Thursday Oct. 22, 2015

Rodrigo y Gabriela "The Soundmaker", Tesoro "Malaguena",  Beth Daunis & Domingo DeGrazia "I'm not sure what the title is".  The latter two groups are local groups.

The Experiment #2 reports were returned in class today.  You are allowed to revise those reports if you want to, it's not required.  The due date for revised reports is Thu., Nov. 5.  You only need to rework sections where you want to earn additional credit but please return the original report with your revised report.

"Midterm" grade summaries were also distributed today.  More about them at the end of today's notes.

My goal now is to have the Causes of the Seasons and the revised Expt. #1 reports graded by this time next week.

A new Optional Assignment was handed out in class.  The assignment is due at the start of class next Tuesday (Oct. 27) (you should have the assignment done before coming to class).

We worked through three humidity variable example problems in class today.  A 4th problem has been inserted into these online notes.  The goal here is for you to become a little more familiar with the humidity variables and see what causes them to change value.
Humidity example problem #1
There are 4 humidity variables (mixing ratio, saturation mixing ratio, relative humidity, and dew point temperature).  Generally I'll give you values for two of them and you'll need to figure out values for the other two.
Here are the starting conditions for this first problem
Tair = 90 F
r = 6 g/kg
RH = ?
 Td = ?

We start by entering the data we were given


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 the dew point.  The dew point temperature 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



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 squeeze 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).

Humidity example problem #2
Tair = 90 F
r = ?
RH = 50% Td = ?

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 could hold up to 30 g/kg of water vapor is filled to 50% of its capacity.  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 (I added more intermediate temperatures in the table above than we did in class).  Notice how the saturation mixing ratio decreases, the mixing ratio stays constant, and the relative humidity increases as the air is cooled.   In this example the RH reached 100% when the air had cooled to 70 F.  That is the dew point temperature.
What does the difference Tair - Td tell you about the relative humidity?

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 might be to remember the case where there is no difference between the air and dew point temperatures. 
The RH then would be 100%.

Humidity example problem #3
Tair = ?
r = 10.5 g/kg
RH = 50% Td = ?

We skipped this problem in class.  But I've included all the details below just in case you want to try to solve the problem on your own.

You're given the the mixing ratio = 10.5 g/kg and a relative humidity of 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
Humidity example problem #4
Tair = 90 F
r = ?
RH = ?
 Td = 50 F


One of the dew point's jobs is the same as the 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 and vice versa.  Knowing the dew point is equivalent to knowing the mixing ratio.


We enter the two temperatures given on 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%.  So on the 50 F row, we can set the mixing ratio equal to the value of the saturation mixing ratio at 50 F, 7.5 g/kg.  The two have to be equal in order for the RH to be 100%.


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%.
Drying moist air
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 same numbers that we had in Example Problem #4.  We imagine cooling this air to the dew point temperature, 50 F.  While doing that the mixing ratio, r, would stay constant.  Relative humidity would increase and eventually reach 100%.  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).  Mixing ratio will decrease, the relative humidity will remain 100%.  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.  Cooling moist air raises the RH.  Warming moist air, as is being down here, lowers the RH.  Once back at the starting temperature, Point 5, the air now has a RH of only 10%.

Drying moist air is basically wringing moisture from a wet sponge.



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 original shape and size (the air warms back to its original temperature).  The sponge (and the air) will be drier than when you started.


Dry air indoors in the winter
The air indoors in the winter is often quite dry.


In the winter, cold air is brought inside your house or apartment and warmed.  Imagine foggy 30 F air (with a RH of 100% this is a best case scenario, the cold air outdoors usually has a relative humidity less than 100% and is drier). Bringing the air inside and warming it will cause the RH to drop from 100% to 20%..  This can cause chapped skin, can irritate nasal passages, and causes 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 would be essentially 0%.  The RH doesn't get this low because the airplane adds moisture to the air to make to make the cabin environment tolerable.  Still the RH of the air inside the plane is pretty low and passengers often complain of dehydration on long airplane flights.  This may increase the risk of catching a cold (ref)

The rain-shadow effect

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* (see below).  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 but isn't something you need to remember).  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.

*This is topographic lifting, the 4th of 4 processes that can cause air to rise.  The other three were: convergence (surface winds spiraling inward toward a low pressure center will rise), fronts (both warm and cold fronts cause air to rise), and convection (warm air rises).





We can see the effects of a rain shadow 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 mountains.  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.  Temperate rainforests are found in some of these coastal locations.  The line separating the green and yellow on the left side of the precipitation map is the summit, the ridgeline, of the coastal mountain range.

That's the Willamette River valley, 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.  The boundary between yellow/green and the red is the ridgeline of the Cascade Mountains.  Yearly rainfall is generally less than 20 inches per year on the eastern side, the rain shadow side, of the Cascades.    That's not too much more than Tucson which averages about 12 inches of rain a year.

Death valley is found on the downwind side of the Sierra Nevada mountains (source of left image).  The Chihuahuan desert and the Sonoran desert are found downwind of the Sierra Madre mountains in Mexico (source of the right image).  Mexico might be a little harder to figure out because moist air can move into the interior of the country from the east and west.  But there are mountains along both coasts, so some of that moisture will be removed before arriving in the center of the county.

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 traveled 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.   Beginning in early July in southern Arizona we start to get air coming from the south or southeast.  This air can be much moister and leads to development of our summer thunderstorms.

Just as some of the world's driest regions are found on the downwind side (the rain shadow side) of mountain ranges, some of the wettest locations on earth are on the upwind sides of mountains.  There seems to be some debate whether Mt. Wai'ale'ale in Hawaii or Cherrapunji India gets the most rain per year.  Both get between 450 and 500 inches of rain per year.

Measuring humidity with a sling psychrometer

Next in today's potpourri of topics was a short discussion of how you might measure humidity.  One of the ways 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 thermometer measures the air temperature. 

Would the wet thermometer be warmer or colder or the same as the dry thermometer?   You can check it out for yourself - go get one of your hands wet.  Does it feel the same as the dry hand?  You might blow on both hands to increase the evaporation from the wet hand.  I think you'll find the wet hand feels colder.  That's what happens with the wet bulb thermometer.


What could you say about the relative humidity in these two situations (you can assume the air temperature is the same in both pictures).  You would feel coldest on a dry day (the left picture indicates dry air).  The evaporative coolers that many people use in Tucson in the summer work much better (more cooling) early in the summer when the air is dry.  Once the thunderstorm season begins in July and the air is more humid it is hard to cool your house below 80 F.
 
Here are a bunch of details that you can read through if you're so inclined.  My goal is that you understand the basic principle behind a sling psychrometer.  For that I think you can just skip to the summary a few pictures further on. 

Here's the situation on a day with low relative humidity.

The figure 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  Warm water evaporates at a higher rate than cool water (think of a steaming cup of hot tea and a glass of ice tea).

The evaporation is shown as blue arrows because this will cool the thermometer.   The water on the wet thermometer starts out at 80 F and evaporates fairly rapidly.

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.  As the thermometer cools the rate of evaporation will decrease.  The thermometer will continue to cool until the evaporation has decreased enough that it balances the condensation.


The rates of evaporation and condensation are equal.  The temperature will now remain constant.

The figure below shows the situation on a day with higher relative humidity.  There's enough moisture in the air to provide 3 arrows of condensation. 


The rate of evaporation stays the same, the rate of condensation is higher.  The rate of evaporation is still higher than condensation but not by much. 




There'll only be a little cooling before the evaporation is reduced enough to be in balance with condensation.

Here's a visual summary


A large difference between the dry and wet temperatures means the relative humidity is low.  A small difference means the RH is higher.  No difference  means the relative humidity is 100%. 

We saw the same kind of relationship between RH and the difference between air and dew point temperature.

We didn't cover this next section in class.
Wind chill and heat index




Cold temperatures and wind make it feel colder than it really is.   The wind chill temperature tells you how much colder it will feel ( a thermometer would measure the same temperature on both the calm and the windy day).  If your body isn't able to keep up with the heat loss, you can get hypothermia and die.

There's something like that involving heat and humidity.  Your body tries to stay cool by perspiring.  You would feel hot on a dry 105 F day.  You'll feel even hotter on a 105 F day with high relative humidity because your sweat won't evaporate as quickly.  The heat index measures how much hotter you'd feel. The combination of heat and high humidity is a serious, potentially deadly, weather hazard because it can cause heatstroke (hyperthermia). 

Example grade summary
__________

Doe_J
quiz1 -42.0 (170 pts possible) 75.3%
quiz2 -51.0 (175 pts possible) 72.9%                              
writing scores: 0.0 (expt/book report) + 19.5 (1S1P pts)
writing percentage grade: 94.8%
average (no quiz scores dropped): 78.3% + 0.0 = 78.3%     
average (lowest quiz score dropped): 80.2% + 0.0 = 80.2%

*because you haven't completed the experiment or book report yet
(or your report hasn't been graded yet) an average score of 32
was used to compute your writing grade.
__________

The grades shown here are averages for this class. 

The first two items are your scores on the quizzes.  If you did the Upper Level Charts Optional Assignment and chose to have points added to your Quiz #2 score they are included here.  There are two more quizzes left this semester.

The next line would ordinarily show the number of extra credit points that you have earned so far this semester.  I haven't included extra credit points in this grade summary.  I forgot to mention that in class.   I was in a hurry to get these summaries done in time for today's class and there haven't been very many opportunities to earn  extra credit so far this semester.  Extra credit will be included in future grade summaries.

Your writing score and writing percentage grade are next.  This is made up of your experiment report grade (up to 40 pts) and the number of 1S1P pts you've earned so far (this should be 45 pts by the end of the semester).  If you turned in an Expt. #1 or Expt. #2 report the grade summary will show the grade you received (the revised Expt. #1 reports haven't been graded yet).  A score of 0 is shown for everyone else.  If you haven't received credit for an experiment report yet, the computer has assumed an average score of 32 out of 40 points just to show you the effect that your writing grade will have on your overall grade.  The computer has also taken into account the fact that most students haven't earned 45 1S1P pts at this point in the semester (the average student has earned 19.5 points so far).  The writing percentage grade has the same weight as a quiz.  You should end up with a writing percentage grade near 100% by the end of the semester.

Finally two overall averages are computed:
(i) the first doesn't drop any quiz scores.  This is the score that must be 90.0 or above at the end of the semester in order to be exempt from the Final Exam.
(ii) the lowest quiz score is dropped when computing the 2nd average.  If you do have to take the Final Exam, this is the grade you would have going into the Final Exam.  Your overall grade for the class would be based on this average and the score you receive on the Final Exam.

These grade estimates attempt to predict the grade you will end up with at the end of the semester if you keep on doing as you have done so far.  If you're happy with your overall average, you need to keep up the quality of work you have done so far.  If your score is lower than you'd like there is still plenty of time for improvement. 

Finally be sure to check that all of the information on your grade summary is correct.