Thursday Mar. 25, 2010
click here to download today's notes in a more printer friendly format

2 or 3 songs from Vampire Weekend's new CD Contra ("Cousins" and "I Think Ur a Contra" maybe).

The Experiment #3 reports are due next Tuesday.  Try to return your materials this week (either in class or come by my office) so that you can pick up the Supplementary Information handout.  Experiment #4 reports are due on Tuesday April 6.

The revised Expt. #1 reports have been graded and can be picked up in class.

There is a new Optional Assignment on humidity that is due next Tuesday.  Somewhere in today's notes you'll find a link to a hidden Optional Assignment.

We finished class on Tuesday with the first of four humidity problem examples.  We worked through the remaining three at the beginning of class today.  The main reason for doing this is to learn more about the 4 humidity variables (mixing ratio, saturation mixing ratio, relative humidity, and dew point temperature), what they do, and what can cause their values to change. 

Example 2


The work that we did in class is shown above. 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 might be able to figure out the mixing ratio in your head.  Air that is filled to 50% of its capacity could hold up to 30 g/kg.  Half of 30 is 15, that is the mixing ratio.  Or you can substitute into the relative humidity formula and solve for the mixing ratio.


Finally you imagine cooling the air.  The saturation mixing ratio decreases, 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 (20%).  In the 2nd problem the difference between the air and dew point temperatures was smaller (20 F) and the RH was higher (50%).  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 then would be 100%.
Example 3


You're given the mixing ratio (10.5) and the relative humidity (50).  What are the units (g/kg and %).  Here's the play by play solution to the question


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 (80 F air has a saturation mixing ratio of 21)

(3) Then you imagine cooling the air until the RH becomes 100%.  This occurs at 60 F.  The dew point is 60 F.
Example 4
probably the most difficult problem of the bunch.

Here's what we did in class, we were given the air temperature and the dew point temperature.  We were supposed to figure out the mixing ratio and the relative humidity. 


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

We ignore the fact that we don't know the mixing ratio.  We do know that if we cool the 90 F air to 50 F the RH will become 100%.  We can set the mixing ratio equal to the value of the saturation mixing ratio at 50 F, 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%.

Here's a list of some of the important facts and properties of the 4 humidity variables that we have been talking about.  This list wasn't shown in class.

Now we can start to use what we have learned about humidity variables (what they tell you about the air and what causes them to change value) to learn some new things.  The figure below is on p. 87 in the photocopied ClassNotes. 



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 on Monday).  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 3 g/kg, less than half the moisture that it originally did (7.5 g/kg).  Next, Point 4, the 30 F air is warmed back to 90 F, the starting temperature, Point 5.  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 (p. 87 again)

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 (p. 88 in the ClassNotes).



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).  Rising air expands and cools.   Unsaturated air cools 10 C for every kilometer of altitude gain.  This is known as the dry adiabatic lapse rate and isn't something you need to remember.  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 (condensation of water vapor releases latent heat energy, this warming partly offsets the cooling caused by expansion).  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.  Moisture is being removed from the air.

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 disappears.  The sinking unsaturated 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.  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.  The eastern half of Oregon is drier than the western half because air travels from the Pacific up and over the Cascade mountains.  It loses a lot of its moisture on the upslope side of the mountains.

Next in our potpourri of topics today was measuring humidity.  One of the ways of measuring humidity is to use a sling (swing might be more descriptive) psychrometer.


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 thermometer (dry and wet bulb) temperature difference can be used to determine relative humidity and dew point. 

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 water temperature (the 80 F value was just made up in this example).  Warm water evaporates at a higher rate than cool water.

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 (too well sometimes) 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. 

The wet thermometer will cool but it won't cool indefinitely.  We imagine that the wet bulb thermometer has cooled to 60 F.  Because the wet piece of cloth is cooler, the water is evaporating more slowly.  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 (20 F) between the dry and wet bulb thermometers 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 water temperature is still 80 F just as it was 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 (from 4 arrows to 3 arrows) 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.

There won't be any difference in the dry and wet bulb temperatures when the RH=100%.  The rates at which water is evaporating and water vapor is condensing are equal.  That's one of the things that happens when air is saturated.  The dry and wet bulb thermometers would both read 80 F.



A variety of things can happen when you cool air to the dew point and the relative humidity increases to 100%.  Point 1 shows that when moist air next to the ground is cooled to and below the dew point, water vapor condenses onto (or is deposited onto) the ground or objects on the ground.  This forms dew, frozen dew, and frost. 

Air above the ground can also be cooled to the dew point.  When that happens (Point 2 above) it is much easier for water vapor to condense onto something rather than just forming a small droplet of pure water.    In air above the ground water vapor condenses onto small particles in the air called condensation nuclei.  The small water droplets that form are themselves usually too small to be seen with the naked eye.  We can tell they are present (Point 3) because they either scatter (haze or fog) or reflect (clouds) sunlight. 

We'll learn a little bit about the formation of dew and frost today.  Next Tuesday we'll learn more about condensation nuclei and will do a cloud-in-a-bottle demonstration (Point 5) to see the role that cloud condensation nuclei can play in cloud formation.


It might be a little hard to figure out what is being illustrated here.  Point 1 is sometime in the early evening when the temperature of the air at ground level is 65.  During the course of the coming night the air will cool to 35 F.  When the air temperature reaches 40 F, the dew point, the relative humidity reaches 100% and water vapor begins to condense onto the ground.  You would find your newspaper and your car covered with dew the next morning.

The next night is similar except that the nighttime minimum temperature drops below freezing.  Dew forms and first covers everything on the ground with water.  Then the water freezes and turns to ice.  This isn't frost, rather frozen dew.  Frozen dew is often thicker and harder to scrape off your car windshield than frost. 

Now the dew point and the nighttime minimum temperature are both below freezing.  When the RH reaches 100% water vapor turns directly to ice (deposition).  This is frost.

What happens on this night?  Because the nighttime minimum temperature never reaches the dew point and the RH never reaches 100%, nothing would happen.

Class ended a few minutes early so that I could get to a 10 am dental appointment.  I am always impressed by the amount of material that we cover in a period in the T Th class.  This is plenty of material for a class period in my opinion; we should probably always end class around 9 am.  Let me know if you feel strongly one way or the other.  I almost forgot, here's the link to the hidden Optional Assignment.