Thu., Mar. 22, 2012
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The Blue Danube waltz seemed like a good way to celebrate the Spring Equinox and the return of warm weather.

The in-class assignment from Tuesday was returned today.  Everyone received full credit (0.15 pts of extra credit) even if you didn't get all the answers right.  Here are some answers to the questions.

The 1S1P Bonus Assignment report on "Causes of the Seasons" was collected today together with the Expt. #2 revised reports. 

A new take-home Optional Assignment was handed out.  It is due next Thursday (Mar. 29).  The Controls of Temperature assignment is due net Tuesday (Mar. 27).

I hope you saw the graupel (aka "soft hail" or "snow pellets") earlier this week.  If not here's a good picture of some graupel that fell in Catalina AZ (northwest of Tucson).  The photograph was taken by a student in this class.




Now onto the main event for today, some example humidity problems. 

Example 1


Here's what was actually written down in class.   You will have a hard time unscrambling this if you're seeing it for the first time or didn't understand it when we went over it in class.  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 (using the chart on p. 86 in the ClassNotes); 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 (you divide the mixing ratio by the saturation mixing ratio, 6/30, and multiply the result by 100%).  You ought to be able to work out the ratio 6/30 in your head (6/30 = 1/5 = 0.2).  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 (r) 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.  This is probably the most difficult concept to grasp.

(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%.  This is kind of a special point.  You have cooled the air until it has become saturated.  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.  Because water vapor is being taken out of the air (the water vapor is turning into water), the mixing ratio will decrease from 6 to 4.  As you cool air below the dew point, the RH stays constant at 100% and the mixing ratio decreases.

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



Squeezing the sponge and reducing its volume is like cooling moist air and reducing the saturation mixing ratio.  (1) 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.  (2) 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).

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 and understand 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 the mixing ratio = 10.5 g/kg and the relative humidity = 50%.    You need to figure out the air temperature and the dew point temperature.  Here's the play by play solution to the question

(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 above). 

(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.  But one of the things we said about dew point is that it has the same job as mixing ratio - it gives you an idea of the actual amount of water vapor in the air.  This problem will show that if you know the dew point, you can quickly figure out the mixing ratio.  Knowing the dew point is equivalent to knowing the mixing ratio.

Here's what we ended up with 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 45 F the RH will become 100%.  We can set the mixing ratio equal to the value of the saturation mixing ratio at 45 F, 6 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 6 all along.   Once we know the mixing ratio in the 90 F air it is a simple matter to calculate the relative humidity, 20%.

And now for something completely different (actually many of you just turned in a 1S1P report on the Causes of the Seasons so this should be material you're pretty familiar with.  But the Spring Equinox took place earlier this week (it was 05:14 UT on Tuesday Mar. 20 this year which was 10:14 pm Monday night in Tucson).  We can't let a big event like that go unnoticed.

The figure above shows the earth orbiting the sun. 

On or around Dec. 21st, the winter solstice, the north pole is tilted away from the sun.  Note that a small portion of the earth near the N. Pole (north of the Arctic Circle) spends 24 hours in darkness.  Days are less than 12 hours long in the northern hemisphere and the sun is low in the sky.  Both factors reduce the amount of sunlight energy reaching the ground.  That's why it's cold and wintry.

On June 21st, the summer solstice, the north pole is tilted toward the sun.  Now there are 24 hours of sunlight north of the Arctic Circle.  Days are more than 12 hours long in the northern hemisphere and the sun is high in the sky at noon.  A lot more sunlight energy reaches the ground; that's why it is summer.

The equinoxes are a time of transition.  On the equinoxes, the N. Pole still tilted just not toward or away from the sun.  The line separating day and night passes through the pole and the days and nights are each about 12 hours long everywhere on earth (except perhaps at the poles). 

The drawing below shows you what you would see at sunrise (about 6:30 am) on the Spring Equinox here in Tucson (the same would happen on the Fall Equinox).  The sun rises exactly in the east on the equinoxes.  The rest of the year it is a little to the north or south of east.


At noon you would need to look south to see the sun.


The sun reaches its highest point in the sky at noon.  On the equinoxes in Tucson that's almost 60 degrees.  The sun is lower in the sky (34.5 degrees above the horizon) on the winter solstice.  That together with the fact that the days are shorter means much less sunlight energy reaches the ground.  In the summer the days are longer and the sun gets much higher in the sky at noon (81.5 degrees above the horizon, nearly overhead).  Much more sunlight energy reaches the ground and it is much warmer.

The sun passes directly overhead at the equator at noon on the equinoxes.


The sun sets exactly in the west on the equinoxes at about 6:30 pm in Tucson.
This is the 2 pm class.  Most of you are more likely (perhaps) to see the sun set than see the sun rise.  The figure below shows you about what you would see if you looked west on Speedway (from Treat Ave.) at sunset.  In the winter the sun will set south of west, in the summer north of west (probably further south and north than shown here).  On the equinoxes the sun sets exactly in the west.  This is something you should check out for yourself this week before the sun moves noticeably to the north of due west.

Several years ago I positioned myself in the median near the intersecton of Treat and Speedway and pointed my camera west.  I took a multiple exposure photograph of the sun over a 2 or 3 hour period that ended at sunset.  I'll bring the slide photograph to class one of these days.

Something else to note in this figure and something I didn't mention in class.  Note how the sun is changing color.  It changes from a bright yellow white to almost red by the time it sets..  This is due to scattering of sunlight by air.  The shorter wavelengths (violet, blue, green) are scattered more readily than the longer wavelengths.  At sunset the rays of sunlight take a much longer slanted path through the atmosphere and most of the shorter wavelengths are scattered and removed from the beam of sunlight.  All that's left in the beam of light that reaches your eyes are the longer wavelengths: yellow, orange, and red.

If you aren't careful, you can get yourself seriously injured, even killed, on or around the equinoxes.  Here's an article that appeared in the Arizona Daily Star at the time of the equinox last fall (Thu., Sep. 22).

I forgot to come back and discuss the following figure in class.



December 21, the summer solstice, is the shortest day of the year (about 10 hours of daylight in Tucson).  The days have slowly been getting longer since then. The rate of change is greatest at the time of the equinox.

This will continue up until June 21, the summer solstice, when there will be about 14 hours of daylight.  After that the days will start to shorten again as we make our way back to the winter solstice.
There was a very interesting coicidence last semester.  We were covering some of this same material in class on Friday Sep. 23.  There were a few parents in class because it was Parent's Weekend.  I showed these same pictures on that afternoon.  One of the parents came up to the front after class and mentioned having seeing the sun right at the end of 77th St. in New York City around this time of year.  That got me thinking that a picture of sunset at the end of one of the long streets with all the tall buildings might be spectacular.

When I started looking however I found that the major streets in Manhattan aren't oriented EW and NS.  You can see this on a Google map of Manhattan.  77th St. is oriented in more of a NW-SE direction.   So the sun doesn't shine straight down 77th St. at sunrise and sunset on the equinoxes.  I was pretty disappointed but then I stumbled on the this Manhattanhenge map which shows the direction of sunset (the left, west, side of the map) and sunrise (the right, east, side of the map) at various times of the year. 

If you remember that as you move past the Spring Equinox toward summer sunrise move north of east and sunset is north of west.  On May 31 the sun has moved far enough north that it does set right at the west end of 77th St.  Sunset continues to move north up until the summer solstice on June 21.  Then the sunset starts to move back south.  You can again see the sunset at the west end of 77th St. on July 12 and 13.  An article with several Manhattanhenge photographs from the May 31 event appeared in a story on the Business Insider webpage.  That would certainly make a worthwhile field trip in Atmo 170A1 if the semester went that long.  The "henge" part of the name comes from Stonehenge where the rising and setting sun aligns with stones on the solstices.

You can also see the sunrise at the east end of 77th St.  But sunrise has to be in the southeast.  This takes place on Dec. 5 and Jan. 8, just before and just after the winter solstice.

We had a little time for a couple of more topics that begin to apply some of what we have been learning about humidity.

The figure below is on p. 87 in the photocopied ClassNotes.  It explains how you can dry moist air.


At Point 1 we start with some 90 F air with a relative humidity of 25%, fairly dry air.   These are the numbers we ended up with in Example Problem #4.  We imagine cooling this air to the dew point temperature where the relative humidity would reach 100% and a cloud would form (Pt. 2 in the figure above). 

Then we continue to cool the air below the dew point, to 30 F.  Air that is cooled below the dew point finds itself with more water vapor than it can contain.  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).  The air is being warmed back up to 90 F along Path 4.  As it warms the mixing ratio remains constant.  At Point 5, the air now has a RH of only 10%.

Drying moist air is very much like wringing moisture from a wet sponge.  The figure below wasn't shown in class.


You start to squeeze the sponge and it gets smaller.  That's like cooling the air and reducing the saturation mixing ratio, the air's capacity for water vapor.  At first squeezing the sponge doesn't cause anything to happen (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.  This can cause chapped skin, can irritate nasal passages, and cause cat's fur to become charged with static electricity.

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 dehydration on long airplane flights.  The plane's ventilation system probably adds moisture to the air so that it doesn't get that dry.

Next a much more important example of drying moist air (see p. 88 in the photocopied ClassNotes).


We start with some moist but unsaturated air (the RH is about 50%) at Point 1 (the air and dew point temperatures would need to be equal in order for the air to be saturated).  As it is moving toward the right the air runs into a mountain and starts to rise.  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.  So after rising 1 km the air will cool to 10 C which is the dew point.

The air becomes saturated at Point 2 (the air temperature and the dew point are both 10 C).  Would you be able to tell if you were outdoors looking at the mountain?  Yes, you would see a cloud appear. 

Now that the RH = 100%, the saturated air cools at a slower rate than unsaturated air (condensation of water vapor releases latent heat energy inside the rising volume of air, 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 and the value of the mixing ratio (and the dew point temperature) decreases.

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.

I went looking for some good specific illustrations of the rainshadow effect after class.  None of the following figures were shown in class on Thursday.

We can see the effects of a rainshadow illustrated well in the state of Oregon.  The figure above at left shows the topography (here's the source of that map).  Winds generally blow from west to east across the state. 

Coming off the Pacific Ocean the winds first encounter a coastal range of moutains.  On the precipitation map above at right (source) you see a lot of greens and blue on the western sides of the coastal range.  These colors indicate yearly rainfall totals that range from about 50 to more than 180 inches of rain per year.  This is where temperature rain forests are found. 

That's the Willamette River, I think, in between the coastal range and the Cascades.  This valley is somewhat drier than the coast because air moving off the Pacific has lost some of its moisture moving over the coastal range. 

What moisture does remain in the air is removed as the winds move up and over the taller Cascades.  Yearly rainfall is generally less than 20 inches per year on the eastern side, the rainshadow side, of the Cascades.  That's not too much more than Tucson which averages about 12 inches of rain a year.

Here's the best picture of the rain shadow effect I could find (here's the source of the picture).  I didn't show this picture in class either.


The Himalayan mountains stretch across the lower left 1/3 of the picture.  The land below and to the left of the mountains appears somewhat green in the picture.  This is because moist air moving from lower left toward the upper right leaves most of its moisture on this side of the mountain range.  The upper right 2/3rds of the picture, the Tibetan plateau, is in the rain shadow and appears very dry and brown in the photograph.

Most of the year the air that arrives in Arizona comes from the west, from the Pacific Ocean (this changes in the summer).  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.