Thursday Oct. 22, 2009
click here to download today's notes in a more printer friendly format.

Some zydeco music from the Best of Beausoleil CD before class this morning.

The Experiment #2 reports have been graded and were returned in class today.  You are allowed to revise your report; revisions are due on or before Thu., Nov. 5.  Please return your original report with your revised report.

The Experiment #4 materials were distributed in class today.  Experiment #4 reports are due on Tue., Nov. 10.

A new Humidity Optional Assignment was distributed in class.  The assignment is due next Thursday, Oct. 29.  There was also an In-class Optional Assignment today.  You'll find the questions below embedded in the online notes.  You may turn in answers to the questions if you wish at the start of class next Tuesday.
We'll work through 4 humidity example problems today.

Example 1



We were 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 (Td).  You might have a hard time understanding this if you're seeing it for the first time.  The series of steps that we followed are retraced below:

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.  Pay attention to what is happening to the mixing ratio, saturation mixing ratio, and relative humidity as the air cools.

(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 actual amount of water vapor in the air isn't changing, it is just the air's capacity that is decreasing.

Finally at 45 F the RH becomes 100%.   You have cooled the air until it has become saturated.  This is a special point, this is the dew point temperature.

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 of water vapor would condense and form water.  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 (and being turned into water), the mixing ratio will now decrease from 6 to 4.

In many ways cooling moist air is liking squeezing a moist sponge


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 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.

After working through the 4 humidity problem examples you should be able to answer the following questions.  This is the first of two questions that formed part of an In-class Optional Assignment.  If you weren't in class, answer this question & Question 2 below and turn in your work at the beginning of class on Tuesday to receive credit.

Question #1



Question #2

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 didn't have enough time to cover this is class.  I've included it in the online notes because you might be able to use some of what you learn here to answer Question #2 on the In-class Optional Assignment.


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 (see the end of today's notes for more information on this point).  Unsaturated air cools 10 C for every kilometer of altitude gain.  This is known as the dry adiabatic lapse rate.  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.

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 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, there is less evaporation.  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.

The last thing we covered was the formation of dew, frost, and something called frozen dew.  The figures below were redrawn after class to make them clearer.

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.  By the next morning the air has cooled 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.