Wednesday Oct. 17, 2007
Optional Assignment #4 (Controls of Temperature) was collected in class
today. Optional Assignment #5 (Humidity) is due next Monday.
The 1S1P Assignment #2 reports are due 1 week from today
(Wednesday Oct. 24).
The Quiz #2 scores have been corrected. There was a small, 1%
point, rise in the class average. You will be able to see if your
score has changed on a computer generated grade summary that should be
distributed early next week.
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.
In the difference between the air and dew point temperatures was
smaller and the RH was high. 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 would be 100%.
Now on to
Problem #3
This figure was redrawn after class. 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.
(3) Then you
imagine cooling the air until the RH becomes 100%. This occurs at
60 F. The dew point is 60 F.
Problem #4 is probably the most difficult of the bunch.
Here's what we did in class. We were given the air temperature
and the dew point temperature.
We enter the two temperatures onto a chart and look up the
saturation
mixing ratio for each.
Then we know that if we cool the 90 F air to 50 F the RH
will
become
100%. We don't know the mixing ratio or the relative
humidity. But we do know that when we arrive at 50 F the RH will
be 100%. The mixing ratio must be equal to the saturation mixing
ratio value for 50 F air, 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%.
Next we will use what we have learned about humidity
variables (what they tell you about the air and what causes them to
change value) to learn something new.
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). 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 less than half the
moisture (3 g/kg) that it originally did (7.5 g/kg). Next, Point
4, the 30
F air is warmed back to 90 F, the starting temperature. 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.
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 (the figure in
class was redrawn for clarity).
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). 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. 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.
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
evaporates. The sinking 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. The 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.
NOTE: The figure
above illustrates orographic or topographic lifting.
It is one of 4
ways of causing air to rise. We have already run into the other
three in class this semester. They were: convergence
(surface winds spiral into centers of low pressure), convection (warm
air rises), and fronts. Rising air is important because rising
air expands and cools. Cooling moist air raises the relative
humidity and a cloud might form.
Finally we
learned about a simple instrument used to measure humidity, a sling
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 buld temperature difference can be
used to determine relative humidity and dew point (see Appendix D at
the back of the textbook).
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 air temperature (the 80 F
value was just made up in this example).
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.
Note in the bottom left figure 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 between the dry and wet bulb
thermometers (20 F) 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 air temperature is
the same as 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 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 dry and wet bulb thermometers would both read 80 F.