ATMOS 301: Dry Adiabatic Processes

Thursday, October 31, 2013
8:37 p.m.

An adiabatic process is a process in which a material undergoes a change in its physical state without releasing or extracting any heat from the atmosphere. In other words, dq = 0 in the equation below, where dq = change in thermal energy (this is the first law of thermodynamics, check out my previous post for more info on it).

Because of our hydrostatic balance equation,

, and since we've let dq equal 0, we can get the following equation after substitution.

The equation below shows that the temperature decreases 9.8 degrees Celsius or Kelvin (your pick) per kilometer when it is raised at constant pressure without any moisture being released. It inreases at the same rate when it decreases at constant pressure. For a more generalized equation to show what the temperature will be like at any point when risen to a certain height with this process, use the equation below.

But what about using it in terms of pressure?

Well, you substitute from the equation of state, it's a lot uglier to calculate, and in the interest of time (it's now 8:57 p.m. Thursday night), I'm not going to do the proof. I doubt we will have to write it out on a 50 minute test. I think the main thing to take away from using it in terms of pressure is that it is more exact than z-coordinate calculation because you are not assuming the parcel is hydrostatic.

Potential Temperature: (because virtual temperature wasn't good enough)

The potential temperature θ is the temperature of a parcel of air after being brought dry-adiabatically (the process I just described above) to the 1000 mb level. If we know the temperature and pressure of an air parcel, then we know the potential temperature. Here's the relationship between temperature and pressure if we let K = R_d/c_p =~ 0.286.

T_2 = T_1(p_2/p_1)^K

And here's the more specific equation for potential temperature.

θ = T(1000mb/p)^K

If we take the derivative of θ... i.e., find out its rate of change, we get the equation below, which, just like every other equation in this god-forsaken pdf on thermodynamics, is atrociously ugly. The thing that matters is the term on the far right, though. dθ = 0. How nice. This means that the potential temperature is conserved and does not change as a function of height (provided that it doesn't exchange any heat with its environment) This is not true for the actual temperature that is measured by thermometers.

Cool. Now let's add water vapor to the equation. ;)

Charlie

ATMOS 301: The First Law of Thermodynamics

Thursday, October 31, 2013
5:54 p.m.

Our third law that we will concern ourselves with is the first law of thermodynamics. It will be the third of the "big six" we will look at for now. Here's the law.

This law says that the heat added to a certain mass of a gas is equal to its change in internal energy + the work done BY the gas ON the environment.

A classic example of the first law of thermodynamics at work is the piston. In order to make the piston system move, we first need to add heat to it. All or some of this heat Q is used to increase the internal energy of m. The rest of this heat is used to move the piston upwards.

In order to move a piston up, we need to do work on it. To push the piston outward through an external force dx, we set up the following equation, where dW equals the derivative of work and dx equals the derivative of x. For those of you who don't know calculus, derivative means "rate of change."

dW = Fdx

In the case of the piston, we have a cross-sectional area A exposed, so because F=p*A, we plug that in to get dW=p*A*dx = p*dV. Our final equation, dW = p*dV, says that the work done by the substance when its volume increases by a small increment dV is equal to the pressure of the substance multiplied by its increase in volume. If you want to find the total work done, just take the integral from V_1 to V_2.

__________________________________________________________________

I thought that was pretty easy to understand. Then we started getting into some other proofs that looked quite ugly.

Just to review:
 
Work = F*d (force*distance) = ρ*ΔV (density*change in volume)
Volume=A*L(area*length)
α = V/m (volume/mass)


Below are some variations on the ideal gas law. C_v is equal to the specific heat of vapor (water vapor), which is the amount of heat per unit mass required to raise the temperature by one degree Celsius. dq is the change in thermal energy, dT is the change in temperature, dα is the change in specific volume (1/density), and dp is the change in pressure.

We didn't spend a ton of time talking about the first law of thermodynamics, but we did spend a great deal talking about adiabatic processes. Without further ado, let's move on to some of those.

Charlie

ATMOS 301: The Hydrostatic Equation

Thursday, October 31, 2013
3:20 a.m.

I got a lot of sleep today, OK? So I'm allowed to do this. Plus, I don't have class until 1:30 tomorrow.

You ever feel your ears pop in an airplane? That's because you are going into the upper levels of the atmosphere where the pressure is lower. Remember, pressure is equal to force/area, and force is equal to mass*acceleration. When we increase in elevation, all that mass of the atmosphere that was previously weighing down on us is now below us, and the pressure on us is therefore lower. When we undergo this change in elevation and babies of all shapes and sizes begin to cry, we are directly experiencing the effects of the pressure gradient force. And, of course, as I'm sure all of you know, airplanes, while not counteracting the effects of gravity, keep us afloat at 35,000 feet so we can travel from point A to point B. These two forces - the pressure gradient force and the gravitational force - play a huge role in what is called hydrostatic equilibrium.

The gravitational force pulls objects towards the center of the Earth, so theoretically, it should pull all the atmosphere down to the ground. However, the pressure gradient force causes air to flow from high pressure to low pressure, so according to it, the atmosphere should have a homogeneous pressure distribution. So why don't we see either of those outcomes? Well, that's where the hydrostatic equilibrium comes into play.

Here are the individual forces that contribute to hydrostatic equilibrium.

g = gravity (9.8 m/s^2)
ρ = density
A = area
p = pressure
Δ = change in (insert variable here)

Gravitational force downward: g*ρ*A*Δz
Pressure force downward: (p + Δp)*A
Pressure force upward = p*A

Ok, it's now 4:49 a.m. I should sleep. Goodnight.

Alright, 11:45 a.m. and back to work. I have class at 1:30. I'd like to take a shower and eat before then as well. I hope you realize how epic this will be if I get all this done.

Anyway, we were talking about the hydrostatic equilibrium and why the atmosphere is structured the way it is. Since we have these forces that work in opposite directions, they get to a point where they equalize and the general structure of the atmosphere doesn't change. The equation for this is given by:

p*A =  (p + Δp)*A + g*ρ*A*Δz

This equation shows the pressure force upward having an equal magnitude to the sum of the pressure force above the specified level of air downward and the gravitational force downward. We can also restate this as:

Δp/Δz = -ρ*g

And by taking limits and partial derivatives, we end up with our final hydrostatic equation that we all use... an equation so important that it deserves its own picture.

Geopotential Height:

I know what you're thinking. These terms aren't getting any easier to understand. I'm right there with you. Although, I have to admit, I've always seen this term thrown around on the UW mm5/WRF model charts, and I've always wondered what it was but for some reason or another never actually tried to figure it out. Well, now that I have a midterm in 25 hours and 20 minutes, I'm suddenly a lot more interested in it. Imagine that!

The geopotential Φ at any point in the Earth's atmosphere is defined as the work that must be performed to raise 1kg of something to that point. The units are J/kg or m^2/s^2, suggesting that Φ is the gravitational potential per unit mass. The force (in newtons) acting on 1 kg at a given height z above sea level is the same as g, as g is, for our purposes, constant with altitude. The work is given by the integral of gdz from 0 to z, where z is equivalent to the height. The resulting work ends up just being equal to gravitational acceleration (9.81 meters/second^2) multiplied by the height the parcel is lifted. The equation below is just basic calculus.

The Hypsometric Equation:

Would you like fries with that?

Everybody loves a combo. And that's exactly what the hypsometric equation is. It's a conglomerate of our equation of state and our hydrostatic equation. Remember, the equation of state is P=ρR_dT_v and the hydrostatic equation is ∂p/∂z = -ρ*g. Now watch what we do here when we combine the equations. To make things easier to read, I'm just going to take a snapshot of Professor Houze's slide instead of trying to make a bunch of ugly subscripts and fractions.

 Once you have dZ = (R_d/g_o)*T_v*d*ln(p), you take the integral of both sides and you get the hypsometric equation below. For those of you taking the test, I would highly recommend memorizing the equation and what each of the variables mean. It showed up on one of the quizzes and I couldn't remember it, so I tried to try and BS it by using the hydrostatic equation, which is much simpler. It didn't really work, but I managed to scrape a few points. :)

The Scale Height:

In atmospheric science, the scale height is the distance over which a substance decreases by a factor of e (2.71828... the base of natural logarithms). Because the atmosphere is well-mixed below the turbopause (about 105 km), the pressures and densities of individual gases decrease at about the same rate as a function of altitude with a scale height directly proportional to their gas constant R. Since R* is just a universal constant and the only thing that results in R being different for different gases is the apparent molecular weight of the mixture in the denominator, the scale height is inversely proportional to the apparent molecular weight. The average temperature of the troposphere and stratosphere is around -18 degrees Celsius (255 degrees Kelvin), and this gives us a scale height of around 8 km.

To be honest, we took some notes on this in class but I barely understood any of it, and our book only barely touches on it and we don't have any lecture notes on it. So hopefully it won't be on the midterm. If you are interested in researching it by yourself, by all means, go ahead, especially if you can get it done by the morning of November 1, 2013.

Thicknesses and Heights:

Many things in atmospheric science intersect. Fronts intersect. Winds intersect. In our neck of the woods, the Puget Sound Convergence Zone is a classic example of winds intersect. But pressure ALWAYS decreases with height, and because of that, pressure surfaces (imaginary surfaces on which pressure is constant) never intersect.

If we take a look back at our hypsometric equation, check out the term on the left. This difference in height represents the thickness of the atmosphere (usually in decameters) from one pressure level to another. The thickness of the atmosphere is determined by the temperature. Remember, a cold atmosphere is more dense, so the thickness between the different pressure levels will be more compact than it would if the atmosphere was warmer. I feel like the diagram below shows this well.

Here are some common manifestations of this effect that we see in storm systems. Mid-latitude storms (cold-core cyclones) have low heights associated with them because, as the name suggests, the air at their core is cold and therefore more dense. The opposite is true with tropical cyclones, which have very high heights.

All these diagrams, with the exception of Thumper, the Super Cool Ski Instructor, were taken from Professor Houze's lecture on thermodynamics here. Now, onto the first law of thermodynamics!



Charlie

ATMOS 301: Equation of State

Thursday, October 31, 2013
12:54 a.m.

Happy Halloween!

Happy Halloween! To our beloved Professor Houze, I'll give you a free pass on this one, but please don't schedule a midterm on April 21st.

Our professor told us there are six laws that govern the atmosphere. Today, we just finished talking about Newton's 2nd Law, which, as opposed to just being F=ma, is F = 1/(ρ)*gradient(P)*(fk X v)*av. I think. I still have to review that... today's lecture was more confusing than why Rebecca Black ever became famous. Who knows...

Anyway, the first law we went over is the equation of state law, also known as the ideal gas law. It is given by the equation below:

, where

P = Pressure (pascals)
V = Volume (cubic meters)
n = Amount of gas (moles)
T = Temperature of gas (kelvin)
R* = Universal Gas constant (bear with me and pretend there is a star there. I got this equation off Wikipedia and it looked nice)

This can also be written as PV=mRT, where the R has no star and m = mass. The two equations are solved in different ways. This equation is often stated as P=ρRT, where ρ = mass/volume. Sometimes, it is written as Pα=RT, where α = 1/ρ = the specific volume of the gas, i.e., the volume occupied by 1 kg of the gas at pressure P and temperature T. This is the equation we are going to focus on solving.

The gas constant R of a gas is given by R*/M where R* is the universal gas constant (8.314 Joules*Kelvin⁻¹*moles⁻¹) and M is the molecular weight of the gas. Therefore, the gas constant varies with each gas. If you wanted to find out the gas constant of water vapor, you would divide R* by the molecular weight of a mole of water vapor, which would be 1 oxygen + 2 hydrogen = 16 + 2 = 18 grams per mole, and this = .462 Joules*Kelvin⁻¹*gram⁻¹). We want this in SI units, so we multiply by 1000 to turn grams into kilograms, and we get 462, which is our gas constant for water vapor (it's actually closer to 461.51, I just rounded the atomic masses of oxygen and hydrogen)! Water vapor plays a huge role in these gas laws; such a big role, in fact, that it has been given its own variable. The "vapor pressure" of water, or P in the equation of state above when it is solved for water, is commonly known as e. Therefore, eα_v=R_vT. I used the underscores to represent subscripts... all they denote is α and R for water vapor.

Since dry air is a mixture of gases, calculating R for it requires a two-step process. First, we add up the total mass of each gas the atmosphere and divide this sum by the total number of moles in the atmosphere. We then divide R* by molecular weight that is a conglomerate of all the components of air in the atmosphere to get R, which happens to be 287 J/kg for dry air. We do the same thing for moist air, but we add water vapor to the equation.

The reason why we can do this for a mixture of gases is because of Dalton's law of partial pressures, which states that the total pressure exerted by a mixture of gases that do not interact chemically is equal to the sum of the partial pressures of the gases. The partial pressure of a gas is the pressure it would exert at the same temperature as the mixture if it alone occupied all of the volume that the mixture occupies.

While we're at it, let me explain how the equation of state came to be. There were two dudes named Charles and Boyle, and they were heavily involved in the whole industry of making scientific laws that had to do with gases. Boyle discovered that if you keep your m and T constant, P is inversely proportional to V, and Charles discovered that V is directly proportional to T if you keep m and P constant AND that P is directly proportional to T if you keep m and V constant. Some smart fellow/s came around and combined Boyle's and Charles' findings into one beautiful law, and it remains one of the most fundamental laws in physics and chemistry today.

So we got Dalton, Boyle, and Charles. It's like the three musketeers.

Retrieved from http://ownbond.com/what-were-the-real-names-of-the-three-musketeers/

There's another important constant that the equation of state gives us. The ratio of the gas constant of dry air to water vapor (R_d/R_v) is equal to the ratio of the molecular weight of water vapor to dry air (M_w/M_d). This value is equal to 0.622 and is denoted by ε.
 
Virtual Temperature (because regular temperature wasn't good enough)

The equation of state is useful to derive an expression for something my textbook calls "a fictitious temperature" (Wallace & Hobbes, 2006). Sounds useless to me, but turns out it is just the opposite. The virtual temperature of a parcel of air with some moisture in it is the temperature at which a theoretical dry air parcel would have an equivalent pressure and density to the moist parcel. Because moist air is less dense than dry air at the same temperature and pressure, the virtual temperature is always greater than the actual temperature. It usually doesn't exceed it my more than a few degrees though.

Earlier, we solved for R for dry air and were able to form an equation of state suited to it. We weren't able to do that for moist air... until now. There's a long proof that shows that T_v = 1/.622*T, but the main thing to take away from the virtual temperature is that it allows for an equation of state that is always correct no matter how moist the air is. I'll give the equation its own line for ease of reading.

P=ρR_dT_v

, where

ρ = Density (kg/m^3)


R_d = Gas constant for dry air = 287 J/kg
T_v = Virtual temperature

Whew! It's 3:04 a.m. But you know what, this is an effective studying method because I feel like I have to not only study for myself but also put out some quality blogs for my friends... with a serious time deadline. Next!

Charlie

ATMOS 301: Troughs, Ridges, Waves, and Fronts

Wednesday, October 30, 2013
10:40 p.m.

It almost sounds like it could be a Dr. Seuss poem. Troughs, ridges, waves and fronts. Ditches, mountains, surfboards, and stunts.

Anyway, this will be the last of my "light duty" reviews/online lectures/blogs. The next section will be on gas laws and atmospheric thermodynamics. I'll try and write them in a manner that I can understand, and let me tell you, it won't be easy. But before I write one blog, two blog, red blog, blue blog (and possibly more), let's do some more qualitative analyses of atmospheric phenomena.

Ridges and troughs are pretty easy to conceptualize. They represent the shape of the jet stream in the upper atmosphere. In the northern hemisphere, a ridge represents a northern shift in the jet stream caused by a high pressure "pushing" it northward, whereas a trough represents a southern sagging of the jet stream caused by an area of low pressure.

The general flow of the atmosphere in the mid-latitudes in the northern hemisphere is eastward. Because these ridges and troughs travel eastward, they are often called waves. Waves come in all sorts of shapes and sizes, but for simplicity's sake, we divide them into two main categories: short waves and long waves. Generally, the shorter the wave, the faster it moves. I haven't taken any ocean physics classes, but I would like to find out if this same phenomena is true with water waves. I can't remember if/how the length of light or sound waves affects their speed when they are not in a vacuum. The picture below is another one I got from Professor Houze's presentation, and it shows the ridges and troughs quite nicely. The far right hand portion got cut off because of a formatting problem with the PDF.

Retrieved from http://www.atmos.washington.edu/~houze/301/protected/Notes/CompObsMaps.pdf

Here's an example of a 500mb chart from our latest WRF-GFS model here at the UW. Can you distinguish the shortwaves from the longwaves?

Add captValid 07:00 am PST, Tue 05 Nov 2013 - 135hr Fcst: 500mb Heights, Absolute Vorticity. UW WRF-GFS 36km Resolution: Initialized 00z 31 Oct 2013. Retrieved from the UW Pacific Northwest Environmental Forecasts and Observations Website. Model URL: http://www.atmos.washington.edu/~ovens/wxloop.cgi?mm5d1_x_500vor+///3

Remember, pressure is the vertical coordinate here, not height. Using pressure as the vertical coordinate is useful because it helps us get rid of some variable in some equation that I didn't write down. But trust me, it does in fact help get rid of some variable in some equation.

Fronts:

Many people don't know what troughs or ridges or waves are. Even I didn't have the firmest grasp of them for a long time. And I still don't think I do... I plan to study these and their effect on lower atmospheric dynamics at some point in the future over this quarter or over winter break. Whilst working with Steve Pool, we don't usually put fronts on our weather maps because he claims that much of the populace that watches the news doesn't understand what fronts are and just wants to know whether their suit should be of the swimming or survival variety. Most of my friends and family have a general idea of what they are, though.

Fronts mark the warm edge of a zone of strong temperature contrast; they are not simply the boundary between two different air masses. This horizontal temperature gradient is conducive to the formation of a low pressure system. Hopefully you all know what a low pressure system is.

Retrieved from http://www.atmos.washington.edu/~houze/301/protected/Notes/CompObsMaps.pdf

The formation of a low pressure system, known as cyclogenesis, is really complicated, and the above diagram is greatly simplified. Still, it offers a fantastic idea of the general life cycle of a cyclone.

Below is a diagram of the vertical structure of a midlatitude cyclone. The occluded front results because the cold front moves faster than the warm front, and when it does this, the warm air goes away. When I was young and studying this stuff on my own, I learned that there were two types of occlusions: warm occlusions and cold occlusions. It turns out there is only one - the cold occlusion - with cold air on both sides.

Retrieved from http://www.atmos.washington.edu/~houze/301/protected/Notes/CompObsMaps.pdf

Fronts are responsible for clouds and precipitation, but explaining that and a lot of the other intricacies of fronts wouldn't be the best use of my time. If you are interested in that stuff, I recommend clicking on the links above where I am getting the pictures from.

One last thing that I think is important, however, is the relationship between ridges and troughs in the upper atmosphere and fronts at sea level. Fronts tend to form on the right edge of the trough as it transitions into the ridge. The trough and low/front move at the same speed, but remember, not all troughs themselves move at the same speed. Those with shorter wavelengths move faster than those with longer wavelengths.

Shortwave troughs are also commonly superimposed on longwave troughs. Imagine a worm wiggling, and then these wiggles having their own individual wiggles associated with them. That's what the upper atmosphere is like. A worm wiggling its own wiggles.

Now, onto thermodynamics!

Charlie