Thursday Oct. 23, 2014

I was too busy grading lab reports yesterday and going to the dentist to think about music choices.  So I just looked through some past semesters and picked Gotye "Somebody That I Used to Know", The XX "Infinity", and a live version of  "Reckoner" from Radiohead. 

The graded Expt. #2 reports were available for pickup in class today.  You can revise your report if you want to (it isn't required).  The revised reports are due in 2 weeks - by Thu., Nov. 6.  Please return your original report if you turn in a revised report.

Any remaining In-class Optional Assignments and the 1S1P reports on ultraviolet light were collected today.  A new In-class Optional Assignment on Humidity Variables was handed out today and collected at the end of the class period.  If you weren't in class and want to do the assignment you can download it here.  If you complete the assignment and turn it in by the start of class on Tuesday next week you can earn at least partial credit (most likely full credit).

The are a few remaining sets of Expt. #2 and Expt. #3 materials available for checkout.  If you are planning on doing either the Scientific Paper report or a Book Report in place of an experiment, it is time to get started on that work.  Both reports are due by Thu., Nov. 13.  Here's a link to the list of book titles available.

We needed to finish up a few short greenhouse effect-related topics:

A more realistic picture of energy balance on the earth.

The more realistic picture of energy balance on the earth is shown below (the bottom part of the figure below).  The top part of the figure below is our simplified representation of energy balance.  Incidentally the top figure contains the answer to the Optional Assignment assigned at the end of class on Tuesday (you needed to add the 3 bluish arrows being emitted by the atmosphere: one went upward & into space, the other two downward to the ground).
 


In the top figure you should recognize the incoming sunlight (green), IR emitted by the ground that passes through the atmosphere (violet), IR radiation emitted by the ground that is absorbed by greenhouse gases in the atmosphere (orange) and IR radiation emitted by the atmosphere (blue). 

The lower part of the figure is pretty complicated.  It would be difficult to start with this figure and find the greenhouse effect in it.  That's why we used a simplified version.  Once you understand the upper figure, you should be able to find and understand the corresponding parts in the lower figure (I've tried to use the same colors for each of the corresponding parts).

Some of the incoming sunlight (51 units in green) reaches the ground and is absorbed.  19 units of sunlight are absorbed by gases in the atmosphere.  The 30 units of reflected sunlight weren't included in the figure.

The ground emits a total of 117 units of IR light.  Only 6 shine through the atmosphere and go into space.  The remaining 111 units are absorbed by greenhouse gases. 


There were 3 things I wanted you to notice in the bottom figure and try to explain.

(1).  How can the ground be emitting more energy (117 units) than it gets from the sun (51 units).  It is able to achieve energy balance because it also gets energy from the atmosphere (96 units).  That's thanks to the greenhouse effect. 

If you're really paying attention you would notice that 117 units emitted doesn't balance 96 + 51 = 147 units absorbed.  The surface is emitting 117 units but an additional 30 units are being carried from the ground to the atmosphere by conduction, convection, and latent heat (at the far left of the figure).  That brings everything into balance (117 + 30 = 147).  Note how much smaller the energy transport by conduction, convection, and latent heat are compared to radiant energy transport.


(2).  Why are the amounts of energy emitted upward (64 units) and downward (96 units) different? 

One reason might be that the lower atmosphere is warmer than the upper atmosphere (warm objects emit more energy than cold objects).  But I think a bigger part of the explanation is probably that there is more air in the bottom of the atmosphere (the air is denser) than near the top of the atmosphere.  It is the air in the atmosphere that is emitting radiation.  More air =  more emission.

(3).  The ground is receiving more energy from the atmosphere (96 units) than it gets from the sun (51 units)!  Doesn't that seem odd?  I think the main reason for this is that the sun just shines for part of the day.  We receive energy from the atmosphere 24 hours per day, 365 days per year.

The effects of clouds on daytime high and nighttime low temperatures

This can be found on pps 72a & 72b in the ClassNotes (I've rearranged things slightly to try to make it clearer).   And before we go any further ask yourself what you already know about this.  It was fairly cool overnight.  If it had been cloudy last night would it have been colder, warmer or about the same?  It is supposed to get up near 90 F this afternoon with clear skies.  If it clouds over this morning will the high temperature be above 90 F or below 90 F?



Here's the simplified picture of radiative equilibrium again.  You should be able to say something about every arrow in the picture.  The two pictures below show what happens at night when you remove the two green rays of incoming sunlight.



Note first of all that neither picture is in radiative equilibrium.  The picture on the left shows a clear night.  The ground is losing 3 arrows of energy and getting one back from the atmosphere.  That's a net loss of 2 arrows.  The ground cools rapidly and gets cold during the night.

A cloudy night is shown at right.  Notice the effect of the clouds.  Clouds are good absorbers of far infrared radiation (10 μm wavelength).   If we could see 10 μm far IR light, clouds would appear black, very different from what we are used to (actually it would be more complex than that because clouds also emit IR light, the clouds might also glow).  Because of the clouds none of the IR radiation emitted by the ground passes through the atmosphere into space.  It is all absorbed either by greenhouse gases or by the clouds.  Because the clouds and atmosphere are now absorbing 3 units of radiation they must emit 3 units: we'll draw 1 going upward into space, the other 2 downward to the ground. 

There is still a net loss of energy at the ground on the cloudy night but it's smaller, only 1 arrow.  The ground won't cool as quickly and won't get as cold on a cloudy night as it does on a clear night.  That makes for very pleasant early morning bicycle rides this time of the year. 

The next two figures compare clear and cloudy days.



Sunlight is made up of mostly visible and near-IR light.  The fact that clouds are white tells us they reflect visible light.  What about near IR light?  Have a look at the photograph of the tree near the start of the Tue., Oct. 21 notes.  The clouds in both pictures are white, that tells us clouds are good reflectors of both visible and near-IR light.  The effect of this is to reduce the amount of sunlight energy reaching the ground in the right picture.  With less sunlight being absorbed at the ground, the ground won't warm up as much during the day.

It is generally cooler during the day on a cloudy day than on a clear day.

Clouds raise the nighttime minimum temperature and lower the daytime maximum temperature, they reduce the diurnal temperature range.  Here are some typical daytime high and nighttime low temperature values on clear and cloudy days for this time of the year.




Afternoon temperatures will be as much as 10 F above average Friday and this weekend.  This weekend will be a perfect time to conduct Expt. #3.  Don't procrastinate, you never know what the weather will be like by the end of next week.

We can use our simplified representation of radiative equilibrium to understand enhancement of the greenhouse effect and global warming.

But first, a common misconception about the cause of global warming.




Many people know that sunlight contains UV light and that the ozone layer absorbs much of this dangerous type of high energy radiation.  People also know that release of chemicals such as CFCs are destroying stratospheric ozone and letting some of this UV light reach the ground.  That is all correct. 

But then they conclude that it is this additional UV energy reaching the ground that is causing the globe to warm.  This is not correct.  There isn't much UV light in sunlight in the first place (only about 7% of sunlight is UV) and just a portion of that would reach the ground.  The small amount of additional UV light reaching the ground won't be enough to cause global warming.  It will cause cataracts and skin cancer and those kinds of problems but not global warming.

If all 7% of the UV light in sunlight were to reach the ground it probably would cause some warming.  But it probably wouldn't matter because some of the shortest wavelength and most energetic forms of UV light would probably kill us and most other forms of life on earth.


Enhancement of the greenhouse effect and global warming
Here's the real cause of global warming and the reason for concern (this is also the last time you'll see these energy balance pictures)



The figure (p. 72c in the photocopied Class Notes) on the left shows energy balance on the earth without an atmosphere (or with an atmosphere that doesn't contain greenhouse gases).  The ground achieves energy balance by emitting only 2 units of energy to balance out what it is getting from the sun.  The ground wouldn't need to be very warm to do this, only 0 F.

If you add an atmosphere and greenhouse gases, the atmosphere will begin to absorb some of the outgoing IR radiation.  The atmosphere will also begin to emit IR radiation, upward into space and downward toward the ground.  After a period of adjustment you end up with a new energy balance.  The ground is warmer and is now emitting 3 units of energy even though it is only getting 2 units from the sun.  It can do this because it gets a unit of energy from the atmosphere.  This is what I refer to as the beneficial greenhouse effect.  It makes the earth more habitable by raising the average surface temperature to 60 F.

In the right figure the concentration of greenhouse gases has increased even more (due to human activities).  The earth might find a new energy balance.  In this case the ground would be warmer and could be emitting 4 units of energy, but still only getting 2 units from the sun.  With more greenhouse gases, the atmosphere is now able to absorb 3 units of the IR emitted by the ground.  The atmosphere sends 2 back to the ground and 1 up into space.  A new balance is achieved but the earth's surface is warmer.  How much warmer?  That's the big question.


Humidity variables
Now a new block of material on humidity and an introduction to humidity variables.  This topic and the terms that we will be learning are probably new and might be confusing.  That's the reason for this introduction.  We will be mainly be interested in 4 variables:





Your task will be to learn the "jobs" of these variables, their units, and what can cause them to change value. 

An In-Class Optional Assignment was handed out in class.  You were supposed to complete and turn in the assignment at the end of class.  If you weren't in class you can download the assignment and turn it in before the start of class next Tuesday.

Mixing ratio
The bottom half of the figure below can be found on p. 83 in the 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).  A kilogram of air is about one cubic meter of air (about one cubic yard of air).  Mixing ratio 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 (with one exception - 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

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 and depends on the air's temperature; warm air can potentially hold more water vapor than cold air.  It doesn't say anything about how much water vapor is actually in the air (that's the mixing ratio's job).    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 ClassNotes.



The sugar dissolved in tea analogy is still helpful.  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.  And not only that, the amount depends on temperature: 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.  The numbers 15 and 5 came from the table on p. 86.


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.  Three of the open circles have been colored in.  The mixing ratio, r, is 3 g/kg in both cases.

Relative humidity
After looking at the figure above you might be able to figure out what relative humidity is


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, how close the air is to being "saturated" with water vapor.  RH has units of %.

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.  How full the room is 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 of water vapor.


Both the 70 F and the 40 F air 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 capacity, the 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 values of relative humidity.  You could just as easily have two volumes of air with the same relative humidity but different actual amounts of water vapor.

What is the RH good for if it doesn't tell you how much moisture is in the air?  When the RH reaches 100% dew, fog, and clouds form.  RH tells you whether clouds or fog are about to form or not.

Dew point temperature


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.  This is something we learned early in the semester.

The dew point is a temperature and has units of   oF or oC
Second the dew point tells you how much you must cool the air in order to raise the RH to 100% (at which point a cloud, or dew or frost, or fog would form).  This idea of cooling the air until the RH increases to 100% is important and is something we will use a lot.


 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.

We're going to work 4 example problems.  It's by doing these problems that you start to understand how these various variables work and what can cause them to change value.  We'll do just one today.
Humidity example problem #1 (of 4)
In this example you are given an air temperature of 90 F and a mixing ratio value of 6 g/kg.  You are supposed to determine the mixing ratio and the dew point temperature.


Here's something close to what we ended up with in class.  This would be hard enough 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 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).