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:
All matter radiates infrared energy, as we will see in the next lesson. When an air mass radiates this energy, 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.
- 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.
An example of radiative cooling in action is radiation fog, which occurs overnight when Earth's surface and the air near it cool until a fog forms (see figure below).
Assume two air parcels with different temperatures and water vapor partial pressures are at the same total pressure. If these two parcels mix, then the temperature and the water vapor partial pressure 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:
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.
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):
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.
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).
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.
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."
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.
- Find Practice Quiz 5-2 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.
- When you feel you are ready, take Quiz 5-2. You will be allowed to take this quiz only once. Good luck!