Thu., Oct. 18 notes

Thursday Oct. 18, 2007

Optional Assignments #4 (Controls of Temperature) and #5 (Humidity) are both due at the beginning of class next Tuesday. The in-class Optional Assignment was returned in class today (the answers were added to the Tue., Oct. 16 notes)

1S1P Assignment #2 reports are due one week from today (Thu., Oct. 25). All of the Assignment #1 reports have now been graded.

We need to work several humidity problems. By doing these problems you should become more familiar with the humidity variables (mixing ratio, saturation mixing ratio, relative humidity, and dew point temperature). You'll also learn "how they behave" and what can cause each of these variables to change value.

Keep this compilation of saturation mixing ratio values (shown in a table and on a graph) handy, we will use it a lot as we work through the humidity problem examples. Remember that saturation mixing ratio is the maximum amount of water vapor that can be found in air. It is a property of the air and depends on the air's temperature.

The beakers (beakers were also brought to class) are meant to show graphically the relative amounts of water vapor that air at different temperatures can contain.

Now the first of 4 example problems (actually done in class on Tuesday, Oct. 16)

Here is what we actually did in class. You might have a hard time unscrambling this if you're seeing it for the first time. The series of steps that we followed are retraced below:

We're given an air temperature of 90 F and a mixing ratio (r) of 6 g/kg. We're supposed to find the relative humidity (RH) and the dew point temperature.

We start by entering the data we were given in the table. Once you know the air's temperature you can look up the saturation mixing ratio value; it is 30 g/kg for 90 F air. 90 F air could potentially hold 30 grams of water vapor per kilogram of dry air (it actually contains 6 grams per kilogram in this example).

Once you know mixing ratio and saturation mixing ratio you can calculate the relative humidity. The RH is 20%.

The numbers we just figured out are shown on the top line above.

(A) We imagined cooling the air from 90F to 70F, then to 55F, and finally to 45F.

(B) At each step we looked up the saturation mixing ratio and entered it on the chart. Note that the saturation mixing ratio values decrease as the air is cooling.

(C) The mixing ratio doesn't change as we cool the air. The only thing that changes r is adding or removing water vapor and we aren't doing either.

(D) Note how the relative humidity is increasing as we cool the air. The air still contains the same amount of water vapor it is just that the air's capacity is decreasing.

Finally at 45 F the RH becomes 100%. The dew point temperature in this problem is 45 F.

What would happen if we cooled the air further still, below the dew point temperature?

35 F air can't hold the 6 grams of water vapor that 45 F air can. You can only "fit" 4 grams of water vapor into the 35 F air. The remaining 2 grams would condense. If this happened at ground level the ground would get wet with dew. If it happens above the ground, the water vapor condenses onto small particles in the air and forms fog or a cloud. Now because water vapor is being taken out of the air (and being turned into water), the mixing ratio will decrease from 6 to 4. That is why the mixing ratio is changing.

In many ways cooling moist air is liking squeezing a moist sponge (the figure below wasn't shown in class)

Squeezing the sponge and reducing its volume is like cooling moist air and reducing the saturation mixing ratio. At first when you sqeeze the sponge nothing happens, no water drips out. Eventually you get to a point where the sponge is saturated. This is like reaching the dew point. If you squeeze the sponge any further (or cool air below the dew point) water will begin to drip out of the sponge (water vapor will condense from the air).

Now Problem #2 (this is where we actually started in class on Thursday)

Here is what we did in class. Given an air temperature of 90 F and a relative humidity of 50% you are supposed to figure out the mixing ratio (15 g/kg) and the dew point temperature (70 F). The problem is worked out in detail below:

First you fill in the air temperature and the RH data that you are given.

(A) since you know the air's temperature you can look up the saturation mixing ratio (30 g/kg).

(B) Then you can substitute into the relative humidity formula and solve for the mixing ratio (15 g/kg).

Finally you imagine cooling the air. Cooling causes the saturation mixing ratio to decrease, the mixing ratio stays constant, and the relative humidity increases. In this example the RH reached 100% when the air had cooled to 70 F. That is the dew point temperature.

We can use results from humidity problems #1 and #2 to learn a useful rule.

In the first example the difference between the air and dew point temperatures was large (45 F) and the RH was low. In the difference between the air and dew point temperatures was smaller and the RH was high. The easiest way to remember this rule is to remember the case where there is no difference between the air and dew point temperatures. The RH would be 100%.

Now on to Problem #3

This figure was redrawn after class. You are given a mixing ratio of 10.5 g/kg and a relative humidity of 50%. You need to figure out the air temperature and the dew point temperature.

(1) The air contains 10.5 g/kg of water vapor, this is 50%, half, of what the air could potentially hold. So the air's capacity, the saturation mixing ratio must be 21 g/kg (you can either do this in your head or use the RH equation following the steps shown).

(2) Once you know the saturation mixing ratio you can look up the air temperature in a table.

(3) Then you imagine cooling the air until the RH becomes 100%. This occurs at 60 F. The dew point is 60 F.

Problem #4 is probably the most difficult of the bunch.

Here's what we did in class. We were given the air temperature and the dew point temperature.

We enter the two temperatures onto a chart and look up the saturation mixing ratio for each.

Then we know that if we cool the 90 F air to 50 F the RH will become 100%. We don't know the mixing ratio or the relative humidity. But we do know that when we arrive at 50 F the RH will be 100%. The mixing ratio must be equal to the saturation mixing ratio value for 50 F air, 7.5 g/kg.

Remember back to the three earlier examples. When we cooled air to the the dew point, the mixing ratio didn't change. So the mixing ratio must have been 7.5 all along. Once we know the mixing ratio in the 90 F air it is a simple matter to calculate the relative humidity, 25%.

Next we will use what we have learned about humidity variables (what they tell you about the air and what causes them to change value) to learn something new.

At Point 1 we start with some 90 F air with a relative humidity of 25%, fairly dry air (these data are the same as in Problem #4). Point 2 shows the air being cooled to the dew point, that is where the relative humidity would reach 100% and a cloud would form. Then the air is cooled below the dew point, to 30 F. Point 3 shows the 30 F air can't hold the 7.5 g/kg of water vapor that was originally found in the air. The excess moisture must condense (we will assume it falls out of the air as rain or snow). When air reaches 30 F it contains less than half the moisture (3 g/kg) that it originally did (7.5 g/kg). Next, Point 4, the 30 F air is warmed back to 90 F, the starting temperature. The air now has a RH of only 10%.

Drying moist air is like wringing moisture from a wet sponge.

You start to squeeze the sponge and nothing happens at first (that's like cooling the air, the mixing ratio stays constant as long as the air doesn't lose any water vapor). Eventually water will start to drop from the sponge (with air this is what happens when you reach the dew point and continue to cool the air below the dew point). Then you let go of the sponge and let it expand back to its orignal shape and size (the air warms back to its original temperature). The sponge (and the air) will be drier than when you started.

This sort of process ("squeezing" water vapor out of moist air by cooling the air below its dew point) happens all the time. Here are a couple of examples.

In the winter cold air is brought inside your house or apartment and warmed. Imagine 30 F air with a RH of 100% (this is a best case scenario, the cold winter air usually has a lower dew point and is drier). Bringing the air inside and warming it will cause the RH to drop from 100% to 20%.. Air indoors during the winter is often very dry.

The air in an airplane comes from outside the plane. The air outside the plane can be very cold (-60 F perhaps) and contains very little water vapor (even if the -60 F air is saturated it would contain essentially no water vapor). When brought inside and warmed to a comfortable temperature, the RH of the air in the plane will be very close 0%. Passengers often complain of becoming dehydrated on long airplane flights. The plane's ventilation system probably adds moisture to the air so that it doesn't get that dry.

Here's a very important example, the rain shadow effect (the figure in class was redrawn for clarity).

We start with some moist but unsaturated air (RH is about 50%) at Point 1. As it is moving toward the right the air runs into a mountain and starts to rise (see the note below). Unsaturated air cools 10 C for every kilometer of altitude gain. This is known as the dry adiabatic lapse rate. So in rising 1 km the air will cool to 10 C which is the dew point.

The air becomes saturated at Point 2, you would see a cloud appear. Rising saturated air cools at a slower rate than unsaturated air. We'll use a value of 6 C/km (an average value). The air cools from 10 C to 4 C in next kilometer up to the top of the mountain. Because the air is being cooled below its dew point at Point 3, some of the water vapor will condense and fall to the ground as rain.

At Point 4 the air starts back down the right side of the mountain. Sinking air is compressed and warms. As soon as the air starts to sink and warm, the relative humidity drops below 100% and the cloud evaporates. The sinking air will warm at the 10 C/km rate.

At Point 5 the air ends up warmer (24 C vs 20 C) and drier (Td = 4 C vs Td = 10 C) than when it started out. The downwind side of the mountain is referred to as a "rain shadow" because rain is less likely there than on the upwind side of the mountain. The rain is less likely because the air is sinking and because the air on the downwind side is drier than it was on the upslope side.

Most of the year the air that arrives in Arizona comes from the Pacific Ocean. It usually isn't very moist by the time it reaches Arizona because it has travelled up and over the Sierra Nevada mountains in California and the Sierra Madre mountains further south in Mexico. The air loses much of its moisture on the western slopes of those mountains.

NOTE: The figure above illustrates orographic or topographic lifting. It is one of 4 ways of causing air to rise. We have already run into the other three in class this semester. They were: convergence (surface winds spiral into centers of low pressure), convection (warm air rises), and fronts. Rising air is important because rising air expands and cools. Cooling moist air raises the relative humidity and a cloud might form.

A sling psychrometer is a simple instrument that can be used to measure relative humidity and dew point temperature.

A sling psychrometer consists of two thermometers mounted side by side. One is an ordinary thermometer, the other is covered with a wet piece of cloth. To make a humidity measurement you swing the psychrometer around for a minute or two and then read the temperatures from the two thermometers. The dry - wet buld temperature difference can be used to determine relative humidity and dew point (see Appendix D at the back of the textbook).

The figure at upper left shows what will happen as you start to swing the wet bulb thermometer. Water will begin to evaporate from the wet piece of cloth. The amount or rate of evaporation will depend on the air temperature (the 80 F value was just made up in this example).

The evaporation is shown as blue arrows because this will cool the thermometer. The same thing would happen if you were to step out of a swimming pool on a warm dry day, you would feel cold. Swamp coolers would work well on a day like this.

The figure at upper left also shows one arrow of condensation. The amount or rate of condensation depends on how much water vapor is in the air surrounding the thermometer. In this case (low relative humidity) there isn't much water vapor. The condensation arrow is orange because the condensation will release latent heat and warm the thermometer.

Because there is more evaporation (4 arrows) than condensation (1 arrow) the wet bulb thermometer will drop.

Note in the bottom left figure we imagine that the wet bulb thermometer has cooled to 60 F. Because the wet piece of cloth is cooler, there is less evaporation. The wet bulb thermometer has cooled to a temperature where the evaporation and condensation are in balance. The thermometer won't cool any further.

You would measure a large difference between the dry and wet bulb thermometers (20 F) on a day like this when the air is relatively dry.

The air temperature is the same in this example, but there is more water vapor in the air. You wouldn't feel as cold if you stepped out of a pool on a warm humid day like this. Swamp coolers wouldn't provide much cooling on a day like this.

There are four arrows of evaporation (because the air temperature is the same as in the previous example) and three arrows now of condensation (due to the increased amount of water vapor in the air surrounding the thermometer). The wet bulb thermometer will cool but won't get as cold as in the previous example.

The wet bulb thermometer might well only cool to 75 F. This might be enough to lower the rate of evaporation enough to bring it into balance with the rate of condensation.

You would measure a small difference (5 F) between the dry and wet bulb thermometers on a humid day like this one.

There won't be any difference in the dry and wet bulb temperatures when the RH=100%. The dry and wet bulb thermometers would both read 80 F.

We will be discussing many of the phenomena above. They all involve cooling air to (and/or below) the dew point temperature. The air becomes saturated (RH=100%) and water vapor begins to condense.

It turns out that it is much easier for water vapor to condense onto something rather than just forming a small droplet of pure water Near the ground water vapor will condense onto cold objects on the ground (the grass, automobile, and newspaper above). In air above the ground water vapor condenses onto small particles in the air called condensation nuclei. We'll learn a little bit about these today.

The formation of fog and haze, scattering of light, and clouds will all be covered next Tuesday.

In the first example air starts out with a temperature of 65 F early in the evening. It cools to 35 F during the night. When the air reaches 40 F, the dew point, the RH reaches 100%. As the air temperature drops below the dew point and cools to 35 F water vapor will condense onto the ground or objects on the ground (such as an automobile). This is dew.

The dew point is the same but the nighttime minimum temperature is below freezing in the second example. Dew will form again on this night when the air temperature reaches 40 F. Once the air temperature drops below 32 F though the dew will freeze and form frozen dew.

In the third example both the dew point and nighttime minimum temperatures are below freezing. When the air temperature drops below the dew point, water vapor turns directly to ice (deposition) and forms frost. The dew point in this case is sometimes called the frost point.

The air never becomes saturated in the fourth example because the nighttime minimum temperature never cools to the dew point. You wouldn't see anything on this night.

When air above the ground reaches 100% relative humidity it is much easier for water vapor to condense onto small particles in the air called condensation nuclei than to just form a small droplet of water. There are hundreds even thousands of these small particles in every cubic centimeter of air. We can't see them because they are so small.

You can learn why it is so hard to form small droplets of pure water by reading the top of p. 92 in the photocopied class notes.

Water vapor will condense onto certain kinds of condensation nuclei even when the relative humidity is below 100% (again you will find some explanation of this on the bottom of p. 92). These are called hygroscopic nuclei.

A short video showed how water vapor would, over time, preferentially condense onto small grains of salt rather than small spheres of glass.

The start of the video at left showed the small grains of salt were placed on a platform in a petri dish containing water. Some small spheres of glass were placed in the same dish. After about 1 hour small drops of water had formed around each of the grains of salt (shown above at right). The figure above wasn't shown in class.

In humid parts of the US, water will condense onto the grains of salt in a salt shaker causing them to stick together. Grains of rice apparently will keep this from happening and allow the salt to flow freely out of the shaker when needed.