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Lesson 5: Cloud Physics

Overview

Overview

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.

Learning Objectives

By the end of this lesson, you should be able to:

  • identify cloud types
  • describe the essentials for cloud formation
  • on a Koehler curve, explain the behavior of a particle in different supersaturation environments
  • explain the lifecycle of cloud formation through precipitation

Questions?

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.

5.1 Do you recognize these clouds, drops, and snowflakes?

5.1 Do you recognize these clouds, drops, and snowflakes?

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 19th century and has 10 basic cloud types with many minor variations (see figure below).

chart of cloud types: cumulonimbus, cumulus, stratus, stratocumulus, nimbostratus, altostratus, altocumulus, cirrocumulus, cirrostratus, cirrus
Genus classification of clouds by altitude of occurrence.
Credit: Wikipedia [1]

Clouds!

This website contains a nice overview of the cloud types [2] with descriptions and accompanying images. Check out some of their amazing photos!

NOAA and NASA put together this thorough Sky Watcher Chart [3] 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.

Cloud drop sizes. See text below image
Cloud drop sizes and characteristics. D is the typical diameter; n is the typical number per volume of air. Sizes are almost but not quite to scale.
Click here for a text description of the image above

Rain Drop: D~1000μm, n~1L-1

CCN Particle: D~0.1μm, n~1000cm-3

 Haze Drop: D~1μm, n~1000cm-3

Cloud Drop: D~10μm, n~1000cm-3

Drizzle Drop: D~100μm, n~1cm-3

Credit: W. Brune (after Lamb and Verlinde)

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 vapor 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). The liquid water content can be defined as:

LWC= ω L = mass of liquid water volume of air , units=g  m −3 This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.1]

Typical LWC are 0.1- 0.9 g m-3, but a few g m-3 are possible for wetter conditions.

Check Your Understanding

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?

Click for answer.

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.

n cloud V cloud = V rain n cloud = V rain V cloud = 4 3 π ( r rain ) 3 4 3 π ( r cloud ) 3 = ( r rain r cloud ) 3 = ( 500 5 ) 3 = 10 6 This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

So we see that it takes about a million cloud drops to make one raindrop. Thus 109 cloud drops per m3 of cloud should make about 103 raindrops per m3 of cloud. This is about the number per m3 that are observed.

Quiz 5-1: Cloud drops and liquid mass.

  1. Find Practice Quiz 5-1 in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.
  2. When you feel you are ready, take Quiz 5-1. You will be allowed to take this quiz only once. Good luck!

Some clouds exist in regions where the temperature is below 0oC and thus can be made of ice. Ice is simply water in an organized crystalline form. While on the molecular scale the arrangement of water molecules in the ice matrix is the same, the visible shape can vary dramatically. In fact, certain shapes are favored in different temperature regimes (see figure below). Excess vapor density is simply the amount of water vapor (in terms of milligram per cubic meter of air) in excess of ice saturation, not liquid water saturation, which is shown as the dashed line. At temperatures above -3oC, simple hexagonal plates occur. Between -3oC and -8oC, columns form. Between -8oC and -22oC, plates are formed, with the more complex plates occurring with a greater excess of water vapor. At temperatures below -22oC, columns once again dominate and go from simpler to more complex as the excess water vapor increases.

Ice crystal shape for different temperatures and different excess water density levels as described in the text above. See link in caption for text description
Ice crystal shape for different temperatures and different excess water density levels. Excess water vapor is the amount of water vapor above the saturation water vapor amount. This situation can occur as air is lifted and the temperature (and thus saturation vapor pressure) drops faster than the water vapor can deposit on the ice.
Click for a text description of the ice crystals image.

Ice crystal shapes for different temperatures:

Columns (-30 to -23ºC):

Above the liquid-water saturation line: bullet rosettes

Below the liquid-water saturation line: hollow columns, solid columns

Plates (-23 to -8ºC):

Above the liquid-water saturation line: dendrites

Below the liquid-water saturation line: stellar crystals, broad branch plates, sector plates, solid plates

Columns (-8 to -4ºC):

On the liquid-water saturation line: needles

Below the liquid-water saturation line: solid columns

Plates (-4 to 0):

Below the liquid-water saturation line: Solid plates

Credit: W. Brune (after Lamb and Verlinde)

Ice crystal habits as a function of temperature and excess water vapor (i.e., water vapor greater than saturation water vapor).

close up view of a snowflake
A snowflake. Its shape with dendrites indicates that it formed with a lot of excess water vapor and a temperature of about -16oC.
Credit: bkaree1 [4] via flickr

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, 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 -22oC and -8oC and the excess water vapor 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.

Snowflake Safari
Click here for transcript of the Snowflake Safari video.

FLORA LICHTMAN: Sure there's sledding, snowmen, skiing, but a winter storm can also mean safari. KEN LIBBRECHT: You really just need a snowy day. Take a magnifying glass, go out, there's all sorts of different things you can see. FLORA LICHTMAN: That's Ken Libbrecht, the physicist at Caltech who also happens to be a snowflake expert. He's been hunting flakes for years and documenting them before they melt with this microscope camera rig. KEN LIBBRECHT: My travel with that the hard part is getting it through airport security. FLORA LICHTMAN: Snow crystals come in roughly 35 flavors Libbrecht says. Some more common than others of course. KEN LIBBRECHT: Stellar dendrites are pretty common standard sort of shopping mall snowflake with six branches. FLORA LICHTMAN: Then there's the variant fern-like stellar dendrites. KEN LIBBRECHT: ...and they look like a little bitty ferns. FLORA LICHTMAN: Also common are... KEN LIBBRECHT: ..needles, columns. One of my favorites are capped columns. FLORA LICHTMAN: Which look kind of like a satellite or... KEN LIBBRECHT:...two wheels on axle. Unfortunately the most common thing you'll find is just kind of junky looking snow looks like sand. FLORA LICHTMAN: The least common, the ivory-billed woodpecker of snowflakes is big... KEN LIBBRECHT: ...five millimeters in diameter and nicely symmetrical with lots of intricate markings. Those are really gorgeous and hard to find. FLORA LICHTMAN: But you can increase your chances if you seek out snowflake hotspots. KEN LIBBRECHT: Northern Ontario is a good spot. Vermont and Michigan and I have been there. Northern Japan actually is pretty good. I'm anxious to try to Siberia. FLORA LICHTMAN: See certain conditions breed better crystals. KEN LIBBRECHT: The best temperature is around five degrees Fahrenheit. Sometimes though you can see it's really nice crystals just below freezing. FLORA LICHTMAN: Ok a little review of where snowflakes come from. They're born in the clouds. It all starts with a speck of dust or bacterium. KEN LIBBRECHT: Gunk in the air. FLORA LICHTMAN:...and the gunk floats around the cloud. KEN LIBBRECHT:...for half a mile. FLORA LICHTMAN: Picking up water molecules. KEN LIBBRECHT: Then they shuffle around a little bit until they find the right spot to sit in and that the water molecules themselves are lined up in the hexagonal array. That's where the the order is generated. FLORA LICHTMAN: ...and that order is what makes it a crystal. KEN LIBBRECHT:...and as a grows larger the points of the hexagon stick out a little bit in the air so each of the six corners sprouts and arm and that's one of the things we're trying to understand in details how crystals grow. FLORA LICHTMAN: The details of that growth are determined by the microenvironment, the flake encounters, as it travels through the cloud. KEN LIBBRECHT: Humidity is low the crystals grow slow and humidity is high they go fast. FLORA LICHTMAN: In other words of flakes identity is shaped by the environment it grows up in and because two snow crystals aren't likely to follow the exact same path, you're not likely to find two of the exact same flake. Just how environment affects crystal growth is something Librecht studies in the lab, by growing his own snowflakes. KEN LIBBRECHT: We call these designer snowflakes. You can sort of dial-up what you want. FLORA LICHTMAN: Give it the right environment and something to grow on and it'll build itself. KEN LIBBRECHT: A really nice example of how really complicated structures emerged spontaneously not alive test the DNA or anything like that genetic code. It just happens. To understand more about how works will be able to use it for something or at the very least we'll just understand how it works. FLORA LICHTMAN: Happy new year. For Science Friday, I'm Flora Lichtman.

Credit: SciFri [5]

Discussion Activity: Cloud Identification

(3 discussion points)

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.

You will upload each image in its own post. You should include the following in your post:

  • your name
  • the picture's location
  • the picture's date and time
  • your identification of the cloud
  • your reasoning for the identification in a short sentence or phrase.

Copy and paste your picture into the post box. Instructions for how to embed an image in your post can be accessed here [6].

  1. You can access the Cloud Identification Discussion Forum in Canvas.
  2. Post your pictures of clouds with their identification using the format described above.
  3. Keep the conversation going! Comment on at least one other person's post. Your comment should include follow-up questions and/or reasoning for an alternate identification of the clouds in the post.

This discussion will be worth 3 discussion points. I will use the following rubric to grade your participation:

Discussion Activity Grading Rubric
Evaluation Explanation Available Points
Not Completed Student did not complete the assignment by the due date. 0
Student completed the activity with adequate thoroughness. Posting answers the discussion question in a thoughtful manner, including some integration of course material. 1
Student completed the activity with additional attention to defending his/her position. Posting thoroughly answers the discussion question and is backed up by references to course content as well as outside sources. 2
Student completed a well-defended presentation of his/her 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

5.2 What are the requirements for forming a cloud drop?

5.2 What are the requirements for forming a cloud drop?

There are three requirements for forming a cloud drop:

  1. Moisture
  2. Aerosol
  3. Cooling

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 water saturation vapor pressure.

An easy way to remember these key ingredients is to think of a Big MAC.

a McDonald's Big Mac`
A Big MAC. Hungry for some more?
Credit: Cleaveland via flickr

Saturation occurs when e=es, w=ws, and condensation = evaporation. At saturation, RH = e/es ~ w/ws = 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) = Td (as determined from the constant water vapor mixing ratio), or when w = ws.

Let’s define two new variables that are useful in discussing the cloud drop formation.

Often we talk about the saturation ratio:

S= e e s This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.2]

where e is the water vapor pressure and es is the saturation vapor pressure. S < 1 for a subsaturated environment, S = 1 for saturation (condensation = evaporation), and S > 1 for a supersaturated environment.

We also talk about supersaturation:

s=S−1= e e s −1 This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.3]

s = 0 at saturation; s < 0 for a subsaturated environment; s > 0 for a supersaturated environment. Note that s and S are both unitless.

This equation applies only for a flat surface of pure water. When we get into situations where we have curvature or a solute, we need to think about the supersaturation relative to the equilibrium value of e, eeq, which can be different from es. So, depending on the circumstances, eeq can be es (flat liquid water), ei (flat ice), esc (curved liquid water), esol (curved solution), or some combination. We will see that a small supersaturation is actually needed to form clouds.

Check Your Understanding

The relative humidity is 85%. What is the saturation ratio? What is the supersaturation?

Click for answer.

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?

Click for answer.

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?

5.3 How can supersaturation be achieved?

5.3 How can supersaturation be achieved?

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:

Supersaturation Processes 2
Click here for transcript of the Supersaturation Processes 2 video.

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.

Radiative Cooling

All matter radiates energy as electromagnetic waves, as we will see in the next lesson. When an air mass 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 (see figure below). Initially, the air mass is at the position of the blue dot. As the air parcel cools and the temperature drops, the air parcel temperature moves to the left on the diagram but the water vapor pressure does not change. However, because the temperature drops, es drops. When es becomes slightly less than e, a cloud forms.

Summary:

  • e is constant as T decreases.
  • Since es depends only on T, es also decreases until es < e.
  • When es becomes slightly less than e, a cloud forms.
described sufficiently in caption
Water phase diagram for radiative cooling, with an air parcel starting with e and T marked by the blue dot. The horizontal arrow marks the cooling of the air parcel and the downward pointing arrow marks the change in es and T as the parcel cools. When es becomes slightly less than e, a cloud forms.
Credit: W. Brune

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).

fog surrounding trees
Radiation fog. Moist air cools overnight by radiation to space while its water vapor mixing ratio remains roughly constant. However, the saturation vapor pressure drops as the temperature drops. When the saturation vapor pressure drops to be the same as the vapor pressure, a radiation fog forms.
Credit: Montgomery County Planning Commission [7] via flickr

Mixing

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 is going to be a weighted average of the T and e 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 e1, T1, and N1 (number of moles) and parcel 2 with e2, T2, and N2, the e and T of the mixed parcel are given by the equations:

e= N 1 N 1 + N 2 e 1 + N 2 N 1 + N 2 e 2   T= N 1 N 1 + N 2 T 1 + N 2 N 1 + N 2 T 2 or approximately e= M 1 M 1 + M 2 e 1 + M 2 M 1 + M 2 e 2   T= M 1 M 1 + M 2 T 1 + M 2 M 1 + M 2 T 2 This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.4]

where M1 and M2 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 warm mixes into more and more of the drier air, it gets increasingly diluted but the mixed parcel continues to grow. As the amount of environmental air in the mixture increases, the average e and T of the mixed air 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 warmer parcel, as the mixed parcel continues to grow, eventually the e and T will hit the Clausius-Clapeyron curve. As it 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 to the left of the Clausius-Clapeyron curve. However, once the mixed parcel comes to the right of the curve, the cloud will evaporate.

Water phase diagram with two air parcels at (T1,e1) and (T2,e2). described sufficiently in caption
Water phase diagram for mixing, with two air parcels at (T1,e1) and (T2,e2). When the two air parcels mix, the temperature and vapor pressure of the mixed parcel lies along the mixing line between the two parcels. If a small, warm, moist parcel is mixed into a colder, drier environment, then as the warm moist parcel mixes with more environmental air, the size of the mixed air parcel grows and the temperature and vapor pressure follow the mixing line toward the environmental temperature and vapor pressure (Eq. 5.4).
Credit: W. Brune

Example:

Suppose air Parcel 1 has e = 20 hPa, T = 40oC, and N = 40,000 moles; and Parcel 2 has e = 5 hPa, T = 10oC, and N = 80,000 moles. Then using equation 5.4 (top):

e= 40, 000 1 40,000+80,000 20+ 80,000 40,000+80,000 5 =10hPa  T= 40, 000 1 40,000+80,000 40+ 80,000 40,000+80,000 10= 20 o C MathType@MTEF [8]@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 air moves over warm moist ground, say just after rain.

contrails in a blue sky over desert
Contrails of different ages. Look at the fresh contrails. They are very thin and consist of water from the jet engines. However, the older contrails look bigger and more spread out, yet we know that the water from the jet engines is not enough to make such big contrails. This extra water vapor must have come from the atmosphere.
Credit: Mike Lewinski [9] via flickr

Uplift

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 water vapor mixing ratio remains the same, but e drops as p drops, thus reducing the possibility that RH = e/es will reach 100%.
  • The temperature drops in accordance with Poisson’s relations so that es also drops.

The question is “Does e or es drop faster so that eventually e equals es?" It turns out that es drops faster. As a result, in uplifted air, e and es converge at the lifting condensation level (LCL) and a cloud forms just at that level (see figure below).

described sufficiently in caption
Water phase diagram for uplift, with an air parcel starting at the beginning of the arrow. As the parcel ascends both e and T (and thus es) decrease, but es decreases faster than e so that eventually e > es. Once the cloud forms as the line passes into the liquid part of the phase diagram, the water line tries to achieve saturation.
Credit: W. Brune

The arrow on the figure above shows the changes in e and T (and thus es) as an air parcel rises. Once es <= e, then s > 0 and the air parcel is supersaturated. This supersaturated situation is not stable; the water vapor in excess of es forms liquid. As the uplift continues, more water vapor is converted into liquid water and the vapor pressure remains close to es. 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.

A cumulus cloud over the ocean.
A cumulus cloud over the ocean.
Credit: JanneG [10] via pixabay.com

Why is supersaturation required for a cloud drop to form?

I thought that cloud drops formed when w = ws. 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."

Glory: The Cloud makers
Click here for transcript of the Glory: The Cloud Makers video.

[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]

Credit: NASA [11]

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 water vapor equilibrium 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.

Quiz 5-2: Cloud formation essentials.

  1. When you feel you are ready, take Quiz 5-2. You will be allowed to take this quiz only once. Good luck!

5.4 Curvature Effect: Kelvin Effect

5.4 Curvature Effect: Kelvin Effect

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/m2.

diagram of curvature effect, as described in the text above and below

Sketches of the curvature effect. Left is a flat surface of pure water; right is a curved surface of pure water.
Credit: W. Brune

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 ( T )= e s ( T )⋅exp( 2 σ n L ⋅R⋅T⋅ r d )

[5.5]

where esc is the equilibrium vapor pressure over a curved surface of pure water, es is the equilibrium vapor pressure over a flat surface of pure water, both of which are functions of temperature, although esc is also a function of the drop radius, nL is the number of moles per volume of water (55.5 moles L-1). R* is the molar gas constant, σ is the water surface tension, and r is the radius of the drop.

( 2σ n L ⋅R* )=3.3x 10 −7  m  K −1 This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.6]

Since the evaporation is much, much greater over a curved surface, the condensation must also be much, much greater in order to keep condensation = evaporation, which is required for saturation (i.e., equilibrium). Thus, the saturation vapor pressure over a curved surface is much greater than the saturation vapor pressure over a flat surface of pure water.

When we plot this equation, we get the following figure:

radius on the x-axis, e(sc)/e(s) on y-axis. e(sc)/e(s) is higher with smaller radius
Dependence of ratio of the saturation vapor pressure over a curved surface to saturation vapor pressure over a flat surface on the drop radius.
Credit: W. Brune

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)⋅( 1+ 2σ n L ⋅R*⋅T⋅ r d )= e s (T)⋅( 1+ a K r d ), where    a K = 2σ n L ⋅R*⋅T This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.8]

Summary

Cloud drops start as nanometer-size spherical drops, but the vapor pressure required for them to form is much greater than es until they get closer to 10-2 μm 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 esc approaches es.

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.

5.5 Solute Effect: Raoult’s Law

5.5 Solute Effect: Raoult’s Law

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 H2O.

Solute: The chemical that is being dissolved in the solvent.

red dots with a few black dots spaced out and surrounded by red dots
Sketch of a flat liquid surface with a solvent (water, red dots) and a solute (black dots).
Credit: W. Brune

The simplest view of this effect is that solute molecules are evenly distributed in the water (solvent). Therefore, 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 water saturation vapor pressure will be lower. The equilibrium vapor pressure is less than es, 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 be even less.                

We can quantify this equilibrium vapor pressure over a solution with a few simple equations.

The mole fraction is defined as:

χ s = n s n w + n s ≈ n s n L , if  n w > n s

[5.9]
where nw is the number of moles of water in a liter of solution, ns is the number of moles of solute in a liter of solution, and nL is the number of moles per liter of pure water.

As a result, the evaporation rate of the water molecules from the surface decreases. This can be written as Raoult’s Law:

e sol = e s ⋅(1− χ s )
[5.10]

We can approximate Raoult’s Law for a reasonably dilute solution by writing:

χ s ≈ n s n L = i N s n L V drop = i N s n L ( 4π r d 3 3 ) ≡ Bi N s r d 3 , where B= 3 4π n L This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.11]

Ns is the total moles of solute. Note that an “i” was added. This factor is called the Van’t Hoff factor and it 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-, which means that i = 2 in this case.

5.6 Koehler Theory

5.6 Koehler Theory

So, now we can put the pieces together. Both the Kelvin effect and the solute effect are found by multiplying the equilibrium vapor pressure by the effect:

e eq ( n s ,r,T)= e s (T)⋅( 1− χ s )⋅exp( 2σ n L R*Tr ) e eq ( n s ,r,T)≈ e s (T)⋅( 1− Bi N s r d 3 )⋅( 1+ a K r d )= e s (T)⋅( 1+ a K r d − Bi N s r d 3 − a K Bi N s r d 4 ) e eq ( n s ,r,T)≈ e s (T)⋅( 1+ a K r d − Bi N s r d 3 ) This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.12]

This equation is called the Koehler Equation. It gives us the equilibrium vapor pressure that a drop will have for a given drop radius rd. 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/es. We also talked about supersaturation, s = S-1= e/es-1. We define a new saturation ratio and supersaturation for a particle, Sk and sk, where “k” stands for “Koehler.”

sk is defined as follows:

S k = e eq ( n s ,r,T) e s (T) ≈( 1+ a K r d − Bi N s r d 3 ) s k = S k −1= a K r d − Bi N s r d 3 This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.13]
where a K = 2σ n L ⋅R*⋅T and B= 3 4π n L .

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 20oC, ak = 1.1x10-9 m, B = 4.3x10-6 m3 mole-1, and Ns is typically in the range of 10-18 to 10-15 moles.

Koehler curve, which is the supersaturation of a particle as a function of the particle radius. The radius increases as the particle takes on water.
Credit: W. Brune

Interpretation of the figure:

  • sk is the supersaturation of the particle, not the atmosphere.
  • At the smallest rd (~10’s nm), the Raoult solute effect (-1/rd3) is stronger than the Kelvin effect.
  • As rd gets bigger as water is added to the particle, the Kelvin effect (+1/rd) starts to win out over the solute term (-1/rd3). As a result, sk becomes positive.
  • As rd gets very large, then the surface begins to look “flat” and the solution becomes so dilute that the liquid water approaches “pure” and sk approaches 0, which means that eeq approaches es.
  • sc is called critical supersaturation. The particle cannot activate and grow into a cloud drop until the atmospheric supersaturation, s, exceeds sc.
  • sc occurs at the critical radius rc, for activation. Drops that reach this radius can activate and grow into a cloud drop. Those that don’t won’t.
  • The greater Ns, the lower sc, and the greater rc (see next figure).

The video below (1:35) explains the Koehler Curve Equation in more depth:

Koehler Curve Equation
Click here for a transcript of the Koehler Curve Equation video.

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.

Check Your Understanding

same graph as above, larger # of moles of solute is a shorter peak and a more linear decrease
Koehler curve for two drops: Ns=1x10-17 moles (blue solid line), Ns=5x10-17 moles (red dashed line).
Credit: W. Brune

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?

Click for the answer.

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, sk, 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 sk=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 sk=s if it can! Another way to think about sk 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 sk < s, then net condensation will occur and the drop will grow. If sk > 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.

see text below for interpretation. Graph is same curve as above, ambient supersaturation is above (rc,sc) point
Koehler curve for a drop in an environment where the ambient supersaturation is greater than the entire Koehler curve.
Credit: W. Brune

Interpretation of the figure:

  • The ambient supersaturation does not depend on the radius of the particle and so is a straight line on the graph.
  • The basic concept is the particle supersaturation is constrained to the sk curve, the particle grows or shrinks in an effort to have sk = s, that is, to have the particle condensation and evaporation be in equilibrium with the environment’s water vapor.
  • If s>sc, then the condensation of ambient water vapor is greater than the evaporation from the drop no matter what the drop radius is.
  • The particle continues to grow, becomes a critical nucleus, and then continues to add water and grow into a cloud drop.
  • The drop can start at any size and it will grow into a cloud drop as long as the ambient supersaturation is greater than the Koehler curve.

In the second case, the ambient supersaturation intersects the Koehler curve:

see text below for interpretation. Graph is same curve as above, ambient supersaturation is below (rc,sc) point and intersects the curve 2x
Koehler curve for a drop in an environment where the ambient supersaturation intersects the Koehler curve.
Credit: W. Brune

Interpretation of the figure:

  • If s < sc and the drop radius is very small, then the drop grows to a radius on the left side of sc and stops when sk = s.
  • The drops are called haze drops. They do not nucleate to become cloud drops but instead stay a smaller size. You have probably often seen haze on a warm, moist summer day.
  • If a drop finds itself in an environment where s<sk, (as could happen if the drop's air is mixed with drier air), then the evaporation from the drop would exceed the condensation from the environment and the drop would shrink until s=sk.
  • If a drop happens to find itself in an environment where s=sk, the drop can shrink back into a haze drop or grow into a cloud drop, depending on the drop radius. If r < rc, then the drop will shrink until it has a radius for which s=sk. If r > rc and s=sk, then the drop is in unstable equilibrium and can either grow into cloud drops by adding a water molecule or can shrink all the way back to a haze drop with a radius that matches the radius for s=sk.
  • If s = sc (which is a special case of s = sk) then the drop will grow to radius r = rc and then with the random addition of a little water will grow a little bit more so that sk < s. At this point, there will be net condensation and the drop will activate and continue to grow.

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!

Quiz 5-3: How cloud drops form.

  1. Find Practice Quiz 5-3 in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.
  2. When you feel you are ready, take Quiz 5-3. You will be allowed to take this quiz only once. Good luck!

5.7 Vapor Deposition

5.7 Vapor Deposition

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, senv, must be greater than sk for this to happen, but as the drop continues to grow, sk approaches 0 (i.e., eeq approaches es), 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.

  • The growth rate (dmd/dt, where md is the mass of the drop) is proportional to senv – sk. Physically, this statement means that the greater the difference between the supersaturation in the environment and supersaturation at the particle’s surface, the faster water vapor will diffuse and stick on the surface. For instance, if senv equaled sk­, then the evaporation and condensation of water on the particle’s surface would be equal and there would be no mass growth.
  • As water vapor diffuses to the drop and forms water, energy is released (i.e., latent heat of condensation) and this raises the temperature of the cloud drop surface, Tsfc, so that Tsfc > Tenv. But an outward energy flow occurs and is proportional to Tsfc – Tenv. Physically, this statement means that the particle and the air molecules around it are warmed by latent heat release. These warmer molecules lose some of this energy by colliding with the cooler molecules further away from the particle, and warm them by increasing their kinetic energy (see figure below).
arrow 2 cloud drop (mass transport to surface), & arrow away (thermal energy transport), reflux of of mass & thermal energy @ surface
Schematic of the two physical processes in the growth of a cloud drop by vapor deposition. One is vapor deposition and the other is the transfer of condensational heating to the atmosphere;
Credit: W. Brune (after Lamb and Verlinde)
  • When we account for both the flow of water molecules to the cloud drop surface and the flow of energy away from the surface, we can show that:

d m d dt =4π r d ρ l G( T,p )( s env − s k ) This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.14]
where G is a coefficient that is a function of T and p, ρL is the density of liquid water, and the other variables have already been defined. G incorporates the effects of the mass transport of water vapor molecules to the surface and the transport of heat generated on condensation away from the particle surface. As a result, the drop radius grows as the square root of a constant times time.
 
rd = (C time)1/2
[5.15]
growth time (s) on x, drop radius (um) on y, slope curves up and starts to flatten. Point marked (1000, 20)
Growth of a cloud drop by vapor deposition as a function of time. Dashed lines indicate drop size after the typical cloud lifetime.
Credit: W. Brune

Physical explanation:

  • The nucleated cloud drop radius increases fairly rapidly at the beginning, but within minutes slows down because of the square root dependence on time.
  • So, cloud drops can grow to 10–20 μm in 15 or so minutes, but then grow bigger much more slowly.
  • Since a typical cloud only lasts 10s of minutes, it is not possible for cloud drops to grow into rain drops by vapor deposition alone.
  • CCN nucleation followed by vapor deposition can make clouds, but it can’t make them rain.
  • We can develop a similar expression for vapor deposition on ice, but the vapor depositional growth on ice is a little faster than on liquid.

Conclusion:

We need other processes to get cloud drops big enough to form precipitation, either liquid or solid.

5.8 Did you know most precipitation comes from collision-coalescence?

5.8 Did you know most precipitation comes from collision-coalescence?

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

Collisions occur in both cold and warm clouds and can involve either liquid drops or solid particles or both.

  • Collision–Coalescence: Large liquid drop scavenges smaller liquid drops as it falls.
  • Riming: Falling ice collects liquid water, which freezes on its surface.
  • Capture Nucleation: Large liquid drop captures small ice particle, which acts as an ice nucleus and causes the large drop to freeze. The particle that is collected can be either an ice nucleus (IN) or a piece of ice, which also is a good ice nucleus. In either case, the supercooled liquid drop freezes on contact with the IN.
  • Aggregation: Falling snowflake scavenges other snowflakes that aggregate to make a larger snowflake bundle.

For a cloud drop at rest, gravity is the only external force. Once the cloud drop accelerates, 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. For instance, the terminal velocity of a 10 μm radius cloud drop is about 1 mm s–1, while the terminal velocity for a 100 μm drop is about 1 m s–1.

Air flow around a falling particle. Free & distorted streamlines pointed up, drag force pointed up, gravity pointing down arrow
Air flow around a falling particle. The shaded area is the cross sectional area of the particle. Note the movement of air around the particle. Only the air in innermost streamline collides with the particle; the rest goes around it.
Credit: W. Brune (after Lamb and Verlinde)

The growth of a cloud drop into a precipitation drop by collision–coalescence is given by the equation:

d m L dt =Area swept out⋅efficiency of collection⋅velocity difference⋅liquid water content d m L dt = A g ⋅ E c ⋅( v L − v s )⋅LWC d m L dt =π ( r L + r s ) 2 ⋅ E c ⋅( v L − v s )⋅LWC This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.16]
  • mL is the mass of the large drop that is falling,
  • Ag is the geometric cross-sectional area for which collisions between the falling large drop and the many drops below is possible,
  • E­c is the collision-coalescence efficiency (i.e., a collection efficiency), which is the fraction of the actual cross sectional area that is swept out compared to the cross sectional area that is geometrically possible (smaller drops can follow air streamlines and go around the big drop) (see the figure below),
  • vL is the velocity of the large drop and vs is the velocity of the smaller, slower falling drops below,
  • and LWC is the liquid water content.

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-section becomes less than the actual cross-section, 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.

grazing air trajectory around drop, collected drops attached to collector drop, effective cross section much smaller than geometric
Schematic of the maximum possible geometric cross-section of a large and small drop and the actual cross-section due to particles following air streamlines around the big particle.
Credit: W. Brune (after Lamb and Verlinde)

Ec 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). Ec 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.

collision efficiency (%) on y, radius on (x), higher efficiency the smaller the radii
Collision–collection efficiencies for two drops, with the percent collision efficiency on the vertical axis; the ratio of the radius of the small drop, rs, to the radius of the large drop, rL, on the horizontal axis; and lines for individual large drop radii.
Credit: W. Brune (after Rogers and Yau)

Once a collecting drop has reached a radius of a few hundred μm, it is falling fast and its collision–coalescence efficiency is close to 100%. The growth of its radius then has the approximate form:

r d ∝exp(time) This equation is not rendering properly due to an incompatible browser. See Technical Requirements in the Orientation for a list of compatible browsers.

[5.17]

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 they 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 particles. 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 break-up.

How Raindrops are Formed
Click here for transcript of the How Raindrops are Formed.

Now this is a familiar scene. The sun's heat causes water from plants, lakes, and oceans to turn from a liquid to a vapor. High in the atmosphere the water vapor then cools down and condenses from a gas back into a liquid. The liquid water then falls back to the surface in the form of rain, snow, ice, or hail. Water runs off into streams lakes and oceans or is stored in the ground or in snow path. This is the water cycle and it describes our most vital resource moves through the whole earth system, but like most things in our world when we look at the tiny parts that make up the whole we can learn a lot more about the phenomenon. Take the shape of a single raindrop. Small droplets of water in the atmosphere are spherical in shape due to the surface tension or skin of the water molecules. As these droplets grow they become heavier and start to fall through the air. As they fall, the raindrop collides with other drops and continues to get bigger. These larger raindrops fall through the air faster the wind resistance on the underside of the drop causes the bottom of the drop to flatten resulting in a drop looking like a hamburger bun. As the drop continues to fall and grow at some point it becomes too large for the surface tension to hold it together, so the raindrop breaks apart into smaller spiracle drops. Investigating the processes we can't see with the naked eye is nothing new. Science and technology drive each other forward and often lead to insights and discoveries along the way. With the invention of high-speed photography we finally saw the most basic elements of our watery planet in action. Understanding how a tiny raindrop falls through the atmosphere does more than debunk the myth that a raindrop Falls like a teardrop. It actually makes a difference when it comes to measuring precipitation in particular for ground radars. Ground radars look at the sides of the raindrops and then estimate the vertical and horizontal sighs. A heavier, flatter drop allows radars to identify heavier precipitation. In fact the two radars on board the GPM satellite can also measure drop sizes from space and so a more accurate look at rain drops gives us a more accurate look at how global rainfall is shaping up.

Credit [12]

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.

5.9 An Unusual Way to Make Precipitation in Mixed-Phase Clouds

5.9 An Unusual Way to Make Precipitation in Mixed-Phase Clouds

Recall that water can exist in liquid form even when T < 273 K. This supercooled liquid needs ice nuclei (IN) in order to become ice, although at a temperature of about –40oC, it can convert to ice homogeneously.

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 with T < 273K, 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 min in mixed-phase clouds (see figure below).

Cartoon of ice growth in the presence of supercooled liquid drops. See link in caption for text description
Cartoon of ice growth in the presence of supercooled liquid drops
Click for a text description of the cartoon.

Frame 1: The cloud drops were supercool, balancing evaporation and condensation, when one drop bumped into an ice nuclei...

Frame 2: It changed into ice, evaporated less, but the vapor kept condensing on it, the ice grew, and the vapor decreased...

Frame 3: The other drops kept evaporating but with less vapor to condense, they shrank, their water turned to vapor, and the ice kept growing

Frame 4: The other drops evaporate away, and the ice crystal came into balance with less water vapor near it. It's big enough to fall...

Credit: W. Brune

This process, as unusual as it seems, actually works, as can be seen in the figure below!

snowflake
Photograph of an ice crystal growing by the Bergeron effect in a field of small supercooled liquid drops.
Credit: R. Pitter

5.10 Looking at the Whole Cloud

5.10 Looking at the Whole Cloud

Extra Credit Alert!

Announcing a new activity worth one point of extra credit for each lucky winner: Picture of the Week! Here is how to participate:

  1. You take a picture of some atmospheric phenomenon—a cloud, wind-blown dust, precipitation, haze, winds blowing different directions—anything that strikes you as interesting.
  2. Add a short description of the processes that you think are causing your observation. A Word file is a good format for submission.
  3. Use your name as the name of the file. Upload it to Picture of the Week Dropbox in this week's lesson module. To be eligible for the week, your picture must be submitted by 23:59 UT on Sunday of each week.
  4. I will be the sole judge of the weekly winners. A student can win up to three times.
  5. There will be a Picture of the Week dropbox each week through Lesson 11. Keep submitting entries!

Here is an example below:

aerial view of cloud over farmland
Heavy precipitation from a cumulonimbus cloud over Alabama, Summer 2013. This cumulus cloud rose to over 30,000 ft, about 9 km, and heavy afternoon rains fell under this cloud. This cloud is in the "mature" stage of development. This picture was taken from the NASA DC-8 aircraft, which was flying a few thousand feet above the ground in the cloud's inflow region.
Credit: W. Brune

Putting It All Together

We can put all of the processes from this lesson together to look at the lifecycle of a cloud:

diagram of lifecycle of a cloud as described in the text below
The lifecycle of a cloud. This schematic contains essentially all the processes that we have talked about in this lesson.
Credit: W. Brune

The following is a description of convection’s stages of development:

  • Local perturbation in atmospheric density fields, sometimes driven by uneven surface heating or moisture evaporation, starts relative vertical motion
  • Stage 1: “Developing Stage” (also called Cumulus Stage)
    • Updraft dominates center of cloud, cloud drops form and grow
    • Release of latent heat provides the energy for vertical motion and growth
  • Stage 2: “Mature Stage”
    • Downdrafts form in addition to updrafts, causing gust fronts
    • Cloud reaches height so that freezing occurs and precipitation develops
    • Evaporation of precipitation drives downdrafts
  • Stage 3: “Dissipating Stage”
    • Downdrafts only
    • Water mass is removed by sedimentation/evaporation

The video below (2 min.) includes some great time-lapse video of clouds forming and disappearing (No audio). Check it out:

Quiz 5-4: How precipitation forms.

  1. You can take Practice Quiz 5-4 as many times as you like.
  2. When you feel you are ready, take Quiz 5-4. You will be allowed to take this quiz only once. Good luck!

Summary and Final Tasks

Summary

Clouds are shaped and sized by atmospheric motions and mixing with the surrounding air and are composed of either liquid or ice drops depending on their environmental temperature. The basic shapes are stratus or cumulus or mixtures of the two; 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.

Reminder - Complete all of the Lesson 5 tasks!

You have reached the end of Lesson 5! Double-check that you have completed all of the activities before you begin Lesson 6.


Source URL: https://www.e-education.psu.edu/meteo300/node/525

Links
[1] http://en.wikipedia.org/wiki/Cloud
[2] http://www.clouds-online.com/
[3] https://www.weather.gov/jetstream/cloudchart
[4] https://www.flickr.com/photos/bkaree/6850865695/in/photolist-brotG4-b9xz6V-bkj7Dx-9zTrBi-ja4ZGL-4aTM3e-5SyyGu-7pcVYQ-8ZWApw-bdrQwF-bz5epY-5SugrZ-joih25-7kTKYZ-4DjBuF-dUc3Xk-dWxa49-jyok9h-dE15Tb-9kpPNn-5SufP6-5Suf4F-rzJYPe-aTNkdk-rxsogS-dzgRoh-4fJv8w-7k6YZG-i1RX48-rTv7T4-7kTKZe-dzPRSt-Q2knN-4TKYTr-vdsh2-6brtvC-b9xw6i-5SyzbG-5MmEZF-dDUHci-5SufmX-5Syzko-91KLXC-5Sufdv-7rmkwB-4hywbG-hsRyzF-7kTKZn-5SyAr3-dE167q
[5] https://www.youtube.com/user/SciFri
[6] https://community.canvaslms.com/docs/DOC-10700
[7] https://www.flickr.com/photos/75012107@N05/8609938706/in/photolist-e7QbP5-9GLHEQ-csaSmN-4x9Cmb-qS5abP-qgUJto-otURZR-oZQbYW-5FUKqN-cWpbBs-egTyta-4fgqzP-58gAua-a5hdDR-93KzLx-hofR8p-aCkGKb-49Krqp-7Q4HBC-5TSEyd-7HLnf8-ifvbKJ-igksoG-ifvct7-qjcw9S-61ja7t-dQNsKM-ifuQVz-qKXn15-q6TVpa-ifuQze-pFgruE-ifvhrd-8EeXnX-8TVVHU-6Mxtjv-pBgmkS-e3TcJC-kt6F9b-ddajGq-qnVxoe-4FVTLo-5eVcaf-owEPC8-q4b5jr-4hrMUN-ouqQTm-qjnWSd-ifvq72-5WoFTB
[8] mailto:MathType@MTEF
[9] https://www.flickr.com/photos/ikewinski/7170809855/in/photolist-bVEgWF-qvBVDW-nsA6uk-rv3TYi-rDceCr-rGkgeg-5CeaxE-cXgvn1-4Wh1ws-8oZxpt-8TQHn-iuAZ2Y-axw4r-bdsADX-qJedmV-dSsXid-e1KnJ3-dUMY6o-6h7TQe-6XBiYV-6XBiXZ-6XFjfd-6XFje7-6XBiev-6XFitU-kHZCX-9di7p-9CV8nn-e86Rqt-ryE64N-e8cwqf-Gj57k-ebMfPe-duvpHF-cXgr8Q-dVRzm-d8TPNJ-85a4am-qVnqtr-byeZiR-8L8z7k-e8griu-asKKQB-9e3X-CBsXe-ps34Mb-obKT13-dSsX7h-bq4hLo-9CV8tH
[10] http://pixabay.com/en/cumulus-nimbus-thundercloud-240600/
[11] https://www.youtube.com/user/NASAexplorer
[12] https://www.youtube.com/channel/UCcorzrM23aYyvcZTqaRoX0Q