Fall 2007 - Homework #2
Answer
the following questions on a separate sheet of paper. Homework answers squeezed onto this page will not be accepted. If you need to calculate an answer, you must
show your work. To answer question 2,
you will need to refer to the skew-T diagram located under the homework link on
the course web page and the table of saturation mixing ratios provided with an
in-class handout (referenced by temperature in Celsius). For questions 3 and 4, you will need to use the table of saturation
mixing ratios provided on the class web page under the homework link
(referenced by temperature in Fahrenheit).
Use the heat index and wind chill tables (provided in lecture notes) to
help answer questions 5 and 6. Make sure you read and answer all the parts
to each question!
1.
Suppose
you were going to walk from the ocean near Calcutta, India up to the top of
Mount Everest at 8846 meters above sea level.
(Round the elevation to 9000 meters).
We will look at how air temperature, pressure, and density change on
your way up.
|
Elevation (meters) |
Fraction of way up by
altitude |
Air Temperature |
Air Pressure |
Percentage of the
atmosphere below you by weight |
|
0 |
At bottom |
30° C |
1000 mb |
0 % |
|
3000 |
1/3 |
? |
700 mb |
? |
|
6000 |
2/3 |
? |
500 mb |
? |
|
9000 |
At top |
? |
330 mb |
? |
(a)
Estimate
the air temperature at 3000, 6000, and 9000 meters. The information you need to do this is contained on the lecture
page entitled “Vertical variation of temperature, pressure, and density in the
atmosphere.” Do not use the rules
for lifting air parcels to answer this question.
(b)
Compute
the percentage of the atmosphere below 3000, 6000, and 9000 meters (based on
weight).
(c)
Explain
why the rate of decrease of air pressure is not constant with increasing
altitude, i.e., it drops by 300 mb over the first 3000 meters of the climb
(from 0 m to 3000 m), 200 mb over the next 3000 meters of the climb (from 3000
m to 6000 m), and 170 mb over the last 3000 meters of the climb (from 6000 m to
9000 m).
2. You must use the skew-T diagram located under the
homework link on the class web page to answer this question. The data was taken at 0000 UTC (or 00Z) on
September 7, 2006 at Vandenberg AFB in California.
(a) What is the local Tucson date and time when this data
was measured and plotted?
(b) Fill in the missing values in the table below by
reading values from the skew-T chart. Re-write
the table on your own paper. Do not
squeeze answers into the tables below.
(c) Using the table of saturation mixing ratios (in
Celsius), which was provided as part of an in-class handout, compute the relative humidity of the air at 700 mb
and 500 mb. You may round off the air
temperature and dew point temperature to the nearest values in the table.
|
Air Pressure (mb) |
Altitude Above Sea Level (m) |
Air Temperature (°C) |
Dew Point Temperature (°C) |
Wind Direction |
Wind Speed (knots) |
|
200 |
12430 |
-52 |
-62 |
East |
65 |
|
250 |
|
|
|
|
|
|
300 |
|
|
|
|
|
|
400 |
|
|
|
|
|
|
500 |
|
|
|
|
|
|
700 |
|
|
|
---- |
Calm |
|
850 |
|
|
|
|
|
|
1000 |
126 |
14 |
13 |
Southeast |
5 |
3.
On
a day last winter, the following conditions were measured on the UA
campus. You must use the saturation
mixing ratio table (with temperature in Fahrenheit) provided under the homework
link on the class web page to answer this question.
n
At
8 AM: air temperature, T = 45° F; dew point temperature, Td = 25° F.
n
At
11 AM: air temperature, T = 60° F; dew point temperature, Td = 25° F.
n
At
2 PM: air temperature, T = 70° F; dew point temperature, Td
= 25° F.
(a)
Compute
the relative humidity for each of the times/conditions specified above.
(b)
Explain
why the relative humidity changed the way it did from 8 AM through 2 PM. How did the water vapor content in the air
change between 8 AM and 2 PM?
4.
Values
of air temperature and relative humidity are given below for Presque Isle,
Maine and Tucson, Arizona as observed on a day in spring 2004. You must use
the saturation mixing ratio table (with temperature in Fahrenheit) provided
under the homework link on the class web page to answer this question.
Air Temperature
|
35° F |
|
Relative Humidity |
100 % |
|
Weather Conditions |
Rain |
Air
Temperature
|
90° F |
|
Relative Humidity |
25 % |
|
Weather Conditions |
Sunny |
(a)
What
are the approximate dew point temperatures at the two locations?
(b)
Of
these two locations, which has the higher concentration of water vapor in the
air? How do you know? Explain how a desert location with a low
relative humidity can actually have a higher water vapor content than a
location where the relative humidity is 100% with rain falling?
5.
On
a day in summer 2004, the conditions in Tucson, Arizona and Charleston, South
Carolina are given
Tucson, Arizona
|
Air Temperature |
105° F |
|
Relative Humidity |
10 % |
Air
Temperature
|
95° F |
|
Relative Humidity |
50 % |
(a)
Using the heat index chart provided with the course lecture notes, find the heat index for
the two cities. Compare the rate of
heat loss from the human body at these two locations.
6.
On
a day last winter, conditions measured at Flagstaff, Arizona and West
Yellowstone, Montana are given
Air
Temperature
|
0° F |
|
Wind Speed |
20 MPH |
West Yellowstone, Montana
Air
Temperature
|
-10° F |
|
Wind Speed |
5 MPH |
(a)
Using the wind chill chart provided with the course lecture notes, determine the wind chill
equivalent temperature for Flagstaff and West Yellowstone. Compare the rate of heat loss from the human
body at these two locations.
7.
Evaporative
cooling is one of the most ancient and one of the most energy-efficient methods
of cooling a home. It long has been regarded as environmentally
"safe," since the process uses no ozone-depleting chemicals, and
demands one-fourth as much energy as refrigeration during the peak cooling
months of the year. In dry climates such as Tucson, evaporative cooling can be
used to inexpensively cool large homes.
Locally, these devices are often referred to as “swamp coolers”. The most common form of residential
evaporative cooling uses a vertical pad of absorbent cellulose fiber, a system
for delivering water to the pad to keep it soaked with water, and a fan to draw
air through the porous pad. As warm, dry outside air is drawn through the wet
pad, water evaporates into the air, and the air gives up its heat. In other words, energy is removed from the
air in order to evaporate water. Thus, air that has moved through the wet pad
is both cooler and contains more water vapor than the outdoor air.
The drop in temperature depends on how much water
can be evaporated into the air. This is
obviously a function of relative humidity.
When the relative humidity is low, the temperature drop can be
large. However, when the relative
humidity is high, the temperature drop will be small (and the swamp cooler
doesn’t help much).
The wet bulb temperature, which was not discussed
in lecture, is the lowest temperature to which air can be cooled by
evaporating water into it. This is the
theoretical lower limit for the temperature of the air that comes out of an
evaporative cooler.
Explain the following statements:
(a)
When
the relative humidity is 100%, the air temperature, the dew point temperature,
and the wet bulb temperature are identical.
Explain.
(b)
When
the relative humidity is less than 100%, the dew point temperature and the wet
bulb temperature are both lower than the air temperature. Explain.
(c)
When
the relative humidity is less than 100%, the wet bulb temperature will always
be higher than the dew point temperature.
Explain. (Hint: What is happening to the water vapor content
and dew point temperature of the air as it is being evaporatively cooled? At what point does it become impossible to
further cool air by evaporation?)