Tuesday Oct. 19, 2010
click here to download today's notes in a more printer friendly format

Three songs from The Beatles ("I'll Follow the Sun", "Mr Moonrise", and "While My Guitar Gently Weeps") before class.

The quizzes were returned in class today together with a midterm grade summary printout.  The grade summary is discussed at the end of today's notes.

There was an In-class Optional Assignment today.  If you download the assignment, answer the questions, and then turn it in at the start of class on Thursday you can receive at least partial credit for it.

Here's an idea of what we will be covering between now and the next quiz
(with maybe some emphasis on the items in bold)

humidity, measuring humidity, heat index
dew and frost, radiation and steam fog
condensation nuclei and cloud formation
identifying and naming clouds
satellite photographs
precipitation formation and types of precipitation

The following is an introduction to the first item: humidity (moisture in the air).  This topic and the terms that we will be learning and using can be confusing.  That's the reason for this introduction.  We will be mainly be interested in 4 variables: mixing ratio, saturation mixing ratio, relative humidity, and dew point temperature.  Our first job will be to figure out what they are and what they're good for.  Then we see what can cause the value of each variable to change.  You will find much of what follows on page 83 in the photocopied ClassNotes. 



Mixing ratio tells you how much water vapor is actually in the air.  Mixing ratio has units of grams of water vapor per kilogram of dry air (the amount of water vapor in grams mixed with a kilogram of dry air).  It is basically the same idea as teaspoons of sugar mixed in a cup of tea. 



The value of the mixing ratio won't change unless you add water vapor to or remove water vapor from the air.  Warming the air won't change the mixing ratio.  Cooling the air won't change the mixing ratio (one exception is when the air is cooled below its dew point temperature and water vapor starts to condense).  Since the mixing ratio's job is to tell you how much water vapor is in the air, you don't want it to change unless water vapor is actually added to or removed from the air.


Saturation mixing ratio is just an upper limit to how much water vapor can be found in air, the air's capacity for water vapor.  It's a property of air, it doesn't say anything about how much water vapor is actually in the air (that's the mixing ratio's job).  Warm air can potentially hold more water vapor than cold air.  This variable has the same units: grams of water vapor per kilogram of dry air.  Saturation mixing ratio values for different air temperatures are listed and graphed on p. 86 in the photocopied class notes. 


Just as is the case with water vapor in air, there's a limit to how much sugar can be dissolved in a cup of hot water.  You can dissolve more sugar in hot water than in cold water.

 The dependence of saturation mixing ratio on air temperature is illustrated below:


The small specks represent all of the gases in air except for the water vapor.  Each of the open circles represents 1 gram of water vapor that the air could potentially hold.  There are 15 open circles drawn in the 1 kg of 70 F air; each 1 kg of 70 F air could hold up to 15 grams of water vapor.  The 40 F air only has 5 open circles; this cooler air can only hold up to 5 grams of water vapor per kilogram of dry air.

Now we have gone and actually put some water vapor into the volumes of 70 F and 40 F air (the open circles are colored in).  The same amount, 3 grams of water vapor, has been added to each volume of air.  The mixing ratio, r, is 3 g/kg in both cases.


The relative humidity is the variable most people are familiar with, it tells you how "full" the air is with water vapor, how close it is to being filled to capacity with water vapor.

In the analogy (sketched on the right hand side of p. 83 in the photocopied notes) 4 students wander into Classroom A which has 16 empty seats.  Classroom A is filled to 25% of its capacity.  You can think of 4, the actual number of students, as being analogous to the mixing ratio.  The classroom capacity is analogous to the saturation mixing ratio.  The percentage occupancy is analogous to the relative humidity.
The figure below goes back to the volumes (1 kg each) of 70 F and 40 F air that could potentially hold 15 grams or 5 grams, respectively of water vapor.


Both the 70 F and the 40 F air that each contain 3 grams of water vapor.  The 70 F air is only filled to 20% of capacity (3 of the 15 open circles is colored in) because this warm air's saturation mixing ratio is large.  The RH in the 40 F is 60% even though it has the same actual amount of water vapor because the 40 F air can't hold as much water vapor and is closer to being saturated. 

Something important to note: RH doesn't really tell you how much water vapor is actually in the air.  The two volumes of air above contain the same amount of water vapor (3 grams per kilogram) but have very different relative humidities.  You could just as easily have two volumes of air with the same relative humidities but different actual amounts of water vapor.



The dew point temperature has two jobs.  First it gives you an idea of the actual amount of water vapor in the air.  In this respect it is just like the mixing ratio.  If the dew point temperature is low the air doesn't contain much water vapor.  If it is high the air contains more water vapor. 

Second the dew point tells you how much you must cool the air in order to cause the RH to increase to 100% (at which point a cloud, or dew or frost, or fog would form).


If we cool the 70 F air or the 40 F air to 30 F we would find that the saturation mixing ratio would decrease to 3 grams/kilogram.  Since the air actually contains 3 g/kg, the RH of the 30 F air would become 100%.  The 30 F air would be saturated, it would be filled to capacity with water vapor.  30 F is the dew point temperature for 70 F air that contains 3 grams of water vapor per kilogram of dry air.  It is also the dew point temperature for 40 F air that contains 3 grams of water vapor per kilogram of dry air.Because both volumes of air had the same amount of water vapor, they both also have the same dew point temperature.
Now back to the student/classroom analogy


The 4 students move into classrooms of smaller and smaller capacity.  The decreasing capacity of the  classrooms is analogous to the decrease in saturation mixing ratio that occurs when you cool air.  Eventually the students move into a classroom that they just fill to capacity. This is analogous to cooling the air to the dew point.

A short break at this point to look at a couple of video about Experiment #3.  The object of that experiment is to measure the solar irradiance, the energy in sunlight.  Here's a little more information about the experiment (this  link wasn't mentioned in class and it's not something you need to worry about unless you're actually doing Expt. #3)

The rest of the period was spent working out some humidity example problems.  This way you will learn more about the 4 humidity variables; you'll see what they do and what can cause their values to change. 

Example 1


Here is the first sample problem that we worked 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).  A table of saturation mixing ratio values can be found on p. 86 in the ClassNotes.

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 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%.  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.  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.

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 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).

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 worked in class on Monday 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.

Here's a midterm grade summary example (numbers in the example are class averages)


1.    You should find your two quiz scores here.  The quiz percentage grades are used to compute your overall grade; all the quizzes have the same weight.

2.    This is the total number of extra credit points you have earned on the Optional Assignments.  The In-class Optional Assignment from last Friday that is being returned today is not included in the total.  If you have done all the Optional Assignments you should have 1.4 pts (unless you elected to receive a Green Card on one of the assignment instead of extra credit points, in that case you should have 0.9 pts)

3.    If you have turned in an experiment or book report and it has been graded you should see the score here (only the scores on the original Expt. #1 reports have been entered.  The Expt. #2 reports are currently being graded, the Expt. #1 revised reports will be graded after that).  If there is a 0 here, an average grade of 34 out of 40 has been used in the computer to show the effect of the writing on your overall grade.

4.    This shows the total number of 1S1P pts you have earned so far.  You should try to earn 45 1S1P pts by the end of the semester.  There will be 2 more 1S1P assignments (with at least 2 topics per assignment) and 2 or 3 more Bonus Assignments. 

5.    This is the average that needs to be 90.0% or above in order for you to not have to take the final exam.

6.    This is the average with the lowest quiz score dropped.  This is the grade that would be used together with your Final Exam score to determine your overall grade.

The mid term grade estimate gives you an idea of the grade you would receive at the end of the semester if you continue to perform as you have so far.  It is possible for you to significantly raise your grade between now and the end of the semester.  It is also possible, of course, for your grade to drop.