Wednesday Oct. 17, 2012
click here to download today's notes in a more printer friendly format

A song I heard earlier this week while eating lunch, "Somebody That I Used to Know" from Gotye.

Most of the Experiment #2 reports have been graded and were returned in class today.  You can revise the reports if you want to (it isn't required).  Revised reports are due by Wed., Oct. 31.  I should have several reports that were turned in Monday this week graded in time to return on Friday.

I am also planning on handing out midterm grade summaries in class on Friday.

The Causes of the Seasons and Ozone and the Ozone Hole 1S1P reports were collected today.  The 3rd Assignment #2 topic is due next Wednesday.

And here is something I didn't announce  in class, a hidden optional assignment.  Download the assignment if you like and turn it in on Friday if you'd like to earn some extra credit.  I'll also have another, take home, assignment that I will hand out in class on Friday.  The second assignment will be due at the start of class next Wednesday (Oct. 24). 

The main event today was working through four humidity problems.  Hopefully this will increase your understanding of the roles the various humidity variables play and what can cause their values to change. 

Example #1

I gave my notes from class to a student so I'm using notes from a previous class.

But the notes from class are too hard to sort out even if you were in class.  So we'll work through this problem in a more detailed, step-by-step manner.

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 this data in the table.

Anytime you know the air's temperature you can look up the saturation mixing ratio value on a chart (such as the one on p. 86 in the ClassNotes); the saturation mixing ratio 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 g/kg to 4 g/kg.  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.  At first (Path 1 in the figure) 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 (Path 2) water will begin to drip out of the sponge (water vapor will condense from the air).

Example 2

We're given an air temperature of 90 F and a relative humidity of 50%; we'll try to figure out the mixing ratio and the dew point temperature.  Here's something like what we ended up with in class.

 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.  The details of that calculation are shown above at B.

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 g/kg)

(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 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%.

Finally in the last 5 minutes or so I showed a short student made video tape that described Experiment #3.  The object of the experiment is to measure the energy in sunlight arriving at the ground here in Tucson.  The apparatus used is sketched below:

It consists of two pieces of wood connected together with a hinge.  A styrofoam insert fits into one of the pieces of wood and hold a small rectangular piece of metal painted black so that it will absorb sunlight.  There is a hole drilled into the side of the metal block so that a thermometer can be inserted to measure the temperature of the block.  The block can be elevated and turned so that it is pointing straight at the sun (rays of sunlight strike the metal block perpendicularly).  When the block is oriented corectly a small dowel sticking out the front of the apparatus won't cast a shadow.

A couple of photographs of the apparatus are shown above.  You can see the black metal block, the dowel, and the small piece of wire that keeps one of the pieces of wood elevated so that it points at the sun.

In the left photo the apparatus hasn't yet been properly oriented and the dowel is casting a shadow.  At right the apparatus has been turned until the dowel isn't casting a shadow.

After setting up the device you simply the measure the block's temperature and time while the block heats up.  This change of temperature with time data together with the mass, crossectional area of the block and the specific heat of aluminum are enough to estimate the amount of energy striking the block.  Click on this link if you would like to read more about how that is done.