This is the course outline.
Meteorologists devote their lives to one purpose: to forecast the weather. But many other scientists work to build our understanding of the atmosphere, which is the basis for better prediction that meteorologists can use. Atmospheric science is the foundation upon which all meteorology is based.
Just as with all other physical sciences, mathematics is the language of atmospheric science. If you had our first-year meteorology course, you learned about weather forecasting and some of the physical basis behind the forecasts. This course will introduce you to even more physical concepts of atmospheric science and some of the mathematics that describe and quantify those physical concepts.
You have learned some of the mathematics that you need for this course and will be learning more about vectors and vector calculus soon. In this lesson, we will practice some of the mathematics. You will also prepare an Excel workbook that will help you with some of this course’s assessment problems.
By the end of this lesson, you should be able to:
If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.
The atmosphere is amazing, awe-inspiring, frightening, deadly, powerful, boring, strange, beautiful, and uplifting – just a few of thousands of descriptions. So much of our lives depend on the atmosphere, yet we often take it for granted.
Atmospheric science attempts to describe the atmosphere with physical descriptions using words, but also with mathematics. The goal is to be able to write down mathematical equations that capture the atmosphere’s important physical properties (predictability) and to use these equations to determine the atmosphere’s evolution with time (prediction). Predicting the weather has long been a primary focus, but, increasingly, we are interested in predicting climate.
We know quite a lot about the atmosphere. It has taken decades, if not centuries, of careful observation and insightful theory that is based on solid physical and chemical laws. We have more to learn. You could help to advance the understanding of the atmosphere, but you must first understand the physical concepts and mathematics that are already well known. That is a primary purpose of this course – to give you that understanding.
What follows, below, is a series of pictures and graphical images. Each one depicts some atmospheric process that will be covered in this course. Look at these images; you will see them again, each in one of the next ten lessons. Of course, in each observation there are many processes going on simultaneously. In the last lesson, you will have the opportunity to look at an observation and attach the physical principles and the mathematics that describe several processes that are causing the phenomena that you are observing.
We have offered Meteo 300 online several times but use of the Canvas assessment platform is relatively recent. There may still be errors despite our best efforts.
As we go through the course, if you find an error or typo, post it in the Finding Errors Discussion Forum in Canvas. If you are the first to find an error, you will be awarded 0.1 additional points on your final grade. If you aren't the first one to post, you receive no credit for your find, so it's a good idea to read each lesson at the beginning of each week. A student can earn up to 1 point total (i.e., by being the first to find an error ten times). Now, I hope that we have very few errors, but ...
You’ve been told many times that meteorology is a math-intensive field. It is. But for this course, you already know much of the math, and what you haven’t seen, you will see in vector calculus. To get ready for the meteorology and atmospheric science in this course, you will need to refresh your ability to solve simple math problems, including solving simple problems in differential and integral calculus. At the same time, we will remind you about the importance of correctly specifying significant figures and units in your answers to the problems. The goal of this first lesson is to boost your confidence in the math you already know.
Suppose you are asked to solve the following word problem:
In the radar loop, a squall line is oriented in the north-south direction and is heading northeast at 57 km hr^{-1}. In the last frame of the loop, the line is 17 km west of the Penn State campus. You are out running and know that you can make it back to your apartment in 25 minutes. Will you get back to your apartment before you get soaked?
You reason that the line is moving northeast, and thus, at an angle of 45^{o} relative to the east. Therefore, the eastward motion of the squall line is just the velocity times the cosine of 45^{o}. That gives you the eastward speed. You decide to divide the distance by the eastward speed to get the amount of time before the line hits campus. You plug the numbers into your calculator and get the following result:
$\text{time}=\frac{17\text{km}}{\text{(}57\text{km/h)}\cdot \mathrm{cos}\left(\mathrm{45\xba}\right)}=0.42178\text{hours}=25.3070\text{minutes}$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaabshacaqGPbGaaeyBaiaabwgacqGH9aqpdaWcaaqaaiaaigdacaaI3aaabaGaaGynaiaaiEdacqGHflY1ciGGJbGaai4BaiaacohadaqadaqaaiaaisdacaaI1aaacaGLOaGaayzkaaaaaiabg2da9iaaikdacaaI1aGaaiOlaiaaiodacaaIWaGaaG4naiaaicdacaqGGaGaaeyBaiaabMgacaqGUbGaaeyDaiaabshacaqGLbGaae4Caaaa@501D@
According to your calculation, you will make it back with 0.3 minutes (18 seconds) to spare. But can you really be sure that the squall line will strike in 25.3070 minutes? Maybe you should figure out how many significant figures your answer really has. To do that, you need to remember the rules:
Number(s) | Answer | Number of Significant Figures | Reason |
---|---|---|---|
$25+.3$ | 25 | 2 | 25 has only 2 significant figures |
$25\xb70.3$ | 8 | 1 | $25\xb70.3=7.5$ , round to 8 because 0.3 has only 1 significant figure |
$1.5\left({10}^{3}\right)+3.24\left({10}^{2}\right)$ | $1.8\left({10}^{3}\right)$ | 2 | $1.5\left({10}^{3}\right)+0.324\left({10}^{3}\right)=1.824\left({10}^{3}\right)$ , then drop 2 to get $1.8\left({10}^{3}\right)$ |
$1.5\left({10}^{3}\right)+3.86\left({10}^{2}\right)$ | $1.9\left({10}^{3}\right)$ | 2 | $1.5\left({10}^{3}\right)+3.86\left({10}^{2}\right)=1.886\left({10}^{3}\right)$ , round up then drop 2 to get $1.9\left({10}^{3}\right)$ |
$\frac{\left(57.3+6.41\right)}{15.6}$ | 4.08 | 3 | $\frac{63.71}{15.6}=4.0840$ , trim to 3 significant figures to get 4.08 |
$200\left(3.142\right)$ | 600 | 1 | 200. has 3 significant figures; 200 (no decimal point) has 1 but is ambiguous |
$152\left({e}^{-.52}\right)$ | 90 | 2 | number in exponent has only 2 significant figures |
Check out this video (11:23): Unit Conversions & Significant Figures for a brief (1 minute) explanation of those rules! Start watching at 9:14 for the most relevant information. Note a minor error starting at 9:50 in which "60" should actually have a decimal point following the zero.
There are two types of variables – scalars and vectors. Scalars are amount only; vectors also have direction.
Most variables have dimensions. The ones used in meteorology are:
Some constants such as π have no units, but most do.
The numbers associated with most variables have units. The system of units we will use is the International System (SI, from the French Système International), also known as the MKS (meter-kilogram-second) system, even though English units are used in some parts of meteorology.
We will use the following temperature conversions:
$K{=}^{o}C+273.15$
$\left(\frac{5}{9}\right)\left({}^{o}F-32\right){=}^{o}C$
We will use the following variables frequently. Note the dimensions of the variables and the MKS units that go with their numbers.
Type | Variable | Dimensions | MKS Units | Common Unit Name |
---|---|---|---|---|
Scalar | length (x or ...) | L | m | |
area (A) | L^{2} | m^{2} | ||
volume (V) | L^{3} | m^{3} | ||
speed (u, v, w) | L/T | m/s | ||
energy (E) | ML^{2}/T^{2} | kg m^{2}/s^{2} | J = Joule | |
power (P) | ML^{2}/T^{3} | kg m^{2}/s^{3} | W = Watt | |
density (ρ) | M/L^{3} | kg/m^{3} | ||
pressure (p) | M/LT^{2} | kg/ms^{2} | Pa = Pascal | |
electrical potential | ML^{2}/T^{3}A | kg m^{2}/s^{3}A | V = Volt | |
temperature (T) | Θ | K | ||
Vectors | velocity (v) | L/T | m/s | |
momentum (mv) | ML/T | kg m/s | ||
acceleration (a) | L/T^{2} | m/s^{2} | ||
force (F) | ML/T^{2} | kg m/s^{2} | N = Newton |
p = (normal force)/area = (mass x acceleration)/area = ML/T^{2}L^{2} = M/LT^{2}
1 Pa = 1 kg m^{–1} s^{–2}; 1 hPa = 100 Pa = 1 mb = 10^{–3} bar (hPa = hecto-Pascal)
1013.25 hPa = 1.01325 x 10^{5} Pa = 1 standard atmospheric pressure = 1 atm
The knot (kt) is equal to one nautical mile (approximately one minute of latitude) per hour or exactly 1.852 km/hr. The mile is nominally equal to 5280 ft and has been standardized to be exactly 1,609.344 m.
Thus, 1 m/s = 3.6 km/hr ≈ 1.944 kt and 1 kt ≈ 1.151 mph.
surface winds are typically 10 kts ~ 5 m/s
500 mb winds are ~50 kts ~ 25 m/s
250 mb winds are ~100 kts ~ 50 m/s
Kelvin (K) must be used in all physical and dynamical meteorology calculations. Surface temperature is typically reported in ^{o}F (or ^{o}C for METARS) and in ^{o}C for upper air soundings.
$$\text{w=}\frac{{\text{massH}}_{2}\text{O}}{\text{massdryair}}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaabEhacaqGGaGaaeypaiaabccadaWcaaqaceaadbGaaeyBaiaabggacaqGZbGaae4CaiaabccacaqGibWaaSbaaSqaaiaaikdaaeqaaOGaae4taaqaaiaab2gacaqGHbGaae4CaiaabohacaqGGaGaaeizaiaabkhacaqG5bGaaeiiaiaabggacaqGPbGaaeOCaaaaaaa@47B5@
Usually the units for water vapor mixing ratio are g kg^{-1}. In the summer w can be 10 g kg^{-1}; in the winter, it can be 1-2 g kg^{-1}.
Dimensions truly are your friend. Let me give you an example. Suppose you have an equation ax + b = cT, and you know the dimension of b, x (a distance), and T (a temperature), but not a and c. You also know that each term in the equation – the two on the left-hand side and the one on the right-hand side – must all have the same units. Therefore, if you know b, you know that the dimensions of a must be the same as the dimensions of b divided by L (length) and the dimensions of c must be the same as the dimensions of b divided by Θ.
Also, if you invert a messy equation and you're not sure that you didn’t make a mistake, you can check the dimensions of the individual terms and if they don’t match up, it’s time to look for your mistake. Or, if you have variables multiplied or divided in an exponential or a logarithm, the resulting product must have no units.
Always write units down and always check dimensions if you aren’t sure. That way, you won’t crash your spacecraft on the back side of Mars [4]. View the following video (2:42).
Now it's time to to take a quiz. I highly recommend that you begin by taking the Practice Quiz before completing the graded Quiz. Practice Quizzes are not graded and do not affect your grade in any way–except to make you more competent and confident to take the graded Quizzes : ).
Calculus is an integral part of a meteorologist’s training. The ability to solve problems with calculus differentiates meteorologists from weather readers. You should know how to perform both indefinite and definite integrals. Brush up on the derivatives for variables raised to powers, logarithms, and exponentials. We will take many derivatives with respect to time and to distance.
Visit the Khan Academy website that explains calculus with lots of examples [6], practice problems, and videos. You can start with single variable calculus, but may find it useful for more complicated calculus problems.
1.
$$\begin{array}{l}\frac{da}{dt}=-ka\\ \frac{da}{a}=-kdt\\ {\displaystyle \underset{{a}_{o}}{\overset{{a}_{1}}{\int}}\frac{da}{a}}=-{\displaystyle \underset{{t}_{o}}{\overset{{t}_{1}}{\int}}kdt}\\ \mathrm{ln}({a}_{1})-\mathrm{ln}({a}_{0})=-k\left({t}_{1}-{t}_{0}\right)\\ \mathrm{ln}({a}_{1}/{a}_{0})=-k\left({t}_{1}-{t}_{0}\right)\\ {a}_{1}/{a}_{0}={e}^{\left(-k\left({t}_{1}-{t}_{0}\right)\right)}=\mathrm{exp}\left(-k\left({t}_{1}-{t}_{0}\right)\right)\\ {a}_{1}={a}_{0}{e}^{\left(-k\left({t}_{1}-{t}_{0}\right)\right)}={a}_{0}\mathrm{exp}\left(-k\left({t}_{1}-{t}_{0}\right)\right)\end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@B6C6@
2. $p={p}_{o}{e}^{\left(-z/H\right)}\text{\hspace{1em}};\text{\hspace{1em}}{\displaystyle \underset{0}{\overset{\infty}{\int}}pdz=\text{\hspace{0.17em}}\text{?(Dothedefiniteintegral.)}}$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiCaiabg2da9iaadchadaWgaaWcbaGaam4BaaqabaGccaWGLbWaaWbaaSqabeaadaqadaqaaiabgkHiTiaadQhacaGGVaGaamisaaGaayjkaiaawMcaaaaakiaaywW7caGG7aGaaGzbVpaapehabaGaamiCaiaadsgacaWG6bGaeyypa0JaaGjbVlaab+dacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabIcacaqGebGaae4BaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeizaiaabwgacaqGMbGaaeyAaiaab6gacaqGPbGaaeiDaiaabwgacaqGGaGaaeyAaiaab6gacaqG0bGaaeyzaiaabEgacaqGYbGaaeyyaiaabYgacaqGUaGaaeykaaWcbaGaaGimaaqaaiabg6HiLcqdcqGHRiI8aaaa@615B@
$${{\displaystyle \underset{0}{\overset{\infty}{\int}}p\text{\hspace{0.17em}}dz=-H{p}_{o}{e}^{-z/H}|}}_{0}^{\infty}=-H{p}_{o}\left(0-1\right)={p}_{o}H$$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaapehabaGaamiCaiaaykW7caWGKbGaamOEaiabg2da9maaeiaabaGaeyOeI0IaamisaiaadchadaWgaaWcbaGaam4BaaqabaGccaWGLbWaaWbaaSqabeaacqGHsislcaWG6bGaai4laiaadIeaaaaakiaawIa7aaWcbaGaaGimaaqaaiabg6HiLcqdcqGHRiI8aOWaa0baaSqaaiaaicdaaeaacqGHEisPaaGccqGH9aqpcqGHsislcaWGibGaamiCamaaBaaaleaacaWGVbaabeaakmaabmaabaGaaGimaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH9aqpcaWGWbWaaSbaaSqaaiaad+gaaeqaaOGaamisaaaa@55F1@
3. $p={p}_{0}{e}^{\left(-\frac{z}{H}\right)};\frac{1}{p}\frac{dp}{dz}=?$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiCaiabg2da9iaadchadaWgaaWcbaGaaGimaaqabaGccaWGLbWaaWbaaSqabeaadaqadaqaaiabgkHiTmaalaaabaGaamOEaaqaaiaadIeaaaaacaGLOaGaayzkaaaaaOGaai4oamaalaaabaGaaGymaaqaaiaadchaaaWaaSaaaeaacaWGKbGaamiCaaqaaiaadsgacaWG6baaaiabg2da9iaac+daaaa@3F8A@
$$\frac{dp}{dz}=-\frac{1}{H}{p}_{0}{e}^{\frac{-z}{H}}=-\frac{1}{H}p;\frac{1}{p}\frac{dp}{dz}=-\frac{1}{H}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGKbGaamiCaaqaaiaadsgacaWG6baaaiabg2da9iabgkHiTmaalaaabaGaaGymaaqaaiaadIeaaaGaamiCamaaBaaaleaacaaIWaaabeaakiaadwgadaahaaWcbeqaamaalaaabaGaeyOeI0IaamOEaaqaaiaadIeaaaaaaOGaeyypa0JaeyOeI0YaaSaaaeaacaaIXaaabaGaamisaaaacaWGWbGaai4oamaalaaabaGaaGymaaqaaiaadchaaaWaaSaaaeaacaWGKbGaamiCaaqaaiaadsgacaWG6baaaiabg2da9iabgkHiTmaalaaabaGaaGymaaqaaiaadIeaaaaaaa@49A9@
4. $\begin{array}{l}\frac{d\mathrm{ln}(ax)}{dt}=?\text{}\frac{d\mathrm{ln}(ax)}{dt}=\text{}\frac{1}{ax}\frac{d(ax)}{dt}=\text{}\frac{1}{ax}\frac{adx}{dt}=\frac{1}{x}u\text{,whereu=velocity}\\ \end{array}$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaWaaSaaaeaacaWGKbGaciiBaiaac6gacaGGOaGaamyyaiaadIhacaGGPaaabaGaamizaiaadshaaaGaeyypa0Jaai4paiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccadaWcaaqaaiaadsgaciGGSbGaaiOBaiaacIcacaWG4bGaaiykaaqaaiaadsgacaWG0baaaiabg2da9iaabccadaWcaaqaaiaaigdaaeaacaWGHbGaamiEaaaadaWcaaqaaiaadsgacaWG4baabaGaamizaiaadshaaaGaeyypa0ZaaSaaaeaacaaIXaaabaGaamyyaiaadIhaaaGaamyDaiaabccacaqGSaGaaeiiaiaabEhacaqGObGaaeyzaiaabkhacaqGLbGaaeiiaiaabwhacaqGGaGaaeypaiaabccacaqG2bGaaeyzaiaabYgacaqGVbGaae4yaiaabMgacaqG0bGaaeyEaaqaaaaaaa@6C7F@
5. $d(\mathrm{cos}(x))=?\text{}d(\mathrm{cos}(x))=-\mathrm{sin}(x)dx\text{}$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadsgacaGGOaGaci4yaiaac+gacaGGZbGaaiikaiaadIhacaGGPaGaaiykaiabg2da9iaac+dacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaWGKbGaaiikaiGacogacaGGVbGaai4CaiaacIcacaWG4bGaaiykaiaacMcacqGH9aqpcqGHsislciGGZbGaaiyAaiaac6gacaGGOaGaamiEaiaacMcacaWGKbGaamiEaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaaaa@59D0@
Often in meteorology and atmospheric science you will need to manipulate equations that have variables raised to powers. Sometimes, you will need to multiply variables at different powers together and then rearrange your answer to simplify it and make it more useful. In addition, it is very likely that you will need to invert an expression to solve for a variable. The following rules should remind you about powers of variables.
If $a={b}^{x}$ , then raise both sides to the exponent $\frac{1}{x}$ to move the exponent to the other side: ${a}^{\frac{1}{x}}={\left({b}^{x}\right)}^{\frac{1}{x}}={b}^{\frac{x}{x}}=b$
If ${a}^{x}{b}^{y}$ , and you want to get an equation with a raised to no power, then raise both sides to the exponent $\frac{1}{x}$ : ${\left({a}^{x}{b}^{y}\right)}^{\frac{1}{x}}={\left({a}^{x}\right)}^{\frac{1}{x}}{\left({b}^{y}\right)}^{\frac{1}{x}}=a{b}^{\frac{y}{x}}=\text{newconstant}$
This brief video (7:42) sums up these important rules:
Are you ready to give it a try? Solve the following problem on your own. After arriving at your own answer, click on the link to check your work. Here we go:
What does y equal?
$$\begin{array}{l}{x}^{1/b}={\left(a{y}^{b}\right)}^{1/b}={a}^{1/b}{\left({y}^{b}\right)}^{1/b}={a}^{1/b}y\\ y={x}^{1/b}/{a}^{1/b}={\left(\frac{x}{a}\right)}^{1/b}\end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@5F4C@
Now it's time to to take another quiz. Again, I highly recommend that you begin by taking the Practice Quiz before completing the graded Quiz, since it will make you more competent and confident to take the graded Quiz : ).
Meteorologists and atmospheric scientists spend much of their time thinking deep thoughts about the atmosphere, the weather, and weather forecasts. But to really figure out what is happening, they all have to dig into data, solve simple relationships they uncover, and develop new ways to look at the data. Much of this work is now done by programming a computer. Many of you haven’t done any computer programming yet, and for those of you who have, congratulations – put it to good use in this class. For those who are programming novices, we can introduce you to a few of the concepts of programming by getting you to use Excel or another similar spreadsheet program.
To help you learn and retain the concepts and skills that you will learn in this course, you will solve many word problems and simple math problems. For several activities, we give you the opportunity to practice solving particular types of problems enough times until you gain confidence that you can solve those same types of problems on a quiz. That means that you will be solving some types of problems several times and only the numbers for the variables will change. The simplest way for you to solve these problems is to program a spreadsheet to do that repetitive math for you.
Screenshot shows an Excel Spreadsheet
A text box says "put activity number in row 1" and an arrow points to cell A1.
A second text box says "put variable names in row 2" with an arrow pointing to cells A2 and B2
A third text box says "start numbers for variables in row 3" with an arrow pointing to cells A3 and B3
A final text box says "calculations follow variable numbers" with an arrow pointing to C3.
Let’s do a simple example. Suppose we have several boxes, some with different shapes and sizes, and we want to calculate the volume of the boxes and find the total volume. I have put in the names of the variables (with units!) and then the numbers for the length, width, and height of each box type and the total number of each box. To calculate the volume of each box, click on E3 and put an “= a3*b3*c3” in the equation line. Hit enter and it will do the calculation and put the answer in E3. A small square will appear in the lower right corner of E3. Click on this square with the mouse and pull down over the next three rows. Excel will automatically do the calculations for those rows. To calculate the total volume, go to F3 and enter “=d3*e3,” and hit “enter.” Grab the small box and pull down to get the total volume of each type of box. To get the total volume, click on F7, click on “Formulas,” and then “AutoSum,” and finally “Sum.” Excel will show you which cells it intends to sum. You can change this by adjusting the edges of the box it shows. Finally, pay attention to significant digits in your calculated volumes. You can adjust the number of decimal places by highlighting a cell or group of cells and then by clicking on the appropriate icon on the Excel tool bar. Note that an incorrect number of significant figures is displayed in some of the answers--see if you can figure out where.
length (m) | width (m) | height (m) | number of boxes | box volume (m3) | total volume (m3) |
---|---|---|---|---|---|
0.72 | 0.52 | 0.52 | 3 | =A3*B3*C3 | |
0.52 | 0.52 | 0.52 | 4 | ||
1.2 | 0.85 | 1.01 | 1 | ||
0.62 | 0.62 | 0.95 | 2 |
length (m) | width (m) | height (m) | number of boxes | box volume (m3) | total volume (m3) |
---|---|---|---|---|---|
0.72 | 0.52 | 0.52 | 3 | 0.19 | 0.58 |
0.52 | 0.52 | 0.52 | 4 | 0.14 | 0.56 |
1.2 | 0.85 | 1.01 | 1 | 1.03 | 1.03 |
0.62 | 0.62 | 0.95 | 2 | 0.37 | 0.73 |
Hopefully this example is a refresher for most of you. For those who are totally unfamiliar with Excel, please click on the question mark in the upper right of the screen and type in the box “creating your first workbook.” You can also visit Microsoft's help page for additional step-by-step instructions for how to Use Excel as Your Calculator [10]. The best way to learn, after the introduction, is by doing. The Keynote Support website [11] also lists helpful summaries of instructions.
Please follow the instructions above for setting up an Excel workbook. You will be using this workbook to do calculations, plot graphs, and answer questions on quizzes and problems for the rest of the course.
Use the same headings and the same numbers in the first four columns. Enter your own equations and find the box volumes and total volumes.
This assignment is worth 15 points. Your grade will mostly depend upon showing that you set up the workbook, but some additional points will be assigned contingent upon how well you follow the instructions. When your Excel workbook is complete, please do the following:
There is a very good reason that you are taking this class and I am teaching it – all of us are fascinated by the weather, awed by the atmosphere’s power, and passionate about learning more about it. Quite honestly, I can’t imagine a more rewarding career than the one that you are embarking upon or the one that I have. Nothing could be more rewarding than saving lives by making the atmosphere more predictable or by making the perfect prediction. Nothing.
But, do you know what? The best forecasters are the ones who can not only read weather maps, but who also know physically what the atmosphere is doing. The best forecasters know how to translate the physics into mathematics so that hand-waving can be turned into usable numbers. This course will start to make all of these connections between observations and physical cause-and-effect and help us find numerical solutions to questions.
For those of you who are in related disciplines, this course will give you a solid basic understanding of the atmosphere that you can apply in your studies and career, whether it be civil engineering, mechanical engineering, environmental engineering, chemistry, hydrology, or many other fields.
We have now reviewed some important concepts like significant figures and dimensions and units. You will continue to gain confidence in using the differential and integral calculus that you already know. As you go through the course, I want you to look back at the pictures of the atmosphere and imagine which equations are governing the processes that are causing your observations.
You have reached the end of Lesson 1! Make sure that you have completed all of the tasks in Canvas.
The atmosphere is composed of billions of billions of molecules, 10^{43} molecules to be exact. Each molecule is zooming at hundreds of meters per second but colliding with other molecules each billionth of a second, exchanging kinetic energy (½ mv^{2}), momentum (mv), and internal energy (rotations and vibrations). It’s impossible to calculate what all of these molecules will do, so instead we study the energy and energy changes of volumes of molecules and use this information in our forecast models. This study is called thermodynamics. Thermodynamics has some difficult concepts to master, but it is also part of your everyday experience. In this lesson, you will learn the fundamental laws of thermodynamics and see how they describe our atmosphere’s pressure and temperature structure. You will learn why the atmosphere is sometimes stable and what happens when it isn’t.
By the end of this lesson, you should be able to:
If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.
Understanding atmospheric thermodynamics begins with the gas laws that you learned in chemistry. Because these laws are so important, we will review them again here and put them in forms that are particularly useful for atmospheric science. You will want to memorize these laws because they will be used again and again in many other areas of atmospheric science, including cloud physics, atmospheric structure, dynamics, radiation, boundary layer, and even forecasting.
Before you begin this lesson's reading, I would like to remind you of the discussion activity for this lesson. This week's discussion activity will ask you to take what you learn throughout the lesson to answer an atmospheric problem. You will not need to post your discussion response until you have read the whole lesson, but keep the question in mind as you read:
This week's topic is a hypothetical question involving stability. The troposphere always has a capping temperature inversion—it's called the stratosphere. The tropopause is about 16 km high in the tropics and lowers to about 10 km at high latitudes. The stratosphere exists because solar ultraviolet light makes ozone and then a few percent of the solar radiation is absorbed by stratospheric ozone, heating the air and causing the inversion. Suppose that there was no ozone layer and hence no stratosphere caused by solar UV heating of ozone.
Would storms in the troposphere be different if there was no stratosphere to act like a capping inversion? And if so, how?
You will use what you have learned in this lesson about the atmosphere's pressure structure and stability to help you to think about this problem and to formulate your answer and discussions. So, think about this question as you read through the lesson. You'll have a chance to submit your response in 2.6!
The atmosphere is a mixture of gases that can be compressed or expanded in a way that obeys the Ideal Gas Law:
$$pV=N{R}^{*}T$$
where p is pressure (Pa = kg m^{–1} s^{–2}), V is the volume (m^{3}), N is the number of moles, R* is the gas constant (8.314 J K^{–1} mole^{–1}), and T is the temperature (K). Note also that both sides of the Ideal Gas Law equation have the dimension of energy (J = kg m^{2} s^{–2}).
Recall that a mole is 6.02 x 10^{23} molecules (Avogadro’s Number). Equation 2.1 is a form of the ideal gas law that is independent of the type of molecule or mixture of molecules. A mole is a mole no matter its type. The video below (6:17) provides a brief review of the Ideal Gas Law. Note that the notation in the video differs slightly from our notation by using n for N, P for p, and R for R*.
So here I have a tank filled with gas. And these little dots represent some of the gas particles that would be in this tank. The arrows I put in here because all of these particles are in constant random motion. They're like a bunch of hyperactive little kids, running into each other all the time, banging into the sides of the container, and so forth. So we've got this tank of gas. Let's think about the characteristics that we could use to describe it. So one of the things that we could do is we could say what its temperature is. The higher the temperature, remember, the faster these gas particles are moving around, so temperature is very important when we talk about gas. Temperature for gases should always be reported in Kelvin. So we could say, for example, that the temperature of this guy here is 313 Kelvin. That's how hot these gas particles in the sample are. When you talk about gas, another important characteristic is pressure. How hard are these gas particles bouncing against the sides of the tank? How much pressure are they exerting on them? And we could measure these with a pressure gauge or something like that on the top of this tank. We could say, the pressure for this is 3.18 atm. That might be a pressure. And another thing that we spend a lot of time talking about when it comes to gas is volume. And again, I have these letters here that are how each one of these things are abbreviated. Volume, V, volume of this tank might be something like 95.2 liters. And finally, look at these particles that I've drawn. There is a certain amount of gas that's in here. And the amount of gas, which is abbreviated by the little letter n, is usually reported in moles, which is a convenient measure of how much of something we have. So we could say that the amount of gas in this tank is 7.5 moles. Now, whenever we have a sample of gas like this, if it's a tank or it's in a balloon or wherever it is, we can describe-- we can give it these various characteristics. And it turns out that also, for any sample of gas, if we know three of these characteristics, we can figure out what the fourth is. All we need to do is know three. And in order to do that, we use an equation that's a representation of the Ideal Gas Law. And it's written as P times V, pressure times volume, equals n, the amount of gas, times R times T, temperature. I'll get to R in a second. Don't worry about it for right now. It's going to be a number that we know. So let's say, for example, that we didn't know what pressure was, but we still knew the temperature, volume, and the amount of gas. No big deal. We could take the equation, PV equals nRT, and rearrange it. Divide both sides by V. Get rid of the V. And then we'd have P equals nRT divided by V. Plug these values in, and we could figure out what the pressure was. Or let's say that we knew what the pressure was of a particular gas sample. We know what the temperature was in a volume. But we didn't know what the amount of gas was. We don't know how much we had. We could figure out that fourth characteristic by rearranging the Ideal Gas Law for n, canceling out R and T on one side, rearranging it to solve for n. And then we could plug in the pressure, the volume, and the temperature, and we could figure out the amount of gas. So in other words, if we know three of these characteristics, we can always figure out what the fourth is. So you may be asking yourself, so R-- what's R? R is what we call a constant. It's a number that we know ahead of time that doesn't depend on the variables in our problem. The R that I'm going to be using most of the time for the videos is 0.0821 liters times atm divided by Kelvin times moles. Now notice that this is a fraction. It has both a top and a bottom. And it also is not just a number, but it has units. And check this out-- the units on R match the units in my problem. They match the characteristics that I'd be using. So I have liters here, liters here, atm, atm, Kelvin, Kelvin, and moles, moles. You always want the units on R to match the units of the characteristics in your Ideal Gas Problem. So because you always want the units to match, there are also different values of R, although I'm going to be using this mostly for the videos I'm doing. For example, let's say that instead of atm, I was using a pressure that was in millimeters of mercury. In this case, I wouldn't want to use this R here. I'd want to use this R here, so that the units match-- millimeters of mercury here, millimeters of mercury here, and the number's different-- 62.4. So again, that's what I use here. Let's say that instead of millimeters of mercury, my pressure was given to me in kPa. I would then use this value of R so that the units match. I've got kPa here, kPa here, and all the others are the same, so 8.31 for that. Now as I keep saying, in most of the videos that I'm going to be doing, I'm going to be using this top R with atm. But you may be asked by your teacher to use a different R. It's no big deal. That's probably just because they're giving you problems that have different pressure units, and they want the pressure units to match. So don't worry at all if you're using one of these other R's. Setting up and solving the Ideal Gas Law is exactly the same. No matter which of these R's you use, it's just a matter of plugging a different R in at the very end. So no matter which one you're using, you should be able to follow all these lessons, and it should all make sense.
Usually in the atmosphere we do not know the exact volume of an air parcel or air mass. To solve this problem, we can rewrite the Ideal Gas Law in a different useful form if we divide N by V and then multiply by the average mass per mole of air to get the mass density:
$$\rho =\frac{NM}{V}$$
where M is the molar mass (kg mol^{–1}). Density has SI units of kg m^{–3}. The Greek symbol ρ (rho) is used for density and should not be confused with the symbol for pressure, p.
Thus we can put density in the Ideal Gas Law:
$$p=\frac{\rho {R}^{*}T}{M}or\rho =\frac{Mp}{{R}^{*}T}$$
Density is an incredibly important quantity in meteorology. Air that is more dense than its surroundings (often called its environment) sinks, while air that is less dense than its surroundings rises. Note that density depends on temperature, pressure, and the average molar mass of the air parcel. The average molar mass depends on the atmospheric composition and is just the sum of the mole fraction of each type of molecule times the molar mass of each molecular constituent:
$${M}_{average}=\frac{{\displaystyle \sum _{i}{N}_{i}{M}_{i}}}{{\displaystyle \sum _{i}{N}_{i}}}=\frac{{\displaystyle \sum _{i}{N}_{i}{M}_{i}}}{N}={\displaystyle \sum _{i}\frac{{N}_{i}}{N}{M}_{i}}={\displaystyle \sum _{i}{f}_{i}{M}_{i}}$$
where the i subscript represents atmospheric components, N is the number of moles, M is the molar mass, and f is the mole fraction. This video (3:19) shows you how to find the gas density using the Ideal Gas Law. You will note that the person uses pressure in kPa and molar mass in g/mol. Since kPa = 1000 Pa and g = 1/1000 kg, the two factors of 1000 cancel when he multiplies them together and he can get away with using these units. I recommend always converting to SI units to avoid confusion. Also, note that the symbol for density used in the video is d, which is different from what we have used (ρ, the convention in atmospheric science).
Hey guys how do you solve ideal gas law questions involving density? The key is to have a formula or know how to derive the formula on your own. Remember density is mass over volume. Now, the way that mass is found in the ideal gas law equation is in n because the number of moles is the same as mass over molar mass. So, check this set. I'm going to replace n with mass over molar mass, and then I'm going to rearrange for m over v. I'm going to undo division by molar mass on the other side and then I'm going to undo multiplication by RT and bring my V over. Here's a what I mean. P times the molar mass divided by RT gives me mass over volume. Mass over volume is density and so my equation is density equals pressure times molar mass divided by RT. We can now use this equation to find the density of oxygen at 55 Celsius and a hundred and three kilopascals. So let's do it. The density is pressure that's 103 kilopascals times molar mass for oxygen. That's 32 grams per mole. R, now, I'm going to put my volume in liters and I'm going to put my pressure in kilopascals which means the relevant are that I want is 8.314 liters, kilopascals per mole Kelvin and my temperature in Kelvin is the temperature in Celsius plus 273, which gives me 328 Kelvin. And all of these units should cancel out to give me a density unit. Kelvin cancels of Kelvin per moles cancel / moles kilopascals canceled the scale pascals and left with grams per liter. Let's do this on the calculator 103 times 32 divided 8.314 divided 328. That's 1.21 grams per liter. That may not seem like a lot, but remember you're dealing with the gas here. If you have a 1-liter balloon how much is it actually going to weigh? Probably the amount of the rubber plus like a gram or so. This here is the density of oxygen gas at 55 and 103 kilopascals. This is your density formula in terms of the ideal gas law. Be able to use it. Best of luck!
Solve the following problem on your own. After arriving at your own answer, click on the link to check your work.
Let’s calculate the density of dry air where you live. We will use the Ideal Gas Law and account for the three most abundant gases in the atmosphere: nitrogen, oxygen, and argon. The mole fractions of the gases are 0.78, 0.21, and 0.01, respectively. M is the molar mass of air; M = 0.029 kg mol^{–1}, which is just an average that accounts for the mole fractions of the three gases:
$$\begin{array}{l}M=\text{}0.78\text{}{M}_{N}{}_{2}+0.21\text{}{M}_{O}{}_{2}\text{}+0.01\text{}{M}_{A}{}_{r}\text{}\\ =0.78\cdot 0.028+0.21\cdot .032+0.01\cdot 0.040=0.029\text{\hspace{0.17em}}kg\text{\hspace{0.17em}}mo{l}^{-1}\end{array}$$R* = 8.314 J K^{–1} mol^{–1}. Here, p = 960 hPa = 9.6 x 10^{4} Pa and T = 20 ^{o}C = 293 K.
Putting these values into the equation 2.3, we get that the dry air density is 1.1 kg m^{–3}.
What would the density be if the room were filled with helium and not dry air at the same pressure and temperature?
Helium density = (pressure times molar mass of helium)/(Ideal Gas Law constant in SI units times temperature in K) = 0.16 kg m^{–3}
Often in meteorology we use mass-specific gas laws so that we must specify the gas that we are talking about, usually only dry air (N_{2} + O_{2} + Ar + CO_{2} +…) or water vapor (gaseous H_{2}O). We can divide R^{*} by M_{i} to get a mass-specific gas constant, such as R_{d} = R*/M_{dry air}.
Thus, we will use the following form of the Ideal Gas Law for dry air:
$${p}_{d}={\rho}_{d}{R}_{d}T$$
where:
${R}_{d}=\frac{R*}{{M}_{dryair}}=\frac{8.314{\text{JK}}^{-1}{\text{mol}}^{-1}}{0.02897{\text{kgmol}}^{-1}}=287{\text{m}}^{2}{\text{s}}^{-2}{\text{K}}^{-1}=287{\text{Jkg}}^{-1}{\text{K}}^{-1}$ .
M_{dry air} is 0.02897 kg mol^{–1}, which is the average of the molar masses of the gases in a dry atmosphere computed to four significant figures.
Note that p must be in Pascals (Pa), which is 1/100th of a mb (a.k.a, hPa), and T must be in Kelvin (K).
We can do the same procedure for water vapor:
$${p}_{v}\equiv e={\rho}_{v}{R}_{v}T$$
where ${R}_{v}=\frac{R*}{{M}_{watervapor}}=\frac{8.314{\text{JK}}^{-1}{\text{mol}}^{-1}}{0.01802{\text{kgmol}}^{-1}}=461{\text{m}}^{2}{\text{s}}^{-2}{\text{K}}^{-1}=461{\text{Jkg}}^{-1}{\text{K}}^{-1}$ .
Typically e is used to denote the water vapor pressure, which is also called the water vapor partial pressure.
This gas law is used often in meteorology. Applied to the atmosphere, it says that the total pressure is the sum of the partial pressures for dry air and water vapor:
$$p={p}_{d}+{p}_{H2O}={p}_{d}+e$$
Imagine that we put moist air and an absorbent in a jar and screw the lid on the jar. If we keep the temperature constant as the absorbent pulls water vapor out of the air, the pressure inside the jar will drop to p_{d}. Always keep in mind that when we measure pressure in the atmosphere, we are measuring the total pressure, which includes the partial pressures of dry air and water vapor.
So it follows that the density of dry air and water vapor also add:
$$\rho ={\rho}_{d}+{\rho}_{v}$$
Suppose we have two air parcels that are the same size and have the same pressure and temperature, but one is dry and the other is moist air. Which one is less dense?
We can solve this one without knowing the pressure, temperature, or volume. Let’s assume that 98% of the molecules are dry air, which means the remaining 2% are dry air in the first case and water vapor in second case. Dry air is 0.029 kg mol^{–1} and water vapor is 0.018 kg mol^{–1}, so 2% of the moist air is lighter than the 2% of dry air, and when we consider the total air, this means that for the same temperature and pressure, moist air is always less dense than dry air.
Suppose there are two air parcels with different temperatures and water vapor amounts but the same pressure. Which one has a lower density? We can calculate the density to determine which one is lighter, but there is another way to do this comparison. Virtual temperature, T_{v}, is defined as the temperature dry air must have so that its density equals that of ambient moist air. Thus, virtual temperature is a property of the ambient moist air. Because the air density depends on the amount of moisture (for the same pressure and temperature), we have a hard time determining if the air parcel is more or less dense relative to its surroundings, which may have a different temperature and amount of water vapor. It is useful to pretend that the moist parcel is a dry parcel and to account for the difference in density by determining the temperature that the dry parcel would need to have in order to have the same density as the moist air parcel.
We can define the amount of moisture in the air by a quantity called specific humidity, q:
$$q=\frac{{\rho}_{v}}{{\rho}_{d}+{\rho}_{v}}$$
We see that q is just the fraction of water vapor density relative to the total moist air density. Usually q is given in units of g of water vapor per kg of dry air, or g kg^{–1}.
Using the Ideal Gas Law and Dalton’s Law, we can derive the equation for virtual temperature:
$${T}_{v}=T\left[1+0.61q\right]$$
where T and T_{v} have units of Kelvin (not ^{o}C and certainly not ^{o}F!) and q must be unitless (e.g., kg kg^{–1}).
Note that moist air always has a higher virtual temperature than dry air that has the same temperature as the moist air because, as noted above, moist air is always less dense than dry air for the same temperature and pressure. Note also that for dry air, q = 0 and the virtual temperature is the same as the temperature.
Solve the following problem on your own. After arriving at your own answer, click on the link to check your work.
Consider a blob of air (T_{blob} = 25^{ o}C, q_{blob} = 10 g kg^{–1}) at the same pressure level as a surrounding environment (T_{env} = 26^{ o}C and q_{env} = 1 g kg^{–1}). If the blob has a lower density than its environment, then it will rise. Does it rise?
We will use equation 2.10. Remember to convert T from ^{o}C to K and q from g kg^{–1} to kg kg^{–1}!
$$\begin{array}{l}{T}_{vblob}=(25+273)\left[1+0.61\cdot .010\right]=299.8K=26.8{}^{o}C\\ {T}_{venv}=(26+273)\left[1+0.61\cdot .001\right]=299.2K=26.2{}^{o}C\end{array}$$
We see that the blob is less dense than its environment and so will rise. This difference of 0.6 ^{o}C may seem small, but it makes a huge difference in upward motion.
The following are some mistakes that are commonly made in the above calculations:
$$\begin{array}{l}{T}_{vblob}=(25)\left[1+0.61\cdot .010\right]=25.15{}^{o}C\\ {T}_{venv}=(26)\left[1+0.61\cdot .001\right]=26.02{}^{o}C\end{array}$$
We calculate that T_{vblob} < T_{venv}, which is the wrong answer.
$$\begin{array}{l}{T}_{vblob}=(25+273)\left[1+0.61\cdot 10\right]=2115K=1842{}^{o}C\\ {T}_{venv}=(26+273)\left[1+0.61\cdot 1\right]=481K=208{}^{o}C\end{array}$$
We calculate that T_{vblob} > T_{venv}, which is correct in this case, but the numbers are crazy! After you complete your calculations, if the numbers you get just don’t seem right—like these—then you know that you have made a mistake in the calculation. Go looking for the mistake. Don’t submit an answer that makes no sense.
Once we find T_{v}, we can easily find the density of a moist parcel by using equation 2.5, in which we substitute T_{v} for T. Thus,
$${\rho}_{}=\frac{{p}_{}}{{R}_{d}{T}_{v}}$$
This quiz will give you practice calculating the virtual temperature and density using the Excel workbook that you set up in the last lesson.
The atmosphere’s vertical pressure structure plays a critical role in weather and climate. We all know that pressure decreases with height, but do you know why?
The atmosphere’s basic pressure structure is determined by the hydrostatic balance of forces [18]. To a good approximation, every air parcel is acted on by three forces that are in balance, leading to no net force. Since they are in balance for any air parcel, the air can be assumed to be static or moving at a constant velocity.
There are 3 forces that determine hydrostatic balance:
$${F}_{top}=-{p}_{top}A$$
$${F}_{bottom}={p}_{bottom}A$$
$${F}_{weight}=-\rho Vg=-\rho gA\Delta z$$
By balancing these forces, the total force on the fluid is:
$$\sum F={F}_{bottom}+{F}_{top}+{F}_{weight}={p}_{bottom}A-{p}_{top}A-\rho \mathrm{gA}\Delta z$$
This sum equals zero if the air's velocity is constant or zero. Dividing by A,
$$0={p}_{bottom}-{p}_{top}-\rho g\Delta z$$
or:
$${p}_{top}-{p}_{bottom}=-\rho g\Delta z$$
P_{top} − P_{bottom} is a change in pressure, and Δz is the height of the volume element – a change in the distance above the ground. By saying these changes are infinitesimally [23] small, the equation can be written in differential [24] form, where dp is top pressure minus bottom pressure just as dz is top altitude minus bottom altitude.
$$dp=-\rho gdz$$
The result is the equation:
$\frac{dp}{dz}=-\rho g$
This equation is called the Hydrostatic Equation. See the video below (1:18) for further explanation:
Using the Ideal Gas Law, we can replace ρ and get the equation for dry air:
$$\frac{dp}{dz}=-g\frac{p}{{R}_{d}T}or\frac{dp}{p}=-\frac{g}{{R}_{d}T}dz=-\frac{Mg}{{R}^{*}T}dz$$
We could integrate both sides to get the altitude dependence of p, but we can only do that if T is constant with height. It is not, but it does not vary by more than about ±20%. So, doing the integral,
$$p={p}_{o}{e}^{-\raisebox{1ex}{$z$}\!\left/ \!\raisebox{-1ex}{$H$}\right.}where{p}_{o}=surfacepressureandH=\frac{{R}^{*}\overline{T}}{{M}_{air}g}$$
H is called a scale height because when z = H, we have p = p_{o}e^{–1}. If we use an average T of 250 K, with M_{air} = 0.029 kg mol^{–1}, then H = 7.3 km. The pressure at this height is about 360 hPa, close to the 300 mb surface that you have seen on the weather maps. Of course the forces are not always in hydrostatic balance and the pressure depends on temperature, thus the pressure changes from one location to another on a constant height surface.
From equation 2.20, the atmospheric pressure falls off exponentially with height at a rate given by the scale height. Thus, for every 7 km increase in altitude, the pressure drops by about 2/3. At 40 km, the pressure is only a few tenths of a percent of the surface pressure. Similarly, the concentration of molecules is only a few tenths of a percent, and since molecules scatter sunlight, you can see in the picture below that the scattering is much greater near Earth's surface than it is high in the atmosphere.
This quiz will give you practice using the hydrostatic equation to learn interesting and useful properties and quantities of the atmosphere.
Weather involves heating and cooling, rising air parcels and falling rain, thunderstorms and snow, freezing and thawing. All of this weather occurs according to the three laws of Thermodynamics. The First Law of Thermodynamics tells us how to account for energy in any molecular system, including the atmosphere. As we will see, the concept of temperature is tightly tied to the concept of energy, namely thermal energy, but they are not the same because there are other forms of energy that can be exchanged with thermal energy, such as mechanical energy or electrical energy. Each air parcel contains molecules that have internal energy, which when thinking about the atmosphere, is just the kinetic energy of the molecules (associated with molecular rotations and, in some cases, vibrations) and the potential energy of the molecules (associated with the attractive and repulsive forces between the molecules). Internal energy does not consider their chemical bonds nor the nuclear energy of the nucleus because these do not change during collisions between air molecules. Doing work on an air parcel involves either expanding it by increasing its volume or contracting it. In the atmosphere, as in any system of molecules, energy is not created or destroyed, but instead, it is conserved. We just need to keep track of where the energy comes from and where it goes.
Let U be an air parcel’s internal energy, Q be the heating rate of that air parcel, and W be the rate that work is done on the air parcel. Then:
$$\frac{dU}{dt}=Q+W$$
The dimensions of energy are M L^{2} T^{–2} so the dimensions of this equation are M L^{2} T^{–3}.
To give more meaning to this energy budget equation, we need to relate U, Q, and W to variables that we can measure. Once we do that, we can put this equation to work. To do this, we resort to the Ideal Gas Law.
For processes like those that occur in the atmosphere, we can relate working, W, to a change in volume because work is force times distance. Imagine a cylinder with a gas in it. The cross-sectional area of the piston is A. If the piston compresses the gas by moving a distance dx, the amount of work being done by the piston on the gas is the force (pA) multiplied by the distance (dx). W is then pAdx/dt. But the volume change is simply –Adx/dt and so:
$$W=-p\frac{dV}{dt}$$
Reducing a volume of gas (dV/dt < 0) takes energy, so working on an air parcel is positive when the volume is reduced, or dV/dt < 0. Thus:
$$\frac{dU}{dt}=Q-p\frac{dV}{dt}$$
The heat capacity C is the amount of energy needed to raise the temperature of a substance by a certain amount. Thus, $C=\frac{Q}{\frac{dT}{dt}}$ and has SI units of J/K. C depends on the substance itself, the mass of the substance, and the conditions under which the energy is added. We will consider two special conditions: constant volume and constant pressure.
Consider a box with rigid walls and thus constant volume: $\frac{dV}{dT}=0$ . No work is being done and only internal energy can change due to heating.
The candle supplies energy to the box, so Q > 0 and dU/dt > 0. The internal energy can increase via increases in molecular kinetic and potential energy. However, for an ideal gas, the attractive and repulsive forces between the molecules (and hence the molecular potential energy) can be ignored. Thus, the molecular kinetic energy and, hence, the temperature, must increase:
$$\frac{dT}{dt}>0$$
So,
$${Q}_{\begin{array}{c}const\\ volume\end{array}}=\frac{dU}{dt}={C}_{V}\frac{dT}{dt}$$
C_{V}, the constant relating Q to temperature change, is called the heat capacity at constant volume. Heat capacity has units of J K^{-1}.
Remember that ${C}_{v}\frac{dT}{dt}$ is the change in the air parcel’s internal energy.
The heat capacity, C_{V}, depends on the mass and the type of material. So we can write C_{V} as:
$${C}_{V}=mass\cdot {c}_{V}$$
where c_{V} is called the specific heat capacity. The adjective “specific” means the amount of something per unit mass. The greater the heat capacity, the smaller the temperature change for a given amount of heating.
Some specific heat capacity values are included in the table below:
gas | c_{V} (@ 0^{o}C) J kg^{–1} K^{–1} |
---|---|
dry air | 718 |
water vapor | 1390 |
carbon dioxide | 820 |
Solve the following problem on your own. After arriving at your own answer, click on the link to check your work.
Consider a sealed vault with an internal volume of 10 m^{3} filled with dry air (p = 1013 hPa; T = 273 K). If the vault is being heated at a constant rate from the outside at a rate of 1 kW (1,000 J s^{–1}), how long will it take for the temperature to climb by 30 ^{o}C?
The 1^{st }Law can be rewritten as:
$$Q=\frac{dU}{dt}+p\frac{dV}{dt}={C}_{V}\frac{dT}{dt}+p\frac{dV}{dt}=mass\cdot {c}_{V}\frac{dT}{dt}+p\frac{dV}{dt}$$
However, dV/dt = 0 because the vault’s volume isn’t changing. So, we can use the equation, rearrange it and integrate it:
$$Q=mass\cdot {c}_{V}\frac{dT}{dt}$$
How do we find the mass of the air inside the vault? Use the Ideal Gas Law to find the number of moles and then multiply by the mass per mole!
$$mass={M}_{dryair}\cdot N={M}_{dryair}\cdot \frac{pV}{{R}^{*}T}=0.029\cdot \frac{1.013x{10}^{5}\cdot 10}{8.314\cdot 273}=12.9kg$$
$$\Delta t=\frac{mass\cdot {c}_{V}\cdot \Delta T}{Q}=\frac{12.9\cdot 718\cdot 30}{{10}^{3}}=278\mathrm{s}\text{\hspace{0.17em}}(~5\mathrm{min})$$
Often we do not have a well-defined volume, but instead just an air mass. We can easily measure the air mass’s pressure and temperature, but we cannot easily measure its volume. Often we can figure out the heating rate per volume (or mass) of air. Thus:
$$\frac{Q}{mass}\equiv q=\frac{mass\cdot {c}_{V}\frac{dT}{dt}}{mass}={c}_{V}\frac{dT}{dt};\text{\hspace{1em}}q={c}_{V}\frac{dT}{dt}$$
where q is the specific heating rate (SI units: J kg^{–1} s^{–1}).
The atmosphere is not a sealed box and when air is heated it can expand. We can no longer ignore the volume change. On the other hand, as the volume changes, any pressure changes are rapidly damped out, causing the pressure in an air parcel to be roughly constant even as the temperature and volume change. This constant-pressure process is called isobaric.
$$Q=\frac{dU}{dt}+p\frac{dV}{dt}$$
Now the change in the internal energy could be due to changes in temperature or changes in volume. It turns out that internal energy does not change with changes in volume. It only changes due to changes in temperature. But we already know how changes in internal energy are related to changes in temperature from the example of heating the closed box. That is, the internal energy changes are related by the heat capacity constant volume, C_{v}. Thus:
$$Q={C}_{V}\frac{dT}{dt}+p\frac{dV}{dt}$$
Note that when volume is constant, we get the expression of heating a constant volume.
Suppose we pop the lid off the box and now the air parcel is open to the rest of the atmosphere. What happens when we heat the air parcel? How much does the temperature rise?
It’s hard to say because it is possible that the air parcel’s volume can change in addition to the temperature rise. So we might suspect that, for a fixed heating rate Q, the temperature rise in the open box will be less than the temperature rise in the sealed box where the volume is constant because the volume can change as well as the temperature.
Enthalpy (H) is an energy quantity that accounts not only for internal energy but also the energy associated with working. It is a useful way to take into consideration both ways that energy can change in a collection of molecules – by internal energy changes and by volume changes that result in work being done.
$$enthalpy\equiv H=U+pV$$
Enthalpy is the total energy of the air parcel including effects of volume changes. We can do some algebra and use the Chain Rule to write the First Law of Thermodynamics in terms of the enthalpy:
$$Q=\frac{dU}{dt}+p\frac{dV}{dt}=\frac{dU}{dt}+\frac{d\left(pV\right)}{dt}-V\frac{dp}{dt}=\frac{d\left(U+pV\right)}{dt}-V\frac{dp}{dt}=\frac{dH}{dt}-V\frac{dp}{dt}$$
If the pressure is constant, which is true for many air parcel processes, then dp/dt = 0 and:
$$Q=\frac{dH}{dt}$$
In analogy with constant volume process, for a constant pressure process, we can write:
$${Q}_{\begin{array}{c}constant\\ pressure\end{array}}=\frac{dH}{dt}={C}_{p}\frac{dT}{dt}$$
where C_{p} is the heat capacity at constant pressure and c_{p} is the specific heat capacity at constant pressure.
Note that c_{p} takes into account the energy required to increase the volume as well as to increase the internal energy and thus temperature.
What is the difference between c_{p} and c_{v}? You will see the derivation of the relationship, but I will just present the results:
gas | c_{V} (@ 0^{o}C) J kg^{–1} K^{–1} | c_{p} (@ 0^{o}C) J kg^{–1} K^{–1} |
---|---|---|
dry air | 718 | 1005 |
water vapor | 1390 | 1858 |
Since c_{p} > c_{v}, the temperature change at constant pressure will be less than the temperature change at constant volume because some of the energy goes to increasing the volume as well as to increasing the temperature.
$Q={C}_{V}\frac{dT}{dt}+p\frac{dV}{dt}$
$Q={C}_{p}\frac{dT}{dt}-V\frac{dp}{dt}$
$Q=\frac{dU}{dt}+p\frac{dV}{dt}$
$Q=\frac{dH}{dt}-V\frac{dp}{dt}$
It is often useful to express these equations in terms of specific quantities, such as specific volume (α ≡ V/m = ρ^{-1}), specific heat at constant pressure (c_{p} ≡ C_{p}/m), and specific heat at constant volume (c_{V} ≡ C_{V}/m). With these definitions, the first three equations above become:
$q={c}_{V}\frac{dT}{dt}+p\frac{d\alpha}{dt}$
$q={c}_{p}\frac{dT}{dt}-\alpha \frac{dp}{dt}$
$q=\frac{du}{dt}+p\frac{d\alpha}{dt}$
You can figure out which form to use by following three steps:
Consider the atmospheric surface layer that is 100 m deep and has an average density of 1.2 kg m^{–3}. The early morning sun heats the surface, which heats the air with a heating rate of F = 50 W m^{–2}. How fast does the temperature in the layer increase? Why is this increase important?
$$Q=FA=50\text{}W\text{}{m}^{-2}A$$
$${C}_{p}={c}_{p}\rho V={c}_{p}\rho \text{}\Delta z\text{}A$$
$$Q=FA={c}_{p}\rho \text{}\Delta z\text{}A\frac{dT}{dt}$$
$$\frac{dT}{dt}=\frac{FA}{{c}_{p}\text{}\rho \text{}\Delta z\text{}A}=\frac{F}{{c}_{p}\rho \text{}\Delta z}=\frac{50}{1005\text{1.2100}}=4.2x{10}^{-4}K\text{}{s}^{-1}=1.5\text{}K\text{}h{r}^{-1}$$
This temperature increase is important because it is one of the most important factors in determining whether convection will occur later in the day. We will talk more about instability soon.
Here is a video (1:30) explanation of the above problem:
Consider the atmospheric surface layer that is 100 m deep and has an average density of 1.2 kg m^{–3}. It is night and dark and the land in contact with the air is cooling at 50 W m^{–2}. If the temperature at the start of the night was 25 ^{o}C, what is the temperature 8 hours later?
$$Q=FA=-50\text{}W\text{}{m}^{-2}A$$
$${C}_{p}={c}_{p}\rho V={c}_{p}\rho \text{}\Delta z\text{}A$$
$$Q=FA={c}_{p}\rho \text{}\Delta z\text{}A\frac{dT}{dt}$$
$$\frac{dT}{dt}=\frac{FA}{{c}_{p}\text{}\rho \text{}\Delta z\text{}A}=\frac{F}{{c}_{p}\rho \text{}\Delta z}=\frac{-50}{1005\text{1.2100}}=-4.2x{10}^{-4}K\text{}{s}^{-1}=-1.5\text{}K\text{}h{r}^{-1}$$$${\mathrm{}}^{}$$
Since the cooling continues for 8 hours, the total amount of cooling is –1.5 K/hr x 8 hr = 12 K or 12 ^{o}C. Thus, the temperature 8 hours later will be 13 ^{o}C.
This cooling near the surface creates a layer of cold air near the surface with a layer of warmer air above it. The layering of warm air over colder air creates an temperature inversion, which suppresses convection and lock pollutants into the air layer near Earth's surface.
A surprising way to relate the distance between two pressure surfaces to the temperature of the layer between them.
Consider a column of air between two pressure surfaces. If the mass in the column is conserved, then the column with the greater average temperature will be less dense and occupy more volume and thus be higher in altitude. But the pressure at the top of the column is related to the weight of the air above the column, which is constant, and so the upper pressure surface is higher in altitude. If the temperature of the column is lower, then the pressure surface at the top of the column will be lower in altitude.
We can look at this behavior from the point-of-view of hydrostatic equilibrium.
$\frac{\text{d}p}{dz}=-\rho g=-\frac{pg}{{R}_{d}T}$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaalaaapaqaa8qacaqGKbGaamiCaaWdaeaapeGaamizaiaadQhaaaGaeyypa0JaeyOeI0IaeqyWdiNaam4zaiabg2da9iabgkHiTmaalaaapaqaa8qacaWGWbGaam4zaaWdaeaapeGaamOua8aadaWgaaWcbaWdbiaadsgaa8aabeaak8qacaWGubaaaaaa@42C0@
If the temperature is greater, then the change in p with height is less, which means that any given pressure surface is going to be higher.
The difference between any two pressure surfaces is called the thickness.
We can show that the thickness depends only on temperature:
$dz=-\frac{dp}{p}\frac{{R}_{d}}{g}T$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaadsgacaWG6bGaeyypa0JaaiiOamaalaaapaqaa8qacaWGKbGaamiCaaWdaeaapeGaamiCaaaadaWcaaWdaeaapeGaamOua8aadaWgaaWcbaWdbiaadsgaa8aabeaaaOqaa8qacaWGNbaaaiaadsfaaaa@3E3B@
Integrate both sides:
$\underset{{z}_{1}}{\overset{{z}_{2}}{{\displaystyle \int}}}dz=-\underset{{p}_{1}}{\overset{{p}_{2}}{{\displaystyle \int}}}\frac{dp}{p}\frac{{R}_{d}}{g}T$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaawahabeWcpaqaa8qacaWG6bWdamaaBaaameaapeGaaGymaaWdaeqaaaWcbaWdbiaadQhapaWaaSbaaWqaa8qacaaIYaaapaqabaaaneaapeGaey4kIipaaOGaamizaiaadQhacqGH9aqpcaGGGcWaaybCaeqal8aabaWdbiaadchapaWaaSbaaWqaa8qacaaIXaaapaqabaaaleaapeGaamiCa8aadaWgaaadbaWdbiaaikdaa8aabeaaa0qaa8qacqGHRiI8aaGcdaWcaaWdaeaapeGaamizaiaadchaa8aabaWdbiaadchaaaWaaSaaa8aabaWdbiaadkfapaWaaSbaaSqaa8qacaWGKbaapaqabaaakeaapeGaam4zaaaacaWGubaaaa@4C07@
or
${z}_{2}-{z}_{1}=\frac{{R}_{d}}{g}\text{ln}\left(\frac{{p}_{1}}{{p}_{2}}\right)\overline{T}$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaadQhapaWaaSbaaSqaa8qacaaIYaaapaqabaGcpeGaeyOeI0IaamOEa8aadaWgaaWcbaWdbiaaigdaa8aabeaak8qacqGH9aqpcaGGGcWaaSaaa8aabaWdbiaadkfapaWaaSbaaSqaa8qacaWGKbaapaqabaaakeaapeGaam4zaaaacaqGSbGaaeOBamaabmaapaqaa8qadaWcaaWdaeaapeGaamiCa8aadaWgaaWcbaWdbiaaigdaa8aabeaaaOqaa8qacaWGWbWdamaaBaaaleaapeGaaGOmaaWdaeqaaaaaaOWdbiaawIcacaGLPaaaceWGubWdayaaraaaaa@46A3@
where T is the average temperature of the layer between p_{1} and p_{2}. So, the thickness is actually a measure of the average temperature in the layer.
As some of you already know, you can use the thickness between different pressure surfaces to estimate the type of precipitation that will fall - snow, rain, or a mixture. You can check out these resources for some more information and example problems:
Suppose that the 500 mb surface is at 560 dam (decameters, 10s of meters) and the 1000 mb surface is at 0 dam. What is the average temperature of the layer between 1000 mb and 500 mb?
Rearrange equation 2.4 to get an expression in terms of the average temperature and then put all the numbers into the equation to find the average temperature of the layer. Make sure that all the units are correct.
$$\overline{\text{T}}=\frac{10*\left({z}_{2}-{z}_{1}\right)}{\left(\raisebox{1ex}{$287$}\!\left/ \!\raisebox{-1ex}{$9.8$}\right.\right)\mathrm{ln}\left(\raisebox{1ex}{${p}_{1}$}\!\left/ \!\raisebox{-1ex}{${p}_{2}$}\right.\right)}=\frac{10*\left(560-0\right)}{\left(\raisebox{1ex}{$287$}\!\left/ \!\raisebox{-1ex}{$9.8$}\right.\right)\mathrm{ln}\left(\raisebox{1ex}{$1000$}\!\left/ \!\raisebox{-1ex}{$500$}\right.\right)}=276\text{\hspace{0.17em}}K$$So far, we have covered constant volume (isochoric) and constant pressure (isobaric) processes. There is a third process that is very important in the atmosphere—the adiabatic process. Adiabatic means no energy exchange between the air parcel and its environment: Q = 0. Note: adiabatic is not the same as isothermal.
Consider the Ideal Gas Law:
$pV=n{R}^{*}T$
If an air parcel rises, the pressure changes, but how does the temperature change? Note that the volume can change as well as the pressure and temperature, and thus, if we specify a pressure change, we cannot find the temperature change unless we know how the volume changed. Without some other equation, we cannot say how much the temperature will rise for a pressure change.
However, we can use the First Law of Thermodynamics to relate changes in temperature to changes in pressure and volume for adiabatic processes.
I do not expect you to be able to do this derivation, but you should go through it to make sure that you understand all the steps as a way to continue to improve your math skills. Start with the following specific form of the 1^{st} Law for dry air (Equation 2.41) and then assume an adiabatic process (q = 0):
$q={c}_{p}\frac{dT}{dt}-\alpha \frac{dp}{dt}$
$q=0={c}_{p}\frac{dT}{dt}-\alpha \frac{dp}{dt}$
Divide both sides by T:
$\frac{{c}_{p}}{T}\frac{dT}{dt}-\frac{{R}_{d}T}{pT}\frac{dp}{dt}=0=\frac{{c}_{p}}{T}\frac{dT}{dt}-\frac{{R}_{d}}{p}\frac{dp}{dt}={c}_{p}\frac{dln\left(T\right)}{dt}-{R}_{d}\frac{dln\left(p\right)}{dt}$
where we figured out the two terms were just derivatives of the natural log of T and p.
$\frac{d}{dt}({c}_{p}ln\left(T\right)-{R}_{d}ln\left(p\right))=0$
But d/dt = 0 just means that the value is constant:
$({c}_{p}ln\left(T\right)-{R}_{d}ln\left(p\right))=constant$
Divide by c_{p}:
$ln\left(T\right)-\frac{{R}_{d}}{{c}_{p}}ln\left(p\right)=ln\left(T\right)+ln\left({p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}\right)=ln\left(T{p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}\right)=constant$
If the natural log of a variable is constant then the variable itself must be constant:
$T{p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}=constant$
We can rewrite R_{d}/c_{p} as a new term denoted by the Greek letter gamma, γ
$\gamma \equiv \frac{{c}_{p}}{{c}_{v}}=\frac{{c}_{v}+{R}_{d}}{{c}_{v}}=1.4$
$\frac{{R}_{d}}{{c}_{p}}=\frac{\gamma -1}{\gamma}=0.286$
We can use the Ideal Gas Law to get relations among p, V, and T, called the Poisson’s Relations:
$T{p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}=T{p}^{\left(\raisebox{1ex}{$\left(1-\gamma \right)$}\!\left/ \!\raisebox{-1ex}{$\gamma $}\right.\right)}=constant$
${\alpha}^{\gamma}p=constant$
$T{\alpha}^{\gamma -1}=constant$
The Poisson Relation that we use the most is the relation of pressure and temperature because these are two variables that we can measure easily without having to define a volume of air:
$\frac{T}{\theta}={\left(\frac{p}{{p}_{o}}\right)}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}$
or
$\theta =T{\left(\frac{{p}_{o}}{p}\right)}^{\raisebox{1ex}{${R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.}=T{\left(\frac{1000}{p}\right)}^{0.286}$
We call θ the potential temperature, which is the temperature that an air parcel would have if the air is brought to a pressure of p_{o} = 1000 hPa. Potential temperature is one of the most important thermodynamic quantities in meteorology.
Adiabatic processes are common in the atmosphere, especially the dry atmosphere. Also, adiabatic processes are often the same as isentropic processes (no change in entropy).
Air coming over the Laurel Highlands descends from about 700 m (p ~ 932 hPa) to 300 m (p ~ 977 hPa) at State College. Assume that the temperature in the Laurel Highlands is 20 ^{o}C. What is the temperature in State College?
We can find the temperature in State College due only to adiabatic changes by the following equation:
$$\begin{array}{l}{T}_{7}{}_{00}{p}_{7}{{}_{00}}^{-0.286}={T}_{3}{}_{00}{p}_{3}{{}_{00}}^{-0.286}\\ {T}_{300=}{T}_{700}{\left(\frac{{p}_{300}}{{p}_{700}}\right)}^{0.286}=\left(273+20\right){\left(\frac{977}{932}\right)}^{0.286}=297K=24C\end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@745A@
This temperature change is 4 ^{o}C, or 7 ^{o}F just from adiabatic compression.
We can plot adiabatic (isentropic) surfaces in the atmosphere. An air parcel needs no energy to move along an adiabatic surface. Also, it takes energy for an air parcel to move from potential surface to another potential energy surface.
Suppose an air parcel has p = 300 hPa and T = 230 K. How much heating per unit volume of dry air would be needed to increase the potential temperature by 10 K?
The heating raises the temperature, and the amount of heating required depends on the heat capacity, constant pressure, which depends on the mass of air, or the density times the volume. Let's do the calculation for a volume of air; that way we can use the density.
First we need to find the temperature rise that is the same as a potential temperature rise of 10 K at a pressure of 300 hPa.
$d\theta =dT{\left(\frac{1000}{p}\right)}^{0.286}\text{}dT=10{\left(\frac{300}{1000}\right)}^{0.286}=7.1K$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@5831@
Then we need to find the density so that we can calculate the heat capacity:
$$\rho =\frac{p}{{R}_{d}T}=\frac{{\mathrm{3x10}}^{4}}{287\cdot 230}=0.45\text{\hspace{0.17em}}kg\text{\hspace{0.17em}}{m}^{-3}$$
Now we can put it all together:
$$\frac{Q\Delta T}{V}=\frac{\rho \cdot V\cdot {c}_{p}\cdot \Delta T}{V}\Rightarrow \frac{Q\Delta T}{V}=\rho \cdot {c}_{p}\cdot \Delta T=0.45\cdot 1005\cdot 7.1=3.2x{10}^{3}\text{\hspace{0.17em}}J\text{\hspace{0.17em}}{m}^{-3}$$
Note that we were asked to provide the total heating per unit volume, which is just the heating rate times time divided by the unit volume. So the quantity on the left is what we want. Is this heating large? Yes! So it takes a lot of heating or cooling the raise or lower an air parcel potential temperature just 10 K.
The temperature change with change in pressure (and thus change in altitude) is a major reason for weather. For dry air, the main effect is buoyancy. So because the pressure change generally follows the hydrostatic equation, the change in height translates into a change in pressure which translates into a change in temperature due to adiabatic expansion. Note that as the air parcel rises, its pressure quickly adjusts to the pressure of the surrounding air. Thus we can determine the dry adiabatic lapse rate by starting with the Poisson relation between pressure and temperature:
$$T{p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}=constant$$
Take the derivative w.r.t. z:
$$\frac{dT}{dz}{p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)}+T\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right){p}^{\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right)-1}\frac{dp}{dz}=\frac{dT}{dz}+T\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right){p}^{-1}\frac{dp}{dz}=0$$
But we also know from the hydrostatic equation that:
$$\frac{dp}{dz}=-\rho g$$
Substituting –ρg for dp/dz into the equation and rearranging the terms:
$$\frac{dT}{dz}=T\left(\raisebox{1ex}{$-{R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right){p}^{-1}\rho g=-T\left(\raisebox{1ex}{${R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.\right){p}^{-1}g\left(\frac{p}{{R}_{d}T}\right)$$
Γ_{d} is called the dry adiabatic lapse rate. Note that the temperature decreases with height, but the dry adiabatic lapse rate is defined as being positive.
Air coming over the Laurel Highlands descends from about 700 m to 300 m at State College. Assume that the temperature in the Laurel Highlands is 20 ^{o}C. What is the temperature in State College?
We can find the temperature in State College due only to adiabatic changes by using the dry adiabatic lapse rate multiplied by the height change:
$$dT=-{\Gamma}_{d}(300-700)=9.8Kk{m}^{-1}\cdot 0.4={4}^{\circ}C$$
This temperature change is 4.0 ^{o}C, or 7 ^{o}F just from adiabatic compression. This answer is very similar to the answer we obtained using the change in potential temperature.
This quiz is practice calculating energy balance using the First Law of Thermodynamics. Some of the problems will involve potential temperature.
We know that an air parcel will rise relative to the surrounding air at the same pressure if the air parcel’s density is less than that of the surrounding air. The difference in density can be calculated using the virtual temperature, which takes into account the differences in specific humidity in the air parcel and the surrounding air as well as the temperature differences.
In equilibrium, the sum of forces are in balance and the air parcel will not move. The question is, what happens to the parcel if there is a slight perturbation in its vertical position?
For the figure on the left, if the ball is displaced a tiny bit to the left or the right, it will be pulled by gravity and will continue to roll down the slope. That position is unstable. For the figure on the right, if the ball is displaced a little bit, it will be higher than the central position and gravity will pull it back down. It may rock back and forth a little bit, but eventually it will settle down into its original position.
To assess instability of air parcels in the atmosphere, we need to find out if moving the air parcel a small amount up or down causes the parcel to continue to rise or to fall (instability) or if the air parcel returns to its original position (stability).
Now look at some atmospheric temperature profiles. Important: a dry air parcel that is pushed from its equilibrium position always moves along the dry adiabatic lapse rate (DALR) line.
Note that we can also show that if the air parcel is pushed down, it will keep going if the atmospheric (environmental) profile looks like the one on the left and will return to the original position if it looks like the one on the right.
Use the image above to determine the following:
Is the air parcel stable or unstable at each of the points, 1-5?
The red line is the atmospheric temperature profile; the dashed lines are the dry adiabatic lapse rate lines (-9.8 K/km). Consider points 1-5. When an air parcel is pushed up the DALR and its temperature is greater than the atmospheric temperature at the new level, it is warmer and thus less dense. It will continue to rise. When an air parcel is pushed down the DALR and its temperature is less than the atmospheric temperature at that new level, it is colder and thus more dense. It will continue to fall. Both cases are unstable. However, if when an air parcel is pushed up the DALR and its temperature is less than the atmospheric temperature at that new level, it is colder and thus more dense. It will sink back down to its original position and is stable.
Using this thinking, air parcels at points 1, 2, and 5 are stable and at points 3 and 4 are unstable.
We can calculate the acceleration an unstable air parcel will have and, from this, can determine the parcel’s velocity at some later point in time. This acceleration is called buoyancy (B).
$\sum F={F}_{bottom}+{F}_{top}+{F}_{weight}={p}_{bottom}A-{p}_{top}A-{\rho}^{\prime}gAh$
Let’s look at the forces on an air parcel again, like we did to derive the hydrostatic equilibrium. But this time, let’s assume that the parcel has a different density than the surrounding air. We will designate quantities associated with the air parcel with an apostrophe (’); environmental parameters will have no superscript.
If the forces are not in balance, then we need to keep the acceleration that we set to zero in the hydrostatic equilibrium case. We can also divide both sides of the equation by the mass of the air parcel:
$\frac{\sum F}{{\rho}^{\prime}V}=\frac{-\frac{{p}_{top}-{p}_{bottom}}{\Delta z}-{\rho}^{\prime}g}{{\rho}^{\prime}}\to \frac{-\left(-\rho g\right)-{\rho}^{\prime}g}{{\rho}^{\prime}}$
$a\equiv B=\frac{\left(\rho -{\rho}^{\prime}\right)g}{{\rho}^{\prime}}$
where we have used the hydrostatic equilibrium of the environment to replace the expression for the pressure change as a function of height with the density times the acceleration due to gravity.
We can then use the Ideal Gas Law to replace densities with virtual temperatures because the pressure of the parcel and its surrounding air is the same:
$B=\frac{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${T}_{v}$}\right.-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${T}_{v}^{\text{'}}$}\right.\right)g}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${T}_{v}^{\text{'}}$}\right.}=\frac{\left({T}_{v}^{\text{'}}-{T}_{v}\right)}{{T}_{v}}g$
If B > 0, then the parcel accelerates upwards; if B < 0, then the parcel accelerates downwards.
We look at the instability at each point in the environmental temperature profile and can determine Γ_{env} for each point.
Thus,
${T}_{env}={T}_{0}+{\frac{\partial T}{\partial z}|}_{env}\Delta z={T}_{0}-{\Gamma}_{env}\Delta z;{T}_{parcel}={T}_{0}+{\frac{\partial T}{\partial z}|}_{parcel}\Delta z={T}_{0}-{\Gamma}_{d}\Delta z;$
so that:
$B=\frac{\left({T}_{0}-{\Gamma}_{d}\Delta z-{T}_{0}+{\Gamma}_{env}\Delta z\right)}{{T}_{v}}g=\frac{\left({\Gamma}_{env}-{\Gamma}_{d}\right)}{{T}_{v}}g\Delta z$
If Γ_{env} < Γ_{d}, the parcel accelerates downward for positive Δz (statically stable environment).
If Γ_{env} > Γ_{d}, the parcel accelerates upward for positive Δz (statically unstable environment).
We can put this idea of buoyancy in terms of potential temperature.
$\theta =T{\left(\frac{{p}_{o}}{p}\right)}^{\raisebox{1ex}{${R}_{d}$}\!\left/ \!\raisebox{-1ex}{${c}_{p}$}\right.}$
We want to find dθ/dz. Taking the log of both sides of the equation and replacing a dp/dz term with –gρ, we are able to find the following expression for buoyancy in terms of potential temperature:
$B=-g\Delta z\frac{1}{\theta}\frac{d\theta}{dz}$
Remember that no matter what the environmental temperature or potential temperature profiles, a change in height of an air parcel will result in a temperature that changes along the dry adiabat and a potential temperature that does not change at all. As you can see below, the stability of a layer depends on the change in environmental potential temperature with height. Air parcels try to move vertically with constant potential temperature.
Parcels will move to an altitude (and air density) for which B = 0. However, if they still have a velocity when they reach that altitude, they will overshoot, experience a negative acceleration, and then descend, overshooting the neutral level again. In this way, the air parcel will oscillate until its oscillation is finally damped out by friction and dissipation of the air parcel. Note that in the neutral section of vertical profile where potential temperature does not change, it is not possible to determine if an air parcel will be stable or unstable. For instance, if the air parcel in the neutral region is given a small upward push, it will continue to rise until it reaches a stable region.
This quiz provides practice determining stability or instability of an air parcel and in calculating the buoyancy of air parcels.
(3 discussion points)
This week's discussion topic is a hypothetical question involving stability. The troposphere always has a capping temperature inversion–it's called the stratosphere. The tropopause is about 16 km high in the tropics and lowers to about 10 km at high latitudes. The stratosphere exists because solar ultraviolet light makes ozone and then a few percent of the solar radiation is absorbed by stratospheric ozone, heating the air and causing the inversion. Suppose that there was no ozone layer and hence no stratosphere caused by solar UV heating of ozone.
Would storms in the troposphere be different if there were no stratosphere to act like a capping inversion? And if so, how?
Use what you have learned in this lesson about the atmosphere's pressure structure and stability to help you to think about this problem and to formulate your answer and discussions. It's okay to be wrong, as long as you have some solid reasoning to back up your ideas. My goal is to get you all to communicate with each other and think hard about atmospheric science.
This discussion will be worth 3 discussion points. I will use the following rubric to grade your participation:
Evaluation | Explanation | Available Points |
---|---|---|
Not Completed | Student did not complete the assignment by the due date. | 0 |
Student completed the activity with adequate thoroughness. | Student answers the discussion question in a thoughtful manner, including some integration of course material. | 1 |
Student completed the activity with additional attention to defending their position. | Student thoroughly answers the discussion question and backs up reasoning with references to course content as well as outside sources. | 2 |
Student completed a well-defended presentation of their position, and provided thoughtful analysis of at least one other student’s post. | In addition to a well-crafted and defended post, the student has also engaged in thoughtful analysis/commentary on at least one other student’s post as well. | 3 |
We started with the Ideal Gas Law and Dalton’s Law to develop an understanding of the atmosphere’s behavior and atmospheric density, a fundamental driving force for vertical motion in the atmosphere. We saw that composition matters, and developed a new quantity, virtual temperature, that allows us to compare the densities of different moist air parcels. Using these laws along with Newton’s laws, we were able to derive a fundamental property of the atmosphere: the hydrostatic equation, which states that the change in pressure with altitude is proportional to the negative of the density times gravity. We applied a new constraint – the First Law of Thermodynamics, which states that energy is conserved and saw that combining it with the gas laws enabled us to calculate temperature changes and to derive an important atmospheric quantity – the potential temperature. With these concepts, we were able to determine the stability (and instability) of an air parcel. We could also determine the buoyancy of an air parcel, which allows us to calculate the acceleration and thus the velocity of an air parcel after it has accelerated for a while.
You have reached the end of Lesson 2! Be sure to complete the Activities within three days of the end of the lesson. I strongly encourage you to do them as you are going through the lesson.
The atmosphere’s most abundant chemicals are molecular nitrogen (N_{2}), molecular oxygen (O_{2}), and Argon (Ar). These are all only in the gas phase. Water vapor, the next most abundant, can exist as vapor, liquid, or solid. The phase changes of water have a major role in weather and in climate. In the atmosphere, water is always trying to achieve a balance between evaporation and condensation while never really succeeding. In this lesson, you will discover the conditions under which the phases of water are in balance and will see that they depend on only two quantities—the amount of water and the temperature. Equilibrium conditions, often called saturation, are expressed mathematically by the Clausius–Clapeyron Equation. We will see that phase changes of water create weather, including severe weather, and that we can use the 1^{st} Law of Thermodynamics to do many calculations involving situations where there are phase and temperature changes. Combining the Clausius–Clapeyron equation with the equations of thermodynamics, we can construct a diagram called the skew-T. The skew-T is useful in helping us understand both the atmosphere’s temperature structure and the location and behavior of clouds.
By the end of this lesson, you should be able to:
If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.
Up to now, we have dealt with water vapor only as the specific humidity in order to determine the virtual temperature. But there are many ways for us to quantify the amount of water vapor in the atmosphere. The most common are specific humidity, water vapor mixing ratio, relative humidity, and dewpoint temperature.
Specific humidity (q) is the density of water vapor (mass per unit volume) divided by the density of all air, including the water vapor:
$$q=\frac{{m}_{water\_vapor}}{{m}_{all\_air}}=\frac{{\rho}_{water\_vapor}}{{\rho}_{all\_air}}=\frac{{\rho}_{v}}{{\rho}_{}}$$
We have already seen that specific humidity is used to calculate virtual temperature. Specific humidity is unitless, but often we put it in g kg^{–1}.
Water vapor mixing ratio (w) is the density of water vapor divided by the density of dry air without the water vapor:
$$w=\frac{{m}_{water\_vapor}}{{m}_{dry\_air}}=\frac{{\rho}_{water\_vapor}}{{\rho}_{dry\_air}}=\frac{{\rho}_{v}}{{\rho}_{d}}$$
Water vapor mixing ratio is widely used to calculate the amount of water vapor. It is also the quantity used on the skew-T diagram, which we will discuss later in this lesson. Water vapor mixing ratio is unitless, but often we put it in g kg^{–1}.
Since ρ_{d} = ρ – ρ_{v} we can rearrange the equations to get the relationship between w (water vapor mixing ratio) and q (specific humidity):
$$q=\frac{w}{1+w}\text{\hspace{1em}}\text{\hspace{1em}}w=\frac{q}{1-q}$$
The water vapor mixing ratio, w, is typically at most about 40 g kg^{–1} or 0.04 kg kg^{–1}, so even for this much water vapor, q = 0.040/(1 + 0.040) = 0.038 or 38 g kg^{–1}.
Thus, water vapor mixing ratio and specific humidity are the same to within a few percent. But specific humidity is less than the water vapor mixing ratio if the humidity is more than zero.
Here is one example of global specific humidity.
The greatest absolute specific humidity is in the tropics with maximum values approaching 30 g kg^{–1}. The smallest values are at the high latitudes and are close to zero. Why is specific humidity distributed over the globe in this way?
Relative humidity (RH) is another measure of water vapor in the atmosphere, although we must be careful when using it because a low relative humidity may not mean a low water vapor mixing ratio (i.e., at high temperatures) and a high relative humidity might still be quite dry air (i.e., at low temperatures).
According to the World Meteorological Organization (WMO) definition,
$$RH=\frac{w}{{w}_{s}}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamOuaiaadIeacqGH9aqpdaWcaaWdaeaapeGaam4DaaWdaeaapeGaam4Da8aadaWgaaWcbaWdbiaadohaa8aabeaaaaaaaa@3CAC@
where w_{s} is the saturation mixing ratio (the mixing ratio at which RH = 100%). w and w_{s} can both have units of g kg^{–1} or kg kg^{–1}, as long as they are consistent. Relative humidty is usually expressed as a percent. Thus, when w = w_{s}, RH = 1 = 100%. In most problems involving RH, it is important to keep in mind conversions between decimal fractions and percent.
A more physically based definition of the relative amount of moisture in the air is the saturation ratio, S:
$$S=\frac{e}{{e}_{s}}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaam4uaiabg2da9maalaaapaqaa8qacaWGLbaapaqaa8qacaWGLbWdamaaBaaaleaapeGaam4CaaWdaeqaaaaaaaa@3BBC@
where e is the vapor pressure and e_{s} is the saturation vapor pressure. The saturation ratio is used extensively in cloud physics (Lesson 5). To see how RH and S are related, start with the Ideal Gas Law and then do some algebra:
$$\begin{array}{l}e={\rho}_{v}{R}_{v}T,\text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{0.17em}}{p}_{d}={\rho}_{d}{R}_{d}T,\text{\hspace{1em}}\text{\hspace{1em}}\epsilon \equiv \raisebox{1ex}{${R}_{d}$}\!\left/ \!\raisebox{-1ex}{${R}_{v}$}\right.\hfill \\ \text{so}\hfill \\ w=\frac{\epsilon e}{{p}_{d}}=\frac{\epsilon e}{p-e},\text{\hspace{1em}}\text{}\frac{w}{{w}_{s}}=\frac{e}{{e}_{s}}\left(\frac{p-{e}_{s}}{p-e}\right),\text{}RH=S\left(\frac{p-{e}_{s}}{p-e}\right)\hfill \end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@8C59@
where ε = 0.622 is just the molar mass of water (18.02 kg mol^{–1}) divided by the mass of dry air (28.97 kg mol^{–1}). e and e_{s} are typically less than 7% of p, and since e is usually 20%–80% of e_{s}, the difference between the two definitions is usually less than a few percent.
Note that at saturation, you can replace w with w_{s} and e with e_{s} in the equation that relates w to e.
Some processes depend upon the absolute amount of water vapor, which is given by the specific humidity, water vapor mixing ratio, and water vapor pressure, and other processes depend on the relative humidity. For example, the density of a moist air parcel depends on the absolute amount of water vapor. So does the absorption and emission of infrared atmospheric radiation. On the other hand, cloud formation depends on the relative humidity, although the cloud might be kind of wimpy if the absolute humidity is small.
One of the most common indicators of absolute humidity is the dewpoint temperature. We will postpone the discussion of it until after we learn about the relationship between temperature and saturation vapor pressure, e_{s}.
If the density of water vapor is 10.0 g m^{–3} and the density of dry air is 1.10 kg m^{–3}, what is the water vapor mixing ratio and what is the specific humidity?
$$w=\frac{10.0{\text{gm}}^{-3}}{1.10{\text{kgm}}^{-3}}=9.09\text{\hspace{0.17em}}\frac{\text{g}}{\text{kg}};\text{\hspace{1em}}\text{\hspace{1em}}q=\frac{10.0{\text{gm}}^{-3}}{1.10{\text{kgm}}^{-3}+(10.0{\text{gm}}^{-3}/1000{\text{gkg}}^{-1})}=9.01\text{\hspace{0.17em}}\frac{\text{g}}{\text{kg}}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@8503@
If the water vapor mixing ratio is 21 g kg^{–1} and the relative humidity is 84%, what is the saturation water vapor mixing ratio?
$$RH=\frac{w}{{w}_{s}}\text{\hspace{1em}}\to \text{\hspace{1em}}{w}_{s}=\frac{w}{RH}=\frac{21{\text{gkg}}^{-1}}{0.84}=25\text{\hspace{0.17em}}{\text{gkg}}^{-1}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@5DB2@
When you feel you are ready, take Quiz 3-1. This quiz can be found in Canvas. You will be allowed to take this quiz only once. Good luck!
What is vapor pressure? Because of the Ideal Gas Law (Equation 2.1), we can think of vapor pressure e (SI units = hPa or Pa) as being related to the concentration of water vapor molecules in the atmosphere,
$$eV=NR*T\text{\hspace{1em}}\text{and}\text{\hspace{1em}}e=nR*T$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamyzaiaadAfacqGH9aqpcaWGobGaamOuaiaacQcacaWGubGaaGzbVlaabggacaqGUbGaaeizaiaaywW7caWGLbGaeyypa0JaamOBaiaadkfacaGGQaGaamivaaaa@477F@
where n is the number of moles per unit volume (n = N/V).
What makes liquid water different from ice or water vapor? It is actually the weak bonds between water molecules that are called hydrogen bonds. These bonds are 20 times weaker than the bonds between hydrogen and oxygen in the same molecule and can be broken by collisions with other molecules if they are traveling fast enough and have enough kinetic energy to break the bonds. So the differences between vapor, liquid, and ice are related to the number of hydrogen bonds. In vapor, there are essentially no hydrogen bonds between molecules. In ice, each water molecule is hydrogen bonded to four other water molecules. And in liquid, only some of those hydrogen bonds are made and they are constantly changing as the water molecules and clusters of water molecules bump into and slide past each other.
Think about a liquid water surface on a molecular scale. What is happening all the time is that some water molecules in the gas phase are hitting the surface and sticking (i.e., making hydrogen bonds), while at the same time other water molecules are breaking free from the hydrogen bonds that tie them to other molecules in the liquid and are becoming water vapor. The water vapor surface is like a Starbucks, but even busier. We can easily calculate the flux of molecules that are hitting the surface using simple physical principles, although it is harder to calculate the number that are leaving the liquid. Both are happening all the time, although usually the amount of condensation and evaporation aren't the same, so that we usually have net evaporation or net condensation.
In equilibrium, the flux of molecules leaving the surface exactly balances the flux of molecules that are hitting the surface. This condition is called equilibrium, or saturation. We can show that:
$$\frac{\text{condensation}}{\text{evaporation}}=\frac{e}{{e}_{s}}=S\cong RH=\frac{w}{{w}_{s}}\text{\hspace{1em}}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeWaaSaaa8aabaWdbiaabogacaqGVbGaaeOBaiaabsgacaqGLbGaaeOBaiaabohacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaaWdaeaapeGaaeyzaiaabAhacaqGHbGaaeiCaiaab+gacaqGYbGaaeyyaiaabshacaqGPbGaae4Baiaab6gaaaGaeyypa0ZaaSaaa8aabaWdbiaadwgaa8aabaWdbiaadwgapaWaaSbaaSqaa8qacaWGZbaapaqabaaaaOWdbiabg2da9iaadofacqGHfjcqcaWGsbGaamisaiabg2da9maalaaapaqaa8qacaWG3baapaqaa8qacaWG3bWdamaaBaaaleaapeGaam4CaaWdaeqaaaaak8qacaaMf8oaaa@5BB9@
Thus, when S = 1, e = e_{s}, RH is approximately 100%, and w is approximately w_{s}. Condensation and evaporation are in balance. These two processes are going on all the time, but sometimes there can be more evaporation than condensation, or more condensation than evaporation, or evaporation equaling condensation. However, water is always trying to come into equilibrium.
So we know that the amount of water in vapor phase determines the condensation rate and thus e. So what determines e_{s}? We will see next that e_{s} depends on only one variable: temperature!
We can derive the equation for e_{s} using two concepts you may have heard of and will learn about later: entropy and Gibbs free energy, which we will not go into here. Instead, we will quote the result, which is called the Clausius–Clapeyron Equation,
$$\frac{1}{{e}_{s}}\frac{d{e}_{s}}{dT}=\frac{{l}_{v}}{{R}_{v}{T}^{2}}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaalaaabaGaaGymaaqaaiaadwgadaWgaaWcbaGaam4CaaqabaaaaOWaaSaaaeaacaWGKbGaamyzamaaBaaaleaacaWGZbaabeaaaOqaaiaadsgacaWGubaaaiabg2da9maalaaabaGaamiBamaaBaaaleaacaWG2baabeaaaOqaaiaadkfadaWgaaWcbaGaamODaaqabaGccaWGubWaaWbaaSqabeaacaaIYaaaaaaaaaa@4182@
where l_{v} is the enthalpy of vaporization (often called the latent heat of vaporization, about 2.5 x 10^{6} J kg^{–1}), R_{v} is the gas constant for water vapor (461.5 J kg^{–1} K^{–1}), and T is the absolute temperature. The enthalpy of vaporization (i.e., latent heat of vaporization) is just the amount of energy required to evaporate a certain mass of liquid water.
What is the physical meaning? The right-hand side of [3.9] is always positive, which means that the saturation vapor pressure always increases with temperature (i.e., de_{s}/dT > 0). This positive slope makes sense because we know that as water temperature goes up, evaporation is faster (because water molecules have more energy and thus a greater chance to break the bonds that hold them to other water molecules in a liquid or in ice). At saturation, condensation equals evaporation, and since evaporation is greater, condensation must be greater as well. Much of the higher condensation comes from having more water vapor molecules hitting the liquid surface, which according to the Ideal Gas Law, means that the water vapor pressure is higher.
The temperature sensitivity of e_{s} is quite high. Plugging in the appropriate values to the right side of [3.9] yields 0.07 K^{–1}, which means that the saturation vapor pressure increases by 7% for every 1 K increase in temperature. This high sensitivity has profound implications for weather and climate.
Separating variables (e_{s} and T) in [3.9] and integrating, assuming that l_{v} is a constant with temperature (it is not quite constant!), yields:
$${e}_{s}={e}_{so}\text{\hspace{0.17em}}\mathrm{exp}\left(\frac{{l}_{v}}{{R}_{v}{T}_{o}}\right)\mathrm{exp}\left(\frac{-{l}_{v}}{{R}_{v}T}\right)$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadwgadaWgaaWcbaGaam4CaaqabaGccqGH9aqpcaWGLbWaaSbaaSqaaiaadohacaWGVbaabeaakiaaykW7ciGGLbGaaiiEaiaacchadaqadaqaamaalaaabaGaamiBamaaBaaaleaacaWG2baabeaaaOqaaiaadkfadaWgaaWcbaGaamODaaqabaGccaWGubWaaSbaaSqaaiaad+gaaeqaaaaaaOGaayjkaiaawMcaaiGacwgacaGG4bGaaiiCamaabmaabaWaaSaaaeaacqGHsislcaWGSbWaaSbaaSqaaiaadAhaaeqaaaGcbaGaamOuamaaBaaaleaacaWG2baabeaakiaadsfaaaaacaGLOaGaayzkaaaaaa@4F84@
Generally T_{o} is taken to be 273 K and e_{so} is then 6.11 hPa.
What does the plot of this equation look like?
What happens between vapor and ice? The same methods can be applied and the same basic equations are obtained, except with a different constant:
$${e}_{si}={e}_{so}\text{\hspace{0.17em}}\mathrm{exp}\left(\frac{{l}_{s}}{{R}_{v}{T}_{o}}\right)\mathrm{exp}\left(\frac{-{l}_{s}}{{R}_{v}T}\right)$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadwgadaWgaaWcbaGaam4CaiaadMgaaeqaaOGaeyypa0JaamyzamaaBaaaleaacaWGZbGaam4BaaqabaGccaaMc8UaciyzaiaacIhacaGGWbWaaeWaaeaadaWcaaqaaiaadYgadaWgaaWcbaGaam4CaaqabaaakeaacaWGsbWaaSbaaSqaaiaadAhaaeqaaOGaamivamaaBaaaleaacaWGVbaabeaaaaaakiaawIcacaGLPaaaciGGLbGaaiiEaiaacchadaqadaqaamaalaaabaGaeyOeI0IaamiBamaaBaaaleaacaWGZbaabeaaaOqaaiaadkfadaWgaaWcbaGaamODaaqabaGccaWGubaaaaGaayjkaiaawMcaaaaa@506C@
where e_{si} is the saturation vapor pressure for the ice vapor equilibrium and l_{s} is the enthalpy of sublimation (direct exchange between solid water and vapor = 2.834 x 10^{6} J kg^{–1}).
Equations for e_{s} and e_{si} that account for variations with temperature of l_{v} and l_{s}, respectively, can be found in Bohren and Albrecht (Atmospheric Thermodynamics, Oxford University Press, New York, 1998, ISBN 0-19-509904-4):
$$\begin{array}{l}{e}_{s}={e}_{so}\mathrm{exp}\left[\left(6808\text{K}\right)\left(\frac{1}{{T}_{o}}-\frac{1}{T}\right)-5.09\mathrm{ln}\frac{T}{{T}_{o}}\right]\hfill \\ \hfill \\ {e}_{si}={e}_{so}\mathrm{exp}\left[\left(\text{6293K}\right)\left(\frac{1}{{T}_{o}}-\frac{1}{T}\right)-0.555\mathrm{ln}\frac{T}{{T}_{o}}\right]\hfill \\ \hfill \\ \text{where}\text{\hspace{0.17em}}{T}_{o}=273\text{K},\text{\hspace{1em}}{e}_{so}=6.11\text{hPa}\hfill \end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaauaabaqafeaaaaqaaabbaaaaaaaaIXwyJTgapeGaamyza8aadaWgaaWcbaWdbiaadohaa8aabeaak8qacqGH9aqpcaWGLbWdamaaBaaaleaapeGaam4Caiaad+gaa8aabeaak8qaciGGLbGaaiiEaiaacchadaWadaqaamaabmaabaGaaGOnaiaaiIdacaaIWaGaaGioaiaabccacaqGlbaacaGLOaGaayzkaaWaaeWaa8aabaWdbmaalaaapaqaa8qacaaIXaaapaqaa8qacaWGubWdamaaBaaaleaapeGaam4BaaWdaeqaaaaak8qacqGHsisldaWcaaWdaeaapeGaaGymaaWdaeaapeGaamivaaaaaiaawIcacaGLPaaacqGHsislcaaI1aGaaiOlaiaaicdacaaI5aGaciiBaiaac6gadaWcaaWdaeaapeGaamivaaWdaeaapeGaamiva8aadaWgaaWcbaWdbiaad+gaa8aabeaaaaaak8qacaGLBbGaayzxaaaapaqaaaqaa8qacaWGLbWdamaaBaaaleaapeGaam4CaiaadMgaa8aabeaak8qacqGH9aqpcaWGLbWdamaaBaaaleaapeGaam4Caiaad+gaa8aabeaak8qaciGGLbGaaiiEaiaacchadaWadaqaamaabmaabaGaaeOnaiaabkdacaqG5aGaae4maiaabccacaqGlbaacaGLOaGaayzkaaWaaeWaa8aabaWdbmaalaaapaqaa8qacaaIXaaapaqaa8qacaWGubWdamaaBaaaleaapeGaam4BaaWdaeqaaaaak8qacqGHsisldaWcaaWdaeaapeGaaGymaaWdaeaapeGaamivaaaaaiaawIcacaGLPaaacqGHsislcaaIWaGaaiOlaiaaiwdacaaI1aGaaGynaiGacYgacaGGUbWaaSaaa8aabaWdbiaadsfaa8aabaWdbiaadsfapaWaaSbaaSqaa8qacaWGVbaapaqabaaaaaGcpeGaay5waiaaw2faaaWdaeaaaeaapeGaae4DaiaabIgacaqGLbGaaeOCaiaabwgacaqGGaGaaGPaVlaadsfapaWaaSbaaSqaa8qacaWGVbaapaqabaGcpeGaeyypa0JaaGOmaiaaiEdacaaIZaGaaeiiaiaabUeacaGGSaGaaGzbVlaadwgapaWaaSbaaSqaa8qacaWGZbGaam4BaaWdaeqaaOWdbiabg2da9iaaiAdacaGGUaGaaGymaiaaigdacaqGGaGaaeiAaiaabcfacaqGHbaaaaaa@9770@
Simply put, the dewpoint temperature is the temperature at which the atmosphere’s water vapor would be saturated. It is always less than or equal to the actual temperature. Mathematically,
$$w(T)={w}_{s}({T}_{d})$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEhacaGGOaGaamivaiaacMcacqGH9aqpcaWG3bWaaSbaaSqaaiaadohaaeqaaOGaaiikaiaadsfadaWgaaWcbaGaamizaaqabaGccaGGPaaaaa@3C79@
which means that the water vapor pressure at some temperature T (not multiplied by T) equals the water vapor saturation pressure at the dewpoint temperature, T_{d}. So we see that because w_{s} depends only on T_{d} at a given pressure, T_{d} is a good method for designating the absolute amount of water vapor.
We can draw the phase diagram for water. There are three equilibrium lines that meet at the triple point, where all three phases exist (e_{s} = 6.1 hPa; T = 273.14 K). Along the line for e_{s}, vapor and liquid are in equilibrium, and evaporation balances condensation. Along the line for e_{si}, vapor and ice are in equilibrium and sublimation equals deposition. Along the line for e_{sm}, liquid and ice are in equilibrium and melting balances fusion.
Is it possible to have water in just one phase? Yes!
The simplest case is when all the water is vapor, which occurs when the water vapor pressure is low enough and the temperature (and thus saturation vapor pressure) is high enough that all the water in the system is evaporated and in the vapor phase.
Let’s think about what it would take to have all the water in the liquid phase. Suppose we have a vertical cylinder closed on one end and a sealed piston at the other end. The whole cylinder is immersed in a constant-temperature water bath so that we can hold the cylinder and its contents at a fixed temperature (i.e., isothermal). Initially we fill the cylinder with liquid water and have a small volume of pure water vapor at the top. If we set the bath temperature to, say, 280 K and let the system sit for a while, the vapor will become saturated, which is on the e_{s} line. For isothermal compression, in which energy is removed from the system by the bath in order to keep the temperature constant, a push on the piston will slightly raise the vapor pressure above e_{s} and there will be net condensation until equilibrium is obtained again. If we continue to slowly push in the piston, eventually all the cylinder’s volume will be filled with liquid water and the cylinder will contain only one phase: liquid. If we continue to push the piston and the bath keeps the temperature constant, then the water pressure will increase.
In the atmosphere, ice or liquid almost always has a surface that is exposed to the atmosphere and thus there is the possibility that water can sublimate or evaporate into this large volume. Note that the presence or absence of dry air has little effect on the condensation and evaporation of water, so it is not the presence of air that is important, but instead, it is the large volume for water vapor that is important.
Conditions can exist in the atmosphere for which the water pressure and temperature are in the liquid or sometimes solid part of the phase diagram. But these conditions are unstable and there will be condensation or deposition until the condensation and evaporation or sublimation and deposition come into equilibrium, just as in the case of the piston above. Thus, more water will go into the liquid or ice phase so that the water vapor pressure drops down to the saturation value. When the water pressure increases at a given temperature to put the system into the liquid region of the water phase diagram, the water vapor is said to be supersaturated. This condition will not last long, but it is essential in cloud formation, as we will see in the lesson on cloud physics.
Note also that the equilibrium line for ice and vapor lies below the equilibrium line for supercooled liquid and vapor for every temperature. Thus e_{si} < e_{s} for every temperature below 0 ^{o}C because l_{s} > l_{v} in the Clausius–Clapeyron Equation. This small difference between e_{si} and e_{s} can be very important in clouds, as we will also see in the lesson on cloud physics.
When a cloud drop evaporates, the energy to evaporate it must come from somewhere because energy is conserved according to the 1^{st} Law of Thermodynamics. It can come from some external source, such as the sun, from chemical reactions, or from the air, which loses some energy and thus cools. Thus, temperature changes and phase changes are related, although we can think of phase changes as occurring at a constant temperature. The energy associated with phase changes drives much of our weather, especially our severe weather, such as hurricanes and deep convection. We can quantify the temperature changes that result from phase changes if we have a little information on the mass of the air and the mass and phases of the water.
In the previous lesson, we said that all changes of internal energy were associated with a temperature change. But the phase changes of water represent another way to change the energy of a system that contains the phase-shifter water. So often we need to consider both temperature change and phase change when we are trying to figure out what happens with heating or cooling.
For atmospheric processes, we saw that we must use the specific heat at constant pressure to figure out what the temperature change is when an air mass is heated or cooled. Thus the heating equals the temperature change times the specific heat capacity at constant pressure times the mass of the air. For dry air, we designate the specific heat at constant pressure as c_{pd}. For water vapor, we designate the specific heat at constant pressure as c_{pv}. So for example, the energy required to change temperature for a dry air parcel is c_{pd} m ΔT = c_{pd} ρV ΔT, where c_{pd} is the specific heat capacity for dry air at constant pressure. If we have moist air, then we need to know the mass of dry air and the mass of water vapor, calculate the heat capacity of each of them, and then add those heat capacities together.
For liquids and solids, the specific heat at constant volume and the specific heat at constant pressure are about the same, so we have only one for liquid water (c_{w}) and one for ice (c_{i}).
For phase changes, there is no temperature change. Phase changes occur at a constant temperature. So to figure out the energy that must be added or removed to cause a phase change, we only need to know what the phase change is (melting/freezing, sublimating/depositing, evaporating/condensing) and the mass of water that is changing phase. So, for example, the energy needed to melt ice is l_{f} m_{ice}.
The following tables provide numbers and summarize all the possible processes involving dry air and water in its three forms.
Dry air c_{pd} |
Water vapor c_{pv} |
Liquid water c_{w} |
Ice c_{i} |
---|---|---|---|
1005 | 1850 | 4184 | 2106 |
Vaporization @ 0 ^{o}C l_{v} |
Vaporization @ 100 ^{o}C l_{v} |
Fusion @ 0 ^{o}C l_{f} |
Sublimation @ 0 ^{o}C l_{s} |
---|---|---|---|
2.501 x 10^{6} | 2.257 x 10^{6} | 0.334 x 10^{6} | 2.834 x 10^{6} |
Dry air | Water vapor | Liquid water | Ice |
---|---|---|---|
c_{pd} m_{d} |
c_{pv} m_{v} |
c_{w} m_{liquid} |
c_{i} m_{ice } |
vapor→liquid | liquid→vapor | vapor→ice | ice→vapor | liquid→ice | ice→liquid |
---|---|---|---|---|---|
l_{v} m_{vapor} | l_{v} m_{liquid} | l_{s} m_{vapor} | l_{s} m_{ice} | l_{f} m_{liquid} | l_{f} m_{ice} |
To solve energy problems you can generally follow these steps:
Knowing how to perform simple energy calculations helps you to understand atmospheric processes that you are observing, and to predict future events. Why is the air chilled in the downdraft of the thunderstorm? When will the fog dissipate? When might the sun warm the surface enough to overcome a near-surface temperature inversion and lead to thunderstorms? We can see that evaporating, subliming, and melting can take up a lot of energy and that condensing, depositing, and freezing can give up a lot of energy. In fact, by playing with these numbers and equations, you will see how powerful phase changes are and what a major role they play in many processes, particularly convection.
With the elements in the tables above, you should be able to take a word problem concerning energy and construct an equation that will allow you to solve for an unknown, whether the unknown be a time or a temperature or a total mass.
In the atmosphere, these problems can be fairly complex and involve many processes. For example, when thinking about solar energy melting a frozen pond, we would need to think about not only the solar energy needed to change the pond from ice to liquid water, but we would also need to consider the warming of the land in which the pond rests and the warming of the air above the pond. Further, the land and the ice might absorb energy at different rates, so we would need to factor in the rates of energy transfer among the land and the pond and the air.
So we can make these problems quite complex, or we can greatly simplify them so that you will understand the basic concepts of energy required for temperature and phase changes. In this course, we are going to solve fairly simple problems and progress to slightly more complicated ones. Let’s look at a few examples. I will give you some examples and then you can do more for Quiz 3-3.
A small puddle is frozen and its temperature is 0 ^{o}C. How much solar energy is needed to melt all the ice? Assume that m_{ice} = 10.0 kg.
$${{\displaystyle \int}}^{\text{}}Qdt={l}_{f}{m}_{ice}=\left(0.334\times {10}^{6}{\text{Jkg}}^{-1}\right)\text{\hspace{0.17em}}\left(10.0\text{kg}\right)=3.34\times {10}^{6}\text{\hspace{0.17em}}\text{J}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@688A@
To put this amount of energy into perspective, this energy is equivalent to a normal person walking at about 4 mph for 2 hours (assuming the person burns 400 calories per hour, which is really 400 kilocalories per hour in scientific units).
Now let’s assume that the ice is originally at –20.0 ^{o}C. Now we have to both raise the temperature and melt the ice. If we don’t warm the ice, some of it will simply refreeze. Our equation now becomes:
$$\begin{array}{l}{{\displaystyle \int}}^{\text{}}Qdt={l}_{f}{m}_{ice}+{c}_{i}{m}_{ice}\Delta T\\ =\left(0.334\times {10}^{6}{\text{Jkg}}^{-1}\right)\text{\hspace{0.17em}}\left(10.0\text{kg}\right)+\left(2106{\text{Jkg}}^{-1}{\text{K}}^{-1}\right)\left(10.0\text{kg}\right)\text{\hspace{0.17em}}\left(0.0--20.0\right)\text{K}\\ =3.76\times {10}^{6}\text{J}\end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@91CC@
We see that the amount of energy required increased by about 25%. Most of the energy is still required to melt the ice, not change the temperature.
Now let’s assume that the solar heating rate is constant at 191 W m^{–2} and that the area of the puddle is 2.09 m^{2}. How long does it take the sun to raise the temperature of the ice and then melt it?
$$\begin{array}{l}{{\displaystyle \int}}^{\text{}}Qdt=\frac{\overline{\Delta Q}}{\Delta A}A\text{\hspace{0.17em}}\Delta t={l}_{f}{m}_{ice}+{c}_{i}{m}_{ice}\Delta T\hfill \\ \Delta t=\frac{{l}_{f}{m}_{ice}+{c}_{i}{m}_{ice}\Delta T}{\frac{\overline{\Delta Q}}{\Delta A}A}=\frac{3.76\times {10}^{6}\text{J}}{\left(191{\text{Jm}}^{-2}{\text{s}}^{-1}\right)\text{\hspace{0.17em}}\left(2.09{\text{m}}^{2}\right)}=9.42\times {10}^{3}\text{\hspace{0.17em}}\text{s}=2.6\text{\hspace{0.17em}}\text{h}\hfill \end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@A230@
We could now assume that the source of heating is not the sun but instead is warm air passing over the puddle. If the temperature of the air is 20.0 ^{o}C and we assume that its temperature drops to 0.0 ^{o}C after contacting the ice, what is the mass of air that is required to warm the ice and then melt it?
$$\begin{array}{l}{{\displaystyle \int}}^{\text{}}Qdt={c}_{pd}{m}_{air}\Delta {T}_{air}={l}_{f}{m}_{ice}+{c}_{i}{m}_{ice}\Delta {T}_{ice}\hfill \\ {m}_{air}=\frac{{l}_{f}{m}_{ice}+{c}_{i}{m}_{ice}\Delta {T}_{ice}}{{c}_{pd}\Delta {T}_{air}}=\frac{3.76\times {10}^{6}\text{J}}{\left(1005{\text{Jkg}}^{-1}{\text{K}}^{-1}\right)\text{\hspace{0.17em}}\left(20.0\text{K}\right)}=187\text{\hspace{0.17em}}\text{kg}\hfill \end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@A663@
See this video (2:28) for further explanation:
The skew-T vs –lnp diagram, often referred to as the skew-T diagram, is widely used in meteorology to examine the vertical structure of the atmosphere as well as to determine which processes are likely to happen.
Check out this video to learn the basics of reading skew-T diagrams (1:23):
You may know a little about the skew-T from previous study, but for those who did not take a previous course or who need a refresher, there are many useful websites that can help you understand the skew-T and how to use it. Two useful resources are the following:
Weatherprediction.com Review of Skew-T Parameters [32]
Introduction to Mastering the Skew-T Diagram Video [33]
In this video (1:24) I will show you how the skew-T relates to a cumulus cloud:
Here's a picture of a mature cumulus cloud over the ocean. We can see the cloud base here, the vertical growth, and the cloud top here. Above and below the cloud is clear air. We can imagine what temperature and dew point are radius on record if you were to launch one from below the cloud. Initially we would see a temperature decrease, probably close to the dry adiabatic lapse rate of 10 degrees c per kilometer. We will see the dew point decrease slightly relative to the temperature which is skewed to 45 degrees on the Skew-T diagram. At cloud base temperature and dew point are about the same. Inside the cloud the temperature and dew point stay together along the moist adiabat, which is a temperature decrease of about six degrees c per kilometer. Remember that the relative humidity is about 100% in the clouds. The air above the cloud is likely stable, which is why the cloud's height is limited. Stable air has a lapse rate that is less than the adiabatic lapse rate. In addition the dew point likely drops off because the middle to upper troposphere tends to be drier than the lower troposphere. When you look at an upper air sounding you can often pick out where the clouds are by looking at where the temperature and dew point get close together.
First, familiarize yourself with all of the lines. Look at a radiosonde ascent [34], such as the one from the National Center for Atmospheric Research Research Applications Laboratory [35] (type of plot: GIF of skew-T). The atmospheric sounding lines are temperature (solid red line) and dewpoint temperature (solid green line). These lines are plotted on a grid that shows –lnp on the vertical axis (horizontal blue lines) and temperature on the horizontal axis (blue lines tilting up and to the right—which is where "skew" comes from). –lnp is used because it varies nearly linear with height and the skewing of the temperature lines allows soundings to be shown without going out of bounds of the graph.
Here [36] is a blank skew-T diagram that you can play with and practice on. The diagram has an electronic pen that allows you to draw on it. This diagram will be used in Practice Quiz 3-4 and Quiz 3-4, so become familiar with it.
Please also note the following:
See the video below (1:19) for further explanation:
Let's see how to find the lifting condensation level, the LCL. The LCL is the level where a cloud will form if the air mass near a surface is pushed upward. We have two quantities that are conserved when lifted. The potential temperature, or theta, and the water vapor mixing ratio, w. As the air parcel is pushed up, then w goes up the constant w line and theta goes up the dry adiabat, which is the constant theta line. Note that both the temperature and the dew point temperature are changing and getting closer to each other as the air parcel ascends. When the two lines meet, the relative humidity is 100%, a cloud forms, and this is the lifting condensation level. Once the LCL has been reached and the cloud is formed, any further ascent will be in the cloud. The air parcel temperature will follow the moist adiabat, which is less than the dry adiabat. Because as water condenses it gives up its energy to warm the air a little bit. If the air were pushed down, its temperature would follow the moist adiabat, as long as it was above the LCL. But below the LCL, it will follow the dry adiabat. And the water vapor mixing ratio will follow the constant w line.
When the air parcel is in a cloud, ascent causes a temperature decrease while the air remains saturated (i.e., w = w_{s}, RH = 100%). Since w_{s} decreases, the amount of water in the vapor phase decreases while the amount in the liquid or solid phase increases, but the total amount of water is constant (unless it rains!). As water vapor condenses, energy is released into the air and warms it a little bit. Thus, the lapse rate of the moist adiabat (curved dot-long-dash green lines tilting toward the upper left) is less than the lapse rate of the dry adiabat (9.8 K/km).
As long as it doesn’t rain or snow, an air parcel will move up and down a moist adiabat as long as it is in a cloud and will move up and down a dry adiabat when w < w_{s} below the LCL.
The following is a summary for air parcel ascent and descent:
The following video (1:43) discusses the process of adiabatic cooling and heating.
We've all seen clouds build up on one side of the mountain and then on the other side just dissipate into blue sky. Maybe you want to know why that happens. Like most natural events, this one has an impressive scientific term attached to it. It's called adiabatic cooling and heating, and occurs because of changes in air pressure. Here's some time lapse video that shows what happens. Basically, as a parcel of air encounters a mountain, it is forced upward. As air pressure decreases with altitude, the air parcel expands. Expansion causes air to cool. When the air cools to its so-called dew point, the water vapor in the air condenses and becomes visible as a cloud. If there's enough moisture and the adiabatic cooling is strong enough, it rains or snows. Essentially the opposite occurs on the other side of the mountain. The cool air sinks and compresses. Compression results in increased temperature. When temperature rises above the dew point, the cloud dissipates into invisible water vapor. In Wyoming, especially in winter, most of the moisture-laden air masses come from the Pacific, approaching our mountains from the west. So as adiabatic cooling occurs, more rain and snow is dumped on west-facing slopes. As warmer, drier air descends on the eastern slopes, it accounts for another famous phenomenon of the plains, the so-called Chinook winds. So we've looked at clouds from both sides now. Knowing why they form and disappear does not diminish their beauty. But if it weren't for our mountains and the dynamic processes that occur, one, we would be a much drier place, and frankly, much less interesting. I'm Tom Hill from the University of Wyoming Cooperative Extension Service exploring the nature of Wyoming.
There are other potential temperatures that are useful because they are conserved in certain situations and therefore can help you understand what the atmosphere is doing and what an air parcel is likely to do.
Virtual potential temperature is the potential temperature of virtual temperature, where density differences caused by water vapor are taken into account in the virtual temperature by figuring out the temperature of dry air that would have the same density:
$${\theta}_{v}={\left(\frac{{p}_{o}}{p}\right)}^{\raisebox{1ex}{$(\gamma -\mathrm{1)}$}\!\left/ \!\raisebox{-1ex}{$\gamma $}\right.}{T}_{v}={\left(\frac{{p}_{o}}{p}\right)}^{\raisebox{1ex}{$(\gamma -\mathrm{1)}$}\!\left/ \!\raisebox{-1ex}{$\gamma $}\right.}T\left(1+0.61q\right)$$
This quantity is useful when comparing the potential temperatures (and thus densities) of air parcels at different pressures.
The wet-bulb temperature is the temperature a volume of air would have if it were cooled adiabatically while maintaining saturation by liquid water; all the latent heat is supplied by the air parcel so that the air parcel temperature, when it descends to 1000 hPa, is less than its temperature would be had it descended down the dry adiabat.
The wet bulb temperature at any given pressure level is found by finding the LCL and then bringing the parcel up or down to the desired pressure level on the moist adiabat.
The wet bulb potential temperature, Θ_{w}, is the wet bulb temperature at p = 1000 hPa.
How can we use the wet bulb potential temperature? The wet bulb potential temperature is conserved, meaning it does not change, when an air mass undergoes an adiabatic process, such as adiabatic uplift or descent. If we consider large air masses that acquire similar temperature and humidity, then this entire air mass can take on the same wet bulb potential temperature. Colder, drier air masses will have a lower Θ_{w}. The Θ_{w} of this air mass can change if a diabatic process occurs, such as the warming of a cold air mass as it moves over warmer land, or the cooling of an air mass by radiating to space during the night, but these processes can sometimes take days. So an 850-mb map of Θ_{w} is one indicator of air masses and the fronts between air masses.
See the video below (:32) for further explanation:
Let's see how to find the wet-bulb potential temperature on the Skew-T. The first step is to find the LCL. Once we find the LCL, then we have a saturated air parcel. And it's temperature is the wet-bulb temperature. To find the wet-bulb potential temperature, we simply follow the moist adiabat down to a pressure of 1,000 millibar. We see that the wet-bulb potential temperature is about 19 C, while the potential temperature is about 34 C.
The equivalent potential temperature is the temperature that an air parcel would have if it were lifted adiabatically until all of the vapor were condensed and removed, and then brought to 1000 hPa adiabatically. For example, let's say you have an air parcel with some water vapor in it below the LCL. To find the equivalent potential temperature, you would (1) lift the parcel to the LCL along a dry adiabat, (2) lift the parcel along the moist adiabat all the way to the stratosphere so that all the water vapor condensed into liquid, and (3) bring the parcel down to 1000 hPa along the dry adiabat. Equivalent potential temperature accounts for the effects of condensation or evaporation on the change in the air parcel temperature.
Every 1 g/kg (g water vapor to kg of dry air) causes Θ_{e} to increase about 2.5 K. So, a moist air parcel with w = 10 g kg^{-1}, which is not uncommon, will have Θ_{e} that is 25 K greater than Θ.
Approximately,
$${\theta}_{e}\approx \theta +\frac{{l}_{v}w}{{c}_{p}}$$
where Θ is the potential temperature, l_{v} is the latent heat of vaporization, w is the water vapor mixing ratio, and c_{p} is the specific heat capacity at constant pressure.
How can we use the equivalent potential temperature? The equivalent potential temperature, Θ_{e} is conserved when an air parcel or air mass undergoes an adiabatic process, just like the wet bulb potential temperature, Θ_{w}, is. Note also the total amount of water in vapor, liquid, and ice form is also conserved during adiabatic processes. The total amount of water is quantified using the total water mixing ratio, w_{t}, which is defined in the same way that the water vapor mixing ratio (w) is defined except that it also includes liquid and solid water in the numerator. So, if we look at Θ_{e} and w_{t}, we can learn a lot about the history of an air parcel. These conserved quantities are very useful to understand the history of air parcels around clouds. For example, if Θ_{e} changes but w_{t} is constant, then the air parcel was either heated or cooled by a non-adiabatic process. On the other hand, if both Θ_{e} and w_{t} change proportionally, then two air parcels with different initial values for Θ_{e} and w_{t} have mixed. On a larger, more synoptic scale, gradients in Θ_{e} can be used to indicate the presence of fronts.
Another use of Θ_{e} is as an indicator of unstable air. Air parcels that have higher Θ_{e} tend to be unstable. Thus regions of high-Θ_{e} air are regions where thunderstorms might form if the surface heating is great enough to erase a temperature inversion.
See the video (1:01) below for further explanation:
Let's see how to find the equivalent potential temperature, called theta-e, on this Skew-T. The equivalent potential temperature is the potential temperature that an air parcel would have if all this water vapor were converted to liquid water, thus warming the air. And then the liquid water was removed. To find theta-e, we find the LCL. Go up to the moist adiabat until it's parallel with the dry adiabat. And then go down the dry adiabat that matches the moist adiabat until we reach the pressure of 1,000 millibar. In this case, theta-e is about 330 Kelvin, or 57 degrees C. Note that the lines aren't marked with such high temperatures. But we can determine which temperature this line represents by looking at the 360 Kelvin dry adiabat. And then counting one, two, three lines over, where the lines are in intervals of 10 K.
Now we can begin to understand the reasons for the troposphere’s typical temperature profile. The atmosphere is mostly transparent to the incoming solar visible radiation, so Earth’s surface warms, and thus warms and moistens the air above it. This warm, moist air initially rises dry adiabatically, and then moist adiabatically once a cloud forms. Different air masses with different histories and different amounts of water mix and the result is a typical tropospheric temperature profile that has a lapse rate of (5–8) K km^{-1}.
If atmospheric temperature profiles were determined only by atmospheric moisture, drier air masses would have lapse rates that are more like the dry adiabatic lapse rate, in which case we would expect that the skies would have fewer, thinner clouds. Moister air masses would have lapse rates that are closer to the moist adiabatic lapse rate, resulting in a sky filled with clouds at many altitudes.
But many processes affect the temperature of air at different altitudes, including mixing of air parcels, sometimes even from the stratosphere, and rain and evaporation of rain. Exchange of infrared radiation between Earth’s surface, clouds, and IR-absorbing gases (i.e., water vapor and carbon dioxide) also plays a major role in determining the atmosphere’s temperature profile, as we will show in the lesson on atmospheric radiation.The resulting atmospheric profiles can have local lapse rates that can be anywhere from less than the dry adiabatic lapse rate to greater than the moist adiabatic lapse rate. Look carefully at the temperature profile below. You will see evidence of many of these processes combining to make the temperature profile what it is.
If we average together all of these profiles over the whole year and over the whole globe, we can come up with a typical tropospheric temperature profile. According to the International Civil Aviation Organization (Doc 7488-CD, 1993), the standard atmosphere has a temperature of 15 ^{o}C at the surface, a lapse rate of 6.5 ^{o}C km^{–1} from 0 km to 11 km, a zero lapse rate from 11 km to 20 km, and a lapse rate of –1 ^{o}C km^{–1} from 20 km to 32 km in the stratosphere (i.e., temperature increases with height). Even though this standard profile is a good representation of a globally averaged profile, it is unlikely that such a temperature profile was ever seen with a radiosonde.
Combining knowledge of stability along with the knowledge of moist processes enables us to understand the behavior of clouds in the atmosphere. The following picture of water vapor released from a cooling tower at the Three-Mile Island nuclear reactor near Harrisburg, PA shows the water vapor quickly condensing to form a cloud. The cloud ascends, but then reaches a level at which its density matches the density of the surrounding air. The cloud then stops ascending and begins to spread out.
Water vapor is a key atmospheric constituent that is essential for weather. There are many ways to express and measure the amount of atmospheric water vapor—specific humidity, water vapor mixing ratio, partial pressure, relative humidity, and dewpoint temperature—and these are all related and can be used interchangeably, although some provide more physical insight than others depending on the question being asked. Water’s most important characteristic in the atmosphere is that it can change phases between vapor, liquid, and ice. In the atmosphere, water is either in the vapor phase or trying to establish an equilibrium between vapor and liquid or vapor and ice. The equilibria conditions are given by the Clausius–Clapeyron equation, which shows that the equilibrium (a.k.a. saturation) water vapor pressure depends only on the temperature. Water phase changes pack a big energy punch and drive weather events. We can calculate the atmospheric temperature changes resulting from phase changes and then see that these temperature changes greatly affect the buoyancy of air parcels and therefore their vertical motion.
A good way to visualize atmospheric vertical structure and behavior is the skew-T diagram. With it, we can readily deduce atmospheric properties and predict what weather is likely to happen if solar heating causes some air near the surface to ascend. Some of the most important properties found using the soundings on the skew-T are the lifting condensation level, the potential temperature, and the equivalent potential temperature. The behavior of a typical sounding on the skew-T shows that the troposphere’s thermal structure is caused in large part by adiabatic ascent and descent, although we will see later that absorption and emission of infrared radiation by water vapor and carbon dioxide also have a hand in shaping the temperature vertical profile.
You have reached the end of Lesson 3! Make sure you have completed all of the activities before you begin Lesson 4.
The atmosphere consists mostly of dry air—mostly molecular nitrogen (78%), molecular oxygen (21%), and Argon (0.9%)—and highly variable amounts of water vapor (from parts per million in air to a few percent). Now we will consider gases and particles in the atmosphere at trace levels. The most abundant of the trace gases in the global atmosphere is carbon dioxide (~400 parts per million, or 400 x 10^{-6}), but there are thousands of trace gases with fractions much less than a few parts per million. Some, particularly the reactive hydroxyl (OH) radical, are important even though their abundance is less than 1 part per trillion (10^{–12}). The atmosphere also contains small particles with sizes from nanometers (10^{–9} m) to microns (10^{–6} m) coming from many sources. These trace gases and particles are as important to atmospheric structure and weather as are nitrogen, oxygen, and water vapor and they also play a huge role in human and ecological health and global climate. In this lesson we will examine the atmosphere’s composition and its changes over time. The atmosphere is continually inundated with surface emissions of gases and particles (and some from space) but it has chemical mechanisms to clean itself. We will see how two atmospheric pollutants—ozone and small particles—are produced. In later lessons, we will see that without these chemical processes and particles, there would be no clouds and, thus, no real weather.
By the end of this lesson, you should be able to:
If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.
The major gases that comprise today's atmosphere are listed in the table below. In most studies of atmospheric composition, the volume mixing ratio is used to specify the amount of a gas. The volume mixing ratio (also called the molar mixing ratio) of a gas is the number of moles of the gas divided by the number of moles of air. For example, 78 moles of every 100 moles of air is nitrogen, so nitrogen's volume mixing ratio is 0.78.
Constituent |
Molecular Mass (g mol^{–1}) |
Volume Mixing Ratio (mol mol^{–1}) |
Role in the Atmosphere |
---|---|---|---|
nitrogen (N_{2}) | 28.013 | 0.7808 | transparent; provides heat capacity and momentum; exchanged with biomass; decomposed in combustion |
oxygen (O_{2}) | 31.998 | 0.2095 | transparent except in the extreme ultraviolet; provides some heat capacity and momentum; exchanged with life; source of important reactive gases like ozone |
argon (Ar) | 39.948 | 0.0093 | no role |
carbon dioxide (CO_{2}) | 44.010 | 0.000385 (385 ppmv) | transparent in visible; absorbs infrared light (i.e., contributes to global warming); exchanged with life; product of combustion |
neon (Ne) | 20.183 | 0.0000182 | no role, but makes colorful glowing signs |
water vapor (H_{2}O) | 18.015 | 2 x 10^{–6} to 0.05 | transparent in visible; absorbs infrared light (i.e., contributes to global warming); exchanges with liquid and solid forms; exchanges with life; product of combustion |
aerosol particles | varies | 0–500 µg m^{–3} (note different units) | essential for cloud formation; interact with visible and infrared light; exchange with surfaces and life |
methane (CH_{4}) | 16.04 | 0.00000182 (1820 ppbv) | transparent in visible; absorbs in infrared (i.e., contributes to global warming); exchange with life; source of CO_{2} and H_{2}O |
ozone (O_{3}) | 48.00 | 0.01–10 x 10^{–6} (10 ppbv to 10 ppmv) | transparent in visible; absorbs in UV and infrared; reactive and source of more reactive gases |
particles | varies | 0–100’s µg m^{–3} (note different units) | absorb and scatter light; act as CCN and IN (see below) |
Key features of the gases include their compressibility (i.e., ability to expand or shrink in volume), their transparency in the visible, their momentum, and their heat capacity. Key greenhouses gases—those that absorb infrared radiation and hence warm the planet—are water vapor, carbon dioxide, methane, and ozone. Water vapor has the additional important feature of exchanging with liquid and solid phases in the atmosphere and on Earth’s surface. The most important properties of small particles include their ability to dissolve in water in order to be cloud condensation nuclei (CCN) or to maintain a lattice structure similar to ice in order to be ice nuclei (IN), as well as their ability to absorb and scatter sunlight. These properties depend completely on the particle size and composition. Most atmospheric gases participate in the atmosphere's chemistry, which is initiated by sunlight, as you will soon see.
The amount of a gas is typically specified in one of three different ways. You have already been introduced to the first, the volume mixing ratio, in the table above. For gases with relatively large fractions like nitrogen, oxygen, and argon, we use percent to indicate this fraction. For minor gases like carbon dioxide and ozone, we use parts per million (10^{–6}) by volume (ppmv) or parts per billion (10^{–9}) by volume (ppbv). The second is the mass mixing ratio, which is the mass of a chemical species divided by the total mass of air. You have already encountered this ratio with the specific humidity. The third way to specify the amount of a gas is the concentration, which is the number of molecules per unit volume.
It is straightforward to convert between volume mixing ratio and concentration. For a species X, to convert from a volume mixing ratio, notated χ_{X}, to a concentration, notated [X], use the Ideal Gas Law to find the number of total molecules in a cm^{3} and then multiply by χ_{X}, expressed as a fraction. For example, let p = 960 hPa, T = 296 K, and χ_{X} = 60 ppbv, then the concentration can be calculated as follows:
Here we have used the Boltzmann constant k, which is simply the universal gas constant divided by Avogadro's number.
Since the Earth was formed more than 4 billion years ago, the atmosphere has changed profoundly. A wide variety of geochemical and ecological (fossil) evidence indicates that oxygen levels rose dramatically about 2 billion years ago. Such evidence also indicates that carbon dioxide levels were much higher earlier in Earth's history, which allowed the Earth to be at a habitable temperature despite the fact that the output from the Sun was much lower (about 25%) compared to today, resolving the so-called "Faint-Young-Sun Paradox." Below we discuss changes in atmospheric composition over the past 800,000 years, the past few hundred years, and the past several decades.
We have learned a lot about how atmospheric composition has changed from measurements of gases in bubbles trapped in ice cores. The figure below, which shows such measurements from an Antarctic ice core, reveals that two key greenhouse gases, CO_{2} and CH_{4}, underwent large, rapid variations over the past 400,000 years. The variations are also periodic, with a rapid decline followed by a more gradual increase every 100,000 years or so, and are in roughly in phase with the temperature and out of phase with ice volume. The periodic variations reflect the coming and going of the ice ages. Similar cycles have been measured as far back as 800,000 years for CO_{2} and CH_{4} and other fossil evidence suggests that the current period of ice ages that we are in now began about 2.6 million years ago. These changes in CO_{2} and CH_{4} were driven ultimately by changes in the Earth's orbit and axis of rotation, which led to changes in the amount of solar radiation received at various latitudes during various seasons. Summer solar radiation at high northern latitudes is particularly important (bottom curve below), because it regulates how the large northern hemisphere ice sheets grow. These changes in solar radiation led to changes in Earth's temperature, ocean circulation, and other processes that influence atmospheric CO_{2} and CH_{4}, which amplified the changes in Earth's temperature.
The Holocene period began about 12,000 years ago, at the end of the last ice age, and marks a period of relative stability in climate and atmospheric gas concentrations. During this time, the ice core data reveal that levels of CO_{2} and CH_{4} in the atmosphere were relatively constant, at about 280 ppmv and 650 ppbv, respectively. Then, as shown below, levels of both gases, as well as another greenhouse gas, nitrous oxide (N_{2}O), increased rapidly about 200 years ago. These increases, which coincide with the industrial revolution, were due to anthropogenic activity, including the burning of fossil fuels and enhanced deforestation and agriculture. The increases in these three greenhouse gases are the primary cause of the warming of the Earth by about 1 ^{o}C over the past century.
Ice core data reveal many other interesting changes in the atmosphere that occurred during the last 200 years or so, particularly in the Northern Hemisphere. For example, within the ice itself, one can see increases in the amount of nitrate and sulfate [39], which are produced ultimately from the combustion of fossil fuels. These constituents are the key components of acid rain and, indeed, data from the same ice core also reveal an increase in acidity.
Changes in the composition of the atmosphere over the past several decades primarily reflect changes in human activity.
As fossil fuel emissions have increased over recent decades, so has the growth rate of atmospheric CO_{2}, as indicated by the concave-upward curvature in the figure below. The growth rate has approximately doubled from about 1 ppmv per year in the 1960s to about 2 ppmv per year in the 2000s. According to the Global Carbon Project [40], 86% of the anthropogenic CO_{2} emissions during 2009–2018 were from fossil fuel burning and 14% were from land-use change (e.g., deforestation). However, CO_{2} injected to the atmosphere from human activity does not stay there. 44% of the emissions from human activity during 2009–2018 accumulated in the atmosphere, 29% were absorbed by terrestrial ecosystems, 23% were absorbed by the ocean, and 4% is unaccounted for (Global Carbon Project). Superimposed on the accelerating trend over the past few decades is an annual cycle in which CO_{2} declines during Northern Hemisphere summer and rises during most of the rest of the year. This cycle reflects photosynthesis (an atmospheric CO_{2} sink) and respiration (an atmospheric CO_{2} source) of terrestrial ecosystems in the Northern Hemisphere, where most land is present. Note that the current increase to above 400 ppmv now extends well above any other time in the past half 800,000 years, at least, when CO_{2} varied between about 180 and 280 ppmv.
The changes in methane over the past several decades are more complex than those of carbon dioxide. The figure below shows that the methane growth rate (the slope of the curve) gradually declined from the 1980s to roughly zero by the mid 2000s. Since then, the growth rate has increased, though it is still not as high as it was in the 1980s. A detailed budget of methane has been developed by the Global Carbon Project [40] and indicates that for 2008–2017 about 60% of the methane sources were anthropogenic (mainly from agriculture, waste, fossil fuel production and use, and biomass burning) and 40% were natural (mainly from wetlands, inland waters, geological sources, ocean, termites, wild animals, permafrost, and vegetation). The main sink of methane is oxidation by OH, which we will discuss later in this lesson, as well as consumption within soils. Despite our basic understanding of methane sources and sinks, we do not know why the methane growth rate has changed in the way it has over the past several decades. This is an important unresolved problem, because methane levels are very high currently. In 2019, the global-average methane mixing ratio was more than 1860 ppbv, which is well above values over the last 800,000 years, at least, which varied between about 400 and 700 ppbv.
Numerous lines of evidence show that water vapor is increasing in the atmosphere. Surface specific humidity measurements, which have been made mostly in the Northern Hemisphere, show clear trends, as indicated in the figure below. Data from satellites indicate that precipitable water (the total amount of water that is in a column from the surface to the top of the atmosphere) increased 1.49% per decade from 1988 to 2017 (Mears et al., 2005) [42]. These trends are generally consistent with expectations of a warming world and the Clausius–Clapeyron equation. Variability from year to year is also consistent with temperature change. For example, the increases in temperature resulting from the very large 1997–1998 El Niño event were accompanied by large increases in surface specific humidity and total precipitable water.
There are trends and variations in many of the other trace gases as well, such as chlorofluorcarbons, which are decreasing in abundance due to emissions reductions. There are other trace gases that increase as the sun rises and decrease as it sets and are heavily involved in atmospheric chemistry. We will talk about these gases next.
Hundreds of different trace gases have been measured in the atmosphere and perhaps thousands more have yet to be measured. Many of these are volatile organic compounds (VOCs). Volatile means that the compound may exist in the liquid or solid phase but that it easily evaporates. Organic means that the compound contains carbon but is not carbon dioxide, carbon monoxide, or carbides and carbonates found in rocks. There are also other chemicals like the nitrogen oxides (e.g., nitric oxide (NO), nitrogen dioxide (NO_{2}), nitric acid (HNO_{3})), sulfur compounds (e.g., sulfur dioxide (SO_{2}), sulfuric acid (H_{2}SO_{4})) and halogen compounds (e.g., natural methyl chloride (CH_{3}Cl), human-made chlorofluorocarbons (CCl_{2}F_{2})). If we pay attention, we can often smell and identify many of these chemicals, even at trace levels, although some, like methane, carbon monoxide (CO), and chlorofluorocarbons, are odorless. We enjoy smelling the VOCs emitted by trees in a forest—aah, that fresh pine smell—but we hold our nose to escape the smells of a stagnant swamp.
In addition to these thousands of chemicals that are emitted into the atmosphere every day, there are also some very reactive compounds that are created by atmospheric chemistry and play the important role of cleaning the atmosphere of many gases. The most important reactive gases are ozone (O_{3}) and hydroxyl (OH). We will focus the discussion of atmospheric chemistry on these two.
Earth’s atmosphere is an oxidizing environment. This term means what you think it would: gases that are emitted into the atmosphere react in a way that increases their oxygen content. Gases that contain oxygen tend to be “stickier” on surfaces and more water soluble, which means that they stick when they hit a surface or they can be readily taken up in clouds and rain drops and be deposited on Earth’s surface. We call gases hitting the surface and sticking “dry deposition” and gases being taken up in precipitation and rained out “wet deposition.”
Let’s consider a natural gas that is very important in our lives, methane, also known as natural gas. More and more methane is being extracted from below Earth’s surface and used to run our electrical power plants, heat our homes, cook our food, and, increasingly, to run our transportation vehicles. Methane is a simple molecule, CH_{4}, in which each of carbon’s four bonds is made with a hydrogen atom. Energy comes from heating methane to high enough temperatures that cause it to react, giving off energy as more stable molecules are formed. In complete combustion, each methane molecule is converted into one CO_{2} molecule and two H_{2}O molecules. In the process, four oxygen atoms or two oxygen molecules are consumed.
This same process occurs in the atmosphere, but at much lower temperatures and at a much slower rate. In both cases, the first step in the methane oxidation sequence is the reaction with the hydroxyl radical (OH). In water, hydroxyl loses an electron and is ionized (OH^{–}), but in the atmosphere, hydroxyl is not ionized. We call OH a free radical because it has an odd number of electrons (eight for oxygen and one for hydrogen). Any gas with an odd number of electrons is reactive because the electrons want to be paired up in molecules because that makes them more stable.
Often, combustion is inefficient, resulting in the formation of carbon monoxide (CO). Examples include forest fires, humans burning fields to clear them for planting, poorly tuned vehicles, inefficient industrial processes, and other human-caused processes. The primary way that CO is removed from the atmosphere is by reacting with atmospheric OH. It takes a while for CO to be removed from the atmosphere by the reaction with OH, so that satellite instruments can track CO plumes as they emerge from their sources and flow around the world.
Before we tackle this question, let’s first look at where ozone (O_{3}) comes from. We will start with stratospheric ozone (a.k.a. good ozone because it blocks solar UV that harms humans, other animals, agriculture, and ecosystems) and then eventually we will consider tropospheric ozone (a.k.a. bad ozone, which is the ozone that hurts our health when we breathe it and that damages plants and their fruit).
For this week's discussion activity, I would like you to think about which trace gas is the most important and why. By trace gas I mean a gas with a mixing ratio of less than 20 ppm in the atmosphere. Defend your choice. Use information from this lesson as well as other sources (credit them, please!) to describe the qualities of this gas that make you think that it is the most important trace gas. Then read the choices of your classmates and respond to their choices and follow-up with further questions and/or analysis.
This discussion will be worth 3 discussion points. I will use the following rubric to grade your participation:
Evaluation | Explanation | Available Points |
---|---|---|
Not Completed | Student did not complete the assignment by the due date. | 0 |
Student completed the activity with adequate thoroughness. | Student answers the discussion question in a thoughtful manner, including some integration of course material. | 1 |
Student completed the activity with additional attention to defending their position. | Student thoroughly answers the discussion question and backs up reasoning with references to course content as well as outside sources. | 2 |
Student completed a well-defended presentation of their position, and provided thoughtful analysis of at least one other student’s post. | In addition to a well-crafted and defended post, the student has also engaged in thoughtful analysis/commentary on at least one other student’s post as well. | 3 |
Ozone is ozone no matter where it is in the atmosphere. Good ozone is good only because it is in the stratosphere where we cannot breathe it (see figure below). Bad ozone also absorbs solar ultraviolet light, but it is down near Earth's surface where we can breathe it. For UV protection, we are interested in the total number of ozone molecules between us and the Sun. 90% of ozone molecules are in the stratosphere and 10% are in the troposphere—some down near Earth's surface where we can breathe them. There are important issues affecting human and ecological health for both good ozone and bad ozone. For good ozone, the most important issues are the reduction of ozone globally, the Antarctic Ozone Hole, and Arctic ozone loss that is caused by chlorofluorocarbons. Reduced ozone means more solar UV gets to the ground causing more skin cancer. For bad ozone, the most important issues include the production of too much ozone in cities and nearby regions that is caused by too many pollutants from traffic, industrial processes, power generation, and other human activities. Increased ozone means more people have respiratory and heart problems. Let's look at both the good and the bad, starting with the stratospheric ozone.
To get the total amount of ozone between us and the Sun, we simply add up the ozone amount starting at the surface and going up to the top of the ozone layer. Note how much more ozone there is in the stratosphere. At higher latitudes, the bottom of the stratospheric ozone layer is at approximately 10–12 km. Recall the following image from Lesson 2:
The process of stratospheric ozone formation starts with ozone (O_{3}), which is made by ultraviolet sunlight in the stratosphere (but not the troposphere, as we shall see). The two reactions are:
$${O}_{2}+UV\text{}\to \text{}O+O$$
$$O+{O}_{2}+{N}_{2}\to \text{}{O}_{3}+{N}_{2}$$
Note that N_{2} doesn't really react in this last chemical equation, but instead, simply bumps into the O_{3} molecule as it is being formed and stabilizes it by removing some of the energy from O_{3}. We call O_{3} an oxidant because it can react with some compounds and oxidize them.
This O_{3} can be broken apart by ultraviolet light to make O_{2} and O. Usually O combines with O_{2} to form O_{3 }in this way: O + O_{2} + N_{2} → O_{3} + N_{2}, so nothing really happens, except that the solar energy that breaks apart the O_{2} ends up as extra energy for the O_{3} and for the colliding N_{2} and, as a result, ends up warming the air. Sometimes O collides with O_{3} and reacts as follows: O + O_{3 }→ O_{2} + O_{2}. Putting all of the reactions together, we can see the chemical lifecycle of ozone in the stratosphere. This set of reactions was proposed in the 1930s by Chapman:
O_{2} + UV → O + O | production |
2(O + O_{2} + N_{2 }→ O_{3 }+ N_{2}) | cycling |
O_{3} + UV → O_{2} + O | |
O + O_{3 }→ O_{2 }+ O_{2} | loss |
Net: UV → higher T |
These four reactions could produce the basic characteristics of the ozone layer as it was in the 1940s through the 1970s. The basic ingredients are UV radiation and O_{2}, which can help to explain why ozone reaches a maximum concentration at a certain height in the atmosphere–that is, why there is an ozone layer at all. At very high levels in the atmosphere (e.g., in the mesosphere), there is plenty of UV but too little O_{2} (simply because pressure decreases with height). On the other hand, at very low levels in the atmosphere (the troposphere), there is plenty of O_{2} but very little UV (because of absorption by O_{2} above). It's only in the stratosphere that there is enough of both UV and O_{2} to make plenty of ozone.
While Chapman's theory predicted the existence of an ozone layer, the theory produced peak ozone levels that were 50 milliPascals (mPa), not the 25–30 mPa seen in the first figure above. Thus, the measured levels of stratospheric ozone were about half of those predicted by Chapman's theory—it was a real puzzle. However, in the 1970s, scientists proposed new sets of reactions by other gases that accomplished the same results as the loss reaction shown above. A famous example involved chlorine, which comes mostly from human-made chlorofluorocarbons (CFCs):
CFCs + UV → product + Cl | production |
Cl + O_{3 }→ ClO + O_{2} | cycling |
ClO + O → Cl + O_{2} | |
Cl + CH_{4 }→ HCl + CH_{3} | loss |
Net: O_{3} + O → O_{2} + O_{2} |
During the cycle, chlorine (Cl) and chlorine monoxide (ClO) aren’t destroyed but instead are just recycled into each other. With each cycle, two ozone molecules are lost (one directly and a second because O almost always reacts with O_{2} to form O_{3}). This cycle can run for hundreds of thousands of times before Cl gets tied up in HCl. So ClO and Cl levels of tens of parts per trillion of air (10^{–12}) are able to destroy several percent of the few parts per million of O_{3}. Sherry Rowland and Mario Molina figured this cycle out and wrote a paper about it in 1974. They received a Nobel Prize in Chemistry in 1995 for this work. When catalytic cycles involving chlorine, nitrogen oxides, and OH are included with the theory, the agreement between the theory and the measurements gets much better.
Note that the total ozone amount at midlatitudes is greater than the amount in the tropics. This should seem strange to you because the solar UV that is part of the Chapman mechanism is strongest in the tropics. Why do you think that total ozone is distributed this way?
ANSWER: In addition to the production and destruction processes described above, the ozone distribution in the stratosphere is due to the motion of air. Air comes from the troposphere into the stratosphere mostly in the tropics and then slowly moves to middle and high latitudes, where it sinks and re-enters the troposphere. This air motion is known as the Brewer–Dobson circulation. Even though most ozone is made in the tropical stratosphere, the production process is relatively slow, and hence ozone abundance in the tropical stratosphere is quite low. As the air moves poleward, the production process adds ozone to the air, leading to relatively high ozone abundance at midlatitudes.
The low ozone over Antarctica above is the Antarctic Ozone Hole; the video below (:31) entitled "Ozone Minimums With Graph" (from NASA) shows changes in ozone concentration between 1979 and 2013. Video is not narrated:
The Antarctic ozone hole is an extreme example of the destructive power of chlorine catalytic cycles. Different catalytic cycles dominate the ozone destruction over Antarctica and, to a lesser extent, the Arctic. But, when aided by chemistry on the surfaces of naturally occurring polar stratospheric clouds, all the Cl in the form of HCl is liberated so that the polar catalytic cycles are able to destroy a few percent of the ozone per day in a plug the size of Antarctica from an altitude of 12 km all the way up to 20 km.
Fortunately, the amount of chlorine being injected into the stratosphere is decreasing due to the Montreal Protocol, the world’s first international global environmental treaty.
The atmosphere's oxidation capacity is its ability to clean itself of all of the gases that are emitted into it. What does stratospheric ozone have to do with the atmosphere’s oxidation capacity, which mostly occurs in the troposphere and mostly by the atmosphere's PAC-MAN, hydroxyl (OH)? It turns out that natural dynamic processes actually pull air down from the stratosphere and mix it into the troposphere, eventually mixing some of this ozone to Earth’s surface. This naturally occurring surface ozone provides a baseline value for near-surface ozone, but ozone pollution is more than ten times greater than this baseline in cities. Ozone is both sticky on surfaces and fairly reactive in the atmosphere. It is lost both by depositing on surfaces and through being chemically destroyed by reactions in the atmosphere.
The following chemical sequences are the humble beginnings of the atmosphere's PAC-MAN. OH is generated throughout the stratosphere and troposphere by a two-step reaction sequence. The first step is:
$${O}_{3}+\text{}UV\text{}\to \text{}{O}_{2}+O*$$
where O* is an excited-state oxygen atom that has extra chemical energy. O* can lose this extra energy by colliding with N_{2} and O_{2}, but it can also collide with a water molecule to make two OH molecules:
$$O*+{H}_{2}O\text{}\to \text{}OH\text{}+OH$$
OH is very reactive. You can think of OH as being water that has had a hydrogen taken away and wants it back. There are other sources for OH, but this one is the most important globally. OH reacts with many other atmospheric constituents. In fact, it is so reactive, that its lifetime in the atmosphere is less than a second.
Another important oxidant is nitric oxide (NO). It comes from combustion (power plants, internal combustion engines, fires) or lightning. In cities, the NO mixing ratio is tens of ppbv during morning rush hour and a bit smaller during evening rush hour, but there is typically about a ppbv around during the day. In very remote areas, the levels of NO are a hundred times less. NO can react with many chemicals, but a particularly important reaction is with O_{3}:
$$NO+{O}_{3}\to \text{}N{O}_{2}+{O}_{2}$$
which forms nitrogen dioxide, NO_{2}. NO_{2} is not very stable:
$$N{O}_{2}+UV\text{}\to \text{}NO+O$$
but the O reacts immediately with O_{2} to form ozone:
$$O+{O}_{2}+{N}_{2}\to \text{}{O}_{3}+{N}_{2}$$
If an NO_{2} molecule is produced, then an O_{3} molecule will be produced during the day when the sun is out. Note that if we think of these three reactions as a cycle, no ozone was either created or destroyed because it is destroyed in [4.5] and created in [4.2].
Methane is a volatile organic compound (VOC). Methane oxidation is a good model for what happens to all of the volatile organic compounds that you smell every day and all the ones that you can’t smell. I am not going to show you the entire reaction sequence. Instead, here are just a few steps.
The first step is the reaction between methane and hydroxyl:
$$C{H}_{4}+\text{}OH\text{}\to \text{}C{H}_{3}+\text{}{H}_{2}O$$
Note that water vapor is made and CH_{3} is a radical because it has 6 + 3 = 9 protons and, therefore, 9 electrons. Just as for most other VOCs and some other trace emissions, the reaction with OH is the main way methane is removed from the atmosphere. Otherwise, it would build up to high abundance.
CH_{3} is very reactive. It combines with O_{2}:
$$C{H}_{3}+\text{}{O}_{2}+\text{}{N}_{2}\to C{H}_{3}{O}_{2}+\text{}{N}_{2}$$
If there is any NO around, the following reaction happens:
$$C{H}_{3}{O}_{2}+\text{}NO\text{}\to \text{}C{H}_{3}O\text{}+\text{}N{O}_{2}$$
followed by:
$$C{H}_{3}O\text{}+\text{}{O}_{2}\to \text{}C{H}_{2}O\text{}+\text{}H{O}_{2}$$
and:
$$H{O}_{2}+\text{}NO\text{}\to \text{}OH\text{}+\text{}N{O}_{2}$$
The chemical CH_{2}O is formaldehyde. Some of you may have encountered it in high school chemistry or biology and so may be familiar with the smell. You also see that we got the OH molecule back.
Ultimately, formaldehyde gets broken down to CO and the net reaction of methane oxidation is:
$$C{H}_{4}+\text{}2\text{}NO\text{}+\text{}3\text{}{O}_{2}\to \to \to \to \text{}C{O}_{2}+\text{}2\text{}{H}_{2}O\text{}+\text{}2\text{}N{O}_{2}$$
Remember that NO_{2} is easily broken apart by the UV sunlight that reaches Earth’s surface, so we can take this reaction sequence a step further and show that in the presence of sunlight, reactions [4.6] and [4.2] give:
$$N{O}_{2}+\text{}UV\text{}+\text{}{O}_{2}\to \text{}NO\text{}+\text{}{O}_{3}$$
or
$$C{H}_{4}+\text{}5{O}_{2}\to \to \to \to \to \text{}C{O}_{2}+\text{}2\text{}{H}_{2}O\text{}+\text{}2\text{}{O}_{3}$$
In this final chemical equation, we do not see OH, HO_{2}, NO, or NO_{2}, yet they are essential to the formation of ozone. They are catalytic, which means that they are neither created nor destroyed in the reaction sequence, but instead are simply recycled between OH and HO_{2} and between NO and NO_{2}. There are other reactions that destroy these reactive chemicals by producing other chemicals that are much less reactive and sticky, a main one being:
$$OH\text{}+\text{}N{O}_{2}+\text{}{N}_{2}\to \text{}HN{O}_{3}+\text{}{N}_{2}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9Gqpi0dc9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeeaaaaaaaaigBHn2Aa8qacaWGpbGaamisaiaabccacqGHRaWkcaqGGaGaamOtaiaad+eapaWaaSbaaSqaa8qacaaIYaaapaqabaGcpeGaey4kaSIaaeiiaiaad6eapaWaaSbaaSqaa8qacaaIYaaapaqabaGcpeGaeyOKH4QaaeiiaiaadIeacaWGobGaam4ta8aadaWgaaWcbaWdbiaaiodaa8aabeaak8qacqGHRaWkcaqGGaGaamOta8aadaWgaaWcbaWdbiaaikdaa8aabeaaaaa@4405@
where HNO_{3} is nitric acid, a very sticky and water-soluble chemical. However, each OH that is produced can typically oxidize more than ten methane molecules before it reacts with NO_{2} to form nitric acid. And as reaction [4.13] shows, each time methane is completely oxidized, two O_{3} molecules are produced. That's a lot of chemical steps to remember, but I don't want you to necessarily remember them. I want you to see that the process started with a reaction of OH with a volatile organic compound (in this case methane) and that in the subsequent reactions, the product molecules had more and more oxygens attached to them. This process is why we say that the atmosphere is an oxidizing environment.
Ozone is a different sort of pollutant from others because it is not directly emitted by a factory or power plant or vehicle but instead is produced by atmospheric chemistry.
Three ingredients are needed to make ozone pollution: volatile organic compounds (VOCs) (like methane); nitric oxide (NO from combustion); and sunlight. When we say this, we assume that we already have some ozone and water to provide the OH to get the reactions started. Every VOC goes through an oxidation process that is similar to the methane oxidation reaction sequence. In the methane oxidation sequence, steps [4.9] and [4.11] make NO_{2}, which in the presence of sunlight makes ozone through step [4.6] followed by step [4.2]. Voila! Ozone is formed from methane oxidation in the presence of nitrogen oxides and sunlight. Now imagine the thousands of volatile organic compounds in the atmosphere and realize that all of them—both anthropogenic and natural—can participate in the production of ozone pollution. Now you have seen the sequence of chemical reactions that produce tropospheric ozone.
Let's look at a video (3:14) entitled "Ground Level Ozone: What Is It?" that explains ozone production without getting into the gory details of the chemistry.
We're all pretty familiar with what O2 is. I hope so. You need to breathe it to live. Yes, O2 is oxygen, that life-giving gas, but what is O3? O3 is another gas essential to our survival but it's definitely not for breathing. O3 is ozone high up in the stratosphere. It's made naturally and absorbs harmful ultraviolet rays from the Sun. Without it life as we know it wouldn't, couldn't exist. We need the ozone layer in the stratosphere. We want it, we rely on it. But don't get too used to singing ozone's praises. High ozone levels at lower altitudes, what we call the troposphere, where we live and breathe or anything but natural and beneficial. In fact, down here it turns out to be a toxic atmospheric pollutants. Yep, you heard me right. ground-level ozone primarily exists due to human activities that burn fossil fuels. Transportation, our power and industrial plants, and other activities expel nitrogen oxides and hydrocarbons. When those compounds interact with sunlight, voila, ozone is created a contributor to smog. that's why I ozone levels increase during the summer months when sunlight is abundant. Yes, smog love summer just like many of us. We run, bike, hike, fish, play, stroll, oh yeah, and breathe. Yes, the fact that more people are outside when it's warmer makes us particularly vulnerable to Ozone's harmful impacts. Ozone is a harmful oxidant when we inhale it it's like getting a sunburn inside your lungs and it can be particularly serious for the young, old, active, and those with respiratory conditions at any age. And it's not just humans that are vulnerable ozone harms plants, crops, and agricultural yield interfering with pretty important processes like well, photosynthesis and even our economy. To make matters worse ozone production increases with higher temperatures which are occurring more frequently with climate change. The EPA sets national ambient air quality standards for several pollutants in the United States including ground level ozone. When a county is out of compliance they need to know what can be done to improve air quality. and let's not forget that air pollution is a global comments. air pollution is shared from surrounding cities states and also country's halfway around the world. What can we do, what are we willing to do to improve current levels? Drive less, carpool, avoid car idling, set your home's thermostat higher in the summer and lower in the winter, avoid gas powered lawn & garden tools on severe ozone days. There's a lot to do and lots to know about air quality knowing more about the sources and contributors to ozone and other atmospheric pollutants will help us chart our course.
Ozone pollution is bad for the health of people, crops, and forests. Ozone can react with some types of VOCs, including types that make up our lungs, and breathing it can cause serious health problems and even death. Ozone reacts with the VOCs that make up plants and stunts their growth and damages their fruit. The Clean Air Act from the 1970s has dramatically decreased the levels of air pollution in the United States, including ozone. The EPA can take the credit for much of the progress against air pollution in the United States. But there is still a ways to go and the progress may be reversed due to effects of climate change. Since ozone pollution increases at higher temperatures, the increases in global temperatures could actually reverse the steady progress in ozone reduction and ozone pollution could once again increase, unless volatile organic compounds and nitrogen oxides are reduced even more.
Now you can see why OH is called the PAC-MAN of the atmosphere. But how can we tell how long it will take for OH to remove from the atmosphere some trace gas like methane? Let’s look at an equation for the budget of methane. It is produced in the atmosphere by all the emissions from cows and wetlands. It is removed from the atmosphere by reactions with OH [4.7]. The rate of removal, that is the change in the methane concentration with time, is always proportional to the amount of the two reactants, in this case, CH_{4} and OH. So, the change in methane with time is given by the balance between methane production and methane loss by reaction with OH:
$$\frac{d[C{H}_{4}]}{dt}=production-{k}_{OH+CH4}[OH][C{H}_{4}]$$
where k_{OH+CH4} is the reaction rate coefficient (units: cm^{3} molecule^{–1} s^{–1}) and [OH] and [CH_{4}] are the concentrations of OH and CH_{4} (units: molecules cm^{–3}). Note that the production is positive and increases CH_{4} with time while the loss is negative and decreases CH_{4} with time.
We use [OH] to indicate the concentration of OH (molecules cm^{–3}), which is quite different from the OH mixing ratio, which is usually expressed in ppt (10^{–12}). Note that 1 ppt ~ 2.4 x 10^{7} molecules cm^{–3} for typical surface conditions. See the video below (1:47) entitled "Rate Equation" for further explanation:
Let me explain equation 4.15, which is a rate equation for methane. A rate equation is just a differential equation. The change of something with respect to time equals the production rate of something, minus the fraction of something that is lost each unit of time, multiplied by the amount of something. Note that the loss rate of something is always proportional to something. That something can be anything. It does not have to be a chemical concentration. It could be the amount of milk in your refrigerator, or the number of socks in your drawer, both of which tend to disappear over time. And equation 4.15 is the methane concentration, which has units of molecules per centimeter cubed. The production rate is in units of molecules per centimeter cubed per second. Remember, each term of the equation must have the same units. The last term is the loss rate. The reaction rate coefficient has units of centimeter cubed per molecule per second, but when we multiply it by the OH concentration, we get a product that has units of per second, which is a frequency. Now, OH varies from almost 0 at night, to a peak value at midday. However, we can take an average OH to find the average loss rate of methane. Note that if we assume that the production rate suddenly goes to 0, then we find a very simple equation, which has an exponential solution. We designate the time that it takes the exponential factor to go to minus 1 as a lifetime, which is just the inverse of a loss frequency.
How can we find out what the lifetime of methane is? We assume that the production suddenly stops and equals 0. Then [4.15] becomes:
$$\begin{array}{l}\frac{d[C{H}_{4}]}{dt}=-{k}_{OH+CH4}[OH][C{H}_{4}]\\ \frac{d[C{H}_{4}]}{[C{H}_{4}]}=-{k}_{OH+CH4}[OH]\text{\hspace{0.17em}}dt\end{array}$$
k_{OH+CH4 }is the reaction rate coefficient for this reaction. Assume that OH is constant. Because OH is generated mostly from sunlight, it follows the sunshine and is greatest near midday and is very small at night. However, we assume that the OH concentration is the average over the day and night in order to assign it a constant value. Now integrate both sides of the equation:
$$\begin{array}{l}{\displaystyle \underset{{[C{H}_{4}]}_{o}}{\overset{[C{H}_{4}]}{\int}}\frac{d[C{H}_{4}]}{[C{H}_{4}]}}={\displaystyle \underset{0}{\overset{t}{\int}}-{k}_{OH+C{H}_{4}}[OH]\text{\hspace{0.17em}}dt}\\ \mathrm{ln}\left([C{H}_{4}]\right)-\mathrm{ln}\left({[C{H}_{4}]}_{0}\right)=-{k}_{OH+C{H}_{4}}\overline{[OH]}\text{\hspace{0.17em}}t\\ \mathrm{ln}\left(\frac{[C{H}_{4}]}{{[C{H}_{4}]}_{0}}\right)=-{k}_{OH+C{H}_{4}}\overline{[OH]}\text{\hspace{0.17em}}t\\ \text{takeexponentialofbothsides}\\ \frac{[C{H}_{4}]}{{[C{H}_{4}]}_{0}}={e}^{-{k}_{OH+C{H}_{4}}\overline{[OH]}\text{\hspace{0.17em}}t}\\ [C{H}_{4}]={\left[C{H}_{4}\right]}_{0}{e}^{-{k}_{OH+C{H}_{4}}\overline{[OH]}t}\end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOabaeqabaWaa8qCaeaadaWcaaqaaiaadsgacaGGBbGaam4qaiaadIeadaWgaaWcbaGaaGinaaqabaGccaGGDbaabaGaai4waiaadoeacaWGibWaaSbaaSqaaiaaisdaaeqaaOGaaiyxaaaaaSqaaiaacUfacaWGdbGaamisamaaBaaameaacaaI0aaabeaaliaac2fadaWgaaadbaGaam4BaaqabaaaleaacaGGBbGaam4qaiaadIeadaWgaaadbaGaaGinaaqabaWccaGGDbaaniabgUIiYdGccqGH9aqpdaWdXbqaaiabgkHiTiaadUgadaWgaaWcbaGaam4taiaadIeacqGHRaWkcaWGdbGaamisamaaBaaameaacaaI0aaabeaaaSqabaGccaGGBbGaam4taiaadIeacaGGDbGaaGjbVlaadsgacaWG0baaleaacaaIWaaabaGaamiDaaqdcqGHRiI8aaGcbaGaciiBaiaac6gadaqadaqaaiaacUfacaWGdbGaamisamaaBaaaleaacaaI0aaabeaakiaac2faaiaawIcacaGLPaaacqGHsislciGGSbGaaiOBamaabmaabaGaai4waiaadoeacaWGibWaaSbaaSqaaiaaisdaaeqaaOGaaiyxamaaBaaaleaacaaIWaaabeaaaOGaayjkaiaawMcaaiabg2da9iabgkHiTiaadUgadaWgaaWcbaGaam4taiaadIeacqGHRaWkcaWGdbGaamisamaaBaaameaacaaI0aaabeaaaSqabaGcdaqdaaqaaiaacUfacaWGpbGaamisaiaac2faaaGaaGjbVlaadshaaeaaciGGSbGaaiOBamaabmaabaWaaSaaaeaacaGGBbGaam4qaiaadIeadaWgaaWcbaGaaGinaaqabaGccaGGDbaabaGaai4waiaadoeacaWGibWaaSbaaSqaaiaaisdaaeqaaOGaaiyxamaaBaaaleaacaaIWaaabeaaaaaakiaawIcacaGLPaaacqGH9aqpcqGHsislcaWGRbWaaSbaaSqaaiaad+eacaWGibGaey4kaSIaam4qaiaadIeadaWgaaadbaGaaGinaaqabaaaleqaaOWaa0aaaeaacaGGBbGaam4taiaadIeacaGGDbaaaiaaysW7caWG0baabaGaaeiDaiaabggacaqGRbGaaeyzaiaabccacaqGLbGaaeiEaiaabchacaqGVbGaaeOBaiaabwgacaqGUbGaaeiDaiaabMgacaqGHbGaaeiBaiaabccacaqGVbGaaeOzaiaabccacaqGIbGaae4BaiaabshacaqGObGaaeiiaiaabohacaqGPbGaaeizaiaabwgacaqGZbaabaWaaSaaaeaacaGGBbGaam4qaiaadIeadaWgaaWcbaGaaGinaaqabaGccaGGDbaabaGaai4waiaadoeacaWGibWaaSbaaSqaaiaaisdaaeqaaOGaaiyxamaaBaaaleaacaaIWaaabeaaaaGccqGH9aqpcaWGLbWaaWbaaSqabeaacqGHsislcaWGRbWaaSbaaWqaaiaad+eacaWGibGaey4kaSIaam4qaiaadIeadaWgaaqaaiaaisdaaeqaaaqabaWcdaqdaaqaaiaacUfacaWGpbGaamisaiaac2faaaGaaGjbVlaadshaaaaakeaacaGGBbGaam4qaiaadIeadaWgaaWcbaGaaGinaaqabaGccaGGDbGaeyypa0ZaamWaaeaacaWGdbGaamisamaaBaaaleaacaaI0aaabeaaaOGaay5waiaaw2faamaaBaaaleaacaaIWaaabeaakiaadwgadaahaaWcbeqaaiabgkHiTiaadUgadaWgaaadbaGaam4taiaadIeacqGHRaWkcaWGdbGaamisamaaBaaabaGaaGinaaqabaaabeaalmaanaaabaGaai4waiaad+eacaWGibGaaiyxaaaacaWG0baaaaaaaa@E0FD@
So we see that methane decreases exponentially with time.
The lifetime of an atmospheric constituent is defined as the time it takes that constituent to decrease to e^{–1} = 0.37 of its initial value once production stops. So the lifetime of methane in the atmosphere is the value of t when k_{OH+CH4}[OH]t = 1, or:
$$\tau =\frac{1}{{k}_{OH+CH4}[OH]}$$
where τ indicates the lifetime. k_{OH+CH4} = 3 x 10^{–15} cm^{3} molecule^{–1} s^{–1} and [OH] ~ 10^{6} molecules cm^{–3}, so:
$$\tau =\frac{1}{3x{10}^{-15}\text{\hspace{0.17em}}{10}^{6}}=3{\mathrm{x10}}^{8}\text{\hspace{0.17em}}\text{seconds~10years}$$
This reaction rate coefficient is fairly low. Other VOCs have reaction rate coefficients that are typically hundreds to hundreds of thousands of times greater, so the lifetime of most VOCs is hours to days.
The atmospheric lifetime of a gas is very important for determining how far a gas can travel from its source. Some trace gases have lifetimes of hours, so unless they are made by atmospheric chemistry, they can't travel more than a few tens of kilometers from their sources. Other gases have much longer lifetimes; methane is a good example with its 10-year lifetime. In 10 years, it can travel from its sources to most anywhere around the globe, even to the stratosphere. NASA measures the amounts of several gases from space. An excellent NASA website for accessing these satellite data and having it plotted as global maps is the Center for Trace Gas Data & Information Website at the NASA Goddard Space Flight Center's Earth Sciences Distributed Active Archive Center (GES DISC) [43].
This concept of atmospheric lifetime is very important. For instance, what if an industry is spewing a chemical into the atmosphere that is toxic at a certain concentration in the atmosphere? Then it is important to know if that chemical is removed in less time than it takes to become toxic or if it is going to continue to build up at toxic levels and not leave the atmosphere for a long, long time. If the chemical's atmospheric lifetime is hundreds to thousands of years, then maybe we shouldn’t let that industry dump that chemical into the air.
Now it is your turn to solve some problems concerning the atmospheric lifetimes and rates of change for atmospheric constituents in a quiz.
Now that you've learned about the atmosphere’s gas-phase composition, it is time to look at its particle composition. We are interested in atmospheric particles for several reasons:
Atmospheric aerosol is most obvious to us on warm and muggy summer days. Under these conditions, there are lots of aerosol particles and they absorb water and swell up to a size that is quite efficient at scattering sunlight. The following picture was taken over Maryland on a flight between Washington Dulles airport and State College airport. Above the fair-weather cumulus clouds is blue sky in the free troposphere. Below the clouds is the atmospheric boundary layer, which is filled with aerosol that has been well-mixed by warm, moist air parcels rising and stirring the boundary layer air. The haze is so thick that it is a little hard to see the ground.
Atmospheric particles come from many different sources. Good cloud condensation nuclei (CCN) must be small particles, so that they do not settle too fast, and must be hydrophilic, which means that water can stick. They can be either soluble (i.e., dissolvable in water), or insoluble, but most are soluble.
Most particles originate from emissions from Earth’s surface. Primary aerosols are emitted directly from the source, although the smaller ones start off as hot gases that rapidly condense to form particles even before they leave the smokestack or tailpipe. Secondary aerosols are gaseous emissions that are converted to aerosol particles by chemical reactions in the atmosphere. Some of these become CCN. This process is often called gas-to-particle conversion. Most CCN are secondary aerosols.
The sources of aerosols are both natural and anthropogenic (human-made). Seaspray, volcanoes, forests, and forest fires, as well as gas-to-particle conversion of naturally occurring gases such as sulfur dioxide (SO_{2}) and some naturally occurring VOCs, such as α-pinene (which gives the pine smell) are important natural particle sources. Industry, power plants, using fires to clear cropland, transportation, and gas-to-particle conversion of anthropogenic SO_{2} and numerous other gas emissions are important anthropogenic particle sources.
Note that we must pay attention not only to the aerosol sources but also the aerosol sinks, as shown in the diagram below.
The different sources make particles of different sizes. The typical size distribution (i.e., number of particles in a volume of air, plotted as a function of size) has bumps in it, with more particles at some sizes than at others, as seen in the diagram below. Reading these bumps tells us a lot about how the particles were made.
The nucleation mode (there are other designations for this) includes particles that are made by gas-to-particle conversion. A low-volatility vapor is one that will condense onto particles or other surfaces when its vapor pressure exceeds its low saturation vapor pressure. This situation is analogous to water.
The coarse mode includes particles made by mechanical processes. The hydrophilic coarse particles can be CCN, but they settle out pretty fast.
The accumulation mode includes particles that are usually made when nucleation particles collide and stick (called coagulation) or when gases accumulate on a nucleation mode particle. The accumulation mode particles neither settle fast nor coagulate, so they tend to hang around in the atmosphere for a few weeks. They make pretty good CCN.
PM2.5 is a particle size designation that means “Particle matter smaller than 2.5 µm in diameter.” Another common term is PM10, which is "particle matter smaller than 10 µm in diameter." PM2.5 particles are the ones that are most important for human health and climate, and, in many cases, cloud formation, because of their longer lifetime in the atmosphere.
Secondary particles start with the emission of VOCs or sulfur compounds, which react mainly with OH to start a sequence of reactions. These reactions tend to add oxygen to the molecules, which chemically makes them stickier (with a lower saturation vapor pressure) and more water soluble, which is just what is needed to make them better cloud condensation nuclei.
For particles that start as gaseous sulfur compounds, such as sulfur dioxide (SO_{2}), the reaction sequence starts with OH and the reaction product is sulfuric acid, a compound that has a very low vapor pressure and is very sticky:
$$S{O}_{2}+\text{}OH\text{}\to \to \text{}{H}_{2}S{O}_{4}$$
Sulfuric acid is easily taken up into cloud drops and raindrops and then can be deposited on Earth’s surface when it rains. The good news is that the rain cleans the atmosphere. The bad news is that the rain is very acidic and has earned the name “acid rain” because of its harmful effects on forests and on buildings, memorials, and statues.
If sulfur sources are upwind of an area, the particles in that area will contain some sulfur. But almost all atmospheric particles also contain some organic compounds and sometimes particles are mainly made up of carbon-containing organic compounds. Some of these organic particles are primary, but most of the small ones are made by gas-to-particle conversion, which is just a simple way to say the volatile organic compounds react in the atmosphere with OH or O_{3} to form less volatile organic compounds that become aerosol particles. The chemicals in these particles can continue to oxidize, thus making them even better CCN.
We can demonstrate gas-to-particle conversion of a VOC that is often emitted into the atmosphere by trees. This compound is limonene and also comes from oranges. In the video (4:47) below entitled "Demonstration of Gas-to-Particle Conversion," I will use orange peel to demonstrate this effect.
Atmospheric composition, even of trace gases, has a huge influence on weather and climate. Carbon dioxide is the most abundant trace gas – its mixing ratio is 400 ppm and growing, but other trace gases are also emitted into the atmosphere. The atmosphere cleans itself of these gases by atmospheric chemistry, which oxidizes the gas emissions and produces new chemicals that contain oxygen and so are stickier and more water soluble.
These new chemicals can be removed from the atmosphere either by hitting surfaces and sticking or by being taken up in clouds or rain drops and precipitated to the ground. The main oxidant is hydroxyl (OH), which is made with ozone, UV sunlight, and water vapor and starts the removal sequence by reacting with gas emissions. In these reaction sequences ozone pollution is produced if the pollutant nitric oxide (NO) is also present.
This pollutant ozone is nearby and is harmful to human health and agriculture. Stratospheric ozone, on the other hand, shields Earth from harmful UV, and is made a completely different way – by the breaking apart of O_{2} to produce O, which reacts readily with O_{2} to form O_{3}. Some of this stratospheric ozone is then transported to Earth, but at levels much lower than pollutant levels. Methane oxidation is an example of the VOC reactions that produce ozone and particles.
An important concept is the atmospheric lifetime of gases and particles. This can be determined by solving a simple linear differential equation. The methane lifetime was shown to be about 10 years.
Particles have many natural and anthropogenic sources; some are emitted directly from the sources (primary particles) and some are produced by atmospheric chemistry (secondary particles). Particles affect human health, visibility, scattering and absorption of light, and are essential for cloud formation, as will be seen in the next lesson on cloud physics.
You have reached the end of Lesson 4! Double-check that you have completed all of the activities before you begin Lesson 5.
Clouds and precipitation are integral to weather and can be difficult to forecast accurately. Clouds come in different sizes and shapes that depend on atmospheric motions, their composition, which can be liquid water, ice, or both, and the temperature. While clouds and precipitation are being formed and dissipated over half the globe at any time, their behavior is driven by processes that are occurring on the microscale, where water molecules and small particles collide. We call these microscale processes “cloud microphysics” and microphysics is the focus of this lesson. Three ingredients are required for the formation of clouds: moisture, aerosol, and cooling. If any one of these is missing, a cloud will not form. Over eighty years ago, a simple hypothesis was developed to explain the formation of clouds. This hypothesis has been thoroughly tested and validated and is now called Koehler Theory. We will learn the elements of Koehler Theory and how to use them to determine when clouds will form and when they will not, becoming only haze. Clouds do not automatically precipitate. In fact, most clouds do not. We will learn about the magic required for precipitation to form. Thus, cloud formation through precipitation is a series of microsteps, each of which is necessary, but not sufficient, to achieve precipitation.
By the end of this lesson, you should be able to:
If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.
Clouds have fascinated people for millennia, but it wasn’t until 1802 that Luke Howard first classified clouds with the terms that are used today. His classification scheme was formalized later in the 19^{th} century and has 10 basic cloud types with many minor variations (see figure below).
This website contains a nice overview of the cloud types [45] with descriptions and accompanying images. Check out some of their amazing photos!
NOAA and NASA put together this thorough Sky Watcher Chart [46] that describes a wide variety of cloud formations.
Cloud physics goes beyond the classification of clouds to determine the actual physical and chemical mechanisms that create clouds and cause their evolution over time. There are two aspects of cloud physics. One is the physics on the cloud scale, which is tens to hundreds of meters in size. This physics is driven in part by behavior in the cloud’s environment, such as the wind shear or the location of a front, and determines the evolution of the cloud and the cloud’s size and shape. All of this action, however, is not possible without the physics that is occurring on the microscale, which is less than a few centimeters in size.
This lesson deals mostly with the physics that occurs on the microscale and is often called cloud microphysics. Now that you are familiar with the concepts of thermodynamics and water vapor, we are ready to look at the fundamentals of cloud microphysics. To understand cloud-scale physics will require an understanding of atmospheric dynamics and turbulence, which are introduced in later lessons of this course.
A cloud is defined as a (visible) suspension of small particles in the atmosphere. For a water cloud, there are a number of types of particles that we are interested in.
Note the wide range in size, volume, and number of particles in the figure above. The smallest, the cloud condensation nuclei (CCN), can have rather little water and are made up of substances to which water can attach (called hydrophilic, water loving). The other particles grow by adding water molecules but still contain the original CCN upon which they formed.
We can specify the amount of water that is in liquid form by using the liquid water content (LWC), which is defined as:
$$LWC={\omega}_{L}=\frac{mass\text{\hspace{0.17em}}of\text{\hspace{0.17em}}liquid\text{\hspace{0.17em}}water}{volume\text{\hspace{0.17em}}of\text{\hspace{0.17em}}air},\text{\hspace{1em}}units=g\text{\hspace{0.17em}}{m}^{-3}$$
Typical values of LWC are 0.1–0.9 g m^{–3}, but a few g m^{–3} are possible for wetter conditions.
A cloud drop is typically 5 µm in radius, while a raindrop, which comes from a collection of cloud drops, is typically 0.5 mm (500 µm) in radius. How many cloud drops does it take to make a raindrop?
ANSWER: Find the volume of the cloud drop and the volume of the raindrop and then find out how many times bigger the raindrop is. The answer is the number of cloud drops it takes to make a raindrop.
$$\begin{array}{l}{n}_{cloud}{V}_{cloud}={V}_{rain}\\ {n}_{cloud}=\frac{{V}_{rain}}{{V}_{cloud}}=\frac{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\pi {\left({r}_{rain}\right)}^{3}}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\pi {\left({r}_{cloud}\right)}^{3}}={\left(\frac{{r}_{rain}}{{r}_{cloud}}\right)}^{3}={\left(\frac{500}{5}\right)}^{3}={10}^{6}\end{array}$$
So we see that it takes about a million cloud drops to make one raindrop. Thus 10^{9} cloud drops per m^{3} of cloud should make about 10^{3} raindrops per m^{3} of cloud. This is about the number per m^{3} that are observed.
Ice crystal habits as a function of temperature and excess water vapor (i.e., water vapor greater than saturation water vapor).
The next time it snows, catch snowflakes on a cold surface and take a good look at them. Their shape will tell you a lot about the environment in which they were formed. In State College, Pennsylvania, we often see plates with broad branches and sometimes we see dendrites, telling us that the snowflakes were formed at altitudes in the cloud where the temperature was between –22 ^{o}C and –8 ^{o}C and the excess vapor density was large.
The following video (3:52) entitled "Snowflake Safari" gives a simple explanation of snowflake formation and shows some nice pictures of different snowflake shapes.
It's time to look up at the sky to observe the clouds. During the next week, take pictures of clouds and identify the clouds in the pictures. Try to focus on just one cloud type per image. Submit an image that depicts at least one cloud type.
This discussion will be worth 3 discussion points. I will use the following rubric to grade your participation:
Evaluation | Explanation | Available Points |
---|---|---|
Not Completed | Student did not complete the assignment by the due date. | 0 |
Student completed the activity with adequate thoroughness. | Student answers the discussion question in a thoughtful manner, including some integration of course material. | 1 |
Student completed the activity with additional attention to defending their position. | Student thoroughly answers the discussion question and backs up reasoning with references to course content as well as outside sources. | 2 |
Student completed a well-defended presentation of their position, and provided thoughtful analysis of at least one other student’s post. | In addition to a well-crafted and defended post, the student has also engaged in thoughtful analysis/commentary on at least one other student’s post as well. | 3 |
There are three requirements for forming a cloud drop:
If any one of these three is missing, a cloud cannot form. We have talked about moisture and aerosol and now need to consider ways that the air can be cooled. The air needs to be cooled so that the water vapor pressure initially equals and then exceeds the saturation water vapor pressure.
An easy way to remember these key ingredients is to think of a Big MAC.
Saturation occurs when e = e_{s}, w = w_{s}, and condensation = evaporation. At saturation, RH = e/e_{s} ~ w/w_{s} = 1, or in terms of percent, 100%. When we find the lifting condensation level (LCL) on a skew-T, we are finding the pressure level at which T (as determined from the dry adiabat) = T_{d} (as determined from the constant water vapor mixing ratio), or when w = w_{s}.
Two variables that are useful in discussing the cloud drop formation are the saturation ratio and the supersaturation. In Lesson 3, we introduced the saturation ratio:
$$S=\raisebox{1ex}{$e$}\!\left/ \!\raisebox{-1ex}{${e}_{s}$}\right.$$
where e is the water vapor pressure and e_{s} is the saturation vapor pressure. S < 1 for a subsaturated environment, S = 1 for a saturated environment (condensation = evaporation), and S > 1 for a supersaturated environment. A second useful variable is the supersaturation:
$$s=S-1=\raisebox{1ex}{$e$}\!\left/ \!\raisebox{-1ex}{${e}_{s}$}\right.-1$$
s < 0 for a subsaturated environment, s = 0 for a saturated environment, and s > 0 for a supersaturated environment. Note that s and S are both unitless.
The above equations apply only for a flat surface of pure liquid water. When we get into situations where the water has a curved surface (as in a cloud drop), contains a solute, or is in solid form, we need to think about the saturation ratio and the supersaturation relative to the equilibrium value of e, e_{eq}, which can be different from e_{s}. So, depending on the circumstances, e_{eq} can be e_{s} (flat liquid water), e_{sc }(curved liquid water), e_{sol} (curved solution), e_{i} (flat ice), or some combination. We will see that a small supersaturation is actually needed to form clouds.
The relative humidity is 85%. What is the saturation ratio? What is the supersaturation?
ANSWER: S = 0.85 and s = 0.85 – 1 = –0.15
The relative humidity is 102%. What is the saturation ratio? What is the supersaturation?
ANSWER: S = 1.02 and s = 0.02
Note that it is possible to have the relative humidity be greater than 100%, which makes the supersaturation positive. This condition can't last long because condensation will exceed evaporation until they become equal. But how can supersaturation happen?
Three basic mechanisms for cooling the air are RUM: Radiation, Uplift, and Mixing.
Radiation and mixing happen at constant pressure (isobaric); uplift happens at constant energy (adiabatic). Let’s consider these three cases in more detail. A good way to show what is happening is to use the water phase diagram. The video (3:15) entitled "Supersaturation Processes 2" below will explain these three processes in greater detail:
Clouds will not form unless the air becomes supersaturated, meaning that it's relative humidity is slightly greater than 100%. Or put it another way, it's supersaturation is greater than 0%. Let's look at the three ways that supersaturation can be achieved, radiative cooling, mixing, and adiabatic ascent. We can use the water phase diagram of water vapor on the y-axis versus temperature on the x-axis to examine these processes. Supersaturation means that the environment moves from the all-vapor part of the phase diagram into the all-liquid part by crossing the equilibrium line, which is given by the Clausius Clapeyron equation. I will mention only the essentials for each process, what changes and what stays the same. For radiative cooling, the water vapor pressure stays the same, but the temperature drops. And because the saturation vapor pressure depends only on temperature, the saturation vapor pressure also drops. The saturation vapor pressure decreases until it gets equal to and then a little less than the vapor pressure. And then the supersaturation above 0. The next process is mixing. Mixing clouds usually form when unsaturated, warm, moist air from a source is mixed into the unsaturated, colder, drier environmental air. As the warm, moist air mixes with the colder, drier air, the temperature and vapor pressure of the moist air parcel becomes the average of the temperature and vapor pressure of the moist, warm air parcel multiplied by the number of moles and the temperature and vapor pressure of the cold, dry environmental air multiplied by the number of moles, all this divided by the total number of moles. As the air parcel mixes with more environmental air, the parcel's temperature and vapor pressure move along the mixing line between the two initial air parcel states. If this line crosses the equilibrium line and goes into the liquid part of the phase diagram, supersaturation becomes greater than 0 and the cloud forms. If the air parcel continues to entrain the dry air, continues along the mixing line, and it may eventually cross the equilibrium line back into the vapor region, ant the cloud will evaporate. Contrails are one example of a mixing cloud. The contrail length tells you something about what the temperature and environmental pressure of the environmental air must be. The third process is adiabatic ascent. As an air parcel ascends, it's pressure and temperature drop. Because the water vapor mixing ratio is constant until a cloud forms, the drop in the pressure means a drop in the water vapor pressure. At the same time, a drop in the temperature means a drop in the saturation vapor pressure, which depends only on temperature. So vapor pressure and saturation vapor pressure are both dropping. However, in adiabatic ascent, the saturation vapor pressure drops faster than the vapor pressure, and eventually, they become equal. And then supersaturation becomes greater than 0, and the cloud forms.
All matter radiates energy as electromagnetic waves, as we will see in the next lesson. When an air parcel radiates this energy (mostly in the infrared part of the spectrum), it cools down, but the amount of water vapor does not change.
We can understand this process by using the water phase diagram in the figure below. Consider a situation in which an air parcel is undersaturated, which is represented by the blue dot. As the air parcel emits radiation, the air parcel cools but does not change its vapor pressure. Hence, the blue dot moves to the left on the diagram. However, because the temperature drops, e_{s} drops. When e_{s} becomes slightly less than e, a cloud forms.
An example of radiative cooling in action is radiation fog, which occurs overnight when Earth's surface and the air near it cool until a fog forms (see figure below).
Assume two air parcels with different temperatures and water vapor partial pressures are at the same total pressure. If these two parcels mix, then the temperature and the water vapor partial pressure are going to be weighted averages of the T and e, respectively, of the two parcels. The weighting is determined by the fraction of moles that each parcel contributes to the mixed parcel. Mathematically, for parcel 1 with e_{1}, T_{1}, and N_{1} (number of moles) and parcel 2 with e_{2}, T_{2}, and N_{2}, the e and T of the mixed parcel are given by the equations:
$$\begin{array}{l}e=\frac{{N}_{1}}{{N}_{1}+{N}_{2}}{e}_{1}+\frac{{N}_{2}}{{N}_{1}+{N}_{2}}{e}_{2}\text{\hspace{1em}}\text{\hspace{1em}}T=\frac{{N}_{1}}{{N}_{1}+{N}_{2}}{T}_{1}+\frac{{N}_{2}}{{N}_{1}+{N}_{2}}{T}_{2}\\ or\text{\hspace{0.17em}}approximately\\ e=\frac{{M}_{1}}{{M}_{1}+{M}_{2}}{e}_{1}+\frac{{M}_{2}}{{M}_{1}+{M}_{2}}{e}_{2}\text{\hspace{1em}}\text{\hspace{1em}}T=\frac{{M}_{1}}{{M}_{1}+{M}_{2}}{T}_{1}+\frac{{M}_{2}}{{M}_{1}+{M}_{2}}{T}_{2}\end{array}$$
where M_{1} and M_{2} are the masses of the air parcels. On the phase diagram, these give straight lines for different proportions of the mixed parcel being from parcel 1 (0% to 100%) and parcel 2 (100% to 0%), as in the figure below.
Note that both of these two air parcels are unsaturated. So how does a cloud form? Think about a single warm, moist air parcel mixing into the environment of colder, drier air. As the parcel mixes into more and more of the environmental air, it gets increasingly diluted but the mixed parcel grows in size. As the amount of environmental air in the mixed parcel increases, the average e and T of the mixed parcel decreases to be closer to the environmental values and the mixed parcel’s e and T follow a mixing line. Starting in the upper right near the initial warm parcel, as the mixed parcel continues to grow, eventually the e and T will hit the Clausius–Clapeyron curve. As the mixed parcel continues to push into the liquid portion of the phase diagram and become supersaturated, a mixing cloud will form. The cloud will stay as long as the mixed e and T put the parcel above the Clausius–Clapeyron curve. However, once the mixed parcel comes below the curve, the cloud will evaporate.
Suppose air Parcel 1 has e = 20 hPa, T = 40 ^{o}C, and N = 40,000 moles; and Parcel 2 has e = 5 hPa, T = 10 ^{o}C, and N = 80,000 moles. Then using equation 5.4 (top):
$$\begin{array}{l}e=\frac{40,{000}_{\mathrm{}}}{40,000+80,000}20+\frac{80,000}{40,000+80,000}5\text{\hspace{1em}}=10\text{hPa}\text{\hspace{1em}}\\ T=\frac{40,{000}_{\mathrm{}}}{40,000+80,000}40+\frac{80,000}{40,000+80,000}10={20}^{o}C\end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@849B@
There are many good examples of mixing clouds. One is a jet contrail; a second is your breath on a cold day; a third is a fog that forms when cold, dry air moves over warm, moist ground, say just after rain.
The uplift of air can lead to cloud formation, as we know from the skew-T. Uplift is generally the same as adiabatic ascent. This adiabatic ascent can be driven by convection, by a less dense air mass overriding a more dense one, or by air flowing up and over a mountain. The following happens:
The question is “Does e or e_{s} drop faster?" It turns out that e_{s} drops faster. As a result, in uplifted air, e and e_{s} converge at the lifting condensation level (LCL) and a cloud forms just at that level (see figure below).
The arrow on the figure above shows the changes in e and T (and thus e_{s}) as an air parcel rises. Once e_{s} <= e, then s > 0 and the air parcel is supersaturated. This supersaturated situation is not stable; the water vapor in excess of e_{s} forms liquid. As the uplift continues, more water vapor is converted into liquid water and the vapor pressure remains close to e_{s}. All convective clouds, that is clouds with vertical extent, form this way. An example of adiabatic uplift is a cumulus cloud, as seen in the figure below.
I thought that cloud drops formed when w = w_{s}. Why is supersaturation required for a cloud drop to form?
To answer this question, we need to look through a microscope at the nanometer scale, which is the scale of molecules and small particles. You all know that cloud condensation nuclei are needed for clouds to form, but do you know why? Watch the following video (3:16) entitled "Glory: The Cloud Makers."
[music playing] NARRATOR: Aerosols are suspended throughout Earth's atmosphere, and the tiny, varied particles play a mysterious role in human induced climate change. Just like people, every aerosol particle is unique. Sometimes aerosols occur naturally, from things like volcanoes, but they can also originate from human activity. Aerosols are short-lived, but have an active lifetime! In just a short expanse of time, particles can change their size and composition and even travel across vast oceans. Aerosols are difficult to study, and one important new area of research involves how these particles impact clouds. Without aerosols, clouds could not exist. MICHAEL MISHCHENKO: An aerosol particle can serve as a cloud condensation nucleus. NARRATOR: The introduction of too many aerosols will modify a cloud's natural properties. MICHAEL MISHCHENKO: The more aerosol particles we have in the atmosphere, the more cloud droplets we can have. NARRATOR: Clouds play an important role in regulating Earth's climate; aerosol-rich clouds become bigger, brighter, and longer lasting. Aerosols impact clouds in other ways. Some aerosol particles primarily reflect solar radiation and cool the atmosphere, and others absorb radiation, which warms the air. When aerosols heat the atmosphere, they create an environment where clouds can't thrive. The suppression of clouds leads to further warming of the atmosphere by solar radiation. Researchers are still working to understand the role of these curious particles. MICHAEL MISHCHENKO: We need to study the distribution of particles globally, and the only way to do that is from satellites. NARRATOR: New tools will soon help scientists study aerosols. The Aerosol Polarimetry Sensor, or APS, is among a suite of instruments onboard NASA's upcoming Glory mission. The APS will provide a global dataset of aerosol distribution with unprecedented accuracy and specificity. Unique data from the Glory mission, along with NASA's fleet of Earth observing satellites, will help researchers investigate the intricacies of Earth's changing climate. [music playing] [wind blowing]
In the atmosphere, relative humidity rarely rises much above 100% because small aerosol particles act as Cloud Condensation Nuclei (CCN). Two effects most strongly determine the amount of supersaturation that each particle must experience in order to accumulate enough water to grow into a cloud drop. The first is a physical effect of curvature on increasing the equilibrium water vapor pressure; the second is a chemical effect of the aerosol dissolving in the growing water drop and reducing its vapor equilibrium pressure. You will learn about these two effects in the next two sections of this lesson.
Let’s look at curvature effect first (see figure below). Consider the forces that are holding a water drop together for a flat and a curved surface. The forces on the hydrogen bonding in the liquid give a net inward attractive force to the molecules on the boundary between the liquid and the vapor. The net inward force, divided by the distance along the surface, is called surface tension, σ. Its units are N/m or J/m^{2}.
If the surface is curved, then the amount of bonding that can go on between any one water molecule on the surface and its neighbors is reduced. As a result, there is a greater probability that any one water molecule can escape from the liquid and enter the vapor phase. Thus, the evaporation rate increases. The greater the curvature, the greater the chance that the surface water molecules can escape. Thus, it takes less energy to remove a molecule from a curved surface than it does from a flat surface.
When we work through the math, we arrive at the Kelvin Equation:
$${e}_{sc}\left(T\right)={e}_{s}\left(T\right)\cdot \mathrm{exp}\left(\frac{2\sigma}{{n}_{L}\cdot R*\cdot T\cdot {r}_{d}}\right)$$
where e_{sc} is the equilibrium vapor pressure over a curved surface of pure water, e_{s} is the equilibrium vapor pressure over a flat surface of pure water, σ is the water surface tension, n_{L} is the number of moles of liquid water unit per unit volume, R* is the universal gas constant, and r_{d} is the radius of the drop. Note that e_{s} is a function of temperature while e_{sc} is a function of temperature and drop radius. Because σ and n_{L} are relatively weak functions of temperature and R* is constant, it is useful to combine them as follows:_{ }
$$\left(\frac{2\sigma}{{n}_{L}\cdot R*}\right)=3.3x{10}^{-7}\text{\hspace{0.17em}}\mathrm{m\; K}\text{\hspace{0.17em}}{}^{\mathrm{}}$$
where we have used 0 ^{o}C values for σ (0.0756 J m^{–2}) and n_{L} (5.55 x 10^{4} mol m^{–3}).
Since the evaporation is greater over a curved surface than over a flat surface, at equilibrium the condensation must also be greater over a curved surface than over a flat surface in order to keep condensation equal to evaporation, which is required for saturation (i.e., equilibrium). Thus, the saturation vapor pressure over a curved surface is greater than the saturation vapor pressure over a flat surface of pure water.
When we plot this equation, we get the following figure:
Note the rapid increase in equilibrium vapor pressure for particles that have radii less than 10 nm. Of course, all small clusters of water vapor and CCN start out at this small size and grow by adding water.
The Kelvin Equation can be approximated by expanding the exponential into a series:
$${e}_{sc}(T)={e}_{s}(T)\cdot \left(1+\frac{2\sigma}{{n}_{L}\cdot R*\cdot T\cdot {r}_{d}}\right)={e}_{s}(T)\cdot \left(1+\frac{{a}_{K}}{{r}_{d}}\right),\text{\hspace{1em}}where\text{\hspace{0.17em}}{a}_{K}=\frac{2\sigma}{{n}_{L}\cdot R*\cdot T}$$
Cloud drops start as nanometer-size spherical drops, but the vapor pressure required for them to form is much greater than e_{s} until they get closer to 10 nm in size. The Kelvin effect is important only for tiny drops; it is important because all drops start out as tiny drops and must go through that stage. As drops gets bigger, their radius increases and e_{sc} approaches e_{s}.
So, is it possible to form a cloud drop out of pure water? This process is called homogeneous nucleation. The only way for this to happen is for two molecules to stick together, then add another, then another, etc. But the radius of the nucleating drop is so small that the vapor pressure must be very large. It turns out that drops probably can nucleate at a reasonable rate when the relative humidity is about 440%. Have you ever heard of such a high relative humidity?
So, the lesson here is that homogeneous nucleation is very unlikely because of the Kelvin effect.
On the other hand, the atmosphere is not very clean either. There are all kinds of dirt and other particles in the atmosphere. Some of these are hydrophilic (i.e., they like water) and water-soluble (i.e., they dissolve in water). So let’s see what the effect of soluble CCN might be on the water evaporation rate for a flat water surface. We’ll then put the curvature and the solute effects together.
First, here are some important definitions:
Solvent: The chemical that another chemical is being dissolved into. For us, the solvent is H_{2}O.
Solute: The chemical that is being dissolved in the solvent.
The simple view of this effect is that solute molecules are evenly distributed in the water (solvent) and, therefore, that some solute molecules occupy surface sites that would otherwise be occupied by water molecules. Thus, the solute prevents water molecules from evaporating from those sites. Adding more solute means that more surface sites would be occupied by solute molecules and water vapor would have even less opportunity to break hydrogen bonds and escape the liquid. The real view is more complicated by the electrostatic interactions between water and solute molecules that cause an attraction between water and solute molecules, but the basic result is the same as the simple view.
Because the evaporation rate is lowered, that means that there will be net condensation until the water vapor flux to the surface matches the water vapor flux leaving the surface. When equilibrium is established between the lower evaporation and condensation, the condensation will be less, which means that the saturation vapor pressure will be lower. The equilibrium vapor pressure is less than e_{s}, which, remember, is the saturation vapor pressure over a flat surface of pure water. As the amount of solute is increased, the equilibrium vapor pressure of the solution will decrease further.
We can quantify this equilibrium vapor pressure over a solution with a few simple equations.
The mole fraction is defined as:
$${\chi}_{s}=\frac{{n}_{s}}{{n}_{w}+{n}_{s}}\approx \frac{{n}_{s}}{{n}_{L}},\text{\hspace{1em}}if\text{\hspace{0.17em}}{n}_{w}>{n}_{s}$$
Raoult’s Law relates the equilibrium vapor pressure of the solution (e_{sol}) to that of pure water and to the mole fraction:
We can approximate Raoult’s Law for a reasonably dilute solution by writing:
$${\chi}_{s}\approx \frac{{n}_{s}}{{n}_{L}}=\frac{i{N}_{s}}{{n}_{L}{V}_{drop}}=\frac{i{N}_{s}}{{n}_{L}\left(\raisebox{1ex}{$4\pi {r}_{d}{}^{3}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}\equiv \frac{Bi{N}_{s}}{{r}_{d}{}^{3}},\text{\hspace{1em}}where\text{\hspace{0.17em}}B=\frac{3}{4\pi {n}_{L}}$$
In the above equations, V_{drop} is the volume of a water drop, N_{s} is the total moles of solute, and i is called the Van’t Hoff factor, which accounts for the splitting of some solutes into components when they dissolve. An example is salt, NaCl, which splits into two ions in solution, Na^{+} and Cl^{–}; in this case, i = 2.
So, now we can put the pieces together. The equilibrium vapor pressure for a water drop containing a solute is simply the triple product of the saturation vapor pressure for pure water over a plane surface, the curvature effect, and the solute effect:
$$\begin{array}{c}{e}_{eq}\left({n}_{s},r,T\right)={e}_{s}\left(T\right)\cdot \left(1-{\chi}_{s}\right)\cdot \text{exp}\left(\frac{2\sigma}{{n}_{L}\cdot R*\cdot T\cdot {r}_{d}}\right)\\ \approx {e}_{s}\left(T\right)\cdot \left(1-\frac{Bi{N}_{s}}{{r}_{d}{}^{3}}\right)\cdot \left(1+\frac{{a}_{K}}{{r}_{d}}\right)\\ ={e}_{s}\left(T\right)\cdot \left(1+\frac{{a}_{K}}{{r}_{d}}-\frac{Bi{N}_{s}}{{r}_{d}{}^{3}}-\frac{{a}_{K}Bi{N}_{s}}{{r}_{d}{}^{4}}\right)\\ \\ {e}_{eq}\left({n}_{s},r,T\right)\approx {e}_{s}\left(T\right)\cdot \left(1+\frac{{a}_{K}}{{r}_{d}}-\frac{Bi{N}_{s}}{{r}_{d}{}^{3}}\right)\end{array}$$
This equation is called the Koehler Equation. It gives us the equilibrium vapor pressure that a drop will have for a given drop radius r_{d} and given number of moles of solute N_{s}. We can see that there are two competing effects: the curvature or Kelvin effect that depends on the inverse of the drop radius, and the solute or Raoult effect that depends on the inverse of the drop radius cubed.
Remember we talked about the saturation ratio, S = e/e_{s}. We also talked about supersaturation, s = S – 1= e/e_{s} – 1. We define a new saturation ratio and supersaturation for a particle, S_{k} and s_{k}, where “k” stands for “Koehler.”
s_{k} is defined as follows:
$$\begin{array}{l}{S}_{k}=\frac{{e}_{eq}({n}_{s},r,T)}{{e}_{s}(T)}\approx \left(1+\frac{{a}_{K}}{{r}_{d}}-\frac{Bi{N}_{s}}{{r}_{d}{}^{3}}\right)\\ {s}_{k}={S}_{k}-1=\frac{{a}_{K}}{{r}_{d}}-\frac{Bi{N}_{s}}{{r}_{d}{}^{3}}\end{array}$$
Equation 5.13 is the form of the Koehler Equation that is most often used. Remember, this equation applies to individual drops. Each drop has its own Koehler curve because each drop has its own amount of solute of a given chemical composition. No two Koehler curves are alike, just as no two snowflakes are alike.
Let's look at a graph of this equation. At 20 ^{o}C, a_{K} = 1.1 x 10^{–9} m, B = 4.3 x 10^{–6} m^{3} mole^{–1}, and N_{s} is typically in the range of 10^{–18} to 10^{–15} moles.
Interpretation of the figure:
The video below (1:35) explains the Koehler Curve Equation in more depth:
Koehler theory is at the heart of cloud microphysics. It deals with two competing processes. One raises the equilibrium saturation vapor pressure above the [INAUDIBLE] saturation vapor pressure of the flat surface of your water. This is called the Kelvin effect, or the curvature effect. And the second process lowers the equilibrium vapor pressure. This is called Raoult's Law or the solute effect. Review the previous two sections of this lesson if you've forgotten these two effects. We can approximate the curvature effect as a constant over the drop radius and we can approximate the solute effect as the negative a constant over the drop radius cubed. Together they give us supersaturation for a drop. How do these two give us the Koehler curve? The curvature effect goes as a positive inverse of the radius. But the solute effect goes as the inverse of the radius cubed is negative. And at small r it is greater negative than the curvature effect is positive. As a drop gets bigger, then the curvature effect becomes more important. And then the drop equilibrium supersaturation follows the curvature effect. Note that each drop has its own Koehler curve. Supersaturation of the environment, which can be positive by radiative heating, cooling, mixing, [INAUDIBLE] descent, determines what will happen to the drop.
Note that the supersaturation is less than 0.2% for the smaller particle and less than 0.1% for the larger particle. As cooling occurs, which one will activate first?
ANSWER: The larger particle, because it has a lower critical supersaturation.
To see what happens to the drop in the atmosphere, we need to compare the Koehler curve for each drop to the ambient supersaturation of the environment. The Koehler curve is the equilibrium supersaturation, s_{k}, for each drop and it varies as a function of the drop size. The ambient supersaturation, s, is the amount of water vapor available in the environment. When s_{k} = s, the drop is in equilibrium with the environment. The drop will always try to achieve this equilibrium condition by growing (condensing ambient water vapor) or shrinking (evaporating water) until it reaches the size at which s_{k} = s if it can! Another way to think about s_{k} is that it is telling us something about the evaporation rate for each drop size and temperature. For the drop to be in equilibrium with the environment, the condensation rate of atmospheric water vapor must equal the evaporation rate of the drop. If s_{k} < s, then net condensation will occur and the drop will grow. If s_{k} > s, then net evaporation will occur and the drop will shrink.
Let's look at two cases. In the first case, the ambient supersaturation is always greater than the entire Koehler curve for a drop.
Interpretation of the figure:
In the second case, the ambient supersaturation intersects the Koehler curve:
Interpretation of the figure:
You can imagine all kinds of scenarios that can happen when there is a distribution of cloud condensation nuclei of different sizes and different amounts of solute. Drops with more solute have lower values for the critical supersaturation and therefore are likely to nucleate first because, in an updraft, the lower supersaturation is achieved before the greater supersaturation. So you can imagine the larger CCN taking up the water first and taking up so much water that the ambient supersaturation drops below the critical supersaturation for the smaller CCN. As a result, the larger CCN nucleate cloud drops while the smaller CCN turns into haze. The moral of the story? If you're a CCN, bigger is better!
The growth of the cloud drop depends initially on vapor deposition, where water vapor diffuses to the cloud drop, sticks, and thus makes it grow. The supersaturation of the environment, s_{env}, must be greater than s_{k} for this to happen, but as the drop continues to grow, s_{k} approaches 0 (i.e., e_{eq} approaches e_{s}), so smaller amounts of supersaturation still allow the cloud drop to grow. Deriving the actual equation for growth is complex, but the physical concepts are straightforward.
$\frac{d{m}_{d}}{dt}=4\pi {r}_{d}{\rho}_{l}G\left(T,p\right)\left({s}_{env}-{s}_{k}\right)$
Physical explanation:
Conclusion:
We need other processes to get cloud drops big enough to form precipitation, either liquid or solid.
There are two types of processes for growth into precipitation drops: warm cloud processes and cold cloud processes. In warm clouds, the processes all involve only liquid drops. In cold clouds, the processes can involve only solid particles, as well as mixed phases (both supercooled liquid and ice). Some of the most important processes involve collisions between drops, whether they be liquid or solid.
Collisions occur in both cold and warm clouds and can involve either liquid drops or solid particles or both.
For a cloud drop at rest, gravity is the only external force. Once the cloud drop starts to fall, then the air resistance forms another force called drag, which is a function of the velocity.
In less than a second, the particle reaches a fall speed such that the drag force exactly balances the gravitational force and the velocity becomes constant. This velocity is called the terminal velocity. Because the gravitational force depends on the volume of the drop, it goes as the cube of the drop radius. In contrast, drag acts on the surface of the drop, and so it depends on the drop area and goes as the square of the drop radius (times the velocity). Setting the gravitational and drag forces equal to other and then solving for the terminal velocity, it is easy to show that the terminal velocity should vary linearly with drop radius. Measurements bear this linear relationship out. For instance, the terminal velocity of a 50 μm radius drop is about 0.3 m s^{–1}, while the terminal velocity for a drop 10 times larger (500 μm radius) is about 4 m s^{–1}, which just a little more than a factor-of-10 increase.
The growth of a cloud drop into a precipitation drop by collision–coalescence is given by the equation:
$$\begin{array}{l}\frac{d{m}_{L}}{dt}=Area\text{\hspace{0.17em}}swept\text{\hspace{0.17em}}out\cdot efficiency\text{\hspace{0.17em}}of\text{\hspace{0.17em}}collection\cdot velocity\text{\hspace{0.17em}}difference\cdot liquid\text{\hspace{0.17em}}water\text{\hspace{0.17em}}content\\ \frac{d{m}_{L}}{dt}={A}_{g}\cdot {E}_{c}\cdot ({v}_{L}-{v}_{s})\cdot LWC\\ \frac{d{m}_{L}}{dt}=\pi {\left({r}_{L}+{r}_{s}\right)}^{2}\cdot {E}_{c}\cdot ({v}_{L}-{v}_{s})\cdot LWC\end{array}$$
The figure below provides a good conceptual picture of collision–coalescence. The collector drop must be falling faster than the smaller collected drop so that the two of them can collide. As the air streamlines bow out around the drop, they carry the smaller drops with them around the drop, and the effective cross-sectional area becomes less than the actual cross-sectional area, which is simply the cross-sectional area of a disk with a radius that is the sum of the radii of the large collector drop and the smaller collected drops. As drops get bigger, they have too much inertia to follow the air streamlines, thus making the collision more likely.
E_{c} is small for 10 μm drops, so by a random process, some drops become bigger than others and begin collecting smaller drops (see figure below). E_{c} increases as the radius of the falling drop increases. When the larger falling drop gains a radius of more than 100 μm, its collision–coalescence efficiency is very good for all smaller drops down to sizes of about 10–20 μm.
Once a collecting drop has reached a radius of a few hundred μm, it is falling fast (v_{L} >> v_{s}) and its collision–coalescence efficiency is close to 100%. Now take the following steps to rewrite Equation 5.16: (1) E_{c} equal to 1, (2) v_{L} >> v_{s}, (3) v_{L} = constant x r_{L}, (4) m_{L} = 4ρ_{l}πr_{L}^{3}/3 , and (5) solve for dr_{L}/dt. Once you take these steps, you can show that dr_{L}/dt is proportional to r_{L}. That is, the bigger the drop gets the faster it grows. Separating variables (r_{L} and t) and integrating from r_{L} = 0 at t = 0 to arbitrary values of r_{L} and t reveals that r_{L} increases exponentially with time:
$${r}_{d}\propto \mathrm{exp}(time)$$
With the constant of proportionality between the terminal velocity and drop radius set at 0.8 x 10^{–3} s^{–1} and LWC = 1 g m^{–3}, it can be shown that a drop can grow from 50 μm to 1000 μm by collision–coalescence in only 25 minutes. So, the activated cloud drops grow to 10–20 μm by the slow growth of vapor deposition (square root of time). Then when collision–coalescence starts and produces a few big drops, they can grow exponentially with time.
Smaller drops are typically spherical. Once these drops get to be above a mm in radius, they become increasingly distorted, with a flattened bottom due to drag forces, and they look a little like the top half of a hamburger bun. They can be further distorted so that the middle of the bun-shape gets pushed up by the drag forces so that the drop takes on a shape resembling an upside down bowl.
Eventually the drops break up, either by getting thin enough in the middle that they break into pieces or by colliding with other drops so hard that filaments or sheets of liquid break off to form other drops. These processes create a whole range of sizes of drops. Thus rain consists of drops that have a wide spectrum of sizes. The following video (2:50) entitled "How Raindrops are Formed" starts with a simplified view of the atmosphere's water cycle, but then shows examples of a falling drop, collision–coalescence, and cloud-drop breakup.
For riming, capture nucleation, and aggregation, there are similar equations with terms similar to those in Equation 5.16—an area swept out, a collection efficiency, the relative velocity, and the liquid or solid mass concentration of the smaller drops or ice. These are typically a bit more complicated if the ice is not spherical, but the concepts are the same. These ice collision–coalescence processes are able to produce ice particles big enough to fall, and if these particles warm as they pass through the warm part of the cloud, they can turn into liquid rain. A significant fraction of rain in the summer can come from ice collision–coalescence processes above the freezing line in the clouds.
Recall that water can exist in liquid form even below the freezing point. This supercooled liquid needs ice nuclei (IN) in order to become ice, although at a temperature of about –40 ^{o}C, the liquid can freeze homogeneously (without IN).
Recall from Lesson 4 that the vapor pressure over supercooled liquid water is greater than the vapor pressure over ice at the same temperature. So, if an ice particle is introduced into air that contains liquid water below the freezing point, the ambient vapor pressure in equilibrium with the liquid will be greater than the saturation vapor pressure of the ice. The ice will grow, but this uptake of water vapor will cause the ambient water vapor pressure to be less than the saturation vapor pressure for the liquid drops and the liquid drops will have net evaporation. This process will continue so that the ice grows at the expense of the liquid drops, which will shrink. The transfer of water is not by the liquid drops colliding with the ice crystal; the transfer of water comes from the liquid drops evaporating water to make water vapor and then that water vapor diffusing over to the ice, where it condenses. This process is called the Bergeron–Findeisen Process, and is a way that precipitation-sized drops can be formed in about 40 minutes in mixed-phase clouds (see figure below).
This process, as unusual as it seems, actually works, as can be seen in the figure below!
Announcing a new activity worth one point of extra credit for each lucky winner: Picture of the Week! Here is how to participate:
Here is an example below:
We can put all of the processes from this lesson together to look at the lifecycle of a cloud:
The following is a description of convection’s stages of development:
The video below (2 min.) includes some great time-lapse video of clouds forming and disappearing (No audio). Check it out:
Clouds are shaped and sized by atmospheric motions and mixing with the surrounding air and are composed of either liquid drops, ice crystals, or both, depending on the temperature. The basic shapes are stratus, cumulus, and cirrus or combinations thereof; the altitudes define low, middle, and high clouds.
Understanding clouds requires looking at individual cloud drops through a microscope. Cloud drops form when there is sufficient moisture, aerosol to act as Cloud Condensation Nuclei (CCN), and cooling air. This cooling air becomes supersaturated with water vapor by radiative cooling (e.g., valley fog), uplift (e.g., cumulus convection), or mixing (e.g., contrail). Each CCN particle requires supersaturation to grow into a cloud drop as a competition takes place between a curvature effect (tiny particles have higher saturation vapor pressure than flat surfaces) that inhibits water uptake, while a solute effect (the particle dissolving in liquid water) enhances water uptake. Once the atmosphere has cooled enough to achieve supersaturation greater than the critical supersaturation for a CCN particle, that particle can take on enough water to continue growing large enough to become a cloud drop.
Initially, the drop grows by vapor deposition, but this process slows down as the square root of time, so that the formation of raindrops is not possible within the typical 30-minute lifetime of a cloud. Other processes are at work. In a warm cloud, where all the drops are liquid, collisions and coalescence of drops, with occasional breakup, exponentially increases the size of the drops as they fall. In a cold cloud, precipitation drops can grow either by riming of ice with supercooled liquid drops or by collisions and aggregation of ice particles or by vapor deposition from supercooled liquid to ice.
You have reached the end of Lesson 5! Double-check that you have completed all of the activities before you begin Lesson 6.
Atmospheric radiation plays a critical role in life on Earth and in weather. Without solar heating, Earth would be a dead frozen ball hurtling through space. Luckily, the energy that Earth receives from solar radiation is sufficient to produce liquid water on its surface, thus enabling life to thrive. In this lesson we will look at solar radiation and its changes over time. Radiation is just another form of energy and can be readily converted into other forms, especially thermal energy, which is sometimes called "heat." In this lesson, we will use the word "radiation" to mean all electromagnetic waves, including ultraviolet, visible, and infrared radiation. We will introduce some unfamiliar terms like "radiance" and "irradiance" and will be careful with our language to prevent confusion.
An important concept in studying atmospheric radiation is that all objects emit and absorb radiation. For a perfect emitter, the radiation emitted by an object, called the irradiance, is determined by the Planck function, which depends only on temperature and wavelength. The higher the temperature, the greater the radiation emitted at all wavelengths and the shorter the wavelength of the peak energy. The Sun emits mainly in the visible while Earth and its atmosphere emit mainly in the infrared. No object is really a perfect emitter at every wavelength; the unitless number emissivity measures how good or poor an emitter is. At each wavelength, a good emitter is a good absorber.
How well an object absorbs at different wavelengths of radiation, called its absorptivity, depends on its chemical composition and the rules of quantum mechanics. As a result, some absorption is strong and some is weak; some is in sharp lines while some is in broad features in the wavelength spectrum; some is in the UV, particularly due to O_{2} and O_{3}, little is in the visible, and much absorption, including broad bands and sharp lines, occurs in the infrared, particularly by H_{2}O, CO_{2}, and O_{3}.
The radiation that is not absorbed by a gas, liquid or solid is either transmitted or scattered. The amount of transmitted radiation depends on the absorption and scattering cross sections of the gas, liquid, or solid components of matter, so that the larger the cross section and the distance through the matter, the less radiation passes through the matter. The decay of the transmitted light with distance through the matter is exponential, as is described by Beer’s Law.
Earth’s surface and atmospheric gases can emit and absorb radiation at the same wavelengths. Most of Earth’s emissions are at infrared wavelengths, whether the emission be from the surface, clouds, or atmospheric gases.
Scattering of atmospheric radiation complements absorption and is even more difficult to track through the atmosphere than absorption is. The wavelength of the radiation and the size, shape, and composition of the scattering particle together determine the scattering efficiency and scattering pattern. Many of the skies that we remember best are due to the scattering and absorption of sunlight.
By the end of this lesson, you should be able to:
If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.
Everything radiates—the Sun, the Earth, the atmosphere, and you. The energy provided by the Sun is reused in the Earth system to provide the energy that drives weather and climate. But ultimately, the infrared radiation radiated by Earth into space must balance the solar visible radiation coming into the Earth system. From the point-of-view of the Earth system, we are most concerned about how atmospheric radiation interacts with matter. Matter is simply molecules and atoms and the structures that they build, such as the air, the clouds, the Earth, and the Sun.
When radiation encounters matter, three things can happen. The radiation can be transmitted through matter; it can be absorbed by the matter; it can be scattered by the matter. One of these three things must happen, so we can sum them up to one:
$$a+t+s=1$$
where t is the transmissivity, the fraction of the radiation that is transmitted; a is the absorptivity, the fraction absorbed; and s is the reflectivity, the fraction that is scattered or reflected.
Scattering and reflecting are related but are different because reflection is scattering in a particular direction whereas scattering tends to go in a range of directions.
Here is another chance to earn one point of extra credit: Picture of the Week!
Solar radiation drives the Earth system and makes life possible. Solar radiation is absorbed and then put to use to increase the surface temperature, to change the phase of water, and to fuel atmospheric chemistry. The uneven distribution of solar radiation on Earth’s surface drives atmospheric dynamics.
The total amount of solar energy per unit time and unit area, also called the solar irradiance, is 1361 W m^{–2} at the top of the atmosphere (Stephens et al., 2012, Nature Geoscience 5, p. 691). It is distributed unevenly over Earth’s surface. That distribution changes over the course of the seasons (see next two figures). The seasons result primarily from the Earth's rotation axis not being perpendicular to the plane of the Earth's orbit around the Sun.
Earth’s spin and its orbit around the sun are not constant, but instead, change with time, like a spinning top. The orbit’s eccentricity (i.e., how different it is from circular) varies with a 100,000-year period. The tilt of Earth’s rotation axis with respect to a line perpendicular to the plane of Earth's orbit, which is called the obliquity, varies from 22.1^{o} to 24.5^{o} during a 41,000-year cycle. The obliquity is currently 23.4^{o} and decreasing. Finally, the precession of Earth’s orbit, which is the orientation of the rotation axis with respect to Earth’s orbital position, also varies with a period of about 26,000 years, although the orbit itself is also rotating around the sun, so that the effective period of precession is about 21,000 years. These motions, when taken together, slowly and periodically change the distribution of solar irradiance on Earth’s surface and are described by the Milankovitch Theory (see next two figures). The changes in solar radiation ultimately lead to very large changes in climate and greenhouse gas concentrations, particularly the ice ages, which occur about every 100,000 years. The greenhouse gas changes (methane and carbon dioxide, in particular) were presented in Lesson 4.
Changes in Earth’s orbit and spin are not the only ways that solar irradiance changes—the Sun’s energy output also changes. It has been increasing slightly (0.05–0.10%) over the past 300 years and varies by another ~0.1% over the course of the 11-year solar cycle. The ultraviolet (200–300 nm) irradiance has increased by about 3% in the past 300 years and varies by ~1.5% between solar maximum and solar minimum. This increased UV leads to greater stratospheric ozone production, which increases stratospheric heating, leading to the poleward displacement of the stratospheric meridional wind.
We can think of radiation either as waves or as individual particles called photons. The energy associated with a single photon is given by $$
The energy of a single photon that has the wavelength λ is given by:
$$E\text{}=\text{}\frac{hc}{\lambda}=\text{}\frac{1.986\times \text{}{10}^{-16}{\text{Jnmphoton}}^{-1}}{\lambda}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamyraiaabckacqGH9aqpcaqGGcWaaSaaaeaacaWGObGaam4yaaqaaiabeU7aSbaacqGH9aqpcaqGGcWaaSaaa8aabaWdbiaaigdacaGGUaGaaGyoaiaaiIdacaaI2aGaey41aqRaaeiOaiaaigdacaaIWaWdamaaCaaaleqabaWdbiabgkHiTiaaigdacaaI2aaaaOGaaeiOaiaabQeacaqGGaGaaeOBaiaab2gacaqGGcGaaeiCaiaabIgacaqGVbGaaeiDaiaab+gacaqGUbWdamaaCaaaleqabaWdbiabgkHiTiaaigdaaaaak8aabaWdbiabeU7aSbaaaaa@5AA2@
Note that as the wavelength of light gets shorter, the energy of the photon gets greater. The energy of a mole of photons that have the wavelength λ is found by multiplying the above equation by Avogadro's number:
$${E}_{m}\text{}=\text{}\frac{hc{N}_{A}}{\lambda}=\text{}\frac{1.196\text{}\times \text{}{10}^{8}{\text{Jnmmol}}^{-1}}{\lambda}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamyra8aadaWgaaWcbaWdbiaad2gaa8aabeaak8qacaqGGcGaeyypa0JaaeiOamaalaaabaGaamiAaiaadogacaWGobWdamaaBaaaleaapeGaamyqaaWdaeqaaaGcpeqaaiabeU7aSbaacqGH9aqpcaqGGcWaaSaaa8aabaWdbiaaigdacaGGUaGaaGymaiaaiMdacaaI2aGaaeiOaiabgEna0kaabckacaaIXaGaaGima8aadaahaaWcbeqaa8qacaaI4aaaaOGaaeiiaiaabQeacaqGGaGaaeOBaiaab2gacaqGGcGaaeyBaiaab+gacaqGSbWdamaaCaaaleqabaWdbiabgkHiTiaaigdaaaaak8aabaWdbiabeU7aSbaaaaa@5A32@
In the lesson on atmospheric composition, you saw how solar UV radiation was able to break apart molecules to initiate atmospheric chemistry. These molecules are absorbing the energy of a photon of radiation, and if that photon energy is greater than the strength of the chemical bond, the molecule may break apart.
Consider the reaction O_{3} + UV → O_{2} + O*. If the bond strength between O_{2} and O* (i.e., excited state oxygen atom) is 386 kJ mol^{–1}, what is the longest wavelength that a photon can have and still break this bond?
ANSWER: Solve for wavelength in equation [6.2b]
$$\lambda =\text{}\frac{1.196\text{}\times \text{}{10}^{8}{\text{Jnmmol}}^{-1}}{{E}_{m}}\text{}=\frac{1.196\text{}\times \text{}{10}^{8}{\text{Jnmmol}}^{-1}}{386\text{}\times \text{}{10}^{3}{\text{Jmol}}^{-1}}=309\text{nm}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaeq4UdWMaeyypa0JaaeiOamaalaaapaqaa8qacaaIXaGaaiOlaiaaigdacaaI5aGaaGOnaiaabckacqGHxdaTcaqGGcGaaGymaiaaicdapaWaaWbaaSqabeaapeGaaGioaaaakiaabckacaqGkbGaaeiOaiaab6gacaqGTbGaaeiOaiaab2gacaqGVbGaaeiBa8aadaahaaWcbeqaa8qacqGHsislcaaIXaaaaaGcpaqaa8qacaWGfbWdamaaBaaaleaapeGaamyBaaWdaeqaaaaak8qacaqGGcGaeyypa0ZaaSaaa8aabaWdbiaaigdacaGGUaGaaGymaiaaiMdacaaI2aGaaeiOaiabgEna0kaabckacaaIXaGaaGima8aadaahaaWcbeqaa8qacaaI4aaaaOGaaeiOaiaabQeacaqGGcGaaeOBaiaab2gacaqGGcGaaeyBaiaab+gacaqGSbWdamaaCaaaleqabaWdbiabgkHiTiaaigdaaaaak8aabaWdbiaaiodacaaI4aGaaGOnaiaabckacqGHxdaTcaqGGcGaaGymaiaaicdapaWaaWbaaSqabeaapeGaaG4maaaakiaabckacaqGkbGaaeiOaiaab2gacaqGVbGaaeiBa8aadaahaaWcbeqaa8qacqGHsislcaaIXaaaaaaakiabg2da9iaaiodacaaIWaGaaGyoaiaabckacaqGUbGaaeyBaaaa@817C@
The Sun emits radiation from X-rays to radio waves, but the irradiance of solar radiation peaks in the visible wavelengths (see figure below). Common units of irradiance are Joules per second per m^{2} of surface that is illuminated per nm of wavelength (e.g., between 300 nm and 301 nm), or W m^{–2} nm^{–1} for the plot below. These units are the units of spectral irradiance, which is also simply called irradiance, but as a function of wavelength. To get the total irradiance in units of W m^{–2}, the spectral irradiance should be integrated over all the wavelengths.
Note the following for the solar spectrum:
For solar wavelengths at which the absorptivity is high, the solar irradiance at sea level is small. Note that the big absorbers of infrared irradiance are water vapor, carbon dioxide, and ozone.
Recall that molecules have a wide range of speeds and thus a wide range of energies. The Maxwell–Boltzmann Distribution, which gives the distribution of molecules as a function of energy, is given approximately by the equation:
$$f(E)=\frac{2\sqrt{E}}{\sqrt{\pi}{\left(kT\right)}^{3/2}}\mathrm{exp}\left(-E/kT\right)$$
where f is the probability that a molecule has an energy within a small window around E, T is the absolute temperature, and k is the Boltzmann constant. The above equation, when integrated over all energies, gives the value of 1.
The functional form of this distribution is shown below:
All objects—gas, liquid, or solid—emit radiation. If we think of radiation as photons, we would say that these photons have a distribution of energies, just like molecules do. However, photons cannot have continuous values of photon energy; instead, the photon energy is quantized, which means that it can have only discrete energy values that are different by a very very small amount of energy. When this quantized distribution is assumed, then the distribution of spectral irradiance leaving a unit area of the object’s surface per unit time per unit wavelength interval into a hemisphere is called the Planck Distribution Function of Spectral Irradiance:
$${P}_{e}(\lambda )=\frac{2\pi h{c}^{2}}{{\lambda}^{5}}\frac{1}{\mathrm{exp}(hc/\lambda \text{\hspace{0.05em}}k\text{\hspace{0.05em}}T)-1}$$
where h is Planck’s constant, c is the speed of light, k is the Boltzmann constant (1.381 x 10^{–23} J K^{–1}), T is the absolute temperature, and λ is the wavelength. The integral of this function over all wavelengths leads to the Stefan–Boltzmann Law Irradiance, which gives the total radiant energy per unit time per unit area of the object’s surface emitted into a hemisphere.
P_{e}(λ)/π is called the Planck Distribution Function of Spectral Radiance and commonly has units of W steradian^{–1} m^{–2} nm^{–1} and is often denoted by the letter I. A steradian is just a unit solid angle. Just as a radian for a circle is a length of the circle's arc that is equal to the circle's radius, a steradian is the area on a sphere's surface that is equal to the sphere's radius squared. There are 2π radians along a circle and 4π steradians (abbreviated sr) over a sphere. For some of this discussion we will use the spectral irradiance P_{e}(λ) and consider the radiation from any area on the sphere to be emitted not in a single direction but into the full hemisphere of directions. Later on, when we start talking about absorption, we will need to think about the irradiance in very specific directions, in which case we will use radiance. Be mindful of the difference between irradiance and radiance.
The spectral irradiance is the amount of energy that is emitted from or falling on a unit area in space per unit time per unit wavelength (W m^{–2} µm^{–1}). So the m^{2} in this case indicates a surface area on which energy is leaving or falling. To get the Sun’s spectral irradiance at the top of Earth’s atmosphere, we must multiply the spectral irradiance emitted by the Sun's surface by the Sun's surface area to get the total energy emitted per unit time and per unit wavelength by the Sun and then divide by the surface area of the sphere that has a radius equal to the Earth–Sun distance to get the energy per time per unit wavelength per unit area of the surface located at the top of Earth's atmosphere.
All the radiation that is emitted from the Sun’s surface continues to move outward at the speed of light until it hits objects, but there are very few objects between the Sun and Earth, except occasionally the Moon. Thus, in the absence of any objects, the total amount of solar irradiance striking the planets decreases as the square of the distance between the center of the Sun and the surface of the planet.
The Planck distribution function spectral irradiance for an object at a temperature of 5777 K (the Sun's surface temperature) is shown in the figure below. Note the rapid rise at the shorter wavelengths, the peak value, and then the slower decline at longer wavelengths. Look at the peak value of the irradiance, which is about 25 million W m^{–2} nm^{–1}! That's a lot of energy being radiated from the Sun's surface, but of course the Earth is 150 million km away from the Sun and thus intercepts less than half a billionth of the solar irradiance.
The peak of this distribution as a function of wavelength can be found by taking the derivative of P_{e}(λ) with respect to wavelength, setting the value equal to 0, and solving for the wavelength. The result is the Wien Displacement Law:
$${\lambda}_{max}=\frac{2898\text{\hspace{0.17em}}\mu \text{m}\text{\hspace{0.17em}}\text{K}}{T}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaeq4UdW2damaaBaaaleaacaWGTbGaamyyaiaadIhaaeqaaOWdbiabg2da9maalaaapaqaa8qacaaIYaGaaGioaiaaiMdacaaIWaGaaGPaVlabeY7aTjaab2gacaaMc8Uaae4saaWdaeaapeGaamivaaaaaaa@46FA@
For the sun with a photospheric temperature of about 5780 K, λ_{max} ~ 0.500 μm or 500 nm, which is the color green. However, for Earth with a mid-tropospheric temperature of about 260 K, the peak wavelength is closer to 11 μm, well into the infrared (see below).
If the Planck distribution function spectral irradiance is integrated over all wavelengths, then the total irradiance emitted into a hemisphere is given by the Stefan–Boltzmann Law:
$${F}_{s}=\sigma \text{\hspace{0.05em}}{T}^{4}$$
where σ is called the Stefan–Boltzmann constant (5.67 x 10^{–8} W m^{–2} K^{–4}). F_{s} has SI units of W m^{–2}, where the m^{2} refers to the surface area of the object that is radiating.
The Stefan–Boltzmann law (total) irradiance applies to an object that radiates according to the Planck distribution function spectral irradiance. If we look at the figure below, we see that the solar spectrum at the top of the atmosphere is similar to the Planck distribution function but does not follow it perfectly. However, the Planck distribution function with the same total irradiance as the sun has a temperature of 5777 K, as in the second figure.
Clouds radiate. Assume two spherical clouds, one with a radius of 100 m and a temperature of 275 K and a second with a radius of 100 m and a temperature of 230 K. Assuming that they both radiate according to the Planck distribution function, calculate the emission for each cloud in W m^{–2} and in W. Which cloud is radiating more total energy and by how much?
ANSWER:
Cloud T (K) | Cloud radius (m) | F_{s} (W m^{–2}) | F_{s} x 4πR_{c}^{2} (W) |
---|---|---|---|
275 | 100 | 324 | 4.1 x 10^{7} |
230 | 100 | 159 | 2.0 x 10^{7} |
The warmer cloud is radiating about twice as much energy as the cooler cloud. These little clouds are radiating quite a lot of energy in all directions, but some of it is down toward Earth’s surface. If we make the simple assumption that half the radiation goes up and the other half goes down, the amount of energy radiated toward Earth’s surface per second is approximately 10 million W. If the clouds are not too far from the surface, this downward radiation could contribute a few hundred W m^{–2} of heating at Earth’s surface. Thus clouds can act like additional heat sources for Earth’s surface, keeping its temperature higher than it would be on a clear night. The image below is an infrared photograph of the sky above Ogden, Utah. Infrared radiation detected by the camera has been converted to temperature, with higher temperatures indicating more infrared emission.
Remember that when radiation encounters matter it may be absorbed or transmitted or scattered (including reflected). For an object acting as a perfect Planck distribution function, it must absorb all radiation completely with no scattering and no transmission. Some objects absorb very well at some wavelengths but not at others. For instance, water vapor absorbs little visible radiation but absorbs infrared radiation at some wavelengths very well.
At the same time, the sun, like other objects, does not radiate perfectly according to the Planck distribution function spectral irradiance, but instead radiates at a fraction of it at some wavelengths. This fraction, which goes from 0 to 1, is called the emissivity and is denoted by ε. How is an object’s emissivity related to its absorptivity?
Kirchhoff’s Law states that at any given wavelength, an object’s emissivity ε is equal to its absorptivity, that is:
$$\epsilon \left(\lambda \right)=\alpha \left(\lambda \right)$$
Thus, if an object has some wavelengths at which radiation is scattered or reflected, then the object will have an emissivity less than 1 at the wavelength, and the fraction that is absorbed will be equal to the emissivity at each wavelength.
Thus, when we integrate the Planck distribution function spectral irradiance over wavelength to obtain the irradiance emitted by the object, it first has to be multiplied by the wavelength-dependent emissivity, thus leading to the modified form of the Stefan–Boltzmann law:
$$F=\epsilon \text{\hspace{0.17em}}\sigma \text{\hspace{0.05em}}{T}^{4}$$
where we understand that ε is some form of averaged emissivity.
Watch the following video (1:07), where the Stefan–Boltzmann Law is described in greater detail:
Some typical average emissivities are listed in the table below. These are emissivities averaged over all wavelengths. At any particular wavelength, the emissivity may be greater or less than the average.
Material | Emissivity, ε |
---|---|
ice | 0.97 |
pure water | 0.96 |
snow | 0.8–0.9 |
trees (oak, beech, maple, pine) | 0.97–0.98 |
grass | 0.98 |
soil | 0.93 |
aluminum foil | 0.03 |
asphalt | 0.88–0.94 |
What about gases? Gases absorb and thus emit like all other matter. To know more about the emissivity of all objects, we need to know more about the absorption of objects.
The atmosphere absorbs a significant amount of radiation in the infrared but rather little in the visible. Also, we see that gases absorb strongly at some wavelengths and not at others. Why is this?
To answer this question, we need to look at the configurations of the electrons that are zooming around atoms and molecules. More than 100 years ago, scientists began using prisms to disperse the light from the sun and from flames containing different elements. While the sun gave the colors of the rainbow, the flames had light in very distinct lines or bands. This puzzle was finally resolved a little more than 100 years ago with the invention of quantum mechanics, which basically says that the electrons zooming around atoms and molecules and the vibrations and rotations of molecules can have only discrete energies that are governed by rules of conservation of angular momentum.
The following bulleted list is a crash course in absorption by the electrons in atoms and molecules. Refer to the figure below the box.
The absorption cross section, σ, varies significantly over the width of the absorption line. So it is possible for all the radiation to be absorbed in the middle of the line but very little absorbed in the “wings.”
Atoms and molecules can absorb radiation (a photon) only if their structure has an energy difference between levels that matches the photon’s energy (hc/λ). Otherwise, the atom or molecule will not absorb the light. Once the molecule has absorbed the photon, it can either lose a photon and go back to its original lower energy level; or it can break apart if the photon energy is greater than the chemical bond holding the molecule together; or it can collide with other molecules, such as N_{2} or O_{2}, and transfer energy to them while it goes back to its lower energy level. Collisions happen often, so the energy of the absorbed photon is often transferred to thermal energy.
Note that Earth's outgoing infrared irradiance is limited to a few atmospheric "windows" and the irradiance at all other wavelengths is strongly absorbed, mostly by water vapor, but also by carbon dioxide, ozone, nitrous oxide, methane, and other more trace gases that aren't shown in the figure above.
Gases, liquids, and solids can all absorb radiation. Since we are most interested in gases, we will use a gas to develop the equation that we need for absorption.
Suppose we have a volume of uniformly distributed gas that absorbs. Let's suppose that there is a beam of radiation that passes into this medium. Let's narrow this beam down so that all the radiation is traveling in essentially the same direction, so now we are interested in the irradiance that is traveling in a narrow range of angles, which is called the radiance, which is denoted by I and has units of W m^{–2} steradian^{–1} nm^{–1}. Refer back to Lesson 6.5 [76] to refresh your memory about what steradians are.
In an infinitesimal slab of that medium, a certain fraction dI of the radiation with radiance I (W m^{–2}) is absorbed. dI is equal to the absorption cross section, σ (m^{2}), of a single molecule or particle multiplied by the number density, n (# m^{–3}) of absorbers, the length of the light-path through the slab, ds (m), and the amount of radiance I itself :
Note that both I and σ are functions of wavelength. If there is a beam of radiation with a certain wavelength, then the chance of absorbing the radiation depends on the cross-sectional area of each molecule or particle, which quantifies the efficiency with which the molecule absorbs the radiation with that particular wavelength. The absorption cross sections are largest if the energy of the radiation's wavelength matches the transition energy of an electron, atom, or molecule.
Integrating this expression over the total thickness of the volume, we get the Beer–Lambert Law:
$$I\left(\lambda \right)={I}_{o}\left(\lambda \right)\mathrm{exp}\left[-\sigma \left(\lambda \right)ns\right]$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamysamaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaacqGH9aqpcaWGjbWdamaaBaaaleaapeGaam4BaaWdaeqaaOWdbmaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaaciGGLbGaaiiEaiaacchadaWadaqaaiabgkHiTiabeo8aZnaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaacaWGUbGaam4CaaGaay5waiaaw2faaaaa@4DF0@
where I_{o} is the radiance at the front edge of the path and I is the radiance at distance s along the path. Sometimes, we substitute a variable κ_{a}, called the absorption coefficient, for the product σn. In fact, you will see the absorption coefficient reported many ways.
When σns = 1, I(λ)/I_{o}(λ) = exp(–1) = 0.368.
For a uniform gas, the optical depth equals σns and is often given by the symbol τ (tau). If the optical depth gets large, then very little radiation gets through.
Ozone has an absorption cross section of 1.0 x 10^{–19} cm^{2} at a wavelength of 310 nm. If the average ozone concentration in the ozone layer is 2 x 10^{12} molecules cm^{–3} and the thickness of the ozone layer is 25 km, what is the fraction of sunlight that gets through when the sun is directly overhead?
ANSWER:
$$\frac{I\left(\lambda \right)}{{I}_{o}\left(\lambda \right)}=\mathrm{exp}\left(\mathrm{\u20131}\times {10}^{-19}{\text{cm}}^{2}\cdot 2\times {10}^{12}{\text{cm}}^{-3}\cdot 25\times {10}^{5}\text{cm}\right)=0.6$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@6A4A@
At 310 nm, 60% of the solar UV gets through to Earth’s surface. Yet at 290 nm, where σ = 1.09 x 10^{–18} cm^{2}, only 0.4% of the solar UV gets through.
Once we know the absorption coefficients of all of the absorbing molecules between the sun and the air volume of interest and the number densities of the absorbers, we can determine how much radiation will be absorbed at every wavelength.
The angle of the Sun with respect to the zenith (directly overhead) is called the solar zenith angle (SZA) and is 0^{o} overhead and 90^{o} on the horizon. The SZA affects the total path through absorbers, and thus is important. We often make an assumption that the atmosphere is a parallel plane (see figure below).
The result is that we can write $ds=\frac{1}{\mathrm{cos}\left(SZA\right)}dz=\mathrm{sec}\left(SZA\right)dz$ in the differential form of Beer's Law:
$$\frac{dI\left(\lambda \right)}{I\left(\lambda \right)}=-\sigma \left(\lambda \right)n\text{sec}\left(\text{SZA}\right)dz$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeWaaSaaa8aabaWdbiaadsgacaWGjbWaaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaaWdaeaapeGaamysamaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaaaaGaeyypa0JaeyOeI0Iaeq4Wdm3aaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaiaad6gacaqGGaGaae4CaiaabwgacaqGJbWaaeWaa8aabaWdbiaabofacaqGAbGaaeyqaaGaayjkaiaawMcaaiaadsgacaWG6baaaa@5168@
which gives:
$$I\left(\lambda ,z\right)={I}_{0}\left(\lambda \right)\mathrm{exp}\left[-\text{sec}\left(\text{SZA}\right){{\displaystyle \int}}^{\text{}}\sigma \left(\lambda \right)n\left(z\right)dz\right]$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamysamaabmaapaqaa8qacqaH7oaBcaGGSaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadMeapaWaaSbaaSqaa8qacaaIWaaapaqabaGcpeWaaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaiGacwgacaGG4bGaaiiCamaadmaabaGaeyOeI0Iaae4CaiaabwgacaqGJbWaaeWaa8aabaWdbiaabofacaqGAbGaaeyqaaGaayjkaiaawMcaa8aadaqfGaqabSqabeaacaaMb8oaneaapeGaey4kIipaaOGaeq4Wdm3aaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaiaad6gadaqadaWdaeaapeGaamOEaaGaayjkaiaawMcaaiaadsgacaWG6baacaGLBbGaayzxaaaaaa@5DB3@
Often, the integral is called the optical thickness, or optical path, where s_{1} is one point along the path and s_{2} is another. The exponential of the optical thickness, factoring in the sec(SZA), equals exp[–sec(SZA) τ(s_{1},s_{2})] and is called the transmittance, often designated with the symbol t, where τ is the optical depth.
$$\tau ({s}_{1},{s}_{2})\equiv {\displaystyle \underset{{s}_{1}}{\overset{{s}_{2}}{\int}}\sigma (s)\cdot n(s)\cdot ds}$$
For matter that absorbs and does not scatter, the absorptivity, a = 1 – t. Note that in this case the emissivity ε = a = 1 – t.
Usually, but not always, σ is a function of altitude because it often is a function of pressure and temperature, which vary for different altitudes.
Here is a video (2:03) further explaining Beer's Law:
Absorption and scattering apply to infrared radiation, just as they do to UV, visible, and microwave radiation. However, all matter that is at Earth-like temperatures is radiating in the infrared but isn’t hot enough to radiate significantly in the visible.
Remember that the peak in the Planck distribution function for a body at T = 300 K is given by Wien's Law:
$${\lambda}_{\mathrm{max}}=\frac{2898\text{\hspace{0.17em}}\mu \text{m}\text{\hspace{0.17em}}\text{K}}{T}=\frac{2898\text{\hspace{0.17em}}\mu \text{m}\text{\hspace{0.17em}}\text{K}}{300\text{K}}=10\text{\hspace{0.17em}}\mu \text{m}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaeq4UdW2damaaBaaaleaapeGaciyBaiaacggacaGG4baapaqabaGcpeGaeyypa0ZaaSaaa8aabaWdbiaaikdacaaI4aGaaGyoaiaaiIdacaaMc8UaeqiVd0MaaeyBaiaaykW7caqGGaGaae4saaWdaeaapeGaamivaaaacqGH9aqpdaWcaaWdaeaapeGaaGOmaiaaiIdacaaI5aGaaGioaiaaykW7cqaH8oqBcaqGTbGaaGPaVlaabccacaqGlbaapaqaa8qacaaIZaGaaGimaiaaicdacaqGGaGaae4saaaacqGH9aqpcaaIXaGaaGimaiaaykW7cqaH8oqBcaqGTbaaaa@5D95@
To examine how extinction and emission work together, consider the case of an absorbing atmosphere with no scattering at a fixed wavelength. Assume that there is an absorbing gas between us and some radiation source which has a radiance I. Then I will be reduced as it travels through the absorbing gas:
$$\begin{array}{l}d{I}_{absorption}\text{=}-\sigma \text{\hspace{0.17em}}n\text{\hspace{0.17em}}I\text{\hspace{0.17em}}ds=-{\kappa}_{a}\text{\hspace{0.17em}}I\text{\hspace{0.17em}}ds\hfill \\ {\kappa}_{a}\equiv \text{absorption}\text{\hspace{0.17em}}\text{coefficient,}\text{\hspace{1em}}{\text{SIunits:m}}^{-\text{1}}\text{\hspace{0.17em}}\hfill \end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@84E7@
But, at the same time, the layer with width ds is emitting:
$$d{I}_{emission}\text{=}{\kappa}_{a}\text{\hspace{0.17em}}{P}_{e}\text{\hspace{0.17em}}ds$$
Normally, we would multiply P_{e} by the emissivity for that wavelength. But because the absorptivity = emissivity for any given wavelength and type of matter by Kirchhoff’s Law, we can replace the emissivity, ε, with the absorptivity, κ_{a}ds. Thus:
$$\begin{array}{l}dI=d{I}_{absorption}+d{I}_{emission}=-{\kappa}_{a}\text{\hspace{0.17em}}I\text{\hspace{0.17em}}ds+{\kappa}_{a}\text{\hspace{0.17em}}{P}_{e}\text{\hspace{0.17em}}ds={\kappa}_{a}\text{\hspace{0.17em}}\left({P}_{e}-I\right)\text{\hspace{0.17em}}ds\hfill \\ \begin{array}{l}\text{or}\\ \frac{dI}{ds}={\kappa}_{a}\text{\hspace{0.17em}}\left({P}_{e}-I\right)\text{\hspace{0.17em}}\end{array}\hfill \end{array}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@8546@
This equation is called Schwarzschild’s Equation. Please watch the video below (2:29).
At every infrared wavelength, the change in the irradiance is due to the absorption of irradiance coming from some source, F, and the emission of irradiance by the absorbing gas itself. If the source is relatively strong (i.e., it produces a spectral irradiance greater than the Planck distribution function spectral irradiance from the gas), then the change in F will decrease mainly according to the Beer–Lambert Law. However, if the source is relatively weak, then the irradiance that gets to us will be due mostly to the gas itself.
Let’s look at this problem another way. Suppose we have an infrared sensor and are pointing it at the Sun. The Sun radiates some energy in the infrared, but at some wavelengths the water vapor and carbon dioxide are also radiating and the sensor may be picking up their irradiance. So what is the infrared sensor seeing?
Break the atmosphere into very thin layers, so thin that the temperature and amount of water vapor and carbon dioxide is almost constant over the thickness of each layer. Start with the layer closest to the sensor. It radiates both down to the sensor but also up to the layers above it. Since this first layer is just above us, the sensor measures the infrared radiation from it.
The next layer may be at a different temperature, but it too is radiating both down and up. Some of the irradiance in this second layer might get absorbed in the first layer depending upon what the first layer’s absorptivity is. If this absorptivity is a lot, then very little of the infrared radiation from the second layer will get through the first layer, according to the Beer–Lambert Law, but at the same time, the emissivity of the first layer will be great, so the irradiance of the first layer will be close to that of the Planck distribution function spectral irradiance at the layer’s temperature. As a result, the sensor will see mostly the first layer and not the second. If, on the other hand, the first layer has low absorptivity and thus low emissivity, then irradiance from the first layer to the sensor will be weak and irradiance from the second layer will get to the sensor along with irradiance from the first layer.
If we add a third layer, it too will be radiating the same amount of energy both down and up. Some or all of its irradiance might be absorbed in the first and second layers, so depending on the absorptivity of the first two layers, the sensor may or may not see any irradiance from the third layer.
You can see how we can build a model of atmospheric radiation by stacking individual layers with distinct amounts of water vapor and carbon dioxide on top of each other, one at a time, and as we add each one, use the Beer–Lambert Law to determine how much radiation from one layer is absorbed by the layers below or above it.
Besides being absorbed or transmitted, radiation can be scattered. The scattering can be by particles of all sizes and by molecules. We can talk about the extinction of radiation by a particle that both absorbs and scatters.
$$\text{extinction=absorption+scattering,or}{\sigma}_{e,\lambda}={\sigma}_{a,\lambda}+{\sigma}_{s,\lambda}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@6EA4@
When we have absorption and scattering by particles or molecules, we simply replace the absorption cross section with the extinction cross section in Equations 6.9 to 6.13 (see section 6.10).
Scattering is messy!!!! While absorption simply removes photons, or energy, from a beam of radiation, scattering can redirect the beam of radiation into other directions, where it can be absorbed or scattered some more. Furthermore, the intensity of the scattered radiation in different directions depends strongly on the radiation's wavelength and on the size, shape, and composition of the “particle,” where we can consider a molecule to be a particle.
A very important parameter is called the size parameter:
$$x\equiv \frac{2\pi r}{\lambda}$$
where r is the radius of a spherical particle and λ is the radiation’s wavelength.
The effectiveness of scattering depends on the size parameter, which separates different types of scattering. For x > 2000, we have geometric optics. Geometric optics is when the particles get so big relative to the wavelength that we can calculate the interaction of radiation with them using optical principles (think of a lens or prism). For x = 0.2 to 2000, we have Mie scattering. Mie scattering is when the particle size is comparable to the size of the radiation wavelength, such as many atmospheric particles and light, from UV to microwave. For x between 0.002 and 0.2, we have Rayleigh scattering. Rayleigh scattering occurs for particles that are smaller than the radiation’s wavelength. This is the regime for molecules for the UV into the visible and for drizzle and raindrops for radar. Let’s look at its properties.
Without going through the math or the reasoning, I will state only that the scattering cross section for a molecule, particle, or raindrop in the Rayleigh regime (x << 1) is proportional to the sixth power of the particle radius and fourth power of the radiation wavelength.
$${\sigma}_{s}\propto \frac{{r}^{6}}{{\lambda}^{4}}$$
In the Rayleigh regime, this scattering is equally strong in the backward and forward directions, with slightly less scattering in other directions (see figure above).
This strong wavelength dependence of scattering by small particles includes molecules at the wavelengths of visible radiation. This wavelength dependence of scattering leads to the preferential scattering of blue solar radiation compared to red solar radiation, which contributes to the blue of the sky in directions away from the sun and the deep orange-to-red colored sun as it sets. The image below shows both effects in one picture—a yellow-orange sunset below the cloud and a bright blue sky above the cloud.
For radar wavelengths, the r^{6} dependence of scattering by cloud and rain drops leads to much stronger back scattering of the radar beam with increasing drop size, thus allowing for estimates of raindrop sizes and thus rainfall rates.
See below for a video (2:09) explaining scattering in greater depth:
When you feel you are ready, take Quiz 6-4. You will be allowed to take this quiz only once. Good luck!
This lesson provided you with a basic understanding of atmospheric radiation. You should now be familiar with the following basic concepts:
We will use many of these concepts in the next lesson on Applications of Atmospheric Radiation, in which we will consider two cases important to the atmosphere and weather: Earth’s radiant energy balance and Earth observed in the infrared by satellites in space.
You have reached the end of Lesson 6! Double-check that you have completed all of the activities before you begin Lesson 7.
Now that you are familiar with the principles of atmospheric radiation, we can apply them to help us better understand weather and climate. Climate is related to weather, but the concepts used in predicting climate are very different from those used to predict weather.
For climate, we need to understand the global energy budget, which is comprised of solar radiation coming into the Earth’s atmosphere and infrared radiation leaving the atmosphere to go into space. We will see that, when averaged over the Earth and over sufficient time, the energy associated with infrared radiation emitted to space by the Earth’s surface and atmosphere essentially always balances the energy associated with solar radiation absorbed by the Earth’s surface and atmosphere. By increasing atmospheric concentrations of CO_{2} and other greenhouse gases during the industrial era we have slightly perturbed this balance such that less infrared radiation is currently leaving the Earth system as compared to solar radiation being absorbed by it. This leads to additional energy being deposited into the Earth system that has been exhibited, in part, as a rise in surface air temperatures. At Earth’s surface the energy budgets of both downwelling solar and downwelling longwave radiation at short (second to minute to hour) timescales depends strongly on the presence of gases that absorb, emit, and scatter radiation in the atmosphere. Thus, Earth’s local surface temperature is exquisitely sensitive to the amounts and radiative properties of those gases and particles. We will do some simplified radiation calculations to show you how the Earth’s atmosphere affects the surface temperature.
For weather, we make predictions using models that consist of the equations of thermodynamics, motion, and microphysics. We initialize the models with observations and then let the model calculate the air motions going into the model future, thus giving weather forecasts. The models are good, but not so good that they can run for many days and continue to make accurate forecasts. So periodically, the models are adjusted by adding more observations, a process called data assimilation, in order to correct them and keep the forecasts accurate. Increasingly, satellite observations are being assimilated into the models to improve weather forecasts.
Satellite instruments observe atmospheric radiation: both visible sunlight scattered by Earth’s surface, clouds, and aerosols; and infrared radiation emitted by Earth’s surface and many of its atmospheric constituents. What the satellites measure depends on the wavelengths at which they collect radiation coming up to them. Typically, satellites observe in different wavelength bands, some of which cover wavelengths at which water vapor absorption is much stronger than for others. Taken together, the radiation in these different bands tells us much about the atmosphere’s temperature and moisture structure, which is just the kind of information that the models need to assimilate. You will learn how to interpret satellite observations of atmospheric radiation in support of applications such as vertically resolved temperature and moisture retrievals.
By the end of this lesson, you should be able to:
If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.
Let’s use what you learned in Lesson 6 to examine two applications of atmospheric radiation. The first application involves the role of atmospheric radiation and greenhouse gases in Earth’s climate. The second application is the interpretation of upwelling infrared radiation spectra measured by satellite instruments in space with an eye on improving weather forecasting. These two applications use the principles of atmospheric radiation in quite different ways, but understanding both is critical to you becoming a competent meteorologist or atmospheric scientist.
Earth’s atmosphere is essentially always in radiative energy balance, which is also called radiative equilibrium. By this, I mean that, when averaged over the whole Earth, the total amount of solar radiation energy per second that is absorbed by the Earth’s surface and atmosphere is about equal to the total amount of infrared radiation energy per second that leaves the Earth’s surface and atmosphere to go into space. There can be periods when this balance is not exact because changes in atmospheric or surface composition can alter the absorption or scattering of radiation in the Earth system. It can take a little while for all of the temperatures of all of the Earth system's parts to adjust, but if the changes stop, the Earth system will adjust its temperatures to come back into balance. Right now we are in a period where atmospheric CO_{2} concentrations are increasing due to industrialization, the outgoing infrared radiation is slightly less than the incoming absorbed solar radiation, and the Earth system's temperatures are adjusting (by increasing) to try to bring the outgoing infrared radiation into balance with the incoming absorbed solar radiation. For most of the following discussion, we will use this concept of radiative equilibrium even though the current balance is not exact.
Always keep in mind that atmospheric radiation moves at the speed of light and that all objects are always radiating. Moreover, as soon as an object absorbs radiation and increases its temperature, its emitted radiation will increase. Thus energy is not “trapped” in the atmosphere and greenhouse gases do not “trap heat.” We will see instead that greenhouse gases act like another radiation energy source for Earth’s surface.
Before we do any calculations, let's summarize how different parts of the Earth system affect visible and infrared radiation (see table below). Earth's surface either absorbs or scatters both visible and infrared radiation, while the atmosphere mostly transmits the visible radiation, with a little scattering; and the atmosphere mostly absorbs infrared radiation, with a little transmission. Clouds, an important part of the Earth system, strongly absorb infrared radiation and both scatter and absorb visible radiation.
Earth’s surface | atmosphere | clouds | ||||
---|---|---|---|---|---|---|
visible | IR | visible | IR | visible | IR | |
absorptivity | large | opaque | tiny | large | large | opaque |
emissivity | large | large | tiny | large | large | large |
scattering (reflectivity) | large | large | moderate | none | large | small |
transmissivity | none | none | large | small | none | none |
Watch this video (52 seconds) to learn more:
Here is another chance to earn 0.2 points of extra credit: Picture of the Week!
Let’s first look at the general energy balance—the radiative equilibrium—of the Earth system (see figure below). The solar irradiance is essentially composed of parallel radiation beams (or radiances) that strike half the globe. At the same time, outgoing infrared radiation is emitted to space in all directions from both the sunlit and dark sides of the globe. At the top of the atmosphere, the difference of the incoming solar radiation energy minus the amount of solar radiation energy that is scattered back to space (this difference being the amount of solar radiation energy absorbed by the Earth system) must balance the emitted infrared radiation energy for radiative equilibrium to hold. The total amount of solar radiation energy striking Earth per second is equal to the solar irradiance, F (W m^{–2}), times the Earth’s cross sectional area, $\pi {R}_{Earth}{}^{2}({\text{m}}^{2})$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaeqiWdaNaamOua8aadaWgaaWcbaWdbiaadweacaWGHbGaamOCaiaadshacaWGObaapaqabaGcdaahaaWcbeqaa8qacaaIYaaaaOGaaiiOaiaacIcacaqGTbWdamaaCaaaleqabaWdbiaaikdaaaGccaGGPaaaaa@4372@. Some of the solar radiation energy is reflected by clouds, aerosols, snow, ice, and the land surface back to space and is not absorbed, hence does not contribute energy to raise Earth’s temperature. The fraction that is reflected is called the albedo, and we can account for it by subtracting the albedo from 1 and multiplying $F\text{}\pi {R}_{Earth}{}^{2}$ times the difference: $F\text{}\pi {R}_{Earth}{}^{2}\left(1-a\right)$. The albedo has been estimated to be 0.294 (Stephens et al., 2012, Nature Geoscience 5, p. 691). On the other hand, Earth and its atmosphere radiate in all directions and the radiation can be described by the Stefan–Boltzmann Law, which, recall, is the integral of the Planck function over all wavelengths. Thus the emitted infrared energy per unit area (or emitted infrared irradiance) out the top of the atmosphere is $\sigma {T}_{top}{}^{4}(W\text{}{m}^{-2})$, where we have assumed an emissivity of 1 for the atmosphere at all emitted infrared radiation wavelengths. To get the total energy we must multiply this irradiance by the Earth’s total surface area, $4\pi {R}_{Earth}{}^{2}$. The top of the atmosphere is at an altitude of ~50–100 km above the surface, compared to Earth’s radius of 6400 km, so we will ignore this small difference.
See the video (1:37) below for a more detailed explanation:
Equating the solar radiation energy absorbed by the Earth system to the infrared radiation energy emitted by the Earth system to space gives the equation:
But what is the temperature at the top of the atmosphere, T_{top}? Put in the values F = 1361 W m^{–2}, a = 0.294, and σ = 5.67 x 10^{–8} W m^{–2} K^{–4}. Therefore,
The temperature at the top of the atmosphere is 255 K, which equals –18 ^{o}C or 0 ^{o}F. It is substantially less than Earth’s average surface temperature of 288 K, which equals 15 ^{o}C or 59 ^{o}F. This top-of-the-atmosphere temperature is the same as what the Earth’s surface temperature would be if Earth had no atmosphere but had the same albedo. It is clear from these calculations that the atmosphere, modeled with an emissivity (and hence absorptivity) of 1 over all emitted infrared radiation wavelengths, is creating a difference between the temperature at the top of the atmosphere and the temperature at Earth’s surface.
In particular, let’s look at only the vertical energy balance averaged over the entire globe. We will think of everything in terms of the SI units of irradiance (or energy per second per unit area), which is W m^{–2}. Consider two idealized cases first before examining the actual atmosphere: (1) the no-atmosphere model and (2) the infrared-opaque model.
Let’s build a simple, flat atmosphere with all solar and infrared radiation energy moving only vertically (see figure below). For both models we assume:
For the opaque-infrared model, we make the following additional assumptions:
Please watch the following video (2:18)
In the no-atmosphere model, the only radiating bodies are the Sun and the Earth. (By the way, if Earth had a pure nitrogen atmosphere, the results would be very similar to the no-atmosphere scenario.) The solar radiation passes through the altitude levels where a stratosphere and troposphere would be and the fraction 1 – a of the radiation is absorbed by the Earth’s surface. We assume that Earth’s albedo is still 0.294 so that 0.706, or 70.6%, of the solar radiation is absorbed at the surface with the rest reflected back to space. The Earth’s surface radiates infrared radiation energy back out to space with no absorption at the levels where the stratosphere and troposphere would be. The surface temperature in this model is such that the infrared radiation energy leaving the surface balances the incoming solar radiation energy absorbed by the surface. In terms of the arrows in the figure, there is one down arrow and one up arrow at every level.
model | no atmosphere |
atmosphere transparent in visible opaque in infrared |
||
---|---|---|---|---|
interface | down arrows | up arrows | down arrows | up arrows |
space–stratosphere | 1 | 1 | ||
stratosphere–troposphere | 2 | 2 | ||
troposphere–surface | 3 | 3 | ||
space–surface | 1 | 1 |
So what would the temperature at Earth’s surface be if there was no atmosphere? Equation [7-2] applies to the no-atmosphere case and hence the Earth with no atmosphere has a surface temperature of 255 K. This temperature is the same as the radiating temperature at the top of our Earth with an atmosphere whose absorptivity (and hence emissivity) is 1 at all emitted infrared radiation wavelengths. The surface would be so cold that any water on it would freeze and stay frozen.
Now consider the opaque-infrared model. The atmosphere in this model is identical to the atmosphere modeled in the derivation of Equation [7-2]. Now, however, we will be paying attention to the temperature of Earth’s surface under such an atmosphere. As before, this atmosphere is transparent to all solar radiation energy coming down to Earth’s surface and is opaque to all infrared radiation. “Opaque” means that the infrared radiation is completely absorbed over very short distances (i.e., the absorptivity and emissivity are 1, and the absorption optical depth is great, so by Beer’s Law, very little infrared radiation is transmitted). The atmosphere itself is strongly emitting in all directions, both up and down, and the only infrared radiation that does not get absorbed is that emitted out the top of the stratosphere to space.
We know that the infrared radiation leaving the Earth system must come close to balancing the solar radiation absorbed by the Earth system. Otherwise, the temperatures of Earth’s surface and atmosphere would adjust until this condition was true. So, we will assume radiative equilibrium. Our model is a two-layer model—an upper layer and a lower layer—with a solid Earth beneath them. We are assuming that each layer is at a constant temperature and absorbs all infrared radiation energy impinging on it, and then emits infrared radiation out its top and its bottom in equal amounts (because the layer emits infrared radiation energy in both directions equally). The amount of infrared radiation energy emitted by the layer is determined by its temperature only because its emissivity is set to 1 at all infrared wavelengths. Thus between the upper layer and space, we have one arrow going down and one arrow going up: the outgoing emitted infrared radiation energy exactly balances the incoming solar radiation energy that is absorbed.
The upper layer thus also emits one arrow of infrared radiation down. So, at the interface between the upper and lower layer, the solar radiation and the upper layer's infrared radiation are going down (two arrows). To be in radiative equilibrium there must be enough upwelling infrared radiation from the lower layer to equal the incoming solar radiation energy that is absorbed and the downward infrared radiation emitted by the upper layer (two arrows). But that means that the lower layer must also be emitting the same amount of infrared radiation down to Earth’s surface (two arrows).
At Earth’s surface, there is the incoming solar radiation energy that is absorbed and the tropospheric downward emitted infrared radiation, equivalent to three times the incoming solar radiation energy that is absorbed. Thus Earth’s surface must be radiating upwelling infrared radiation energy equivalent to this incoming energy to maintain radiative equilibrium. So, in this simple model Earth’s surface is radiating three times the energy that the model without the atmosphere does. But to emit this larger amount of radiation the surface must be much warmer than the surface in the model without an atmosphere. We can calculate the surface temperature that would be required using equation [7-2], but adding the downward emitted infrared radiation energy from the troposphere to the solar radiation energy. One way to look at this situation is that the lower layer is providing a source of radiation energy at the Earth’s surface in addition to the solar radiation energy.
Mathematically, we can account for this extra energy near Earth's surface by simply multiplying the solar radiant energy by an IR multiplier, multiplier_{IR} = 3, in equation [7-2]:
This temperature (336 K = 63 ^{o}C = 145^{ o}F) is deadly and much higher than Earth’s actual surface temperature, 288 K. So this model also fails to simulate the real Earth. The no-atmosphere model is too cold while the model with a two-layer, infrared-opaque atmosphere is too hot. So we can guess that something in between might be just right.
Indeed this is the case! If you look at the infrared absorption spectrum in Lesson 6, you will recall that there are some wavelengths at which all the infrared is absorbed and others, called windows, at which only a small fraction of the infrared radiation is absorbed. So, we find that a mix of total absorption, partial absorption, and no absorption at various wavelengths gives an atmosphere that allows Earth’s surface to radiate much radiation directly to space at some wavelengths but not at other wavelengths, where troposphere absorption is strong. But a large absorptivity implies a large emissivity so that at those wavelengths for which there is strong absorption there is also emission; however, given that the troposphere is cooler than the surface, the troposphere emits less upwelling infrared radiation energy than it absorbs from the warmer surface underneath. But irrespective of wavelength, emission by the troposphere is downwards as well as upwards, and provides another radiation energy source to heat Earth’s surface. This is called the greenhouse effect, which is poorly named because a greenhouse warms the Earth by suppressing heat loss by convection whereas the troposphere warms the Earth by emitting infrared radiation.
A study by Kiehl and Trenberth (1997, Bulletin of the American Meteorological Society 78, p. 197) determined the contributions to the greenhouse effect. It was shown that 81% of the greenhouse effect is due to greenhouse gases and 19% is due to clouds. Of the greenhouse effect resulting from gases, 60% is contributed by water vapor, 26% by carbon dioxide, and 14% by ozone, nitrous oxide, and methane. Though clouds are an important contributor to the greenhouse effect, they are actually more effective at reflecting solar radiation back to space: clouds cause a greenhouse warming of 30 W m^{–2} but a reflective cooling of 50 W m^{–2} for a net radiative cooling of 20 W m^{–2} (National Research Council, 2008 [81]).
In parts of the spectrum where water vapor, carbon dioxide and other gases absorb more weakly, the atmosphere is less opaque. However, if the amounts of these gases are increased, then they will absorb more strongly and thus start emitting more strongly, thus increasing the radiation emitted by the atmosphere to the surface and thus increasing the surface temperature in order for the surface to come into radiative equilibrium. Remember that the energy going out of the top of the atmosphere is still essentially the same as the solar radiation energy coming into the atmosphere that is absorbed. In a sense, by adding carbon dioxide and other greenhouse gases to the atmosphere, we are moving Earth’s surface temperature from being closer to the lower value of the no-atmosphere model to being closer to the higher value of the infrared-opaque model.
In summary, the greenhouse effect has a very dramatic impact on Earth's climate by warming the surface by 33 ^{o}C. The greenhouse effect is mostly natural and keeps the Earth habitable. Recall from Lesson 6 that the greenhouse effect probably kept the Earth at a habitable temperature early in Earth's history, when the energy output of the Sun was much lower. Despite water vapor being the most important greenhouse gas, it is condensable and thus has an extremely short lifetime in the atmosphere (about 8 days, easily calculated as the global total precipitable water divided by the global precipitation rate). It is the longer-lifetime greenhouse gases, particularly carbon dioxide, that ultimately have the bigger impact on climate. In fact, climate model simulations show that when carbon dioxide is completely removed from the atmosphere, the Earth's surface temperature drops by more than 30 ^{o}C within 30 years, consistent with our simple estimations of the greenhouse effect (Lacis et al., 2010, Science 330, pg. 356). The large temperature drop is possible because the initial cooling due to carbon dioxide reduces the amount of water vapor in the atmosphere, as expected from the Clausius–Clapeyron equation. A positive feedback, called water-vapor feedback, then ensues, leading to further cooling and even lower water vapor, until after 30 years, the water vapor content of the atmosphere is only 10% of its initial value. Water vapor thus follows carbon dioxide, which is why Penn State glaciologist and climate scientist, Richard Alley, refers to carbon dioxide as the "control knob" [82] of the Earth's climate.
The real atmosphere's energy balance includes not only radiation energy but also energy associated with evaporation and convection (see figure below). However, the atmosphere is still very close to total energy balance at each level.
First, let’s go through each set of arrows to see what is happening. The average solar irradiance at the top of the atmosphere is 340.2 W m^{–2}, which we will represent as being 100 units and then compare all other energy amounts to it.
At each level, the amount of energy going down must equal the amount of energy going up. Thus, at the top of the stratosphere, 100 units cross into the stratosphere from space, and to balance this downward energy are 30 units of reflected solar irradiance upward to space and 70 units upward emitted infrared radiation that makes it to space. At the top of the troposphere, the downwelling of 97 units of solar irradiance and 5 units of infrared irradiance is balanced by the upwelling of 30 units of reflected solar irradiance and 72 units of infrared irradiance. At Earth’s surface, the downward fluxes of solar irradiance (50 units) and infrared irradiance (89 units) balance the upward fluxes of 110 units infrared irradiance, the 24 units of latent heat, and the 5 units of sensible heat.
In reality, the Earth’s surface and atmosphere are not in simple radiative equilibrium, but are instead in radiative–convective equilibrium. Furthermore, the atmosphere is in radiative–convective equilibrium globally, but not locally (see figure below). The absorbed solar irradiance is much greater near the equator than the poles because that is where the surface is most perpendicular to the incoming solar irradiance. The radiative and convective net upward energy transport is greatest at the equator as well (because Earth’s surface is warmer there than at the poles). Overall, there is significant net incoming radiation energy between 30^{o}S and 30^{o}N latitude and a net outgoing radiation energy poleward of 30^{o} in both hemispheres.
This uneven distribution of incoming and outgoing radiation results in a flow of energy from the tropics to the poles (see figure below). It unleashes forces that cause warm air to move poleward and cold air to move equatorward. The poleward motion of warmer air, coupled with the Coriolis force that curves moving air to the right in the Northern Hemisphere and to the left in the Southern Hemisphere, causes the atmosphere’s basic wind structure, and thus its weather. We'll talk more about these forces and the resulting motion in the next few lessons when we discuss atmospheric motion (kinematics) and the forces (dynamics) that cause the motion that results in weather.
A second application of the principles of atmospheric radiation is satellite remote sensing (see figure below).
The visible channel (0.55–0.75 μm) records reflected sunlight radiances, where whiter shades are more reflected light and darker shades are less, just like in a black-and-white photograph. Land reflects more light than oceans and lakes; clouds and snow cover reflect more light than land. The visible channel goes dark at night.
The infrared window channel (10.2–11.2 μm) is over a wavelength band where the cloud-free atmosphere is transparent. As a result, it primarily records infrared radiation emitted from Earth’s surface and clouds, with emission and absorption by the gases in the atmosphere playing a secondary role. In the figure above, the greater the surface temperature (and hence the greater the radiance or radiation energy according to Equation [6-5]), the darker the shading. Thus clouds tops, which are at higher altitudes and thus colder, appear brighter.
The water vapor channel (6.5–7.0 μm) covers a strong water vapor absorption band. Thus, radiation energy at this wavelength is strongly absorbed and the radiation energy recorded by the satellite for this channel must originate from the top of the highest moist layer. Within the moist layer, the absorptivity at this wavelength is effectively 1 and it is only near the top of the moist layer that the absorption optical thickness becomes small enough that the radiation energy can escape to space and be recorded by the satellite. Note that the higher the top of the moist layer, the lower the temperature and the less radiance recorded by the satellite. Lower radiances (and hence higher, colder moist layers) are given whiter shading; darker shading is given to higher radiances (and hence lower, warmer moist layers).
A few remarks on the water vapor channel. Even the driest column of air will have enough water vapor to absorb all 6.5–7.0 μm infrared radiation emitted from Earth’s surface and just above Earth's surface. Therefore, all the radiation energy at these wavelengths recorded by the satellite comes from atmospheric water vapor at least a kilometer or more above the surface.
Second, in a drier column, some of the radiation energy emitted by water vapor at lower altitudes will not be absorbed by the water vapor above, thereby making it to space. Because lower-altitude water vapor has a higher temperature than the water vapor above, it emits a greater amount of infrared radiation than the overlying water vapor. Therefore, as a column dries and there is less high-altitude water vapor, the water vapor channel radiance recorded by a satellite will go up in value (or become darker) in the water vapor image.
Thus, brighter shades indicate emissions from higher altitudes and lower temperatures; darker shades indicate emissions from lower altitudes and thus higher temperatures. In no case, however, is the Earth's surface or the water vapor just above the Earth's surface observed. So whiter shades indicate more water vapor in a column at higher altitudes and can be used as a qualitative indicator of air moisture and as a tracer of atmospheric motion because the amount of moisture does not change significantly on daily time scales.
In Lesson 6, we derived an equation (Schwarzschild’s equation) for the change in radiance as a function of path between an infrared source and an observer:
where I is the directed beam of radiation (or radiance) along the path from the object to the observer, s is the distance along that path, P_{e} is the Planck function radiance at the temperature of the air (really the greenhouse gases in the air) along the path, and κ_{a} is the absorption coefficient of the air along the path.
Let’s apply this equation to the point-of-view of an Earth-observing satellite. Define $\tau $ (tau) as the optical path between the satellite $(\tau =0)$ and some arbitrary point along the optical path given by $\tau $ . We are not using Earth’s surface as the zero point as we often do, but instead, we are using the satellite as the zero point and letting the distance, s, and thus the optical path, change from there. The change in the optical path equals:
where the negative sign indicates that τ increases as s decreases.
Integrating both sides from the satellite to some distance s from the satellite:
To make it easier to understand what is going on, we will switch variables in [6.16] from actual distance s to optical path τ because it is the optical path, not the actual distance, that determines what the satellite detects.
This equation can be integrated to give the radiance observed by the satellite at an optical depth ${\tau}_{i}$ looking down at Earth:
So, what does this mean?
We have neglected scattering in these equations. Molecular scattering is insignificant at infrared and longer (for example, microwave) wavelengths. Cloud particle and aerosol scattering is important at visible and near-infrared (1–4 μm) wavelengths, but less so at thermal infrared (4–50 μm) wavelengths, where absorption dominates. In the thermal infrared, water clouds have an absorptivity, hence emissivity, close to 1 and emit according to the Planck function (equation 6.4).
Let's look back at a figure from Lesson 6 (reproduced below) and focus on terrestrial radiation in the atmosphere (the right side of the figure). We can see at which wavelengths the greenhouse gases in the atmosphere—mostly water vapor and carbon dioxide—absorb and at which wavelengths radiation mostly passes through the atmosphere. Note that much of Earth's infrared irradiance is absorbed by the atmosphere. However, there is a "window," extending from about 8 to 13 μm, in which most of the radiation from the Earth's surface passes through the atmosphere into space. One exception in this window is a fairly narrow band of absorption around 9.6 μm that is due to ozone.
Satellites observe radiance from both Earth's surface and from the atmosphere at different pressure levels. The example below shows the infrared spectrum observed by a special satellite, which captures the details of the full infrared spectrum. In contrast, weather satellites, such as GOES, look in only selected wavelength bands. The special satellite, in this case, observed the Earth under clear skies over the Western Pacific Ocean. Hence, the radiance observed between about 8 and 13 μm came from Earth's surface (the ocean, in this case) and had a temperature of about 295 K, or 22 ^{o}C. Hence, a GOES weather satellite in the IR band (10.2–11.2 μm, indicated in the figure below), would see the ocean surface. At wavelengths lower than 8 μm, note that the radiance is coming from sources that are colder. Specifically, the radiance is coming from water vapor with a temperature of about 260 K between 7 and 8 μm and 240 K between 6 and 7 μm. The temperature difference is due to the increase in water vapor absorptivity as wavelength decreases from 8 to 6 μm (see lower panel in figure above). Because lower temperatures are related to higher altitudes, the special satellite observed water vapor at lower altitudes near 8 μm and higher altitudes near 6 μm. Thus, satellites can observe radiance from different altitudes in the atmosphere by using different wavelengths. In this case, a GOES weather satellite in the water vapor band (6.7–7.0 μm, indicated in the figure below), would be seeing the atmosphere at an altitude where the temperature is about 250 K.
Another example from the figure below is the strong carbon dioxide and water vapor absorption near 15 μm. At wavelengths near 13 μm, the satellite is observing radiance mostly from CO_{2} and H_{2}O from lower in the atmosphere because the emissivity of CO_{2} is less at those wavelengths. At wavelengths nearer 15 μm, the CO_{2} emissivity is much greater and the satellite is observing CO_{2} and H_{2}O radiance from temperatures below 220 K and therefore much higher in the atmosphere, actually at the tropopause. Note the very narrow spike right in the middle of this strongly absorbing (and thus emitting) CO_{2} absorption band. Why does the temperature go up? Answer: In this most-strongly absorbing part of the band the satellite is seeing the CO_{2} radiance coming from the stratosphere, which is warmer than the tropopause. Note that the CO_{2} and H_{2}O at lower altitudes are emitting in the 15 μm band, but all of that radiance is being absorbed; only the layer that has no significant absorption above it can be observed by the satellite.
Watch the following video (2:46) on infrared spectrum analysis:
Look at another scene, which is the top of a thunderstorm in the tropical western Pacific. Remember that reasonably thick clouds are opaque in the infrared and therefore act as infrared irradiance sources that radiate at the temperature of their altitude. The cloud's radiance was equivalent to Planck distribution function irradiance with a temperature of 210 to 220 K. These temperatures occur at an altitude just below the tropical tropopause, which means that this storm cloud reached altitudes of 14–16 km. Note that in the middle of the 15 μm CO_{2} absorption band the satellite observed only the CO_{2} in the stratosphere (there is essentially no water vapor in the stratosphere). We know this because the radiance temperature is higher and the absorption is so strong that the radiance must be coming from higher altitudes closer to the satellite.
Let’s put all of this together.
As I said earlier, by observing the CO_{2} radiance at different wavelengths, the satellite can be sampling CO_{2} radiance from different altitudes (see figure below). The top panel is the radiance from 12 to 18 μm centered on the strong 15 μm CO_{2} absorption band. Look at the wavelengths marked 1 through 4. The bottom left panel in the figure shows the absorptivity from the top of the atmosphere to a given pressure level as a function of pressure level at these four wavelengths. Note that for the most strongly absorbed wavelength, 1, the radiance of all the CO_{2} and H_{2}O below a pressure level of about 150 hPa is completely absorbed. Thus, very little of the radiation received by the satellite comes from below this pressure level. On the other hand, very little of the radiance received from the satellite comes from above the 0.1 hPa pressure level because the absorptivity (and hence emissivity) there is zero. Thus, the radiance reaching space must primarily come from between the 150 and 0.1 hPa pressure levels. The panel on the lower right shows the relative contribution of each pressure level to the radiance that reaches space. For wavelength 1, we see that almost all radiance comes from the stratosphere (between about 100 and 1 hPa).
Look at equation 7.6 to see that the absorption of lower layers is exponential so that there are no sharp layers that emit radiance at each wavelength, but instead, the radiance the satellite observes at any wavelength comes from a band that has soft edges. If we look at the wavelength at 2, 3, and 4, we see that the CO_{2} and H_{2}O radiance comes from further down in the atmosphere. For wavelength 4, the satellite is observing radiance from Earth's surface as well as from the CO_{2} and H_{2}O below about 500 hPa, whereas for the wavelength marked 3, the radiance is only slightly from Earth's surface and mostly from CO_{2} and H_{2}O in the middle troposphere.
This week's discussion topic asks you to reflect on the impact of this lesson's material on your own thinking. Please answer the following question:
How has studying this lesson altered your thoughts about greenhouse gases and climate change?
If it has not, say why not.
Your posts need not be long, but they should tie back to the material in Lesson 7 (also Lessons 4 and 6) as well as other sources.
This discussion will be worth 3 discussion points. I will use the following rubric to grade your participation:
Evaluation | Explanation | Available Points |
---|---|---|
Not Completed | Student did not complete the assignment by the due date. | 0 |
Student completed the activity with adequate thoroughness. | Student answers the discussion question in a thoughtful manner, including some integration of course material. | 1 |
Student completed the activity with additional attention to defending their position. | Student thoroughly answers the discussion question and backs up reasoning with references to course content as well as outside sources. | 2 |
Student completed a well-defended presentation of their position, and provided thoughtful analysis of at least one other student’s post. | In addition to a well-crafted and defended post, the student has also engaged in thoughtful analysis/commentary on at least one other student’s post as well. | 3 |
Two applications of the theory of atmospheric radiation have been presented. The most important concepts used are:
For climate, these principles mean that water vapor, carbon dioxide, and other gases radiate energy to Earth’s surface, keeping it warmer than it would be if the atmosphere did not have these gases. This downward infrared radiation is the greenhouse effect, a natural phenomenon that has been enhanced by human activity, mainly fossil fuel burning. Clouds contribute to the greenhouse effect, but they actually do more cooling by reflecting visible radiation from the sun back to space.
For satellite infrared observations, some wavelength bands are in windows, so that the satellites see radiation from Earth’s surface. Other bands are completely absorbed by water vapor or carbon dioxide, so that the infrared getting to the satellite comes from the top of the water vapor column. Clouds are opaque in the infrared, so the satellite sees their tops, which are radiating at the temperature of that altitude.
You have reached the end of Lesson 7! Double-check that you have completed all of the activities before you begin Lesson 8.
In previous lessons, we were able to explain physical and chemical processes using only algebra and differential and integral calculus. Thermodynamics, moist processes, cloud physics, atmospheric composition, and atmospheric radiation and its applications can all be quantified (at this level of detail) with fairly simple mathematics. However, more math skill is required to understand and quantify the dynamics of the atmosphere.
This lesson introduces you to the math and mathematical concepts that will be required to understand and quantify atmospheric kinematics, which is the description of atmospheric motion; and atmospheric dynamics, which is an accounting of the forces causing the atmospheric motions that lead to weather. Weather is really just the wind (the motion of air in the horizontal and the vertical) and the consequences of the wind. Wind, which has both direction and speed, is best described using vectors.
The Earth is a spinning, slightly squashed sphere. The atmosphere is a tenuous thin layer on this orb, so from a human’s limited view, the Earth appears to be flat. For some applications, a simple Cartesian coordinate system, with three dimensions in the x, y, and z directions, seems like a good way to mathematically describe motion. For processes that occur on the larger scale, where the Earth’s curvature is noticeable, we must resort to using coordinates that are natural for a sphere.
The way wind direction is described sprang out of wind observations, and is now firmly implanted in the psyche of every weather enthusiast: easterly, northerly, westerly, and southerly. This wind convention, however, is quite different than that used in the equations that govern atmospheric motion, which are the basis of weather forecast models. Here we will see that a conversion between the two conventions is straightforward but requires some care.
Finally, we will see that movement of air can either be described by an observer at a fixed location (called the Eulerian framework) or by someone riding along with a moving parcel of air (called the Lagrangian framework). These two points-of-view are very different, but we will see that they are related to each other by advection, which is just the movement of air with different properties (such as temperature, pressure, and humidity) from a place upwind.
With this math and these concepts you will be ready to take on atmospheric kinematics and dynamics.
By the end of this lesson, you should be able to:
If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.
Here is another chance to earn one point of extra credit: Picture of the Week!
In your first calculus class you learned about derivatives. Suppose we have a function f that is a function of x, which we can write as f(x). What is the derivative of f with respect to x?
What about a new function that depends on two variables, h(x,y)? This function could, for example, give the height h of mountainous terrain for each horizontal point (x,y). So what is the derivative of h with respect to x? One way we determine this derivative is to fix the value of y = y_{1}, which is the same as assuming that y is a constant, and then take the ordinary derivative of h with respect to x. In a sense, we are taking a slice through the mountain in the x-direction at a fixed value of y = y_{1}. Thus,
This is called the partial derivative of h with respect to x. It’s pretty easy to determine because we do not need to worry about how y might depend on x.
Let $h={\left(x-3\right)}^{2}\mathrm{cos}\left(y\right)$ . What is the partial derivative of h with respect to x?
$$\frac{\partial h}{\partial x}=\frac{\partial \left({\left(x-3\right)}^{2}\mathrm{cos}(y)\right)}{\partial x}=2\left(x-3\right)\mathrm{cos}(y)$$
We can also find the partial derivative of h with respect to y. Can you do this?
$$\frac{\partial h}{\partial y}=\frac{\partial \left({\left(x-3\right)}^{2}\mathrm{cos}(y)\right)}{\partial y}=-{\left(x-3\right)}^{2}\mathrm{sin}(y)$$
So you can see that the $\partial h/\partial x$ may be different for each value of y and $\partial h/\partial y$ may be different for each value of x. Thus, even if you are not entirely familiar with partial derivatives and their notation, you can see that they are no different from ordinary derivatives but you take the derivative with respect to just one variable at a time.
(7:29)
Remember that a scalar has only a magnitude while a vector has both a magnitude and a direction. The following video (12:33) makes this difference clear.
Typically the vectors used in meteorology and atmospheric science have two or three dimensions. Let’s think of two three-dimensional vectors of some variable (e.g., wind, force, momentum):
Sometimes we designate vectors with bold lettering, especially if the word processor does not allow for arrows in the text. When Equations [8.3] are written with vectors in bold, they are:
Yet another notation uses parentheses and commas, as in A = (A_{x}, A_{y}, A_{z}). Be comfortable with all of these notations.
In the equations for vectors, A_{x} and B_{x} are the magnitudes of the two vectors in the x (east–west) direction, for which $\stackrel{\rightharpoonup}{i}$or i is the unit vector; A_{y} and B_{y} are the magnitudes of the two vectors in the y (north–south) direction, for which $\stackrel{\rightharpoonup}{j}$ or j is the unit vector; and A_{z} and B_{z} are the magnitudes of the two vectors in the z (up–down) direction, for which $\stackrel{\rightharpoonup}{k}$ or k is the unit vector. Unit vectors are sometimes called direction vectors.
Sometimes we want to know the magnitude (length) of a vector. For example, we may want to know the wind speed but not the wind direction. The magnitude of $\stackrel{\rightharpoonup}{A}$ , or A, is given by:
We often need to know how two vectors relate to each other in atmospheric kinematics and dynamics. The two most common vector operations that allow us to find relationships between vectors are the dot product (also called the scalar product or inner product) and the cross product (also called the vector product).
The dot product of two vectors A and B that have an angle $\beta $ between them is given by:
The dot product is simply the magnitude of one of the vectors, for example A, multiplied by the projection of the other vector, B, onto A, which is just $B\text{}cos\beta $. Thus, if A and B are parallel to each other, then their dot product is AB. If they are perpendicular to each other, then their dot product is 0. The dot product is a scalar and therefore has magnitude but no direction.
Also note that the unit vectors have the following properties:
Note that the dot product of the unit vector with a vector simply selects the magnitude of the vector's component in that direction ($\overrightarrow{i}\xb7\overrightarrow{A}={A}_{x}$) and that the dot product is commutative $(\overrightarrow{A}\xb7\overrightarrow{B}=\overrightarrow{B}\xb7\overrightarrow{A})$ .
Equation [8.4] can be rearranged to yield an expression for $\mathrm{cos}\beta $[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaci4yaiaac+gacaGGZbGaeqOSdigaaa@3ADE@ in terms of the vector components and vector magnitudes:
The cross product of two vectors A and B that have an angle $\beta $ between them is given by:
The magnitude of the cross product is given by:
where $\beta $ is the angle between A and B, with $\beta $ increasing from A to B.
Note that the cross product is a vector. The direction of the cross product is at right angles to A and B, in the right hand sense. That is, use the right-hand rule (have your hand open, curl it from A to B, and A x B will be in the direction of your right thumb). The magnitude of the cross product can be visualized as the area of the parallelogram formed from the two vectors. The direction is perpendicular to the plane formed by vectors A and B. Thus, if A and B are parallel to each other, the magnitude of their cross product is 0. If A and B are perpendicular to each other, the magnitude of their cross product is AB.
The following video (2:06) reminds you about the right-hand rule for cross products.
It follows that the cross products of the unit vectors are given by:
Note finally that $\overrightarrow{A}\times \overrightarrow{B}=-\overrightarrow{B}\times \overrightarrow{A}$.
We sometimes need to take derivatives of vectors in all directions. For that we can use a special vector derivative called the Del operator, $\overrightarrow{\nabla}$.
Del is a vector differential operator that tells us the change in a variable in all three directions. Suppose that we set out temperature sensors on a mountain so that we get the temperature, T, as a function of x, y, and z. Then $\overrightarrow{\nabla}$T would give us the change of T in the x, y, and z directions.
The Del operator can be used like a vector in dot products and cross products but not in sums and differences. It does not commute with vectors and must be the partial derivative of some variable, either a scalar or a vector. For example, we can have the following with Del operator and a vector A:
In meteorology and other atmospheric sciences, we mostly use the standard x, y, and z coordinate system, called the local Cartesian coordinate system, and the spherical coordinate system. Let’s review some of the main points of these two systems.
The local Cartesian coordinate system applies to three dimensions (as seen in the figure below). The convention is simple:
Unit vectors (length 1 along standard coordinates) are i (east); j (north); k (up).
Often we will consider motion in two dimensions as being separate from motions in the vertical. We usually denote the horizontal with a subscript H; for example, r_{H} = i x + j y, where r_{H} is a horizontal distance vector.
This coordinate system works well over relatively small scales on Earth, perhaps the size of an individual state, where Earth's curvature is not important. This is why the qualifier "local" is used in the name of the coordinate system. This coordinate system does not work so well for large-scale motion on Earth, which is spherical.
When we place the Cartesian coordinate system on a sphere, note that x always points toward the east, y always points toward the north, and z always points up along the direction of Earth’s radius (as seen in the figure above).
Life would be much easier if the Earth were flat. We could then use the local Cartesian coordinate system with no worries. But the Earth is very nearly a perfect sphere, which implies that to accurately describe motion, we must take the Earth’s spherical shape into account.
We use the following terms:
Note that 1^{o} of latitude is always 111 km or 60 nautical miles, but 1^{o} of longitude is 111 km only at the equator. It is smaller in general and equal to 111 km x cos($\Phi $). Note that 1 nautical mile = 1.15 miles.
To find the horizontal distance between any two points on Earth's surface, we first need to find the angle of the arc between them and then we can multiply this angle by Earth's radius to get the distance. To find the angle of the arc, $\Delta \sigma $ , we can use the Spherical Law of Cosines:
where the latitude and longitude of the two points are ${\Phi}_{1},\text{}{\lambda}_{1}and\text{}{\Phi}_{2},\text{}{\lambda}_{2}$ , respectively and $\Delta \lambda =|{\lambda}_{1}-{\lambda}_{2}|$ is the absolute difference between the longitudes of the two points. Note that the angle of the arc must be in radians, where $2\pi \text{}radians\text{}=\text{}{360}^{o}$. To find the distance, simply multiply this arc angle by the radius of the Earth, 6371 km.
Show that 1^{o} of latitude = 111 km distance.
Distance = 6371 km * (1/360)*2π = 111.2 km
In summary, we will use a local Cartesian coordinate system when our scales of interest are not too large (synoptic scale or smaller), but will need to use spherical coordinates when the scale of interest is larger than synoptic scale.
For another explanation of these two systems, visit this Coordinate Systems website [88].
Three different vertical coordinates are used in meteorology and atmospheric science: height, pressure, and potential temperature.
We have already introduced the vertical coordinate z, which is a height, usually in m or km, above the Earth's surface in the local Cartesian coordinate system; z is related to r in spherical coordinates through r = a + z, where a is the Earth's radius. The vertical coordinate z is the most commonly used in meteorology and in any process that involves getting off the ground, such as flight. Often pilots talk about flight levels, which are measured in hundreds of feet. So, flight level 330 is about 10 km altitude.
Another useful vertical coordinate is pressure, which decreases with height. Pressure is often a useful vertical coordinate in dynamics calculations. To a good approximation, pressure falls off exponentially with height, p = p_{o}exp(-z/H), as we learned in Lesson 2, so that ln(p) is fairly linear with height. We’ll get into this in greater detail later. For now, consider the following table of typically used pressure heights:
altitude (km) | altitude (kft) | pressure level (hPa or mb) |
---|---|---|
0 | 0 | 1000 |
1.5 | 4.4 | 850 |
3.0 | 9.9 | 700 |
5.5 | 18.3 | 500 |
A third important vertical coordinate is potential temperature, $\theta $ (Equation 2.58). This quantity is the temperature that an air parcel would have if it were brought to a pressure of 1000 hPa without any exchange of heat with its surroundings. This vertical coordinate has a nice property: air parcels tend to move on surfaces of constant potential temperature because moving on such a surface requires no energy. This coordinate is particularly useful in the stratosphere, where the rapid increase with altitude tends to keep air motion stratified.
Meteorologists use terms such as northeasterlies and southerlies when they describe winds. These terms designate directions that the winds come from. But when we think about the dynamic processes that cause the wind, we use the conventions for direction that are common in mathematics and in coordinate systems like the Cartesian coordinate system. The conversion between the two conventions—math and meteorology—is not simple. However, we will show you a simple way to do the conversion (see the second figure below).
The wind vector is given by U = i u + j v + k w. The wind vector points to the direction the wind is going.
The subscript “H” will be used to denote horizontal vectors, such as the horizontal velocity, U_{H} = i u + j v (though note that sometimes the symbols V , v_{H}, and v will be used to denote the horizontal velocity). The magnitude of U_{H} is U_{H} = (u^{2} + v^{2})^{1/2}. The math wind angle, $\alpha $, is the angle of the wind relative to the x-axis, so that tan($\alpha $) = v/u and the angle increases counterclockwise as the direction moves from the eastward x-axis ($\alpha $ = 0^{o}) to the northward y-axis ($\alpha $ = 90^{o}) .
The meteorology wind convention is often used in meteorology, including station weather plots [90]. The wind vector points to the direction the wind is coming from. The angle is denoted by delta, $\delta $, which has the following directions:
direction wind is coming from | angle $\delta $ |
---|---|
north (northerlies or southward) | 0^{o} |
east (easterlies or westward) | 90^{o} |
south (southerlies or northward) | 180^{o} |
west (westerlies or eastward) | 270^{o} |
Meteorology angles, designated by $\delta $, increase clockwise from the north (y) axis. Math angles, designated by $\alpha $ , increase counterclockwise from the east (x) axis.
In the diagram on the left, the wind is southwesterly, the meteorology angle (measured clockwise from the north or y-axis) δ = 225^{o }, and the math angle (measured counterclockwise from the east or x-axis) $\alpha \text{}=\text{}{45}^{o}$. If the wind is northerly (southward), the wind vane points to the north, the wind blows to the south, δ = 0^{o} and α = 270^{o}. If the wind is westerly (eastward), δ = 270^{o} and α = 0^{o}.
Note that in all cases, we can describe the relationship between the math and the meteorology angles as:
$$math\text{}angle\text{}=\text{}27{0}^{o}-\text{}meteorology\text{}angle$$
When the meteorology angle is greater than 270^{o}, the math angle will be negative but correct. However, to make the math angle positive, simply add 360^{o}.
Drawing a figure like those shown in the figure above often helps when you are trying to do the conversion. The following video (2:17) explains the conversion between meteorology and math wind angles using the figure above.
The gradient of a variable is just the change in that variable as a function of distance. For instance, the temperature gradient is just the temperature change divided by the distance over which it is changing: $\Delta T/\Delta distance$. The gradient is a vector and thus has a direction as well as magnitude.
Consider the surface temperature contour plot from NOAA for 8 September 2012 in the figure above. A strong temperature variation is draped across the eastern and southern US from New York down to Texas. How do we quantify this temperature variation? First we have to specify where we want to measure the temperature gradient. Then we simply need to choose isotherms on either side of the point, take the difference between the isotherms, figure out how far apart they are in horizontal distance, and divide the temperature change between the isotherms by the distance between the isotherms. The direction for the gradient is perpendicular to the isotherms and pointing from the lower temperatures to the higher temperatures. It’s pretty easy to figure out where the gradient vector points just by quick examination, but it is a little harder to figure out what the gradient magnitude and actual direction are.
Now watch this video (2:12) on finding distances:
Mathematically, if we know the algebraic expression for the temperature change, such that T = T(x,y), we can find the gradient by using the del operator, which is also called the gradient operator.
Recall the del operator:
If we are looking only at changes in x and y, then we can define a horizontal del operator:
At any point, we can determine the gradient of the temperature:
Note that this quantity has dimensions of $\theta /L$ and a magnitude and a direction. The gradient direction is always normal to the isolines and pointing in the direction of an increase. We can define the normal vector, which is just the unit vector in the direction of the increasing temperature. We will call this normal vector n.
We can calculate a gradient for every point on the map, but to do this we need to know the change in the temperature over a distance that is centered on our chosen point. One approach is to calculate the gradients in the x and y directions independently and then determine the magnitude by:
and the direction by:
We can program a computer to do these calculations.
However, often we just want to estimate the gradient. The gradient can be determined by looking at the contours on either side of the point and computing the change in temperature over the distance. These partial derivatives can be approximated by small finite changes in temperatures and distances, so that $\partial $ is replaced by $\Delta $ in all places in these equations. We can calculate gradients by using “centered differences” as shown in the figures below.
We then calculate the magnitude with Equation [8.12] and the direction with Equation [8.13], where we replace the partial derivatives with the small finite differences in all places in these equations.
The magnitude and direction are:
When you calculate the arctangent, keep in mind that the tangent function has the same values every 180^{o} or every π radians. If you get an answer for the arctangent that is 45^{o}, how do you know whether the angle is really 45^{o} or 45^{o} + 180^{o} = 225^{o}? The gradient vector always points towards higher values, so always choose the angle so that the gradient vector points toward the warmer air. Alternatively, if ∂T/∂x is greater than 0, choose the value of the arctangent between –90^{o} and 90^{o}, whereas if ∂T/∂x is less than 0, add 180^{o}.
Recap of the process for calculating the temperature gradient:
Now watch this video (3:52) on finding gradients:
A word about finding gradients in the real world. Sometimes the centered differences method is difficult to apply because the gradient is too much east–west or north–south. For instance, in the temperature map at the beginning of this section, the x-gradient is hard to determine by the centered difference method in the Oklahoma panhandle and the y-gradient is hard to determine in central Pennsylvania because in both cases, the temperature hardly changes those directions. In these cases, you could say that the gradient in that direction equals 0, but then your computer program might have a hard time finding the arctangent. One way around this problem is to put in a very small number for the gradient in that direction, say 1 millionth of your typical gradient numbers, to do the calculation.
A second word about finding gradients in the real world. When you are finding temperature gradients from a temperature map, it is sometimes hard to determine the temperature gradient at some locations because the isotherms are not evenly spaced and can be curvy. Don't despair! Use your best judgment as to what the gradients are. Check your answers for the magnitude and direction of the temperature gradient vector by estimating the magnitude and direction by eyeballing the normal to the isotherms at that location and pointing the gradient vector to the warmer air. If your calculated direction is very different from your eyeballed value by more than, say, 45^{o}, check your math.
Suppose you were driving underneath an eastward-moving thunderstorm at the same speed and in the same direction as the storm, so that you stayed under it for a few hours. From your point-of-view, it was raining the entire time you were driving. But from the point-of-view of people in the houses you passed, the thunderstorm approached, it rained really hard for twenty minutes, and then stopped raining.
The people in houses formed a network of observers, and if they talked to each other, they would find that the thunderstorm moved rapidly from west to east in a path that rained on some houses and missed others. This point-of-view is called the Eulerian description because it follows the thunderstorm through a fixed set of locations. You, on the other hand, followed along with the thunderstorm; any changes you saw were due to changes in the thunderstorm’s intensity only. Your point-of-view was a Lagrangian description because you stayed with the storm.
We can now generalize these ideas to any parcel of air, not just a thunderstorm. An air parcel is a blob of air that hangs more-or-less together as it moves through the atmosphere, meaning that its mass and composition are conserved (i.e., not changing) as it moves. It has a fairly uniform composition, temperature, and pressure, and has a defined velocity.
Air parcels do not live forever. They are just air moving in air, so they change shape as they bump into other air parcels and they mix until they disappear. Larger air parcels tend to live longer than smaller air parcels. Even though air parcels do not live forever, they are very useful concepts in explaining the differences between the Eulerian and Lagrangian frameworks.
Eulerian Framework:
Lagrangian Framework:
Here's an interesting fact. Radiosonde measurements are considered Eulerian because they measure properties like temperature and humidity at a fixed location. However, the horizontal velocity is measured by a radiosonde’s lateral movement with the horizontal wind and so is actually a Lagrangian measurement. A cup anemometer or sonic anemometer is an example of a Eulerian method of measuring wind velocity.
In the lesson, I presented an example of a driver in a car that happened to be traveling at the same speed as a rainstorm so that to the driver it was raining the entire time, while to the observers in the houses that the driver passed, the rain showers were brief. We could have given a second case in which there was widespread rain so that both the car driver and the house occupants observed constant rainfall for several hours. In the first case, the advection exactly matched the local rainfall change wherever the driver was, so that the driver was in rain constantly, but the house occupants saw rain only for a short while. In the second case, there was no gradient in the rainfall rate, so even if the storm was moving, neither the driver nor the house occupants observed a change in rainfall.
Think of one event or phenomenon from the Eulerian and Lagrangian points-of-view, one in which the event or phenomenon looks very different from the two points-of-view. These events do not have to be weather related. Be creative.
This discussion will be worth 3 discussion points. I will use the following rubric to grade your participation:
Evaluation | Explanation | Available Points |
---|---|---|
Not Completed | Student did not complete the assignment by the due date. | 0 |
Student completed the activity with adequate thoroughness. | Student answers the discussion question in a thoughtful manner, including some integration of course material. | 1 |
Student completed the activity with additional attention to defending their position. | Student thoroughly answers the discussion question and backs up reasoning with references to course content as well as outside sources. | 2 |
Student completed a well-defended presentation of their position, and provided thoughtful analysis of at least one other student’s post. | In addition to a well-crafted and defended post, the student has also engaged in thoughtful analysis/commentary on at least one other student’s post as well. | 3 |
Changes over time in properties, such as temperature and precipitation, can be expressed in Lagrangian and Eulerian frameworks, and often the changes are different in the two frameworks (as in the precipitation change for the thunderstorm example). We can express these changes mathematically with time derivatives. Suppose we have a scalar, R, which could be anything, but let’s make it the rainfall rate. It is a function of space and time:
To find the change in the rainfall rate R in an air parcel over space and time, we can take its differential, which is an infinitesimally small change in R:
where dt is an infinitesimally small change in time and dx, dy, and dz are infinitesimally small changes in x, y, and z coordinates, respectively, of the parcel.
If we divide Equation [8.15] by dt, this equation becomes:
where dx/dt, dy/dt, and dz/dt describe the velocity of the air parcel in the x, y, and z directions, respectively.
Let’s consider two possibilities:
Case 1: The air parcel is not moving. Then the change in x, y and z are all zero and:
So, the change in the rainfall rate depends only on time. $\frac{\partial R}{\partial t}$ is called the Eulerian or local time derivative. It is the time derivative that each of our weather observing stations record.
Case 2: The air parcel is moving. Then the changes in its position occur over time, and it moves with a velocity, $\overrightarrow{U}=\overrightarrow{i}u+\overrightarrow{j}v+\overrightarrow{k}w$ , where:
A special symbol is given for the derivative when you follow the air parcel around. It is called the substantial derivative or total derivative and is denoted by:
Mathematically, we can express this equation in a more general way by thinking about the dot product of a vector with the gradient of a scalar as we did in an example of the del operator:
where the second term on the right hand side is called the advective derivative, which describes changes in rainfall that are solely due to the motion of the air parcel through a spatially variable rainfall distribution.You should be able to show that equation [8.19] is the same as equation [8.18].
We can rearrange this equation to put the local derivative on the left.
The term on the left is the local time derivative, which is the change in the variable R at a fixed observing station. The first term on the right is the total derivative, which is the change that is occurring in the air parcel as it moves. The last term on the right, $-\overrightarrow{U}\xb7\overrightarrow{\nabla}R$, is called the advection of R. Note that advection is simply the negative of the advective derivative.
To go back to the analogy of the thunderstorm, the change in rainfall that you observed driving in your car was the total time derivative and it depended only on the change in the intensity of the rain in the thunderstorm. However, for each observer in a house, the change in rainfall rate depended not only whether the rainfall within the thunderstorm was changing with time (which would depend, for example, on the stage of the storm) but also on the movement of the thunderstorm across the landscape.
R can be any scalar. Rainfall rate is one example, but the most commonly used are pressure and temperature.
Equation [8.20] is called Euler’s relation and it relates the Eulerian framework to the Lagrangian framework. The two frameworks are related by this new concept called advection.
Let’s look at advection in more detail, focusing on temperature.
We generally think of advection being in the horizontal. So often we only consider the changes in the x and y directions and ignore the changes in the z direction:
So what’s with the minus sign? Let’s see what makes physical sense. Suppose T increases only in the x-direction so that:
If u > 0 (westerlies, blowing eastward), then both u and $\frac{\partial T}{\partial x}$ are positive so that temperature advection is negative. What does this mean? It means that colder air blowing from the west is replacing the warmer air, and the temperature at our location is decreasing from this advected air. Thus $\frac{\partial T}{\partial t}$ should be negative since time is increasing and temperature is decreasing due to advection.
If the temperature advection is negative, then it is called cold-air advection, or simply cold advection. If the temperature advection is positive, then it is called warm-air advection, or simply warm advection.
Some examples of simple cases of advection show these concepts (see figure below). When the wind blows along the isotherms, the temperature advection is zero (Case A). When the wind blows from the direction of a lower temperature to a higher temperature (Case B), we have cold-air advection. When the wind blows as some non-normal direction to the isotherms, then we need to multiply the magnitude of the wind and the temperature gradient by the cosine of the angle between them. We can estimate the temperature advection by doing what we did for the gradient, that is, replace all derivatives and partial derivatives with finite $\Delta s$ .
When the isotherms with the same temperature difference are further apart on the map (see figure below), then the horizontal temperature advection will be less than when the isotherms are closer together, if the wind velocity is the same in the two cases.
In summary, to calculate the temperature advection, first determine the magnitude and the direction of the temperature gradient. Second, determine the magnitude and direction of the wind. The advection is simply the negative of the dot product of the velocity and the temperature gradient.
Watch this video (2:20) on calculating advection:
You now have all of the math tools that you will need to understand the lessons on kinematics and dynamics. For some of you, the concept of partial derivatives is new, but you see how easy it is to apply. We have reviewed vectors and a little vector calculus that you will need in the next few lessons. Dot products and cross products will appear frequently in the next few lessons, so make sure that you are comfortable with them and their applications. We introduced a new operator, the “del” or “gradient” operator, which is essential for describing weather observations and how atmospheric properties vary in space. There are two major coordinate systems that are used to describe atmospheric motion: local Cartesian and spherical. And there are three different coordinates that are used to define the vertical distance starting at Earth’s surface and rising radially: height z (m); pressure or logarithm of pressure p (hPa); and potential temperature $\theta $ (K). You will become skilled in converting between the meteorological wind directions shown on weather maps and in station weather plots and the wind direction needed to do calculations of wind motion and its effects. We introduced the concepts of the Eulerian and Lagrangian frameworks and showed that they are related by advection of a scalar, which is simply the dot product of the wind velocity and the gradient of the scalar.
Once you successfully complete the activities in this lesson, you will be ready to learn about atmospheric kinematics (the description of air movement) and atmospheric dynamics (the study of why air moves the way that it does).
You have reached the end of Lesson 8! Double-check that you have completed all of the activities before you begin Lesson 9.
The study of kinematics provides a physical and quantitative description of our atmospheric motion, while the study of dynamics provides the physical and quantitative cause-and-effect for this motion. This lesson discusses kinematics.
When we look at weather in motion from a satellite, we see very complicated swirls and stretching that evolve over time. We can see the same types of motions on a much smaller scale by observing swirling leaves. These complex motions can be ascribed to combinations of just five different types of atmospheric motion. Quantifying these motions with mathematics, without assigning a cause to the motion, is the focus of this lesson on kinematics.
By the end of this lesson, you should be able to:
If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.
Streamlines are lines that are everywhere parallel to the velocity vectors at a fixed time. They consider the direction of the velocity but not the speed. Sometimes more streamlines are drawn to indicate greater speed, but this is not usually done. Streamlines generally change from one time to the next, giving us “snapshots” of the motion of air parcels. For maps of wind observations for a fixed time, we often look at streamlines. On a map of streamlines, you will see that the lines aren’t always straight and don’t always have the same spacing. Confluence is when streamlines come together. Diffluence is when they move apart.
Trajectories are the actual paths of the moving air parcels, and indicate both the direction and velocity of air parcels over time. Convergence is when the velocity of the air slows down in the direction of the streamline. Divergence is when the velocity of the air speeds up in the direction of the streamline. We will talk more about convergence/divergence later, but for now, you should understand that convergence/divergence come from changes in velocity while confluence/diffluence come from changes in spacing between streamlines.
Confluence/diffluence and convergence/divergence are illustrated in the figure below:
Here is another chance to earn one point of extra credit: Picture of the Week!
The water vapor image from the GOES 13 satellite, above, indicates different air masses over the United States. As we know from Lesson 7, the water vapor image actually shows the top of a column of water vapor that strongly absorbs in the water vapor channel wavelengths, but it is not a bad assumption to think that there is a solid column of moister air underneath the water vapor layer that is emitting and is observed by the satellite. In a single snapshot, it is not possible to see what happens to the air parcels over time. But if we look at a loop, then we can see the air parcels moving and changing shape as they move.
Visit this website to see a loop [95]. Pick any air parcel with more water vapor in the first frame and then watch it evolve over time. What does it do? Maybe it moves; it spins; it stretches; it shears; it grows. Maybe it does only a few of these things; maybe it does them all.
We can break each air parcel’s complex behavior down into a few basic types of flows and then mathematically describe them. We will just describe these basic motions here and show how they lead to weather.
Assume that we have an air parcel as in the figure above. We focus on motion in the two horizontal directions to aid in the visualization (and because most motion in the atmosphere is horizontal) but the concepts apply to the vertical direction as well. If the air parcel is moving and does not change its orientation, shape, or size, then it is only undergoing translation (see figure below).
The air parcel can do more than just translate. It can undergo changes relative to translation, and its total motion will then be a combination of translation and relative motion. Let’s suppose that different parts of the air parcel have slightly different velocities. This situation is depicted in the figure below.
If we consider very small differences dx and dy, then we can write u and v at point (x_{o} + dx,y_{o} + dy) as a Taylor series expansion in two dimensions:
We see that u(x_{o},y_{o}) and v(x_{o},y_{o}) are the translation, and the relative motion is expressed as gradients of u in the x and y directions and gradients of v in the x and y directions.
There are four gradients represented by the four partial derivatives. Each can be either positive or negative for each partial derivative.
$\frac{\partial v}{\partial x}$ is the following change in velocity in the x direction:
Note that a partial derivative is positive if a positive value is becoming more positive or a negative value is becoming less negative. Similarly, a negative partial derivative occurs when a positive value is becoming less positive or a negative value is becoming more negative. Be sure that you have this figured out before you go on.
Watch this video (2:38) for further explanation:
Generally, air velocities change with distance in such a way that more than one partial derivative is different from zero at any time. It turns out that any motion of an air parcel is a combination of five different motions, one being translation, which we have already discussed, and four of which can be represented by pairs of partial derivatives of velocity. Of these four, one is a deformation of the air parcel, called stretching, which flattens and lengthens the air parcel. A second is another deformation of the air parcel, called shearing, which twists the air parcel in both the x and y directions. A third is pure rotation, called vorticity. A fourth enlarges or shrinks the parcel without changing its shape, called divergence. Let’s consider each of five types of air motion alone, even though more than one is often occurring for an air parcel.
Translation simply moves the air parcel without stretching it, shearing it, rotating, or changing its area. There are no partial derivatives of velocities involved with translation.
For the remaining four cases, we will provide examples in which the motion (stretching, shearing, vorticity, and divergence) has a positive value. We could have provided examples in which the motion has negative value, but the conclusions would be the same.
Stretching deformation is represented by $\frac{\partial u}{\partial x}-\frac{\partial v}{\partial y}$. u gets more positive as x gets more positive and u gets more negative as x gets more negative (so that the derivative is always positive), making the parcel grow in the x direction. In the other direction, v gets more negative as y gets more positive and v gets more positive as y gets more negative (so that the derivative is always negative), making the parcel shrink in the y direction (see figure below). However, the total area of the air parcel will remain the same if $\partial u/\partial x=\partial v/\partial y$. Shown in the figure is positive stretching deformation; negative stretching deformation occurs when the parcel is stretched in the y direction.
Shearing deformation is represented by $\frac{\partial v}{\partial x}+\frac{\partial u}{\partial y}$. In this case, v gets more positive as x gets more positive and v gets more negative as x gets more negative, resulting in the air parcel part at lower x getting pushed towards lower y, and the air parcel part at higher x getting pushed towards higher y. At the same time, u gets more positive as y gets more positive and u gets more negative as y gets more negative, resulting in the air parcel part at lower y getting pushed to lower x and the air parcel part at higher y getting pushed to higher x (see figure below). The total area of the air parcel remains the same after the shearing occurs. Shearing deformation is positive when the air parcel stretches in the southwest/northeast direction and contracts in the southeast/northwest direction (as in the figure below). Shearing deformation is negative when the parcel stretches in the southeast/northwest direction and contracts in the southwest/northeast direction.
As the two figures above show, both stretching and shearing deformation cause stretching along the axis of diffluence and contraction along the axis of confluence, with the two axes at right angles to each other. These deformations result in weather fronts. In both cases, these motions cause some parts of the air parcel to move away from each other and some parts of the air parcel to move towards each other. The air coming together is called frontogenesis.
Vorticity is represented by $\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\equiv \zeta $ . Vorticity is special, and because it is special, it is represented by a Greek lower-case letter, zeta (ζ). In this case, the air parcel does not get distorted if $\partial v/\partial x=-\partial u/\partial y$ and does not change area. It simply rotates (see figure below).
This difference in partial derivatives may look familiar to you.
Vorticity is actually a vector that follows the right-hand rule. Your fingers curve in the direction of the flow and your thumb is the vorticity vector. Here we are discussing only the vertical component of the vorticity. In a right-handed coordinate system, counter-clockwise flow in the x-y plane will result in your thumb pointing in the positive z direction. Hence, vorticity is positive if the rotation is counter-clockwise and is negative if the rotation is clockwise. In the Northern Hemisphere, low-pressure systems are typically characterized by counter-clockwise flow and thus have positive vorticity whereas high-pressure systems are typically characterized by clockwise flow and thus have negative vorticity. The vorticity definition is the same in the Southern Hemisphere (with counter-clockwise flow being positive and clockwise flow being negative), but low-pressure systems usually have clockwise flow and high-pressure systems usually have counterclockwise flow. Vorticity is an important quantity because low- and high-pressure systems are responsible for a lot of weather.
Divergence is represented by $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\equiv \delta $ . Divergence is also special, and because it is special, it is represented by a Greek lower-case letter, delta ($\delta $ ). When the divergence is positive, the air parcel grows (i.e., its area increases) (see figure below). If the divergence is negative, then the air parcel shrinks (i.e., its area decreases). Strictly speaking, $\delta $ is the horizontal divergence because it describes a change in parcel area projected onto a horizontal plane. Adding ∂w/∂z to the horizontal divergence gives the 3-D divergence.
The divergence can be written in vector notation:
Watch this video (1:56) for further explanation:
We see that divergence is positive when the parcel area grows and is negative when it shrinks. We call growth “divergence” and shrinking “convergence.” We wish to know whether air parcels come together (converge) or spread apart (diverge) or if the parcel area increases with time (divergence) or decreases with time (convergence).
Let’s see how divergence in the horizontal two dimensions is related to area change. We can do a similar analysis that relates divergence in three dimensions to a volume change, but we will stay with the two-dimensional case because it is easier to visualize and also has important applications. Consider a box with dimensions Δx and Δy. Different parts of the box are moving at different velocities (see figure below).
The box's area, A, is given by:
$$\begin{array}{l}A=\Delta x\text{\hspace{0.17em}}\Delta y\hfill \\ \frac{dA}{dt}=\frac{d(\Delta x\text{\hspace{0.17em}}\Delta y)}{dt}=\Delta x\text{\hspace{0.17em}}\frac{d(\Delta y)}{dt}+\Delta y\frac{d(\Delta x\text{\hspace{0.17em}})}{dt}=\hfill \\ \text{\hspace{1em}}\text{\hspace{1em}}\text{\hspace{1em}}\Delta x\left[v(y+\Delta y)-v(y)\right]+\Delta y\left[u(x+\Delta x)-u(x)\right]\hfill \\ \hfill \\ \text{divideby}A\text{=}\Delta x\text{\hspace{0.17em}}\Delta y\hfill \\ \hfill \\ \frac{1}{A}\frac{dA}{dt}=\frac{v(y+\Delta y)-v(y)}{\Delta y}+\frac{u(x+\Delta x)-u(x)}{\Delta x}\hfill \\ \hfill \\ \text{Let}\Delta y\to \text{0,}\Delta x\to \text{0}\hfill \\ \hfill \\ \frac{1}{A}\frac{dA}{dt}=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}={\overrightarrow{\nabla}}_{H}\u2022{\overrightarrow{U}}_{H}\hfill \end{array}$$[2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@E7CE@
So we see that the fractional change in the area is equal to the horizontal divergence. Note that the dimension of divergence is time^{–1} and the SI unit is s^{–1}.
We can do this same analysis for motion in three dimensions to get the equation:
$$\frac{1}{V}\frac{dV}{dt}=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}={\overrightarrow{\nabla}}_{H}\xb7{\overrightarrow{U}}_{H}+\frac{\partial w}{\partial z}=\overrightarrow{\nabla}\xb7\overrightarrow{U}$$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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aagaWca8qacqWIpM+zceWGvbWdayaalaaaaa@62C1@
where V is the parcel volume. Thus, the 3-D divergence is just the fractional rate of change of an air parcel’s volume.
Suppose that an air parcel has an area of 10,000 km^{2} and it is growing by 1 km^{2} each second. What is its divergence?
$\frac{\Delta A}{\Delta t}=1{\text{km}}^{2}{\text{s}}^{-1}$ , so $\left(\frac{1}{A}\right)\left(\frac{\Delta A}{\Delta t}\right)=\left(\frac{1}{{10}^{4}}{\text{km}}^{2}\right)\left(1{\text{km}}^{2}{\text{s}}^{-1}\right)={10}^{-4}{\text{s}}^{-1}$ .
Suppose that an air parcel has a area of 10,000 km^{2} and has a divergence of –10^{–4} s^{–1}. Is the air parcel growing or shrinking?
$\text{divergence=}\delta =\left(\frac{1}{A}\right)\left(\frac{\Delta A}{\Delta t}\right)$ , or $\frac{\Delta A}{A}=\delta \Delta t=\left(-{10}^{-4}{\text{s}}^{-1}\right)\left(1\text{s}\right)=-{10}^{-4}$ . The air parcel is shrinking.
Check out this video (1:33) for further explanation:
Our goal here is to relate horizontal convergence and divergence to vertical motion. If vertical motion is upward, then the uplifted air will cool, clouds will form, and it might rain or snow. If vertical motion is downward, then the downwelling air will warm by adiabatic descent, clouds will evaporate, and it will become clear.
To find out what will happen, we need to go back to a fundamental law of mass conservation [96], which we will derive in detail in Lesson 10. Here we simply quote the result:
where $\rho $ is the density and D/Dt is the total derivative.
For divergence,$\overrightarrow{\nabla}\u2022\overrightarrow{U}>0$ , volume increases and density must decrease to conserve mass.
For convergence,$\overrightarrow{\nabla}\u2022\overrightarrow{U}<0$ , volume decreases and density must increase to conserve mass.
However, to good approximation, density does not change with time for any given horizontal surface. Sure, density decreases exponentially with height, but for each height level, the density at that level is fairly constant.
So, to a good approximation:
and because we can separate out the horizontal and vertical components of divergence:
we see that:
Thus, horizontal divergence is compensated by vertical convergence and horizontal convergence is compensated by vertical divergence.
Horizontal divergence gives a decrease in vertical velocity with height.
Horizontal convergence gives an increase in vertical velocity with height.
Now, in the troposphere, the vertical velocity is close to zero (w ~ 0) at two altitudes. The first is Earth’s surface, which forms a solid boundary that stops the vertical wind. The second is the tropopause, above which the rapid increase in stratospheric potential temperature strongly inhibits vertical motion from the troposphere (see two figures below), so much so, that we can say that the vertical wind must be ~ 0 at the tropopause.
These processes can be summarized in the following table:
plane | process | surface area change | ∂w/∂z | w |
---|---|---|---|---|
surface | convergence | decrease | + | up |
surface | divergence | increase | – | down |
aloft | convergence | decrease | + | down |
aloft | divergence | increase | – | up |
Let’s now consider the effect that divergence/convergence aloft has on surface convergence/divergence (see figure below).
Divergence aloft is associated with rising air throughout the troposphere, which is associated with low pressure and convergence at the surface.
Convergence aloft is associated with sinking air throughout the troposphere, which is associated with high pressure at the surface and thus divergence at the surface.
So, starting at the surface, the vertical velocity becomes more positive with height when there is surface convergence, reaches some maximum vertical velocity, and then becomes less positive with height again toward the divergence aloft.
Similarly, starting again at the surface, the vertical velocity becomes more negative with height when there is surface divergence, reaches some maximum negative velocity, and then becomes less negative with height again near convergence aloft.
Now watch this video (3:52) on horizontal divergence:
We have shown that convergence and divergence aloft near the tropopause is related to surface highs and lows. Now it's your turn to find some examples. Go to a source of information about surface pressure and upper-air winds and pick out some regions that show this relationship. One good source is the Penn State e-Wall [97], for which you can use the "U.S. Satellite Overlays."
What are typical values of the vertical velocity caused by convergence or divergence and how do they vary with height? The vertical velocity, w, is typically too small to measure by a radiosonde. But we can estimate w from the convergence/divergence patterns:
Note that this equation just gives the derivative of the vertical velocity, not the vertical velocity itself. So to find the vertical velocity, we must integrate both sides of the equation over height, z.
Integrate this equation from the surface (z = 0) to some height z:
where we have assumed that w(0) is equal to zero, which is true if the surface is horizontal. Equation [9.6] gives the kinematic vertical velocity.
To a good approximation, it has been determined that divergence/convergence for horizontal flow at large scales (e.g., synoptic scales, ~1000 km) varies linearly with altitude.
where δ_{s} is the surface divergence and b is a constant. Substituting this expression for the horizontal divergence into Equation [9.6], we get:
The trick is to find b using some other information. To find b, note that the derivative $\frac{\partial w}{\partial z}$ must equal 0 at some level because w must be 0 at both Earth's surface and the tropopause while, in general, w is non-zero elsewhere. If, for example, the derivative is negative near Earth's surface, it must become positive at the tropopause in order for w to go to zero at the tropopause. Therefore, at some level in between being negative in the lower troposphere and positive in the upper troposphere, the derivative must be zero. We call this level the level of nondivergence, z_{LND}, and can use it to find an expression for w as a function of z by setting the divergence equal to zero in Equation [9.7]:_{ }
The large-scale surface divergence typically has a value of 10^{–5} s^{–1}. The large-scale level of nondivergence is typically about 5000 m. So, a typical value for b is:
So, for typical large-scale surface divergence:
At $z\text{}=\text{}{z}_{LND}=5000\text{m},w({z}_{LND})=-2.5{\text{cms}}^{-1}=\left(-2.5\times {10}^{-5}{\text{kms}}^{-1}\right)\left(86400{\text{sday}}^{-1}\right)=\text{}-2.2{\text{kmday}}^{-1}$[2]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@886C@ (see figure below).
The result is that w is only a few cm s^{–1}. In a day, the air mass can rise or fall only a few kilometers. Compare this vertical motion dictated by large-scale convergence and divergence to the vertical motion in the core of a powerful thunderstorm (horizontal scale of a few km), where the vertical velocities can be many m s^{–1}. This simple model is called the bowstring model because the shape of the vertical velocity looks like a bowstring that is fixed at two points but can vary as a parabola in between.
Kinematics describes the behavior of atmospheric motion but not the cause. Streamlines provide snapshots of that motion and trajectories show where individual air parcels actually go. All atmospheric motion in the horizontal is made up of one or more of five distinct types of motion: translation, stretching deformation, shearing deformation, vorticity, and divergence. Stretching and shearing deformation lead to the formation or the dissolution of surface weather fronts. Vorticity describes the counter-clockwise rotation around low pressure (in the Northern Hemisphere) and clockwise rotation around high pressure (in the Northern Hemisphere) and is thus associated with much of weather. Divergence/convergence aloft leads to vertical winds that connect to convergence/divergence at the surface, and through this mechanism, air motion aloft communicates with air motion at the surface.
This lesson showed the mathematics necessary to quantify all of these processes. So besides identifying streamline confluence/diffluence, you practiced quantifying the five flow types from weather maps of streamlines with wind vectors. Finally, you calculated the vertical wind and its direction (up or down) based on the divergence/convergence of the winds aloft.
You have reached the end of Lesson 9! Double-check that you have completed all of the activities before you begin Lesson 10.
To forecast the weather, we use numerical weather prediction models that are based on mathematical expressions for conservation of energy, mass, and momentum. Climate prediction models are based on the same conservation laws. Conservation simply means that the amount of a quantity such as total energy, mass, or momentum remains constant even though the forms of that quantity may change. Conservation of energy is described by the 1^{st} Law of Thermodynamics, which was discussed in Lesson 2; conservation of mass and conservation of momentum are discussed in this lesson. The total mass of an air parcel is constant, but density and volume may change. The conservation of momentum is based on Newton's 2^{nd} Law and involves forces that can change momentum. The conservation of momentum is relatively simple when cast in an inertial (non-accelerating) reference frame because there are only three real forces that really matter for atmospheric motion: gravity, the pressure gradient force, and friction. But when cast on a rotating Earth, we need to add apparent forces to this equation in order to compensate for the fact that an air parcel on Earth is always accelerating as the Earth rotates. We end up adding the apparent forces—Coriolis force and centrifugal force—to the real forces to get an equation of motion whose predictions we can readily match with our observations.
Some atmospheric motion occurs with air masses and waves that are thousands of kilometers across, while other motion, such as tornadoes, is at most a few kilometers across. Sometimes air flows in a straight line; sometimes it flows around ridges and troughs. In all of these different cases, the most important forces are different, allowing the momentum conservation equation to be simplified in different ways. We will discuss these different conditions and show how you can determine the wind velocity from knowledge of the balance of the most important forces and thus determine the impact of air motion at upper levels on the air motion near Earth’s surface. Finally, we will describe why midlatitude winds are westerly.
By the end of this lesson, you should be able to:
If you have any questions, please post them to the Course Questions discussion forum. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.
Scientists like things that are conserved. There are good reasons for this. First, if something is conserved, that means we can always count on it being the same no matter what happens. Second, when we write down the equation for the conserved quantity, we can use that equation to understand how the equation’s variables will change with differing conditions. For example, in Lesson 2, we were able to use the First Law of Thermodynamics (a.k.a., conservation of energy) along with the Ideal Gas Law to derive the equation for potential temperature, which is very useful for understanding and calculating the vertical motion of air parcels.
In atmospheric dynamics, we like three conservation laws:
So, let’s step back and look at the mass of an air parcel, which equals the density times the volume of the parcel:
$m=\rho V$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaad2gacqGH9aqpcaGGGcGaeqyWdiNaamOvaaaa@3897@
In a parcel, the mass is conserved, and since m = ρV,
$\frac{D}{Dt}\left(m\right)=\frac{D}{Dt}\left(\rho V\right)=0$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaalaaapaqaa8qacaWGebaapaqaa8qacaWGebGaamiDaaaadaqadaWdaeaapeGaamyBaaGaayjkaiaawMcaaiabg2da9maalaaapaqaa8qacaWGebaapaqaa8qacaWGebGaamiDaaaadaqadaWdaeaapeGaeqyWdiNaamOvaaGaayjkaiaawMcaaiabg2da9iaaicdaaaa@4235@
Apply the product rule to Equation [10.2]:
$V\frac{D\rho}{Dt}+\rho \frac{DV}{Dt}=0$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbiaadAfadaWcaaWdaeaapeGaamiraiabeg8aYbWdaeaapeGaamiraiaadshaaaGaey4kaSIaeqyWdi3aaSaaa8aabaWdbiaadseacaWGwbaapaqaa8qacaWGebGaamiDaaaacqGH9aqpcaaIWaaaaa@406A@
Divide both sides by ρV:
$\frac{1}{\rho}\frac{D\rho}{Dt}+\frac{1}{V}\frac{DV}{dt}=0$
Recall that the specific rate of change in parcel volume is equal to the divergence (Equation 9.4) and so we can write:
$\frac{1}{V}\frac{DV}{dt}=\overrightarrow{\nabla}\u2022\overrightarrow{U}$ $$
Rearranging the equation gives us an expression for the conservation of mass:
$\frac{1}{\rho}\frac{D\rho}{Dt}+\overrightarrow{\nabla}\u2022\overrightarrow{U}=0$
This equation is for the conservation of mass in a continuous fluid (i.e., the fluid particles are so small that the air parcel behaves like a fluid). It is also called the Equation of Continuity. Physically, this equation means that if the flow is converging ($\overrightarrow{\nabla}\u2022\overrightarrow{U}<0$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaaFiaabaGaey4bIenacaGLxdcacaGGIaYaa8HaaeaacaWGvbaacaGLxdcacqGH8aapcaaIWaaaaa@3B2C@ ), then the density must increase ($\frac{D\rho}{Dt}$[2]@5@5@+=faaagCart1ev2aqaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCG4uz3bqee0evGueE0jxyaibaieYlNi=xH8yiVC0xbbL8FesqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbbG8FasPYRqj0=yi0dXdbba9pGe9xq=JbbG8A8frFve9Fve9Ff0dc9Gqpi0dmeaabaqaciGacaGaaeqabaWaaeaaeaaakeaaqaaaaaaaaaWdbmaalaaapaqaa8qacaWGebGaeqyWdihapaqaa8qacaWGebGaamiDaaaaaaa@3779@ >0). Note that in Lesson 9.5 [99]we said that density doesn’t change much at any fixed pressure level, which is how we were able to relate horizontal divergence/convergence with vertical ascent/descent. What did change was the vertical size of the air parcel as the horizontal size increased or decreased. The total mass, however, remained the same.
Newton’s 2^{nd} Law, F = ma, applies to a mass with respect to the inertial coordinate system of space. But we are interested in motion with respect to the rotating Earth. So, to apply Newton’s 2^{nd} Law to Earth’s atmosphere, our mathematics will need to account for the forces of Earth’s rotating coordinate system:
$$\overrightarrow{F}=m\overrightarrow{a}={{\displaystyle \sum}}^{\text{}}\text{(realforces)}+{{\displaystyle \sum}}^{\text{}}\text{(apparentforces)}=m\frac{D\overrightarrow{U}}{Dt}$$[2]@5@5@+=faaahmart1ev3aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=ribbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@60E8@
where the first set of forces are real forces and the second set are apparent (or effective) forces that will be used to correct for using a coordinate system attached to a rotating Earth.
When we use the word “specific” as an adjective describing a noun in science, we mean that noun divided by mass. So, specific force is F/m = a, acceleration. In what follows, we will use the terms “force” and “acceleration” interchangeably, assuming that if we say “force,” we mean “force/mass,” which is acceleration. At this point, you should be able to check the units—if there is no “kg,” then obviously we are talking about accelerations.
$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{D\overrightarrow{U}}{Dt}$
Here is another chance to earn one point of extra credit: Picture of the Week!
There are three real forces important for atmospheric motion:
Hence we can sum these real forces:
$\sum \overrightarrow{{F}_{a}}=\overrightarrow{{F}_{g}}+\overrightarrow{{F}_{p}}+\overrightarrow{{F}_{f}}$
We put the subscript "a" on these forces to indicate "absolute" because they are true in an inertial reference frame. Thus, in the absolute reference frame,
$$\frac{{D}_{a}{\overrightarrow{U}}_{a}}{Dt}=\frac{{{\displaystyle \sum}}^{\text{}}{}_{}}{}$$