Monday Oct. 17, 2011
click here to download today's notes in a more printer friendly format

You heard two songs from Lhasa de Sela ("Con Toda Palabra" and "La Confession").  I'm trying to improve my French so listening to songs that are sung in French is educational.

The in-class Optional Assignment from last Friday has been graded and was returned in class.  A new Optional Assignment is now online.  Even if you don't normally do the Optional Assignments I would encourage you to do this one.  It will give you some practice with the humidity variables that we learned about last Friday and with the humidity type problems that we'll be doing in class today.

Experiment #4 materials should be available in class on Wednesday.  I'm still hoping to be able to hand out grade summaries sometime this week (most likely it will be Friday).

Here's a review of the humidity variables that were introduced last Friday.
mixing ratio (r)
actual amount of water vapor in the air
saturation mixing ratio (rs)
maximum amount of water vapor that can be in the air (depends on temp.)
relative humidity (RH)
how close is the air to being "filled" to capacity or saturated with water vapor
dew point  (Td)
1. gives an idea of the actual amount of water vapor in the air
2. cooling the air to the dew point raises the RH to 100%

Most 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, how they "behave", and what can cause their values to change. 

We spend maybe 10 minutes reviewing some material that was stuck onto the online notes from last Friday's class.  The objective there was to explain why there is an upper limit to the amount of water vapor that can be found in air and why that limit depends on temperature.

Here's a little bit shorter explanation.

The first thing we need to realize is that warm water will evaporate more rapidly than cool water.  You probably know that already.  If a cup of iced tea were set next to a cup of hot tea you probably be able to tell which was which by just looking at them.  You wouldn't need to touch or taste the tea or look for ice cubes in the iced tea.

You might notice that one of the cups of tea was steaming (the cup on the right above).  This would be the hot tea.  You're not actually seeing water vapor.  Rather water vapor is evaporating so quickly that it is saturating the air above.  The air isn't able to accomodate that much water vapor and some of it condenses and forms a cloud of steam.  That's what you see.

Now we'll redraw the picture and cover both cups so that water vapor can begin to buildup in the air above the water in both cups. 

Arrows represent the different rates of evaporation.  One arrow is shown evaporating from the cup of cold water.  The warmer water at right is evaporating 3 times more rapidly.  We've arbitrarily assigned rates of evaporation of 10 and 30 to the water in the two cups.

Water vapor will start to buildup in the air above each cup.  And, even though it has just evaporated, some of the water vapor will condense and rejoin the water at the bottom of each cup.  Let's just assume that 1% of the water vapor molecules will condense. 

The water vapor concentration in each glass will increase until it reaches a point where
water evaporation rate = water vapor condensation rate

for the cup of cold water
10 = 0.01 x water vapor concentration

The 0.01 is 1% expressed in decimal form.  Solving this equation gives you a water vapor concentration of 1000.  The air is saturated when you reach this point and the RH = 100%.

The saturation water vapor concentration in the air in the warm cup would be 3000.  And again the relative humidity would be 100%.

The fact that the rates of evaporation and condensation are equal when air is saturated (RH = 100%) is something we'll be using later when we study the formation of precipitation.  Here's a picture of how that would look inside a cloud.

The air inside the cloud is saturated.  The rate of evaporation from the cloud droplet (2 green arrows) is balanced by an equal rate of condensation (2 orange arrows).  The RH = 100%.  The cloud droplet won't grow any bigger or get any smaller.

Here's something to test your understanding of this material.


What information can you add to this picture?  Is the water in one of the glasses warmer than the other?  Is there more water vapor in the air in one of the glasses than the other?  Is the relative humidity in each glass more than 100%, less than 100% or is it equal to 100%.  The rates of evaporation and condensation aren't equal in either glass, so the pictures will change with time.  What will the glasses look once they have reached equilibrium?  Think about this for a while and then click here for the answers and some explanation.

Now onto the main event for today, some example humidity problems.  I got the impression that this was very confusing when you saw it for the first time in class.  Hopefully the more detailed explanations below will help.

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 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 (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 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.  Now 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.  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 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 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%.

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.