Tuesday Oct. 18, 2011
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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.
Quiz #2 has been graded and was returned in class today.
A few Expt. #3 materials were available again today. Also because
we are a little short of materials if you can collect your data in time
to be able to return your materials by Tuesday next week (Oct. 25)
you'll receive a Green Card.
Experiment #4 materials should be
available on Thursday.
A Tuesday, Nov. 8, due date has been set for the Scientific Paper report. That was
an option you could choose instead of performing one of the experiments.
Two new Optional
Assignments are now available. They're both due next Tuesday
(Oct. 25). The topic of the first assignment is controls of temperature.
There
is
some
online
reading that you should do first (also some optional reading that
you can choose to read or not). The second assignment will give
you some practice with humidity problems, something we will begin
covering today.
Finally,
there was also an in class assignment today. If you'd like to
download the
assignment and turn it in at the beginning of class on Thursday you
can earn at at least partial credit.
I showed a short student produced video about Experiment #3.
The video is on VHS tape. I don't have a digital version so I
can't put it online.
Also we took a short detour so that I could mention the "required"
online reading about the Controls of Temperature and the associated
Optional Assignment.
We spent the majority of the class period on an introduction to
the next major topic we will be covering: 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 their "jobs" are and what can cause them to
change value . 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. You can think of it as just a number: when the value is
large there's more water vapor in the air than when the value is
small. But it's not a difficult concept to grasp. 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's 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
(unless
the air is
cooled below its dew point temperature and water
vapor starts to condense but in that case the air is losing water
vapor). 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 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 photocopied class
notes.
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. 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. 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 of water vapor.
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.
Now onto some example humidity
problems. This material can be confusing when
you see 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%.
