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.

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).

Genus classification of clouds by altitude of occurrence.

Credit: Wikipedia

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 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 ~ 1000 μm, n ~ 1 L^–1

CCN Particle: D ~ 0.1μm, n ~ 1000 cm^–3

 Haze Drop: D ~ 1μm, n ~ 1000 cm^–3

Cloud Drop: D ~ 10 μm, n ~ 1000 cm^–3

Drizzle Drop: D ~ 100 μm, n ~ 1 cm^–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 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:

Typical values of LWC are 0.1–0.9 g m^–3, but a few g m^–3 are possible for wetter conditions.

Some clouds exist in regions where the temperature is below 0 ^oC 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 excess vapor density 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 –3 ^oC, simple hexagonal plates occur. Between –3 ^oC and –8 ^oC, columns form at low excess vapor densities and needles at high excess vapor densities. Between –8 ^oC and –22 ^oC, plates are formed, with more complex forms occurring at a higher excess vapor densities. At temperatures below –22 ^oC, columns once again dominate and go from simpler to more complex forms as the excess vapor density increases.

Ice crystal shape for different temperatures and different excess vapor density levels. Excess vapor density is the density of water vapor above the saturation water vapor density. Positive values of excess vapor density can occur as air is rapidly lifted and cooled, causing the saturation vapor pressure to drops faster than the vapor pressure itself, due to the finite amount of time it takes to deposit water vapor onto the ice.

Click for a text description of the ice crystals image.

Ice crystal shapes for different temperatures:

Columns (–30 to –23 ºC):

Above the liquid-water saturation line: bullet rosettes

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

Plates (–23 to –8 ºC):

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 –4 ºC):

On the liquid-water saturation line: needles

Below the liquid-water saturation line: solid columns

Plates (–4 to 0 ºC):

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).

A snowflake. Its shape with dendrites indicates that it formed with a lot of excess water vapor and a temperature of about –16 ^oC.

Credit: bkaree1 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, 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 ^oC and –8 ^oC 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.

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 saturation water vapor pressure.

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

A Big MAC. Hungry for some more?

Credit: Big Mac Snip by Ian Burt is licensed under CC BY 2.0.

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:

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 < 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.

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:

Radiative Cooling

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.

Summary:

• e is constant as T decreases.
• Since e_s depends only on T, e_s also decreases until e_s < e.
• When e_s becomes slightly less than e, a cloud forms.

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 e_s and T as the parcel cools. When e_s becomes slightly less than e, a cloud forms.

Credit: W. Brune © Penn State is licensed under CC BY-NC-SA 4.0

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).

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 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 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 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:

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 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.

Water phase diagram for mixing with an initially warm, moist air parcel at (T₁, e₁) in a cooler, drier environment at (T₂, e₂). When the air parcel mixes with the environment, the temperature and vapor pressure of the mixed parcel lies along the mixing line between the parcel and the environment. Thus, as the warm, moist parcel is mixed into the colder, drier environment, 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: Credit: Fog over the Weaver Farm in Limerick by Mongomery County Planning Commission's Photostream is licensed under CC BY-SA 2.0

Example:

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

$$\begin{array}{c}e=\frac{40,{000 }_{\mathrm{ }}}{40,000+80,000 }20+\frac{80,000 }{40,000+80,000 }5=\mathrm{10 }\text{hPa}\\ T=\frac{40,{000 }_{\mathrm{ }}}{40,000+80,000 }40+\frac{80,000 }{40,000+80,000 }10={20}^{o}C \end{array}$$

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.

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: Credit: Contrails by Mike Lewinski is licensed under CC BY 2.0

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/e_s will reach 100%.
• The temperature drops in accordance with Poisson’s Relations (Lesson 2) so that e_s also drops.

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).

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 e_s) decrease, but e_s decreases faster than e so that eventually e > e_s and a cloud forms. As the parcel continues to rise, it continues to cool and water vapor continues to condense so that the parcel stays close to saturation and hence travels along the saturation vapor pressure curve.

Credit: W. Brune © Penn State is licensed under CC BY-NC-SA 4.0

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.

A cumulus cloud over the ocean.

Credit: Cumulus cloud by JanneG from Pixabay is licensed under the Simplified Pixabay License

Why is supersaturation required for a cloud drop to form?

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."

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 Goddard. "NASA/Glory: The Could Makers." YouTube. January 28, 2011.

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.

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!

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.

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

Credit: W. Brune © Penn State is licensed under CC BY-NC-SA 4.0

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:

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: 

where we have used 0 ^oC values for σ (0.0756 J m^–2) and n_L (5.55 × 10⁴ 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:

Dependence of the ratio of the saturation vapor pressure over a curved surface to the saturation vapor pressure over a flat surface on the drop radius.

Credit: W. Brune © Penn State is licensed under CC BY-NC-SA 4.0

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:

Summary

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 H₂O.

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

Sketch of a flat liquid surface with a solvent (water, red dots) and a solute (black dots).

Credit: W. Brune © Penn State is licensed under CC BY-NC-SA 4.0

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:

where n_w is the number of moles of water per unit volume of solution, n_s is the number of moles of solute per unit volume of solution, and n_L is the number of moles of water per unit volume of pure water.

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:

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:

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:

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 ^oC, a_K = 1.1 x 10^–9 m, B = 4.3 x 10^–6 m³ mole^–1, and N_s 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 © Penn State is licensed under Penn State is licensed under CC BY-NC-SA 4.0

Interpretation of the figure:

• s_k is the supersaturation of the particle, not the atmosphere.
• At the smallest r_d (~10s of nm), the Raoult solute effect (–1/r_d³) is stronger than the Kelvin effect.
• As r_d gets bigger as water is added to the particle, the Kelvin effect (+1/r_d) starts to win out over the solute term (–1/r_d³). As a result, s_k becomes positive.
• As r_d gets very large, then the surface begins to look “flat” and the solution becomes so dilute that the liquid water approaches “pure” and s_k approaches 0, which means that e_eq approaches e_s.
• s_c is called critical supersaturation. The particle cannot activate and grow into a cloud drop until the atmospheric supersaturation, s, exceeds s_c.
• s_c occurs at the critical radius r_c, for activation. Drops that reach this radius can activate and grow into a cloud drop. Those that don’t won’t.
• The greater N_s, the lower s_c, and the greater r_c (see next figure).

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

Check Your Understanding

Koehler curve for two drops: N_s = 1 x 10^–17 moles (blue solid line), N_s = 5 x 10^–17 moles (red dashed line).

Credit: W. Brune © Penn State is licensed under CC BY-NC-SA 4.0

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, 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.

Koehler curve for a drop in an environment where the ambient supersaturation is greater than the entire Koehler curve.

Credit: W. Brune © Penn State is licensed under CC BY-NC-SA 4.0

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 s_k curve, the particle grows or shrinks in an effort to have s_k = s, that is, to have the particle condensation and evaporation be in equilibrium with the environment’s water vapor.
• If s > s_c, 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:

Koehler curve for a drop in an environment where the ambient supersaturation intersects the Koehler curve.

Credit: W. Brune © Penn State is licensed under CC BY-NC-SA 4.0

Interpretation of the figure:

• If s < s_c and the drop radius is very small, then the drop grows to a radius on the left side of s_c and stops when s_k = 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 < s_k, (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 = s_k.
• If a drop happens to find itself in an environment where s = s_k, the drop can shrink back into a haze drop or grow into a cloud drop, depending on the drop radius. If r < r_c, then the drop will shrink until it has a radius for which s = s_k. If r > r_c and s = s_k, 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 = s_k.
• If s = s_c (which is a special case of s = s_k) then the drop will grow to radius r = r_c and then with the random addition of a little water will grow a little bit more so that s_k < 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!

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.

• The growth rate (dm_d/dt, where m_d is the mass of the drop) is proportional to s_env – s_k. Physically, this statement means that the greater the difference between the supersaturation in the environment and supersaturation at the drop’s surface, the faster water vapor will diffuse and stick on the surface. For instance, if s_env equaled s_k­, then the evaporation and condensation of water on the drop’s surface would be equal and there would be no mass growth.
• As water vapor diffuses to the drop and forms liquid water, energy is released (i.e., latent heat of condensation) and this raises the temperature of the cloud drop surface, T_sfc, so that T_sfc > T_env. But an outward energy flow occurs and is proportional to T_sfc – T_env. Physically, this statement means that the drop 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 drop, and warm them by increasing their kinetic energy (see figure below).

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:

where G is a coefficient that is a function of T and p, ρ_l 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. Plugging in m_d = ρ_l4πr_d³/3 to 5.14, taking the derivative, and solving for dr_d/dt, it is easy to show (try it!) that dr_d/dt is proportional to 1/r_d. That is, the bigger the drop gets the slower it grows. Separating variables (r_d and t) and integrating from r_d = 0 at t = 0 to arbitrary values of r_d and t reveals that r_d is proportional to the square root of time:

where the constant C = 2G(s_env – s_k). The figure below shows how r_d varies with time for a typical value of C, 4 x 10^–13 m² s^–1.

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. For conditions shown in the figure above, it would take 30 days to reach the diameter of a typical rain drop (1000 μm).
• 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.

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 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.

Air flow around a falling drop. The shaded area is the cross sectional area of the drop. Note the movement of air around the drop. Only the air within the innermost streamlines collides with the drop; 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:

• m_L is the mass of the large drop that is falling,
• A_g is the geometric cross-sectional area for which collisions between the falling large drop and the many drops below are possible,
• E­_c 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),
• v_L is the velocity of the large drop and v_s 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-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.

Schematic of the maximum possible geometric cross-sectional area of a large and small drop and the actual cross-sectional area due to small drops following air streamlines around the large drop.

Credit: W. Brune (after Lamb and Verlinde)

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.

Collision–coalescence efficiency E_c for large drops of radius r_L (individually labeled red lines) as a function r_s/r_L, where r_s is the radius of the small drop.

Credit: W. Brune (after Rogers and Yau)

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_lπr_L³/3 , 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:

With the constant of proportionality between the terminal velocity and drop radius set at 8 x 10³ 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 ^oC, 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).

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!

Photograph of an ice crystal growing by the Bergeron-Findeisen Process in a field of small supercooled liquid drops.

Credit: R. Pitter

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:

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 © Penn State is licensed under CC BY-NC-SA 4.0

Putting It All Together

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

The lifecycle of a cloud. This schematic contains essentially all the processes that we have talked about in this lesson.

Credit: W. Brune © Penn State is licensed under CC BY-NC-SA 4.0

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

• Local perturbation in atmospheric density fields, sometimes driven by uneven surface heating or 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
  • Liquid water 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

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.

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.