In this class we will be focusing upon the type of clouds most commonly seen in the earth's atmosphere. These are a collection of liquid water drops and/or ice suspended or falling very slowly in the atmosphere.

How do clouds form? Previously we defined R.H. = 100% as the point at which the rate of condensation of water equals the rate of evaporation (also called saturation point) over a FLAT surface of PURE water. There are, however, two important complications when referring to cloud drops in the atmosphere.

CONDENSATION:

CURVATURE

The first complication is the fact that drops do not have flat surfaces, but instead curved surfaces. Due to the surface tension of water, it is easier for liquid water molecules to evaporate if the surface is curved than if the surface is flat. Therefore, though the rate of evaporation is equal to the rate of condensation at the saturation point (RH=100%) over a flat surface of water, the evaporation rate is greater than the condensation rate over a drop (curved surface). As a matter of fact the RH would have to be much larger than 100% for the evaporation rate to equal the condensation rate over the surface of a drop. For example, a drop of radius 0.01 micrometers would need a RH value of over 112% for evaporation to equal condensation and a drop of 1.0 micrometer radius would need RH=100.12% for equilibrium. This effect of curvature decreases with the radius of the drop and it is not a factor for an infinitely large drop (a flat surface). So, one would need very high RH values if cloud drops formed just by liquid water molecules sticking together. This process involving only water molecules coming together to form a drop is called HOMOGENOUS NUCLEATION.

However, we don't notice in the atmosphere that relative humidity are much above 100%. Why is that? There are particles in the atmosphere, such as dust and dirt, sea salt, ash from combustion, that serve as sites upon which condensation can occur. This is just like the process of dew forming upon the grass or the ground where these surfaces provide places for the water to condense. The particles in the atmosphere that serve as condensation sites are called Cloud Condensation Nuclei (CCN). The process of cloud drops forming upon CCN is called heterogeneous nucleation. Many of these particles attract water (call hygroscopic) and allow drops to form at RH less than or equal to 100%. These CCN can also dissolve in the droplet, and then we have a solution. This solution reduces the evaporation rate (since it is proportional to the number of liquid water molecules at the surface and this is reduces when there is a solution). Thus, compared to pure water, the air would be super-saturated over the solution if the same air was saturated over the pure water. That is to say in an atmosphere that is saturated over a pure water surface (rate of evaporation equals the rate of condensation) that as we dissolve CCN in the water to create a solution, the evaporation rate will decrease, but the condensation rate will remain the same. So, this imbalance between the two rates means that the liquid water would gain molecules. This means that a solution droplet will grow in size when a pure droplet would remain the same size. This allows droplets to form at RH < 100%.

In order for a droplet of water to freeze, enough water molecules must collect together within the droplet to form a nucleus for freezing that is large enough to survive and grow. This is similar to the droplet formation from water vapor. Because of this need for a nucleus to start the freezing process, pure water droplets of just a few micrometers in size will not freeze until a temperature of about -40 C. This process of freezing that only involves water is called homogeneous freezing and is due to the chance aggregation of a sufficient number of water molecules in the droplet.

Just like the process of condensation, where cloud condensation nuclei aid in the formation of cloud droplets in the atmosphere, the freezing of water in the atmosphere is aided by freezing nuclei. However, these are not as plentiful in the atmosphere and the requirements upon the form of the nuclei are greater than in the case of condensation. So, it is possible to have droplets of liquid water in the atmosphere at temperatures below 0 C or 32 F.

This has the effect of allowing clouds to be composed of both liquid and solid forms of water at the same time. At temperatures between 0 and -4C a cloud consists almost solely of liquid droplets because there are no natural freezing nuclei that operate in that temperature range. In fact, little ice is observed at temperatures above -10C. Between -10C and -30C clouds are seen to contain a mixture of both droplets and crystals. At temperatures less than -30C the cloud is mostly ice crystals.

Don't forget that the air must also be very near saturation for the ice crystal to form, just like the case of droplets forming near RH=100%. The one difference is that air over an ice surface is actually super-saturated at RH=100%. This fact plays a big role in the formation of most of the precipitation around the world.

Mechanisms of Cloud Formation:

To form a cloud, air must be saturated with water vapor. This can be achieved 3 different ways:

1. Cool the air until SMR=AMR and air temperature = dew point.
1. Buoyant ascent (convection)
2. Forced ascent over terrain (Orographic)
3. Forced ascent along a front
4. Forced ascent due to convergence (low pressure)
5. Contact with the earth's surface (radiation and advection fog)
6. Adiabatic expansion and cooling due to a rapid local reduction in pressure (funnel clouds)
2. Add water vapor until AMR=SMR and air temperature = dew point.
1. Water vapor from the earth's surface
2. Water vapor from evaporating precipitation
3. Mix parcels of air with different properties (cool air and warm moist air). The result is air with a temperature in between and a dew point in between the two.

First we must start with a concept of a bubble of air or as meteorologist call it an air parcel.  This parcel has the following characteristics:
 

1.  Air parcel is free to expand and contract in response to the surrounding pressure.
2. Air parcels expand as they rise in altitude into lower pressure and contract as they sink into higher pressure below.
3. Work is done BY a rising (expanding) parcel and thus decreases the internal energy of the parcel. As the internal energy decreases, so does the temperature.
4. Work is done ON a sinking (compressing) parcel and thus increasing the internal energy of the parcel. As the internal energy increases, so does the temperature.
5. These temperature changes in 3) and 4) are NOT the result of removing heat from or adding heat to the parcel. This process is called an adiabatic process because no heat is added.

10 C/km or 5.5 F/1000ft.

This cooling rate is called the Dry Adiabatic Lapse Rate.

Dry --> unsaturated (RH < 100%)

Adiabatic --> no heat is added to the parcel

Lapse Rate --> there is a lapse (decline) in the temperature as the altitude of the parcel increases

As an example, if air here in Tucson is at 86F and is lifted from an altitude of 2500ft to an altitude of 8500ft (near the top of Mt. Lemon), the air would cool;

5.5 F/1000ft x (8500ft - 2500ft)

5.5 F/1000ft x (6000ft)

5.5 F * 6 = 33F

33 F degrees to about 53 F.

On thing to note is that as this dry or unsaturated air parcel is lifted, the water vapor content is unchanged. So, the dew point remains the same.

Saturated Pseudo Adiabatic Process

Now at some point the rising parcel of air will cool so much that the temperature is equal to the dew point. Then we have a state of saturation. If the parcel continues to rise beyond this point, the air will continue to expand and cool, thus the saturation mixing ratio (SMR) will also decrease. Last time we said that water droplets form when the RH is approximately equal to 100% or saturation mixing ratio is approximately equal to the actual mixing ratio (AMR). So, we should not expect to see the RH increase much above 100% as the parcel cools, but remain at 100% or remain saturated. To do this as the SMR decreases, the AMR must also decrease. Therefore, water vapor must be taken out of the air. This causes the cloud droplets to nucleate (form) and grow via condensation. This condensation process releases latent heat that warms the parcel.

So as the saturated parcel rises, we have a competition between the cooling due to expansion and heating due to condensation. Since the expansion cooling is greater than the condensation heating, the parcel continues to cool, but at a much slower rate than when it was unsaturated (dry).

What would determine the rate of saturated cooling? Remember that expansion cooling is nearly constant with altitude. How might the condensation heating vary? It will depend upon the amount of condensation. Remember that the saturation mixing ratio did NOT change linearly with temperature, but slowly with cool temperatures and quickly with warm temperatures. So, the condensation heating will be larger for warmer temperatures than for cooler temperatures. The logic is;

warm saturated parcel ---> rapid condensation ---> slow cooling

cold saturated parcel ---> slow condensation ---> rapid cooling

This causes the rate of cooling to vary between:

4 C/km or 2.2 F/1000ft

to

9 C/km or 4.9 F/1000ft

we will use an average value of 3.3 F/1000ft for this saturated pseudo adiabatic lapse rate.

Saturated: RH=100%

Pseudo Adiabatic: heat is added during condensation (not quite adiabatic)

Lapse Rate: temperature lapse (decline) with increasing altitude

Summary:

1) When a rising parcel reaches saturation, latent heat is released, which slows the cooling rate. The altitude at which this saturation occurs is referred to as the condensation level (CL). Here the air temperature is equal to the dew point. This can often be seen in the atmosphere as the altitude of the base of a cloud. You'll note that the base of the cloud is fairly flat and the base altitude is much the same for all the clouds in the field.

2) This saturated cooling rate is less than the dry cooling rate.

3) Variations in the saturated cooling rate are due to variations in the rate of condensation.

4) In a warm parcel condensation is rapid, thus much latent heat is released, and the parcel cools slowly.

5) A cold parcel contains less water, so the rate of condensation is less and the parcel cools rapidly (almost as quickly as the dry rate).

Strictly speaking, it is not really true in the atmosphere, but we will assume in this class that any water that condenses falls out of the parcel (it rains).

Let?s return to our example of the parcel of air lifted from the surface here in Tucson to near the top of Mt. Lemon. If the dew point was 53 F (which would be more like summer time conditions), and at 8500 ft the air temperature, then the air would be saturated at this altitude and the condensation level would be 8500 ft. If the air was lifted an additional 1000ft to 9500ft, the temperature would decrease at a the saturated rate of 3.3 F/1000ft. Thus the temperature of the parcel at 9500ft would be

53 F - 3.3 F/1000ft x 1000ft

53 F- 3.3 F

49.7 F

What would be the dew point of the air parcel at 9500ft? Remember that the parcel was lifted from 8500ft to 9500ft via a saturated process, so the RH=100% during this process. So, the dew point would be equal to the air temperature. The dew point would be 49.7 F.

Now, if the air parcel comes back down in altitude from 9500ft to 2500ft, what would the temperature and dew point of the air parcel be at 2500 ft? As the parcel decreases in altitude, the pressure around it will increase causing the parcel to warm. If the parcel started at a state of saturation and was then warmed, it would no longer be saturated, but unsaturated. The RH would then be less than 100%. If the parcel is unsaturated, then it warms at a rate of 5.5 F/1000ft. So the total warming from 9500ft to 2500ft would be

5.5 F/1000ft x (9500ft - 2500ft)

5.5 F/1000ft x 7000ft

5.5 F x 7

38.5 F

The air temperature at 2500 ft, after being lifted initially from 2500ft to 8500ft then on to 9500ft and back down to 2500ft would be

49.7 F + 38.5 F= 88.2 F

or 2.2 F warmer than when it began. As the dew point does not change during this dry adiabatic process, the dew point of the parcel now at the surface would be 49.7 F or 3.3 F less than when it started. So, the parcel is warmer and dryer than when it began.
 
 
 
 

Atmospheric Stability:

Stability of the atmosphere refers to whether or not a parcel of air if forced to rise, will continue to rise on its own (unstable) or sink back down to the level it started at (stable). To determine this we must first look at a principle put forth many years ago.

Archimedes's Prinicple:

An object is subject to an upward force when it is immersed in a liquid. The force is equal to the weight of the liquid displaced. The net force acting upon the object is the difference between the downward weight of the object due to gravity and the upward buoyant force due to the weight of the displaced liquid.

Weight of object = Density of object x Volume of object x Acceleration of gravity

Buoyant force =Weight of displaced liquid = Density of liquid x Volume of object x Acceleration of gravity

Net force = Buoyant force - Weight of object

= density of liquid x volume of object x acceleration of gravity -

density of object x volume of object x acceleration of gravity

= (density of liquid - density of object) x volume of object x acceleration of gravity

So, if the density of the liquid is greater than the density of the object, then the net force will be in the upward direction and the object will rise.

If the density of the liquid is less than the density of the object, then the net force will be in the downward direction and the object will sink.

For example, if we took a balloon with a volume of 1 cubic meter and filled it with Helium gas with a density of 0.07 kg/cubic meter and placed it in the atmosphere which has a density of about 1kg/cubic meter what would happen. Well the density of the air is about 14 times that of the Helium, so it would have a net force in the upward direction and would rise.

If the same balloon was filled with pure oxygen with a density of 1.10 kg/cubic meter, the balloon would have a net force in the downward direction and it would sink since it is more dense than the air.
 
 
 
 
 
 

So, if we were to return to our parcel of air and wanted to know if it would rise or sink we must compare the density of the parcel to the air around it (the liquid in which it is immersed). Previously we learned how the temperature of the parcel will change with altitude and we can calculate that. Also remember that Charles? Law tells us how we can relate the temperature and density of a gas at the same pressure. This will allow us to compare the density of the parcel and the surrounding air, which are at the same pressure. The gist of the law said that warm air is less dense than cold air.

D1 * T1 = D2 * T2

or

D1/D2 = T2/T1
 
 

So, we just need to compare the temperature of the parcel and the surrounding air to see which direction the net force will be.

For example;

1) If the parcel is colder than the surrounding air, then the parcel will be denser and sink. In this case the atmosphere is said to be stable.

2) If the parcel is warmer that the surrounding air, then the parcel will be less dense and will rise. In this case the atmosphere is said to be unstable.

3) If the parcel is the same temperature as the surrounding air, then the densities will be the same and the parcel will remain where it is. In this case the atmosphere is said to be neutral.