plume will change with the wind speed, but not the concentration within the plume.

Source would be the rate at which pollutant is being added to the atmosphere at the power plant. 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.

Lapse Rate [00:27:10]

Let's?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.

In the troposphere, 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.?That's just an average. Sometimes it's very different than that, sometimes it's even positive. For example, in an inversion,?a layer where the temperature increases with height,?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. 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. 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)=-g/Cp?air?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.?

Cp air is the heat capacity of constant pressure for air. You could look it up in a table of physical constants.?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.?

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.

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.

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.

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,?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. 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. 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.?